- #1
demonelite123
- 216
- 0
Suppose f(z) is analytic in a region R including the point z0. Prove that f(z) = f(z0) + f'(z0)(z-z0) + η(z-z0) where η ~> 0 as z ~> z0.
this is actually a lemma my book proves first before actually proving L'Hospital's rule. I understood how they used the lemma to prove the rule but i don't really understand the logic in proving this lemma. my book did:
Let [f(z) - f(z0)]/(z-z0) - f'(z0) = η so that f(z) = f(z0) + f'(z0)(z - z0) = η(z-z0).
Then, since f(z) is analytic at z0, we have as required:
lim (z ~> z0) of η = lim (z ~> z0) of [f(z) - f(z0)]/(z-z0) - f'(z0) = f'(z0) - f'(z0) = 0.
i don't understand how f(z) = f(z0) + f'(z0)(z - z0) = η(z-z0). shouldn't it be f(z) = η(z-z0) + f'(z0)(z - z0) + f(z0) since they let [f(z) - f(z0)]/(z-z0) - f'(z0) = η?
this is actually a lemma my book proves first before actually proving L'Hospital's rule. I understood how they used the lemma to prove the rule but i don't really understand the logic in proving this lemma. my book did:
Let [f(z) - f(z0)]/(z-z0) - f'(z0) = η so that f(z) = f(z0) + f'(z0)(z - z0) = η(z-z0).
Then, since f(z) is analytic at z0, we have as required:
lim (z ~> z0) of η = lim (z ~> z0) of [f(z) - f(z0)]/(z-z0) - f'(z0) = f'(z0) - f'(z0) = 0.
i don't understand how f(z) = f(z0) + f'(z0)(z - z0) = η(z-z0). shouldn't it be f(z) = η(z-z0) + f'(z0)(z - z0) + f(z0) since they let [f(z) - f(z0)]/(z-z0) - f'(z0) = η?
… + and = are the same key on most keyboards! 
