sequence /series convergence of 2^2n 3^(1-n) this step in the proof is confusing me:

$$\sum_1^\infty {\frac{4^{n}}{3^{n-1}}}\qquad \longrightarrow \qquad\sum_1^\infty 4\left(\frac43\right)^{n-1}$$

please explain how/why this happened?

cheers,

gregg

The $n^\text{th}$ term was rewritten by pulling out a factor of $4$ from the numerator.  Maybe seeing a couple of extra steps will help:

$$\frac{4^n}{3^{n-1}}=\frac{4\cdot4^{n-1}}{3^{n-1}}=4\frac{4^{n-1}}{3^{n-1}}=4\left(\frac{4}{3}\right)^{n-1}$$

4^n is really 4 * 4 * 4 * 4 …. 4^n
so by taking one of those 4's out of 4^n, it becomes 4^n-1


An elementary proof for a series involving logarithm I want to prove that

$$ \sum_{x=1}^{\infty} \frac{\ln x}{x^2} \leq 1$$

in an elementary way, but haven't found such a proof yet. Would you please share with us if you find any elementary proof of the inequality?

Also, is there any elementary method to find upper bounds for series involving logarithms?

An alternative way is to exploit creative telescoping. One may prove the inequality
$$ \forall n\geq 2,\qquad \frac{\log n}{n^2}\leq \frac{\log(n+1)+1}{n}-\frac{\log(n+2)+1}{n+1}\tag{1} $$
and deduce from it that
$$ \sum_{n\geq 1}\frac{\log n}{n^2} \leq \frac{\log 2}{4}+\frac{\log(4)+1}{3} < 1 \tag{2}$$
since $\log(2)<\frac{8}{11}$. This can be shown via
$$ 0\leq \int_{0}^{1}\frac{x^3(1-x)^3}{1+x}\,dx = \frac{111}{20}-8\log(2) \tag{3}$$
and $\frac{111}{160}<\frac{8}{11}.$

Have you tried using a taylor expansion for ln(x), Then dividing this by x^2. It should always be 1/x^n for each term in that expansion.


Combinatorics Homework Problem: $n$ People Sit Down. What is the Probability that $m$ Seats are Occupied? From Y.A. Rozanov. Probability Theory: A Concise Course: Chapter 1. Problem 6.

Suppose $n$ people sit down at random and independently of each other in an auditorium containing $n+k$ seats. What is the probability that m seats specified in advance $(m \le n)$ will be occupied?

My Solution:

First, there are $ n+k \choose n $ ways to choose n seats from n+k seats. That is the denominator.

What I want to find is the number of ways the m seats can be occupied by n people where there are more people than m and n+k total seats.

I start by thinking of the m seats as baskets that can be filled with n balls. What is the number of ways to fill m baskets with n balls? The first basket can be filled n different ways. The second basket can be filled n-1 different ways. The mths basket can be filled n-m+1 different ways. Therefore, by the usual rule for sampling without replacement, all the m baskets can be filled $\frac{n!}{m!*(n-m)!)}$ different ways. But the other n+k-m baskets (these are the "seats" leftover from the n+k total after m have been selected) can be filled also. The question now is, how many ways can these be filled given that that all the m seats are filled? There are n-m left over balls. And there are n+k-m left over baskets. So I believe these can be filled $n+k-m \choose n-m$ ways.

So my answer is:

$\frac{\frac{n!}{m!*(n-m)!)}{n+k-m \choose n-m}}{ n+k \choose n }$

I'm not confidant in this answer. :(

Sorry that I re-expressed it in terms of baskets instead of seats, but that makes it easier for me to think about.

The denominator is right. There are $\binom{n+k}{n}$ ways to choose $n$ seats, and all choices are equally likely.

We have $m$ seats specified in advance. Call them the marked seats. How many ways can seats be chosen so that all the $m$ marked seats are chosen?

We have no choice about the $m$ seats, they must be occupied. So our only freedom is in choosing $n-m$ seats from the  $n+k-m$ unmarked seats. There are $\binom{n+k-m}{n-m}$ ways to do this. Now divide.

Remark: You identified the denominator correctly, and got a term $\binom{n+k-m}{n-m}$ in the numerator. But you had an additional term that should not be there.

Here' how I understand it:
The probability of the first of those m seats to be occupied is 1/(n+k), the second 1/(n+k-1), the third 1/(n+k-2), the mth 1/(n+k-m+1). Then the probability of all m seats being occupied is 1 / (n+k)!/m!= m!/(n+k)!


Show that $f(x,y) = \frac{x^3+y^3}{x-y}$ when $x \neq y$ is discontinuous at the origin Show that the following function is discontinuous at the origin
  $(x,y)=(0,0)$

  $$
\begin{equation}
f(x,y) =
\begin{cases}
\frac{x^3+y^3}{x-y} &x\neq y\\
0  &x = y
\end{cases}
\end{equation}
$$

Now I am taking $x=my\ (m \neq 1)$ and getting $f(x,y)$ as

$$\lim_{(x,y) \to (0,0)}f(x,y)=\lim_{x \to 0}\frac{x^3+m^3x^3}{x-mx} = \lim_{x \to 0}\frac{x^2+m^3x^2}{1-m}$$

which tends to zero showing that the function is continuous. But we need to show discontinuance, I am not getting any other substitution for $y$ that will prove discontinuity.

which tends to zero showing that the function is continuous

No, the existence of the limit of $f(x,y)$ as $(x,y)$ approaches $(0,0)$ along one path does not show that the function is continuous at $(0,0)$. To prove continuity, you need to prove this limit exists (and has the same value) along every path to $(0,0)$.

Thus to disprove continuity, you only need to show the non-existence of the limit, or the existence of a limit with a value different from $f(0,0)$, along a single path to $(0,0)$. You already tested all linear paths, so try something that's not linear.

It might be helpful to use long division to write:
$$\frac{x^3+y^3}{x-y} = x^2+xy+y^2+\frac{2y^3}{x-y}, \quad x \neq y.$$
The $x^2+xy+y^2$ portion is continuous at $(0,0)$, so the $\frac{2y^3}{x-y}$ portion is the discontinuous portion. Notice that the numerator is cubic in $y$, while the denominator is linear in $x$ and $y$, so (as you found) if $y$ is a linear function of $x$ then the numerator will approach $0$ faster than the denominator. To get a limit other than $0$, you want the numerator to approach zero no faster than the denominator, which suggests trying the path $y=mx^{1/3}$, or something that goes even slower like $y=mx^{1/4}$.

If $n\in\mathbb N$, then$$f\left(\frac1n,\frac1n+\frac1{n^3}\right)=-1-\left(1+\frac1{n^2}\right)^3$$and therefore $\lim_{n\to\infty}f\left(\frac1n,\frac1n+\frac1{n^3}\right)=-2$.


Is $\mathbb{C}$ equal to $\mathbb{R}^2$? Complex numbers are usually formally defined as pairs of real numbers. Although there are operations on $\mathbb{C}$, such as complex multiplication, which are not found in operations usually applied to $\mathbb{R}^2$, the sets themselves seem to be the same. Each consists of pairs of real numbers.

So is it okay to say that $\mathbb{C} = \mathbb{R}^2$? It seems formally correct, but something doesn't feel quite right about it.

The answer is both yes, and no.

$\Bbb C$ and $\Bbb R^2$ are both sets with the same cardinality, and they have a very natural bijection between them which preserves a lot of nice properties. So much that we can almost say that these two sets are the same for a lot of purposes.

But these two carry very different structure as a natural structure. $\Bbb C$ is a field and $\Bbb R^2$ is not (because pointwise multiplication does not form a field).  One can even argue that formally $\Bbb C$ is in fact $\Bbb R[x]/(x^2+1)$, and not $\Bbb R^2$, and one would be at least partially correct.

Personally, I'd support the "no" answer more than the "yes" answer. And here's why. We often like to think about $\Bbb R$ as a subset of $\Bbb C$. Namely $x\in\Bbb C$ is a real number if and only if $\overline x=x$. But $\Bbb R$ is not a subset of $\Bbb R^2$, instead there is an obvious embedding $x\mapsto(x,0)$, but still real numbers are generally not ordered pairs of real numbers (you can even notice that this approach takes $\Bbb C$ as sort of a primitive notion, and not quite as $\Bbb R[x]/(x^2+1)$ as others might see it).

Although, as I said, it depends on how you define things, because "formally" things can be done in plenty of different ways.

If you want to be super formal, you'd say that there exists an equivalence relation between $C$ and $R^2$ defined as $x+yi \sim (x, y)$. It's a very useful and intuitive relationship (in fact one of the reasons it's useful is because it's intuitive), but as you note there are operations that usually only make sense in one or other of the sets, and you'd have to explicitly define that operation in the other set and then show that it's preserved across the relation.


How to calculate weighted average based on recency I am trying to estimate the unit cost of a product based on the unit cost on past purchase orders. For example, let's say I have the following purchase order history:

PO 1: January 1, 2019   -    Quantity 10   -   Unit cost $10
PO 2: March 1, 2019     -    Quantity 20   -   Unit cost $12
PO 3: July 1, 2019      -    Quantity 30   -   Unit cost $16

On August 1, 2019 I receive a product back from a customer. I do not know which PO the product originally came from. I would like to estimate the unit cost of that one product by calculating a weighted average based on quantity for all of the unit prices from past POs. That part seems straight forward.

However, I would also like to ensure that the most recent PO has more weight than POs in the past.

Does anyone have a recommendation for how to do this?

I use a backwards halving algorithm to do this. It requires running sums, but can be done in a spreadsheet, so not so bad...
Pick a time period for the first halving (say 365 days). For POn, the weight will be:
Weight = 2 ^ (POdate - Now) / 365
Now, take the value of POn and multiply it by the weight to get the weighted value:
WeightedValue = POvalue * Weight
Finally, take the sum of the weighted values and divide it by the sum of weights (hard to show in text). This value will provide a recency-biased average.


Solve $5^{x/2}-2^x=1$ I tried to think of it a lot but didn't get any breakthrough. I was trying with substitution method, but things were not fitting in.

Actual answer is 2

Let $x=2u$, then the equation becomes:
$$5^u-2^{2u}=1\Rightarrow 5^u=4^u+1^u$$
Divide both sides by $5^u$,
$$1=\left(\dfrac{4}{5}\right)^u+\left(\dfrac{1}{5}\right)^u$$
Easily observe that $u=1$ is one of the answer. However, as $\left(\dfrac{4}{5}\right)^u+\left(\dfrac{1}{5}\right)^u$ is strictly decreasing, so there must only have at most one answer. Therefore the answer is $u=1$, which means that $x=2$

Rewrite,

$$5^{x/2}-2^x=1$$

as

$$5^{x/2}=4^{x/2} + 1^{x/2}$$

Since $5=4+1$ and they all have the same exponential, the exponential must be 1,

$$\frac x2=1$$

which yields

$$x=2$$


Lipschitz-constant gradient implies bounded eigenvalues on Hessian I've read in a few places that if we have a Lipschitz gradient

$$\|\nabla f(x) - \nabla f(y)\|\leq L\|x-y\|,\, \forall x,y, $$
we can equivalently say $\nabla^2f\preceq LI.$ But I'm having a hard time showing this. (Equivalently, I want to show $z^T \nabla^2f(x)z\leq z^TLIz=Lz^Tz,\forall\, x,z $.)

This is not true as stated. For example, the function $f(x)=x|x|$ on the real line has Lipschitz gradient, but is not twice differentiable.
Also, the function $f(x)=-x^4$ satisfies $f''\le LI$ with $L=0$, but its gradient is not Lipschitz continuous.

The two properties are equivalent for functions that are convex and twice differentiable. For such functions, $\nabla^2 f$ is a positive semidefinite matrix, so its norm is its largest eigenvalue. Hence,
$$\nabla^2 f \preceq LI \iff \|\nabla^2 f\|\le L \iff \|\nabla f(x)-\nabla f(y)\|\le L\|x-y\|$$
where the last equivalence is based on the mean value theorem.

It is equivalent for $||\bigtriangledown ^2f(x)||_2 \leq L$ where $||\bigtriangledown ^2f(x)||_2$ means the maximum singular value.


Dual Bundles' Local Trivializations Confusion $$\newcommand{\R}{\mathbb R}$$
Let $\pi:E\to M$ be a rank $k$ smooth vector bundle over a smooth manifold $M$. I will below describe how to form the dual bundle, wherein lies my question.

Let $E_p=\pi^{-1}(p)$ for each $p\in M$, and define $E^*=\bigsqcup_{p\in M}E_p^*$. Further define $\pi^*:E^*\to M$ as $\pi(E^*_p)=\{p\}$ for all $p\in M$.

The main part is to define the local trivializations. For each local trivialization $\Phi:\pi^{-1}(U)\to U\times \R^k$ of $E$ over $U$, define $\Phi^*:\pi^{*-1}(U)\to U\times \R^{k*}$ as
$$\Phi^*(\omega)=(p, (\Phi|_{E_p})^{-t}\omega)$$
for all $\omega\in E_p^*$, and all $p\in U$. (The '$t$' denotes the transpose).

The problem with the above definition is that a local trivialization is supposed to map $\pi^{*-1}(U)$ to $U\times \R^k$ and not $U\times \R^{k*}$.
Now I know that $\R^k$ can be canonical identified with $\R^{k*}$ provided we have an inner product on $\R^k$.

But then different inner products will lead to different $\Phi^*$ and consequently different transition functions.

Is there a preferred inner product with respect to which the identification of $\R^k$ with $\R^{k*}$ is made?

The identification of $V$ with $V^*$ via a scalar product is done for vector spaces in which there is no preferred basis. This is not the case of $\mathbb R ^n$ which has the usual basis (in terms of which it is constructed, after all) that allows you to define the identification $e_i \mapsto e^i$ where $e^i (v) = e^i (\sum v^j e_j) =v^i$. So, the idea is that for $\mathbb R^n$ identifies with $(\mathbb R^n)^*$ purely algebraically, with no need for a scalar product.

The fibers $p^{-1}(\{x\})$ of a vector bundle $p : E \to X$ are just finite-dimensional vector spaces (say, over $\mathbb{R}$ here). No bases are chosen. So it is "wrong" to write $\mathbb{R}^n$ here. $\mathbb{R}^n$ is not "the" $n$-dimensional vector space, but it is the "the" $n$-dimensional vector space with a chosen basis.

The dual of the trivial vector bundle $X \times V \to X$ (where $V$ is a finite-dimensional vector space) is defined by the trivial vector bundle $X \times V^* \to X$. Any isomorphism $X \times V \to X \times W$ of vector bundles over $X$ is induced by a unique isomorphism $V \to W$ of vector spaces. This may be dualized to an isomorphism $W^* \to V^*$ of vector spaces. The inverse of this is an isomorphism of vector spaces $V^* \to W^*$, which thus corresponds to an isomorphism $X \times V^* \to X \times W^*$ of vector bundles over $X$.

Thus, when $E \to X$ is any vector bundle, glued from trivial vector bundles, along isomorphisms on the intersections, we may also glue the corresponding dual trivial vector bundles together and obtain the dual vector bundle $E^* \to X$.

Notice that $E^* = \coprod_{x \in X} (E_x)^*$ is wrong (or at least, utterly confusing); we only have $|E^*| = \coprod_{x \in X} |(E_x)^*|$, where $|-|$ means the underlying set here.


Why is $a^{-1}$ used instead of $\frac{1}{a}$ in Calculus by Spivak? In Calculus by Michael Spivak, why did he use the form $a^{-1}$ instead of $\frac{1}{a}$ to explain the multiplicative inverse (reciprocal) property? Is there a conceptual reason why the former is preferred rather than the latter? I know they are the same thing but I can't help but feel those who are not strong on proofs would have an easier time seeing the solution to problem 1(i) if it was used in that way rather than what was given. Any insight as to why his style is used or preferred in analysis is greatly appreciated.

Because the notation $\frac1a$ suggests that this is the number that you get when you divide $1$ by $a$. And, yes, it is! But one should use that notation only after it was proved that $a^{-1}{}{}{}{}{}{}$ is indeed what you get when you divide $1$ by $a$.

The question here is, how do we define exponentiation for negative integer powers? Given that our definition for $n\in\mathbb{N}$ is $$x^n=\underbrace{x\cdot ...\cdot x}_{n\text{ times}}$$
With the definition $x^0:=1$.
The property that we want exponentiation to satisfy is the equation $x^{a+b}=x^ax^b$. If we take this property as an axiom, then we can see
$$x^{1-1}=x^0=1$$
But on the other hand,
$$x^{1-1}=x^1x^{-1}=x\cdot x^{-1}$$
So we have
$$x\cdot x^{-1}=1$$
Hence we conclude
$$x^{-1}=\frac{1}{x}.$$
We can now extend this to any negative integer that we want. First write
$$x^{-n}=x^{\sum_{i=1}^n (-1)}$$
Then using the defining property of exponentiation,
$$=\prod_{i=1}^n x^{-1}={(x^{-1})}^n$$
We can go further! If we take as another axiom $x^{ab}={(x^a)}^b={(x^b)}^a$ (which we definitely want, since our previous work demands that $x^{mn}=x^m\cdot ...\cdot x^m$, $n$ times) then we can define exponentiation for reciprocals of natural numbers. The question is, what is $x^{1/n}$ ? Well,
$$x=x^1=x^{n/n}$$
Now using our other defining property of exponentiation,
$$x^{n/n}=(x^{1/n})^n=x$$
If we now define $\sqrt[n]{}$ as the inverse operation of exponentiation, we can say that
$$\sqrt[n]{(x^{1/n})^n}=\sqrt[n]{x}$$
But by the way we have defined this operation,
$$x^{1/n}=\sqrt[n]{x}$$
So in words, $x^{1/n}$ is the (I won't worry about uniqueness for now) number $a$ such that $a^n=x$. Now using the properties of exponentiation, we can extend this to any rational power:
$$x^{p/q}=\sqrt[q]{x^p}=\left(\sqrt[q]{x}\right)^p$$
Then of course we can extend this to any real number. Let $r\in\mathbb{R}$ be an irrational number. Let $\{p_n\},\{q_n\}$ be sequences of integers such that the sequence
$$r_n=\frac{p_n}{q_n}\to r\text{  as  }n\to\infty$$
We can then say
$$x^r:=\lim_{n\to\infty} x^{p_n/q_n}$$
While this definition is not particularly computationally efficient, it is the most intuitive.
Hopefully this serves as a good overview of why the notation is the way that it is and how, I think, exponentiation was defined in antiquity before fancy tools like $e^x$ and whatnot.


Must a function that 'preserves r.e.-ness' be computable itself? Does there exist a non-recursive function (say, from naturals to naturals) such that the inverse of every r.e. set is r.e.?

If yes, how to construct one?
If no, how to prove that?
Any References?

For any sequence $\mathcal{B} = \langle B_i : i \in \mathbb{N}\rangle$ of subsets of $\mathbb{N}$, there is a $\mathcal{B}$-cohesive set $A$: an infinite set such that, for any $B_i \in \mathcal{B}$, either there are only finitely many elements of $A$ in $B_i$, or all but finitely many elements of $A$ are in $B_i$. The set $A$ is called "cohesive" because, although it is infinite, none of the sets $B_i$ is able to split it into two infinite pieces.

Apply that fact with $\mathcal{B}$ containing all the r.e. sets to obtain a cohesive set $A$. Let $f$ be the function that enumerates $A$ in increasing order.   First, $A$ is not computable, or even r.e., because if $A$ was r.e. then we could use that enumeration to effectively enumerate "every other element" of $A$, which would give us an r.e. set that splits $A$ into two infinite pieces, which is impossible.  Moreover, because $A$ is not computable, $f$ is not computable - if we could compute $f$ then we could enumerate $A$ in increasing order, which would imply $A$ is computable.

Now let $B$ be any r.e. set. Then, because $A$ is cohesive for r.e. sets, one of two things must happen:

All but finitely many elements of $A$ are in $B$. In this case $f^{-1}(B)$ is cofinite, and hence is computable.
Only finitely many elements of $A$ are in $B$. In this case, $f^{-1}(B)$ is a finite set, and is thus computable.

Either way $f^{-1}(B)$ is not only r.e., it is computable.

I guess that you have meant image instead of inverse...

In that case in my opinion the answer is that such a function exists. Since you can construct a function $\phi$ as follows: for numbers representing valid pairs (code of a Turing machine, input) return 1, if the machine halts on the input, 0 otherwise. For all other numbers, return 0. This function is clearly not recursive. However, for all r.e. sets, the image is either $\{0\}$, or $\{0,1\}$, which are both r.e.


Estimate the decrease in the period of the satellite to the nearest one-hundredth hour.... According to Kepler's Third Law, the period T (in days) of a satellite moving in a circular orbit x mi above the surface of the earth is given by $T=.0588(1+\frac{x}{3959})^{\frac{3}{2}}$ Suppose that a communications satellite is moving in a circular orbit 28,000 mi above the earth's surface. Because of friction, the satellite drops down to a new orbit 27,500 mi above the earth's surface. Estimate the decrease in the period of the satellite to the nearest one-hundredth hour.

So this is what I think I have, not sure whether I'm right or wrong. $ΔT=T'dx$ So $dx=-500$ and $T'=.0588(\frac{3}{2}(1+\frac{x}{3959})^{.5}*\frac{1}{3959}$ So $T'(28,000)=.000063298$ and then I multiply $.000063298(-500)=-.03(rounded)$ but it's wrong and I suppose I'm approaching the problem wrong. A precise step by step solution would be greatly appreciated, thanks.

Could you just sub in T(28000) - T(27500)?
Edit: After a quick read of linear approximation, it looks like what you're doing is right. Did you check your units though?


What is the best calculus book for my case? My professor of calculus said that he will not adopt books like "stewart" or "thomas" because they are too easy for a physics undergrad course.

He said that Apostol is a great book for our case, but some people told me that this book focus a lot in the historical concept and was not good to them.

I know that my professor likes extreme rigour in math, so do i.

In the words of Apostol itself (preface):

Some people insist that the only way to really understand calculus is to start off with a thorough treatment of the real-number system and develop the subject step by step in a logical and rigorous fashion

What books follow this principle of learning calculus?

Thank you.

Spivak's Calculus and Calculus on Manifolds (for multivariable calculus) are pretty standard rigorous calculus texts. Rudin's Principle's of Mathematical Analysis is standard for a first analysis course, but may be too abstract for a physics course. If you don't like Apostol but still want the mathematical rigor, these are good alternatives.

There's also Strang's free calculus text here.

For more calculations, Hubbard's Vector Calculus book works as well.

Edit: Yes, it does depend on where you are from, since American students (such as myself) wouldn't study Spivak in our first year (to be fair, I actually never studied Spivak's Calculus, just his Calculus on Manifolds for a freshman advanced math course but this was far from the norm and only for math majors, not physics majors) and it is considered to be one of the best rigorous calculus textbooks as far as I'm aware. Seeing as the OP's professor mentions Apostol (although not mentioning which Apostol book he means, either his two Calculus books or his more advanced Mathematical Analysis), I think Spivak's Calculus books are a good alternative to Apostol's Calculus books, although they do not contain any differential equations theory or as extensive a treatment of linear algebra as Apostol (Calculus on Manifolds includes a bit of linear algebra in as far as it allows him to discuss total derivatives and multilinear forms).

All this said, I'd honestly need a bit more information on the course in order to give a proper alternative. There are far too many advanced "rigorous" vector calculus textbooks to choose from for such a general recommendation, so I gave what are considered by most people to be fairly standard references.

There are other books of course:

Fitzpatrick's Advanced Calculus: Again, little in the way of
linear algebra or differential equations, but I think he explains
integration much better than Spivak and a bit more drawn out explanations, which I think is nice if you're new to the material.
Boas's Mathematical Methods in
the Physical Sciences: This book probably has more material than you
need (PDEs, integral transforms, basic complex variables). However,
I'd say probably the first 400 pages is a great alternative to
Apostol, and as the title suggests it's good for a physics course.
Stroyan's Foundations of Infinitesimal Calculus: Haven't read this one to be honest, but seems pretty good based on the table of contents, although it does seem a bit short.
Protter's Basic Elements of Real Analysis: A nice springer undergraduate text which is an abstract version of Stewart's Calculus, but not quite as rigorous as others on this list. For more rigor but using a similar style, he also has a book called A First Course in Real Analysis, which I also like.

These are my best "general recommendations". Honestly, we do need me information to give good recommendations (does your class have a website/what is the class on/etc.).

Hope this helps.

You probably mean "math" undergrad course. In your case, you should check "About Vector by Hoffman" and "div grad curl and all that by schey." But of them are cheap books, and they are pretty abstract, but they are only use for Cal 3. From cal 1-2, I used steward, which is pretty basic though.


Having two points of a square and only a compass, how to find the remaining two? I remember being presented a mathematical puzzle some years back that I still can't solve. The problem is defined as follows:

We have two points on a plane, and using only a compass, how do we find other two points, so that all four of them would be vertices of a square?

I'm not sure whether the first two points were supposed to be vertices of the same edge of a square or not, so solutions to both variants are welcome.

Here is a description of one of the constructions:

Given the two points $\color{maroon}0$ and $\color{maroon}d$ that form one side of the square:

Construct the five $\color{gray}{\text{gray}}$ circles of common radius $\color{maroon}{od}$ shown in the
diagram and locate the points  $\color{maroon}a$, $\color{maroon}b$,
and $\color{maroon}c$.
Construct the two $\color{maroon}{\text{maroon}}$ circles: one with
center $\color{maroon}c$ and radius $\color{maroon}{cb}$, and one
with center $\color{maroon}d$ and radius $\color{maroon}{ad}$. Note
that $\color{maroon}{ad}$ and $\color{maroon}{cb}$ have the same
length.
Locate the point of intersection $\color{pink}e$ of the two maroon
circles.
Construct the $\color{pink}{\text{pink}}$ circles of radius $\color{pink}{oe}$ at centers
$\color{maroon}c$ and $\color{maroon}d$.
The point of intersection,  $\color{pink}f$, of the pink circles  is a vertex of the
square.
Draw the $\color{darkgreen}{darkgreen}$ circle centered at $\color{pink}f$ of radius $\color{maroon}{od}$.
The point of intersection,  $\color{darkgreen}g$, of the darkgreen
circle with the gray circle centered at $\color{maroon}d$ is the
final vertex of the square.

Justification of step 5:

From step 1., $\color{maroon}{aboc}$ is a rhombus with common side length $  l(\color{maroon}{co})$.
Since the point $\color{pink}e$ is equidistant to the points $\color{maroon}c$ and $\color{maroon}d$ , the segment $\color{pink}e\color{maroon}o$ is perpendicular to the segment $\color{maroon}{cd}$. Since $\color{pink}f$ is equidistant to the points $\color{maroon}c$ and $\color{maroon}d$, the points $\color{pink}e$, $\color{pink}f$, and $\color{maroon}o$ are colinear and the segment $of$
is perpendicular to  the segment $\color{maroon}od$.

We need to show that $l(fo)=l(co)$.
Proceeding with some abuse of notation:

Considering the rhombus $aboc$, we have
$$
cb^2+ao^2=2(ac^2+co^2 );
$$
or, since $ao=co=ac$,
$$
cb^2=3co^2.
$$
Since $cb=ce$, we have
$$\tag{1}
ce^2=3co^2
$$
Considering now the right triangle $ceo$:
$$\tag{2}
ce^2=eo^2+co^2.
$$
Combining equations $(1)$ and $(2)$:
$$\tag{3}
eo^2=2co^2.
$$

Considering now the right triangle $cfo$:
$$\tag{4}
cf^2=fo^2+co^2
$$
since $cf=oe$:
$$\tag{5}
oe^2=fo^2+co^2
$$
From $(3)$ and $(5)$ now, we finally obtain
$$
 fo=co,
$$
as desired.

There's an easier solution with a collapsible compass.
Let the final square side (distance between two initial points) be equal to 1.

First, draw three small circles with the radius of 1
Then draw tree large circles with the radius of sqrt3.
Now all you need is to find sqrt2. That will be outer crossing of the large circles. Draw two medium circles centered in your initial points.
The crossing of the medium circles with the first two small ones are the top points you need.

See diagram


Proving that $\liminf (x_n)+\liminf (y_n)\le \liminf (x_n+y_n)$ I know that $\limsup (x_n)+\limsup (y_n)\ge \limsup (x_n+y_n)$, but I'm not sure how to show that $\liminf (x_n)+\liminf (y_n)\le \liminf (x_n+y_n)$. Please help me.

Start with the inequality for $\limsup$ that you know, and use the hint of RRL.
\begin{align}
\limsup (-x_n) + \limsup (-y_n) &\ge \limsup (-x_n-y_n)\\
-\liminf x_n - \liminf y_n &\ge -\liminf (x_n+y_n)\\
\liminf x_n + \liminf y_n &\le \liminf (x_n+y_n)
\end{align}

Fix $n$ for the moment and let $l \ge n$. Since $\inf_{ k \ge n} x_k \le x_l$ and $\inf_{ k \ge n} y_k \le y_l$, we see that
$\inf_{ k \ge n} x_k + \inf_{ k \ge n} y_k \le x_l + y_l$. Hence
$\inf_{ k \ge n} x_k + \inf_{ k \ge n} y_k \le \inf_{ k \ge n} ( x_k + y_k)$.

Since these are non decreasing sequences, we have
$\inf_{ k \ge n} x_k + \inf_{ k \ge n} y_k \le \liminf_{ k } ( x_k + y_k)$ and
hence $\liminf_{ k } x_k + \liminf_{ k } y_k \le \liminf_{ k } ( x_k + y_k)$.


Given (p ∧ q), use the Fitch system to prove (q ∨ r) I'm having trouble solving a Fitch Style Proof and I was hoping someone would be able to help me.

Premises: p ∧ q

Goal: q ∨ r

Would be an assumption (p ∧ q => q ∨ r) in first step a correct one?

Then, q ∨ r Implication Elimination: 2, 1

Why is it not correct?

Thank you very much for any help!

HINT

Your premise says $p \land q$ ... what does this mean in plain English?  It means that both $p$ and $q$ are true. Now: are you interested in $p$ being true? Would $p$ help you to get to your goal $q \lor r$? How about $q$?

Here is a proof using a Fitch-style proof checker:

The first line is the premise. The second line is derived from the premise use conjunction elimination ($\land$E). The last line is derived from the second using disjunction introduction ($\lor$I).

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/


How to prove periodicity Modulo b We have the sequence

$$
x_n=a^n \mod{b},
$$
where $a$ and $b$ are positive integers. How to show that it's periodic? It is intuitively clear but I have no clue how to prove it rigorously from first principles...

Thanks in advance

Note first that periodic here means there are $K$, $P > 0$ such that $a^{i} = a^{i+P}$ for $i > K$.

One begins with showing that there are $m > n$ such that $a^{m} =  a^{n}$. Then for all $i > n$ we have
$$
a^{i} = a^{i - n + n} = a^{i - n} a^{n} = a^{i - n} a^{m} =
a^{i + (m - n)}.
$$
So you can take $K = n$ and $P = m - n$.

Using the fact that
\begin{align}
a^{\varphi(b)}\equiv 1 \mod b
\end{align}
and the fact that for any $n$
\begin{align}
n = \varphi(b)k+r
\end{align}
for some $0\leq r<\varphi(b)$ we have the desired result. Here $\varphi(b)$ is the Euler Totient function.


Prove that $\tan x =\sin x + 1$ have only one solution in $(-\frac{\pi}{2},\frac{\pi}{2})$ Prove that $\tan x = \sin x + 1$ have only one solution in $(-\frac{\pi}{2},\frac{\pi}{2})$

Well, I believe I need to show that there's a solution. so I need to find x which solves $\tan x = \sin x + 1$.

Then, I need to derivate it and show that the derivation either bigger than zero, or smaller than zero, Thus - there will be no solutions.

So in intuition way, I know how to solve it, but practically I'm having difficulties. Can you guys help with the steps I've shown?

I believe It's Rolle's rule. Because we just studied it :)

let me rewrite the equation as $$\tan t = \sin t + 1.\tag 1$$  i will use $x = \cos t, y = \sin t.$ then $(1)$ becomes $$\frac y x = y + 1 \to y = \frac x{1-x}\tag 2 $$

the solution to $(1)$ are the points where the hyperbola (2) cuts the unit circle in the first quadrant. this is easy to see because the  branch of the hyperbola going through $(0,0)$ can only cut the unit circle at one in the first quadrant and the second time at a pint in the third quadrant.

$\tan(x)$ is a bijection from $]-\pi/2, \pi/2[$ to $]-\infty, \infty[$ as a continous and increasing function (derivative >0). $\sin(x)+1$ is increasing, continuous function: this is a bijection from $]-\pi/2, \pi/2[$ to $]0,2[$.

hence, the two graphs must have a unique common point as $\sin(x)+1 \in [0, 2|$. To have an interval, you apply the Bolzano's theorem:

if f is continuous on [a,b] and f(a)< 0 and f(b)>0 then, there is $\alpha$ where $f(\alpha)=0$. $\alpha$ is unique if f is a bijection on [a,b] (example : f monotone).

here $f(x)=\tan(x)-\sin(x)-1$. we have $\alpha=1.08264952...$


If the line $ax+by +c = 0$ touches the circle $x^2+y^2 -2x=\frac{3}{5}$ and is normal to $x^2+y^2+2x-4y+1=0$, what is (a,b)? I have a question that goes:

If the line $ax+by +c = 0$ touches the circle $x^2+y^2 -2x=\frac{3}{5}$ and is normal to $x^2+y^2+2x-4y+1=0$, what is (a,b)?

So what I tried was I know that since the line is normal to the 2nd circle, so it must pass through the center of the second circle which is $(-1,2)$.

So from that I got that $$-a+2b+c=0$$

But I cant really find any other equations here that would help, I tried differentiation the curves but I dont have the point of contact so can't really do anything there.

I also know that the tangent to the circle $x^2+y^2+2gx+2fy+c = 0$ at $(a,b)$ is $ax+by+(a+x)g+(b+y)f +c = 0$

I don't know how to proceed, can someone help?

By sketching a graph in mind, we see that the line $ax+by+c=0$ is just a line that is passing through $(-1,2)$ (the center of the circle $x^2+y^2+2x-4y+1=0$) and is tangent to the circle $x^2+y^2-2x=3/5$.

Therefore we looking for tangent lines from point $(-1,2)$ to the circle $x^2+y^2-2x=3/5$.

The line passing through $(-1,2)$ with slop $m$ is
$$y=m(x+1)+2.$$
The intersects points of the line and the circle $x^2+y^2-2x=3/5$ is found by
$$x^2+(m(x+1)+2)^2-2x=3/5$$
simplifying:
$$(1+m^2)x^2+[2m(m+2)-2]x+(m+2)^2-3/5=0\tag{*}$$
Since we need the line $y=m(x+1)+2$ to touch the circle, then the discriminant of $(*)$ must be zero:
$$(2m(m+2)-2)^2-4(1+m^2)((m+2)^2-3/5)=0$$
which simplifies to
$$\frac{-4}{5}(m+3)(m+\frac{1}{3})=0$$
i.e. $m=-3$ or $m=-1/3$. Thus the tangent lines are
$$y=-3(x+1)+2\quad\text{and}\quad y=-\frac13(x+1)+2$$

Another way.

Let $y=mx+n$ be an equation of the tangent.

Thus, $$mx-y+n=0,$$ which since equations of our circles they are
$$(x+1)^2+(y-2)^2=4$$ and $$(x-1)^2+y^2=\frac{8}{5},$$ we obtain that the point $(-1,2)$ is placed on the line $y=mx+n$

and the distance from $(1,0)$ to the line is equal to $\sqrt{\frac{8}{5}}.$

Thus,
$$-m-2+n=0$$ and
$$\frac{|m\cdot1-1\cdot0+n|}{\sqrt{m^2+1}}=\sqrt{\frac{8}{5}},$$ which gives
$$\frac{|2m+2|}{\sqrt{m^2+1}}=\sqrt{\frac{8}{5}}$$ or
$$5(m+1)^2=2(m^2+1).$$
Can you end it now?


Simplifying Algebraic Expression Under A Square Root This is part of a problem for calculating the length of a curve. Unfortunately, I'm stuck on a pretty basic algebra concept.

Solutions for my problem say that:

$\sqrt{\frac{1}{2t} + 1 + \frac{t}{2}} = \frac{\sqrt{t} + \frac{1}{\sqrt{t}}}{\sqrt{2}}$

I cannot understand how they got this at all.

The most I can simplify the original expression is as

$\sqrt{\frac{1}{2}(\frac{1}{t} + 2 + t)} = \sqrt{1/2}\sqrt{\frac{1}{t} + 2 + t}  = \frac{1}{\sqrt{2}}\sqrt{\frac{1}{t} + 2 + t}$

I can't see how they get simplify the algebraic expression under the square root.

Note that $$t + 2 + \frac{1}{t} = t + 2\cdot t\cdot \frac{1}{t}+ \frac{1}{t}=  \left(\sqrt{t} + \frac{1}{\sqrt{t}}\right)^2$$

Put over a common denominator: $$\sqrt{\frac{1}{2t} + 1 + \frac{t}{2}} =\sqrt{\frac{1 + 2t + t^2}{2t}}$$

Factor the top term: $$\sqrt{\frac{(1+t)^2}{2t}}$$

Take the square root: $$\frac {1+t}{\sqrt{2t}}$$

Divide top and bottom by $\sqrt t$:  $$\frac {\frac 1{\sqrt{t}} + \sqrt{t}}{\sqrt 2}$$


Properties of Closure Definitions:

$A'$ is the set of all accumulation or limit points.

$\bar{A} = A \cup A'$ - this is known as the closure of $A$.

Let $A$ be a subset of $\mathbb{R}$. A point $p\in\mathbb{R}$ is an accumulation point of $A$ if and only if every open set $G$ containing $p$ contains a point of $A$ different from $p$.

Prove or disprove: $(\overline{A})^{c} = \overline{(A^{c}})$

proof: This is true since we know what the compliment of the closure set is open.

I am not sure if I am right, assume already that I have proven that the compliment of the closure set is open.

The statement is not true. If $A$ is the set of rationals, then $(\bar{A})^c = (\Bbb R)^c = \emptyset$, but $\overline{A^c} = \bar{\Bbb I} = \Bbb R$, where $\Bbb I$ is the set of irrationals.

This is not generally true. Furthermore, we can make the following claim:

Let $X$ be a connected topological space and $A\subseteq X, A\neq\emptyset, A\neq X$. Then $\left(\overline{A} \right)^c\neq\overline{A^c}$.

Assume the contrary. Then $\left(\overline{A}\right)^c$ is open, but since $\overline{A}=\left(\overline{A}\right)^{c^c}=\left(\overline{A^c}\right)^c$, we also have that $\overline{A}$ is open. Then $X$ is the union of two disjoint nonempty open sets, which contradicts $X$ being connected.


Show that the limit $\lim_{z \to i}{\frac{1-|z|}{i-z}}$ does not exist. Show that the limit $\lim_{z \to i}{\frac{1-|z|}{i-z}}$ does not exist.

I tried by putting $z =ix $, where $x \to 1$ and got that the limit is $\frac{1}{i}$, but can't think of another example.

Hint:
$$\lim_{z \to i}{\frac{1-|z|}{i-z}}= \lim_{s \to 0}{\frac{|s+i|-1}{s}}.$$
Choose $s$ real number then the limit will be 0.

Choose some $k\in\mathbb{C}$ and $\epsilon\in\mathbb{R}$ and substitute $z=i+\epsilon k$ then we have the limit equal to
$$\begin{align}
\lim_{\epsilon\to0}\frac{1-|i+\epsilon k|}{i-(i+\epsilon k)}
&=\lim_{\epsilon\to0}\frac{1-\sqrt{(\Re{(\epsilon k)})^2+(1+\Im{(\epsilon k)})^2}}{-\epsilon k}\\
&=\lim_{\epsilon\to0}\frac{1-\sqrt{1+(\Re{(\epsilon k)})^2+2\Im{(\epsilon k)}+(\Im{(\epsilon k)})^2}}{-\epsilon k}\\
&=\lim_{\epsilon\to0}\frac{1-(1+\frac12((\Re{(\epsilon k)})^2+2\Im{(\epsilon k)}+(\Im{(\epsilon k)})^2+O(\epsilon^3))}{-\epsilon k}\\
&=\lim_{\epsilon\to0}\frac{-\frac12(\Re{(\epsilon k)})^2-\Im{(\epsilon k)}-\frac12(\Im{(\epsilon k)})^2+O(\epsilon^3)}{-\epsilon k}\\
&=\lim_{\epsilon\to0}\frac{-\frac12(\epsilon\Re{(k)})^2-\epsilon\Im{(k)}-\frac12(\epsilon\Im{(k)})^2+O(\epsilon^3)}{-\epsilon k}\\
&=\lim_{\epsilon\to0}\frac{-\frac12\epsilon^2\Re^2{(k)}-\epsilon\Im{(k)}-\frac12\epsilon^2\Im^2{(k)}+O(\epsilon^3)}{-\epsilon k}\\
&=\lim_{\epsilon\to0}\frac1k\left(\frac12\epsilon\Re^2{(k)}+\Im{(k)}+\frac12\epsilon\Im^2{(k)}+O(\epsilon^2)\right)\\
&=\frac1k\Im{(k)}\\
\end{align}$$
Where $\frac1k\Im{(k)}$ is an arbitrary complex number, so the limit does not exist.


Odds of getting specific color of Jelly Beans in a handful? I have a bag of jelly beans with approx 1190 Jelly Belly's in it. There are 50 different flavors. Assuming the amount of Jelly Belly's per flavor are equal (so, 23.8 of each bean):

If I pull 6 Jelly Beans from my unopened bag, what are the odds that 3 of them will be the same color?

I'm asking because I got a bag of jelly beans for Christmas and got 3 of the same color in the handful I just pulled out... and it's been many many years since my statistics class in college.

Thank you for the help. It's driving me batty and nobody at work cares about my Jelly Belly question except me =(

I would have tried to figure this out on my own as it seems very easy, but I don't even know where to begin looking.

Let's pretend the bag is very large, so drawing one bean of a flavor doesn't change the probabilities.  There are $50^6$ possible draws, all the same probability.  The ways to get exactly $3$ of a flavor can be counted as $\binom 63=15$ choices for which $3$ will match times $50$ choices for which flavor to match times $49^3$ for the other $3$.  There is a tiny error as we double count the case you get two three of a kinds.  So the chance is $\frac {15*50*49^3}{50^6}=352947/62500000 = 0.005647152$ or about $1$ in $177$.

The accepted answer is incorrect (logic is right) but 6C3 = 20 (not 15) , hence the final answer would be .0077 not .0056

Also another way to look at this is plain old binomial probability : nCr p^r (1-p)^n-r
Here we have 6 tries (n=6), to get 3 success (same color, r=3 ) , probability of  success won't change in any meaningful way as beans are too many (1190) .
Calculating p (probability of success is bit tricky ) Its 24/1190 ( as 23.8 are of same color) .
So for a one specific color

Pr(x=3 same color) = 6C3*(24/1190)^3 (1166/1190)^3 = .000154.
But this could be true for any of the 50 flavor possible , so 50*0.000154 = .0077


Mathematicians who overcame academic failure to achieve success Does anyone have any story of mathematicians who overcame "academic failure" or setbacks to achieve success later as a result of their perseverance? This is a soft question, that hopefully can inspire aspiring students.

New edit: I am especially interested in academic failure (e.g. failure of exams, failure in proof/ wrong proof, failure in getting academic jobs).  This is to narrow down the question so it is not too broad. There is another related question on blind/disabled mathematicians which is very good: Who are some blind or otherwise disabled mathematicians who have made important contributions to mathematics?

My ideal accepted answer is a relatively less well known answer (so that we all learn something new), supported by factual evidence (e.g.  a hyperlink to a page or a quote).

Some that I can list are:

1) Zhang Yitang, who worked in Subway (arguably a sort of a setback) but later proved a result related to the Twin Prime Conjecture.

2) Robion Kirby, who failed his oral Ph.D. qualifying examination (http://www-groups.dcs.st-and.ac.uk/history/Biographies/Kirby.html) but later proved the "torus trick".

Thanks! (Hope this question is on topic for Math Stackexchange..)

The analytic number theorist Hua Loo-Keng overcame abject poverty, handicappedness, political persecution; for more information refer to one of his faithful biography.

Charles Hermite overcame much too, but in different aspects; he failed nearly every math exam that he was to take.

To supplement, the analytic number theorist Chen Jing-Run, the man closest to Goldbach conjecture (who proved "1 + 2") and a student of Hua Loo-Keng, overcame unemployment, poverty, and political persecution; the serial political persecution done unto him was even more violent. But he survived anyway.

Grigori Perelman who possibly proved a generalization of the Poincare Conjecture - the Thurston Geometrization Conjecture  ( has anyone claimed to show any flaw in the proof yet? ).

That being said, it seems that it's not very uncommon with odd behaviour amongst great mathematicians. They have a huge uphill struggle. Anything they discover could overshine or put others' work in the shadows with all the nasty responsive behaviours which that could provoke in any community. Especially if they are also a political nuisance (like Galois was).


How to prove that continuous function do not necessarily preserve cauchy sequences I am trying to construct a proof that continuous function do not
preserve Cauchy sequences
Every proof I can find is disprove by counter example, which is great
but these counter examples cannot be extend to how we can fix the
failure of Cauchy preservation through uniform continuity.

  Prove: Let $(x_n)$ be a Cauchy sequence on a metric space $(M,d)$, let
  $f: (M,d_M) \mapsto (N,d_N)$ be a continuous function, then $(f(x_n))$
  may not be Cauchy.

(Note: I worded the problem myself which maybe a little bit awkward, feel free to point out a good fix)

(Use definition + by contradiction)

Let $(x_n)$ be a Cauchy sequence on $(M,d_N)$. Then $(x_n)$ is Cauchy if $\forall \epsilon > 0, \exists N \in \mathbb{N}$, such that $\forall n, m \geq N, d_M(x_n, x_m) < \epsilon$

Let $f$ be a continuous function such that  $f: (M,d_M) \mapsto (N,d_N)$. Then given $x_o \in M, \forall \epsilon>0, \exists \delta(\epsilon,x_o) > 0 \text{ s. t. } \forall x \in M, d_M(x,x_o) < \epsilon \implies d_N(f(x), f(x_o)) < \delta $

Suppose $(f(x_n))$ is Cauchy on $(N,d_N)$, then $\forall \epsilon > 0, \exists N' \in \mathbb{N}$, such that $\forall n', m' \geq N', d_N(f(x_{n'}), f(x_{m'})) < \epsilon$

At this point I thought I would be notice some inconsistencies between the definition thus revealing a problem but I am not confident with my conclusions...

Can we say that we reach a contradiction because $f$ cannot be
continuous since it isn't true for $\forall x$ since $(x_n)$ is only
Cauchy for $n \geq N$?
Can we say that since $M$ is not complete, therefore $(x_n)$ may fail
to convergence, thus $f(x_n)$ is not defined as $n \to \infty$?

What is the problem here exactly?

If $d$ is a complete metric on $X$ and $f:X\to X$ is continuous then any Cauchy sequence $(p_n)_{n\in N}$ has a limit point $p$,  and the continuity of $f$ requires that $(f(p_n))_{n\in N}$ converges to $f(p).$

So to find an exception we must use  an incomplete metric.

Example. Let $X$ be the positive reals with the usual metric $d(x,y)=|x-y|$. Let $f(x)=1/x.$ The image of the Cauchy sequence $(1/n)_{n\in N}$ is the sequence $(n)_{n \in N}.$

It is not necessary that the metric be unbounded. In the above example, replace the metric $d$ with the equivalent metric $e(x,y)=\min (1, |x-y|).$

The statement says that $(f(x_n))$ may not be Cauchy.

The keyword here is "may not", hence trying to prove by contradiction (in the manner you described) is not possible, as the negation of "may not" is still "may not".


Calculate the probability with a finite arithmetic progression We have a finite arithmetic progression $ a_n $, where $ n \geq 3 $ and its  $r\neq 0 $.
We draw three different numbers. We have to calculate the probability, that
these three numbers in the order of drawing will create another arithmetic progression.

My proposition :

$ \Omega={n!\over (n-3)!}$

$ \mathbf{A}= {n \choose 3} \cdot 2$

But I think that my way of thinking about way of counting $ \mathbf{A}$ is incorrect.

Any suggestions how can I count it? Thanks in advice, Kuba!

$\Omega = n(n-1)(n-2)$

We're interested in triples whose elements differ by $r$, $2r$ ... up to $\lceil n/3\rceil r$
Notice, that if a triple $a_i,a_k,a_l$ is OK, then so is $a_l,a_k,a_i$.
My idea was to "stick together" those desired elements that form each triple and calculate their quantity. To be precise: there are $2(n-2)$ triples whose elements differ by r, $2(n-4)$ triples whose elements differ by $2r$ .... and $2(n-2\lceil n/3\rceil)$ triples whose elements differ by $\lceil n/3\rceil r$. Then:
$P(A) = \frac{2(n-2+n-4...+n-2 \lceil n/3\rceil)}{n(n-1)(n-2)}$

Let the sequence be $a_1,a_2,a_3,...a_n$. Then $|\Omega | = {n\choose 3} = {n(n-1)(n-2)\over 6}$

and each good 3-sequence is uniquely defined by it's middle term and the difference $d$. So if middle term is $a_m$ then the difference can be $d,2d,...,(m-1)d$.

So if we have $n=2k$, then we have $$1+2+...+(k-1)+(k-1)+...+2+1 = k(k-1)$$
good 3-sequences. So the probability is $$P = {6k(k-1)\over 4k(2k-1)(k-1)} = {3\over 2(n-1)}$$

If $n=2k+1$ then we have $$1+2+...+(k-1)+k+(k-1)+...+2+1 = k(k-1)+k=k^2$$
good 3-sequences. So the probability is $$P = {6k^2\over 2(2k+1)k(2k-1)} = {3(n-1)\over 2n(n-2)}$$


Area enclosed by curve $f(x,y)$ defined implicitly Find area bounded by the curve
  $$5x^2+2y^2+6xy+7x+6y+6=0$$

I can find the area using integration for curves defined explicitly in $x$ or $y$. I have no idea how to do this.

The hint.

By the affine transformation with  determinant $\Delta$ write the equation of this ellipse in the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the needed area is equal to $\frac{\pi ab}{\Delta}$

I got the following.
$$\left(2x+y+\frac{1}{2}\right)^2+\left(x+y+\frac{5}{2}\right)^2=\frac{1}{2}.$$
Can you end it now?

just like Michael Rozenberg said, you can use the completing the square approach to make the equation take the form $a^{2}x^{2}+b^{2}y^{2}=a^{2}b^{2}$ the the result follows.


In how many ways can you pick $m$ cats from $n$ cats if $2$ given cats must never be together at the $m$-group? In how many ways can you pick $m$ cats from $n$ cats if $2$ given cats must never be together at the $m$-group?

Is it $C(n,m)-C(n-2,m-2)$,  $C(n-2,m)+C(n-1,m)+C(n-1,m)$ or something else?

The first answer is correct, the second is wrong.

The first formula amounts to (all combos) - (combos with both the cats together)

The second formula tries to compute (neither cat in combo) + (cat 1 in combo) + (cat 2 in combo), but fails to see that the 2nd and 3rd terms do not guarantee the specific cat actually being in the combo.

The correct expression would be $\dbinom{n-2}{m} + 2\dbinom11\dbinom{n-2}{m-1}=  \dbinom{n-2}{m} + 2\dbinom{n-2}{m-1}$

Both answers are true. To see the equality you can use $\dbinom{n}{k}=\dbinom{n-1}{k-1}+\dbinom{n-1}{k}$.


Does $\sum a_n$ converge if $a_n = \sin( \sin (...( \sin(x))...)$ Does $\sum a_n$ converge if $a_n = \sin( \sin (\cdots( \sin(x))\cdots)$, $\sin$ being applied n times and $x \in (0, \pi/2)$?

What about $\sum a_n^r$ for $r \in \mathbb{R^+}$?

Classicaly, $\displaystyle a_n=\sqrt\frac{3}{n}+o\left(\sqrt\frac{3}{n}\right)$

The series $\sum a_n$ is therefore divergent.

Using the previous asympotical estimate, $\sum a_n^r$ converges iff $1/2-r> 1$

Proof of the first claim:

Notation: Given two sequences $u_n$ and $v_n$, we note $u_n\sim v_n$ iff $ \lim_{n\to \infty}\frac{u_n}{v_n}=1$

Now, back to the asymptotics:

$$\begin{align} \frac{1}{a_{n+1}^2}-\frac{1}{a_n^2}=\frac{1}{\sin(a_n)^2}-\frac{1}{a_n^2}&=\frac{1}{a_n^2-\frac{a_n^4}{3}+o(a_n^4)}-\frac{1}{a_n^2}\\&=\frac{1}{a_n^2}\left(\frac{1}{1-\frac{a_n^2}{3}+o(a_n^2)}-1\right)\\&=\frac13+o(1)  \\\end{align}$$

By Cesaro mean theorem, $\displaystyle \frac{1}{n}\sum_{k=1}^n \left(\frac{1}{a_{k+1}^2}-\frac{1}{a_k^2}\right)=\frac13+o(1)$

Hence $\displaystyle a_n^2=\frac{1}{\frac{n}{3}+o(n)}=\frac{3}{n}\left(1+o(1)\right)$

Hence $\displaystyle a_n=\sqrt\frac{3}{n}+o\left(\sqrt\frac{3}{n}\right)$

The estimate can be refined as $$a_n=\sqrt\frac{3}{n}-\frac{3\sqrt{3}}{10}\frac{\log(n)}{n\sqrt{n}}+o\left(\frac{\log(n)}{n\sqrt{n}}\right)$$

Proof of the estimate on partial sums

Lemma: Let $u_n$ and $v_n$ be sequence such that $u_n\sim v_n$, and $\forall n\in \mathbb N,v_n\geq 0$. If $\sum_{n\geq 1}v_n$ diverges, so does $\sum_{n\geq 1}v_n$ and $\sum_{k=1}^nu_k \sim \sum_{k=1}^nu_k$

We've proven so far that $\displaystyle a_n\sim \sqrt\frac{3}{n}$.

Hence, by the lemma, $\displaystyle S_n \sim \sum_{k=1}^n \sqrt\frac{3}{k}$

To estimate the last sum, an integral method comes in handy: indeed

$$\forall n\geq 2, \int_n^{n+1}\frac{1}{\sqrt{t}}dt \leq \frac{1}{\sqrt{n}} \leq \int_{n-1}^n \frac{1}{\sqrt{t}}dt$$

Summing these inequalities yields $$\sum_{k=1}^n \sqrt\frac{3}{k} \sim 2\sqrt{3n} $$

Hence $S_n \sim 2\sqrt{3n}$

Since $sin(x)$ has values only in $[0,1]$, the function $sin(sin(x))$ has values on $[sin(0),sin(1)]$. Therefore $a_{n+1}<a_n$. Estimating the upper bound:

$|\sum a_n| \leq \sum |a_n| \leq \sum sin(sin(...(1)))$ (used triangle inequality and sine is applied $n-1$ times).

Moreover it holds $sin(x) \leq x$ such that (especially for small arguments in the sine) $sin(sin(...(x))) \leq x$. Therefore the upper bound can be estimated:

$|\sum a_n| \leq \sum 1= \infty$.

Because the multiple sines decreasing slowly, this series diverges. If the $a_n$ are exponentiated, the $a_n^r$ become smaller when $r$ become larger; it remains the upper bound in order $x^r$ for $x \leq 1$ and when summing over it, the sum will diverge as in the case for $r=1$.


Transformation to polar coordinates in an integral Suppose that the domain of integration for a double integral is: $\{(x,y), - \infty < x \le a, -\infty < y \le a \}$. If I want to do a change of variable (to polar coordinates), how do I express the new domain? I first thought it is $0 \le \theta \le 2\pi$, $0 \le r \le \alpha \sqrt 2$, but it seems incorrect

I think the easiest thing you could do is make a change of variables $(x,y)\to(x',y')$, such that $x' = x - a, y' = y = a$, this allows you to easily make the transformation to polar, and just worry about the region $\{(x',y'),\ x' \leq 0, y' \leq 0\}$ which could correspond to $\pi \leq \theta \leq \frac{3\pi}{2},\ 0 \leq r < \infty$.

In the space you originally defined, you constrained yourself to a rectangle (because $x$,$y$ were constrained individually). If you are interested in constraining to a circle (making a direct constraint on radius), you would have the space $R=\{(x,y): \{\pi/2 \leq \theta \leq 2\pi,-\infty< x \leq 0,-\infty< y\leq 0\}\cup\{0\leq \theta\leq \pi/ 2,\ x^2 + y^2 \leq \alpha\}$
Not sure what you want to do with that extra quadrant.

domain/range

$0 \le \theta \le 2\pi$ and $- \infty \le r \le \alpha \sqrt 2$.


Convergence of $\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p$ Prove that $$\sum_{n=0}^{\infty}  \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p$$ converges if $p>1$ and diverges when $0<p\leqslant1$.

My attempt:

Since the terms of the series $\sum_{n=0}^{\infty} \frac1n$ are monotonically decreasing and positive, I applied Cauchy's Condensation theorem, i.e, if $\sum a(n)$ and $\sum2^na(2^n)$ will converge and diverge together.
Applying this I got, $\sum\left(1-\frac{1}{2^n}\right)^p$, but I am stuck now. Don't know how to proceed further.

On one hand since $\frac {1}{n}\le \frac 1n\left(1+\frac 12+\ldots+\frac 1n\right)$ it is not difficult to see that for $p>0$ we have

$$\sum_{n=0}^{\infty} \frac {1}{n^p}\le\sum_{n=0}^{\infty}  \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p$$ so the divergence statement from Riemann series which diverges for $p\le 1$

On the other hand it follows from Hardy inequality that: for $p>1$
$$\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p\le \left(\frac{p}{p-1}\right)^p\sum_{n=0}^{\infty} \frac {1}{n^p}<\infty$$
 the converges blatantly follows from the Riemann series on the right hand side which converges only for $p>1.$

In $$\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p$$

$$1+\frac 12+\ldots+\frac 1n=O(log(n))$$

so , given sum equals

$$\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p=\sum_{n=0}^{\infty}\left(\frac{log(n)}{n}\right)^p$$

Here, $O(n)>>O(log(n))$ ,

so convergence of given sum depends on convergence of $$\sum_{n=0}^{\infty}\left(\frac{1}{n}\right)^p$$

which converges if $p>1$ and diverges when $0<p\leq1$.

For case $p=1$

given sum equals $$\sum_{n=0}^{\infty}\left(\frac{log(n)}{n}\right)$$  which is divergent as

$$\sum_{n=0}^{\infty}\left(\frac{1}{n(log^p(n))}\right)$$

converges if $p>1$


Why is 'abuse of notation' tolerated? I've personally tripped up on a few concepts that came down to an abuse of notation, and I've read of plenty more on stack exchange.  It seems to all be forgiven with a wave of the hand.  Why do we tolerate it at all?

I understand if later on in one's studies if things are assumed to be in place, but there are plenty of textbooks out there assuming certain things are known before teaching them.  This is a very soft question, but I think it ought to be asked.

I doubt I could put it better than this:

  "The student of mathematics has to develop a tolerance for ambiguity. Pedantry can be the enemy of insight." - Gila Hanna

I also highly recommend Terence Tao's article describing the "pre-rigorous", "rigorous", and "post-rigorous" stages of a mathematician's development.

I believe it was proper mathematical notation that have birth to my love of mathematics, and I all now a mathematics professor. Also, may we be reminded that mathematics is a language, the most concise language today is truly universal.


A goat tied to a corner of a rectangle A goat is tied to an external corner of a rectangular shed measuring 4 m by 6 m.  If the goat’s rope is 8 m long, what is the total area, in square meters, in which the goat can graze?

Well, it seems like the goat can turn a full circle of radius 8 m, and a rectangular shed's diagonal is less than 8m (actually √52), and so shouldn't it be just 6 x 4 = 24 sq metre? The answer says it is 53 pi, and I have no clue why it is so or why my way of solving doesn't work.

Updated:
Oh, and the only area given is that of the shed's. How can I know the full area in which the goat can actually graze on?


Number of solutions $(x_1)(x_2)(x_3)(x_4) = 2016$ Having some trouble wrapping my head around this one:

  find the number of solutions to the equation $(x_1)(x_2)(x_3)(x_4) = 2016$, where $(x_i)$s are integers that are not necessarily positive.

Here's what I've been thinking so far:

Let's bring $2016$ down to its prime factorized form, such that we have $2016 = 2^5\times3^2\times7^1$.

Because the solution may include negative integers, note that a pair or 4 integers will form positive numbers, therefore we have $C(4,2)$ OR $C(4,4)$ ways of doing this.

Let's also note that the solution may include 1s in $3, 2, 1$, or $0$ positions for $(x_i)$, such that we have $C(4,3), C(4,2), C(4,1)$, or $C(4,0)$.

Now I'm thinking that we should look at each particular element of the prime factorization, such that we have elements $2, 2, 2, 2, 3, 3, 3, 7$ that are not entirely distinct. We should now think of the number of ways we can fill in $8$ elements into each of these $1, 2, 3$, or $4$ positions depending on the number of 1s we have.

And this is where I am having a little bit of trouble thinking through the solution. I'm thinking I could check the number of distinct integers that I could form with $2^5\times3^2\times7^1$ and then go from there? But I'm having trouble thinking through this methodology as well.

Thoughts?

I will assume that you want the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ such that $x_1x_2x_3x_4=2016$.

First let us find the number of ordered quadruples of positive integers.

We want to distribute the five $2$'s between $x_1,x_2,x_3,x_4$.

By Stars and Bars there are $\binom{5+4-1}{4-1}$ ways to do this.

For each such choice we need to distribute the two $3$'s, and then the one $7$. I will assume you know now how to complete this part. Call the resulting number $N$.

To take account of negative possibilities, we could have all negative, giving another $N$ choices, or two positive and two negative, giving another $\binom{4}{2}N$ choices.

The solution is hidden in
2^5*3^2*7
We have to select 4 factor whose multiplication is 2016

So for selecting 4 factor there are
C(8,2)waysways

&you have to also include negative no
Cases


What is wrong with the following u-substitution? We will calculate $\displaystyle\int^{2 \pi}_0 x \, dx$.  Let $u=\sin (x)$, and observe that $\sin(2 \pi)=0$ and $\sin(0)=0$.  We also have that $\frac{du}{dx}=\cos(x)=\sqrt{1-u^2}$. Hence,
$$
\int^{2 \pi}_0 x \, dx=\int^0_0 \frac{\sin^{-1}(u)}{\sqrt{1-u^2}} \, du = 0.
$$
This is very obviously wrong, but I am not sure how to explain the error formally.

Edit:
Thanks for the responses and in particular the link below to the related problem!  The error is indeed caused by the substitution $x=\sin^{-1}(u)$.  The integration is performed over $[0,2 \pi]$ which is outside the range of the $\sin^{-1}$ function.

Remark
The error is slightly better disguised when calculating $\displaystyle\int^1_{-1}\frac{2x}{1+x^2} \, dx.$

Let $u(x)=1+x^2$, and observe that $u(1)=u(-1)=2$.  Then since $dx=\frac{1}{2x} du$, we have that
$$
\int^1_{-1} \frac{2x}{1+x^2} \, dx =  \int^2_2 \frac{1}{u} \, du=0.
$$
This time, no trigonometric substitution is used, but it is still an incorrect proof for the same reason as above.  A correct proof can be obtained by using the fact that $x \mapsto \displaystyle\frac{2x}{1+x^2}$ is odd.

This example is more disturbing because the procedures above are entirely intuitive and yield the correct result. It seems to me that students when taught integration by substitution of definite integrals should also be taught that great care be exercised in checking the range of integration, particularly when the (apparent) substituting function is not invertible in that range.

Contrary to the OP's Remark/claim, the integration-by-substitution solution provided for the example $\displaystyle\int^1_{-1}\frac{2x}{1+x^2} \, \mathrm{d}x\,$ is actually perfectly valid—and not disturbing—despite the substitution function $u=1+x^2$ not being injective (and certainly not invertible either) on $[-1,1]$: integration by substitution does not intrinsically require the substitution function to be invertible/bijective or monotonic—or even injective—on the interval of integration.
Integration by substitution (change of variable):
If $g'$ is integrable on $[a,b]$ and $f$ is continuous on $g[a,b]$,
then $$\int_a^bf(g(x))g'(x)\mathrm{d}x=\int_{g(a)}^{g(b)}f(u)\mathrm{d}u.$$
In this example: Let $g(x)=1+x^2,\,\,f(u)=\frac1u,\text{ and }\,a=-1,b=1$.
Then $f(g(x))=\frac1{1+x^2},\,\,g'(x)=2x,\text{ and }\,g(a)=g(b)=2$.
The above theorem is validly applicable here
$$\int^1_{-1}\frac{2x}{1+x^2}\,\mathrm{d}x=\int_2^2\frac1u\,
   \mathrm{d}u=0,$$ yet $g(x)$ isn't injective on $[-1,1].$

However, splitting up the original integral such that the
substitution function is injective on each subinterval does
generally prevent the substitution process from running into
"glitches" (due to misapplication of the integration-by-substitution
theorem).
So, for the OP's main example:
Let $u=\sin x\,$ so that $\,\mathrm{d}u=\cos x \,\mathrm{d}x$.
Now, for $x\in[0, \frac{\pi}{2}]$, \begin{align}\cos x &=
   \sqrt{1-u^2},\\x&=\arcsin u;\end{align} for
$x\in[\frac{\pi}{2},\frac{3\pi}{2}]$, \begin{align}\cos x &=
   -\sqrt{1-u^2},\\x&=\pi -\arcsin u;\end{align} for $x\in[\frac{3\pi}{2}, 2\pi]$, \begin{align}\cos x &=
   \sqrt{1-u^2},\\x&=2\pi +\arcsin u.\end{align}
Therefore \begin{align}\int^{2\pi}_0 x
   \,\mathrm{d}x&=\int^\frac{\pi}{2}_0 x \,\mathrm{d}x +
   \int^\frac{3\pi}{2}_\frac{\pi}{2} x \,\mathrm{d}x +
   \int^{2\pi}_\frac{3\pi}{2} x \,\mathrm{d}x\\&=\int^1_0 \frac{\arcsin
   u}{\sqrt{1-u^2}}\,\mathrm{d}u + \int^{-1}_1 \frac{\pi-\arcsin
   u}{-\sqrt{1-u^2}} \,\mathrm{d}u + \int^0_{-1} \frac{2\pi + \arcsin
   u}{\sqrt{1-u^2}} \,\mathrm{d}u\\&=\pi\int^1_{-1}
   \frac{\,\mathrm{d}u}{\sqrt{1-u^2}}+2\pi\int^0_{-1}
   \frac{\,\mathrm{d}u}{\sqrt{1-u^2}}\\&=\pi\bigg[ \arcsin
   u\bigg]_{-1}^1+2\pi \bigg[\arcsin
   u\bigg]_{-1}^0\\&=2\pi^2.\end{align}

It shoud be obvious that the integral must be evaluated in the progressive direction, otherwise some parts of the summation are deducted rather than added. To achieve this, the transformation must be monotonous.


Function whose limit does not exist at all points There are functions which are discontinuous everywhere and there are functions which are not differentiable anywhere, but are there functions with domain $\mathbb{R}$ (or "most" of it) whose limit does not exist at every point? For example, $ f:\mathbb{R}\to\mathbb{N}, f(x) = $ {last digit of the decimal representation of $x$}. Is this even a valid function?

As suggested in the comments define $f:\mathbb{R}\rightarrow \mathbb{R}$ by $f(x)=0$ if $x$ is irrational and $f(x)=1$ if $x$ is rational. Let's prove it doesn't have a limit at any point. Let $y\in \mathbb{R}$, suppose $y$ is irrational and that there exists $L$ the limit of $f$ at $y$. Then given any $\epsilon<1/2$, there is $\delta>0$, such that $x\in (y-\delta, y+\delta)\setminus\{y\}$ implies $|f(x)-L|<1/2$. Now, since there exists $x_0,x_1\in (y-\delta, y+\delta)\setminus\{y\}$ such that $x_0$ is rational and $x_1$ is irrational we get that $1=|1-0|=|f(x_0)-f(x_1)|\leq |f(x_0)-L|+|f(x_1)-L|<1/2+1/2=1$, a contradiction. So the limit at any irrational $y$ does not exists. The same argument applies to any rational $y$, so the limit doesn't exists at every point.

Now f(x) is define f:R→R by f(x)=x if x is rational and f(x)=-x if x is irrational. How to prove it limt (x approach to a) f(x) doesn't exist at a belongs to R with a not equal to 0.

Here limit doesn't go to constant value. How to prove it by using above theory.


Are there 3 trig functions or are there 6 trig functions? In my algebra class we are being taught that there are only the 3 basic trig functions (cosine, sine, and tangent). But my friend who is 2 math grade levels ahead of me is saying that there is 6 trig functions (cosine, sine, tangent, cotangent, secant, and cosecant). Does anyone know why we are being taught differently and which one of us is correct?

It depends on how you look at it I guess, but:
$$\cot(x) = \frac{1}{\tan(x)}$$
$$\csc(x) = \frac{1}{\sin(x)}$$
$$\sec(x) = \frac{1}{\cos(x)}$$
So the three "extra" functions your friend told you about are just derived from the three you know. But if that's the rule, then two of the ones you know,
$$\cos(x) = \sin\left(\frac{\pi}{2} - x\right)$$
$$\tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{\sin{x}}{\sin\left(\frac{\pi}{2} - x\right)}$$
are also just derived functions. Hence we would say there is only one trigonometric function, for example $\sin{x}$.
(As others have mentioned, this statement works even counting hyperbolic functions, because of properties like $\cosh(x) = \cos(ix)$ and so on, or using $e^{i\theta} = \cos{\theta} + i\sin{\theta}$. But since you don't appear to be at this level of math yet, I won't go into detail about that.)
Bottom line: We only need one trigonometric function, but for practical reasons, there are more.

Actually there is hyperbolic, elliptical, cardioid, integral, cubic, quartic, and many more. It can be viewed as curve functions.


Math behind rotation in MS Paint For those who don't know, MS Paint only has the options to rotate an image by right angles. To carry out an arbitrary rotation ($\theta^\circ$), the following hack is suggested:

  Horizontal skew by $\theta$
  Vertical Stretch by $\displaystyle \frac{1}{\cos^2\theta}$
  Vertical Skew by  $-\theta$
  Horizontal and Vertical Stretch by $\cos \theta$

I would like to know how this works.

(I guess one way would be to show that multiplying the above operations represented as matrices should return the $2$D rotation matrix, but I do not know what are the matrix representations for the intermediate steps.)

I think the following matrix product gives an explanation.

The first step is given by the matrix
$$
S_1=\left(\begin{array}{cc}1&\tan\theta\\0&1\end{array}\right).
$$

The second step corresponds to
$$
D_1=\left(\begin{array}{cc}1&0\\0&\frac{1}{\cos^2\theta}\end{array}\right).
$$

The second shearing operation corresponds to
$$
S_2=\left(\begin{array}{cc}1&0\\-\tan\theta&1\end{array}\right).
$$

At this point we have done $S_2D_1S_1$ that after a little bit of matrix manipulation and using the identity $\sin^2\theta +\cos^2\theta=1$ becomes
$$
S_2D_1S_1=\left(\begin{array}{cc}1&\tan\theta\\-\tan\theta&1\end{array}\right).
$$

The last step amounts to a scalar multiplication by $\cos\theta$ and gives us the familiar rotation matrix
$$
\cos\theta S_2D_1S_1=\left(\begin{array}{rc}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{array}\right).
$$

Edit: I got to use this as an exercise in a freshman course that just ended. I was actually a bit unhappy with that youtube video, because it doesn't show the transitions from one step to the other. To fix this problem I made an animation, where the four steps are done gradually: in steps one and three I continuously move the shear (skew) parameter from zero to the desired value. Similarly in steps two and four the scaling parameter changes continuously from one to the proper value. All the steps are done sequentially.

It is worth noting that you can accomplish rotation with only a single trip to the Resize and Skew menu.
Paint applies the operations in the order: Resize, Skew Horizontal, Skew Vertical.

As a sequence of matrix transforms acting on a column vector, that is:

$$(\begin{bmatrix}1&0\\K_v&1\end{bmatrix}(\begin{bmatrix}1&K_h\\0&1\end{bmatrix}(\begin{bmatrix}S_h&0\\0&S_v\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix})))$$

Where $K_v$ and $K_h$ are the vertical and horizontal skew factors, and $S_v$ and $S_h$ are likewise the scale factors.

Because matrix multiplication is associative, let's remove the parentheses and multiply these transforms from left to right to give an equivilent single-matrix transform:
$$\begin{bmatrix}S_h&K_hS_v\\S_hK_v&K_hK_vS_v+S_v\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$$

Setting this matrix equal to the rotation matrix, $\begin{bmatrix} \cos(θ) & \sin(θ) \\ -\sin(θ) & \cos(θ) \end{bmatrix}$, we get a system of equations which we can solve for our scale and skew factors:
$$S_h = \cos(θ)$$
$$K_h = \sin(θ)cos(θ)$$
$$S_v =1/\cos(θ)$$
$$K_v = -\tan(θ)$$

These variables describe factors in an affine transformation, but paint wants "% Resizing" and "degrees skew", so input:
$$\%R_h = 100*S_h=100*\cos(θ)$$
$$\%R_v = 100*S_v=100/\cos(θ)$$
$$DegSk_h = \arctan(K_h)=\arctan(\sin(θ)\cos(θ))$$
$$DegSk_v = \arctan(K_v) = \arctan(-\tan(θ))=-θ$$

As an operational note: Paint will automatically resize your selection after a skew if the skew transformation would put the selection outside of the canvas, so make sure you have plenty of room around your selection.


Showing that $H$ is a normal subroup of $A_4$ Consider the subgroup

$H$={$(1),(12)(34),(13)(24),(14)(23)$}

How would I show that $H$ is a normal subgroup of $A_4$?

If I perform a few computations such as:
$$(132)^{-1}(12)(34)(132)=(123)(12)(34)(123)=(14)(23)$$
$$(243)^{-1}(12)(34)(243)=(234)(12)(34)(243)=(13)(24)$$
$$(124)^{-1}(12)(34)(124)=(142)(12)(34)(124)=(12)(34)$$

,then we see that an element of the form $\beta^{-1}\alpha\beta$, where $\beta \in A_4$ and $\alpha \in H$, remains in $H$.

But because it's tedious to show all the other cases, how would one prove this without showing all the cases?

Hint: Conjugation preserves the cycle type (this spares you quite some computation) and here you even only have one non-trivial cycle type.

The elements of $H$ are exactly the solutions in $A_4$ to the equation $x^2=e$. Therefore, if $H$ is a sugbgroup of $A_4$, then it is a characteristic subgroup.


Definite Integral of Polynomial of Sine and Cosines I was wondering if there is a way to compute definite integral
\begin{align}
I(m,n) :=\int_{0}^{\pi} \sin^m (\theta) \cdot \cos^{n}(\theta) \ \mathrm{d} \theta
\end{align}
in general for integer-valued $m$ and $n$. This problem arises when I try to compute integrals of trigonometric functions over a high dimensional sphere.

In fact I am more interested in the asymptotic order of this integral. I was wondering how $I(m,n)$ behaves when $n$ goes to infinity. For example, what is the limit of $\lim _{n\rightarrow \infty} I(2, n)$?

I'd try to expand
$$\cos^n(x) = \frac{(e^{ix}+e^{-ix})^n}{2^n} = \frac{e^{-inx}(e^{2ix}+1)^n}{2^n} = \frac{e^{-inx}}{2^n}\sum_{k=0}^n {n\choose k} e^{2ikx}$$
$$\sin^m(x) = \frac{(e^{ix}-e^{-ix})^m}{(2i)^m} = \frac{e^{-imx}(e^{2ix}-1)^m}{(2i)^m} = (-1)^m\frac{e^{-imx}}{(2i)^m}\sum_{l=0}^m {m\choose l} (-1)^l e^{2ilx}$$

$$\cos^n(x) \sin^m(x) =\frac{(-1)^m}{2^{n+m}i^m}\sum_{l=0}^m \sum_{k=0}^n {m\choose l} {n\choose k} (-1)^l e^{\textstyle i(2l+2k-n-m)x}  $$

for $a \in \mathbb{Z}^*$ we know that $\int_0^\pi e^{2 i a x} dx = 0$ so that

$$\int_0^\pi \sin^m(x) \cos^n(x) dx = \pi \frac{(-1)^m}{2^{n+m}i^m}\sum_{(l,k) \in E_{n,m}} {m\choose l} {n\choose k} (-1)^l $$

where $E_{n,m} = \left\{ (l,k) \ | \ 2l+2k=n+m, l \in \{0 \ldots m\}, k \in \{0 \ldots n\} \right\}$

this also tells us that the result is non-zero only when $m$ and $n$ are even (if $m$ is odd the result is imaginary and if $m+n$ is odd $E_{n,m}$ is empty)


How to formalize "It doesn't matter how places the brackets" Let $X$ be a set and $X\times X\to X: (a, b)\mapsto a\cdot b$ an associative operation on $X$. Now one can prove by induction that it doesn't matter how one places the brackets in a product $x_1\cdot x_2\cdot\text{ }\dots\text{ } \cdot x_n$, the product always evaluates to the same value. This result justifies the notation of a product $x_1\cdot x_2\cdot\text{ }\dots\text{ } \cdot x_n$ without any brackets. But how can one formalize it?

One nice way to formalize the general associative law is via binary trees. For us, a tree of operations will be a finite tree, with a root node, where each node is either a leaf (no successors) or has exactly two successors (a left successor and a right successor), and each leaf is labelled with an element of $X$.

The idea is that each leaf of the tree represents an input, and each non-leaf node is an instance of the operation $*$ applied to the values associated to the $*$-product of the left and right sub-trees of that node. So, for instance, $(a*b)*c$ is represented by a tree with one split at the root, and then the left side splits once more; and the leaves are labelled $a, b, c$, respectively.

A vauation of a tree of operations is a map $m$ from the nodes to $X$, satisfying $m(\sigma)$ is the label already on $\sigma$ - if $\sigma$ is a leaf - and $m(\sigma)=m(S_{left}(\sigma))*m(S_{right}(\sigma)),$ where $S_{left}$ and $S_{right}$ are the left and right successors of $\sigma$, if $\sigma$ is not a leaf.  We can prove by induction that each tree $T$ of operations has exactly one valuation; call this "$v(T)$."

Note that the leaves of a tree of operations are ordered in a natural way: one leaf $\sigma$ is before another leaf $\tau$ if it is "to the left" of $\tau$, that is, when $\rho$ is the furthest node of the tree above both $\sigma$ and $\tau$ we have that $\sigma$ goes left from $\rho$ while $\tau$ goes right. The general associative law is then:

  If $T_1, T_2$ are two trees of operations with the same leaf-labels occurring in the same left-right order, then $v(T_1)$ and $v(T_2)$ assign the same element of $X$ to the root of $T_1$ and $T_2$, respectively.

This is called the associative law which says precisely: $$\forall x,y,z [(x*(y*z))=((x*y)*z))] $$

That it extends to arbitrary finite products can then proven using induction.

Some more details: you will prove that every expresention is equivalent to the one with brackets on left e.g. to $(((x*y)*z)*w))$ (every expresion uses each variable only once). Then you can define complexity $C:\mathrm{EXP}\to\mathbb{N}$ of an expression recursively as : $C(x)=0$, and for composed expressions $C(e*f)=C(e)+C(f)+\mathrm{length}(f)-1$, where $e,f$ are expressions and $\mathrm{length}(f)$ is a number of variables in $f$.

Examples:
 $C(x*y)=C(x)+C(y)+1-1=0$,

$C((x*y)*z))=C(x*y)+C(z)+1-1=0$,

but $C(x*(y*z))=C(x)+C(y*z)+2-1=1$,

and use induction on the complexity. Calculate that any one application of associativity decreases the complexity.


How can we count in base eleven? We work in base 10. We know base 3 would count 1, 2, 10, 11, 12, 20. How would we count in base 11?

You would have to add a symbol for $10$, such as $A$. Then it would be
$$\begin{array}{|c|c|c|c|c|c|c|}
\hline
\text{Base 10} & 1 & 2 & 3 & \dots & 9 & 10 & 11 \\ \hline
\text{Base 11} & 1 & 2 & 3 & \dots & 9 & A & 10 \\ \hline
\end{array}$$

You know how counting in Hexadecimal is done right? It could be done similarly with just the A included.Leave out the other alphabets.


Square root of negative numbers If:

$$a = \sqrt{ b^2 - b }$$

The problem I have is that for values of:

$0 < b < 1$

the result of:

$b^2 - b$

Is a negative number which gives rise to an error on Excel and my calculator.

I understand that negative numbers don't have square roots (I read it on Wikipedia at least), so how do I solve this for values of $b$ less than 1?

Thanks! :)

Mathematicians have defined a new number, called $i$, such that $i^2=-1$ (it's not really that new). Commonly $i$ is called an imaginary number. If you're familiar with coordinate geometry like the Cartesian plane, the complex numbers are very similar. Every complex number has the form $a+bi$ for some $a,b\in\mathbb{R}$ so we can plot complex numbers (that is, numbers that have a real part and an imaginary part) as pairs (a,b) where we view the typical $x$-axis as the real part and the $y$-axis as the imaginary axis. If you have a number like $\sqrt{-64}$, you can simplify it by pulling out the $-1$ as an $i$. That is,
$$
\sqrt{-64}=i\sqrt{64}=\pm8i
$$
Complex numbers have lots of interesting properties. I recommend checking out the wikipedia page on complex numbers for more information.

Specifically to answer your question, if $b^2-b<0$, there are no solutions over the real numbers. You need to use complex numbers in order to find solutions.

Svenkat's and Google answers:
"one cannot take the square root of negative numbers"
"Negative numbers don't have real square roots"
"you can't take the square root of a negative number"
It is very difficult to break 16th century scholastic dogma, e.g. imaginary numbers, used in taking the square root of negative numbers. Nobody wants to discuss or in other words - taboo.
If we are consistent, we should accept to take square roots only from positive numbers. The result of multiplication of the same numbers is always a positive number.
If we take the square root of a negative number, we should place the negative sign before the radical sign/radix -√. Details: https://globid.eu/


Does the category framework permit new logics? It appears to me that a topos permits a broader concept of subsets than the yes/no decission of a characteristic function in a set theory setting. Probably because the subobject classifier doesn't have to be {0,1}.

But I wonder, aren't all the multivalued logics also part of/can be modeled in set theory? Is there some new logic coming in with topoi which weren't there before? Did it just help discovering new ideas? Fuzzy stuff etc. are all existent in "conventional set theory mathematics" already, right?

Category theory itself, and thus topos theory, can be formalized in set theory(*). So, in a weak sense, everything in topos theory is already "in" set theory. Of course, set theory can be formalized in topos theory using the category Set, and so set theory is also "in" topos theory. The real question that matters is which formalization is useful for a particular purpose. For some purposes, topos theory provides a useful framework to the people who use it, and they prefer this framework over the equivalent framework where everything is rephrased in terms of set theory.

The key point about formalization in set theory is that category theory, and topos theory, are formal axiomatic systems, and any axiomatic system can be studied using set theory as a metatheory.

(*): There is a minor issue that some things in topos theory may use axioms that appear to be large cardinal axioms from the point of view of set theory. But this is not an impediment to formalizing things in set theory if we simply assume the necessary large cardinal axioms.

No, you can't model all multivalued logics in set theory.  Set theory models classical propositional logic, but it does not model a logic where say the principle of contradiction fails and its negation fail also.  All formal theorems of any multivalued logic exist within classical logic in the sense that if A comes as a formula in multivalued logic, it will also happen in classical logic and thus can get modeled by set theory (the converse does seem to hold for some multivalued logics, but hardly all that many of them).  But, the domain of truth values differs for a formula in a multivalued logic than in classical logic.

Fuzzy stuff does NOT exist withing conventional set theory mathematics.  The axiom of extensionality does not hold for fuzzy sets.


State 2 values of $x$ for which the value of $3x^2+4x-14$ is a perfect square. State 2 values of $x$ for which the value of $3x^2+4x-14$ is a perfect square.

I can't seem to factor it and I'm really lost. Does using the quadratic equation for it work? Help I need it for grade 10 quadratics test.

$$(1)\quad3x^2+4x-14=a^2$$ $$\implies 3x^2+4x-(14+a^2)=0\implies x=\dfrac{-4\pm\sqrt{16-12(-14-a^2)}}{6}$$

This does not need to be an integer. This says that for $(1)$ to be true, just choose an $a$ here and it will give you an $x$ value that satisfies. For example, $a=1$: $$x=\dfrac{-4\pm\sqrt{16-12(-15)}}{6}=\dfrac{-4\pm\sqrt{196}}{6}=\frac{10}{6},-3$$
Plugging either one of these values into the original equation will give you $a=1$.

3x^2+4x-14=0.
Let q and w be 2 roots of the equation.
q+w=4,qw=-42
(q-w)^2=(q+w)^2-4qw
So q-w=√184=2√46
On solving q+w=4and q-w=2√46
We get w=q={4+2√46}÷\2
Now put the value of w andq and solve it.on verifying q+w is not equal with 4 . slight deviations will come as we have taken imaginary roots.


"such that" logical symbol So, in the definition of what is a square root,

  $\sqrt{x}$ are all numbers $y$ such that $y×y=x$.

are there any logical mathematical symbols so that the above definition can be written using logical operators only, and no natural language?

Where can I get some introductory or reference material on all such logical symbols?

update:
I noticed, some time after asking the question that the definition of square root I am giving is wrong. The square root of $x$ is to defined to be the non-negative number $y$ that satisfies $y*y=x$. But the question was about notation, not square roots, so I am leaving it as it stands due to some answers using the supplied (erroneous) definition.

You could write this in a few different ways... I'm not sure what you're asking, so let me show you a couple.

For one, you could define the condition $y\in\text{Sqrt}(x)$, rather than the set itself:
$$
y\in\text{Sqrt}(x)\Leftrightarrow y^2=x
$$

The following two are commonly used in set definitions:
$$
\text{Sqrt}(x)=\{y\mid y^2=x\}\qquad \text{or}\qquad \text{Sqrt}(x)=\{y:\ y^2=x\}
$$

I also see people use (and have used myself) "s.t." as an abbreviation for such that in formulas.

I ALSO have seen a backwards ∈ symbol for "such that."  I saw it in logical notation for the definition of the limit of a function. M. Del Nero


Why does a transcritical bifurcation occur when $ab = 1$? From Nonlinear Dynamics and Chaos by Strogatz:

Show that the first-order system $\dot{x} = x(1-x^{2})-a(1-e^{-bx})$ undergoes a transcritical bifurcation at $x=0$ when the parameters $a,b$ satisfy a certain equation, to be determined. (This equation defines a bifurcation curve in the $(a,b)$ parameter space.) Then find an approximate formula for the fixed point that bifurcates from $x=0$, assuming that the parameters are close to the bifurcation curve.
Solution: Note that $x=0$ is a fixed point for all $(a,b)$. This makes it plausible that the fixed point will birfurcate transcritically, if it bifurcates at all. For small $x$, we find
\begin{align*}
1-e^{-bx} &= 1-[1-bx+\dfrac{1}{2}b^{2}x^{2} + O(x^{3})] \\
&= bx - \dfrac{1}{2}b^{2}x^{2} + O(x^{3})
\end{align*}
and so
\begin{align*}
\dot{x} &= x - a(bx - \dfrac{1}{2}b^{2}x^{2})+O(x^{3}) \\
&=(1-ab)x + (\dfrac{1}{2}ab^{2})x^{2}+O(x^{3}).
\end{align*}
Hence a transcritical bifurcation occurs when $ab=1$; this is the equation for the bifurcation curve. The nonzero fixed point is given by the solution of $1-ab+(\dfrac{1}{2}ab^{2})x \approx 0,$ that is $$x^{*} \approx \dfrac{2(ab-1)}{(ab)^{2}}.$$

After looking through the solution given in my textbook, I still don't understand why a transcritical bifurcation occurs when $ab=1$?

The transcritical bifurcation occurs when $r=0$ in the normal form $\dot{y}=ry\pm y^2$. You can notice that if you rescale $x$ and $t$ to eliminate the quadratic term you will obtain the transcritical normal form with $r=(1-ab)/\sqrt{2a}b$. The zero of $r$ are given by $ab=1$ as shown in the textbook.


greatest common divisor for polynomials Let $k$ be a field and $f(x), g(x)\in k [x]$. The greatest common divisor is defined to the monic common divisor having largest degree.  In this definition, can the condition $k$ being a field be weakened to $k$ being a commutative ring? That is, is it well-defined due to the uniqueness?

For example, the gcd of $2x+1$ and $(2x+1)x$ in $\mathbb Q[x]$ is $x+\frac{1}{2}$, which is not in $\mathbb Z[x]$.

So we cannot just replace $k$ with any commutative ring. This is really because of the monic condition on the gcd; we can divide by the leading entry when working in $k$, but this is not always possible in a general commutative ring.


In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together? In the below solved problem, every thing is okay, but if we have $4$ consonants then why we are giving $5!$? and is this a combination problem? how to distinguish?

Question:
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

Answer:
The word 'OPTICAL' contains $7$ different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, $5$ letters can be arranged in $5! = 120$ ways.
The vowels (OIA) can be arranged among themselves in $3! = 6$ ways.
Required number of ways $= (120*6) = 720$.

There are $4$ consonants and $1$ group of vowels, so there are $5$ elements to permute. Yes, this is a combinatorial problem because we are counting the number of possibilities that satisfy certain conditions.

For this problem I use the multinomial coefficient.
We have 7 letters in total 3 of which are vowels and 4 are consonants
The number of vowels combinations are 3! and the number of consonants combinations are 4! (I've used the multinomial coefficient)
So by the end we've 3!*4! ways


Find a sequence $(p_n)_{n\in N}$ of discrete density Let $F:\mathbb{R}\to [0,1]$ be a repartition function and $D=\{x\in \mathbb{R}: F(x)\neq F(x^-)\}\subset \mathbb{R}$. I have to find a sequence $(p_n)_n$ of discrete density in $\mathbb{R}$ such that $\lim_{n\to\infty}F_n(x)=F(x)\ \forall x\in\mathbb{R}\setminus D$, where $F_n$ is the repartition function associated to $p_n$, (i.e. $F_n(x)=\sum_{y\in\mathbb{R}:y\le x}p_n(y)$).

I use the fact that $\{X\le x\}=\bigcap_{n\in N}\{X\le x_n\}$. Indeed if $X(\omega)\le x$, we necessarily have $X(\omega)\le x_n$ because $x\le x_n$ and vice versa, since the sequence of events $\{X\le x_n\}$ is decreasing, $\{X\le x_{n+1}\}\subseteq\{X\le x_n\}$, so $F(x)=P(X\le x)=\lim_{n\to\infty}P(X\le x_n)=\lim_{n\to\infty}F(x_n)$. But this is a sequence of events and not of discrete density. So my question is: is it possible to use this fact to find the density sequence?

Let $F_n (t)=P\{[nX] \leq nt\}$ where $[x]$ is the largest integer not exceeding $x$ and $X$ is random variable whose distribution is $F$. Since $\frac {[nX]} n \to X$ it follows that $F_n(t) \to F(t)$ at all continuity points $t$ of $F$, i.e., at all points of $\mathbb R \setminus D$. Also the distribution of $\frac {[nX]} n $ is discrete since it takes only countable many values.


Counting subsets of a given set Could anyone be kind in helping me with this simple question?

What is the number of subsets of $$\{1,2,3,\cdots, n\}$$ containing k elements?

Thanks.

It is the sum of number of ways of choosing $r$ elements from n elements, where $r$ ranges from $0$ to $n$. This is because $r = 0$ is the null set, $k = 1$ is all $nC1$ subsets of cardinality $1$, and so on...

Just imagine yourself picking all subsets of size 0, then size 1, ..., till size n.

This is also the sum of the nth row of Pascal's Triangle, which is also $2^n$.

Permutations do not count, because the order of a set does not matter.

The number of subsets in a set is given by $2^n$ where n is the number of elements in the set . It also includes the empty set as a subset.


Is "The empty set is a subset of any set" a convention? Recently I learned that for any set A, we have $\varnothing\subset A$.

I found some explanation of why it holds.

  $\varnothing\subset A$ means "for every object $x$, if $x$ belongs to the empty set, then $x$ also belongs to the set A". This is a vacuous truth, because the antecedent ($x$ belongs to the empty set) could never be true, so the conclusion always holds ($x$ also belongs to the set A). So $\varnothing\subset A$ holds.

What confused me was that, the following expression was also a vacuous truth.

  For every object  $x$, if $x$ belongs to the empty set, then $x$ doesn't belong to the set A.

According to the definition of the vacuous truth, the conclusion ($x$ doesn't belong to the set A) holds, so $\varnothing\not\subset A$ would be true, too.

Which one is correct? Or is it just a convention to let $\varnothing\subset A$?

There’s no conflict: you’ve misinterpreted the second highlighted statement. What it actually says is that $\varnothing$ and $A$ have no element in common, i.e., that $\varnothing\cap A=\varnothing$. This is not the same as saying that $\varnothing$ is not a subset of $A$, so it does not conflict with the fact that $\varnothing\subseteq A$.

To expand on that a little, the statement $B\nsubseteq A$ does not say that if $x\in B$, then $x\notin A$; it says that there is at least one $x\in B$ that is not in $A$. This is certainly not true if $B=\varnothing$.

I'm also learning about sets, but the way that I've begun to see things like this is as follows:

I think about what a nice system needs, and a few things spring to mind. I need an addition, the idea of pushing two things together. How about a concept of negatives? That could be useful. I could add these negatives to some of these positives, and eventually I'm going to come across a situation where my negatives meets my positives and what do I have? That's the null set.

Think about 'zero' logically, this concept of nullness, but more than just the number zero, but its function, and realize why it fits soundly within our understanding of mathematics. Say we were to not have this zero, the what would happen? This concept of zero is such an obvious idea in numbers that when we get zero in more bizarre systems like sets, the concept of zero along this feels odd, because 0 is just a number after all.

But in sets your '0' still retains the set construct, but it represents the analogous function that 0 holds in numbers.


I Don't Understand Error Bounds I understand they're supposed to give us a limit on how off our approximation of an integral can be, but I don't understand how the formula gives that.

What does the second derivative have to do with anything? Why are the constants 12 and 24 put in front of $n^2$? I'm really looking for an explanation as to why this formula gives us error bounds.

Here is a derivation of the error bound for the midpoint rule.

The midpoint rule is

$$\int_{a}^{b}f(x)dx \approx \frac{b-a}{n}\sum_{j=1}^{n}f(\bar{x}_j),$$

where $x_j = a + j(b-a)/n$ and $\bar{x}_j = (x_{j-1}+x_j)/2.$

The absolute error is

$$E = \left|\int_{a}^{b}f(x)dx - \frac{b-a}{n}\sum_{j=1}^{n}f(\bar{x}_j)\right| = \left|\sum_{j=1}^{n}\int_{x_{j-1}}^{x_{j}}[f(x)-f( \bar{x}_j)]dx \right|.$$

Using Taylor's theorem,

$$f(x) = f(\bar{x}_j)+f'(\bar{x}_j)(x-\bar{x}_j) + \frac1{2}f''(\xi_x)(x-\bar{x}_j)^2,  $$

where $\xi_x$ is between $x$ and $\bar{x}_j$ and $|f''(\xi_x)| \leq K$.

Substituting and integrating we get

$$E =  \left|\sum_{j=1}^{n}\frac1{2}\int_{x_{j-1}}^{x_{j}}f''( \bar{x}_j)(x-\bar{x}_j)^2dx \right| \\ \leq \frac1{2}\sum_{j=1}^{n}\int_{x_{j-1}}^{x_{j}}|f''( \bar{x}_j)|(x-\bar{x}_j)^2dx  \leq \frac{K}{24}\sum_{j=1}^{n}(x_j-x_{j-1})^3 = \frac{K}{24}\sum_{j=1}^{n}\left(\frac{b-a}{n}\right)^3\\= \frac{K(b-a)^3}{24n^2}$$

These formulas are methods for approximating integrals.

Because the actual answer isn't the answer obtained by using simpson's rule, trapezoidal rule, etc.

We allocate for that difference by using a formula to calculate that difference.

We use this difference to then see how close our value is to the accepted value.

This is useful application in both physics and mathematics.


Intuitive explanation for how could there be "more" irrational numbers than rational? I've been told that the rational numbers from zero to one form a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? There is always a rational between two irrationals, and always an irrational between two rationals, so it seems like it should be split pretty evenly.

I think the most intuitive explanation I have heard is to considering writing down a rational number in decimal form.  This means that either it is a repeating decimal or a terminating decimal, for example $2.3737\overline{37}$ or $0.42$, which we will write as $0.4200\bar{0}$. Now, consider the probability of randomly writing down a number.  So you have ten options every time you go to place a digit down. How likely is it that you will just "happen" to get a repeating decimal or a decimal where you only have zeros after a certain point?  Very unlikely. Well those unlikely cases are the rational numbers and the "likely" ones are the irrational.

This is the most intuitive explanation I can think of. It's long, but hopefully easy to follow.

Consider a natural number, for instance $123$. This natural number is implied by our mathematical notation system to have the decimal point after the last digit, with no fractional part ($123 = 123.00...0$). However, we can take this integer number and specify that it is the stream of digits of a non-integer, by specifying that there is a decimal point at any arbitrary single position of the stream. For instance, we could transform $123$ into $12.3$, or $1.23$, or $0.123$, or $0.0123$. We do this all the time, in what's called scientific or engineering notation; $123 = 1.23*10^2$. This is also a convenient way to store these decimal quantities in computers; binary floating-point arithmetic is based on this system (using a fixed-length binary mantissa and a base-2 integer exponent). We can also use the integer negatives of the natural numbers in the same way. In fact, the ability to construct a rational number in this way is derivable* from the true definition of a rational number.

The integer mantissa (and exponent) used to construct a rational can be arbitrarily large, but both are countable numbers; theoretically, if you started counting from 0, you'd eventually reach the number used (though it may take you longer than the universe has existed or will continue to exist to do so).

Now, with irrational numbers, this changes. The mantissa is no longer any one natural number, but the concatenation of an infinite number of unique natural numbers. Some irrationals are defined as such, using a formula to define each number to be concatenated (e.g. Champernowne's constant, $0.123456789101112131415161718192021...$). For most, however, the construction is not formulaic, but it holds that any irrational number can be constructed by concatenating a subset of the natural numbers (order independent), then placing a decimal point somewhere.

The upshot is that there are only as many rational numbers as there are naturals (there's a step I'm missing to formally prove that; the exponential construction I'm using is a good start, but without further restrictions to the allowable values, it trivially allows for infinite combinations of an integer mantissa and integer exponent to produce the same rational. Cantor did it much more rigorously). However, there are as many real numbers as there are subsets of the naturals, because by appending the digits of any unique natural number to any concatenation of other numbers, and sticking a decimal place in arbitrarily, you get a whole new real number. The set of all subsets of a set is known as the power set, and its cardinality is given by $|P(A)| = 2^{|A|}$. The cardinality of the naturals is the trans-infinite number $\aleph_0$ (aleph-null or aleph-zero), and so the cardinality of the reals is $\beth_1 = 2^{\aleph_0}$ (beth-one).

The cardinality of the set of irrational numbers is the cardinality of the set of real numbers, minus the cardinality of the set of rationals (which is the same as those of the integers and the naturals), which using the trans-infinite numbers is $\beth_1 - \aleph_0$, similar to the notation of the set itself as $\bar{\mathbb Q} = R-Q$ (the set difference of the rationals from the reals, also known as the set complement of the rationals over the reals).

* - A rational number's official definition is a value that can be represented as a fraction with integer terms. For any natural number $x$, $10^x$ is a natural number (because the set of natural numbers is closed over multiplication), and $10^{-x} = 1/10^x$ and so by definition it is also rational as both terms of the fraction are naturals (which are integers). Any integer mantissa $m$, multiplied by 10 to the power of any integer $x$, is thus either $(m*10^x)/1$ or $m/10^x$ for exponents $x>0$ and $x<0$ respectively (with the trivial case of an integer $m$, including 1, being $m=m*10^0$), and so any number constructed in this way is rational as all terms of the fraction are always integers. Not all rationals can be constructed by using an integer power of 10, but by allowing any natural number base it becomes trivial to express any rational in this same exponential form; $a/b = a*b^{-1}$ for all $a \in \mathbb Q, b \in \mathbb N$.


The number of edge cut sets simultaneously containing two different edges of complete graph Let $a$, $b$ be two different edges in the complete graph $K_n$. Try to find the number of edge cut sets that simultaneously contain $a$ and $b$.

Here, an edge cut of a connected graph $G$ is a set $S$ of $G$'s edges such that $G-S$ is disconnected and $G-S'$ is connected for any proper subset $S'$ of $S$.

I try to calculate based on whether $a$, $b$ have a common vertex, but I don't know the exact answer.

We get an edge cut whenever we partition the vertices of $K_n$ into two parts, and take all the edges going between the parts.

Let's consider two cases. First, suppose that edges $a$ and $b$ share an endpoint: they are edges $uv$ and $uw$ for some vertices $u,v,w$. Then we need to put $u$ on one side of the cut and $v,w$ on the other. There are $n-3$ other vertices in $K_n$; each of them can go on $u$'s side of the cut, or on $v$'s side. So there are $2^{n-3}$ possible cuts.

Second, suppose that edges $a$ and $b$ share no endpoints: they are edges $uv$ and $wx$ for some vertices $u,v,w,x$. There are two ways for both edges to be included in the cut: we could put $u,w$ on one side and $v,x$ on the other, or we could put $u,x$ on one side and $v,w$ on the other. Either way we do it, there are $n-4$ other vertices, each of which can go on one of two possible sides. So there are $2^{n-4}$ possible cuts either way, and again there are $2^{n-3}$ possible cuts altogether.

I see that there are $2$ edge cut sets and they are either cutting all the edges adjacent to $a$ or all those adjacent to $b$ (each has $n-1$ edges)

Any other edge cutting that will disconnect $a$ and $b$ has more than $n-1$ edges.