5) The Lagrange's polynomial for the function f (x) when f(0) = 4,f(1) = 3,f(2)= 6 is
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Your data pairs (xi,fi), 0≤i≤3 are (−1,−6),(0,2),(1,−2) and (2,6). The 3-degree polynomial P3(x) that you would get using Lagrange's interpolation is
P3(x)=∑i=03fiLi(x)
where
Li(x)=∏j=0j≠i3x−xjxi−xj
are the basis polynomials. For example,
L0(x)=x(x−1)(x−2)(−1)(−1−1)(−1−2).
Similarly find L1(x),L2(x),L3(x) and then calculate P3(x).
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answered Apr 9 '17 at 12:37
Prajwal Kansakar
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I would like to propose an alternative way to solve the problem. Coefficients of Lagrange interpolation polynomial can be found if one uses a determinant form of Lagrange interpolation presented in "Beginner's guide to mapping simplexes affinely", section "Lagrange interpolation" (you may check for concrete example in "Workbook on mapping simplexes affinely"). General formula looks as follows
f(x)=(−1)det⎛⎝⎜⎜⎜⎜⎜⎜0xnxn−1⋯1f0xn0xn−10⋯1f1xn1xn−11⋯1⋯⋯⋯⋯⋯fnxnnxn−1n⋯1⎞⎠⎟⎟⎟⎟⎟⎟det⎛⎝⎜⎜⎜⎜xn0xn−10⋯1xn1xn−11⋯1⋯⋯⋯⋯xnnxn−1n⋯1⎞⎠⎟⎟⎟⎟.
EXAMPLE
Let's consider your case: f(−1)=−6, f(0)=2, and f(2)=6. Previous equation should be written as
f(x)=(−1)det⎛⎝⎜⎜⎜0x2x1−6(−1)2−112020162221⎞⎠⎟⎟⎟det⎛⎝⎜(−1)2−1102012221⎞⎠⎟=2+6x−2x2.
Now it is easy to check that
f(−1)f(0)f(2)=2−6−2(−1)2=−6,=2+0−0=2,=2+12−2(2)2=61
Your data pairs (xi,fi), 0≤i≤3 are (−1,−6),(0,2),(1,−2) and (2,6). The 3-degree polynomial P3(x) that you would get using Lagrange's interpolation is
P3(x)=∑i=03fiLi(x)
where
Li(x)=∏j=0j≠i3x−xjxi−xj
are the basis polynomials. For example,
L0(x)=x(x−1)(x−2)(−1)(−1−1)(−1−2).
Similarly find L1(x),L2(x),L3(x) and then calculate P3(x).
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answered Apr 9 '17 at 12:37
Prajwal Kansakar
5,12822 gold badges1717 silver badges2828 bronze badges
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I would like to propose an alternative way to solve the problem. Coefficients of Lagrange interpolation polynomial can be found if one uses a determinant form of Lagrange interpolation presented in "Beginner's guide to mapping simplexes affinely", section "Lagrange interpolation" (you may check for concrete example in "Workbook on mapping simplexes affinely"). General formula looks as follows
f(x)=(−1)det⎛⎝⎜⎜⎜⎜⎜⎜0xnxn−1⋯1f0xn0xn−10⋯1f1xn1xn−11⋯1⋯⋯⋯⋯⋯fnxnnxn−1n⋯1⎞⎠⎟⎟⎟⎟⎟⎟det⎛⎝⎜⎜⎜⎜xn0xn−10⋯1xn1xn−11⋯1⋯⋯⋯⋯xnnxn−1n⋯1⎞⎠⎟⎟⎟⎟.
EXAMPLE
Let's consider your case: f(−1)=−6, f(0)=2, and f(2)=6. Previous equation should be written as
f(x)=(−1)det⎛⎝⎜⎜⎜0x2x1−6(−1)2−112020162221⎞⎠⎟⎟⎟det⎛⎝⎜(−1)2−1102012221⎞⎠⎟=2+6x−2x2.
Now it is easy to check that
f(−1)f(0)f(2)=2−6−2(−1)2=−6,=2+0−0=2,=2+12−2(2)2=6
Step-by-step explanation: