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Show that discrete analogy of partial integral formula as below. $$ \sum\limits_{k=0}^{n-1} y(k)\Delta x(k)=x(n)y(n)-\sum\limits_{k=0}^{n-1}x(k+1) \Delta y(k) + c $$

I have tried as below and I stuck.

\begin{align*} \sum\limits_{k=0}^{n-1} y(k)\Delta x(k)&= \sum\limits_{k=0}^{n-1} y(k)\left(x(k+1)-x(k)\right)\\ &=\sum\limits_{k=0}^{n-1} \left(x(k+1)y(k)-x(k)y(k)\right)\\ &=\sum\limits_{k=0}^{n-1} x(k+1)y(k)- \sum\limits_{k=0}^{n-1} x(k)y(k)\\ \end{align*}

Now I can't expand it in form

$$=x(n)y(n)-\sum\limits_{k=0}^{n-1}x(k+1) \Delta y(k) + c.$$

Anyone can give me hint?

Ongky Denny Wijaya
  • 2,298

1 Answers1

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\begin{align*} \sum\limits_{k=0}^{n-1} y(k)\Delta x(k)&= \sum\limits_{k=0}^{n-1} y(k)\left(x(k+1)-x(k)\right)\\ &=\sum\limits_{k=0}^{n-1} \left(x(k+1)y(k)-x(k)y(k)\right)\\ &=\sum\limits_{k=0}^{n-1} \left(x(k+1)y(k)-x(k+1)y(k+1)+x(k+1)y(k+1)-x(k)y(k)\right)\\ &=\sum\limits_{k=0}^{n-1} \left(-x(k+1)(y(k+1)-y(k))+x(k+1)y(k+1)-x(k)y(k)\right)\\ &=-\sum\limits_{k=0}^{n-1} x(k+1)\Delta y(k)+ \sum\limits_{k=0}^{n-1} \left(x(k+1)y(k+1)-x(k)y(k)\right) \\ &=-\sum\limits_{k=0}^{n-1} x(k+1)\Delta y(k)+ \left(x(n)y(n)-x(0)y(0)\right). \end{align*} Let $c=-x(0)y(0)$, we have \begin{align*} \sum\limits_{k=0}^{n-1} y(k)\Delta x(k)&=x(n)y(n)-\sum\limits_{k=0}^{n-1} x(k+1)\Delta y(k) +c. \end{align*}

Ongky Denny Wijaya
  • 2,298