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Find an algebraic function

  1. Jul 1, 2017 #1
    1. The problem statement, all variables and given/known data
    Snap1.jpg

    2. Relevant equations
    Algebraic Fuction:
    $$P_0(x)y^n+P_1(X)y^{n-1}+...+P_{n-1}(x)y+P_n(x)=0$$
    ##P_i(x)~## are polinomials, ##~P_0(x)\neq 0##

    3. The attempt at a solution
    $$\left( x^{2/3} \right)^3 (1-x)+\left( x^{2/3} \right)^2(x)=x^2-x^3+x^{5/3}$$
    Any integer exponent can't cancel the ##~x^{5/3}##
     
  2. jcsd
  3. Jul 1, 2017 #2

    fresh_42

    Staff: Mentor

    I assume, here is ##y=f(x)##. So what is ##y^3##?
     
  4. Jul 1, 2017 #3
    $$y=x^{2/3} \rightarrow~y^3=\left( x^{2/3} \right)^3=x^2$$
     
  5. Jul 1, 2017 #4

    fresh_42

    Staff: Mentor

    Yes, and now you need to write it ##...=0## to see what ##P_0(x)## and ##P_3(x)## are.
     
  6. Jul 1, 2017 #5
    $$P_0(x)x^2+x^{\frac{5}{3}}+P_3(x)=0$$
    P0 and P3 are dependent on each other, i have to choose one:
    $$\Rightarrow~P_0(x)=-\frac{-P_3(x)-x^{5/3}}{x^2}-x^{-\frac{1}{3}}$$
    No polinomial will result
     
  7. Jul 1, 2017 #6

    fresh_42

    Staff: Mentor

    Why so complicated? It is far, far easier than this. You already have ##y^3-x^2= 0##. All you have to do is to write down the coefficients of ##y^3,y^2,y^1,y^0##.
     
  8. Jul 1, 2017 #7
    I don't have ##~y^3-x^2= 0##. I assume you use only: ##~\left( x^{2/3} \right)^3 (1-x)=x^2-x^3##, or, maybe you have made a mistake, it's: ##~\left( x^{2/3} \right)^3 (1-x)=y^3-x^3##
    The coefficient of y3 is, as i chose: ##~(1-x)##
    $$(1-x)y^3+P_1(x)y^2+P_2(x)y-x^3=0$$
    Is it the equation? i arbitrarily chose 3d degree and the coefficient P0=1-x
     
    Last edited: Jul 1, 2017
  9. Jul 1, 2017 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Why did you choose ##1-x##? Why not just ##1##?
     
  10. Jul 1, 2017 #9

    fresh_42

    Staff: Mentor

    What about:
    You only have to compare ##y^3-x^2= 0## with ##P_0(x)y^3+P_1(x)y^2+P_2(x)y+P_3(x)=0##.
    Now what are the choices for the ##P_i##?
     
  11. Jul 1, 2017 #10
    $$1\cdot y^3+0\cdot y^2+0\cdot y-x^2=0$$
     
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