## Exercise 2.5

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) =Ā (exp x) * (exp r)

Notice that exp rx has value 1 when x = 0, and has derivative r exp rx. 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 rx = (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.