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Kay [80]
3 years ago
15

Find the half range Fourier sine series of the function

Mathematics
1 answer:
worty [1.4K]3 years ago
5 0
The full range is -\pi (length 2L=2\pi), so the half range is L=\pi. The half range sine series would then be given by

f(x)=\displaystyle\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L=\sum_{n\ge1}b_n\sin nx

where

b_n=\displaystyle\frac2L\int_0^Lf(x)\sin\dfrac{n\pi x}L\,\mathrm dx=\frac2\pi\int_0^\pi(\pi-x)\sin nx\,\mathrm dx

Essentially, this is the same as finding the Fourier series for the function

\begin{cases}g(x)=\begin{cases}\pi-x&\text{for }0

Integrating by parts yields

b_n=\dfrac2\pi\left(\dfrac\pi n-\dfrac{\sin n\pi}{n^2}\right)=\dfrac2n

So the half range sine series for this function is simply

f(x)=\displaystyle\sum_{n\ge1}\frac{2\sin nx}n
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If the legs of a right triangle are x, (2x - 1) and its hypotenuse is (2x + 1), what is the value of x?​
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