TITLE: Find limit of $x_n =n2^{-n}, n\in\mathbb{N}$
QUESTION [0 upvotes]: By writing out the first few terms of the sequence, I see that it is a decreasing sequence (monotonic non-increasing) and want to show it converges to 0. But I don't know how to manipulate the expression and I want to find an $N$ and an $\epsilon$ to prove the limit exists formally.

REPLY [0 votes]: We would like to show that for every $\phi > 0$ satisfaing $\frac{n}{2^N} < \phi$ there exists $n_0$ that for all $n > n_0$ holds $\frac{n}{2^N} < \phi$. But $\frac{2^n}{n} > \frac{n^2}{n} = n > \frac{1}{\phi}$ so it is sufficient to take $n_0 = \lfloor{\frac{1}{\phi}}\rfloor$.


TITLE: Is the space of solutions to the Queens Domination Problem connected?
QUESTION [11 upvotes]: A configuration of queens on an 8 by 8 chessboard (or n by n if you like) is a queen domination if every square on the board lies in the same row, column, or diagonal as at least one of the queens. The Queens Domination Problem is to find the minimum number of queens necessary for a queen domination. A solution to the Queens Domination Problem is a queen domination using the minimum number of queens.
For the 8 by 8 chessboard, brute force has shown that 5 queens is the minimum number. More discussion about that here.
The adjacency relation on the set of solutions to the Queens Domination Problem is defined as follows: we say solutions are adjacent if they differ by only the placement of one queen. For example: C3, E4, D5, B6, F4 is adjacent to C3, E4, D5, B6, F2.

How many equivalance classes are there in the equivalance relation generated by adjacency?

In other words, starting with one solution, can we reach any other solution by moving one queen at a time, such that the result of each move is itself a solution?

REPLY [11 votes]: Revised and correct:
589 equivalence classes.
http://www.math.ucsd.edu/~etressle/classes.txt
It seems that the pairs (#vertices,#components) are (1,388) (2,100) (3,40) (4,34) (5,20) (18,4) (20,2) (3804,1) – damiano 15 hours ago
1: Class I.D. 2, 4, 5, 6, 7,..., 589.
2: Class I.D. 22, 29, 35, 47, 48,..., 579.
3: Class I.D. 28, 33, 38, 49, 58,..., 574.
4: Class I.D. 13, 90, 95, 126, 141,..., 576.
5: Class I.D. 32, 37, 62, 89, 97,..., 395.
18: Class I.D. 17, 31, 185, 303.
20: Class I.D. 3, 75.
3804: Class I.D. 1.


TITLE: Line integral calculation for a line segment
QUESTION [0 upvotes]: Compute the line integral $\int_y e^z dz$ where $y$ is the line segment from $0$ to $z_0$.

Since there is only one variable z here, I can directly compute $\int_y e^z dz=\int_0^{z_0} e^z dz=e^{z_0}-1$. I assume $z_0$ means a fixed point. Am I missing something or is this it?

REPLY [1 votes]: It is not that "there is only one variable $z$ there" you can directly do so, but that when $f$ has a primitive $F$, i.e., $F'(z)=f(z)$ in the domain (that contains the path $\gamma$),
$$
\int_\gamma f(z)=F(B)-F(A)
$$
where $A$ is the initial of $\gamma$ and $B$ the final point of $\gamma$.
In general, you can't calculate $\int_\gamma f(z)dx$ in this way. Consider for instance $f(z)=\frac{1}{z}$.


TITLE: Locus of a particular geometric situation
QUESTION [0 upvotes]: Two fixed points S(a,b) and T(c,d) are 4 units apart and are on the same side of a moving line L. If perpendicular is distance of S and T from line L are p1, p2 respectively, then p1 and p2 satisfy
p1+3p2=k where K is a constant number.
  It is given that the line L always touches a fixed circle C.

Find the.centre and radius of the circle C in terms of k and coordinates of S AND T.
An attempt at a solution
I took a general line y=mx+c and tried to solve manually for
p1+3p2=k . However I got mingled up with the variables and the locus did not seem like a circle.
I was looking for a hint or an alternate solution to the problem.

REPLY [1 votes]: You’re given that there’s a common tangent circle for all possible lines $L$ that meet the criteria in the problem, which greatly simplifies your task: all you really have to do is find this circle’s center and radius.
Observe first that, by symmetry, the center of $C$ must be colinear with $S$ and $T$: for any line $L$ that meets the criteria, its reflection in the line through $S$ and $T$ does, too. I’ll assume that $k$ is meant to be some arbitrary positive constant.
Place $T$ at the origin and let $S=(0,4)$. I’ll use $r_S$ and $r_T$ for the respective distances to $L$ to make it easier to keep track of which is which. The line $L$ is a common tangent to the circles with centers $S$ and $T$ and radii $r_S$ and $r_T$, respectively. For $S$ and $T$ to be on the same side of $L$, it must be one of the common outer tangents of these two circles.

Consider first the case $r_S=r_T$. In this case, $L$ is one of the two vertical lines $x=\pm\frac k4$, therefore $C$’s radius is equal to $\frac k4$. Now let $r_S\ne r_T$. Using similar triangles, we have for $L$’s $y$-intercept $b$ $${b \over r_T} = {b-4 \over r_S} \tag 1$$ so $$b = {4 r_T \over r_T-r_S} = {4 r_T \over 4 r_T - k}.$$ $C$ is also tangent to $L$. Applying similar triangles again using its center $(0,c)$ we have $${c-b \over k/4} = {-b \over r_T}, \tag 2$$ therefore $$c = {4r_T-k \over 4r_T}b=1.$$ There is a second tangent circle to $L$ with radius $\frac k4$ centered on the $y$-axis, but for that circle $c=-{k+4r_T\over k-4r_T}$ (you can compute this by squaring both sides of eqn. (2)), which we can reject because it depends on $r_T$, so isn’t common to all possible $L$.
Finally, the general solution is obtained from this special case via a rigid motion of the plane. This doesn’t affect a circle’s radius, therefore $C$’s center is one-fourth of the way from $T$ to $S$, i.e., at $\frac14 S+\frac34T$ and its radius is equal to $\frac k4$. I expect that the way this solution reflects the ratios of $r_S$ and $r_T$ to $k$ is not coincidental and that there’s some more clever way to obtain this result using those ratios. Indeed, if you generalize this problem by letting $S$ and $T$ be any points any distance apart and require that $k=r_S+\frac{1-\lambda}\lambda r_T$ for some fixed $\lambda\in(0,1)$, the common tangent circle will have center $\lambda S+(1-\lambda)T$ and radius $\lambda k$.


TITLE: What would be the graph of $P(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)\cdots$ (with infinitely-many factors)?
QUESTION [3 upvotes]: Suppose A polynomial $P(x)$, such that,
$$P(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)\cdots$$
having infinite roots. Where $x_1$, $x_2$, $x_3$, $x_4$, $\ldots$ are roots of the polynomial.
My question is what would be the graph of this function? Will it be just the $x$-axis or there will some other curve?

REPLY [2 votes]: When you have something like this you have to deal with infinite series and radius of convergence and the result is very sensitive to the way in which you define your objects analytically.
I think a good example for this comes from number theory and the theory of partitions. In there the following equation (due to Euler) is proven as a formal expansion:
$\prod_{k = 1}^{\infty}(1 - x^k) = 1 + \displaystyle\sum_{j = 1}^{\infty}(-1)^j(x^{j(3j-1)/2} + x^{j(3j+1)/2})$.
This is a formality between series, without reference to actual values of x, and is obtained byt multiplying the product and collecting powers of $x$ (the proof of it is a different story, and it's called pentagonal number theorem).
Nevertheless you can study both sides and study radius of convergence. For example, with the right hand side, if $|x| \le r < 1$ then we have
$\displaystyle\sum_{j = 1}^{\infty}\left|(-1)^j(x^{j(3j-1)/2} + x^{j(3j+1)/2})\right| \le 2\displaystyle\sum_{j = 1}^{\infty}r^{j(3j-1)/2} \le 2(1 + r  + r^2 + ...) = \dfrac{2}{1 - r}$,
which proves this functions converges absolutely and uniformly in compacts of $(-1, 1)$. This means there exists a continuous function to which this series converges and so it makes sense to graph it, but it might not be the graph of a polynomial and it only makes sense in the region of convergence.
Analogously you can check what happens with the product and obtain a similar convergence phenomenon (by expanding the products into polynomials you can basically repeat the argument above). You can see the first products here:

If instead you do it with the polynomials you get (I did it for k = 4):

These are starting to look similar and it indeed seems some function is at the limit indeed.
Notice how all these graphs go through (0, 1) since in both the product and the series version, in the terms that form the sequence these graphs all go thorugh there. On the other hand, in the product version all go through $(\pm, 0)$ since these are zeros of all finite products, meanwhile in the series version it doesnt have those zeroes yet, it just approaches to it in the limit.
Finally, notice how outside $[-1, 1]$ the graphs do not seem to go nowhere, and that makes sense since  in there the behaviour at $\pm\infty$ dominates and the degree of the polynomial changes the behaviour.


TITLE: Find point farthest on $(x-5)^2+(y+1)^2+(z-7)^2=69$ from point $(2, 3, -6)$
QUESTION [1 upvotes]: So, I have the equation of a sphere $$(x-5)^2+(y+1)^2+(z-7)^2=69$$ and a point, $P(2, 3, -6)$. I am supposed to find the point on the sphere farthest away from $P$.
I tried to figure out how to use parametric equations to represent the sphere and maximize that using the distance formula. However, they are multiple variables in this equation, and this led me to nowhere.

REPLY [0 votes]: The center of the sphere is $C(5,-1,7)$. The line passing through $C$ and $P(2,3,-6)$ has parametric equations
$$
\begin{cases}
x=5+(2-5)t&x=5-3t\\
y=-1+(3-(-1))t&y=-1+4t\\
z=7+(-6-7)t&z=7-13t\\
\end{cases}
$$
Substitute in the equation of the sphere
$$(x-5)^2+(y+1)^2+(z-7)^2=69$$
$$(5-3t-5)^2+(-1+4t+1)^2+(7-13t-7)^2=69$$
$$194 t^2 = 69\to t=\pm\sqrt{\frac{69}{194}}$$
the point on the sphere farthest away from $P$ is:
$$\left(3 \sqrt{\frac{69}{194}}+5,-2 \sqrt{\frac{138}{97}}-1,13 \sqrt{\frac{69}{194}}+7 \right)$$


TITLE: Proving the inradius (r) for an equilateral triangle
QUESTION [0 upvotes]: I am trying to prove the basic properties of an equilateral triangle given in the image. I can do the first three, but am stumped at the fourth one.
Any hints?

REPLY [0 votes]: Hint:
The medians in a triangle  intersect at the centroid, which is located on each median at the $2/3$ of  the median length from the corresponding vertex.


TITLE: A Normal distribution inequality
QUESTION [0 upvotes]: Denote by $\phi(z)$ and $\Phi(z)$ the density and distribution of a standard Normal variable (mean=0, sd=1), respectively. It appears that for $z > 0$,
$$
\frac{\phi(z)}{1 - \Phi(z)} < z + \frac{1}{z}
$$
Is there a proof (or disproof) of this?

REPLY [1 votes]: A similar question has been answered Here
Anyway, this is an alternative proof:
Let's rewrite your inequality in the following way
$$\frac{1-\Phi(z)}{\phi(z)}>\frac{z}{1+z^2}$$
$$1-\Phi(z) > \phi(z)\frac{z}{1+z^2}$$
Let's define the following function:
$g(z)=1-\Phi(z)-\phi(z)\frac{z}{1+z^2}$
As:

$g(0)=\frac{1}{2}>0$
$g'(z)=-2 \frac{\phi(z)}{(1+z^2)^2}<0$ for every z
$g(\infty) \rightarrow 0$

thus $g(z)$ is always positive... and this immediately proves what you are requested to do


TITLE: How to show that: $ \gcd \Big(2^{2^a}+1 , 2^{2^b}+1\Big)=1$
QUESTION [2 upvotes]: How to show that:
$$ \gcd     \Big(2^{2^a}+1 , 2^{2^b}+1 \Big)=1$$

REPLY [1 votes]: We know $(x-y)\mid(x^c-y^c)$ where $x,y,c$ are natural numbers.
Putting $x=a^{2^n},b=-1, c=2^{m-n}$
$$\{a^{2^n}-(-1)\}\mid \{(a^{2^n})^{2^{m-n}}-(-1)^{2^{m-n}}\}$$ if $m>n$ where $a,m,n$ are natural numbers.
$$\implies (a^{2^n}+1)\mid (a^{2^m}-1)$$   if $m>n$
Now, $$(a^{2^m}-1,a^{2^m}+1)=(a^{2^m}-1,a^{2^m}+1-(a^{2^m}-1))=(a^{2^m}-1,2)$$ is $2$ if $a$ is odd, is $1$ if $a$ is even
So, $$(a^{2^n}+1,a^{2^m}+1)$$ will be $1$ or $2$ accordingly.