Solution to polynomial of unknown degree


by Jhenrique
Tags: degree, polynomial, solution, unknown
Jhenrique
Jhenrique is offline
#1
Dec2-13, 04:50 PM
P: 427
Dear!

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

Thx!
Phys.Org News Partner Mathematics news on Phys.org
Hyperbolic homogeneous polynomials, oh my!
Researchers help Boston Marathon organizers plan for 2014 race
'Math detective' analyzes odds for suspicious lottery wins
glappkaeft
glappkaeft is offline
#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
Jorriss is offline
#3
Dec2-13, 05:49 PM
P: 1,025
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
pwsnafu is offline
#4
Dec2-13, 05:56 PM
Sci Advisor
P: 779

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
Mark44 is offline
#5
Dec3-13, 12:36 AM
Mentor
P: 21,063
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
Jhenrique is offline
#6
Dec3-13, 06:02 AM
P: 427
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
Mentallic is offline
#7
Dec3-13, 06:54 AM
HW Helper
P: 3,436
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
Citan Uzuki is offline
#8
Dec3-13, 07:14 AM
PF Gold
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
SteamKing is offline
#9
Dec3-13, 07:21 AM
HW Helper
Thanks
P: 5,582
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
Mentallic is offline
#10
Dec3-13, 07:46 AM
HW Helper
P: 3,436
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
Jhenrique is offline
#11
Dec3-13, 09:53 AM
P: 427
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
jackmell is offline
#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.


Register to reply

Related Discussions
Third degree polynomial General Math 4
Solution of a polynomial with degree 25 Precalculus Mathematics Homework 10
How approximate a sextic polynomial to a lower degree polynomial General Math 6
Degree 3 Polynomial Precalculus Mathematics Homework 5
Polynomial with 3 unknown variables Precalculus Mathematics Homework 7