Closed form solution for P(x) = exp(x)

[Monomer]

A student at my Mum's work sent this email out and Mum sent it to me thinking I'd know how to solve it. She's sweet, but forgets that it's been more than a decade since I studied maths.

 

So is there anyone at Hypo that can help out with this one?

 

 

I am trying to find a closed form solution for the following equation:

 

P(x) = exp(x),

 

where P(x) is a polynomial of degree d >0.

 

Does anyone know the solution for this? Any help will be appreciated.

 

[Qfwfq]

Can't give an exact proof without looking back through textbooks but I'd say no, for a polynomial P(x) of finite degree.

 

The only sensible way to treat an equality between the two is by an expansion of the exponential, which is a polynomial of infinite degree. Expanding from 0 is easy:

 

[math]e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}[/math]

 

Now writing the equation as difference equals 0 means altering a finite subset of the series coefficients. There's not a general method for finding the zeros of the new "polynomial" although there certainly is in the case of degree 0 because the equation's original form is trivial to solve. A full proof would be slightly more complicated but I think degree 0 is the only finite one for which there's a solution.

[Monomer]

Thanks for the reply Qfwfq!

 

I'll pass it on to the student.

[Pyrotex]

Can't give an exact proof without looking back through textbooks but I'd say no, for a polynomial P(x) of finite degree....

Maybe it depends on what you want the polynomial for. If you're looking for the "zeros" for P(x), then it appears hopeless. But if you just want a polynomial that has nearly the same shape in the region 0 < x < 1, the same area under the curve, the same slope, etc, then a close approximation may be to just take the first m terms of the expansion, for m between 4 and 8.

[Monomer]

Thanks for your input Pyrotex!

 

I'll pass that info on too.

[Qfwfq]

Yes Pyro, solving the equation would mean finding the zeros of [math]e^x - P(x)[/math] so I'd say it's hopeless except for brute number crunching.

[Nootropic]

The Lambert W-function, or product log, actually provides a closed form for the solution of this equation in certain instances. Check it out: Lambert W function - Wikipedia, the free encyclopedia

[Qfwfq]

Well I suppose that with:

...in certain instances.

you mean those of the form [imath]P(x)=\frac{\alpha}{x}[/imath] or even [imath]P(x)=\alpha x[/imath] with a bit of manipulation. Even for [imath]P(x)=\alpha x+\beta[/imath] I don't seem to get Lambert equals a constant, I get:

 

[math]xe^x=\frac{1}{\alpha+\frac{\beta}{x}}[/math]

 

Of course one could always define the Gilbert W-squared function [imath]w^2e^w[/imath] and the Dilbert W-cube function etc. and, for any polinomial P(x), one couild define the Gollywooglesmurfbert [imath]P^{-1}(w)[/imath] function as:

 

[math]f(w)=\frac{e^w}{P(w)}[/math]

 

and of course in each case you need to know the inverse of the defined function.

[Monomer]

Here's one reply you might be interested in:

 

I don't believe there is a general closed form solution. For P(x)=0, general closed form solutions only exist up to d=4 [1, Chapter 4]. For P(x)=exp(x), the solution for d=0 is trivial. For d=1, we we have a*x+b=exp(x). Doing a graph, we can see that there can be one, two, or no real solutions, depending on the value of a and b. We can express exp(x) as an infinite polynomial, but we know this does not have a general closed form solution. Given the difficulty of finding a solution even for d=1, I very much doubt a general solution exists.

 

Your best bet is to solve the equation numerically. Newton-Raphson's (or Newton's) method is probably the most useful.

 

p_{n+1} = p_n - f(p_n)/f'(p_n), n >=0

 

where f(x) = P(x) - exp(x), and f'(x) = dP(x)/dx - exp(x).

 

[1] N. Jacobson, "Basic algebra I, Second Ed.," W. H. Freeman and Company, New York, 1985.

[Nootropic]

Certain instances, where p(x) = x. For larger degree polynomials, we may be out of luck, unless we use the suggested method of approximating the solution with the mclaurin series for the exponential function. Or a more efficient method, while not exact, would be to use newton's method. Which is certainly plausible, and can be used to good degree of accuracy.

[Qfwfq]

I don't see why McLaurin would be more advantageous than Taylor. While it's obvious that one may find numerical values with these or Newton etc. the title of this thread was "Closed form solution", that's why I haven't mentioned these as being actual answers.

 

:evil: