1.5. Series expansionNonlinear functions can be nasty, but polynomials are the “nicest” version. A
series expansion is a systematic approach to turn a general nonlinear
equation into a polynomial. Suppose we have some function \(f(x)\) that we want
to approximate as:
(1.1) \[\begin{equation}
f(x) = \sum_{n=0}^{\infty} c_n(x-x_0)^n
\end{equation}\]
To determine the coefficients \(c_n\) , consider the values of f and its
derivatives at \(x_0\)
(1.2) \[\begin{align}
f(x) &= c_0 + c_1(x-x_0) + c_2(x-x_0)^2 + c_3(x-x_0)^3 + \cdots & f(x_0) &= c_0 \\
f'(x) &= c_1 + 2 c_2(x-x_0) + 3 c_3(x-x_0)^2 + \cdots & f'(x_0) &= c_1 \\
f''(x) &= 2 c_2 + 6 c_3(x-x_0) + \cdots & f''(x_0) &= 2 c_2 \\
f'''(x) &= 6 c_3(x-x_0) + \cdots & f''(x_0) &= 6 c_3
\end{align}\]
The pattern of matching derivatives gives rise to the Taylor series .
Taylor series
The Taylor series for \(f(x)\) about \(x_0\) is given by
(1.3) \[\begin{equation}
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}\cdot(x-x_0)^n
\end{equation}\]
where \(f^{(n)}\) is the n th derivative of f . A Taylor series about \(x=0\) is
sometimes called a Maclaurin series.
Example: Taylor series
Compute the Taylor series for \(f(x) = \ln x\) about \(x_0 = 2\) .
(1.4) \[\begin{align}
f(x) &= \ln x & f(2) &= \ln 2 \\
f'(x) &= \frac{1}{x} & f'(2) &= \frac{1}{2} \\
f''(x) &= -\frac{1}{x^2} & f''(2) &= -\frac{1}{4}
\end{align}\]
Hence,
(1.5) \[\begin{equation}
f(x) \approx \ln(2) + \frac{1}{2}(x-2) - \frac{1}{2}\cdot\frac{1}{4} (x-2)^2 + \cdots
\end{equation}\]
In general, we could determine the infinite series and general expressions for
coefficients, but for practical purposes, we will usually stop at either first
or second order polynomials since that is easier to work with.
Example: Taylor series
Find the two-term Taylor series for
(1.6) \[\begin{equation}
f(x) = xe^{2 x^2}
\end{equation}\]
about \(x = 0\) .
We will evaluate the required derivatives. This will be a series of
product rule
The value of the function is \(f(0) = 0\) . The first derivative is:
(1.7) \[\begin{align}
f'(x) &= x(e^{2x^2}4x) + e^{2x^2} \\
&= (4x^2 + 1)e^{2x^2}
\end{align}\]
so \(f'(0) = 1\) . The second derivative is:
(1.8) \[\begin{align}
f''(x) &= (4x^2+1)(e^{2x^2}4x)+8xe^{2x^2} \\
&= (16x^3+12x)e^{2x^2}
\end{align}\]
so \(f''(0) = 0\) . This means we need to keep going! The third derivative is:
(1.9) \[\begin{align}
f'''(x) &= (16x^3+12x)(e^{2x^2}4x)+(48x^2+12)e^{2x^2} \\
&= (64x^4+96x^2+12)e^{2x^2}
\end{align}\]
so \(f'''(0) = 12\) .
Putting it all together,
(1.10) \[\begin{equation}
f(x) \approx 0 + 1 \cdot x + \frac{1}{2} \cdot 0 \cdot x^2
+ \frac{1}{6} \cdot 12 \cdot x^3 + \cdots = x + 2x^3 + \cdots
\end{equation}\]
1.5.1. Combining seriesSome of these calculations can sometimes be avoided by combining series
together. For example, the Taylor series for \(e^x\) is well known
(1.11) \[\begin{equation}
e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \cdots
\end{equation}\]
The argument of e in f is \(2x^2\) , which is also zero when x is zero, so we
can substitute it in the known series to obtain
(1.12) \[\begin{equation}
e^{2x^2} \approx 1 + 2x^2 + \frac{(2x^2)^2}{2} + \cdots
\end{equation}\]
Finally, inserting this series into f and expanding
(1.13) \[\begin{align}
f(x) &= x e^{2x^2} \approx x (1 + 2x^2 + 2x^4 + \cdots) \\
&= x + 2x^3 + \cdots
\end{align}\]
When doing this process, it can be helpful to keep track of dropped terms using
\(O\) -notation
(1.14) \[\begin{align}
e^x &= \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + O(x^3) \\
e^{2x^2} &= 1 + 2x^2 + 2x^4 + O(x^6) \\
xe^{2x^2} &= x + 2x^3 + O(x^5)
\end{align}\]
1.5.2. Skill builder problemsExpand the following to two terms:
\(f(x)=\cos x\) at \(x=0\)
The function and its derivatives at \(x=0\) are:
(1.15) \[\begin{align}
f(x) &= \cos(x) & f(0) &= 1 \\
f'(x) &= -\sin(x) & f'(0) &= 0 \\
f''(x) &= -\cos(x) & f''(0) &= -1
\end{align}\]
So:
(1.16) \[\begin{equation}
\cos x \approx 1-\frac{x^2}{2}
\end{equation}\]
\(f(x)=\sin x\) at \(x=0\)
The function and its derivatives at \(x=0\) are:
(1.17) \[\begin{align}
f(x) &= \sin(x) & f(0) &= 0\\
f'(x) &= \cos(x) & f'(0) &= 1 \\
f''(x) &= -\sin(x) & f''(0) &= 0 \\
f^{(3)}(x) &= -\cos(x) & f^{(3)}(0) &= -1
\end{align}\]
So:
(1.18) \[\begin{equation}
\sin x \approx x-\frac{x^3}{6}
\end{equation}\]
\(f(x)=e^x\) at \(x=0\)
The function and its derivatives at \(x=0\) are:
(1.19) \[\begin{align}
f(x)&=e^x & f(0) &= 1\\
f'(x)&=e^x & f'(0) &= 1
\end{align}\]
So:
(1.20) \[\begin{equation}
e^x \approx 1+x
\end{equation}\]
\(f(x)=\ln(1+x)\) at \(x=0\)
The function and its derivatives at \(x=0\) are:
(1.21) \[\begin{align}
f(x)&=\ln(1+x) & f(0) &= 0\\
f'(x)&=\frac{1}{1+x} & f'(0) &= 1\\
f''(x)&=-\frac{1}{(1+x)^2} & f''(0) &= -1
\end{align}\]
So:
(1.22) \[\begin{equation}
\ln(1+x) \approx x-\frac{x^2}{2}
\end{equation}\]
\(f(x)=\dfrac{1}{1+x}\) at \(x=0\)
The function and its derivatives at \(x=0\) are:
(1.23) \[\begin{align}
f(x)&=\frac{1}{1+x} & f(0) &= 1\\
f'(x)&=-\frac{1}{(1+x)^2} & f'(0) &= -1
\end{align}\]
So:
(1.24) \[\begin{equation}
\frac{1}{1+x} \approx 1-x
\end{equation}\]
\(f(x)=\cos(4x)\) at \(x=0\)
Recall:
(1.25) \[\begin{equation}
\cos x \approx 1-\frac{x^2}{2}
\end{equation}\]
So:
(1.26) \[\begin{equation}
\cos(4x) \approx 1-\frac{(4x)^2}{2} = 1-8x^2
\end{equation}\]
\(f(x)=\cos(x-\pi)\) at \(x=\pi\)
Recall:
(1.27) \[\begin{equation}
\cos(x) \approx 1-\frac{x^2}{2}
\end{equation}\]
So:
(1.28) \[\begin{equation}
\cos(x-\pi) \approx 1-\frac{(x-\pi)^2}{2}
\end{equation}\]
about \(x_0 = \pi\) .
\(f(x)=e^x \sin x\) at \(x=0\)
Recall:
(1.29) \[\begin{align}
e^x &\approx 1+x \\
\sin(x) &\approx x-\frac{x^3}{6}
\end{align}\]
So:
(1.30) \[\begin{equation}
e^x \sin x \approx (1+x+\cdots)(x-\frac{x^3}{6}+\cdots) = x+x^2
\end{equation}\]
\(f(x)=\cos x\) at \(x=\pi\)
The function and its derivatives at \(x=\pi\) are:
(1.31) \[\begin{align}
f(x) &= \cos(x) & f(\pi) &= -1 \\
f'(x) &= -\sin(x) & f'(\pi) &= 0 \\
f''(x) &= -\cos(x) & f''(\pi) &= 1
\end{align}\]
So:
(1.32) \[\begin{equation}
\cos x \approx -1+\frac{1}{2}(x-\pi)^2
\end{equation}\]
at \(x = \pi\) .
\(f(x)=7x^2-6x+1\) at \(x=2\)
The function and its derivatives at \(x=2\) are:
(1.33) \[\begin{align}
f(x)&=7x^2-6x+1 & f(2) &= 17\\
f'(x)&=14x-6 & f'(2) &= 22
\end{align}\]
So:
(1.34) \[\begin{equation}
f(x) \approx 17+22(x-2)
\end{equation}\]