Direct way to calculate nth term of cyclically repeating function?

  • Thread starter ktoz
  • Start date
  • #1
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3

Main Question or Discussion Point

"cyclic functions" may not be the correct term, but I don't know what else to call them. (only have basic high school math training, no calculus) Here's what I'm looking for.

Given the following series of functions

f(a) = b (step 1)
f(b) = c (step 2)
f(c) = d (step 3)
...
f(n) = a (step n)

Is there a general way to rework these so that you can directly calculate the value at steps 1,2,3 etc rather than having to iterate through all the steps from a to x, something like this?

f(0) = a
f(1) = b
f(2) = c
f(3) = d
...
f(n) = a

This problem arose from a computer program I'm working on and it would be much more efficient to directly calculate the n'th term rather than having to iterate through them all.

Thanks for any help
 

Answers and Replies

  • #2
Tide
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How about using the mod (modulo) function?
 
  • #3
153
3
Example?

Tide said:
How about using the mod (modulo) function?
Here's a little more background. Using the Mandelbrot generating function Z1 = Z0^2 + c yeilds either:

Escape set - values not in set
Chaotic set - values don't escape but also don't settle into a repeating set of values
Convergent set - values converge to either a single value or a repeating set of values

Convergent values take the form
f(a) = b
f(b) = c
f(c) = d
...
f(n) = a

With the above in mind, how would I apply your modulo suggestion?

Thanks

Ken
 
  • #4
Tide
Science Advisor
Homework Helper
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That's a little different than what I thought you were asking in your original post. I'm afraid you're stuck with iterating.
 

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