$\lim_{x\rightarrow+\infty} x^{\alpha+\beta}P(|X|>x)=0\Leftrightarrow\lim_{x\rightarrow+\infty}x^{\alpha}E||X|^{\beta}\cdot1_{\{|X|>x\}}|=0$ I have met with a probability problem which I have no idea to deal with. It says:

Let $\alpha>0$, $\beta\geq0$, prove:

$\lim_{x\rightarrow+\infty} x^{\alpha+\beta}\mathbb{P}(|X|>x)=0\Leftrightarrow\lim_{x\rightarrow+\infty}x^{\alpha}\mathbb{E}||X|^{\beta}\cdot1_{\{|X|>x\}}|=0$.

The sufficiency is easy to prove. However, I have no ideas how to prove the necessity. Can you give me some advice? Any hints would be appreciated.

Thanks for your time and patience.

You can simplify the statement by writing $Y$ for $|X|^{\beta}$ and repalcing $x^{\beta} $ by $t$. The statement then becomes the following: $t^{1+s}P(Y>t) \to 0$ iff $t^{s}EYI_{Y>t} \to 0$ as $t \to \infty$.[Here $s=\frac {\alpha} {\beta}$. I will leave the case $\beta=0$ to you].  As noted by you the íf'part follows easily by Chebychev's Inequality. For the ónly if'part write $t^{s}EYI_{Y>t} \to 0$ as $t^{s}\int_t^{\infty} P(Y>u)du$ which is less than or equal to $t^{s}\int_t^{\infty} \epsilon u^{-1-s}du=\epsilon$ for $t$ sufficiently large.

Suppose
$\lim_{x\to \infty} x^{\alpha+\beta}P(|X|>x) = 0$.
Note that
\begin{align}
E[|X|^\beta \cdot 1_{|X|>x}]
&= P(|X|>x)E[|X|^\beta | |X| > x] \\
&= x^\beta P(|X|>x) E[|X/x|^\beta | |X/x|>1] \\
&\le x^\beta P(|X|>x) E[|X|^\beta], \quad \text{for large $x$}.
\end{align}
Note that for large $x$, $|X/x| < |X|$, and thus
$|X/x|^\beta < |X|^\beta$.

Assuming $E[|X|^\beta] < \infty$,
$$
\lim_{x \to \infty} x^\alpha E[|X|^\beta \cdot 1_{|X|>x}]  = 0.
$$


Intuition concerning Schwartz kernels of Operators Consider a (for example differential) operator $A$ acting on an appropriate function space over a smooth compact manifold without boundary.

Using the Schwartz kernel $K(x,y)dy$ of the operator, its action can be described as
$$
Af(x) = \int_M K(x,y)f(y)dy\,.
$$

Sometimes one may also have families of operators $A_t$ depending on a parameter which may for example describe time.

I was wondering whether there are "intuitive ideas" behind the kernel $K(x,y)$, for example
1) what does its value at the point $(x,y)$ tell us about the action of the operator?
2)How can I translate something like a discontinuity into some insight regarding the behaviour of the abstract operator? (e.g. kernels of inverses to differential operators, which are pseudodifferential operators, are discontinuous at the diagonal)

Many thanks for your insight!

A large part of modern harmonic analysis is about inferring properties of the operator $A$ like being bounded from $L^p$ to $L^q$, from properties of the kernel $K(x,y)$ and in particular its behavior near the diagonal. See, e.g., the theory of Calderon-Zygmund operators.

Someone will likely correct me, but I wouldn't expect too much. $\{K(x,y)\}_{(x,y)}$ is the continuous version of the matrix representation of the operator. And usually you don't get much information about an operator by looking at the entries of its matrix in some basis.


Difficult integration problem (unique headline) So I found this difficult integral at Physicsforum (I'm quite sure you are familiar with this forum) and I've been at it for about an hour now and it's quite frustrating because I can't get the answer.

$$\int \frac{4x^5 -1}{x^5 + x +1} dx $$

I would really love it if someone could help me with this

Taking the question as it is, it is a rational function and therefore cen be integrated by partial fractions. Fist divide the denominator into the numerator to get a fraction where the numerator has a smaller degree. Integrating the quotient gives $4x$. Now the problem is to factor the denominator. However it does not have any rational roots. It is of odd degree so has at least one real root, so irrational quantities will be an essential part of the answer. The other question is how many factors are linear and how many complex. You can work this out by graphing the function (maxima and minima).
Each linear factor gives a logarithm and each quadratic a inverse tangent. So you should be able to determine the general form of the integral, but not the specific constants involved.


Set intersection theory $W = \{a,b,c\}, Z = \{W,\emptyset\}$

What is $Z\cap W$?

I'm not sure how to work this out as my thoughts would be
$$Z = \{W,\emptyset\}  = \{\{a,b,c\},\emptyset\}$$
Which means $Z \cap W = \{a,b,c,\{a,b,c\},\emptyset\}$

Or should it just be
$$Z = \{W,\emptyset\} = \{a,b,c,\emptyset\}$$
Which means $Z \cap W = \{a,b,c,\emptyset\}$

Or should it just be
$$Z = \{W,\emptyset\}$$
Which means $Z\cap W = \{W,\emptyset,a,b,c\}$

Any thoughts would be appreciated!

The elements of $W$ are: $a,b$ and $c$

The elements of $Z$ are: $W$ and $\emptyset$

Can you see any element both in $W$ and $Z$? If yes, those elements are the answer. If there is no such element, then the answer is $\emptyset$.

Intersection = available in both the sets. So,

Z intersection W = W $ \rightarrow $ {a,b,c}


Find the value of $S$ in term of $k$ (telescoping sums) Let $k=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+\cdots+\frac{1}{2549\times2550}$.

Find the value of $S=\frac{1275}{1276}+\frac{1276}{1277}+\frac{1277}{1278}+\cdots+\frac{2548}{2549}$ in term of $k$.

I've tried to write the terms of $S$ in the following way:

\begin{align}
\frac{1275}{1276} & =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{1275\times1276} \\
\frac{1276}{1277} & =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{1276\times1277} \\
& \,\,\,\vdots \\
\frac{2548}{2549} & =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{2548\times2549}
\end{align}

But it's just not enough, because in $k$ we don't have $\frac{1}{2\times3}$, $\frac{1}{4\times5}$, etc.

Can you please help me solve the problem? Thanks!

Your sums are $k_{1275}$ and $S_{1275}$ where $$\cases{k_n=\sum_{i=1}^n\frac1{(2i-1)(2i)}\\S_n=\sum_{i=n}^{2n-2}\frac{i}{i+1}}$$

You can prove using induction that
$$\forall n\geq 2,S_n=n-1+\frac1{2n}-k_n.$$

Note that your sum is $$\sum_{n=1}^{1275}{\frac{1}{2n(2n-1)}}$$
We can rewrite this using partial fractions as:
$$\sum_{n=1}^{1275}{\bigg[\frac{1}{2n-1}-\frac{1}{2n}\bigg]}$$
and use telescoping from there.


Prove 3 vectors are collinear I am asked to prove A(2,4), B(8,6), C(11,7) are collinear using vectors.

I can work AB by subtracting A from B and BC by subtracting B from C in vector form.

I can say that BC = 2AB.

But I don't understand why this proves they are collinear.

If $A=(2,4), \; B=(8,6), \; C=(11,7)$, then
$$
\vec{AB} = B - A = (6,2) \quad ; \quad
\vec{BC} = C - B = (3,1)
$$
So $\vec{AB} = 2 \vec{BC}$, which is different from your conclusion.

Anyway, this says that $\vec{AB}$ and $\vec{BC}$ are parallel, since one is just a multiple of the other. So, to get from $B$ to $C$, you go in the same direction you went to get from $A$ to $B$ (no turn is required). This means that $A$, $B$, $C$ must all lie on the same line.

Two vectors are colinears (and so your the end points) if, and only if there one is multiple of other.


How to evaluate $\lim_{n \to \infty} \sum_{i=1}^n \arcsin{(u \cdot i/n)}$ without Riemann-sums How to evaluate the limit $$\lim_{n \to \infty} \sum_{i=1}^n \dfrac{1}{n}\arcsin{(u \cdot i/n)}$$
Where $0 < u \leq 1$. There is a closed form expression, but I just don't know how to find it.

Without Riemann-summs.

Note that

$$\dfrac{1}{n}\arcsin{(u \cdot i/n)}=i\cdot \left(\frac{u}{n^2}+o\left(\frac1{n^2}\right)\right)$$

then

$$\sum_{i=1}^n \dfrac{1}{n}\arcsin{(u \cdot i/n)}=\left(\frac{u}{n^2}+o\left(\frac1{n^2}\right)\right)\sum_{i=1}^n i =\frac{u}{n^2}\frac{n(n+1)}2+o\left(\frac1{n^2}\right)\frac{n(n+1)}2=$$

$$=\frac{u}{2}\frac{n(n+1)}{n^2}+\frac{o\left(\frac1{n^2}\right)}{\frac1{n^2}}\frac{n(n+1)}{2n^2}\to \frac u 2$$


Profit or Loss Justification What is the justification of the following two sentences:

1.If an article is sold at a gain of 35% then its selling price is 135% of Cost price.

If an article is sold at a loss of 35% then its selling price is 65% of cost price.

Remember that the base value is $1$ times of its original value, which translates to $100$% of the original value. We can say this because by definition, $100$ lots of $1$ percent make up a whole part (equal to $1$).

Therefore, a percentage gain of $35$% results in $100$% + $35$% = $135$%,
and a percentage loss of $35$% results in $100$% - $35$% = $65$%.

I am not sure what justification you want here but If ideally you consider all the other expenses to be null(i. e transporting, maintaining, etc.)  or consider the cost price to include all the other expenses only then are the above statements justified...


Maximum term of (a + b) ^ n I would like a demonstration of the fact below.

Being given real numbers a and b (nonzero) and a positive integer n, the order p, that occupies the maximum term (in absolute value)
 of the development of power (a+b)^ n, according to decreasing powers of a is given by:

p = 1 + integer part of [|b|(n+1)/(|a|+|b|)]

When n is integer, there are maximum two terms: those of order p and p-1.

|a| and |b| are the modules of the numbers a and b, respectively.

Already, very grateful.

Paulo Argolo

Rio de Janeiro, Brazil

Let $f(k)=C_n^k |a|^{n-k} |b|^k$.

Then $f(k+1)/f(k)=\frac{|b|}{|a|} \frac{n-k}{k+1} > 1$ iff $k<\frac{|b|(n+1)}{(|b|+|a|)}-1$, so $f(k)$ is increasing until $k=I\left(\frac{|b|(n+1)}{(|b|+|a|)}\right)$ where $I$ denotes the integral part.

Every positive integer is of the form 3k-2,3k-1, or 3k.

If n is of the form 3k-2, then ak has the maximum value
If n is of the form 3k-1, then ak = a[k+1] have the maximum value.
If n is of the form 3k, then a[k+1] has the maximum value.

It's not hard to prove each of those.  Write out the terms
with the factorials and powers of 2 for positive j<=k and prove:
a[k-j] < ak {and/or a[k+1]}  and ak {and/or a[k+1]} > a[k+1+j]

Edwin McCravy


Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$ Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$

My work so far:

1) $$y =\frac{2x+9}{\sqrt{-x^2+4x+12}+\sqrt{-x^2+2x+3}} $$

2) I used a derivative and found the answer ($y=\sqrt3$ at $x=0$). Is there any other way?

Rewriting the function that way only complicates things.

First let's determine the domain:
\begin{cases}
-x^2+4x+12\ge0\\
-x^2+2x+3\ge0
\end{cases}
reduces to $-1\le x\le3$.

The derivative is
$$
y'=
\frac{-x+2}{\sqrt{-x^2+4x+12}}-
\frac{-x+1}{\sqrt{-x^2+2x+3}}
$$
(which is undefined at $-1$ and $3$) and we want to see where it vanishes, that is,
$$
(2-x)\sqrt{-x^2+2x+3}=(1-x)\sqrt{-x^2+4x+12}
$$
We need either $-1<x<1$ or $2<x<3$ so that both terms are either positive or negative. Now we can square safely:
$$
(2-x)^2(-x^2+2x+3)=(1-x)^2(-x^2+4x+12)
$$
becomes $12x^2-16x=0$, so the only critical point is $0$ (because $4/3$ doesn't satisfy the above limitations).

We also have
$$
f(-1)=\sqrt{7},\qquad
f(0)=\sqrt{3},\qquad
f(3)=\sqrt{15}
$$
Since maxima and minima are at critical points or at the extremes of the domain, we can conclude.

find the first derivative $dy/dx$ and set it equal to zero. Study the variation in case you are looking for a "local" minimum.


Is this a correct characterisation of natural equivalence? In the category of functors of type $\mathcal{C}\rightarrow \mathcal{D}$ an arrow $\psi: F \Rightarrow G$ is defined as a collections of $\left(\psi_X: FX \rightarrow GX| X:\mathcal{C} \right)$ of compenents in $\mathcal{D}$, such that $\psi_Y \circ FX = GY \circ \psi_X$ and in case all the components are invertible, $\psi$ is called a natural equivalence or natural isomorphism of $F$ and $G$. My question is whether or not:

  There exists a natural equivalence of $F,G: \mathcal{C}\rightarrow \mathcal{D}$, iff $FX \simeq GY \mbox{ in } \mathcal{D} \mbox{ whenever } X \simeq Y \mbox{ in }\mathcal{C}$

in other words:
$$F\simeq G \mbox{ in } \mathcal{C}^{\mathcal{D}} \Leftrightarrow FX \simeq GY \mbox{ in } \mathcal{D} \mbox{ whenever } X \simeq Y \mbox{ in } \mathcal{C}$$

Thanks

$FX\simeq GX$ for all objects $X$ doesn't imply that $F\simeq G$ as functors.

For example, let $\mathcal{C}$ be the category with two objects $X$ and $Y$, and only one morphism other than the identity morphisms, say $\alpha:X\to Y$. Then there are two functors $F$ and $G$ from $\mathcal{C}$ to (say) the category of real vector spaces with $FX=GX=FY=GY=\mathbb{R}$, $F\alpha=\operatorname{id}_{\mathbb{R}}$ and $G\alpha=0$, which are not isomorphic as functors.

The answer is no.

From left to right:

For all $X \simeq Y \mbox{ in } \mathcal{C}$

$$F\simeq G \mbox{ in } \mathcal{C}^{\mathcal{D}} \Rightarrow FX \simeq FY \simeq GY \mbox{ in } \mathcal{D} $$

From right to left however:

For all $X \mbox{ in } \mathcal{C}$

$$ X \simeq X \mbox{ in } \mathcal{C}  \Rightarrow FX \simeq GX \mbox{ in } \mathcal{D}$$

but this is not enough to imply that $ F\simeq G \mbox{ in } \mathcal{C}^{\mathcal{D}}$
as Jeremy Rickard shows above


Sequence And Series - How To Find The Limit Of A Converging Sequence With No Common Ratio I have the following series
3/2, 4/3, 5/4, 6/5....

the nth term could be expressed as n+2/n+1.

I cant seem to work out how to get the limit of this function as most of the books describe the sum to infinity as a/1 -r.

With a = 3/2 and being the first term, I am confused as to what r should equal.  As this sequence does not seem to have a common ratio.  I thought that r would be the common ratio.

Maybe I am going down the wrong line of thinking so please clarify how to find the limit and the r variable in the equation.

Many Thanks

Suppose that the $n$-th term of a certain sequence is $\dfrac{n+2}{n+1}$.  Note that
$$\frac{n+2}{n+1}=1+\frac{1}{n+1}.$$
 As $n\to\infty$, the $\dfrac{1}{n+1}$ part approaches $0$, so our limit is $1$.

Or else we can divide top and bottom by $n$, obtaining
$$\frac{n+2}{n+1}=\frac{1+\frac{2}{n}}{1+\frac{1}{n}}.$$
As $n\to\infty$, $1+\dfrac{2}{n}\to 1$, and $1+\dfrac{1}{n}\to 1$, so our limit is $1$.

Remark: One can get useful information from the calculator. If you are interested in finding the limit as $n\to\infty$ of $a_n$, you can calculate $a_n$ for a few largish $n$. For example, ca;culate $\dfrac{n+2}{n+1}$ for $n=1000$ and for $n=10000$. It will now be plausible that the limit might be $1$. And once one knows that the "answer" should be $1$, it becomes easier to show that it is $1$.

for a seaquency Un..a limit L is such that for its value there is a positive integer N such that n>N..so a sequency only has this limit L if there are still some integer values of n falling in this lane


Pedagogy: How to cure students of the "law of universal linearity"? One of the commonest mistakes made by students, appearing at every level of maths education up to about early undergraduate, is the so-called “Law of Universal Linearity”:

$$ \frac{1}{a+b} \mathrel{\text{“=”}} \frac{1}{a} + \frac{1}{b} $$

$$ 2^{-3} \mathrel{\text{“=”}} -2^3 $$

$$ \sin (5x + 3y) \mathrel{\text{“=”}} \sin 5x + \sin 3y$$

and so on.  Slightly more precisely, I’d call it the tendency to commute or distribute operations through each other. They don't notice that they’re doing anything, except for operations where they’ve specifically learned not to do so.

Does anyone have a good cure for this — a particularly clear and memorable explanation that will stick with students?

I’ve tried explaining it several ways, but never found an approach that I was really happy with, from a pedagogical point of view.

I think this is a symptom of how students are taught basic algebra. Rather than being told explicit axioms like $a(x+y)= ax+ay$ and theorems like $(x+y)/a = x/a+y/a,$ students are bombarded with examples of how these axioms/theorems are used, without ever being explicitly told: hey, here's a new rule you're allowed to use from now on. So they just kind of wing it. They learn to guess.

So the solution, really, is to teach the material properly. Make it clear that $a(x+y)=ax+ay$ is a truth (perhaps derive it from a geometric argument). Then make it clear how to use such truths: for example, we can deduce that $3 \times (5+1) = (3 \times 5) + (3 \times 1)$. We can also deduce that $x(x^2+1) = xx^2 + x 1$. Then make it clear how to use those truths. For example, if we have an expression possessing $x(x^2+1)$ as a subexpression, we're allowed to replace this subexpression by $x x^2 + x 1.$ The new expression obtained in this way is guaranteed to equal the original, because we replaced a subexpression with an equal subexpression.

Perhaps have a cheat-sheet online, of all the truths students are allowed to use so far, which is updated with more truths as the class progresses.

I think that, if you teach in this way, students will learn to trust that if a rule (truth, whatever) hasn't been explicitly written down, then its either false, or at the very least, not strictly necessary to solve the problems at hand. This should cure most instances of universal linearity.

I would say that the main causation, the underlying disease, is the fact that, here in the U.S. we often have too many students, all crammed into a small room, that lack a love for math. Students are worried about grades, not the inner machinations of mathematics. Teach them to love maths and they ought to take care of the rest on their own.


Probability of printing characters A printing machine capable of printing $n$ characters $A_1,A_2,\dotsc,A_n$ is operated by feeding electronic impulse. There is a distinct impulse designed to produce each character. One impulse is chosen at random and fed into the machine twice and found that

                1.A_1 is printed twice
                2.First A_1 is printed than A_2

Suppose $p$ is the probability of printing the right character.Then find the probability that the impulse designed to produce $A_1$ was fed into the machine.

Here's an answer to (1), (2) is very similar. First note that each impulse is capable of producing any character, I assume with a probability of $\frac{1-p}{n-1}$. So the probability of impulse k producing character k is $p^2$. k then not k needs breaking down into component parts, so for example (k,1) is $p\frac{1-p}{n-1}$, and similarly (1,1) is $\frac{1-p}{n-1}\times\frac{1-p}{n-1}$. We can re-sum these to prove the answer is $1$. Now we only need to add the probability the impulse is $k=1$ and it fires correctly twice  ($p^2$), plus the impulse is not $k=1$ and it misfires twice: $n-1 \times\frac{1-p}{n-1}\times\frac{1-p}{n-1} = (1-p)\times\frac{1-p}{n-1} = \frac{1-2p+np^2}{n-1}.$


How to prove this inequality $\sum_{i=1}^{n}\left(x^k_{i}\ln{x_{i}}\ln{\frac{x_{i}}{n}}\right)\le 0$ Let $x_{i}\ge 0$ for $i\in\{1,2,\cdots,n\}$ and $x_{1}+x_{2}+\cdots+x_{n}=n$ for $n\ge 3$

  Show that for all strictly positive integers $k\ge2$ the following inequality holds :
  $$\sum_{i=1}^{n}x^k_{i}\ln{x_{i}}\ln{\dfrac{x_{i}}{n}}\le 0$$

We consider
$$f(x)=x^k\ln{x}\ln{\dfrac{x}{n}}$$
Then
$$f'(x)=kx^{k-1}\ln{x}\cdot\ln{\dfrac{x}{n}}+x^{k-1}\ln{\dfrac{x}{n}}+\dfrac{x^{k-1}}{n}\ln{x}$$
$$\Longrightarrow
f''(x)=k(k-1)x^{k-2}\cdot\ln{x}\cdot\ln{\dfrac{x}{n}}+kx^{k-2}\ln{\dfrac{x}{n}}+\dfrac{kx^{k-2}}{n}\ln{x}+(k-1)x^{k-2}\ln{\dfrac{x}{n}}+\dfrac{x^{k-2}}{n}+\dfrac{k-1}{n}x^{k-2}\ln{x}+\dfrac{x^{k-2}}{n}.$$
Unfortunatly I can't know  the sign of $f''(x)$ because I want to use Jensen's Inequality to prove it.

So how can we prove this inequality ?

Firstly, you diffrentiated it wrongly:
$$f(x)=x^k\ln{x}\ln{\dfrac{x}{n}}$$
$$f'(x)=kx^{k-1}\ln{x}\ln{\dfrac{x}{n}}+x^{k-1}\ln{\dfrac{x}{n}}+x^{k-1}\ln{x}$$
$$f'(x)=x^{k-1}\left(k\ln x\ln\frac xn+\ln \frac xn+\ln x\right)$$
$$f''(x)=(k-1)x^{k-2}\left(k\ln x\ln\frac xn+\ln \frac xn+\ln x\right)+x^{k-1}\left(\frac  kx\ln\frac xn+\frac kx\ln x+\frac1x+\frac1x\right)$$
$$f''(x)=(k-1)x^{k-2}\left(k\ln x\ln\frac xn+\ln \frac xn+\ln x\right)+x^{k-2}\left(k\ln\frac xn+k\ln x+2\right)$$
$$f''(x)=x^{k-2}\left(k(k-1)\ln x\ln\frac xn+(k-1)\ln\frac xn+(k-1)\ln x+k\ln\frac xn+k\ln x+2\right)$$
$$f''(x)=x^{k-2}\left(k(k-1)\ln x\ln\frac xn+(2k-1)\ln\frac {x^2}n+2\right)$$

As $k>2$, $x^{k-2}>0$ Also $x<n$.See this graph:

I think you need to look at  $\ln\frac{x_i}{n}$ in your equation. Since, $$x_1 + x_2 + \dots + x_n = n$$
therefore $$x_i < n$$
and notice that $$\ln\frac{x_i}{n} = \ln(x_i) - \ln(n) \implies \ln\frac{x_i}{n} < 0$$
hence the expression $${x_i}^k\ln{x_i}\ln\frac{x_i}{n} < 0$$
as ${x_i}^k$ and $\ln{x_i}$ are going to be positive. Hence the sum of all negative numbers is going to be less than zero.


Find the limit : $\lim\limits_{n\rightarrow\infty}\int_{n}^{n+7}\frac{\sin{x}}{x}\,\mathrm dx$ I have this exercise I don't know how to approach :

  Find the limit : $$\lim_{n\rightarrow\infty}\int_{n}^{n+7}\frac{\sin x}{x}\,\mathrm dx$$

I can see that with $n\rightarrow\infty$ the area under the graph of this function becomes really small as $\sin{x} \leq 1$ so $\dfrac{\sin{n}}{n}\rightarrow_{\infty}0$ but can I get something from it?

Hint: $\def\abs#1{\left|#1\right|}$
$$\abs{\int_n^{n+7}\frac{\sin x} x \, dx}\le \int_n^{n+7}\frac{\abs{\sin x}}{x}\, dx\le \frac 1n \int_n^{n+7}\abs{\sin x}\, dx $$

No I don't think that there is a lot that can be got out of it. For example the result is not yet sufficient condition for convergence of the integral from $0$ to $\infty$. However, the Riemann improper integral of the first kind
\begin{equation}
\int_0^\infty \textrm{sinc}(x/\pi) dx
\end{equation}
where $\textrm{sinc}(x) = \Bigg\{ \begin{eqnarray} & \frac{\sin(\pi x)}{\pi x} & , \ \ x \neq 0 \\ &1&, \ \ x = 0 \end{eqnarray}$
converges as we know. Define
\begin{equation}
a_n = \int_0^n \textrm{sinc}(x/\pi) dx \ .
\end{equation}
Now $\{a_n\}_{n=1}^\infty$ is a Cauchy-sequence. The integral under the limit can be expressed as follows.
\begin{eqnarray}
\int_n^{n+7} \frac{\sin(x)}{x} dx & = & \int_n^{n+7} \textrm{sinc}(x/\pi) dx = \int_0^{n+7} \textrm{sinc}(x/\pi) dx - \int_0^n \textrm{sinc}(x/\pi) dx \\
\\
\\
\\
& = & a_{n+7} - a_n \rightarrow 0
\end{eqnarray}
as $n \rightarrow \infty$.

Without the knowledge that the integral from $0$ to $\infty$ converges we can proceed as follows. We first show that $a_n$ is a Cauchy-sequence. We may assume $m \leq n$. Otherwise we can apply $|a_n-a_m| = |a_m-a_n|$ and interchange the order of $m$ and $n$ in the first term of the following calcualtion. We have
\begin{eqnarray}
|a_n-a_m| & = & \bigg| \int_0^n \textrm{sinc}(x/\pi)dx - \int_0^m \textrm{sinc}(x/\pi)dx \bigg| = \bigg| \int_m^n \textrm{sinc}(x/\pi)dx \bigg| \\
& = & \bigg| \int_m^n \frac{\sin(x)}{x}dx \bigg| = \bigg| \bigg|_m^n \frac{-\cos(x)}{x} - \int_m^n -\frac{-\cos(x)}{x^2} dx \bigg| \\
& = & \bigg|-\frac{\cos(n)}{n}+\frac{\cos(m)}{m}-\int_m^n \frac{\cos(x)}{x^2} dx \bigg| \\
& \leq & \bigg|-\frac{\cos(n)}{n} \bigg| + \bigg| \frac{\cos(m)}{m} \bigg| + \bigg| \int_m^n \frac{\cos(x)}{x^2} dx \bigg| \\
& \leq & \bigg|\frac{\cos(n)}{n} \bigg| + \bigg| \frac{\cos(m)}{m} \bigg| + \int_m^n \bigg| \frac{\cos(x)}{x^2} \bigg| dx \\
& \leq & \frac{|\cos(n)|}{|n|} + \frac{|\cos(m)|}{|m|} + \int_m^n \frac{|\cos(x)|}{|x^2|} dx \\
& \leq & \frac{1}{n} + \frac{1}{m} + \int_m^n \frac{1}{x^2} dx \leq \frac{1}{n} + \frac{1}{m} + \bigg|_m^n -\frac{1}{x} \\
& \leq & \frac{1}{n} + \frac{1}{m} - \frac{1}{n} + \frac{1}{m} \leq \frac{1}{n} + \frac{1}{m} + \frac{1}{n} + \frac{1}{m} = \frac{2}{n} + \frac{2}{m} \rightarrow 0
\end{eqnarray}
as $m, \ n \rightarrow \infty$. Hence $a_n$ is a Cauchy-sequence. We can then calculate as above after the definition of $a_n$ and obtain the desired result
\begin{equation}
\lim_{n \rightarrow \infty} \int_n^{n+7} \frac{\sin(x)}{x} dx = 0 \ .
\end{equation}
The existence of the integral from $0$ to $\infty$ can be now established by calculating the distance $|a_M-a|$ where $a$ is the limit of the Cauchy-sequence. For the next integer $m$ from $M$ we have
\begin{equation}
|a_M-a| = |a_M-a_m+a_m-a| \leq |a_M-a_m| + |a_m-a|
\end{equation}
the second term goes to $0$ by definition of limit and the first can be estimated as you thought in your question. Hence
\begin{equation}
a = \lim_{M \rightarrow \infty} a_M = \lim_{m \rightarrow \infty} \int_0^M \textrm{sinc}(x/\pi) dx = \int_0^\infty \textrm{sinc}(x/\pi) dx \ .
\end{equation}
Hopefully you got some perspective to the integration of $\frac{\sin(x)}{x}$.

Then the exercise. We calculate using the previous estimate
\begin{eqnarray}
\Bigg|\int_n^{n+7} \frac{\sin(x)}{x} dx \Bigg| \leq \frac{1}{n+7} + \frac{1}{n} - \frac{1}{n+7} + \frac{1}{n} = \frac{2}{n} \rightarrow 0 \ ,
\end{eqnarray}
as $n \rightarrow \infty$. Note that we succeed to get rid of the $7$ in the question. Another approach is to estimate
\begin{eqnarray}
\sup_{x \in [0,7]} \Bigg|\frac{\sin(x+n)}{x+n} - 0 \Bigg| \leq \sup_{x \in [0,7]} \frac{1}{n} = \frac{1}{n} \rightarrow 0 \ ,
\end{eqnarray}
as $n \rightarrow \infty$. Hence $\frac{\sin(x+n)}{x+n}$ converges uniformly to $0$. Hence
\begin{eqnarray}
\lim_{n \rightarrow \infty} \int_n^{n+7} \frac{\sin(x)}{x} dx = \lim_{n \rightarrow \infty} \int_0^{7} \frac{\sin(x+n)}{x+n} dx = \int_0^7 \lim_{n \rightarrow \infty} \frac{\sin(x+n)}{x+n} dx = \int_0^7 0 dx = 0 \ ,
\end{eqnarray}
where the order of integration and limit is changed by uniform convergence of functions defined on the compact set $[0,7]$.


Adherent point and limits I am reading Terence Tao Analysis I. I am confused in his remark 9.3.11. Can someone clarify this comment. If someone could also give me an example of a set that doesn't have any adherent point (This is probablyI don't understand remark 9.3.11 given in the image where the confusion arises)?

You are misunderstanding Tao. What he is saying is that he will not defined $\lim_{x\to x_0}f(x)$ when $x_0$ is not an adherent point of $X$. For instance, if $X=[0,1]$, he does not define the meaning of $\lim_{x\to2}f(x)$ (note that $2$ is not an adherent point of $[0,1]$).

An adherent point to a set is either an accumulation point or an isolate point.
If $ a$ is isolate , then we cannot define $\lim_{x\to a}f(x)$ .
For example, if $ D_f=E=[0,1]\cup \{2\}$, then $ 2$ is adherent but not an accumulation. So, it is not intetesting to speak about $\lim_{x\to 2}f(x)$.
Each element of a set is adherent to this set. Every non empty set contains adherent points.


Unsolved Problems due to Lack of Computational Power I was recently reading up about computational power and its uses in maths particularly to find counterexamples to conjectures. I was wondering are there any current mathematical problems which we are unable to solve due to our lack of computational power or inaccessibility to it.

What exactly am I looking for?

Problems of which we know that they can be solved with a finite (but very long) computation?

(e. g. NOT the Riemann hypothesis or twin prime conjecture)

I am looking for specific examples.

Goldbach's weak conjecture isn't a conjecture anymore, but before it was proved (in 2013), it had already been proved that it was true for every $n>e^{e^{16\,038}}$. It was not computationally possible to test it for all numbers $n\leqslant e^{e^{16\,038}}$ though.

Prime gaps - https://en.wikipedia.org/wiki/Prime_gap - of maximum known merit are found by increasing CPU (and GPU) computational power on the gapcoin network. See: https://gapcoin.club

  ... So the difficulty will simply be the length of the prime gap?

  Not exactly. The average length of a prime gap with the starting prime
  p, is log(p), which means that the average prime gap size increases
  with lager primes. Then, instead of the pure length, we use the merit
  of the prime gap, which is the ratio of the gap's size to the average
  gap size.

  Let p be the prime starting a prime gap, then m = gapsize/log(p) will
  be the merit of this prime gap.

  Also a pseudo random number is calculated from p to provide finer
  difficulty adjustment.

  Let rand(p) be a pseudo random function with 0 less than rand(p) less
  than 1. Then, for a prime gap starting at prime p with size s, the
  difficulty will be s/log(p) + 2/log(p) * rand(p), where 2/log(p) is
  the average distance between a gap of size s and s + 2 (the next
  greater gap) in the proximity of p.

  When it actually comes to mining, there are two additional fields
  added to the Blockheader, named “shift” and “adder”. We will calculate
  the prime p as sha256(Blockheader) * 2^shift + adder. As an additional
  criterion the adder has to be smaller than 2^shift to avoid that the
  PoW could be reused. ...

Source: gapcoin.org

...

https://web.stanford.edu/~tonyfeng/bounded_gaps.pdf

  ... (Twin prime conjecture) ...

  ... This  is  a  just  a  special  case  of  a  far-reaching  conjecture
  of  Hardy  and  Lit- tlewood  describing  the  frequency  of  prime
  gaps  of  any  sizes.  The  Hardy- Littlewood conjecture predicts not
  only how often twin primes occur, but also how often any finite
  tuple of the form ( n
  + h 1 , n
  + h 2 , . . . , n
  + h k ) consists en- tirely of prime numbers. But even though analytic number theorists have be- lieved for many years that they know the
  answers to these questions, progress towards proving the existence of
  small gaps between primes has been slow. As recently as 2005 , the
  problem of establishing infinitely many bounded gaps be- tween primes
  was considered by many mathematicians to be “hopeless” ...

...

Further,

List of unsolved problems in mathematics:
https://en.wikipedia.org/wiki/List_of_unsolved_problems_in_mathematics


Confusion in Argand plane When I say a argand plane , in my textbook it says the plane having a complex number assigned to each of its point. I interpreted a meaning for this statement down below , let me know if it’s correct
Second is does it mean that when I talk about a plane. It is like x and z axis is one plane , x and y axis , y and z axis . So , it has to be 2D.  x,y and z axis all together can’t be an argand plane. Is this correct ?
Confusion in ordered pairs :

So , there is no squaring of numbers done here like R ^2 as written . That is my confusion.

The expression “Argand plane” simply means the set $\Bbb C$ of complex numbers, when we see it as a plane, that is, when we identify each complex number $a+bi$ ($a,b\in\Bbb R$) with the point $(a,b)\in\Bbb R^2$. So, yes, it is a $2$-dimensional thing.

It says Argand plane, not any volume, is n't it?
The Argand diagram  contains points  $z= x+iy$  and lines in the  the complex plane.


For which $a$ and $b$ is this matrix diagonalizable? For which $a$ and $b$ is this matrix diagonalizable?

$$A=\begin{pmatrix} a & 0 & b \\ 0 & b & 0 \\ b & 0 & a  \end{pmatrix}$$

How to get those $a$ and $b$? I calculated eigenvalues and eigenvectors, but don't know what to do next?

That is a symmetric matrix: it is always diagonalizable over a field with characteristic$\;\neq2\;$. Not only that: it is orthogonally diagonalizable.

Matrix A has 3 eigenvalues a-b, a+b and b. In order for them to be distinguishable, following conditions should be met. $$b \ne 0$$ $$a \ne 0$$ $$a \ne 2b$$


$\lim_{n\to\infty} \frac{(\log n)^n}{n^{\log n}}$ $$\lim_{n\to\infty} \frac{(\log n)^n}{n^{\log n}}$$

This limit is equal to $+\infty$ but I don't understand why. I know that every exponential function with a base greater than 1 grows to infinity faster than any polynomial. But is this true only for a fixed base exponential that doesn't depend on the variable?

Or is it that, approaching positive infinity, the function at the numerator behaves like $a^n, a > 1$ in terms of comparison with a polynomial?

Sorry if this is a silly question.

By taking the logarithm
$$\log\left(\frac{(\log n)^n}{n^{\log n}}\right)=n\log\log n-\log^2 n$$

and then rewriting
$$n\log\log n-\log^2 n= n\left(\log\log n-\frac{\log^2n }{n}\right)$$
In the parenthesis, the second term goes to $0$. What happens to the first?


Number of polynomials of degree less than 4 satisfying 5 points Let polynomial $P(x)$ have the property that $P(1),$ $P(2),$ $P(3),$ $P(4)$ and $P(5)$ are equal to $1$, $2$, $3$, $4$, $5$ in some order. How many possibilities are there for the polynomial $P,$ given that the degree of $P$ is strictly less than $4$?

By Lagrange interpolation, given the values of $P(1)$, $P(2)$, $P(3)$, $P(4)$, and $P(5)$, there is a unique polynomial of degree $\leq 4$ taking those values.  So the question is, for which permutations of the numbers $1$ through $5$ will the resulting $P(x)$ actually have degree $<4$?  To determine this, we can look at the actual explicit formula for Lagrange interpolation and see what the coefficient of $x^4$ will be in terms of our five values.  That formula is
$$\begin{align*}P(x)=P(1)\frac{(x-2)(x-3)(x-4)(x-5)}{(1-2)(1-3)(1-4)(1-5)}&+P(2)\frac{(x-1)(x-3)(x-4)(x-5)}{(2-1)(2-3)(2-4)(2-5)}\\
&+P(3)\frac{(x-1)(x-2)(x-4)(x-5)}{(3-1)(3-2)(3-4)(3-5)}\\
&+P(4)\frac{(x-1)(x-2)(x-3)(x-5)}{(4-1)(4-2)(4-3)(4-5)}\\
&+P(5)\frac{(x-1)(x-2)(x-3)(x-4)}{(5-1)(5-2)(5-3)(5-4)},
\end{align*}$$
so the coefficient of $x^4$ will be $$\frac{P(1)}{24}-\frac{P(2)}{6}+\frac{P(3)}{4}-\frac{P(4)}{6}+\frac{P(5)}{24}=\frac{P(1)-4P(2)+6P(3)-4P(4)+P(5)}{24}.$$  So we get a polynomial of degree $<4$ iff $$P(1)+6P(3)+P(5)=4(P(2)+P(4)).$$

Now we just have some casework to consider.  If $P(3)=1$, the RHS will be at least $4(2+3)=20$ and the LHS is at most $5+6+4=15$, so there are no solutions.  Since the problem is symmetric with respect to conjugating by $x\mapsto 6-x$, there are no solutions with $P(3)= 5$ either.

If $P(3)=2$, then $P(1)+P(5)$ is divisible by $4$, which means $P(1)$ and $P(5)$ must be $1$ and $3$ (in some order) or $3$ and $5$ (in some order).  The equation will hold in the second case but not the first case.  This gives four different solutions, since you can swap the two values of $P(1)$ and $P(5)$, and also the two values of $P(2)$ and $P(4)$.  Again, by symmetry there are four more solutions if $P(3)=4$.

Finally, suppose $P(3)=3$.  Then $P(1)+P(5)$ must be $2$ mod $4$, so $P(1)$ and $P(5)$ must be $1$ and $5$ (in some order) or $2$ and $4$ (in some order).  Both cases work, and each gives four solutions.  Thus there are eight solutions total if $P(3)=3$.

Taking all the cases together, then, we find there are sixteen different solutions.

Some closing remarks:  It's not a coincidence that the coefficients of the values of $P$ in the equation we got are binomial coefficients; you can see this by noticing that the denominator in the Lagrange interpolation term with $P(n)$ is $\pm(n-1)!(5-n)!$ (grouping the positive and negative factors together), and this generalizes if you replace $5$ by another positive integer.  Still, it would be nice to have a more conceptual explanation for why we're getting binomial coefficients.  It would also be nice to have a more conceptual explanation for how to count the solutions (in particular, something that would generalize if you replaced $5$ by any positive integer).

There are only one polynomial of degree 1 . There are three cases

  Case 1 the degree of the polynomial is 1
  If we let the degree of the polynomial is 1 then let the polynomial is P(x)=ax+b . Then put the values & solve it. The polynomial became P(x)=x .
  Case 2 the degree of the polynomial is 2.
  If the degree of the polynomial is 2 then let the polynomial is $P(x)=ax^2+bx+c$ . Again solve this you get the previous polynomial P(x)=x
  Case 3 the degree of the polynomial is 3
  Similarly you get the polynomial P(x)=x .
  So there exist only one polynomial.


Question on Differential Equations of a Family of Circles Q. The differential equation of the family of circles with fixed radius 5 units and centre on the line y = 2 is?

My Attempt:
I can write the equation of a random circle, satisfying the condition mentioned in the question, as:
$$(x – h)^2 + (y – 2)^2 = 25$$

Differentiating, with respect to x,

$$2(x – h) + 2(y – 2) \frac{dy}{dx}= 0$$

I am not sure how to proceed further.

Can you eliminate the arbitrary constant $h$ from your system of equations? That will give you the answer.

Rearrange your equation:

$$2(y-2)\frac {dy}{dx}=-2(x-h)$$

$$\frac {dy}{dx}=-\frac{(x-h)}{(y-2)}$$


An inequality with determinant Let $A$ be an invertible $n \times n$ matrix such that all the elements of $A$ and $A^{-1}$ are integers. Prove that if all the eigenvalues of $A$ are real numbers then:
$$|\det(A + A^{-1})| \ge 2^n$$
I've already known that if $A$ is a diagonalizable matrix with eigenvalues $\lambda_1$, $\lambda_2$,... $\lambda_n$ then we can prove that $|\det(A + A^{-1})| = |(\lambda_1 + \lambda_1^{-1})...(\lambda_n + \lambda_n^{-1})| \ge 2^n$ but I still cannot solve the problem, could somebody please give me some help? Thank you.

$A$ may not be diagonalizable, but it is triangularizable over $\mathbb C$. Since you know that all the eigenvalues of $A$ are real numbers, you get the same eigenvalues when considering $A$ as a matrix with complex entries, thus $A$ is similar to some $$ T=\left(
    \begin{array}{ccccc}
    \lambda_1                                   \\
      & \lambda_2             &   & \huge *\\
      &               & \ddots                \\
      & \large 0 &   & \lambda_{n-1}            \\
      &               &   &   & \lambda_n
    \end{array}
    \right)$$
that is $A=PTP^{-1}$ where $P\in Gl_n(\mathbb C)$. $T^{-1}$ is upper triangular as well, with the form
$$ T^{-1}=\left(
    \begin{array}{ccccc}
    \lambda_1^{-1}                                    \\
      & \lambda_2 ^{-1}            &   & \huge **\\
      &               & \ddots                \\
      & \large 0 &   & \lambda_{n-1} ^{-1}           \\
      &               &   &   & \lambda_n^{-1}
    \end{array}
    \right)$$
so that $A^{-1}=PT^{-1}P^{-1}$, which yields $|\det(A + A^{-1})| = |(\lambda_1 + \lambda_1^{-1})...(\lambda_n + \lambda_n^{-1})|$.

\begin{eqnarray*}
\lambda+ \frac{1}{\lambda} =\left( \sqrt{\lambda} - \frac{1}{\sqrt{\lambda}} \right)^2+2 \geq 2.
\end{eqnarray*}


Number of moves necessary to solve Rubik's cube by pure chance Suppose, random moves are made to solve Rubik's cube. A move consists of
a $90$-degree-rotation of some side. The starting position is also random.

What is $E(X)$, where $X$ is the number of moves until the cube is solved ?
How many moves must be made, that the probability that the cube is solved, exceeds $99$% ?

Each edge has 2 faces, providing for $12!\times 2^{12}$ possibilities.

Each vertex has 3 faces, providing for $8!\times 3^8$ possibilities.

Therefore if you dismantle the cube it can be reassembled in 519,024,039,293,878,272,000 possible states.

Thanks to Gerry Myerson for this: According to Singmaster's Notes on Rubik's Magic Cube, the total permutation of the edge and vertex cubes must be even.  Then, if you orient all but one of the edges, the remaining one is forced, and similarly for the vertices, so only 1 in 12 possible assemblies of the cube is actually a "solvable" or "arrivable at by rotation" state.

Dividing the above number by 12 therefore confirms Turner and Gold's result (which i haven't seen first hand) that the number of states "arrivable at by rotation" is $43,252,003,274,489,856,000$.

Therefore any given random move has probability $1/43,252,003,274,489,856,000$ of completing the cube.

For such high $n$ the binomial will approximate the normal distribution, so the expected number of moves will be the mean.

I think Wolfram Alpha struggled with the number-crunching in my last answer so I've solved by substitution:

$0.5=\binom{n}{0}p^0q^n$ where $q=(1-1/43,252,003,274,489,856,000)$

$0.5=(1-1/43,252,003,274,489,856,000)^n$

$n\approx 3×10^{19}$ moves is the expected number of moves to solve it.

Repeating the substitution for 0.01:

$n\approx 1.992*10^{20}$ moves is the expected number of moves to reach 99% confidence.

Use of the binomial in this answer makes the technically incorrect assumption that the probability of solving on any particular move is not related to the probability of solving on a move n steps earlier. (i.e. there is no serial correlation). However as an experienced cube user I know that the cube very quickly diverges from its current position given random movements and therefore the effect of serial correlation will be very low.

This divergence is clearly illustrated by the fact that it follows from Rokicki, Kociemba, Davidson, and Dethridge's 2010 result that it takes at most 20 Singmaster moves to take the cube to any obtainable state you wish.  Contrast that 20 moves with the expected number of random moves to solve the cube and you get an idea how minimal serial correlation will be.

The center squares of each face don't actually move independently of each other, they rotate but since each orientation of that square is indistinguishable its the other three, there are $4^6= 32,768$ acceptable finished solutions to the cube, not just 1, considering only the permutations of the pieces relative to each other, which is perhaps not obvious to everybody.

Once solved it can also be held with a choice of 6 faces facing the floor and each of those can be rotated into 4 unique positions, making, depending how you permutate your possibilities, 786,432 possible "correct" solutions.

To eliminate this, if we fix the axes and the centre-faces, which is governed by the manufacturing of the cube anyway, hold one face to the floor, don't rotate any centre squares and don't move any of the axes, we're left with 12 "edges cubes" and 8 "vertex cubes".

Each edge has 2 faces, providing for $12!\times 2^{12}$ possibilities.

Each vertex has 3 faces, providing for $8!\times 3^8$ possibilities.

Therefore ignoring the for all intents and purposes identical, centre squares and orientations of the cube, the cube has 519,024,039,293,878,272,000 possible states.

The probability of solving it on any given random move is therefore 1 in 519 billion billion.

Achieving a probability of $99\%$ is therefore a simple binomial probability... and you may have to correct this bit as it's 25 years since I did  maths:

Our desired result is most simply thought of as the probability of "not having no successes", i.e. $1-P(x=0)$

$0.99=1-\binom{n}{0}p^0q^n$ where $q=(1-1/519024039293878272000)$

$0.01=(1-1/519024039293878272000)^n$

$n=6.4812×10^{14}$ moves, which I'm reliably informed by Wolfram Alpha is 32 moves per red blood cell in your body.

I should add to this answer that it is of course fatally flawed in that it makes no allowance for the fact that the positions of the cube are serially correlated and therefore if one starts in a position which is "more distant" from complete, it is more likely one will cycle repeatedly through those distant permutations than move through permutations having an average likelihood of success.

This has the effect of making "fast" solutions faster on average and "slow" solutions slower than the binomial estimator and therefore the 99% confidence interval, being a distant interval, will typically take a little longer than the $6.4812\times 10^{14}$ moves estimated.  However if you require an accurate answer which makes allowance for that, I suspect you will be waiting for a very long time!

This approximating a normal distribution for large $n$, the estimated number of moves required to solve it will be approximately half of the 99% interval, say $3.2\times 10^{14}$ moves.


How to prove that a torus has the same volume as a cylinder (with the height equal to the torus' perimeter) I want to find the volume of a torus with a given thickness and a given radius.

Let r be the radius of a circle with its midpoint at $M(0|b)$ ($b \geq r$). Now I want to rotate this circle about the x-axis, that is to say about a circular path which has the length $2 \pi \cdot |b|$. So I thought I'd simply integrate:

$V = \int\limits_0^{2 \pi \cdot |b|} \pi r^2 dz = 2 \pi \cdot r^2 \cdot |b|$, which turns out to be the correct result.

However, I don't find it trivial that the volume of this torus is the same as the volume of a cylinder with the corresponding height. I read the article on Wikipedia about the torus and it said that this was due to Cavalieri's theorem, which to my mind doesn't really have a lot to do with the torus vs. the cylinder...

Is there some easy way to prove that a torus has the same volume as a cylinder with the height equal to the torus' perimeter?

(i) Slice the torus into a million near-disks.
(ii) Rotate every second disk through $180^\circ$.
(iii) Stick them all together again.
You get a near-cylinder whose height is nearly $2\pi b$. Now let a million tend to infinity.

Perhaps I don't understand the question, but if you have a cylinder, you can imagine bending it so that the top and bottom are connected. Then you have a torus.


Equation to calculate best liked comments In my website, users can either "like" or "dislike" a posted comment. I want to put a link to sort comments by liking such that the most liked ones becomes on top.

Of course I cannot just sort by the number of likes only. I have to subtract the dislikes. But what if the difference (likes - dislikes) are the same. e.g. 10 - 8 = 2 and 5 - 3 = 2. I think in this case, the comment of 10 likes and 8 dislikes has to come before the 5 likes and 3 dislikes comment.

So is there an equation that you feed it the number of likes and number of dislikes and then it gives you a meaningful rating number that I can sort with?

You have to define what criterion will say that a pair $(L_1,D_1)$ for comment 1 is better than $(L_2,D_2)$ for comment 2.  One way is just to subtract, so $(L_1,D_1) \geq (L_2,D_2)$ if $L_1-D_1 \geq L_2-D_2$.  But on stackexchange there are many more upvotes than downvotes, so maybe you want $(L_1,D_1) \geq (L_2,D_2)$ if $L_1-10*D_1 \geq L_2-10*D_2$ or some such.  Maybe you want to compare on $\frac{L-D}{L+D}$.  There are many choices, and you need to consider your audience and their behavior to select one.  Any such function will map (L,D) to some number, which you can then sort.

Here's what governs (or at least used to govern) the Reddit "best" comment sorting:
http://www.evanmiller.org/how-not-to-sort-by-average-rating.html
The lower bound of the Wilson score confidence interval represents an estimate of "at least" how good a comment should be.
However, if you want to entertain people into conversations and also show "new", "untested" comments on top, you might want to consider the upper bound (replace the "+/-" with a "+").
That way, the comments are sorted by the optimistic potential they have ("at most"), given current votes.
So, use the lower bound to see proven-good comments on top, and the upper bound if you want new comments (of unknown quality) to be above.


How can I generate "random" curves? In game programming (my profession) it is often necessary to generate all kinds of random things, including random curves (for example, to make a procedural island or for an agent to follow some path).

For one dimensional things, we usually use some random generator that generates say floats (which are really rational numbers) in some range and that is good enough as an approximation for most purposes. (Even though we cannot generate actual real numbers within a range uniformly, we can get a good sense of it with the rationals that we DO generate.)

When it comes to 2D things, though, things are very murky. For example, suppose I want to generate curves uniformly between two points (say, all curves bounded in a box, and say, additional requirements such as that the curves are differentiable).

The way we usually do it is to generate random parameters for some specific type of curve - say a Bezier curve, but this is not (AFAI can see) uniform in the original requirements - i.e. some curves that fit the bill will be more likely than others.

Is this even a sensible question to ask?

And if so, is there a way to generate curves (to a decent approximation) so that they are uniform within the parameters? (bounded and smooth)?

It "feels" like there are too many curves; it is strictly true with real numbers too... but there we have that rationals can be close enough for practical purposes; but with 2D things it seems less clear that any possible real curve is "close enough" to a "rational curve".

So I guess the main question is... if we have a set of "all curves", whether we can find a way to generate another set of approximations so that each "real" curve is close enough to our approximation.

Or: is there a mapping from "approximations to reals" to "approximations of continuous, differentiable, bounded curves between two points".... (that preserves uniformity, at least intuitively)?

Or: is there a notion of distribution of (bounded differential) curves? (And way to pick items from it).

Edit: I am more interested in the theoretical possibilities. I know LOTS of ways to generate curves... I am particular interested in generating curves without some kind of bias, and whether this even makes sense to want.

Edit: @pjs36 pointed out the curve may be arbitrary long. I don't mind additional restrictions to prevent pathological curves. Restrictions like "not longer than $x$", or not self-crossing.

(1) A version of your question was explored here: "Generating Random Curves with Fixed Length and Endpoint Distance."
I'd say not definitively answered.

(2) You might look at this recent paper:
Igor Rivin, "Random space and plane curves," arXiv:1607.05239, 2016. It relies on random trigonometric series:
"our class is precisely the class of random tame knots."

(3) A recent theoretical paper:
Kemppainen, Antti, and Stanislav Smirnov. "Random curves, scaling limits and Loewner evolutions." The Annals of Probability 45.2 (2017): 698-779.
"[W]e show that a weak estimate
on the probability of an annulus crossing implies that a random curve arising
from a statistical mechanics model will have scaling limits and those will be
well-described by Loewner evolutions with random driving forces."

Fig.1: arXiv version abstract.

If You want to draw irregular closed shape (e.g. asteroid or island), then the simplest way is to use polar coordinates - rotate the polar axis using random-size steps from 0..360, change the polar radius also using random steps, both in certain limits min..max -


initial value problem: y'' + 4y = f(t) , y(0)= y'(0)=0. f(t) = { 0 if t 3} Solve the initial value problem:

$$y'' + 4y = f(t) , y(0)= y'(0)=0. $$
where
$$ f(t) = \begin{cases} 0 &t < 3 \\ t & t > 3\end{cases} $$

I've solved for the homogeneous equation, $y'' + 4y = 0,$ to get the general solution,
$y = A\cos(2t) + B \sin(2t)$ for some arbitrary constants $A$ and $B$.

But, now where do I go next?

Thanks!

A related problem. In fact, you need to solve these two differential equations

$$ \begin{cases} y'' + 4y = 0 &t < 3 \\ y'' + 4y = t & t > 3\end{cases} .$$


Does Monty Hall logic apply to this real world situation? I recently posted a tweet claiming I had encountered a real life Monty Hall dilemma. Based on the resulting discussion, I'm not sure I have.

The Scenario

I have 3 tacos (A,B,C) where tacos A and C are filled with beans, and taco B is filled with steak.
I have no foreknowledge of the filling of any tacos.
My wife only knows that taco C is filled with beans.
My wife and I both know that I want steak.
After I pick taco A, my wife informs me taco C is filled with beans.
I switch my pick from taco A to taco B, thinking the logic behind the Monty Hall problem is relevant to my choice.

Edit for clarity

Timing:  The contents of taco C were not revealed to me until after I had made my selection of taco A.
My knowledge of what my wife knew:  When she told me the contents of taco C, I knew that she had previously opened taco C.  I also knew that she had no other knowledge of the contents of the other tacos.

Questions

Even though my wife does not know the fillings of all the tacos, does her revealing that taco C is definitively not the taco I want after I've made my initial selection satisfy the logic for me switching (from A to B) if I thought it would give me a 66.6% chance of getting the steak taco?
If this is not a Monty Hall situation, is there any benefit in me switching?

No, this is not a Monty Hall problem.  If your wife only knew the contents of #3, and was going to reveal it regardless, then the odds were always 50/50/0.  The information never changed.  It was just delayed until after your original choice.  Essentially, you NEVER had the chance to pick #3, as she would have immediately told you it was wrong. (In this case, she is on your team, and essentially part of the player).  #3 would be eliminated regardless: "No, not that one!"

Imagine you had picked #3.  Monty Hall never said, "You picked a goat.  Want to switch?"

If he did, the odds would immediately become 50/50, which is what we have here.

Monty always reveals the worst half of the 2/3 you didn't select, leaving the player at 33/67/0.

You can turn this into a Monty Hall problem:

Your wife knows all the fillings (say, for example, she prepared them).

You pick A.

She picks C, cuts it open, revealing beans.  It does not matter which taco is labled A or B or C.  Since she knows all, she can always pick one with beans after your pick, since there are two with beans, just like Monty's goats.

Now you have the situation where initially your probability was 1/3 steak and 2/3 beans.

The new situation is that your original pick is the steak with probability 1/3 and the unpicked taco is the steak with probability 2/3 ... however ...

Probabilities only make sense over many samples (many plays of the same game). Like the Monty Hall game, you only get to play once, so it really does not matter at all - this is the Monty Hall Myth.

The logic only applies to Monty, because he is the only one playing all the games.  If every player switched, over time, Monty would lose more often (66.6% of the time), and in consequence every player would have a better chance.

But you do not control the other players.  The odds are, many will not switch, based on 'tells' or other non-probabilistic intuition. That impairs your odds, since your play is a singleton - one event of many.  In your singleton event, the only thing you know for sure is that one has beans and the other has steak, and this circumstance will always arise, since Monty knows all and there are two goats, so he can always find one.

In your singleton event, Monty shifts your odds from 33:67 to 50:50, and it does not matter whether you switch.


solving logarithmic equation $\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x}$ Solve for $x$

$$\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x} $$

Assume that $\log$ is the natural logarithm and $x > 1$ (Otherwise, it would not make sense the power in LHS). Since $\log \sqrt{x} = \dfrac{\log x}{2}$, we can simplify the equation as $$\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x}  \implies \left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 3)} = 1$$
$$\implies \left(\log^2x + 2\log x - 3\right)\log\left(\dfrac{\log x}{2}\right) = 0.$$

Thus, $$\log\left(\dfrac{\log x}{2}\right) = 0 \implies x = e^{2}.$$ Or,
$$\log^2x + 2\log x - 3 = 0 \implies \log x = -1\pm2  \implies x = e ~~\text{and}~~x = e^{-3}.$$

$\frac{\log x}{2}=\log{\sqrt{x}}$

Using this,

$( \log{\sqrt{x}})^{(\log x)^2+2\log x-2}= \log{\sqrt{x}}$

$\implies (\log x)^2+2\log x-2=1$

$\implies (\log x+3)(\log x-1)=0$

$\implies x=e^{-3}$ or $e$

Again,

$\log{\sqrt{x}}=1$ gives a solution.

$\implies x=e^2$ is also a solution


Would it be possible to extract all the points in the area and still have positive area? I'm reading Apostol's Calculus. Given the axioms

I got a silly doubt. A point is a measurable set with zero area, according to the axioms. Suppose I have a square with $a(S)=16$ and a point with $a(P)=0$, then I can use $(3)$ from the axioms above and subtract one from another.

Wouldn't that imply that I could apply $(3)$ repeatedly, remove all the points in $S$ and still have $a(S)=16$ even when $S$ has no points and should have $a(S)=0$?

No, because no matter how many times you apply (3) removing one point, you'll only have removed finitely many points.

I'm completely out of my league here, so bear with me. But I recently viewed a video from vsauce that seems to suggest that theoretically what you describe might actually be possible and would end up creating The Banach–Tarski Paradox


Discuss the convergence of the sequence whose $n$-th term is given by $a_n= \frac{n!}{n^n}$ Discuss the convergence of the sequence whose $n$-th term is given by

$$a_n= \frac{n!}{n^n}$$

I'm just a little confuse on how this can be solved because the factorial is confusing me.

Put

$$a_n:=\frac{n!}{n^n}\implies\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e<1$$

thus the infinite positive series $\;\sum a_n\;$  converges, so $\;a_n\xrightarrow[n\to\infty]{}\;?$

$$a_n  = \frac{{n!}}{{n^n }} \cong \frac{{\sqrt {2\pi n} }}{{e^n }} \Rightarrow \mathop {\lim }\limits_{n \to  + \infty } a_n  \cong \mathop {\lim }\limits_{n \to  + \infty } \frac{{\sqrt {2\pi n} }}{{e^n }} = 0
$$


Interval of convergence and integration of a power series The $\arctan(x)$ can be expanded as a MacLaurin series starting from the integral

$$\arctan(x) = \int \frac{1}{1 + x^2} \mathrm{d}x$$

and using

$$\frac{1}{1 + x^2} = \sum_{n = 0}^{\infty} (-1)^n x^{2n}$$

as suggested in this answer. This series converges for $x \in (-1,1)$, but, after integration, it can be shown that the resulting MacLaurin series

$$\arctan(x) = \sum_{n = 0}^{\infty} (-1)^n \frac{x^{2n + 1}}{2n + 1}$$

converges for $x \in \left[ -1, 1 \right]$.

The integration of a series is possible only when the series is evaluated within its interval of convergence, $x \in (-1,1)$: the MacLaurin series for $x = \pm 1$ shouldn't even be considered, because the above result for $\arctan(x)$ should not be available!

Why instead, here two more points ($x = \pm 1$) can be added to the interval of convergence?

Integration does not change the radius of convergence of a series.

What are then the differences between the interval of convergence and the radius of convergence? Do $(-1,1)$ and $\left[ -1, 1 \right]$ correspond to the same radius of convergence?

I read also this question, answer and comments am I am not familiar with Cauchy-Hadamard Radius Formula.

A comment here states (given an interval of convergence $(a - R, a + R)$):

  The issue of convergence at the points $x= a ± R$ is independent of the convergence within the interval $(a−R,a+R)$.

My questions above essentially are: why?

On the request of asker I am writing a full blown answer based on my comments.

Let us then note that a series of the form $$\sum_{n=0}^{\infty} a_nz^n$$ is called a power series in variable $z$. Here $a_n, z$ are complex numbers.

A fundamental feature of such a series is the existence of a non-negative real number $R$ called the radius of convergence and it has the following properties:

The series $\sum_{n=0}^{\infty} a_nz^n$ converges absolutely for $|z|<R$.
The series $\sum_{n=0}^{\infty} |a_nz^n|$ diverges to $\infty $ for $|z|>R$.

There exist special and useful cases when $R=\infty$ and then the series converges everywhere in the complex plane. Functions represented by such series are called entire functions.

The region $|z|<R$ is called the circle of convergence. In case of real variable $z$ this gets replaced by interval of convergence $(-R, R) $.

A power series can be always be integrated as well as  differentiated term by term to lead to another power series with the same radius of convergence.

The convergence of a power series at boundary points $|z|=R$ requires special analysis on a case by case basis and there are no general theorems without any additional hypotheses to confirm convergence or divergence at these points.

Some textbooks define a power series in the form $\sum a_n(z-a) ^n$ and to handle these we can just replace $z$ with $z-a$ everywhere in previous part of this answer.

In general, you won't be able to say things about endpoints of an interval after integrating.

In this case, however, we can just check that the sum converges at the endpoints directly; if $x = \pm 1$, then $x^{2n+1} = x$, so the sum becomes
$$x \sum_{n=0}^\infty \frac{(-1)^n}{2n+1},$$
which converges because $\frac{(-1)^n}{2n+1}$ is an alternating series with $\frac{1}{2n+1}$ decreasing and going to $0$.


How to find the limit of $\frac{(1-x)(1-2x)(1-3x)...(1-85x)-1}{x}$? I am struggling to find out how to correctly write the steps to solve the limit of this series:
$$\lim_{x\to 0}=\frac{(1-x)(1-2x)(1-3x)...(1-85x)-1}{x}$$
I know, that the result is -3655 and I also know why: Since the limit approaches zero, powers higher than $x^1$ will be zero, and elements without $x$ will cancel out $(1-1=0)$. Therefore, I only care about elements that have $x^1$, which can be simplified to
$$-x(1+2+3+4+...+85)$$
and that gives us the formula $n(n+1)/2$, hence the result. The $x$ will then cancel out with the one in the denominator, leaving me with $-3655$.
What I do not know is how to solve this problem properly, that is using proper steps and formation.
Can anyone provide any solutions or tips?

FIRST METHOD :
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ defined for all $x \in \mathbb{R}$ by
$$f(x)=(1-x)(1-2x)(1-3x)...(1-85x)$$
What you want to calculate is
$$\lim_{x \rightarrow 0} \frac{f(x)-1}{x} = \lim_{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}$$
By definition of the derivative, you deduce $$\lim_{x \rightarrow 0} \frac{f(x)-1}{x} = f'(0)$$
which can be calculated, if you are courageous.
SECOND METHOD :
The function $f$ is polynomial, and its constant coefficient is $1$. That means that you can write $f(x)=1+\sum_{k=1}^{85} a_k x^k$. Then you see that
$$\frac{f(x)-1}{x}  = \sum_{k=1}^{85} a_k x^{k-1}$$
so the limit when $x$ tends to $0$ is equal exactly to $a_1$, i.e. the coefficient of $x$ in the polynomial function $f$. It is easy to see that this coefficient is equal to
$$a_1=- \sum_{k=1}^{85} k = -3655$$


Let p(x) be a quadratic polynomial such that distinct reals a and b ,. p(a)=a and p(b)=b then find the roots of. p[p(x)]-x=0 Well a and b are surely two of the roots which I found easily but I cannot find the other.
I used the approach that
P(x)=A(x-a)(x-b)+x
then I found p[p(x)]-x from which I can easily see that a and b are it's roots but how can I find the other two.

Note that $p(x)=x$ is a quadratic equation, hence, there exist at most two distinct real numbers that satisfy it. You have $p(a)=a$ and $p(b)=b$, hence, $a,b$ are these numbers. So, in your equation $p(p(x))=x$, the only roots are exactly the solutions of the (quadratic) equations $p(x)=a$ and $p(x)=b$. Since we know that $p(a)=a$ and $p(b)=b$, $a,b$ are two solutions.

Note that, as you have observed, $p(x)=C(x-a)(x-b)+x$, $C\neq0$. So:

$$p(x)=a\Leftrightarrow C(x-a)(x-b)+x-a=0\Leftrightarrow(x-a)(Cx-Cb+1)=0,$$

so, $x=a$ (we have found that) or $x=\dfrac{Cb-1}{C}$.

Similarly, we can find that the other two roots are $x=b$ or $x=\dfrac{Ca-1}{C}$.


sum of numbers formed by permutations I have digits 2,3,4,5. I have been asked to find the sum of all 4 digits the numbers that can be formed using these digits without repetition such that all are included in the number.

Can someone help me solve this quickly?

The answer is 93324.

Fixing 5 in the units digit, you get $3!=6$ numbers. So 5 comes 6 times in the units digit. Same goes for 2,3,4

hence sum of the units digits = $6(2+3+4+5)=84$

Same argument goes for ten's place, hundreds place and so on. hence the sum=

$84 \times 1000+84 \times 100+84 \times 10+84=84 \times 1111=93324$

For the 2 you get:
$2000 \cdot 3! + 200 \cdot 3! + 20 \cdot 3! + 2 \cdot 3!$
The 1st term corresponds to 2 being at position 1,
the 2nd term corresponds to 2 being at position 2 and so on.

Then you have the same pattern for 3,4,5.

So you get: $(2222 + 3333 + 4444 + 5555) \cdot 6 = 93324$


Find five prime factors of $3^{140 }- 1$ I tried to simplify $3^{140}$ but I couldn't go past $81^{35}$, any help would be greatly appreciated.

$$3^{140}-1=(3^{70}+1)(3^{70}-1)=(3^{70}+1)(3^{35}+1)(3^{35}-1)$$
and from $a-b\mid a^7-b^7$, the number
$ 3^{35}\pm1$ is a multiple of $3^5\pm1$. These are of feasible size and give us $2, 11, 61$ as prime factors. Also, $3^{35}\pm1$ is a multiple of $3^7\pm1$, which are still of manageable size and give us $547$ and $1093$.

Alternatively, note that $3^{p-1}\equiv 1\pmod p$ for $p\ne 3$, hence looking for small primes with $p-1\mid 140$ is helpful. This way you find $p=2, 5, 11, 29, 71$ very easily.


Adding sine wave with different amplitude and different frequency How to express the following equation into a sine wave equation ?

$$S(t) = 4 + 3 \sin 100 \pi t + 5 \sin 200 \pi t$$

I understand that, for the sine wave with same frequency and different amplitude,

I can use the formula

$$\sin (\omega t)+ A_{2}\sin (\omega t)=(A_{1}+A_{2})\sin (\omega t)$$

but how to add sine wave that has different frequency ?

Thanks!

You can't add two sines of different frequencies.
This is a fundamental limit which much of math and physics takes advantage of.

Its also the basis of the fourier transform.

To put it simply, sines of different frequencies are orthogonal.


Continuity of a function defined by an integral Ok, Here's my question:

  Let $f(x,y)$ be defined and continuous on a $\le x \le b, c \le y\le d$, and $F(x)$ be defined >by the integral $$\int_c^d f(x,y)dy.$$ Prove that $F(x)$ is continuous on $[a,b]$.

I think I want to show that since $f(x,y)$ is continuous on $[a,b]$, I can use proof by contradiction to get $F'(x)$ continuous on $[a,b]$, which would then imply that $F(x)$ is continuous. But How do I go about setting this up? Any hints would be great. Thank you in advance. Also, this is my first attempt to format everything properly, So I'm sorry if this didn't post properly.

Here is a start

$$ F(x+h)-F(x) = \int_c^d f(x+h,y)dy -  \int_c^d f(x,y)dy  $$

$$ \implies |F(x+h)-F(x)|\leq  \int_c^d |f(x+h,y)- f(x,y)|dy. $$

Now, use the assumptions you have been given to finish the proof. Note that, $f$ is continuous on a compact set.

Consider $f(x,y)=y^{x-1}$ then $F(x)$ defined above is $\frac{y^{x}}{x}$. The definition of continuous requires that the function equal it's limit at the point in question among other things. Clearly $\frac{y^{x}}{x}$ is not defined and thus not equal to it's limit at the point $x = 0$. Yet $f(0,y)=\frac{1}{y}$ has no problem for suitable $y$. Thus the statement you set out to prove is in fact false.

Note If additionally you knew that the $\frac{\partial }{\partial x}F(x,y)$ was continuous, only then could you conclude $F(x)$ continuous on an interval $[a,b]$ as then you would know that both $F(x,y)$'s partials (one of them is $f(x,y)$) where continuous, thus $F$ differentiable and hence continuous.


Diameter of union is at most the sum of diameters, when the sets intersect Prove that $d (A \cup B) \leq d(A)+ d(B)$, given that   $ A \cap B \neq \varnothing$.

Here $d$ stands for the diameter of the set. Please note that my knowledge is limited to metric spaces only, with no knowledge of topology at all.

My attempt:

I'm trying to use the relation that $A \subset B \Rightarrow d(A) \leq d(B)$

I prove this result in the following way:

If $a,b \in A$, then  $a,b \in B$, so
 { $d(a,b) : a,b \in A$ }$ \subset$ {$d(a,b) : a,b \in B$}

So, sup { $d(a,b) : a,b \in A$ }$ \subset$ sup {$d(a,b) : a,b \in B$}

Is this part correct?

Now, using this result:

$ A \subset A \cup B$ and $ B \subset A \cup B$

So $d(A) \leq d(A \cup B)$ and $d(B) \leq d(A \cup B)$

Adding, $d(A)+ d(B) \leq 2d(A \cup B)$

This is where I'm stuck.

Consider two points $x,y$ in $A\cup B$. If they are both in $A$, then $d(x,y)\leq d(A)$. If they are both in $B$, then $d(x,y)\leq d(B)$. Otherwise, suppose $x\in A$ and $y\in B$. Let $z$ be a point in $A\cap B$. Then by the triangle inequality we have
$$d(x,y)\leq d(x,z)+d(z,y)\leq d(A)+d(B)$$
Taking sups we obtain that $d(A\cup B)\leq d(A)+d(B)$.

You need $A\cap B\ne \varnothing$. What happens if $A$ and $B$ are just two different points?

The more general inequality (any $A$ and $B$) should have $d(A,B)$ in it. See here for detail. In this case, $A\cap B\ne \varnothing$ implies $d(A,B)=0$.


Simplify $\frac{\sqrt{x^4 +3 x^2}}{x}$ The question states to simplify $$\frac{\sqrt{x^4 +3 x^2}}{x}$$where x is negative.

and part b is Let g(x) be the result from the simplification. Are f and g two equal functions? Y/N?

Much help is appreciated as i am struggling.

The answer is $\sqrt{x^2 + 3}$, and I don't know yet how to display so nicely...


Limits of an increasing function Let $a<b$ and let $f: [a,b] \to R$ be an increasing function.
I'm asked to provide an example where $\lim_{x\to a^{+}}f(x)$ does not equal
$$\inf\{\,f(x) \; | \;x \in [a,b]\,\},$$
and I'm having trouble understanding what exactly is going on in this situation and was wondering if anyone can enlighten me.
Background
Prior to this question I was asked to prove that there exists an $S$ in $R$ such that
$$ S = \inf\{\,f(x)\; |\; x \in (a,b]\,\}.$$
I used the fact that $F$ is an increasing function and showed $F$ is bounded from below and therefore must have an infimum.
Then I was asked to prove that the $\lim_{x\to a^{+}}f(x)$ exists, and is equal to S.
I showed this by definition of limit, using the fact that if $F$ is bounded from below, then there exists $L$ such that $L$ is less than or equal to $F(x)$.
I have a feeling my professor wants me to use the things I showed above to find the example to the initial question, but I just can't connect the dots…

This is a picture of what can happen at the endpoint (what needs to happen, in fact).

Remember to keep your counterexamples as simple as possible!
For the limit part ($L$ is the unfilled circle), you can perhaps use the sequential definition of limit, with the version of the completeness axiom that applies to monotonic sequences.

The big difference between the two infimums is that one includes $a$ while the other doesn't. To take advantage of it we use cases. For example,
$$
f(x) = \begin{cases}g(x), &x \neq a\\
-1 &x=  a\end{cases}
$$
What will happen if $g(x) > -1\ \forall x \in (a,b]$?


Big theta notation question can someone please explain to me the big theta notation and big omega and also

How i can show that $$ 3x+7\quad \text{is}\quad \Theta (x); $$

I don't really get how growth of functions works.

The definition that I'm accustomed to for $O$, $\Theta$, and $\Omega$ are the following:

$f(x)$ is $O(g(x))$ if there is some positive real number $C$ such that $$|f(x)| \leq C|g(x)|$$ for all $x>N$, where $N$ is some real number.

$f(x)$ is $\Omega(g(x))$ if there is some positive real number $C$ such that $$|f(x)| \geq C|g(x)|$$ for all $x>N$, where $N$ is some real number.

$f(x)$ is $\Theta(g(x))$ if $f(x)$ is both $O(g(x))$ and $\Omega(g(x))$ (but these can be with different constants!).

So $x$ is $O(x^2)$, but not $\Omega(x^2)$, because you can't find an $N$ to satisfy the condition.

But $3x+7$ is $O(x)$ ($C=5$, $N=4$, for example) and $\Omega(x)$ ($C=1$, $N=0$), so $3x+7$ is also $\Theta(x)$.

If the following limit exists:
$$
  \lim \frac f g = L
$$
and $L>0$ and $L<+\infty$ then $f \in \Theta(g)$. This applies in your case.


Is Rayo's number really that big? I was reading about large numbers, and came across Rayo's Number which is defined to be the smallest integer that is not nameable by any expression in the language of set theory that contains less than $10^{100}$ symbols.

Now, my question is: Is this number really that large?

If we pick some "random" number $2091580284...384901284021$ with $10^{100}$ digits, wouldn't it be non-nameable with less than a googol symbols? Wouldn't this number be bigger or equal to Rayo's Number?

As some of the comments already mentioned, you misquoted the definition. The correct definiton (quoting Googology Wiki) is:

  the smallest positive integer bigger than any finite positive integer named by an expression in the language of first order set theory with a googol symbols or less

So while there are only approximately $(10^{100})^{(10^{100})}$ possible expressions, and only a very small fraction of them actually name a number, Rayo's number can be very large.

We don't exactly know how large, but there are good heuristic arguments that the busy beaver function $\Sigma$ can be implemented in a few million symbols. Given that $\Sigma$ grows much faster then $\mathrm{TREE}$ and all other computable functions, Rayo's number is much larger than  $(10^{100})^{(10^{100})}$ .

No, a number like 10^(10^100) is much smaller than Rayo(10^100), because there are not a Googol digits, but a Googol symbols in first order set-theory, and it is pretty efficient to write down big numbers such as TREE(3), which can be expressed with MUCH LESS than a Googol symbols, you don't need anything like a Googol symbols to express TREE(3),and if we can say that :
              Rayo's Number > TREE(3)

We can certainly say that :
              Rayo's Number > Any number with 10^100 digits

So Rayo's Number is really that big.


Rotation of general conic equation - angle $\theta$ quadrant I need to identify the conic represented by the equation
  $$9x^2 -6xy +y^2 - 40x -20y + 75=0$$

The book provides the solution as below: $\tan 2\theta = -3/4$

Then says take $\tan \theta = 3$. I reached this point as well.
However it then calculates

$\sin \theta = -3/{\sqrt {10}}$

$\cos \theta = -1/{\sqrt {10}}$

I was solving it assuming $\theta$ is in first quadrant, however solution shows it lies in third quadrant but not sure how it arrived at that conclusion.

Please help if I am missing something.

The tangent function has period $\pi$, and the tangent of the double angle has period $\dfrac\pi2$. So you are free to take $\theta$ in any quadrant.

Whatever the choice, you will reduce to a parabola, and

$$9x^2-6xy+y^2=(3x-y)^2,$$ is the same as $$(-3x+y)^2.$$

Since the discriminant vanishes it is a parabola. The presence of $xy$ term  indicates that its axis of symmetry is parallel to neither axis, but inclined at $\theta$ to $x-$ axis.

Evaluation of $\theta $ from $ 2 \theta$ gives two  values of arctan function value separated by $ \pi$ from square root of quadratic equation solution, two inclinations one each for the tangent and symmetry axis corresponding to  your calculated values.

EDIT1:

I try to elucidate fully; the given parabola equation
$$ 9x^2-6xy + y^2 -40x-20y +75=0    \tag1$$
To get rid of $xy$ term the slope (= rotation) of the axis of symmetry is $3$ ( given by OP ) and slope of tangent at vertex is its negative reciprocal $\dfrac{-1}{3}$

So let the equation of tangent be

$$ y= \frac{-x}{3} +p \tag 2$$

where $p$ is the $y-$ axis intercept yet to be calculated.

Eliminate $y$ from above to get

$$ x^2(9+2+1/9) + x(-6p -2p/3-40+20/3) + (p^2-20p+75) = 0 \tag3$$
simplifies to

$$ x^2(100/9) + x(-20p/3 -100/3) + (p^2-20p+75) = 0 \tag4$$

Condition for double root of tangent is that its discriminant should vanish.

$$ \left(\frac{10p+50}{3} \right)^{2}= \frac{100}{9} (p^2-20p+75)=0 \rightarrow p=\frac{5}{3} \tag5 $$

Plug 5) into 4) and simplify to get equation of tangent line as

$$x+3y=5 \tag6$$

Again plug 5) into 1)  and simplify

$$ [(10p +50))^2 = 100/9(p^2-2p+75) \tag7$$

or

$$p= \frac{ 5}{3} \tag8 $$

plugging it into 6) gives

$ x_t=(2,2)$ and corresponding $y_t= (1,1) $ as it is a double (tangent/ coincident) point.

Thus the vertex point is

$$(2,1) \tag 9 $$

Finally axis of symmetry is given by a straight line of given slope through known point $V$as

$$ y=3x - 5 \tag {10} $$

The above are all graphically depicted for the parabola with its tangent at vertex and normal which is also the axis of symmetry.


Show that $\Bbb Q$ is not a finitely generated $\Bbb Z$-algebra. Show that $\Bbb Q$ is not a finitely generated $\Bbb Z$-algebra.

I know that $\Bbb Q$ is not a finitely generated $\Bbb Z$-module. From here how can I conclude that $\Bbb Q$ is not a finitely generated $\Bbb Z$-algebra?

Please help me in this regard.

Thank you in advance.

Hint:

If we had $\mathbf Q=\mathbf Z\biggl[\dfrac{m_1}{n_1},\dots,\dfrac{m_k}{n_k} \biggr]$, we could as well write
$$\mathbf Q=\mathbf Z\biggl[\frac1N \biggr],\ \text{where }\;N=\operatorname{lcm}(n_1,\dots,n_k).$$
Consider a prime $p$ not dividing $N$. Can $\dfrac1p$ be a polynomial in
$\dfrac 1N$?

Suppose $\mathbb Q$ is finitely generated as a $\mathbb Z$-module (in this case, $\mathbb Z$-algebra). By the Structure Theorem for modules (in this case, it is the theorem on finitely-generated abelian groups) and the fact that $\mathbb Q$ is torsion free, $\mathbb Q\cong \mathbb Z^r$ for some $r\geq 0$. Let $\varphi:\mathbb Q\rightarrow \mathbb Z^r$ be an isomorphism. Since $\varphi$ is a $\mathbb Z$-module homomorphism, if $n\in \mathbb Z$ is nonzero, then $\varphi(1)=\varphi\left(n\cdot \frac{1}{n}\right)=n\varphi\left(\frac{1}{n}\right)$. Let $\varphi(1)=(a_1,\ldots,a_r)$. Take $n=\mbox{gcd}(a_1,\ldots,a_r)+1$. Then $\varphi\left(\frac{1}{n}\right)\notin\mathbb Z^r$, a contradiction.


a question about the Collatz conjecture (relation of smallest number in a cycle and minimal cycle-length) I've only done a bit of research on the current findings, not sure if anyone here can answer this.

Q1: I just haven't been able to find,

  has it been shown yet that a it is impossible for a loop to exist,
  and for a counter example to exist does it have to grow to infinity?

Q2: I've seen that all numbers up to 2^60 have been checked, and using this information I'm able to show that no loops exist of length 101 or less (as a lower bound).

  Has something like this been shown before? I just can't seem to find much information on it.

It has neither been shown that a loop other than the trivial $\{4,2,1\}$ does not exist, nor that there exists a trajectory that grows to infinity.

Reformulate the Collatz iteration in a compact form as

(1) $v_n = z_n + 2^{2n+1-r}m_r$

for any input $u \in 2\mathbb{N}+1 \not \equiv 0\pmod{3}$ such that $u=3m_r+r$ where $r=1$ if $u \equiv 1\pmod{3}$ or $r=2$ if $u \equiv 2\pmod{3}$ and $z_n = \sum_{i=1}^{n} 2^{i-1}$ for $n \in \mathbb{N}$. Iterating (1) by one step for each $n \in \mathbb{N}$ yields an infinite set $\{v_n\}_{n \in \mathbb{N}}$, termed sibling set, arising from parent $u$. Suppose we consider $u$ and $v_n$ as vertices connected by edge $(u, v_n)$ from $u$ to $v_n$ for every $n \in \mathbb{N}$. Then starting with $u_0=1$, repeatedly iterating (1) constructs an arborescence graph $G=(V,E)$ that is the union of infinitely many sibling sets, with root at $r_0$, vertex set $V$ and edge set $E$, as shown here. Every path of any length from the root in $G$ represents an odd Collatz sequence in reverse. It can be shown that every vertex in $G$ is unique and that $V=2\mathbb{N}+1$, i.e. $V$ exactly covers the set of odd natural numbers. Since $G$ is weakly connected, then we infer that: (i) only the trivial loop exists (but is necessarily ignored in $G$) and (ii) every odd Collatz sequence always converges to $1$ (hence, cannot diverge). These results, subject to further scrutiny, answer both questions under Q1 above.


Prove that the triangle is isosceles, based on the intersection of two line segments connecting vertices to sides Given $AB=AC$ and $DE=DF$, prove $DB=DC$

I have no idea how to solve it using elementary way, can somebody help me?

Assume that $DB>DC$, then $\angle BCD>\angle CBD$. So $\alpha>\beta$. From the following picture

we see that $AE<AF$. Thus, in $\triangle AEF$ we have $x>y$. But $x+\alpha=y+\beta$ is a contradiction.

Given that DE=DF, a parallel line can be drawn through EF which is parallel to BC. as AB=AC angleABC=angleACB,THUS following that angleFEA=angleEFA WHICH implies EA=FA .THUS due to all of these angleDBF=angleDCE and thus DB=DC .


Incremental averageing Is there a way to incrementally calculate (or estimate) the average of a vector (a set of numbers) without knowing their count in advance?

For example you have a = [4 6 3 9 4 12 4 18] and you want to get an estimate of the average but you don't have all the values at hand so you want to have a running average without keeping all the previous values available.

You need to keep at least two pieces of information: the number of terms so far and the running mean (or something equivalent to it).

Let's suppose the $n$th component of the vector is $a_n$ and the running mean up to this is $m_n$ so $$m_n= \frac{1}{n}\sum_{i=1}^n a_i.$$

Starting with $m_0=0$, you can use the obvious single pass $$m_n = \frac{(n-1)m_{n-1}+a_n}{n}$$ but precision errors are likely to be smaller if you use the equivalent $$m_n = m_{n-1}  + \frac{a_{n}-m_{n-1}}{n}.$$

Here is the general solution to the problem. We start calculating average from (m+1)-th element up to (n+1)-th element of the incoming data.
Giving that:

$a_{n}$ is the n-th element of incoming data,
$a_{m}$ is the element from which we start averaging,
$\widehat{a_{n-m}}$ is the avarage from m-th to n-th element,
$\widehat{a_{n+1-(m+1)}}$ is the avarage from (m+1)-th to (n+1)-th element

So, if we initially have $a_{m}, a_{m+1}, a_{m+2}, \ldots, a_{n}$, an average of $n-m$ elements can be easily calculated. That is
$$
\widehat{a_{n-m}} = \frac{a_{m}+a_{m+1}+a_{m+2}+\ldots+a_{n}}{n-m}
$$

Next, when the $n+1$ elementh comes, we want to obtain average calculated of the same number of elements that is from $a_{m+1}$ to $a_{n+1}$:
$$
\widehat{a_{(n+1)-(m+1)}} = \frac{a_{m+1} + a_{m+2} + \ldots + a_{n} + a_{n+1}}{(n+1) - (m+1)}
$$

Then from first equation
$$
\widehat{a_{n-m}}(n-m) - a_{m} = a_{m+1}+a_{m+2}+\ldots+a_{n}
$$

Substituting this to equation 2
$$
\widehat{a_{(n+1)-(m+1)}} = \frac{\widehat{a_{n-m}}(n-m) - a_{m} + a_{n+1}}{n-m}
$$

In order to dynamicaly calculate average for new element we need previous average $\widehat{a_{n-m}}$, first element of the previous average $a_{m}$, number of the elements we include in the average $n-m$.
I hope you will find this solution inspiring. Please write if you find any errors in the above reasoning.


Square free and congruence modulo n I am trying to show that if $a^n\equiv a\pmod n$ for all integers $a$ then $n$ is square free.

I have an idea to start with the contradiction that suppose $n=p^2m$ for some prime $p$, then n does not divide $a^{p^2m}-a$ for some integer $a$. Any hints would be appreciated thanks.

Suppose that $n$ is divisible by $p^2$. Let $a=p$. Since $p^n$ is divisible by $p^2$, the number $p^2$ does not divide $p^n-p$.

Your claim is equivalent to the saying that $\{0\} \subset \mathbb{Z}/n\mathbb{Z}$ is a radical ideal in $\mathbb{Z}/n\mathbb{Z}$ only if n is squarefree. This equivalence follows because, for any commutative ring R, an ideal $I \subset R$ is radical if and only if $rad(I) = I$. What exactly is $rad(I)$ (aka the radical of I), you might ask?
\begin{equation}
rad(I) = \{r \in R : r^n \in I \,\,\,for\,\, some\,\, n \in \mathbb{Z}^+\}
\end{equation}
Notice that this is precisely what you're trying to prove, since in this case, $R = \mathbb{Z}/n\mathbb{Z}$ and $I = n\mathbb{Z}$. Observe that the trivial ideal in this R is precisely $n\mathbb{Z}$. If the trivial ideal is radical, then for any $t \in  n\mathbb{Z}$, we have $t^n \in n\mathbb{Z}$. It is easy to see that this same property holds for all the additive cosets of $n\mathbb{Z}$ whenever this property holds.

Below I will freely denote classes mod n using their representatives by enclosing them in brackets. It is clear that we are working in $\mathbb{Z}/n\mathbb{Z}$.

Proceeding to the proof of my original claim, suppose that n is squarefree. Then by the fundamental theorem of arithmetic, $n = p_1 \cdot ...\cdot p_k$ for unique prime factors $p_1, ..., p_k \in \mathbb{Z}^+$. First, note that we have the containment $\{0\} \subset rad(\{0\})$ since $[0] = [0]^1 \in \{0\}$. Now, let $[r] \in rad(\{0\})$ be arbitrary. Then, $[r]^m = [0]$ for some $m \in \mathbb{Z}^+$ since $[0]$ is the only element in the trivial ideal. Equivalently, $r^m = tn$ for some $t \in \mathbb{Z}$. Furthermore, by the fundamental theorem of arithmetic, $r=q_1^{f_1} \cdot ...\cdot q_h^{f_h}$ for unique prime factors $q_1, ..., q_h \in \mathbb{Z}^+$ and $f_1, ..., f_h \in \mathbb{Z}^+$. It follows that
$$\begin{align}
q_1^{mf_1} \cdot ...\cdot q_l^{mf_h} &= (q_1^{f_1} \cdot ...\cdot q_h^{f_h})^m \\
&= r^m \\
&= tn \\
&= t \cdot p_1 \cdot ...\cdot p_k
\end{align} $$
Since $p_i\,|\,t \cdot p_1 \cdot ...\cdot p_k$ for each $i \in \{1,...,k\}$, it must also be the case that $p_i\,|\,q_1^{mf_1} \cdot ...\cdot q_h^{mf_h}$ for each $i \in \{1,...,k\}$. Furthermore, because each $p_i$ is a unique prime, $p_i\,|\, q_j^{mf_j}$ for some unique $j \in \{1,...,h\}$. Thus, $p_i\,|\, q_j$. Since $p_i$ and $q_j$ are prime for every choice of $i$ and $j$, it follows that $p_i > 1$ and therefore $p_i = \pm q_j$ since the only units in $\mathbb{Z}$ are $\pm 1$. Without loss of generality, suppose that $p_i = q_j$ since the unit multiple will make no difference once we finally consider classes mod n. Thus, we have
$$ \begin{align}
r &= q_1^{f_1} \cdot ...\cdot q_h^{f_h} \\
&= p_1^{f_1} \cdot ... \cdot p_k^{f_k} \cdot q_{k+1}^{f_{k+1}} \cdot ... \cdot q_h^{f_h} \\
&= p_1 \cdot p_1^{f_1-1} \cdot ... \cdot p_k \cdot p_k^{f_k-1} \cdot q_{k+1}^{f_{k+1}} \cdot ... \cdot q_h^{f_h} \\
&= (p_1 \cdot ...\cdot p_k) \cdot p_1^{f_1-1} \cdot ... \cdot p_k^{f_k-1} \cdot q_{k+1}^{f_{k+1}} \cdot ... \cdot q_h^{f_h} \\
&= n \cdot p_1^{f_1-1} \cdot ... \cdot p_k^{f_k-1} \cdot q_{k+1}^{f_{k+1}} \cdot ... \cdot q_h^{f_h} \\
\end{align} $$
Denote $x = n \cdot p_1^{f_1-1} \cdot ... \cdot p_k^{f_k-1} \cdot q_{k+1}^{f_{k+1}} \cdot ... \cdot q_h^{f_h}$ for convenience. Hence, $r = nx$ and it follows that $[r] = [0]$. Therefore, $[r] \in \{0\}$, and since $[r]$ was chosen arbitrarily, we have the containment $rad(\{0\}) \subset \{0\}$. Thus, $rad(\{0\}) = \{0\}$ and $\{0\}$ is a radical ideal in $\mathbb{Z}/n\mathbb{Z}$. Q.E.D.


Real life examples for eigenvalues / eigenvectors There are already good answers about importance of eigenvalues / eigenvectors, such as this question and some others, as well as this Wikipedia article.

I know the theory and these examples, but now in order to do my best to prepare a course I'm teaching, I'm looking for ideas about good real life examples of usage of these concepts.

Do you know some good simple real-life examples (in economics or data analysis or anything else), in which  the usage of eigen values/vectors is a crucial tool?

Here are just some of the many uses of eigenvectors and eigenvalues:

Using singular value decomposition for image compression. This is a note explaining how you can compress and image by throwing away the small eigenvalues of $AA^T$. It takes an $8$ megapixel image of an Allosaurus, and shows how the image looks after compressing by selecting $1$,$10$,$25$,$50$,$100$ and $200$ of the largest singular values.
Deriving Special Relativity is more natural in the language of linear algebra. In fact, Einstein's second postulate really states that "Light is an eigenvector of the Lorentz transform." This document goes over the full derivation in detail.
Spectral Clustering. Whether it's in plants and biology, medical imaging, buisness and marketing, understanding the connections between fields on Facebook, or even criminology, clustering is an extremely important part of modern data analysis. It allows people to find important subsystems or patterns inside noisy data sets. One such method is spectral clustering which uses the eigenvalues of a the graph of a network. Even the eigenvector of the second smallest eigenvalue of the Laplacian matrix allows us to find the two largest clusters in a network.
Dimensionality Reduction/PCA. The principal components correspond the the largest eigenvalues of $A^TA$ and this yields the least squared projection onto a smaller dimensional hyperplane, and the eigenvectors become the axes of the hyperplane. Dimensionality reduction is   extremely useful in machine learning and data analysis as it allows one to understand where most of the variation in the data comes from.
Low rank factorization for collaborative prediction. This what Netflix does (or once did) to predict what rating you'll have for a movie you have not yet watched. It uses the SVD, and throws away the smallest eigenvalues of $A^TA$.
The Google Page Rank algorithm. The largest eigenvector of the graph of the internet is how the pages are ranked.

Let me give you a direct answer. In application eigen values can be:
1- Control Field: eigen values are the pole of the closed loop systems, if there values are negative for analogue systems then the system is stable, for digital systems if the values are inside the unit circle also the system is stable.

2- Mechanical system: eigen values are the natural frequency and the eigen values are the mode shapes.


How do I get the derivative of a summation Sorry if this has been asked before. I'm fairly new to this.
How do I find the derivative of a summation?

$$f(x)= \sum_{i=1}^{10} \frac{10x^2}i$$

what is the derivative of $f(x)$?

$f(x) = 10x^2/1+10x^2/2+\cdots + 10x^2/10$ Now termwise differentiate.


X is reflexive, so X is Weakly complete. Plese, help for this questions

"X is reflexive, so X is Weakly complete."

Thank you.

A google search should help you out here: http://people.math.gatech.edu/~bwick6/teaching/math6338/math6338_hw4.pdf

Specifically see attached.  I've posted pics of the solution from the site given above.


Can this expression $(\neg B \land \neg D) \lor (\neg A \land B \land C) \lor (A \land C \land D)$ be further simplified? I have assignment for computer architecture where I have to simplify a big boolean function:

  f(a, b, c, d) = a'b'c'd + a'bcd' + abcd + a'bcd + a'b'cd' + ab'cd' +
  ab'c'd' + ab'cd + a'b'c'd'

Now I arrive at this point

  $(\neg B \land \neg D) \lor (\neg A \land B \land C) \lor (A \land C \land D)$

but it still seems too complicated, but I am not able to simplify it further, can someone help me?

As far as I can see no further simplification is needed.

Besides, note that your formula

$$(\neg B \land \neg D) \lor (\neg A \land B \land C) \lor (A \land C \land D)$$

is in disjunctive normal form already (this does not imply that no simplification is needed though).

Good work!

You can use Karnaugh maps to simplify this function.
The result is:
$${A'B'C'+ AB'D' + A'BC + A'CD' + ACD}$$


Is there a way to figure out the $n$ such that $n!$ ends with exactly $k$ zeroes? Given $n$, I can find the number of zeroes at the end of the decimal representation of $n!$ by
$$
\sum_{i=1}^\infty\left\lfloor\frac{n}{5^i}\right\rfloor.
$$

Is there a way to reverse this? That is, given $k$, is there a way to find out how many $n$ exist such that $n!$ has exactly $k$ zeroes at the end of its decimal representation besides making educated guesses and checking them?

There's a way to almost reverse this. I find it easier to think about this in terms of quinary (base $5$) representations. So your formula says that the number of zeroes in the decimal representation of $n!$ is the sum of all right-shifted and truncated versions of the quinary representation of $n$. That means that the $m$-th digit $a_m$ of the quinary representation of $n$ contributes

$$
\sum_{i=0}^{m-1}5^i=\frac{5^m-1}4
$$

zeroes, so we have

$$
4k=\sum_m a_m5^m-\sum_ma_m=n-\sum_ma_m\;.
$$

So $n$ is basically $4k$, and then you have to adjust a bit because the sum of the quinary digits gets subtracted.

This is wrong:

Well, a number is going to have as many 0's at the end as it has factors of 10 in it, and since $n!$ is going to have an abundance of 2's, you only need to look for 5's. So every time you hit a $n$ that is divisible by 5, $n!$ gains another 0. So there are 5 n such that $n!$ has k 0's. Specifically, $n = 5k,\, 5k+1,\, 5k+2,\, 5k+3,\, 5k+4$

At least I'm pretty sure I'm right, but I have no experience with number theory.

Why it is wrong: 25! has 2 5s, so does 50!, 75!; 125! has 3 etc. ie, Problems with higher powers of 5. But you will have either 5 or 0 n for any k.


Variance of binomial distribution Why for $X\sim B(n,p)$ is $Var(X)=np(1-p)$?

$Var(X)=\sum x_i^2 p_i -(\sum x_i p_i)^2=\sum_{r=0}^n r^2 \binom{n}{r}p^r(1-p)^{n-r}+( \sum_{r=0}^n r \binom{n}{r}p^r(1-p)^{n-r} )^2$

In my short-sightedness, I don't see any viable ways to derive the variance from this.

An easier way is to recognize that $X = Y_1 + Y_2 + \cdots Y_n$ where $Y_k$ are independent Bernoulli random variables with parameter $p$. For a Bernoulli random variable $Y_k$, we have $$\text{Var}(Y_k) = p(1-p)$$ Since $Y_k$ are independent, we have that
$$\text{Var}(X) = \text{Var}(Y_1) + \text{Var}(Y_2) + \cdots + \text{Var}(Y_n) = np(1-p)$$

To go the direct way, we need to first evaluate couple of summations.

We will evaluate the sums $$\sum_{k = 0}^n k \mathbb{P}(X=k) \text{ and }\sum_{k = 0}^n k^2 \mathbb{P}(X=k)$$
First note that $\mathbb{P}(X=k) = \dbinom{n}k p^k (1-p)^{n-k}$. Hence, $$\sum_{k = 0}^n k \mathbb{P}(X=k) = \sum_{k = 0}^n k \dbinom{n}k p^k (1-p)^{n-k}$$
Note that $$k \dbinom{n}k = \dfrac{n!}{(n-k)! (k-1)!} = n \dbinom{n-1}{k-1}$$ Hence,
\begin{align}
\sum_{k = 0}^n k \mathbb{P}(X=k) & = \sum_{k = 1}^n n \dbinom{n-1}{k-1} p^k (1-p)^{n-k} = np \sum_{k = 1}^n \dbinom{n-1}{k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}\\
& = np \left( p + (1-p)\right)^{n-1} = np
\end{align}
Similarly,
\begin{align}
k^2 \dbinom{n}k & = k \dfrac{n!}{(n-k)! (k-1)!} = n k \dbinom{n-1}{k-1}\\
& = n (k-1) \dbinom{n-1}{k-1} + n \dbinom{n-1}{k-1}\\
& = n(n-1) \dbinom{n-2}{k-2} + n \dbinom{n-1}{k-1}
\end{align}
Hence,
\begin{align}
\sum_{k = 0}^n k^2 \mathbb{P}(X=k) & = \sum_{k = 2}^n n(n-1) \dbinom{n-2}{k-2} p^k (1-p)^{n-k} + \sum_{k = 1}^n n \dbinom{n-1}{k-1} p^k (1-p)^{n-k}\\
& = n(n-1)p^2 + n p
\end{align}
Hence,
\begin{align}
\text{Var}(X) & = \sum_{k = 0}^n k^2 \mathbb{P}(X=k) - \left(\sum_{k = 0}^n k \mathbb{P}(X=k) \right)^2\\
& = n(n-1)p^2 + n p - (np)^2 = n^2p^2 - np^2 + np - n^2 p^2\\
& = np(1-p)
\end{align}

Replace $1-p$ by $q$, then replace each factor of $r$ by $p\partial/\partial p$.


Proving the length of angle bisector How do I prove that a triangle with sides a, b, c, has an angle bisector (bisecting angle A) is of length:

$$\frac{2 \sqrt{bcs(s-a)}}{b+c}$$

I have tried using the sine and cosine rule but have largely failed. A few times I have found a way but they are way too messy to work with.

Assuming the usage of Trigonometry is allowed,

Let $AD$ be the bisector of $\angle BAC$

$\triangle ABC=\frac12bc\sin A $

$\triangle ABD=\frac12\cdot c\cdot AD\sin\frac A2$ and $\triangle ADC=\frac12\cdot b\cdot AD\sin\frac A2$

$\triangle ABC=\triangle ABD+\triangle ADC$

$\sin A=2\sin\frac A2\cos\frac A2$

As $\displaystyle 0<A<\pi,0<\frac A2<\frac\pi2\implies \cos\frac A2>0\implies \cos \frac A2=+\sqrt{\frac{1+\cos A}2}$ as $\cos A=2\cos^2\frac A2-1$

Use $\displaystyle\cos A=\frac{b^2+c^2-a^2}{2bc}$  and $2s=a+b+c$

The length of the angle bisector of a triangle is (2bc*cosa/2)/b+c
Just prove this using cosine rule and you will be done because $\cos a= \sqrt{s(s-a)bc}$


Differential equation $\left(x^2+xy\right)y'=x\sqrt{x^2-y^2}+xy+y^2$ I am not sure which type of differential equation this falls into:

$$\left(x^2+xy\right)y'=x\sqrt{x^2-y^2}+xy+y^2$$ any hints? P.S. I first tried reornazing it so I have $y'$ alone, and hoping that I would get a homogeneous equation, but no such luck.

Divide $x^2$ from both sides and then let $z=\dfrac{y}x$.

Note that $y'=z+xz'$.

The new equation becomes $(1+z)y'=\sqrt{1-z^2}+z+z^2$.

$y'=\sqrt{\dfrac{1-z}{1+z}}+z$

$x\dfrac{\mathrm dz}{\mathrm dx}=\sqrt{\dfrac{1-z}{1+z}}$

$\dfrac{\mathrm dx}{x}=\sqrt{\dfrac{1+z}{1-z}}\ \mathrm dz$

$\ln(x)=\sqrt{\dfrac1{1-z}}\left((z-1)\sqrt{z+1}+2\sqrt{1-z}\sin^{-1}\left(\dfrac{\sqrt{1+z}}{\sqrt2}\right)\right)+C$

And... it gets too complicated.


Random walk in the plane A particles moves in $\mathbb{R^2}$ started at the origin. At each stage $i (i = 1, 2, ...)$, the particles would move, independently of all the stages before, one of the four directions North, East, South and West 1 unit, with probability $1 \over 4 $ each.Let $T_n$ be the distance from the origin just after $n$ steps. What is $E(T_n)$. I tried define 4 $1\times 2$matrice with only $1,0,-1$ in all entries. The use technique similar to simple random walk in 1 dimension and we know that for a $1\times 2$ matrice $(x,y)$, the distance from origin =$\sqrt {x^2+y^2}$ but i am not sure how to like the 1 D to 2 D. As in 2 D there are 4 possible direction.

An explanation is given here.

Using that formula you get for the 2D case for the distance from the origin just after n steps:

$$\sqrt{2n} \dfrac{\Gamma(\frac{3}{2})}{\Gamma(\frac{1}{2})}=\sqrt{\frac{n}{2}}$$


How do I find the area of this region? A square with edge length 2 cm has semicircles drawn on each side.
Find the total area of the shaded region.

Here is an image of the diagram shown :

Please show your work in pictures, numbers, words, anything. (Try to keep it to a Grade 8 level too)

The square has area $4$.

The top and bottom white region together have area $4$ minus $2$ times the area of the semi circle, that is, $4-2\times \pi r^2/2=4-\pi$ (since $r=1$).

So, dividing by $2$, the top blue region has area $2-\pi/2$.

If you remove this area from the area of the top semi circle you obtain the area of the two top red parts.

Double it to obtain the red area.

A layman's (mine) approach to solving it:

There are 4 half-circles contained in the square. The 4 half-circles have the same area as two whole circles with the same radius. The radius (half the length of the diameter, which is the length of a side of the cube) is 1 and the area of a circle is $\pi r^2$, so the area of the four half-circles is $2 ( \pi * (1)^2)$, which is $2\pi$. The square can only hold an area of $4$ and all its space is used, so the excess area of the circles' area must be contained in overlap. The the area not able to be contained be the square without overlap is given by $2\pi - 4$.

This, of course, only works out nicely because ALL the area inside the square is being used. Were part of the square unused, you would have to add the area of the unused portion into the calculation of the overlap.