# Recursive formula find closed something

• Feb 27th 2010, 05:17 PM
Henryt999
Recursive formula find closed something
Hi!
There is this problem on recursive formulas I do not understand.
The problem is:
examine the recursive: $\displaystyle a_{n+1}=\frac{a_n}{1+a_n}$ where $\displaystyle a_0 =\frac{1}{2}$ and $\displaystyle n=0,1,2,3,4...$
Find a closed formula of the type $\displaystyle a_n = f(n)$ where $\displaystyle f(n)$is a expression that depends on n but not on for example $\displaystyle a_{n-1}$

I can probably do it if I would understand what they want, closed formula????
No idéa...where to start..
• Feb 27th 2010, 05:22 PM
NonCommAlg
Quote:

Originally Posted by Henryt999
Hi!
There is this problem on recursive formulas I do not understand.
The problem is:
examine the recursive: $\displaystyle a_{n+1}=\frac{a_n}{1+a_n}$ where $\displaystyle a_0 =\frac{1}{2}$ and $\displaystyle n=0,1,2,3,4...$
Find a closed formula of the type $\displaystyle a_n = f(n)$ where $\displaystyle f(n)$is a expression that depends on n but not on for example $\displaystyle a_{n-1}$

I can probably do it if I would understand what they want, closed formula????
No idéa...where to start..

they want you to prove that $\displaystyle a_n=\frac{1}{n+2}.$
• Feb 28th 2010, 01:32 AM
HallsofIvy
Quote:

Originally Posted by Henryt999
Hi!
There is this problem on recursive formulas I do not understand.
The problem is:
examine the recursive: $\displaystyle a_{n+1}=\frac{a_n}{1+a_n}$ where $\displaystyle a_0 =\frac{1}{2}$ and $\displaystyle n=0,1,2,3,4...$
Find a closed formula of the type $\displaystyle a_n = f(n)$ where $\displaystyle f(n)$is a expression that depends on n but not on for example $\displaystyle a_{n-1}$

I can probably do it if I would understand what they want, closed formula????
No idéa...where to start..

This problem tells you what a "closed formula" is! It "is an expression that depends on n" only. In other words, what you have always thought of, before, as a formula.

If $\displaystyle a_0= \frac{1}{2}$, then $\displaystyle a_1= \frac{\frac{1}{2}}{1+ \frac{1}{2}}$$\displaystyle = \frac{\frac{1}{2}}{\frac{3}{2}}= \frac{1}{2}\frac{2}{3}= \frac{1}{3}. Then \displaystyle a_2= \frac{\frac{1}{3}}{1+ \frac{1}{3}}= \frac{\frac{1}{3}}{\frac{4}{3}}$$\displaystyle = \frac{1}{3}\frac{4}{3}= \frac{1}{4}$.

Now do you see where NonCommAlg got his formula?

Of course you will need to prove that is correct. I recommend proving it by induction. Recursive formulas are ideally suited for induction.