Paper 4, Section I, H

Topics in Analysis
Part II, 2019

Show that π\pi is irrational. [Hint: consider the functions fn:[0,π]Rf_{n}:[0, \pi] \rightarrow \mathbb{R} given by fn(x)=xn(πx)nsinx.]\left.f_{n}(x)=x^{n}(\pi-x)^{n} \sin x .\right]