Need help with a recurion formula

1. May 15, 2005

skrying

Hi everyone,

Right now my class is studying recursion formulas - er, when the next answer is based off the answer before it, if that makes any sense? I am having a hard time understanding this.

A picture of the attached problem is included...Thanks for any and all help for this problem.

skrying

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2. May 15, 2005

dextercioby

That is a geometrical progression.The "n+1" term is gotten from the "n"-th by dividing through 2.

It converges to 0.What else...?

Daniel.

3. May 15, 2005

saltydog

You have:
$$a_n=\frac{1}{2^n}$$

That's just an expression for the n'th term. Something like:

$$a_n=2a_{n-1}+3a_{n-2}$$

would be a recursive formula defining the n-th term as a function of the n-1 term and the n-2 term.

Last edited: May 15, 2005
4. May 15, 2005

skrying

Thanks Dexter and Salty

:rofl: Thank you dexter and salty for your input. It does make more sense when you put it that way. I think I am just making all this harder than it really is perhaps?

Thanks, Skrying

5. May 15, 2005

skrying

How far can this recursion formula go?

I had another quick question (and probably a dumb one) since math of any kind is not my forte'....how far does or can this formula continue..?

Thanks!

6. May 15, 2005

OlderDan

What Dexter was saying is

$$a_n = \frac{1}{2^n}$$

can be rewritten. Since

$$a_{n-1} = \frac{1}{2^{n-1}}$$

$$a_n = \frac{1}{2^n} = \frac{1}{2 \bullet 2^{n-1}}=\frac{1}{2} a_{n-1}$$

That is the recursion formula, and it goes on forever. There is no limit on the value of n.

7. May 15, 2005

skrying

Thank you OlderDan

Thanks OlderDan...that looks like the same way some of my classmates were working on the problem. I really appreciate your help!

Thanks Skrying!