Maclaurin Series Formula

A Maclaurin series is a function that has expansion series that gives the sum of derivatives of that function. The Maclaurin series of a function

\begin{array}{l} f (x) \end{array}

up to order n may be found using Series

\begin{array}{l} [f, x, 0, n] \end{array}

It is a special case of Taylor series when x = 0. The Maclaurin series is given by

$f (x) = f (x_{0}) + f^{'} (x_{0}) (x - x_{0}) + \frac{f ” (x_{0})}{2!} (x - x_{0})^{2} + \frac{f ”^{'} (x_{0})}{3!} (x - x_{0})^{3} + \dots . .$

The Maclaurin series formula is

$f (x) = \sum_{n = 0}^{\infty} \frac{f^{n} (x_{0})}{n!} (x - x_{0})$

Where,
f(x_o), f’(x_o), f’‘(x_o)……. are the successive differentials when x_o = 0.

Function	Maclaurin Series
$\begin{array}{l} e^{x} \end{array}$	$\begin{array}{l} \sum_{k = 0}^{\infty} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \dots . . \end{array}$
$\begin{array}{l} s i n x \end{array}$	$\begin{array}{l} \sum_{k = 0}^{\infty} (- 1)^{2} = \frac{x^{2 k + 1}}{(2 k + 1)!} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \frac{x^{7}}{7!} + \dots . . \end{array}$
$\begin{array}{l} c o s x \end{array}$	$\begin{array}{l} \cos x = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} x^{2 n}}{(2 n)!} = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \dots \end{array}$
$\begin{array}{l} \frac{1}{1 - x} \end{array}$	$\begin{array}{l} \sum_{k = 0}^{\infty} x^{k} = 1 + x + x^{2} + x^{3} + \dots . (i f - 1 < x < 1) \end{array}$
$\begin{array}{l} l n (1 + x) \end{array}$	$\begin{array}{l} \ln (1 + x) = \sum_{n = 1}^{\infty} (- 1)^{n + 1} \frac{x^{n}}{n} = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \dots \end{array}$

Question 1: Expanding

\begin{array}{l} e^{x} \end{array}

: Find the Maclaurin Series expansion of

\begin{array}{l} f (x) = e^{x} \end{array}

Solution:

Recalling that the derivative of the exponential function is

\begin{array}{l} f^{'} (x) = e^{x} \end{array}

In fact, all the derivatives are

\begin{array}{l} e^{x} \end{array}

\begin{array}{l} f^{'} (0) = e^{0} = 1 \end{array}

\begin{array}{l} f ” (0) = e^{0} = 1 \end{array}

\begin{array}{l} f ”^{'} (0) = e^{0} = 1 \end{array}

We see that all the derivatives, when evaluated at x = 0, give us the value 1.

Also, f(0)=1, so we can conclude the Maclaurin Series expansion will be simply:

\begin{array}{l} e^{x} \approx 1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \frac{1}{24} x^{4} + \frac{1}{120} x^{5} + \dots . \end{array}

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