Suppose f has period 2 and f(x) = x for 0≤x&lt;2. Find the fourth-degree Fourier polynomial and graph it on 0≤x&lt;2

Step 1 The given function f(x)=x,0≤x&lt;2 Since we know the general expression of Fourier series for f on [−b2,b2] is a0+∑k=1∞akcos⁡(2πkxb)+∑k=1∞bksin⁡(2πkxb) Where constant expressed as a0=1b∫−b2b2f(x)dx,ak=2b∫−b2b2f(x)cos⁡(kx)dx, for k&gt;0 bk=2b∫−b2b2f(x)sin⁡(kx)dx, for k&gt;0 Now the value of constant a0=12[∫02xdx]=12[x22∣02]=1 Furthermore a1=42[∫02xcos⁡πxdx]=2[(xsin⁡πxπ+cos⁡πxπ2)∣02]=0 And b1=42[∫02xsin⁡πxdx]=2[(−xcos⁡πxπ+sin⁡πxπ2)∣02]=−4π Therefore the fourier polynomial of degree 1 is given by

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2021-08-09

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

Answer & Explanation

Brighton

Skilled2021-08-10Added 103 answers

Step 1
The given function
$f (x) = x, 0 \leq x < 2$
Since we know the general expression of Fourier series for f on $[- \frac{b}{2}, \frac{b}{2}]$ is
$a_{0} + \sum_{k = 1}^{\infty} a_{k} \cos (\frac{2 π k x}{b}) + \sum_{k = 1}^{\infty} b_{k} \sin (\frac{2 π k x}{b})$
Where constant expressed as
$a_{0} = \frac{1}{b} \int_{- \frac{b}{2}}^{\frac{b}{2}} f (x) d x, a_{k} = \frac{2}{b} \int_{- \frac{b}{2}}^{\frac{b}{2}} f (x) \cos (k x) d x$ , for $k > 0$
$b_{k} = \frac{2}{b} \int_{- \frac{b}{2}}^{\frac{b}{2}} f (x) \sin (k x) d x$ , for $k > 0$
Now the value of constant
$a_{0} = \frac{1}{2} [\int_{0}^{2} x d x] = \frac{1}{2} [\frac{x^{2}}{2} ∣_{0}^{2}] = 1$
Furthermore
$a_{1} = \frac{4}{2} [\int_{0}^{2} x \cos π x d x] = 2 [(\frac{x \sin π x}{π} + \frac{\cos π x}{π^{2}}) ∣_{0}^{2}] = 0$
And
$b_{1} = \frac{4}{2} [\int_{0}^{2} x \sin π x d x] = 2 [(\frac{- x \cos π x}{π} + \frac{\sin π x}{π^{2}}) ∣_{0}^{2}] = - \frac{4}{π}$
Therefore the fourier polynomial of degree 1 is given by

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