0

How to construct generating function such that $$g(x) = \sum_{n=0}^\infty \frac{n!}{(2n)!} t^n $$

Mula Ko Saag
  • 2,177
  • 1
  • 23
  • 45
  • 1
    Woops!!, it seems to be related to $e^{x^2}$. http://mathworld.wolfram.com/DoubleFactorial.html should have googled property :/?? – Mula Ko Saag Jul 17 '15 at 17:33

3 Answers3

2

It might be better to consider the exponential generating function in this case. \begin{align} a_{n} &= \frac{n!}{(2n)!} \\ f(t) &= \sum_{n=0}^{\infty} a_{n} \, \frac{t^{n}}{n!} = \sum_{n=0}^{\infty} \frac{t^{n}}{(2n)!} = \cosh(\sqrt{t}). \end{align}

Linear generating function: \begin{align} g(t) &= \sum_{n=0}^{\infty} a_{n} \, t^{n} \\ &= \sum_{n=0}^{\infty} \frac{(1)_{n} \, (1)_{n} \, t^{n}}{n! \, (1)_{2n}} = \sum_{n=0}^{\infty} \frac{(1)_{n} \, t^{n}}{n! \, \left(\frac{1}{2}\right)_{n}} = {}_{1}F_{1}\left(1; \frac{1}{2}; \frac{t}{4}\right) \\ &= 1 + \frac{\sqrt{\pi \, t}}{2} \cdot e^{\frac{t}{4}} \cdot erf\left(\frac{\sqrt{t}}{2}\right). \end{align}

Equation (21) of MathWorld is as follows: \begin{align} \sum_{n=0}^{\infty} \frac{t^{2n+1}}{(2n+1)!!} &= \frac{1}{t} \, \sum_{n=0}^{\infty} \frac{2^{n+1} \, (n+1)! \, t^{2n+2}}{(2n+2)!} \\ &= \frac{1}{t} \, \sum_{n=1}^{\infty} \frac{2^{n} \, n! \, t^{2n}}{(2n)!} \\ &= \frac{1}{t} \, \left[ -1 + g(2 t^{2}) \right] \\ &= \sqrt{\frac{\pi}{2}} \cdot e^{\frac{t^{2}}{2}} \cdot erf\left(\frac{t}{\sqrt{2}}\right) \end{align}

Leucippus
  • 26,329
  • no no .. you are making too complicated, I don't understand hypergeometric funtions. could you simply provide hints (even on comments) to arrive at eq (21) on this page http://mathworld.wolfram.com/DoubleFactorial.html – Mula Ko Saag Jul 17 '15 at 17:48
  • @MulaKoSaag equation (21) has been demonstrated – Leucippus Jul 17 '15 at 18:00
1

$$g(x)=1+e^{x/4}\frac{\sqrt{\pi x}}{2} \mbox{erf}(\sqrt{x}/2).$$

Alex R.
  • 32,771
1

There's no easy way, I'm afraid... My suggestion would be to write the well-known Taylor series expansions for both the exponential and error function, where that of the latter can be found by integrating the former term by term, and then employ the dreadful Cauchy product formula for their multiplication.

Lucian
  • 48,334
  • 2
  • 83
  • 154
  • Sorry, it's $$\int_0^\infty e^{-x} \frac{x^{-1/2}}{1+2x}dx$$. I didn't put it correctly. :( any hints? I got this result form equation (2) http://mathworld.wolfram.com/DoubleFactorial.html and Reflection formula. – Mula Ko Saag Jul 17 '15 at 18:40
  • 1
    @MulaKoSaag: Substitute $x=t^2$, and then insert an extra parameter a in the exponent. Notice that the new integral satisfies the differential equation $I(a)-2I'(a)=\sqrt{\dfrac\pi a}$ – Lucian Jul 17 '15 at 18:57