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

In summary: This is actually what I get by using a series solution for f.In summary, the conversation discusses the evaluation of the series \sum{\frac{x^n}{n!}e^{cn^2}} and its equivalence to finding a modified exponential function f(x) such that \frac{d^{n}f(0)}{dx^{n}} = \frac{e^{cn^{2}}}{n!}. The conversation also mentions the possibility of expressing f(x) as e^{g(x)} for some function g(x) and provides a possible solution for f'(x). The topic of numerical stability and a closed form solution for f(x) is also brought up.
  • #1
donifan
12
0
Hello, I am trying to evaluate the series

[tex]\sum{\frac{x^n}{n!}e^{cn^2}}[/tex]

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


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

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?
 
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  • #2
I can't answer your question. However, warning! c > 0 may be a problem!
 
  • #3
No biggie, x<0. To avoid confusion, let's write it better like

[tex]
\sum{\frac{(-1)^nt^n}{n!}e^{cn^2}}
[/tex]

I also know from the physical process that:

[tex]\displaystyle\lim_{t\to\infty} f(t)=0[/tex]

Thanks!
 
  • #4
I'm not sure whether you can express this particular function as some sort of exponential such as [tex]e^{g(x)}[/tex] for some g(x) or not but if you take the derivative it doesn't seem likely that you can. I came up with [tex]f'(x) = e^cf(e^{2c}x)[/tex].
 
  • #5
By expanding the exponential as a second series, assuming we can swap the order of summations, I get

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

The derivative should work out to a polynomial times an exponential. Not sure if that polynomial can be worked out explicitly.
 
  • #6
Thanks guys!

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

[tex]
\frac{d^{n}f(0)}{dx^{n}} = e^{cn^{2}}
[/tex]


So far I was able to check that [tex] f^{'}(x) = e^cf(e^{2c}x) [/tex] (Very nice work Wizlem). I even tried making the substitution [tex] u=e^{2c}x [/tex] so that

[tex] \frac{df}{du}=e^{c}f [/tex]

which can be easily solved (with [tex] f(0)=1 [/tex]) to [tex] f(x)=exp(e^{c}x) [/tex]. 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.
 
  • #7
donifan said:
[tex] \frac{df}{du}=e^{c}f [/tex]

which can be easily solved (with [tex] f(0)=1 [/tex]) to [tex] f(x)=exp(e^{c}x) [/tex]. 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 [itex]df/du = e^{-c}f[/itex], 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.
 
  • #8
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).
 

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

1. What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms, where each term is a derivative of the function evaluated at a specific point.

2. How is a Taylor series related to an exponential function?

A Taylor series can be used to approximate any function, including exponential functions. If a Taylor series is expanded around the point x=0, and the function is an exponential function, the resulting series will look very similar to the original exponential function.

3. What is the significance of the coefficient in a Taylor series?

The coefficient in a Taylor series is multiplied by the corresponding power of x in the series. This coefficient represents the value of the derivative of the function at the point around which the series is expanded. In the case of an exponential function, the coefficient will be the value of the function's derivative at x=0.

4. How can a Taylor series be used to approximate an exponential function?

By using more and more terms in the Taylor series, the approximation becomes more accurate. For an exponential function, using more terms will result in the series more closely resembling the original function. However, it should be noted that the series will only converge to the original function within a certain interval or range of x values.

5. Are there any limitations to using a Taylor series to approximate an exponential function?

Yes, there are limitations. As mentioned before, the series will only converge to the original function within a certain interval or range of x values. Also, the series will only be an accurate approximation for a finite number of terms, and may diverge at certain values of x. Additionally, the series may not be an accurate representation of the original function if the function has singularities or discontinuities.

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