how i solve a polynomial equation with exponents that are not integers?

For example: 0,32567*x^(-0,05673)+0,2223*x^(0,7319)+...+0,00863*x^(-0,000657)=0

Answer 1

[John D'Errico]

It is a general nonlinear equation, not a polynomial. Calling it that does not make it so, even though it looks vaguely like one.

You find the root of any such problem using fzero, IF a root exists. Note that when x is negative there will be issues, as a negative number raised to a non-integer power will not yield a real number. So using your example...

fun = @(x) 0.32567*x.^-0.05673 + 0.2223*x.^0.7319 + 0.00863*x.^-0.000657;
fun(-1)
ans =
      0.18114 +    0.10812i

As it turns out, this function is a bit boring along the positive real line. Ok, it is VERY boring. With all positive coefficients, there will never be a positive real solution. We need something with both positive and negative coefficients in the mix to make it interesting. (Of course, in context of polynomials, I think Descartes had something to say here.)

fun = @(x) -0.723 + 0.32567*x.^-0.25673 + 0.8223*x.^0.7319 -   0.363*x.^1.657;
ezplot(fun,[0,2])
grid on

It appears this function has three roots on the positive real line. ALWAYS plot your problems when you can. Plots give you a great deal of intuition about what is happening.

format long g
x = fzero(fun,[eps,2])
x =
          1.37482509521088

fun(x)
ans =
      2.22044604925031e-16

So close enough. Of course, this is only one of the roots. By a careful search for sign changes, we can force fzero to find the other roots. But roots cannot work on problems like this.

Or, we can throw the symbolic TB at it, which in this case, happens to find the second positive real root.

syms x
fsym = -0.723 + 0.32567*x.^-0.25673 + 0.8223*x.^0.7319 - 0.363*x.^1.657;

xs = solve(fsym)
xs =
0.22525988470397199278453292791815

[George]

Thank you very much John. You were very enlightening.

[John D'Errico]

I'm realizing I should have added an example where a solution exists that is findable by fzero. Next I could look at what happens in the complex plane.

[Bjorn Gustavsson]

Enlighten me John, what did Descartes have to say about this?

[John D'Errico]

Descartes rule of signs. which can tell you the maximum number of zeros of a polynomial function.

While I won't attempt to make any conclusions about whether it would hold in general, it certainly applies in the simple case. I.e., if all coefficients of the function are positive, then there can NEVER be any real positive roots.

Answer 2

[Walter Roberson]

Those are not polynomial equations, by definition.

[George]

Yes you are right. I wrote it like this in order to see it.

[George]

So has anyone an aswer on this?