Maclaurin Series for the Exponential Function

The Maclaurin series expansion of the exponential function is: $$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!} + o(x^n) $$

Proof

Let’s start from the Maclaurin series formula:

$$ f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!} \cdot x^k + o(x^n) $$

In this case, f(x) = e^x.

For k = 0

Since f⁽⁰⁾(x) = f(x) = e^x:

$$ \frac{f^{(0)}(0)}{0!} \cdot x^0 = \frac{e^0}{0!} \cdot x^0 = \frac{1}{1} \cdot 1 = 1 $$

Thus, the first term of the Maclaurin series is:

$$ e^x = 1 + o(x^0) $$

$$ e^x = 1 + o(1) $$

For k = 1

The first derivative of e^x is again e^x:

$$ \frac{f^{(1)}(0)}{1!} \cdot x^1 = \frac{D[e^0]}{1!} \cdot x = \frac{e^0}{1} \cdot x = x $$

So the Maclaurin series up to the first-order term is:

$$ e^x = 1 + x + o(x^1) $$

For k = 2

The second derivative of e^x is still e^x:

$$ \frac{f^{(2)}(0)}{2!} \cdot x^2 = \frac{D^{(2)}e^0}{2!} \cdot x^2 = \frac{1}{2!} \cdot x^2 = \frac{x^2}{2} $$

Therefore, the Maclaurin series up to the second-order term is:

$$ e^x = 1 + x + \frac{x^2}{2} + o(x^2) $$

For k = 3

The third derivative of e^x remains e^x:

$$ \frac{f^{(3)}(0)}{3!} \cdot x^3 = \frac{D^{(3)}e^0}{3!} \cdot x^3 = \frac{x^3}{3!} $$

Hence, the Maclaurin series up to the third-order term is:

$$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + o(x^3) $$

For k = 4

The fourth derivative of e^x is again e^x:

$$ \frac{f^{(4)}(0)}{4!} \cdot x^4 = \frac{D^{(4)}e^0}{4!} \cdot x^4 = \frac{x^4}{4!} $$

Thus, the Maclaurin series up to the fourth-order term is:

$$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + o(x^4) $$

For k = 5

The fifth derivative of e^x is still e^x:

$$ \frac{f^{(5)}(0)}{5!} \cdot x^5 = \frac{D^{(5)}e^0}{5!} \cdot x^5 = \frac{x^5}{5!} $$

Hence, the Maclaurin series up to the fifth-order term is:

$$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + o(x^5) $$

Graphically, this looks like:

And so forth.