Maclaurin series degree 6 involving substitution

Question

So I am aware that you can perform operations on taylor series, such as integration, differentiation, etc. However, I am not sure of when exactly one is allowed to substitute values into another taylor series. For example:

Find the degree 6 taylor polynomial of:

$f(x) = e^{\sin(x^2)} $ about $x=0$

So what exactly would be the process? Clearly they don't expect you to differentiate $f(x)$ 6 times under exam conditions, so I thought maybe you could find the degree 6 taylor polynomial of $e^x$:

$P_{6,0}(e^x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 + \frac{1}{120}x^5 + \frac{1}{720}x^6$

And then making a $\sin(x^2)$ substitution for x giving:

$P_{6,0}(x) = 1 + x + \frac{1}{2}\sin(x^2)^2 + \frac{1}{6}\sin(x^2)^3 + \frac{1}{24}\sin(x^2)^4 + \frac{1}{120}\sin(x^2)^5 + \frac{1}{720}\sin(x^2)^6$

But I don't suppose this is right, and I don't know why. Even so, I know this can't always be done. Any tips or points are appreciated. Thanks in advance.

Answer 1

Since$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\cdots\text{ and }\sin(x^2)=x^2-\frac{x^6}{3!}+\cdots,$$you have that $P_{6,0}(x)$ is$$1+\left(x^2-\frac{x^6}{3!}\right)+\frac{\left(x^2-\frac{x^6}{3!}\right)^2}{2!}+\frac{\left(x^2-\frac{x^6}{3!}\right)^3}{3!}+\frac{\left(x^2-\frac{x^6}{3!}\right)^4}{4!}+\frac{\left(x^2-\frac{x^6}{3!}\right)^5}{5!}+\frac{\left(x^2-\frac{x^6}{3!}\right)^6}{6!}$$minus those monomials whose degree is greater than $6$. Note, furthermore, that every monomial obtained when you expand$$\frac{\left(x^2-\frac{x^6}{3!}\right)^4}{4!}+\frac{\left(x^2-\frac{x^6}{3!}\right)^5}{5!}+\frac{\left(x^2-\frac{x^6}{3!}\right)^6}{6!}$$has degree greater than $6$, and therefore they don't matter.

Answer 2

You can expand $\sin(x^2)$ and then use the exponential property $\exp(a+b)=\exp(a)\exp(b)$ to see that we have

\begin{align}e^{\sin(x^2)}&=\exp\left(\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}x^{4k+2}\right)\\&=\prod_{k=0}^\infty\exp\left(\frac{(-1)^k}{(2k+1)!}x^{4k+2}\right)\\&=\prod_{k=0}^\infty\sum_{n=0}^\infty\frac1{n!}\left(\frac{(-1)^k}{(2k+1)!}x^{4k+2}\right)^n\end{align}

from which you can collect coefficients. Written out to the $x^6$ term,

\begin{align}e^{\sin(x^2)}&=\prod_{k=0}^\infty\sum_{n=0}^\infty\frac1{n!}\left(\frac{(-1)^k}{(2k+1)!}x^{4k+2}\right)^n\\&=\left[\sum_{n=0}^\infty\frac1{n!}x^{2n}\right]\left[\sum_{n=0}^\infty\frac{(-1)^n}{n!6^n}x^{6n}\right]\left[1+\mathcal O(x^{10})\right]\\&=\left[1+x^2+\frac12x^4+\frac16x^6+\mathcal O(x^8)\right]\left[1-\frac16x^6+\mathcal O(x^{12})\right]\left[1+\mathcal O(x^{10})\right]\end{align}

$$e^{\sin(x^2)}=1+x^2+\frac12x^4+\mathcal O(x^8)$$