Question

Polynomial interpolation using linear systems?? Suppose we wish to find a quadratic polynomial of the form y=a_2(x^2) + a_1(x) + a_0, where a_2 a_1 and a_0 are scalars that passes through the points (-1,1),(2,4) and (p,1). Here p is a constant. Write the augmented matrix?

Answer 1

Oay I don't get how you can find a_0

Answer 2

thinking. joe?

Answer 3

ok not thinking. looking it up

Answer 4

why yes! when i see "polynomial interpolation" i think this:
\[f(-1)\frac{(x-2)(x-p)}{-1-2)(-1-p)}+f(2)\frac{(x+1)(x-p)}{2+1)(2-p)}+f(p)\frac{(x+1)(x-2)}{(p+1)(p-2)}\]

Answer 5

oops forgot a left parenthesis there twice >.>

Answer 6

find the least squares line (quadratic)

Answer 7

http://en.wikipedia.org/wiki/Polynomial_interpolation

Answer 8

looks like you set up a matrix equation right? vandermonde?

Answer 9

set up the standard matrix and use the normal equation

\[(A^TA)^{-1}A^Tb\]

Answer 10

in my book it says [1 x_1 x_1^2.... x_1^(n-1);1 x_2 x_2^2.... x_2^(n-1)] etc. and they used it for cubic polynomial

Answer 11

can I use the same matrix? but the augmented matrix doesn't have the first column being fully 1's

Answer 12

correct...that is the standard matrix that is needed

Answer 13

no but the answer for this question doesn't have a column of 1's

Answer 14

ok help me understand. because what i read had a column of 1's

Answer 15

\[A=\left[\begin{matrix}1 & -1 & 1 \\ 1 & 2 & 4\\ 1 & p & p^2\end{matrix}\right]\]

Answer 16

the answer is [1 -1 1 1; 4 2 1 4; p^2 p 1 1 ]?

Answer 17

same thing ...I am finding \[a_0+a_1x+a_2x^2\]

your matrix finds \[a_2x^2+a_1x+a_0\]

Answer 18

hmm...looks like you have some extra stuff there

Answer 19

an extra 1,4 and 1 which you don't need to find a quadratic polynomial

Answer 20

that just comes from the:
f(-1) = 1
f(2) = 4
f(p) = 1

right?

Answer 21

Oh would that come from the y

Answer 22

ohhhhhhhhh that make sense

Answer 23

I get it now thanks for your help! but one question does it really not matter where the columns are placed

Answer 24

like how the 1 column was the first and in this question it placed it in the third column

Answer 25

multiply \[\left[\begin{matrix}1 & -1 & 1 \\ 1 & 2 & 4\\ 1 & p & p^2\end{matrix}\right]\left[\begin{matrix}a_0 \\a_1\\ a_2 \end{matrix}\right]\]

that gives the form \[a_0+a_1x+a_2x^2\]

Answer 26

now compute ... \[(A^TA)^{-1}A^Tb\]

b being the vector consisting of the y-values

Answer 27

Okay my only question was why my answer key put it in a_2x^2 + a_1x + a_0

Answer 28

@joe

yes...but his matrix had it in twice which is not correct.

Answer 29

that way is fine..it is really the same form

Answer 30

No I understand what joe is saying, it make sense, okay thanks for your help everyone

Answer 31

I get

\[\left(\frac{2}{p-2}+2\right)x^2+\frac{p-1}{p-2}x-\frac{1}{p-2}\]

Answer 32

my other question is why is it infinitely solutions when p=-1

Answer 33

the matrix when reduced is [1 0 0 -1/(p-2); 0 1 0 (p-1)/(p-2); 0 0 1 2(p-1)/(p-2)]

Answer 34

if be was -1, then the first and third row of the matrix would be the same, and it would be singular.

Answer 35

same problem with 2

Answer 36

ohh I that what it means by infinitely many I thought it was when 1 =1 or something

Answer 37

so if two rows are the same then the solutions are infinite

Answer 38

then the matrix will not have full rank and therefore

\[A^TA\] will not be invertable

Answer 39

okay make sense, thanks again! I'm actually liking this stuff b/c it make sense now haha

Answer 40

cool

Answer 41

btw, that formula i put at the top also gives that solution you got by calculating:
\[(A^TA)^{-1}A^Tb\]
just thought i would point that out lol

Answer 42

after you simplify it and whatnot.

Answer 43

is that the Lagrange polynomial

Answer 44

i think it is..i caught a random glimpse of it in a book once. Let me see if i can find it....

Answer 45

I'm pretty sure it is...I'd be a little surprised if it always gave the least squares solution

Answer 46

yeah, its called Lagrange Interpolation in this book.

Answer 47

if we had more than 3 points then it would be different

Answer 48

i remember reading this section and not understanding a lot of it >.< it just came to mind because of the word "interpolation" lol >.<