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Given the function f(x) = {0, -pi le x < 0 \\ 4, 0 le x < pi. Find the first three nonzero terms...

Question:

Given the function {eq}f(x) = \begin{cases} 0,& -\pi \le x < 0 \\ 4,& 0 \le x < \pi \end{cases} {/eq}. Find the first three nonzero terms of the Fourier series expansion of this function

Fourier Series:

The trigonometric series {eq}\displaystyle \frac{a_0}{2}+\sum_{n=1 }^{\infty }\left \{ a_n \ \cos nx+b_n \ \sin nx \right \} {/eq} is the Fourier series representation of {eq}f(x) {/eq} on {eq}(-\pi, \pi). {/eq}


{eq}a_0, \ a_n {/eq} and {eq}b_n {/eq} determine the coefficients of the series.

To find those coefficients, we can use the following formulas:

{eq}\displaystyle a_0=\frac{1}{\pi }\int_{-\pi }^{\pi }f(x) \ dx {/eq}


{eq}\displaystyle a_n=\frac{1}{\pi }\int_{-\pi }^{\pi }f(x) \ \cos nx \ dx {/eq}


{eq}\displaystyle b_n=\frac{1}{\pi }\int_{-\pi }^{\pi }f(x) \ \sin nx \ dx {/eq}

Answer and Explanation: 1

Given {eq}f(x) = \begin{cases} 0,& -\pi \le x < 0 \\ 4,& 0 \le x < \pi \end{cases} {/eq}

Here {eq}f(x) {/eq} is defined on {eq}(-\pi,\pi). {/eq}

Thus the Fourier series of {eq}f(x) {/eq} is {eq}\boldsymbol{\displaystyle f(x)=\frac{a_0}{2}+\sum_{n=1 }^{\infty }\left \{ a_n \ \cos nx+b_n \ \sin nx \right \}}. {/eq}


Now:

{eq}\begin{align*} a_0 &=\frac{1}{\pi }\int_{-\pi }^{\pi }f(x) \ dx & \left [ \text{Function is defined in two ways on two intervals, so we need to split the integral} \right ]\\ \\ &=\frac{1}{\pi }\left ( \int_{-\pi }^{0}0 \ dx+ \int_{0}^{\pi }4 \ dx \right ) & \left [ \text{First integral vanishes} \right ]\\ \\ &=\frac{1}{\pi }\left [ 4x \right ]_{0}^{\pi }\\ \\ &=\frac{4}{\pi }\left [ \pi-0 \right ] & \left [ \text{Applying the limits of integration} \right ]\\ \\ &=4. \end{align*} {/eq}


{eq}\begin{align*} a_n &=\frac{1}{\pi }\int_{-\pi }^{\pi }f(x) \ \cos nx \ dx\\ \\ &=\frac{1}{\pi }\left ( \int_{-\pi }^{0}0 \ \cos nx dx+ \int_{0}^{\pi }4 \ \cos nx \ dx \right ) & \left [ \text{First integral vanishes} \right ]\\ \\ &=\frac{4}{\pi }\left [ \frac{\sin nx }{n} \right ]_{0}^{\pi } & \left [ \because \displaystyle \int \cos ax \ dx=\frac{ \sin ax}{a}+c \right ]\\ \\ &=\frac{4}{n \pi }\left [ \sin n \pi-\sin 0 \right ] & \left [ \text{Applying the limits of integration} \right ]\\ \\ &=0 & \left [ \because \sin n\pi=0 \right ]. \end{align*} {/eq}



{eq}\begin{align*} b_n &=\frac{1}{\pi }\int_{-\pi }^{\pi }f(x) \ \sin nx \ dx\\ \\ &=\frac{1}{\pi }\left ( \int_{-\pi }^{0}0 \ \sin nx dx+ \int_{0}^{\pi }4 \ \sin nx \ dx \right ) & \left [ \text{First integral vanishes} \right ]\\ \\ &=\frac{4}{\pi }\left [ \frac{-\cos nx }{n} \right ]_{0}^{\pi } & \left [ \because \displaystyle \int \sin ax \ dx=\frac{- \cos ax}{a}+c \right ]\\ \\ &=\frac{-4}{n \pi }\left [ \cos n \pi-\cos 0 \right ] & \left [ \text{Applying the limits of integration} \right ]\\ \\ &=\frac{-4}{n \pi }\left [ (-1)^n-1 \right ] & \left [ \because \cos n\pi=(-1)^n \right ]\\ \\ b_n&=\frac{4}{n \pi }\left [ 1-(-1)^n \right ]. \end{align*} {/eq}


Applying the above coefficients in the series, we obtain:

{eq}\begin{align*} f(x) &=\frac{4}{2}+\sum_{n=1 }^{\infty }\left \{ 0 \ \cos nx+\frac{4}{n \pi }\left [ 1-(-1)^n \right ] \ \sin nx \right \}\\ \\ &=2+\frac{4}{ \pi }\sum_{n=1 }^{\infty }\left \{ \frac{1}{n }\left [ 1-(-1)^n \right ] \ \sin nx \right \}\\ \\ &=2+\frac{4}{ \pi } \left \{ \frac{1}{1 }\left [ 1-(-1) \right ] \ \sin x+ \frac{1}{2 }\left [ 1-1 \right ] \ \sin 2x+ \frac{1}{3 }\left [ 1-(-1) \right ] \ \sin 3x+ \frac{1}{4 }\left [ 1-1 \right ] \ \sin 4x+... \right \} & \left [ \text{Expand sum notation} \right ]\\ \\ &=2+\frac{4}{ \pi } \left \{ 2 \sin x+0+ \frac{2}{3 } \sin 3x+ 0+... \right \}\\ \\ &=2+\frac{8}{ \pi } \left \{ \sin x+ \frac{1}{3 } \sin 3x+ ... \right \}. \end{align*} {/eq}


Therefore the first three non-zero terms are {eq}{\color{Blue} {\displaystyle 2, \ \frac{8}{ \pi } \sin x, \ \frac{8}{3 \pi } \sin 3x.}} {/eq}


Learn more about this topic:

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Evaluating Definite Integrals Using the Fundamental Theorem

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Chapter 16 / Lesson 2
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The fundamental theorem of calculus makes finding your definite integral almost a piece of cake. See how the definite integral becomes a subtraction problem after applying the fundamental theorem of calculus.


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