MATH 285 E1 F1 GRADED HOMEWORK SET 5 DUE WEDNESDAY NOVEMBER 5 IN LECTURE This time the homework has just one part Please staple your homework together and put your name and section on it Thank you 1 15 points from 1 p 190 Expand x3 and x in Fourier sine series valid when x and hence find the value of the sum of the series 1 1 1 sin x 3 sin 2x 3 sin 3x 3 sin 4x 2 3 4 for all values of x 1 Since both x and x3 are odd functions the constant term and all of the cosine terms of their Fourier series vanish so the Fourier series contains only sine terms as the problem statement says Again using the odd symmetry the sine coefficients are computed by the integrals Z Z 2 2 3 x sin nx dx x sin nx dx 0 0 Let s consider the function x first this was actually done in lecture We use integration by parts Z Z x sin nx dx 1 n x cos nx 1 n cos nx dx 1 n x cos nx 1 n2 sin nx C Plugging in the limits of integration 0 and gives Z 2 bn x sin nx dx 0 2 1 n cos n 1 n2 sin n 1 n 0 cos 0 1 n2 sin 0 2 2 1 n cos n 1 n 1 n Thus on the interval x we have the equality x X bn sin nx 2 n 1 X 1 n 1 n 1 n sin nx 1According to Whittaker and Watson 1 this problem was on an exam at Jesus College Cambridge in 1902 1 2 GRADED HOMEWORK 5 Now let s consider the function x3 We reuse the symbol bn to denote the sine coefficients of x3 We need to integrate x3 sin nx and for this we use integration by parts recursively Z Z 3 3 x sin nx dx x 1 n cos nx 3x2 1 n cos nx dx Z 3 1 n x cos nx 3 n x2 cos nx dx Z 2 2 Z x cos nx dx x 1 n sin nx Z 2x 1 n sin nx dx Z 1 n x2 sin nx 2 n x sin nx dx x sin nx dx 1 n x cos nx 1 n2 sin nx C Putting it together Z x3 sin nx dx 1 n x3 cos nx 3 n 1 n x2 sin nx 2 n 1 n x cos nx 1 n2 sin nx C Rather than multiplying this out let s just see what happens when we plug in 0 and If we plug in 0 the the terms x3 cos nx x2 sin nx x cos nx and sin nx all become zero so the whole expression is zero If we plug in the terms x2 sin nx and sin nx become zero so we are left with 1 n 3 cos n 3 n 2 n 1 n cos n 1 n 3 1 n 6 n3 1 n Thus Z 2 3 2 bn x sin nx dx 1 n 1 1 n 3 1 n 6 n3 0 1 n 1 2 n 2 1 n 12 n3 Thus for x we have x3 X 1 n 1 2 n 2 1 n 12 n3 sin nx n 1 This completes the first part of the problem The second part is to find the sum of the given series Call that P function f x We notice that this series may be written using notation as X f x 1 n 1 1 n3 sin nx n 1 GRADED HOMEWORK 5 3 The sine coefficients of x are Pn 1 n 1 2 n those of x3 are Qn 1 n 1 2 n 2 1 n 12 n3 and those of f x are Rn 1 n 1 1 n3 There is a relationship between these coefficients Qn 2 Pn 12Rn and thus a relationship between the corresponding functions x3 2 x 12f x Thus for x f x 1 12 2 x x3 The function f x repeats periodically with period 2 and it is in fact continuous 2 5 points Find the Fourier cosine series of the function f t 1 t defined on the interval 0 t 1 Recall that the cosine series is defined by taking the Fourier series of the even extension of period 2L 2 The Fourier cosine coefficients are Z Z 1 2 L a0 f t dt 2 1 t dt 2 t t2 2 10 2 1 1 2 1 L 0 0 2 an L Z 0 L n t f t cos dt 2 L Z 1 1 t cos n t dt 0 The integral is done by parts Z Z 1 1 sin n t 1 sin n t dt 1 t cos n t dt 1 t n n 1 1 1 t sin n t cos n t C n n 2 1 1 1 an 2 1 t sin n t cos n t n n 2 0 1 1 1 1 1 1 sin n cos n 1 0 sin 0 cos 0 2 n n 2 n n 2 2 2 cos 0 cos n 1 1 n 2 n n 2 Thus an 0 if n is even and an 4 n 2 if n is odd The cosine series is then f t a0 X 1 4 X 1 an cos n t 2 cos n t 2 2 n2 n 1 n odd 4 GRADED HOMEWORK 5 3 10 points Let f t be the periodic function of period 2 defined on the interval 0 t 2 by the formula f t t2 The Fourier series of this function is 4 4 X 1 4X1 2 cos n t sin n t 3 n2 n n 1 n 1 a 5 points Describe precisely the value of the sum of the Fourier series at every value of t including at the points of discontinuity of the original function f t The function f t is continuous except at the points t 2k where k is an integer At each such point the limit from the left is 22 4 while the limit from the right is 02 0 So the Fourier series converges to 4 0 2 2 at these points At any point t not of the form 2k where k is an integer the Fourier series converges to s2 where s is the number in the interval 0 s 2 obtained by adding a multiple of 2 to t If x denotes the greatest integer less than or equal to x then s t 2 t 2 In other words 2 t 2k 4 X 1 4X1 4 cos n t sin n t 2 t 6 2k 3 2 n2 n t 2 t 2 n 1 n 1 b 5 points Suppose we differentiate the Fourier series term byterm Show that the resulting series does not converge to anything …
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