This question is an exact duplicate of:

I've this expression to simplify:

$$\frac {1}{2^{3N}}\frac{(2m\pi)^{3N/2}}{h^{3N}\sqrt{\frac{2\pi3N}{2}}\left(\frac{3N}{2}\right)^{3N/2}e^{-3N/2}}U^{3N/2}\frac{V^N}{\pi^{3N}}$$

and I'm supposed to end with:


But whenever I try I couldn't reach it


marked as duplicate by Community Jan 18 '18 at 9:29

This question was marked as an exact duplicate of an existing question.

  • 1
    $\begingroup$ Are you sure the term in the denominator under the square root is correct? $\endgroup$ – Rohan Jan 18 '18 at 7:56
  • $\begingroup$ Yes. $\sqrt{\frac{2\pi3N}{2}}\left(\frac{3N}{2}\right)^{3N/2}e^{-3N/2}$ has to be the Stirling approximation of $\Gamma(3N/2+1)$ $\endgroup$ – Lo Scrondo Jan 18 '18 at 8:00
  • $\begingroup$ @LoScrondo The two expressions are not equivalent equal as written. Check for example the exponent of $\,\pi\,$, which doesn't match between them. If there are some other known relations between the variables besides what's posted, then you'll need to spell those out. Same goes if by "simplify" you maybe meant some sort of asymptotic equivalence, instead, which your accept vote seems to indicate. $\endgroup$ – dxiv Jan 18 '18 at 8:15

We have: $$I = \frac {1}{2^{3N}}\frac{(2m\pi)^{3N/2}}{h^{3N}\color{red}{\sqrt{\frac{2\pi3N}{2}}}\left(\frac{3N}{2}\right)^{3N/2}e^{-3N/2}}U^{3N/2}\frac{V^N}{\pi^{3N}}$$

$$I = \frac{2^{3N/2}m^{3N/2}\pi^{3N/2}e^{3N/2}U^{3N/2}V^N\times 2^{3N/2}}{2^{3N}h^{3N}3^{3N/2}N^{3N/2}\pi^{3N}}$$

$$I= \frac{2^{3N}}{2^{3N}} \times \frac{\pi^{3N/2}}{\pi^{3N}}\times \frac{(me)^{3N/2}}{3^{3N/2}h^{\color{green}{2}\times 3N/2}N^{3N/2}} \times U^{3N/2}\times V^N$$

Simplify it and we are done.

Unfortunately, I have to say that though the Stirling approximation to $\Gamma(3N/2+1)$ is correct, I never used that term: $\sqrt{\dfrac{2\pi3N}2}$, but I still get the required answer. Maybe, you just give the original expression without any simplification and we will be able to deduce from it.

  • $\begingroup$ You mean I have to edit the question I've posted, o I've to open another one (many many thanks indeed)? $\endgroup$ – Lo Scrondo Jan 18 '18 at 8:14
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    $\begingroup$ Just post another one stating a reference to this question. $\endgroup$ – Rohan Jan 18 '18 at 8:15
  • $\begingroup$ I've posted the new one here $\endgroup$ – Lo Scrondo Jan 18 '18 at 8:29

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