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Basic ability multiplication problem?

  1. May 20, 2007 #1
    Hello! I am just invastigating the cubic function and proving my conclusion formally.

    I am finding the formula for the tangent to the function at the point equal to the average of two different roots.

    The function is:
    P(x) = q(x-a)(x-b)(x-c);
    after multiplication:
    P(x) = q (x^3 + (-a-b-c)x^2 + (ab+bc+ac)x -abc)

    I know the leading coefficient of the tangent:
    [tex]
    \frac{-q}{4}(a-b)^{2}
    [/tex]
    And I know that it is calculated properly (checked on a few examples).

    Now, I am calculating the other variable of the linear function of the tangent using the fact that the value at that point must be the same for both functions:
    P(x)=P'(x)*x + z

    [tex]
    P(\frac{a + b}{2})=P\text{'} (\frac{a+b}{2})\ast(\frac{a+b}{2})+z
    \linebreak
    [/tex]
    [tex]
    q((\frac{a + b}{2})^{3}+(-a-b-c)(\frac{a + b}{2})^{2}+(\mathit{ab}+\mathit{bc}+\mathit{ac})(\frac{a + b}{2})-\mathit{abc})=(\frac{-q}{4}(a-b)^{2})\ast(\frac{a+b}{2})+z
    \linebreak
    [/tex][tex]
    z=q(\frac{(a^{3}+3\mathit{ba}^{2}+3\mathit{ab}^{2}+b^{3})}{8}-\frac{(a^{3}+b^{3}+3\mathit{ab}^{2}+3\mathit{ba}^{2}+\mathit{ca}^{2}+\mathit{cb}^{2}+2\mathit{abc})}{4}+\frac{(\mathit{ba}^{2}+\mathit{ca}^{2}+\mathit{ab}^{2}+\mathit{cb}^{2})}{2}+\frac{(a^{3}+\mathit{ba}^{2}-\mathit{ab}^{2}+b^{3})}{8})\\
    \linebreak
    [/tex][tex]
    z=q((\frac{a + b}{2})^{3}+(-a-b-c)(\frac{{a + b}}{2})^{2}+(\mathit{ab}+\mathit{bc}+\mathit{ac})(\frac{a + b}{2})-\mathit{abc}+\frac{(a-b)^{2}}{4})\ast (\frac{a+b}{2})\\
    \linebreak
    [/tex][tex]
    z=q(\frac{(2\mathit{ba}^{2}+2\mathit{ca}^{2}+2\mathit{cb}^{2}-4\mathit{abc})}{8})=q(\frac{(c(a-b)^{2}+\mathit{ba}^{2})}{4})
    [/tex]

    The problem is... that it does not work as it should... I mean, the result is probably wrong. I checked it a few times and I am always getting the same result.

    Could any of you try to multiply the equation and see whether this result is really wrong or the problem is not there?

    Thank you for help!
    Greetings,
    Theriel
     
    Last edited: May 20, 2007
  2. jcsd
  3. May 20, 2007 #2
    Because there were no replies I decided to learn Latex... and I wrote the whole equation. Maybe now somebody can help me and check where the error is?

    P.S. I am sorry for the line breaking however it is not working properly, I do not know why...
     
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