5
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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?

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    $\begingroup$ 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. $\endgroup$
    – Braindead
    Jan 4, 2015 at 18:39
  • $\begingroup$ @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. $\endgroup$
    – renu
    Jan 5, 2015 at 8:23

1 Answer 1

5
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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 $2e$.)

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  • $\begingroup$ Nice! Also thanks for clearing up my earlier misreading of the question. $\endgroup$ Jan 5, 2015 at 7:40
  • $\begingroup$ @mrf thankx sir, now my doubt regarding this type of problem is clear. $\endgroup$
    – renu
    Jan 5, 2015 at 8:27

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