Derivatives and Integrals Derivatives and Integrals

Introduction

Taylor Series:

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


Maclaurin Series (Taylor Series centered at
a=0
):
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n

Common Maclaurin Series

\begin{align*} e^x &= \sum_{n=0}^{\infty} \frac{x^n}{n!} \\\\ \cos x &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \\\\ \sin x &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \\\\ \frac{1}{1-x} &= \sum_{n=0}^{\infty} x^n \quad \text{for } |x| < 1 \end{align*}

Deriving New Series

We can use differentiation and integration of known power series to find the taylor series representations of other functions. These operations are done term-by-term within the radius of convergence (for converging series).

\begin{align*} f(x) &= \sum_{n=0}^{\infty} a_n x^n \\\\ f'(x) &= \frac{d}{dx} \left[ \sum_{n=0}^{\infty} a_n x^n \right] = \sum_{n=1}^{\infty} n a_n x^{n-1} \\\\ \int f(x) \, dx &= \int \left[ \sum_{n=0}^{\infty} a_n x^n \right] dx = \sum_{n=0}^{\infty} \frac{a_n}{n+1} x^{n+1} + C \end{align*}

Examples

Example 1: Derivation
Find the Maclaurin series for

f(x) = \cos(x)
by differentiating the Maclaurin series for
\sin(x)
.
\begin{align*} \sin(x) &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \\ \frac{d}{dx} \left[ \sin(x) \right] &= \frac{d}{dx} \left[ \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \right] \\ \cos(x) &= \sum_{n=0}^{\infty} \frac{d}{dx} \left[ \frac{(-1)^n x^{2n+1}}{(2n+1)!} \right] \\ &= \sum_{n=0}^{\infty} \frac{(-1)^n (2n+1) x^{2n}}{(2n+1)!} \\ &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \end{align*}
The Maclaurin series for
\cos(x)
is
\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}
.

Example 2: Derivation
Find the Maclaurin series for
f(x) = \frac{1}{(1-x)^2}
by differentiating the Maclaurin series for
\frac{1}{1-x}
.
\begin{align*} \frac{1}{1-x} &= \sum_{n=0}^{\infty} x^n \\ \frac{d}{dx} \left[ \frac{1}{1-x} \right] &= \frac{d}{dx} \left[ \sum_{n=0}^{\infty} x^n \right] \\ \frac{1}{(1-x)^2} &= \sum_{n=0}^{\infty} \frac{d}{dx} \left[ x^n \right] \\ &= \sum_{n=0}^{\infty} n x^{n-1} \\ &= \sum_{n=1}^{\infty} n x^{n-1} \quad \text{Exclude } n=0 \text{ as } a_0 = 0x^{-1}=0 \\ &= \sum_{n=0}^{\infty} (n+1) x^{n} \quad \text{Re-index } n \to n+1 \end{align*}
The Maclaurin series for
\frac{1}{(1-x)^2}
is
\sum_{n=0}^{\infty} (n+1) x^{n}
.

Example 3: Derivation
Find the Maclaurin series for
f(x) = \cosh(x)
by differentiating the Maclaurin series of
\sinh(x)
.
\begin{align*} \sinh(x) &= \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \\ \frac{d}{dx} \left[ \sinh(x) \right] &= \frac{d}{dx} \left[ \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \right] \\ \cosh(x) &= \sum_{n=0}^{\infty} \frac{d}{dx} \left[ \frac{x^{2n+1}}{(2n+1)!} \right] \\ &= \sum_{n=0}^{\infty} \frac{(2n+1) x^{2n}}{(2n+1)!} \\ &= \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \end{align*}
The Maclaurin series for
\cosh(x)
is
\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}
.

Example 4: Integration
Find the Maclaurin series for
f(x) = -\ln(1-x)
by integrating the Maclaurin series for
\frac{1}{1-x}
.
\begin{align*} \frac{1}{1-x} &= \sum_{n=0}^{\infty} x^n \\ \int \frac{1}{1-x} \, dx &= \int \left[ \sum_{n=0}^{\infty} x^n \right] dx \\ -\ln(1-x) &= \sum_{n=0}^{\infty} \int x^n \, dx \\ &= \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} + C \\\\ \text{To find C, set } x&=0 \\ -\ln(1-0) &= 0 = \sum_{n=0}^{\infty} \frac{0^{n+1}}{n+1} + C = C \\ C &= 0 \\\\ -\ln(1-x) &= \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} \\ &= \sum_{n=1}^{\infty} \frac{x^{n}}{n} \quad \text{Re-index } n \to n-1 \end{align*}
The Maclaurin series for
-\ln(1-x)
is
\sum_{n=1}^{\infty} \frac{x^{n}}{n}
.

Example 5: Integration
Find the Maclaurin series for
f(x) = \arctan(x)
by integrating the Maclaurin series for
\frac{1}{1+x^2}
.
\begin{align*} \frac{1}{1+x^2} &= \sum_{n=0}^{\infty} (-1)^n x^{2n} \\ \int \frac{1}{1+x^2} \, dx &= \int \left[ \sum_{n=0}^{\infty} (-1)^n x^{2n} \right] dx \\ \arctan(x) &= \sum_{n=0}^{\infty} \int (-1)^n x^{2n} \, dx \\ &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} + C \\\\ \text{To find C, set } x&=0 \\ \arctan(0) &= 0 = \sum_{n=0}^{\infty} \frac{(-1)^n 0^{2n+1}}{2n+1} + C = C \\ C &= 0 \\\\ \arctan(x) &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \end{align*}
The Maclaurin series for
\arctan(x)
is
\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}
.

Example 6: Integration
Find the Maclaurin series for
e^x
by integrating the Maclaurin series for
e^x
.
\begin{align*} e^x &= \sum_{n=0}^{\infty} \frac{x^n}{n!} \\ \int e^x \, dx &= \int \left[ \sum_{n=0}^{\infty} \frac{x^n}{n!} \right] dx \\ &= \sum_{n=0}^{\infty} \int \frac{x^n}{n!} \, dx \\ e^x &= \sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!} + C \\\\ \text{To find C, set } x&=0 \\ e^0 &= 1 = \sum_{n=0}^{\infty} \frac{0^{n+1}}{(n+1)!} + C = C \\ C &= 1 \\\\ e^x &= 1 + \sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!} \\ &= 1 + \sum_{n=1}^{\infty} \frac{x^{n}}{n!} \quad \text{Re-index } n \to n-1 \\ &= \frac{x^{0}}{0!} + \sum_{n=1}^{\infty} \frac{x^{n}}{n!} \\ &= \sum_{n=0}^{\infty} \frac{x^{n}}{n!} \\ \end{align*}
The Maclaurin series for
e^x
is
\sum_{n=0}^{\infty} \frac{x^{n}}{n!}
.