4,712 reputation
917
bio website motls.blogspot.com
location Czech Republic
age 40
visits member for 3 years, 5 months
seen Aug 4 at 12:45

Hi, I am a string theorist and a publicist.


Jun
10
comment Question about direct sum of function space
Dear @vonjd, your problem already starts with ${\mathbb R}^3$: the first sentence says that $V$ itself are functions from $U$ which is a subset of ${\mathbb R}^3$. $V$ itself are functions that take value in ${\mathbb R}$ and $V\oplus V\oplus V$ are functions from ${\mathbb R}^3$ to another ${\mathbb R}^3$. But functions from ${\mathbb R}$ never appear in your problem at all so I don't understand in what sense it would be "natural". Real numbers and their 3rd power are equally natural but only the latter appear in your problem. The tripling only affects the value of the function not the domain
Jun
10
awarded  Enlightened
Jun
10
comment Finding power series representation of $ \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt {1 - k^2\sin^2{x}}}\;{dx}$
It's just a small digit "2" that someone missed, but you didn't. Maybe they thought it was a mark for a footnote. ;-)
Jun
10
revised Finding power series representation of $ \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt {1 - k^2\sin^2{x}}}\;{dx}$
added 329 characters in body; added 1 characters in body
Jun
10
comment Finding power series representation of $ \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt {1 - k^2\sin^2{x}}}\;{dx}$
Oh, I eventually noticed that it's actually the same thing. In that case, I am sure that your result is correct. In the original formula, the whole big ratio in the parentheses should be squared, otherwise it's correct.
Jun
10
awarded  Nice Answer
Jun
10
answered Finding power series representation of $ \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt {1 - k^2\sin^2{x}}}\;{dx}$
Jun
9
comment A radius for which the inverse of a function is well defined (Inverse Function Theorem)
The four derivatives $\partial f_{1,2} / \partial (x,y)$ are all nonzero at $(1/2,0)$ and the matrix is non-singular over there, so it's clear that in a vicinity of the point, the inverse exists.
Jun
9
comment A radius for which the inverse of a function is well defined (Inverse Function Theorem)
Maybe they just want an approximate formula for the inverse? I can't imagine how the formulae could simplify. Or maybe you're allowed to use some symbols inside? Something that makes the solution "non-explicit"?
Jun
9
answered Question about direct sum of function space
Jun
9
answered A radius for which the inverse of a function is well defined (Inverse Function Theorem)
Jun
9
comment A radius for which the inverse of a function is well defined (Inverse Function Theorem)
I see, thanks, it's a comma. ;-)
Jun
9
comment A radius for which the inverse of a function is well defined (Inverse Function Theorem)
Sorry, I may be missing something, but how can you invert a function of two variables? Your function is from $R^2$ to $R$, so it cannot be a simple function, so it can't be (fully) inverted. The inverse function would have to be from $R$ to $R^2$, right?
Jun
9
comment Finding number of matrices whose square is the identity matrix
Hi, every $9\times 9$ matrix $A$ may be brought into the standard form $A=CDC^{-1}$ for a $D$ which is either diagonal or has the Jordan blocks on the diagonal. That's a basic result in algebra. In this form, $A^2 = CDC^{-1}CDC^{-1} = CD^2 C^{-1}$. It should be equal to $I = CC^{-1}$ which implies $D^2=I$. So $D$ has to have $\pm 1$ eigenvalue and one may check that the nondiagonal Jordan blocks would fail to produce $D^2=1$, too.
Jun
9
revised Finding number of matrices whose square is the identity matrix
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Jun
9
answered Finding number of matrices whose square is the identity matrix
Jun
8
answered What's the value of $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$?
Jun
6
revised Bounding ${(2d-1)n-1\choose n-1}$
added 199 characters in body
Jun
6
revised Bounding ${(2d-1)n-1\choose n-1}$
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Jun
6
answered Bounding ${(2d-1)n-1\choose n-1}$