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Is it true that $\sum_{k=1}^{m+n} \binom {m+n}k k^m (-1)^k=0 , \forall m,n\in \mathbb N$ ?

I feel that it is true because if we define $H_1 (x,r)=rx(1+x)^{r-1}$ , and

$H_{m+1}(x,r)=x \dfrac d {dx} {H_m (x,r)}$ , then $H_m (x,r)=\sum_{k=1}^{r} \binom {r}k k^m x^k$ and I have noticed , for first few $m$

that we can write $H_m (x,r)=\sum_{i=1}^m f(x,r,i)(1+x)^{r-i}$ , for some function $f$ so I thought that if I

let $r>m$ , then I should have $H_m (-1,r)=0$ i.e. $H_m (-1,m+n)=0$ ,(since $m+n>n$) but I am not totally sure .

Note:- $ 0\notin \mathbb N$

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  • $\begingroup$ For you, does $0\in \mathbb{N}$? $\endgroup$ – user119228 Feb 17 '14 at 23:31
  • $\begingroup$ @Julien: No , $0 \notin \mathbb N$ $\endgroup$ – Souvik Dey Feb 18 '14 at 11:58
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We must be assuming that $0\not\in\mathbb{N}$ since this is false for $m+n=1$.

Since $m\gt0$, this is the same as $$ \sum_{k=0}^{m+n}\binom{m+n}{k}k^m(-1)^k=0\tag{1} $$ which is the $m+n$ order forward difference of the function $k^m$. Each forward difference decreases the degree of a polynomial by one. Since the order is greater than the degree, the forward difference is $0$.

Note that we can write any polynomial as a linear combination of combinatorial polynomials of the same degree and lower. To be precise, $$ \newcommand{\stirtwo}[2]{\left\{{#1}\atop{#2}\right\}} k^m=\sum_{j=0}^m\binom{k}{j}\,\stirtwo{m}{j}j!\tag{2} $$ where $\stirtwo{m}{j}$ is a Stirling Number of the Second Kind. Plugging $(2)$ into $(1)$ gives $$ \begin{align} &\sum_{k=0}^{m+n}\sum_{j=0}^m\binom{m+n}{k}\binom{k}{j}\,\stirtwo{m}{j}j!(-1)^k\\ &=\sum_{j=0}^m\stirtwo{m}{j}j!\sum_{k=0}^{m+n}(-1)^k\binom{m+n}{k}\binom{k}{j}\\ &=\sum_{j=0}^m\stirtwo{m}{j}j!\sum_{k=0}^{m+n}(-1)^k\binom{m+n}{j}\binom{m+n-j}{k-j}\\ &=\sum_{j=0}^m\stirtwo{m}{j}j!\binom{m+n}{j}\sum_{k=0}^{m+n-j}(-1)^{k+j}\binom{m+n-j}{k}\\ &=\sum_{j=0}^m\stirtwo{m}{j}j!\binom{m+n}{j}(-1)^j(1-1)^{m+n-j}\\ &=0\tag{3} \end{align} $$ since $m+n-j\ge n\gt0$ in all terms of the sum.

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  • $\begingroup$ No , $ 0 \notin \mathbb N$ $\endgroup$ – Souvik Dey Feb 18 '14 at 11:59
  • $\begingroup$ @SouvikDey: yes, that is what I concluded (and I prefer). I just wanted to make that explicit since "There is no universal agreement about whether to include zero in the set of natural numbers". $\endgroup$ – robjohn Feb 18 '14 at 12:27
  • $\begingroup$ robjohn: True , and I like your method. $\endgroup$ – Souvik Dey Feb 18 '14 at 12:30
  • $\begingroup$ Perhaps this remark was for my comment..I had never heard about Stirling Number. Nice approach. (+1). $\endgroup$ – user119228 Feb 18 '14 at 13:57
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\sum_{k = 1}^{m + n}{m + n \choose k}k^{m}\pars{-1}^{k} = 0:\ {\Large ?} \,,\quad\forall\ m,n\ \in\ \mathbb N}$

$$ \pars{1 - z}^{m + n}=\sum_{k = 1}^{m + n}{m + n \choose k}z^{k}\pars{-1}^{k} $$

$$ \pars{z\,\partiald{}{z}}^{m}\pars{1 - z}^{m + n} =\sum_{k = 1}^{m + n}{m + n \choose k}k^{m}z^{k}\pars{-1}^{k} $$

$$ \!\!\!\!\!\color{#00f}{\large\sum_{k = 1}^{m + n}{m + n \choose k}k^{m}\pars{-1}^{k}} =\lim_{z \to 1^{-}}\bracks{\pars{z\,\partiald{}{z}}^{m}\pars{1 - z}^{m + n}} =\lim_{z \to 0^{-}}\partiald[m]{\bracks{\pars{1 - \expo{z}}^{m + n}}}{z} =\color{#00f}{\large 0} $$

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