Central Forces: Spring-2026
HW 01 (SOLUTION): Due W1 D3: Math Bits

1. Syllabus and Schedule S1 5530S

   Find the course syllabus and schedule on Canvas. Read through them carefully and bring your questions to the second day of class.

2. Compare Fourier Series to Particle-in-a-Box S1 5530S Consider the function: \begin{equation*} f(x)=\begin{cases} 0&0<x<\frac{L}{3}\\ D & \frac{L}{3}<x<\frac{2L}{3}\\ 0&\frac{2L}{3}<x<L \end{cases} \end{equation*}

   1. You can think of this function as periodic, with period \(L\). Expand this function in a Fourier series.

      Plot an approximation containing eight nonzero terms.

      Make a histogram of your coefficients. Be thoughtful about the horizontal axis of your histogram.

      This function is even, so I'm expecting that all the coefficients in front of sine functions (\(b_n\)'s) to be zero: \begin{align*} b_n &= \frac{2}{L} \int_0^L \sin\left(\frac{2n \pi x}{L}\right) f(x) dx\\ &= \frac{2}{L}\Bigg[ \int_0^{L/3} (0) dx + \int_{L/3}^{2L/3} D\sin\left(\frac{2n \pi x}{L}\right)dx + \int_{2L/3}^L (0) dx \Bigg]\\ &= \frac{D}{n\pi} \Bigg[ \cos\left(\frac{2n\pi}{3} \right) -\cos\left(\frac{4n\pi}{3} \right)\Bigg]\\ &= 0 \end{align*}

      Convince yourself that this is zero by looking at the unit circle: \begin{align*} \cos\left(\frac{2\pi}{3} \right) -\cos\left(\frac{4\pi}{3}\right) &= 0 \\ \cos\left(\frac{4\pi}{3} \right) -\cos\left(\frac{8\pi}{3}\right) &= 0 \\ \cos\left(\frac{6\pi}{3} \right) -\cos\left(\frac{12\pi}{3}\right) &= 0 \\ \end{align*}

      and so on. Now, now I'll look at the coefficients of the cosine terms: \begin{align*} a_0 &= \frac{2}{L} \int_0^L f(x) dx\\ &= \frac{2}{L}\Bigg[ \int_0^{L/3} (0) dx + \int_{L/3}^{2L/3} D dx + \int_{2L/3}^L (0) dx \Bigg]\\ &= \frac{D}{3} \end{align*} \begin{align*} a_n &= \frac{2}{L} \int_0^L \cos\left(\frac{2n \pi x}{L}\right) f(x) dx\\ &= \frac{2}{L}\Bigg[ \int_0^{L/3} (0) dx + \int_{L/3}^{2L/3} D\cos\left(\frac{2n \pi x}{L}\right) dx + \int_{2L/3}^L (0) dx \Bigg]\\ &= \frac{D}{n\pi} \Bigg[ -\sin\left(\frac{2n\pi}{3} \right) +\sin\left(\frac{4n\pi}{3} \right)\Bigg]\\ \end{align*}

      So, for: \begin{align*} a_1 &= -\frac{D\sqrt{3}}{\pi} \\ a_2 &= \frac{D\sqrt{3}}{2\pi} \\ a_3 &= 0 \\ a_4 &= -\frac{D\sqrt{3}}{4\pi} \\ a_5 &= \frac{D\sqrt{3}}{5\pi} \\ a_6 &= 0 \\ a_7 &= -\frac{D\sqrt{3}}{7\pi} \\ a_8 &= \frac{D\sqrt{3}}{8\pi} \\ a_9 &= 0 \\ a_{10} &= -\frac{D\sqrt{3}}{10\pi} \\ \end{align*}

      

   2. Extra Credit Challenge You can also think of this function as representing a quantum particle in a box of size \(L\). Expand this function in quantum particle in a box energy eigenstates.

      Plot an approximation containing eight nonzero terms.

      Make a histogram of your coefficients. Be thoughtful about the horizontal axis of your histogram.

      Now I'm going to approximate this function using the energy eigenstates of the infinite square well: \begin{align*} \left|{f}\right\rangle = \sum_{n=0}^{\infty} c_n \left|{n}\right\rangle \end{align*}

      where \(\left|{n}\right\rangle \) are the energy eigenstates of the infinite square well (ISW). In wavefunction language, this is: \begin{align*} f(x) = \sum_{n=0}^{\infty} c_n \sqrt{\frac{2}{L}} \sin \left(\frac{n \pi x}{L}\right) \end{align*}

      So, now I have two different ways to write the same function!: \begin{align*} &\mbox{Piecewise: } f(x) =\begin{cases} 0&0<x<\frac{L}{3}\\ D & \frac{L}{3}<x<\frac{2L}{3}\\ 0&\frac{2L}{3}<x<L \end{cases}\\ &\mbox{ISW Energy Eigenstates: } f(x) = \sum_{n=0}^{\infty} c_n \sqrt{\frac{2}{L}} \sin \left(\frac{n \pi x}{L}\right)\\ \end{align*}

      To find the expansion coefficients for the ISW Energy Eigenstate form, I'll take advantage of the orthonormality of the energy eigenstates \(\left|{n}\right\rangle \) and do inner products: \begin{align*} c_n &= \left\langle {n}\middle|{f}\right\rangle \\ &= \int_0^{L} \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right) f(x) dx\\ \end{align*}

      where I will use the piecewise definition of \(f(x)\) inside the integral. This seems circular - using \(f(x)\) to calculate the coefficients for representing \(f(x)\) - but I need information about the function in order to figure out the unique expansion coefficients for this function. Now I'll plug in my original \(f(x)\) and integrate: \begin{align*} c_n &= \int_0^{L/3} (0) dx + \int_{L/3}^{2L/3} D\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right) dx + \int_{2L/3}^{L} (0) dx\\ &= \int_{L/3}^{2L/3} D\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right) dx\\ &= -\frac{DL}{n\pi}\sqrt{\frac{2}{L}}\cos\left(\frac{n\pi x}{L}\right)\Bigg|_{L/3}^{2L/3} \\ &= -\frac{D\sqrt{2L}}{n\pi}\Big[\cos \left(\frac{2n\pi}{3}\right)- \cos \left(\frac{n\pi}{3}\right) \Big] \\ \end{align*}

      Because this Center Step function is symmetric in my well, I expect only contributions from terms that are symmetric in the well - only odd \(n\). All the even \(n\) terms are assymmetric in the well. Plugging in for integer \(n\): \begin{align*} c_1 &= \frac{D\sqrt{2L}}{\pi} \\ c_2 &= 0 \\ c_3 &= -\frac{2D\sqrt{2L}}{3\pi} \\ c_4 &= 0 \\ c_5 &= \frac{D\sqrt{2L}}{5\pi}\\ c_6 &= 0 \\ c_7 &= \frac{D\sqrt{2L}}{7\pi}\\ c_8 &= 0 \\ c_9 &= -\frac{2D\sqrt{2L}}{9\pi}\\ c_{10} &= 0\\ c_{11} &= \frac{D\sqrt{2L}}{11\pi}\\ c_{12} &= 0\\ c_{13} &= \frac{D\sqrt{2L}}{13\pi}\\ c_{14} &= 0\\ c_{15} &= -\frac{2D\sqrt{2L}}{15\pi}\\ \end{align*}

      As expected, all the \(n=\) even terms are zero. The spectrum looks like:

      

      The approximation looks like: \begin{align*} f(x) &\approx \frac{D\sqrt{2L}}{\pi} \sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right) -\frac{2D\sqrt{2L}}{3\pi} \sqrt{\frac{2}{L}}\sin\left(\frac{3\pi x}{L}\right)\\ & +\frac{D\sqrt{2L}}{5\pi} \sqrt{\frac{2}{L}}\sin\left(\frac{5\pi x}{L}\right) +\frac{D\sqrt{2L}}{7\pi} \sqrt{\frac{2}{L}}\sin\left(\frac{7\pi x}{L}\right) \\ & -\frac{2D\sqrt{2L}}{9\pi} \sqrt{\frac{2}{L}}\sin\left(\frac{9\pi x}{L}\right) +\frac{D\sqrt{2L}}{11\pi} \sqrt{\frac{2}{L}}\sin\left(\frac{11\pi x}{L}\right) \\ & +\frac{D\sqrt{2L}}{13\pi} \sqrt{\frac{2}{L}}\sin\left(\frac{13\pi x}{L}\right) -\frac{2D\sqrt{2L}}{15\pi} \sqrt{\frac{2}{L}}\sin\left(\frac{15\pi x}{L}\right) \\ \end{align*}

      

   3. Extra Credit Challenge: Briefly reflect on the similarities and differences between the previous two cases. You may find it interesting to plot your approximations from \(x=-2L\) to \(x=2L\)

      I have three different ways to write the same function!: \begin{align*} &\mbox{Piecewise: } f(x) =\begin{cases} 0&0<x<\frac{L}{3}\\ D & \frac{L}{3}<x<\frac{2L}{3}\\ 0&\frac{2L}{3}<x<L \end{cases}\\ &\mbox{Fourier Series: } f(x) = \frac{a_0}{2} + \sum_{n=0}^{\infty} a_n \cos \left(\frac{2 \pi n x}{L}\right) + \sum_{n=0}^{\infty} b_n \sin \left(\frac{2 \pi n x}{L}\right)\\ &\mbox{ISW Energy Eigenstates: } f(x) = \sum_{n=0}^{\infty} c_n \sqrt{\frac{2}{L}}\sin \left(\frac{n \pi x}{L}\right)\\ \end{align*}

      This problem has me expanding function my piecewise-defined \(f(x)\) in two completely different sets of basis functions: Fourier Series and the energy eigenstates of the infinite square well.
      

      The Fourier expansion includes a constant term (\(n=0\)) and cosine terms - no sine terms.
      

      The ISW energy basis only includes sine terms - no constant term and no cosine terms.
      

      Therefore, even though it's tempting to think of the ISW energy basis and the Fourier functions to be the same, they really aren't.
      

      If I include 8 terms each, the approximations look pretty similar, but I needed \(n=15\) in order to get 8 non-zero terms for the ISW energy basis. For the Fourier Series, I only need \(n=10\).

      I'll plot what my expansion looks like outside of \(0<x<L\):

      

      Notice that the Fourier approximation repeats the same, but the ISW energy approximation alternates signs.