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X^3/(e^x-1) series

  1. May 30, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the power series of x^3/(e^x-1) and keep terms through x^4

    2. Relevant equations

    Taylor series

    f(x) ~ f(a) + f '(a)*x + f ''(a)/2!*x^2...

    3. The attempt at a solution

    Taking the first term f(a) around the point 0 shows 0/0 so I discarded that and moved on. But this happens in each of the terms for the series. To try and double check this I punched it into Wolfram Alpha which spit out

    around x=0:
    x^2-x^3/2+x^4/12-x^6/720+x^8/30240+O(x^9)

    I am confused how wolfram alpha obtained this and where I am going wrong. Thank you for your help.
     
  2. jcsd
  3. May 30, 2009 #2

    Cyosis

    User Avatar
    Homework Helper

    Mathematica takes the limit as x tends to 0 for every term. That said you're being asked to find the power series, are you certain they want you to calculate the Taylor series?

    You can find the power series by comparing it to a geometric series.
     
  4. May 30, 2009 #3
    You need to take the limit for x to zero if you want to use the formula for the Taylor expansion. This is not an efficient way to compue the Taylor series. A more efficient method is to substitute the Taylor expansion for e^x and then perform the division. Alternatively, you can use the power series of the geometric series.

    So, we have:

    x^3/[exp(x) - 1] =

    x^3/[x + x^2/2 + x^3/6 + x^4/24 + ...] =

    x^2/[1+x/2 + x^2/6 + x^3/24 +...]


    If you use the geometric series power expansion, you use that:

    1/(1+y) = 1 - y + y^2 - y^3 + ...

    and substitute

    y = x/2 + x^2/6 + x^3/24 + ...

    in here.

    If you want to do a division, you can do long division. You can also write the expansion in undetermined form:


    x^2/[1+x/2 + x^2/6 + x^3/24 +...] = a x^2 + b x^3 + c x^4 + ...

    (note that we can immediately tell that the series expansion will start with x^2).

    If you multiply both sides by 1+x/2 + x^2/6 + x^3/24 +... and equate the coefficients of the powers of x you can solve for a, b , recursively. This procedure is equivalent to long division.


    Then there exist a much faster way to do division of power series that is based on the Newton-Raphson technique of findng zeroes of nonlinear equations. This works as follows. In case of division of ordinary numbers, we need to compute 1/y where y is some given number. Then this amounts to solving the equation

    1/x - y = 0

    for x.

    Newton-Raphson then gives you successive approximants x_n that satisfy the recursion:

    x_{n+1} = x_{n} - (1/x_{n} - y)/(-1/x_{n}^2) =

    2x_n - y x_{n}^2

    Since this algorithm does not contain any divisions, it is a bona fide division algorithm.

    Now, this recursion can also be used for power series. If y is not a number but a power series and x_0 is 1/y correct to zeroth order, then the algorithm will yield the x_n as the expansion of 1/y correct to
    order 2^n-1. The number of correct expansion coefficients doubles at each iteration, while in case of long divsion you only get one coefficient per step.
     
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