What Are Solutions for Universal Linear Recursions?

In summary: That's why its characteristic polynomial is quadratic.So it doesn't seem to me that your formulas and the ones at Wiki are in any way in conflict. They're just different ways of looking at things. I'm sure that if you followed their method through, you would get exactly the same thing that you have.In summary, we discussed the use of recursive formulas in various number systems and their applications in different fields, such as the Fibonacci series and the Mandelbrot series. We also mentioned the existence of Benet formulas for the Fibonacci numbers and how linear, homogeneous recurrence relations can be turned into non-recursive formulas by obtaining solutions for the characteristic polynomial. Lastly, we compared the different approaches for solving inhomogeneous linear recurrence relations and concluded
  • #1
coolul007
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I may have reinvented the wheel, here is a pdf on my observations
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  • #2
coolul007 said:
Summary:: I may have reinvented the wheel, here is a pdf on my observations

https://www.testsite.cocoams.org/linrec.pdf
Looks ok so far (re-opened).

You could try to prove why your closed formulas are correct, consider why there are repetitions in other number systems to a different basis, or what happens if ##a\in \mathbb{R}.##
 
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  • #3
I haven't got a clue how to prove my formulas, they are correct. My aha moment was seeing the rep digits in the different bases. As for the reals, I'm a positive integer kind of guy. I also looked to see if there was anything that may be useful for the 3N+1 conjecture. So far nothing. Thank you for your comments.
 
  • #4
Recursive formulas occur everywhere. Some examples:
  • The Fibonacci series: [itex]x_{n+1}=x_{n}+x_{n-1} [/itex]
  • The Mandelbrot series: [itex]z_{n+1}=z_{n}^{2}+c [/itex] where c is a given constant
  • Factorials: n! = n⋅(n-1)!, (n integer >0, 0!=1)
  • Square roots: [itex]x_{n+1}=\frac{1}{2}(x_{n}+\frac{a}{x_{n}}) [/itex] (for a>0)
Other recursives inlude the Sierpinski curves and the Hilbert curves.
 
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  • #5
Svein said:
Recursive formulas occur everywhere. Some examples:
  • The Fibonacci series: [itex]x_{n+1}=x_{n}+x_{n-1} [/itex]
  • The Mandelbrot series: [itex]z_{n+1}=z_{n}^{2}+c [/itex] where c is a given constant
  • Factorials: n! = n⋅(n-1)!, (n integer >0, 0!=1)
  • Square roots: [itex]x_{n+1}=\frac{1}{2}(x_{n}+\frac{a}{x_{n}}) [/itex] (for a>0)
Other recursives inlude the Sierpinski curves and the Hilbert curves.
There already exists a Benet formula for the Fibonacci numbers.
 
  • #6
coolul007 said:
There already exists a Benet formula for the Fibonacci numbers.
Linear, homogeneous recurrence relations (sequences which can be defined using the form ##a_n=k_1 a_{n-1} + k_2 a_{n-2} + ... k_m a_{n-m}##) can be turned into non-recursive formulae by obtaining solutions for the characteristic polynomial.

The motivation for the technique is straightforward. Suppose that a solution takes the form ##x^n## for some value x. Then the recurrence states that ##x^m=k_1 x^{m-1} + k_2 x^{m-2} + ... k_m##. This is the characteristic polynomial for the recurrence. It will normally have m solutions. Any of those solutions will have the property that ##x^n## solves the recurrence. Any linear combination of those (##a x_1^n + b x_2^n + ...##) will also solve the recurrence. You plug in initial values of the sequence to figure out which linear combination you need. You do have to worry about complex solutions. Those end up being sines and cosines.

Or one can consult Wikipedia where they describe the exact same thing and more.
 
  • #7
jbriggs444 said:
Linear, homogeneous recurrence relations (sequences which can be defined using the form ##a_n=k_1 a_{n-1} + k_2 a_{n-2} + ... k_m a_{n-m}##) can be turned into non-recursive formulae by obtaining solutions for the characteristic polynomial.

The motivation for the technique is straightforward. Suppose that a solution takes the form ##x^n## for some value x. Then the recurrence states that ##x^m=k_1 x^{m-1} + k_2 x^{m-2} + ... k_m##. This is the characteristic polynomial for the recurrence. It will normally have m solutions. Any of those solutions will have the property that ##x^n## solves the recurrence. Any linear combination of those (##a x_1^n + b x_2^n + ...##) will also solve the recurrence. You plug in initial values of the sequence to figure out which linear combination you need. You do have to worry about complex solutions. Those end up being sines and cosines.

Or one can consult Wikipedia where they describe the exact same thing and more.
It seems these are solutions for a particular recursion, what I have supplied is a universal formula for all positive linear recursions.
 
  • #8
coolul007 said:
It seems these are solutions for a particular recursion, what I have supplied is a universal formula for all positive linear recursions.
As I understand your terminology, your "positive linear recursion" would be what we would call an "inhomogeneous first order linear recurrence relation".

It is "inhomogeneous" because of the constant term.
It is "linear" because the contribution from prior terms is a fixed multiple of the terms. Not of their squares or anything fancier.
It is "first order" because it only goes back one term.

As I recall from many years ago (it's been about 40 years now since I studied these), the first move you make when attacking an inhomogeneous recurrence is to make it homogeneous with a change of variables. You can then solve the homogeneous recurrence and change variables back.

The same Wiki article I linked to above covers inhomogenous linear recurrence relations as well. It uses a different technique from the one that I recall.

https://en.wikipedia.org/wiki/Recur...currence_relations_with_constant_coefficients

Fibonacci is an example of a second order homogenous linear recurrence relation. It goes back two terms and has no constant term.
 
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Related to What Are Solutions for Universal Linear Recursions?

1. What is linear recursion?

Linear recursion is a programming technique where a function calls itself repeatedly until a specific condition is met. It involves breaking down a larger problem into smaller sub-problems, and solving each sub-problem until the original problem is solved.

2. How is linear recursion different from iteration?

Linear recursion and iteration are both ways to solve problems through repetition, but they differ in their approach. In linear recursion, the function calls itself, while in iteration, a loop is used to repeat a set of instructions. Linear recursion is often more elegant and intuitive, while iteration can be more efficient in terms of time and space complexity.

3. What is the base case in linear recursion?

The base case in linear recursion is the condition that stops the function from calling itself and allows the program to exit the recursive loop. It is the simplest form of the problem that can be solved without further recursion.

4. What are the advantages of using linear recursion?

Linear recursion can simplify complex problems by breaking them down into smaller, more manageable sub-problems. It also allows for elegant and concise code, making it easier to understand and maintain. Additionally, linear recursion can be used to solve problems that are difficult or impossible to solve using iteration.

5. What are some common applications of linear recursion?

Linear recursion is commonly used in computer science and mathematics to solve problems such as searching, sorting, and data structures. It is also used in various fields of science, such as biology and physics, to model patterns and behaviors. Additionally, many programming languages use linear recursion to implement their own internal functions and algorithms.

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