Find Taylor series generated by e^x centered at 0.

Lo.Lee.Ta.
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1.
a. Find Taylor series generated by ex2 centered at 0.

b. Express ∫ex2dx as a Taylor series.

2. For part a, I just put the value of "x2" in place of x in the general form for the e^x Taylor series:

ex: 1 + x + x2/2! + x3/3! + ...

ex2: 1 + x2 + x4/2! + x6/3! + ...


For part b, I just took the integral of the Taylor series for ex2:

= 0 + 2x + 1/2*4x3 + 1/6*6x5 + ...

= 2x + 2x3 + x5 + ...

Is this the right way to go about this?
Thanks! :)
 
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HI Lo.Lee.Ta.! :smile:

a. looks fine
Lo.Lee.Ta. said:
For part b, I just took the integral …

Looks like the derivative to me :redface:
 
#O_O AGH! I did take the derivative! Thanks! ha

So, it should be: x + 1/3(x3) + 1/10(x5) + 1/42(x7) + ...

Is this right?

Thanks! :)
 
Don't forget the constant of integration ... Otherwise, it looks fine.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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