《高数双语》课件section 7-6.pptx

1、Fourier Series of Other Forms1Fourier Expansions of Periodic Functions with Period 2L2Suppose a periodic function f(x)satisfies Dirichlet conditions on the interval L,L and has period 2L.Set,Lxt ,.t ()().Lf xftg t Then,g(t)is a periodic function with period 2 and satisfies Dirichlet conditions on th

2、e interval ,.01()cossin2nnnag tantbnt We have011();()cos(1);1()sin(1).nnag t dtag tntdtnbg tntdtn where3Fourier Expansions of Periodic Functions with Period 2LThe trigonometric seriesThe constants a0,an and bn are the Fourier coefficients of whose coefficients are determined by.LxLf.is called the Fo

3、urier series of the function f over the intervalDefinition(Fourier series of a function with period 2L)01cossin2nnnan xn xabLL 011();()cos(1);1()sin(1).LLnLLLnLn xaf x dxaf xdxnLLLn xbf xdxnLL 4Convergence TheoremTheorem(Dirichlets theorem)Assume that the function f is piecewise monotone on the inte

4、rval and is continuous except for a finite number of discontinuous points of the firsttype.Then the Fourier series of the function f must converge on the interval and its sum function is,LxL(),(0)(0)(),2(0)(0),2f xxf xf xS xxfLf LxL if is a point of continuity;if is a point of discontinuity;if.,LxLI

5、n this case we also say the Fourier series is the of f.0111();()cos;()sin.LLLnnLLLn xn xaf x dxaf xdxbf xdxLLLLL5Fourier Expansions of Periodic Functions with Period 2LSolution Find the Fourier series for the function20,()1,01,0,12.xxf xxx and the sum of the series.20102201111()()1222LLaf x dxf x dx

6、xdxdxL022111;2222x 6Fourier Expansions of Periodic Functions with Period 2L20,()1,01,0,12.xxf xxx Solution(continued)0111();()cos;()sin.LLLnnLLLn xn xaf x dxaf xdxbf xdxLLLLL22012011()cos()cos2211coscos2222LnLn xn xaf xdxf xdxLLn xn xxdxdx 2221(1cos)sin;2nnnn 22012011()sin()sin2211sinsin2222LnLn xn

7、xbf xdxf xdxLLn xn xxdxdx 1(1cos2cos).2nnn 7Fourier Expansions of Periodic Functions with Period 2LSolution(continued)0221211;(1cos)sin;(1cos2cos).222nnnnaanbnnnn The Fourier series for the given function is012211cossin2121(1cos)sincos4221(1cos2cos)sin.22nnnnnan xn xabLLnn xnnnnn xnn 8Fourier Expans

8、ions of Periodic Functions with Period 2LSolution(continued)20,()1,01,0,12.xxf xxx yxO264-6-2-41-2(),42,4,411,42,1()0,1,2,4.21,412f xxkkkxkS xkxkxk 9Integrals of Even and Odd FunctionsEven function:0()2()LLLg x dxg x dx Odd function:()0LLg x dx Because of these two rules,even and odd extensions of a

9、 functionThe following results also hold for even and odd functions.are convenient to use.1.The product of two even functions is even.2.The product of an even functions with an odd function is odd.3.The product of two odd functions is even.Fourier Expansions of Periodic Functions with Period 2L10Cor

10、ollary Suppose that the function f satisfies Dirichlet conditions on the interval-L,Land has period 2L.Let .(1)While f(x)is an odd function,its Fourier expansion iswhere(2)While f(x)is an even function,its Fourier expansion iswhere1|()(0)(0)2Cxf xf xf x1()sin()nnn xf xbxCL 02()sin.Lnn xbf xdxLL 01()

11、cos()2nnan xf xaxCL 00022();()cos.LLnn xaf x dxaf xdxLLL 11Fourier Expansions of Periodic Functions with Period 2L Find the Fourier Expansion for f,where f is a periodic function with period 6 and defined on interval-3,3 as follows1,30;()103.xf xx Solution033000;0;2232sincos(1cos)3333nnaan xn xbdxnn

12、n 0,221(1).4,21(21)nnknknk 12Fourier Expansions of Periodic Functions with Period 2LSolution(continued)14(21)()sin,(,).(21)3kkxf xxAk Since f(x)is continuous except|3,0,1,2,Ax xk kwe have Find the Fourier Expansion for f,where f is a periodic function with period 6 and defined on interval-3,3 as fol

13、lows1,30;()103.xf xx Find the Fourier Expansion for f,where f is a periodic function with period 4 and defined on interval-2,2 as follows13Fourier Expansions of Periodic Functions with Period 2L1,|;()20|2.xf xx yxO4+4-4+-4-1/(2)-44-2214Fourier Expansions of Periodic Functions with Period 2LSolution0

14、000022110;();22221121cossinsin.222222Lnnbaf x dxdxLn xn xnadxnn So,111()sincos.422nnn xf xn Find the Fourier Expansion for f,where f is a periodic function with period 4 and defined on interval-2,2 as follows1,|;()20|2.xf xx 15Fourier Expansions of Functions Defined on 0,LQuestion:How than can we ca

15、lculate a Fourier series expansion for f on 0,L?To do so,we extend the function so that it is defined over the symmetric intervalHow,though,do we define the extension of f for 0?Lx.LxL0LxThe answer is that we define the extension to be any function over we choose as long as the extension and its der

16、ivative are piecewise continuous(in order to satisfy the Dirichlet conditions).LOx16Anyway,there are two special extensions that are particularly useful and whoseFourier Expansions of Functions Defined on 0,LFourier coefficients are especially easy to calculate;extensions of f.LOxthese are the even

17、and odd-L1()sinnnn xf xbL Question:How than can we calculate a Fourier series expansion for f on 0,L?17Integrals of Even and Odd FunctionsThe graph of an odd function is symmetric about the origin.The graph of an even function is symmetric about the y-axis.This observation can make the integrals of

18、even and oddfunctions over intervals symmetric about the origin relatively easy to calculate.18Even Extension:Fourier Cosine SeriesWe define the even extension of f by requiring that is specified for the interval 0.xL()yf x Suppose that the function()(),.fxf xLxLabout()yf x Graphically,we obtain the

19、 even extension by reflecting The even extension of a function is illustrated in the following.the y-axis.19Therefore,if we used the even extension for a function f,we obtain theEven Extension:Fourier Cosine SeriesFourier coefficients01()cos.2nnan xf xaL 0012()()LLLaf x dxf x dxLL 012()cos()cos,LLnL

20、n xn xaf xdxf xdxLLLL EvenThe Fourier series of f is1()sin0.LnLn xbf xdxLL Odd20Fourier Cosine Seriesis the cosine series01()cos,2nnan xF xaL where00022(),()cos.LLnn xaf x dx af xdxLLL LxLof an even function F on the interval Suppose we extend the function f from interval 0,L to-L,0)so that the exte

21、nded function F is an even function,that is(),0,()()0.f xxLF xfxLx This extension is called the even extension.In this case,The Fourier expansion 01()cos.2nnan xf xaL The Fourier expansion of the function f on the interval 0,L is 21Fourier Cosine Series Find the Fourier cosine series for the functio

22、n1,0,2()0,.2xf xx Solution22Fourier Cosine SeriesFor the Fourier cosine series,we select the evenSolution(continued)The Fourier coefficients are.xextension of the function over/2021x 2sin.2nn /202dx 002()af x dx 02()cosnn xaf xdx /202cosnxdx Therefore,we have the Fourier cosine expansion112()sincos.

23、22nnf xnxn 0000122()cos,(),()cos.2LLnnnan xn xf xaaf x dxaf xdxLLLL Finish.23Fourier Cosine SeriesNext figure shows the Fourier cosine approximation f as n varies up to1,5 and 20 terms.24Odd Extension:Fourier Sine Seriesspecified for the interval 0.xLConsider again a function()yf x We defined the od

24、d extension of f by requiring that()(),.fxf xLxL about()yf x Graphically,we obtain the even extension by reflecting The odd extension of a function is illustrated in the following.the origin.25Therefore,if we used the odd extension for a function f,we obtain theOdd Extension:Fourier Sine SeriesFouri

25、er coefficients1()sin.nnn xf xbL 01()0LLaf x dxL 1()cos0,LnLn xaf xdxLL OddThe Fourier series of f is012()sin()sin.LLnLn xn xbf xdxf xdxLLLL Even26Fourier Sine Seriesis the sine seriesLxL an odd function F on the interval 1()sin,nnn xF xbL where02()sin.Lnn xbf xdxLL Suppose we extend the function f

26、from interval(0,L to-L,0 so that the extended function F is an odd function,that is(),0,()00,()0.f xxLF xxfxLx This extension is called the odd extension.In this case,The Fourier expansionThe Fourier expansion of the function f on the interval(0,L is1()sin.nnn xf xbL 27Fourier Sine Series Find the F

27、ourier sine series for the function1,0,2()0,.2xf xx Solution28Fourier Sine SeriesFor the Fourier sine series,we select the oddSolution(continued)The Fourier coefficients are.x extension of the function over 21cos.2nn 02()sinnn xbf xdx /202sinnxdx Therefore,we have the Fourier sine expansion12()1cossin.2nnf xnxn 012()sin,()sin.Lnnnn xn xf xbbf xdxLLL Finish.29Fourier Sine SeriesNext figure shows the Fourier sine approximation f as n varies up to1,5 and 20 terms.