# Maclaurin Series

A **Maclaurin series** is a **power series** that allows one to calculate an approximation of a function \(f(x)\) for input values close to zero, given that one knows the values of the successive derivatives of the function at zero. In many practical applications, it is equivalent to the function it represents.

An example where the Maclaurin series is useful is the sine function. The definition of the sine function does not allow for an easy method of computing output values for the function at arbitrary input values. On the other hand, it is easy to calculate the values of \(\sin(x)\) and all of its derivatives when \(x=0\). The Maclaurin series allows one to use these derivative values at zero to calculate precise approximations of \(\sin(x)\) for inputs close to but not equal to zero. The Maclaurin series is used to create a polynomial that matches the values of \(\sin(x)\) and a chosen number of its successive derivatives when \(x=0\). The resulting polynomial matches the sine curve closely.

A Maclaurin series can be used to approximate a function, find the **antiderivative** of a complicated function, or compute an otherwise uncomputable sum. Partial sums of a Maclaurin series provide polynomial approximations for the function.

A Maclaurin series is a special case of a ** Taylor series**, obtained by setting \(x_0=0\). The Maclaurin series of a function \(f\) is therefore the

**series**

\[\sum_{n=0}^{\infty}f^{(n)}(0)\frac{x^{n}}{n!}=f(0)+f'(0)x+\dfrac{f''(0)}{2!}x^2+\cdots +\dfrac{f^{(k)}(0)}{k!}x^k+\cdots.\]

#### Contents

## Derivation

Intuitively, it makes sense that if an infinite series is equal to a function within a certain interval, then the values of their derivatives should be equal as well. Since a power series has easily expressible derivatives at \(x = 0\), it turns out the series can be expressed entirely in terms of the values of its derivatives.

Suppose \(f(x) = \sum_{n=0}^\infty a_n x^n\) converges for all \(x \in (- \epsilon, \, \epsilon)\), where \(\epsilon\) is some small positive number. Define the series \(g(x) = \sum_{n=0}^\infty (n+1) a_{n+1} x^n\) obtained by differentiating \( f(x) \) term by term. Then the following argument shows that \( g(x) \) also converges on \( (- \epsilon, \, \epsilon):\)

Given \( |x| < \epsilon, \) choose \( |x| < \rho < \epsilon \) and define \( r = \frac{\rho}{\epsilon} < 1.\) It follows that

\[ \big| (n+1) a_{n+1} x^{n} \big| = \left | (n+1) a_{n+1} \frac{x^{n}}{\rho^n} \rho^{n} \right | = \frac{n+1}{\rho} r^n \big| a_{n+1} \rho^{n+1} \big|. \]

The sum \( \sum\limits_{n=0}^{\infty} | a_{n+1} \rho^{n+1} | = \sum\limits_{n=1}^{\infty} | a_{n} \rho^{n} |\) converges since \( \rho < \epsilon.\) Since \( r < 1, \) the sum \( \sum\limits_{n=0}^{\infty} (n+1)r^n \) also converges by the ratio test. It follows that \( \lim\limits_{n \to \infty} (n+1) r^n = 0,\) so \( (n+1) r^n \) is necessarily bounded above, say by \( M.\) This implies \( |(n+1) a_{n+1} x^{n} | \leq \frac{M}{\rho} | a_{n+1} \rho^{n+1} |. \) By the comparison test, \(g(x)\) converges for \( |x| < \epsilon.\)

Therefore, the function \(g(x)\) exists and is equal to the derivative of \(f(x)\) for all \(x \in (-\epsilon, \, \epsilon)\).

It follows from induction that the \(n^\text{th}\) derivative of the power series converges in \((-\epsilon, \, \epsilon)\) and is equal to the \(n^\text{th}\) derivative of \(f\). Then, for \(x = 0\),

\[f^{(n)}(0) = \displaystyle\sum_{k=0}^\infty a_{k+n} (k+n) \cdot (k+n-1) \cdots (k+1) 0^k = a_n (n) \cdot (n-1) \cdots (1) \implies a_n = \frac{f^{(n)}(0)}{n!}. \]

Therefore, any power series equal to \(f(x)\) in an open interval of the origin is of the following form. This power series is defined to be the **Maclaurin series**:

\[ f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n\]

A function \(f\) is defined for all real numbers and satisfies

\[f(0) = 1,\quad f^{(1)}(0) = 0,\quad f^{(2)}(0) = 0,\quad f^{(3)}(0) = -1,\quad f^{(4)}(0) = 0,\]

and, in general, \(f^{(k)}(0) = (-1)^{k}\) if \(k\) is divisible by 3, and \(f^{(k)}(0) = 0\) otherwise.

What is the Maclaurin series of \(f?\)

**Clarification:** In the answer choices, \(!\) denotes the factorial function. For example, \(8! = 1\times2\times3\times\cdots\times8 \).

## Interval and Radius of Convergence

The **interval of convergence** is the open, closed, or semiclosed range of values of \(x\) for which the Maclaurin series converges to the value of the function; outside the domain, the Maclaurin series either is undefined or does not relate to the function. The **radius of convergence** is half the length of the interval; it is also the radius of the circle in the complex plane within which the series converges.

Convergence may be determined by a variety of methods, but the ratio test tends to provide an immediate value \(r\) for the radius of convergence. The interval of convergence may then be determined by testing the value of the series at the endpoints \(-r\) and \(r\).

Find the interval and radius of convergence for the series \( \sum_{k=1}^{\infty} \frac{x^k}{k} \).

Use the ratio test to note that the series will converge only if \(x\) satisfies

\[ \lim_{k \to \infty} \left| \frac{\hspace{2mm} \frac{x^{k+1}}{k+1}\hspace{2mm} }{\frac{x^k}{k}}\right| < 1 \implies \lim_{k \to \infty} \left| \frac{kx}{k+1} < 1 \right| \implies \left| x \right| < 1. \]

The preliminary interval of convergence is \( -1 < x < 1 \), but the series could possibly converge at the "endpoints" \(x = -1\) and \(x = 1\) as well.

If \( x = 1 \), then the series becomes \( \sum_{k = 1}^{\infty} \frac{1}{k} \), which is the harmonic series. This diverges.

If you put in \( x = -1 \), you get \( \sum_{k=1}^{\infty} \frac{(-1)^k}{k}\). This is the alternating harmonic series, which converges by the alternating series test.

Therefore, the interval of convergence is \([-1, \, 1)\), and the radius of convergence is \(1\). \(_\square\)

It may help to note that for simple functions, like the one above, \(1\) and \(\infty\) are common radii of convergence.

## Frequently Used Maclaurin Series

Most Maclaurin series expressible in terms of elementary functions can be determined through the composition and combination of the following functions:

Function | \(\hspace{10mm}\) Maclaurin Series | \(\hspace{10mm}\) Interval of Convergence |

\(\dfrac{1}{1-x}\) | \(\hspace{10mm}\) \(\displaystyle \sum_{k=0}^{\infty} x^k\) | \(\hspace{10mm}\) \(-1 < x < 1\) |

\(e^x\) | \(\hspace{10mm}\) \(\displaystyle \sum_{k=0}^{\infty} \dfrac{x^k}{k!}\) | \(\hspace{10mm}\) \(-\infty < x < \infty\) |

\(\ln(1+x)\) | \(\hspace{10mm}\) \(\displaystyle \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}x^k}{k}\) | \(\hspace{10mm}\) \(-1 < x \leqslant 1\) |

\(\sin x\) | \(\hspace{10mm}\) \(\displaystyle \sum_{k=0}^{\infty} \dfrac{(-1)^k x^{2k+1}}{(2k+1)!}\) | \(\hspace{10mm}\) \(-\infty < x < \infty\) |

\(\cos x\) | \(\hspace{10mm}\) \(\displaystyle \sum_{k=0}^{\infty} \dfrac{(-1)^k x^{2k}}{(2k)!}\) | \(\hspace{10mm}\) \(-\infty < x < \infty\) |

Determine the Maclaurin series for \(f(t) = \arctan t\).

Note that \(f(t) =\int_0^t \frac{1}{1 + u^2} \, du.\) Using the substitution \(x = -u^2\) from the table above, it follows that

\[f(t) = \int_0^t \sum_{k=0}^\infty (-u^2)^k \, du = \sum_{k=0}^\infty \int_0^t (-1)^k u^{2k} \, dt = \sum_{k = 0}^\infty \frac{(-1)^k}{2k + 1} t^{2k + 1}.\ _\square\]

## Taylor Series

Main article: Taylor series

The Maclaurin Series is a Taylor series centered about 0. The Taylor series can be centered around any number \(a \) and is written as follows:

\[\sum_{n=0}^{\infty}f^{(n)}(a)\frac{(x-a)^{n}}{n!}=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+\cdots +\dfrac{f^{(k)}(a)}{k!}(x-a)^n+\cdots.\]

## See Also

**Cite as:**Maclaurin Series.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/maclaurin-series/