Checking convergence of Gaussian integrals

[mitochan]

So you are prepared to write down $c_n=...$.

 

[JD_PM]

So you are prepared to write down $c_n=...$.

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*}

 

[mitochan]

No summation and take a look at factor $2^n$.

 

[JD_PM]

No summation and take a look at factor $2^n$.

Oops my bad. It can be simplified further

\begin{equation*}
c_n = \frac{1}{2} \sum_{n=0}^N\frac{(-)^n}{n!}\frac{1}{(4!)^n} \frac{(4n)!}{2^{3n}(2n)!}
\end{equation*}

 

[mitochan]

Do it more carefully.

 

[JD_PM]

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*}

 

[mitochan]

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.

 

[JD_PM]

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.

 

[JD_PM]

OK Got it, thanks for your patience!

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*}

 

[mitochan]

Thanks. So we can proceed to investigate the series converges or diverges.

 

[JD_PM]

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?

 

[mitochan]

If you do not mind please show your result.

 

[JD_PM]

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 $R= + \infty$

 

[mitochan]

I calculated it
\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
Please check our results.

 

[JD_PM]

I calculated it
\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
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

$$R=0$$

 

[JD_PM]

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*}

 

[mitochan]

Let me confirm you say

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

Why don't you try changing integral paremeters
\frac{u^4}{4!}=t
and do a similar job with b) ?

 

[JD_PM]

@mitochan I see how to complete the exercise, thank you for your help and patience