]> 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.