How to find 4th degree polynomial equation from given points?

Question

I am trying to find 4th degree polynomial equation from given points. I do not own a graphing calculator so this task is very difficult for me to solve. So far I would out what points I need. The points are $$(-2, 3), (-8, -5), (-11.5, -1), (3, -5), (9.5, 2), (-10, -3), (-5, -3), (1, -3), (5.5, -3)$$ I know that the format of the equation should be $P(x)=ax^4+bx^3+cx^2+dx+e$. I tried to find the way to get the equation but so far all of them require a calculator. I really need help with this problem. Any help would be appreciated.

Answer 1

Lagrange Interpolation is the technique most commonly used for this purpose. Just take 5 points and plug them into Lagrange's formula (which I think is best articulated in the Examples section of the Wiki page). Assuming all of your points do indeed lie on a polynomial of degree 4, you will get that polynomial via Lagrange interpolation.

Answer 2

Before doing anything, I felt that there is a serious problem with the data for which you want to find a fourth degree polynomial supposed to fit them. $$\left( \begin{array}{cc} x & y & \text{note} \\ -2. & +3. \\ -8. & -5. & * \\ -11.5 & -1. \\ 3. & -5. & *\\ 9.5 & +2. \\ -10. & -3. & **\\ -5. & -3. & **\\ 1. & -3. & **\\ 5.5 & -3. & ** \end{array} \right)$$

In order to go (more or less) through the data points, you would need a much higher degree.

For sure, since there are $9$ data points, a polynomial of degree $8$ will make a perfect fit but any lower degree will do a quite poor job.

In any manner, the problem has to be treated using multilinear regression.

Using a fourth degree polynomial, the predicted values would be $$\left( \begin{array}{cc} x & y & y_{calc} \\ -2. & +3. & -0.25\\ -8. & -5. & -4.20 \\ -11.5 & -1. & -0.85\\ 3. & -5. & -3.16\\ 9.5 & +2. & +2.31\\ -10. & -3. & -3.75\\ -5. & -3. & -1.88\\ 1. & -3. & -1.33\\ 5.5 & -3. & -4.90 \end{array} \right)$$

Answer 3

What you need is a linear least-squares fit. Don't be deceived by the name, "linear" means linearity in the polynomial coefficients $a,b,c,d,e$, not linearity in $x$. If you insist on solving it by hand, you'll need to solve a $5\times 5$ set of equations, which is time-consuming but not impossible.

Answer 4

@Alex : There is no need for a calculator to draw 9 points on a graph.

You wrote : I know that the format of the equation should be $P(x)=ax^4+bx^3+cx^2+dx+e$.

Well, from what do you know this ? Is it in the wording of your exercise ? You should have given more information. Without one cannot show you the better way to solve your problem.

If it is asked for finding a good but approximate fit of the fourth degree polynomial curve to the points, the regression method is recommended. It can be computed without calculator, but this would be very tiresome. I will not discuss more about the methods to apply because several answers were already given on this subject.

If it is asked for finding a fourth polynomial curve passing exactly on the 9 points, it is impossible. It is obvious just looking at the graph. Nevertheless, a proof is shown below :

We see that four points have the same value $y=-3$. Changing of function $Y(x)=y(x)+3$ shows that the four points are at the four roots of the function $Y(x)=c(x+10)(x+5)(x-1)(x-5.5)$.

So, the four points are exactly on the curve
$$y(x)=-3+c(x+10)(x+5)(x-1)(x-5.5)$$
To make a fifth point $(x_5\:,\:y_5)$ exactly on the curve : $$c=\frac{y_5+3}{(x_5+10)(x_5+5)(x_5-1)(x_5-5.5)}$$ The equation of the fourth degree polynomial is : $$y(x)=-3+(y_5+3)\frac{(x+10)(x+5)(x-1)(x-5.5)}{(x_5+10)(x_5+5)(x_5-1)(x_5-5.5)} $$ The figure below shows the five cases : On each one, they are five points exactly on the curve and of course four remaining points far from the curve.