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seen Mar 13 at 4:40

Feb
9
awarded  Commentator
Feb
9
comment Matrix diagonalization
If you need $c \neq 0$, you can WLOG set $c=1$ and scale the other parameters accordingly.
Jan
24
comment $1 + 1 + 1 +\cdots = -\frac{1}{2}$
@Tunk-Fey: I didn't calculate these numbers myself, Mathematica (or any other serious CAS) does it for you.
Jan
24
comment $1 + 1 + 1 +\cdots = -\frac{1}{2}$
@Tunk-Fey: there's no easy explanation. If you know the story about analytically continuing through different patches, it's not difficult to write a computer program to estimate these coefficients. But the precise numbers only follow from serious computations; typically it involves using an integral representation for the $\zeta$ function, contour integrals etc. Another starting point can be the reflection formula for the gamma function.
Jan
23
awarded  Nice Answer
Jan
23
awarded  Yearling
Jan
23
comment $1 + 1 + 1 +\cdots = -\frac{1}{2}$
Although methods like that video seem attractive (because they're not "fancy") it's better to learn the real methods (for example zeta functions). When you need to work with naively divergent series, you should understand how regulators work and in what sense your result is meaningful. In the video, the guy writes down $1 - 1 + 1 -1 + \ldots =$ something, but give an solid argument why that must be true.
Jan
23
answered $1 + 1 + 1 +\cdots = -\frac{1}{2}$
Sep
22
awarded  Editor
Sep
22
revised The inverse Fourier transform of $1$ is Dirac's Delta
added 77 characters in body
Sep
20
comment The inverse Fourier transform of $1$ is Dirac's Delta
This is meant as a constructive remark, but you might want to re-read your argument. We write $f$ as the inverse Fourier transform of the box function of height 1 (or $1/2\pi$) and width $L$, calculate what $f$ looks like in real space, transform back to $k$-space and conclude. If you are confident that the box function is the FT of the Dirac delta, you don't need all these intermediate steps.
Sep
20
answered The inverse Fourier transform of $1$ is Dirac's Delta
Jul
12
answered Heaviside step function squared
Jun
15
comment Infinitesimal $SO(N)$ transformations
I don't fully agree with the calculation. Why repeat the index $i$ twice but sum over $j$ explicitly? It's much cleaner to write $$(R \cdot {}^tR)_{ik} = (\delta_{ij} + \theta_{ij})(\delta_{jk} + \theta_{kj}) = \delta_{ik} + (\theta_{ik} + \theta_{ki}) + \mathrm{O}(\theta^2)$$ to reach the desired conclusion.
Jun
15
comment Infinitesimal $SO(N)$ transformations
You are on the right track. Note that ${}^t \theta_{ij} = \theta_{ji}$ and please be careful with (dummy) indices. You should never write anything like $\delta_{ii}$ in a covariant calculation.
Mar
2
awarded  Supporter
Feb
7
comment Good examples of Ansätze
@Emilio: I wouldn't say that the Ansatz you give is "of course unjustified." You could probably develop a theorem which states that all solutions of certain families of differential equations are given by exponentials. However, it would be a completely trivial statement once you understand how the find the right $k$.
Jan
16
awarded  Teacher
Jan
16
comment Use Taylor Series Expansion in Calculating Integral
You've made a mistake somewhere. $I$ should not depend on $x$, but the expression in the last line does.
Jan
16
answered General solution of $\frac{d^{2}u}{d\rho^{2}}=\frac{l\left(l+1\right)}{\rho^{2}}u$