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Let | Let | ||
| − | <math>\frac{f(z)-f(a)}{z-a} - f'(a) = | + | <math>\frac{f(z)-f(a)}{z-a} - f'(a) = E_f(z).</math> |
Then | Then | ||
| − | <math>f(z)= f(a)+f'(a)(z-a)+ | + | <math>f(z)= f(a)+f'(a)(z-a)+E_f(z)(z-a).</math> |
| + | |||
| + | Note that if <math>f(z)</math> is complex differentiable at <math>a</math>, then | ||
| + | |||
| + | <math>\lim_{z\to a} E_f(z)=0.</math> | ||
| + | |||
| + | Next, let | ||
| + | |||
| + | <math>\frac{g(w)-g(A)}{w-A} - g'(A) = E_g(w),</math> | ||
| + | |||
| + | where <math>A=f(a).</math> | ||
| + | |||
| + | Now | ||
| + | |||
| + | <math>g(w)= g(A)+g'(A)(w-A)+E_g(w)(w-A).</math> | ||
| + | |||
| + | To prove the Chain Rule, let <math>w=f(z)</math> in the formula above and write out the difference quotient for <math>g(f(z))</math>. The limit should become obvious. You will need to use the fact that <math>f</math> complex differentiable at <math>a</math> implies that <math>f</math> is continuous at <math>a</math> in order to deduce that <math>E_g(f(z))</math> tends to zero as <math>z</math> tends to <math>a</math>. | ||
Revision as of 08:06, 13 January 2010
Homework 1
Hint for complex chain rule:
Let
$ \frac{f(z)-f(a)}{z-a} - f'(a) = E_f(z). $
Then
$ f(z)= f(a)+f'(a)(z-a)+E_f(z)(z-a). $
Note that if $ f(z) $ is complex differentiable at $ a $, then
$ \lim_{z\to a} E_f(z)=0. $
Next, let
$ \frac{g(w)-g(A)}{w-A} - g'(A) = E_g(w), $
where $ A=f(a). $
Now
$ g(w)= g(A)+g'(A)(w-A)+E_g(w)(w-A). $
To prove the Chain Rule, let $ w=f(z) $ in the formula above and write out the difference quotient for $ g(f(z)) $. The limit should become obvious. You will need to use the fact that $ f $ complex differentiable at $ a $ implies that $ f $ is continuous at $ a $ in order to deduce that $ E_g(f(z)) $ tends to zero as $ z $ tends to $ a $.
