]> Exercise 2.1
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## Exercise 2.1

Figure out the series for $exp ⁡ x$ ; and prove it  to be so.

Solution:

The power series expansion of $exp ⁡ x$ about 0 has the form

$exp ⁡ x = a 0 + a 1 x + a 2 x 2 + ⋯$

When x is near 0, $exp ⁡ x$ is near 1. This implies $a 0 = 1$ .

The derivative of $exp ⁡ x$ is itself and so is also near 1 when $x$ is near 0.

Differentiating the series we find

$d exp ⁡ x d x = a 1 + 2 a 2 x + 3 a 3 x 2 + ⋯$

This allows us to identify $a 1 = a 0 , 2 a 2 = a 1 , 3 a 3 = a 2$ , from the fact that the coefficients of each power of $x$ must be the same here and in the previous expression for $exp ⁡ x$ and in general $j a j = a j − 1$ .

This allows us to identify $a 1 = 1 , a 2 = 1 2$ and in general $a j = 1 j !$ . We conclude that the series for $exp ⁡ x$ is the sum from $j = 0$ to infinity of $x j j !$ , which we write as

$exp ⁡ x = ∑ j = 0 ∞ x j j !$
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