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!}
.