[PDF] Lecture notes on Fourier Series(C8-unit-4)





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Lecture notes on Fourier Series(C8-unit-4)

by

Shilpa PatraDepartment of Mathematics

Narajole Raj College

West bengal, India

Periodic Functions

Denition

A function f(x) is said to be periodic if there exists a numberP>0 such that f(x + P) = f(x) for every x. The smallest such P is called the period of f(x). Geometrically, a function f is periodic with period P if the graph of f is invariant under translation in the x-direction by a distance of P.Example sin x and cos x are periodic with period 2.If L is a xed number, then sin( 2xL ) and cos(2xL ) have period L.The period of sin nx or cos nx, where n is a positive integer, is 2n .Lecture notes on Fourier Series(C8-unit-4)

Trigonometric Functions with period

Lecture notes on Fourier Series(C8-unit-4)

Introduction: Fourier series

When the French mathematician Joseph Fourier (1768{1830) was trying to solve a problem in heat conduction, he needed to express a function as an innite series of sine and cosine functions: f(x) =a0+1X n=1(ancosnx+bnsinnx) (0.1) Earlier, Daniel Bernoulli and Leonard Euler had used such series while investigating problems concerning vibrating strings and astronomy. The series in Equation 0.1 is called a trigonometric series or Fourier series and it turns out that expressing a function as a Fourier series is sometimes more advantageous than expanding it as a power series. In particular, astronomical phenomena are usually periodic, as are heartbeats, tides, and vibrating strings, so it makes sense to express them in terms of periodic functions.

Lecture notes on Fourier Series(C8-unit-4)

Fourier series

We start by assuming that the Fourier series converges and has a continuous function f(x) as its sum on the interval [;], that is, f(x) =a0+1X n=1(ancosnx+bnsinnx)x(0.2) Our aim is to nd formulas for the coecientsanandbnin terms of f. Recall that for a power seriesf(x) =Pcn(xa)nwe found a formula for the coecients in terms of derivatives:cn=f(n)(a)n!. Here we use integrals. If we integrate both sides of Equation 0.2 and assume that it's permissible to integrate the series term-by-term, we get f(x)dx= a 0dx+ 1 X n=1(ancosnx+bnsinnx)dx = 2a0+1X n=1a n cosnx dx+1X n=1b n sinnx dx:Lecture notes on Fourier Series(C8-unit-4)

Determining Fourier coecienta0But

cosnx dx= 0 = sinnx dx: So, f(x)dx= 2a0; and solving fora0gives a 0=12 f(x)dx:(0.3) To determineanforn1 we multiply both sides of Equation 0.2 by cosmx(where m is an integer andm1) and integrate term-by-term fromto: f(x)cosmx dx= h a 0+1X n=1(ancosnx+bnsinnx)i cosmx dx =a0 cosmx dx+1X n=1a n cosnxcosmx dx+1X n=1b n sinnxcosmx dx:Lecture notes on Fourier Series(C8-unit-4) Determining Fourier coecientsanandbnA simple calculation shows that sinnxcosmx dx= 0for all m;n: cosnxcosmx dx=(

0 ifn6=m;

ifn=m:

So we get,

f(x)cosmx dx=am Solving foram, and then replacing m by n , we have a n=1 f(x)cosnx dx n= 1;2;3;:(0.4) Similarly, if we multiply both sides of Equation 0.2 by sinmxand integrate fromto, we get b n=1 f(x)sinnx dx n= 1;2;3;:(0.5)Lecture notes on Fourier Series(C8-unit-4)

Remark:

We have derived Formulas (0.3), (0.4) , and (0.5) assuming f is a continuous function such that Equation (0.2) holds and for which the term-by-term integration is legitimate. But we can still consider the Fourier series of a wider class of functions:A piecewise continuous functionon [a,b] is continuous except perhaps for a nite number of removable or jump discontinuities.

Lecture notes on Fourier Series(C8-unit-4)

Fourier Series

Denition 1

Let f be apiecewise continuous functionon [;]. Then theFourier seriesof f is the series a 0+1X n=1(ancosnx+bnsinnx) where the coecientsanandbnin this series are dened by a 0=12 f(x)dx a n=1 f(x)cosnx dx bn=1 f(x)sinnx dx and are called theFourier coecientsof f. We write f(x)a0+1X n=1(ancosnx+bnsinnx)Lecture notes on Fourier Series(C8-unit-4) Note: Notice in the above Denition 1 that we are not saying is equal to its Fourier series i.e., we useand not =. Later we will discuss conditions under which that is actually true. For now we are just saying that associated with any piecewise continuous function on [;] is a certain series called a Fourier series.Example 1 Find the Fourier coecients and Fourier series of the square-wave function f dened by f(x) =(

0 ifx<0;

1 if 0x< and f(x+ 2) =f(x);

so f is periodic with period 2.Lecture notes on Fourier Series(C8-unit-4)

Solution:

Using the formulas for the Fourier coecients in Denition 1 , we have a 0=12 f(x)dx=12 0

0dx+12

0

1dx= 0 +12() =12

and, forn1 a n=1 f(x)cosnx dx=1 0 0dx+1 0 cosnx dx= 0: b n=1 f(x)sinnx dx=1 0 0dx+1 0 sinnx dx

0 if n is even

2nif n is odd:Lecture notes on Fourier Series(C8-unit-4)

Continued..

Solution

The Fourier series of f is therefore

a 0+1X n=1(ancosnx+bnsinnx) 12 +2 sinx+23sin3x+25sin5x+ Since odd integers can be written asn= 2k1, where k is an integer, we can write the Fourier series in sigma notation as 12 +1X k=12(2k1)sin(2k1)x: In Example 1 we found the Fourier series of the square-wave function, but we don't know yet whether this function is equal to its Fourier series.

Lecture notes on Fourier Series(C8-unit-4)

Theorem 1

Fourier Convergence Theorem

If f is a periodic function with period 2and f andf0are piecewise continuous on [;], then the Fourier series (1) is convergent. The sum of the Fourier series is equal f(x) to at all numbers x where f is continuous. At the numbers x where f is discontinuous, the sum of the Fourier series is the average of the right and left limits, that is 12 [f(x+) +f(x)]:If we apply the Fourier Convergence Theorem to the square-wave function in Example 1, we get 12 +1X k=12(2k1)sin(2k1)x=( f(x) ifx6=n 12 ifx=n:Lecture notes on Fourier Series(C8-unit-4)

FUNCTIONS WITH PERIOD 2L

If a function f has period other than 2, we can nd its Fourier series by making a change of variable. Suppose f(x) has period 2L, that is f(x+ 2L) =f(x) for all x. If we lett=xL and g(t) =f(x) =f(Lt then, as you can verify, g has period 2andx=Lcorresponds to t=. The Fourier series of g is a 0+1X n=1(ancosnt+bnsinnt) where a 0=12 g(t)dt a n=1 g(t)cosnt dt;bn=1 g(t)sinnt dt:

If we now use the Substitution Rule withx=Lt

, thent=xL dt=Ldx, and we have the following:Lecture notes on Fourier Series(C8-unit-4)

Fourier Series of a function with Period 2L

Denition 2

Let f be apiecewise continuous functionon [L;L]. Then theFourier seriesof f is the series a 0+1X n=1h a ncosnxL +bnsinnxL i where a 0=12L L

Lf(x)dx

and, forn1, a n=1Lquotesdbs_dbs11.pdfusesText_17
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