- Admin
- #26

- Jan 26, 2012

- 4,055

This might be wrong so I am hoping someone can check it and comment if necessary. The end result is correct but there is one spot where I am unsure, so read with scrutiny and skepticism.

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We are given \(\displaystyle \frac{s^n}{n!}\frac{d^n}{dq^n}(e^{-q^2})\) and want to try to match it to this form: \(\displaystyle \sum_{n=0} ^ {\infty} \frac {f^{(n)}(a)}{n!} (x-a)^n\)

There is one term that they share which represents the nth derivative at $a$. These must match up so $f(a)=e^{-q^2}$. There are two things we need to know however, $f(x)$ and $a$. There are perhaps various combinations to try but the one that comes to mind first is $f(x)=e^{-x^2}$ and $a=q$.

Looking at $s^n$ it seems that this should match up with $(x-a)^n$ so we find that $x-a=s$ and since $a=q$ that is equivalent to $x-q=s$. Finally that gives us that $x=s+q$. If we plug that into $f(x)=e^{-x^2}$ then we have our result.