1、Fourier Series12The Fourier SeriesWhen investigating the problem of heat conduction in a long thin insulated rod,Joseph Fourier needed to express a function as a trigonometric series 三角级数三角级数.()f xcoefficients and ()for which0,naanb1n Generally,if is defined on the interval ,we need to know the()f x
2、x nnnaf xanxbnx01()cossin2 (1)3The Fourier SeriesNotice that the interval is symmetric about the origin.xThis equation called a Fourier series for on the interval .()f x(,)01()cossin2nnnaf xanxbnx Assume that is expressible as the trigonometric series given by()f xthe Equation(1).01212,.a a ab bWe w
3、ant to find a way to calculate the coefficients4Orthogonality of the system of trigonometric functionsIt is easy to prove that definite integral of the product of any two different functions of1,cos,sin,cos2,sin2,cos,sin,xxxxnxnxsquare of any function of the system over is not equal to zero.,over th
4、e interval is zero;and the definite integral of the,on .,The key to the calculations is the definite integral 01()cossin2nnnaf xanxbnx 5Orthogonality of the system of trigonometric functions5.0,sinsin,mnmxnxdxmn 1.1 cos0nxdx 6.212dx 2.1 sin0nxdx 3.0,coscos,mnmxnxdxmn 4.cossin0mxnxdx Trigonometric In
5、tegrals6Orthogonality of the system of trigonometric functionsOrthogonality正交性),.a b()()0,baf xg x dx then f and g are said to be orthogonal on If functions f and g are both integrable over the interval and,a b,.a bSuppose that is a sequence of functions on()nfxdifferent functions of the sequence ar
6、e orthogonal on()nfx,.a bsystem of orthogonal functions on()nfxand2()0(1,2,),bnafx dxn then the sequence is called a ,a bIf any two7Fourier SeriesCalculation of a0operations for integration and summation can be interchanged to obtainWe integrate both sides of Equation(1)from to(2)011()cossin.2nnnnaf
7、 x dxdxanxdxbnxdxTherefore,000().22aa xf x dxdxa For every positive integer n,the last two integrals on the right-hand side of Solving for a0 yieldsequation(2)are zero.01().af x dx and assume that the 01()cossin2nnnaf xanxbnx 8Fourier Series Calculation of amWe multiply both sides of Equation(1)byco
8、s,0,mx m the result from:to (3)011()coscos2coscossincos.nnnnaf xmxdxmxdxanxmxdxbnxmxdx and integrate 01()cossin2nnnaf xanxbnx 9Fourier Series 1()cos.maf xmxdx Therefore,can be further reduced toThe first integral on the right-hand side of Equation(3)is zero and the equation()coscoscos.mmf xmxdxamxmx
9、dxa Calculation of am 01()cossin2nnnaf xanxbnx 10Fourier Series Calculation of bmWe multiply both sides of Equation(1)bysin,0,mx m tothe result from:and integrate(4)011()sinsin2cossinsinsin.nnnnaf xmxdxmxdxanxmxdxbnxmxdx 01()cossin2nnnaf xanxbnx 11Fourier Series Thus,the equation(4)can be further re
10、duced to1()sin.mbf xmxdx Therefore,()sinsinsin.mmf xmxdxbmxmxdxb 011();()cos(1);1()sin(1).nnaf x dxaf xnxdxnbf xnxdxn Calculation of bm 01()cossin2nnnaf xanxbnx 12Fourier SeriesThe trigonometric seriesThe constants a0,an and bn are the Fourier coefficients of whose coefficients are determined by.xf.
11、is called the Fourier series of the function f over the interval 01cossin2nnnaanxbnx 011();()cos(1);1()sin(1).nnaf x dxaf xnxdxnbf xnxdxn 13Finding a Fourier Expansion Find the Fourier series of the function1,0,(),0.xf xxx SolutionThe piecewise continuous functionFrom the last definition we have01()
12、af x dx 00111.2dxxdx 0111();()cos;()sin.nnaf x dxaf xnxdx bf xnxdx14Finding a Fourier ExpansionSolution(continued)201cosnxn 1()cosnaf xnxdx 0011coscosnxdxxnxdx 000111sinsinsinxnxnxnxdxnnn 21(cos1)nn 2(1)1.nn cos(1)nn 0111();()cos;()sin.nnaf x dxaf xnxdx bf xnxdx Find the Fourier series of the functi
13、on1,0,(),0.xf xxx 15Finding a Fourier ExpansionSolution(continued)1()sinnbf xnxdx 0011sinsinnxdxxnxdx (1)(1)1.nn 2111(1)1(1)(1)1cossin.24nnnnnxnxnnTherefore,the Fourier series of the given function isFinish.0111();()cos;()sin.nnaf x dxaf xnxdx bf xnxdx Find the Fourier series of the function1,0,(),0
14、.xf xxx 16Finding a Fourier Expansionterms is given in the following figure.A graph of the Fourier series approximations as n varies up to 1,5,and 20 17Convergence of Fourier SeriesAssume that the function is piecewise monotone on the interval and is continuous except for a finite number of disconti
15、nuous points of the firsttype.Then the Fourier series of the function fmust converge on the interval fand its sum function is,x(),(0)(0)(),2(0)(0),2f xxf xf xS xxffx if is a point of continuity;if is a point of discontinuity;if.,x18Convergence of Fourier Series(0)(0)011.222ffthe Fourier Series conve
16、rges,xin the intervalFor every 0 x converges to the average valuethe function has a jump0,x At().f xtoThe functionsatisfies the conditions of1,0,(),0.xf xxx Theorem of Convergence of Fourier Series.discontinuity and the Fourier series 19Find the Fourier Series of a FunctionSuppose that f is a period
17、ic function with period 2,andas follows:(,Find its Fourier expansion.0,0,(),0.xf xxx is defined in the interval Sawtooth wave 3 320Find the Fourier Series of a Function222,1(1)10,nnnnn odd,even.Obviously,f is piecewise continuous and monotone.By the Dirichlet theorem it can be expanded in a Fourier
18、series.Solution000111()sinsincoscosnbf xnxdxxnxdxxnxnxdxn 0011(),2af x dxxdx 000111()coscossinsinnaf xnxdxxnxdxxnxnxdxn 1(1).nn 21converges to f at all other points in Find the Fourier Series of a FunctionHence according to the Dirichlet theorem,we know that the Fourier series of f,that is the serie
19、s on the right sideSolution(continued)3212sin221()cossincos3sin342332(1)cos(21)sin;4(21)kkxf xxxxxkxkxkk (,),x for,x andwhen 1()().22ff Thus,the of the last equation converges to(1,3,5,)xnn 2 atFourier series of f converges to(,).Finish.22Find the Fourier Series of a FunctionyxO32-3-21Suppose that f
20、 is a periodic function with period 2,andas follows:,)Find its Fourier expansion.0,0,()1,0.xf xx is defined in the interval 23Find the Fourier Series of a FunctionObviously,f satisfies the Dirichlet conditions.By theDirichlet theorem it can be expanded in a Fourier series.Solution0011()1,af x dxdx 0
21、0111()coscossin0naf xnxdxnxdxnxn 2,11(1)0,nnnnn odd,even.00111()sinsincosnbf xnxdxnxdxnxn Hence,the Fourier series of f is112sin(21).221kkxk 24Find the Fourier Series of a FunctionSolution(continued)According to the Dirichlet theorem we know that thesum function of the Fourier series on the interval
22、 is,0,0,1(),0,21,0.xS xxx In the interval it is(,)(),()0,1,2,1.2f xxnS xnxn The Fourier expansion of the function f is112sin(21)(),(,).221kkxf xxk Finish.25Find the Fourier Series of a FunctionyxO1121k 3k 20k -terms is given in the following figure.A graph of the Fourier series approximations as k varies up to 1,3,and 20Suppose that f is a periodic function with period 2,andas follows:,)Find its Fourier expansion.is defined in the interval Find the Fourier Series of a Function263,0,()3,0.xxf xxx yxO32-3-23
