Static Fields: Fall-2026
HW 03 Practice (SOLUTION): Due W2 D5
- Power Series Practice
Problems like this were covered in Theoretical Mechanics. We will use these techniques in Static Fields. It's here for you to review, especially if you were not in Theoretical Mechanics.
- Calculate the \(n=0, 1, 2, 3, 4\) coefficients of the power series for \(\cos{z}\) expanded around \(z=\pi\). Using these coefficients, find a power series approximation for this function.
First, the form of the answer will be: \[ f(z) = \sum_{n=0}^{\infty} c_n (z-z_0)^n\]
where \(c_n\) are expansion coefficients and \(z_0\) is the point I'm expanding the function around.
In this case, since I'm expanding around \(\pi\):
\[ \cos z = c_0 + c_1 (z-\pi) + c_2 (z-\pi)^2 + c_3(z-\pi)^3 + c_4(z-\pi)^4+\dots\]
Since I'm asked for the fourth order expansion, I'll keep only terms up to the fourth power.
\[ \cos z \approx c_0 + c_1 (z-\pi) + c_2 (z-\pi)^2 + c_3(z-\pi)^3 + c_4(z-\pi)^4\]
I use the approximation sign \(\approx\) when I want to emphasize that I have done an approximation. I use the \(=\) sign and \(\dots\) at the end when I want to indicate that there are more terms that I am not writing in the exact power series.
Now, to calculate the coefficients: \begin{eqnarray*} c_0 &=& \frac{1}{0!}f(z=z_0) = \cos\pi = -1 \\ c_1 &=& \frac{1}{1!}\left.\frac{df}{dz}\right|_{z=z_0} = -\sin \pi = 0 \\ c_2 &=& \frac{1}{2!}\left.\frac{d^2f}{dz^2}\right|_{z=z_0} = -\frac{1}{2}\cos \pi = \frac{1}{2} \\ c_3 &=& \frac{1}{3!}\left.\frac{d^3f}{dz^3}\right|_{z=z_0} = \frac{1}{6}\sin \pi = 0 \\ c_4 &=& \frac{1}{4!}\left.\frac{d^4f}{dz^4}\right|_{z=z_0} = \frac{1}{24}\cos \pi = -\frac{1}{24} \\ \end{eqnarray*} (Don't forget the \(1/n!\).)
So, the approximation I end up with is:
\[ \cos z \approx -1 + \frac{1}{2} (z-\pi)^2 +-\frac{1}{24}(z-\pi)^4\]
This formula is the correct form of the power serie approximation. Do NOT multiply out and regroup the terms (do Not "FOIL") or you are undoing all the work you did to find the power series.
- Plot both the original function and your approximation.
Here is the code in Mathematica and the graph it produces.
- For what values of \(z\) is your approximation “good”?
When I plot this approximation against the cosine function, I see that the expansion approximates the cosine function reasonably well between about \(\pi/2\) and \(3\pi/2\). Inside this region, I can not even distinguish the graphs from each other. Beyond that region, the function and its approximation rapidly start to differ.
- Calculate the \(n=0, 1, 2, 3, 4\) coefficients of the power series for \(\cos{z}\) expanded around \(z=\pi\). Using these coefficients, find a power series approximation for this function.
- Series Notation 1
Try this problem if you need some practice with summation notation for series.
Write out the first four nonzero terms in the series:
\[\sum\limits_{n=0}^\infty \frac{1}{n!}\]
\begin{eqnarray*} \sum\limits_{n=0}^\infty \frac{1}{n!} &=& \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...\\ &\approx&1 + 1 + \frac{1}{2} + \frac{1}{6} \end{eqnarray*} Note that \(0!=1\) by definition.
\[\sum\limits_{n=1}^\infty \frac{(-1)^n}{n!}\]
\begin{eqnarray*} \sum\limits_{n=1}^\infty \frac{(-1)^n}{n!} &=& \frac{(-1)^1}{1!}+\frac{(-1)^2}{2!}+\frac{(-1)^3}{3!}+\frac{(-1)^4}{4!}+...\\ &\approx& -1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \end{eqnarray*} Note that this series starts at \(n=1\).
- \begin{equation}
\sum\limits_{n=0}^\infty {(-2)^{n}\,\theta^{2n}}
\end{equation}
\begin{eqnarray*} \sum\limits_{n=0}^\infty {(-2)^{n}\,\theta^{2n}} &=&(-2)^0\theta^0+(-2)^1\theta^2+(-2)^2\theta^4+(-2)^3\theta^6+\,\dots\\ &\approx&1-2\theta^2+4\theta^4-8\theta^6 \end{eqnarray*}
- Series Notation 2
Try this problem if you need some practice with summation notation for series.
Write (a good guess for) the following series using sigma \(\left(\sum\right)\) notation. (If you only know a few terms of a series, you don't know for sure how the series continues.)
\[1 - 2\,\theta^2 + 4\,\theta^4 - 8\,\theta^6 +\,\dots\]
\[\sum\limits_{n=0}^\infty {(-2)^{n}\,\theta^{2n}}\]
\[\frac14 - \frac19 + \frac{1}{16} - \frac{1}{25}+\,\dots\]
\[\sum\limits_{n=2}^\infty \frac{(-1)^{n}}{n^2}\]