Calculating Taylor Series for e^(x^2) around x=0

dantheman57
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Homework Statement


Find the Taylor series of e^(x^2) about x=0


Homework Equations



Taylor Series = f(a) +f'(a)(x-a) + (f''(a)(x-a)^2)/2 ...

The Attempt at a Solution



So, the first term is pretty obvious. It's e^0^2, which is zero.

The second term is what got me. (e^x^2)'=2x*(e^x^2), so at zero that is zero. Multiply by x, still zero. But the answer key says the second term is x^2. I really cannot understand this.

Thanks!
 
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dantheman57 said:
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So, the first term is pretty obvious. It's e^0^2, which is zero.

My bad. e^0^2 is one. Typo.
 
Keep going. It's the f'' term that is the second term.
 
Thank you so much!
 
You can also take the Taylor series of ex and then just fill in x2.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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