A power series
centred at \(c\) may be written:
\[
f(x) = \sum_{n=0}^{\infty} a_n (x-c)^n
\]
When \(c = 0\), the power series becomes:
\[
\sum_{n=0}^{\infty} a_n x^n
= a_0
+ a_1 x
+ a_2 x^2
+ a_3 x^3
+ a_4 x^4
+ \ldots
\]
Let
\[
f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots
\]
Differentiate and evaluate at \(x = 0\):
\[
f'(x) = a_1 + 2a_2 x + 3a_3 x^2 + \cdots
\]
\[
f'(0) = a_1
\]
Differentiate again:
\[
f''(x) = 2a_2 + 2 \cdot 3a_3 x +3 \cdot 4a_4 x^2 + \cdots
\]
\[
f''(x) = 2a_2 + 6a_3 x + 12a_4 x^2 + \cdots
\]
\[
f''(0) = 2a_2
\]
And again:
\[
f^{(3)}(x) = 6a_3 + 2 \cdot 12a_4 x + \cdots
\]
\[
f^{(3)}(x) = 6a_3 + 24a_4 x + \cdots
\]
\[
f^{(3)}(0) = 6a_3
\]
And again:
\[
f^{(4)}(x) = 24a_4 + \cdots
\]
\[
f^{(4)}(0) = 24a_4
\]
Continuing this pattern gives:
\[
a_n = \frac{f^{(n)}(0)}{n!}
\]
Substituting back into the original series:
\[
f(x) =
\frac{f(0)}{0!}
+ \frac{f'(0)}{1!}x
+ \frac{f''(0)}{2!}x^2
+ \frac{f^{(3)}(0)}{3!}x^3
+ \cdots
\]
Giving **Maclaurin’s Series**:
\[
f(x) = \sum_{n=0}^{\infty}
\frac{f^{(n)}(0)}{n!} x^n
\]
Which has valid values for x when
\[
\lim_{n\to\infty}
\left|
\frac{u_{n+1} x}{u_n}
\right|
\lt1
\]
\[
\lim_{n\to\infty}
\left|
\frac{
f^{(n+1)}(0)\, x^{\,n+1}
}{
(n+1)!}
\;\times\;
\frac{n!}{
f^{(n)}(0)\, x^{\,n}}
\right|
\lt 1
\]
\[
\lim_{n\to\infty}
\left|
\frac{
x\, f^{(n+1)}(0)
}{
(n+1)}
\;\times\;
\frac{1}{f^{(n)}(0)}
\right|
\lt 1
\]
\[
\lim_{n\to\infty}
\left|
\frac{
x\, f^{(n+1)}(0)
}{
(n+1)\, f^{(n)}(0)}
\right|
\lt 1
\]
Example
Find the Maclaurin expansion of
\[
f(x) = (1+x)^n
\]
Differentiate repeatedly:
\[
\text{Differentiate}
\]
\[
f(x) = (1+x)^n
\]
\[
f'(x) = n(1+x)^{\,n-1}
\]
\[
f''(x) = n(n-1)(1+x)^{\,n-2}
\]
\[
f'''(x) = n(n-1)(n-2)(1+x)^{\,n-3}
\]
\[
f^{iv}(x) = n(n-1)(n-2)(n-3)(1+x)^{\,n-4}
\]
\[
\text{etc}
\]
Evaluate at \(x = 0\):
\[
\text{set } x = 0
\]
\[
f(0) = 1
\]
\[
f'(0) = n
\]
\[
f''(0) = n(n-1)
\]
\[
f'''(0) = n(n-1)(n-2)
\]
\[
f^{iv}(0) = n(n-1)(n-2)(n-3)
\]
\[
\text{etc}
\]
\[
f(x) =
\frac{f(0)}{0!}
+ \frac{f'(0)}{1!}x
+ \frac{f''(0)}{2!}x^2
+ \frac{f^{(3)}(0)}{3!}x^3
+ \cdots
\]
Thus:
\[
(1+x)^n
= 1 + nx + \frac{n(n-1)}{2!}x^2
+ \frac{n(n-1)(n-2)}{3!}x^3 + \cdots
\]
This is the Binomial Series.
Example
Find the Maclaurin expansion of
\[
f(x) = (1-x)^n
\]
\( f(x) = (1 - x)^n = (1 + (-x))^n \)
\[
(1-x)^n
= 1 - nx + \frac{n(n-1)}{2!}x^2
- \frac{n(n-1)(n-2)}{3!}x^3 + \cdots
\]
Example
Use the Maclaurin Series to find a series for
\[
f(x) = \sin x
\]
Differentiate
\[
f(x) = \sin x
\]
\[
f'(x) = \cos x
\]
\[
f''(x) = -\sin x
\]
\[
f'''(x) = -\cos x
\]
\[
f^{(iv)}(x) = \sin x
\]
Derivatives:
\[ f(0) = 0 \]
\[ f'(0) = 1 \]
\[ f''(0) = 0 \]
\[ f^{(3)}(0) = -1 \]
\[ f^{(4)}(0) = 0 \]
\[ f^{(5)}(0) = 1 \]
\[
f(x) =
\frac{f(0)}{0!}
+ \frac{f'(0)}{1!}x
+ \frac{f''(0)}{2!}x^2
+ \frac{f^{(3)}(0)}{3!}x^3
+ \cdots
\]
\[
f(x) =
\frac{0}{0!}
+ \frac{1}{1!}x
+ \frac{0}{2!}x^2
+ \frac{-1}{3!}x^3
+ \cdots
\]
Thus:
\[
\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!}
- \frac{x^7}{7!} + \cdots
\]
Example
Use the Maclaurin Series to find a series for
\[
f(x) = (1+x)^{1/4}
\]
\[
f(x) = (1 + x)^{1/4}
\]
\[
\text{Differentiate}
\]
\[
f(x) = (1 + x)^{1/4}
\]
\[
f'(x) = \frac{1}{4}(1 + x)^{-3/4}
\]
\[
f''(x) = \frac{1}{4}\cdot\frac{-3}{4}(1 + x)^{-7/4}
\]
\[
f'''(x) = \frac{1}{4}\cdot\frac{-3}{4}\cdot\frac{-7}{4}(1 + x)^{-11/4}
\]
\[
f^{iv}(x) = \frac{1}{4}\cdot\frac{-3}{4}\cdot\frac{-7}{4}\cdot\frac{-11}{4}(1 + x)^{-15/4}
\]
\[
\text{etc}
\]
Derivatives:
\[
\text{set } x = 0
\]
\[
f(0) = 1
\]
\[
f'(0) = \frac{1}{4}
\]
\[
f''(0) = \frac{1}{4}\cdot\frac{-3}{4}
= -\frac{3}{16}
\]
\[
f'''(0) = \frac{1}{4}\cdot\frac{-3}{4}\cdot\frac{-7}{4}
= \frac{21}{64}
\]
\[
f^{iv}(0) = \frac{1}{4}\cdot\frac{-3}{4}\cdot\frac{-7}{4}\cdot\frac{-11}{4}
= -\frac{231}{256}
\]
\[
f(x)
= \frac{f(0)}{0!}
+ \frac{f'(0)\,x}{1!}
+ \frac{f''(0)\,x^2}{2!}
+ \frac{f'''(0)\,x^3}{3!}
+ \frac{f^{iv}(0)\,x^4}{4!}
+ \ldots
\]
\[
(1 + x)^{1/4}
= 1
+ \frac{1}{4}x
- \frac{3x^2}{16\cdot 2!}
+ \frac{21x^3}{64\cdot 3!}
- \frac{231x^4}{256\cdot 4!}
+ \ldots
\]
\[
= 1
+ \frac{1}{4}x
- \frac{3x^2}{32}
+ \frac{21x^3}{384}
- \frac{231x^4}{6144}
+ \ldots
\]
So
\[
(1+x)^{1/4}
= 1 + \frac{1}{4}x
- \frac{3}{32}x^2
+ \frac{7}{128}x^3
- \frac{77}{2048}x^4 + \cdots
\]
Example
Find the Maclaurin expansion of
\[
f(x) = e^x
\]
All derivatives equal \(e^x\), so at \(x=0\):
\[
f^{(n)}(0) = 1
\]
\[
f(x)
= \frac{f(0)}{0!}
+ \frac{f'(0)\,x}{1!}
+ \frac{f''(0)\,x^2}{2!}
+ \frac{f'''(0)\,x^3}{3!}
+ \frac{f^{iv}(0)\,x^4}{4!}
+ \ldots
\]
Thus:
\[
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}
+ \frac{x^4}{4!} + \cdots
\]
\[
e^{x}
= \sum_{n=0}^{\infty} \frac{x^{n}}{n!}
\]
\[
= 1
+ \frac{x}{1!}
+ \frac{x^{2}}{2!}
+ \frac{x^{3}}{3!}
+ \frac{x^{4}}{4!}
+ \ldots
\]
\[
e^{ax}
= \sum_{n=0}^{\infty} \frac{(ax)^{n}}{n!}
\]
\[
= \sum_{n=0}^{\infty} a^{n}\,\frac{x^{n}}{n!}
\]
\[
= 1
+ \frac{ax}{1!}
+ \frac{(ax)^{2}}{2!}
+ \frac{(ax)^{3}}{3!}
+ \frac{(ax)^{4}}{4!}
+ \ldots
\]
\[
e^{ax+b}
= e^{ax}\,e^{b}
\]
\[
= \sum_{n=0}^{\infty} a^{n} e^{b}\,\frac{x^{n}}{n!}
\]
\[
= e^{b}
+ \frac{e^{b}(ax)}{1!}
+ \frac{e^{b}(ax)^{2}}{2!}
+ \frac{e^{b}(ax)^{3}}{3!}
+ \ldots
\]
\[
\ln(1+x)
= \sum_{n=0}^{\infty}
(-1)^{\,n+1}\,
\frac{x^{\,n}}{n}
\]
\[
= x
- \frac{x^{2}}{2}
+ \frac{x^{3}}{3}
- \frac{x^{4}}{4}
+ \frac{x^{5}}{5}
- \ldots + \ldots
\]
\[
\ln(1 - x)
= - \sum_{n=0}^{\infty} \frac{x^{\,n}}{n}
\]
\[
= -x
- \frac{x^{2}}{2}
- \frac{x^{3}}{3}
- \frac{x^{4}}{4}
- \frac{x^{5}}{5}
- \ldots
\]
\[
\text{provided } -1 \lt x \lt 1
\]
\[
\sin x
= \sum_{n=0}^{\infty}
\frac{(-1)^{\,n}}{(2n+1)!}\,x^{\,2n+1}
\]
\[
= x
- \frac{x^{3}}{3!}
+ \frac{x^{5}}{5!}
- \frac{x^{7}}{7!}
+ \frac{x^{9}}{9!}
- \ldots + \ldots
\]
\[
\cos x
= \sum_{n=0}^{\infty}
\frac{(-1)^{\,n}}{(2n)!}\,x^{\,2n}
\]
\[
= 1
- \frac{x^{2}}{2!}
+ \frac{x^{4}}{4!}
- \frac{x^{6}}{6!}
+ \frac{x^{8}}{8!}
- \frac{x^{10}}{10!}
+ \ldots
\]
\[
(a + x)^{r}
= a^{r}
+ r\,a^{\,r-1}x
+ \frac{r(r-1)}{2!}\,a^{\,r-2}x^{2}
+ \frac{r(r-1)(r-2)}{3!}\,a^{\,r-3}x^{3}
+ \ldots
\]
\[
\text{provided } -a \lt x \lt a
\]
\[ (1+x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k =1 + nx + \frac{n(n-1)}{2!}x^2
+ \frac{n(n-1)(n-2)}{3!}x^3 + \cdots\]
\[ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \]
\[ \arctan x = x - \frac{x^3}{3} + \frac{x^5}{5}
- \frac{x^7}{7} + \cdots \]
Example
Find the first five terms of a power series for \(e^{3x+6}\).
\[
e^{ax+b}
= e^{b}
+ \frac{e^{b}(ax)}{1!}
+ \frac{e^{b}(ax)^{2}}{2!}
+ \frac{e^{b}(ax)^{3}}{3!}
+ \frac{e^{b}(ax)^{4}}{4!}
+ \ldots
\]
\[
e^{3x+6}
= e^{6}
+ \frac{e^{6}(3x)}{1!}
+ \frac{e^{6}(3x)^{2}}{2!}
+ \frac{e^{6}(3x)^{3}}{3!}
+ \frac{e^{6}(3x)^{4}}{4!}
+ \ldots
\]
\[
= e^{6}
+ 3e^{6}x
+ \frac{9e^{6}x^{2}}{2}
+ \frac{27e^{6}x^{3}}{6}
+ \frac{81e^{6}x^{4}}{24}
\]
\[
= e^{6}
+ 3e^{6}x
+ \frac{9e^{6}x^{2}}{2}
+ \frac{9e^{6}x^{3}}{2}
+ \frac{27e^{6}x^{4}}{8}
\]
Example
Find the first four terms of a power series for \(\cos(3x)\).
\[
\cos x
= \sum_{n=0}^{\infty}
\frac{(-1)^{\,n}}{(2n)!}\,(3x)^{\,2n}
\]
\[
= 1
- \frac{(3x)^{2}}{2}
+ \frac{(3x)^{4}}{4}
- \frac{(3x)^{6}}{6}
+ \ldots
\]
\[
= 1
- \frac{9x^{2}}{2}
+ \frac{81x^{4}}{4}
- \frac{729x^{6}}{6}
+ \ldots
\]
\[
= 1
- \frac{9x^{2}}{2}
+ \frac{81x^{4}}{4}
- \frac{243x^{6}}{2}
+ \ldots
\]
Example
Express
\(
\sin\!\left(3x + \frac{\pi}{4}\right)
\)
as a power series in \(x\):
\[
\sin\!\left(3x + \frac{\pi}{4}\right)
= \sin(3x)\cos\!\left(\frac{\pi}{4}\right)
+ \cos(3x)\sin\!\left(\frac{\pi}{4}\right)
\]
\[
= \frac{1}{\sqrt{2}}\,\sin(3x)
+ \frac{1}{\sqrt{2}}\,\cos(3x)
\]
\[
= \frac{1}{\sqrt{2}}
\sum_{n=0}^{\infty}
\frac{(-1)^{\,n}}{(2n+1)!}\,(3x)^{\,2n+1}
\;+\;
\frac{1}{\sqrt{2}}
\sum_{n=0}^{\infty}
\frac{(-1)^{\,n}}{(2n)!}\,(3x)^{\,2n}
\]
\[
= \frac{1}{\sqrt{2}}
\sum_{n=0}^{\infty}
\frac{(-1)^{\,n}}{(2n+1)!}\,(3x)^{\,2n+1}
\;+\;
\frac{(-1)^{\,n}}{(2n)!}\,(3x)^{\,2n}
\]
\[
= \frac{1}{\sqrt{2}}
\sum_{n=0}^{\infty}
\frac{(-1)^{\,n}}{(2n)!}\,(3x)^{\,2n}
\left(
1 + \frac{3x}{2n+1}
\right)
\]
Example
Calculate \(\sin(0.6)\) correct to five decimal places.
\[
\sin x
= x
- \frac{x^{3}}{3!}
+ \frac{x^{5}}{5!}
- \frac{x^{7}}{7!}
+ \frac{x^{9}}{9!}
- \ldots + \ldots
\]
\[
\sin(0.6)
= 0.6
- \frac{0.6^{3}}{3!}
+ \frac{0.6^{5}}{5!}
- \frac{0.6^{7}}{7!}
+ \frac{0.6^{9}}{9!}
- \ldots + \ldots
\]
\[
= 0.6
- \frac{0.216}{6}
+ \frac{0.07776}{120}
- \frac{0.027994}{5040}
+ \frac{0.6^{9}}{9!}
- \ldots + \ldots
\]
\[
= 0.6
- 0.036
+ 0.000648
- 0.00000555428571428571
+ \ldots
\]
\[
0.6 - 0.036 = 0.564
\]
\[
0.564 + 0.000648 = 0.564648
\]
\[
0.564648 - 0.00000555428571428571
= 0.56464244571429
\]
\[
0.56464244571429 + 0.00000002777142857143
= 0.56464247348571
\]
\[
\sin(0.6) = 0.56464 \text{ (5 dp)}
\]
© Alexander Forrest