vladimir69
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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.
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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