TITLE: Nth term of 1, 2, 4, 5, 7, 8, 9, 11, 12 ............?
QUESTION [0 upvotes]: How to derive Nth term of 1, 2, 4, 5, 7, 8, 9, 11, 12 ............?
I tried a lot but can't find it.
This sequence is known as Beatty sequence.
It's formula for nth term is a(n) = $$\lfloor(n * \sqrt2)\rfloor$$.
I want to know it's derivation.

REPLY [2 votes]: Of course you can't prove that any sequence with first nine terms being $1, 2, 4, 5, 7, 8, 9, 11, 12$ is beatty sequence. Indeed, you can find infinitely many sequence that start with these as the first nine terms. So, yeah, beatty sequence has these nine terms but the converse is not true.


TITLE: Show that a group of order $1755=3^3\times 5\times 13$ must have a normal Sylow $13$ subgroup or a normal Sylow $5$ subgroup.
QUESTION [2 upvotes]: Question: Show that a group of order $1755=3^3\times 5\times 13$ must have a normal Sylow $13$ subgroup or a normal Sylow $5$ subgroup.
My thoughts: $n_{13}\equiv 1\pmod{13}$ and $n_{13}\mid  135$, so $n_{13}=1$ or $27$.  If $n_{13}=1$, then we have a normal Sylow $13$ subgroup of $G$, and we're done.  If $n_{13}=27$, then let $Q\in{\rm Syl}_{13}(G)$.  As $G$ acts on $Q$ by conjugation, we induce a homomorphism $\phi:G\rightarrow S_{27}$, where $|G:N_G(Q)|=27$, thus $|N_G(Q)|=5\times 13$.  Since $13\neq 1\pmod{5}$, we have that $N_G(Q)$ is cyclic thus abelian.  Let $R\in{\rm Syl}_5(N_G(Q))$, thus $R\trianglelefteq N_G(Q)$.  From here, I am not quite sure how to show that $R$ must then be normal in $G$.
Any help is greatly appreciated!
Thank you.

REPLY [2 votes]: Here is a proof. First, as you noticed, $n_{13}=27$ so there are $12\cdot 27=324$ elements of order $13$ in $G$. Next $n_5$ divides $351$ and $n_5\equiv 1 \pmod{5}$, hence $n_5=351$. Then the number of elements of order $5$ is $351\cdot 4=1404$. The number of elements of order $5$ or $13$ is then $1728$.
There are then at most $1755-1728=27$ elements whose orders are powers of $3$, including $1$. Hence $n_3=1$ and the group $G$ has one normal Sylow 3-subgroup $N$. For every $5$-Sylow subgroup $S$, $SN/N$ is a $5$-Sylow subgroup of $G/N$. If $S'\ne S$ is another Sylow $5$-subgroup, then $SN/N\ne S'N/N$. So $G/N$ has $1404$ elements of order $5$, a contradiction because $|G/N|<1404$.


TITLE: Is there a function $f''(0)$ exists, $f'$ is not continuous on $(-\delta,\delta)$
QUESTION [5 upvotes]: Is  there a function $f\colon(-\delta,\delta)\to\Bbb R$ satisfying the folowing conditions(real number $ \delta\gt0$)?
(i) $f$ is  differentiable on $(-\delta,\delta)$;
(ii) the second derivative of $f$ exists  at $0$, that is $f''(0)$ exists
(iii) there is a sequence $\{x_n\}$, $-\delta\lt x_n\lt \delta$, $\lim\limits_{n\to\infty}x_n=0$,  such that $f'$ is not continuous  at all $x_n$.

I can't construct such  a  example
Any help will be appreciated!

REPLY [8 votes]: The standard way to get a function that is differentiable everywhere (with bounded derivative) but whose derivative has a discontinuity point is
$$ g(x) = \begin{cases} 0 & x=0 \\ x^2\sin(1/x) & x\ne 0 \end{cases} $$
Now select your $x_n$s and define
$$ f(x) = x^2 \sum_{k=1}^\infty \frac{g(x-x_k)}{2^k} $$
This places a discontinuity of $f'$ at each $x_k$, and the overall factor of $x^2$ squeezes the range of $f'$ around $0$ enough to make sure $f''(0)$ exists.


TITLE: Affine transformation matrix doing translation
QUESTION [0 upvotes]: Greetings All
I have some test matlab code which can scale and do rotation but the
translation (tl value) doesn't seem to be working.
I expected the entire object to be moved over x=2 and y=0 but it
doesn't move any ideas?
clf; clear all
rp=[0 4 2 -1 0; 0 1 -4 -3 0;0 0 0 0 0] %x-y values
scalenum=2 %scale factor
tx=2;ty=0; %translate values
%
for theta=0:45:90
    tl=[1 0 tx; 0 1 ty;0 0 1] %translation
    rot=[cosd(theta) -sind(theta) 0; sind(theta) cosd(theta) 0;0 0 1];
%rotation
    scale=[scalenum 0 0;0 scalenum 0;0 0 1] %scale
    rt=tl*rp %new translation
    r=rot*rp %new rotation of points
    rs=scale*r %new scale with rotation
    hold off
    plot(rt(1,:),rt(2,:))
axis([-8 8 -8 8])
grid on
pause(.5)
end;

REPLY [0 votes]: I'm highly suspicious of
rp=[0 4 2 -1 0; 0 1 -4 -3 0;0 0 0 0 0] %x-y values

If you want to do translation you don't need (x, y): you need (x, y, w), where w is the reciprocal scaling factor. So all your points are at infinity, and translating them isn't going to do much.
rp=[0 4 2 -1 0; 0 1 -4 -3 0;1 1 1 1 1] %(x,y,w) values

should get you somewhere.


TITLE: Power series development of inverse function
QUESTION [0 upvotes]: Problem:

Let $G=\{z\in \Bbb{C} : | \text{Im }z|< \pi/4, \text{Re }z> \pi/4 -1 \}$ and $f:G\to f(G), z\mapsto z e^z$.
Prove that $f$ is injective on $G$ and compute the first three coefficients $a_0, a_1, a_2$ of the power series development in $0$ of $f^{-1}:f(G)\to G$.

I have proved that $f$ is injective on $G$ but I don't see a way of computing $a_0, a_1, a_2$. How do we go about doing that? Do we try to find an explicit form of the inverse function?
Thankful for any hints.

REPLY [0 votes]: $(f^{-1}(z))'=\frac {1}{f'(f^{-1}(z))}$ and from there you can find the other coefficients...


TITLE: What is the definition of coefficient?
QUESTION [1 upvotes]: How to define coefficient? And what does it mean when someone says: put the $i$th term to be the coefficient of $x^i$

REPLY [2 votes]: It means the constant multiplying the term $x^i$.
E.g., If f(x) = $x^3 + 2x^2 + x + 9$, then the coefficient of $x^3 = 1\cdot x^3$ is $1$, the coefficient of the term $x^2$ is $2$, the coefficient of the term $x = x^1$ is $1$, and the coefficient of the term $x^0 = 1$ is $9$.
So "put the $i$th term to be the coefficient of $x^i$" would require knowledge of the domain of $i$. Suppose the $4$th term is $3$. Then we would multiply $x^4$ by $3$, to get $3x^4$.
Note: In a polynomial such as $4x^2 - 1$, the coefficient of the "absent" term "$x$" can be said to be $0$.


TITLE: What is solution to this integration of $sec^{2}x$, limits of integration being from $\pi/6$ to $\pi/2$
QUESTION [0 upvotes]: What is solution to this integration of $sec^{2}x$, limits of integration being from $\pi/6$ to $\pi/2$. The pi/2 screws the integration. What should i do ?
This is part of integration of double integral over a region, in cartesian coordinated no problem arose, but when i changed to polars, this integral showed up
Original question is
http://i.imgur.com/6QN2CJV.png
I solved this with cartesian, i will be thankful if someone does it in polar coordinates

REPLY [2 votes]: Let $\displaystyle I = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\sec^2 xdx = \left[\tan x\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}} = \lim_{a\rightarrow \frac{\pi}{2}}\tan (a)-\tan \left(\frac{\pi}{6}\right).$