# Maybe I'll Get It This TIME

• Apr 30th 2006, 06:24 PM
Ranger SVO
Maybe I'll Get It This TIME
One more time and maybe it will sink in.

D(0) = 5

D(1) = 2.25 = ((1/4)(D(0))+1

D(2) = 1.5625 = ((1/4)(D(1))+1

D(3) = 1.390625 = ((1/4)(D(2))+1

D(n) = ????

This is the final question on an assesment due Thursday. An explination is just as important as the answer. Every step is important so that I might explain the prosses correctly on the assesment

• Apr 30th 2006, 06:46 PM
ThePerfectHacker
Quote:

Originally Posted by Ranger SVO
One more time and maybe it will sink in.

D(0) = 5

D(1) = 2.25 = ((1/4)(D(0))+1

D(2) = 1.5625 = ((1/4)(D(1))+1

D(3) = 1.390625 = ((1/4)(D(2))+1

D(n) = ????

This is the final question on an assesment due Thursday. An explination is just as important as the answer. Every step is important so that I might explain the prosses correctly on the assesment

Notice ya got,
$\displaystyle D_0=5$
$\displaystyle D_1=\frac{1}{4}D_0+1$
$\displaystyle D_2=\frac{1}{4}D_1+1=\frac{1}{4}\left( \frac{1}{4}D_0+1 \right) +1$=$\displaystyle \frac{1}{4^2}D_0+\frac{1}{4}+1$
$\displaystyle D_3=\frac{1}{4}D_2+1=\frac{1}{4}\left( \frac{1}{4^2}D_0+\frac{1}{4}+1 \right)$=$\displaystyle \frac{1}{4^3}D_0+\frac{1}{4^2}+\frac{1}{4}+1$
Notice da pattern?
$\displaystyle D_n=\frac{1}{4^n}D_0+\frac{1}{4^{n-1}}+...+\frac{1}{4}+1$
But, da series,
$\displaystyle 1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{n-1}}$
is geometric.
Thus, its sum iz,
$\displaystyle \frac{1-(1/4)^n}{1-1/4}=\frac{4}{3}-\frac{4}{3}\cdot \left( \frac{1}{4} \right)^n=\frac{4}{3}-\frac{1}{3\cdot 4^{n-1}}$
Thus, you are left with,
$\displaystyle \frac{1}{4^n}D_0+\frac{4}{3}-\frac{1}{3\cdot 4^{n-1}}$
But, $\displaystyle D_0=5$
Thus,
$\displaystyle \frac{5}{4^n}+\frac{4}{3}-\frac{1}{3\cdot 4^{n-1}}$
• May 3rd 2006, 06:10 PM
Ranger SVO
It took a little rearranging but I see what I was sopposed to.

It is a decaying (decreasing) exponential function with a limit of 4/3.
Its rate of decay is 1/4^x with a base of 44/3.

On this assesment the answer was not as important as what I needed to see. Below is a small part of that assesment

"Both are graphs of decreasing exponential functions. Both functions have been transformed up. The horizontal asymptote represents the limit. It also represents how much the function has been shifted up.
Y1 = 2 + 6 * (1/2)x
Y2 = (4/3) + (44/3) * (1/4)x
So L = d + c * ( 1- r )t"

Thank you very much !! :)