How to Solve for f in a Nonlinear Difference Equation?

In summary, the conversation revolves around finding a solution for the equation \varepsilon(x) = e^{\varepsilon(x-1)} with the initial condition \varepsilon(0) = 1. The main goal is to consider x as a number in the interval [0,1]. Various approaches and suggestions are proposed, including defining \varepsilon(x) as an integral and using floor functions. Matt's suggestion of defining \varepsilon(x) = e^{e^{\vdots^e}} with \lfloor x \rfloor e's in the expression is discussed and eventually another approach is proposed using a Taylor series expansion. The conversation ends with a reference to a website that has a solution for the
  • #1
phoenixthoth
1,605
2
I'm posting this under analysis because of the method I'm thinking about using to solve it.

Here is the equation:
[tex]\varepsilon \left( x\right) =e^{\varepsilon \left( x-1\right) }[/tex]
and the initial condition is
[tex]\varepsilon \left( 0\right) =1[/tex].

My main goal is to consider x as a number in [0,1] but we can start with x being a natural number.

Here's what I want to try now. I'm pretty sure [tex]\varepsilon[/tex] is not elementary so let's assume it's of the form
[tex]\varepsilon \left( x\right) =\int_{0}^{\infty }f\left( t,x\right) dt[/tex].

Now my question is how do i solve for f?
 
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  • #2
eps(x) is just repeated exponentiation, isn't it?

eps(1) = e^{eps(0)} = e

eps(2) =e^{eps(1)}= e^e

eps(n) = e^{e^[e^...^{e}...} wit n e's in the expression.

Note ine general the answer is

eps(x) = e^{e^...e^{eps(floor(x)))...} with floor (x) e's in the expression.
 
  • #3
Note ine general the answer is

eps(x) = e^{e^...e^{eps(floor(x)))...} with floor (x) e's in the expression.

Thanks for the feedback but I was hoping to get a solution esp(x) where I can let x be a non-integer. My guess is the integral might be helpful in doing this, if it's like the gamma function.
 
  • #4
Matt's suggestion,

[tex]\varepsilon (x) = e^{e^{\vdots^e}}[/tex]

with [itex]\lfloor x \rfloor[/itex] [itex]e[/itex]'s in the expression works, doesn't it?

For example,

[tex]\varepsilon (0.5) = 1[/tex]

[tex]\varepsilon (1.5) = e[/tex]

[tex]\varepsilon (2.5) = e^e = e^{\varepsilon (1.5)}[/tex]

[tex]\varepsilon (3.5) = e^{e^e} = e^{\varepsilon (2.5)}[/tex]
 
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  • #5
Data said:
Matt's suggestion,

[tex]\varepsilon (x) = e^{e^{\vdots^e}}[/tex]

with [itex]\lfloor x \rfloor[/itex] [itex]e[/itex]'s in the expression works, doesn't it?

For example,

[tex]\varepsilon (0.5) = 1[/tex]

[tex]\varepsilon (1.5) = e[/tex]

[tex]\varepsilon (2.5) = e^e = e^{\varepsilon (1.5)}[/tex]

[tex]\varepsilon (3.5) = e^{e^e} = e^{\varepsilon (2.5)}[/tex]

yeah, it works but that's about as satisfying as defining 2.5! to be 2!.
 
  • #6
Alright, in that case try this:

First define [itex]\langle x \rangle = n[/itex] where [itex]n[/itex] is the greatest integer strictly smaller than [itex]x[/itex] (ie. [itex]x \not {\in} \mathbb{N} \Longrightarrow \langle x \rangle = \lfloor x \rfloor[/itex], but [itex] x \in \mathbb{N} \Longrightarrow \langle x \rangle = \lfloor x \rfloor -1[/itex]). Then let

[tex]\varepsilon (x) = e^{e^{\vdots^{e^{e^{x - \langle x \rangle}}}}}[/tex]

where [itex]e[/itex] appears [itex]\langle x \rangle + 1[/itex] times. Thus

[tex] \varepsilon(0) = 1[/tex]

[tex]\varepsilon(0.2) = e^{\frac{1}{5}}[/tex]

[tex] \varepsilon(0.5) = \sqrt{e}[/tex]

[tex] \varepsilon(1) = e = e^{\varepsilon(0)}[/tex]

[tex]\varepsilon(1.2) = e^{e^\frac{1}{5}} = e^{\varepsilon(0.2)}[/tex]

[tex]\varepsilon(1.5) = e^{\sqrt{e}} = e^{\varepsilon (0.5)}[/tex]

[tex]\varepsilon(2) = e^e = e^{\varepsilon (1)} [/tex]

[tex]\varepsilon(2.5) = e^{e^{\sqrt{e}}} = e^{\varepsilon(1.5)}[/tex]

etc.

I think you will find this function is actually pretty well-behaved... interesting!
 
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  • #7
sorry, i didn't mean floor(x) inside, I mean x - floor(x) ie the "bit after the decimal place"

that is there is no remotely unique solution unless you specify eps(x) for all x in (0,1]
 
  • #8
I'm working on finding another way.

Let f(x)=EXP(x) and note that p=W(-1) is a complex fixed point of f. Then I expanded a taylor series for the nth iterate of f about p and I got, where g is the nth iterate of f,
[tex]g\left( x\right) =p+p^{n}\left( x-p\right) +\frac{p^{n}\left( p^{n}-1\right) }{2\left( p-1\right) }\left( x-p\right) ^{2}+O\left( x-p\right) ^{3}[/tex].

Then, let n=a real number.

1. I can't find a nice formula for the series
2. I'm not sure the outputs for real x are real (they should be)
 
  • #9
Data said:
I think you will find this function is actually pretty well-behaved... interesting!

Well...it's not even C^2 ...does somebody have any proof of the existence of an analytic solution ?
 
  • #10
Any function from (0,1] to R yields a solution, which will be smooth on the domain R\Z if it is smooth on (0,1). If you wish to make it smooth at all points you'll need one such that it is infinitely differentiable at the the integers. Whether or not that can be done I do not wish to guess.

If we think about it as a functional equation on C: exp(f(z)) =f(z+1), then I would be even more reluctant to state there was an analytic solution.
 
  • #11
Focusing on the half-iterate of e^x now...

I rewrote the series above, which is centered at -W(-1), a fixed point of e^x, to be centered at 0. Lo and behold, the imaginary coefficients seem to be small on all coefficients (all of the five i calculated at least!).

Here's the real part of the series for f, where ff(x)=e^x:

f(x) = 0.50998 + 0.8876 x + 0.1720 x^2 + 0.0275 x^3 +0.0222 x^4 - 0.0018 x^5.

Now the last term IS wrong because all derivatives should be positive. I think it came out negative to compensate for the truncation.

Try graphing f(f(x)), f(x), x, and e^x on your CAS. You see that f(f(x)) is close to e^x for x in [-2,2] (at least) and the f(x) is between e^x and y=x which is to be expected.

So is there a real analytic solution with a positive radius of convergence?

And what is the formula for the nth derivative of f at 0?

I'm pretty sure (:rolleyes:) that f(0)=1/2 exactly and that f'(0)=Sqrt(e)/2 but all I could get for the second derivative was this:
f''(1/2)*e/4 + 2 f''(0) / Sqrt(e) = 1. I need f''(1/2) to get f''(0).

Hmm...
 
  • #12
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Related to How to Solve for f in a Nonlinear Difference Equation?

What is a nonlinear difference equation?

A nonlinear difference equation is a mathematical equation that describes a relationship between a variable and its successive values. Unlike linear difference equations, which have a constant rate of change, nonlinear difference equations have a variable rate of change.

What is the difference between a linear and a nonlinear difference equation?

The main difference between linear and nonlinear difference equations is the rate of change. In linear difference equations, the rate of change is constant, while in nonlinear difference equations, the rate of change can vary. This means that the relationship between the variable and its successive values is more complex in nonlinear difference equations.

What are some real-world applications of nonlinear difference equations?

Nonlinear difference equations are used in many fields, including physics, biology, economics, and engineering. Some examples of real-world applications include population growth models, weather forecasting, and stock market analysis.

How are nonlinear difference equations solved?

There is no general method for solving nonlinear difference equations, as the approach will depend on the specific equation. However, some common techniques include numerical methods, such as Euler's method or the Runge-Kutta method, and analytical methods, such as substitution or iteration.

What are the limitations of using nonlinear difference equations?

Nonlinear difference equations can accurately model many real-world phenomena, but they also have some limitations. For example, they may be sensitive to initial conditions and small changes in parameters, making long-term predictions less reliable. Additionally, they may not accurately capture all aspects of a complex system, leading to potential errors in predictions.

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