0
$\begingroup$

I have the following function: $$f(x) = (\frac{x!}{\sqrt[]{\pi}} * (\frac{x}{e})^{-x})^6$$

I want to get the following result using Taylor expansion at Infinity: $$8 x^3+4 x^2+x+\frac{1}{30}-\frac{11}{240x}+O((\frac{1}{x})^{\frac{3}{2}})$$

I have managed to get this result using Mathematica and the following command but I couldn't do it by myself:

Series[((x!/Sqrt[Pi] *((x)/E)^-(x))^6), {x, Infinity, 1}]

I've tried many different ways and failed. Any help would be great. Thanks.

$\endgroup$
  • $\begingroup$ Is $x$ supposed to be an integer in your question? (or is the factorial a Gamma function "in disguise"?) $\endgroup$ – Clement C. Jan 4 '16 at 11:10
  • $\begingroup$ @ClementC, Yes, x is supposed to be an integer. $\endgroup$ – Eyal Godovich Jan 4 '16 at 11:12
  • 2
    $\begingroup$ I'd start with the asymptotic series for $n!$ that extends the Stirling approximation, and work from there. $\endgroup$ – vonbrand Jan 4 '16 at 11:37
0
$\begingroup$

You are looking for an asymptotic expansion on $x$ for the expression, so the number of terms you get depends on the number of terms you specify on Stirling's approximation as vonbrand says. E.g. For the general Stirling approximation on $k$ terms in parentheses (see here):

$$\Gamma_k(x+1)\sim S_k(x)=\sqrt{2\pi x}\left(\frac{x^x}{e^x}\right)\cdot\left(1+\frac{1}{12x}+\frac{1}{288x^2}-\frac{139}{51840x^3}-\cdots\right)$$

Your expression is:

$$\left(\frac{\Gamma(x+1)}{\sqrt{\pi}}\left(\frac{e}{x}\right)^x\right)^6\,\,\,\,(1)$$

Substituting $S_1(x)$ in (1) simplifies it to: $8x^3$

Substituting $S_2(x)$ in (1) simplifies it to: $8x^3+4x^2+\text{stuff}$

Substituting $S_5(x)$ in (1) simplifies it to: $8x^3+4x^2+x+\frac{1}{30}-\frac{11}{240x}+\text{stuff}$

So for an asymptotic correct to $k$ terms you need to substitute at least $S_k(x)$ into your expression.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.