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Is there an easier way to do a Taylor expansion of $e^{u^2+u}$ than do derivatives or substitute and then use Newton's binomial?

For example, expanding until the $4$th term:

$$e^{u^2+u}=1+u^2+u+ \frac{(u^2+u)^2}{2!} + \frac{(u^2+u)^3}{3!}+\frac{(u^2+u)^4}{4!}$$ $$e^{u^2+u}=1+u+\frac{3}{2}u^2+\frac{7}{6}u^3+\frac{25}{24}u^4+H.O.T.$$

I thought of using the $n$th derivative formula on $e^{u^2}\cdot e^u$ $$(f(x)\cdot g(x))^{(n)}=\sum_{n=0}^n{{n\choose k}f^{k}g^{n-k}}.$$

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2 Answers 2

Perhaps an easier way is this:

Note first that $e^{u+u^2}=e^ue^{u^2}$. Now, we know power series for both of these functions: $$ e^u=\sum_{n=0}^{\infty}\frac{u^n}{n!},\qquad\text{and}\qquad e^{u^2}=\sum_{n=0}^{\infty}\frac{u^{2n}}{n!}. $$ From here, we use convolution: in the power series for $e^ue^{u^2}$, the coefficient by $u^k$ is precisely $$ \sum_{n=0}^{k}\left([u^n]e^{u^2}\right)\left([u^{k-n}]e^{u}\right), $$ where $[u^n]f(u)$ denotes the coefficient by $u^n$ in the power series expansion for $f$.

Now, $[u^n]e^{u^2}=0$ for all odd $n$; for even $n=2m$, the coefficient is $$ [u^n]e^{u^2}=[u^{2m}]e^{u^2}=\frac{1}{m!}. $$ On the other hand, the coefficient by $u^{k-2m}$ in $e^u$ is $$ [u^{k-2m}]e^u=\frac{1}{(k-2m)!}. $$ So, we have $$ [u^k]e^{u+u^2}=\sum_{n=0}^{k}\left([u^n]e^{u^2}\right)\left([u^{k-n}]e^u\right)=\sum_{m=0}^{\lfloor k/2\rfloor}\left([u^{2m}]e^{u^2}\right)\left([u^{k-2m}]e^u\right), $$ so that $$ [u^k]e^{u+u^2}=\sum_{m=0}^{\lfloor k/2\rfloor}\frac{1}{m!}\cdot\frac{1}{(k-2m)!}. $$ From here, you can of course try to do some simplification by breaking in to the cases where $k$ is even or odd, but I'll leave that to you.

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You could also multiply $$ e^u \cdot e^{u^2} = \sum_{n,k=0,1,\dots} \frac{u^{k} u^{2n}}{k! n!} = \sum_{n=0,2,\dots} u^n \sum_{k=0}^{n/2} \frac{1}{(2n-k)!k!} + \sum_{n=1,3,\dots} u^n \sum_{k=0}^{(n-1)/2} \frac{1}{(2n-k)!k!}.$$

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