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I was running into a strange identity that is

Given ${x_1},{x_1},...,{x_n}$ and ${y_1},{y_1},...,{y_n}$ are all real number.

Denote ${c_k} = {y_1} + {y_2} + {y_3} + ... + {y_k}$ where $1 \le k \le n$

Proof that

${x_1}{y_1} + {x_2}{y_2} + ...{x_n}{y_n} = \left( {{x_1} - {x_2}} \right){c_1} + \left( {{x_2} - {x_3}} \right){c_2} + ... + \left( {{x_{n - 1}} - {x_n}} \right){c_n} + {x_n}{c_n}$

By plug in some number, I was able to come up with some case but I am not sure how to proof this identity for the general case

For $n=2$, we have:

${a_1}{b_1} + {a_2}{b_2} = \left( {{a_1} - {a_2}} \right){b_1} + {a_2}\left( {{b_1} + {b_2}} \right)$

For $n=3$, we have:

${a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3} = \left( {{a_1} - {a_2}} \right){b_1} + \left( {{a_2} - {a_3}} \right)\left( {{b_1} + {b_2}} \right) + {a_3}\left( {{b_1} + {b_2} + {b_3}} \right)$

My country call this identity as Abel's expansion but I was unable to determine whether if this naming is correct or not.

Edit: I have finally found the name, this process is called Abel transformation https://en.wikipedia.org/wiki/Summation_by_parts

Tuong Nguyen Minh
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    As you say, this trick is sometimes called Abel summation. It's also known as partial summation due to analogy with integration by parts. – Angina Seng Aug 03 '20 at 05:40
  • So if you look up "Abel summation" or "partial summation", you may find a proof, Tuong, and then you can report back to us. Or you may find that it has already been explained in a previous ppost on this site, and you could just link to it. – Gerry Myerson Aug 03 '20 at 06:40
  • I think your formula has some incorrect indices underneath "Proof that". For $n=1$ if you use that formula, we get that $x_1y_1=x_1y_1+x_2y_2$ – C Squared Aug 03 '20 at 07:22

1 Answers1

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We could prove this by induction.

Givens: $(x_1,x_2,...,x_n)$, $(y_1,y_2,...,y_n)$, and $c_k=y_1+y_2+...+y_k$ for $1\leq k\leq n$. Show that $$\sum_{k=1}^n x_ky_k=c_{n}x_{n+1}+\sum_{k=1}^{n} (x_k-x_{k+1})c_k $$ Base case $n=1$: $$\begin{align*}x_1y_1 &= c_1x_2 + (x_1-x_2)c_1\\ &= (y_1)(x_2)+(x_1-x_2)(y_1)\\&=x_2y_1+x_1y_1-x_2y_1\\&=x_1y_1\end{align*}$$

Induction Hypothesis: Assume $$\sum_{k=1}^{j} x_ky_k=c_{j}x_{j+1}+\sum_{k=1}^{j} (x_k-x_{k+1})c_k $$ is true for some integer $j$. We need to show that it is true for $j+1$. $$ \begin{align*}\sum_{k=1}^{j+1} x_ky_k &=x_{j+1}y_{j+1}+\sum_{k=1}^{j} x_ky_k\\&= x_{j+1}y_{j+1}+c_jx_{j+1}+\sum_{k=1}^{j}(x_k-x_{k+1})c_k\\&= x_{j+1}(c_j+y_{j+1})+ \sum_{k=1}^{j}(x_k-x_{k+1})c_k \\&= x_{j+1}c_{j+1}+\sum_{k=1}^{j}(x_k-x_{k+1})c_k\\&= (x_{j+2} +x_{j+1}-x_{j+2})c_{j+1} +\sum_{k=1}^{j}(x_k-x_{k+1})c_k\\&= x_{j+2}c_{j+1} +(x_{j+1}-x_{j+2})c_{j+1} + \sum_{k=1}^{j}(x_k-x_{k+1})c_k\\&= x_{j+2}c_{j+1} + \sum_{k=1}^{j+1}(x_k-x_{k+1})c_k\end{align*}$$

By induction, we are done. As for the name, I believe it was given in the comments.

C Squared
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