Finding Coefficients for A=1 in Problem

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The discussion revolves around finding the coefficients a_k in the equation A = ∑_{k=0}^{∞} ∑_{m=-N}^{+N} a_k cos(mkπ/(N+1)) cos(mjπ/(N+1)) = 1. Participants explore various approaches, including recurrence relations and specific values for a_k, but struggle with the complexity of the inner sums. The consensus suggests that a_0 should equal 1 while other a_k terms may be zero, although there is uncertainty about how to derive these coefficients effectively. The challenge lies in ensuring that the inner sums yield meaningful results for probabilities, as the values of j affect the outcomes. Ultimately, the participants seek a clearer method to determine the a_k terms while maintaining the function's dependence on m and n.
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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 don't 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.
 
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Is j is a constant here?

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Malay
 
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.
 
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.
 
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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 don't 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.
 
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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.
 
vladimir69 said:
HallsofIvy: i hope it is that simple but i don't 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.
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.
 
ok so what's the verdict
a_i = 0 for i > 0?
a_0 = ?
 
i don't 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
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.
 
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  • #11
"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 can't 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
 
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  • #12
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, ...
 
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  • #13
ok thanks Orthodontist that's what i was looking for
but how did you work out that the a_i's must sum to 1?
 
  • #14
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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