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Math Help - cubics

  1. #1
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    cubics

    Prove that for a cubic of the form a(x-b)^3+d then there exists at least one set of real solutions \{a,b,d\} such that:

    y_1=a(x_1-b)^3+d

    y_2=a(x_2-b)^3+d

    y'_1 = 3a(x_1-b)^2
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  2. #2
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    Quote Originally Posted by usagi_killer View Post
    Prove that for a cubic of the form a(x-b)^3+d then there exists at least one set of real solutions \{a,b,d\} such that:

    y_1=a(x_1-b)^3+d

    y_2=a(x_2-b)^3+d

    y'_1 = 3a(x_1-b)^2
    Strictly speaking, what you want to prove here is NOT true! If x_1= x_2 and y_1\ne y_2, the first two conditions are incompatible. I am going to assume that x_1\ne\x_2.

    Since y'= 3a(x- b)^2 that third condition follows from the first so we really have only two conditions. That means we are free to choose one of the parameters to be whatever we want. Choosing b= x_1 makes the first condition y_1= d. With b= x_1 and d= y_1, the second condition becomes y_2= a(x_2- x_1)^3+ y_1 which gives a= \frac{y_2-y_1}{(x_2- x_1)^3} as long as x_1\ne x_2.

    If we have x_1= x_2 and y_1= y_2, then we really only have one condition. Choose b= x_1 and d= y_1. a now can be any number at all.
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  3. #3
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    Sorry x_1 \neq x_2 and y_1 \neq y_2
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  4. #4
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    hey this comes under theory of equations,right??? i expanded (x-b)^3 and got a cubic. if the roots be y1,y2,y3 then i got the following:
    y1+y2+y3=-3b
    y1y2+y2y3+y3y1=-3b^2
    y1y2y3=d/a-b^3

    as evident there should be 3 roots but i simply couldnot find y1,y2,y3.there seems to be some kind of a proof not by usual methods.
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