5

I just have answer of this question which is 6, but I don't know how to arrive at this answer. Please anyone help me solve this. How does one calculate the value of this function?

Braindead
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renu
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    Are you familiar with the Cauchy integral formula or the Cauchy estimate? It can be shown that $|(e^f)''(0)|$ must be less than $2e$ using this method. Let me know if you aren't sure and I can elaborate. – Braindead Jan 04 '15 at 18:39
  • @Braindead yes sir,I am familiar with Cauchy Integral Formula and with help of this I got my answer also. And thankx sir for giving time to this problem. – renu Jan 05 '15 at 08:23

1 Answers1

5

Let $g(z) = \exp(f(z))$. By Cauchy's integral formula, $$ g''(0) = \frac{2!}{2\pi i} \int_{|z|=1} \frac{g(z)}{z^{3}}\,dz $$ so by the "ML"-inequality and the estimate $|g(z)| \le e$, we have $$ |g''(0)| \le \frac{2!}{2\pi} \cdot 2\pi \max_{|z|=1} \frac{g(z)}{z^{3}} \le 2e. $$

In particular, $|g''(0)| < 6$. (All the other options are smaller in modulus than $2e$.)

mrf
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