# Find a Fixed Cubic

• Jun 5th 2008, 08:02 AM
ThePerfectHacker
Find a Fixed Cubic
I am working on a problem, the remaining step I need to do is very computational.
I am mathematically software challenged, can some one find the solution to this problem:

Find $a,b,c,d,e,f,g,h$ (real numbers) and with $a,e\not = 0$ so that if in the function $f(x) = \tfrac{ax^3+bx^2+cx+d}{ex^3+fx^2+gx+h}$ we replace $x$ by $\tfrac{ - x - \sqrt{3}}{\sqrt{3}x-1}$ then we end up with the same function, i.e. $f(x) = f\left( \tfrac{-x-\sqrt{3}}{\sqrt{3}x-1} \right)$.
• Jun 10th 2008, 08:27 AM
mathceleb
Take a look at this. I think I understand your question. I'm holding x as 1 for both sides. I take the value of the 2nd function using x=1, and then plug that in as the x value to the main F(x). Do you want me to write something to loop through a-h to find your target values?
• Jun 10th 2008, 05:39 PM
ThePerfectHacker
Thank you, but like I said I am mathematically software challenged I do not even have Excel.
• Jun 10th 2008, 07:23 PM
Mathstud28
Quote:

Originally Posted by ThePerfectHacker
Thank you, but like I said I am mathematically software challenged I do not even have Excel.

It says
a=-1
b=-20
c=-30
d=-243
e=-2
f=6
h=8
• Jun 10th 2008, 09:03 PM
ThePerfectHacker
Quote:

Originally Posted by Mathstud28
It says
a=-1
b=-20
c=-30
d=-243
e=-2
f=6
h=8

Okay I will try that. Can you check that, if you have computer-algebra system on your computer. I am too lazy to.
• Jun 10th 2008, 09:11 PM
Mathstud28
Quote:

Originally Posted by ThePerfectHacker
Okay I will try that. Can you check that, if you have computer-algebra system on your computer. I am too lazy to.

I don't, I'm sorry (Crying)
• Jun 11th 2008, 07:37 AM
ThePerfectHacker
I just tried graphing this function and it does not work.
I used $g=0$, because you never told me what value is, I assumed it was zero.
• Jun 11th 2008, 07:51 AM
mathceleb
Sorry gentleman, I should have described better. The numbers on that file are starting points, not the answer.

My question to you was do you want me to set x = 1 for both equations. After that, my thought was run my program to come up with a combination for each to match equations.

If that's what you want, I'll program that.
• Jun 11th 2008, 07:59 AM
ThePerfectHacker
Quote:

Originally Posted by mathceleb
If that's what you want, I'll program that.

I am looking for coefficients that when I substitute the linear rational function and after simplifying denominator and all that stuff I end up with the same expression as I started. So it seems this is a linear algebra problem with 8 variables.
• Jun 15th 2008, 08:33 PM
mathceleb
Quote:

Originally Posted by ThePerfectHacker
I am looking for coefficients that when I substitute the linear rational function and after simplifying denominator and all that stuff I end up with the same expression as I started. So it seems this is a linear algebra problem with 8 variables.

Ok, that makes sense. Let me take a look and see if I can code this for you. My "day job" is really busy this week, so I'll see what I can do this week after hours.
• Jun 15th 2008, 09:23 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Thank you, but like I said I am mathematically software challenged I do not even have Excel.

Here is the screen capture. As we see with these coefficients the right side does not equal the right side, they are only an example.

RonL
• Jun 16th 2008, 07:25 AM
ThePerfectHacker
Thank you for the function, but the problem got solved a few days ago. I should have posted on this thread the answer but I forgot to.
• Jun 16th 2008, 07:30 AM
mathceleb
Quote:

Originally Posted by ThePerfectHacker
Thank you for the function, but the problem got solved a few days ago. I should have posted on this thread the answer but I forgot to.

Oh, great news! I'll cease work on this then.