Monday, September 15, 2008

Proofs and Even/Odd Functions

Here is the second part of my chat discussion. I have edited this some, but I hope the essence of the questions and answers.

[Student]: ok, so im still iffy on how to structure proofs, ill know the assumption (obviosly) and the conclusion, but I'm unsure how to structure the premises and justify them
[Professor]: Well, basic ideas include starting the proof with your assumptions.
These are the basic statements you know are true.
[Student]: right

[Professor]: The very last line of the proof should match the conclusion. (and it shouldn't appear earlier) The hard part is what comes in between :-)
[Professor]: But seriously, usually, you take a look at your conclusion and see what type of statement it requires.

[Professor]: If it is an equation that is needed, then you can usually start with one side of the equation and then see what you can put on the other side that you know must be true. (Usually using a definition or algebra)

[Professor]: Then you see how you can use your assumptions to create a statement that leads to your conclusion.

After turning to an actual problem, we started to get into more specifics. The problem was stated as: Suppose that f1 and f2 are odd functions. Prove that f1*f2 is an even function.

[Professor]: What is given knowledge?
[Student]: f1 and f2 are both odd and g1 and g2 are even
[Professor]: What is desired? (conclusion)
[Student]: f1 * f2 is even
[Professor]: What does it mean that f1 is odd?
[Student]: that f(-x) = -f(x)
[Professor]: Make the f into f1, and correct.
[Student]: ah ok ya
[Professor]: So in the actual proof, one line would be...
f1 is an odd function
[Professor]: The next line would interpret...
f1(-x) = -f1(x) for all x in domain of f1
[Professor]: Repeat these two lines for f2.
[Professor]: So far, we have simply restated facts based on the assumptions and our knowledge of odd functions.
[Student]: and that is all stuff that is under the given part? rite
[Professor]: In the proof, it doesn't belong to a "given" part per se, but the reason on the right hand side (if in tabular form) would be that the statement was "Given"
[Professor]: So our proof now has 4 statements.

[Professor]: Now, (scratch work) look at what you need to show: f1*f2 is even. What does that mean?
[Student]: ok ya i have no idea, i feel like it would involve one of the multiplicative properties involving f1 and f2 but i dont really know where to start
[Professor]: What does it mean if I said that G is an even function? g(x) = g(-x) My name was G, so it would be G(x)=G(-x).
[Professor]: But the name could be Brian: Brian(-x)=Brian(x). Voila! I [Brian] am even.
[Professor]: f1*f2 is the name of a function. And for f1*f2 to be even, you need: f1*f2(-x) = f1*f2(x)
[Professor]: It just happens that the name looks like a formula.
[Student]: theres noit a step in between there?
[Student]: to justify how multiplying the functions is ok
[Professor]: Well, there are still steps. But we need to know what we are aiming for.
[Student]: ah
[Professor]: You know what f1 and f2 are. They are odd functions. But exactly what is this new function that we call f1*f2?
[Student]: idk f1(x) * f2(x)
[Professor]: Exactly!
[Student]: or f1(-x)
[Professor]: No
[Student]: * f2(-x)
[Student]: no ok
[Professor]: So we know f1*f2(x) = f1(x)*f2(x)
[Student]: yep
[Professor]: And it looks like you were in the middle of the thought: f1*f2(-x) = f1(-x)*f2(-x)
[Student]: ya
[Professor]: Okay, things are starting to come together.
[Student]: sort of
[Professor]: We want to show f1*f2(-x)=f1*f2(x)
[Student]: substitution?
[Professor]: Yep. Back to the proof. We need to use substitution to get our final result. But we do it one step at a time.

[Professor]: I'll type the left hand side, you type the right hand side.
[Professor]: f1*f2(-x) = ?
[Student]: f1(-x) * f2(-x)
[Professor]: Good. But f1(-x) = ? and f2(-x) = ?
[Student]: f1(x) * f2(x)
[Student]: oh
[Professor]: And finally, f1(x)*f2(x) = ?
[Student]: f1*f2(x)
[Professor]: Precisely. Putting those together, we now know f1*f2(-x) = f1*f2(x).
[Professor]: And that means ...
[Professor]: f1 * f2 is even
[Professor]: Q.E.D. :-)
[Student]: sooo i guess i then assemble all of those steps or just some of them
[Professor]: You can leave out the parts that I said belonged as scratch work.
[Professor]: There were four lines associated with the given information. Then I said "back to the proof" and there were probably four more lines. That is the proof.
[Student]: ic ok
[Student]: ugh, this is def not my fav stuff, i cant believe i haven't learned this before

In summary, here is our proof:
f1 is an odd function
f1(-x) = -f1(x) for x in domain
f2 is an odd function
f2(-x) = -f2(x) for x in domain
f1*f2(-x) = f1(-x) * f2(-x)
= (-f1(x))*(-f2(x))
= f1(x)*f2(x)
= f1*f2(x)
So f1*f2(-x) = f1*f2(x) for x in domain
f1*f2 is an even function

Now, back to the discussion. Another problem dealt with composition. That is, students were to prove: If f1 is an odd function and g1 is an even function, then g1•f1 is an even function. Here is where part of that discussion went.

[Student]: so how do the g1(f1(x)) differ, i guess proving that f1(x) is odd, and then proving then that g1(x) is odd because of f1(x)?
[Professor]: One key point is that the definition of odd or even --- f(-x) = -f(x) and g(-x)=g(x) --- is that the x is simply a place-holder and the same statement would be true regardless of what is in the place of the x.
[Professor]: So for example f(-(x+2)) = -f(x+2), where the "x" was actually the formula x+2.
[Student]: right
[Professor]: Now, be more specific on your question related to composition.
[Student]: so once i prove that f1 is even then it would be the same justification for g1
[Professor]: Careful! f1 is odd (from the given information), so you can't prove it is even.
[Student]: well, opposite justification
[Student]: I'm still not sure of the question.
[Professor]: g1 alone is known to be even. There is nothing to prove about that.
[Student]: ok
[Student]: timeout, so all id have to do is justify g1 as being even regardless of f1?
[Professor]: Why? What are your steps?
[Professor]: And what are you trying to prove?
[Student]: so its given that g1 is even and f1 is odd and we're trying to prove that g1(f1(x)) is even
[Professor]: Technically, g1•f1 is even.
[Professor]: g1(f1(x)) is a value, not a function.
[Professor]: So you need to show g1•f1(-x) = g1•f1(x).
[Student]: wait im confused
[Student]: ok
[Student]: i get it
[Student]: couldnt you just subsitute -x for f1(-x)
[Professor]: No for what you said. f1(-x) and -x are different, so they don't substitute.
[Professor]: But I don't think that is what you were thinking. Try to restate.
[Student]: you could justify g1(f1(x)) = g1(f1(x)) with g(x)=g(-x)
[Student]: -f1(x)*
[Professor]: I think you're on the right track. But be careful that you go step by step.
[Professor]: What are all the steps?
[Professor]: I'll start it off... g1•f1(-x) = ?
[Student]: thats where im confused, do you need a a step for that?
[Professor]: Yes. You must relate the function (g1•f1) to the rest of the formulas.
[Student]: g1(f1(-x))= g1(f1(x) )
[Professor]: What justifies that? The -x belongs to f1, not g1
[Student]: g1(-f1(x)) def of odd fcn
[Professor]: So don't skip that step
[Student]: so whats after that then?
[Professor]: Well, why don't you summarize the statements so far. Start again with: g1•f1(-x) = ...
[Student]: g1(f1(-x)) = g1(-f1(x)) b/c of the def of odd fcns
[Professor]: Very good. Now, what does g1 do when you have g1(-[anything])?
[Student]: g1(-x)=-g1(x) ? but that would make it odd
[Professor]: So use the fact that g1 is even. g1(-x)=?
[Student]: g1(x)
[Professor]: So g1(-U) = g1(U) or g1(-f1(x)) = g1(f1(x))
[Student]: k
[Professor]: It doesn't matter what appears.
[Professor]: g1(-[stuff]) = g1([stuff])
[Professor]: So you left off at: g1 o f1(-x) = g1(f1(-x)) = g1(-f1(x)) = ... (finish it off)
[Student]: g1(-f1(x))=g1(f1(x))
[Professor]: And how does that relate to g1•f1?
[Student]: g1(f1(x)) = even
[Professor]: not equal even. g1(f1(x)) = g1• f1(x).
[Professor]: Recall, you are working with the function g1•f1. You need to show g1•f1(-x) = g1•f1(x).
[Student]: i just got a little confused
[Student]: whats the diff between g1(f1(x)) = g1•f1(x)
[Student]: ah i c nev mind
[Professor]: same value but only g•f can be called the name of the function
[Student]: ok
[Student]: so g1•f1(x) = g1•f1(-x)
[Professor]: That would be the final line to show that g1•f1 is an even function.
[Student]: ok that makes sense

And that leads us to our second proof:
f1 is an odd function
f1(-x) = -f1(x) for x in domain
g1 is an even function
g1(-x) = g1(x) for x in domain
g1•f1(-x) = g1(f1(-x))
= g1(-f1(x))
= g1(f1(x)
= g1•f1(x)
So g1•f1(-x) = g1•f1(x) for x in domain
g1•f1 is an even function

Do you know how to do the justifications of each line now?

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