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Exercise 2.1.16

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Similar to the previous exercise we have $T (\sum_{i = 0}^{n} a_{i} x^{i}) = \sum_{i = 0}^{n} i a_{i} x^{i - 1}$ . It’s onto since $T (\sum_{i = 0}^{n} \frac{a_{i}}{i + 1} x^{i + 1}) = \sum_{i = 0}^{n} a_{i} x^{i}$ . But it’s not one-to-one since $T (1) = T (2) = 0$ .

jephianlin

2011-06-27 00:00

Exercise 2.1.16

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