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917
bio website motls.blogspot.com
location Czech Republic
age 40
visits member for 3 years, 6 months
seen Aug 4 at 12:45

Hi, I am a string theorist and a publicist.


Aug
9
comment Plis, What is the orthogonality conditions for associated legendre polynomials with both two different indexes
If some indices are equal, the results are mathworld.wolfram.com/AssociatedLegendrePolynomial.html
Aug
2
comment Why does $1+2+3+\dots = -\frac{1}{12}$?
Dear @AxelBoldt, $0+1+2+3+4+\dots $ is also demonstrably and always equal to $-1/12$, and it may be shown by pretty much the same proof. And $1+1=2$. If all entries except for a finite number are zero, then one adds a finite number of terms which always has uncontroversial rules.
Mar
6
comment Why does $1+2+3+\dots = -\frac{1}{12}$?
Sure, it's a completely analogous sum. Just like $1+2+3+\dots$ quantifies the ground state energy in 1+1 dimensions (of a string), $1+1+1+\dots$ quantifies the ground state's charge (of a system of free fermions, for example). It's really the reason why the two degenerate states $|0\rangle$ and $c_0|0\rangle$ have ghost numbers $\pm 1/2$, for example.
Feb
10
comment Why does $1+2+3+\dots = -\frac{1}{12}$?
It might be but it is a wrong conclusion. The key property of this $\infty$, which may be subtracted by a counterterm, is that it is independent of all independent variables, so it is formally a scale-invariant constant, and $0$ is the number to be associated with it. The fact that the sum of integers equals $-1/12$ is perhaps the most important fact about mathematics among the laymen - and among mathematicians without breadth and depth.
Jan
24
comment $1 + 1 + 1 +\cdots = -\frac{1}{2}$
Dear @Pedro, I am saying that $\zeta(2k+1)$ are transcendental numbers because I am ready to bet all my wealth on the validity of the claim. After all, even $\pi^n$ are transcendental numbers for rational nonzero values of $n$ so even zeta of even positive integers are (provably) transcendental. But the zetas of odd positive integers are (I claim without a rigorous proof) much more transcendental than the powers of $\pi$. I never claimed that I had a rigorous proof of the transcendentality but if you want to claim that I said something incorrect, you should have a proof which you don't have.
Jan
24
comment $1 + 1 + 1 +\cdots = -\frac{1}{2}$
Nope, it's just believed by me and almost everyone that they're transcendental. The much "easier" proof that zeta(3) is irrational came just recently, in 1978, see en.wikipedia.org/wiki/Ap%C3%A9ry's_theorem
Jan
23
comment $1 + 1 + 1 +\cdots = -\frac{1}{2}$
Nice, +1. An alternative calculation of $\zeta(0)=-1/2$ appears early in this article: motls.blogspot.com/2014/01/… - See my comment under the OP's question for 4 more articles about the topic if you wish...
Jan
23
comment $1 + 1 + 1 +\cdots = -\frac{1}{2}$
Five more articles with 8 different ways to compute the sum of integers (most of the methods extend to the "sum of ones", too) etc.: motls.blogspot.com/2007/09/… motls.blogspot.com/2011/07/… motls.blogspot.com/2014/01/… motls.blogspot.com/2014/01/… motls.blogspot.com/2014/01/…
Jan
11
comment Why does $1+2+3+\dots = -\frac{1}{12}$?
"Incorrect", "not right", and "opposite of right" are the same thing, also known as "synonyma".
Sep
20
comment The inverse Fourier transform of $1$ is Dirac's Delta
Well, I think you would also have to discuss that the negative regions of $(\sin x)/x$ don't combine to other distributions like $\delta'$ etc. But otherwise OK, I upvote it.
Jun
25
comment Fourier transform of a signal sequence?
I would be sort of surprised if there is an analytic formula for the Fourier transform of this contrived function.
May
24
comment Grover Algorithm Orthogonal vectors
Sorry, I made a mistake. I meant that $\omega$ and $s'$ are orthogonal. My answer is hopefully fixed. I don't understand the question "how I will be able to define $\sqrt{N}|\omega\rangle$". It's just multiplication, right? One should know how to multiply vectors by numbers.
Dec
9
comment Rotation matrix in terms of axis of rotation
It's just a conjugation of the simple matrix for a rotation around the $z$-axis, which is effectively just 2 x 2 matrix, by another rotational matrix that rotates the North pole to the point $V$, which is a product of rotation in the theta-direction and the phi-direction to get where you need to get. Conjugation by $U$ is $URU^{-1}$ where the product is matrix product. It's possible ineffective to write these things without matrices so if you don't know matrices, this is a reason to learn them. At any rate, it's not really physics, it's linear algebra and geometry and a basic one.
Dec
2
comment Kummer's Equation
Dear Alex, the equation for $L_n$ is nothing else than a coefficient in front of $t^n$ in the Taylor expansion of the equation for $g$ with respect to $t$. You just compare the terms term-by-term. Please just expand my equation for $g$ as a Taylor expansion in $t$ and don't ask any more questions.
May
23
comment A general pattern to find the roots of the classical lie algebras
Sorry, your updated question got very confusing. I was answering your original question. The new question talks about "levels" etc. There aren't levels in ordinary Lie algebras. The roots are very simple objects and be sure I could easily enumerate all of them for all 7 classes of the Lie algebras, A,B,C,D,E,F,G. They're just the non-negative integral combinations of the simple roots that have the right length (either the same as simple roots or, in non-simply-laced groups, sqrt2 or sqrt3 times longer). ... If you have an algebra, the roots are just defined as the eigenvalues under the Cartan.
Nov
5
comment How do I determine the measure for a volume integral?
Sorry, this was the integral of $r^2=x^2+y^2+z^2$. If I calculate the correct $r_z^2$, the squared distance from the $z$-axis, it's only $x^2+y^2$ so I only get two terms out of three. Due to the spherical symmetry, it's $2/3$ of the result above, so that the moment of inertia is $(2/5)MR^2$ for a solid ball.
Nov
5
comment How do I determine the measure for a volume integral?
Let me calculate the moment of inertia of a ball of radius $R$ here. I choose spherical coordinatets so $dV=r^2\,dr\,d\Omega$. The integral over $d\Omega$ gives $4\pi$, the integral of $r^4dr$ (there was $r^2$ from the moment) from $0$ to $R$ gives $R^5/5$: I can factorize them here. So the result is $4\pi \rho R^5/5=3/5 MR^2$. Less symmetric shapes are more complicated but it's pure maths.
Nov
5
comment How do I determine the measure for a volume integral?
Also, your question as well as your comments below my answer make it spectacularly self-evident that you don't understand what either of these symbols mean, in a striking contradiction with your assertion.
Nov
5
comment How do I determine the measure for a volume integral?
Dear @jnm2, the first equation in your comment above isn't a "magic transformation": it's a totally trivial identity. $\int$ is the integral sign but it's still a sum of infinitesimal pieces. The infinitesimal pieces are proportional to $dV$ and one may choose cubic shapes of $dV$ to make it clear that it may be written as $dx\,dy\,dz$. I don't understand what's your problem. If you can't understand this simple thing about the volume integral, you should give up studying physics quantitatively. Also, I wrote you how you can calculate the triple integral for any shape you mentioned.
Jul
9
comment True, false, or meaningless?
I see, so $\forall i\in \{\}: T(i)$ holds for any $T(i)$ because there is no counterexample in an empty set for which $T(i)$ would fail - it holds for everyone (all zero of them). On the other hand, $\exists i\in\{\}: T(i)$ is always untrue because there doesn't exist any $i$ in an empty set that has a property - whatever property - because there's nothing in an empty set even without adjectives.