Exercise 3.F.14

Answers

Proof. _(a)_ Let $p \in P (ℝ)$ . Then

\begin{aligned} T^{'} (φ) (p) & = φ \circ T p \\ = φ (x^{2} p + p^{''}) \\ = (2 x p + x^{2} p^{'} + p^{''}) |_{x = 4} \\ = 8 p (4) + 16 p^{'} (4) + p^{'''} (4) \end{aligned}

_(b)_

\begin{aligned} (T^{'} (φ)) (x^{3}) & = φ \circ T (x^{3}) \\ = φ (x^{5} + 6 x) \\ = \int_{0}^{1} x^{5} + 6 x \\ , d x \\ = (\frac{x^{6}}{6} + 3 x^{2}) |_{0}^{1} \end{aligned}

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celiopassos

2017-10-06 00:00