Shortcut to taylor series of f, given taylor series of g

[54stickers]

So, I have the series of g(x) = e^{(x-1)^{2}} = 1 + (x-1)^{2} + \frac{(x-1)^{4}}{2} + \frac{(x-1)^{6}}{6} + ... + \frac{(x-1)^{2n}}{n!}

and I am asked to find the series of f(x) = \frac{e^{(x-1)^{2}}-1}{(x-1)^{2}} for x \neq 1 and f(1) = 1. The Taylor series is centered about x = 1

I am told that there is an easy way to do this, but I don't see it.

any help would be appreciated, thanks

 

[Mark44]

So, I have the series of g(x) = e^{(x-1)^{2}} = 1 + (x-1)^{2} + \frac{(x-1)^{4}}{2} + \frac{(x-1)^{6}}{6} + ... + \frac{(x-1)^{2n}}{n!}

and I am asked to find the series of f(x) = \frac{e^{(x-1)^{2}}-1}{(x-1)^{2}} for x \neq 1 and f(1) = 1. The Taylor series is centered about x = 1

I am told that there is an easy way to do this, but I don't see it.

any help would be appreciated, thanks

1. What do you get for e^{(x - 1)2} - 1?
2. What do you get if you divide the result from #1 by (x - 1)²?

Is this a homework problem?

 

[54stickers]

I see it now, thanks

It's an old AP problem that I am doing to prepare for the test