# Doubt in Ivan Niven's proof of irrationality of pi.

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?

• Is a strict upper bound. The maximum isn't required. And your supposed maximum us highly suspicious. – Martín-Blas Pérez Pinilla Mar 17 '14 at 11:56
• @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. – DiffeoR Mar 17 '14 at 12:18
• @DiffeoR, true. I overlooked the symmetry of $f$. – Martín-Blas Pérez Pinilla Mar 17 '14 at 12:44
• 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. – Mobius Mar 20 '14 at 5:16
• DiffeoR and Martín-BlasPérezPinilla, please check out the answer below. As Martín-Blas Pérez Pinilla said, it is a strict upper bound. – Mobius Mar 20 '14 at 5:24

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 $a$.