Trying to solve any equation ever with Recurrences

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Discussion Overview

The discussion revolves around the exploration of solving equations using recurrence relations. Participants examine specific examples, including a quadratic function and an exponential equation, to illustrate the proposed method of setting one side of an equation equal to x and iteratively calculating the other side. The scope includes theoretical approaches and numerical techniques, with an emphasis on convergence behavior.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant proposes a method to solve equations by rearranging them into the form x = f(x) and using iterative calculations to find solutions.
  • Examples provided include the quadratic equation f(x) = x^2 - 3x + 2, which converges to its roots through iterative substitution.
  • A more complex equation, 4x = e^(0.5x), is analyzed, with the participant demonstrating how it can yield a numerical approximation through recursion.
  • Another participant notes that the methods described resemble generic numerical techniques, such as Newton's method, and emphasizes that these approaches are approximations rather than exact solutions.
  • Concerns are raised about the validity of converging to infinity in certain cases, with a challenge to the idea that two functions can intersect at infinity.
  • Suggestions are made to explore the logistic map as a potential application of the proposed method.

Areas of Agreement / Disagreement

Participants express differing views on the validity of certain convergence behaviors, particularly regarding the concept of converging to infinity. While some participants acknowledge the iterative method's potential, there is no consensus on its applicability or correctness across all equations discussed.

Contextual Notes

Limitations include the potential for divergence in certain cases and the dependence on the specific form of the equations used. The discussion also highlights the need for a deeper understanding of the mathematical principles involved in the proposed methods.

Who May Find This Useful

This discussion may be of interest to those exploring numerical methods, iterative techniques for solving equations, and theoretical approaches in mathematics and engineering.

Gackhammer
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So, I am trying to make up this theory of trying to solve any equation ever using recurrences... Ill show you what I mean

Consider the quadratic function -
f(x) = x^2 - 3x + 2

Well, obviously you could do this the easy way and do factoring and figure out x = 2 or 1... no big deal right?

But what if we try a different way... albeit harder and more complicated... but Itll make a bit more sense later in this post

So what if we do this

x^2 - 3x + 2 = 0
x(x) - 3x + 2 = 0
x(x) = 3x-2
x = (3x-2)/x

So what does this mean?

Well, plug in a random number into your calculator... say 10...
Now do (3x-2)/x, and youll get 2.8...
now (if your using a TI-84 like me), plug in ((3*Ans)-2)/Ans
So Ans = 2.8, this should yield 2.28
Now Ans = 2.28, this should yield 2.125
From 2.125, you get 2.0588...

Basically, If you do this an infinite amount of times, it will converge to 2... which is one of the solutions of the equation...

Now say you rearrange the equation differently...
x^2 -3x+2 = 0
x(x) - 3x + 2 = 0
(x-3)x = -2
x = (-2)/(x-3)

If you do the same method as above, it will converge to 1 for any initial value you put in for x, which is the other zero for the quadratic equation...So... why am I wasting my time with this?

Well, consider this equation

4x = e^(.5x)

Well... idk how to solve this equation... but what if we do the thing I just did, where we set one side of the equation to x then infinitely compute the other side...

x = e^(.5x)
ln(4x) = .5x
2ln(4x) = x

If you pick 20 for the first x, 2ln(4x) will yield 8.76, then 7.11, then 6.69, and so on, until it converges to a number 6.523371369...

Is this right? well
4(x) = e^(.5x)
4(6.523371369) = e^(.5(6.523371369))
26.09 = e^(3.26)
26.09 = 26.09

So... that's the solution... by using an infinite recursion of setting one side of the equation to x, then we will get some sort of solution...

Now... this doesn't always work... for example, say we did (from the previous equation)

4x = e^(.5x)
x = (e^(.5x))/4

If you do this, then x will converge to infinity. Now you may think that this is wrong... but I feel this is right, since they both technically (intersect) at infinity. In theory, 4(infinity) = e^(infinity)... so is it really wrong? I don't believe so...Anyway... my theory is that you can solve a solution for an equation when you set the equation to the form of
x = f(x)
Using an infinite recursive calculation.Now, I really don't know how to dive more into this theory, since I think this involves math that's super crazy (more than I know). I am a computer engineer, so I don't have the biggest arsenal of theoretical math at my disposal. I do know that this might have to do something with damping, since some of the recurrences will have an overdamped, underdamped, and critically damped response to converging to the solution... but... i don't know...
So... any thoughts on this? Any suggestions on how to move forward with this? Or has someone done this before?

And no, this does not have anything to do with homework, this is just fun theoretical stuff I am doing on my own and I would like input/other ideas to help me move forward with this. If this post is deemed by the admins to not belong here, then Ill delete this post (no questions asked) and move it to the Homework Section (or wherever they tell me to put it). All I want is some outside input for this.
 
Last edited:
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A lot of these methods are pretty generic numerical techniques (see e.g. Newton's method). Note that you're not really "solving", only approximating solutions.
 
Gackhammer said:
Now... this doesn't always work... for example, say we did (from the previous equation)

4x = e^(.5x)
x = (e^(.5x))/4

If you do this, then x will converge to infinity. Now you may think that this is wrong... but I feel this is right, since they both technically (intersect) at infinity. In theory, 4(infinity) = e^(infinity)... so is it really wrong? I don't believe so...

You don't converge to infinity, you diverge to infinity.
Two real valued functions cannot intersect at infinity.
And "in theory" (i.e. in real analysis), 4 times infinity is not defined. Neither is e to the power of infinity.
You are also using the term "technically" is very non-technical manner.

Anyway... my theory is that you can solve a solution for an equation when you set the equation to the form of
x = f(x)
Using an infinite recursive calculation.

Try it with the logistic map.
 
pwsnafu said:
Try it with the logistic map.

if x := f(x) gets you in a cycle,

x := k f(x) + (1-k)x with k < 1 will probably work. Use a smaller k if there are still problems
 
Gackhammer said:
So, I am trying to make up this theory of trying to solve any equation ever using recurrences... Ill show you what I mean

Consider the quadratic function -
f(x) = x^2 - 3x + 2

Well, obviously you could do this the easy way and do factoring and figure out x = 2 or 1... no big deal right?

But what if we try a different way... albeit harder and more complicated... but Itll make a bit more sense later in this post

So what if we do this

x^2 - 3x + 2 = 0
x(x) - 3x + 2 = 0
x(x) = 3x-2
x = (3x-2)/x

So what does this mean?

Well, plug in a random number into your calculator... say 10...
Now do (3x-2)/x, and youll get 2.8...
now (if your using a TI-84 like me), plug in ((3*Ans)-2)/Ans
So Ans = 2.8, this should yield 2.28
Now Ans = 2.28, this should yield 2.125
From 2.125, you get 2.0588...

Basically, If you do this an infinite amount of times, it will converge to 2... which is one of the solutions of the equation...

Now say you rearrange the equation differently...
x^2 -3x+2 = 0
x(x) - 3x + 2 = 0
(x-3)x = -2
x = (-2)/(x-3)

If you do the same method as above, it will converge to 1 for any initial value you put in for x, which is the other zero for the quadratic equation...


So... why am I wasting my time with this?

Well, consider this equation

4x = e^(.5x)

Well... idk how to solve this equation... but what if we do the thing I just did, where we set one side of the equation to x then infinitely compute the other side...

x = e^(.5x)
ln(4x) = .5x
2ln(4x) = x

If you pick 20 for the first x, 2ln(4x) will yield 8.76, then 7.11, then 6.69, and so on, until it converges to a number 6.523371369...

Is this right? well
4(x) = e^(.5x)
4(6.523371369) = e^(.5(6.523371369))
26.09 = e^(3.26)
26.09 = 26.09

So... that's the solution... by using an infinite recursion of setting one side of the equation to x, then we will get some sort of solution...

Now... this doesn't always work... for example, say we did (from the previous equation)

4x = e^(.5x)
x = (e^(.5x))/4

If you do this, then x will converge to infinity. Now you may think that this is wrong... but I feel this is right, since they both technically (intersect) at infinity. In theory, 4(infinity) = e^(infinity)... so is it really wrong? I don't believe so...


Anyway... my theory is that you can solve a solution for an equation when you set the equation to the form of
x = f(x)
Using an infinite recursive calculation.


Now, I really don't know how to dive more into this theory, since I think this involves math that's super crazy (more than I know). I am a computer engineer, so I don't have the biggest arsenal of theoretical math at my disposal. I do know that this might have to do something with damping, since some of the recurrences will have an overdamped, underdamped, and critically damped response to converging to the solution... but... i don't know...



So... any thoughts on this? Any suggestions on how to move forward with this? Or has someone done this before?

And no, this does not have anything to do with homework, this is just fun theoretical stuff I am doing on my own and I would like input/other ideas to help me move forward with this. If this post is deemed by the admins to not belong here, then Ill delete this post (no questions asked) and move it to the Homework Section (or wherever they tell me to put it). All I want is some outside input for this.

I've learned this before what you found is actually called iteration...
the idea is that as you keep on feeding in the values generated by f(x) in x=f(x), you will converge to the solution.

However the sequence won't always converge (i.e it will diverge from the solution). As you might have noticed, there are different ways of isolating x from the equation. But the values will only converge to the solution if |f'(x)|<1 near the solution.

Therefore, you must try different equations of the form x=f(x). Then you find the gradient/slope of f'(x) near the root. If it lies between -1 and 1, it will converge. Also you can modify the equation such that the slope of f(x) near the solution is close to zero; then the iteration will converge more rapidly!
 
What does this have to do with Newton's method? It looks to me like the Banach fixed point theorem at work here.
 
  • #10
So, just wanted to make some other interesting points (on the damping point i mentioned in the first post)

So here's the equation x^2 -4x-9, with an input of 1.

For the recurrence equation x = \frac{9+4x}{x}, here is the plot of number of recurrences (n = 1,2,3) and the value it produces

PFPic1_zpsa1a4b692.png


(its in Excel, I am lazy)

As you can see, there is a damped curve, which is underdamped (I hope I got that right, been a while since I took Circuits for RLC stuff)

Anyway, as n approaches infinity, it reaches one of the zeros for the equation of 5.6ish

Now, if we manipulate the recurrence function to be x = \frac{9}{x-4}, here is the plot

PFPic2_zps0f763f22.png


Also underdamped... and gets to the -1.6 zero of the quadratic function



So... how about x = ln(x+3), can we find some overdamped or critically damped?

Well, here is that recurrence function with two different inputs

PFPic3_zps012a922b.png


PFPic4_zps92efa890.png



Im not sure if this is critically damped or overdamped...

I thought this was interesting to note, as (and I am not an expert on damping systems), we could probably describe this in a Second ODE...
 
  • #11
Vargo said:
What does this have to do with Newton's method? It looks to me like the Banach fixed point theorem at work here.

It looked similar to Newton's method- converges on a 'fixed point' in some cases, divergent in others. It also looked similar to several root finding algorithms for the same reason.

Never heard of a fixed point theorem before now, although it sounds like an excellent term for describing several things I've seen through the years.
 

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