# Power Series Expansions

1. Jun 6, 2014

### Zondrina

I had a quick question about an expansion. Wolfram and maple have not been very useful in verifying the series.

Could I do these:

$f(x) = e^{-3x^2} = e^{-3}e^{-3(x^2-1)} = \sum_{n=0}^{∞} \frac{e^{-3}(-3)^n}{n!} (x^2-1)^n$

$f(x) = e^{-3x^2} = e^{-12}e^{-6(\frac{x^2}{2}-2)} = \sum_{n=0}^{∞} \frac{e^{-12}(-6)^n}{n!} (\frac{x^2}{2}-2)^n$

$f(x) = e^{-3x^2} = e^{-27}e^{-9(\frac{x^2}{3}-3)} = \sum_{n=0}^{∞} \frac{e^{-27}(-9)^n}{n!} (\frac{x^2}{3}-3)^n$

My overall hunch from this pattern centered about a=k:

$f(x) = e^{-3x^2} = e^{-3k^2}e^{-3k(\frac{x^2}{k}-k)} = \sum_{n=0}^{∞} \frac{e^{-3k^2}(-3k)^n}{n!} (\frac{x^2}{k}-k)^n$

and for any $\alpha$ centered about a=k:

$f(x) = e^{-\alpha x^2} = e^{-\alpha k^2}e^{-\alpha k(\frac{x^2}{k}-k)} = \sum_{n=0}^{∞} \frac{e^{-\alpha k^2}(-\alpha k)^n}{n!} (\frac{x^2}{k}-k)^n$

EDIT: Also one more, if I replace $x^2$ by $x^n$ in the above:

$f(x) = e^{-\alpha x^n} = e^{-\alpha k^n}e^{-\alpha k(\frac{x^n}{k}-k)} = \sum_{n=0}^{∞} \frac{e^{-\alpha k^n}(-\alpha k)^n}{n!} (\frac{x^n}{k}-k)^n$

2. Jun 6, 2014

### jbunniii

I'm not sure exactly what you're trying to do, so my comment may be off base. But the series
$$\sum_{n=0}^{∞} \frac{e^{-3}(-3)^n}{n!} (x^2-1)^n$$
is not centered at $x=1$, at least not in the usual sense of that phrase, because the polynomials $(x^2 - 1)^n$ are not centered at $x=1$. They are centered at (symmetric around) $x=0$.

A series which is centered about $x=1$ would be of the form
$$\sum_{n=0}^\infty a_n (x-1)^n$$

3. Jun 6, 2014

### Zondrina

I was trying to find a general expression without having to find $f^{(n)}(a)$. It seemed to work for (center a=1):

$f(x) = e^{-3x} = e^{-3} e^{-3(x-1)} = \sum_{n=0}^{∞} \frac{e^{-3}(-3)^n}{n!} (x-1)^n$

I see that won't work now though for higher powers of $x$ as I would produce series that are not centered around the desired point.

Is there a way I could adapt this little trick to higher powers?