Recent content by julian

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...

Say ##i=1,2,\cdots n##, and consider the object ##T_{i_i i_2 \cdots i_k}## that is symmetric under the interchange of any pair of indices. The number of independent components, ##N##, is equal to the number degree-##k## monomials in ##n## variables ##x_1, x_2, \dots, x_n##, i.e., expressions of...

The approach is thorough, though perhaps a bit more detailed than necessary. (1) This symmetry applies to ##i## and ##j##. For each fixed ##k##, ##T_{ijk}## is a symmetric 3×3 matrix in indices ##i,j##. The number of independent components in a symmetric 3×3 matrix is 6. Since ##k## can...

Just to clarify — the expression you gave for the integrating factor wasn’t correct, so I was pointing that out.

It should be ##\int \ln x dx = x \ln x -x +C##, so ##e^{\int \ln x dx} = K x^x e^{-x}##.

Alternatively, following the suggestion made by @Paul Colby. Note, \begin{align*} \Gamma(s)=\frac{1}{s}\Gamma(s+1)=\frac{1}{s(s+1)}\Gamma(s+2)=\frac{1}{s(s+1)(s+2) \cdots (s+n)}\Gamma(s+n+1) \end{align*} What does that tell us about the poles of ##\Gamma (s)^2##? Here are useful formula for...

My initial thought was to approach this using complex contour integration due to the structure of the inverse transform. However, there appears to be a possibly more straightforward way to derive the inverse Mellin transform. Consider \begin{align*} \Gamma (s)^2 = \int_0^\infty \left(...

They are mentioned briefly in Arfken, where it is noted that substituting ##t=\ln x## and ##i \omega = s -c## (possibly a typo — perhaps it should be ##i \omega = s -ic##?) into the Fourier transform and its inverse leads to \begin{align*} G(s) = \int_0^\infty x^{s-1} F(x) dx \qquad \text{the...

Your question could have been phrased more clearly. Try evaluating your integral by first expressing it as ##\sum_{n=0}^\infty \int_n^{n+1} [x] x^{s-1} dx## and then writing out the resulting series. That should make the connection to ##\zeta (-s)## clear—if that's what you're aiming for.

Indeed — as they say, "after a rather lengthy calculation."