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Given recurrence:

${d_{n+1}=n(d_n+d_{n-1}) \\ d_0=1, \ d_1=0}$

I was able to calculate the EGF:

$${\sum_{n=0}^{\infty} d_n\frac{x^n}{n!} = \frac{e^{-x}}{|1-x|}}$$

Where the absolute value can be (?) dropped.

Knowing this, normally I would use the Taylor's theorem about 0. So I need to find the n-th derivative of the right part. That's where the problem begins, as these derivatives are not trivial.

How can I find the n-th derivative of the right part? Is there a better way to solve the problem using an EGF?

Umbra
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  • Hint: Look at the first few derivatives of $f(x)g(x)$ – Empy2 Dec 13 '23 at 12:25
  • @Empy2 In derivatives, how do the polynomials next to the e^-x relate to each other? The denominator is trivial. Still, I do not get it. – Umbra Dec 13 '23 at 13:18
  • For example, the third derivative is $\frac{d^3f}{dx^3}g+3\frac{d^2f}{dx^2}\frac{dg}{dx}+3\frac{df}{dx}\frac{d^2g}{dx^2}+f\frac{d^3g}{dx^3}$. So find the derivatives of the two parts separately, then combine them using Pascal's Triangle. – Empy2 Dec 13 '23 at 13:26
  • Don't combine the terms until after you have set $x=0$ – Empy2 Dec 13 '23 at 13:28

1 Answers1

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The way of calculating this EGF can be found here.

${(\frac{1}{1-x})}^{(n)} = \frac{n!}{(1-x)^{n+1}}$

${(e^{-x})}^{(n)} = (-1)^{n}e^{-x}$

Both easy seen, can be proven by induction.

By the general Leibniz rule:

$$ \begin{align} (fg)^{(n)} &= \sum_{k=0}^{n} \binom{n}{k} \frac{(n-k)!}{(1-x)^{n+1-k}} (-1)^{k}e^{-x} \\ &=\frac{e^{-x}n!}{(1-x)^{n+1}} \sum_{k=0}^{n} \frac{(x-1)^{k}}{k!} \end{align} $$

By providing $0$.

$$ (fg)^{(n)}(0) = n!\sum_{k=0}^{n} \frac{(-1)^{k}}{k!} \stackrel{by \space def.}{=} {!n} $$

From the Taylor's theorem:

$$ d_{n} = {!n} $$

Umbra
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