P4 rightarrow P3 be the linear transformation defined by T (p (x)) = d/dx p{x) for all p (x) P4. What is K (T)? Let q (x) = 2 + 3x - 2a:2 - x3. Find one solution, any solution p (x) P4 to T (p (x)) = q (x). (As an aside, your method of solution probably can be generalized to show that T is surjective, but you don\'t have to do that. ) What is T-1 (q (x))?
Solution
a) K(T) = {p(x)| T(p(x))=0} = {p(x)| p\'(x) = 0} = {p(x)| p(x) = constant} = P 0
b)T(p(x)) = q(x)
p\'(x) = 2 + 3x - 2x 2 - x 3
p(x) = 2 + 3x - 2x 2 - x 3 dx = C + 2x + (3/2)x 2 - (2/3)x 3 - (1/4)x 4
a particular would be C = 0: p(x) = 2x + (3/2)x 2 - (2/3)x 3 - (1/4)x 4
to show that T is surjective is to note that for any polynomial in P 3 you can always integrate it to find a polynomial in P 4 . Note because of the constant of integration T is not injective.
c) T(p(x)) = q(x)
p(x) = T -1 (q(x)) = C + 2x + (3/2)x 2 - (2/3)x 3 - (1/4)x 4
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P4 rightarrow P3 be the linear transformation defined by T (p (x)) - d.docx
1. P4 rightarrow P3 be the linear transformation defined by T (p (x)) = d/dx p{x) for all p (x) P4.
What is K (T)? Let q (x) = 2 + 3x - 2a:2 - x3. Find one solution, any solution p (x) P4 to T (p (x))
= q (x). (As an aside, your method of solution probably can be generalized to show that T is
surjective, but you don't have to do that. ) What is T-1 (q (x))?
Solution
a) K(T) = {p(x)| T(p(x))=0} = {p(x)| p'(x) = 0} = {p(x)| p(x) = constant} = P 0
b)T(p(x)) = q(x)
p'(x) = 2 + 3x - 2x 2
- x 3
p(x) = 2 + 3x - 2x 2
- x 3
dx = C + 2x + (3/2)x 2
- (2/3)x 3
- (1/4)x 4
a particular would be C = 0: p(x) = 2x + (3/2)x 2
- (2/3)x 3
- (1/4)x 4
to show that T is surjective is to note that for any polynomial in P 3 you can always integrate it to
find a polynomial in P 4 . Note because of the constant of integration T is not injective.
c) T(p(x)) = q(x)
p(x) = T -1
(q(x)) = C + 2x + (3/2)x 2
- (2/3)x 3
- (1/4)x 4
