- #1
theriel
- 27
- 0
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
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: