How Do Recursion Formulas Work in Mathematics?

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Recursion formulas in mathematics define a sequence where each term is based on previous terms, such as a_n = 2a_{n-1} + 3a_{n-2}. A specific example discussed is a_n = 1/2^n, which can be expressed recursively as a_n = (1/2) a_{n-1}. The formula can continue indefinitely, with no limit on the value of n. Understanding these concepts can simplify the learning process for students struggling with recursion. Overall, the discussion emphasizes the importance of grasping the relationship between terms in recursive sequences.
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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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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.
 
skrying said:
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.
 
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Thanks Dexter and Salty

:smile: 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
 
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!
 
skrying said:
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.
 
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!
 
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