Space filling curve with locality for odd column/row count First of all: I don't have a mathematical background, I'm looking for something I can eventually implement in software.

What I'm looking for is a space filling cure with locality, like the hilbert curve, but one instead that can also handle odd number of rows/pixels. E.g. hilbert curve requires powers of two (2x2, 256x256, etc) and I would like one that also works with 81x81 or 59x59, you get the drift.

Comment

Actually the minimum square that can contain an iteration of Hilbert's curve has an odd number of pixels on each side:

### ###
# # # #
# ### #
#     #
### ###
  # #
### ###

So I assume you are actually thinking of a grid of cells, and one long path through all the cells exactly once each, such that the displacement between two points on the path is bounded by a constant times the grid side length times the fraction of the total number of cells traversed.

Solution

I don't see an easy way to use Hilbert's curve, but there is another space-filling curve that is easy to use. Here is the basic structure (O represents an endpoint and a line represents a connection):

O  .  O-O
 \ . / . \
  O.O  .  O
..|.|.....|
  O.O  .  O
 / . \ . /
O  .  O.O
|.....|.|..
O  .  O.O
 \ . / . \
  O-O  .  O

Notice that the whole region is divided into $3 \times 3$ smaller regions of roughly equal size, and if each can be completed using a suitable path with endpoints at opposite corners, then connecting those $9$ paths according to the the above construction will produce a suitable path for the whole region. But it is impossible to find such a path for any region with length and width both even, and to avoid that case we would need to first use a reduction to make either the length or width odd, and then ensure that on each division the $9$ smaller regions are also of the same nature, which is possible because the odd side can always be divided into $3$ odd parts unless it is $1$.

Reduction

Let $m,n$ be the dimensions of the original grid

If $m$ is odd or $n$ is odd:

  The above solution works

If $m,n$ are both even:

  $m,n$ can each be divided into $2$ odd parts of roughly equal size

  Divide the original grid into $2 \times 2$ regions accordingly and use the following structure:

  O-O
 / . \
O  .  O
|.....|
O  .  O
 \ . /
  O-O

  Now each small region has odd length and width, and the above solution works on each of them

Base cases

The base cases for the above recursive solution are all trivial, being of dimensions $(1,k)$ for some integer $k$.

Proof of locality

This is left as an exercise, but intuitively the recursive structure ensures that to move between any two nonadjacent regions at any level requires traversing an entire region at that level.

I do not have a strong mathematical background neither, but I think that the following trick would work: you have to consider a $59 \times 59$ grid as a multilevel refined grid. Here is an example how to achieve a $5 \times 5$ grid:

Then you put a multilevel hilbert curve (or a multilevel z-curve: see image) on this grid and you have your SFC on an odd column/row count. I know that your grid is regular, but you use the irregular grid (square with different sizes) to do your numbering, and you map this irregular grid to your original grid keeping the same numbering.


Show that $2\cos \frac{2\pi}{7}$ satisfies $x^3+x^2-2x+1$ Show that $2\cos \frac{2\pi}{7}$ satisfies $x^3+x^2-2x+1$

I need to use Galois Theory. I know about the identity $\cos 3\theta = 4\cos^3 -\cos \theta$, but I can't fit things into the polynomial.

Consider $\frac{x^7-1}{x-1}=\Phi_7(x)=1+x+x^2+x^3+...+x^6$

It has his zeros in $\exp(\frac{2\pi i k}{7})$ with $k \in [0,1]$

If $\alpha$ is a root of $\Phi_7$ then $$\left(\alpha^3 +\frac{1}{\alpha^3}\right)+\left(\alpha^2 +\frac{1}{\alpha^2}\right)+\left(\alpha+\frac{1}{\alpha^1}\right)+1=0$$
$$\left(\alpha^2 +\frac{1}{\alpha^2}\right)=\left(\alpha +\frac{1}{\alpha}\right)^2-2$$
$$\left(\alpha^3 +\frac{1}{\alpha^3}\right)-\left(\alpha +\frac{1}{\alpha}\right)\left(\alpha^2-1 +\frac{1}{\alpha^2}\right)=\left(\alpha^3 +\frac{1}{\alpha^3}\right)-3\left(\alpha +\frac{1}{\alpha}\right)$$

Consider $x=\left(\alpha +\frac{1}{\alpha}\right)$ and you obtain:
$$x^3-3x+x-2+x^2+1=0$$

At the end you obtain $$x^3+x^2-2x-1=0$$

When $\alpha=\exp\left(\frac{2\pi i}{7}\right)$ you get $x=2\cos\left(\frac{2\pi}{7}\right)$


Find $n$ and $k$ such that maximum element is minimum Given $a_1, a_2, a_3, \ldots, a_m \in \mathbb {Z}$.
How do I find $n \in \mathbb Z, k \in \mathbb N$ such that
$$\max \{|n - a_1|, |n+k-a_2|, |n+2k-a_3|,...\}$$ is minimum?

The original problem was given $a_1 =2,\ a_2=-3,\ a_3 = -1,\ a_4 =-5$. But I don't know how to solve it, please show me a way or give me a hint.

I'll describe one algorithmic approach to this problem:

Start by imagining we have already determined the best choice of $k$. The problem can then be viewed as trying to find the optimal value of $n$, minimizing $\max\limits_i |n-b_i|$ with $b_i=a_i-(i-1)k$.

It's not too difficult to see that if $b_L$ is the least and $b_G$ the greatest of $b_i$, the optimal choice of $n$ is equal to $\frac{b_L+b_G}{2}$, rounded to the nearest integer. The corresponding value of the maximum is then equal to $\left\lceil\frac{b_G-b_L}{2}\right\rceil$.

This offers one simple (although inefficient) approach to finding the optimal values of $k$ and $n$; just start with $k=1$ and keep increasing it; computing the corresponding optimal value of $n$ in each case, until we find the answer. When can we stop, though?

One useful observation we can make says that with increasing $k$, $L$ never increases and $G$ never decreases and they eventually reach $L=1$ (since $b_1$ does not depend on $k$ at all while all the other $b_i$ keep increasing) and $G=m$ ($b_m$ grows at the fastest pace, so it'll eventually outgrow all the others).

Furthermore, every increase of $k$ by one adds $(G-L)$ to the numerator of $\left\lceil\frac{b_G-b_L}{2}\right\rceil$; as long as neither $G$, nor $L$ changes. Whenever $G$ and/or $L$ changes, the quantity $(G-L)$ actually increases. This means that as $k$ increases, the full expression (maximum of the absolute values) will decrease, reach the optimum and then start to increase again. The optimum point is thus either at $k=1$ (since that's the minimum permitted value) or when $(G-L)$ becomes non-negative. Moreover, the only interesting values of $k$ are those when either $G$ or $L$ changes.

In order to find them, we can look at all pairs $1\leq i<j\leq m$ and calculate $k_{i,j}=\frac{a_j - a_ i}{j-i}$; the value of $k$ when $G$ could possibly change from $i$ to $j$ or $L$ change from $j$ to $i$. Then, we can just try to check the positive integers closest to each of these values, calculate the corresponding $n$ and the corresponding value we're looking for -- and report the smallest one (of course, checking $k=1$ too; due to $k$ being restricted to positive integers). This brings the calculation down to $O(m^2\log m)$ operations (the logarithmic factor is related to sorting the values so that they are processed in the correct order).

This latter part could actually be optimized even further by finding only the "relevant" of the $k_{i,j}$, which there are at most $m$ of and thus reduce the total complexity to linear. The magic word would be "convex hull"

if there's no pattern in your input sequence $a_1$, $a_2$, ... , the formula you gave up there is the best it gets. you can characterize some depending on whether the seq. is (monotonically/-) increasing/decreasing for one.


Combinatorial proof for an identity I'm trying to show the equality below using a combinatorial argument. For the left hand side;
     suppose there are n lands to build houses that can hold 3 families             at once. First it counts the ways of building houses with n choose k in summation and then for each case of houses it puts 3 family inside those k houses. If I algebraically change right hand side getting a slightly longer equation, I can show that they are counting the same thing. However, I should be able to do so without changing the right hand side with algebraic manipulations. Can someone help me find the story of the right hand side?

$$\sum_{k=1}^{n} k^3{n \choose k}=n^2(n+3)2^{(n-3)}$$

There are 3 cases.

Case 1: The families occupy $1$ house.

There are $n$ ways to choose the land to build the occupied house.

For the remaining $n-1$ lands, a house can be built or not to be built, so there are $2^{n-1}$ ways.

Hence, there are $n\cdot 2^{n-1}$ ways for Case 1.

Case 2: The families occupy $2$ houses.

There are $n$ ways to choose the land to build the occupied house for $2$ families and then there are $n-1$ ways to choose the land to build the occupied house for $1$ family.

Then there are ${3 \choose 2}=3$ ways to choose $2$ families for the occupied house for $2$ families and the remaining $1$ family for the occupied house for $1$ family.

For the remaining $n-2$ lands, a house can be built or not to be built, so there are $2^{n-2}$ ways.

Hence, there are $n(n-1)\cdot 3\cdot 2^{n-2}=3n(n-1)\cdot 2^{n-2}$ ways for Case 2.

Case 3: The families occupy $3$ houses.

There are $n$ ways to choose the land to build the occupied house for the $1$st family, then there are $n-1$ ways to choose the land to build the occupied house for the $2$nd family and there are $n-2$ ways to choose the land to build the occupied house for the $3$nd family.

For the remaining $n-3$ lands, a house can be built or not to be built, so there are $2^{n-3}$ ways.

Hence, there are $n(n-1)(n-2)\cdot 2^{n-3}$ ways for Case 3.

Combining the 3 cases

\begin{align*}
\text{The total number of ways} & =n\cdot 2^{n-1}+3n(n-1)\cdot 2^{n-2}+n(n-1)(n-2)\cdot 2^{n-3} \\
& =n[4+6(n-1)+(n-1)(n-2)]\cdot 2^{n-3} \\
& =n^2(n+3)\cdot 2^{n-3} \\
\end{align*}

I think of this expression probabilistically.Let's define a Random variable $X$,which counts the number of heads obtained when an unbiased Coin is thrown n times in a succession independently.So $\mathbb{P}(X=k)=\binom{n}{k}(\frac{1}{2})^n$.So if you multiply the LHS by $(\frac{1}{2})^{n}$ it becomes   $\mathbb{E}(X^{3})$.And we know that $\mathbb{E}(X^{3})=\frac{n^{2}(n+3)}{8}$.Then you get the desired RHS.


How to calculate probabilities for football match? I am looking for the acceptible methods to calculate probability chances "1 x 2" on football matches. I read a lot of articles, but there are no useful information that can help me. I started to investigate probability theory myself and I freezed after some calculations.
Let imagine we have match between Chelsea and Liverpool. I understand how calculate probabilitly for each team. I took last ten Chelsea's home matches and last ten Liverpool's away matches. After simple calculations(win/10; draw/10; lose/10) I received:

P(Chelsea win at home) = 0,5;
P(Chelsea draw at home)=0,3;
P(Chelsea lose at home) = 0,2;

P(Liverpool win away) = 0,2;
P(Liverpool draw away)=0,6;
P(Liverpool lose away) = 0,2;

But I could now understand how to use this calculated data to predict match Chelsea - Liverpool. I tries to apply sum and multiplication theorems, but there could be incredible result, for exmaple over 1. I definetly understand that this calculation would be very approximatly and that there are a lot of parameters that affect on football result, but I think that it could be useful to predict match. Thank you for any help!

Probability of Chelsea winning would still be 0.55. It is not dependent on any other variable.  it means that,

  For any match, Chelsea probability to win is 0.55. Because in a season they win 55% of their matches.

What you have to do is to normalize the data. A simple way for it:

Add the probabilities which show the same result, to each other. In you example, you have to add Chelsea wins to Liverpool losses, Chelsea draws to Liverpool draws and Chelsea losses to Liverpool wins.

What you'll have is:

Chelsea Win: $0.55+0.25=0.80$

Draw: $0.35+0.55=0.90$

Liverpool Win: $0.10+0.20=0.30$

Then, divide those possibilities to two, to normalize the probabilities. Why two? Because you have two teams.

Chelsea Win: $0.40$
Draw: $0.45$
Liverpool Win: $0.15$

Of course, in real world it's not that simple as you have to calculate last 5 games form, injured and banned players and the statistics when they're not playing, last game score and date, games between those two teams in last 10 years etc.

For instance, in Turkey, Galatasaray hasn't won against Fenerbahçe away in last 15 years. Even Galatasaray was way ahead champion, they just couldn't win against Fenerbahçe. So, when calculating the probable result of Fenerbahçe - Galatasaray game, you have to consider that curse. So, in any game, you can't only look at their whole season form, or only last 5 games, or only "something". You have to consider each effect, and this is what makes the calculation complex.


On Multiplication,Division,Addition,Subtraction I was asked if what is the basis of MDAS rule.
(multiplication, division, addition, subtraction).
Will these operations be followed at all times, or can it be changed in some cases?

It equally can be DMAS, DMSA and MDAS. The point is multiplication and division always have higher precedence over addition and subtraction unless there are parantheses. Evaluating a subtraction before or after an addition doesn't change the value of the result. Same thing for multiplication and division.

If you are trying to compare MDAS against DMAS then they are both equivalent as long as you replace division with inverse of multiplication i.e. replace $x/y$ with $x \times\frac{1}{y}$

For instance: $4/2\times3$ using MDAS rule AND (not replacing $x/y$ with $x \times\frac{1}{y}$) , would be like
$$4/2\times3 = 4/6 = 2/3$$
which is not equal to the answer obtained by using DMAS rule $4/2\times3 = 2\times3 = 6$

But if you replace $x/y$ with $x \times\frac{1}{y}$ then solving $4/2\times3$ would be like
$$4/2\times3 = 4\times\frac{1}{2}\times3 = \frac{12}{2} = 6$$
which is the answer obtained by using DMAS rule i.e. $6$

same goes for Addition and Subtraction operations, that is considering Subtraction as inverse of Addition $a-b = a+({-x})$


Restricted number of k non-consecutive combinations out of n Assume we need to choose $k$ numbers out of $[1...n]$ so no two numbers are consecutive. I know the number of such combinations is given by$\binom {n-k+1}{k}$. Assume the numbers are given by $r_1<r_2...<r_k$ when the difference between $r_{i+1}$ and $r_i$ is always higher than $2$. Now, I want to restrict the combinations so that for $p=1$ $(r_{i+1},r_i)$ pairs, the difference between them will be at least $t=3$ (and not at least 2).

Example: for $p=1$ and $t=3$ The number of combinations is given by $\binom {n-k+1}{k}$-$(n-2k+2)$ because we subtract all the combinations in which all the differences are exactly 2.

How can it be generalized to any $p$ and/or any $t$?

For $t=3$ I thought about counting the combinations by $\binom {n-k+1-p}{k}$ so we can insert the $p$ "dummies" in each combination but it seems it doesn't work and doesn't coincide with the example for $p=1$.

Thank you

We will show the particular case when $p=k-1$.

Any selection of $k$ numbers from $\{1,2,\dots ,n-(t-1)(k-1)\}$ gives you a suitable combination when you add $t-1$ to the second smallest element, $2(t-1)$ to the third smallest element$\dots$ $(t-1)(k-1)$ to the largest alement. Using the inverse process (substracting instead of adding)  on a suitable sequence will always give you a set in $\{1,2,3,\dots ,(n-(k-1)(t-1)\}$

Table[Table[Binomial[n - 2 (k - 1), k], {k, 0, Floor[(n + 2)/3]}], {n,
    0, 10}] // Grid

Table[Table[
   Length[Select[Tuples[{0, 1}, n],
     FreeQ[#, {_, 1, 0, 1, }] && FreeQ[#, {, 1, 1, _}] &&
       Count[#, 1] == k &]], {k, 0, Floor[n + 2]/3}], {n, 0,
   10}] // Grid

nn = 10; CoefficientList[
  Series[1/(1 - x)/(1 - y x^3/(1 - x)), {x, 0, nn}], {x, y}] // Grid

These three Mathematica codes return the same triangular array of numbers, T(n,k): the number of size k subsets of {1,2,...,n} such that the difference between any two elements is at least 3.


Parametric Equations for a $2$-torus I know that for a torus (with one hole) the parametric equations describing it are $x= (c + a\cos v)\cos u, y= (c + a\cos v)\sin u, z= a\sin v$, where $c$ is the radius from the center of the hole to the center of the torus tube and $a$ is the radius of the tube. My question is what would be the corresponding parametric equations describing a $2$-torus, i.e. a torus with $2$ holes. How about an $n$-torus?

A few words concerning nomenclature: The surface of a donut is a $2$-torus, or a closed surface of genus $1$. A donut with $2$ holes is not a $2$-torus but a closed surface of genus $2$. An $n$-torus for arbitrary $n\geq1$ is the manifold obtained from ${\mathbb R}^n$ by identifying points whose coordinates differ by integers.

A $1$-torus is nothing else but the circle $S^1$. The map
$$u:\quad\phi\mapsto(\cos\phi,\sin\phi)\qquad (-\infty<\phi<\infty)$$
used to ”parametrize" $S^1$ is actually a covering map: It is locally (i.e., for short $\phi$-intervals) a diffeomorphism, but to each point $z\in S^1$ belong infinitely many $\phi$'s with $u(\phi)=z$.

In the same way the parametrization you gave for the $2$-torus is a covering map: Each point $(x,y,z)$ on the torus is produced infinitely many times. When you want to compute the area of the torus you have to restrict this map to $[0,2\pi]\times[0,2\pi]$, even though it is defined on all of ${\mathbb R}^2$.

Now the closed surfaces of genus $\geq2$: Here no elementary parametrization is possible. In the theory of Riemann surfaces it is shown that such a surface $S$ possesses a "universal" covering map $\pi: \ D\to S$ $\ (D$ is the unit disk in the $z$-plane), where again each point $p\in S$ is produced infinitely many times. Here the different $z$'s that produce the same point $p\in S$ are not related by translations $z\mapsto z+2\pi j +2i\pi k$, as in the case of the $2$-torus, but by a group of Moebius transformations.

$0$ holes, $3$ donuts
$$\Big(\big(-\cos(\theta)^2+\sin(\theta)^2\cos(\phi)^2+\sin(\theta)^2\sin(\phi)^2+h+\sin(\theta)\big)^2+\\
\big(-\sin(\theta)-h+2\cos(\theta)\sin(\theta)\big)\big(\sin(\theta)+h-2\cos(\theta)\sin(\theta)\big)\text,\\
2\sin(\theta)\cos(\phi)\big(1+h-2\cos(\theta)\big)\\
\big(-\cos(\theta)^2+\sin(\theta)^2\cos(\phi)^2+\sin(\theta)^2\sin(\phi)^2+h+\sin(\theta)\big)\text,\\
2\sin(\theta)\sin(\phi)\big(1+h-2\cos(\theta)\big)\\
\big(-\cos(\theta)^2+\sin(\theta)^2\cos(\phi)^2+\sin(\theta)^2\sin(\phi)^2+h+\sin(\theta)\big)\Big)$$
$h$ is the height.
$3$ donut:


Calculate a Ranking Average? I got stuck to find a fair formula to calculate the average ranking of the items that I found after consecutive searches, look:

If I calculate the simple average of the item2 for example I get 1,33 as a result, not even nearly close to an "average" ranking :P
any ideas?

What are the values in the table? Are they rankings under the three searches?  Then you can just average (or sum-gets the same thing and removes the division) the numbers, rank them, and report the result.  $1.33333$ for item $2$ is not bad, actually.  It will probably be worst (if low numbers are bad), which is exactly what you want-it was very low by all three rankings.  You would still have to deal with items that are not ranked by all three searches.  Ties in one search are no problem, and you may have ties in the final ranking in this approach.

The arithmetic mean is a parametric descriptive statistic that presumes normality in the data.  Ordinal data (rank-ordered data) does not meet the assumptions of normality, and you cannot use an arithmetic mean to describe the central tendency—only the median.


Why can't the ratio test be used for geometric series? The ratio test says that, for $a_k\neq 0$, if
$$\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=L$$
exists, then if $0\leq L <1$, then $\sum_k a_k$ converges. If $L>1$, it diverges.

The notes I'm reading say that it's inadmissible to use the ratio test to test for convergence of a geometric series. I can't see why this should be the case.

Say we have some geometric series $\sum_kar^k$. Then
$$\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\to\infty}\frac{\left|ar^{k+1}\right|}{\left|ar^k\right|}=|r|.$$
So the ratio test tells us that the geometric series converges for $|r|<1$, and diverges for $|r|>1$, which is exactly what we get by using the formula
$$\sum_{k=1}^n ar^k=a\left(\frac{1-r^{n+1}}{1-r}\right).$$

What is an example that demonstrates why the ratio test is inadmissible for a geometric series?

The usual proof of the ratio test is to compare the series to a geometric series.  If $$\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \alpha < 1,$$ then we have
$$ |a_{n+1}| < |a_n| \alpha $$
for all sufficiently large $n$.  It then follows from an induction argument that
$$ |a_{n+k}| < |a_n| \alpha^k $$
for $n$ sufficiently large, which means that
$$ \sum_{j=1}^{\infty} |a_j|
= \sum_{j=1}^{n-1} |a_j| + \sum_{j=n}^{\infty} |a_j|
\le \sum_{j=1}^{n-1} |a_j| + \sum_{k=0}^{\infty} |a_n| \alpha^k.$$
The first series has only finitely many terms and is therefore finite, and the second series is geometric and therefore converges by some other argument.  From this, it follows that if
$$ \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| < 1,
\qquad\text{then}\qquad
\sum_{j=1}^{\infty} a_j $$
converges (by the limit comparison test, for example).  The other inequality in the ratio test can be argued by noting that the general term does not go to zero, and the uncertainty at 1 can be argued by considering (for example) the harmonic and alternating harmonic series.

The key point is that we prove that the ratio test implies the convergence of series by comparison to a convergent geometric series.  Since the proof of the ratio test relies on this convergence, it is circular to argue that a geometric series converges by the ratio test (unless, of course, you have another proof for the ratio test that doesn't use the convergence of geometric series).

While one could use the ratio test to establish the convergence of a geometric series (there is nothing stopping us!), it is typically poor style to rely on circular arguments as it can (potentially) lead to overlooking important hypotheses or exceptional cases.  This is particularly important in a pedagogical setting, when students may not be entirely cognizant of the line of reasoning that lead up to a result (it is hard to keep track of all of the lemmata, theorems, and proofs that lead up to a result if it is the first time that you have had to deal with them).

Moreover, I don't see why one would want to use the ratio test to show that a geometric series converges.  Basically no computation is needed to show that a geometric series converges, while a couple of computational steps are needed in order to invoke the ratio test.

Once we have a valid proof of the ratio test, then it can be applied to any series if it satisfies the hypotheses, including geometric series. It does not matter how the ratio test was proven. This is in no way circular reasoning even if the proof used results about geometric series. The idea of circular reasoning is a logical fallacy, but that is not what is being done here. We are not assuming what we are trying to prove. In fact, if we are allowed to assume statement $A$ because it is given to us, then the proof of $A$ is immediate from our given assumption. This is not circular reasoning because we are explicitly allowed to use statement $A$ which is somehow given to us to use.

The OP gave a link to course notes that state 'We cannot prove that a geometric series $\sum r^k$ is convergent/divergent by applying the Ratio Test. Why not?" That's a convincing argument, isn't it? Despite the hypotheses for "D'Alembert's Ratio Test" are satisfied by any positive geometric series.


What is the difference and relationship between the binomial and Bernoulli distributions? How should I understand the difference or relationship between binomial and Bernoulli distribution?

A Bernoulli random variable has two possible outcomes:  $0$ or $1$.  A binomial distribution is the sum of independent and identically distributed Bernoulli random variables.

So, for example, say I have a coin, and, when tossed, the probability it lands heads is $p$.  So the probability that it lands tails is $1-p$ (there are no other possible outcomes for the coin toss).  If the coin lands heads, you win one dollar.  If the coin lands tails, you win nothing.

For a single coin toss, the probability you win one dollar is $p$.  The random variable that represents your winnings after one coin toss is a Bernoulli random variable.

Now, if you toss the coin $5$ times, your winnings could be any whole number of dollars from zero dollars to five dollars, inclusive. The probability that you win five dollars is $p^5$, because each coin toss is independent of the others, and for each coin toss the probability of heads is $p$.

What is the probability that you win exactly three dollars in five tosses?  That would require you to toss the coin five times, getting exactly three heads and two tails. This can be achieved with probability $\binom{5}{3} p^3 (1-p)^2$.  And, in general, if there are $n$ Bernoulli trials, then the sum of those trials is binomially distributed with parameters $n$ and $p$.

Note that a binomial random variable with parameter $n = 1$ is equivalent to a Bernoulli random variable, i.e. there is only one trial.

A Bernoulli random variable $X$ is a random variable that satisfies $P(X=1)=p$, $P(X=0)=1-p$. A canonical example is a coin flip which has $p=1/2$. In fact, you can think of a Bernoulli random variable is just a weighted coin, which comes up $1$ with some probability and $0$ otherwise. A binomial random variable with parameters $n,p$ is what you get when you count the number of $1$'s (successes) that come up in a string of $n$ independent Bernoulli random variables, each with parameter $p$. Another way to say this is that a binomial random variable is the sum of independent and identically distributed Bernoulli random variables.


Limit calculation, calculus I am trying to find

$$  \lim_{x\to 0} \frac{e^{\alpha x} -e^{\beta x}}{\sin(\alpha x) + \sin(\beta x) }$$

but dont have a clue where to start, could someone give me a hint please ?

Since $\alpha \neq -\beta$

$$e^{\alpha x} = 1 + \alpha x + o(x)$$
$$e^{\beta x} = 1 + \beta x + o(x)$$
$$\sin(\alpha x) = \alpha x + o(x)$$
$$\sin(\beta x) = \beta x + o(x)$$
So
$$ \frac{e^{\alpha x} -e^{\beta x}}{\sin(\alpha x) + \sin(\beta x) } = \dfrac{1 + \alpha x - 1- \beta x +o(x)}{\alpha x + \beta x +o(x) } = \dfrac{\alpha - \beta + \dfrac{o(x)}{x}}{\alpha + \beta + \dfrac{o(x)}{x}} \to \dfrac{\alpha - \beta}{\alpha + \beta}$$

Alternatively, use

$$\frac{e^{\alpha x} -e^{\beta x}}{\sin(\alpha x) + \sin(\beta x) }
=\frac{\dfrac{e^{\alpha x} - 1}{\alpha x}\alpha-\dfrac{e^{\beta x} - 1}{\beta x}\beta}{\dfrac{\sin(\alpha x)}{\alpha x }\alpha + \dfrac{\sin(\beta x)}{\beta x}\beta }$$

and $\lim \dfrac{e^{\alpha x} - 1}{\alpha x} = \lim \dfrac{e^{\beta x} - 1}{\beta x} = \lim \dfrac{\sin(\alpha x)}{\alpha x } = \lim \dfrac{\sin(\beta x)}{\beta x} = 1 $

Using Wolfram Alpha:

lim (e^(alpha x)-e^(beta x))/(sin(alpha x)+sin(\beta x)) as x->0

Answer:

(alpha-beta)/(alpha+beta)

http://www.wolframalpha.com/input/?i=lim+%28e%5E%28alpha+x%29-e%5E%28beta+x%29%29%2F%28sin%28alpha+x%29%2Bsin%28%5Cbeta+x%29%29+as+x-%3E0


What does it mean for a polynomial with integer coefficients to have a root in $p$-adic integers? I have just begun delving into $p$-adic number theory. I was wondering, given a poynomial $f(x)$ with integer coefficients, what does it mean when we say, $f(x)$ has a root in $\mathbb{Z}_2$, for instance.

If $f(x) = f_0 + f_1 x + \ldots + f_d x^d \in \mathbb{Z}[x]$, and $f(r) = 0$ over $\mathbb{Z}_2$ then $$f_0 + f_1 r + \ldots + f_d r^d \equiv 0 \bmod 2.$$ Equivalently, $f(x) \equiv (x-r) g \bmod 2$ for some polynomial $g(x) \in \mathbb{Z}_2[x].$


Dividing open domains in $\mathbb R^2$ in parts of equal area (i) Let $A$ be an open, bounded and connected subset of $\mathbb R^2$.
  Prove that for every given direction $d$, there exists a line parallel
  to $d$ that divides $A$ in two parts with the same area.

  (ii) Show that if $A,B$ are open, bounded and connected subsets of
  $\mathbb R^2$ there exists a line which divides both $A$ and $B$ in
  two parts of equal area.

I'm stuck and I do not know how to begin. What should I do? The first statement if obvious if $A$ is a ball, but I do not know how to treat the general case.

Thanks for your help.

Both of these rely on the intermediate value theorem.

(i) Define one direction normal to $d$ as up. Draw a line $l_0$ "below" $A$, and one line $l_1$ "above" $A$ (possible because $A$ is bounded) (both parallel to $d$),and call the distance between them $r$. You can define a function $f$ from the interval $[0, 1]$ into $\mathbb R$ taking a number $x$ to the area below the line parallel to $d$ and distance $rx$ above $l_1$. We now have $f(0) = 0$, $f(1) = |A|$, and it is continuous, so somewhere it has to equal $|A|/2$. That's where you put your line.

To solve (ii) we need to know that the line is unique. We therefore need to show that $f$ is monotone, and that at the point where $f(x)= |A|/2$, it is strictly increasing.

The first part is simple, since we only move the line in one direction, points of $A$ can only go from being over the line to being under the line as $x$ increases, so the calculated area cannot decrease.

The second part is a bit more tricky. I'll show the "top part", the bottom part is completely analogus. Given the point $x$ where $f(x) = |A|/2$, for any $\epsilon > 0$ (and small enough that $x+\epsilon < 1$) we must show that $f(x+\epsilon) > f(x)$. Since $A$ is open and connected there will be a point $p$ on the line corresponding to $x$ with $p\in A$ which is the centre of a ball contained in $A$ with radius less than $r\epsilon$. Half this ball will be above the $x$-line, but below the $x+\epsilon$-line, thus $f$ has increased between these two points.

We now know that the line is unique, and can carry on with (ii).

(ii) For each possible direction, find the line that divides $A$ into equal parts by (i). Remark that opposite directions yield the same line, since opposite directions are still parallell. Now pick a direction $d$. Define a direction normal to $d$ as up, and now define a function $g$ from the unit interval to $\mathbb R$ as follows: For a number $x$, turn the direction $d$ (and the orientation "up") a total of $\pi x$ radians counterclockwise and let the value of the function be how much area of $B$ is above the line with that direction, dividing $A$ in half. The key remark here is that $f(0) + f(1) = |B|$, since they are both calculated from the same direction $d$, but with the orientation "up" in opposite direction. You therefore have a function from the unit interval to $\mathbb R$ with $f(0) = |B|/2 + a$ and $f(1) = |B|/2 - a$ for some real number $a$. Thus it has to be $|B|/2$ somewhere in between, and you then have your line.

Under the given conditions, the sets have a centre of gravity.  One usually learns in multivariate calculus which integrals are used to compute them and the basic property that any line through them splits the areas as you require.

Then for (i) take a line through the centre, parallel to the given direction.

For (ii) take a line through both centres.


D. Williams 'Probability with Martingales' query I don't understand the logic of the paragraph highlighted in red at all. The 'uniqueness lemma' he speaks of states that if two measures agree on a so-called $\pi$-system then they agree on the sigma algebra generated by that system.

So is he saying that the two measures in question, $\mathcal{L}_X$ and $\mathbb{P}$, agree on $\pi(\mathbb{R})$? If so, I don't understand why. I'm generally really confused by the number of sigma algebras in this page (we have $\mathcal{F}$, $\mathcal{B}$, $\sigma(X)$ - why so many and what is happening?) and what everything has a function for.

The steps might be really simple for most people like Williams but I can't understand how that lemma relates to this? If anyone can provide some insight into understanding this I would be really grateful.

The first part is saying that if

$$(\Omega, \mathscr F, \mathbb P)$$ is a probability space

then

$$(\mathbb R, \mathscr B, \mathcal L)$$

is a probability space because $\mathcal L$ is a probability measure on (\mathbb R, \mathscr B).$

This can be shown by proving

$\mathcal L(\mathbb R) = 1$:

$P(X \in \mathbb R) = P(\Omega) = 1$

$\mathcal L$ is $\sigma$-additive:

$\mathbb P(X \in B)$ for $B \in \mathscr B$ is $\sigma$-additive because $\mathbb P(A)$ for $A \in \mathscr F$ is $\sigma$-additive

What the second part is saying is that

$$\mathcal L_X(B) := P(X \in B)$$

is determined uniquely by

$$F_X(t) := P(X \le t) = P(X \in (-\infty,t])$$

In other words, if we had some formula for $P(X \in (-\infty,t])$ for any $t \in \mathbb R$ (or equivalently $P(X \in B)$ for any $B \in \mathbb \pi(\mathbb R)$) then we would be able to compute $P(X \in B)$ for any $B \in \mathscr B$, the latter of which is more difficult to compute than the former.

Why? Well as with any proof for uniqueness suppose we have two random variables $X$ and $Y$ with the same distribution function. Is it possible that

$$P(X \in B) = P(Y \in B), \forall B \in \pi(\mathbb R)$$

but

$$\exists B \in \mathscr B \ \text{s.t.} \ P(X \in B) \ne P(Y \in B)$$

?

No. By the uniqueness lemma,

since $$F_X(t) = F_Y(t), t \in \mathbb R$$

or equivalently

$$P(X \in (-\infty,t]) = P(Y \in (-\infty,t]), t \in \mathbb R$$

or equivalently

$$P(X \in B) = P(Y \in B), B \in \pi(\mathbb R)$$

we have that

$$P(X \in B) = P(Y \in B), B \in \sigma(\pi(\mathbb R)) = \mathscr B$$


How to prove $\lim_{n \to \infty} \frac{\log(n)(n^{\log(n)})} {a^n} ＝0, $ where $ |a|>1$ How to prove $$\lim_{n \to \infty} \frac{\log(n)\left(n^{\log(n)}\right)} {a^n} ＝0,\;  \text{ where } \;|a|>1$$

I don't know how to employ L'Hospital here.

I don't know how to employ L'Hospital here.

Then, don't...
$$
\left|\frac{\log(n)\left(n^{\log(n)}\right)} {a^n}\right| ＝\exp\left(\log\log n+(\log n)^2-\alpha n\right)$$
with $\alpha=\log|a|\gt0$, and
$$
\log\log n+(\log n)^2-\alpha n\sim-\alpha n\to-\infty.
$$

Let $a_n:=\log(n)n^{\log(n)}$.

First compute $\lim_n \frac{\log(n+1)}{\log(n)}=1$. We can use L'Hospital for this (It gets reduced to $\lim_n \frac{n}{n+1}=1$).

From this we get that $\lim_n [\log(n)]^{1/n}=1$ as well. This time by applying L'Hospital for sequences (Stoltz's theorem) which implies that if $\lim \frac{r_{n+1}}{r_n}=L$ then $\lim_n r_n^{1/n}=L$.

Now we compute $\lim_n n^{\log(n)/n}=\lim_{n}e^{\frac{\log^2(n)}{n}}=e^{\lim_n\frac{\log^2(n)}{n}}=1$. This we get by applying L'Hospital to the limit in the exponent $\lim_n\frac{\log^2(n)}{n}=0$, since by L'Hospital it gets reduced to $\lim_n\frac{2\log(n)}{n}$, which gets reduced to $\lim_{n}\frac{2}{n}=0$ by another application of L'Hospital.

Therefore, since $\lim_n[\log(n)]^{1/n}=1$ and $\lim_nn^{\log(n)/n}=1$, we get that $$\lim_n a_n^{1/n}=1.$$

Therefore, the radius of convergence of $\sum_n a_nx^n$ is at least $1$. Therefore $\sum_n \frac{a_n}{a^n}$ converges. Hence $$\lim_n\frac{a_n}{a^n}=0.$$


Complex variable problem I would appreciate some help about the following problem.

  Given $z\in\mathbb{C}$ and
  $$\left\vert {1+z\over 1-z}\right\vert =1 $$
  show that $z$ is purely imaginary.
  Deduce that the roots of the equation
   $$2{(1+z)^5}-(1+{\sqrt 3}i){(1-z)}^5=0 $$
  are purely imaginary, then write down the five distinct roots of the above equation in $z$ .

A slightly different approach is to notice that $f(z)=\frac{1-z}{1+z}$ is an involutive map:
$$\forall z\in\mathbb{C}\setminus\{-1\},\quad f(f(z))=z $$
hence $f(z)$ gives a bijection between the punctured plane and itself. It is simple to notice that $f$ maps the imaginary axis into the unit circle minus the point $-1$ in a bijective way, hence the inverse image of the unit circle minus the point $-1$ is the imaginary axis.

The second equation is equivalent to
$$ \left(\frac{1-z}{1+z}\right)^{5} = \exp\left(-\frac{\pi i}{3}\right) $$
and every solution lies on the imaginary axis by the previous lemma. A solution, for instance, is given by $z=\frac{1-\exp(-\pi i/15)}{1+\exp(-\pi i/15)}=i\tan\frac{\pi}{30}$.

$$|1+z|^2=|1-z|^2$$
$$2Re(z)=-2Re(z)$$
$$Re(z)=0$$
Hence $z$ is purely imaginary.

You can also see this geometrically: If $z$ is equidistant from both $1$ and $-1$, then it has to lie on imaginary axis.


How to compute $\int_0^1 \cos^2{\pi x}\ln \Gamma(x)dx$ Problem
Find the exact value of
$$\int_0^1 \cos^2{\pi x}\ln \Gamma(x)dx$$
My Attempt
\begin{align*}
I&=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)dx\\
2I&=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)dx+\int_0^1 \cos^2{(\pi (1-x))}\ln \Gamma(1-x)dx \tag{Using reflection}\\
2I&=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)+\cos^2{(\pi (1-x))}\ln \Gamma(1-x)dx\\
2I&=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)+\cos^2{\pi x}\ln \Gamma(1-x)dx\\
2I&=\int_0^1 \cos^2{(\pi x)}[\ln \Gamma(x)+\ln \Gamma(1-x)]dx\\
2I&=\int_0^1 \cos^2{(\pi x)}\left[\ln\left(\frac{\pi}{\sin(\pi x)}\right)\right]dx\tag{Using Euler's reflection formula}
\end{align*}
And here is where I am stuck, I have no idea how to evaluate the integral. Any help is appreciated.

Consider reduce the power of cosine term:
$$
\begin{aligned}
\int_0^1\cos^2(\pi x)\log\Gamma(x)\mathrm dx
&=\frac12\left[\int_0^1\cos(2\pi x)\log\Gamma(x)\mathrm dx+\int_0^1\log\Gamma(x)\mathrm dx\right]
\end{aligned}
$$
For the first part, by the property that
$$
\int_a^bf(x)\mathrm dx=\frac12\int_a^b[f(x)+f(a+b-x)]\mathrm dx
$$
we have
$$
\begin{aligned}
\int_0^1\cos(2\pi x)\log\Gamma(x)\mathrm dx
&=\frac12\int_0^1\cos(2\pi x)\log[\Gamma(x)\Gamma(1-x)]\mathrm dx \\
&=\frac12\int_0^1\cos(2\pi x)\log\left[\pi\over\sin(\pi x)\right]\mathrm dx \\
&={\log\pi\over2}\int_0^1\cos(2\pi x)\mathrm dx-\frac12\int_0^1\cos(2\pi x)\log\sin(\pi x)\mathrm dx \\
&=-\frac12\int_0^1\cos(2\pi x)\log\sin(\pi x)\mathrm dx
\end{aligned}
$$
In fact, by the Fourier expansion of log sine
$$
\log\sin\theta=-\log2-\sum_{k=1}^\infty{\cos(2k\theta)\over k}
$$
we can deduce
$$
\begin{aligned}
-\frac12\int_0^1\cos(2\pi x)\log\sin(\pi x)\mathrm dx
&={\log2\over2}\int_0^1\cos(2\pi x)\mathrm dx+\frac12\sum_{k=1}^\infty\frac1k\int_0^1\cos(2\pi x)\cos(2\pi kx)\mathrm dx \\
&=\frac14
\end{aligned}
$$
For the last part, by Raabe's integral we have
$$
\int_0^1\log\Gamma(x)\mathrm dx=\frac12\log(2\pi)
$$
All things combined, we obtain
$$
\int_0^1\cos^2(\pi x)\log\Gamma(x)\mathrm dx=\frac18+{\log(2\pi)\over4}
$$

According to Maple, your last integral is
$$ \frac{1}{4} + \frac{\ln(2\pi)}{2}$$
In fact, $\cos^2(\pi x) \ln(\pi/\sin(\pi x))$ has a rather complicated closed-form antiderivative (involving dilog)


Why Markov matrices always have 1 as an eigenvalue Also called stochastic matrix. Let

$A=[a_{ij}]$ - matrix over $\mathbb{R}$

$0\le a_{ij} \le 1 \forall i,j$

$\sum_{j}a_{ij}=1 \forall i$

i.e the sum along each column of $A$ is 1. I want to show $A$ has an eigenvalue of 1. The way I've seen this done is that $A^T$ clearly has an eigenvalue of 1, and the eigenvalues of $A^T$ are the same as those of $A$. This proof, however, uses determinants, matrix transposes, and the characteristic polynomial of a matrix; none of which are particularly intuitive concepts. Does anyone have an intuitive, alternate proof (or sketch of proof)?
My goal is to intuitively understand why, if $A$ defines transition probabilities of some Markov-chain, then $A$ has an eigenvalue of 1. I'm studying Google's PageRank algorithm.

I just encounter the same problem as you do, here is my solution.

I upload my solution by picture

When we calculate eigen values and eigen vectors:
Given an eigenvalue Y, and eigenvector associated with Y is a non zero vector x
such that:
(A-YI)x=0

Now in markov chain a steady state vector ( when effect multiplying or any kind of linear transformation on prob state matrix yield same vector)  :

qp=q where p is prob state transition matrix
this means Y = 1  as an eigen vector of P'
which also means

There is an unique probability vector q which is an eigenvector associated with eigenvalue 1.


Example of Non-separable stochastic process. This question is related to the link:

https://www.encyclopediaofmath.org/index.php/Separable_process

The link provided a basic definition of separable Stochastic process. I felt all the process under study seems to be separable. Is it true? Is there any example of  Non-separable stochastic process which is of some importance?

Let's consider the following setting: $([0,1],\mathscr L, \mu_L)$ with Lebesgue Set $\mathscr L$ and $t\in[0,1]$. Define the stochastic process as$Y_t=\mathbb 1_{\{\omega\}}(t)$, where $\{\omega\}$ is a singleton. This process just take value 1 at $t=\omega$.(for example one possible sample path is: For $\omega=1$. $Y_t(\omega)=1$ if $t=1$ and $Y_t(\omega)=0$ if $t \neq 1$).

Now we want to show it is not separable:

Consider the close set $A=\{1\}$, the set $\{\omega\in\Omega,Y_t(\omega)\in \{1\}\}$ becomes $\{\omega \in \Omega ,\omega=t\}$ and the union $ \cup_{t\in T\cap O}\{\omega\in\Omega,Y_t(\omega)\in \{1\}\}$ for some open interval $O$ is just the set ${T\cap O}$. Similar for a countable set $S_1,S_2,...\subset T$. The union $ \cup_{t\in (S_j)_{j\in\mathbb N} \cap O}\{\omega\in\Omega,Y_t(\omega)\in \{1\}\}$ is just $(S_j)_{j\in\mathbb N}\cap O$. The difference between the two set is not a subset of null set.

The advantage that we are dealing with separable process is that we can operate with countable operations (Union and common sigma-Algebra operations). This simplify our problem for stochastic process in continous time a lot. There is a theorem by Loeve, that under certain condition one can always find a equivalent seperable process.

Any process that is right continuous and has left limits is separable,

the poisson jump process is also separable. Just notice that you can know the value of $X(t,\omega)$ by looking at a sequence of rational numbers $q_n \downarrow t$.

To get a non separable process you need to consider a process that has more irregularity, say that is not right continuous. This might occur if you have a subset that is considered Bad and you tele-transport it to infinity as it enters the Bad region. This often happens when we are studying meta-stability, but still you can modify your process (ignoring the times you spent in the bad region - this is called the trace of the process).

You could also find a non separable process if it took value on a non separable space (it can't be a real valued process )

One often deals with separable process since it is a good property when we are dealing with denumerable set operations (as it is the case for a $\sigma$-algebra).


An unfair dice question Unfair die of 12 faces. The number 11 has a chance of  40% while other
  faces are equally likely. If someone and you are playing a game and
  whoever is closer to the answer will win, what is your strategy? Also, do
  you start first or second?

So I believe the best strategy is to pick $8$ because the expected value of this dice is given by $8.05454546$

Also, I think it's best to go second because my opponent can just choose the bigger part of my numbers. For example, if I choose $11$, they can choose $10$, and they'll have $6/11$ chance to win.

Is this right? I don't know if there's a better strategy.

The answer is 10

If you pick 10, and your opponent picks 9, they will win if the dice is 1-9. That means there is a 9/11 * 60% = 49% chance of winning. However, if your opponent picks 11, then there is a 1/11 * 60% + 40% = 45% chance of winning. Either way, you've won because your opponent is always under 50%


Losing less than $100 in a game of chance. In a game you win \$10 with probability $\frac{1}{20}$ and lose \$1 with probability $\frac{19}{20}$. Approximate the probability that you lost less than \$100 after the first 200 games. How will this probability change after 300 games?

Since I need to show both winning and losing together, I have come up with $$X_n=10S_n+(n-S_n)$$ where $X_n$ is the amount of winnings, and $S_n$ is the number of times you win the game. I also have $$E(X_n)=10, Var(X_n)=9.5$$ But when I want take $P(X_n\ge-100)$ I can never get a realistic answer.

Hint:

Let $X_i\sim \begin{pmatrix}
0 & 1\\
 19/20 & 1/20 \\
\end{pmatrix} $
and
$S=X\cdot10-(n-X)\cdot1$, where $X=\sum_{i=1}^nX_i$

It's easy to show that $E(X_i)=\frac{1}{20}$ and $Var(X_i)=\frac{19}{400}$.

Calculate $E(S)$ and $Var(S)$ and
then apply central limit theorem.


Find $a$, $b$, and $c$ so $a \mid bc ~$ but $a \nmid b \space $ and $a \nmid c$ I'm working out of the text "Elementary Number Theory" by James K. Strayer and I've run into the following problem:

  Find integers $a$, $b$, and $c$ such that $a \mid bc ~$ but $a \nmid b \space $ and $a \nmid c$

When I first saw this problem I thought it was weird because I thought for sure I've seen a theorem somewhere that says that this would never happen. The only thing is that usually if a question isn't valid they say "Prove or Disprove", which makes me think I'm overlooking the obvious. Is there an easy solution to this problem that I'm overlooking? If not, how do you prove its falsity. Thanks

Hint: Consider integers $b, c,\;\; b\neq 1, c\neq 1$ such that $a = bc$. Then then clearly $a = bc\nmid b$ and $a = bc\nmid c,\;$ while, of course, $a = bc\mid bc$.

EXist.
let a = 6, b=3 and c=4 then b.c=3.4=12.
we see that 6 divides 12 but, 6 not divides 3 and 6 not divides 4.


Find $\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \to \min$ I'm trying to solve
\begin{align*}
&\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \to \min \\
&x + y + z = 1 \\
&x, y, z > 0,
\end{align*}
using only inequalities.
How can i solve it? I used am-gm, tried to "break" terms, but got only after all $(xy + yz + xz) = \frac{1}{2}$

You can use Cauchy-Schwarz:
$$\sum\dfrac{x^2}{y}\sum y\geq(x+y+z)^2$$
or an even simpler one is:
$$\sum\dfrac{x^2}{y}-\sum y = \sum\left(\dfrac{x^2}{y} - 2x+y\right)=\sum\dfrac{(x-y)^2}{y}\geq 0.$$

By Cauchy-Schwarz we get$$\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \geq \frac{(x+y+z)^2}{x+y+z}=1$$


Error on Wikipedia about Dilations Looking at Wikipedia here, the definition of a scale factor has two definitions of a scale factor which are clearly not equivalent.

Is definition $(1.)$, or comment $(2.)$ the correct one of a "scale factor" in this context? In other words, can the ratio of magnification, dilation factor, scale factor, or similitude ratio be negative?

It states:

  $(1.)$ "In Euclidean geometry, a homothety of ratio $\lambda$ multiplies distances between points by $|\lambda|$ and all areas by $\lambda^2$." The next sentence then clearly refers to and defines $|\lambda|$ to be the "scale factor."

However, this is a problem as it later describes $\lambda$ to be the scale factor (i.e. where $\lambda<0$):

  $(2.)$ "The image of a point $(x, y)$ after a homothety with center $(a, b)$ and scale factor $\lambda$ is given by $(a + \lambda(x − a), b + \lambda(y − b))$."

I finally found the answer! I appreciate the time others took to help me. Looking on p. 117 (i.e. p.8 of 23 on the pdf) the definition of "ratio of similitude" is given which is the positive real number of a transformation on a metric space of which a dilation is being scaled by. Thus, the first number must be referring to $|\lambda|$ when talking about the ratio of magnification or dilation factor or scale factor or similitude ratio.

Thus, $(2.)$ is wrong and really should say the "ratio" instead of "scale factor."


Examples of finite nonabelian groups. Can anybody provide some examples of finite nonabelian groups which are not symmetric groups or dihedral groups?

Here's a bunch of examples I've collected from my notes, group theory texts, and various places around the Internet.  This has become somewhat of a treatise, but nonetheless, I hope you and others enjoy them.

Familiar generalizations.

Generalized dihedral groups, denoted $\mathcal{D}(A)$ or $\operatorname{Dih}(A)$, are formed by letting an involution (element of order $2$) act on an arbitrary abelian group $A$ by inversion.  So, $D_{2n}=\operatorname{Dih}(\mathbb{Z}/n\mathbb{Z})$.
Another sister of the dihedral groups are the semidihedral groups.  These are all $2$-groups which behave just a tiny bit different from their dihedral counterparts (see the presentation).
The quaternion group $Q_8$ (not to be confused with the related-but-not-identical quaternion algebra) is a Hamiltonian group of order $8$.  It's made of three normal subgroups $\langle i \rangle$, $\langle j \rangle$, and $\langle k \rangle$, which intersect at the order $2$ center $Z(Q_8)=\langle -1 \rangle$. $Q_8$ is the first counterexample you should try whenever you think anything might be true. (one example)
The generalized quaternion groups, denoted $Q_{4n}$ for $n\geq 2$, naturally extend the group presentation of $Q_8$ to higher orders. The important thing about generalized quaternion groups is that they have a unique element of order $2$, whereas all other $p$-groups with unique elements of order $p$ are cyclic.

These can be most easily understood as the quotient of the semidirect product $\mathbb{Z}_{2^n}\rtimes \mathbb{Z}_4$ by the subgroup $\langle(2^{n-1},2)\rangle$.
You can also realize $Q_{4n}$ as the subgroup of $\mathbb{H}^\times$ generated by $\cos(\pi/n)+\mathbf{i}\sin(\pi/n)$ and $\mathbf{j}$, where $\mathbb{H}$ denotes the (division) ring of Hamiltonian quaternions (with generators $\mathbf{i},\mathbf{j},$ and $\mathbf{k}$, under the usual relations).  I found this construction very interesting

I'll stop here for a moment to note that the dihedral groups, semidihedral groups, and generalized quaternion groups constitute all $2$-groups of maximal class, which makes them super important to know for people who study $p$-groups.

We can even generalize the generalized quaternions.  Replacing $\mathbb{Z}_{2^n}$ in the above definition with an arbitrary cyclic group $A$ of even order $n$ gives us the dicyclic groups, denoted $\operatorname{Dic}(A)$ or $\operatorname{Dic}_n$.  We can get even further into outer space and define the generalized dicyclic groups $\operatorname{Dic}(A,y)$ by allowing $A$ to be any abelian group with even order.  (For generalized dicyclic groups one must specify which element $y\in A$ of order $2$ behaves like the unique element of order $2$ in the generalized quaternion group.  You can read about that here.)

Matrix groups.

$\operatorname{GL}_n(\mathbb{F}_q)$ is the group of linear transformations of an $n$-dimensional vector space over a finite field $\mathbb{F}_q$.  The special linear group $\operatorname{SL}_n(\mathbb{F}_q)$ is the kernel of the homomorphism $\operatorname{det}:\operatorname{GL}_n(\mathbb{F}_q)\rightarrow \mathbb{F}_q^{\hspace{1pt}\star}$.
$\operatorname{PGL}_n(\mathbb{F}_q)$ and $\operatorname{PSL}_n(\mathbb{F}_q)$ are their projective counterparts, which essentially removes effects of scalars from $\operatorname{GL}_n(\mathbb{F}_q)$ and $\operatorname{SL}_n(\mathbb{F}_q)$ to focus on their structure.  $\operatorname{PSL}_n(\mathbb{F}_q)$ is simple for $n\geq 3$ or $q\geq 4$ (which means that $\operatorname{SL}_n(\mathbb{F}_q)$ is quasisimple).
General semilinear groups, written $\Gamma\!\operatorname{L}_n(\mathbb{F}_q)$, are split extensions of $\operatorname{GL}_n(\mathbb{F}_q)$ by the Galois group of $\mathbb{F}_q$ over its prime subfield. These have projective counterparts too.
Unitriangular matrix groups (upper triangular matrices with $1$'s on the diagonal) and symplectic groups are other good matrix groups to know.

Scary groups.

The sporadic groups are the scariest things which lurk outside of the happy little bubble that is solvable groups.  This is where the Monster lives.  In most cases, understanding the structure of these groups requires some esoteric machinery, so I would avoid these until you're really good at group theory.

By the way, any nonabelian simple group is an alternating group, a $\operatorname{PSL}$ or other group of Lie type, or a sporadic group.  Figuring that out took about $110$ years of focused sorcery, which is fun to read about if you like math history.

Frobenius groups.

Frobenius groups are made of two parts: a normal subgroup $N$, called the Frobenius kernel, which is acted upon by a subgroup $K$, the Frobenius complement, with the defining property that $K$ acts on $N$ fixed point freely - that is, $n^k\ne n$ for all non-identity elements $n\in N,k\in K$.  One example is $S_3$, for which the kernel is generated by the $3$-cycles and the complement is generated by (your choice of) a $2$-cycle. The structure of Frobenius groups is well understood - as it turns out, $N$ must be nilpotent, and the Sylow subgroups of $K$ must be either cyclic or generalized quaternion.

Frobenius groups are also the subject of a major theorem in representation theory which proves an alternative definition is equivalent to the standard one.  I personally don't find this alternative definition particularly enlightening, but the proof of this theorem demonstrates how powerful representation theory can be.

$2$-Frobenius groups are groups $G$ which have some subgroup $F$ which is a Frobenius group with kernel $K$ which is normal in $G$ for which $G/K$ is also a Frobenius group.  An example of this is $S_4$, where $F=A_4$ (the subgroup $K$ generated by the permutions $(12)(34)$ and $(13)(24)$ are acted upon fixed point freely by $\langle (123) \rangle$) and $G/K\cong S_3$.

Frobenius and $2$-Frobenius groups are important for element orders.  Whenever a solvable group $G$ lacks an element of order $pq$, where $p$ and $q$ are primes dividing $|G|$, the Hall $\{p,q\}$ subgroups of $G$ (subgroups of order $p^aq^b$ with index coprime to their order) are either Frobenius or $2$-Frobenius.

$2$-Frobenius groups were also important in the proof of the Odd Order theorem, which derived a contradiction by showing that a non-solvable group with odd order would have to contain certain $2$-Frobenius groups in a way that was not possible.

Zassenhaus groups are somewhat related to Frobenius groups in that they are defined by counting fixed points of their action on a set.  (I'll include more info on these soon.)

Cool families of $p$-groups.

$p$-groups are groups of prime power order $p^n$.  We already saw some families of $2$-groups in the first section; here, we will discuss families of $p$-groups of where $p$ is not necessarily even.

A special group is a $p$-group $G$ in which the center, derived, and Frattini subgroups coincide ($Z(G)=G^\prime=\Phi(G)$).  If these subgroups are cyclic of order $p$, we call the group extraspecial. The combination of these properties gives rise to some very interesting representation-theoretic behavior. In special groups, the commutator map from $G/Z(G)\times G/Z(G)\rightarrow G^\prime$ satisfies $[g,hk]=[g,h][g,k]$, so in extraspecial groups this may be considered as bilinear form over $\mathbb{F}_p$ (which, as it turns out, is skew-symmetric, alternating, and non-singular).  If $p$ happens to be $2$, we can additionally show that the squaring map is a quadratic form.  You can read more about this in $\oint 3.10$ of Wilson's Finite Simple Groups, which can be seen in a preview on the web.  Note also that it is particularly easy to classify the conjugacy classes of extraspecial groups, so they often make good testing grounds for conjectures in representation theory.

Some favorites from my toolbox.

$M_{16}$, the group with the coolest name, is the smallest example of a group with two distinct isomorphic characteristic subgroups, and thus another great counterexample group.  $M_{16}$ is pretty close to being abelian - it is just $\mathbb{Z}_2\times \mathbb{Z}_8$ with a little "twist."
The octahedral group is the symmetries of an octahedron.  I'm partial to the binary octahedral group $2\mathcal{O}$, which has order $48$ and is constructed by replacing certain elements of $\operatorname{GL}_2(\mathbb{F}_3)$ with scalar multiples in $\operatorname{GL}_2(\mathbb{F}_9)$. (I recently heard from a coauthor that Marty Isaacs refers to this group "fake $\operatorname{GL}(2,3)$", which I think is hilarious.)
$\Gamma$ groups are the affine transformation version of $\Gamma\operatorname{L}_n(\mathbb{F}_q)$.  I like taking small subgroups of one dimensional $\Gamma$ groups when I want a very nonabelian group in which I can still easily write out arithmetic.
My second favorite matrix group is the Valentiner group, isomorphic to $\operatorname{PGL}_3(\mathbb{F}_4)$.  My favorite sporadic group is the Thompson group.
I'm a big fan of making Frobenius groups of the form $\mathbb{Z}_{p}\rtimes \mathbb{Z}_q$ (and $2$-Frobenius groups of the form $(\mathbb{Z}_{p}\rtimes \mathbb{Z}_q)\rtimes \mathbb{Z}_r$), where $p,q,r$ are primes (necessarily) satisfying $p\equiv 1 \pmod q$ (and $p\equiv 1 \pmod {qr})$.  Need an easy nonabelian group of order $21$?  Boom, done.
Another particularly nice, easy class of Frobenius groups are those of the form $C_p\rtimes C_{p-1}$, where $p$ is an odd prime, also known as the holomorph of $C_p$, $\operatorname{Hol}(C_p)$.  These groups arise naturally in Galois theory as the Galois group of $\mathbb{Q}\left(2^{1/p},\xi\right)$, the splitting field of $x^p-2$ over $\mathbb{Q}$. (Here, $\xi$ is a $p^\text{th}$ root of unity.)  In this case, the most obvious set of generators is $$\sigma:\left\{\begin{array}{l}2^{1/p}\mapsto \xi 2^{1/p}\\ \xi\mapsto \xi\end{array}\right.\hspace{15pt}\text{and}\hspace{15pt}\tau:\left\{\begin{array}{l}2^{1/p}\mapsto 2^{1/p}\\ \xi \mapsto \xi^m\end{array}\right.$$ where $m$ has order $p-1$ mod $p$.
For an abstraction of the previous example, check out Z-groups, groups with all Sylow subgroups are all cyclic.  These are related to extraspecial groups and Zassenhaus groups, and tend to be good examples in character theory.  Like most things with well-understood character theory, Z-groups came up a lot in the papers leading up to the Classification theorem.
Double and triple covers of alternating groups show up as counterexamples in finite group theory often.

Make your own.

You can take wreath products of any of the above nonabelian groups to get very strange looking groups indeed.  (Just the other day $(S_p\wr S_p)\wr S_p$ came up in chat.)
If you know something about the automorphisms of a group $G$, you can take semidirect products $G\rtimes H$ with other groups $H$ (usually those that embed into $\operatorname{Aut}(G)$).  In particular, holomorphs can be an easy way to make new groups with easy arithmetic.
If two groups $H$ and $K$ have isomorphic central subgroups $Z_1\leq Z(H)$ and $Z_2\leq Z(K)$, you can take the (external) central product of $H$ and $K$, written $H{\small\text{ Y }} K$.  This is done by identifying points in $Z_1$ with the corresponding points in $Z_2$, then taking the quotient of $H\times K$ by this diagonal central subgroup. (Explicitly, if $\theta:Z_1\rightarrow Z_2$ is an isomorphism, let $Z=\{(z,\theta(z)^{-1}):z\in Z_1\}$, and $H{\small\text{ Y }} K=\left(H\times K\right)/Z$.)  There's a rich theory behind central products, and you can make some nifty things happen with them. (example)
If you want to give a group theorist a headache, take a direct product with one component from each of the above categories, wreath it with itself, and ask about the structure of its Sylow $2$-subgroup.


Why is $\frac{1}{\frac{1}{X}}=X$? Can someone help me understand in basic terms why $$\frac{1}{\frac{1}{X}} = X$$

And my book says that "to simplify the reciprocal of a fraction, invert the fraction"...I don't get this because isn't  reciprocal by definition the invert of the fraction?

$1/x$ is, by definition, the number that you multiply $x$ by to get $1$.

Similarly, $1/\left(1/x\right)$ is the number that you multiply $1/x$ by to get $1$.

But wait a sec: we just learned in the first sentence that that number is $x$.

Simply because:

$$
\frac{1}{\frac{1}{X}}=\frac{1}{(\frac{X}{1})^{-1}}=\frac{1}{1}\frac{X}{1}=\frac{1X}{1}=X
$$

And yes, the reciprocal of a fraction, is the inverted fraction...


Complex number: calculate $(1 + i)^n$. I have to solve the following complex number exercise: calculate $(1 + i)^n\forall n\in\mathbb{N}$ giving the result in $a + ib$ notation.

Basically what I have done is calculate $(1 + i)^n$ for some $n$ values.
$$(1 + i)^1 = 1 + i$$
$$(1 + i)^2 = 2i$$
$$(1 + i)^3 = - 2 + 2i$$
$$\boxed{(1 + i)^4 = - 4}$$
$$(1 + i)^5 = (1 + i)^4\cdot(1 + i)^1 = (-4)\cdot(1 + i) = - 4 - 4i$$
$$(1 + i)^6 = (1 + i)^4\cdot(1 + i)^2 = (-4)\cdot2i = - 8i$$
$$(1 + i)^7 = (1 + i)^4\cdot(1 + i)^3 = (-4)\cdot(- 2 + 2i) = 8 - 8i$$
$$(1 + i)^8 = (1 + i)^4\cdot(1 + i)^4 = (-4)\cdot(-4) = (-4)^2 = 16$$
We can write $n = 4\cdot q + r$ (Euclidean division), so we have:
$$(1 + i)^n = (1 + i)^{(4\cdot q + r)} = ((1 + i)^4)^q\cdot(1 + i)^r = (-4)^q\cdot(1 + i)^r$$
Finally if you want to calculate say... $(1 + i)^n$ for $n = 625$ you have:
$$625 = 4\cdot156 + 1\Rightarrow q = 156, r = 1$$
$$(1 + i)^{625} = (-4)^{156}\cdot(1 + i)^1 = (-4)^{156} + (-4)^{156}i$$
What other approach would you suggest? Mine works, but you have to find $q$ and $r$ in order to do the calculation, and I think it is not "calculate" technically speaking which was what the exercise is asking (although I am not sure what they mean by calculate).

Write $1+i$ in modulus argument notation ($\rho e^{i \theta}$). Then $(1+i)^n = \rho^n e^{i n \theta}$ will be pretty easy to compute.

How about setting L = lim_n/rarrow (1 + i) ^n


Is there a finitist semantics for transfinite mathematics? I'm sympathetic to the Aristotelian view that potential infinity makes sense while actual (completed) infinity doesn't. However, I also find transfinite set theory to be fascinating, and I'm under the impression that this domain of mathematics is founded upon the idea of performing calculations with completed infinities. So for a while I felt like I had to "pick sides", so to speak.
Upon reflection, however, this doesn't really make sense. Formalists and fictionalists ignore or deny the reality of mathematical objects without ignoring/denying mathematics as an activity, so why should my ontological reservations matter when it comes to working with "higher infinities"? Clearly there is something interesting and systematic going on in this domain.
It seems like my difficulty with transfinite set theory is merely semantic: I don't think it makes sense to talk about adding one to $\omega$ if the symbol refers to a completed infinity, but there's nothing wrong with adding one to $\omega$ as long as $\omega$ means something else. My question here is basically: what are my options for that "something else"?
(As an example: one thing I'm wondering is if it's possible to understand transfinite mathematics as operating on potential infinities instead of actual infinities.${}^1$ Perhaps we could understand $\omega$ not as a completed object but as an endless process.${}^2$ Is there a "process" interpretation of transfinite set theory?)
Edit: What about coalgebras? Is there a coalgebraic version of transfinite set theory that treats all completed infinities as incomplete processes?

${}^1$ Potentially relevant: Can we formally distinguish between actual and potential infinities?
${}^2$ What would it mean to "add" and "multiply" processes? Can we make sense of the countable/uncountable distinction in a process context?

The hyperreal numbers do what I believe you are trying to do.  While there are many different notations, I prefer the one that uses $\epsilon$ for a standard marker for an infinitesimal, and $\omega$ as a standard marker for an infinity, such that $\epsilon = \frac{1}{\omega}$.
This makes infinities that act semantically basically like uncompleted infinities.  You can do arithmetic with them in the ordinary sense, just treating $\omega$ and $\epsilon$ as if they were variables, but then using the "standard part" function (std(x)) to reduce it to a real if you need.
In this system, $2\omega + 1$ is a different value than $\omega - 1$.  They can also be put into ratio with one another and cancel: $\frac{5\omega}{10\omega} = \frac{1}{2}$.
You can even do things like $\frac{\omega^2 + 2}{\omega^2}$ which yields $1 + \frac{2}{\omega^2} = 1 + 2\epsilon^2$, whose standard part is just $1$.
This also allows you to assign values to certain divergent series, essentially treating them as if they were convergent to a hyperreal value.  For instance, $\frac{1 + 3 + 5 + \ldots}{1 + 2 + 3 + \ldots} \simeq 2$.


Why don't previous events affect the probability of (say) a coin showing tails? Why doesn't a previous event affect the probability of (say) a coin showing tails?

Let's say I have a fair and unbiased coin with two sides, heads and tails.

For the first time I toss it up the probabilities of both events are equal to $\frac{1}{2}$.
This much is intuitive and clear to me.

Now suppose that I toss it up $1000000000$ times and the scores are,

$501000000$ Heads

$499000000$ Tails

Now, for the $1001000000^{th}$ toss, shouldn't the probability of a tail coming up be greater than that of heads showing up?

I have seen many books which say that even for the $1001000000^{th}$ toss, the probabilities of both events are equal to $\frac{1}{2}$.

This seems wrong to me since the same books affirm that if a coin is tossed a large number of times, the quantity $\frac{heads}{tails}$ will approach $1$.

I know this is very elementary and naive, yet I had only superficially studied probability and I hope you all will bear with me.

My Objections with some of the top-voted answers

  It isn't that future flips compensate for the imbalance, it is that there are so many of them it doesn't matter.

I don't get this statement. What exactly does the second sentence mean? Moreover, if what you said is true then, the following comment by a user should be wrong,

  Law of large numbers

So these are contradicting each other I feel. Please bear with my lack of knowledge.

The assumption for a coin is that is has no memory.  That means that the chance of heads is the same on every toss.  For a fair coin, that chance is $\frac 12$ regardless of the history.  If you toss $100$ times and get heads every time (very unlikely, but it could happen)  the most probable event after a million tosses (including the $100$ you already did) is $500050$ heads and $499950$ tails.  It isn't that future flips compensate for the imbalance, it is that there are so many of them it doesn't matter.  Look how close the head/tail ratio would be to $1$ at that point.

The probability will always be 1/2 for each coin toss because the event physically has no tie to the previous throws.  It will approach 1 because, as you approach infinity, the difference between the number of heads and tails will become smaller.  The difference between heads and tails will become smaller because the probably for each are equal to each other.  So, when you have 10000000000003 heads vs 10000000000017 tails, you get 0.99999999999860000000000238 or very close to 1.  Just keep going infinitely and you will get closer and closer to 1.


What are some examples of when Mathematics 'accidentally' discovered something about the world? I do not remember precisely what the equations or who the relevant mathematicians and physicists were, but I recall being told the following story. I apologise in advance if I have misunderstood anything, or just have it plain wrong. The story is as follows.

  A quantum physicist created some equations to
  model what we already know about sub-atomic particles. His equations
  and models are amazingly accurate, but they only seem to be
  able to hold true if a mysterious particle,
  currently unknown to humanity, exists.

  More experiments are run and lo and
  behold, that 'mysterious particle' in actual fact exists! It was found to be a quark/dark-matter/anti-matter, or something of the sort.

What similar occurrences in history have occurred, where the mathematical model was so accurate/good, that it 'accidentally' led to the discovery of something previously unknown?

If you have an answer, could you please provide the specific equation(s), or the name of the equation(s), that directly led to this?

I can recall one other example.

  Maxwell's equations predicted the existence of radio waves, which were
  then found by Hertz.

The planet Neptune's discovery was an example of something similar to this. It was known that Newtons's Equations gave the wrong description of the motion of Uranus and Mercury. Urbain Le Verrier sat down and tried to see what would happen if we assumed that the equations were right and the universe was wrong. He set up a complicated system of equations that incorporated a lot of ways contemporary knowledge of the universe could wrong, including the number of planets, the location and mass of the planets, and the presences of the forces other than gravity. He would eventually find a solution to the equations where the dominating error was the presence of another, as of yet undetected, planet. His equations gave the distance from the sun and the mass of the planet correctly, as well as enough detail about the planet's location in the sky that it was found with only an hour of searching.

Mercury's orbit's issues would eventually be solved by General Relativity.

I did a quick scan and saw no mention of Riemannian manifolds here. Historically, it might've been the exact high-precision mathematical machinery that Einstein needed; from this  this wikipedia link:

  Albert Einstein used the theory of Riemannian manifolds to develop his
  general theory of relativity. In particular, his equations for
  gravitation are constraints on the curvature of space.


Computing the solution of a recurrence relation Let $y_k=\dfrac{16}{3}y_{k-1}-\dfrac{5}{3}y_{k-2}$

with $y_0=1$ and $y_1=\dfrac{1}{2}$ for $k \geq 2, k \in \mathbb{N}$.

To find the solution $\lbrace y_k \rbrace_{k \in \mathbb{N_0}}$ I used:

Firstly, I computed it for $k=2$:

$y_2=\dfrac{16}{3}\dfrac{1}{2}-\dfrac{5}{3}1=1$

What has to be done next? Induction isn't helpful here.

I don't see how can it be found for $\lbrace y_k \rbrace_{k \in \mathbb{N_0}}$.


Finding the minimum and the maximum when speaking of relations I'm trying to figure out how to find the minimum and the maximum when speaking of relations. Consider $A=(2,3,\ldots,13)$. Also consider the following relation $R$:
$$
(x,y)\in R \Leftrightarrow x|y
$$
where $x,y\in A$. I need to find the minimum and the maximum of the relation. What does it mean? How do I find them?

$
\newcommand{\p}{\preccurlyeq}
$
A "maximal" element of a set $A$ with partial order $\p$ is an $M \in A$ such that there is no $a \in A$ where both $M \p a$ and $M \ne a$. Likewise, a minimal element of $(A,\p)$ is an $m \in a$ such that no $a \in A$ satisfies $a \p m$ and $a \ne m$ together.
Consequently, $a \in A$ is not maximal if there exists $b \in A$, $b\ne a$, such that $a \p b$. Similarly, $a$ is not minimal if $b \p a$ but $b \ne a$.
So, run through the list invalidating candidates for maxima first:

$\not \exists x \in A$ such that $x \mid 2$ (except itself)
$\not \exists x \in A$ such that $x \mid 3$ (except itself)
$2 \mid 4$
$\not \exists x \in A$ such that $x \mid 5$ (except itself)
$3 \mid 6$
$\not \exists x \in A$ such that $x \mid 7$ (except itself)
$2 \mid 8$
$3 \mid 9$
$2 \mid 10$
$\not \exists x \in A$ such that $x \mid 11$ (except itself)
$2 \mid 12$
$\not \exists x \in A$ such that $x \mid 13$ (except itself)

Thus the maxima are the primes in $A$ -- which makes sense, for the divisibility relation.
We can similarly go through the list again and eliminate candidates for minima:

$2 \mid 4$
$3 \mid 6$
$4 \mid 8$
$5 \mid 10$
$6 \mid 12$
$7,8,\cdots,13$ only divide themselves in $A$

Thus the maxima are $7, 8,9,10,11,12$.

Note that there is no singular "greatest maximum"/"least minimum" for this relation. If there were, we would not have multiple maximal/minimal elements.

$2|4$ and no element $x\in A$ other than $2$ divides $2$, thus $ 2$ is the minimum of $ A$ according to the relation $ R$.

no element of $A$ divides $ 13$ but $2|12 $ and $12$ does not divide any other element of $A$. its follows that $ 12$ is the maximum of $A$.


Why is the period of sin(2x) π? In my textbook, it is stated that the period of $\sin(2x)$ is $\pi$. The author justifies this using a mathematical statement which I cannot understand.
He writes that, since $\sin(2x) = \sin(2x+2\pi) = \sin(2(x+\pi))$ the period of $\sin(2x)$ is $\pi$.

Though my intuition tells me that the period of $\sin(2x)$ is $\pi$, I just cannot understand this reasoning. To me the period of $\sin(2x)$ appears to be to $2\pi$ since $\sin(2x)=\sin(2x+2\pi)$. I would be very thankful if someone could explain this reasoning to me.

By the definition of a period of a real function $f$: $$T=\inf_{t \in \mathbb R}\{t\mid \forall x\in \mathbb R:f(x)=f(x+t)\}$$
We can deduce from that the period of $f(x)=\sin(2x)$:
$$T_{\sin(2x)}=\inf_{t \in \mathbb R}\{t\mid \forall x\in \mathbb R:\sin(2x)=\sin(2x+2t)\}=\pi$$

The period of a function $f$ is (informally) the smallest value of $k$ (if any) so that $f(x + k) = f(x)$ for all $k$.

The period of $\sin()$ is $2\pi$ as you no doubt accept.  We'll take that as a given.

$f(x) = \sin(2x)$ is a different function.

$f(x + \pi) = \sin (2(x+\pi)) = \sin (2x + 2\pi) = \sin (2x) = f(x)$.  So the period of $f$ is $\pi$ or smaller.  (It isn't smaller.  If $f(x + k) = f(x)$ then $\sin(2x + 2k) = \sin 2x$ so the period of $\sin $ would be $2k$ or smaller.  So $2k$ is not smaller than $2\pi$.)

Your confusion lies in you think we are adding $2\pi$ to $2x$ into the argument of $\sin$ it makes the period $2\pi$.  True; it makes the period of SINE(x) $2\pi$. But we are sticking the $2\pi$ into $\sin ()$; we are not sticking it into $\sin (2x)$. We are only sticking $\pi$ into ...

... okay, look at this:  $\sin (x)$ can be written as $\sin( [\backslash stick input here/])\backslash$ and $\sin(2x)$ can be written as $\sin(2\times [\backslash stick input here/])$.  And $a + b$ can be written as $\text {a is the main thing} ---\text{ b is tacked on for the ride}$ or $a --tackon-- b$.

So $\sin( [\backslash 2x/])\backslash = \sin( [\backslash 2x -tackon- 2\pi/])\backslash $. So the period is $2\pi$

But $\sin(2x [\backslash x -tackon- \pi/])=\sin(2x [\backslash x/]--tackon-- 2\pi)$

$= \sin([\backslash 2\times x/]--tackon-- 2\pi)=$

$\sin([\backslash 2\times x --tackon-- 2\pi/])=$

$\sin ([\backslash 2\times x /]) = \sin(2\times [\backslash x /])$.

So the period of $\sin(2\times [\backslash put input here /]$ is $\pi$.


When finding relative minimum/maximum, what is the point of using the second derivative test? Why use the second derivative test over the first derivative test when finding maxima and minima if it's uncertain what $f''(x) = 0$ is? Why not just always use the first derivative since we need to take the first derivate either way?

Here's what I tell my students:

The first derivative test is usually harder to use, but will always give you an answer (at least for reasonable functions);
The second derivative test is usually easier to use, but it may not be applicable (if it is not a stationary point) and it may be inconclusive.

Why do I say that? To use the second derivative test, you just need to plug in a number and evaluate. That's straightforward. To use the first derivative test, you need to analyse the sign of the first derivative on a neighborhood of the point and figure out whether it changes signs as it goes through the point. This can be harder to do than just plugging in a single point.

As you say, you always already have the first derivative handy. But you may also need to have the second derivative handy: for example, if you are doing a full analysis of the function, looking for intervals in which the function is increasing and decreasing, and intervals in which the function is concave up and concave down, you will also have the second derivative handy.

It's a tradeoff, then:

For the First Derivative Test, you can always use it (stationary points and points where the first derivative does not exist), will almost always give you an answer either way (detecting critical points that are not relative extremes), and does not require any further derivatives to be calculated. But it requires a sign analysis, which may not be trivial (or possible).
For the Second Derivative, you can't always use it, and even if you can use it, it may be inconclusive and it will never detect critical points that are not relative extremes, and it requires you to calculate another derivative. But it is usually much easier to actually use, as it only requires figuring out the sign at a single point.

But more generally: it's always good to have more than one way to do something, as different ways may provide strengths and weaknesses that the other methods don't have, or provide a different kind of insight.

Because a zero of the derivative has no sign ! Hence it cannot tell a minimum from a maximum.

And what about $f(x)=x^3$ ?


Non-compactness of the resolvent Consider a complete non-compact Riemannian manifold $M$ and the resolvent of the Laplacian $(-\Delta + \lambda I)^{-1}$. It is known that the resolvent is in general not a compact operator. I am trying to understand why not. I realize that the proof in the compact setting does not follow through for the non-compact case because of the failure of compact Sobolev embedding. I would be happy to see why the resolvent is not compact even for $M = \mathbb{R}^n$, and a justification for general non-compact $M$ would be excellent. Thanks.

For $M=\mathbb R^n$ take a smooth function with compact support $\phi\in C_0^\infty(M)$.

Then
$$
\phi = (-\Delta+\lambda I)^{-1}((-\Delta+\lambda I)\phi)
$$
obviously. Now consider translations $\phi_k(x):=\phi(x+kv)$ for $v\in \mathbb R^n$, $k\in \mathbb N$. Take the norm of $v$ large enough such that the supports of different $\phi_k$'s are empty.

The set
$$
X:=\{ (-\Delta+\lambda I)\phi_k,\ k\in\mathbb N\}
$$
is bounded in $L^2$, but
$$
 (-\Delta+\lambda I)^{-1}X
$$
is not relatively compact in $L^2$.

The simplest way to show that for a non-compact manifold the resolvent does not have to be compact is to give a counterexample. Following your (very good) choice of the manifold $M=\mathbb{R}^n$, lets show that $(-\Delta - \lambda I)^{-1}$ is not compact. For simplicity, assume $n=2$.

Assume $\vec{\mathbf{x}}, \vec{\mathbf{y}} \in \mathbb{R}^2$. Then the  fundamental solution $\mathit{\Phi}$  of 2D Laplacian $\Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}  \ $  takes form

$$
\mathit{\Phi}\big(\vec{\mathbf{x}}, \vec{\mathbf{y}}\big) =- \frac{1}{2\pi}\ln \big|\vec{\mathbf{x}} - \vec{\mathbf{y}} \big| =
- \frac{1}{4\pi}\ln \Big( (x_1 - x_2)^2 + (y_1-y_2)^2 \Big). \label{1}\tag{1}
$$

Recall that the fundamental solution of a linear PDE, (which is equivalent to the Green's function) is the kernel of resolvent of the corresponding differential operator.

   For example, see  this link, page 4.

Denote the resolvent $R_\lambda = (-\Delta - \lambda I)^{-1}$. Then the resolvent of 2D Laplacian takes form
$$
R_\lambda[u](\mathbf{x}) = \int_{\mathbb{R}^2} \mathit{\Phi}\big(\mathbf{x}, \mathbf{y}\big) u (\mathbf{y}).\label{2}\tag{2} $$
Substituting $\eqref{1}$ into $\eqref{2}$, we get

$$
R_\lambda[u](\mathbf{x}) = - \frac{1}{2\pi}  \int_{\mathbb{R}^2} \bigg( \ln \big|\vec{\mathbf{x}} - \vec{\mathbf{y}} \big| \!\cdot\! u (\mathbf{y}) \bigg)\,\mathrm{d}^2 \mathbf y.\label{3}\tag{3}
$$
or, equivalently,
$$
R_\lambda[u]\big(x_1, x_2\big) =
- \frac{1}{4\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty \!\bigg(\! \ln\! \Big( (x_1 - x_2)^2 + (y_1-y_2)^2 \Big)\! \cdot\! u\big(x_1,x_2\big)\! \bigg)dy_1  dy_2, \label{4}\tag{4}
$$
where
$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \in \mathbb{R}^2$ and $\mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \in \mathbb{R}^2$.

Now, when we have the explicit expression  \eqref{3}, it is easy to see that  the resolvent $R_\lambda$ is not compact, since the integral kernel $\mathit{\Phi}\not\in L^2\big( \mathbb{R}^2\times  \mathbb{R}^2\big) $.


Find area of the triangle ABC, given the coordinates of vertices in plane $A, B$ and $C$ are the points $(7,3), (-4,1)$ and $(-3,-2)$ respectively. Find the area of the triangle $ABC$.

I've worked out the lengths of each side of the triangle which are $AB=5\sqrt5$, $BC=\sqrt10$ and $AC=5\sqrt5$.

I know that the formula for the area of a triangle is $\frac12hb$ but when I checked the solutions the answer to the area of this triangle is $17\frac12$.

I do not understand how this answer is achieved.

Since you have obtained the length of each side, using Heron's Formula is a natural way to find the area.  Let's consider the approach suomynonA suggested in the comments. Consider the figure below.

We can find the area of $\triangle ABC$ by subtracting the sum of the areas of the three right triangles $ABD$, $ACF$, and $BCE$ from the area of rectangle $ADEF$.  I will leave the details of the calculations to you.

I need anwer .A(6,2),B(2,4) and C(8,-4) are three in a plane in the line BC and Q is a point in the plane, such that PAQB is a rectangle. Find the (I) equation of the sides AP and BP. (ii) the coordinate of the points P (iii) the area of the rectangle


Is the relation $R = \{ ( 1,2 ) \}$ on $A = \{ 1,2,3,4 \}$ not reflexive, not symmetric, antisymmetric and transitive? Given $R = \{ ( 1,2 ) \}$ over $A = \{ 1,2,3,4 \}$:

$R$ is not reflexive because there's no $(1,1)$ and $(2,2)$.

$R$ is not symmetric because there isn't $( 2,1 )$.

$R$ is anti-symmetric because there isn't $( 2,1 )$.

And $R$ is transitive because there isn't a $(2,x)$ element in $R$.

If  $(1,2) \in R \land (2,x) \in R$ is false, then

$(1,2) \in R \land (2,x) \in R \to (1,x) \in R$ is true,

and therefore $R$ is transitive.

Is this correct?

The relation is not reflexive, since $(a,a) \not \in R$ for every $a \in \{1,2,3,4\}$.

The relation is not symmetric, since while $(1,2) \in R$, the reversal $(2,1) \not \in R$.

The relation is anti-symmetric in a vacuous sense: since $(1,2) \in R$ and is the only ordered pair in $R$, there is no "violation" in the sense that since there is no pair of related pairs $(a,b),(b,a) \in R$ with $a \ne b$ to begin with. Thus, anti-symmetry holds trivially.

The relation is transitive, for similar reasons. If $R$ were not transitive, then there would be pairs $(2,x),(x,y) \in R$ for which $(1,y) \not \in R$. However, no $(2,x) \in R$ for any $x$, so there is no issue there. Vacuously, in a sense, $R$ thus is transitive.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.

Not reflexive since $(1,1,),\ldots,(4,4)$ are missing, and not symmetric since $(2,1)$ is missing, but both antisymmetric and transitive since premises of implications (in their definitions) are not fulfilled.


Proving convergence of a recursively defined sequence Using a computer it is easy to see that the sequences defined by letting $a_1=1$, $a_2=m$, and
$$a_n=\frac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}$$
converges to $\frac{3m}{m+2}$. I would very much like to know how to prove this. I don't even know where to start.

Hint: The following is I think a natural starting "move."

Let $a_n=\dfrac{1}{x_n}$.  When the smoke clears you get a nice linear recurrence.

Continuation: Making the suggested substitution, and simplifying, we obtain the recurrence $x_n=\dfrac{1}{2}\left(x_{n-1}+x_{n-2}\right)$. We have to be a little careful, since there will be trouble if $a_{n-1}+a_{n-2}=0$. But for example if $m\gt 0$ then all terms are safely positive.

There are many ways to solve the recurrence. For example we can use a general technique for solving homogeneous linear recurrences with constant coefficients, and write down the characteristic polynomial $2x^2-x-1=0$, which has the roots $1$ and $-\dfrac{1}{2}$.

Thus the general solution of the recurrence is $x_n=A+B\left(-\frac{1}{2}\right)^n$. Putting $x_1=1$ and $x_2=\dfrac{1}{m}$ we find that $A=\dfrac{m+2}{3m}$, and  $B=\dfrac{4-2m}{3m}$. Thus
$$x_n=\frac{m+2}{3m} + \left(-\frac{1}{2}\right)^n \frac{4-2m}{3m}.$$
As $n\to\infty$, $x_n\to \dfrac{m+2}{3m}$.

Remark: Because the numbers are quite simple, we do not need to use the "general" procedure to solve the recurrence. More generally, consider the recurrence
$$y_{n}=py_{n-1}+(1-p)y_{n-2},$$
which comes up in some probability calculations.
This may be rewritten as
$$y_n-y_{n-1}=(-1)(1-p)(y_{n-1}-y_{n-2}).$$
If we let $z_n=y_{n}-y_{n-1}$, we get $z_n=(-1)(1-p)z_{n-1}$. So the sequence $(z_n)$ is a geometric sequence with common ratio $-(1-p)$. Now we can easily find a general expression for $z_n$, and hence for $y_n$.

Let $A_n=\frac{a_n}{a_{n-1}}$, then we have $A_n=\frac{2}{1+A_{n-1}}$, it is easy to find that $\lim\limits_{n\to\infty }A_n$ exists
then we have $\lim\limits_{n\to\infty }\frac{a_n}{a_{n-1}}=\frac{\sqrt{2}-1}{2}<1$, we can use the $\epsilon- N$ to illustrate the result......


Proof by induction: $2^{n} < n!$ Prove that $2^{n} < n!$ $\forall$ n > 4

$n=5:$ $$2^{5}<5!$$

$$32 < 120$$

This is true.

Now, after knowing it worked for $n$ we need to show it works for every other, so $n+1$:

$$2^{n+1} < (n+1)!$$

$$2^{n} < \frac{(n+1)!}{2}$$

We know from beginning that $2^{n} < n!$

So we replace it with it here and then show:

$$n! < \frac{(n+1)!}{2}$$

$$n! < \frac{n!\cdot(n+1)}{2}$$

$$1<\frac{n+1}{2}$$

Task say for all $n > 4$ so the thing on the right side will really be greater than $1$.

I hope everything is ok?

Edit: The possible-duplicate-link didn't help me because I'm not really looking for a solution to the task. I'm rather interested in knowing if MY proof is correct.

I would simply note that for sufficiently large $n$, $2$ < $n$ + $1$.  Then by induction, when $2^n$ < $n$!, $2^n$ ${\times}$ $2$ < $n$! ${\times}$ $2$ and $n$! ${\times}$ $2$ < $n$! ${\times}$ $n + 1$, so by transitivity $2^n$ ${\times}$ $2$ < $n$! ${\times}$ $n + 1$.

Are you stating that if $a<b$ and $a<c$ then $b<c?$ But you have the correct answer. I'd do it in this way.

Assume its true that it's true that $2^n<n!=>\frac{2^n}{n!}<1$ then for $n+1$ we have to prove that:

$2^{n+1}<(n+1)!$

$2^n<n!(n+1) * 1/2$

$\frac{2^n}{n!}<(n+1)*1/2$

And then state that as $\frac{2^n}{n!}<1$ and $\frac{n+1}{2}>1$ for any $n>4$ Therefore the inequality is true.


Let $A,B,C,D$ lie on a circle, $AD$ and $BC$ meet at $E$. The circle through $A,C,E$ have centre $O$. Prove that $BD$ is perpendicular to $EO$. Let A,B,C,D lie on a circle, where AD and BC meet at E. Let the circle through A,C,E have centre O. Prove that BD is perpendicular to EO.

I saw this question posted earlier but it was put on hold and now I can't find it anymore... I thought I'd repost with my solution. Not sure about the rules on that; Let me know if I have to take it down.

Are there any other ways to prove this? There was an answer to the old post regarding defining a circle through B,E,G, with BE as the diameter, and by Thales theorem this would prove that the angle at G is 90 degrees. But how can you prove that BE would be the diameter of the circle through B,E,G without assuming angleBGE to be 90 degrees?

Picture by someone who answered the original post

You can also use Reim's theorem:

https://mathsolympian.wordpress.com/2014/08/02/reims-theorem/

Then by Reim we have that tangent $EE$ on circle $ACE$ is parallel to $DB$. But since tangent $EE$ is perpendicular to radius $EO$ we are done.

I used circle theorems (which I just realised I forgot to name exactly).


How to prove $ E(|X-Y|) \le E(|X+Y|)$ when $X,Y$ are i.i.d variables? Let $X$ and $Y$ be independent random variables having the same distribution and the finite mathematical expectation. How to prove the inequality $$ E(|X-Y|) \le E(|X+Y|)?$$

After a little inspection, we see that
$$
E(|X+Y|-|X-Y|) = 2E[Z(1_{XY\geq 0}-1_{XY<0})]
$$
where $Z = \min(|X|,|Y|)$.

Remember that for any non-negative random variable $T$,
$$
E(T) = \int_0^\infty P(T>t)\,dt.
$$

We apply this with $T=Z\,1_{X \geq 0, Y\geq 0}$, $T=Z\,1_{X < 0, Y< 0}$ and $T=Z\,1_{X \geq 0, Y< 0}$. Since $\{Z > t\} = \{|X| > t\}\cap\{|Y| > t\}$, we obtain

$$
E(Z \,1_{X \geq 0,Y\geq 0}) = \int_0^\infty P(X > t)P(Y > t)\,dt = \int_0^\infty P(X > t)^2\,dt
$$

$$
E(Z\, 1_{X < 0, Y < 0}) = \int_0^\infty P(X < -t)P(Y < - t)\,dt = \int_0^\infty P(X < -t)^2\,dt
$$

$$
E(Z\,1_{X \geq 0, Y< 0}) = E(Z\,1_{\{X < 0, Y \geq 0\}}) = \int_0^\infty P(X > t)P(X < -t)\,dt
$$

So finally,
$$
E(|X+Y|-|X-Y|) = 2\int_0^\infty (P(X>t)-P(X<-t))^2\,dt \geq 0
$$

Remark 1. The inequality is an equality if and only if the distribution of $X$ is symetric, that is $P(X > t) = P(X < -t)$ for any $t \geq 0$.

Remark 2. When $|X|=1$ a.s. the inequality is nothing but the semi-trivial fact that if $X$ and $Y$ are independent with same distribution, then $P(XY \geq 0) \geq \dfrac{1}{2}$.

Remark 3. It is worthwile to mention a nice corollary : $E(|X+Y|) \geq E(|X|)$. The function $x \mapsto |x|$ is convex hence $|X| \leq \frac{1}{2}(|X+Y|+|X-Y|)$. Taking expectations we find
$$
\Bbb E(|X+Y|-|X|) \geq \frac{1}{2}\Bbb E(|X+Y|-|X-Y|) \geq 0.
$$
Furthermore, there is an equality if and only if $X=0$ a.s.

E[x] is a linear operator.

This means E[X + Y] = E[X] + E[Y]

Also, E[X - Y]  = E[X] - E[Y]

The statement will be true when $E[Y] \ge 0$


Why is gradient the direction of steepest ascent? $$f(x_1,x_2,...x_n):\mathbb{R}^n \rightarrow \mathbb{R}$$
The definition of the gradient is
$$ \frac{\partial f}{\partial x_1}\hat{e}_1 +\  ... +\frac{\partial f}{\partial x_n}\hat{e}_n$$

which is a vector.

Reading this definition makes me consider that each component of the gradient corresponds to the rate of change with respect to my objective function if I go along with the direction $\hat{e}_i$.

But I can't see why this vector (defined by the definition of the gradient) has anything to do with the steepest descent.

Why do I get maximal value again if I move along with the direction of gradient?

Each component of the gradient tells you how fast your function is changing with respect to the standard basis.  It's not too far-fetched then to wonder, how fast the function might be changing with respect to some arbitrary direction?  Letting $\vec v$ denote a unit vector, we can project along this direction in the natural way, namely via the dot product $\text{grad}( f(a))\cdot \vec v$.  This is a fairly common definition of the directional derivative.

We can then ask in what direction is this quantity maximal?  You'll recall that $$\text{grad}( f(a))\cdot \vec v = |\text{grad}( f(a))|| \vec v|\text{cos}(\theta)$$

Since $\vec v$ is unit, we have $|\text{grad}( f)|\text{cos}(\theta)$, which is maximal when $\cos(\theta)=1$, in particular when $\vec v$ points in the same direction as $\text{grad}(f(a))$.

I think the most intuitive explanation is the following: Draw one vector in the X direction. Draw another vector in the Y direction. Make each vector any length you want. Make X longer than Y or Y longer than X or make them the same length. Doesn't matter. Now draw the vector that represents the sum of those two vectors, using the rectangle method. This is key to the intuition. You have to draw a rectangle. The resulting vector (representing the sum of the X-direction vector and the Y-direction vector) goes from one corner of the rectangle to the opposite corner. Remember that the length of the resulting vector represents the magnitude, which in this context represents slope. Now look at the drawing and ask yourself: is there any vector within this rectangle, starting at the origin, that is longer than the diagonal one? No. Any vector you draw that starts at the origin and goes to another side of the rectangle is shorter than the diagonal one. Of course, you can draw a vector at the origin that is longer than the diagonal one, but only if that vector leaves the rectangle. But if it leaves the rectangle, then it's no longer operating within the constraint of the given X vector and Y vector; it's working with some different X vector and Y vector. I hope that helps. It took me a while to understand this intuitively and the image of a diagonal within a rectangle finally switched the lamp on.


Calculate limits without l'Hôpital I am trying to calculate the following limits
$$ \lim_{x\to\infty}(2x+1) \ln \left(\frac{x-3}{x+2}\right) $$
and
$$ \lim_{x\to1}\frac{x}{3x-3}\ln(7-6x) $$
I can't use l'Hôpital's rule, so I am not sure how to solve it.

By continuity of the logarithm,
$$ \lim_{x\to\infty} \ln \left(\frac{x-3}{x+2}\right)^{2x+1}
= \lim_{x\to\infty} \ln \left(1-\frac5{x+2}\right)^{2(x+2)-3}\\
=\ln((e^{-5})^2)-3\lim_{x\to\infty}\ln \left(1-\frac5{x+2}\right).$$

$$\lim_{x\rightarrow \infty }\left ( 2x+1 \right )\ln \left ( \frac{x-3}{x+2} \right )=\lim_{x\rightarrow \infty }\frac{\frac{\mathrm{d} }{\mathrm{d} x}\ln \left ( \frac{x-3}{x+2} \right )}{\frac{\mathrm{d} }{\mathrm{d} x}\frac{1}{\left ( 2x+1 \right )}}=-\lim_{x\rightarrow \infty }\frac{5\left ( 2x+1 \right )^2}{2\left ( x-3 \right )\left ( x+2 \right )}=-\frac{5}{2}\lim_{x\rightarrow \infty }\frac{\left ( 2x+1 \right )^2}{\left ( x-3 \right )\left ( x+2 \right )}=-\frac{5}{2}\cdot 4\lim_{x\rightarrow \infty }\frac{2x+1}{2x-1}=-\frac{5}{2}\cdot 4=-10$$
$$\lim_{x\rightarrow 1 }\frac{x}{3x-3}\ln \left ( 7-6x \right )=\lim_{x\rightarrow 1 }\frac{\ln \left ( 7-6x \right )}{3x-3}=\lim_{x\rightarrow 1 }\frac{\frac{\mathrm{d} }{\mathrm{d} x}\ln \left ( 7-6x \right )}{\frac{\mathrm{d} }{\mathrm{d} x}3x-3}=\lim_{x\rightarrow 1 }\frac{2}{6x-7}=-2$$


Find $\lim \limits_{x\to 8} {\frac{64-x^2}{8-x}}$ I'm making a bunch of limits exercises and I found me stuck with this limit:

$$\begin{align}
\lim_{x\to 8} {\frac{64-x^2}{8-x}}
\end{align}$$

It gives me an indetermination of

$$\begin{align} \frac{0}{0} \end{align}$$

So, the next step is to factorize. I used the Ruffini's Rule.

The solution is $16$. And my result was $-16$, something is wrong :(

Any help?

To use Ruffini's Rule, you need to use x-8, not 8-x. So, multiply both numerator and denominators by -1 one and then apply Ruffini's rule. You will get the correct answer, which is +16.
Another trick for futur reference, is to compute the derivatives of both numerators and denominators and then compute the limit. In this case, you will get -2x divided by -1, which gives you 2x. When you replace x with 8, you get the answer +16.

Note: I saw that you said Ruffini's rule gives you -x-8, I think that is where your error is: when you multiply by -1, you should get -(8-x) = x-8. Same applies to the numerator.

$$
\frac{64 - x^2}{8-x} = 8 + x \;,
$$
so, the limit as $x \rightarrow 8$ is of course $16$.


Help with a certain proof For all $x,y \in \mathbb{R} - \{0\}$, $(xy)^{-1}=x^{-1}y^{-1}$.

I was wondering how I could solve this.

$\mathbb R$ is a field, so nonzero elements are invertible, then
$$xy\cdot x^{-1}y^{-1}=xy\cdot y^{-1}x^{-1}=x\cdot1\cdot x^{-1}=x\cdot x^{-1}=1$$
So $(xy)^{-1}=x^{-1}y^{-1}$.


Prove that sequence $a_n/n$ is convergent for $a_n = 1+\frac15+\frac19... + \frac1{4n-3}$ Show that the sequence $\langle a_n\rangle$ where

$a_n = 1+\frac{1}{5}+\frac{1}{9}... + \frac{1}{4n-3}$

does not converge, whereas $\langle b_n\rangle$ defined by $b_n = \frac{1}{n}  a_n$ converges to 0.

My Approach: I was able to prove that $\langle a_n\rangle$ does not converge by showing that it is not a Cauchy sequence but not able to prove that $\langle b_n\rangle$ is converging sequence. Is there a theorem related to this?.

I completely misread the question at first. Let's try again. We know that for $k\geq1$, we have $4k-3\geq k$, and hence
$$
\sum_{k=1}^n\frac{1}{4k-3}\leq \sum_{k=1}^n\frac{1}{k}.
$$
The sum on the right gives the harmonic numbers, $H_n$, which are approximated asymptotically by $\ln(n)$ (given by a Riemann approximation of $\int_1^n\frac{1}{x}dx$). Thus since $$\lim_{n\to\infty}\frac{\ln(n)}{n}=0,$$
we may apply the squeeze theorem to
$$0\leq \frac{1}{n}\sum_{k=1}^n\frac{1}{4k-3}\leq \frac{C+\ln(n)}{n}$$
(where $C$ is an arbitrarily large, fixed constant to make the inequality hold). Thus we have $b_n\to0$.


When a periodic function is squared (or cubed, and so on...) does it always lose its periodicity? For instance $$\sin^{2}\left(-\frac{\pi}{6}\right) = \sin^{2}\left(\frac{\pi}{6}\right)$$

i.e., $\sin^2 (x)$ is an even function and loses the $2\pi$-periodicity of $\sin x $.

Is this true in general?

Does $\tan^2 x$ lose the $\pi$-periodicity of $\tan x$?  The $\tan^2 x$ function still blows up wherever $\cos^2 x$ is equal to zero - I am currently studying $\tan^2 x$'s singularities to try to understand a solution to a problem I've been working on.

If $f : \mathbb R \to \mathbb R$ is periodic with period $T$, i.e., $f(t + T) = f(t)$ for all $t$, then certainly $f^n$ is also periodic of period $T$ as $$f^n(t) = (f(x))^n = (f(t + T))^n = f^n(t+T)$$

However $f^n$ can also pick up other shorter periods, as your example of $f = \sin$ illustrates: $\sin$ has period $2\pi$ as does $\sin^2$; but $\sin^2$ also has period $\pi$. The minimum period of $\sin^2$ of $\pi$ is therefore less than the minimum period of $2\pi$ for $\sin$.

$\tan$ has period $\pi$, as does $\tan^2$. What $\tan^2$ does not have is a minimum period less the minimum period of $\tan$.

Let $f$ be a periodic function of (minimum) period $T$ and let $g(x)=\left(f(x)\right)^2$.

$\textbf{Proposition:}$ $g$ have a (minimum) period of $\frac{T}{2}$ iff $f(x+\frac{T}{2})=\pm f(x)$

$\textbf{Proof:}$ Left to the reader.


Has the 3x3 magic square of all squares entries been solved? It is my understanding that it has not yet been determined if it is possible to construct a $3$x$3$ magic square where all the entries are squares of integers.  Is this correct?  Has any published work been done on this problem?

The existence or not of a non-trivial integer 3x3 magic square of squares is STILL a unsolved problem.

The quoted reference to Kevin Brown's web pages only discusses an extremely special configuration of numbers, which does not exist. The page does NOT claim to prove non-existence for all possible magic squares.

If you are interested in this topic you should consult the web-site

http://www.multimagie.com/

which gives lots of details and references.

There is a claimed proof of nonexistence by J. C. Ferreira, but is it correct? https://arxiv.org/abs/1506.06621


Find the number of subsets? (There is no proper solution for the duplicate question, please provide proper proof ) The number of subsets {1,2,…,n} with odd cardinality is ?

(Is it asking about number of subsets, we have to take power set then?)

For $n>0$ There are half subset containing an odd  and half containing an even number of elements

The set of subsets has $2^n$ elements so the number of subsets with odd cardinality

is half $2^n$ that is $2^{n-1}$

Hope this helps