We'll find the interpolating polynomial passing through the three points ( 1 , − 6 ) {\displaystyle (1,-6)} , ( 2 , 2 ) {\displaystyle (2,2)} , ( 4 , 12 ) {\displaystyle (4,12)} , using the Vandermonde matrix .
For our polynomial, we'll take ( 1 , − 6 ) = ( x 0 , y 0 ) {\displaystyle (1,-6)=(x_{0},y_{0})} , ( 2 , 2 ) = ( x 1 , y 1 ) {\displaystyle (2,2)=(x_{1},y_{1})} , and ( 4 , 12 ) = ( x 2 , y 2 ) {\displaystyle (4,12)=(x_{2},y_{2})} .
Since we have 3 points, we can expect degree 2 polynomial.
So define our interpolating polynomial as:
p ( x ) = a 2 x 2 + a 1 x + a 0 {\displaystyle p(x)=a_{2}x^{2}+a_{1}x+a_{0}} .
So, to find the coefficients of our polynomial, we solve the system p ( x i ) = y i {\displaystyle p(x_{i})=y_{i}} , i ∈ { 0 , 1 , 2 } {\displaystyle i\in \{0,1,2\}} .
( x 0 2 x 0 1 x 1 2 x 1 1 x 2 2 x 2 1 ) ∗ ( a 2 a 1 a 0 ) = ( y 0 y 1 y 2 ) {\displaystyle \left({\begin{array}{ccc}x_{0}^{2}&x_{0}&1\\x_{1}^{2}&x_{1}&1\\x_{2}^{2}&x_{2}&1\\\end{array}}\right)*\left({\begin{array}{c}a_{2}\\a_{1}\\a_{0}\end{array}}\right)=\left({\begin{array}{c}y_{0}\\y_{1}\\y_{2}\end{array}}\right)} In order to solve the system, we will use an augmented matrix based on the Vandermonde matrix, and solve for the coefficients using Gaussian elimination . Substituting in our x {\displaystyle x} and y {\displaystyle y} values, our augmented matrix is:
( 1 2 1 1 − 6 2 2 2 1 2 4 2 4 1 12 ) {\displaystyle \left({\begin{array}{ccc|c}1^{2}&1&1&-6\\2^{2}&2&1&2\\4^{2}&4&1&12\end{array}}\right)}
Then, using Gaussian elimination,
( 1 1 1 − 6 4 2 1 2 16 4 1 12 ) ⇒ ( 1 1 1 − 6 0 − 2 − 3 26 0 − 12 − 15 108 ) ⇒ ( 1 1 1 − 6 0 − 2 − 3 26 0 0 3 − 48 ) ⇒ ( 1 1 0 10 0 2 0 − 22 0 0 1 − 16 ) ⇒ ( 1 0 0 − 1 0 1 0 11 0 0 1 − 16 ) {\displaystyle \left({\begin{array}{ccc|c}1&1&1&-6\\4&2&1&2\\16&4&1&12\end{array}}\right)\Rightarrow \left({\begin{array}{ccc|c}1&1&1&-6\\0&-2&-3&26\\0&-12&-15&108\end{array}}\right)\Rightarrow \left({\begin{array}{ccc|c}1&1&1&-6\\0&-2&-3&26\\0&0&3&-48\end{array}}\right)\Rightarrow \left({\begin{array}{ccc|c}1&1&0&10\\0&2&0&-22\\0&0&1&-16\end{array}}\right)\Rightarrow \left({\begin{array}{ccc|c}1&0&0&-1\\0&1&0&11\\0&0&1&-16\end{array}}\right)}
Our coefficients are a 2 = − 1 {\displaystyle a_{2}=-1} , a 1 = 11 {\displaystyle a_{1}=11} , and a 0 = − 16 {\displaystyle a_{0}=-16} . So, the interpolating polynomial is
p ( x ) = − x 2 + 11 x − 16 {\displaystyle p(x)=-x^{2}+11x-16} .
Adding a point
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Now we add a point, ( 3 , − 10 ) = ( x 3 , y 3 ) {\displaystyle (3,-10)=(x_{3},y_{3})} , to our data set and find a new interpolation polynomial with this method.
Since we have 4 points, we will have degree 3 polynomial.
Thus our polynomial is p ( x ) = a 3 x 3 + a 2 x 2 + a 1 x + a 0 {\displaystyle p(x)=a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}} ,
and we get the coefficients by solving the system p ( x i ) = y i {\displaystyle p(x_{i})=y_{i}} .
Constructing our augmented matrix as before and using Gaussian elimination, we get:
( 1 3 1 2 1 1 − 6 2 3 2 2 2 1 2 4 3 4 2 4 1 12 3 3 3 2 3 1 − 10 ) ⇒ ( 1 1 1 1 − 6 0 − 4 − 6 − 7 50 0 − 48 − 60 − 63 396 0 − 18 − 24 − 26 152 ) ⇒ ( 1 1 1 1 − 6 0 − 4 − 6 − 7 50 0 0 12 21 − 204 0 0 3 11 12 − 73 ) {\displaystyle \left({\begin{array}{cccc|c}1^{3}&1^{2}&1&1&-6\\2^{3}&2^{2}&2&1&2\\4^{3}&4^{2}&4&1&12\\3^{3}&3^{2}&3&1&-10\end{array}}\right)\Rightarrow \left({\begin{array}{cccc|c}1&1&1&1&-6\\0&-4&-6&-7&50\\0&-48&-60&-63&396\\0&-18&-24&-26&152\end{array}}\right)\Rightarrow \left({\begin{array}{cccc|c}1&1&1&1&-6\\0&-4&-6&-7&50\\0&0&12&21&-204\\0&0&3&{\frac {11}{12}}&-73\end{array}}\right)}
⇒ ( 1 1 1 1 − 6 0 − 4 − 6 − 7 50 0 0 12 21 − 204 0 0 0 1 4 − 22 ) ⇒ ( 1 1 1 0 82 0 − 4 − 6 0 − 566 0 0 12 0 1644 0 0 0 1 − 88 ) ⇒ ( 1 1 0 0 − 55 0 − 4 0 0 256 0 0 1 0 137 0 0 0 1 − 88 ) {\displaystyle \Rightarrow \left({\begin{array}{cccc|c}1&1&1&1&-6\\0&-4&-6&-7&50\\0&0&12&21&-204\\0&0&0&{\frac {1}{4}}&-22\end{array}}\right)\Rightarrow \left({\begin{array}{cccc|c}1&1&1&0&82\\0&-4&-6&0&-566\\0&0&12&0&1644\\0&0&0&1&-88\end{array}}\right)\Rightarrow \left({\begin{array}{cccc|c}1&1&0&0&-55\\0&-4&0&0&256\\0&0&1&0&137\\0&0&0&1&-88\end{array}}\right)}
⇒ ( 1 0 0 0 9 0 1 0 0 − 64 0 0 1 0 137 0 0 0 1 − 88 ) {\displaystyle \Rightarrow \left({\begin{array}{cccc|c}1&0&0&0&9\\0&1&0&0&-64\\0&0&1&0&137\\0&0&0&1&-88\end{array}}\right)}
Therefore, our polynomial is:
p ( x ) = 9 x 3 − 64 x 2 + 137 x − 88 {\displaystyle p(x)=9x^{3}-64x^{2}+137x-88} .