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Series formula

  1. Mar 26, 2005 #1


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    I'm trying to get an expression for the nth term in this series:

    [tex]c_1 = 1[/tex]

    [tex]c_n = \sum_{k=1}^{n-1} \frac{c_n}{(n-k)!}[/tex]

    For example:

    [tex]c_2 = 1/1! = 1[/tex]

    [tex]c_3 = 1/2! +1/1! = 3/2[/tex]

    [tex]c_4 = 1/3! + 1/2! + 3/(2\cdot 1!) = 13/6[/tex]


    Since the factor in front of each term is different in each series, you can't express [itex]c_n[/itex] just in terms of [itex]c_{n-1}[/itex]. I have no idea how to start.
  2. jcsd
  3. Mar 29, 2005 #2


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    First, that is supposed to be ck inside the sum. It's too late to edit it.

    Is no one replying because this is too hard or too easy? If it helps, this was found when I was trying to derive a formula for

    [tex]\sum_{n=1}^{\infty} \frac{1}{n^{2k}} [/tex]

    I know this involves the Bernoulli numbers, so maybe they are involved in the series from my first post.
    Last edited: Mar 29, 2005
  4. Mar 29, 2005 #3
    [tex] \sum_{n=1}^\infty \frac{1}{n^{2k}} = \zeta(2k), \; \forall k > \frac{1}{2}[/tex]


    Is there a particular [itex]k[/itex] that you want to evaluate it for?
    Last edited: Mar 29, 2005
  5. Mar 29, 2005 #4


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    I was using fourier series to find an explicit formula. Namely, I started with a sawtooth wave with fourier coefficients proportional to 1/n. Then I normalized it so the integral over one period was one, which meant that:

    [tex]\sum_{n=1}^{\infty} |c_n|^2 = 1[/tex]

    From which I got the first sum (k=1). Then I integrated to get a series with 1/n2,l fixed the constant so there were only sinusoidal terms in the fourier series, and renormalized. I correctly got the formula for k = 1, 2, and 3. Then trying to find the general form, which is made complicated by the integrating constants that depend on the constants from previous iterations, I ran into a series very similar to the one from the first post. I was just curious if there is a closed form for that expression, or if you have to resort to more complicated methods.
    Last edited: Mar 29, 2005
  6. Mar 30, 2005 #5
    Maybe this will help, although I don't surely know cause I haven't tried it yet, but I recall a method I used to solve a similar problem once.


    [tex] f(x) := c_1 + c_2x + c_3x^2 + c_4x^3 + c_5x^4 + .....[/tex]

    where [tex]c_k[/tex] are the co-efficients you originally defined.

    Then expand the inverse of [tex]f(x)[/tex] as a power series, call this new power series [tex]g(x)[/tex], say. I'm assuming the co-efficients of [tex]g(x)[/tex] might be easier to recognise. If you have maple or mathematica this can be easily done.

    Then the [tex] c_n [/tex] co-efficients of [tex]f(x)[/tex] can be calculated by knowing the n'th derivitive of [tex]1/g(x)[/tex] , evaluated at the point zero.

    If you can calculate it, I'd be interested to know what the power series of [tex]g(x)[/tex] is.
    Last edited: Mar 30, 2005
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