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In the proof, how do we get the upper limit for $f(x) \sin{x}$ as $\pi^n \cdot \frac{a^n}{n!}$ ?

I thought $f(x) \sin{x}$ would be maximum at $x=\pi/2$ when its value would be: $$\pi^n \cdot \frac{a^n}{2^{2n}n!}$$

Where am I going wrong?

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    $\begingroup$ Is a strict upper bound. The maximum isn't required. And your supposed maximum us highly suspicious. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 17 '14 at 11:56
  • $\begingroup$ @Martín-BlasPérezPinilla : The maximum evaluated by Moibus is correct. As $f(x)$ is a symmetric function on $[0,\pi]$ attaining maximum at $x = \frac{\pi}{2}$, and $\sin{x}$ also attains maximum at $x = \pi/2$, hence their product attains maximum also at $x = \pi/2$. So now we can just put $x = \pi/2 = a/2b$. But your observation is indeed correct. One upvote from me. $\endgroup$ – DiffeoR Mar 17 '14 at 12:18
  • $\begingroup$ @DiffeoR, true. I overlooked the symmetry of $f$. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 17 '14 at 12:44
  • $\begingroup$ Thanks to naslundx, Git Gud and Sabyasachi for editing the question and making it more readable. I was not familiar with LaTeX. Now I have used it in the answer below. $\endgroup$ – Mobius Mar 20 '14 at 5:16
  • $\begingroup$ DiffeoR and Martín-BlasPérezPinilla, please check out the answer below. As Martín-Blas Pérez Pinilla said, it is a strict upper bound. $\endgroup$ – Mobius Mar 20 '14 at 5:24
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The upper limit of $f(x) \sin{x}$ mentioned in the proof is not the least upper bound.

The least upper bound is $$\pi^n \cdot \frac{a^n}{2^{2n}n!}$$ as mentioned in the question.

Given $f(x) = \frac{x^n.(a-bx)^n}{n!}$ a general argument for getting the upper bound mentioned in the proof would be that the maximum value of $x$ is $\pi$ and the maximum value of $(a-bx)$ is $a$.

After posting this question here, I found an explanation of the limits of the proof in Presh Talwalkar's blog

I posted my doubts in the comments section and Presh was kind enough to respond and clarify my doubt.

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