# Trying to solve a Taylor series problem

I have a Taylor series problem, well more precisely a Maclaurin series.

I am trying to find convergence of: $f(x) = e^{x^3} + e^{{2x}^3}$

Okay here goes:

$$f'(x) = 3xe^{x^3} + 6x e^{{2x}^3}$$ $$f''(x) = 9x^2e^{x^3} + 3e^{x^3} + 36x^2e^{{2x}^3} + 6e^{{2x}^3}=e^{x^3}(9x^2+3) + e^{{2x}^3}(36x^2+6)$$ $$f'''(x) = (9x^2+3)(3xe^{x^3}) + 18xe^{x^3} + (36x^2 + 6)(6xe^{2x^3}) + 72xe^{2x^3}$$ $$=27xe^{x^3}(x^2 + 1) + 108xe^{2{x^3}}(2x^2 + 1)$$

Now I don't see a pattern, but I can note at $a=0$ we have:

$$f(0)=2$$ $$f'(0)=0$$ $$f''(0)=9$$ $$f'''(0)=0$$

I am not sure where this is going, or if this is the right way to attack the problem. Any advice?

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You mis-differentiated $x^3$, that yields $3x^2$, not $3x$. The right way to attack the problem is to take the expansion of $e^z$, substitute in $z = x^3$ resp $z = 2x^3$, and add. – Daniel Fischer Jun 5 '14 at 12:34
What is the problem - finding the expansion at $x=0$? – mvw Jun 5 '14 at 12:36
@DanielFischer Oh...my... I can't believe that happened... But yes that second part of the comment makes sense. – Tony Jun 5 '14 at 12:36

To supplement the other answer, here's how you would determine convergence using the ratio test:

We would like to test the convergence of the sum $$\sum_{n=1}^\infty a_{3n} x^{3n} = \sum_{n=1}^\infty \frac{1 + 2^n}{n!}x^{3n}$$ In order to do so, we need to determine the value of $$\limsup_{n \to \infty} \left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right| = \lim_{n \to \infty} \frac{(1 + 2^{n+1})x^{3(n+1)}}{(n+1)!} \cdot \frac{n!}{(1 + 2^n)x^{3n}}$$ Simpifying, this becomes $$\lim_{n \to \infty} \frac{x^3}{n+1} \cdot \frac{1+2^{n+1}}{1+2^n}$$ Noting that this limit becomes zero for all $x$ and noting that $0<1$, we conclude that this sum converges for all $x$ by the ratio test.

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Use the familiar Maclaurin expansion for the exponential:

$$e^t=1+t+\frac{t^2}{2!}+\dots=\sum_{n=0}^\infty \frac{t^n}{n!}.$$ If you plug in $t=x^3$ you get $$e^{x^3}=\sum_{n=0}^\infty \frac{x^{3n}}{n!}.$$ If you plug in $t=2x^3$ you similarly get $$e^{2x^3}=\sum_{n=0}^\infty 2^n \frac{x^{3n}}{n!}$$ Try adding these two expansions.

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The sum of two series, is the series of their sum, is this correct? – Tony Jun 5 '14 at 12:38
@user151558 Correct! you should find that the coefficient of $x^{3n}$ is $\frac{1+2^n}{n!}$. – user1337 Jun 5 '14 at 12:38
How do I then find where it converges to $f(x)$? – Tony Jun 5 '14 at 12:48
Use the ratio test? This should have infinite radius of convergence, since the function is entire (in the sense of complex analysis) – Steven Gubkin Jun 5 '14 at 13:09
@StevenGubkin So it is exact at $a=0$, so at $n=1$? – Tony Jun 5 '14 at 13:10