Wednesday, April 28, 2010

Graph from a graph of f ' (x)

First, pay attention: the graph provided on the assignment is the graph of the derivative 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.)

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.
• f '(x) = + and f ''(x) = + means f(x) looks like Quadrant IV (incr, conc. up)
• f '(x) = - and f ''(x) = + means f(x) looks like Quadrant III (decr, conc. up)
• f '(x) = + and f ''(x) = - means f(x) looks like Quadrant II (incr, conc. down)
• f '(x) = - and f ''(x) = - means f(x) looks like Quadrant I (decr, conc. down)
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.

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).

Sums of Geometric Sequences

The second problem on the project introduces a new closed form for a sum:

k=1n [A ρk] = A ρ (ρn-1)/(ρ-1).

Unfortunately, too many of you are still intimidated simply by the symbols that are used.

The formula for the geometric sequence, A ρk, 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:
2/3 (k=1), 2/9 (k=2), 2/27 (k=3), 2/81 (k=4), etc.

The summation is simply the sum of these values:
k=1n [2 (1/3)k] = 2/3 + 2/9 + 2/27 + 2/81 + ... + 2/3n.
The closed form gives a formula answering the value of this sum:
2(1/3)[(1/3)n-1]/[(1/3) - 1] = (2/3)*[(1/3)n-1]/(-2/3) = 1 - (1/3)n.

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.

Populations, Birth Rates, and Death Rates

This 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 biological sense.)

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.

Suppose you know f '(x). Then A(x) = ∫0x 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) + ∫0x f '(z) dz.

This will always work, even if I don't start the integral at 0: f(x) = f(a) + ∫ax f '(z) dz. Written another way, it looks like the first Fundamental Theorem of Calculus: f(x) - f(a) = ∫ax f '(z) dz.
In other words, the 2nd FTC implies that every function is its starting value plus the integral of its rate of change.

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, ∫02 b(t) dt can only be estimated with a single rectangle while ∫04 b(t) dt would involve two rectangles. The idea of the Riemann sum is that we choose b(tk*) as one of our data points (either on the left or right).

More specifically, on the interval [0,2] (k=1), we can either use t1*=0 so that b(t1*)=100 or use t1*=2 so that b(t1*)=135. In the first case, the rectangle for k=1 contributes b(t1*)Δt = 200, while the second case leads to a contribution of b(t1*)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.

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.

Friday, February 5, 2010

Exponential Functions

I'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:
• bx is not just a formula "b to the power x" but is a new function, which I'm asking you to call expb(x).
• The properties of exponents like bx+y=bx by and (bx)y=bxy become properties of the exponential functions.
expb(x+y)=expb(x)*expb(y)
expb(xy)= [expb(x)]y

• Logarithms are the inverse functions of exponential functions.
expb(logb(x))=x
logb(expb(x))=x

In formula representation, these are written as follows:
blogb(x)=x
logb(bx)=x
• Whenever you see a formula with an exponential, say b3x-2, you should be able to think in both formula and function modes interchangeably.
b3x-2 = b3x b-2 = (b3)x b-2
b3x-2 = expb(3x-2)
The first mode allows us to recognize that b3x-2t is actually of the form Aqx where A=b-2 and q=b3. 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.
• There is a special base e that is most important because the corresponding exponential function is its own derivative.
exp'e(x)=expe(x)
d/dx[ex]=ex

The natural exponential is written without a base. The corresponding inverse function is called the natural logarithm, ln(x).
exp(ln(x))=x
ln(exp(x))=x

In formula representation, these are written as follows:
eln(x)=x
ln(ex)=x
The x in these formulas, as always, is a placeholder. So any number or formula could be used in place of x.
• Any exponential can be written in terms of the natural exponential. The key is to use the properties of exponents.
b = eln(b)
bx = [eln(b)]x = eln(b) x
Another way to think of this is using composition of inverse functions: exp(ln( ))
bx = exp(ln(bx))
ln(bx) = x ln(b)
bx = exp(x ln(b)) = ex ln(b)

That is, by replacing bx by eln(b) x, we can write any exponential A bx in the form A ekx where k is the number ln(b).
• Derivatives of exponentials use the basic property that exp'(x) = exp(x), or d/dx[ex] = ex. Usually, this also requires the chain rule: d/dx[eu] = eu u'.
d/dx[e2x] = e2x (2) = 2e2x (u = 2x)
d/dx[e-3x] = e-3x (-3) = -3e-3x (u = -3x)
d/dt[e(t2-5t)] = e(t2-5t) (2t-5) = (2t-5)e(t2-5t) (u = t2-5t)

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.