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Solution to polynomial of unknown degree

by Jhenrique
Tags: degree, polynomial, solution, unknown
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Jhenrique
#1
Dec2-13, 04:50 PM
P: 686
Dear!

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

Thx!
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glappkaeft
#2
Dec2-13, 05:40 PM
P: 82
You have sevens unknowns and one equation.

Could you solve 0 = a + b + c + d + e + f + g + h?
Jorriss
#3
Dec2-13, 05:49 PM
P: 1,066
Quote Quote by Jhenrique View Post
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
#4
Dec2-13, 05:56 PM
Sci Advisor
P: 834
Solution to polynomial of unknown degree

Quote Quote by Jorriss View Post
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
#5
Dec3-13, 12:36 AM
Mentor
P: 21,314
Quote Quote by Jhenrique View Post
Dear!

Is possible to solution a polynomial of kind y = Ax^a + Bx^b ?
Quote Quote by glappkaeft View Post
You have sevens unknowns and one equation.
I count six: A, a, B, b, x, and y.
Jhenrique
#6
Dec3-13, 06:02 AM
P: 686
omg, my question is simple!

I would like to know if is possible to isolate the x variable in equation
[tex]f(x)=ax^{\alpha}+bx^{\beta}[/tex]
Mentallic
#7
Dec3-13, 06:54 AM
HW Helper
P: 3,542
Quote Quote by Jhenrique View Post
omg, my question is simple!

I would like to know if is possible to isolate the x variable in equation
[tex]f(x)=ax^{\alpha}+bx^{\beta}[/tex]
In general, no. If [itex]\alpha[/itex] and [itex]\beta[/itex] 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
#8
Dec3-13, 07:14 AM
P: 274
Yes, it's possible, and quite easy. Let us suppose that [itex]\alpha > \beta[/itex]. Let [itex]\zeta = e^{2\pi i/(\alpha - \beta)}[/itex] be a primitive root of unity. Then the polynomial factors as:
[tex]ax^\beta \prod_{k=1}^{\alpha - \beta} (x - \left( \frac{b}{a} \right)^{1/(\alpha - \beta)} \zeta^k)[/tex]
SteamKing
#9
Dec3-13, 07:21 AM
Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 6,559
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
Mentallic
#10
Dec3-13, 07:46 AM
HW Helper
P: 3,542
Quote Quote by Citan Uzuki View Post
Yes, it's possible, and quite easy. Let us suppose that [itex]\alpha > \beta[/itex]. Let [itex]\zeta = e^{2\pi i/(\alpha - \beta)}[/itex] be a primitive root of unity. Then the polynomial factors as:
[tex]ax^\beta \prod_{k=1}^{\alpha - \beta} (x - \left( \frac{b}{a} \right)^{1/(\alpha - \beta)} \zeta^k)[/tex]
He's not looking to factor the polynomial but rather to find the inverse [itex]f^{-1}(x)[/itex], I think...
Jhenrique
#11
Dec3-13, 09:53 AM
P: 686
Quote Quote by Mentallic View Post
He's not looking to factor the polynomial but rather to find the inverse [itex]f^{-1}(x)[/itex], I think...
Yeah!
jackmell
#12
Dec4-13, 03:58 AM
P: 1,666
Quote Quote by Jhenrique View Post
omg, my question is simple!

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