Interdependent Recursive Equations

Click For Summary
SUMMARY

The discussion focuses on solving interdependent recursive equations in terms of the variable m, with constants Rbbase, Rtbase, and Rcc. The user Himmel suggests using numerical calculations with tools like R, Python, or Matlab to derive patterns from the sequences generated by the equations. The conversation highlights the simplification of the recursion formulas through division, leading to expressions involving K_m and S_m. Ultimately, the goal is to establish direct expressions for R_m and K_m using these recursive relationships.

PREREQUISITES
  • Understanding of recursive equations and their properties
  • Familiarity with numerical computation tools such as R, Python, or Matlab
  • Knowledge of mathematical induction for proof validation
  • Basic algebraic manipulation skills, particularly with fractions and ratios
NEXT STEPS
  • Explore numerical methods for solving recursive equations using R
  • Learn about mathematical induction and its application in proving recursive formulas
  • Investigate the behavior of sequences generated by recursive equations in Python
  • Study the simplification techniques for complex algebraic expressions in recursive contexts
USEFUL FOR

Mathematicians, data scientists, and engineers working with recursive equations, as well as anyone interested in numerical analysis and algorithm development.

Himmel
Messages
1
Reaction score
0
Hi, I have a system that I am trying so solve in terms of m, and have two recursive equations:
O5Igc.png


The problem for me is that each recursion is dependent on the value from the other! I know that they are both solvable, however I have no idea what approach I could take to express each only in terms of m. Rbbase, Rtbase and Rcc are all constants, and if it helps Rbbase is proportional to Rtbase.

Cheers, Himmel
 

Attachments

  • O5Igc.png
    O5Igc.png
    26.4 KB · Views: 1,436
Physics news on Phys.org
With sequences like this, where there is obvious way in at first, a good way to develop intuitions that may point towards solutions is to first make numerical calculations of the sequence for a fair number of elements, say twenty. Assume values for the three constants and then use a calculator like R, Python or Matlab to calculate the first twenty values of the two sequences. Then change the constants a few times and see what changes. Hopefully a pattern will start to emerge, suggesting formulas that give values of the sequences in terms of only m and the three constants. For any candidate formula you think of, project the first twenty items using that, and see if they match. If they do, you might then set out to prove that the formula is correct for all m. Induction may well be a handy way to do such a proof.

I also notice that both recursion formulas simplify significantly if you divide both numerator and denominator by the numerator, giving you expressions of the form ##\frac{1}{1/R_{bbase}+1/\textrm{messy factor 1}}## and ##\frac{1}{1/R_{tbase}+1/\textrm{messy factor 2}}## where the 'messy factors' are the second factors in the numerators.
 
Last edited:
Let ##K_m = \frac{R_{bm}}{R_{tm}}##. Let ##R_{cc}=C## to avoid overusing R. Let ##D=R_{b0}## so Rb is gone and drop the index "t" on the remaining ##R_{tx}##.

Then $$R_{m} = \frac{
R_{0} \cdot \left( R_{m-1}(K_{m-1}+1) + \frac{C}{K_{m-1}} \right)
}{
R_{0} + \left( R_{m-1}(K_{m-1}+1) + \frac{C}{K_{m-1}} \right)
}$$
and
$$K_m R_{m} = \frac{
D \cdot \left( R_{m-1}K_{m-1} + CK_{m-1} \right)
}{
D + \left( R_{m-1}K_{m-1} + CK_{m-1} \right)
}$$
Define ##S_m = \frac{1}{R_m}##. And look at m+1 because it shortens the equation. Now
$$S_{m+1} = S_0 + \frac{1}{ \frac{1+K_m}{S_m}+\frac{C}{K_m} }$$
and
$$\frac{S_{m+1}}{K_{m+1}} = \frac{1}{D} + \frac{1}{\frac{K_m}{S_m} + C K_m}$$
I'm still not sure how to find direct expressions, but it looks much nicer already.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K