# Difference Equations

• Jun 9th 2009, 05:43 PM
Roam
Difference Equations
Here's my question:
http://img8.imageshack.us/img8/3861/91929979.gif

The answer to this probelm is:

\$\displaystyle
S_{n} = 0.076 S_{n-1}
\$

I don't see how to get this answer, here's my attempt:

Equation has to be of the form \$\displaystyle S_{n} = k S_{n-1}\$, where k is some constant. I don't understand how they've got k=0.076

• Jun 10th 2009, 07:10 AM
the_doc
Clearly this should be:

\$\displaystyle S_n = 0.976 \, S_{n-1}\$

so either you've mistyped it or the person who set the question mistyped it putting a zero in place of where the 9 should have been!
• Jun 10th 2009, 11:45 AM
Roam
Quote:

Originally Posted by the_doc
Clearly this should be:

\$\displaystyle S_n = 0.976 \, S_{n-1}\$

so either you've mistyped it or the person who set the question mistyped it putting a zero in place of where the 9 should have been!

Oh, ok, but how did you get 0.976?
• Jun 10th 2009, 11:48 AM
Random Variable
Quote:

Originally Posted by Roam
Oh, ok, but how did you get 0.976?

\$\displaystyle S_{n} - S_{n-1} = -0.024S_{n-1} \$
• Jun 10th 2009, 01:12 PM
Roam
I'm sorry but I can't follow what you've done, could you please explain a little more?
• Jun 10th 2009, 01:19 PM
the_doc
Say after \$\displaystyle (n-1)\$ years the amount is \$\displaystyle S_{n-1}\$ so then after one more year you lose 2.4 % which is \$\displaystyle 0.024\$ of \$\displaystyle S_{n-1}\$ so

the amount lost \$\displaystyle = 0.024 S_{n-1}\$ so then after \$\displaystyle n\$ years we have the amount \$\displaystyle S_n\$ given by the amount at the start of the year, \$\displaystyle S_{n-1}\$, minus the amount lost so

\$\displaystyle S_n = S_{n-1} - 0.024 S_{n-1}\$,

\$\displaystyle = (1-0.024) S_{n-1}\$,

\$\displaystyle = 0.976 S_{n-1}\$.

Any better?
• Jun 10th 2009, 01:21 PM
skeeter
Quote:

... How did you get 0.976?
100% - 2.4% = 97.6%