Can't recognize the series. Can somebody take a look?

Question

I am solving a question as part of which I got the below mentioned series. I tried a lot but couldn't recognize this.

$$ ^m C_m m^n - {^m C_{m-1} (m - 1)^n} + \cdots \pm {^mC_1} 1^n $$

I am sure it's expansion of some famous series.

Can somebody help?

Answer 1

If you define $(Df)(x)=f(x+1)-f(x)$ then your expression is $$ (D^m f)(x)=\binom{m}{m} f(x+m)-\binom{m}{m-1}f(x+m-1)\pm…+(-1)^{m-1}\binom{m}1f(x+1)+(-1)^m\binom{m}{0}f(x+0) $$ where $f(x) = x^n$ and evaluated at $x=0$, in total it is an $m$-th derivative approximated by an $m$-th order difference quotient of step size 1.

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• If $f$ is a polynomial of degree $n$, then $D^mf$ is a polynomial of degree $n-m$ if that is non-negative, or else the zero polynomial.
• If $n
• If $n=m$ and $f(x)=x^n$, then $(D^mf)(x)=m!$.

Answer 2

We use the notation $\binom{m}{j}$ instead of $^mC_j$ and we also use the coefficient extraction operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write e.g. \begin{align*} n![z^n]e^{jz}=j^n \end{align*}

We obtain for $m\geq 1$ \begin{align*} \sum_{j=1}^m\binom{m}{j}(-1)^{m-j}j^n &=\sum_{j=1}^m\binom{m}{j}(-1)^{m-j}n![z^n]e^{jz}\\ &=n![z^n]\sum_{j=1}^m\binom{m}{j}\left(e^z\right)^j(-1)^{m-j}\\ &=n![z^n]\left((e^z-1)^m-1\right)\\ &=n![z^n]\left(\left(z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right)^m-1\right)\\ &=\begin{cases} -1&\qquad n=0\\ 0&\qquad 1\leq n < m\\ m!&\qquad n=m\\ n^m&\qquad n\geq m \end{cases} \end{align*}