Taylor Series Substitution $e^{x^2-1}$

Question

If I'm using substitution to find a Taylor series about $x=1$ for $e^{x^2-1}$, but I'm given the Maclaurin series for $e^x$, how come the fact that the Taylor series is about $x=1$ doesn't matter when computing the Taylor series? If it were about $x=0$, it would be: $$1+(x^2-1)+(x^2-1)^4/2+(x^2-1)^6/6+....$$

and if it were about $x=1$ it would still be $$1+(x^2-1)+(x^2-1)^4/2+(x^2-1)^6/6+....$$

Shouldn't the Taylor series be dependent on the center ($a$), since each term contains the nth derivative at $a$? ($e^1$ is different from $e^0$, for example). Also, $(x-a)^k$, is also a term, which would be $x^k$ for $0$ and $(x-1)^k$ for $1$. Why are the taylor series the same for $a=0$ and $a=1$?

Answer 1

As I mentioned in the comment, a direct substitution does not give you a power series in $x$. The series you have in your question is not a Taylor series at all.

If you want to get a Taylor series of $e^{x^2-1}$ centered at $x=0$, this is not that bad: $e^{x^2 -1 } = \dfrac{e^{x^2}}{e}$, and you get:

$$e^{x^2 -1 } = \frac{1}{e} \sum_{n=0}^\infty \dfrac{(x^2)^n}{n!} = \frac{1}{e} \sum_{n=0}^\infty \dfrac{x^{2n}}{n!}$$

This in fact, is a power series in $x$, because it's written as the sum of powers of $x$. Therefore, it is a Taylor series centered at 0.

Note that in order to get the Taylor series, I manipulated the exponent so that there is a power of $x$ in the exponent.

For $x=1$, this is a bit more difficult, but the idea is the same. See if you can rewrite $x^2-1$ so that it is in terms of powers of $x-1$ only.

Answer 2

To get the series around $x=1$ start by rewriting

$\exp(x^2-1) = \exp[(x-1)^2] \exp[2(x-1)] = \left(\sum_{m=0}^\infty \frac{(x-1)^{2m}}{m!}\right)\left(\sum_{n=0}^\infty \frac{2^n(x-1)^n}{n!}\right)$

Now by multiplying out the two sums we get the desired Taylor series

$\exp(x^2-1) = \sum_{k=0}^\infty c_k (x-1)^k$

where

$c_{k} = \sum_{2m + n = k} \frac{1}{m!}\frac{2^n}{n!}$

Here the sum is over all pairs of integers $n,m\geq 0$ such that $2m + n = k$.

For the even terms this gives us

$c_{2k} = \sum_{m + n = k} \frac{1}{m!}\frac{4^n}{(2n)!} = \sum_{n = 0}^k \frac{1}{(k-n)!}\frac{4^{n}}{(2n)!}$

and for the odd terms we get

$c_{2k+1} = \sum_{m + n = k} \frac{2}{m!}\frac{4^n}{(2n+1)!} = \sum_{n = 0}^k \frac{2}{(k-n)!}\frac{4^n}{(2n+1)!}$