Problem 165 Find the Fourier cosine series of the function f x x x \u03c0 2 \u03c0 x \u03c0 2

Problem 165 find the fourier cosine series of the

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Problem 16.5 Find the Fourier cosine series of the function f ( x ) = x, 0 x π 2 π - x, π 2 x π Problem 16.6 Find the Fourier cosine series of f ( x ) = x on the interval [0 , π ] . Problem 16.7 Find the Fourier sine series of f ( x ) = 1 on the interval [0 , π ] . Problem 16.8 Find the Fourier sine series of f ( x ) = cos x on the interval [0 , π ] . Problem 16.9 Find the Fourier cosine series of f ( x ) = e 2 x on the interval [0 , 1] . Problem 16.10 For the following functions on the interval [0 , L ], find the coefficients b n of the Fourier sine expansion. (a) f ( x ) = sin ( 2 π L x ) . (b) f ( x ) = 1 (c) f ( x ) = cos ( π L x ) .
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126 SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS Problem 16.11 For the following functions on the interval [0 , L ], find the coefficients a n of the Fourier cosine expansion. (a) f ( x ) = 5 + cos ( π L x ) . (b) f ( x ) = x (c) f ( x ) = 1 0 < x L 2 0 L 2 < x L Problem 16.12 Consider a function f ( x ) , defined on 0 x L, which is even (symmetric) around x = L 2 . Show that the even coefficients ( n even) of the Fourier sine series are zero. Problem 16.13 Consider a function f ( x ) , defined on 0 x L, which is odd around x = L 2 . Show that the even coefficients ( n even) of the Fourier cosine series are zero. Problem 16.14 The Fourier sine series of f ( x ) = cos ( πx L ) for 0 x L is given by cos πx L = X n =1 b n sin nπx L , n N where b 1 = 0 , b n = 2 n ( n 2 - 1) π [1 + ( - 1) n ] . Using term-by-term integration, find the Fourier cosine series of sin ( πx L ) . Problem 16.15 Consider the function f ( x ) = 1 0 x < 1 2 1 x < 2 (a) Sketch the even extension of f. (b) Find a 0 in the Fourier series for the even extension of f. (c) Find a n ( n = 1 , 2 , · · · ) in the Fourier series for the even extension of f. (d) Find b n in the Fourier series for the even extension of f. (e) Write the Fourier series for the even extension of f.
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17 SEPARATION OF VARIABLES FOR PDES 127 17 Separation of Variables for PDEs Finding analytic solutions to PDEs is essentially impossible. Most of the PDE techniques involve a mixture of analytic, qualitative and numeric ap- proaches. Of course, there are some easy PDEs too. If you are lucky your PDE has a solution with separable variables. In this chapter we discuss the application of the method of separation of variables in the solution of PDEs. 17.1 Second Order Linear Homogenous ODE with Con- stant Coefficients In this section, we review the basics of finding the general solution to the ODE ay 00 + by 0 + cy = 0 (17.1) where ab, and c are constants. The process starts by solving the character- istic equation ar 2 + br + c = 0 which is a quadratic equation with roots r 1 , 2 = - b ± b 2 - 4 ac 2 a . We consider the following three cases: If b 2 - 4 ac > 0 then the general solution to ( 17.1 ) is given by y ( t ) = Ae - b - b 2 - 4 ac 2 a t + Be - b + b 2 - 4 ac 2 a t . If b 2 - 4 ac = 0 then the general solution to ( 17.1 ) is given by y ( t ) = Ae - b 2 a t + Bte - b 2 a t . If b 2 - 4 ac < 0 then r 1 , 2 = - b 2 a ± i 4 ac - b 2 2 a and the general solution to ( 17.1 ) is given by y ( t ) = Ae - b 2 a t cos 4 ac - b 2 2 a t + Ae - b 2 a t sin 4 ac - b 2 2 a t.
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128 SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS 17.2 The Method of Separation of Variables for PDEs
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