Solution to polynomial of unknown degree

[Jhenrique]

Dear!

Is possible to solution a polynomial of kind y = Ax^a + Bx^b ?

Thx!

 

[glappkaeft]

You have sevens unknowns and one equation.

Could you solve 0 = a + b + c + d + e + f + g + h?

 

[Jorriss]

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?

 

[pwsnafu]

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.

 

[Mark44]

Dear!

Is possible to solution a polynomial of kind y = Ax^a + Bx^b ?

You have sevens unknowns and one equation.

I count six: A, a, B, b, x, and y.

 

[Jhenrique]

omg, my question is simple!

I would like to know if is possible to isolate the x variable in equation
$$f(x)=ax^{\alpha}+bx^{\beta}$$

 

[Mentallic]

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.

 

[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:
$$ax^\beta \prod_{k=1}^{\alpha - \beta} (x - \left( \frac{b}{a} \right)^{1/(\alpha - \beta)} \zeta^k)$$

 

[SteamKing]

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's_Last_Theorem

 

[Mentallic]

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...

 

[Jhenrique]

He's not looking to factor the polynomial but rather to find the inverse $f^{-1}(x)$, I think...

Yeah!

 

[jackmell]

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.