From: israel@math.ubc.ca (Robert Israel) Subject: Re: Properties of sin(1/x) Date: 22 Nov 2001 22:30:33 GMT Newsgroups: sci.math Summary: Estimating Fourier coefficients of sin(1/x) In article , Vijay Kannan wrote: >Right now I am working on a calculation that I need estimates for the >Fourier coefficients of these functions. >For example both sin(1/x) and x*sin(1/x) are L_2 on an interval about >the origin. So I know that their Fourier coefficients < M/sqrt(n) for >some constant M. But there must be a better estimate since M/sqrt(n) >is not quite good enough to guarantee that these functions are L_2. The fact that a function is in L_2 does _not_ guarantee that its Fourier coefficients are bounded by M/sqrt(n). For example, take f(x) = sum_{j=1}^infinity 1/j sin(2^j x) which has |c_n| = const/log(n) for infinitely many n. It is true that the Fourier coefficients of sin(1/x) are bounded by M/sqrt(n), and I don't know if it's possible to get a better estimate. You can do this: |int_0^a sin(1/x) sin(n x) dx| <= a |int_a^Pi sin(1/x) sin(n x) dx| = (integrating by parts) |(-1/n sin(1/x) cos(n x))|_a^Pi - int_a^Pi cos(1/x)cos(n x)/(n x^2) dx| <= C/n + 1/(n a) Take a = 1/sqrt(n) to get the estimate. On the other hand, for x sin(1/x), with two integrations by parts: |int_0^a x sin(1/x) sin(n x) dx| <= a^2/2 |int_a^Pi x sin(1/x) sin(n x) dx| <= C1/n + C2/(a^2 n^2) and with a = 1/sqrt(n) again we get a bound of M/n. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ============================================================================== From: israel@math.ubc.ca (Robert B. Israel) Subject: Re: Properties of sin(1/x) Date: 25 Nov 2001 15:01:04 -0800 Newsgroups: sci.math israel@math.ubc.ca (Robert Israel) wrote in message news:<9tjua9$ffk$1@nntp.itservices.ubc.ca>... > It is true that the Fourier coefficients of sin(1/x) are bounded by > M/sqrt(n), and I don't know if it's possible to get a better estimate. It is possible to approximate these coefficients using the method of stationary phase. I'll suppose we're working on the symmetric interval [-Pi, Pi]. Since sin(1/x) is odd, only the sine series is needed. The coefficients are b_n = 1/Pi int_{-Pi}^Pi sin(1/x) sin(n x) dx = 1/Pi int_0^Pi [cos(-1/x + n x) - cos(1/x + n x)] dx = 1/Pi Re(c_n - d_n) where c_n = int_0^Pi exp(-i/x + i n x) dx and d_n = int_0^Pi exp(i/x + i n x) dx I claim that c_n = O(1/n) while d_n ~ sqrt(Pi) exp(2 i sqrt(n) + i Pi/4) n^(-3/4) so b_n ~ Pi^(-1/2) cos(2 sqrt(n) - 3 Pi/4) n^(-3/4) Using the change of variables x = n^(-1/2) t, c_n = n^(-1/2) int_0^(n^(1/2) Pi) exp(i n^(1/2) (-t + 1/t)) dt d_n = n^(-1/2) int_0^(n^(1/2) Pi) exp(i n^(1/2) (t + 1/t)) dt Note that h(t) = -t + 1/t is decreasing on (0,infinity), with h'(t) = -1 - 1/t^2. By the Second Mean Value Theorem for Integrals, there is x between 1 and b = n^(1/2) Pi such that int_1^b 1/h'(t) cos(n^(1/2) h(t)) h'(t) dt = 1/h'(1) int_1^x cos(n^(1/2) h(t)) h'(t) dt + 1/h'(b) int_x^b cos(n^(1/2) h(t)) h'(t) dt = 1/h'(1) n^(-1/2) (sin(n^(1/2) h(x)) - sin(n^(1/2) h(1)) + 1/h'(b) n^(-1/2) (sin(n^(1/2) h(b)) - sin(n^(1/2) h(x)) (and similarly with cos replaced by sin) so int_1^b exp(i n^(1/2) h(t)) dt = O(n^(-1/2)). For the integral from 0 to 1, we use the change of variables s = 1/t, h(s) = -h(t), so for 0 < r < 1 there is x between 1 and 1/r such that int_r^1 cos(n^(1/2) h(t)) dt = int_1^{1/r} cos(n^(1/2) h(s)) ds/s^2 = 1/h'(1) int_1^x cos(n^(1/2) h(s)) h'(s) ds + r^2/h'(1/r) int_x^{1/r} cos(n^(1/2) h(s)) h'(s) ds so int_0^1 exp(i n^(1/2) h(t)) dt = O(n^(-1/2)) also. Thus c_n = O(1/n). Similar arguments can be used for the parts of the integral of d_n for x from 0 to 1/2 and 3/2 to n^(1/2) Pi. Then for the integral from 1/2 to 3/2, we can use the Stationary Phase Theorem (see Marsden and Hoffman, Basic Complex Analysis, Theorem 7.2.10). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2