# Simplifying an Infinite Composite Recursion in a System of Equations

Hello, I'm working with a system of equations that has an infinite recursion function, and am wondering if its possible to simplify or remove the recursion in terms of the other functions in the system. Any insight into the framework or family of this system is appreciated.

Given two functions P(x) and C(x), define a 3rd function A(x), such that:

A(x) = 2x - P(x) - C(x)

Then define an infinitely recursive composite 4th function Q(x), such that:

Q(x) = ...A(x) +C(A(x)+C(A(x)+C(A(x))))

Can Q(x) be simplified/solved to remove the recursion and/or shown in terms of a non recursive function of P(x) and C(x)?

I did notice that it's almost stating that Q(x) = A(x) +C(Q(x)), and was going to try and approach it that way, but that's not quite right, seeing as the recursion is on the front end and not terminating on the inside. Or does that represent it correctly? In which case how do I get the Q(x) out from the input of C(x)? Is it solely dependent on the form of C(x) itself, or removable by some other method?

Also, I was able to think of Q(x) as a series of equations, Qn(x), such that:

Q1(x) = C(A(x))
Q2(x) = C(A(x)+Q1(x))
Q3(x) = C(A(x)+Q2(x)) ... or in general,
Qn(x) = C(A(x)+Qn-1(x))

in which case, the infinite recursion is represented as Q(x), but I didn't really know where to go from here.

Anyways, thanks for any insight into this system.

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mfb
Mentor
If Q(x) is well-defined, the recursion process has to converge. If P and C are not completely crazy (what do you know about them?), this implies Q(x) = A(x) +C(Q(x)). If this equation has multiple solutions, you have to identify the right one.

Thanks for your help. Reading what you asked, and thinking about what I was trying to describe, I think I can make it clearer, and see where I may have miss stated Q(x) from the original iterations. The original and correct recursive function I am working with is:

Iteration 1: 2x - P(x) - C(x)
Iteration 2: 2x - P(x) - C(x) + C(2x - P(x) - C(x))
Iteration 3: 2x - P(x) - C(x) + C(2x - P(x) - C(x)) + C(2x - P(x) - C(x) + C(2x - P(x) - C(x)))
Iteration 4: 2x - P(x) - C(x) + C(2x - P(x) - C(x)) + C(2x - P(x) - C(x) + C(2x - P(x) - C(x))) + C(2x - P(x) - C(x) + C(2x - P(x) - C(x)) + C(2x - P(x) - C(x) + C(2x - P(x) - C(x))))
... and so on

This is the same as:

Iteration 1, Iteration 1 + C(Iteration 1), Iteration 2 + C(Iteration 2), Iteration 3 +C(Iteration3),....

I then defined Iteration 1 as A(x). What I'm trying to find, loosely speaking, is a closed form or limit, explicit or implicit, for the full infinite Iteration.

Lastly, it's funny that you questioned P and C. C(x) and P(x) are not easily invertible, if at all, and the above recursion came about as a method to avoid trying to find the Inverse of P(x). They are Sumations of Trig Functions where sometimes a partial sum is used, and other times, an infinite sum is used. In either case the Inversion gets challenging as it is not known whether the trig sums I am using have a closed form. That was another question, I may eventually create a thread for, but I figured I'd investigate this angle first. Thanks again.

mfb
Mentor
If P and C are continuous, that would be a good start, as that gives Q(x) = A(x) +C(Q(x)).

I'm working this problem in a few locations, and am updating the other sources respectively. A response and my reply from one of these sources has led to the correct statement of the recursion, additional information, and has me leaning toward the function being non-reducible. I have included the two relevant posts below. Of course, if anyone knows better, just let me know, thanks.

Post 1:
Ah, I see. That is much different than what I thought.

So, the recursion is actually, $$Q_1(x) = A(x), Q_n(x) = Q_{n-1}(x) + C(Q_{n-1}(x)).$$ This recursion is only well defined if the function converges pointwise to a limit function: $$\lim_{n \to \infty} Q_n(x) = Q(x).$$

If the limit function exists, then, in the limit, $$\lim_{n \to \infty} Q_n(x) = \lim_{n \to \infty} Q_{n-1}(x) + C(Q_{n-1}(x))$$ implies $$Q(x) = Q(x) + C(Q(x)),$$ which then implies $$C(Q(x)) = 0.$$ So, you know that Q(x) is a root of C(x) for all x. If that is not true, then I doubt any limit function exists. (Also, I am assuming that C(x) is continuous over all x).

If C(x) is not continuous, but the limit function of $$Q_n(x)$$ exists, then $$\lim_{n \to \infty} C(Q_n(x)) = 0.$$ So, now you need to check if $$C(Q_n(x))$$ converges, and if it does, does it converge to the zero function?

Post 2:

Nice, after looking over my initial reply I also saw the correct sectioning of the recursion, and you confirmed it.

$$Q_n(x) = Q_{n-1}(x) + C(Q_{n-1}(x))$$ is correct.

Your statements on the limits also proved insightful, as I can speak to some of the implications.

First, I agree that, $$\lim_{n \to \infty} Q_n(x) = \lim_{n \to \infty} Q_{n-1}(x) + C(Q_{n-1}(x))$$ implies $$Q(x) = Q(x) + C(Q(x)),$$ but only if Q and C are acting determinately. Sometimes the indeterminate limit is different than working with the limit of the parts depending on the functions. To that end, I'm not exactly sure. Temporarily shelving that possibility, and assuming the determinate case, then like you said, $$C(Q(x)) = 0,$$ and that Q(x) is a root of C(x) for all x, however, I do know some things conceptually about C(x) and Q(x).

Namely that:
C(Q(x)) = 0 only when x=1 or x=2.
C(x) as it is, is only continuous across the Integers, but not the Reals, it only has values at integers.
C(x) has 3 roots, at x=1, x=2, or x=3.

I think I may be able to alter C(x) to still serve its purpose and be continuous across the reals, or and also even add non-integer roots, however I don't think I would gain anything from that.

Finally, treating C(x) in its non-continuous form, and assuming the limit function of $$Q_n(x)$$ exists, you stated that $$\lim_{n \to \infty} C(Q_n(x)) = 0.$$ However, that limit equals $$Q(x) - x - 1,$$ which again, only equals 0 when x=1 or x=2.

So, overall, I'm generally thinking the recursion can't be truncated. Back to the chalkboard. Thanks again for your insights, they helped.