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...

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- Sep 14th 2008, 06:51 PMbotanicSimple 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, 10:13 PMbkarpuz
The values you gave come out froum the solution of the following difference equation:

$\displaystyle f(x)=f(x-1)+100*(x+1)\text{ for }n\in\mathbb{N}\text{ with the initial condition }f(0)=0.$

Then, we have

$\displaystyle f(x)-f(x-1)=100*(x+1).$

By summing up from $\displaystyle i=1$ to $\displaystyle x$, we get

$\displaystyle \sum\limits_{i=1}^{x}\big[f(i)-f(i-1)\big]=100\sum\limits_{i=1}^{x}(i+1).\qquad(*)$

Note that left-hand side of (*) is a telescopic series, hence we deduce that

$\displaystyle \Rightarrow f(x)-f(0)=100\sum\limits_{i=1}^{x}(i+1)$

...................._$\displaystyle =100\Bigg(\sum\limits_{i=1}^{x}i+\sum\limits_{i=1} ^{x}1\Bigg)$

...................._$\displaystyle =100\Bigg(\frac{x(x+1)}{2}-x\Bigg),$

or simply, we have

$\displaystyle \Rightarrow f(x)=50x(x+3)$

for all $\displaystyle x\in\mathbb{N}$ since $\displaystyle f(0)=0$.

I hope this is what you want. (Wink)