 # Suppose f has period 2 and f(x) = x for 0 \leq x&lt.2. Find the fourth-degree Fourier polynomial and graph it on 0 \leq x&lt.2. Line 2021-08-09 Answered

Suppose f has period 2 and f(x) = x for $0\le x<2$. Find the fourth-degree Fourier polynomial and graph it on $0\le x<2$

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Step 1
The given function
$f\left(x\right)=x,0\le x<2$
Since we know the general expression of Fourier series for f on $\left[-\frac{b}{2},\frac{b}{2}\right]$ is
${a}_{0}+\sum _{k=1}^{\mathrm{\infty }}{a}_{k}\mathrm{cos}\left(\frac{2\pi kx}{b}\right)+\sum _{k=1}^{\mathrm{\infty }}{b}_{k}\mathrm{sin}\left(\frac{2\pi kx}{b}\right)$
Where constant expressed as
${a}_{0}=\frac{1}{b}{\int }_{-\frac{b}{2}}^{\frac{b}{2}}f\left(x\right)dx,{a}_{k}=\frac{2}{b}{\int }_{-\frac{b}{2}}^{\frac{b}{2}}f\left(x\right)\mathrm{cos}\left(kx\right)dx$, for $k>0$
${b}_{k}=\frac{2}{b}{\int }_{-\frac{b}{2}}^{\frac{b}{2}}f\left(x\right)\mathrm{sin}\left(kx\right)dx$, for $k>0$
Now the value of constant
${a}_{0}=\frac{1}{2}\left[{\int }_{0}^{2}xdx\right]=\frac{1}{2}\left[\frac{{x}^{2}}{2}{\mid }_{0}^{2}\right]=1$
Furthermore
${a}_{1}=\frac{4}{2}\left[{\int }_{0}^{2}x\mathrm{cos}\pi xdx\right]=2\left[\left(\frac{x\mathrm{sin}\pi x}{\pi }+\frac{\mathrm{cos}\pi x}{{\pi }^{2}}\right){\mid }_{0}^{2}\right]=0$
And
${b}_{1}=\frac{4}{2}\left[{\int }_{0}^{2}x\mathrm{sin}\pi xdx\right]=2\left[\left(\frac{-x\mathrm{cos}\pi x}{\pi }+\frac{\mathrm{sin}\pi x}{{\pi }^{2}}\right){\mid }_{0}^{2}\right]=-\frac{4}{\pi }$
Therefore the fourier polynomial of degree 1 is given by