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 $a=1$?
Best Answer
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.