Checking convergence of Gaussian integrals

In summary, the conversation discusses the computation of the integral Z(λ) and its range of convergence. The first part of the conversation explores the use of Gaussian integral formulas and Taylor expansions to find the expression for Z(λ). The second part talks about completing the square and using a change of variables to simplify the integration process. Eventually, the conversation leads to finding the explicit form of Z(λ) by expanding e^(-λx^4/4!) and integrating it term by term.
  • #36
mitochan said:
Do it more carefully.

\begin{align*}
c_n &= \frac{1}{\sqrt{2}}\frac{(-)^n}{n!}\frac{1}{(4!)^n}(2)^{2n}\frac{1}{\sqrt{2}} \frac{(4n)!}{2^{4n}(2n)!} \\
&= \frac{1}{2} \frac{(-)^n}{n!}\frac{1}{(4!)^n} \frac{(4n)!}{2^{2n}(2n)!}
\end{align*}
 
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  • #37
Thanks. I got c0=1 in post #21 and your result shows c0=1/2. I should appreciate it if you would check factor 2 missing or duplicating.
 
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  • #38
Let me summarize what we are doing. At #21 you obtained (let me plug the solution of the integral)

\begin{align*}
c_n &= \frac{(-)^n}{\sqrt{2\pi}}\frac{2}{(4!)^n n!}\int_{0}^{+\infty}x^{4n} e^{-\frac{x^2}{2}}dx=\frac{(-)^n}{\sqrt{\pi}}\frac{2^{2n+1}}{(4!)^n n!}\int_{0}^{+\infty}x^{4n} e^{-x^2}dx \\
&= \frac{(-)^n 2^{2n+1}}{(4!)^n n!} (\sqrt{2})^{4n - 1} \frac{(4n)!}{2^{4n}(2n)!}.
\end{align*}

And I got

\begin{align*}
c_n &= \frac{1}{\sqrt{2}}\frac{(-)^n}{n!}\frac{1}{(4!)^n}(2)^{2n}\frac{1}{\sqrt{2}} \frac{(4n)!}{2^{4n}(2n)!} \\
&= \frac{1}{2} \frac{(-)^n}{n!}\frac{1}{(4!)^n} \frac{(4n)!}{2^{2n}(2n)!}
\end{align*}

It indeed seems I am missing a ##2## factor. I will recheck my computation.
 
  • #39
OK Got it, thanks for your patience! :smile:

JD_PM said:
Indeed (I would say it cannot be simplified any further)

\begin{equation*}
c_n = \frac{1}{\sqrt{2}} \sum_{n=0}^N\frac{(-)^n}{n!}\frac{1}{(4!)^n} (\sqrt{2})^{4n - 1} \frac{(4n)!}{2^{4n}(2n)!}
\end{equation*}

I dropped a two (i.e. we have ##\int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx = 2 (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right)## and not ##\int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx = (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right)##).

So now my answer matches yours i.e.

\begin{align*}
c_n &= \frac{2}{\sqrt{2}}\frac{(-)^n}{n!}\frac{1}{(4!)^n}(2)^{2n}\frac{1}{\sqrt{2}} \frac{(4n)!}{2^{4n}(2n)!} \\
&= \frac{(-)^n}{n!}\frac{1}{(4!)^n} \frac{(4n)!}{2^{2n}(2n)!}
\end{align*}
 
  • #40
Thanks. So we can proceed to investigate the series converges or diverges.
 
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  • #41
mitochan said:
Thanks. So we can proceed to investigate the series converges or diverges.

I did this section some days ago via ratio test and got that the series was convergent, with radius of convergence ##R=\infty##

Do you mind if we advance or do you want me to show it explicitly?
 
  • #42
If you do not mind please show your result.
 
  • #43
mitochan said:
If you do not mind please show your result.

Using the ratio test we get

\begin{align*}
\lim_{n \to \infty} \left|\left( c_{n + 1} \frac{1}{c_n}\right)\right| &= \frac{(-)^{n+1}}{(n+1)!} \frac{1}{(4!)^{n+1}} \frac{\left[4(n+1) \right]!}{2^{2(n+1)}\left[ 2(n+1)\right]!} \\
&\times \frac{n!}{(-)^n}(4!)^n \frac{2^{2n}(2n)!}{(4n)!} \\
&= \lim_{n \to \infty} \left(\frac{1}{4! (n+1)}\right) \\
&= 0 < 1
\end{align*}

Hence ##R= + \infty##
 
  • #44
I calculated it
[tex]\frac{c_{n+1}}{c_n}=\frac{-1}{4*4!}\frac{(4n+4)(4n+3)(4n+2)(4n+1)}{(n+1)(2n+2)(2n+1)}=\frac{-1}{4!}\frac{(4n+3)(4n+1)}{(n+1)}=-\frac{2}{3}\frac{(1+3/4n)(1+1/4n)}{1+1/n}n[/tex]
Please check our results.
 
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  • #45
mitochan said:
I calculated it
[tex]\frac{c_{n+1}}{c_n}=\frac{-1}{4*4!}\frac{(4n+4)(4n+3)(4n+2)(4n+1)}{(n+1)(2n+2)(2n+1)}=\frac{-1}{4!}\frac{(4n+3)(4n+1)}{(n+1)}=-\frac{2}{3}\frac{(1+3/4n)(1+1/4n)}{1+1/n}n[/tex]
Please check our results.

You are right, my bad! (I have to acknowledge that I rushed, my apologies).

Let me be more careful

\begin{align*}
c_{n + 1} \frac{1}{c_n} &= \frac{(4n+4)!}{(n+1)!(4!)^{n+1} 2^{2n+2}(2n+2)!}\times \\
&\times \frac{n! (4!)^n 2^{2n}(2n)!}{(4n)!} \\
&= \frac{(4n+3)(4n+1)}{24(n+1)} \\
&= \frac{n}{24}\left( \frac{(4+3/n)(4+1/n)}{(4+1/n)} \right)
\end{align*}

Where, to get to the first equality, I used factorial properties and simplified

$$(4n+4)! = (4n+4)(4n+3)(4n+2)(4n+1)(4n)!$$

$$(n+1)! = (n+1)(n)!, \qquad (2n+2)! = (2n+2)(2n+1)(2n)!$$

So the series diverges!

\begin{equation*}
\lim_{n\to \infty} \left( \frac{n}{24}\left( \frac{(4+3/n)(4+1/n)}{(4+1/n)} \right) \right) = \infty
\end{equation*}

And the radius of convergence is given by

$$R=0$$
 
  • #46
Regarding c). Let me check if I am on the right track

I would introduce a change of variables ##u := \lambda^{1/4} x## and expand one of the exponentials only so that ##Z(\lambda)## becomes

\begin{align*}
Z(\lambda) &= \frac{1}{\sqrt{2\pi}} \lambda^{-1/4} \int_{-\infty}^{\infty} du \exp\left( -\frac{u^2}{2!\sqrt{\lambda}}-\frac{1}{4!}u^4\right) \\
&= \frac{1}{\sqrt{2\pi}} \lambda^{-1/4} \sum_{n=0}^{\infty} \int_{-\infty}^{\infty} du e^{-u^4/4!} \left( \frac{-u^2}{2! \sqrt{\lambda}} \right)^n \frac{1}{n!}
\end{align*}Mmm so all would boil down to computing an integral of the form

\begin{align*}
\int_{-\infty}^{\infty} du \ e^{-u^4/4!} u^{2n}
\end{align*}
 
  • #47
Let me confirm you say

[tex]d_n=\frac{1}{\sqrt{2\pi}}\frac{(-)^n}{n!}\int_{-\infty}^{+\infty} e^{-\frac{u^4}{4!}}\frac{u^{2n}}{2^n} du[/tex]

Why don't you try changing integral paremeters
[tex]\frac{u^4}{4!}=t[/tex]
and do a similar job with b) ?
 
Last edited:
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  • #48
@mitochan I see how to complete the exercise, thank you for your help and patience :smile:
 
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