Finding coefficients

1. Jul 12, 2006

vladimir69

this problem has been bugging me for a few weeks now and here it is
$$A=\sum_{k=0}^{\infty}\sum_{m=-N}^{+N}a_k\cos(\frac{mk\pi}{N+1})\cos(\frac{mj\pi}{N+1})=1$$
what i am trying to do is find the $a_k$ terms.
i've tried a couple of different ways but there just seems to be too many different possibilities to be able to find the $a_k$ terms.
like for example i found that $\sum_{m=-N}^{+N}\cos(\frac{mk\pi}{N+1})\cos(\frac{mj\pi}{N+1})$
=N (if k=j)
=1 (if k+j) is odd)
=-1 (if k+j is even)
=2N+1 (if k=j=0)
i have been told there is nothing wrong with the above solution but from that i still dont know how to find a_k
i also suggested solving a recurrence relation as suggest by another guy but that proved to be unsuccessful also as it just turned into a giant mess.
if it helps i am trying to calculate probabilities so $A$ should be returning probabilities but that is not the major problem at the moment i just want the a_k terms so i can work out some numerical values.

Last edited: Jul 13, 2006
2. Jul 14, 2006

MalayInd

Is j is a constant here?

Keep Smiling
Malay

3. Jul 14, 2006

HallsofIvy

Am I missing something here? Doesn't a0= 1, ai= 0 if i is not 0 work?

In that case both $cos\left(\frac{m(k\pi)}{N+1}\right)$ and $cos\left(\frac{mj\pi}{N+1}\right)$ (what is j?) are 1 for m= 0, the other values don't matter and the sum on the right reduces to the single term 1.

4. Jul 14, 2006

0rthodontist

If you want to find the inner sum disregarding the ak's, solving the sequence should not be a mess. Back in the original thread
https://www.physicsforums.com/showthread.php?t=125068
all you have to do to finish solving it is find
$$a_n = a_{n-1}+2cos(\theta n)$$
where you know that the particular solution is of the form
$$C_1 sin(\theta n) + C_2 cos(\theta n)$$
and you know that the homogeneous solution is a constant function, adding another coefficient C3. To sum it up,
$$a_n = C_1 sin(\theta n) + C_2 cos(\theta n) + C_3$$
Just find some actual values of the sequence and use them to solve for the C's.

Last edited: Jul 14, 2006
5. Jul 14, 2006

vladimir69

firstly, thanks for the replies
MalayInd: j is an integer which can take on any value you choose, which is troublesome for me because $B=\sum_{m=-N}^{+N}\cos(\frac{mk\pi}{N+1})\cos(\frac{mj\pi}{N+ 1})$ takes on different values depending on the value of j (which can be either odd or even)

HallsofIvy: i hope it is that simple but i dont follow how you went about your answer, and in particular "the other values don't matter " part. would your solution change whether j=1 or j=2 or j=3 for example? if not then i think this is what i am after but its magic to me how you got it - i'm a bit slow yes.
Orthodontist: yeah i solved that recurrence thing to get
$$a_n=2+\frac{\sin(\theta)\sin(n\theta)}{\cos(\theta)-1}-\cos(n\theta)$$
$$\theta = \frac{(k+j)\pi}{N+1}$$
and then my sum is something like,
$$\sum_k a_k((N+1)\delta_{k,j} + a_N) = 1$$ (1)
from that i am unable to determine the a_k terms. i was hoping i would get a $\delta_{k,j}$ popping out of the sum $B$ to be able to find the a_k terms but the above sum (1) is too difficult for me.

this may help you understand where i am coming from
i originally found a function
$$p(m,n)=\sum_k a_k \cos ^n (\theta) \cos(m\theta)$$
where $p(m,0)=\delta(m)$ where
$$\delta(m) = 1$$ if m=0
$$\delta(m)=0$$ otherwise
in an attempt to find a_k i multiplied both sides by
$$\cos(\frac{mj\pi}{N+1})$$
so hopefully that made it a bit clearer.

the solution still eludes me.

Last edited: Jul 14, 2006
6. Jul 15, 2006

0rthodontist

I'm not sure what HallsofIvy means by the other terms not mattering, but using the same idea, letting the inner sum except for ak equal X for k = 0, then you can let a0 = 1/X and let all the other a's be 0.

7. Jul 15, 2006

HallsofIvy

I was wrong! For some reason, I got in my head that the inner sum was from -k to k rather than -N to N. InTHAT case, the first term, with k=0, the inner sum would also just be m=0 so both cosines would just give 1. "The other terms don't matter" because they are all multiplied by ai= 0.

Of course, since the inner sum is always from -N to N whatever k is, that's not true.

8. Jul 15, 2006

vladimir69

ok so whats the verdict
a_i = 0 for i > 0?
a_0 = ?

9. Jul 17, 2006

vladimir69

i dont see how you got that the a_i's are 0
Orthodontist do you mean this?
$$a_0=\frac{1}{\sum_{m=-N}^{+N}\cos(\frac{mj\pi}{N+1})}$$

10. Jul 17, 2006

0rthodontist

No, I meant
$$a_0 = \frac{1}{\sum_{m=-N}^{+N}\cos(\frac{mk\pi}{N+1})\cos(\frac{mj\pi} {N+1})}$$
The thing is that your equation does not completely determine the ai's--for example, you could let the ai's be such that the inner sums are arranged in a geometric sequence summing to 1.

Last edited: Jul 17, 2006
11. Jul 17, 2006

vladimir69

"The thing is that your equation does not completely determine the ai's"
i see.

originally i had
$$p(m,n)=\sum_k a_k \cos ^n (\frac{k\pi}{N+1}) \cos(\frac{mk\pi}{N+1})$$
i think if i used the value for a_0 you have there that would make p no longer a function of m or n. do you know how i would go about getting some sort of a_i's (even if u cant completely determine them) so that p was a function of m and n ? because if u set a_k = 0 for k>0 then pop in k=0 into the following
$$\cos ^n (\frac{k\pi}{N+1}) \cos(\frac{mk\pi}{N+1})$$ = 1, and if i am not wrong that means m and n do the disappearing act.
i have a hazy idea it might have something to do with what you were talking about with the geometric sequence. why do the a_i's have to sum to 1?

thanks for your help

Last edited: Jul 17, 2006
12. Jul 17, 2006

0rthodontist

That looks like it would be a function of m and n to me, specifically one that is constant in n.

The inner sums (not the a's) would have to sum to 1--it was just an example. You can pick any infinite sequence that sums to 1, and choose your a's so that the inner sums match that sequence. If you don't want p to be constant in n, the simplest choice is to let that sequence be 0, 1, 0, 0, ...

Last edited: Jul 17, 2006
13. Jul 17, 2006

vladimir69

ok thanks Orthodontist thats what i was looking for
but how did you work out that the a_i's must sum to 1?

14. Jul 17, 2006

0rthodontist

I said, they don't--the inner sums do because you have stated that they have to when you asked the original question.

Maybe I should clarify:

$$a_1 = \frac{1}{\sum_{m=-N}^{+N}\cos(\frac{mk\pi}{N+1})\cos(\frac{mj\pi} {N+1})}$$
where k = 1, all other ai's are 0.

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