tag:blogger.com,1999:blog-41562010201887143812014-10-02T23:46:37.527-07:00Walton's JMU Math BlogA blog for math courses that I teach at James Madison University.Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.comBlogger38125tag:blogger.com,1999:blog-4156201020188714381.post-52057078652489215142013-01-03T13:38:00.000-08:002013-10-08T10:58:36.452-07:00Oh yeah! I've got infinity plus one!Earlier this week, my brother Chris sent me a Facebook question posed by my preschool-aged nephew: "What is infinity minus infinity? What is zero minus infinity?" I'm sure most of us at some point engaged in the one-upmanship game of making bigger numbers than our sibling or friend.<br /><br />Chris: 20<br /><br />Brian: 21<br /><br />Chris: 100<br /><br />Brian: 101<br /><br />Chris: 1000<br /><br />Brian: 1001<br /><br />And then your brother makes the leap!<br /><br />Chris: Infinity!<br /><br />Brian: Infinity + 1!<br /><br />Chris: That's just infinity. There's nothing bigger than infinity. I win!<br /><br />No fair! What's up with this? Up to the point where we leap into the infinite, we are dealing with numbers. As it happens, infinity is not a number. Some people like to say, "Infinity is not a number, it's a concept." But this isn't very helpful at all. What good does it do to say it's a concept? What is a concept anyway? Or for that matter, what is a number?<br /><br />Another of my Facebook friends relatively recently introduced me to a <a href="http://ed.ted.com/lessons/how-big-is-infinity">TED video</a> that explores the cardinality idea of infinity. That is, cardinality is about sizes of sets, and natural numbers (positive integers) or counting numbers are exactly the type of numbers that measure cardinality. The idea of infinity in relation to cardinality corresponds to the idea that you have an unending number of elements in a set.<br /><br />Bizarre things happen in the cardinality sense of infinity. For example, if you take all of the positive integers (1, 2, 3, 4, ...) and double each number (2, 4, 6, 8, ...), we still have the same number of elements in the set (since we just manipulated each object). But our new set happens to be a subset of the original. When dealing with finite sets, a proper subset always has <b>fewer</b> objects than the original set. But we just saw an example where a proper subset has an <b>equal</b> number of objects (infinitely many). So here, the phrase "number of objects" does not represent an actual number, but represents the concept of cardinality.<br /><br />Alternatively, we could have started with all of the positive integers (1, 2, 3, 4, ...) and then just deleted every other number in the list. This also gives us (2, 4, 6, 8, ...) since the odd numbers were all removed. This is one way of showing that an infinite set taken away from an infinite set can still be infinite. This is an example where ∞ - ∞ = ∞. On the other hand, if we start with (1, 2, 3, 4, ...) and then take away the infinite collection of numbers (11, 12, 13, 14, ....), we are left with (1, 2, 3, 4, 5, 6, 7, 8, 9, 10). This would be an example where ∞ - ∞ = 10. In fact, by choosing how many numbers we want to leave and then just deleting all of the rest, it is easy to create examples where ∞ - ∞ = n for any integer represented by n.<br /><br />The point here is that a cardinality interpretation of subtraction (removing elements from a set, like taking candy pieces out of a pile) reveals that the cardinality of ∞ (not a number) does not follow ordinary arithmetic rules. The infinite does that; it breaks the ordinary rules we are comfortable with when dealing with finite things.<br /><br />In mathematics, we say ∞ - ∞ is <b>indeterminate</b> because the result actually depends on how the subtraction takes place.<br /><br />A cardinality interpretation of 0 - ∞ does not actually make sense. Interpreting 0 - ∞ first requires an extension of the idea of numbers to negative numbers. For example, my children want to tell me that 2 - 5 = 0 because if you start with 2 candies and try to take 5, you don't have any left. But at first they don't realize that there are 3 candies that you never got to take.<br /><br />We could introduce the idea of borrowing (loans). Suppose that I have 2 candies in my bowl and my daughter wants to eat 5 candies. If I want to give her 5 candies, then I can give her my 2 candies but then I'll need to get 3 more candies from someone else. This puts me in debt for 3 candies (-3) which I might represent by little paper IOUs. I have 3 IOUs. I can pay off the IOUs when I obtain candies. For each candy I receive, I pay off an IOU.<br /><br />This is the idea of addition extended to all integers. -5 + 2 corresponds to have 5 IOUs and 2 candies. The 2 candies pay off 2 IOUs, leaving me the same as if I had only 3 IOUs to begin with. So -5 + 2 = -3. Subtraction is normally thought of as actually taking candies away. That is 5 - 3 corresponds to having 5 candies and taking 3 away, leaving only 2. The extended idea of subtraction is to think of having 5 candies and using 3 IOUs: 5 - 3 = 5 + -3. Taking away the candies is equivalent to redeeming IOUs.<br /><br />That is, once we start dealing with negative numbers, subtraction is really about adding negatives (redeeming IOUs). The value -∞ simply means that we have an infinite number (there's that cardinality idea again) of IOUs. So 0 - ∞ is really the idea 0 + -∞, which means we start with a pile of 0 candies and an infinite number of IOUs. Since we can't redeem any of the IOUs, we still have infinitely many. That is, 0 - ∞ = -∞. By the same argument, 5 - ∞ = -∞ and ∞ - 5 = ∞.<br /><br />In addition, this gives us a way of extending our earlier idea of subtraction for ∞ - ∞ to end with <b>any possible answer</b> from -∞ to ∞. The extended cardinality approach to ∞ - ∞ would mean that we have infinitely many candies and infinitely many IOUs. If we put the candies and IOUs each in some order (an interesting philosophical question is if this is always possible), then we can make a choice on how we redeem our candies.<br /><br />We might redeem every IOU but occasionally (or regularly) skip some of the candies. For example, the first IOU could take the 1st candy, the 2nd IOU takes the 3rd candy, the 3rd IOU takes the 5th candy, and so on, leaving candies 2, 4, 6, .... This would correspond to ∞ - ∞ = ∞. Or we could redeem only some of the IOUs but use all of the candies. For example, the 1st candy redeems the 1st IOU, the second candy redeems the 3rd IOU, and so on, leaving infinitely many IOUs unredeemed. This corresponds to ∞ - ∞ = -∞. Or we could leave the first 12 IOUs unredeemed and then redeem each subsequent IOU by the subsequent candies, corresponding to an example of ∞ - ∞ = -12.<br /><br />It is essential to see that when we deal with arithmetic involving ∞ or -∞, we are not dealing with numbers. For cardinality, we are actually dealing with correspondences between sets. ∞ - ∞ does not adequately describe the correspondence. But 12 - 5 is adequate because every finite correspondence results in the same final answer. But for infinite correspondences, the nature of the correspondence itself makes a difference. Indeterminate really is indeterminate. <br /><br /><b>Edit</b>: I had a colleague point out to me that there is a second sense of infinite using <b>ordinal numbers</b>. In this setting, it is possible to talk about "Infinity plus one" as having meaning that is actually distinct from "Infinity". For now, you might want to look at the <a href="http://en.wikipedia.org/wiki/Ordinal_number">wikipedia page</a>, but I'm working on my own response as well. (October 8, 2013)Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-85516362324595203542012-11-20T14:02:00.000-08:002012-11-20T14:12:42.694-08:00Derivatives, Velocity, and AccelerationCalculus can be viewed as the study of rates of change of quantities. The most familiar rate of change in our ordinary experiences is velocity as the rate of change of position. As we ride our skateboards, bicycles, and cars, we understand that high velocities mean our position is changing quite rapidly; and we understand that when our velocity is zero, we are standing still.<br /><br />So imagine the experience of a perfect rocket-car that experiences no friction and has no brakes. The only way to change its velocity is with rocket blasters that are installed on either end of the vehicle. Consider the following trajectory, illustrated as an animation, and repeating in a loop. A timer (16 seconds) is also shown to provide a measurable sense of time.<br /><br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-Zn6TDdjBfiY/UKvVUfrzAdI/AAAAAAAAAE0/3qbFcHtI_Dw/s1600/ex1_rocket_tracks.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-Zn6TDdjBfiY/UKvVUfrzAdI/AAAAAAAAAE0/3qbFcHtI_Dw/s1600/ex1_rocket_tracks.gif" /></a></div><br />The following table describes when and which direction the rockets are firing.<br /><br /><table border="1" style="text-align: center;"><tbody><tr> <th>Time Interval</th> <th>Direction of Rocket</th> <th>Magnitude</th></tr><tr><td>[0,1)</td> <td>None</td> <td>0</td> </tr><tr><td>(1,3)</td> <td>Right</td> <td>Moderate</td> </tr><tr><td>(3,5)</td> <td>None</td> <td>0</td> </tr><tr><td>(5,7)</td> <td>Left</td> <td>Large</td> </tr><tr><td>(7,8)</td> <td>None</td> <td>0</td> </tr><tr><td>(8,12)</td> <td>Right</td> <td>Small</td> </tr><tr><td>(12,16]</td> <td>None</td> <td>0</td> </tr></tbody></table><br />In addition, the rocket-car is driving along a track which has positions marked so that we can think of the position as a variable (in meters). That is, we can think of examining the relation between the variables of time and position. The following table provides the position of the rocket-car as recorded at each second. <br /><br /><table border="1" style="text-align: center;"><tbody><tr> <th>time (s)</th> <td>0</td> <td>1</td> <td>2</td> <td>3</td> <td>4</td> <td>5</td> <td>6</td> <td>7</td></tr><tr> <th>position (m) </th> <td> 0.0 </td> <td> 0.0 </td> <td> 1.0 </td> <td> 4.0 </td> <td> 8.0 </td> <td>12.0 </td> <td>14.0 </td> <td>12.0 </td></tr><tr></tr><tr> <th>time (s)</th> <td>8</td> <td>9</td> <td>10</td> <td>11</td> <td>12</td> <td>13</td> <td>14</td> <td>15</td></tr><tr> <th>position (m) </th> <td> 8.0 </td> <td> 4.5 </td> <td> 2.0 </td> <td> 0.5 </td> <td> 0.0 </td> <td> 0.0 </td> <td> 0.0 </td> <td> 0.0 </td></tr></tbody></table><br /><div>A table is only useful for a coarse overview of the relation between the variables. For example, we can see that between <i>t</i>=2 s and <i>t</i>=3 s, the position went from <i>x</i>=1.0 m to <i>x</i>=4.0 m. We can compute an <i>average velocity</i> over this time interval:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-Dhkj2aWycY8/UKvbzB0B48I/AAAAAAAAAFE/7hLV-5vYtxw/s1600/ex1_ave_velocity.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-Dhkj2aWycY8/UKvbzB0B48I/AAAAAAAAAFE/7hLV-5vYtxw/s1600/ex1_ave_velocity.png" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">So the <i>average</i> velocity on this interval was 3.0 m/s. However, this was during one of the intervals the rocket was firing. It would have been going slower at the beginning of the interval and faster at the end. The table does not give enough information for us to estimate the actual velocities.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">Another representation of the relation between time and position is with a graph.</div><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-Ee3abtAW2Uw/UKvdrWyfK_I/AAAAAAAAAFM/jRwqABj2E7Y/s1600/ex1_pos_time_graph.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="203" src="http://3.bp.blogspot.com/-Ee3abtAW2Uw/UKvdrWyfK_I/AAAAAAAAAFM/jRwqABj2E7Y/s320/ex1_pos_time_graph.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">The derivative defines a <i>new</i> variable, <i>dx</i>/<i>dt</i>. Although this looks like a fraction, we should really think of the entire symbol as the name of our new variable, the derivative. This variable measures the <b>rate of change</b> of position <i>x</i> with respect to time <i>t</i>. On the graph, this corresponds to the <b>slope</b> of the graph at each point.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">For example, if I were to consider the point at <i>t</i>=2 s and <i>x</i>=1 m, we might draw the tangent line and measure the slope. This slope is the derivative, which for this point corresponds to <i>dx</i>/<i>dt</i>=2.0 m/s. Notice that the derivative is a <b>velocity</b>, which is precisely what the derivative measures for position with respect to time. So we will use the variables <i>v</i> and <i>dx/dt</i> interchangeably.</div><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-eRQj7TWONPY/UKviY4ICCGI/AAAAAAAAAFk/WQk0DQLEDH8/s1600/ex1_pos_time_graph_tangent.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="203" src="http://3.bp.blogspot.com/-eRQj7TWONPY/UKviY4ICCGI/AAAAAAAAAFk/WQk0DQLEDH8/s320/ex1_pos_time_graph_tangent.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">We could compute the derivative for <i>every</i> point of the original relation. This new variable could be added to our table.</div><br /></div><table border="1" style="text-align: center;"><tbody><tr> <th>time (s)</th> <td>0</td> <td>1</td> <td>2</td> <td>3</td> <td>4</td> <td>5</td> <td>6</td> <td>7</td></tr><tr> <th>position (m) </th> <td> 0.0 </td> <td> 0.0 </td> <td> 1.0 </td> <td> 4.0 </td> <td> 8.0 </td> <td>12.0 </td> <td>14.0 </td> <td>12.0 </td></tr><tr> <th>velocity (m/s) </th> <td> 0.0 </td> <td> 0.0 </td> <td> 2.0 </td> <td> 4.0 </td> <td> 4.0 </td> <td>4.0 </td> <td>0.0 </td> <td>-4.0 </td></tr><tr> <th>time (s)</th> <td>8</td> <td>9</td> <td>10</td> <td>11</td> <td>12</td> <td>13</td> <td>14</td> <td>15</td></tr><tr> <th>position (m) </th> <td> 8.0 </td> <td> 4.5 </td> <td> 2.0 </td> <td> 0.5 </td> <td> 0.0 </td> <td> 0.0 </td> <td> 0.0 </td> <td> 0.0 </td></tr><tr> <th>velocity (m/s) </th> <td> -4.0 </td> <td> -3.0 </td> <td> -2.0 </td> <td> -1.0 </td> <td> 0.0 </td> <td> 0.0 </td> <td> 0.0 </td> <td> 0.0 </td></tr></tbody></table><br />But it would be better, just as with our original relation, to consider the relation as a graph. The figure below illustrates both graphs one above the other.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-1xT_3tHFB6I/UKvl6IFWZjI/AAAAAAAAAF0/LZvfWvO7Aho/s1600/ex1_pos_vel_time_graph.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="http://1.bp.blogspot.com/-1xT_3tHFB6I/UKvl6IFWZjI/AAAAAAAAAF0/LZvfWvO7Aho/s320/ex1_pos_vel_time_graph.png" width="301" /></a></div><br />Notice that the velocity graph itself <i>also</i> has a slope at (nearly) every point. The slope of a velocity graph is also a rate of change, measuring how the rate of change of velocity (m/s) with respect to time (s). Consequently, the derivative <i>dv</i>/<i>dt</i>, which is called <b>acceleration</b> has units (m/s)/s, or more directly m/s<sup>2</sup>. (If velocity was measured in miles per hour, then acceleration might be measured as mph/s.) We can use the variable name <i>a</i>=<i>dv</i>/<i>dt</i>.<br /><br />Because our velocity is defined as <i>piecewise linear</i>, the acceleration will be <i>piecewise constant</i>. On intervals where the velocity was constant, the acceleration will be zero. <br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-qgI1P3uQiig/UKvuFlLz5GI/AAAAAAAAAGE/XDEDyVl2VQU/s1600/ex1_accel_time_graph.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="185" src="http://2.bp.blogspot.com/-qgI1P3uQiig/UKvuFlLz5GI/AAAAAAAAAGE/XDEDyVl2VQU/s320/ex1_accel_time_graph.png" width="320" /></a></div><br />Notice that the acceleration is directly related to our original discussion of the rocket-blasters. When the rockets are blasting to the right, the acceleration is positive; when the rockets are blasting to the left, the acceleration is negative. In fact, this is exactly the idea behind Newton's second law of mechanics, <i>F</i>=<i>ma</i>. The rocket's thrust corresponds to the force <i>F</i>. Newton's law simply states that the acceleration is proportional to the force. The mass, which measures inertia, provides the proportionality constant. If we had the exact same rockets and the car had twice the mass, then the acceleration would be cut in half.<br /><br />Because our rocket-car does not have brakes, the only way to slow down is to turn on the rockets in the opposite direction of motion. In the language of derivatives, the acceleration must have the opposite sign from the velocity. If the acceleration is the same sign as the velocity, then the effect of acceleration is an increase in the <i>speed</i> (the magnitude of velocity).<br /><br />We close this reading by considering the effect of acceleration on the graph of position. Slowing down corresponds to making the slope closer to horizontal. Speeding up corresponds to making the slope steeper. So, let us look at the original graph of position as a function of time, but marking the graph with different colors, depending on how the car is accelerating.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-DkoIeXsOAyI/UKv3CYTjCTI/AAAAAAAAAGU/JIn5_120T3Q/s1600/ex1_pos_time_graph_sliced.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="210" src="http://2.bp.blogspot.com/-DkoIeXsOAyI/UKv3CYTjCTI/AAAAAAAAAGU/JIn5_120T3Q/s320/ex1_pos_time_graph_sliced.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-DxgpN148Q3M/UKv3CqURBvI/AAAAAAAAAGY/2t84BW8GOEE/s1600/ex1_vel_accel_signs.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="134" src="http://3.bp.blogspot.com/-DxgpN148Q3M/UKv3CqURBvI/AAAAAAAAAGY/2t84BW8GOEE/s320/ex1_vel_accel_signs.png" width="320" /></a></div>When the velocity and acceleration are opposite, I have marked the graph in red. Notice that this is when the graph is becoming closer to horizontal (left-to-right). When the velocity and acceleration are the same direction, I have marked the graph in green. Notice that this is when the graph is becoming steeper (left-to-right). When there is no acceleration, I have marked the graph in purple (straight lines).<br /><br />To visualize this relative to the original animation, I have color-coded the timer and have placed a colored-flag on the car. Try to connect the information in the graph to the visualization of the moving rocket-car.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-mb9mtoXzL28/UKv9tc0QUTI/AAAAAAAAAGs/PS-KCqmI08Y/s1600/ex1_rocket_tracks_flag.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-mb9mtoXzL28/UKv9tc0QUTI/AAAAAAAAAGs/PS-KCqmI08Y/s1600/ex1_rocket_tracks_flag.gif" /></a></div><br /><b>Concavity</b> is the word that describes how a graph bends. When the second derivative (acceleration) is positive, the graph will be <b>concave up</b>. On our graph, this corresponds to the <i>first</i> green segment, <i>t </i>∈ (1,3), and the <i>last</i> red segment, <i>t </i>∈ (8,12). When the second derivative is negative, the graph will be <b>concave down</b>. This corresponds to the middle red and green segments, <i>t </i>∈ (5,7).Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-51176964390470810142012-11-13T09:46:00.000-08:002012-11-13T10:07:49.376-08:00Great Circles and Non-Euclidean GeometryA few weeks ago, a few of my children were playing on the computer with <a href="http://www.google.com/earth/index.html">Google Earth</a>. They wanted to see the aerial photographs of some our previous residences. Then one of them wanted to measure exactly how far away we moved, when we moved from <a href="http://www.cityofmlt.com/">Mountlake Terrace, Washington</a>, to <a href="http://www.harrisonburgva.gov/">Harrisonburg, Virginia</a>. So he drew a path connecting our previous address to our current address.<br /><br />What jumped out at me was not the distance, but the heading—90.4 degrees, or almost exactly due east. I was puzzled. Wasn't Virginia at a lower latitude than Washington? So how could we be exactly east of Virginia. Then I realized that Google Earth is calculated distance using great circles.<br /><br />Here is a picture of the path. Notice how the path bends.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-AVL1hKvDd68/UKJj8McXPrI/AAAAAAAAADo/PymMpYJdBnU/s1600/GreatCircle_WAtoVA.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="http://1.bp.blogspot.com/-AVL1hKvDd68/UKJj8McXPrI/AAAAAAAAADo/PymMpYJdBnU/s1600/GreatCircle_WAtoVA.png" width="315" /></a></div><br />A great circle is a path on the surface of a sphere such that the center of the sphere is the center of the circle. My first instinct when I read the heading was 90 degrees was that the path followed a line of constant latitude. That is, I would follow the compass heading of "East", keeping the north pole always exactly to my left. The image (coming from <a href="http://commons.wikimedia.org/wiki/File:Sphere_filled_blue.svg">Wikimedia Commons</a>) below illustrates these lines as being "parallel" to the equator. It also illustrates lines of constant longitude, which go from pole to pole.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://upload.wikimedia.org/wikipedia/commons/7/70/Sphere_filled_blue.svg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="200" src="http://upload.wikimedia.org/wikipedia/commons/7/70/Sphere_filled_blue.svg" width="200" /></a></div>The equator and lines of constant longitude are examples of great circles. But the lines of constant latitude (other than the equator) are not, because their centers are shifted away from the center of the sphere.<br /><br />How can you get other great circles? One way is to choose any point on the surface of the sphere. Then take a line and go through the center of the earth to the diametrically opposite point. (<b><a href="http://www.etymonline.com/">Etymology</a></b>: The word "diametrically" comes from the word "diameter;" great circles always have the same diameter as the sphere on which they are constructed.) These two points will be on longitude lines exactly 180 degrees apart, which together form a great circle. You can then hold your two points constant and rotate that great circle around the axis you created by joining the two points.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-e2RWZEherTg/UKKHAt4NezI/AAAAAAAAAEI/-vxU6HpODWA/s1600/great_circle.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-e2RWZEherTg/UKKHAt4NezI/AAAAAAAAAEI/-vxU6HpODWA/s1600/great_circle.gif" /></a></div><b>Trivia</b> (also known as a <b>Cool Math Fact</b>): Any great circle containing a point must include its diametrically opposite point.<br /><br /><h3>Non-Euclidean Geometry</h3>Now, let's talk about distance. If you take <i>any</i> two points on the surface of the sphere, we want to find the shortest path between the two points. In the geometry of the Greek mathematician and geometer Euclid, the shortest distance between two points is a line. But what happens when we are constrained to paths on the surface of a sphere?<br /><br />Imagine that you actually want to measure distance. You might do this by taking a piece of string (on the earth, this would be a very long string) and lay it down along your path. On this string, you could mark units of measure (like a measuring tape). When you arrive at your destination, you can just read off the distance of your path. Wildly wandering paths would clearly involve a longer distance.<br /><br />Given your path, you might want to see if you can improve it. You could do this by trying to pull your string tighter, and sliding it around on the surface to see if you can use any less string to reach your point. We have to imagine that the surface offers no friction, so that the string naturally slides toward a better path if it is available. A path that can not be improved corresponds to an optimal path.<br /><br />If we did our experiment of minimal paths on a flat table, then the optimal paths would follow Euclid's prediction of what we think of as straight lines. But on a sphere, optimal paths <b>always follow great circles</b>. The idea of non-Euclidean geometry is to consider lines not by our usual sense of straightness (whatever we think that means) but in terms of minimal paths.<br /><br />In geometry, points and lines are basic elements, meaning they can not be defined from more elementary objects. Euclid's geometry, which is our classical geometry based on what we think of as straight lines, is based on the premise that given any two points, you can use a straight edge (ruler) to draw a straight line. And given any finite line segment, you can extend this line indefinitely. But where does this infinite straight edge come from?<br /><br />In non-Euclidean geometry, all we ever get are short straight edges. We can only extend a line a little bit at a time, although we mathematically assume that we can do this perfectly, without any errors in aligning the straight edge to the rest of the line.<br /><br />In Euclidean geometry, using a short straight edge repeatedly is identical to using an infinite straight edge. But if we do our geometry on a curved surface (instead of a perfectly flat infinitely large plane), then using our short straight edge follows a path that <i>locally</i> looks straight. But it has to follow the curving of the surface so that it does not look anything like Euclid's sense of a straight line. On the surface of a sphere, a straight line that is formed by extending a short line segment with our short straight edge over and over again will be a great circle.<br /><br /><h3>Where is this used?</h3>Great circles are used to plan shortest paths for airline flights. Talking about distance "as a bird flies" corresponds to paths along a great circle.<br /><br />But non-Euclidean geometry has an even more fundamental role in understanding physics and light. One of the basic principles of how light works is that it follows a path of least time (not distance). The idea of refraction (why a stick looks like it bends when immersed in water) is a direct consequence of the fact that light travels through different substances at different speeds (slower through water than through air). The amount of bending of the light's path can be exactly calculated by finding the path over which light takes the least time to traverse.<br /><br />Curiously, sometimes there are multiple paths over which the light minimizes its time. For example, the speed of light is faster in less dense air. If air is heated, it expands and decreases its density. So there are times when light can find a path with a greater physical distance but less time by choosing a path through warmer, less dense air. Ripples that form a mirage on a hot road in front of you are evidence of these multiple paths.<br /><br />Thinking about how light travels actually led to <a href="http://www.dummies.com/how-to/content/einsteins-general-relativity-theory-gravity-as-geo.html">Einstein's development of relativity</a>. His theory of general relativity considers how light would need to travel if it was in a box that was accelerating. By imagining that such a box is indistinguishable from a box that is subject to gravity, Einstein realized that light's path needs to bend as it passes through a gravitational field.<br /><br />But the best way to think about these paths is by considering that light is actually always traveling in a straight line, just that the space in which it travels has a constantly changing sense of what is straight. Further, Einstein realized that <a href="http://www.pbs.org/wgbh/nova/physics/spacetime-lemonade.html">space and time are inseparably connected</a>. That is, light is moving in what is called the <b>space–time continuum</b>. Gravity distorts this space–time continuum, and the way this is studied is through non-Euclidean geometry.<br /><br />By the way, the answer to our original question: 2,257 miles.Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-23611738126209946432011-09-14T21:07:00.000-07:002011-09-14T21:09:13.212-07:00Proof by Induction and SummationsThe two previous blog entries introduced the <a href="http://waltonsjmumathblog.blogspot.com/2011/09/mathematical-induction.html">idea of the Principle of Mathematical Induction</a> followed by a discussion of a <a href="http://waltonsjmumathblog.blogspot.com/2011/09/mathematical-induction-part-2.html">typical example of a proof by induction</a>. Be sure you read those two entries before you look at this one. This blog post extends these ideas by talking about how proof by induction applies to summations.<br /><br />(I apologize right now about the formatting of summation notation, or sigma notation. I do not know how to get the math to look like math in a blog setting.)<br /><br />Recall that the second condition of the PMI is that an arbitrary statement S(n) in the chain of statements being proved will guarantee that the next statement S(n+1) in the chain is also true. In a proof by induction, this step always involves using a recursive relation between something in the sentence S(n) with a corresponding object in the sentence S(n+1).<br /><br />For summations, this recursive relation is always that a summation for statement S(n+1) exactly corresponds to the summation appearing in statement S(n) with some additional term(s).<br /><br />For example, think back to our motivating example of a chain of statements:<br />S(1): 1 = 1(2)/2<br />S(2): 1+2 = 2(3)/2<br />S(3): 1+2+3 = 3(4)/2<br />S(4): 1+2+3+4 = 4(5)/2<br />S(5): 1+2+3+4+5 = 5(6)/2<br />...<br />Notice that the sum appearing on the left hand side of any given sentence appears in the sum on the very next sentence, but one more term has been added. Using parentheses to emphasize where the sum from the previous sentence appears in the new sentence, here is the same chain:<br /><br />S(1): 1 = 1(2)/2<br />S(2): (1)+2 = 2(3)/2<br />S(3): (1+2)+3 = 3(4)/2<br />S(4): (1+2+3)+4 = 4(5)/2<br />S(5): (1+2+3+4)+5 = 5(6)/2<br />...<br /><br />Summation (Sigma) notation gives us a handy way to describe sequences of sums like we see in the chain above. The <b>terms</b> of the sum follow a simple pattern. In this example, the pattern is the sequence of terms (1, 2, 3, 4, 5, ...). This sequence can be explicitly described as a<sub>k</sub> = k for k=1, 2, 3, .... In the symbolism (notation) of summation notation, we write:<br /><br />Σ<sup>1</sup><sub>k=1</sub>(k) = 1<br />Σ<sup>2</sup><sub>k=1</sub>(k) = 1+2<br />Σ<sup>3</sup><sub>k=1</sub>(k) = 1+2+3<br /> or<br />Σ<sup>n</sup><sub>k=1</sub>(k) = 1+2+3+...+n<br />(Notice that the formula in the parentheses (k) is the explicit formula of the sequence of terms in terms of the index which is listed at the bottom of Σ along with the first index of the sequence used in the sum. The index of the last term in the sum appears at the top of Σ.)<br /><br />The pattern that we described above to illustrate the recursive relation between the sums of consecutive sentences in the chain of statements we wish to prove can also be written in terms of summation notation:<br />Σ<sup>n+1</sup><sub>k=1</sub>(k) = Σ<sup>n</sup><sub>k=1</sub>(k) + (n+1)<br />When writing a proof by induction, we use this recursive relation.<br /><br /><b>Example:</b> Prove that Σ<sup>n</sup><sub>k=1</sub>(k) = n(n+1)/2 for n=1, 2, 3, ...<br /><br />In the proof, we will write S(n) to represent the sentence: Σ<sup>n</sup><sub>k=1</sub>(k) = n(n+1)/2.<br /><br /><b>Proof</b>:<br />(1) <i>We first need to think about the sentence where n is replaced by 1. That is, we need to prove that </i>Σ<sup>1</sup><sub>k=1</sub>(k) = 1(1+1)/2.<br />Σ<sup>1</sup><sub>k=1</sub>(k) = 1 (Interpretation of summation notation)<br /><span style="font-size: x-small;">(<i>We are creating a true statement involving the summation symbol that appears on the left side of the equation we are proving. The sequence of terms is 1, 2, 3, 4, 5, ..., and the sum starts <b>and </b>ends with the 1st term.</i>)</span><br />1(1+1)/2 = 1(2)/2 = 1 (Algebra)<br /><span style="font-size: x-small;">(<i>We now create an equation involving the formula on the right side of the equation we are proving, and we just showed it has the same value as the summation symbol.</i>)</span><i><br /></i><br />So Σ<sup>1</sup><sub>k=1</sub>(k) = 1(1+1)/2.<br /><span style="font-size: x-small;">(<i>We obtained evidence above that the summation and the formula both represented the same value, so we conclude that they are equal.</i>)</span><br /><br />(2) <i>We now need to show that S(n) implies S(n+1) for n=1, 2, 3, .... We will assume S(n), i.e. Σ<sup>n</sup><sub>k=1</sub>(k) = n(n+1)/2. Using the recursion connecting the sums, we will show S(n+1), i.e., Σ<sup>n+1</sup><sub>k=1</sub>(k) = (n+1)((n+1)+1)/2.</i><br /><br />Assume Σ<sup>n</sup><sub>k=1</sub>(k) = n(n+1)/2 for some n=1, 2, 3, ...<br />Σ<sup>n+1</sup><sub>k=1</sub>(k) = Σ<sup>n</sup><sub>k=1</sub>(k) + (n+1). (Recursive relation on summation of terms a<sub>k</sub> = k)<br />Σ<sup>n+1</sup><sub>k=1</sub>(k) = n(n+1)/2 + (n+1) (Substitution: summation replaced by formula)<br />Σ<sup>n+1</sup><sub>k=1</sub>(k) = n(n+1)/2 + 2(n+1)/2 (Find common denominator)<br />Σ<sup>n+1</sup><sub>k=1</sub>(k) = (n<sup>2</sup>+n+2n+2)/2 (Distribution and adding fractions)<br />Σ<sup>n+1</sup><sub>k=1</sub>(k) = (n<sup>2</sup>+3n+2)/2 (Combining terms)<br /><span style="font-size: x-small;">(<i>Notice that each equation was based on the previous equation. The key step was when we replaced Σ<sup>n</sup><sub>k=1</sub>(k) by the formula n(n+1)/2 that was provided by the assumed hypothesis. At this stage of the proof, we have a formula representing the value of the summation symbol with n+1 that is the left hand side of the equation in the sentence S(n+1). We now need to check the formula that appears in the right hand side of our sentence.</i>)</span><i></i><br />(n+1)((n+1)+1)/2 = (n+1)(n+2)/2 (Arithmetic: 1+1=2)<br />(n+1)((n+1)+1)/2 = [n(n+2)+1(n+2)]/2 (Distributive law)<br />(n+1)((n+1)+1)/2 = [n<sup>2</sup>+2n + n+2]/2 (Distributive law again.)<br />(n+1)((n+1)+1)/2 = [n<sup>2</sup>+3n+2]/2 (Combining terms) <br /><span style="font-size: x-small;">(<i>Notice that each equation was based on the previous equation. Notice that the second and third steps (distributive law) are the formal steps in the idea commonly taught as FOIL when multiplying to binomial expressions together. The key is that we took the formula with n replaced by n+1 of the right hand side and discovered that it equals the same formula that we found for the summation above.</i>)</span><br />So we have Σ<sup>n+1</sup><sub>k=1</sub>(k) = (n+1)((n+1)+1)/2.<br />Consequently, S(n) implies S(n+1) for n=1, 2, 3, ....<br /><br />By the PMI (since S(1) is true and S(n) implies S(n+1)), we have S(n) is true for n=1, 2, 3, ...<br />♦Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-77361350476887751142011-09-14T20:17:00.000-07:002011-09-14T20:17:03.786-07:00Mathematical Induction (part 2)My <a href="http://waltonsjmumathblog.blogspot.com/2011/09/mathematical-induction.html">previous blog post</a> introduced the basic idea of <b>what</b> mathematical induction is about. This post focuses more on the mechanics of writing a proof.<br /><br />A proof is the mathematical form of argument or persuasion. A proof consists of a sequence of logical statements, each of which is shown to be a <b>true</b> sentence based only on information previous to that sentence. Examples of sentences that might appear in a proof are <b>equations</b> or <b>inequalities</b> for which there is clear reason that it is true, and <i>never</i> based on what we hope is true or will later show is true.<br /><br />For example, suppose I needed to prove (x+1)<sup>2</sup>=x<sup>2</sup>+2x+1. In a proof, I can not write down this equation as the first statement because it is not something I know (yet). Instead, I can write down equations known to be true based on basic principles:<br /><ul><li>(x+1)<sup>2</sup> = (x+1)(x+1) (meaning of power 2)</li><li>(x+1)(x+1) = x(x+1) + 1(x+1) (distributive law)</li><li> x(x+1) + 1(x+1) = x<sup>2</sup>+x + x+1 (distributive law again)</li><li>x<sup>2</sup>+x + x+1 = x<sup>2</sup>+2x+1 (collecting like terms)</li></ul>Notice that each of the sentences is a statement of equality. Although they to suggest that the "=" sign is being used in a computational sense (i.e., it looks like it is being used to say apply a rule), we really are seeing these sentences as <b>declaring</b> that the result of applying a rule <b>demonstrates</b> that the sentence is actually true. (This is a subtle distinction that you need to fight your mind until it sinks in.) Technically, our proof is not yet complete. Each sentence on its own is a complete, true sentence. However, we need to end by stating that the sentence we were trying to prove is actually true:<br /><ul><li> (x+1)<sup>2</sup>=x<sup>2</sup>+2x+1 (equivalence of equal quantities, or substitution)</li></ul>Now, you should read the above paragraphs as illustrating the ideas of a proof in that it illustrates how sentences (equations) are listed as statements that are demonstrated to be true. But so far, we have not dealt with the idea of implication.<br /><br />Recall that the Principle of Mathematical Induction (PMI) involves verifying the two conditions:<br /><ol><li>Show that the first statement in the chain, which we call S(1), is true.</li><li>Show that if any single statement in the chain, which we call S(n), is true, then this implies the next statement, which will be S(n+1), is also true.</li></ol>So the structure of <b>every </b>proof by induction involves two subproofs (showing S(1) is true; showing S(n) implies S(n+1)) followed by an application of the PMI. The subproofs are the method we verify that the two conditions of the PMI have been satisfied.<br /><br />Here is what to look for in the subproofs.<br /><ol><li>S(1) is almost always very easy to show. However, you still need to be clear that you follow the pattern of a proof described above.</li><li>Showing S(n) implies S(n+1) will consist of <b>assuming</b> that S(n) is true <i>for some n</i> (don't forget that this is a specific but unspecified value, so you must leave n or k, depending on the label being used, as a symbol and not an actual number). Then the proof will almost always rely on some type of <b>recursive</b> relation between a quantity involved in the statement S(n) and a similar quantity in the statement S(n+1).</li><li>After the subproofs are complete, you must <b>invoke</b> the PMI and then declare that the statements are true for <b>every </b>n=1, 2, 3, ....</li></ol><b>Example</b>: Given a sequence defined by x1=3 and the recursive relation x<sub>k+1</sub>=x<sub>k</sub>+2 for k=1, 2, 3, ..., prove that x<sub>k</sub>=1+2k for k=1, 2, 3, ....<br /><br />Notice what the chain of statements we are trying to prove will be. The sentence S(k) is the statement x<sub>k</sub>=1+2k, where x<sub>k</sub> is the value of the sequence defined recursively, and 1+2k is just a formula involving k. It is very important to remember that the "=" sign in this equation is <b>not</b> defining the value of the sequence, but is simply stating that the two quantities (the sequence value and the formula value) happen to always be equal. <br /><br />Everything in italics is not part of the proof.<br /><br /><b>Proof:</b><br /><b>(1)</b> <i>We first write a subproof that S(1) is true: </i>x<sub>1</sub>=1+2(1). <i>To do this, we need to create a sequence of equations that we <b>know</b> are true based on the symbols themselves, and not the equation above.</i><br /><br />x<sub>1</sub>=3 (Given) <br /><span style="font-size: x-small;">(<i>We looked at the given information and saw this was provided.</i>)</span><br />1+2(1) = 1+2 = 3 (Arithmetic) <br /><span style="font-size: x-small;">(<i>We needed an equation involving </i>1+2(1)<i> so we wrote down an equation that was based on the rules of evaluating this formula.</i>) </span><br />So x<sub>1</sub>=1+2(1). (Equivalence) <br /><span style="font-size: x-small;">(<i>Earlier, we showed </i>x<sub>1</sub>=3<i> and then we showed </i>1+2(1)=3<i>, so this means the two quantities are equal. This ends the subproof since we just finished writing the statement corresponding to </i>S(1)<i> being true.</i>)</span><br /><br /><b>(2)<i> </i></b><i>We next write a subproof that S(k) implies S(k+1) for k=1,2,3,...</i> <i>To do this, we will start by <b>assuming</b> S(k) is true for some unspecified k. Using the recursion to compute </i>x<sub>k+1</sub> <i>we will demonstrate that S(k+1) is also true. Note: S(k+1) is the statement x<sub>k+1</sub>=1+2(k+1)</i>.<br /><br />Assume x<sub>k</sub>=1+2k is true for some k=1, 2, 3, ....<br />x<sub>k+1</sub>=x<sub>k</sub>+2 (Given recursive definition of sequence)<br /><span style="font-size: x-small;">(<i>The statement we are trying to prove involves the symbol </i>x<sub>k+1</sub><i> so we need to create equations based on that symbol.</i>)</span><br />x<sub>k+1</sub>=(1+2k)+2 (Substitution: x<sub>k</sub>=1+2k from assumed hypothesis)<br /><span style="font-size: x-small;">(<i>This is the key step: we use the recursive connection between </i>x<sub>k+1</sub><i> and</i> x<sub>k</sub> <i>to establish a formula for </i>x<sub>k+1</sub>.)</span><br />x<sub>k+1</sub>=3+2k (Algebra) <br /><span style="font-size: x-small;">(<i>We now have a simple formula for </i>x<sub>k+1</sub><i> and now we turn our attention to the other half of the statement S(k+1), namely the formula</i> 1+2(k+1).)</span><br />1+2(k+1) = 1 + 2k + 2 (Distributive law)<br /><span style="font-size: x-small;">(<i>We are creating a true equation involving the formula by considering the results of applying mathematical laws.</i>)</span><br />1+2(k+1) = 3+2k (Algebra from above: 1+2=3) <br /><span style="font-size: x-small;">(<i>Our target should always be that the formula with k+1 in place of k will match whatever we found by the recursion to compute a value for </i>x<sub>k+1</sub>.)</span><br />So x<sub>k+1</sub>=1+2(k+1) (Equivalence or Substitution)<br />That is, S(k) implies S(k+1) for k=1, 2, 3,.... <br /><span style="font-size: x-small;">(<i>This ends the second subproof because we just finished writing the statement we were trying to prove in this part. We are now ready to invoke the PMI.</i>)</span><br /><br />Since S(1) is true and S(k) implies S(k+1) for k=1, 2, 3, ..., the Principle of Mathematical Induction guarantees S(k) is true for every k=1, 2, 3, .... That is, x<sub>k</sub>=1+2k for k=1, 2, 3, ...<br />♦<br /><ul></ul>Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-817283519829475962011-09-14T19:14:00.000-07:002011-09-14T20:18:57.168-07:00Mathematical InductionI actually posted on this very topic <a href="http://waltonsjmumathblog.blogspot.com/2008/10/mathematical-induction.html">a few years ago.</a> However, I have refined how I think about doing proofs by mathematical induction. And so I am writing one more time.<br /><br />First, you might find it interesting to look at the Wikipedia article on <a href="http://en.wikipedia.org/wiki/Mathematical_induction">mathematical induction</a>.<br /><br />The Principle of Mathematical Induction (PMI) is an axiom that describes the set of natural numbers, which is N = {1, 2, 3, 4, ...}. From one point of view, the PMI says that if S is a set that (1) contains the number 1 and (2) guarantees that whenever any number n is in the set, then n+1 is also in the set, then the set S contains all of N. From the perspective of proofs, however, we are really interested in showing that an infinite chain of logical statements is true.<br /><br />Before we proceed with the idea of the PMI, let me be clear about a logical statement. A logical statement is a well-constructed sentence that is definitively true or false. Two well-known but easily misunderstood examples of logical statements are <b>equations</b> and <b>inequalities</b>. An equation "A=B" is a logical statement that declares two quantities (A and B) are equal (have the same value). An inequality "A<b" (a="" 1="2"" `truth="" a="" b). ="" certain="" declares="" example,="" for="" have="" is="" less="" logical="" natural="" of="" order="" quantities="" statement. ="" statement="" than="" that="" the=""><b>false</b>, but the statement itself is logically complete. (Here logical does not mean `makes sense' but it means `can be decided between True or False'.)</b"><br /><br />Related to this issue is a common misunderstanding by students that "=" is like an operation that means "has the value" or "find the answer" as it is often used on a calculator. This is especially important in a proof, where each statement (usually an equation) must be clearly true rather than a statement of what you hope or what you are trying to calculate.<br /><br />Okay, back to the Principle of Mathematical Induction. This principle is about dealing with an infinite chain of logical sentences. For example, consider the following chain of statements. S(1) is the label for the first sentence, S(2) is the label for the second sentence, and so on:<br /><br />S(1): 1 = 1(2)/2<br />S(2): 1+2 = 2(3)/2<br />S(3): 1+2+3 = 3(4)/2<br />S(4): 1+2+3+4 = 4(5)/2<br />S(5): 1+2+3+4+5 = 5(6)/2<br />S(6): 1+2+3+4+5+6 = 6(7)/2<br />...<br />(the pattern continues so that if n is any number n=1,2,3,..., we have):<br />S(n): 1+2+3+...+n = n(n+1)/2<br /><br />On the left of each sentence is a summation. On the right of each sentence is a simple formula involving n. Using summation notation, we would have written this sentence:<br />S(n): Σ<sup>n</sup><sub>k=1</sub> k = n(n+1)/2<br /><br />The pattern described by S(n) looks like a single sentence, but it is important to remember that it is describing the entire infinitely long chain of logical sentences. If you add up the values on the left side of any single sentence and compare it to the value of the formula on the right, you will see that the answers are the same. But this only verifies the formulas that you actually check.<br /><br />The Principle of Mathematical Induction is a tool to prove that the <b>entire chain</b> of sentences are all true. The PMI is much like an infinite chain of dominoes (see the earlier linked wikipedia article). To knock over a chain, if you knock them down one at a time, you'll never finish. But if you show that knocking <b>any</b> single domino down will guarantee the next domino also falls <b>and </b>you show that you can knock down the first domino, then this is enough to guarantee that the entire chain will fall down.<br /><br />So here is the PMI:<br /><blockquote>Suppose S(n), n=1, 2, 3, ..., is a chain of logical statements and that (1) S(1) is true and (2) S(n) implies S(n+1) for any n=1, 2, 3, ..., then S(n) is true for every n=1, 2, 3, ....</blockquote>Notice that there are two conditions to use PMI:<br /><ol><li> We must verify that S(1) (the first sentence) is true.</li><li>We must show that the sentence S(n+1) is true whenever the previous sentence S(n) is true.</li></ol>A proof by induction is the process whereby we verify these two conditions and then apply the PMI to conclude that the entire chain is true.<br /><br />To learn more about writing the proofs and an example with commentary, please continue by reading the <a href="http://waltonsjmumathblog.blogspot.com/2011/09/mathematical-induction-part-2.html">next blog post</a>.Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-68444399517460258652011-03-08T09:44:00.000-08:002011-03-09T06:15:28.487-08:00How We Learn MathematicsI was reading the following paper: D. Breidenbach, E. Dubinsky, J. Hawks, and D. Nichols, "Development of the Process Conception of Function," <span style="font-style: italic;">Educational Studies in Mathematics</span>, <span style="font-weight: bold;">23</span>: 247-285, 1992.<br /><br />Quote Dubinsky (1989): "A person's mathematical knowledge is her or his tendency to respond to certain kinds of perceived problem situations by constructing, reconstructing and organizing mental processes and objects to use in dealing with the situations."<br /><br />"Applying this point of view to mathematics (or any other subject) consists of determining the nature of the specific processes and objects that are constructed and how they are organized when one studies mathematics"<br /><br />Ways of thinking about functions:<br /><ul><li>prefunction - does not understand any real ways of using function concepts</li><li>action - repeatable mental or physical manipulation (e.g., plug in numbers and calculate); static; one step at a time</li><li>process - think of function as a single dynamic transformation</li></ul>I then found another article: A. Sfard and L. Linchevski, "The Gains and the Pitfalls of Reification: The Case of Algebra," <span style="font-style: italic;">Educational Studies in Mathematics,</span> <span style="font-weight: bold;">26</span> (2/3), 191-228, 1994 [Learning Mathematics: Constructivist and Interactionist Theories of Mathematical Development]<br /><br />This article proceeds with the view that in mathematics, there is a duality in mathematical constructs being a process or an object. That is, conceive of things operationally (process) or structurally (object). Historical examples include the expansion of number systems: positive to negative (operational: subtraction as adding a negative to structural: negative numbers as objects), and real to complex (i=sqrt(-1) as an operational convenience to an actual object)<br /><br />An included reference suggests finding another article: Kieran, C.: 1992, 'The learning and teaching of school algebra', in D. A. Grouws (ed.), The Handbook of Research on Mathematics Teaching and Learning, Macmillan, New York, pp. 390-419. I'll have to see if I can find this one, as it is cited for the sentence, "[reification] was also used to introduce some order into the quickly growing bulk of findings about algebraic thinking."<br /><br />Interesting phrase: "the ability to grasp the structural aspect is not easy to achieve" and "those crucial junctions in the development of mathematics where a transition from one level to another takes place are the most problematic."<br /><br />Another interesting way to think about how mathematics is organized: (1) Logical, or the way it fits together; (2) Historical, or the way in which it was developed; and (3) Cognitive, or the processes in which people learn.<br /><br />Modes of Algebra<br />1.1) Algebra as Generalized Arithmetic: The Operational Phase<br />-- solve for the unknown, but not using symbols (grade school algebraic thinking)<br />-- rhetoric algebra<br />-- principally reversing processes<br />1.2) Algebra as Generalized Arithmetic: The Structural Phase<br />1.2.1) algebra of a fixed value (unknown)<br />-- Notational convenience, but treat variable as a fixed value<br />:::: becomes a mental challenge to think of formula as both a process and result<br />:::: example given: 2+3 represents process, 5 represents result. But x+3 represents both, no separate "result"<br />:::: compare to the challenges of new number types required to think about division, subtraction, and extracting square roots<br />** Nice comment: "Once we manage to overcome this difficulty, it is quickly forgotten. ... Our eyes are easily blinded by habit and by our own ontological beliefs. Nevertheless, much evidence for the difficulty of reification may also be found in today's classroom, provided those who listen to the students are open-minded enough to grasp the ontological gap between themselves and the less experienced learners."<br />1.2.2) Functional algebra (of a variable)<br />:::: View formula as object<br />:::: Parameters represented as symbols not numbers.<br />2) Abstract Algebra<br /><br />Give examples of interview questions. Students at early stages of thinking think about formulas as recipes for computations (process) but do not perceive them as valid objects. "The equality sign is interpreted as a 'do something signal' (Behr et al 1976; Kieran 1981)"<br /><br />Here's something I see all the time in calculus classes: "It [the = symbol] serves here as a 'run' command. When treated in this way, the equality symbol looses [sic] the basic characteristics of an equivalence predicate: it stops being symmetrical or transitive. Indeed, young children seem to have no qualms about solving word problems with the help of a chain of non-transitive equalities. For instance, when asked 'How many marbles do you have after you win 4 marbles 3 times and 2 marbles 5 times?', the child would often write: 3*4=12+5*2=12+10=22."<br /><br />Equations of the form 2x-3 = 11 can be interpreted as a formula whose result is 11 (which can be solved by inverse operations); equations of the form 2x-3=5x-9 appear to be two different formulas, and inverse operations do not make sense.Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-37927258182935272032010-04-28T19:57:00.001-07:002010-04-28T20:07:05.365-07:00Graph from a graph of f ' (x)First, pay attention: the graph provided on the assignment is the <span style="font-weight: bold;">graph of the derivative</span> f '(x) and not the graph of f. So you can't look at the picture and say that because the graph you are looking at is increasing that f '(x) is positive; if the graph is increasing, then that means f '(x) is increasing, and not f(x). (This is useful information, but you just need to think about what it does say.)<br /><br />Second, the number line sign analysis summaries will help identify the shape of the graph. Imagine taking the unit circle and breaking it up according to quadrants. The signs of f '(x) and f ''(x) determine which of these four basic shapes the graph is most like.<br /><ul><li>f '(x) = + and f ''(x) = + means f(x) looks like Quadrant IV (incr, conc. up)<br /></li><li>f '(x) = - and f ''(x) = + means f(x) looks like Quadrant III (decr, conc. up)</li><li>f '(x) = + and f ''(x) = - means f(x) looks like Quadrant II (incr, conc. down)</li><li>f '(x) = - and f ''(x) = - means f(x) looks like Quadrant I (decr, conc. down)</li></ul>The graph is just formed by taking these shapes and putting them end-to-end. You wouldn't actually use the entire portion of the unit circle because we probably don't want vertical tangents like the unit circle has. The circle just helps us remember the basic shape. The points where we join the shapes together will probably be inflection points (concavity changes) or extreme values.<br /><br />However, sign analysis does not tell us the heights of any points. The problem gives only one point: f(0) = 1. The rest of the points of interest (especially the local extreme values) can be found by thinking about the information relating to the areas of the graph of f '(x). (Again, think about the Fundamental Theorem of Calculus).Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-183734197392582862010-04-28T19:47:00.001-07:002010-04-28T19:57:03.450-07:00Sums of Geometric SequencesThe second problem on the project introduces a new closed form for a sum:<br /><br />∑<sub>k=1</sub><sup>n</sup> [A ρ<sup>k</sup>] = A ρ (ρ<sup>n</sup>-1)/(ρ-1).<br /><br />Unfortunately, too many of you are still intimidated simply by the symbols that are used.<br /><br />The formula for the geometric sequence, A ρ<sup>k</sup>, is like an exponential, except the power is an integer variable rather than a continuous variable like x. For example, if A=2 and ρ=1/3, we have terms that are increasing powers of (1/3) times 2:<br />2/3 (k=1), 2/9 (k=2), 2/27 (k=3), 2/81 (k=4), etc.<br /><br />The summation is simply the sum of these values:<br />∑<sub>k=1</sub><sup>n</sup> [2 (1/3)<sup>k</sup>] = 2/3 + 2/9 + 2/27 + 2/81 + ... + 2/3<sup>n</sup>.<br />The closed form gives a formula answering the value of this sum:<br />2(1/3)[(1/3)<sup>n</sup>-1]/[(1/3) - 1] = (2/3)*[(1/3)<sup>n</sup>-1]/(-2/3) = 1 - (1/3)<sup>n</sup>.<br /><br />So when you write down the Riemann sum for the integral in question, you need to look at using the properties of exponentials so that the Riemann sum looks just like a sum of a geometric sequence. You should identify a factor that does not involve k, and this is A. You should identify the other factor as some number raised to the k power. Then you can use the closed form.Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-23192340955405553042010-04-28T19:25:00.000-07:002010-04-28T19:46:56.884-07:00Populations, Birth Rates, and Death RatesThis entry is a general assist for my class working on a project. Suppose you knew the rate at which births are occurring (call it a function of time, b(t)) and you knew the rate at which deaths are occurring (a function d(t)). If the only way the population changes is through births and deaths, then if P(t) is the function describing the size of the population in time, then P'(t) = b(t) - d(t). (It is still your job to explain why this makes <span style="font-weight: bold;">biological</span> sense.)<br /><br />Okay, now for the general principle. Anytime you know the rate of change of a quantity, you can always get back to the original quantity through a definite integral (assuming the rate of change is continuous, anyway). This is the heart of the 2nd Fundamental Theorem of Calculus. Not using P and t as variables (so that you have at least something to translate), here is the basic idea.<br /><br />Suppose you know f '(x). Then A(x) = ∫<sub>0</sub><sup>x</sup> f '(z) dz is an antiderivative of f '. But so is f(x) since that is where f '(x) comes from. That is f(x) = A(x) + C for some constant. In particular, A(0) = 0, so C=f(0). That is, f(x) = f(0) + ∫<sub>0</sub><sup>x</sup> f '(z) dz.<br /><br />This will always work, even if I don't start the integral at 0: f(x) = f(a) + ∫<sub>a</sub><sup>x</sup> f '(z) dz. Written another way, it looks like the first Fundamental Theorem of Calculus: f(x) - f(a) = ∫<sub>a</sub><sup>x</sup> f '(z) dz.<br />In other words, the 2nd FTC implies that <span style="font-weight: bold;">every</span> function is its starting value plus the integral of its rate of change.<br /><br />Now, for our population problem, we don't actually know the rate of change completely; we only know the value at specific points. So instead of computing an integral (to get an exact value), we will approximate the integral using a Riemann sum. We are restricted to using the table data, so Δt=2 is forced upon us. For example, ∫<sub>0</sub><sup>2</sup> b(t) dt can only be estimated with a single rectangle while ∫<sub>0</sub><sup>4</sup> b(t) dt would involve two rectangles. The idea of the Riemann sum is that we choose b(t<sub>k</sub>*) as one of our data points (either on the left or right).<br /><br />More specifically, on the interval [0,2] (k=1), we can either use t<sub>1</sub>*=0 so that b(t<sub>1</sub>*)=100 or use t<sub>1</sub>*=2 so that b(t<sub>1</sub>*)=135. In the first case, the rectangle for k=1 contributes b(t<sub>1</sub>*)Δt = 200, while the second case leads to a contribution of b(t<sub>1</sub>*)Δt = 270. The average (midpoint) of these two values (200+270)/2 = 235 is the estimate that would come from using the trapezoid sum. We do this for each of the 8 intervals between our data points, for both the births and the deaths.<br /><br />By considering our estimates for the number of births and deaths in each of the intervals ([0,2], [2,4], [4,6], etc.), we can produce an estimate of the new population at each of the times (2, 4, 6, 8, etc.). By thinking about what estimates lead to the largest predicted population, we get an upper limit (i.e. bound) for our estimate --- no population consistent with this data can ever go above that value. Similarly, we can choose those estimates to create a lower bound. The true population will be somewhere in between.Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-68782843706886347812010-02-05T05:32:00.000-08:002010-02-05T07:30:11.973-08:00Exponential FunctionsI'm getting feedback that exponential functions are giving you extra trouble. I'd appreciate getting feedback to help know how I can clarify the concepts. Here is a summary of some of the key concepts that I'm wanting you to understand:<br /><ul><li>b<sup>x</sup> is not just a formula "b to the power x" but is a new function, which I'm asking you to call exp<sub>b</sub>(x).<br /></li><li>The properties of exponents like b<sup>x+y</sup>=b<sup>x</sup> b<sup>y</sup> and (b<sup>x</sup>)<sup>y</sup>=b<sup>xy</sup> become properties of the exponential functions.<br /><div style="text-align: center;"> exp<sub>b</sub>(x+y)=exp<sub>b</sub>(x)*exp<sub>b</sub>(y)</div><div style="text-align: center;"> exp<sub>b</sub>(xy)= [exp<sub>b</sub>(x)]<sup>y</sup></div></li><br /><li>Logarithms are the inverse functions of exponential functions.<br /><div style="text-align: center;">exp<sub>b</sub>(log<sub>b</sub>(x))=x<br />log<sub>b</sub>(exp<sub>b</sub>(x))=x</div><br />In formula representation, these are written as follows:<br /><div style="text-align: center;">b<sup>log<sub>b</sub>(x)</sup>=x<br />log<sub>b</sub>(b<sup>x</sup>)=x</div></li><li style="text-align: left;">Whenever you see a formula with an exponential, say b<sup>3x-2</sup>, you should be able to think in both formula and function modes interchangeably.<br /><div style="text-align: center;">b<sup>3x-2</sup> = b<sup>3x</sup> b<sup>-2</sup> = (b<sup>3</sup>)<sup>x</sup> b<sup>-2</sup></div><div style="text-align: center;">b<sup>3x-2</sup> = exp<sub>b</sub>(3x-2)</div><div style="text-align: left;">The first mode allows us to recognize that b<sup>3x-2</sup>t is actually of the form Aq<sup>x</sup> where A=b<sup>-2</sup> and q=b<sup>3</sup>. The second mode is useful to remind us that we really have a composition when we need to compute a derivative or when dealing with inverse functions.</div></li><li style="text-align: left;">There is a special base e that is most important because the corresponding exponential function is its own derivative.<br /><div style="text-align: center;">exp'<sub>e</sub>(x)=exp<sub>e</sub>(x)<br /></div><div style="text-align: center;">d/dx[e<sup>x</sup>]=e<sup>x</sup></div><br />The natural exponential is written without a base. The corresponding inverse function is called the natural logarithm, ln(x).<br /><div style="text-align: center;">exp(ln(x))=x<br />ln(exp(x))=x</div><br />In formula representation, these are written as follows:<br /><div style="text-align: center;">e<sup>ln(x)</sup>=x<br />ln(e<sup>x</sup>)=x</div>The x in these formulas, as always, is a placeholder. So any number or formula could be used in place of x.</li><li style="text-align: left;">Any exponential can be written in terms of the natural exponential. The key is to use the properties of exponents.<br /><div style="text-align: center;">b = e<sup>ln(b)</sup><br />b<sup>x</sup> = [e<sup>ln(b)</sup>]<sup>x</sup> = e<sup>ln(b) </sup><sup>x</sup><br /></div>Another way to think of this is using composition of inverse functions: exp(ln( ))<br /><div style="text-align: center;">b<sup>x</sup> = exp(ln(b<sup>x</sup>))<br />ln(b<sup>x</sup>) = x ln(b)<br />b<sup>x</sup> = exp(x ln(b)) = e<sup>x ln(b)</sup></div><br />That is, by replacing b<sup>x</sup> by e<sup>ln(b) </sup><sup>x</sup>, we can write any exponential A b<sup>x</sup> in the form A e<sup>kx</sup> where k is the number ln(b).</li><li style="text-align: left;">Derivatives of exponentials use the basic property that exp'(x) = exp(x), or d/dx[e<sup>x</sup>] = e<sup>x</sup>. Usually, this also requires the chain rule: d/dx[e<sup>u</sup>] = e<sup>u</sup> u'.<br /><div style="text-align: center;">d/dx[e<sup>2x</sup>] = e<sup>2x</sup> (2) = 2e<sup>2x</sup> (u = 2x)<br />d/dx[e<sup>-3x</sup>] = e<sup>-3x</sup> (-3) = -3e<sup>-3x</sup> (u = -3x)<br />d/dt[e<sup>(t<sup>2</sup>-5t)</sup>] = e<sup>(t<sup>2</sup>-5t)</sup> (2t-5) = (2t-5)e<sup>(t<sup>2</sup>-5t)</sup> (u = t<sup>2</sup>-5t)<br /></div></li></ul><br />So this was a moderately long list. Perhaps just re-reading it helped you understand something better. Or perhaps you realize something is still confusing. Please post comments explaining why you are finding problems challenging or perhaps explaining what helped you suddenly understand what you had missed earlier.Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com8tag:blogger.com,1999:blog-4156201020188714381.post-67370431463217791012009-11-17T12:11:00.001-08:002009-11-17T12:16:56.582-08:00Concerns about DerivativesOkay, I have just finished grading HW #4. This assignment had some differentiation rules. But many of you are not comprehending the purpose of a derivative rule.<br /><br />For example, we had the special rule for squares of functions:<br /><br />d/dx[f<sup>2</sup>(x)] = 2 f(x) f '(x)<br /><br />This means that any time there is a formula squared and you need to take its derivative, you can apply this rule.<br /><br />(2x+3)<sup>2</sup> corresponds to the function f(x)=2x+3 being squared. So since f '(x) = 2:<br /><br />d/dx[(2x+3)<sup>2</sup>] = 2 (2x+3) (2) = 4(2x+3)<br /><br />Similarly, (x<sup>2</sup>-4x+5)<sup>2</sup> corresponds to the function f(x) = x<sup>2</sup>-4x+5 being squared, and f '(x)=2x-4<br /><br />d/dx[(x<sup>2</sup>-4x+5)<sup>2</sup>] = 2 (x<sup>2</sup>-4x+5) (2x-4)<br /><br />You need to be an expert at identifying the <span style="font-weight: bold;">form</span> of an expression in order to apply appropriate rules of differentiation, and next semester, rules of integration (anti-differentiation).Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-58420715009704360782009-09-03T08:31:00.000-07:002009-09-03T08:49:48.764-07:00Science and MathematicsThe other day, I had my students respond to a question about how mathematics relates to science. <br /><br />In class, I had pointed out that mathematical definitions are very precise while scientific measurements can be rather messy. A mathematician has a very precise meaning when they say that two variables are proportional or have a linear relation. But when we get real data, even if they do not satisfy these precise meanings, we still gain significant information about the relation and might even say that the measured variables are proportional or linear. Unfortunately, many students seemed to think I was looking for a repeated discussion of this point.<br /><br />Science can be thought of as the study of the physical world through the scientific method. Essentially, we make observations on what happens in the world (whether that be physical, chemical or biological interactions) and want to understand why that is happening as well as to predict what will happen in the future. In order to do this, scientists propose various hypotheses based on their observations (and past accumulated scientific experience) and then test those hypotheses. Experiments quite often include quantitative measurements, and part of the prediction is often to propose relationships between independent variables (the variables in treatments and control) and the dependent variables (outcomes). Experience may support a hypothesis or falsify the hypothesis, but it never can prove a hypothesis.<br /><br />Knowledge based on patterns that we predict will continue, but which we can support but never prove, is called inductive knowledge. Science is an example of inductive knowledge.<br /><br />Mathematics can be thought of as the study of structures that satisfy very specific rules. We have properties of arithmetic, algebra, and calculus. We establish specific axioms that describe our basic assumptions about the structures and then use logical argument to deduce the behavior of more complicated constructions. We might look at examples to see what ideas might be true or false, and in this sense mathematics can also take advantage of inductive knowledge. However, the objective in mathematics is not just to suppose that a pattern will continue; the objective is to determine conclusively if it must continue. We seek for proofs (that it is true) or counterexamples (break the pattern).<br /><br />Knowledge based on basic assumptions (axioms) and logical argument that determines conclusively what must follow from these assumptions is called deductive knowledge. mathematics is an example of deductive knowledge.<br /><br />Models form a connection between mathematics and science. Data often appear to follow a general trend, even in the presence of the noise of messy observation. A mathematical model takes that messiness and forms an abstract clean relationship that mathematics can work with. Based on the deductive approach of mathematics, we can often establish consequences of the assumed model form. We then apply those consequences as hypotheses in our scientific framework. The predictions from the deductive approach provide the predictions that can be used to falsify these hypotheses.Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com1tag:blogger.com,1999:blog-4156201020188714381.post-55296825187175022912009-08-24T10:36:00.000-07:002009-08-24T10:49:08.204-07:00First Reading AssignmentSince I am not sure when Blackboard is going to be available for the class (I procrastinated asking for the two sections to be merged into a single section), here is the reading assignment and preparation for classwork for Wednesday.<div><br /></div><div><b>Reading Assignments:</b><br /><div><br /></div><div><a href="http://www.math.duke.edu/education/calculustext/index.html">Online Calculus Textbook</a>: Read Sections 1.1 and 1.2 (link below). These sections emphasize the idea that variables (which represent physical quantities) can be related, as independent and dependent variables. We want to think about related quantities throughout this semester. This text specifically asks for the web-browser <a href="http://www.mozilla.com/en-US/firefox/firefox.html">Firefox</a> version 3.0 or later (this is to render formulas correctly).</div><div><ul><li><a href="http://www.math.duke.edu/education/calculustext/Chapter1/Section1-1/Chapter1-1-1M.xhtml">Section 1.1</a></li><li><a href="http://www.math.duke.edu/education/calculustext/Chapter1/Section1-2/Chapter1-2-1M.xhtml">Section 1.2</a></li></ul></div><div>Come prepared to class having prepared answers for Problems 3, 4, 6, 7, 10 from <a href="http://www.math.duke.edu/education/calculustext/Chapter1/Section1-2/Chapter1-2-6M.xhtml">Section 1.2 Problems</a>. We will discuss these problems but will not turn them in.</div><div><br /></div><div><a href="http://www.nap.edu/catalog.php?record_id=10126">How Students Learn</a>: This book prepared by the National Academy of Sciences is actually written for teachers to focus on making courses better suited for students to learn. The first 12 pages introduce three concepts that students should be aware of in their own learning. The link below goes to the first page, and then follow the links to read through page 12.</div><div><ul><li>How Students Learn, <a href="http://books.nap.edu/openbook.php?record_id=10126&page=1">page 1</a> through page 12.</li></ul></div></div>Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-53028598233404358692009-08-24T10:25:00.000-07:002009-08-24T10:36:54.665-07:00Why is math fun? Why is math hard?Today in class, I tried to help break the ice and reduce some of the anxiety related to taking a university mathematics course (Math 231). I asked students for examples of why they might find mathematics fun and why they might find mathematics hard. Here are some of the responses.<div><br /></div><div>Why fun?</div><div><ul><li>It's fun when you struggle with a concept and then it finally clicks and you understand.</li><li>It's fun to see mathematics actually being applied to a real problem.</li><li>It's fun when you are able to spot trends and make predictions based on data.</li><li>It's fun why you really understand why instead of just the "required" steps.</li><li>When you understand, it becomes easy.</li><li>It can be a lot like a game or solving a puzzle.</li><li>It's fun to develop things logically.</li></ul><div>Why hard?</div><div><ul><li>Later material builds on earlier material, so missing something early is permanent hardship.</li><li>It can be really hard when the teacher goes too fast.</li><li>It can be really hard when the teacher is unclear, especially if they can't give alternate ways of thinking about an idea.</li><li>It can be hard if the teaching style is very different from your learning style.</li><li>There are so many formulas, it can be overwhelming to try to memorize them.</li><li>The theorems, rules and definitions are full of little details.</li><li>It can be difficult to understand the many conceptual ideas that interact.</li><li>Learning related technology can be challenging.</li><li>Only one answer, so you can't fake it.</li><li>Very hard to cram for exams.</li><li>It can be really hard to find a little (stupid) mistake when proofing your work.</li><li>Lots of homework, and problems can take a lot of time.</li></ul><div>I'd welcome more comments, including examples of when you found mathematics especially exciting or examples of how your relationship with mathematics soured. Feel free to post a comment.</div></div></div>Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-71241554164623759802009-01-13T12:55:00.001-08:002009-01-13T12:56:42.857-08:00A New Semester --- Two ClassesThis semester I am teaching a mathematical models in biology (Math/Bio 342) as well as the first semester of Calculus with Functions (Math 231). So I expect to have entries for both of these courses showing up.Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-38862768856163718992008-12-03T08:08:00.000-08:002008-12-03T08:26:34.363-08:00Values, Equations, and TheoremsThere is some confusion about theorems. For example, consider the Mean Value Theorem: If <span style="font-style: italic;">f</span> is continuous on [<span style="font-style: italic;">a</span>,<span style="font-style: italic;">b</span>] and differentiable on (<span style="font-style: italic;">a</span>,<span style="font-style: italic;">b</span>), then there is some value <span style="font-style: italic;">c</span>∈(<span style="font-style: italic;">a</span>,<span style="font-style: italic;">b</span>) so that <span style="font-style: italic;">f'</span>(<span style="font-style: italic;">c</span>)=(<span style="font-style: italic;">f</span>(<span style="font-style: italic;">b</span>)-<span style="font-style: italic;">f</span>(<span style="font-style: italic;">a</span>))/(<span style="font-style: italic;">b</span>-<span style="font-style: italic;">a</span>).<br /><br />Some students think that the ratio (<span style="font-style: italic;">f</span>(<span style="font-style: italic;">b</span>)-<span style="font-style: italic;">f</span>(<span style="font-style: italic;">a</span>))/(<span style="font-style: italic;">b</span>-<span style="font-style: italic;">a</span>) is the Mean Value Theorem. But it is not; it is just a value that is called the average rate of change of <span style="font-style: italic;">f</span> between <span style="font-style: italic;">a</span> and <span style="font-style: italic;">b</span>.<span style="font-style: italic;"><span style="font-style: italic;"></span></span> You can compute this value as long as both <span style="font-style: italic;">f</span>(<span style="font-style: italic;">a</span>) and <span style="font-style: italic;">f</span>(<span style="font-style: italic;">b</span>) exist. It has nothing to do with derivatives or continuity.<br /><br />Other students think that the equation <span style="font-style: italic;">f'</span>(<span style="font-style: italic;">c</span>)=(<span style="font-style: italic;">f</span>(<span style="font-style: italic;">b</span>)-<span style="font-style: italic;">f</span>(<span style="font-style: italic;">a</span>))/(<span style="font-style: italic;">b</span>-<span style="font-style: italic;">a</span>). This is closer to the truth, but still is incorrect. First of all, what is c? Second, this statement may not be true. For example, suppose that f(x)=-1 if x<0 and f(x)=1 if x>0. Suppose that a=-2 and b=+2. Then the ratio <span style="font-style: italic;"></span>(<span style="font-style: italic;">f</span>(<span style="font-style: italic;">b</span>)-<span style="font-style: italic;">f</span>(<span style="font-style: italic;">a</span>))/(<span style="font-style: italic;">b</span>-<span style="font-style: italic;">a</span>) is equal to 1/2. But <span style="font-style: italic;">f'</span>(<span style="font-style: italic;">x</span>)=0 everywhere except at x=0, where <span style="font-style: italic;">f'</span>(0) does not exist.<br /><br />Even closer is to say that <span style="font-style: italic;">f</span>'(<span style="font-style: italic;">c</span>)=(<span style="font-style: italic;">f</span>(<span style="font-style: italic;">b</span>)-<span style="font-style: italic;">f</span>(<span style="font-style: italic;">a</span>))/(<span style="font-style: italic;">b</span>-<span style="font-style: italic;">a</span>) for some c between a and b. This is actually the conclusion of the Mean Value Theorem. It requires the entire statement, particularly the statement that the equation is true <span style="font-weight: bold;">for some c</span> and that the value c must be <span style="font-weight: bold;">between a and b</span>.<br /><br />But my example above provides an example where the conclusion of the Mean Value Theorem is false. That does not mean that the Mean Value Theorem itself is false. After all, it is a theorem, and that means that it has been proved to be true <span style="font-weight: bold;">always</span>. The part that is missing is the hypothesis for the theorem. The conclusion can only be guaranteed to be true using the theorem if the hypotheses are all satisfied. In this case, you must also check (or give a reason why) the function f is continuous and differentiable on the interval from a to b, including the endpoints for continuity.<br /><br />Similarly, ∫<sub><span style="font-style: italic;">a</span></sub><sup><span style="font-style: italic;">b</span></sup> <span style="font-style: italic;">f</span>(<span style="font-style: italic;">x</span>) <span style="font-style: italic;">dx</span>/(<span style="font-style: italic;">b</span>-<span style="font-style: italic;">a</span>) computes the average value of a function f on an interval [<span style="font-style: italic;">a</span>,<span style="font-style: italic;">b</span>]. The value can be computed anytime the function is integrable over the interval [<span style="font-style: italic;">a</span>,<span style="font-style: italic;">b</span>]. The Mean Value Theorem for Integrals has nothing to do (in principle) with this calculation.<br /><br />However, if <span style="font-style: italic;">f</span> is continuous on [<span style="font-style: italic;">a</span>,<span style="font-style: italic;">b</span>], then <br /><span style="font-style: italic;">f</span>(<span style="font-style: italic;">c</span>)=∫<sub><span style="font-style: italic;">a</span></sub><sup><span style="font-style: italic;">b</span></sup> <span style="font-style: italic;">f</span>(<span style="font-style: italic;">x</span>)<span style="font-style: italic;"> dx</span>/(<span style="font-style: italic;">b</span>-<span style="font-style: italic;">a</span>)<br />for some <span style="font-style: italic;">c</span>∈(<span style="font-style: italic;">a</span>,<span style="font-style: italic;">b</span>). This entire statement comprises the Mean Value Theorem for Integrals. The hypothesis that must be verified to use the theorem is that f is continuous on [<span style="font-style: italic;">a</span>,<span style="font-style: italic;">b</span>]. The conclusion is that you are guaranteed that<br /><span style="font-style: italic;">f</span>(<span style="font-style: italic;">c</span>)=∫<sub><span style="font-style: italic;">a</span></sub><sup><span style="font-style: italic;">b</span></sup> <span style="font-style: italic;">f</span>(<span style="font-style: italic;">x</span>) <span style="font-style: italic;">dx</span> / (<span style="font-style: italic;">b</span>-<span style="font-style: italic;">a</span>)<br />for at least one value <span style="font-style: italic;">c</span> between <span style="font-style: italic;">a</span> and <span style="font-style: italic;">b</span>.Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-26936716707032570232008-12-02T10:19:00.000-08:002008-12-02T10:31:17.420-08:00Limit or Function Evaluation?I've noticed that some students are perplexed about when they use a limit or function evaluation. I presume that the cause of this confusion is that students have learned that you evaluate a limit by plugging in a value. But this is only because nearly all functions that they work with are continuous.<br /><br />You use a limit evaluation when you need to know what the value of the function <span style="font-weight: bold;">should</span> be by using information <span style="font-weight: bold;">from the side</span> of the point of interest. When using a limit, you must use limit notation: lim<sub>x→c</sub> f(x). Then you use the appropriate rules of limits to evaluate (and hopefully, the function is continuous).<br /><br />You use a function evaluation when you need to know the value of the function <span style="font-weight: bold;">at an actual point</span>. There is no limit involved, just function evaluation. You just use function notation, say f(c), and compute the value defined by the function.<br /><br />For example, suppose you are calculating an instantaneous rate of change as the limit of an average rate of change. The average rate of change only makes sense when the interval of interest includes two points (endpoints of an interval). The instantaneous rate of change is found by seeing what the value of the average rate of change does when the two points move closer to each other, or more particularly, as the second point approaches the first point.<br /><br />On the other hand, suppose that you know the derivative, which is itself a function. Then the instantaneous rate of change is calculated by function evaluation using the derivative function. (The limit was already used to create this new function.)<br /><br />In particular, I have noticed this problem when dealing with finding extreme values of a function. When the interval of interest is an open interval, we are acting as though the domain does not include the endpoints. So, with this restricted domain, evaluation of the function is not possible (since the points are not in the domain). So we must use the information about the function immediately adjacent to the endpoints, and this is with a limit. In this context, the value found in the limit is not achieved at the endpoint, although it might be achieved somewhere else in the domain.<br /><br />On the other hand, if the interval is a closed interval, then the endpoints are included in the restricted domain. If the function is continuous at these points, then evaluating the function directly is appropriate, since the point is in the domain.<br /><br />Final remark, if the function is discontinuous at some point in the interval, you must also check the limits at that point for consideration when looking for extreme values.Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-85065400103618043742008-11-17T17:59:00.000-08:002008-11-17T18:23:41.842-08:00Differential Equations Project TipsAs some questions are asked more regularly, I thought I'd provide some general discussion here.<div><br /></div><div>(1) Start with the proposed form of X(t). Compute X'(t) and X''(t) based on that form. Then use those calculations to discover when X'' + k/m X = 0.</div><div><br /></div><div>(2) The question "mean physically about the mass on a spring" is not asking you to think about the mass (as in measurement) but is asking you to think about what the statement X(0)=1 means about the state of the mass at time t=0 and what the statement X'(0)=0 means about the state of the mass at time t=0.</div><div><br /></div><div>(3) X(0) is a constant and has derivative of d/dt[X(0)] = 0. Recall that dX/dt > 0 implies that X is increasing, dX/dt < 0 implies that X is stationary (instantaneous rate = 0)<div><br /></div><div>(4) An arbitrary quantity A is proportional to some other quantity B if it is always that case that A = k B for some constant value k (the constant of proportionality). Now interpret the statements to identify what pieces of the equation are proportional to what.</div><div><br /></div><div>(5) Although you have studied e<sup>x</sup> in precalculus, you will not use the logarithm at all in this work. Instead, I just want you to consider some function that has the special property that exp' = exp. (This sentence is analogous to sin'=cos and cos'=-sin.) However, you do need to think about the chain rule: X(t) = A exp(rt) (Since the argument is not simply t, you must use the chain rule.) This problem is exactly analogous to Step 1.</div><div><br /></div><div>(6) You will get something like</div><div style="text-align: center;">dX/dt = "formula involving X and a, b, and m"</div><div>X is increasing when "formula" > 0, decreasing when "formula" < 0, and stationary when "formula" = 0. So use your skills with algebra (think sign analysis) to find conditions when these are the case.<div><br /></div><div>(7) You need to understand the relationship between a rate of change and an actual change. To understand this as well as possible, see the section we skipped in Chapter 3 (last section). But what you essentially need is that we will follow the tangent line for the time increment Δt. How much change is there when the rate of change and the duration of time are both known?</div><div><br /></div><div>(8) The new version of Excel has some unanticipated differences from what I had when I wrote the project. The labels are not assigned from a menu anymore. Instead of the 3-step process that is described, you just click in the label field in the header section of Excel and type in the new label and then hit enter.</div><div><br /></div><div>The calculations you see in the first few lines should exactly match your hand calculations in part (7). </div><div><br /></div><div>Do not print the spreadsheet (it takes WAY too many pages). That is why I ask you to submit your spreadsheet on Blackboard as part of the project.</div><div><br /></div><div>(9) I must receive a print out of the graph -- hand drawn figures are not acceptable. Ideally, this entire project report would be typed (perhaps using Equation Editor for the equations), with the figures naturally fitting in.</div><div><br /></div><div>(10) Make hypotheses and test your hypotheses.</div></div></div>Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com1tag:blogger.com,1999:blog-4156201020188714381.post-38572805958808427752008-11-17T17:35:00.000-08:002008-11-19T08:14:08.157-08:00Exponential Project TipsAs some questions are asked more regularly, I thought I'd provide some general discussion here.<div><br /></div><div>(1) exp is the name of the function, just as sin and cos are names of functions. From calculus, you learn that sin'=cos and cos'=-sin. This step shows that exp_b' = ln b * exp_b. (That is, it leaves the function alone except for a constant multiple. (But be careful where the chain rule is needed!)</div><div><br /></div><div>(2) You do not need to use the limit definition (epsilons and deltas). Instead, for perhaps the easiest solution, you should think about how to finish the statement:</div><div>lim b^x = lim [(b^x-1)/x ... ]</div><div>That is, if you start with (b^x-1)/x, what do you do to that expression to leave only b^x. Then use elementary limit rules to compute your resulting limit.</div><div><br /></div><div>(3) One method is to use the method of substitution for limits (change of variables) and then use an identity for the function so that the result of Step 2 is applied --- this method mimics what is done to show that sin x is continuous everywhere. A second method is to use a general theorem that makes continuity an obvious conclusion of the results from Step 1.</div><div><br /></div><div>(4) You must start with a statement like:</div><div style="text-align: center;">ln (1/b) = lim_{x → 0} [(1/b)^x - 1]/x</div><div>There are two easy approaches: (1) Find a common denominator to rewrite this as a simple fraction before continuing or (2) Think of (1/b)^x as b to some appropriate power and then use a limit substitution.</div><div><br /></div><div>(5) Since you do not know the derivative of ln x, it is incorrect to use the Mean Value Theorem applied to the logarithm. Instead, you should apply the MVT to the function exp_b(x) on an interval so that b^a and b^b are incredibly easy and where it is clear which value is larger (so that you know if the average rate of change is positive or negative). You may use the fact that b^x is positive for all values of x.</div><div><br /></div><div>(6) The function fb(x) is a linear function. You should write it in slope-intercept form (e.g., mx+b).</div><div><br /></div><div>(7) Do not attempt to solve the equation fb(x) = exp_b(x). There is one obvious solution from the definition: x=0. But the formulas themselves do not explain where there would not be more solutions. Instead, you should define a function (perhaps g) so that</div><div style="text-align: center;">g(x) = exp_b(x) - fb(x).</div><div>You know that g(0) = 0. You need to show that g(x)>0 for all x ≠ 0. My hint suggested Rolle's theorem, but I have since found that the Mean Value Theorem helps even more. Use the Mean Value Theorem to show that for x>0, the average rate of change between 0 and x must be positive. What about x<0? x="0?"><div><br /></div><div>(8) You may not use a limit form of the type b<sup>∞</sup>. You may take a limit of the function fb(x) because that is of a form we know how to work with. Then you should use the result of (7) to conclude what the limit of exp_b(x) must be.</div><div><br /></div><div>(9) and (10) put all of the previous steps together to perform analysis similar to Sections 4.2 and 4.4 to understand the graph.</div></div>Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-787581304701339842008-10-24T13:06:00.001-07:002008-10-24T13:29:59.697-07:00Spotlight: Math GamesThe other day before class, I introduced a little game called Sprouts. I found the rules summarized at the <a href="http://www.maa.org/mathland/mathland_4_7.html">MAA website</a>. There is also a nice discussion on the <a href="http://www.sciencenews.org/sn_arc97/4_5_97/mathland.htm">Science News website</a>. I find it interesting that such a simple game can be analyzed using mathematical properties.<br /><br />One of my favorite "math" games is a game called Eleusis. This game was invented <a href="http://www.logicmazes.com/games/eleusis/">Robert Abbott</a> as an analogy of the scientific method. So perhaps we should call this a "science" game. The game is played with a deck (or multiple decks) of playing cards. I turn over the first card and then think of a pattern that would start with that card. Now, the remaining players take turns attempting to choose a card from their hands that they believe would be a valid next card in the pattern. If they are correct, I leave it there. If they are incorrect, I move the card out of the sequence and below the card they tried to follow (for future reference).<br /><br />The goal for the players is to eventually arrive at a hypothesis that they believe explains the pattern. By playing cards, they attempt to critically assess whether their hypothesis is a complete explanation of the pattern. This mimics the scientific method because we see patterns in how nature functions, and through experiment we attempt to see if controlled efforts are consistent with or contradict our acting hypotheses.<br /><br />Try it out? Let me know how the game goes.Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-79774810807940905262008-10-10T08:58:00.000-07:002008-10-10T09:38:29.616-07:00Mathematical InductionThe principle of <a href="http://en.wikipedia.org/wiki/Mathematical_induction">mathematical induction</a> is a topic that our textbook unfortunately skips over. It is used when we want to prove a rule that applies to positive integers. Often, it is the argument that is needed when you want to say, "See! It works the same way for this case and that case, so the pattern will just keep repeating." But to say that a pattern keeps repeating is exactly what we should attempt to make more precise.<br /><br />The natural numbers are the set of all positive integers: 1, 2, 3, 4, .... It is the dot-dot-dot that creates the problem. Using "..." attempts to tell us that the pattern continues. But what exactly is the pattern? For the natural numbers, the pattern is that you just add one to the previous number. So here is one way of describing the natural numbers, and it is what motivates the principle of mathematical induction.<br /><ul><li>1 is in the set</li><li>For every number that is in the set, call it n, we also have n+1 in the set.</li></ul>We could restate this using an implication:<br /><ul><li>1 is in the set</li><li>If n is in the set, then (n+1) is in the set.</li></ul>And that is what we do for all applications of mathematical induction. We provide a starting point (such as 1 is in the set). Then we establish an implication that if a statement is true for one value (n is in the set), then it must also be that the statement is true for the next value (n+1 is in the set).<br /><br />Here is an example from our past that should have used induction.<br /><br /><span style="font-weight: bold;">Theorem:</span> x<sup>n</sup> is continuous for n=1, 2, 3, ....<br /><br /><span style="font-weight: bold;">Scratchwork:<br /></span>Before proving this statement, we should think how we might attempt this without induction. Well, f(x) is really a product x<sup>n</sup>= x x x ... x, where there are n factors of x. (See how the "..." allows us to hand-wave our notation?) Well, we know that the limit of each factor x will just go to the value c, so the limit must be lim<sub>x → c</sub> f(x) = c<sup>n</sup>. That use of "..." keeps us from clearly stating how we used the limit of a product, other than again referring to a pattern: "Use the limit of a product n-1 times." The use of induction makes this precise.<br /><br /><span style="font-weight: bold;">Proof:<br /></span><span>We prove by induction.<br />1) First, we show that f(x) = x<sup>1</sup> = x is continuous. (This is the starting point)<br />But this is already known: </span>lim<sub>x → c</sub> x = c.<br />So f is continuous at any point c.<br />The statement is true when n=1.<br />2) Second, we assume that f(x) = x<sup>n</sup> is continuous and now show that this implies that g(x) = x<sup>n+1</sup> is also continuous. (This is the inductive step)<br />So assume that f is continuous.<br />g(x) = x f(x) (Relate the new function in terms of what is assumed)<br />So using the limit of a product:<br />lim<sub>x → c</sub> g(x) = lim<sub>x → c</sub> x f(x) = c f(c) = c c<sup>n</sup> = c<sup>n+1</sup> = g(c)<br />So x<sup>n+1</sup> is continuous whenever x<sup>n</sup> is continuous.<br /><br />So by induction, since the statement is true for n=1, and whenever the statement is true for one value n it is also true for the next value n+1, the statement is true for all integers starting with 1.<br /><span style="font-weight: bold;">(End of Proof)</span><br /><br />Sometimes induction is compared to reaching different rungs on a ladder. The first statement is what allows you to climb onto the first step. The inductive implication is what says that if you have already reached one rung, then you can move to the next rung. Putting the two together, you first climb on the ladder's first rung. Then you know that you can climb from the first to the second rung, from the second rung to the third rung, from the third to the fourth, and so on forever. The implication, <a href="http://www.phrases.org.uk/meanings/50500.html">in one fell swoop</a>, justifies climbing each step from the previous. The principle of mathematical induction replaces the uncertainty in "..."Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-823339272729526182008-10-06T07:37:00.000-07:002008-10-06T08:00:51.886-07:00Intermediate Value TheoremA theorem is a statement that is always true because it has been proved. Theorems are usually stated as implications. That is, they usually are stated as "If [something is true], then [something else is true]." However, this does not mean that the hypothesis (what appears as [something is true]) is actually true. Nor does it mean that the conclusion (the statement instead of [something else is true]) is true. It means that you are guaranteed to know that the conclusion is true whenever the hypothesis is true.<br /><br />When applying a theorem, it is your task to establish that the hypothesis is true. Then, by stating the theorem, you are allowed to state that the conclusion is also true.<br /><br />Here is an example using the Intermediate Value Theorem. Recall that the theorem states that if you have a function <span style="font-style: italic;">f</span> that is continuous on a closed interval [a,b] (where a and b can be any numbers with a < b), then for any y-value C between the values f(a) and f(b), you are guaranteed to be able to find a value x such that a < x < c and f(x) = C.<br /><br />Here is a hypothetical situation. My car holds 12 gallons of gasoline. (That is not the hypothetical part -- I have actually filled the tank :-) I have installed an automated gas-tank tracking system that records the amount of gas as a function of the car's mileage. (Yep, that's the hypothetical part) If you ask me how much gas I had when the car was at 97,034 miles, then I can tell you it had exactly 5.93 gallons of gasoline.<br /><br />Last week, I filled up my tank when the car was at 98,012 miles. This morning, I checked my car and it now records the tank as having 1.45 gallons and 98,143 miles. (All figures are also hypothetical, including mileage) So here is a question: will I actually be able to identify a mileage on the car when between that last fill up and today when the car contained exactly 4.7 gallons?<br /><br />Hmm. Let's see. Imagine that we use the variable x to represent the mileage on the car. Also, let f be a function that measures the gallons in the car f(x) when the mileage is x. We know that f(98,012) = 12 and f(98,143) = 1.45. So C=4.7 is between f(98,012) and f(98,143). What does the Intermediate Value Theorem say?<br /><br />Now, before you go on, I need to tell you a story. On Friday, I needed to mow the lawn. My backyard is pretty large, so it takes a while. Funny thing! I ran out of gas. I knew I had recently filled the car, so I found my gas siphon and pumped a gallon out of the car's tank and into my gas can. Phew! Glad that was available! Finished the lawn with nary a problem.<br /><br />So what did you answer?<br /><br />(Extra credit toward quiz grade if you answer correctly this week by e-mail: waltondb at jmu dot edu)Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-63641297020558556972008-10-02T12:59:00.000-07:002008-10-02T13:43:58.252-07:00Common Limit IssuesPart of the challenge of mathematics is learning the language of mathematics. Mathematics is meant to be spoken, kind of like poetry. But in my class, I'm not wanting a <a href="http://www.cinquain.org/">cinquain</a>. Scattered thoughts that are related, but not directly connected in sentences, do not a coherent message provide.<br /><br />So, here are the top ten problems in limits:<br /><br />1. Dropping "lim" suddenly. If you have two expressions f(x) = g(x) where g(x) is a simplified version of f(x) (cancelled something), you should write lim f(x) = lim g(x) and <span style="font-weight: bold;">NOT</span> lim f(x)=g(x). That "lim" doesn't apply to both sides of the equal sign.<br /><br />2. Writing "lim" too many times. Just because you don't want to forget to write "lim" doesn't mean you write it in front of everything. You keep writing "lim" while you massage the formula into a form where you can decide the limit. As soon as you are allowed to "plug-in" the value, you have just "taken the limit" and you should stop writing "lim".<br /><br />Example: f(x) = (x^2-4)/(x-2) and g(x) = x+2. We know that f(x)=g(x) for x ≠ 2. So<br /><div style="text-align: center;">lim<sub>x → 2</sub>(x^2-4)/(x-2) = lim<sub>x → 2</sub>(x+2) = 2+2 = 4.<br /><div style="text-align: left;"><br />3. Lonely "lim". "lim" is not simply an abbreviation for the word "the limit". It is an operator wanting to do something to a formula. It needs a formula next to it at all times. It is without a formula. So when students write "lim = 3", clearly intending to say, "the limit is 3", they are really saying, "the limit of is 3". Of <span style="font-weight: bold;"><span style="font-weight: bold;">what</span></span>? And that is the problem. The limit is lonely and has nothing to act on.<br /><br />4. Stopping at a limit form. Just because you see a zero (0) in the denominator in the limit form does not mean you are done. If the limit has form 0/0, you must try to factor and cancel. If the limit has form L/0, you must identify the sign of the function to decide whether it is going to +&infty; or -&infty;.<br /><br /></div></div>5. Writing "=" for undefined values. (Don't do that!) Use a limit form notation to indicate that the denominator is 0 or terms go to infinity.<br /><br />6. Piecewise using x=<span style="font-style: italic;">a</span>. A piecewise function that has a formula when x=a is a distractor for limits. Remember, a limit always determines what the function would predict <span style="font-style: italic;">if you came from the sides</span>. So a limit never checks at x=a.<br /><br />7. Writing f(a) instead of lim f(x). This is another piecewise function issue. To check the value predicted by the two sides, you need to <span style="font-style: italic;">say</span> you are checking the sides (meaning limit). So you must write that it is a limit.<br /><br />8. Using rules for x going to infinity at a. When x goes to infinity, we can ignore any terms that look like 1/x since those terms go to zero as x goes to infinity. However, when x goes to a, those terms are still numbers other than zero. Don't just forget about them (and don't even factor out the dominant terms).<br /><br />9. Step-by-step when not required. Unless I explicitly ask you to show that the limit has a value using the elementary limit rules, you should just compute the limit. You don't need to spend time showing the step-by-step justification unless asked.<br /><br />10. No work at all. Often you can see the limit from a graph (say on a calculator). But you need to show a reason on the paper based on mathematics that gives your answer. At the very least, say you looked at a calculator to motivate your answer.Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0tag:blogger.com,1999:blog-4156201020188714381.post-42631365980970556182008-10-02T05:23:00.000-07:002008-10-02T12:33:58.097-07:00Epsilons, Deltas and Limits... Oh My!Yes, <a href="http://en.wikipedia.org/wiki/Toto_%28Oz%29">Toto</a>, writing proofs of limits can be as scary as the wicked witch from the east! But do not fear, with the right direction, we can squash those problems with ease.<br /><br />The first step is to realize that we are proving a limit based on its definition. Suppose we need to prove the statement written in its general form:<br /><div style="text-align: center;">lim<sub><span style="font-style: italic;">x</span> → <span style="font-style: italic;">a</span></sub> <span style="font-style: italic;">f</span>(<span style="font-style: italic;">x</span>) = L<br /></div>Notice that this really is saying that when x is a value close to a, the value of f(x) is close to L. The mathematical statement of this says:<br /><div style="text-align: center;">"For any ε > 0, there exists a value δ > 0 so that if 0 < |<span style="font-style: italic;">x</span>-<span style="font-style: italic;">a</span>| < δ then |<span style="font-style: italic;">f</span>(<span style="font-style: italic;">x</span>)-<span style="font-style: italic;">a</span>| < ε." <div style="text-align: left;"><br />Now just because the <a href="http://en.wikipedia.org/wiki/Scarecrow_%28Oz%29">Scarecrow</a> is flapping in the breeze, we don't need to be afraid of this complicated looking formula. Our task is to be able to find a formula for δ in terms of ε so that once you know that the value of <span style="font-style: italic;">x</span> is within δ of the value <span style="font-style: italic;">a</span> (δ says <span style="font-style: italic;">how</span> close), then the value <span style="font-style: italic;">f</span>(<span style="font-style: italic;">x</span>) is within ε of the value L.<br /><br />To reach that fabled <a href="http://en.wikipedia.org/wiki/Wizard_%28Oz%29">wizard</a> of mathematics called a proof, we just need to follow the <a href="http://en.wikipedia.org/wiki/Yellow_brick_road">yellow brick road</a> outlined below. The proof will always take a form involving four steps corresponding to the four parts of the definition:<br /><br />1. For any ε > 0: We need to create a proof that works for any ε > 0. So that we have a value to work with, we start with any ε with the requirement ε > 0. So the first statement of the proof is something like, "Given ε > 0" or "Suppose ε > 0" or "Let ε > 0."<br /><br />2. ... there exists δ > 0 such that: The second step is that we need to provide a recipe for how to provide δ > 0 that will make the rest of the statement true. Unfortunately, by the time we reach this step of the proof, we don't yet know what the right recipe is. Personally, I just write, "Let δ=____" and leave enough space to fill in later.<br /><br />3. ... if 0 < |x-a| < δ ...: We are starting to prove an implication (if...then... statement). We are successful if we can show the conclusion is true whenever the hypothesis is true. So, to accomplish this, we assume that the hypothesis <span style="font-weight: bold;">is</span> true and see what happens. I write, "Assume 0 < |x-a| < δ."<br /><br />4. ... then |f(x)-L| < ε: This is the conclusion of the implication. And this is also the hardest part of the proof. In the middle of completing the work in this proof, we will discover the recipe needed for δ. At that time, we can go back and fill in the missing pieces.<br /><br />So the 4th step is the hard one. Don't be cowardly like the <a href="http://en.wikipedia.org/wiki/Cowardly_Lion">lion</a> and give up; there is a method to this as well. For the polynomials that we work with, the value of |f(x)-L| will <span style="font-weight: bold;">always</span> factor into something of the form:<br /><div style="text-align: center;">|f(x)-L| = |x-a| |"stuff"|<br /><div style="text-align: left;">We know from step 3 that |x-a|< δ. We want |"stuff"| to be less or equal to a number, which for convenience in discussion we'll call k. Once we find that number k, then we know: <div style="text-align: center;">|f(x)-L| < δ k <div style="text-align: left;">So part of our recipe will be to make sure that δ≤ε/k. If the recipe requires no other parts, we can even just use δ=ε/k. With this knowledge, we will have found:<br /><div style="text-align: center;">|f(x)-L| < δ k ≤ (ε/k)k = ε <div style="text-align: left;"><br />All would be well, except that the <a href="http://www.math.jmu.edu/%7Ewalton/">wicked professor of the west</a> hasn't yet told you how to find k. Let's step back a moment to the |"stuff"| factor. If you don't, I'll send my <a href="http://en.wikipedia.org/wiki/Winged_Monkeys">winged monkeys</a> to bring you back :-). When we're lucky (and f(x)=mx+b), the factor is already a number. But for any other problem, there will be a formula that still involves x. In these cases, without knowing more about x, we won't know how big the extra "stuff" can become. In order to keep a handle on this "stuff" we are going to require for our recipe that δ itself never gets too large. For the simplest cases, we can require δ ≤ 1. And this means that we can take advantage of knowing that x will be between a-1 and a+1.<br /><br />In a general problem, we could find the largest value for |"stuff"| using values of x between a-1 and a+1. This might take a bit of work. But for f(x) that is quadratic, "stuff" is going to be another linear looking term. We want that "stuff" to involve x-a, so use x = (x-a) + a.<br /><br />For example, when a=2, the term x+1 can be rewritten x+1 = (x-2)+2+1 = (x-2)+3. The awesome <a href="http://en.wikipedia.org/wiki/Triangle_inequality">Triangle Inequality</a> then tells us:<br /><div style="text-align: center;">|x+1| = |(x-2)+3| ≤ |x-2| + 3<br /><div style="text-align: left;">But we know that |x-2|<δ and we required δ ≤ 1 for our recipe. So |x+1|< 4.<br /><br />For another example, suppose that f(x)=x<sup>2</sup>-x, a=3, and L=6. We assume 0<|x-3|<δ and rewrite <div style="text-align: center;">|f(x)-L| = |x<sup>2</sup>-x-6| = |x-3||x+2|<br /><div style="text-align: left;">We know |x-3|<δ and we need to find a number k so that |x+2|≤k. Since x is in "stuff", we require δ ≤ 1 and use the triangle inequality:<br /><div style="text-align: center;">|x+2| = |(x-3)+3+2| ≤ |(x-3)|+5 < δ+5 ≤6 <div style="text-align: left;">So |x+2| < 6 (This is our value k=6). Thus we also want to use δ = ε/6 in our recipe. Both requirements are taken care of by the formula δ = min(1, ε/6). So now we know<br /><div style="text-align: center;">|x<sup>2</sup>-x-6| = |x+2||x-3| < 6 δ ≤ 6(ε/6) = ε</div><div style="text-align: left;"><br />We have arrived at the <a href="http://en.wikipedia.org/wiki/Emerald_City">emerald city</a> of our desire and proved the limit statement<br /><div style="text-align: center;">lim<sub><span style="font-style: italic;">x</span> → 3</sub> x<sup>2</sup>-x = 6.<br /><div style="text-align: left;"><br />But the proof needs to be in the right order:<br />Given ε>0.<br />Let δ = min(1, ε/6).<br />Assume 0 < |x-3| < δ. |x<sup>2</sup>-x - 6| = |x-3||x+2|<br />|x+2| = |x-3 + 5|<br />|x+2| ≤ |x-3| + 5<br />|x+2| < δ + 5 and δ ≤ 1 so |x+2| < 6<br />|x<sup>2</sup>-x-6| = |x-3||x+2| < 6δ and δ ≤ ε/6<br />So |x<sup>2</sup>-x-6| < ε<br />Thus, for all ε>0, there exists δ>0 so that if 0 < |x-3| < δ, then |(x<sup>2</sup>-x)-6| < ε.<br />Therefore, lim<sub><span style="font-style: italic;">x</span> → 3</sub> x<sup>2</sup>-x = 6</div></div></div></div></div></div> </div></div></div></div></div></div></div></div></div></div></div>Brian Waltonhttp://www.blogger.com/profile/01474965246427735513noreply@blogger.com0