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Exercise 3.1.7

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Rule 8 is broken: If $c f (x)$ is defined to be the usual $f (cx)$ then $(c_{1} + c_{2}) f =$ $f ((c_{1} + c_{2}) x)$ is not generally the same as $c_{1} f + c_{2} f = f (c_{1} x) + f (c_{2} x)$

mathchakra

2022-01-16 13:18

Exercise 3.1.7

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