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Exercise 3.1.8

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If $(f + g) (x)$ is the usual $f (g (x))$ then $(g + f) x$ is $g (f (x))$ which is different. In Rule 2 both sides are $f (g (h (x))) .$ Rule 4 is broken because there might be no inverse function $f^{- 1} (x)$ such that $f (f^{- 1} (x)) = x$ . If the inverse function exists it will be the vector $- f$ .

mathchakra

2022-01-16 13:19

Exercise 3.1.8

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