Question

(4 points) Let x

0
​
=−α;x
1
​
=0; and x
2
​
=α. Compute the interpolation polynomial for f(x)=−x
3
using Lagrange Interpolation: p
n
​
(x)=∑
i=0
n
​
f
i
​
l
i
​
(x)=∑
i=0
n
​
(f(x
i
​
)∏
k=0
k

=i
​

n
​

x
i
​
−x
k
​

x−x
k
​

​
)

Answer 1

Interpolation polynomial for
`f(x) = -x^3 is p(x) = -2α^2x.`.To compute the interpolation polynomial for f(x) = -
`x^3`using Lagrange Interpolation, we need to find the polynomial that passes through the given points (x0, f(x0)), (x1, f(x1)), and (x2, f(x2)).

Given x0 = -α, x1 = 0, and x2 = α, we can substitute these values into the Lagrange Interpolation formula:
p(x) = f(x0) * L0(x) + f(x1) * L1(x) + f(x2) * L2(x)
where L0(x), L1(x), and L2(x) are the Lagrange basis polynomials.
L0(x) = (x - x1) * (x - x2) / (x0 - x1) * (x0 - x2)
= (x - 0) * (x - α) / (-α - 0) * (-α - α)
=
`x * (x - α) / (2α^2)`
L1(x) = (x - x0) * (x - x2) / (x1 - x0) * (x1 - x2)
= (x - (-α)) * (x - α) / (0 - (-α)) * (0 - α)
=
`(x + α) * (x - α) / (2α^2)`
L2(x) = (x - x0) * (x - x1) / (x2 - x0) * (x2 - x1)
= (x - (-α)) * (x - 0) / (α - (-α)) * (α - 0)
=
`x * (x + α) / (2α^2)`
Substituting these values into the interpolation polynomial formula, we have:
p(x) = f(x0) * L0(x) + f(x1) * L1(x) + f(x2) * L2(x)
=
`(-x0^3) * L0(x) + (-x1^3) * L1(x) + (-x2^3) * L2(x)`
=
`α^3 * L0(x) + 0 * L1(x) + (-α)^3 * L2(x)`
=
`α^3 * (x * (x - α) / (2α^2)) + (-α)^3 * (x * (x + α) / (2α^2))`
=
`α^3 * (x^2 - αx) / (2α^2) + (-α)^3 * (x^2 + αx) / (2α^2)`
=
`α * (x^2 - αx) + (-α) * (x^2 + αx)`
=
`αx^2 - α^2x - αx^2 - α^2x`
=
`-2α^2x`
Therefore, the interpolation polynomial for
`f(x) = -x^3 is p(x) = -2α^2x.`