Re: trigonometric series


"Stephen Montgomery-Smith" <stephen@xxxxxxxxxxxxxxxxx> wrote in message
news:29c6h.279221$1i1.144462@xxxxxxxxxxxx
rancid moth wrote:
By steadily, I assume you mean it is a non-increasing sequence.


yes correct. i mean a_n >= a_n+1 and a_n->0 as n->oo


and so by Dirichlet's test sum ( n=1, oo) a_n * sin(n*x) converges.
However, my question is on what domain does it converge? is it true
only for 0<x<2pi or for all real x ? I ask because im having difficulty
seeing the difference between the convergence of sum ( n=1, oo) a_n *
sin(n*x) and sum ( n=1, oo) a_n * cos(n*x)

You know it converges at x=0 or x=2pi, because there it is zero.


Of course - stupid of me! for any behaved a_n as stated above, it must
converge at the boundaries because of this fact. This wont nessecarily
be the case for the cosine series.


The question I think you are intending is - does it converge to a
continuous or bounded function?


In part yes. I was wondering how one could show what domain x could take
to satisfy the convergence.


I know it is not necessarily continuous, because if you put a_n=1/n it is
a saw tooth function which is clearly discontinuous at 0. My guess is
that there are examples to show it need not be bounded.


For example a_n = 1/n, the series would converge for all x, because of
the fact that the terms are zero at that boundary. This wouldnt be the
case for the cosine series, for there the series would diverge!

Thus if one were to look at sum ( n=1, oo) a_n * cos(n*x)

and using the test (Dirichlet) then i would think we would _have_ to say
that the series converges for all x other than even multiples of pi ?
would that be correct? whereas we dont have that restriction with the
sine series. We cant just shift the boundary either because of the fact
that

| sum (n=1,N) sin(n*x) | < csc (x/2) for 0<x<2pi

and so it would seem that the series is, pardon the jargon, stuck within
the 2*pi blocks because of this fact -and it this that causes the
differnece in the convergence properties of the sine and cosine series.
I think now i have it.

thank you.



Anyway, it does look like an interesting question.

A long time ago I was interested in similar questions about the Fourier
transform. It is more or less the same question, if you suppose that f(x)
is non-increasing, positive for x>0. You have the two cases - is f an
even function or an odd function? The Dirichlet test becomes rather
simpler (it is just integration by parts). You quickly come to realise
that the odd case is a lot easier to handle than the even case. The
calculations are almost identical to the ones you did.

Yes, I suppose it becomes a lot less mysterious when one thinks about the
fact that ultimately the series will be used to express a function on a
(here anyway) finite domain. you then have a choice on how you could start
that representation (using these oscillatory functions) by making the terms
in the series either equal to zero at both ends, or not. one choice would
obviously have benefits over the other, given that one reduces the
possibility of divergence at the end points, and smooths the behaviour in
the middle via an alternating series, while the other does the same in the
middle, though has little influence on its behaviour at the end points.
That is the convergence at the endpoints becomes reliant upon the
coefficients in the series.

It's nice how Dirichlet's test shows this.


.

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