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State and prove these fundamental properties ie, expressions for exp (x + r) and for exp rx. (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: a₀ = 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.