# Solution to polynomial of unknown degree

by Jhenrique
Tags: degree, polynomial, solution, unknown
 P: 280 Dear! Is possible to solution a polynomial of kind y = Ax^a + Bx^b ? Thx!
 P: 81 You have sevens unknowns and one equation. Could you solve 0 = a + b + c + d + e + f + g + h?
P: 1,019
 Quote by Jhenrique Dear! Is possible to solution a polynomial of kind y = Ax^a + Bx^b ? Thx!
I assume you are asking if it is possible to find the zeroes of y(x) = Ax^a + Bx^b, where a,b are real numbers?

P: 764

## Solution to polynomial of unknown degree

 Quote by Jorriss I assume you are asking if it is possible to find the zeroes of y(x) = Ax^a + Bx^b, where a,b are real numbers?
If it's a polynomial then a and b are natural.
Mentor
P: 20,440
 Quote by Jhenrique Dear! Is possible to solution a polynomial of kind y = Ax^a + Bx^b ?
 Quote by glappkaeft You have sevens unknowns and one equation.
I count six: A, a, B, b, x, and y.
 P: 280 omg, my question is simple! I would like to know if is possible to isolate the x variable in equation $$f(x)=ax^{\alpha}+bx^{\beta}$$
HW Helper
P: 3,420
 Quote by Jhenrique omg, my question is simple! I would like to know if is possible to isolate the x variable in equation $$f(x)=ax^{\alpha}+bx^{\beta}$$
In general, no. If $\alpha$ and $\beta$ are both less than 5, then yes. If they're both less than three, then it's a quadratic or simpler and hence easily done by the quadratic formula. In most other cases it's either difficult or impossible.
 PF Gold P: 274 Yes, it's possible, and quite easy. Let us suppose that $\alpha > \beta$. Let $\zeta = e^{2\pi i/(\alpha - \beta)}$ be a primitive root of unity. Then the polynomial factors as: $$ax^\beta \prod_{k=1}^{\alpha - \beta} (x - \left( \frac{b}{a} \right)^{1/(\alpha - \beta)} \zeta^k)$$
 HW Helper Thanks P: 5,208 In mathematics, as in other areas of science, just because a question is 'simple', it does not necessarily follow that the solution will be 'simple'. For example, consider Fermat's Last Theorem: http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem
HW Helper
P: 3,420
 Quote by Citan Uzuki Yes, it's possible, and quite easy. Let us suppose that $\alpha > \beta$. Let $\zeta = e^{2\pi i/(\alpha - \beta)}$ be a primitive root of unity. Then the polynomial factors as: $$ax^\beta \prod_{k=1}^{\alpha - \beta} (x - \left( \frac{b}{a} \right)^{1/(\alpha - \beta)} \zeta^k)$$
He's not looking to factor the polynomial but rather to find the inverse $f^{-1}(x)$, I think...
P: 280
 Quote by Mentallic He's not looking to factor the polynomial but rather to find the inverse $f^{-1}(x)$, I think...
Yeah!
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 Quote by Jhenrique omg, my question is simple! I would like to know if is possible to isolate the x variable in equation $$f(x)=ax^{\alpha}+bx^{\beta}$$
Yes if the exponents are positive integers. Consider the general algebraic function, ##y(x)## written implicitly as:

$$f(x,y)=a_1(x)+a_2(x)y+a_3(x)y^2+\cdots+a_n(x)y^n=0$$

with ##a_i(x)## polynomials. In your case we would simply have:

$$f(x,y)=x-ay^{\alpha}-by^{\beta}=0$$

Then by Newton-Puiseux's Theorem, we can compute power series representations of the various branches (solutions) of ##y(x)## having the form:

$$y_d(x)=\sum_{n=-p}^{\infty} c_n\left(x^{1/d}\right)^n$$

with radii of convergences extending at least to the nearest singular point of ##f(x,y)## and often further than that.

Do a search for "Newton Polygon" if you're interested in knowing how to compute these "Puiseux" series.

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