Recent content by julian

@Rick16 asked how can the sum ##1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots##be determined? In the following proof, the auxiliary function may not be an obvious choice, but the calculations are straightforward. The proof uses familiar differentiation and integration rules, together with...

You can prove ##\ln (1-z)## converges everywhere on the unit circle except at ##z=1## by Dirichlet's test: Fix ##z## in the unit circle, i.e. ##|z|=1##. If ##\{ a_n \}## are real numbers and ##\{ b_n \}## complex numbers such that: (i) ##a_1 \geq a_2 \geq \cdots## (ii) ##\lim_{n \rightarrow...

We can obtain the relevant sum directly by considering the factorisation of ##\cos x## instead: \begin{align*} \cos x &=\left(1-\frac{x^2}{\pi^2/4}\right)\left(1-\frac{x^2}{9\pi^2/4}\right)\left(1-\frac{x^2}{25\pi^2/4}\right) \cdots \\...

This is Euler's classic proof of the Basel problem that @.Scott was referring to in post #2. It appears in the book introduced by @mathwonk in another thread: Euler’s factorisation argument for ##sin⁡x## was initially heuristic. A fully rigorous foundation came in the 19th century, especially...

The particular derivation I gave requires knowledge of complex analysis and is quite lengthy. I therefore thought that it illustrated why deriving the result is not straightforward. However, I had forgotten that there is a substantially more elementary proof using Fourier series, which are...

Extracting relevant bits from other thread: We can write \begin{align*} \sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = \sum_{n=1}^\infty \dfrac{1}{n^2} - \sum_{n=1}^\infty \dfrac{1}{(2n)^2} \nonumber \\ & = (1 - 2^{-2}) \sum_{n=1}^\infty \dfrac{1}{n^2} \nonumber \\ & = \frac34 \sum_{n=1}^\infty...

In another thread I provide a derivation of the general closed formula for ##\zeta(2k)## in terms of Bernoulli numbers: \begin{align*} \zeta(2k) = \sum_{n=1}^\infty \frac{1}{n^{2k}} = (-1)^{k+1} (2 \pi)^{2k} \dfrac{B_{2k}}{2 (2k)!} \end{align*} From which you can obatin your series as...

Write \begin{align*} n^A &= \begin{pmatrix} n^0\\ n^1 \end{pmatrix}, & k^A &= \begin{pmatrix} k^0\\ k^1 \end{pmatrix}. \end{align*} If \begin{align*} m^{AB} = n^A k^B, \end{align*} then the matrix representing ##m^{AB}##, with row index ##A## and column index##B##, is \begin{align*}...

We wish to maximise ##\langle L_x \rangle^2 + \langle L_y \rangle^2 + \langle L_z \rangle^2##. Write \begin{align*} | \psi_0 \rangle = \sum_{m=-l}^l a_m |l,m\rangle \end{align*} I get \begin{align*} & | \langle L_+\rangle |^2 + \langle L_z\rangle^2 = \nonumber \\ & = \hbar^2\, \left|...

To show that ##\sin (\frac{n^2}{n+a} x)## has partially bounded sums for ##x \in [1,2]##: Write \begin{align*} \frac{n^2}{n+a} = n - a + \frac{a^2}{n+a} \end{align*} Then \begin{align*} \sin (\frac{n^2}{n+a} x) = \sin (n - a + r_n) x , \quad where \quad r_n = \frac{a^2}{n+a} \end{align*}...

To show that ##\psi_a^0## and ##\psi_b^0## span the space, we must prove they are linearly independent. Suppose \begin{align*} \kappa \psi_a^0 + \zeta \psi_b^0 = 0. \end{align*} Applying ##A## gives \begin{align*} \kappa \mu \psi_a^0 + \zeta \nu \psi_b^0 = 0. \end{align*} Subtracting ##\mu...

If ##\gamma = \mu##, then you don’t necessarily have ##\langle \psi_a^0 , \psi_\gamma (0) \rangle = 0##.

I think this is the fourth edition: https://archive.org/details/landau-l.-d.-lifshitz-e.-m.-course-of-theoretical-vol-2/page/n3/mode/2up

I believe these notes cover Proposition 4.4.2 - in the author's notes it appears as Proposition 4.3.7. I spoke with the author about his notes, but that was likely over 10 years ago. Things are a bit hectic at the moment, so I won’t be able to revisit this right now.

The "stars and bars" thing: Consider the example where ##T_{ijk}## is symmetric under the interchange of any pair of indices, and the indices take the values ##1,2,3,4##. For example, all these are the same: \begin{align*} T_{112} = T_{121} = T_{211} . \end{align*} These can all be represented...