Question

write half range sine series of the function f(x)=x, in 0<x<2.​

Answer 1

Answer:

$\boxed{\displaystyle f(x) = \dfrac{-4}{\pi}\left(\dfrac{-\sin(\pi x)}{1} + \dfrac{\sin(2\pi x)}{2} \cdotp\cdotp\cdotp\cdotp\cdotp\cdotp}$

Step-by-step explanation:

For half range sine series,

$\displaystyle f(x) = \sum_{n=1}^\infty b_n \sin\left(\dfrac{n\pi x}{2}\right)$

where, $b_n$ can be given as:

$\displaystyle b_n = \dfrac{2}{2}\int_0^2x\sin\left(\dfrac{n\pi x}{2}\right) dx$

Using By-parts, i.e ILATE,

$\displaystyle b_n = \left[-x\dfrac{\cos\left(\frac{n\pi x}{2}\right)}{\frac{n\pi}{2}} + \dfrac{\sin\left(\frac{n\pi x}{2}\right)}{\frac{n^2\pi^2}{4}}\right]_0^2$

$\displaystyle b_n = \left[-2\dfrac{\cos\left(n\pi\right)}{\frac{n\pi}{2}} + \dfrac{\sin\left(n\pi\right)}{\frac{n^2\pi^2}{4}}\right] - 0$

$b_n = \dfrac{-2(-1)^n}{\frac{n\pi}{2}} = \dfrac{-4}{n\pi}(-1)^n$

Now,

$\displaystyle f(x) =\dfrac{-4}{\pi}\sum_n^\infty\dfrac{(-1)^n}{n}\sin\left(\dfrac{n\pi x}{2}\right)$

Required half range sine series is:

$\boxed{\displaystyle f(x) = \dfrac{-4}{\pi}\left(\dfrac{-\sin(\pi x)}{1} + \dfrac{\sin(2\pi x)}{2} \cdotp\cdotp\cdotp\cdotp\cdotp\cdotp}$