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[http://www.math.purdue.edu/~bell/MA530/hwk1.pdf HWK 1 problems] | [http://www.math.purdue.edu/~bell/MA530/hwk1.pdf HWK 1 problems] | ||
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| + | Hint for complex chain rule: | ||
| + | |||
| + | Let | ||
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| + | <math>\frac{f(z)-f(a)}{z-a} - f'(a) = E(z).</math> | ||
| + | |||
| + | Then | ||
| + | |||
| + | <math>f(z)= f(a)+f'(a)(z-a)+E(z)(z-a).</math> | ||
Revision as of 06:31, 13 January 2010
Homework 1
Hint for complex chain rule:
Let
$ \frac{f(z)-f(a)}{z-a} - f'(a) = E(z). $
Then
$ f(z)= f(a)+f'(a)(z-a)+E(z)(z-a). $
