How do you find the general expression for the k^{th} derivative of an exponetial function with a function in the exponent?

Question

I'm looking for a general expression for the function $\frac{\delta^k}{\delta \mu^k}[e^{n\mu + \mu^2}]_{\mu=0}$

I was thinking I could use the taylor expansion coefficients, but the function in the exponent is throwing me off.

Edit 1: Here's my newest attempt

Using the Taylor series coefficients we have $e^{n\mu + \mu^2} = \sum_{j=0}^{\infty} \frac{(n+\mu)^j}{j!}\mu^j$

So now we can evaluate our expression above $\frac{\delta^k}{\delta\mu^k}\left [ e^{n\mu + \mu^2} \right]_{\mu=0}=(n+\mu)^k|_{\mu=0}=n^k$

Answer 1

To understand what is going on, write out a few terms of each series: $$e^{n\mu+\mu^2} = \left( \frac{(n\mu)^0}{0!} + \frac{(n\mu)^1}{1!} + \frac{(n\mu)^2}{2!} + \cdots \right)\left( \frac{\mu^0}{0!} + \frac{\mu^2}{1!} + \frac{\mu^4}{2!} + \cdots \right).$$ If you were to expand and collect like exponents in $\mu$, then you could see how this would give you the coefficient of $\mu^j$, $j = 0, 1, 2, \ldots$. This suggests looking for the set of all nonnegative integers $k, m$ such that $j = k + 2m$. If $j$ is odd, then $k$ must be odd; if $j$ is even, then $k$ is also even. So this is what LutzL is referring to by "distinguish the even and odd case."

For example, if $j = 5$, then we could get this through the choices $$(k,m) = (1,2), (3,1), (5,0)$$ only. If $j = 8$, then the choices are $$(k,m) = (0,4), (2,3), (4,2), (6,1), (8,0).$$ So if we let $$e^{n\mu + \mu^2} = \sum_{j=0}^\infty a_j \frac{\mu^j}{j!} = \sum_{i=0}^\infty a_{2i} \frac{\mu^{2i}}{(2i)!} + a_{2i+1} \frac{\mu^{2i+1}}{(2i+1)!},$$ then $$a_{2i} \frac{\mu^{2i}}{(2i)!} = \sum_{t=0}^i \frac{(n \mu)^{2t}}{(2t)!} \frac{\mu^{2(i-t)}}{(i-t)!} = \mu^{2i} \sum_{t=0}^i \frac{n^{2t}}{(2t)! (i-t)!}.$$ Notice how the $\mu^{2i}$ term is constant with respect to the summation and therefore factors out, which confirms the correctness of the internal sum. So we have $$a_{2i} = (2i)! \sum_{t=0}^i \frac{n^{2t}}{(2t)!(i-t)!} = \frac{d^{2i}}{d\mu^{2i}}\left[ e^{n\mu + \mu^2} \right]_{\mu = 0}.$$ A similar approach applies to the odd derivatives, $j = 2i+1$.

Answer 2

$$ e^{nμ+μ^2}=e^{nμ}·e^{μ^2}=\sum_{m=0}^\infty \frac{(nμ)^m}{m!}·\sum_{j=0}^\infty\frac{μ^{2j}}{j!} $$ Now order by exponents. It may be useful to distinguish the even and odd case.

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The coefficient of degree $k$ is composed of all products of terms with $k=m+2j$. This starts with $j=0$ and $k=m$ and decreases $m$ in steps of 2. Or simply $m=k-2j$. $$ c_k=\sum_{j=0}^{\lfloor k/2\rfloor}\frac{n^{k-2j}}{(k-2j)!}·\frac1{j!} $$