How do we get the following solution $$\overrightarrow{h}(t)=\underline{R}(t, \overrightarrow{\xi}(\overrightarrow{x}))\overrightarrow{h}(0)$$ (where $\underline{R}(t, \overrightarrow{\xi}(\overrightarrow{x}))$ is the matrix of transformation as for $t$ around the axis $\overrightarrow{\xi}(\over...
as a rule of thumb, logs are negligible compared to powers, so test for the convergence of x^(-a) and then try to transplant that result into your situation
and then for the case a = 1, the comparison test with 1/x^a doesn't help us because the integral from 1 to infty of 1/x diverges because it's the log x
which sums to infinity
if you want the background, my book includes this question (math.stackexchange.com/q/1308866/93114) (I didn't ask the math SE question) and it stumped me.
the hint in the book matches the answer, i.e. a = 1 and -2 < b < -1, but I'm having trouble confirming that
it seems pretty clear to me that I need a function of the form (x^(-a) |log x|^b)^(1/p), because then when q = p, it reduces to the function x^(-a) |log x|^b, which is what I was trying to show above, or at least a piece of it
you know $U(15)$ has order $\varphi(15)=(3-1)(5-1)=8$. if it had an element of order $8$, it'd be cyclic isomorphic to $C_8$ which has a unique element of order $2$. verify that $U(15)$ has more than one element of order $2$, hence cannot be iso to $C_8$.
the elements of $H\times K$ of order 2 are of the form $(h,k)$ where $h\in H$ and $k\in K$ are both either identity or order 2 (and not both the identity)
so if $h\in H$ and $k\in K$ have order 2, then $(h,e_K)$ and $(e_H,k)$ both have order 2
say $S\subset \Bbb R^{\Bbb N}$ is the set of all strictly positive sequences of reals. Supposedly the sequence $(1, 1, ...), (1/2, 1/2, ...), (1/3, 1/3, ...)$ of elements of $S$ doesn't converge to $(0, 0, 0, ...)$. I can't see why not though.
@anon, tell me a nice/systematic way to show that $\mathrm{Aut}(\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z) \cong S_3$. I have no clue where to start apart from listing homs to find which are isos. I think you said something about considering generators, but I can't remember properly.
suppose $[0,1]=\{x_1,x_2,x_3,\cdots\}$ and $x_1=0.x_{11}x_{12}x_{13}\cdots$ and $x_2=x_{21}x_{22}x_{23}\cdots$ and so forth are the decimal expansions. then construct $y=0.y_1y_2y_3\cdots$ with $y_i\ne x_{ii}$ for all $i$, so that $y\ne x_1,x_2,\cdots$ hence $y$ is not in the list.
indeed if $a,b$ are any two nonidentity elements of $K_4$ (this group is called Klein four) then $K_4=\langle a,b\rangle=\langle a\rangle\times\langle b\rangle$
So if $c,d$ is any other pair, we have an isomorphism $\langle a\rangle\times\langle b\rangle\to \langle c\rangle\times\langle d\rangle$
which we then can note is an automorphism $K_4\to K_4$
there are exactly 6 ordered pairs of distinct elements, so there are at least 6 distinct isomorphisms here
which is an upper bound too, so these are all automorphisms
(that's the abstract way to go about the problem)
you can of course just directly verify that all six permutations of nonidentity elements are automorphisms
Just starting a textbook called 'Basic topology' by Armstrong, and it seems $\Bbb E$ is defined as a Euclidean space, is this just alternative naming to $\Bbb R$?
I mean $\Bbb R$ usually has standard Euclidean structure
$\Bbb Z/p\Bbb Z$ is a field when $p$ is prime, since $p\Bbb Z $ is a maximal ideal of $\Bbb Z$ when $p$ is prime, etcetc. It's an abelian group under addition for non-prime $p$, it's an abelian group under multiplication when $p$ is prime - else it is a monoid under multiplication
Linear algebra question: I know that $A = QQ^{T}$ ($A$, $Q$ square matrices) is positive definite if and only if $Q$ is invertible for every choice of $Q$. Since the product of invertible matrices is invertible, would it be safe for me to say that $A = QQ^{T}$ is invertible if and only if $A$ is positive definite?
i'm not getting this. then 1 can go to any element, which makes the number of automorphisms . . . p? it should be $p-1$, right? and even if I get the number of automorphisms, how do I show that they are all $\cong C_{p-1}$?
But I only have to show that $Aut(C_p) \cong C_{p-1}$. I don't have to show that $Aut(C_p) \cong (Z/pZ)^*$. $(Z/pZ)^*$ is cyclic is a different exercise (the one I posted a pic of right now).
Showing ${\rm Aut}(\Bbb Z/p\Bbb Z)\cong(\Bbb Z/p\Bbb Z)^\times$ is easy, but showing $(\Bbb Z/p\Bbb Z)^\times\cong\Bbb Z/(p-1)\Bbb Z$ is harder and is basically the content of that image you just posted
I know that $A = QQ^{T}$ ($A$, $Q$ square matrices) is positive definite if and only if $Q$ is invertible for every choice of $Q$. Since the product of invertible matrices is invertible, would it be safe for me to say that $A = QQ^{T}$ is invertible if and only if $A$ is positive definite?
your usual mental picture of a torus (a donut) is not flat, since at any point it's got nonzero principal curvatures. but there are flat tori if you consider it abstractly or else embed it into higher dimensions.
hi - off topic: as a conformal map, does it make sense to take the square root of the plane (to get to the upper half plane) or does it make more sense to take the square root of the slit plane (with a slot from $[0,\infty)$?
@anon I have a plane slit from $[0,10]$. I want to get to the upper half plane. Can I take square root there or should I extend the slit to $[0,\infty)$?
@anon I thought it was $[0,\infty)$. Isn't that what is used in the last step of the answer to <http://math.stackexchange.com/questions/1162754/conformal-mappings-dealing-with-slits> ?
sorry, don't know how to post links correctly here
I know that $A = QQ^{T}$ ($A$, $Q$ square matrices) is positive definite if and only if $Q$ is invertible for every choice of $Q$. Since the product of invertible matrices is invertible, would it be safe for me to say that $A = QQ^{T}$ is invertible if and only if $A$ is positive definite?
