# Simple exponential function

• Sep 14th 2008, 07:51 PM
botanic
Simple exponential function
I have 4 values 200,500,900,1400 lets say there x and y increases by 1 each time. I need the formula for this so i can program it.

I figured out that
(100*(x+1))
is the difference between the numbers just i cant get the rest...
• Sep 14th 2008, 11:13 PM
bkarpuz
Quote:

Originally Posted by botanic
I have 4 values 200,500,900,1400 lets say there x and y increases by 1 each time. I need the formula for this so i can program it.

I figured out that
(100*(x+1))
is the difference between the numbers just i cant get the rest...

The values you gave come out froum the solution of the following difference equation:
$f(x)=f(x-1)+100*(x+1)\text{ for }n\in\mathbb{N}\text{ with the initial condition }f(0)=0.$
Then, we have
$f(x)-f(x-1)=100*(x+1).$
By summing up from $i=1$ to $x$, we get
$\sum\limits_{i=1}^{x}\big[f(i)-f(i-1)\big]=100\sum\limits_{i=1}^{x}(i+1).\qquad(*)$
Note that left-hand side of (*) is a telescopic series, hence we deduce that
$\Rightarrow f(x)-f(0)=100\sum\limits_{i=1}^{x}(i+1)$
...................._ $=100\Bigg(\sum\limits_{i=1}^{x}i+\sum\limits_{i=1} ^{x}1\Bigg)$
...................._ $=100\Bigg(\frac{x(x+1)}{2}-x\Bigg),$
or simply, we have
$\Rightarrow f(x)=50x(x+3)$
for all $x\in\mathbb{N}$ since $f(0)=0$.

I hope this is what you want. (Wink)