Question

By using the sine series for f(x)=1 in 0<x<π. show that π²/8=1+1/3²+1/5²+1/7²+...​

Answer 1

Answer:

$\pi^{2} / 8 &=1+1 / 3^{2}+1 / 5^{2}+1 / 7^{2}+\ldots \ldots .\end{aligned}$

Step-by-step explanation:

Given: sine series

To find:

$\pi^{2} / 8=1+1 / 3^{2}+1 / 5^{2}+1 / 7^{2}+1 / 9^{2}+\ldots \ldots$

Solution:

By using the sine series to $f(x)=1 in 0 < x < \pi$ show that

$\pi^{2} / 8=1+1 / 3^{2}+1 / 5^{2}+1 / 7^{2}+1 / 9^{2}+\ldots \ldots$

Sine series is $f(x)=\Sigma b_{n} \sin n x$

$\begin{aligned}b_{n} &=(2 / \pi) \pi \int_{0} f(x) \sin n x d x \\&=(2 / \pi) \pi \int_{0}(1) \sin n x d x=(2 / \pi)(-\cos n x / n)_{0} \pi \\&=(-2 / n \pi)[\cos n \pi-1]=(-2 / n \pi)\left[(-1)^{n}-1\right] \\&=-4 / n \pi \quad \quad \text { if } n \text { is odd }\end{aligned}$

Then, the sine series is

$1=(4 / \pi) \sin x+(4 / 3 \pi) \sin 3 x+(4 / 5 \pi) \sin 5 x+(4 / 7 \pi) \sin 7 x+\ldots \ldots \ldots$

$c \int_{0}[f(x)]^{2} d x=c / 2\left[b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+b_{4}^{2}+b_{5}^{2}+\ldots .\right]$

$\begin{aligned}&c \int_{0}[1]^{2} d x=\pi / 2\left[(4 / \pi)^{2}+(4 / 3 \pi)^{2}+(4 / 5 \pi)^{2}+(4 / 7 \pi)^{2}+\ldots \ldots \ldots\right] \\&{[x]_{0}^{\pi}=(\pi / 2)\left(16 / \pi^{2}\right)\left[1+1 / 3^{2}+1 / 5^{2}+1 / 7^{2}+\ldots \ldots \ldots\right]}\end{aligned}$

$\begin{aligned}{[x]_{0}{ }^{\pi} } &=(\pi / 2)\left(16 / \pi^{2}\right)\left[1+1 / 3^{2}+1 / 5^{2}+1 / 7^{2}+\ldots \ldots . . .\right] \\\pi &=(\pi / 2)\left(16 / \pi^{2}\right)\left[1+1 / 3^{2}+1 / 5^{2}+1 / 7^{2}+\ldots \ldots \ldots\right] \\\pi^{2} / 8 &=1+1 / 3^{2}+1 / 5^{2}+1 / 7^{2}+\ldots \ldots .\end{aligned}$

Hence it is proved