Solving the Taylor Series for e^(-x^2): Is it the Same at x=0?

[sparkle123]

Do you just replace the x's with (x-3)'s? Since e^(-x^2) is defined as the taylor series though, it seems like the answer should be the same as the series about x=0.
Thanks!


P.S. does anyone know how to resize images? :$

 

[vela]

It's probably easiest just to crank out the first 3 derivatives and use the Taylor series formula.

 

[sparkle123]

Thanks, but how would you do that?

 

[jhae2.718]

The Taylor series of a polynomial about x=a is defined by:
<br /> \sum_{n=0}^\infty \frac{f^{(n)}(a)(x-a)^n}{n!}<br />

I would start by finding the Taylor series for e^x about x=3, which will not be the same as the Maclaurin series.

 

[vela]

Thanks, but how would you do that?

Where exactly are you getting stuck?

 

[sparkle123]

Is this right?
Taylor series for e^x about x=3:
e^x+e^x(x-3)+e^x(x-3)^2 / 2 + ... e^x(x-3)^n / n!
=e^3+e^3(x-3)+e^3(x-3)^2 / 2 + ... e^3(x-3)^n / n!

Taylor series for e^(-x^2) about x=3:
e^(-x^2) + e^(-x^2)(x-3) + e^(-x^2)(x-3)^2 /2 + ... e^(-x^2)(x-3)^n / n!
=(e^-9) + (e^-9)(x-3) + (e^-9)(x-3)^2 / 2 + ... + (e^-9)(x-3)^n / n!

Taylor series for int(e^(-x^2)) about x=3:
I don't know how to integrate:
e^(-x^2) + e^(-x^2)(x-3) + e^(-x^2)(x-3)^2 /2 + ... e^(-x^2)(x-3)^n / n!

 

[vela]

The first series is correct, but the other two aren't.

What's your background in calculus? If you don't know how to integrate and differentiate, you won't be able to do this problem.

 

[sparkle123]

D: I'm taking ap calc in 12 days
could you please show me how to do it?

 

[vela]

Actually, I misspoke. You really only need to know how to differentiate.

It's against forum policy to give out solutions, but I can tell you you need to know: how to differentiate an exponential, the chain rule, the product rule, and the fundamental theorem of calculus. That'll get you the derivatives. Then you just have to use the formula for the Taylor series, which jhae2.718 has already provided you.

 

[eczeno]

Do you just replace the x's with (x-3)'s? Since e^(-x^2) is defined as the taylor series though, it seems like the answer should be the same as the series about x=0.
Thanks!


P.S. does anyone know how to resize images? :$

hi,

your first idea is incorrect, you cannot just substitute (x-3) into the series. luckily, your second idea is correct!

no idea about images.

cheers