]> Exercise 2.5

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.