Coefficients of power series $\sin(x e^x)$

Question

I'm trying to find the first couple of coefficients of the power series expansion of $\sin (x e^x)$. The answers are given, but I'm not sure how to derive them.

What I've got so far:

$\sum_{n=0} ^{\infty} \frac{(-1)^n}{ (2n+1)!} * (x * \sum_{k=0} ^{\infty} \frac{x^k}{k!})^{2n+1}$

$\sum_{n=0} ^{\infty} \frac{(-1)^n}{ (2n+1)!} * (\sum_{k=0} ^{\infty}c_k ^{2n+1} x^{k+1})$

Now first question: is this right? Should $c_k$ be to the power of $2n+1$?

Then rewrite and reorder to

$\sum_{k=0} ^{\infty} (\sum_{n=0} ^{\infty} \frac{(-1)^n}{ (2n+1)!} * c_k ^{2n+1})x^{n+1}$

If this is right and the right way to calculate the coefficients, I have trouble calculating them (that $d_0$ is zero I can see, seeing how $c_0 ^n$ is zero for all positive integers. But $d_1$ (first coefficient) is already quite a bit trickier for me; is $c_1^3 = 3$, corresponding to the three ways to get 1, given three elements? The answer for the coefficient $d_1$ is (according to my book) 1, but $(-1)\ 3! * 3 $ definitely isn't one.

I'm not even sure if this is the right way to calculate the coefficients; any help is greatly appreciated.

Answer 1

Using the power series for $e^x$ \begin{eqnarray*} xe^x = x+x^2+ \frac{x^3}{2} + \cdots \end{eqnarray*} The first couple of terms in the power series for $\sin x$ are \begin{eqnarray*} \sin x = x- \frac{x^3}{6} + \cdots \end{eqnarray*} Now substitute ... \begin{eqnarray*} \sin ( x e^{x})= x+x^2+ \frac{x^3}{2} + \cdots- \frac{1}{6} \left(x+x^2+ \frac{x^3}{2} + \cdots \right)^3+ \cdots \end{eqnarray*} We already have the first two terms ... but we shall calculate the third ... to be on the safe side \begin{eqnarray*} \sin ( x e^{x})= x+x^2+ \left( \frac{1}{2}-\frac{1}{6}\right) x^3+ \cdots \end{eqnarray*} So the coefficients of the first two (non zero) terms in the expansion are $\color{red}{1}$ & $\color{red}{1}$ ... & $\color{green}{\frac{1}{3}}$ if you want the third.

Answer 2

Because the power series of the sine function and the power series of the exponential function have infinite radius of convergence we knows that the power series of $f(x):=\sin(xe^x)$ have infinite radius of convergence, so it coincides with it MacLaurin series, that is

$$f(x)=\sin(xe^x)=\sum_{k=0}^\infty a_kx^k=\sum_{k=0}^\infty\frac{f^{(k)}(0)}{k!}x^k$$

then $a_0=f(0)=0$, $a_1=f'(0)=1$, etc...

Answer 3

If functions in some function-composition $h(x)=g(f(x))$ have a Carlemanmatrix, and especially if that matrices are triangular, then the description of the coefficients of the resulting composition is remarkably simple for visual imagination, because function-composition resolves then to matrix-multiplication of the associated Carlemanmatrices.

Since this is the case here, see first the Carlemanmatrix for $f(x)=x \exp(x)$ (note all matrices are thought as of infinite size and only the leading 8x8 segment is shown here):

$$ \Large F= \small \begin{bmatrix} 1 & . & . & . & . & . & . & . \\ 0 & 1 & . & . & . & . & . & . \\ 0 & 1 & 1 & . & . & . & . & . \\ 0 & 1/2 & 2 & 1 & . & . & . & . \\ 0 & 1/6 & 2 & 3 & 1 & . & . & . \\ 0 & 1/24 & 4/3 & 9/2 & 4 & 1 & . & . \\ 0 & 1/120 & 2/3 & 9/2 & 8 & 5 & 1 & . \\ 0 & 1/720 & 4/15 & 27/8 & 32/3 & 25/2 & 6 & 1 \end{bmatrix} $$ The characteristic of Carleman matrices is that in the consecutive columns are the coefficients of the formal powerseries of $f(x)^0 (=1), f(x), f(x)^2, ...$.
Next the Carlemanmatrix of $g(x)=\sin(x)$ looks like $$ \Large G=\small \begin{bmatrix} 1 & . & . & . & . & . & . & . \\ 0 & 1 & . & . & . & . & . & . \\ 0 & 0 & 1 & . & . & . & . & . \\ 0 & -1/6 & 0 & 1 & . & . & . & . \\ 0 & 0 & -1/3 & 0 & 1 & . & . & . \\ 0 & 1/120 & 0 & -1/2 & 0 & 1 & . & . \\ 0 & 0 & 2/45 & 0 & -2/3 & 0 & 1 & . \\ 0 & -1/5040 & 0 & 13/120 & 0 & -5/6 & 0 & 1 \end{bmatrix} $$ With this $h(x)=g(f(x))$ is expressible by $ H = F \cdot G$ having the coefficients of the formal powerseries for $h(x)$ in its second column. I show this here as an explicite matrix-multiplication-scheme: $$\Tiny \begin{array} {r|rrrrrrrr|r|rrrrrrrr|} \hline & . & . & . & . & . & . & . & . & & 1 & . & . & . & . & . & . & . \\ & . & . & . & . & . & . & . & . & & . & 1 & . & . & . & . & . & . \\ & . & . & . & . & . & . & . & . & & . & . & 1 & . & . & . & . & . \\ & . & . & . & . & . & . & . & . & G= & . & -1/6 & . & 1 & . & . & . & . \\ & . & . & . & . & . & . & . & . & & . & . & -1/3 & . & 1 & . & . & . \\ & . & . & . & . & . & . & . & . & & . & 1/120 & . & -1/2 & . & 1 & . & . \\ & . & . & . & . & . & . & . & . & & . & . & 2/45 & . & -2/3 & . & 1 & . \\ & . & . & . & . & . & . & . & . & & . & -1/5040 & . & 13/120 & . & -5/6 & . & 1 \\ \hline & 1 & . & . & . & . & . & . & . & & 1 & . & . & . & . & . & . & . \\ & . & 1 & . & . & . & . & . & . & & . & 1 & . & . & . & . & . & . \\ & . & 1 & 1 & . & . & . & . & . & & . & 1 & 1 & . & . & . & . & . \\ F= & . & 1/2 & 2 & 1 & . & . & . & . & H= & . & 1/3 & 2 & 1 & . & . & . & . \\ & . & 1/6 & 2 & 3 & 1 & . & . & . & & . & -1/3 & 5/3 & 3 & 1 & . & . & . \\ & . & 1/24 & 4/3 & 9/2 & 4 & 1 & . & . & & . & -7/10 & . & 4 & 4 & 1 & . & . \\ & . & 1/120 & 2/3 & 9/2 & 8 & 5 & 1 & . & & . & -7/10 & -88/45 & 2 & 22/3 & 5 & 1 & . \\ & . & 1/720 & 4/15 & 27/8 & 32/3 & 25/2 & 6 & 1 & & . & -16/35 & -136/45 & -83/30 & 20/3 & 35/3 & 6 & 1 \end{array} $$ finding, that the powerseries of $h(x)$ as calculated in the second column of $H$ begins with $$\small h(x) = 1 x + 1 x^2 + \frac13 x^3 - \frac13 x^4 - \frac7{10}x^5 - \frac7{10} x^6 - \frac{16}{35} x^7 + O(x^8) $$

(Note: this is nothing else than that in the answer of @DonaldSplutterwit; only shown in a memorizable (and generalizable) scheme)