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You have four points, so for them to fit the polynomial exactly, you need it to at least have degree three. Anything more you'll have an infinite number of possibilities that will have all data points fit, and anything less then chances are you'll only be able to get a least squares approximation.If p(x) is a polynomial such that p(0)=5 ,p(1)=4 ,p(2)=9,p(3)=20 ,

the minimum degree it can have

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- Jan 26, 2012

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My approach: start with a straight line and see if it fits exactly. If not, try a quadratic. If that doesn't work, try a cubic. As Prove It has pointed out, a cubic (with four arbitrary constants) will definitely work. But you might be able to get by with fewer depending on where the points are.

Hello, jacks!

If [tex]p(x)[/tex] is a polynomial such that: .[tex]p(0) = 5,\;p(1) = 4,\;p(2) = 9,\;p(3) = 20,[/tex]

. . the minimum degree it can have is __.

Plotting the four points, a parabola

The general parabola is: .[tex]p(x) \:=\:ax^2 + bx + c[/tex]

Use the four point to construct a system of equations:

[tex]\begin{array}{ccccccc} p(0) = 5: & a(0^2) + b(0) + c &=& 5 \\

p(1) = 4: & a(1^2) + b(1) + c &=& 4 \\

p(2) = 9: & a(2^2)+ b(2) + c &=& 9 \\

p(3) = 20: & a(3^2) + b(3) + c &=& 20 \end{array}[/tex]

Solve the system: .[tex]a = 3,\;b = \text{-}4,\;c = 5[/tex]

Hence: .[tex]p(x) \;=\;3x^2 - 4x + 5[/tex]

The minimum degree of [tex]p(x)[/tex] is

- Jan 29, 2012

- 1,151

Then polynomial, passing through four given points, will have degree atYou have four points, so for them to fit the polynomial exactly, you need it to at least have degree three. Anything more you'll have an infinite number of possibilities that will have all data points fit, and anything less then chances are you'll only be able to get a least squares approximation.

Really? I would have thought that there would be an infinite number of solutions to, say, four equations in five unknowns, which is what you would get if you substituted the four points into a general polynomial of degree 4...Then polynomial, passing through four given points, will have degree atmostthree, not "at least". It is quite possible that the four points happen to lie on a parabola (which is apparently the case here) or even on a straight line.

- Jan 26, 2012

- 890

Four equations in five unknowns is what you will end up with when you try to fit a quartic, and we know you can always do that but the solution is not unique.Really? I would have thought that there would be an infinite number of solutions to, say, four equations in five unknowns, which is what you would get if you substituted the four points into a general polynomial of degree 4...

Fitting a cubic \(p(x)=a+bx+cx^2+dx^3\) will give you four equations in four unknowns, and as long as there is no degeneracy will have a solution. In this case the equations are:

\[ \left[\begin{array}{cccc} 1&0&0&0 \\ 1&1&1&1 \\ 1 & 2 & 4 & 8 \\ 1&3&9&27 \end{array} \right] \left[ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right]=\left[ \begin{array}{c} 5 \\ 4 \\ 9 \\ 20 \end{array} \right]\]

Which may be solved using your favourite method of solving linear equations to give:

\[ \left[ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right]=\left[ \begin{array}{c} 5 \\ -4 \\ 3 \\ 0 \end{array} \right]\]

Which corresponds to the polynomial:

\[ p(x)=5-4x+3x^2+0x^3=5-4x+3x^2 \]

We may note that this method would produce the required solution whateve the degree of the ploynomial was. And it works because the fitting cubic is unique and all polynomials of lower degree are cubics for the purposes of fitting to the data.

CB

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