How Do Recursion Formulas Work in Mathematics?

[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

 

[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.

 

[saltydog]

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

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.

 

[skrying]

Thanks Dexter and Salty

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

 

[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!

 

[OlderDan]

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!

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.

 

[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!