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A3: This question illustrates how different bases for spaces of polynomials can help solv- ing mathematical problems. In particular, we look at the use of Lagrange polynomials for polynomial interpolation. Let be the space of polynomials of degree at most two. (a) We define the mapping T: P2R3 by evaluating a given polynomial f i.e P2 at 12,, T(f) = f(2) f(3) Show that this is a linear transformation. (b) Consider the bases B b, b2, bs1,t, and G9929s), where ?20?3)(t-2)(t-3) 933 1)-2) (t - 1)(t - 2) For a polynomial f e P2 we can consider its coordinate vectors [ls relative to B and f relative to G. Determine matrices A, B such that, for any f E P2, we have (c) Choosing a basis on P2 allows us to represent an element f E P2 as a vector in R3, by looking at the coordinate vector MB. This gives us a linear transformation R3-+ P. by considering the mapping [fls + T(f). It follows that there is a matrix Mg such that, for all f E P2, we have Determine Ms and Mg (d) If you are given values a,o2.c R and are asked to compute a polynomial f of degree at most 2 such that f(1f(2)2.f (3)-c, describe in terms of the mapping T from (a) the problem in terms of vector equations. Which basis would you use to solve the problem? B or G?

Answer 1

given P₂={c₀+c₁t+c₂t²: c₀,c₁,c₂ $\in$ R²} be the space of polynomials of degree atmost 2 }

then the transformation is given by T(f)=(f₁,f₂,f₃)

part(a) prove that it is the linear transformation

solution:- take f and g $\in$ P₂ and $\alpha \in$ R

then T( $\alpha$ f+g)=( $\alpha$ f₁+g₁, $\alpha$ f₂+g₂, $\alpha$ f₃+g₃)

=( $\alpha$ f₁, $\alpha$ f₂, $\alpha$ f₃)+(g₁,g₂,g₃)

= $\alpha$ (f₁,f₂,f₃)+(g₁,g₂,g₃)

= $\alpha$ T(f)+T(g)

hence proved that it is a linear transformation