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Find a polynomial P3 such that {po, P1, P2, P3} (see Ex- ercise 11) is an orthogonal basis for the subspace P3 of P4. Scale the polynomial p3 so that its vector of values is (-1,2,0, -2,1).

Sagot :

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​

To know more about equation here.

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