Wikipedia:Reference desk/Archives/Mathematics/2006 November 5

November 5

exponentials

"If the population of rabbits (or anything) is initially 2000 and decays at a rate of 10% per yr, how long does it take for the population to reach 300.."

These sort of questions (I just fabricated that) are confusing. You can of course do the math and work out that it takes 18.005986 years for this to be ... rabbits? Since the number of rabbits is always a whole number, the continuous aspect of exponential decay can only refer to a statistical expectation, and the 10% can only be an estimate. As is described in our article on percentages, it is common to round percentages anyway, so 10% may stand for any value from 0.095 to 0.105, giving a range from 17.1 to 19.0 years.  --Lambiam^Talk 08:45, 5 November 2006 (UTC)

find a recurrence relation for the number of ways to climb n stairs if you can go 1 step, 2 or 3...

for problems like these, why is the answer just additive without any exponential terms?

For example, the problem : "find a recurrence relation for the number of bit strings of length n with 3 consecutive 000's" has a 2^(n-3) in the solution.

when is it just additive and when is it additive with an &cd=6

it says that the stair problem is just additive

http://www.comp.nus.edu.sg/~cs1232/pdfs/tut2s.pdf

shows that the bit string problem is additive with the 2^(n-3) at the end.

what i need to know is, given one of these strange word problems, how can I decide if the relation has an exponetial term in it or if it's just simple additive? Does that makes sense?

I think that I can see what you mean. In the stair climbing problem let $a_{n}$ be the number of ways of climbing n steps. If n is at least 3 then there are 3 cases:

1. Climb 1 step in the first leap. Then we must climb $n-1$ further steps and there are $a_{n-1}$ ways of doing this.
2. Climb 2 steps in the first leap. Then we must climb $n-2$ further steps and there are $a_{n-2}$ ways of doing this.
3. Climb 3 steps in the first leap. Then we must climb $n-3$ further steps and there are $a_{n-3}$ ways of doing this.

Note that in each case we reduce the problem to a simpler problem of the same type and we get the recurrence relation $a_{n}=a_{n-1}+a_{n-2}+a_{n-3}$, which is "simply additive".

In general, the recurrence relation will be "simply additive" if the problem can be reduced to several simpler problems of the same type. Also, if the relation is not "simply additive" then the extra terms need not be exponential. We could have a recurrence relation such as $a_{n}=a_{n-1}+n$.

I hope I have been of some assistance. --Matthew Auger 21:55, 5 November 2006 (UTC)

thank you for the response, it is much appreciated. this makes more sense. Thanks.