Tuesday, November 15, 2016

real analysis - Behaviour of sumuncos(ntheta+a) when thetaoverpi is rational and when irrational




I am trying to show that if θ is not equal to 0 or a multiple of 2π, and if u0,u1,u2,... be a sequence such that un0 steadily, then the series uncos(nθ+a) is convergent and also if the limit of un is not zero, but un is still monotonic, the sum of the series is oscillatory if θπ is rational, but that, if θπ is irrational, the sum may have any value between certain bounds whose difference is lim.






Solution Attempt
Based on the upper bound of the partial sums are bounded as shown by Julián Aguirre
below and based on G.H.Hardy's test for uniform convergence which states that if, in a given domain, \left| \sum _{n=1}^{p}a_{n}\left( z\right) \right| \leq k where a_{n}\left(z\right) is real and k is finite and independent of p and z, and if f_{n}\left( z\right) \geq f_{n+1}\left( z\right) and f_{n}\left( z\right)\rightarrow 0 uniformly as n\rightarrow \infty, then \sum _{n=1}^{\infty }a_{n}\left( z\right) f_{n}\left( z\right) converges uniformly.



Since u_n\rightarrow 0 and the sequence u_0,u_1,u_2,... is monotonically decreasing.
We also observed that the partial sum is bounded so the series converges uniformly.

Hence the series must converge.



I am unsure how to attack the second part of the problem, to be more specific how the cos function varies when the parameter part is irrational. Any help would be much appreciated.


Answer



The hypothesis do not say anything about the convergence of the series \sum u_n. There is nothing you can say about the absolute convergence of the series, and in particular, the ratio test is useless.



The series converges by Dirichlet's test. It is enough to show that the partial sums \sum_{k=1}^n\cos(n\,\theta+a) are bounded for \theta\ne2\,k\,\pi:
\sum_{k=1}^n\cos(n\,\theta+a)=\Re\Bigl(\sum_{k=1}^ne^{(n\,\theta+a)i}\Bigr)=\Re\Bigl(e^{ai}\sum_{k=1}^ne^{n\theta i}\Bigr)=\Re\Bigl(e^{ai}\frac{e^{(n+1)\theta i}}{e^{\theta i}-1}\Bigr),

\Bigl|\sum_{k=1}^n\cos(n\,\theta+a)\Bigr|\le\frac{1}{|e^{\theta i}-1|}.



For the second question, if \lim_{n\to\infty}u_n=u, then u_n-u converges monotonically to 0. Writing
\sum_{k=1}^nu_n\cos(n\,\theta+a)=u\sum_{k=1}^n\cos(n\,\theta+a)+\sum_{k=1}^n(u_n-u)\cos(n\,\theta+a).
By the previous argument, \sum_{k=1}^n(u_n-u)\cos(n\,\theta+a) converges; let S be its sum. We see that it is enough to study the behaviour of
\sum_{k=1}^n\cos(n\,\theta+a)=\Re\Bigl(e^{ai}\frac{e^{(n+1)\theta i}}{e^{\theta i}-1}\Bigr)=\frac12\csc\frac\theta2\sin\Bigl(\bigl(n+\frac12\bigr)\theta+a\Bigr).
If \theta/\pi is irrational, then the values of
\sin\Bigl(\bigl(n+\frac12\bigr)\theta+a\Bigr)=\sin\Bigl(\frac{(2\,n+1)\pi}{2}\frac{\theta}{\pi}+a\Bigr)
are uniformly distributed in [-1,1], and the values of \sum_{k=1}^n\cos(n\,\theta+a) are uniformly distributed in
\Bigl[S-\frac{u}2\csc\frac\theta2,S+\frac{u}2\csc\frac\theta2\Bigr].

On the other hand, if \theta/\pi=p/q is rational, then
\sin\Bigl(\bigl(n+\frac12\bigr)\theta+a\Bigr)=\sin\Bigl(\frac{(2\,n+1)p\pi}{2\,q}+a\Bigr)
is a periodic sequence.


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