How Do You Find the Maclaurin Series for e^(x^3)?

NickMusicMan
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I am studying for an exam, and I am trying to figure out:

if you have something like e^(x^3), can you simply substitute x^3 into the M-series for e^x and get the M-series for e^(x^3)? Or would you have to cube the whole e^x series? I have encountered mixed responses to this question.

This is from a practice problem but I am wondering, in general how can you handle ones like f(g(x)), where f has a known expansion.

Thanks,

-NN
 
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You can simply substitute x^3 in for x, but remember that the radius of convergence needs to be taken into account. I.e, if f(x) converges for |x| < 2, then f(x^3) would converge for |x^3| < 2.

Try it out though. You can compute the Taylor series from the definition, and then by substituting. They should be equal.
 
NickMusicMan said:
I am studying for an exam, and I am trying to figure out:

if you have something like e^(x^3), can you simply substitute x^3 into the M-series for e^x and get the M-series for e^(x^3)? Or would you have to cube the whole e^x series? I have encountered mixed responses to this question.
Cubing ex doesn't give you ex3. It gives you e3x.
 
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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