Exercise 2.1.15

Answers

We actually have

T (\sum_{i = 0}^{n} a_{i} x^{i}) = \sum_{i = 0}^{n} \frac{a_{i}}{i + 1} x^{i + 1} .

Hence by detailed check we know it’s one-to-one. But it’s not onto since no function have integral $= 1$ .

jephianlin

2011-06-27 00:00