MATLAB: Polynomial to the power of polynomial.

polynomialroots

So I wanted to check the roots of the eqn (x^2-7*x+11)^(x^2-13*x+42)=0.

I got one root as 2. How do i get the other values?

Best Answer

2 is not a root.

There are three ways that an expression A raised to another expression B can be 0.

One way is if A is 0 at a location that B is not also 0. You can solve this by finding the roots of A and cross checking whether they are also roots of B.

A second way is if the A has a value with absolute value strictly less than 1 while B is positive infinity. This cannot occur with pure polynomials: they always go to infinity only at +/- infinity so at any location where B is infinite, A would be as well.

A third way is if A is absolute value strictly greater than 1 and B is negative infinity. This cannot occur in this particular case as B cannot be negative infinity, but it could occur with other polynomial B.

We can see that only the first case applies for these expressions, so find the roots of A as usual. 2 is not one of them.

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MATLAB: How i solve a polynomial equation with exponents that are not integers

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

MATLAB: How to get the rightmost root in the interval

How about finding the intersection points of the given curve and the line y = 0 ? You can use this to get intersection points.

https://in.mathworks.com/matlabcentral/fileexchange/22441-curve-intersections?focused=5165138&tab=function

Best Answer

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MATLAB: How i solve a polynomial equation with exponents that are not integers

MATLAB: How to get the rightmost root in the interval

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