F(x) of a taylor series that looks a lot like an exponential

[donifan]

Hello, I am trying to evaluate the series

\sum{\frac{x^n}{n!}e^{cn^2}}

where c is a constant. I think this problem is equivalent to find f(x) such that


\frac{d^{n}f(0)}{dx^{n}} = \frac{e^{cn^{2}}}{n!}

I believe this must be a modified exponential since for c=0, it reduces to f(x)=e^x (also because I have plotted the solution). I have tried many things, however I still can't find the form of f(x). Any ideas?

 

[mathman]

I can't answer your question. However, warning! c > 0 may be a problem!

 

[donifan]

No biggie, x<0. To avoid confusion, let's write it better like

<br /> \sum{\frac{(-1)^nt^n}{n!}e^{cn^2}}<br />

I also know from the physical process that:

\displaystyle\lim_{t\to\infty} f(t)=0

Thanks!

 

[Wizlem]

I'm not sure whether you can express this particular function as some sort of exponential such as e^{g(x)} for some g(x) or not but if you take the derivative it doesn't seem likely that you can. I came up with f&#039;(x) = e^cf(e^{2c}x).

 

[Mute]

By expanding the exponential as a second series, assuming we can swap the order of summations, I get

f(x) = \sum_{m=0}^\infty \frac{c^m}{m!} \left( x\frac{d}{dx}\right)^{2m} e^x

The derivative should work out to a polynomial times an exponential. Not sure if that polynomial can be worked out explicitly.

 

[donifan]

Thanks guys!

First of all, I had a mistake in the first post (nothing about the series). The equivalent problem must be:

<br /> \frac{d^{n}f(0)}{dx^{n}} = e^{cn^{2}}<br />


So far I was able to check that f^{&#039;}(x) = e^cf(e^{2c}x) (Very nice work Wizlem). I even tried making the substitution u=e^{2c}x so that

\frac{df}{du}=e^{c}f

which can be easily solved (with f(0)=1) to f(x)=exp(e^{c}x). This however does not work as it does not reproduce all the derivatives of f(x) (actually only the first one). I am missing something there (probably st quite fundamental).

About the double expansion (which I think is equivalent to a Poisson expansion of f), I am not sure about the numerical stability especially when using many terms. x in my problem expands over 70 orders of magnitude (up to 10^50). That's is the reason why I am trying to come up with a closed form for f.

 

[Mute]

\frac{df}{du}=e^{c}f

which can be easily solved (with f(0)=1) to f(x)=exp(e^{c}x). This however does not work as it does not reproduce all the derivatives of f(x) (actually only the first one). I am missing something there (probably st quite fundamental).

I believe that you should get df/du = e^{-c}f, no? When you changed variables, you didn't seem to account for the du/dx = e^(2c) term that should get produced on the left hand side, which results in an e^(-c) instead of e^(+c) on the RHS.

 

[donifan]

Sorry about the typo, that's what I meant. The solution for f is actually for the right equation (the one with the minus in the RHS).