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Let $k_1=2$ and $k_2=3$ and $\mu_1=1$ and $\mu_2=3$. I would like to evaluate the following sum \begin{align} S=&\sum_{i=1}^{2} \sum_{j=1}^{k_i}\\ \times& \sum_{\begin{matrix} n_1+n_2=k_i-j\\ n_i=0 \end{matrix}}^{}\prod_{\begin{matrix} l=1\\ l\neq i \end{matrix}}^{2} \binom{k_l+n_l-1}{n_l}\frac {u_l^{k_l}} {(\mu_l-\mu_i)^{k_l+n_l}} \end{align} My question is how we get $n_1$ and $n_2$ and evaluate the above sum? What is the meaning of this summation and it name where only $k_j$ and $\mu_j$ are known? $$ \sum_{ \begin{matrix} n_1+n_2+\cdots n_n=k_i-j\\ n_i=0 \end{matrix}}^{} $$ Does any one have work with this sum or have some examples about it? This is independent part, I found the same sum in Leibniz formula which is the derivation of product of $m$ functions given by $$ (v_1v_2v_3\cdots v_m)^n=\sum_{ \begin{matrix} t_1+t_2+\cdots t_m=n\\ t_1,t_2,\cdots t_m\geq0 \end{matrix}}^{}\frac {n!} {t_1!t_2!\cdots t_m!}v_1^{(t_1)}v_2^{(t_2)}\cdots v_m^{(t_m)} $$ Just how we can found all $t_i$ or $n_i$ and the sum?

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We have the pleasant situation that the index range of $i$ is small, namely $i\in\{1,2\}$. So, expanding the outer sum gives two terms only. The product with index $l$ is in fact a single factor, since $l$ is also either $1$ or $2$, but complementary to $i$.

We obtain \begin{align*} \color{blue}{\sum_{i=1}^2}&\color{blue}{\sum_{j=1}^{k_i}\sum_{{n_1+n_2=k_i-j}\atop{n_i=0}}\prod_{{l=1}\atop{l\ne i}}^2 \binom{k_l+n_l-1}{n_l}\frac{u_l^{k_l}}{(\mu_l-\mu_i)^{k_l+n_l}}}\\ &=\sum_{j=1}^{k_1}\sum_{{n_1+n_2=k_1-j}\atop{n_1= 0}}\prod_{{l=1}\atop{l\ne 1}}^2 \binom{k_l+n_l-1}{n_l}\frac{u_l^{k_l}}{(\mu_l-\mu_1)^{k_l+n_l}}\tag{1}\\ &\qquad+\sum_{j=1}^{k_2}\sum_{{n_1+n_2=k_2-j}\atop{n_2= 0}}\prod_{{l=1}\atop{l\ne 2}}^2 \binom{k_l+n_l-1}{n_l}\frac{u_l^{k_l}}{(\mu_l-\mu_2)^{k_l+n_l}}\\ &=\sum_{j=1}^{k_1}\sum_{n_2=k_1-j} \binom{k_2+n_2-1}{n_2}\frac{u_2^{k_2}}{(\mu_2-\mu_1)^{k_2+n_2}}\tag{2}\\ &\qquad+\sum_{j=1}^{k_2}\sum_{n_1=k_2-j} \binom{k_1+n_1-1}{n_1}\frac{u_1^{k_1}}{(\mu_1-\mu_2)^{k_1+n_1}}\\ &=\sum_{j=1}^{k_1}\binom{k_1+k_2-j-1}{k_1-j}\frac{u_2^{k_2}}{(\mu_2-\mu_1)^{k_1+k_2-j}}\tag{3}\\ &\qquad+\sum_{j=1}^{k_2}\binom{k_1+k_2-j-1}{k_2-j}\frac{u_1^{k_1}}{(\mu_1-\mu_2)^{k_1+k_2-j}}\\ &=\sum_{j=1}^{2}\binom{4-j}{2-j}\frac{u_2^3}{2^{5-j}}+\sum_{j=1}^{3}\binom{4-j}{3-j}\frac{u_1^2}{(-2)^{5-j}}\tag{4}\\ &=\left(\binom{3}{1}\frac{1}{2^4}+\binom{2}{0}\frac{1}{2^3}\right)u_2^3\\ &\qquad+\left(\binom{3}{2}\frac{1}{(-2)^4}+\binom{2}{1}\frac{1}{(-2)^3}+\binom{1}{0}\frac{1}{(-2)^2}\right)u_1^2\tag{5}\\ &\,\,\color{blue}{=\frac{1}{16}\left(u_1^2+5u_2^3\right)} \end{align*}

Comment:

  • In (1) we expand the outer sum which gives two terms with $i=1$ and $i=2$.

  • In (2) we substitute $l=2$ in the first term and $l=1$ in the second term. We also substitute $n_1=0$ in the first term and $n_2=0$ in the second term.

  • In (3) we substitute $n_2$ in the first term and $n_1$ in the second term.

  • In (4) we set the values with $k_1,k_2,\mu_1$ and $\mu_2$.

  • In (5) we expand the sums.

Note: The sum with index range \begin{align*} \sum_{{t_1+t_2+\cdots+t_m=n}\atop{t_1,t_2,\ldots,t_m\geq 0}} \end{align*} has different character than the sum above, since here we sum over all $m$-tupel $(t_1,t_2,\ldots,t_m)$ with non-negative integral components which sum up to $n$. For small $n$ and $m$ all valid $m$-tupel can be determined manually. Greater values of $m$ or $n$ will usually need assistence by a computer.

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  • $\begingroup$ Ok prof, thank you so much. $\endgroup$ – Monir Sep 18 '18 at 23:05
  • $\begingroup$ @Monir: You're welcome. $\endgroup$ – Markus Scheuer Sep 19 '18 at 6:16

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