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June 27th, 2018, 04:37 PM  #1 
Newbie Joined: Jun 2018 From: Belgrade, Serbia Posts: 4 Thanks: 0  Please help with this interesting sum proof
Hi, I am trying to prove the following identity. I obviously can't see something here. $\displaystyle \sum_{1}^{n}\frac{\prod_{r=1}^{n1}(y_kz_r)}{\prod_{1<=r<=n, r\ne k}(y_ky_r)}=1 $ Where $\displaystyle y_1,y_2,...y_n$ are different numbers and $\displaystyle z_1,z_2,...z_{n1}$ is set of any numbers. I obviously can't see something. Any hints? Last edited by skipjack; June 28th, 2018 at 02:43 PM. 
June 27th, 2018, 04:48 PM  #2  
Senior Member Joined: Sep 2015 From: USA Posts: 2,266 Thanks: 1198  Quote:
Last edited by skipjack; June 28th, 2018 at 02:44 PM.  
June 27th, 2018, 09:19 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,114 Thanks: 1909 
$\displaystyle {\Large\sum_{k=1}^{n}}\frac{\displaystyle {\small \prod_{\large r=1}^{\large n1}}(y_kz_r)}{\displaystyle {\small \prod_{\large1\leqslant r\leqslant n, r\ne k}}(y_ky_r)}=1 $

June 28th, 2018, 02:36 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,665 Thanks: 651 
It has the look of Lagrange interpolation.

July 1st, 2018, 04:15 PM  #5 
Newbie Joined: Jun 2018 From: Belgrade, Serbia Posts: 4 Thanks: 0  I finally did it
Hi, I think I did it. Actually, least common multiple of all sum members is $\displaystyle \prod_{i=1}^{n1}\prod_{k=i+1}^{n}(y_iy_k)$. In numerator, we have the sum with member with ordinal number i is $\displaystyle (1)^{i+1} \prod_{k=1}^{n1}(y_iz_k)\prod_{1\le k\le n1,k \ne i}\prod_{k+1 \le r \le n, r \ne i}(y_ky_r)$ For shorter notation, let's have $\displaystyle \pi(i) = \prod_{1\le k\le n1,k \ne i}\prod_{k+1 \le r \le n, r \ne i}(y_ky_r)$ When we develop the the product, we get $\displaystyle (y_i^{n1} + \sum_{k=1}^{n1}((1)^{k}y_i^{k1}\sum_{r=1}^{\binom{n1}{k}}C(k,r)))\cdot \pi(i)$ where C(k,r) is rth combination of k elements on array $\displaystyle (z_1,z_2,...,z_{n1})$ When we get back to the sum in the numerator, we get $\displaystyle \sum_{i=1}^{n}y_i^{n1}\cdot \pi(i) + \sum_{k=1}^{n1}(1)^{k}\sum_{r=1}^{\binom{n1}{k}}C(k,r)\cdot \sum_{i=1}^{n}y_i^{k1}\cdot \pi(i)$ As we now got multipliers of zarray (their combinations) we have to prove that every $\displaystyle \sum_{i=1}^{n}y_i^{r}\cdot \pi(i)= \sum_{i=1}^{n}y_i^{r}\prod_{1\le k\le n1,k \ne i}\prod_{k+1 \le r \le n, r \ne i}(y_ky_r)=0$ for r < n1. Approach used here uses the fact that all elements have the same polynomial degree and greatest degree of particular variable is n2. Thus for each element (possible repeatable combination of yarray) we will have its negative pair as 1 is equally combined as all yarray elements. Therefore result has to be zero. When we look $\displaystyle \sum_{i=1}^{n}y_i^{n1}\cdot \pi(i)$ We see that it can't be zero as $\displaystyle y_i^{n1}$ can't be found in other sum elements where as maximal degree of particular element there is n2. However we can look for "similar" one that has $\displaystyle y_l^{n1}y_y{n2}$ and the same products in continuation. We then $\displaystyle y_l^{n2}y_i^{n2}(y_iy_l)(rest \; of \; product)$ We now halved number of sum elements. This can be repeated recursively n1 times and then we get one product that is $\displaystyle \prod_{i=1}^{n1}\prod_{k=i+1}^{n}(y_iy_k)$ which gives what is wanted. Last edited by skipjack; July 2nd, 2018 at 07:41 PM. 
July 2nd, 2018, 12:02 PM  #6 
Newbie Joined: Jun 2018 From: Belgrade, Serbia Posts: 4 Thanks: 0  I finally did it
For some reason my yesterday reply is not published. I didn't actually finished the proof as I illustrated idea behind the proof. Waiting for moderator to publish it so I can finalize it.

July 2nd, 2018, 02:43 PM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond  
July 3rd, 2018, 11:30 AM  #8 
Newbie Joined: Jun 2018 From: Belgrade, Serbia Posts: 4 Thanks: 0  And now to get to the point
Please first note that I made mistake by not including $\displaystyle (1)^{k}$ and $\displaystyle (1)^{i+1}$ in sums with $\displaystyle y_i$ elements which you will see below. In divisor from the beginning we have $\displaystyle \prod_{i=1}^{n1}\prod_{k=i+1}^{n}(y_iy_k) = (1)^{n1}det(V_{n1})$ Where V is Vandermonde matrix of order of n1 (n x n dimensions). $\displaystyle \pi(i) = (1)^{n2}det(V_{n2}(i))$ where V(i) is Vandermonde matrix of order of n2 where $\displaystyle y_i$ is missing. Then the sum $\displaystyle \sum_{i=1}^{n}(1)^{i}y_{i}^{k1}\cdot \pi(i)$ is determinant of matrix which has $\displaystyle (1)^{n2}V(i)$ as cofactor and vector $\displaystyle (y_1^{k1}, ...,y_n^{k1})$ where $\displaystyle 1 \le k \le n1$ so it is less or equal to n2. This means that in V(i) there is already such column of values $\displaystyle \begin{bmatrix} 1& y_1& \cdots & y_1^{k1}& y_1^{k1}& \cdots& y_1^{n2}&\\ \vdots & \vdots& \vdots& \vdots& \vdots& \vdots& \vdots& \\ 1& y_n& \cdots& y_n^{k1}& y_n^{k1}& \cdots& y_n^{n2}& \end{bmatrix} $ And determinant of each such matrix is zero. In $\displaystyle \sum_{i=1}^{n}(1)^{i+1}y_i^{n1}\cdot \pi(i)$ we have similar situation with V(i) as cofactors. However we now have vector $\displaystyle (y_1^{n1}, ...,y_n^{n1})$ as matrix column so we are getting Vandermonde matrix of order n1. Therefore we have $\displaystyle \sum_{i=1}^{n}(1)^{i+1}y_i^{n1}\cdot \pi(i)=(1)^{n1}det(V)$ At the end $\displaystyle \frac{(1)^{n1}det(V)}{(1)^{n1}det(V)}=1$ 

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