Exercise 2.1

]> Exercise 2.1

Home \| 18.013A \| Chapter 2 \| Section 2.1		Tools Glossary Index Up Previous Next

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} + \dots

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

\frac{d \exp x}{d x} = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + \dots

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 $a_{1} = 1, a_{2} = \frac{1}{2}$ and in general $a_{j} = \frac{1}{j!}$ . We conclude that the series for $\exp x$ is the sum from $j = 0$ to infinity of $\frac{x^{j}}{j!}$ , which we write as

\exp x = \sum_{j = 0}^{\infty} \frac{x^{j}}{j!}