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Finding coefficients

  1. Jul 12, 2006 #1
    this problem has been bugging me for a few weeks now and here it is
    [tex]A=\sum_{k=0}^{\infty}\sum_{m=-N}^{+N}a_k\cos(\frac{mk\pi}{N+1})\cos(\frac{mj\pi}{N+1})=1[/tex]
    what i am trying to do is find the [itex]a_k[/itex] 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 [itex]a_k[/itex] terms.
    like for example i found that [itex]\sum_{m=-N}^{+N}\cos(\frac{mk\pi}{N+1})\cos(\frac{mj\pi}{N+1})[/itex]
    =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 [itex]A[/itex] 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. jcsd
  3. Jul 14, 2006 #2
    Is j is a constant here?

    Keep Smiling
    Malay
     
  4. Jul 14, 2006 #3

    HallsofIvy

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

    In that case both [itex]cos\left(\frac{m(k\pi)}{N+1}\right)[/itex] and [itex]cos\left(\frac{mj\pi}{N+1}\right)[/itex] (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.
     
  5. Jul 14, 2006 #4

    0rthodontist

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    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
    [tex]a_n = a_{n-1}+2cos(\theta n)[/tex]
    where you know that the particular solution is of the form
    [tex]C_1 sin(\theta n) + C_2 cos(\theta n)[/tex]
    and you know that the homogeneous solution is a constant function, adding another coefficient C3. To sum it up,
    [tex]a_n = C_1 sin(\theta n) + C_2 cos(\theta n) + C_3[/tex]
    Just find some actual values of the sequence and use them to solve for the C's.
     
    Last edited: Jul 14, 2006
  6. Jul 14, 2006 #5
    firstly, thanks for the replies
    MalayInd: j is an integer which can take on any value you choose, which is troublesome for me because [itex]B=\sum_{m=-N}^{+N}\cos(\frac{mk\pi}{N+1})\cos(\frac{mj\pi}{N+ 1})[/itex] 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
    [tex]a_n=2+\frac{\sin(\theta)\sin(n\theta)}{\cos(\theta)-1}-\cos(n\theta)[/tex]
    [tex]\theta = \frac{(k+j)\pi}{N+1}[/tex]
    and then my sum is something like,
    [tex] \sum_k a_k((N+1)\delta_{k,j} + a_N) = 1[/tex] (1)
    from that i am unable to determine the a_k terms. i was hoping i would get a [itex]\delta_{k,j}[/itex] popping out of the sum [itex]B[/itex] 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
    [tex]p(m,n)=\sum_k a_k \cos ^n (\theta) \cos(m\theta)[/tex]
    where [itex]p(m,0)=\delta(m)[/itex] where
    [tex]\delta(m) = 1[/tex] if m=0
    [tex]\delta(m)=0[/tex] otherwise
    in an attempt to find a_k i multiplied both sides by
    [tex]\cos(\frac{mj\pi}{N+1})[/tex]
    so hopefully that made it a bit clearer.

    the solution still eludes me.
     
    Last edited: Jul 14, 2006
  7. Jul 15, 2006 #6

    0rthodontist

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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.
     
  8. Jul 15, 2006 #7

    HallsofIvy

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    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.
     
  9. Jul 15, 2006 #8
    ok so whats the verdict
    a_i = 0 for i > 0?
    a_0 = ?
     
  10. Jul 17, 2006 #9
    i dont see how you got that the a_i's are 0
    Orthodontist do you mean this?
    [tex]a_0=\frac{1}{\sum_{m=-N}^{+N}\cos(\frac{mj\pi}{N+1})}[/tex]
     
  11. Jul 17, 2006 #10

    0rthodontist

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    No, I meant
    [tex]a_0 = \frac{1}{\sum_{m=-N}^{+N}\cos(\frac{mk\pi}{N+1})\cos(\frac{mj\pi} {N+1})}[/tex]
    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
  12. Jul 17, 2006 #11
    "The thing is that your equation does not completely determine the ai's"
    i see.

    originally i had
    [tex]p(m,n)=\sum_k a_k \cos ^n (\frac{k\pi}{N+1}) \cos(\frac{mk\pi}{N+1})[/tex]
    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
    [tex]\cos ^n (\frac{k\pi}{N+1}) \cos(\frac{mk\pi}{N+1})[/tex] = 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
  13. Jul 17, 2006 #12

    0rthodontist

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    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
  14. Jul 17, 2006 #13
    ok thanks Orthodontist thats what i was looking for
    but how did you work out that the a_i's must sum to 1?
     
  15. Jul 17, 2006 #14

    0rthodontist

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    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:

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