1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the general form of a Struve Function

  1. Nov 25, 2011 #1
    I am given the following series which the problem states is known as a Struve function:

    x -x^3/(1^2 3^2)+x^5/(1^2 3^2 5^2)-x^7/(1^2 3^2 5^2 7^2) + ...

    I am trying to figure out the expression of the series. I have thought about this for some time. I am sure I am missing the obvious. It looks a little like the expansion of Sin[x] in powers of x. Anyway, I have worked at the numerator as follows:

    -1^k x^(2k+1)

    I cannot figure out for the life of me an expression that will give the denominator of the terms in the series (odd numbers squared and multiplied, i.e. the first term's denominator is one squared, then the next term has the denominator one squared times three squared, etc.).

    If anyone could help me think about this it would be greatly appreciated. Thanks so much! I have attached a .nb file to run in Mathematica if it helps. It is the second question.

    -EJ
     

    Attached Files:

  2. jcsd
  3. Nov 26, 2011 #2

    I like Serena

    User Avatar
    Homework Helper

    What about (2k+1)!^2?
     
  4. Nov 26, 2011 #3
    Wow. Pretty obvious. Maybe I shouldn't be doing so late at night. Thank you very much!
     
  5. Nov 26, 2011 #4
    Also didn't realize that you need two factorial symbols "!!" rather than the usual one symbol for Mathematica to recognize and properly calculate the equation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook