# Finding the general form of a Struve Function

• eclayj
In summary, the conversation is about a series known as Struve function and the speaker is trying to find the expression for the series. They have tried using the expansion of Sin[x] and the numerator -1^k x^(2k+1) but cannot figure out the denominator (odd numbers squared and multiplied). With help from another person, they realize that the denominator is (2k+1)!^2 and that Mathematica requires two factorial symbols to properly calculate the equation.

#### eclayj

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

#### Attachments

• Johnson_HW_3.06.3.nb
119.5 KB · Views: 391

Wow. Pretty obvious. Maybe I shouldn't be doing so late at night. Thank you very much!

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.