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 Jun 4 awarded Yearling 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.