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Taylor Series and Maclaurin Series Explained: Derivations of Maclaurin Series of 10 Essential Functions

Taylor Series and Maclaurin Series

Table of Contents

Introduction

Calculus, the language of change, equips us with powerful tools to analyze and understand how things move, grow, and interact. Imagine a scenario where you need to determine the future position of a speeding car or calculate the area under a curved line. While these might seem daunting tasks, calculus offers a solution through the concept of functions. Functions represent relationships between quantities, and in some cases, these relationships can be quite intricate.

This is where Taylor series and Maclaurin series come in as heroes! These remarkable tools allow us to represent even the most complex functions as a sum of simpler terms – polynomials. Think of it as taking a challenging equation and breaking it down into a series of easy-to-understand building blocks. By harnessing the power of Taylor and Maclaurin series, we can not only gain deeper insights into a function’s behavior but also unlock efficient ways to approximate function values and solve challenging problems.

What are Taylor Series and Maclaurin Series?

Ever wondered how to represent a seemingly complex curve using a familiar and well-behaved object – a polynomial? Buckle up, because Taylor series and Maclaurin series are about to become your secret weapon!

Taylor Series

Let’s say you have a well-behaved function, \(f(x)\), and a specific point, \(a\), within its domain. The Taylor series of \(f(x)\) centered at a represents \(f(x)\) as an infinite sum of terms. These terms involve the function’s derivatives evaluated at \(a\), all scaled by appropriate factors.

Mathematically, the Taylor series looks like this:

\begin{align} f(x) &= f(a) + \frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n+\cdots \\[0.2cm] &=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n \end{align}

Here, \( f(a) \) represents the value of the function at the point \( a \), \( f'(a) \) represents the first derivative of \( f(x) \) evaluated at \( a \), \( f”(a) \) represents the second derivative, and so on. The term \( (x-a)^n \) are powers of \( x-a\) raised to increasing integer exponents, and \( n! \) deonotes the factorial of \( n \).

Key Point: The closer \(x\) is to the center point ‘\(a\)’, the better the approximation of \(f(x)\) by the terms included in the series (up to a certain number of terms). In essence, the Taylor series provides a local polynomial approximation of the function around the point \(a\).

Macluarin Series

The Maclaurin series is a specific type of Taylor series where the center point, \(a\), is zero (x = 0). In other words, it’s the Taylor series centered at zero.

\begin{align} f(x) &= f(0) + \frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\cdots+\frac{f^{(n)}(0)}{n!}x^n+\cdots \\[0.2cm] &=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n \end{align}

Importance: Maclaurin series are particularly useful because many fundamental mathematical functions (like sine, cosine, and exponential) have well-defined Maclaurin series expansions. These series often converge for a wide range of \(x\) values, making them powerful tools for various calculations.

In essence, Taylor and Maclaurin series offer a powerful way to break down complex functions into simpler polynomial building blocks, allowing us to estimate function values, analyze function behavior, and solve intricate problems in calculus.

Common Taylor/Maclaurin Series Expansions

\begin{align} 1. \quad e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots = \sum_{n=0}^{\infty} \frac{x^{n}}{n!} \end{align}
\begin{align*} & \text{To derive the Maclaurin series for } e^x, \text{ we start by finding the first few derivatives:} \\ & f(x) = e^x \\ & f'(x) = e^x \\ & f''(x) = e^x \\ & f'''(x) = e^x \\ & \\ & \text{Evaluating these derivatives at } x = 0: \\ & f(0) = e^0 = 1 \\ & f'(0) = e^0 = 1 \\ & f''(0) = e^0 = 1 \\ & f'''(0) = e^0 = 1 \\ & \\ & \text{Now, we can use the Maclaurin series formula:} \\ & e^x = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \\ & \\ & \text{Plugging in the values, we get:} \\ & e^x = 1 + \frac{1}{1!}x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \cdots \\ & \\ & \text{Simplified, this becomes:} \\ & e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \\ \end{align*}
\begin{align} 2. \quad \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \end{align}
\begin{align*} & \text{To derive the Maclaurin series for } \sin(x), \text{ let's start by finding the first five derivatives:} \\ & f(x) = \sin(x) \\ & f'(x) = \cos(x) \\ & f''(x) = -\sin(x) \\ & f'''(x) = -\cos(x) \\ & f^{(4)}(x) = \sin(x) \\ & f^{(5)}(x) = \cos(x) \\ & \\ & \text{Evaluating these derivatives at } x = 0: \\ & f(0) = \sin(0) = 0 \\ & f'(0) = \cos(0) = 1 \\ & f''(0) = -\sin(0) = 0 \\ & f'''(0) = -\cos(0) = -1 \\ & f^{(4)}(0) = \sin(0) = 0 \\ & f^{(5)}(0) = \cos(0) = 1 \\ & \\ & \text{Now, we can use the Maclaurin series formula:} \\ & \sin(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 + \cdots \\ & \\ & \text{Plugging in the values, we get:} \\ & \sin(x) = 0 + \frac{1}{1!}x + \frac{0}{2!}x^2 + \frac{-1}{3!}x^3 + \frac{0}{4!}x^4 + \frac{1}{5!}x^5 + \cdots \\ & \\ & \text{Simplified, this becomes:} \\ & \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \\ \end{align*}
\begin{align} 3. \quad \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \end{align}
\begin{align*} & \text{To derive the Maclaurin series for } \cos(x), \text{ let's start by finding the first five derivatives:} \\ & f(x) = \cos(x) \\ & f'(x) = -\sin(x) \\ & f''(x) = -\cos(x) \\ & f'''(x) = \sin(x) \\ & f^{(4)}(x) = \cos(x) \\ & f^{(5)}(x) = -\sin(x) \\ & \\ & \text{Evaluating these derivatives at } x = 0: \\ & f(0) = \cos(0) = 1 \\ & f'(0) = -\sin(0) = 0 \\ & f''(0) = -\cos(0) = -1 \\ & f'''(0) = \sin(0) = 0 \\ & f^{(4)}(0) = \cos(0) = 1 \\ & f^{(5)}(0) = -\sin(0) = 0 \\ & \\ & \text{Now, we can use the Maclaurin series formula:} \\ & \cos(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 + \cdots \\ & \\ & \text{Plugging in the values, we get:} \\ & \cos(x) = 1 + \frac{0}{1!}x + \frac{-1}{2!}x^2 + \frac{0}{3!}x^3 + \frac{1}{4!}x^4 + \frac{0}{5!}x^5 + \cdots \\ & \\ & \text{Simplified, this becomes:} \\ & \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \\ & \\ & \text{In summation notation:} \\ & \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \end{align*}
\begin{align} 4. \quad \tan(x) = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \cdots \end{align}
\begin{align*} & \text{To derive the Maclaurin series for } \tan(x), \text{ let's start by finding the first few derivatives:} \\ & f(x) = \tan(x) \\ & f'(x) = \sec^2(x) \\ & f''(x) = 2\sec^2(x) \tan(x) \\ & f'''(x) = 2\sec^4(x) + 4\sec^2(x) \tan^2(x) \\ & f^{(4)}(x) = 12\sec^4(x) \tan(x) + 8\sec^2(x) \tan^3(x) \\ & f^{(5)}(x) = 12\sec^6(x) + 36\sec^4(x) \tan^2(x) + 8\sec^2(x) \tan^4(x) \\ & \\ & \text{Evaluating these derivatives at } x = 0: \\ & f(0) = \tan(0) = 0 \\ & f'(0) = \sec^2(0) = 1 \\ & f''(0) = 0 \\ & f'''(0) = 2 \\ & f^{(4)}(0) = 0 \\ & f^{(5)}(0) = 12 \\ & \\ & \text{Now, we can use the Maclaurin series formula:} \\ & \tan(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 + \cdots \\ & \\ & \text{Plugging in the values, we get:} \\ & \tan(x) = 0 + \frac{1}{1!}x + 0x^2 + \frac{2}{3!}x^3 + 0x^4 + \frac{12}{5!}x^5 + \cdots \\ & \\ & \text{Simplified, this becomes:} \\ & \tan(x) = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \cdots \\ \end{align*}
\begin{align} 5. \quad e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n!} \end{align}
\begin{align*} & \text{To derive the Maclaurin series for } e^{-x}, \text{ let's start by finding the first few derivatives:} \\ & f(x) = e^{-x} \\ & f'(x) = -e^{-x} \\ & f''(x) = e^{-x} \\ & f'''(x) = -e^{-x} \\ & f^{(4)}(x) = e^{-x} \\ & \\ & \text{Evaluating these derivatives at } x = 0: \\ & f(0) = e^0 = 1 \\ & f'(0) = -e^0 = -1 \\ & f''(0) = e^0 = 1 \\ & f'''(0) = -e^0 = -1 \\ & f^{(4)}(0) = e^0 = 1 \\ & \\ & \text{Now, we can use the Maclaurin series formula:} \\ & e^{-x} = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \cdots \\ & \\ & \text{Plugging in the values, we get:} \\ & e^{-x} = 1 + \frac{-1}{1!}x + \frac{1}{2!}x^2 + \frac{-1}{3!}x^3 + \frac{1}{4!}x^4 + \cdots \\ & \\ & \text{Simplified, this becomes:} \\ & e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \cdots \\ & \\ & \text{In summation notation:} \\ & e^{-x} = \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n!} \end{align*}
\begin{align} 6. \quad \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots = \sum_{n=1}^{\infty} \frac{{(-1)^{n-1} x^n}}{n} \end{align}
\begin{align*} & \text{To derive the Maclaurin series for } \ln(1+x), \text{ let's start by finding the first few derivatives:} \\ & f(x) = \ln(1+x) \\ & f'(x) = \frac{1}{1+x} \\ & f''(x) = -\frac{1}{{(1+x)}^2} \\ & f'''(x) = \frac{2}{{(1+x)}^3} \\ & f^{(4)}(x) = -\frac{6}{{(1+x)}^4} \\ & f^{(5)}(x) = \frac{24}{{(1+x)}^5} \\ & \\ & \text{Evaluating these derivatives at } x = 0: \\ & f(0) = \ln(1+0) = 0 \\ & f'(0) = \frac{1}{1+0} = 1 \\ & f''(0) = -\frac{1}{{(1+0)}^2} = -1 \\ & f'''(0) = \frac{2}{{(1+0)}^3} = 2 \\ & f^{(4)}(0) = -\frac{6}{{(1+0)}^4} = -6 \\ & f^{(5)}(0) = \frac{24}{{(1+0)}^5} = 24 \\ & \\ & \text{Now, we can use the Maclaurin series formula:} \\ & \ln(1+x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 + \cdots \\ & \\ & \text{Plugging in the values, we get:} \\ & \ln(1+x) = 0 + \frac{1}{1!}x - \frac{1}{2!}x^2 + \frac{2}{3!}x^3 - \frac{6}{4!}x^4 + \frac{24}{5!}x^5 + \cdots \\ & \\ & \text{Simplified, this becomes:} \\ & \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots \\ & \\ & \text{In summation notation:} \\ & \ln(1+x) = \sum_{n=1}^{\infty} \frac{{(-1)^{n-1} x^n}}{n} \end{align*}
\begin{align} 7. \quad \ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \cdots = -\sum_{n=1}^{\infty}\frac{{x^n}}{n} \end{align}
\begin{align*} & \text{To derive the Maclaurin series for } \ln(1-x), \text{ let's start by finding the first few derivatives:} \\ & f(x) = \ln(1-x) \\ & f'(x) = -\frac{1}{1-x} \\ & f''(x) = \frac{1}{{(1-x)}^2} \\ & f'''(x) = -\frac{2}{{(1-x)}^3} \\ & f^{(4)}(x) = \frac{6}{{(1-x)}^4} \\ & f^{(5)}(x) = -\frac{24}{{(1-x)}^5} \\ & \\ & \text{Evaluating these derivatives at } x = 0: \\ & f(0) = \ln(1-0) = 0 \\ & f'(0) = -\frac{1}{1-0} = -1 \\ & f''(0) = \frac{1}{{(1-0)}^2} = 1 \\ & f'''(0) = -\frac{2}{{(1-0)}^3} = -2 \\ & f^{(4)}(0) = \frac{6}{{(1-0)}^4} = 6 \\ & f^{(5)}(0) = -\frac{24}{{(1-0)}^5} = -24 \\ & \\ & \text{Now, we can use the Maclaurin series formula:} \\ & \ln(1-x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 + \cdots \\ & \\ & \text{Plugging in the values, we get:} \\ & \ln(1-x) = 0 - \frac{1}{1!}x + \frac{1}{2!}x^2 - \frac{2}{3!}x^3 + \frac{6}{4!}x^4 - \frac{24}{5!}x^5 + \cdots \\ & \\ & \text{Simplified, this becomes:} \\ & \ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \cdots \\ & \\ & \text{In summation notation:} \\ & \ln(1-x) = -\sum_{n=1}^{\infty} \frac{{x^n}}{n} \end{align*}
\begin{align} 8. \quad \sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \end{align}
\begin{align*} & \text{To derive the Maclaurin series for } \sinh(x), \text{ let's start by finding the first few derivatives:} \\ & f(x) = \sinh(x) \\ & f'(x) = \cosh(x) \\ & f''(x) = \sinh(x) \\ & f'''(x) = \cosh(x) \\ & f^{(4)}(x) = \sinh(x) \\ & f^{(5)}(x) = \cosh(x) \\ & \\ & \text{Evaluating these derivatives at } x = 0: \\ & f(0) = \sinh(0) = 0 \\ & f'(0) = \cosh(0) = 1 \\ & f''(0) = \sinh(0) = 0 \\ & f'''(0) = \cosh(0) = 1 \\ & f^{(4)}(0) = \sinh(0) = 0 \\ & f^{(5)}(0) = \cosh(0) = 1 \\ & \\ & \text{Now, we can use the Maclaurin series formula:} \\ & \sinh(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 + \cdots \\ & \\ & \text{Plugging in the values, we get:} \\ & \sinh(x) = 0 + \frac{1}{1!}x + \frac{0}{2!}x^2 + \frac{1}{3!}x^3 + \frac{0}{4!}x^4 + \frac{1}{5!}x^5 + \cdots \\ & \\ & \text{Simplified, this becomes:} \\ & \sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \\ & \\ & \text{In summation notation:} \\ & \sinh(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \end{align*}
\begin{align} 9. \quad \cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \end{align}
\begin{align*} & \text{To derive the Maclaurin series for } \cosh(x), \text{ let's start by finding the first few derivatives:} \\ & f(x) = \cosh(x) \\ & f'(x) = \sinh(x) \\ & f''(x) = \cosh(x) \\ & f'''(x) = \sinh(x) \\ & f^{(4)}(x) = \cosh(x) \\ & f^{(5)}(x) = \sinh(x) \\ & \\ & \text{Evaluating these derivatives at } x = 0: \\ & f(0) = \cosh(0) = 1 \\ & f'(0) = \sinh(0) = 0 \\ & f''(0) = \cosh(0) = 1 \\ & f'''(0) = \sinh(0) = 0 \\ & f^{(4)}(0) = \cosh(0) = 1 \\ & f^{(5)}(0) = \sinh(0) = 0 \\ & \\ & \text{Now, we can use the Maclaurin series formula:} \\ & \cosh(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 + \cdots \\ & \\ & \text{Plugging in the values, we get:} \\ & \cosh(x) = 1 + \frac{0}{1!}x + \frac{1}{2!}x^2 + \frac{0}{3!}x^3 + \frac{1}{4!}x^4 + \frac{0}{5!}x^5 + \cdots \\ & \\ & \text{Simplified, this becomes:} \\ & \cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \\ & \\ & \text{In summation notation:} \\ & \cosh(x) = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \end{align*}
\begin{align} 10. \quad \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + x^5 + \cdots = \sum_{n=0}^{\infty} x^n \end{align}
\begin{align*} & \text{To derive the Maclaurin series for } \frac{1}{1-x}, \text{ let's start by finding the first few derivatives:} \\ & f(x) = \frac{1}{1-x} \\ & f'(x) = \frac{1}{{(1-x)}^2} \\ & f''(x) = \frac{2}{{(1-x)}^3} \\ & f'''(x) = \frac{6}{{(1-x)}^4} \\ & f^{(4)}(x) = \frac{24}{{(1-x)}^5} \\ & f^{(5)}(x) = \frac{120}{{(1-x)}^6} \\ & \\ & \text{Evaluating these derivatives at } x = 0: \\ & f(0) = \frac{1}{1-0} = 1 \\ & f'(0) = \frac{1}{{(1-0)}^2} = 1 \\ & f''(0) = \frac{2}{{(1-0)}^3} = 2 \\ & f'''(0) = \frac{6}{{(1-0)}^4} = 6 \\ & f^{(4)}(0) = \frac{24}{{(1-0)}^5} = 24 \\ & f^{(5)}(0) = \frac{120}{{(1-0)}^6} = 120 \\ & \\ & \text{Now, we can use the Maclaurin series formula:} \\ & \frac{1}{1-x} = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 + \cdots \\ & \\ & \text{Plugging in the values, we get:} \\ & \frac{1}{1-x} = 1 + \frac{1}{1!}x + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 + \frac{24}{4!}x^4 + \frac{120}{5!}x^5 + \cdots \\ & \\ & \text{Simplified, this becomes:} \\ & \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + x^5 + \cdots \\ & \\ & \text{In summation notation:} \\ & \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \end{align*}
\begin{align} 11. \quad \frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - x^5 + \cdots = \sum_{n=0}^{\infty} {(-1)^n x^n} \end{align}
\begin{align*} & \text{To derive the Maclaurin series for } \frac{1}{1+x}, \text{ let's start by finding the first few derivatives:} \\ & f(x) = \frac{1}{1+x} \\ & f'(x) = -\frac{1}{{(1+x)}^2} \\ & f''(x) = \frac{2}{{(1+x)}^3} \\ & f'''(x) = -\frac{6}{{(1+x)}^4} \\ & f^{(4)}(x) = \frac{24}{{(1+x)}^5} \\ & f^{(5)}(x) = -\frac{120}{{(1+x)}^6} \\ & \\ & \text{Evaluating these derivatives at } x = 0: \\ & f(0) = \frac{1}{1+0} = 1 \\ & f'(0) = -\frac{1}{{(1+0)}^2} = -1 \\ & f''(0) = \frac{2}{{(1+0)}^3} = 2 \\ & f'''(0) = -\frac{6}{{(1+0)}^4} = -6 \\ & f^{(4)}(0) = \frac{24}{{(1+0)}^5} = 24 \\ & f^{(5)}(0) = -\frac{120}{{(1+0)}^6} = -120 \\ & \\ & \text{Now, we can use the Maclaurin series formula:} \\ & \frac{1}{1+x} = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 + \cdots \\ & \\ & \text{Plugging in the values, we get:} \\ & \frac{1}{1+x} = 1 - \frac{1}{1!}x + \frac{2}{2!}x^2 - \frac{6}{3!}x^3 + \frac{24}{4!}x^4 - \frac{120}{5!}x^5 + \cdots \\ & \\ & \text{Simplified, this becomes:} \\ & \frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - x^5 + \cdots \\ & \\ & \text{In summation notation:} \\ & \frac{1}{1+x} = \sum_{n=0}^{\infty} {(-1)^n x^n} \end{align*}

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