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Exercise 2.5

State and prove these fundamental properties ie, expressions for exp ⁡ ( x + r ) and for exp ⁡ r x . (Hint: what value do they have at x = 0 ? What are their derivatives? Deduce their series from these statements and identify them.)

Solution:

exp ⁡ ( x + r ) has value exp ⁡ r at argument 0 and is its own derivative by the chain rule. By the logic used to get the series for exp ⁡ x we obtain: a 0 = exp ⁡ r , and the relation between the a 's is exactly as before. We can deduce that each a is exp ⁡ r multiplied by its value in the previous case, which gives us

exp ⁡ ( x + r ) = ( exp ⁡ x ) * ( exp ⁡ r )

Notice that exp ⁡ r x has value 1 when x = 0 , and has derivative r exp ⁡ r x . Also notice that, for values of r for which it is defined, ( exp ⁡ x ) r has the same value at x = 0 and the same derivative (use the power rule and the chain rule). (We conclude that these functions are the same thing: their difference is 0 and has 0 derivative everywhere, which means it never changes from 0)

So we have

exp ⁡ r x = ( exp ⁡ x ) r

for all values of r for which we have a definition for the latter expression. This allows us to define the second expression to be the first everywhere else.

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