Assigning a Piecewise Function in a ProofDate: 01/11/2001 at 14:24:25 From: Mahesh Bhat Subject: Chain rule proof Hi, As I was trying to teach my niece the chain rule, I came across a step which I have trouble understanding. I'll narrate the proof until the step with which I have the problem. Proof: lim f o g(x) - f o g(c) lim f(g(x)) - f(g(c)) (f o g)'(c) = x->c ------------------- = x->c ------------------ x - c x - c We must show that the limit on the right hand side has the value f'(g(c)).g'(x). Note that if we knew that g(x) - g(c) is not equal to zero for all x near c, x not equal to c we could write: f(g(x)) - f(g(c)) f(g(x)) - f(g(c)) g(x) - g(c) ------------------ = ----------------- . ----------- x - c g(x) - g(c) x - c and show that the first factor on the right has limit f'(g(c)) and the other has g'(c). The problem is that g(x) - g(c) could be zero even if x is not equal to c. (Now the step I do not understand.) =================================== We introduce a function: / | f(u) - f(g(c)) | -------------- - f'(g(c)), when u <> g(c) e(u) = { u - g(c) | | 0, when u = g(c) \ How can we say that e(u) is 0 when u = g(c)? The only way I thought that could happen is if the numerator of the first term is (u - g(c)).f'(g(c)) = f(u) - f(g(c)). Even so how can I be sure that the f(u) - f(g(c)) can have (u - g(c)) and f'(g(c)) as its factors? Can we guarantee that this manipulation will always be true? Even though we have introduced e(u), how can we have a rule that conveniently says e(u) = 0 when u = g(c)? I have seen other proofs, but I would like to understand this step. When I was in school I probably just accepted this step and went on. Thank you, Mahesh Date: 01/11/2001 at 14:32:25 From: Doctor Schwa Subject: Re: Chain rule proof >How can we say that e(u) is 0 when u = g(c)? This is simply the DEFINITION we have given e(u). Depending on the value of u, it has different definitions. It's like saying: f(x) = {x^2 when x is positive, x when x is negative}. There's no *reason* f has to be defined that way; that's just the definition. >The only way I thought that could happen is if the numerator of the >first term is (u - g(c)).f'(g(c)) = f(u) - f(g(c)). Even so how can I >be sure that the f(u) - f(g(c)) can have (u - g(c)) and f'(g(c)) as >its factors? Can we guarantee that this manipulation will always be >true? So you see there's no manipulation to be done. It's just the definition of a new function. We choose to define it any way we want to. We define it this particular way to patch up the hole at g(x) = g(c). Thanks for your excellent question and your careful attention to detail. That's an important part of mathematics! You're right that on seeing it for the first time most people probably just accept this step on faith, memorize it, or ignore it... but for future mathematicians, this kind of thinking is important. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/