how to solve trigonometric inequalities? how does one solve trigonometric inequalities? Is there a method to this or is every solution done ad hoc?

simple equations of the type: $cos3x \leq  0$  when: $0\leq x \leq 2π$

The attempt at a solution: equating $cos 3x = 0$  yields $$ π /6  + 2\frac13πk\leq x \leq 2π -π /6 - πk /3 $$ as a general solution...what happens next?

There is a sort of general method. Take your example: $\cos 3x \leq 0$ for $0 \leq x \leq 2 \pi$.

Like you did, first solve the equality $\cos 3x = 0$. You cannot always do this exactly, but here we can. We get
$$
3x = \frac{1}{2}\pi + 2 k \pi \quad\vee\quad 3x = \frac{3}{2} \pi + 2 k \pi
$$. By drawing a graph (or looking at the derivative of $\cos 3x$ at those points), we know that the inequality holds for $3x$ between them. So $$
\frac{1}{2}\pi + 2 k \pi \leq 3x \leq \frac{3}{2}\pi + 2 k \pi
$$
It is important here that you get a representation for the interval. In your example solution, you gave representations for the endpoints of the interval with different meanings for $k$ on the left-hand side and the right-hand side. Your thing works for $k = 0$, but for $k = 10$ the left endpoint is to the right of the right endpoint.

Dividing by 3 we get
$$
\frac{1}{6}\pi + \frac{2}{3} k \pi \leq x \leq \frac{1}{2}\pi + \frac{2}{3} k \pi
$$

This is a general solution for all of the real line. To confine ourselves to values of $x$ in $[0, 2\pi]$, we try different values of $k$ and see if the resulting interval lies within $[0, 2\pi]$. In this case, that holds for $k = 0, 1, 2$.

We get:
$$
k = 0 \;\rightarrow\; x \in [\frac{1}{6} \pi, \frac{1}{2} \pi] \\
k = 1 \;\rightarrow\; x \in [\frac{5}{6} \pi, \frac{7}{6} \pi] \\
k = 2 \;\rightarrow\; x \in [\frac{3}{2} \pi, \frac{11}{6} \pi]
$$

The answer will be the union of these three intervals.

I use Nghi H Nguyen's method to visually solve trig inequalities by the number unit circle.
Solve: $ F(x) = \cos (3x) < 0$
There are totally 6 end points at: $ \frac{\pi}{6}, \frac{\pi}{2}, 5\frac{\pi}{6}, 7\frac{\pi}{6}, 3\frac{\pi}{2} $ and $11\frac{\pi}{6} $that divide the unit circle into $6$ equal arc lengths.
Use the check point method to find the sign status of arc length $(\frac{\pi}{6}, \frac{\pi}{2})$.
Select $x = \frac{\pi}{3}.$
We get : $F(x) = \cos 3x = \cos \pi = -1. $Then, $f(x) < 0 $ inside this arc length.
Color it blue. Use endpoint's property to easily find other sign status of $F(x)$ for other arc lengths.
We have, starting from end point $\frac{\pi}{6}$ : Blue, red, blue, red, blue, red.
We get the solution set of $F(x) = \cos 3x < 0 $ (blue) that are the 3 open intervals:
$(\frac{\pi}{6}, \frac{\pi}{2})$ and $(5\frac{\pi}{6}, 7\frac{\pi}{6})$ and $(3\frac{\pi}{2}, 11\frac{\pi}{2})$


Best martingale for sequence of "dozen" bets at roulette game Jim goes the Casino to play roulette. He only makes
“dozen” bets at each spin ; his probability of winning is therefore
$\frac{1}{3}$ every time (to simplify, we neglect the effect of the
zeros in the roulette wheel), so that if his stake is $x$ units of currency, when
he wins he is returned thrice that amount, the expectation of his gain
is $0$. He has also decided to stop playing and leave the casino as soon
as he has achieved a positive gain (however small), and has brought with
him a total sum of $M$ units of currency. Finally, he decides what
amounts he bets according to a precise rule : when he has lost $k$ units
of currency ($1 \leq k < M$) he bets $f(k)$ units, where $f$ is a function.
We call $f$ a martingale. Note that $f$ must satisfy
$k+f(k)\leq M$ for every $k$ (Jim cannot bet more than he already has).

This is a typical Markov Chain situation, with $M+2$ vertices
$v_{+},v_0,v_1,\ldots,v_{M}$ (where $v_j$ corresponds to the point
where Jim has lost $k$ units, and $v_{+}$ means he has some positive gain). If
we denote by $\pi(k)$ the probability for Jim to eventually win from a position
where he has lost $k$ units of currency, we have

$$
\pi(k)=\frac{1}{3}\pi(k-f(k))+\frac{2}{3}\pi(k+f(k))
$$

Then, the question is, what is the best strategy (i.e. which $f$ maximizes
$\pi(0)$) ?

After looking at a few numerical examples up to $M=20$, it would seem
that there is a unique optimal $f$, satisfying $f(k)=1$ when $k$ is even
and $f(k)=\lceil \frac{k}{2} \rceil$ when $k$ is odd and $k\leq \lfloor \frac{M+1}{3}\rfloor$ (what happens for the other values of $k$ is unclear).

Does anyone know more about how to solve this exercise ?

To maximize the probability of a positive sum, you want each individual bet to give you a total net positive sum if you win.  Then your probability of winning is $1 - ({2\over 3})^N$, where $N$ is the number of bets you make.  To maximize $N$, bet the smallest amount each time that will produce winning results.

To that end, you would bet: $1$ unit on the first bet.  If you win, stop.  If you lose, then bet $1$ again, and continue with $2, 3, 4, 6, 9, 14, \ldots$ units.  The pattern is:
$$
a_i = a_{i-1} \times 1.5
$$
Round fractions down, then up, alternately.  Each $a_i$ is thus the smallest integer greater than half what you have already lost.

If you follow that betting pattern, then if you win any spin, you will end up with either $1$ or $2$ units more than you started with.

If you started with $M = 40$ units, then you can make up to 8 bets, the last being $14$.  You have $(2/3)^8 \approx 4\%$ chance of losing all $40$ units, and a $96\%$ chance of winning $1$ or $2$ units.

I don't know mathematics. I play only single dozen.  I wait for a dozen not occurred 5 times.  Then start betting on that dozen.  Progression is as follows:

4

4

8

12

20

32

52

84

130

200

310

480

I have seen maximum 15 games not occurred and lost.  But 16th game must win. You need to increase bet as above. Lot of patience required.


Arriving at the value of Limits According to the formal definition of limits,

Let $f(x)$ be a function defined on an open interval $D$ that contains $c$, except possibly at $x=c$. Let $L$ be a number. Then we say that

$$\lim_{x \to c} f(x) = L$$
if, for every $\varepsilon>0$, there exists a $\delta>0$ such that for all $x\in D$ with

$$0 < \left| x - c \right| < \delta$$
we have

$$\left| f(x) - L \right| < \varepsilon$$

Suppose now we have a function $f(x)=x$. Now if we want to calculate the limit at 0, we just replace x by the value 0. What decides that it is correct and is the only possible limit? We could as easily have chosen 1 as our limit still satisfy the formal definition by staying within the error limits. Why is this so?

We could as easily have chosen 1 as our limit still satisfy the formal definition by staying within the error limits.

Um, no.  Set $\epsilon = \frac{1}{2}$.  Let $\delta > 0$ be given, and set $x=\min\left\{\frac{\delta}{2},\frac{1}{2}\right\}$. Then $|x| < \delta$ but $$|f(x) - 1| = |x-1| \geq \frac{1}{2}$$

We have shown there exists some $\epsilon >0$ such that for all $\delta > 0$ there exists $x$ with $|x-0| < \delta$ but $|f(x) - 1| \geq \epsilon$.  This is the negation of the statement that $\lim_{x\to a} f(x) = 1$.  That is, the limit cannot be 1.

You do not simply "replace $x$ by the value $0$". What you do to arrive at the conclusion that $0$ is a limit (but $1$, $\pi$, whatever, are not), is realize that the condition "for $\varepsilon>0$, there exists $\delta>0$ ..." does not hold for $L=1$. For $L=1$ to be a limit, you must show that the condition holds for all $\varepsilon>0$, but you can show that it doesn't hold for $\varepsilon=1$, for example:

Let $\varepsilon=1$. Can I find a $\delta>0$ such that for all $x$ in the interval $D=\left(-\delta,\delta\right)\setminus\left\{0\right\}$ (that is, the interval without the zero) we have $\left|x-1\right|<\varepsilon=1$ ? No we cannot, because all the negative $x$'s have distances greater than $1$ to the supposed limit $L=1$.

P.S.: I am talking all the time about $0$ being "a limit" instead of "the limit" for $f\left(x\right)$ because the definition of "limit" states a condition that a number $L$ must hold in order to be called "a limit". However, you can show (as has been mentioned in the comments to your question) that if some number is "a limit", then it is the only limit.


When does one vector has bigger norm than the other? Let $u,v\in\mathbb{R}^n$ be two probability vectors (i.e., with all components $\geq 0$ and summing up to 1). Then I was wondering what are necessary and sufficient conditions for $$\|u\|_p\leq \|v\|_p \quad \forall p\in[1,\infty).$$

I could check that $u_k\leq v_k$ is a sufficient condition, but it need not be necessary.

we put $f(x)=\|x\|_p^p$ , $C=\{ x=(x_i)_{i=1}^n\in R^n: x\geq0, \sum_{i=1}^nx_i=1\}$ and $\partial C=\{x\in C : x_i=0 ,\quad for\quad some \quad i=1,...,n\}$. $f(x)$ is strictly convex function on convex set $C$, and it has unique point that is called by $x_*$ which $x_{*i}$ are $\frac{1}{n}$ for each $i=1,...,n$.  for all $x\in \partial C$, $g_x(t)=tx_*+(1-t)x \quad t\in [0,1]$. If both $f(x)>f(y)$ and $\|x-x_*\|_1\leq \|y-x_*\|_1$ are true then At least two points exist such as $x_1,\quad y_1$ such that $\|x_1-x_*\|_1\leq \|y_1-x_*\|_1$ and $f(x_1)=f(y_1)$ . this is Contrary to the assumption with strictly convex function  $f(x)$, because $\nabla f(tx_1+(1-t)y_1).(x_1-y_1)=0, \quad t\in(0,1)$.

However, we show $\|x\|_p\leq\|y\|_p$ if and only if $ \|x-x_*\|_1\leq \|y-x_*\|_1 $. thus $max\{ x_{i} \}_{ i = 1 } ^n \leq max\{ y_{i}\}_{i=1} ^n
$ if and only if $\|x\|_p\leq\|y\|_p$ for every $p\in [1,\infty]$.

It can be $ max\{ u_{i} \}_{ i = 1 } ^n \leq max\{ v_{i}\}_{i=1} ^n$.

Proof:

it is clear for $p=1$.

We prove for each $1<p\leq \infty $. We know that $\|.\|_p$ is (strictly for all $p<\infty$) convex function and has an unique minimize on feasible solution ($\{x\in R^n : \Sigma_{i=1}^n x_i=1\} $) which it is common for all $p$. We denote this point with $u^*$. It is clear that if $u$ approch from $v$ via $\|.\|_p  \quad \forall p\in(1,\infty]$ to $u^*$. However, it can be investigated by one of $1<p\leq \infty$.


Densesness of irrational numbers Given $Q$: the aggregate of the rational numbers and $R$: the aggregate of the real numbers and $Qc$: the aggregate of the irrational numbers; Give $x,y \in \mathbb{R} \text{ with } x < y$.

Show that one $n \in \mathbb{N}$ and one $r \in \mathbb{Q}$ hast to exist, so:

$x < r  +  $$\frac{\sqrt 2}{n}$ $< y$

The needed property is
provided by the
Archimedean axiom.
See, for example,
here:
https://en.wikipedia.org/wiki/Archimedean_property

One form states that,
for any real $z > 0$,
there is an integer $n$
such that
$\dfrac1{n} < z$.

Now apply this to
$\dfrac1{n} <(y-x)/2$
and look at the
consecutive multiples
of $\dfrac1{n}$
or $\dfrac{\sqrt{2}}{n}$.

More detail.

Let $k$ be the first multiple of
$\dfrac1{n}$ such that
$\dfrac{k}{n} > x$.
This means that
$\dfrac{k-1}{n} \le x$.
These mean
$k > nx$
and
$k-1 \le nx$
so
$k = 1+\lfloor nx \rfloor
$.

Consider
$w
=\dfrac{k+1}{n}
=\dfrac{k}{n}+\dfrac1{n}
$.

From
$\dfrac1{n} <(y-x)/2$
we get

$\begin{array}\\
w
&=\dfrac{k}{n}+\dfrac1{n}\\
&\lt\dfrac{k}{n}+\dfrac{y-x}{2}\\
&\lt x+\dfrac1{n}+\dfrac{y-x}{2}\\
&\lt x+\dfrac{y-x}{2}+\dfrac{y-x}{2}\\
&=y\\
\end{array}
$

Therefore
$\dfrac{k}{n}$
and
$\dfrac{k+1}{n}$
are two rationals
that are between
$x$ and $y$.

If we specify $n$ by
$\dfrac1{n}
\lt \dfrac{y-x}{m}$,
where $m$ is a positive integer,
we can find
$\dfrac{k+j}{n}$
which are between
$x$ and $y$
for $j = 0$ to $m-1$.

If we choose $m=3$,
then
$\dfrac{k}{n}$,
$\dfrac{k+1}{n}$,
and
$\dfrac{k+2}{n}$,
are all between $x$ and $y$,
so that
$\dfrac{k+\sqrt{2}}{n}$
is between
$x$ and $y$.

Do you know how to prove that, if x< y are two irrational numbers, then there exist an irrational number, r, between them: x< r< y?


How can I calculate the Euclidian displacement of two places on a sphere (earth in this case ) and calculate the I would like to get the formula on how to calculate the distance between two geographical co-ordinates on earth and heading angle relative to True North. Say from New York to New Dehli , I draw a straight line THROUGH THE EARTH  - as they were two points in space. How can I calculate that angle from say New York to New Dehli if I was to draw a straight line through the surface of the earth . What kind of mathematical calculation/formula would be involved in order to do that?

Distance

Convert from spherical coordinates to cartesian ones. If you have latitude $\varphi$ and longitude $\lambda$, you can compute

\begin{align*}
x &= r\cos\varphi\cos\lambda \\
y &= r\cos\varphi\sin\lambda \\
z &= r\sin\varphi
\end{align*}

This assumes that the earth is a sphere of radius $r$, and has the $z$ axis pointing north and the $x$ axis pointing twoards the intersection of the equator and the prime meridian.

Do this conversion for both points, and you have two positions in $\mathbb R^3$. Compute their element-wise difference and take the length of that, by squaring the elements, adding them up and computing the square root.

$$\ell = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$$

Direction

In order to compute a kind of bearing from the difference between two locations, you'd most likely want to orthogonally project the displacement vector onto a plane tangent to the surface of the earth at your current location. To do this, let us compute two tangent directions, as derivatives with respect to latitude and longitude. The first points north, the second east.

\begin{align*}
n = \frac{\partial(x,y,z)^T}{\partial\varphi} &=
\begin{pmatrix}
-\sin\varphi\cos\lambda \\
-\sin\varphi\sin\lambda \\
\cos\varphi
\end{pmatrix}
&
e = \frac{\partial(x,y,z)^T}{\partial\lambda} &=
\begin{pmatrix}
-\cos\varphi\sin\lambda \\
\cos\varphi\cos\lambda \\
0
\end{pmatrix}
\end{align*}

Next you have to normalize these to unit length. $n$ already has unit length, but $e$ has to be rescaled.

\begin{align*}
\hat n &= \frac{n}{\lVert n\rVert} = n = \begin{pmatrix}
-\sin\varphi\cos\lambda \\
-\sin\varphi\sin\lambda \\
\cos\varphi
\end{pmatrix} &
\hat e &= \frac{e}{\lVert e\rVert} = \frac1{\cos\varphi}e = \begin{pmatrix}
-\sin\lambda \\
\cos\lambda \\
0
\end{pmatrix}
\end{align*}

Then you can use dot products to project your displacement vector onto these directions. This gives you two components, the ratio of which corresponds to the tangens of the azimuth. With zero at north and positive angle going east, you want the north-pointing component in the denominator and the east-pointing component in the numerator.

\begin{align*}
\tan\alpha &= \frac{\hat e\cdot(p_2-p_1)}{\hat n\cdot(p_2-p_1)}
\\&= \frac
{-\sin\lambda_1(x_2-x_1)
 +\cos\lambda_1(y_2-y_1)}
{-\sin\varphi_1\cos\lambda_1(x_2-x_1)
 -\sin\varphi_1\sin\lambda_1(y_2-y_1)
 +\cos\varphi_1(z_2-z_1)}
\end{align*}

Note that there are two solutions to this, so you either have to check the signs or use an atan2 function.

If you wanted to, you could perform additional computations to determine the elevation, i.e. the angle between the displacement vector and the tangent plane.

Properties of this bearing

I would assume the difference between magnetic north and true north to be rather small in most places, so I'm not sure how much more detail you can expect from a computation like this, with simplifications like a spherical earth and so on.

The initial bearing chosen in this fashion is identical to the one obtained for a greatcircle course. The reason is because in both of these cases you obtain the bearing by intersecting the tangent plane with the plane spanned by origin, destination and center of the earth. From the USA to Mecca that bearing would be mostly NE, according to Google Earth:

I hope this matches your question. If the answer didn't match the answer you expected, then you should probably confer with the authorities which caused that expectation, to see whether you asked the right question in the first place. The rhumb line you mention as an alternative has certain benefits for simple compass navigation, but is not related to an orthogonal projection of the direct displacement vector, or to the shortest distance either through the earth or along its surface.

Rhumb Line Navigation

Rhumb lines or loxodromes are tracks of constant true course. With the exception of meridians and the equator, they are not the same as great circles. They are not very useful approaching either pole, where they become tightly wound spirals. The formulae below fail if any point actually is a pole.

East-West rhumb lines are special. They follow the latitude parallels and form a closed curve. Other rhumb lines extend from pole-to-pole, encircling each pole an infinite number of times. Despite this, they have a finite length given by pi/abs(cos(tc)) (in our angular units, multiply by the radius of the earth to get it in distance units).

When two points (lat1,lon1), (lat2,lon2) are connected by a rhumb line with true course tc :

lon2-lon1=-tan(tc)(log((1+sin(lat2))/cos(lat2))-
                     log((1+sin(lat1))/cos(lat1)))
          =-tan(tc)(log((1+tan(lat2/2))/(1-tan(lat2/2)))-
                     log((1+tan(lat1/2))/(1-tan(lat1/2))))
          =-tan(tc)*(log(tan(lat2/2+pi/4)/tan(lat1/2+pi/4)))
(logs are "natural" logarithms to the base e.)

The true course between the points is given by:

tc= mod(atan2(lon1-lon2,log(tan(lat2/2+pi/4)/tan(lat1/2+pi/4))),2*pi)
The dist, d between the points is given by:

     if (abs(lat2-lat1) < sqrt(TOL)){
         q=cos(lat1)
     } else {
         q= (lat2-lat1)/log(tan(lat2/2+pi/4)/tan(lat1/2+pi/4))
     }
     d=sqrt((lat2-lat1)^2+ q^2*(lon2-lon1)^2)

This formula fails if the rhumb line in question crosses the 180 E/W meridian. Allowing this as a possibility, the true course tc, and distance d, for the shortest rhumb line connecting two points is given by:

dlon_W=mod(lon2-lon1,2*pi)
  dlon_E=mod(lon1-lon2,2*pi)
  dphi=log(tan(lat2/2+pi/4)/tan(lat1/2+pi/4))
  if (abs(lat2-lat1) < sqrt(TOL)){
     q=cos(lat1)
  } else {
     q= (lat2-lat1)/dphi
  }
  if (dlon_W < dlon_E){// Westerly rhumb line is the shortest
      tc=mod(atan2(-dlon_W,dphi),2*pi)
      d= sqrt(q^2*dlon_W^2 + (lat2-lat1)^2)
  } else{
      tc=mod(atan2(dlon_E,dphi),2*pi)
      d= sqrt(q^2*dlon_E^2 + (lat2-lat1)^2)
      }


Differential equations, solving for a chemical rate of change. Two chemicals $\mathrm{A}$ and $\mathrm{B}$ are put together in a solution where they react to form a compound $\mathrm{X}$. The rate of increase of the mass, $x\,\rm kg$, of $\mathrm{X}$ is proportional to the product of the masses of unreacted $\mathrm{A}$ and $\mathrm{B}$ present at time, $t\,\rm min$. It takes $1\,\rm kg$ of $\mathrm{A}$ and $3\,\rm kg$ of $\mathrm{B}$ to form $4\,\rm kg$ of $\mathrm{X}$. Initially, $2\,\rm kg$ of $\mathrm{A}$ and $3\,\rm kg$ of $\mathrm{B}$ are put together in solution, and $1\,\rm kg$ of $\mathrm{X}$ forms in one minute.

What I deciphered from this is only:

$A + B = Xt$, only true for $t = 1$ (doesn't help at all).
$dx/dt = 1$, when $t = 1$.

How can I set up a differential equation expressing $dx/dt$ as a function of $x$?

$dA/dt= (-1/4)(dx/dt)$
$dB/dt=(-3/4)(dx/dt)$
This gives us with $A=2-(1/4)x$ and $B=3-(3/4)x$
We can interpret that $dx/dt =kab$, for some number $k$.
Now we sub in the values of A and B into the equation dx/dt.
This leaves us with $dx/dt= [3k(8 − x)(4 − x)]/16$


Arc length of a sequence of semicircles. Let $\gamma_1$ be the semicircle above the x axis joining $-1$ and $1$.

Now divide this interval into $[-1,0]$ and $[0,1]$, and trace a semicircle joining $-1,0$ above the $x$-axis and another one joining $0,1$ below the $x$-axis, call this curve $\gamma_2$

Keep subdividing the interval into $2^n$ subintervals, proceding as above, and call the resulting curve $\gamma_n$.

Now, from looking at their graphs, I can say that the arc length of all these curves is $\pi$, but the arc length of the limit, $\gamma$ (which I think should be a straight line connecting $-1$ and $1$) is $2$.

Why does this happen? Shouldn't the arc length of the limit be $\pi$ as well? Does $\gamma_n\to \gamma$ uniformly?

One of my professors said that this is because

$$
\operatorname{Arc}(\gamma_i)=\int_D\sqrt{1+\gamma_i'^2}dx
$$

And that $\gamma_i'$ becomes pretty crazy when $i$ is large (as it's $\pm \infty$ on the edges of the circles).

The problem with this argument, and many other alike, is that the arc length of a curve is not a continuous function of a curve, which means that if sequence of curves $\gamma_i$ converges to a curve $\gamma$ (notion of convergence of functions like that can be made formal, but let me not do that here), then there is no guarantee the sequence of arclengths $l(\gamma_i)$ to converge to the arclength $l(\gamma)$. Indeed, if you look at the integrals $\int_0^1\sqrt{1-\gamma_i'(t)^2}dt$, then you might expect that the expression under the integral converges to $\sqrt{1-\gamma'(t)^2}$, which however isn't the case, because convergent sequence of differentiable functions (I am ignoring the "cusps" where the semicircles meet here; they are important as well, but that's beyond my point right now), then the sequence of derivatives need not be convergent.

The arc length should also be $\pi$. The curve "converges" to a straight line, but is actually an infinite union of semicircles of 0 length. You cannot approximate the arc length like that, it is similar to the fallacious proofs that $\pi=4$. In fact, there are $2^n$ semicircles each with diamater $\frac{1}{2^n}$, then the arc length is $2^n\pi\frac{1}{2^n}=\pi$.


Geometric Expectation problem? Calvin and Hobbes play a match consisting of a series of games, where Calvin has
probability $p$ of winning each game (independently). They play with a “win by two”
rule: the first player to win two games more than his opponent wins the match. Find
the expected number of games played.

Hint: Consider the first two games as a pair, then the next two as a pair, etc.

My approach: This is a geometric distribution related problem because you are trying to find the number of games played until "success" occurs which in this case is winning by two.

So using the hint, I thought about the probability of winning by two. For player 1, the probability of winning two games in a row is $p^2$. For player two, the probability of winning two games in a row is $(1-p)^2$ or $q^2$. So the total probability of winning two games in a row for either player is $p^2+q^2$.

After this, I tried plugging this into the geometric expectation formula which is $E[X]=\frac{1-p}{p}$. But I wasn't getting the right answer.

Solution: $\frac{2}{p^2+q^2}$

Can anyone explain what I did wrong? Thanks!

Expectation of a geometric random variable with parameter $p$ is given by $\frac{1}{p}$. So in this case, we would have an expectation of $\frac{1}{p^2+q^2}$ trials. But each trial corresponds to 2 games, so we get an expected value of $\frac{2}{p^2+q^2}$ games played.

First of all number of games played will be even.
now consider games $1-$2 as box1 , $2-$4 as box two ..........
for probability that any of one of the player wins the both the games in box will be $p^2+q^2$
so the game will stop if any such box comes in the sequence
$X$~first_sucess$(p^2+q^2)$
Expected number of such boxes $$E(X)=1/(p^2+q^2).$$
therfore number of games played is $$2/(p^2+q^2).$$


What exactly is sheafification? I have recently learned about the very BASICS of sheaves, but I was wondering is there an easier definition for sheafification? I could not find anywhere an easier definition for sheafification. I kind of compare it to local rings, where in local rings, you collect the local data at a point for a variety or an algebraic set where some function is defined on that regular set. Am I heading in the right direction? If not, what is the rigorous definition of sheafification and what is some intuition one could use to understand it?
I would much more appreciate the intuition and a clear definition, rather than rigour.

Sheafification is the left adjoint of the inclusion $i: Sh(X) \rightarrow Psh(X)$.
That's the important part anyway, the nitty gritty details of the construction are not so important (constructing the Étale space and whatnot). You should think of the sheafification $\mathcal F_+$ of $\mathcal F$ as identifying sections which are locally equal and adding new "formal" sections which are glued together by old ones in $\mathcal F$ in order to satisfy the sheaf conditions (gluing of sections exists and is unique).
For example: If you have a morphism $\phi:\mathcal F \rightarrow \mathcal G$ where $\mathcal F$ is a presheaf and $\mathcal G$ is a sheaf  and two sections $s,s' \in \mathcal F (U)$ which are locally equal we have that $\phi(s),\phi(s')$ are locally equal (since $\phi$ is compatible with restriction maps) and hence equal because $\mathcal G$ is a sheaf.
Thus it doesn't matter if we identify sections $s$ and $s'$ which are locally equal in $\mathcal F$ if all we care about are maps of $\mathcal F$ into sheaves.
Similarly for formal glued sections, if we have a morphism $\phi:\mathcal F \rightarrow \mathcal G$ then $\phi$ extends uniquely to these formal sections $s$ of $\mathcal F$ over $U$ by simply considering $\{s\mid_{U_i}\}_i$ which are actual sections of $\mathcal F(U_i)$, applying $\phi$ to get $\{\phi(s\mid_{U_i}) \}_i$ which are sections in $ \mathcal G(U_i)$ and then gluing these $\phi(s\mid_{U_i})$ to get a section $s'$ in $\mathcal G(U)$ and then set $\phi(s) = s'$.
We thus see that every morphism $\mathcal F \rightarrow \mathcal G$ gives us a unique morphism $\mathcal F_+ \rightarrow G$. And every morphism $\psi:\mathcal F _+ \rightarrow \mathcal G$ comes from the morphism given by precomposing $\psi$ with the apparent map $\mathcal F \rightarrow \mathcal F_+$.
Anyway, this is enough to say that $Hom(\mathcal F,\mathcal G) = Hom(\mathcal F_+,\mathcal G)$ for sheaves $\mathcal G$ and thus that $(-)_+$ is left adjoint to $i:Sh(X) \rightarrow Psh(X)$.
If this was too handwavy I'm sorry! I'll try and elaborate if you feel like it's necessary.

Question: "If not, what is the rigorous definition of sheafification and what is some intuition one could use to understand it?"
Answer: If $X$ is a scheme and $\mathcal{E},\mathcal{F}$ are quasi coherent $\mathcal{O}:=\mathcal{O}_X$-modules, with $\phi: \mathcal{E}\rightarrow \mathcal{F}$ a map of $\mathcal{O}$-modules you want to form kernels, images, tensor products exterior products etc of the sheaves $\mathcal{E},\mathcal{F}$ and the map $\phi$. In the case of the tensor product $\otimes$ you define the presheaf
$$ S(U):= \mathcal{E}(U)\otimes_{\mathcal{O}(U)} \mathcal{F}(U).$$
The presheaf $S$ is not a sheaf, and the associated sheaf $S^+ := \mathcal{E}\otimes \mathcal{F}$ (see Hartshorne, Section II.5) is a sheaf with the property that for any open affine set $V:=Spec(A)$ with $\mathcal{E}(V):=E, \mathcal{F}(V):=F$ with $E,F$ left $A$-modules, it follows
there is an isomorphism of $A$-modules
$$ S^+(V) \cong E\otimes_A F$$
Hence the sheaf $S^+$ has the correct behaviour and generalize the notion of tensor product of modules. If $\mathcal{E}, \mathcal{F}$ are locally trivial, it follows the tensor product, exterior product, symmetric product etc is again locally trivial. You get compatibility formulas valid for modules: If $\mathcal{E}$ is locally trivial it follows
$$ Hom_{\mathcal{O}}(\mathcal{F}\otimes \mathcal{E}, \mathcal{G}) \cong Hom_{\mathcal{O}}( \mathcal{F}, Hom_{\mathcal{O}}(\mathcal{E}, \mathcal{G})).$$
As you can see this is similar to what happens for $A$-modules. The sheafification process seems "unintuitive" but it has all the correct functorial properties.
Example 1. If you want to define a morphism of sheaves
$$f^+: \mathcal{E}\otimes \mathcal{F} \rightarrow \mathcal{G}$$
where $\mathcal{G}$ is a sheaf, by the universal property of the $S^+$-construction, it is enough to define a morphism of presheaves
$$f: S \rightarrow \mathcal{G}.$$
If $f$ exists, by the universal property you get a unique morphism $f^+: S^+ \rightarrow \mathcal{G}$. And to define a morphism $f$ you can do this using basic properties of the tensor product.
Example 2. As an example you should verify that there is an isomorphism of sheaves $\mathcal{O}(m)\otimes \mathcal{O}(n) \cong \mathcal{O}(m+n)$ on $C:=\mathbb{P}^1$. Let $C:=Proj(k[x_0,x_1])$ and let $U_i:=D(x_i)$. Let $t:=\frac{x_1}{x_0}, s:=\frac{1}{t}$. Let $L_m:=\mathcal{O}(m)$ for an integer  $m$. Let $S(U):=L_m(U)\otimes_{\mathcal{O}_C(U)}L_n(U)$. Define
$$ f_0: S(U_0):=k[t]x_0^m\otimes k[t]x_0^n \rightarrow k[t]x_0^{m+n}$$
by
$$f_0(f(t)x_0^m\otimes g(t)x_0^n):= fgx_0^{m+n}.$$
Do a similar construction for $U_1$ to get a map
$$f_1: S(U_1) \rightarrow k[s]x_1^{m+n}.$$
It follows $f_i$ glue to induce a canonical map
$$f: S^+:= \mathcal{O}(m)\otimes \mathcal{O}(n) \rightarrow \mathcal{O}(m+n)$$
and since $f_0,f_1$ are isomorphisms, it follows $f$ is an isomorphism. Try to generalize to projective space.


Mirror image of the parabola about a tangent The mirror image of the parabola $y^2 = 4x$ in the tangent to the
  parabola at $(1,2)$ $y^2 = 4x$ is?

I have solved this question correctly (by finding the image of the focus and the vertex in the given line and then finding the parabola that possess these parameters) and got the answer to be $(x+1)^2 = 4(y-1)$

In the graph, I have marked all my findings.

I want to know if there's a simpler method to solve this question? But more importantly, what does this reflected parabola signify? It doesn't actually look like a "reflection". For instance, if we find the reflection of any letter with respect to a line mirror, we see that the letter's images are symmetrical about the line but here the parabolas are touching each other and don't look like what we call "reflections". In a nutshell, it contradicts our intuitive sense of "reflection".

Hint:

Find the equation of the tangent at $(1,2)$

Any point on $y^2=4x,$ can be $(t^2,2t)$

Now if the image of $(t^2,2t)$  is $(h,k)$

then $$\left(\dfrac{t^2+h}2,\dfrac{2t+k}2\right)$$ must lie on the tangent.

Eliminate $t$

The mirror image should have a common tangent at $(1,2)$ so the mirror graph is incorrect.

(my earlier comment in error actually for $45^0$ )

After correction of my obvious errors the mirror should have equation like:

$$ 4 (y-1) = (x+1)^2 $$


Why $x^2+3x^4+\omicron(x^4)=\omicron(x^3)$is false? Why $x^2+3x^4+\omicron(x^4)=\omicron(x^3)$is false for $x\to 0$?
I know that:
A. $-\omicron(f)=\omicron(f)$
B. $\omicron(f)+\omicron(f)=\omicron(f)$
C. If g is a function $g*\omicron(f)=\omicron(gf)$
C2. if g is costant$g*\omicron(f)=\omicron(gf)=\omicron(f)$
D. $\omicron(f)^n=\omicron(f^n)$
E. $\omicron(f+\omicron(f))=\omicron(f)$
From C I try to write $\omicron(x^4)=x^4\omicron(1)$ but I don't know what my next move is.

If
$$x^2+3x^4+\omicron(x^4)=\omicron(x^3)$$ then
$$\lim\limits_{x \to 0} \frac{x^2+3x^4+\epsilon(x)x^4}{x^3} = 0$$
where $\lim\limits_{x \to 0} \epsilon(x) =0$.
And this is not the case as
$$\frac{x^2}{x^3}$$ is not bounded around zero.


What is the limit of this sequence involving logs of binomials? Define  $$P_{n} = \frac{\log \binom{n}{0} + \log \binom{n}{1} + \dots +\log \binom{n}{n}}{n^{2}}$$ where $n = 1, 2, \dots$ and $\log$ is the natural log function. Find $$\lim_{n\to\infty} P_{n}$$ Using the property of the $\log$, I am thinking to find the product of all the binomial coefficients. But it gets really messy.

Here's an upper bound
(probably not good enough)
for a start:

From the arithmetic-geometric mean inequality,
$\prod_{i=0}^n \binom{n}{i}
<\left(\dfrac{\sum_{i=0}^n \binom{n}{i}}{n}\right)^n
=\left(\dfrac{2^n}{n}\right)^n
=\dfrac{2^{n^2}}{n^n}
$
so
$(\prod_{i=0}^n \binom{n}{i})^{1/n^2}
<\dfrac{2}{n^{1/n}}
< 2
$
or
$\dfrac{\sum_{i=0}^n \ln \binom{n}{i}}{n^2}
< \ln 2
$.


How many different circular bracelets can be formed from 8 red, 6 blue and 4 yellow beads, always using all available beads? How many different bracelets can be made from 8 red, 6 blue and 4 yellow beads, always using all available beads?
Two bracelets are different when we cannot get the color scheme of the beads of one bracelet by rotate or flipping the other bracelet.
Beads of the same color are the same and indistinguishable from each other, ie. by swapping two beads of the same color, we will not change the bracelet.

First you calculate the anagrams:
$$\frac{(8+6+4)!}{8!6!4!}$$
And then you divide by the number of rotations ($8+6+4$) and the number of flips ($2$).


Spaced repetition algorithms Can someone tell me how the Super Memo spaced repetition algorithms (http://www.supermemo.com/english/algsm11.htm) were calculated?

How does someone go about designing algorithms like this?

Is there a particular branch of mathematics that deals with this type of issue?

I understand the logic of the different variables and why they are included in the algorithm, but not how these elements are formed into a single unit.

I'm interested in it and would like to do some reading on the subject.

Thanks for any help

I do not know exactly how it is calculated, but some of the SuperMemo algorithms are open source. Also, I have used SuperMemo for about nine years and have memorized over 77,000 flashcards with a 95% accuracy. I don't know the exact calculations, but I do know that they work.


The intersection points of $m$ balls in $\mathbf{R}^n$ In $\mathbf{R}^n$ , balls of radii $R_1...R_m$ centered at $A_1...A_m$ have a common point . Prove that if $B_1...B_m$ are points such that $|B_iB_j|\le|A_iA_j|$ , then the balls of radii $R_1...R_m$ centered at $B_1...B_m$ also have a common point .

Consider any of the two circles, suppose the one centered at $B_i$ with radius $R_i$ and that at $B_j$ with radius $R_j$. Name them $b_1$ and $b_2$. Translate the point $A_i$ to $B_i$ and rotate the line $A_iA_j$ so that it coincides with $B_iB_j$. Then it becomes obvious that $b_1$ and $b_2$ has a common point. As we have shown this for any two circles, the result follows.


why the limit of this f(x) when x approach infinity is equal to infinity? why the limit of this f(x) when x approach infinity is equal to infinity?:

$$ \lim_{x\to \infty} \frac{x^2 + 4x + 5}{x-1}$$

$$ \lim_{x\to \infty} \frac{1 + \frac{4}{x} + \frac{5}{x^2}}{\frac{1}{x}-\frac{1}{x^2}} = \infty  $$

I know that all the fraction with X be denominator is equal to Zero but is that mean at the end the $ \lim_{x\to \infty} = \frac{1}{0}$ why it is equal to $ \infty $ ?

$$\lim_{x\to\infty}\left(\frac{x^2+4x+5}{x-1}\right)=$$

The leading term in the denominator of $\frac{x^2+4x+5}{x-1}$ is $x$.

So divide the numerator and denominator by this:

$$\lim_{x\to\infty}\left(\frac{x+4+\frac{5}{x}}{1-\frac{1}{x}}\right)=$$

The expressions $\frac{5}{x}$ and $\frac{1}{x}$ both tend to zero as $x$ approaches $\infty$:

$$\lim_{x\to\infty}\left(\frac{x+4+0}{1-0}\right)=$$
$$\lim_{x\to\infty}\left(\frac{x+4}{1}\right)=$$
$$\lim_{x\to\infty}\left(x+4\right)=$$
$$\lim_{x\to\infty}x=\infty$$

i dont think limit can end with 1/0 = infinity as the answer tough.

heres my approach :

instead divide it by $x^2$  divide it by $x$ instead.

so it will be $\frac{(x+4+\frac{5}{x})}{x-1} $

since $\lim _{x\to infinity} \frac{5}{x}  $ and $\lim _{x\to infinity} \frac{-1}{x}  $ are 0 .

it leaves use with $\lim _{x\to infinity} x+4  $ which is equal to infinity


Why does $\sum_{i=0}^{\infty}\frac{a^i}{i!}=e^a$? Here is a standard identity:

$$\sum_{i=0}^{\infty}\frac{a^i}{i!}=e^a$$

Why does it hold true?

A brief answer: Let's consider the exponential function $e^x$. The definition of $e$ is that $\frac{d}{dx}e^x = e^x$. Now let's assume that $e^x$ can be written as an infinite sum of the form $\sum_{i=0}^{\infty}a_ix^i$. Using the sum rule for derivatives, we have $\sum_{i=0}^{\infty}a_ix^i = \sum_{i=0}^{\infty}\frac{d}{dx}a_ix^i = \sum_{i=1}^{\infty}ia_{i}x^{i-1}$. Therefore, $a_i = \frac{a_{i-1}}{i}$, so starting with the base case of $i=0$, $a_i = \frac{1}{i!}$.

(Warning: variables change from the previous section)

Another way to look at this is to consider the more standard definition of $e$, which is $\lim_{n\to\infty}(1 + \frac{1}{n})^n$. Therefore, $e^a$ can be written as $\lim_{n\to\infty}(1 + \frac{1}{n})^{an}$. Using the binomial theorem, the expression expands to $$\lim_{n\to\infty}(1 + \frac{an}{n} + \frac{an(an - 1)}{2n^2}+ \frac{an(an-1)(an-2)}{6n^3} + \ldots)$$ The lesser powers in the numerators drop out, so the expression becomes $$1 + a + \frac{a^2}{2} + \frac{a^3}{6} + \ldots$$which is $\sum_{i=0}^{\infty}\frac{a^i}{i!}$ in sum notation.

The Taylor series for $e^x$ is

$$
\sum_{k = 0}^{\infty} \frac{x^k}{k!}.
$$

Letting $x = a$, we obtain

$$
e^a = \sum_{k = 0}^{\infty} \frac{a^k}{k!}
$$

as desired.


Conjectured compositeness tests for $N=k \cdot 2^n \pm c$ How to prove that these conjectures are true ?

Definition : $\text{Let}~ P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)~ , \text{where}~ m ~\text{and}~ x ~\text{are nonnegative integers} .$

Conjecture 1 : $ \text{Let} ~N=k\cdot 2^n-c ~\text{such that}~ n>2c ,  k>0 , c>0 ~\text{and}~ c\equiv 3,5 \pmod{8}$

$$\text{Let}~ S_i=P_2(S_{i-1})~ \text{with}~ S_0=P_k(6) , ~\text{thus} $$
$$\text{If}~ N ~\text{is prime then}~ S_{n-1} \equiv -P_{\lfloor c/2 \rfloor}(6) \pmod{N}$$

Searching for counterexample (Pari GP) :

CEk2c(k,c,g)=
{
for(n=2*c+1,g,a=6;
N=k*2^n-c;
my(s=Mod(2*polchebyshev(k,1,a/2),N));
for(i=1,n-1, s=s^2-2);
if(!(s==N-2*polchebyshev(floor(c/2),1,a/2)) && isprime(N),print(n)))
}

Conjecture 2 : $ \text{Let} ~N=k\cdot 2^n-c ~\text{such that}~ n>2c ,  k>0 , c>0 ~\text{and}~ c\equiv 1,7 \pmod{8}$

$$\text{Let}~ S_i=P_2(S_{i-1})~ \text{with}~ S_0=P_k(6) , ~\text{thus} $$
$$\text{If}~ N ~\text{is prime then}~ S_{n-1} \equiv P_{\lceil c/2 \rceil}(6) \pmod{N}$$

Searching for counterexample (Pari GP) :

CEk2c(k,c,g)=
{
for(n=2*c+1,g,a=6;
N=k*2^n-c;
my(s=Mod(2*polchebyshev(k,1,a/2),N));
for(i=1,n-1, s=s^2-2);
if(!(s==2*polchebyshev(ceil(c/2),1,a/2)) && isprime(N),print(n)))
}

Conjecture 3 : $ \text{Let} ~N=k\cdot 2^n+c ~\text{such that}~ n>2c ,  k>0 , c>0 ~\text{and}~ c\equiv 3,5 \pmod{8}$

$$\text{Let}~ S_i=P_2(S_{i-1})~ \text{with}~ S_0=P_k(6) , ~\text{thus} $$
$$\text{If}~ N ~\text{is prime then}~ S_{n-1} \equiv -P_{\lceil c/2 \rceil}(6) \pmod{N}$$

Searching for counterexample (Pari GP) :

CEk2c(k,c,g)=
{
for(n=2*c+1,g,a=6;
N=k*2^n+c;
my(s=Mod(2*polchebyshev(k,1,a/2),N));
for(i=1,n-1, s=s^2-2);
if(!(s==N-2*polchebyshev(ceil(c/2),1,a/2)) && isprime(N),print(n)))
}

Conjecture 4 : $ \text{Let} ~N=k\cdot 2^n+c ~\text{such that}~ n>2c ,  k>0 , c>0 ~\text{and}~ c\equiv 1,7 \pmod{8}$

$$\text{Let}~ S_i=P_2(S_{i-1})~ \text{with}~ S_0=P_k(6) , ~\text{thus} $$
$$\text{If}~ N ~\text{is prime then}~ S_{n-1} \equiv P_{\lfloor c/2 \rfloor}(6) \pmod{N}$$

Searching for counterexample (Pari GP) :

CEk2c(k,c,g)=
{
for(n=2*c+1,g,a=6;
N=k*2^n+c;
my(s=Mod(2*polchebyshev(k,1,a/2),N));
for(i=1,n-1, s=s^2-2);
if(!(s==2*polchebyshev(floor(c/2),1,a/2)) && isprime(N),print(n)))
}

Any hint is appreciated .

First thing I would do is simplify $P_2(x)$
$$P_2(x)=\dfrac{(x-\sqrt{x^2-4})^2+(x+\sqrt{x^2-4})^2}{4}=\dfrac{x^2-2x\sqrt{x^2-4}+x^2-4+x^2+2\sqrt{x^2-4}+x^2-4}{4}=\dfrac{4x^2-8}{4}=x^2-2$$

How did you come up with these conjectures? And why do you think they would work? Because,if we know that, it's way easier to prove them.

This is not an answer, but this would not enter a comment.
Here is a PARI/gp program that implements a unic formula that manages all 4 conjectures, showing that they are related and can be generalized into only one, showing that it looks to be a deep property.
I've found no counter-example with the two verification loops provided after.
You have found something deep, I think. However, it needs a proof now ! ;)

CEk2c(k,c,g)=
{
a=6;
if(c>0,s=1,s=-1;c=-c);
for(n=2*c+1,g,
N=k*2^n+s*c;
e=c%4;if(e!=1,e=-1);
d=((c-e)%8)/4;
A=(-1)^d;
B=(c-A*s)/2;
s0=Mod(2*polchebyshev(k,1,a/2),N);
sn=Mod(A*2*polchebyshev(B,1,a/2),N);
my(s=s0);
for(i=1,n-1,s=Mod(s^2-2,N));
if(s!=sn && isprime(N),print("k: ",k," c: ",c," n: ",n))
)
}

for(k=1,100,for(c=0,50,CEk2c(k,2*c+1,100)))
for(k=1,100,for(c=0,50,CEk2c(k,-(2*c+1),100)))

New conjecture summarizing the 4 conjectures:

$$\text{Let } ~N=k\cdot 2^n+ s \cdot c ~\text{ such that: }$$

$$ n>2c ,  k>0 , c>0 ~\text{ and }~ s = \pm 1 ~\text{ and }~ c \equiv 4 \cdot d ~ \pm 1 \pmod 8 ~\text{ and }~ d=0,1$$

$$\text{Let }~ S_i \equiv P_2(S_{i-1}) \pmod{N} ~ \text{ with } ~ S_0=P_k(6) ~ \text{ , thus: }$$

$$\text{If } ~ N ~\text{ is prime then }~ S_{n-1} \equiv {(-1)}^d \cdot P_{(c - {(-1)^d \cdot s})/2} ~ (6) \pmod{N}$$


What does "If three real numbers $a, x, y$ satisfy the inequalities $a≤x≤a+\frac{y}{n}$ for every integer $n≥1$, then $x=a$" mean? If three real numbers $a, x, y$ satisfy the inequalities $a≤x≤a+\frac{y}{n}$ for every integer $n≥1$, then $x=a$

I saw this theorem in Apostol's Calculus, and even though I understand the proof, I quite don't get what it means, why it's a consequence of the Archimedean property of real numbers (even though I know it's used to prove it), and why it's important to Calculus, as he states. I suppose it has something to do with the inexistence of infinitesimal and infinite numbers that the Archimedean property states, but I still quite don't get it.

Thanks in advance.

The first inequality
says that
$x$ is at least as large as $a$.

If we rewrite the
second inequality as
$x-a
\le \dfrac{y}{n}$,
this says that
the amount $x$ exceeds $a$ by
is less than
$\dfrac{y}{n}$
for any integer $n$.

This is where the
Archimedean axiom comes in.
If the amount $x$ exceeds $a$ by
was greater than zero,
call it $c > 0$,
then
$c \le \dfrac{y}{n}$
for all integers $n$.

This is equivalent to
$n \le yc$.
But Archy says that
for any real
there is a larger integer,
so this can not hold.

Therefore
$x = a$.

You are trying to think too deeply about this.  Subtract $a$ in the inequality and you get $0\leq x-a\leq y/n$ and consider what happens if $x-a>0$. The real number $y/n$ is then positive and becomes less than $x-a$ if $n>y/(x-a)$ and you get a contradiction. Hence $x-a=0$.

Results such as these are trivial and are made complicated only by the fact that they are presented in a book meant for undergraduate students and written using some amount of formalism. If you replace the word "real" in your result with the word "rational" then the statement becomes familiar to any student who has studied fractions and their manipulation. The jump to "real" from "rational" is achieved by noting that all the properties possessed by rational numbers (as far algebraic operations and order relations are concerned) are also possessed by real numbers.


Is multiplication not just repeated addition? If multiplication is not repeated addition, what is it? How do you define multiplication?
All of us had to memorize the multiplication table in elementary school, but how did they come up with it if so many people claim that it is wrong to think of it as repeated addition?
Voodoo magic?

The first thing you should consider is that there are, in some sense, different "hierarchies" of numbers. At each stage, we enlarge the class of numbers, and try to do so in a way that leaves everything we could do before still the same, but that now we can do more.

A common way of doing this is to start with the natural numbers (sometimes called "counting numbers" or positive integers). We start with $1$, $2$, $3,\ldots$.

Here, we do define multiplication as repeated addition. For example, one way to define multiplication is to assume we know how to add, and then define it by saying:
$$\begin{align*}
n\times 1 &= n\\
n\times (k+1) &= (n\times k) + n % need to edit at least 6 characters
\end{align*}$$
Using mathematical induction, we can show this defines multiplication for all positive integers, and that it has the usual properties we know (commutative, so that $n\times k = k\times n$ for all positive integer $n$ and $k$, distributes over the sum, is associative, etc).

Then we have two choices for "expanding our universe of numbers": we can now define negative integers, by considering things that would help us solve all equations of the form $a+x=b$ with $a$ and $b$ positive integers; or we can introduce positive rationals (fractions) by considering all things that would help us solve all equations of the form $ax = b$. Let's do the latter, since that came first historically.

So, we had the positive integers, and we knew how to add and multiply them. Now we are going to have more numbers: now, for every pair of positive integers $a$ and $b$, we will have a number "$\frac{a}{b}$", which is a number that satisfies the property that
$$b\times\left(\frac{a}{b}\right) = a.$$
We also say that $\frac{a}{b}$ is "the same fraction" as $\frac{c}{d}$ if and only if $ad=bc$ (here we are comparing products of positive integers, so that's fine).

We also notice that our old positive integers can also be considered fractions: the positive integer $a$ is a solution to $1x = a$, so $a$ corresponds to the fraction $\frac{a}{1}$.

Now, how do we add two of these numbers? Since $\frac{a}{b}$ represents the solution to $bx=a$, and $\frac{r}{s}$ represents the solution to $sx=r$, then $\frac{a}{b}+\frac{r}{s}$ represents the solution to something; to what? A bit of algebra will tell you that it is the solution to precisely $(bs)x = (as+br)$. So we define
$$\frac{a}{b}+\frac{r}{s} = \frac{as+br}{bs}.$$
There's a bit of work that needs to be done to ensure that if you write the fractions differently, the answer comes out the same (if $\frac{c}{d}=\frac{a}{b}$, and if $\frac{t}{u}=\frac{r}{s}$, does $\frac{cu+td}{du} = \frac{as+br}{bs}$? Yes). And we also notice that if we add positive integers as if they were fractions, we get the same answer we did before:
$$\frac{a}{1} + \frac{c}{1} = \frac{a1+c1}{1} = \frac{a+c}{1}.$$
That's good; it means we are enlarging our universe, not changing it.

How about products? If $\frac{a}{b}$ represents the solution to $bx=a$, and $\frac{r}{s}$ represents the solution to $sy=r$, their product will be the solution to $(bs)z = ar$. So we define
$$\frac{a}{b}\times\frac{r}{s} = \frac{ar}{bs}.$$
And then we notice that it extends the definition of multiplication for integers, since $\frac{a}{1}\times\frac{b}{1} = \frac{a\times b}{1}$.  And we check to see that multiplication and addition still have the properties we want (commutativity, associativity, etc).

(There are other ways to figure out what multiplication of fractions "should be", on the basis of what we want it to do. For example, we want multiplication to extend multiplication of integers, so $\frac{a}{1}\times\frac{b}{1}$ should be $\frac{ab}{1}$; and we want it to distribute over the sum, so we want
$$\frac{a}{1} = \frac{a}{1}\times \frac{1}{1} = \frac{a}{1}\times\left(\underbrace{\frac{1}{b}+\frac{1}{b}+\cdots+\frac{1}{b}}_{b\text{ summands}}\right) = \underbrace{\left(\frac{a}{1}\times\frac{1}{b}\right) + \cdots + \left(\frac{a}{1}\times\frac{1}{b}\right)}_{b\text{ summands}}.$$
So $\frac{a}{1}\times \frac{1}{b}$ should be a fraction which, when added to itself $b$ times, equals $a$; that is, a solution to $bx=a$; that is, $\frac{a}{b}$. And so on).

Then we move on from the positive rationals (fractions) to the positive reals. This is more complicated, as it involves "filling in gaps" between rationals. It is very technical. But what it turns out is that for every real number you can find a sequence of rationals $q_1,q_2,q_3,\ldots$ that get progressively closer to each other and to $r$ (we say the sequence "converges to $r$"); it won't hurt too much if you think of the $q_i$ as being progressive decimal approximations to $r$ (they don't have to be, and ahead of time you don't have any notion of decimal approximation, but you can think of it that way for our purpose). So then the way we define multiplication of real numbers $r$ and $s$ is to find a sequence of rationals $q_1,q_2,q_3,\ldots$ giving the approximation to $r$, and one $p_1,p_2,p_3,\ldots$ giving the approximation to $s$, and we define $r\times s$ to be whatever it is that the sequence
$$p_1\times q_1,\ p_2\times q_2,\ p_3\times q_3,\ \ldots$$
approximates. This ensures that if you take rational numbers and multiply them as if they were reals you get the same thing as if you multiply them as rationals, and likewise for integers.

So multiplication of positive reals is really a series of approximations made up by multiplying rationals; and multiplication of rationals is really a way to codify solutions to certain equations with integers; and it is only multiplication of (positive) integers that really corresponds to "repeated addition".

Finally, once you have the positive reals, you can introduce the negative real numbers. We consider solutions to equations of the form $a+x=b$ with $a$ and $b$ positive real numbers. Some of these already have solutions, some don't. This gives us "zero" and "negative reals". We then extend the definition of multiplication to "zero" and to "the negative reals" in a way that makes sense relative to this definition. Turns out we need to make $0\times r = 0$ for all $r$, and have to respect the "rules of signs" to make sure everything still works. So we define it that way to make sure everything works and what we had before still works exactly the same.

Devlin is profane - do not trust him!

Multiplication is repeated addition ($N = n·u= u+u+u+...$, unit is preserved) , but there is another thing called product.($u×u = u^2$, unit is changed).

Euclid had problems with this issue too, but clearly he was aware of it.
Euclid speaks of product $A\times B$ as plane number, and of product $ A \times B \times C$ as solid number.

René Descartes resolved the issue completely.

These terms is often(almost always) misused, because $|A × B| = |A| · |B|$.


Modelling Dirac Delta Function and Proving its Area Property I have modeled the Dirac delta function using limiting value of the rectangular pulse and now proving its area property. In the proof, I need to interchange the order of integration and limit, which does not look possible. I looked out at the similar problems on math stackexchange but could not find it anywhere. The details about the problem are as follows:

To model the delta function, first I took the rectangular pulse function as:

\begin{equation}
p_n(t) = \frac{1}{n} \ \textrm{for}\ |t| \le \frac{n}{2}\ \textrm{otherwise}\ 0
\end{equation}

Based on this rectangular pulse, we can define the delta function as follows:

\begin{equation}
\delta(t) = \lim_{n \rightarrow 0}{p_n(t)}
\end{equation}

So the above equation precisely captures the behavior of the delta function.

Now, I want to prove the area property of the delta function, which is:

\begin{equation}
\int_{-\infty}^{+\infty}{\delta(t) dt} = 1
\end{equation}

By writing the above equation using the definition of the delta function, it results into:

\begin{equation}
\int_{-\infty}^{+\infty}{\lim_{n \rightarrow 0}{p_n(t)}}dt = 1
\end{equation}

Next, if we look at the integrand of the above integral, it is divergent as the function $\lim_{n \rightarrow 0}{p_n(t)}$ has infinite value, so overall its not integrable. Now, I want to swap the limit and integral operator, which is only possible if the limit is convergent (where it is divergent in this case).

My question is that is there really a way I can swap the order of this limit and integral? If yes then how would I do it?

Note: I also checked out many questions/answers regarding this on math stackexchage, but in all cases they directly take limit outside the integral, i.e., $\lim_{n \rightarrow 0}{\int}_{-\infty}^{+\infty}{p_n(t)} dt$, and it would only be equal to $\int_{-\infty}^{+\infty}{\delta(t) dt}$, if the limits and integral operators can be swapped. Moreover, the integral that I am using is the gauge integral.

I would highly appreciate your earliest response on it.

Thanks in advance.

Since the support of $\delta $ is a point, the natural interpretation of the gauge integral of $\delta $ is that it should be zero. To get a gauge integral you need to calculate Riemann sums; for an interval $(u_i,u_{i+1}) $, you need to calculate $\delta (t_i)\, (u_{i+1}-u_i) $. You now want to get $\infty\times (u_{i+1}-u_i)=1$, and there's no arithmetic that will give you that.
So, the question about exchanging limit and integral does not really apply, since $\delta $ is not actually a function.
The usual approach to avoid this problem is the following. The $\delta$ is never used on its own. It used in the sense that $$\tag1\int_{\mathbb R} f(t)\,\delta(t)\,dt=f(0)$$ for any (good enough, say) function $f$.
One way to achieve $(1)$ is to consider the left-hand-side as the limit
$$
\lim_n \int_{\mathbb R} f(t) p_n(t)\,dt.
$$
This is still not ideal because you are subjecting yourself to the choice of the $p_n$, and in the end what one cares about is the net result, which is the equality $(1)$. The solution to this is to be more abstract, and consider $\delta$ as a linear functional, that to every $f$ assigns the number $f(0)$.
A second approach, that agrees with the previous one, is to consider $\delta$ as a measure, namely $$\delta(E)=\begin{cases} 1,&\ 0\in E\\ \ \\ 0,&\ 0\not\in E\end{cases}$$ Then one can write
$$\tag2
\int_{\mathbb R} f(t)\,d\delta(t)=f(0)
$$
in the sense of Lebesgue integration.

Edit: why $\delta$ "function"?
On the set $C_0(\mathbb R)$, of continuous functions vanishing at infinity, one considers (bounded) linear functionals, which are linear maps $\varphi:C_0(\mathbb R)\to\mathbb R$. By the Riesz-Markov theorem, for any such $\varphi$ there exists a measure $\mu_\varphi$ such that
$$\tag3
\varphi(f)=\int_{\mathbb R} f(t)\,d\mu_\varphi(t),\ \ \ \ f\in C_0(\mathbb R).
$$
If $\mu_\varphi$ is absolutely continuous with respect to Lebesgue measure, by Radon-Nikodym there exists a function $g_\varphi$ such that
$$\tag4
\varphi(f)=\int_{\mathbb R} f(t)\,g(t)\, dt, \ \ \ \ f\in C_0(\mathbb R).
$$
Dirac needed to use the functional $\psi(f)=f(0)$. Doing things "the physicist way", he assumed a $g$ like in $(4)$ would somehow exist, even if not a function, he called it $\delta$ and wrote $(1)$.
Mathematically, what's happening is $(2)$. There is no "$\delta$-function".

Just because the function under the integral contains an infinite value, it does not mean that the function is not integrable. A simple example would be $f(x)=1/\sqrt{|x|}$. When $x\rightarrow 0$ $f$ goes to infinity. You can however integrate in an interval around zero. You can integrate between $-1$ and $1$, and you get $4$.

Since the integration limits do not change with $n$, you can safely move the limit outside of the integration.


$Var(aX) = a^2Var(X)$ but $Var(X+Y) = Var(X) + Var(Y)$. How does this make any sense? Apologies for the less than clear question, I wasn't quite sure how to phrase it.

Say you have a random variable $X ~ N(5, 10^2)$. Say you have another random variable $Y ~ N(5, 10^2)$.

Var(2X) = 4Var(X) = 4*10^2

But:

$Var(X+Y) = Var(X) + Var(Y) = 2\times 10^2$

But now a question arises: what?! X = Y so Var(2X) = Var(X+X) = Var(X+Y) and yet two different answers are reached using the two distinct accepted formulas. How does this make any sense? What's going wrong? It shouldn't matter what we call the random variables, and yet it seems as if it does.

Because in general
$$
\mathbb{V}{\rm ar}(X + Y) = \mathbb{V}{\rm ar}(X) + \mathbb{V}{\rm ar}(Y) + \color{red}{2\operatorname{\mathbb{C}{\rm ov}}(X,Y)}
$$
in your last step you missed to include the covariance of $X$ with $X$

X is distributed identically and independently to $Y$, but $X$ IS NOT $Y$. Let's say $X$ and $Y$ each represent a die roll. If you roll $X$ and $Y$, you might get a $2$ and a $3$, which clearly are not equal. If you compare $X$ to $X$, both are $2$. For any pair of die rolls, the value of $X$ will always be equal to the value of $X$, but will not necessarily be equal to the value of $Y$, and the same goes for any pair of iid $X$ and $Y$.


Critical point of $f(x)=(x^2-1)^2(x-3)^2$ Determine the critical point of $$f(x)=(x^2-1)^2(x-3)^2$$

Firstly, I get the derivative of $f(x)$. I got $=4x(x^2-1)(x-3)^2+2(x-3)(x^2-1)^2$. Any help will be appreciated.

If you are allowed to use critical points calculator, here is solution at
Wolfram|Alpha service.


A unique circle with 3 points proof I have prove the theorem: There is only one circle passing through three given non-collinear points in both geometrical and algebraic ways.
THere is one question that I just have no idea with.
'the accuracy and limitations of this technique"

I dont know what to write...! thank you

In the geometrical way, you can use the fact that in a circle, if you draw a line between any two points in the circle and you bisect that line, the bisected line will go through the center of the circle. So if you have three points, you can draw two bisects and they will intersect at a unique point, which will be the center of your circle. To determine its radius, just compute the distance from your center to any of your three points.

In the algebraic way, you have three points $(x_1, y_1), (x_2, y_2), (x_3,y_3)$ such that
$$
(x_i - a)^2 + (y_i - b)^2 = c^2, \qquad i=1,2,3
$$
which gives you three equations in three unknowns, namely $a,b,c$ (the coordinates of the center $(a,b)$ and the radius $c$). You can use one of the equations to reduce the other two to linear equations and solve the linear system to get the coordinates $a$ and $b$. Once you have those two, any of the three equations will give you $c$.

Hope that helps,

In the algebraic way, you have three points $(x_1,y_1),(x_2,y_2),(x_3,y_3)(x_1,y_1),(x_2,y_2),(x_3,y_3)$ such that
$(x_i−a)^2+(y_i−b)^2=c_2,i=1,2,3$
$(x_i−a)^2+(y_i−b)^2=c_2,i=1,2,3$
which gives you three equations in three unknowns, namely $a,b,c$ (the coordinates of the center $(a,b)(a,b)$ and the radius $c$). You can use one of the equations to reduce the other two to linear equations and solve the linear system to get the coordinates $(a,a)$ and $(b,b)$. Once you have those two, any of the three equations will give you cc.

Hope that helps,


Taylor series for tan x about the point 0 Write down the Taylor series for tan x about the point 0.
Also write down a precise remainder term Rn(x).
finding this much harder than the Taylor series for cos x, arctan x about the point 0.

stuck in this problem from our tutorial sheet of course MA105

You can use long division to divide the Maclaurin expansions of sin(x) over cos(x)

https://www.youtube.com/watch?v=ET3qjWs7g7w This guy explains it in 5mins


What is ⌊0.9 recurring ⌋? For a ceiling and floor function, the number is taken to 0 decimal places. Does this process mean that 0.9 recurring inside a floor function would go to 0? Or would the mathematician take 0.9 recurring to be equal to 1, thus making the answer 1?

And if 0.9 recurring does equal 1, does that mean (by definition) that ⌊1⌋ = 0?

Because $0.\bar{9} = 0.999999$ (recurring) is equal to $1$, the floor function of it is equal to the floor of $1$, and that is $1$.


Hilbert's hotel: why can't I repeat it infinitely many times? I was wondering about the following:

  Suppose a new guest arrives and wishes to be accommodated in the
  hotel. We can (simultaneously) move the guest currently in room 1 to
  room 2, the guest currently in room 2 to room 3, and so on, moving
  every guest from his current room n to room n+1. After this, room 1 is
  empty and the new guest can be moved into that room. By repeating this
  procedure, it is possible to make room for any finite number of new
  guests.

However, if we have an infinite amount of guests, why can't we just say that each guest follows this procedure? Everyone would have to relocate infinitely many times, but what's the problem?

Keep in mind that Hilbert's Hotel is really just an analogy for analyzing countable and uncountable sets, i.e., deciding whether we can construct a bijection from $\mathbb{N}$ to a given set.

If you insist on staying within the analogy, here's the issue with telling everyone to move "infinitely many times." As a hotelier, you need to have a list of who is each room. Similarly, each of your (very accomodating) guests, needs a specific room number to move to. If a guest is in room #1, telling them to move to room #2 (or room #37 or room #123094871230948172 or ...) is a well defined operation. While it may be a long walk, they can get to that room in finite time and know exactly where they are headed.

On the other hand, telling the guest in room #1 to keep walking down the hall until they seem a room that has a number larger than any natural number on the door is ill-defined. There is no way that they could possibly reach this room by walking down the hallway. In effect, you have told them "I don't actually have a room for you, but you are welcome to spend the rest of your life walking down the hallway trying to find one!" This sort of behavior by hotel management is generally frowned upon and leads to poor online reviews of the establishment.

You can accomodate a countably infinite number of new guests in Hilbert's Hotel. Just have each current guest, who is in room number i, move to room 2i. Then all the odd numbered rooms are open and the new guests can move in, and everybody only has to move once.


Transformation matrix for matrix indices to cartesian coordinates In MatLab matrices, the indices are as follows:

(1,1) (1,2) (1,3)
(2,1) (2,2) (2,3)
(3,1) (3,2) (3,3)

This is an example 3x3 matrix. In corresponding cartesian coordinate system, the representation would be:

(-1,1) (0,1) (1,1)
(-1,0) (0,0) (1,0)
(-1,1) (0,-1) (1,-1)

Say, I have any square matrix with dimension-N, where N is odd. I need a generic transformation matrix such that I can get a vector as cartesian coordinates from matrix indices. Does such a function already exist? How should I go ahead in implementing this?

Thanks.

The transformation of indices is the following:

$$
(x,y) = f(i,j) = \left( j-\frac{n+1}{2} ,-i + \frac{n+1}{2}\right) \ .
$$

Here $i$ is the index for the rows, $j$  the one for the columns and $n$ the order of your square matrix.

clear all; clc; close all;

% % % % % % % % Create array with '1' and mark the center with '0' % % % % % % % % %
ones = ones(8,8);
ones(4,5)=2;

[x,y] = size(ones);
for i=1:x
    for j=1:y
        if ones(i,j) == 2;
            index = [i j];
        end

    end
end

newT1 = zeros(x,y);

newT1(index(1,1),index(1,2)) = 5;

[x,y] = size(ones);
for i=1:x
    for j=1:y
            A(i,j) = int2str(i)+","+int2str(j);
    end
end

c = strsplit (A(index(1,1),index(1,2)),',');

% % % % % % % % find Upper and Down limit % % % % % % % % %
uplimit= strsplit(A(1,index(1,2)),',');
dnlimit= strsplit(A(x,index(1,2)),',');

rilimit= strsplit(A(index(1,1),y),',');
lelimit= strsplit(A(index(1,1),1),',');

% % % % % % % % find how many blocks are until the end % % % % % % % % %
up = abs(str2num(c(1,1)) - str2num(uplimit(1,1)));
down = abs(str2num(c(1,1)) - str2num(dnlimit(1,1)));

right = abs(str2num(c(1,2)) - str2num(rilimit(1,2)));
left = abs(str2num(c(1,2)) - str2num(lelimit(1,2)));

% % % % % % % % Create X,Y axis of the cartesians  % % % % % % % % %
for i=1:up
    one = strsplit(A(index(1,1)-i,index(1,2)),',');
    A(index(1,1)-i,index(1,2)) = "0," + int2str(i);
end

for i=1:down
    one = strsplit(A(index(1,1)+i,index(1,2)),',');
    A(index(1,1)+i,index(1,2)) =  "0,"+int2str(-i);
end

for i=1:right
    one = strsplit(A(index(1,1),index(1,2)+i),',');
    A(index(1,1),index(1,2)+i) = int2str(i) + ",0";
end

for i=1:left
    one = strsplit(A(index(1,1),index(1,2)-i),',');
    A(index(1,1),index(1,2)-i) = int2str(-i) + ",0";
end

% % % % % % % % Complete the matrices with the values  % % % % % % % % %

 A(index(1,1),index(1,2)) = "0,0";

 for i=1:up
     for j=1:right
         A(i,y-right+j) = int2str(j) +"," +int2str(up+1-i);
     end
 end

  for i=1:down
     for j=1:right
         A(up+i+1,y-right+j) =int2str(j)+","+int2str(0-i);
     end
  end

for i=1:up
     for j=1:left
         A(i,j) =  int2str(-left-1+j)+","+int2str(up+1-i);
     end
end

for i=1:down
     for j=1:left
         A(up+i+1,j) = int2str(-left-1+j)+","+int2str(0-i);
     end
 end

A

That's my solution for MATLAB :)


Soft question: Unconventional proofs I'm not sure if I understood it correctly, but one of my professors told us that one theorem was proved this way: A mathematician assumed the truth of the Riemann hypothesis and was able to prove a certain mathematical statement. Then a second mathematician assumed the negation of the Riemann hypothesis and was also able to prove the same statement. These two proofs prove that the statement is indeed true. (Does anybody know what this theorem is?)

Another example of such unconventional proof is the proof of the Fermat's last theorem for $n=5$. As I understand it, Sophie Germain showed that if ever there is a solution, one of the integers must be divisible by 5. Dirichlet then proved that if such a solution exists, then the number divisible by 5 must be odd. In the same year, Legendre proved that if such a solution exists, then the number divisible by 5 must be even. Since there are no integers that are simultaneously odd and even, no solution exists.

I also read somewhere that an unconventional way of showing that a set is nonempty is to show that its cardinality is odd (since if the cardinality is odd, it can't be zero).

Do you know of any other very interesting and unconventional proofs that are relatively easy to understand?

One example of a proof along unconventional lines is the following proof that there exist irrational numbers $a$ and $b$ such that $a^b$ is rational.

$\sqrt{2}^\sqrt{2}$ either is, or is not, rational.  If it's rational, we're done.  If not, consider

$(\sqrt{2}^\sqrt{2})^\sqrt{2} = \sqrt{2}^{\sqrt{2} \cdot \sqrt{2}} = \sqrt{2}^2 = 2$

which is rational.  So either way, we have an example of irrational numbers $a$ and $b$ Such that $a^b$ is rational, but we have no clue which is the correct example.

Frankly, it's simpler just to consider $e^{\ln 2}$, but it's still a nice, if highly frustrating, proof.

There are lots of cool examples of unconventional proofs. The criteria for a proof is historically dependent. Very little of the work by Euclid would count as a proof in a modern mathematical journal (same goes for Euler, actually), but what they did was sufficiently rigorous for their time.

So if you look at historical proofs many will strike you as unconventional. For example, look at the Ancient Greek “proofs by exhaustion.” I’d also recommend looking at the original formulation of the solution to the Königsberg bridge problem, which started the field of graph theory.


Second (and higher) derivatives of maps between manifolds I'm trying to understand derivatives of maps between manifolds, and specifically something I read in Dodson and Poston's Tensor Geometry. I'll try to provide as much background as I can for those without the book (for anyone who has the second edition the relevant section is VII.2... pages 166-168 in my copy).

For starters, if $(U, \phi)$ and $(U', \phi')$ are charts on a manifold M (e.g., $\phi: U \subset M \to \mathbb{R}^m$), $u \in U \cap U'$, and $\vec{t}, \vec{t}'$ tangent vectors to $\mathbb{R}^m$, then tangent vectors to $M$ (elements of $T_uM$) are an equivalence class of triples defined by a relation $\sim$ given by

$$
(U,\phi,\vec{t}) \sim (U', \phi', \vec{t}') \iff D_{\phi(u)}(\phi' \circ \phi^{-1})\vec{t} = \vec{t}'.
$$

If $f : M \to N$ is differentiable at $u$, and $(V, \psi)$ is a chart on $N$ (say $\psi : V \to \mathbb{R}^n$) with $f(u) \in V$, then

$$
\begin{align*}
(U,\phi,\vec{t}) &\sim (U', \phi', \vec{t}') \\
&\Rightarrow
D_{\phi(u)}(\psi \circ f \circ \phi^{-1})\vec{t} = D_{\phi(u)}(\psi \circ f \circ \phi'^{-1})\vec{t}' \\
&\Rightarrow
(V,\psi,D_{\phi(u)}(\psi \circ f \circ \phi^{-1})\vec{t}) \sim (V,\psi,D_{\phi(u)}(\psi \circ f \circ \phi'^{-1})\vec{t}')
\end{align*}
$$

so that $f$ induces a well defined map

$$
D_uf : T_uM \to T_{f(u)}N
$$

taking the $\sim$ equivalence class of $(U,\phi,\vec{t})$ to that of $(V,\psi,D_{\phi(u)}(\psi \circ f \circ \phi^{-1})\vec{t})$.

Note that here $D_{\phi(u)}$ is just the regular derivative... as I understand what's going on, we basically create a map $\psi \circ f \circ \phi^{-1} : \mathbb{R}^m \to \mathbb{R}^n$  and take its derivative. The result (in combination with the arbitrary chart $\psi$) uniquely specifies the tangent vector to $N$ for a given tangent vector to $M$ (answering the question: as we move in some direction on $M$, what direction are we moving on $N$?).

The above construction can easily be extended to higher derivatives, by simply taking more derivatives of $\psi \circ f \circ \phi^{-1}$. Dodson and Poston say "In a similar way we can define higher derivatives, with $D^k_uf \in L^k(T_uM;T_{f(u)}N)$". $L^k(T_uM;T_{f(u)}N)$ is the space of multilinear mappings taking in $k$ vectors in $T_uM$ and returning a vector in $T_{f(u)}N$.

Moving towards my questions. Consider a map $g : U \subset \mathbb{R}^2 \to V \subset S^2$. Following the above, $D^2_u(g)(\vec{t},\vec{t}') : T_u \mathbb{R}^2 \times T_u \mathbb{R}^2 \to T_{g(u)}S^2$, i.e. after feeding it two vectors we get a vector tangent to the sphere. This doesn't seem to be what you get if you take derivatives of $\iota \circ g : \mathbb{R}^2 \to \mathbb{R}^3$, i.e., considering the sphere as embedded in $\mathbb{R}^3$ ($\iota$ being the inclusion map). Fixing $\vec{t}$, after the first derivative we have a field of tangent vectors to the sphere at each point $g(x)$ for $x \in U$, and the second derivative I presume looks at how this field changes with position (yielding vectors no longer necessarily tangent to the sphere, since we're using just a regular and not covariant derivative).

My questions: I think my confusion festers somewhere in that last paragraph. Are second derivatives of $g$ and $\iota \circ g$ not the same because the second derivative of the inclusion map is not the inclusion map between the relevant tangent spaces (i.e., chain rule)? Is my picture of the second derivative of $\iota \circ g$ incorrect or misleading? What does $D^2_u(f)$ actually tell us about $f$? Is there some intuitive question that can be formulated like in the first derivative case, which tells us how we move on N under $f$ as we move on M?

Any and all help in untying my knotty brain would be greatly appreciated.

Your doubt is well-placed: Even in the case $f:M\to \mathbb R$ you cannot sensibly define $D^2_u f : T_u M \times T_u M \to \mathbb R$ unless $u$ is a critical point of $f$. Otherwise your definition $$D^2 f(\partial_i, \partial_j)=\frac{\partial^2 f}{\partial x^i \partial x^j} \ \ \text{extended by multilinearity}$$ depends on the coordinate system $x^i$: if we have some other coordinate system $y^\alpha$ then by multilinearity we should have
$$
D^2 f\left(\frac{\partial}{\partial x^{i}},\frac{\partial}{\partial x^{j}}\right) =\frac{\partial y^{\alpha}}{\partial x^{i}}\frac{\partial y^{\beta}}{\partial x^{j}}D^{2}f\left(\frac{\partial}{\partial y^{\alpha}},\frac{\partial}{\partial y^{\beta}}\right)=\frac{\partial y^{\alpha}}{\partial x^{i}}\frac{\partial y^{\beta}}{\partial x^{j}}\frac{\partial^{2}f}{\partial y^{\alpha}\partial y^{\beta}};
$$
but the chain rule gives $$\frac{\partial^{2}f}{\partial x^{j}\partial x^{i}}=\frac{\partial}{\partial x^{i}}\left(\frac{\partial y^{\beta}}{\partial x^{j}}\frac{\partial f}{\partial y^{\beta}}\right)=\frac{\partial y^{\alpha}}{\partial x^{i}}\frac{\partial y^{\beta}}{\partial x^{j}}\frac{\partial^{2}f}{\partial y^{\alpha}\partial y^{\beta}}+\frac{\partial^{2}y^{\beta}}{\partial x^{i}\partial x^{j}}\frac{\partial f}{\partial y^{\beta}}.$$

In Riemannian geometry this is fixed by using the covariant Hessian $$D^2f (X,Y) = XYf - (\nabla_X Y) f;$$ but without the additional structure of a conenction there is no way to think of $k^\text{th}$-order derivatives as taking $k$ tangent vectors.

Instead the general formulation uses higher tangent bundles: the first derivative is a map $TM \to TN$, so the second derivative is a map $TTM \to TTN$. Thus the picture you should have in your head is that the first input (a vector $v \in T_u M$) is the base direction you're moving in, and then the second input (a vector in $T_v TM$) is the direction you're varying that $v$ in.

For the formula of the "double tangent map" (for a map between Banach spaces) see page 100, Exercise 2.4H of "Manifolds, Tensor Analysis, and Applications" by R.H. Abraham, J.E. Marsden, and T.S. Ratiu (Addison Wesley, 1983). This can be generalized to the setting of manifolds.


Computing the Gaussian integral with step functions Say, we are interested in deriving

$$\int_{-\infty}^{\infty}e^{-x^2}=\sqrt{\pi}\tag{1}$$

There are many well known ways to do it, for example:

by polar coordinates
via the gamma function, etc.

After coming across this limit
$$\lim\limits_{n\to\infty} \frac{\sqrt{\pi n}}{2^{2n}}  \binom {2n} {n+\lfloor x\sqrt{n} \rfloor} = e^{-x^2}\tag{2}$$
I wonder how can we derive $(1)$ by using $(2)$ as an approximation via step functions.

Here is a picture of $n=4$:

For every $n\geqslant1$, consider the function $u_n$ defined by
$$
u_n(x)=\frac{\sqrt{\pi n}}{2^{2n}}  \binom {2n} {n+\lfloor x\sqrt{n} \rfloor}.
$$
It happens that, for every $n$,
$$
\int_\mathbb Ru_n(x)\,\mathrm dx=\sqrt\pi.
$$
To see this, note that $u_n(x)=0$ for every $x\lt-\sqrt{n}$ and every $x\geqslant\sqrt{n}$ and that $u_n(x)=u_n(k/\sqrt{n})$ for every $x$ in the interval $[k/\sqrt{n},(k+1)/\sqrt{n})$, for every integer $-n\leqslant k\leqslant n$. The length of each of these $2n+1$ intervals is $1/\sqrt{n}$, hence
$$
\int_\mathbb Ru_n(x)\,\mathrm dx=\sum_{k=-n}^nu_n(k/\sqrt{n})/\sqrt{n}=\sqrt{\pi}\,2^{-2n}\sum_{k=-n}^n{2n\choose n+k}=\sqrt{\pi}\,2^{-2n}\sum_{i=0}^{2n}{2n\choose i}.
$$
By the binomial theorem, the last sum on the RHS is $2^{2n}$ hence we are done.

My hint:
 This is Gauss integration that is famous. Laplace proof equal $\sqrt{\pi}$. Also the technique to calculate like this:
$$I=\int_{-\infty}^{+\infty}e^{-x^2}dx$$
Squared two sides and change variable:

$$I^2=\left ( \int_{-\infty}^{+\infty}e^{-x^2}dx \right )^2=\int_{-\infty}^{+\infty}e^{-x^2}dx\int_{-\infty}^{+\infty}e^{-y^2}dy=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)}dxdy$$

$dxdy$ is the differential surface area in $Oxy$, integrating domain is the whole plane should we switch to polar coordinates:
$$\left\{\begin{matrix} x=r \cos\varphi\\ y=r\sin\varphi \end{matrix}\right.$$
In which
$$\left\{\begin{matrix} \varphi: 0 \rightarrow 2\pi\\ r: 0 \rightarrow +\infty \end{matrix}\right.$$

$$I^2=\int_{0}^{+\infty}\int_{0}^{2\pi}e^{-r^2}rdrd\varphi=\int_{0}^{+\infty}e^{-r^2}rdr\int_{0}^{2\pi}d\varphi=\frac{1}{2}.2\pi=\pi$$

Hence $$I=\sqrt{\pi}$$


Finite rings $R$ in which $x^{25}=x$ holds I want to classify finite rings $R$ in which $x^{25}=x$ for all $x\in R$.

I know Jacobson's Theorem that if $x^n=x$ for all $x\in$ then $R$ is commutative. I don't know how to show the Theorem for the special case $n=25$ by elementary methods.

Furthermore, after showing that $R$ is commutative, is there any Theorem similar to Wedderburn's Theorem to conclude that $R$ is a field?

Your ring is

commutative, by a famous theorem of Jacobson, as mentioned in the comments
reduced, (clearly)
Finite by assumption, hence it is a semisimple, and therefore a product of finitely many fields (Wedderburn), and
All the fields satisfy $x^{24}=1$ on their nonzero elements, which limits the cardinality of each field appearing to at most $25$.

If you run through the cardinalities of finite fields between $2$ and $25$, you'll find that $A=\{2, 3, 4, 5, 7, 9, 13, 25\}$ all will work, while $8, 11, 16, 17, 19$ and $23$ will not. If $q$ is the size of the field, then the nonzero elements are a cyclc group of order $q-1$, and that cyclic group must satisfy $x^{24}=1$, which means $q-1|24$.

So in summary, this proves that any finite product of finite fields with orders in $A$ will satisfy $x^{25}=x$, and conversely every such ring is such a product.

So you see, it is only a field in precisely one of these cases...

Hint: $\forall x,y \in R$, we have $(xy)^{25} = xy$. But $x=x^{25}$ and $y = y^{25}$. Then $(xy)^{25} = x^{25}y ^{25}$. It is enough to show that $xy = yx$.


Does there exist a field $K$ and $x \in K$ with $K(x^{1/4})=K(x^{1/2}) \neq K$ I want to know if there exists a field $K$ and $x \in K$ with $K(x^{1/4})=K(x^{1/2}) \neq K$.

My attempt thus far: if $K(x^{1/2}) \neq K$ then $x^{1/2} \notin K$ so the minimal polynomial of $x^{1/2}$ over $K$ has degree $>1$ and is therefore $X^2-x$. Then $[K(x^{1/2}):K] = 2$ and $\{1, x^{1/2}\}$ is a $K$-basis for $K(x^{1/2})$. Since $K(x^{1/4})=K(x^{1/2})$ we have $x^{1/4} \in K(x^{1/2})$ so there exist $a, b \in K$ such that $x^{1/4} = a + bx^{1/2}$.

Then I tried rearranging and squaring both sides etc to see if I could find something useful but I didn't get anywhere... Any tips? I don't even know whether such a $K$ exists at all!

Yes. $K= \mathbb R$, $x=-1$.

Added: Generalising comments by k.stm and Jyrki Lahtonen, as well as reuns under the OP, I claim that assuming $char(K) \neq 2$, such an $x \in K$ exists if and only if $-1$ is not a square in $K$. The proof, as well as the case $char(K)=2$, are left as exercise.

$s=x^{1/4}$ , $d= x^{1/2}$

s£ k(d)==> s=a+bd ===> s^2 =a^2 +b^2d^2 +2abd ==> d=a^2+b^2x+2adb
===> d( 1-2a) = a^2 +x
d= ( 1-2ab)^(-)  ( a^2+xb^2)  , if 1-2ab#0 ==> d£K but this absurde since  k( d) #k  sorry i have note Latex for better redaction


Multiplication Operator on $L^2( \mathbb{R})$ is not Compact Let $H=L^2(\mathbb{R})$, we define the multiplication operator with a function $h\in H$, ($0\leq h)$ and $||h||_1>0$
.
We have $$T_h:  H \rightarrow H, T_h(f)=h.f$$

Note that $T$ is bounded linear operator.

I want to prove that $T_h$ is not a compact operator.
So If we suppose that $T_h$ is compact, I need to prove for example that the spectral set is non countable. But I can't find a good start.

The idea from MaoWao's answer can also be used to give a direct proof, without any spectral theory.

Since $h$ is not a.e. zero, there is a set $E$ of positive measure on which $h$ is bounded away from zero; say $|h| \ge \epsilon$ on $E$.  As in MaoWao's approach, find a sequence of disjoint sets $E_n \subset E$, each with finite positive measure.  Set $f_n = 1_{E_n} / \sqrt{m(E_n)}$; then the $f_n$ are orthonormal, and in particular they are all in the closed unit ball.  Now verify that the functions $T_h f_n$ are also orthogonal, and $\|T_h f_n\|_{2} \ge \epsilon$.  Conclude from this that $\{T_h f_n\}$ has no convergent subsequence.

So If we suppse that $||h||_1>0$ then we can find the inverse operator for $(T_h-\lambda I) $ for selected $\lambda \in \mathbb{R}$.
So since, $||h||_1>0$, there exists $\alpha>0$ such that $||h||_1>\alpha$.

Now If we take $\lambda \in [0,\alpha]$ the operator $(T_h-\lambda I)$ will be invertible, and it's inverse will be: $\psi=\frac{I}{h-\lambda}$

And then, we come to the conclusion that $[0,\alpha]\subset \sigma(T_h)$ then $T_h$ is not compact.


Finding number of real roots The question is

  Find the number of real roots of the equation
  $(x-1)x(x+1)(x+2)-1=0$

I simply tried to simplify the equation but it transforms into biquadratic which becomes really tough to conclude anything from it.My second approach was :-
$y=(x^2-1)  ;$
$y=\frac{1}{(x^2+3x+2)}$
Then i tried to find their intersection but i failed
Please help to solve the question without using any tools (like wolfram) other than pure mathematics since it is subjective mathematical paper's question which needs a proper solution.

The function is symmetric about $x = -{1 \over 2}$ since the zeroes of $(x+2)(x+1)x(x-1)$ are at $-2,-1,0,$ and $1$. So letting $y = x + {1 \over 2}$ the polynomial is
$$(y-{1 \over 2})(y + {1 \over 2})(y - {3 \over 2})(y + {3 \over 2}) - 1$$
$$ = y^4 - {5 \over 2}y^2 -  {7 \over 16}$$
This is a quadratic equation in $y^2$, so you can find all possible $y$ easily, and from that the number of solutions $x$.


Transformation in PDE Using the transformation $\displaystyle u=\frac{w}{y}$ in the PDE:
$$x\frac{du}{dx} = u + y\frac{du}{dy}$$
 The transformed equation has a solution of the form
"$w=?$".

What is the method to solve such question?

I substituted $\displaystyle u=\frac{w}{y}$ in the equation and then tried to solve. I got:
$$-x\frac{w}{y^2} \frac{dy}{dx} = u - \frac{w}{y}$$
Is this correct?

What you got : $\quad -x\frac{w}{y^2} \frac{dy}{dx} = u - \frac{w}{y}\quad $
 is not correct.

Since you didn't show the detail of your calculus, it's not possible to show you where is the mistake. I suppose that you can find the correct result by yourself.  Or show us what you have done.

For this PDE the method of change of function is very good. Even simpler, use the change $u(x,y)=xv(x,y)$.

For information only, another method method is shown below (method of characteristics) :

The equations of characteristics curves are : $\quad \frac{dx}{x}=-\frac{dy}{y}=\frac{du}{u}\quad\to\quad \begin{cases}xy=c_1 \\ \frac{u}{x}=c_2\end{cases}$

This is valid with independent $c_1$ and $c_2$  on the characteristic curves only. Elsewhere $c_1$ and $c_2$ are dependent : $c_2=f(c_1)$. So, the general solution of the PDE is :

$$u=xf(xy)$$
with any differentiable function $f$.


Solving an obvious problem using Lagrange multipliers/KKT I have the following optimization problem:

$\max_{x,y} xy $ subject to $ x, y \geq 0$ and $y + x^2 \leq 6$.

I know how to solve this problem. By looking at the objective function it is clear that the maximum must lie on $y + x^2 = 6$ and then it can be found as

$\max_x x(6-x^2)$ for $x \in [0, \sqrt{6}]$.

I would like to practice Lagrange multipliers by doing it that way.
So I set up the Lagrangian

$L(x,y, \lambda, \mu, \nu) = xy - \lambda (y + x^2 - 6) + \mu x + \nu y$

and look at the KKT conditions:

$ \frac{\partial L}{\partial x} = y - 2\lambda x + \mu = 0$

$ \frac{\partial L}{\partial y} = x - \lambda + \nu = 0$

$x,y,\lambda, \mu, \nu \geq 0 $

$y + x^2 \leq 6$

$\mu x = \nu y = \lambda (y + x^2 -6) = 0$

I'm able to solve for $x$ and $y$ in terms of the Lagrange multipliers:

$x = \lambda - \nu$ and $y = 2 \lambda^2 - 2 \lambda \nu - \mu$

But after that I'm stuck. I'm not sure how to proceed, and the apparent simplicity of the objective function seems to have vanished.

Should I consider the 8 possible cases from complementary slackness one by one?

(Context: this is an example of the Nash bargaining problem)

You can obtain the solutions by reasoning.

Suppose $\lambda=0$, then $y=-\mu$ and $x=-\nu$. Then $y=\mu=0$ and $x=\nu=0$ due to nonnegativity. Thus we obtain one point that satisfies the KKT conditions.

Now suppose $\lambda>0$, then $y+x^2=6$. Suppose $x=0$, then $y=6$, $\nu=0$, $\lambda=0$, contradiction. So, $x>0$. Suppose $y=0$, then $x=\sqrt{6}$, $\mu=0$, $\lambda=0$, contradiction. So, $y>0$. So, $\mu=\nu=0$. So, $x=\lambda$ and $y=2\lambda x = 2 x^2$. So, $3x^2 = 6$, $x=\sqrt{2}$, $y=4$.

The reason we obtain multiple KKT points is that the problem is not a convex optimization problem. The KKT conditions are therefore necessary but not sufficient. Comparing the objective value for both solutions, the second solution is optimal. If you perform a logarithmic transformation on the objective, the problem becomes convex and the KKT conditions are also sufficient.

We have: $f(x,y) = xy \le x(6-x^2)= 6x-x^3 = g(x)\implies g'(x) = 6-3x^2 = 3(2-x^2)=0 \iff x = \sqrt{2}\implies g''(\sqrt{2})= - 6\sqrt{2} < 0 \implies g_{\text{max}} = g(\sqrt{2})= 4\sqrt{2}= f_{\text{max}}$. Note that $f_{\text{max}}$ occurs at $ x = \sqrt{2}, y = 4$


Finding the least significant digit of a large exponential. I am trying to find the least significant digit of $17^{{17}^{17}}$. I know that I need to use to use the properties of modular arithmetic and mod base 10, but I am not sure how to go about it.

Please provide some hints/first few steps to help me get started.

Start off by looking at $17^n$ mod 10. (In your case, n will end up being $17^{17}$, but that's way too big to calculate yet.)

$17^0$ ends in a $1$, $17^1$ ends in a $7$, $17^2$ is congruent to $7 \times 7$ so it ends in a $9$, $17^3$ likewise is congruent to $9 \times 7$ so it ends in a $3$, and finally $17^4$ is congruent to $3 \times 7$ so it ends in a $1$.

Since $17^0$ and $17^4$ are congruent mod 10, it follows that $17^n$ mod 10 will repeat every time the exponent $n$ goes up by 4.

Therefore, to solve your problem, you now need to calculate the exponent $17^{17}$ mod 4. Then you can use that along with the pattern I just described to get the final answer. Since this is homework, I'll let you calculate $17^{17}$ mod 4 yourself... hint, use the same idea that I used above!

Hint: For distinct primes $p, q,$ and any $a \neq 0
(\mod pq)$, use the formula:
$$a^{(p-1)}(q-1) = 1 (\mod pq)$$


Find the image and preimage of functions Given the function $f(x)=e^{(x-1)^2}, f:\mathbb{R}  → \mathbb{R}$ ,  find $f[[0, +∞)] $.

Given the function $f(x)=2x^2 + 5x - 5, f:\mathbb{R}→\mathbb{R}$ ,  find $f^{-1} [[0, 3]] $.

I have no idea how to solve the first example, so I would be very grateful for explanation.

In the following example I got $f^{-1} [[0, 3]] = [-28,\frac{-5 - \sqrt{65}}{4}]\cup [\frac{-5 + \sqrt{65}}{4}, 28] $. I am not sure if I'm right. Could you tell me if my result is correct?

For the first function observe that it is a composition of $(x-1)^2$ and $e^x$ and since both are continuous and moreover, they are differentiable on $\Bbb R$, the composition i.e. $e^{(x-1)^2}$ is differentiable over $\Bbb R$.

So differentiating, $f^/(x) = 2(x-1)e^{(x-1)^2}$ and $$ f^{(2)}(x) = {4(x-1)^2}e^{(x-1)^2} + 2e^{(x-1)^2}$$.

Now equating, $f^/(x) = 0 $ we obtain that x = 1.

Putting x = 1 at  $ f^{(2)}(x) = {4(x-1)^2}e^{(x-1)^2} + 2e^{(x-1)^2}$ we deduce that $f^{(2)}(x) > 0$, so f has a global minima at x = 1 . At that point , f(1) = 1.

So at all other points $f(x)>1$ and since $f$ is continous on a connected interval, it satisfies the IVP and thus answer to your first question becomes $[1,+\infty)$ .

[Actually, you can consider the graph of $e^{(x-1)^2}$ to be that of $e^{x^2}$ with the origin right-shifted at x = 1.]

For the second question observe that $2x^2 + 5x - 5$ is a polynomial and so it is also continuous and also since the set you're considering the pre-image of is a connected set you just need to solve the quadratic for the boundary points of the inetrval i.e. you just need to solve

$$  2x^2 + 5x - 5 = 0$$ and $$2x^2 + 5x - 5 = 3$$
The roots you'll obtain probably, $$x = \frac{-5 \pm \sqrt{65}}{4}$$ for the first one and $$x = \frac{-5 \pm \sqrt{89}}{4}$$ for the second one. Now take suitable union and you'll obtain answer .

[$\frac{-5 + \sqrt{89}}{4}$ will replace 28 in your answer and $\frac{-5 - \sqrt{89}}{4}$ will replace -28]

For the first example, calculate $f(0)$, see if the function is increasing. If it is, then the image is $[f(0), \infty]$, the function is continuous and explodes to infinit.

For the second one, solve the inequation $0 \leq 2x^{2}+5x−5 \leq 3$. Wolfram says that you are wrong. Calculate $f(28)$ you are going to see that you are wrong.


Why must the power series for the derivative of this be shifted over one unit? I was asked to find the power series for $f(x) = \frac{x}{(1+7x)^2}$ I proceeded by finding the power series for  $g(x) = \frac{1}{1+7x}$, differentiating that series to get $g'(x) = \frac{-7}{(1+7x)^2}$, and mutiplying by $\frac{-x}{7}$ to get an equivalent answer. However, when looking at a solution to the problem, I saw that the power series for the power series for $g'(x)$ was shifted over by one unit to give:

$$\sum_{n=1}^\infty (-4)^nx^{n-1}$$

Why is the power series for $g'(x)$ shifted over here?


Does Arzelà-Ascoli theorem hold for pointwise convergence? Consider the space $C(\left[0,1\right])$ of continuous, real-valued functions on the interval, equipped with the topology of pointwise convergence.

Is it true that a subset $S\subseteq C(\left[0,1\right])$ is compact if and only if it is bounded and equicontinuous?

I guess that the answer is no, as this seems to be a weakened version of Arzelà-Ascoli theorem, which guarantees the validity of the statement when the topology is given by uniform convergence instead. I was trying to find a counterexample but it's not easy to check that a certain subset is or is not compact, especially because pointwise convergence topology is not metrisable, so we cannot use sequential compactness arguments. Can you help me with that? Thank you.

As a starting point we need to take the correct version of Arzela-Ascoli for the compact-open (or in this case also uniform) topology: $S$ is relatively compact (the closure is compact) iff it is pointwise relatively compact, and equicontinuous.

In the weaker pointwise topology, we have more compact sets, and there the criterion is simply: $S$ is relatively compact iff it is pointwise relatively compact: this makes it a subset of a product space of compact intervals of the reals, and thus has a compact closure due to the Tychonoff theorem.

In both cases, to characterize compactness, add closed to the list of conditions.

It might be helpful if you think of C([0,1]) as a subspace of the (compact,hausdorff) product space $R^{[0,1]}$.
- Apologies, this is not correct!


prove existence of the limit of a sequence So I have the following problem:
$ x_0 = 1 , x_1 = 2 , $and $x_{n+1} = 2x_n + x_{n-1} $for $ n \geq 1.$

Show that: $\hspace{2mm} \lim_{n\to \infty} \frac{x_n}{x_{n+1}}   $ exists.

Then show that the Limit is equal to $\sqrt{2}-1$.

For this I thought i could use the fact that $x_n$ is bounded and I thought that it was monotonically falling, but that is not the case, so I ran out of ideas. And I don't know how to calculate the limit... Thank you very much for your help!

Without guessing the limit you may proceed as follows:

Set $q_n = \frac{x_n}{x_{n+1}}$
$$\Rightarrow x_{n+1} = 2x_n + x_{n-1} \Leftrightarrow \frac{1}{q_n} = 2+q_{n-1}\Leftrightarrow q_n = \frac{1}{2+q_{n-1}}$$

Now, it follows
$$|q_{n+1} - q_n| = \left|\frac{1}{2+q_{n}} - \frac{1}{2+q_{n-1}} \right| = \left|\frac{q_{n-1} - q_n}{(\color{blue}{2}+q_{n})(\color{blue}{2}+q_{n-1})}\right|$$ $$< \frac{1}{\color{blue}{2\cdot 2}}\left|q_{n-1} - q_{n} \right| \stackrel{\mbox{see below}}{\Rightarrow} \boxed{(q_n) \mbox{ is convergent}}$$

So, we get for the limit
$$L =\frac{1}{2+L} \Leftrightarrow (L+1)^2=2 \stackrel{L>0}{\Rightarrow}\boxed{L = \sqrt{2}-1}$$

Edit after comment:

Additional info concerning the convergence of the sequence:

$q_{n+1} = q_1 + \sum_{k=1}^n(q_{k+1} -q_k)$
The sums converge (absolutely) as $|q_{k+1} -q_k| < \left( \frac{1}{4}\right)^{k-1}|q_2 - q_1|$ since
$\left|\sum_{k=1}^{\infty}(q_{k+1} -q_k) \right| \leq \sum_{k=1}^{\infty}|q_{k+1} -q_k| < |q_2 - q_1|\sum_{k=1}^{\infty} \left( \frac{1}{4}\right)^{k-1} < \infty$
As $s_n = \sum_{k=1}^n(q_{k+1} -q_k)$ converges, it follows that $q_{n+1} =q_1 + s_n$ converges

Let

$$\lim_{n\to \infty} \frac{x_n}{x_{n+1}} = \lim_{n\to \infty} \frac{x_{n-1}}{x_{n1}} = k$$

Now

$$\lim_{n\to \infty} \frac{x_n}{x_{n+1}} = k$$
$$\lim_{n\to \infty} \frac{x_n}{2x_{n} + x_{n-1}} = k$$

Take $x_n$ out from numerator and denominator
$$\lim_{n\to \infty} \frac{1}{2 + \frac{x_{n-1}}{x_n}} = k$$

Using the first equation

$$ \frac{1}{2 + k} = k$$
$$k^2+2k-1=0$$
This gives two solutions $k=\sqrt{2}-1$ and $k=-\sqrt{2}-1$. Since none of the terms can be negative, we reject . the second solution thereby giving us

$$\lim_{n\to \infty} \frac{x_n}{x_{n+1}} = k = \sqrt2 - 1$$

EDIT - As suggested by the commenter we need to prove that it is a finite limit before we start with the proof. Initially it's a $\frac{\infty}{\infty}$ form as both $x_n$ and $x_{n+1}$ approach $\infty$ as $n$ approaches $\infty$. I'll use a finite upper bound to show that the limit is finite which means it exists.

For any $n$

$$\frac{x_n}{x_{n+1}} =\frac{x_n}{2x_n + x_{n-1}}$$

As $x_{n-1}$ is always a positive quantity
$$\frac{x_n}{x_{n+1}} \leq \frac{x_n}{2x_n}$$
$$\frac{x_n}{x_{n+1}} \leq \frac{1}{2}$$

For any $n$, you can take the last statement to prove the monotonicity as
$$x_{n+1}\geq x_n$$
And since both $x_n$ and $x_{n+1}$ are positive values, the lower bound is $0$. The upper bound along with lower bound and the monotonicity proves that the limit is finite.


Are all Elementary Periodic Functions Trigonometric? Hypothesis:

  Given a non constant differentiable periodic function $f(x),\ \mathbb{R}\to\mathbb{R}$, which has an explicit finite form involving
  only elementary functions. The expression for $f(x)$ must include trigonometric functions, or equivalently, complex exponents.

In other words, we can't construct a periodic function from only roots, real exponents, logarithms and polynomials.

Is there a simple proof or counter example for the above hypothesis?

Partial answer.

  If $f$ is a composition, then the last composed functions can be ignored if they are injective. Thus, for anything of the form $e^{h},h^{2n+1},\ln h,\ldots$, we need just consider $h(x)$.

Proof: Suppose we were to construct a periodic, composite function $f(x)=g\circ h(x)$, where $g$ is injective and thus aperiodic. Then $h$ must have already been periodic since it is trivially $h=g^{-1}\circ f$.

  $f$ cannot be a rational function with an injective function as its variable.

Proof: If $f(x)$ were a periodic polynomial then, for some $x_0$, $f(x)-f(x_0)$ would be a polynomial with infinite roots, $x_0+nP$, where $P$ is a period and $n\in\mathbb{Z}$. This would violate  the fundamental theorem of algebra. This can be generalised to any rational function $r(x)=\frac{a(x)}{b(x)}$ since the polynomial $a(x)-r(x_0)b(x)$ would have infinite roots, $x_0+nP$. We can then generalise it further to $r\circ h(x)$ for any rational function, $r$, with an injective function, $h$, as its variable since it would have at most as many roots as $r(x)$. So all functions of the form $\frac{e^{2x}-3e^x}{e^{2x}+1},\frac{\ln(x)^2+1}{\ln(x)},\ldots$ cannot be periodic.

The question remains for combinations of different elementary functions such as $\frac{e^x+\ln(x)+\ldots}{x+\ldots}$. A difficulty in the problem is that the composition of two aperiodic, elementary functions can form a periodic function. For example, take $g(x)=\cos\sqrt{x}$ and $h(x)=x^2$. So too can the sum or product of two aperiodic, elementary functions. For instance, take $\left(\sin x+e^x\right)+\left(-e^x\right)$ or $\frac{\sin(x)}{e^x}\cdot e^x$.

If we consider $[x]$ to be elementary then $x-[x]$ is also periodic with period $=1$
Other than that, Only trigonometric functions are periodic as they are circular functions which repeat after an interval and all periodic functions are composed of trigonometric functions in some form or other.


Finding the sum of this series: $1+\frac 12 + \frac 13 + \cdots + \frac 1{50}$ I need to find the sum of this series: $1+\frac 12 + \frac 13 + \cdots + \frac 1{50}$

Please help me find the sum of this series.

There is no nice closed formula for partial harmonic sums. However, there is a technique called partial summation used to estimate these types of sums. Partial summation, or Abel summation, is just integration by parts for the Riemann-Stieltjes integral.

You can look up how to approximate sums by the method of partial summation if you are interested. The estimate it provides for partial harmonic sums is

\begin{align}
\sum_{n \leq x}{\frac{1}{n}} &= \log(x) + \gamma + O\left(\frac{1}{x}\right),
\end{align}

where $\gamma \approx 0.5772$ is the Euler-Mascheroni constant. Note that the range of the sum is over all positive integers $n$ at most $x$. The $O(1/x)$ term just means the approximation gets better as we chose larger $x$.

For $x = 50$ this formula tells us

\begin{align}
\sum_{n \leq 50}{\frac{1}{n}} &\approx 4.4892.
\end{align}

If you calculate the sum exactly, the error in the approximation is about $0.00997$.


Prove that if $M$ is rotation matrix, then so is $TMT^{-1}$ Prove that if $M$ is rotation matrix, then so is $TMT^{-1}$

I already proved that the determinant and eigenvalues are the same for $M$ and $TMT^{-1}$, but I'm not sure how to prove they are both rotations.  Do i need to write out the matrix and its components, or is their a rotation equation/relationship I can use to show they are both rotations.

T is any invertible matrix

The claim is false in general. If one takes $$M = \pmatrix{0&-1\\1&0}$$ (anticlockwise rotation by $90^{\circ}$) and $$T = \pmatrix{1&0\\0&\lambda}$$ (dilation by $\lambda$ in the $y$-direction). Then, $$TMT^{-1} \pmatrix{1\\0} = \pmatrix{0\\ \lambda} .$$
In particular, unless $\lambda = \pm 1$, we see that multiplication by $TMT^{-1}$ does not preserve lengths and so is not a rotation matrix.