The equation of the polynomial P3 is p3(t) = 5/6t³− 17/6t
Equation:
In algebra, an equation consists of variable, number and constants.
Given,
Here we need to find the polynomial P3 such that {P0, P1, P2, P3} is an orthogonal basis for the subspace P3 of P4.
And the scale the polynomial p3 so that its vector of values is (-1,2,0, -2,1).
Here we know that the polynomial (before the scaling) p3 is just the difference between t³ and its orthogonal projection on the span of 1,t,t²−2.
Then we showed that the projection is , hence
p3(t) = t³− 17/5t
When here we have the scale this polynomial such that the vector of values at t=−2,−1,0,1,2 is (−1,2,0,−2,1).
Now, we need to find the scalar α, it can be written as,
=> α⋅p3(−2)=−1
=> α⋅p3(−1)=2
=> α⋅p3(0)=0
=> α⋅p3(1)=−2
=> α⋅p3(2)=1
Here from the 4-th equation we have the value of
=> α⋅(1- 17/5) = -2
=> α = 5/6
And it is easy to check that all the equations are satisfied with this α.
Therefore, the required polynomial p3 is
=> p3(t) = 5/6(t³− 17/5t)
=> p3(t) = 5/6t³− 17/6t
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