Taylor series Mostly conceptual

trajan22
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I was just curious why when doing a taylor series like xe^(-x^3) we must first find the series of e^x then basically work it from there, why can't we instead do it directly by taking the derivatives of xe^(-x^3). But doing it that way doesn't give a working taylor series why is this so?
 
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You can do it either way. But the derivatives of xe^(-x^3) get complicated pretty fast. So you may just be doing it wrong. I probably would. It's just a question of choosing the easiest method.
 
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Oh ok I see. I was looking at the wrong answer, I think. Thanks though
 
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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