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 Yearling
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Mar
4
revised Let $\sigma\in S_n$ be a $k$-cycle, $k>1$. Show that $\sigma^j$ (where $j$ is an integer) is a cycle if and only if $j$ is coprime with $k$
latex corrections.
Mar
4
suggested approved edit on Let $\sigma\in S_n$ be a $k$-cycle, $k>1$. Show that $\sigma^j$ (where $j$ is an integer) is a cycle if and only if $j$ is coprime with $k$
Mar
4
answered Show that if $\lim_{k\to\infty} x_k= -\infty$, then $\lim_{k\to\infty} \frac{1}{x_k} = 0$
Mar
4
comment Laurent series of a function around $t=0$
Sorry, it's quite difficult your integral. A more `appropiated' countour integral would be replacing $C$ by $-C$, that is, $C$ starts in $-\infty+i0$, goes to $\varepsilon+0i$, makes the circle and then goes back to $-\infty+i0$. In that case, if we take $0<\varepsilon <x^/2a$, we obtain that the second pole is not inside the countourn and then it is not necessary to compute its resudue. I hope this helps you, though it is not an answer for your question.
Mar
4
comment Laurent series of a function around $t=0$
A pole cannot be on the countourn $C$, thus you must assume $\varepsilon>x^2/2a$.
Mar
4
comment I make correction, i want to know the sum of this series 1*2+2*3+3*4…+99*100
Do you mean $\sum_{n=1}^{990} (1+\frac{n}{10})$? Or maybe $\sum_{n=1}^{981} (1+\frac{n}{10})$ with the last term equal to $99.1=991/10$.
Mar
4
comment Laurent series of a function around $t=0$
Is $x$ a real number? In this case, if $a>0$, then $-x^2/2a$ is on $C$ unless you assume $\varepsilon>x^2/2a$. Am I right?
Mar
4
comment Is $D=\left \{ x\in\mathbb{R}: \left | x \right |\leq 1 \right \}$ a discrete valuation ring?
No, it is not a ring.
Mar
3
revised What the expression of a one-dimensional representation of $H$
I added the word "unitary".
Mar
3
answered Any books on isospectral manifolds?
Mar
3
answered What the expression of a one-dimensional representation of $H$
Mar
3
comment why $3 - \sin{x}$ is always positive?
$\sin(x)\leq 1$ for all $x$, thus $3-\sin(x)\geq2>0$.
Mar
3
awarded  Yearling
Mar
1
comment Proving this $\gcd(2^n-1,3^n+2)=1$ for all postive integers $n$
You are right for 'small' values of $n$, but for $n =22528$ we have 65537, which is also a prime number.
Mar
1
answered Proving this $\gcd(2^n-1,3^n+2)=1$ for all postive integers $n$
Mar
1
answered Differential in Huybrechts
Mar
1
comment Differential in Huybrechts
$f=f_1+if_2$, thus $\frac{\partial f}{\partial z}=\frac{\partial f_1}{\partial z} + i\frac{\partial f_2}{\partial z}= \frac12\left(\frac{\partial f_1}{\partial x}-i\frac{\partial f_1}{\partial y} + i \frac{\partial f_2}{\partial x}+\frac{\partial f_2}{\partial y}\right)$. A similar work with the others, and then working with your matrix, it would appear the usual matrix in differential geometry with basis $\{\frac{\partial }{\partial x}, \frac{\partial }{\partial y}\}$.
Mar
1
awarded  Organizer
Mar
1
revised Differential in Huybrechts
New tags and the basis of the tangent space is corrected
Mar
1
suggested approved edit on Differential in Huybrechts