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Question

The subject of Fourier series is concerned with the representation of a

2π -periodic 2 \pi \text { -periodic }

function f as the following infinite linear combination of the set of functions

{1,sinnx,cosnx}n=1:f(x)=12a0+n=1(ancosnx+bnsinnx).(5.2.5)\{ 1 , \sin n x , \cos n x \} _ { n = 1 } ^ { \infty } : \\ f ( x ) = \frac { 1 } { 2 } a _ { 0 } + \sum _ { n = 1 } ^ { \infty } \left( a _ { n } \cos n x + b _ { n } \sin n x \right) . \quad ( 5.2 .5 )

In this problem, we investigate the possibility of performing such a representation. (a) Use appropriate trigonometric identities, or some form of technology, to verify that the set of functions

{1,sinnx,cosnx}n=1\{ 1 , \sin n x , \cos n x \} _ { n = 1 } ^ { \infty }

is orthogonal on the interval

[π,π].[ - \pi , \pi ].

(b) By multiplying (5.2.5) by cos mx and integrating over the interval

[π,π],[ - \pi , \pi ],

show that

a0=1πππf(x)dx and am=1πππf(x)cosmx dx.a _ { 0 } = \frac { 1 } { \pi } \int _ { - \pi } ^ { \pi } f ( x ) d x \text { and } a _ { m } = \frac { 1 } { \pi } \int _ { - \pi } ^ { \pi } f ( x ) \cos m x\ d x.

(c) Use a similar procedure to show that

bm=1πππf(x)sinmx dx.b _ { m } = \frac { 1 } { \pi } \int _ { - \pi } ^ { \pi } f ( x ) \sin m x\ d x.

It can be shown that if f is in

C1(π,π),C ^ { 1 } ( - \pi , \pi ),

then Equation (5.2.5) holds for each

x(π,π).x \in ( - \pi , \pi ).

The series appearing on the right-hand side of (5.2.5) is called the Fourier series of f, and the constants in the summation are called the Fourier coefficients for f. (d) Show that the Fourier coefficients for the function

f(x)=x,π<xπ,f(x+2π)=f(x),f ( x ) = x , - \pi < x \leq \pi , f ( x + 2 \pi ) = f ( x ),

are

an=0,n=0,1,2,bn=2ncosnπ,n=1,2,\begin{array} { l l } { a _ { n } = 0 , } & { n = 0,1,2 , \ldots } \\ { b _ { n } = - \frac { 2 } { n } \cos n \pi , } & { n = 1,2 , \ldots } \end{array}

and thereby determine the Fourier series of f. (e) Using some form of technology, sketch the approximations to f(x) = x on the interval

(π,π)( - \pi , \pi )

obtained by considering the first three terms, first five terms, and first ten terms in the Fourier series for f. What do you conclude?

Solution

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a) We have the general results

sin(mx)sin(nx)dx=nsin(mx)cos(nx)mcos(mx)sin(nx)m2n2\begin{equation} \int \sin (m x) \sin (n x) \, dx=\frac{n \sin (m x) \cos (n x)-m \cos (m x) \sin (n x)}{m^2-n^2} \end{equation}

cos(mx)cos(nx)dx=msin(mx)cos(nx)ncos(mx)sin(nx)m2n2\begin{equation} \int \cos (m x) \cos (n x) \, dx=\frac{m \sin (m x) \cos (n x)-n \cos (m x) \sin (n x)}{m^2-n^2} \end{equation}

sin(mx)sin(mx)dx=x2sin(2mx)4m\begin{equation} \int \sin (m x) \sin (m x) \, dx=\frac{x}{2}-\frac{\sin (2 m x)}{4 m} \end{equation}

cos(mx)cos(mx)dx=2mx+sin(2mx)4m\begin{equation} \int \cos (m x) \cos (m x) \, dx=\frac{2 m x+\sin (2 m x)}{4 m} \end{equation}

sin(mx)dx=cos(mx)m\begin{equation} \int \sin (m x) \, dx=-\frac{\cos (m x)}{m} \end{equation}

cos(mx)dx=sin(mx)m\begin{equation} \int \cos (m x) \, dx=\frac{\sin (m x)}{m} \end{equation}

cos(mx)sin(nx)dx=msin(mx)sin(nx)+ncos(mx)cos(nx)m2n2\begin{equation} \int \cos (m x) \sin (n x) \, dx=\frac{m \sin (m x) \sin (n x)+n \cos (m x) \cos (n x)}{m^2-n^2} \end{equation}

Now from the above results we can obtain, on the interval (π,π)(-\pi,\pi) the results

ππsin(mx)sin(mx)dx=π\begin{equation} \int_{-\pi }^{\pi } \sin (m x) \sin (m x) \, dx=\pi \end{equation}

ππcos(mx)cos(mx)dx=π\begin{equation} \int_{-\pi }^{\pi } \cos (m x) \cos (m x) \, dx=\pi \end{equation}

ππsin(mx)sin(nx)dx=0\begin{equation} \int_{-\pi }^{\pi } \sin (m x) \sin (n x) \, dx=0 \end{equation}

ππcos(mx)cos(nx)dx=0\begin{equation} \int_{-\pi }^{\pi } \cos (m x) \cos (n x) \, dx=0 \end{equation}

ππsin(mx)dx=0\begin{equation} \int_{-\pi }^{\pi } \sin (m x) \, dx=0 \end{equation}

ππcos(mx)dx=0\begin{equation} \int_{-\pi }^{\pi } \cos (m x) \, dx=0 \end{equation}

ππcos(mx)sin(nx)dx=0\begin{equation} \int_{-\pi }^{\pi } \cos (m x) \sin (n x) \, dx=0 \end{equation}

Then, the set {(1,sin(nx),cos(nx))}n=1\{(1,\sin(nx),\cos(nx))\}_{n=1}^{\infty} is orthogonal on the interval (π,π)(-\pi,\pi).

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