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Ghella [55]
3 years ago
12

Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Mathematics
1 answer:
Zolol [24]3 years ago
3 0
f(x)=\begin{cases}1&\text{for }-7

The Fourier series expansion of f(x) is given by

\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7

where we have

a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx
a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)
a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2

The coefficients of the cosine series are

a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx
a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)
a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}
a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}

When n is even, the numerator vanishes, so we consider odd n, i.e. n=2k-1 for k\in\mathbb N, leaving us with

a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}

Meanwhile, the coefficients of the sine series are given by

b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx
b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)
b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}
b_n=\dfrac{7(-1)^{n+1}}{n\pi}

So the Fourier series expansion for f(x) is

f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7
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