Exercise 11. Isao Sasano

Transcription

1 Exercise Isao Sasano Exercise Calculate the value of the following series by using the Parseval s equality for the Fourier series of f(x) x on the range [, π] following the steps ()-(5). () Calculate the linear combination of the following orthogonal functions that is closest to the function f(x). As for the measure of the distance, use (the half of) the integral of the square of the difference on the range [, π]. {, cos x, sin x, cos x, sin x,..., cos nx, sin nx} () Obtain the Fourier series of f(x) on the range [, π]. (The Fourier series of f(x) x is the limit of the linear combination obtained in () as n goes to infinity.) (3) Normalise the series obtained in (). () Write down the Parseval s equality for the series obtained in (3). (5) Calculate the value of the series Note that in an inner product space L, when the approaximation, in the sense of the least square, of u L by a linear comination of an orthonormal basis {e i i } in L n c e.

2 converges to u in the sense that the norm of the difference converges to 0 as n goes to infinity, the following equation, called Parseval s equality, holds. u c Solution () Assume the following equation holds. (Note: There are no coefficients a 0,..., a n, b,..., b n that satisfy the equation, but it s o.) f(x) n a 0 + (a cos x + b sin x) () Integrate the both sides of the equation () on the range [, π]. f(x)dx a 0dx a 0 π Then we calculate a 0 as follows. a 0 π π 3 π f(x)dx x dx Multiply the both sides of the equation () by cos x and integrate them on the range [, π]. f(x) cos xdx a cos xdx Then we calculate a as follows. a π f(x) cos xdx x cos xdx π {[ ] π sin x x π x sin xdx π a π } sin x x dx [ sin x (since x ] π is 0)

3 Here we calculate the integral x sin xdx. [ ] π π cos x π cos x x sin xdx x dx [x cos cos x x]π (since dx is 0) (π cos π () cos()) (π cos π + π cos π) π cos π π ( ) We resume the calculation of a. a π ( ( ) ) ( ) Multiply the both sides of the equation () by sin x and integrate them on the range [, π]. Then we calculate b as follows. b π f(x) sin xdx b b π f(x) sin xdx sin xdx x sin xdx π 0 (since x sin x is an odd function) So the linear combination that is closest to the function f(x) is 3 π n + ( ) cos x. 3

4 () The Fourier expansion of f(x) x is the limit of the above linear combination as n goes to infinity: 3 π + ( ) cos x. (3) Firstly we calculate the norm of and cos x. (, ( ) π dx dx [ ]π x π cos x (cos x, cos x) cos xdx π So the Fourier series of f(x) x is normalized as follows. 3 π + ( ) cos x 3 π π π + π ) ( ) cos x π π () By the Parseval s equality, we obtain the following equation. f ( ) π 3 π + ( ) ( ) π ()

5 The left hand side of the equation () is calculated as follows. f (f, f) [ ] x 5 π 5 ( π 5 5 π5 f(x) dx x dx 5 (5 5 ) The right hand side of the equation () is calculated as follows. RHS ( ) π 3 π + 9 π π + 9 π5 + 6π ) ( ) ( ) π 6 π So the Parceval s equality for the Fourier series of f(x) x is obtained as follows. 5 π5 9 π5 + 6π (5) We calculate the value of the series 6π. 5 π5 9 π5 8 5 π5 0 5 π5 8 5 π5 5

6 So the value of the series is obtained as follows. 6π 8 5 π5 90 π (Note) The obtained value 90 π is the value of the zeta function ζ(n) when n. The zeta function is given as follows. So ζ() 90 π. ζ(n) n 6