Are the rules of this tournament fair? My daughter just took part to a volleyball tournament and she wonders whether the rules of the tournament were fair or not.

There are 10 teams, gathered into 3 groups: Group 1 with 4 teams and Groups 2 and 3 with 3 teams. Each team plays a match against each other team in the same group. The result of each match being either 2-0 or 2-1, 1-2 or 0-2 (there is no tie in volleyball), it gives each team either 2, 1, -1 or -2 points for the tournament ranking.

After this first series of matches, the teams are redistributed into three new groups as follows: Group A (4 teams: the three winners of each group and the best among the seconds of the three groups), Group B (3 teams: the remaining seconds of each group and the best among the thirds of groups 1, 2 and 3), Group C (the remaining 3 teams).

The key rule explains how to decide which team is best second (respectively best third). The rule consists to add the points obtained by each team and to divide it by the number of matches. For instance, if a team of Group 1 won the first match by 2-1, won the second one by 2-0 and lost the third one by 1-2, this team would get a score of $(1 + 2 - 1)/3 = 2/3$. In case of tie, the total number of points scored in each set can be used.

Question. My daughter had the feeling that the teams in Group 1 (the one with 4 teams) had a slight advantage to end up being in either Group A or Group B. Is this feeling justified?

In a more mathematical setting, assuming that the results of the matches are randomly distributed with equal probability 1/4 for each score $(-2, -1, 1, 2)$, what is the probability for a team of Group 1 (respectively 2 or 3) to end up in Group A (respectively B and C)?

Although my daugther is mainly interested in the case of 10 teams divided into 3 groups, a mathematical argument for the case of $n$ teams divided into $r$ groups would be appreciated.

tl;dr: Group 1 teams each have about a $2$ percent higher probability of advancing to Group A than their Group 2 and 3 counterparts; they each have about a $2$ percent lower probability of moving on to Group B than their Group 2 and 3 counterparts.

If I've understood the situation correctly, Group 1 teams do indeed have an advantage, at least as far as Group A is concerned.  Intuitively, this is easy to understand: Keep in mind that all games are zero-sum.  If one team gains a point, its opponent loses a point.  If one team gains two points, its opponent loses two points.  At the end of round-robin play, the points of the teams in each numbered groups sum up to zero.

This is the crux (the key, if you will).  By symmetry, the second-place teams in the three-team Group 2 and Group 3 will have zero points on average.  By contrast, the second-place team in Group 1 is on the plus side, by as much as the third-place team in Group 1 is on the minus side.  Thus, we should expect that Group 1 will contribute the best second-place team more often that Group 2 or Group 3, even when their extra game is taken into account.

This was borne out by numerical computation.  There are $12$ games in all, with $4$ equally likely results for each.  Accordingly, I simulated round-robin play in all $4^{12} = 16777216$ scenarios.  For each scenario, I kept track of both the group with the best second-best (third-best, resp.) team, as well as the average points-per-game for that team.  In case of a two-way or three-way tie, I considered it equally likely for each Group to contribute the team to Group A (Group B, resp.).

The results were quite interesting.  The average second-best team in Group 1 had about $0.33984$ points per game, as compared with $0$ points per game in Groups 2 and 3.  Accordingly, Group 1 contributed its second-best team to Group A a whopping $0.64924$ of the time—almost two-thirds—as compared with Groups 2 and 3, which contributed their second-best team only $0.17538$ of the time.  (All results have been rounded to five significant digits.)

The results for the best third-place team were similar.  Here, we are comparing the second-worst team in Group 1 with the very worst teams in Groups 2 and 3.  The third-place team in Group 1 had an average points-per-game of $-0.33984$ (the flip side of the second-place team, as expected), whereas the third-place teams in Groups 2 and 3 had an average points-per-game of $-1.2188$.  This led to an almost landslide contribution from Group 1 to Group B of $0.79220$—nearly four-fifths of the time—as compared to Groups 2 and 3 with their contribution of $0.10390$.

Interestingly, the average first-place teams in Groups 2 and 3 had higher points per game than their counterparts in Group 1: $1.2188$ (again, the flip side of the third-place team) as opposed to $1.1055$.  But points-per-game have no bearing on a first-place team.

Counteracting this seeming dominance is the fact that Group 1 has more teams than Groups 2 and 3; that's why the rule was instituted in the first place.  Accounting for that, the average team from Group 1 has a $0.41231$ chance of advancing to Group A, respectively; this compares with $0.39179$ for Groups 2 and 3.  I think the figures for Group B are $0.28574$ for Group 1, and $0.30951$ for Groups 2 and 3, but I haven't confirmed that.

Well, out of the four teams in Group 1, one team that gets the highest score is placed in Group A, but that winning score doesn't matter in relation to the team scores from other groups. For example, the winning score from Group 1 can be 1, and the winning score from Group 2 can be 2, but both teams still go to Group A. In this first case where the winning teams are sent, the teams from Group 1 are at a slight disadvantage, since a team has a 1/4 chance of winning rather than the 1/3 chance teams from other groups face (does this make sense? Group 1 teams have to beat out 3 other teams rather beating out two like the teams from Group 2/3 need to).

Now, we must consider the fourth team that progresses to Group A. As you previously mentioned, the KEY RULE counteracts any advantage teams may gain by playing more/less games by dividing by the number of games played, thus making the probability of the fourth team in group A being from Group 1 1/3 (equal chance). Hence, if I were a team in Group 1, I would have a probability of getting into Group A of (1/4)+(1/3)=(7/12), whereas if I were from Group 2/3, I would have a probability of (1/3)+(1/3)=(2/3)=(8/12). This means that a team originally from Group 2/3 would have a slightly higher chance of going to Group A.

This can be intuitively understood through simple casework: probability of winning in your group and probability of being the "best second". Due to the KEY RULE, the latter is equal for all groups. However, in order to win in your group, it's harder in Group 1 since you have to beat 3 teams, as opposed to 2 other teams in Group 2/3.
 This method can be generalized to n teams r groups through simple substitution of n and r into the probabilities.


When this series converges? $\sum_{n=45}^{\infty}(-1)^n(x^2+2x)^{\log({R(n)})}$ with R(n)=... For n=1,2,.. is $P(n)=n^{10}-\sum_{k=1}^9{\ k  n^k}$ and $Q(n)=\sum_{k=1}^8{\   n^k}$ and is $R(n)=\dfrac{P(n)}{Q(n)}$.
Determine for which values of $x$ the series

$$\sum_{n=45}^{\infty}(-1)^n(x^2+2x)^{\log({R(n)})}$$

is convergent,  distinguishing simple and absolute convergence.

My answer:
Is $a=x^2+2x$. I wrote that for $|a|<1$ converges absolutely. Necessary condition for convergence is also simple that the limit of $(x^2+2x)^{\log({R(n)})}$ is $0$: being that if |a|>1 the limit is different from $0$ get that converges only for $|a|<1$. Is that okay?

Hint: For $x^2 + 2 x > 0$, you'll want to think about alternating series. For $x^2 + 2 x < 0$, things are rather more complicated because $(x^2+2x)^{\log(R(n))}$ is complex, but absolute convergence should be easy to test because $\log(R(n)) = 2 \log(n) + o(1)$.


Working out an algorithm for finding out whether a point is in or outside of a 2D closed polygon I'm having a bit of an issue with the following problem:

Write a brief (1/2 page) design specification document (including pseudo code for the algorithm itself) that describes your approach to solving the following geometric problem:
Given two straight lines in 2D (each line defined by a pair of points), determine whether they intersect.

Explain how you could apply the test in to write an algorithm to find whether a point lies inside or outside a 2D closed polygon consisting of $N$ points joined by straight lines.

As far as I am aware, it is possible to work out whether 2 lines intersect by working out each line equation, and then simultaneously solving those. IE:

2 lines at $(4, 2)$ to $(2, 0)$ and $(0, 4)$ to $(4, 0)$.

Solving simultaneously, the two lines intersect at $(3, 1)$

However, I am not quite sure how I would apply this method into an algorithm that for a 2D polygon as specified in the question.

Any help would be greatly appreciated!

You have one point p1 (x1, y1) that is given. Now, create another point far away from everything (one way to do: you can find maximum y coordinate of the polygon and add 1 to it. That is your y coordinate. Keep the x coordinate as x1. That is p2 = (x1, max(polygon_y)+1 ).

Let the vertices of the given polygon be v1, v2, ... , vn. So the edges are (v1,v2)...(vn,v1). Check how many of these edges intersect with (p1,p2). If it is even then the point is outside the polygon and if it is odd then it is inside.

This problem is really asking you to describe the Convex Hull of the polynomial.

For this, you'll want to arrange the points along the boundary of the polygon and then compute the line segments that border the polygon. The interior of that bounding region is what you're after.


Can synchronized chaotic systems be periodic or stochastic? I am confused regarding the terms chaotic, periodic, and stochastic in the context of coupled chaotic systems:

In general, if chaotic systems are coupled together and synchronized then does the dynamics of the synchronized system to a layman be referred to as still chaotic or periodic?
If the synchronized system becomes desynchronized by removing the chaotic oscillators, then can we say that the dynamics of the desynchronized system is stochastic and there is no chaos? What is the opposite of chaos? Periodicity of randomness?

From the application of chaotic circuits like Chua's chaotic circuit, if the synchronized system is composed of Chua chaotic oscillator and if one of the Chua's oscillator circuit becomes faulty then would it show random dynamics?

What is the opposite of chaos? Periodicity of randomness?

First of all, chaos, periodicity, deterministicity and stochasticity are best used as properties of stand-alone mathematical models¹.
A simple characterisation of these different properties is this:

In a stochastic dynamics, identical initial conditions lead to different outcomes.
A system that is not stochastic is called deterministic.
In modelling, you achieve stochasticity by employing random-number generators.²
This is something you have to explicitly do.
Systems do not become stochastic just like that.
In a chaotic dynamics, identical initial conditions lead to identical outcomes, but similar initial conditions lead to dissimilar outcomes (in general).
In a regular dynamics (i.e., periodic, quasiperiodic or fixed-point), similar initial conditions lead to similar outcomes and identical initial conditions lead to identical outcomes.

These three are clearly distinct.
Chaotic dynamics and regular dynamics are deterministic.
Thinking in terms of opposites is not helpful here.

¹ As to whether reality is any of those: We don’t know, it’s inherently impossible to find out, and even if we knew, this wouldn’t help the least with practical modelling applications.
² You can (and probably will) use pseudorandom number generators here, which are in turn deterministic, but if that affects your model, you are doing something wrong. For all intents and purposes, you should be able to replace a pseudorandom generator with a true random number generator.

  In general if chaotic systems are coupled together and synchronized then does the dynamics of the synchronized system to a layman be referred to as still chaotic or periodic?

If you couple two systems together, you create a new system.
If and only if the original dynamics (and the coupling) are deterministic, the coupled system will be deterministic³.
Thus, if you couple two chaotic systems, the result will be deterministic.
However, in general the resulting dynamics can be chaotic or regular.
For example:

It can exhibit some generalised synchrony, that in turn can either be regular or chaotic.
The systems may not synchronise at all, but exhibit distinct chaotic dynamics.

You can only make statements like this:
If two identical chaotic systems are diffusively coupled, the coupled system will for some initial conditions exhibit a completely synchronised chaotic dynamics, in which the individual systems exhibit their uncoupled dynamics.
However, the same system may already exhibit a different dynamics for different initial conditions.

  If the synchronized system becomes desynchronized by removing the chaotic oscillators, then can we say that the dynamics of the desynchronized system is stochastic and there is no chaos?

If you start with a deterministic system, removing one component cannot possibly make it stochastic.
As to what dynamics you will get, this depends on how exactly your removal looks like, but in general it can be anything.

  From the application of chaotic circuits like Chua's chaotic circuit, if the synchronized system is composed of Chua chaotic oscillator and if one of the Chua's oscillator circuit becomes faulty then would it show random dynamics?

As already mentioned, chaos, periodicity, deterministicity and stochasticity are properties of stand-alone mathematical models.
To translate these to real systems, the question is which kind of model adequately describes the system.
A real system always has a “stochastic component”, i.e., one that we can model only by stochasticity and that is beyond the scope of what we can access, e.g. thermal noise – if such components are negligible for whatever question we want to answer, we can use a deterministic model; otherwise, we cannot.

So, here it depends on what exactly you mean by faulty and how you can adequately model it.
It may be that the original system can be adequately modelled by a deterministic dynamics, while the new system cannot.
If faulty means that the circuit is exhibiting random fluctuations (that you cannot model deterministically), you get a stochastic system.

It doesn't matter wether you are a layman or an academic. In both cases one should use the proper notation and/or define the terms orderly and accurately. Remember a chaotic system can be periodic, if it has pseudorandom properties inside a periodic boundary condition. The question doesn't make much sense that its one or the other.

The opposite of chaos is also non-randomness, but it is a very complicated field to go into. One can say that non-randomness (or non-chaotic systems) have regular patterns that one can easily see or deduce, but sometimes chaotic systems have patterns combined with eachother, and it behaves chaotically that way. There is another thing called Complexity that can be intertwined in these concepts. Simple rules (methods) can produce complex behavior and pseudorandomness. Chua oscillators seems to be real hardware random generators and might be more than pseudorandom. But nobody knows if the real world randomness is pseudorandom. Periodicity is when something repeats, like a pendulum, but if it has a damping factor the state of the pendulum is never equal to a previous state.


How to find a suitable orthogonal matrix Assume $x,y \in \text{R}^n$ are two vectors of the same length, how to prove that there is an orthogonal matrix $A$ such that $Ax=y$?

Thanks for your help.

You can explicitly use a Householder matrix. Normalise $x$ and $y$ so that they become unit vectors (the special case $x=y$ is trivial and is ignored here). Let $u = x-y$ and $Q = I - \frac{uu^\top}{x^\top u}$. Then $Q$ is a real symmetric and orthogonal matrix such that $Qx=y$.

Since every orthogonal matrix $A$ always has inverse, $$ AA^{t} = A^{t}A = I $$, so that if $A$ is orthogonal then any equation $Ax=y$ has a solution $$ x = (...)y $$


For what value of $a$ does the equation $|x(x-4)|=a$ have exactly 3 real solutions? I simplified the equation and got two cases where $$x(x-4)=a$$
and
$$x(x-4)=-a.$$
How do I go after that. I know how you could get $2$ solutions or $4$ solutions, but how do you get $3$ solutions?

Note that we must have $a\geq0$ or the original equation is unsolvable. We then, as you showed have

$$x(x-4)=a$$

and

$$x(x-4)=-a$$

Solving these with the quadratic formula gives

$$x=\frac{4\pm\sqrt{16+4a}}{2}$$
$$x=\frac{4\pm\sqrt{16-4a}}{2}$$

Note that, since $a\geq0$, the first equation always has two solutions. Is there then an $a$ that will make the second only have $1$? You should see that happens when $\sqrt{16-4a}=0$, which tells us $a=4$.

Hint:
Linear equations have one solution
Quadratic equations have two solutions
Hence cubic equations will have 3 solutions


De Morgan's laws in logic and set theory In logic De Morgan's law means $\lnot (A \land B) \Leftrightarrow \lnot A \lor \lnot B$

In set theory De Morgan's law means $(A \cap B)^C = A^C \cup B^C$

I'm surprised that the same idea is true in different fields. Is there any explanation for this phenomenon or is it just a coincidence?

The explanation can be seen in the following chain of equivalences: $$\begin{align}(A \cap B)^C & = \{x\mid x \notin (A \cap B)\}\\ \\ & = \{x\mid \lnot [x\in (A\cap B)]\} \\ \\& = \{x\mid \lnot (x \in A \land x \in B)\} \\ \\ & = \{x\mid \lnot(x \in A)\lor \lnot (x \in B)\}  \\\\& = \{x \mid x \notin A \lor x \notin B\}\\ \\ & = \{x\mid x\in A^C \lor x \in B^C\} \\ \\ &= A^C \cup B^C\end{align}$$

My opinion is that set theory law is just a special case of the logic law. For example, we can note $a\in A$(in set domain) as $\mathcal{A}$(in logic domain) and $a\notin A$ noted as $\lnot\mathcal{A}$.


Finite limit of $f/g$ at infinity implies integrable Let $f,g : [a,\infty)\to \mathbb{R}$ be such that $f/g$ has finite limit at $\infty$.
If either $f$ or $g$ is Lebesgue-integrable then why must the other be also?
I know I have to use the comparison test but I’m not sure how.

$a=1, f(x)=\frac 1 {x^{2}}, g(x) =\frac 1 x$ is a counter-example. However, if the limit of $\frac f g$ is finite and nonzero then the conclusion holds.

Suppose $\displaystyle \lim_{x\,\to\,+\infty} \frac {f(x)} {g(x)} = L \in (0,+\infty)$ and $\displaystyle\int\limits_{[a,+\infty)} |g(x)|\,dx <+\infty.$ Then for some positive number $b$ you have $0 < f(x) < (L+1)g(x)$ whenever $x>b.$ Therefore
$$
\int\limits_{[a,+\infty)} |f(x)|\,dx \le \int\limits_{[a,+\infty)} (L+1)|g(x)|\,dx < +\infty.
$$
Since we assumed $L\ne0$ above, you can apply the same argument with $g/f$ in place of $f/g$ and $1/L$ in place of $L.$ But that doesn't work if $L=0.$
But what if $x\not>b\text{ ?}$ We can't say what happens if $a<x\le b.$


How many 0's are in the end of this expansion? How many $0's$ are in the end of:

$$1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4.... 99^{99}$$

The answer is supposed to be $1100$ but I have absolutely NO clue how to get there.

Any advice?

To find out the number of zeroes at the end of that huge number, we just have to find out the exponent of $10$ in it.
Since $10=2\cdot 5$, we can just find out the exponent of $5$ in that number as the exponent of $2$ will be more than $5$.
All the numbers which have $5$ as their factor will be counted.
These numbers would be
$$5^5,10^{10},15^{15},\ldots,95^{95}$$
We know that all numbers except $25,50$ and $75$ will only contribute themselves as the exponent, whereas the aforementioned numbers, will contribute double of themselves.
So, we have
$$(5+10+\cdots+95)+(75+50+25)=\frac{19}{2}(5+95) + 150=1100$$

The product is:

  [encoded data, 8815 characters]


3D Geometry Question In $3$-dimensional Geometry, if angle made of line segment $OP$ with $X,Y,Z$-axis are in
$1:2:3$, then what is the angle made by line segment with $Y$-axis?

My Solution:

Let $\alpha,\beta$ and $\gamma$ be the angle made by line with $X,Y,Z$-axis, respectively. Then

$$\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma = 1$$

Now given $\alpha = \theta$ and $\beta = 2\theta$ and $\gamma = 3\theta$,

So $\cos^2 \theta+\cos^2 2\theta+\cos^2 3\theta = 1$,
or
$$2\cos^2 \theta+2\cos^2 2\theta+2\cos^2 3\theta = 2,$$
or
$$1+\cos 2\theta+\cos^2 2\theta+1+\cos 6\theta = 2,$$
or
$$1+\cos 2\theta+2\cos^2 2\theta+1+4\cos^3 2\theta-3\cos 2 \theta = 2.$$

Let $\cos 2 \theta = y$, then

$$y+2y^2+4y^3-3y = 0,$$

$$4y^3+2y^2-2y = 0,$$

$$2y(2y^2-y-1) = 0.$$

Maybe I have missed something, but I feel that this method is very tedious.

So could anyone explain me a better method?

Thanks.

say smallest angle is $\theta$

therefore $\theta+2\theta +3\theta= 180^\circ$.

Sum of the angles of a triangle equal $180^\circ$.
therefore $\theta=30^\circ$ (since $6\theta=180^\circ=> \theta=\dfrac{180}{6}=30^\circ$)

therefore angle made with y axis is $60^\circ$.


Maximizing sum of metric function on a set (Adaptation of Hungarian Algorithm) Suppose I have a set of unique elements $A=\{a_1, a_2, ..., a_n\}$. Suppose I also have a metric function $f:A\times A \rightarrow R^+$. I want to choose $k$ elements from $A$ (i.e. $a_{i_1},a_{i_2}, ..., a_{i_k}$, $I=\{i_1,...,i_k\}$) such that it maximizes the sum:

$$S=\sum _ {i,.j\in I, i\neq j}^{}{f(a_i,a_j)}$$

Is there any known algorithm to do something like this? I know the Hungarian Algorithm can be used to solve the case of $k=n$. Can it be adapted for $k<n$?

This problem is NP-hard. The decision problem of max-clique is NP-complete. Deciding whether a clique of size $k$ exists in a graph of $N$ vertices can be translated to your problem as follows. Let $A$ be the vertex set, and let $f(a_i,a_j)=1$ if the vertices are adjacent, $f(a_i,a_j)=0$ otherwise. A clique of size $k$ exists if and only if the optimal value of $S$ is $0.5 k (k-1)$.

This doesn't mean that the problem is not easy for some values of $k$ (such as $k=1$) but that an algorithm that works for any $k$ can probably not solve it in polynomial time.


Sum of $1/[(n + 1)^2] + 1/[(n + 2)^2] + ... + 1/[(2n)^2]$ when n goes to infinity Could someone please help me understand the steps to solve this problem?

For finite $n$ the sum is $\psi ^{(1)}(n+1)-\psi ^{(1)}(2 n+1)$, where $\psi$ is the polygamma function.  Take the limit:

$$\lim\limits_{n \to \infty} \left( \psi ^{(1)}(n+1)-\psi ^{(1)}(2 n+1) \right) = 0.$$


Is the language "substrings of an even-lengthed regular language" also regular? I want to prove that for a regular language $L$ where $\forall w \in L$ the length of $w$ is even, the language containing the first halves of the words of $L$ and the language containing the second halves are both regular.

That is, for the language containing only aabbab, the language containing aab is regular (as is the one containing bab).

I know that concatenation of two regular languages yields a regular language, and that the reverse is not always true. But with this restriction on length, I feel unequipped to answer this question for sure.

For any language $L$, let $L_{\frac12-}$ be the set of first halves of the words in $L$ and $L_{-\frac12}$ be the set of second halves of the words in $L$. Also let $L^R$ be the set of reversals of words in $L$.

If $L$ is regular then so is $L^R$. Since $L_{-\frac12} = ((L^R)_{\frac12-})^R$ it is sufficient to show that if $L$ is regular then so is $L_{\frac12-}$.

Let $M = (Q,\Sigma,\delta,q_0,F)$ be a DFA such that $L(M) = L$.

Construct a DFA $N$ as follows:

The set of states of $N$ is $Q \times P(Q)$ where $P(Q)$ is the power set of $Q$.
The start state of $N$ is $(q_0,F)$.
The set of accept states of $N$ is $\{(q,S) \in Q \times P(Q) \mid q \in S\}.$
Define a function $\gamma : P(Q) \to P(Q)$ by $$\gamma(S) = \{q \in Q \mid {\rm there\ exists\ }  a \in \Sigma {\rm\ such\ that\ } \delta(q,a) \in S\}.$$
The transition function $\bar\delta$ of $N$ is defined by $\bar\delta((q,S),a) = (\delta(q,a), \gamma(S)).$

Then $L(N) = L_{\frac12-}$.

Proof

Suppose $(q_0,S_0),\ldots,(q_n,S_n)$ is the path taken by $N$ on input $w = w_1\cdots w_n$, and $(q_n,S_n)$ is an accept state. Then $q_0,\ldots,q_n$ is the path taken by $M$ on input $w$ and $q_n \in S_n$.

From the definition of $\bar\delta$, there exist $q_{n+1}, \ldots q_{2n} \in Q$ and $v_1,\ldots,v_n \in \Sigma$ such that $q_{n+i} = \delta(q_{n+i-1}, v_i) \in S_{n-i}$, so $q_0,\ldots,q_n,q_{n+1},\ldots q_{2n}$ is the path taken by $M$ on input $w_1\cdots w_nv_1 \cdots v_n$. Also $q_{2n} \in S_0 = F$ so $w$ is the first half of a word accepted by $M$.

Conversely suppose $wv \in L(M)$ and $|w|=|v|=n$. Let $q_0,\ldots,q_{2n}$ be the path taken by $M$ on input $wv$, and let $(q_0,S_0), \ldots, (q_n,S_n)$ be the path taken by $N$ on input $w$. Then $q_{2n} \in F = S_0$ and so it follows by induction on $i$ that $q_{2n-i} \in S_i$ for $0 \leq i \leq n$. In particular, $q_n \in S_n$ so $N$ accpets $w$.

Yes, the language $M$ of even-lengthed strings over a particular alphabet is regular. And the intersection of two regular languages is regular. So the language $L \cap M$ of even-lengthed substrings of the language $L$ is regular.


Reason for setting coefficients of $x$ and $x^2$ equal to zero when roots are infinite. While studying coordinate geometry, I came across a topic to find the equation of asymptote of a standard hyperbola, $$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $$ in that topic, the equation of asymptote was assumed to be $y = mx + c$, and them after some work (forming a quadratic equation having roots that gave $x$ coordinates of the point of intersection of line and hyperbola), the coefficient of $x$ squared and $x$ was set to be zero due to the fact that asymptote never intersects the curve. (So the roots of quadratic that is , the $x$ coordinates of point of intersection will tend to infinity).

I want to know the REASON behind this analogy, although I'm guessing since the sum of roots is given to be the negative of coefficient of $x$ DIVIDED by coefficient of $x$ squared (by Veita's formula), since the sum of roots tends to infinity, the denominator that is, the coefficient of x squared must tend to zero, is my assumed analogy correct? But then, why the coefficient of $x$ is set to zero?

Any help will be greatly appreciated.

EDIT:

Here's the photo of the page of book in which equation of asymptote was derived. (https://i.stack.imgur.com/Via9z.jpg)

  the straight line $y = mx + c \tag 1$ meets the hyperbola in points, whose abscissae are given by the equation $\qquad x^2(b^2 - a^2m^2) - 2a^2 mcx - a^2(c^2+b^2) = 0 \tag 2$ If the straight line be an asymptote, both the roots of $(2)$ must be infinite.  Hence, the coefficients of $x^2$ and $x$ in it must both be zero.

What is going to happen if the coefficients of $x$ and $x^2$ are not zero?

$$\qquad x^2(b^2 - a^2m^2) - 2a^2 mcx - a^2(c^2+b^2) = 0 \tag 2$$

As $x\to \infty$ the right hand side stays at $0$ while the left hand side approaches either infinity or negative infinity depending on the other parameters.

Another way to see the asymptotes is to solve$$ \frac {x^2}{a^2} - \frac {y^2}{b^2} = 1$$  for $y$

Note that $$ y=\pm (b/a)x\sqrt {1-(a/x)^2}$$ which is asymptotic to  $$ y=\pm (b/a)x $$

If you now know why the coefficient of $x^2$ is zero here's an explanation for why coefficient of $x$ is zero

asymptotes are tangents to  the hyperbolas at infinity and therefore has discriminant  equal to zero
i.e. $b^2=4ac$     in equation $ax^2 + bx +c$
      $(a =0)$
     now to satisfy this condition $b=0$


Space of continuous functions with compact support dense in space of continuous functions vanishing at infinity How can we prove that the space of continuous functions with compact support is dense in the space of continuous functions that vanish at infinity?

Presumably your norm is the maximum absolute distance at any point. If I give you a continuous $f$ that vanishes at $\infty$ and an $\epsilon \gt 0$, you need to find a continuous $g$ with compact support that is everywhere within $\epsilon$ of $f$. Since $f$ vanishes at $\infty$, there is some $N$ such that $f(x) \lt \epsilon/2$ whenever $ x \gt N$. So if $g=f$ out to $N$, drops quickly to $0$ and stays there...

Added:  We are trying to show that any continuous function that vanishes at infinity can be approximated closely (within $\epsilon$) by a function with compact support.  The range where $f$ is large (greater that $\epsilon/2$) is some interval $(M,N)$  Our approximating function will agree with $f$ over that range.  We then know that $|f(x)| \lt \epsilon/2$ on $(\infty,M)$ and $(N,\infty)$.  If we let the approximating function go to zero on $(M',M)$ and $(N,N')$ and be zero on $(-infty,M')$ and $(N',\infty)$ it will have compact support and be within $\epsilon$ of $f$ over the whole line.

In other words: Let me denote by $C_c$ the space of continuous function with compact support and $C_0$ the one of continuous function vanishing at infinity.

  $C_c$ dense in $C_0$ if and only if $\forall g \in C_0$ it exists a sequence $g_n\in C_c$ s.t. $g_n\to g$ w.r.t. the supremum-norm

So we can define
\begin{equation}
g_N(x)=g(x) \text{ for } \vert x\vert\leq N \text{ and } g_N(x)=0 \text{ if } \vert x\vert>N
\end{equation}
So $g_N$ is compactly supported.
Thus $\lim_{N\to \infty} \Vert g_N-g\Vert_{\infty}=0$


Cartesian  a+bi conversion For the question in the image,

Question and my answer

I answered this question with (1/root5, -1/root5), which correspond to x and y value in Cartesian coordinate.

However, the correct answer for this question says, 1/5-3i/5, as following picture:

Correct answer for this question

What I actually did is to convert that 1/5-3i/5 into x and y form using x=rcos(theta) and y=rsin(theta), becuz I thought x and y value represents Cartesian coordinate if it is about 2 dimensional.

So my question is, does that form of a+bi really represents Cartesian coordinate?

$a+bi\ $ represents coordinates in the complex plane, where $a$ is the real component, and $b$ is the complex component.

The Cartesian plane can be used to represent any complex number. Your answer was to be given in Cartesian coordinates.

Check this section out.


Finding sum $\frac{1}{2} x^2 + \frac{1}{2} \frac{1}{3} x^3 + \frac{1}{4} \frac{1}{3} x^4 + ... $ I need to find following: for $0 < x < 1$

$$\frac{1}{2} x^2 + \frac{1}{2} \frac{1}{3} x^3 + \frac{1}{4} \frac{1}{3} x^4  + ... $$

My attempt:

I can see that the sum is composed of two infinite sums, one is $\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \sum_{n=1}^{\infty} \left( \frac{1}{n} -
 \frac{1}{n+1} \right)$ (Telescoping ) and another is, $\sum_{n=2}^{\infty} x^n$ (it's a G.P.). How can I use these for solving the sum in question?

Any hints will be appreciated...

Let us set
$$
S(x)=\frac{1}{2} x^2 + \frac{1}{2} \frac{1}{3} x^3 + \frac{1}{4} \frac{1}{3} x^4  + \ldots
$$
then
$$
\frac{dS(x)}{dx}=x + \frac{1}{2}x^2 + \frac{1}{3} x^3  + \ldots.
$$
This is a well-known Taylor series being
$$
   x + \frac{1}{2}x^2 + \frac{1}{3} x^3  + \ldots = -\ln(1-x).
$$
We now note that $S(0)=0$, then we have to solve the simple differential equation
$$
   \frac{dS(x)}{dx}=-\ln(1-x)
$$
with the given initial condition. This yields
$$
  S(x)=(1-x)\ln(1-x)+x
$$
that is the result you were looking for.

Hint
U can make the series into a geometric progression where $$a=\frac{x^2}{2}$$  and $$r=(x/(n+1))/(1/(n-1))$$


Limit of $f$ at $(0,0)$ vs limits of $f(x,kx)$ when $x\to0$ Why if for all $k\in\mathbb R$ we have $f(x,kx)\to\ell$ when $x\to 0$ then this is not enough to conclude that $f(x,y)\to\ell$ if $(x,y)\to (0,0)$?

Indeed, if the limit is the same when we can consider all straight lines that pass through $(0,0)$, logically it should be enough to conclude on the limit, but I found a counter example. So why doesn't it work ? What is the mathematical reason behind this phenomenon?

For a counterexample, let $f(x,y)=\frac{x^2y}{x^4+y^2}$. Then $f(x,kx)=\frac{kx^3}{x^4+k^2x^2}=\frac{kx}{x^2+k^2}\to 0$ as $x\to 0$. But $f(x,x^2)=\frac{x^4}{2x^4}=\frac{1}{2}\not\to 0$

The idea is that $y$ can approach $0$ faster than $x$ can by a LOT, giving issues.

For a function in more than one  variable;

For limit of a  function $f$   to exist at a particular point we have to consider all possible paths for reaching that point and see whether the limit exits or not. What you are considering are only  the straight line paths.

There can be many ways to reach that point


How to find x in this case $\frac{x}{2}-1=e^x$? $$\frac{x}{2}-1=e^x$$

Let say we can write above function as:
$$\frac{x}{2(e^x-1)}=1$$

In Wolframalpha I get a solution $x\approx -1.59362$.

How can I find x?

There is no real solution since

$$e^x\geq 1+x$$

For $x> -4$ we have $e^x\geq 1+x>\frac{x}{2}-1$, so $e^x-\left (\frac{x}{2}-1\right)>0$. This means that they don't intersect for $x\in \left <-4,\infty\right >$.

When $x\leq -4$ we have $e^x>0$ and so $e^x-\left (\frac{x}{2}-1\right )>0$ for $x\in \left <-\infty,-4\right]$. Therefore they don't intersect, and no real solution exists.

$$ \frac{x}{2} - 1 = e^x $$
$$ \frac{x}{2} - 2 = e^x - 1  $$
$$ \frac{x-4}{2} = e^x -1 $$
$$ \frac{x-4}{2(e^x -1)} = 1 $$

  Let say we can write above function as:

You can't do that. Because then $x-4 = x \implies -4 = 0$, which is nonsense.

EDIT:

Assuming you meant:

$$ \frac{x}{2(e^x +1)} = 1 $$

Then you can just plot the function $f(x) = \frac{x}{2(e^x+1)}$ and solve $f(x) = 1$

EDIT 2:

I'm sure you're making a mistake:


lines drawn in a plane Seventeen lines drawn in a plane, with no three concurrent and no two parallel divide the plane into closed regions and open regions. What is the number of closed regions?

I figured out the number of regions is 154 but I am not able to figure out the number of closed regions.

The answer to this question would be 120. As you can see from a chart
number of lines   0 1 2 3 4 5
number of closed  0 0 0 1 3 6

The formula to find the number of closed should be: n-2(n-1)/2
If we input 17 then, 15*16/2 is equal to 120.


Every continuous function $f:ℝ\to ℝ$ can be presented as a sum of a certain power series. I've encountered a statement that I am unable to prove or disprove:

  Every continuous function $f:ℝ\to ℝ$ can be presented as a sum of a certain power series.

Is this statement true? If yes, can somebody prove it?

This isn't even true for smooth functions. The classic example: try to expand
$$
f(x)=\begin{cases}e^{-1/x^2}&x\ne 0\\
0&x=0\end{cases}
$$
in a power series at $x=0$. Convince yourself that $f^{n}(0)$ exists for any $n$ and is zero (these are the coefficients of a hypothetical power series). But the function certainly isn't zero in any positive radius of the origin.

This is not true in all cases. Power series are differentiable almost everywhere but there exist continuous functions which are differentiable nowhere. Now consider the continuous smooth function $f(x)=\exp(-1/x^2)$ with $f(0)=0$. What power series would you like this to be equal to? You should find that the only power series it could be is: $0 + 0x + 0x^2 +\cdots$ but this function is not zero.


relating mole fraction to molality If the mole fraction of water in a HCl solution is 0.78 . calculate the molality of the solution?
 I don't understand how to relate mole fraction to molality.

The mole fraction is the ratio of moles of solute to moles of solvent. A mole fraction of $.78$ means that for every mole of HCl, there are $.78$ moles of water. Molality is the number of moles of solute per mass of solvant. Since HCl has a molar mass of 36.5 g/mol, we would use dimensional analysis to calculate the molality:
$$\frac{0.78 \text{ molH$_2$O}}{1\text{ mol HCl}}\cdot\frac{1 \text{ mol HCl}}{36.5\text{ g HCl}}=.021 \frac{\text{mol H$_2$O}}{\text{g HCl}}.$$
Usually, molality is given in mol/kg, so this is $21$ mol/kg.

Mole fraction of water = 0.78
Mole fraction of HCl = 1 - 0.78 = 0.22
If no. of moles of water = n1 and that of HCl = n2,
n1/(n1 + n2) = 0.78
n2/(n1 + n2) = 0.22

n1/n2 =0.78/0.22 = 3.54

Molality is no. of moles of solute in 1000g of water
No. of moles of water in 1000g of water = 1000/18 = 55.55 = n1
n2 = 55.55/3.54 = 15.69
So, molality = 15.69m


Recurrence Question involving logarithm Can anyone please solve this recurrence

$$T(n)=T(3\sqrt n)+\log n.$$

It came in my paper. I want to know whether the following answer is right or wrong:

My answer: $T(n)=\log^3 n.$

For $n=9$ the recurrence $T(n)=T(3\sqrt n)+\lg n$ gives

$T(9)=T(9)+\lg 9$,

hence $\lg9=0$  ??

Something wents wrong ...


Is this Gram-Scmidt (or an application of) it? I am given a $2\times 2$ matrix
$$\left[ \begin{array}{ccc}
a & 0  \\
0 & b \\
\end{array} \right] $$
where $a,b \in \mathbb{R}$. I was told than an orthnormal basis for the colums of this matrix is $\frac{1}{\sqrt{a}}e_1$ and $\frac{1}{\sqrt{a}} e_2$. How does one show this? At first I thought it was an application of Gram-Schmidt, but I am not sure how to apply it.

There is something incorrect in the statement. I'll first explain why.

In order for a basis to be orthonormal, the basis vectors need to be of length one in addition to being orthogonal.

The notation $e_1$ usually means a vector of all 0's and 1 in the first spot. In general $e_i$ is a vector of all 0's with a 1 in the $i^{th}$ spot.

Therefore the statement that $\frac{1}{\sqrt{a}} e_1 $ and $ \frac{1}{\sqrt{a}} e_2$ are orthnormal is incorrect because each of these vectors have length
$$\|\frac{1}{\sqrt{a}} e_1\|_2= \sqrt{{\frac{1}{\sqrt{a}}}^2+0^2+...+0^2}=\sqrt{\frac{1}{a} }=\frac{1}{\sqrt{a}}$$

which is only 1 when $a=1$. $ \frac{1}{\sqrt{a}} e_2$ is similar.

Since a basis is closed under multiplication, any multiple of a basis element can be treated as the same basis element. For example,
$$ \left(\begin{array}{c}
a\\
0
\end{array}\right) \text { and } \left(\begin{array}{c}
1\\
0
\end{array}\right) $$
can be consider the same because
$$ \left(\begin{array}{c}
a\\
0
\end{array}\right) = a \left(\begin{array}{c}
1\\
0
\end{array}\right). $$

Looking at $$\left[\begin{array}{cc}
a & 0\\
0 & b
\end{array}\right],$$ $$ \left(\left\{ \begin{array}{c}
a\\
0
\end{array}\right\} ,\left\{ \begin{array}{c}
0\\
b
\end{array}\right\} \right) $$ Is an orthogonal basis, but it isn't an orthonormal one. In order to make it orthonormal you need to divide each vector by its length. (In this case, their lengths are just $a$ and $b$.) Thus your orthonornmal basis is
$$ \left(\left\{ \begin{array}{c}
1\\
0
\end{array}\right\} ,\left\{ \begin{array}{c}
0\\
1
\end{array}\right\} \right) .$$

Gram-Schmidt could be used here to get an orthonormal basis. However, since this example is relatively simple, we don't need such a complicated process. Instead, we can get the orthonormal basis using the thought process I described.

If you meant an orthonormal basis for the columns space, and assuming $\;a,b\neq0\;$,  then it could be

$$\left\{\;\binom10\;,\;\;\binom 01\;\right\}$$

Of course, the above is not the only orthonormal basis for that space.


What was the first bit of mathematics that made you realize that math is beautiful? (For children's book) I'm a children's book writer and illustrator, and I want to to create a book for young readers that exposes the beauty of mathematics. I recently read Paul Lockhart's essay "The Mathematician's Lament," and found that I, too, lament the uninspiring quality of my elementary math education.

I want to make a book that discredits the notion that math is merely a series of calculations, and inspires a sense of awe and genuine curiosity in young readers.

However, I myself am mathematically unsophisticated.

  What was the first bit of mathematics that made you realize that math is beautiful?

For the purposes of this children's book, accessible answers would be appreciated.

This wasn't the first, but it's definitely awesome:

This is a proof of the Pythagorean theorem, and it uses no words!

I don't find it beautiful, but I still find the idea expressed by the following something of a psychological curiosity:

How can it be that when some algebraists say "AND" and "OR" they mean exactly the same thing?

OR means this that "false or false" is false, "false or true", "true or false" as well as "true or true" are true, or more compactly:

    F  T
 F  F  T
 T  T  T

AND means this:

    F  T
 F  F  F
 T  F  T

But, since NOT(x OR y)=(NOT x AND NOT y) and NOT(T)=F and NOT(F)=T, OR and AND, to an algebraist, mean exactly the same thing!


Limit of $\sum \limits_{k=1}^{n} \frac{2^kn+2n^2+k}{2^{k+1}n^2+2^kk}$ when $n\to\infty$ I have to show the convergence of the series
$$\lim\limits_{n \to \infty}a_n=\sum \limits_{k=1}^{n} \frac{2^kn+2n^2+k}{2^{k+1}n^2+2^kk}.$$ I am quite sure that the limit is 1.5. I wanted to show this by estimation for another series smaller than $a_n$ and a further one that is bigger than $\lim\limits_{n \to \infty}a_n$.
I have been trying to find this estimation for hours without success. Can someone help me, please?

Fine as they are, the other answers gloss over an important nuance:  In a comparison of (positive) sequences, if $a_n\le b_n$ and $\lim_{n\to\infty}b_n=L$, we cannot conclude that $\lim_{n\to\infty}a_n=L$.  In fact, we cannot even conclude that $\lim_{n\to\infty}a_n$ exists (unless $L=0$).  We need to squeeze $a_n$ between two sequences that both converge to the same limit.

To do this, we begin with

$$
{2^kn+2n^2+k\over2^{k+1}n^2+2^kk}
={2^kn+(2n^2+k)\over2^k(2n^2+k)}
={n\over2n^2+k}+{1\over2^k}
$$

The key is to note that, for $1\le k\le n$,

$${1\over2n+1}={n\over2n^2+n}\le{n\over2n^2+k}\lt{n\over2n^2}={1\over2n}$$

and thus

$${1\over2n+1}+{1\over2^k}\le{2^kn+2n^2+k\over2^{k+1}n^2+2^kk}\lt{1\over2n}+{1\over2^k}$$

From this we see that

$${n\over2n+1}+\sum_{k=1}^n{1\over2^k}=\sum_{k=1}^n\left({1\over2n+1}+{1\over2^k}\right) \le a_n\lt\sum_{k=1}^n\left({1\over2n}+{1\over2^k}\right)={1\over2}+\sum_{k=1}^n{1\over2^k}$$

and now we see that $a_n$ is squeezed between two sequences that both tend to ${1\over2}+1=1.5$.

Factor $n^2$ out of the numerator and denominator to get
    $$a_n = \sum_{k=1}^n\frac{2 + \frac{2^k}{n} + \frac{k}{n^2}}{2^{k+1} + \frac{2^kk}{n^2}}.$$
Making the denominator smaller makes the whole fraction bigger, so you have the comparison
    $$a_n\leq \sum_{k=1}^n\frac{2+\frac{2^k}{n} + \frac{k}{n^2}}{2^{k+1}}.$$
Writing everything over a common denominator and performing some simple algebra, you get
    $$a_n\leq \sum_{k=1}^n\frac{1}{2^k} + \frac{1}{2} + \frac{1}{n^2}\sum_{k=1}^n\frac{k}{2^{k+1}}.$$
The first sum on the right hand side is a geometric series, which converges as $n\to\infty$, the second term is a constant, and for the third we can use that $k\leq 2^k$ when $k\geq 1$ to get the comparison
    $$a_n\leq \sum_{k=1}^n\frac{1}{2^k} + \frac{1}{2} + \frac{1}{2n}.$$
Taking the limit as $n\to\infty$ as using the algebraic limit laws, since each of the three terms is convergent, we get
    $$\lim_{n\to\infty}a_n = \sum_{k=1}^{\infty}\frac{1}{2^k} + \frac{1}{2} + \lim_{n\to\infty}\frac{1}{2n}$$
which will give you 1.5 as you claimed since the geometric series converges to 1 and $\frac{1}{2n}\to 0$ as $n\to\infty$.

EDIT: As pointed out in the comments, this doesn't actually show you what the limit is, rather bounds it above by 1.5, provided it exists.


Every infinite set has a countably infinite subset. Prove every infinite set has a countably infinite subset.
I was wondering if this approach is completely correct for this problem, my approach was to recursively define a function $f: \mathbb{N} \rightarrow$ countably infinite subset of an infinite set.
Attempt:
Suppose $A$ is infinite. We will construct a function $f: \mathbb{N} \rightarrow B$, where $B \subset A$, and $B$ is countably infinite. Since $A \neq \emptyset$, we can choose an element $a_{1} \in A$. Set $f(1)=a_1$. Since $A$ is infinite, choose an element $a_2 \in A-\{a_1\}$.Set $f(2)=a_2.$Assume for each $m<n$, $f(m)$ has been chosen. Since $A$ is infinite there exists in element $a_{n} \in A-\{a_1,...a_{n-1}\}$. Set $f(n)=a_n$. Now that $f$ has been defined inductively, $B=\{a_1,a_2,...\} \subset A$ is countably infinite.

Nerding on Wikipedia I discovered the following facts. A set that admits an injective function from natural numbers is called "Dedekind-infinite". An infinite set is one that is not in bijection with a finite set $\{0, \ldots, n\}$.
Suppose we assume the Zermelo-Frankel (ZF) axioms. Then:

We always have that a Dedekind infinite set is infinite, because there exist no injection from $\mathbb{N}$ to a finite set.

Assume the countable axiom of choice ($AC_{\omega}$). Then your proof can actually be formalized to choose all the values $f(n) $ "together". This is a very subtle point, but you should be aware that generally without the axiom of choice very pathologicsl things can happen, see for example the Banach-Tarski paradox.

Logicians would write this point as $ZF +AC_{\omega} $ proves "infinite $\Rightarrow$ Dedekind-infinite".

If we don't assume the countable axion of choice, surprisingly there exist infinite sets which are not Dedekind-infinite! As far as I understood, this proof is not constructive however, so it is kind of a monster.

In this case, logicians would say that the equivalence "Dedekind-infinite $\Leftrightarrow$ Infinite " is strictly weaker than the countable axiom of choice.


A very elementary change of variable Let $f,g\in C^0(\mathbb R)$ and consider having an equation of the form
$$f(x)g\left(\frac{y-x}{\alpha}\right)=F(x,y),$$
where the right hand side is some function involving $x,y\in\mathbb R$. For the question at hand, the right hand side of the equality doesn't come into it at all; it is only to pose the above equality. What I am interested in, exclusively is the left hand side of the equality.

What is it, precisely and rigorously, which allows me to reconsider the left hand side, equivalently, as

$$f(x)g\left(\frac{y-x}{\alpha}\right)=f(y-\alpha x)g\left(x\right)?$$

On the one hand, it seems like we equate both arguments of the functions in the concatenation on the left hand side and rearrange, via,
$$x=\frac{y-x}\alpha\iff \alpha x=y-x \iff \alpha x - y=-x,$$
to obtain the desired change. But this feels like cheating. What is going on behind this change of variable which allows one to reconsider the left hand side in this way?

Your second displayed equation makes no sense since it uses the same letters $x$, $y$ for different things.

We have a coordinate transformation $\psi: \>{\mathbb R}^2\to{\mathbb R}^2$ defined by
$$\psi:\>(u,v)\mapsto\left\{\eqalign{x&=v-\alpha u\cr y&=v\cr}\right.\ ,\qquad{\rm resp.},\qquad\psi^{-1}:\>(x,y)\mapsto\left\{\eqalign{u&={y-x\over\alpha}\cr v&=y\cr}\right.\quad.$$
We then want to express the function $F:\>{\mathbb R}^2\to{\mathbb R}$ in terms of the new coordinates. This means that we consider the function $$G(u,v):=F\bigl(x(u,v),y(u,v)\bigr)=F\bigl(v-\alpha u, v\bigr)=f(v-\alpha u) g\left({v-(v-\alpha u)\over\alpha}\right)\ ,$$
so that we obtain
$$G(u,v)=f(v-\alpha u) g(u)\ .$$

Nothing is going on behind the scenes: this is just different ways of writing the same thing.  In your initial expression $f(x)g((y-x)/\alpha)$ you have that $f$ and $g$ are both linear functions of $x$ (and, given you've equated it to $F(x,y)$ that g is also a linear function of $y$).  So it doesn't really matter how you express the argument of both functions so long as you don't change the nature of them -- whatever works best for your analysis of the problem can be picked.

It might be clearer if you look at it as a two-step process though.  The first step is to change the variable from $x$ to $z$:
$$ z = \frac{y-x}{\alpha} \Longleftrightarrow x=y-\alpha \cdot z $$
so that your expression (in $z$) becomes $f(y-\alpha z)g(z)$.  Then in the second step you rename $z$ as $x$ (this $x$ is now a 'new' $x$ which is used because you like $x$ as a variable, or you need $z$ for something else, or for whatever reason).


Prove that $10101\ldots01$ can't be a perfect square. Prove that $10101\ldots01$, where the sequence $"01"$ is repeated $k$ times can't be a perfect square for any value of $k$.

All I managed to prove is that the general formula is $1+\frac {100}{99}(100^k-1)$ but I don't know how to continue.

An odd square has to be $\equiv 1 \bmod 8$ because$(2m+1)^2=4m^2+4m+1=4m(m+1)+1$ where $m(m+1)$ is even.  Any number rendered in base $10$ is a multiple of $1000=125×8$ plus its last three digits, thus congruent with the last three digits $\bmod 8$.  But $101 \equiv 5 \bmod 8$, a contradiction.

It can be proved by the conclusions of the following 2 points:$$$$
1.  Suppose we take k = 1,
In this case, 101 is 1 more than 100 (the nearest perfect square). $$$$
similarly, for any value of k, the nearest perfect square will be  1 less than the number. ex. for k=2, 10101 nearest square ->  10000(  $100^2$ ),for k=3, 1010101 -> 1,000,000(  $ 1000^2$ ) $$$$
2. Now Observe the following pattern $$$$
$2^2$ - $1^2$ = 4-1 = 3 $$$$
$3^2$ - $2^2$ = 9-4 =5 $$$$ . $$$$ . $$$$
$(x+1)^2$ - $x^2$ = $x^2$ + 2x +1 - $x^2$ = 2x+1 $$$$
Clearly the difference between square of two consecutive CAN NOT BE 1 for any value of k.
Ex. for k =1 the nearest squares are 100 and 121 and difference is 21.
    for k=2 these are 10000 and 10201 and difference is 201 and so on .

So any for any value of k 10101010....01 can not be perfect square.


Simplification of the sum of all 4-digit palindromic numbers I apologize for the simplicity of this question, if necessary.

I understand that the sum of all 4-digit palindromes is as follows:
$$\sum_{a=1}^{9}(\sum_{b=0}^{9}abba),$$
which can be further expanded upon such that:

\begin{equation}
    \begin{aligned}
\sum_{a=1}^{9}(\sum_{b=0}^{9}abba) &= \sum_{a=1}^{9}\sum_{b=0}^{9}(1001a + 110b) \\
                             &= \sum_{a=1}^{9}[10(1001a) + 110\sum_{b=0}^{9}b].
    \end{aligned}
\end{equation}
The last line of the equation is what's confusing me. Where does the $10$ come from in $10(1001a)$?

I don't use double summation notation often, so I assume it has to do with moving the summation of $b$ inside the equation to make the triangular number formula more apparent, but WHY do we need the $10$?

Thank you for any help.

The 10 comes from the sum of all the palindromes in one digit. Since there are 10 palindromes for every thousand, that is where they come from.

For Example,

The Palindromes for 1000 is 1001 to 1991, so if we were to find the sum, we have to add all the ones, or multiply it by 10


Is there a numeral system that makes both addition and multiplication easy? Decimal positional notation, the system for writing numbers we all use every single day, makes addition very easy by transforming it from a computation to a repeated operation on individual digits (long hand addition). A more formal description would be that if two numbers $a$ and $b$ are presented in decimal notation, long hand addition is a constant-memory, logarithmic time algorithm to obtain the decimal notation of the number $a+b$.

Multiplication (that is, long multiplication) is not so easy in decimal notation, requiring doubly logarithmic memory and logarithmic time while relying heavily on long hand addition as a "subroutine".

It is of course straightforward to invent a numeral system where the roles of the two operations are reversed: Just use the decimal notation of $\log(a)$ to denote the number $a$. Multiplication then becomes as simple as long hand addition in standard notation, because $\log(ab)=\log(a)+\log(b)$ (this is the principle behind the slide rule). Unfortunately, the price one has to pay for this convenience is that addition is suddenly very difficult.

Is there a numeral system that provides a balance between these two extremes, that is, a system for writing numbers in such a way that both addition and multiplication can be done by hand with less effort than decimal long multiplication, but perhaps greater effort than decimal long addition? (Note the focus on still being a system that is of practical use for manual computations; I doubt anyone would be able to execute the Schönhage–Strassen algorithm by hand, even it it has better asymptotic complexity than long multiplication.)

Maybe using primoradic(see :https://oeis.org/wiki/Primorial) . Very difficult to answer here.

Addition : We use the remainders and quotients of Euclidean divisions. The quotients become R.
$Example:4096+1024+512+256+2+1=5891$
$$primes... :11-7-5-3-2-1$$
$$R:1-2-1-2$$
$$4096=(1-8-3-2-2-0)$$
$$1024=(0-4-6-0-2-0)$$
$$0512=(0-2-3-0-1-0)$$
$$0256=(0-1-1-2-2-0)$$
$$0002=(0-0-0-0-1-0)$$
$$0001=(0-0-0-0-0-1)$$

$$5891=(2-6-0-1-2-1)$$

Multiplication :$137\times43=5891$
Here, first I will convert to base 2:
$$137\times43=(10.001.001)_2\times(101.011)_2=(1.011.100.000.011)_2=4096+1024+512+256+2+1=(1:8:3:2:2:0)+(4:6:0:2:0)+(2:3:0:1:0)+(1:1:2:2:0)+(1:0)+1=(2:6:0:1:2:1)$$according to the example given for the addition.
A better example to illustrate multiplication would be calculating the powers of two in primoradic, first by hand, then using a spreadsheet up to $2 ^ {16384}$ for example, or the powers of ten, or the powers of any integer, for example three($1,3,9,...$). See : I'm trying to generalize some simple results about $ 2 ^ n $. It's useful to write them in primoradic (see stub OEIS).. Let us take three. With a pen and a sheet of paper, old fashioned way, using Charles-Ange Laisant's notation in his article of 1888 : 1=(1); $3=3\times1=3\times(1)=(11)$; $9=3\times3=3\times(11)=(111)...$

Remark : 4 = (2:0) = ((1:0):0)...

If you haven't heard of the duodecimal system (or its variations) you might give it a look over. Some (obvious) examples which operate wrt the duodecimal system...

Standard timekeeping.

60 s in a min.
60 min in an h.
24 h in a day.

The imperial system.

A thou = $\frac{1}{12000}$'.
An inch = $\frac{1}{12}$'.
A yard = $3$'.
etc.

1 of the reasons why the duodecimal system is desirable, & why people think that it simplifies hand calculations, has to do w/ $12$ having so many factors $\left(1\;2\;3\;4\;6\;12\right)$. It really shines wrt to division/ratios/proportions, e.g., for a carpenter to perform calculations based on the partitioning of a log into 3 equal pieces, he must consider their lengths to be...

$0.\overline{3}$ $\left[\mathrm{m}\right]$, if the log length was $1$ $\left[\mathrm{m}\right]$.
$4$ $\left[\mathrm{ft}\right]$, if the log length was $12$ $\left[\mathrm{ft}\right]$.


Can $X$ have compact connected components? Let $X$ be a Hausdorff, locally compact but non-compact topological space.
If the (Alexandroff) one-point compactification is connected, can $X$ have compact connected components?

I think I proved the following

  Lemma

  Let $X$ be a Hausdorff space and $C \subset X$ have a compact neighbourhood $K$. Then $C$ is a component of $X$ if and only if $C$ is a component of $K$.

in this answer.

For the present problem this implies a negative answer, since a compact set
in a locally compact Hausdorff space has a compact neighbourhood. Let $C$ be
a compact component of $X$, and let $K$ be some compact neighbourhood of $C$ in
$X$. Applying the lemma one way then shows that $C$ is a component of $K$, and
then applying it the other way shows that $C$ is a component of the
compactification, since $K$ remains a compact neighbourhood of $C$.

Well, no. Assume $X=Y\sqcup C$, where $C$ is compact. Then the 1PC is $X\cup\{\infty\}$ with the topology: $U$ is open iff $U\subset X$ is open in the previous topology or $\infty\in U$ and $U^c$ is compact. Then $C = (U\cup\{\infty\})^c$ is compact, and thus $U\cup\{\infty\}$ is open. $C$ is also open (it was open in $X$). Thus the 1PC of $X$ is not connected.


Why do we take the axiom of induction for natural numbers (Peano arithmetic)? More precisely, when we define the set of natural numbers $\mathbb{N}$ using the Peano axioms, we assume the following:

$0\in\mathbb{N}$
$\forall n\in\mathbb{N} (S(n)\in\mathbb{N})$
$\forall n\in\mathbb{N}(0\neq S(n))$
$\forall m,n (m\neq n\to S(m)\neq S(n))$
If $P(n)$ denotes the fact that $n$ has property $P$, then $\Big(P(0)\wedge \forall n\in\mathbb{N}\big(P(n)\to P(S(n))\big)\Big)\implies \forall n\in \mathbb{N} (P(n))$

I understand that using these axioms we can derive everything about the natural numbers, but I also think it's helpful to know why the axioms were chosen the way they are. So my question is why we choose to accept the axiom of induction ((5.) above), which in a way makes this more of a metamathematical question.

For example in Tao's Analysis I, it says that the axiom of induction keeps unwanted elements (such as half-integers) from entering the set.

Wikipedia says, "Axioms [1], [2], [3] and [4] imply that the set of natural numbers is infinite, because it contains at least the infinite subset $\{ 0, S(0), S(S(0)), \ldots \}$, each element of which differs from the rest. To show that every natural number is included in this set requires an additional axiom, which is sometimes called the axiom of induction. This axiom provides a method for reasoning about the set of all natural numbers."---But I find this tautological: $\mathbb{N}$ is defined as the set of natural numbers so "$n$ is a natural number" means "$n\in\mathbb{N}$", right? So isn't every natural number included in  $\mathbb{N}$ by definition?

Suppose we want to show $\mathbb{N}=\{0,1,2,3,\ldots\}$ using all five of the Peano axioms.

If we let $P(n)$ denote $n\in\{0,1,2,3,\ldots\}$, then $P(0)$ is true. Suppose $n$ is in $\{0,1,2,3,\ldots\}$. Then (informally) the dots indicate that $S(n)$ is in $\{0,1,2,3,\ldots\}$. So $\mathbb{N}\subseteq\{0,1,2,3,\ldots\}$, i.e., our defined set contains no "extra" elements (as in Tao's Analysis I).

Yet I still do not see how to show $\{0,1,2,3,\ldots\}\subseteq\mathbb{N}$ (in order to complete the "proof" that $\mathbb{N}=\{0,1,2,3,\ldots\}$) without just assuming it. (I think this is what the Wikipedia article was doing(?))

Thanks in advance for any help and I apologize if this kind of question is unsuitable for this site.

You are trying to deduce something about the importance of 5th axiom only from the axioms itself, which is impossible.
The axioms by itself do not carry any significance. What is important is that the set defined by the axioms is (a) unique and (b) isomorphic to some real-world object (real-world natural numbers, as in "two apples" and "three oranges").

Giving away the fifth axiom means that:

a) Peano axioms no longer define a set of natural numbers, as there could be two non-isomorphic (and even not of the same cardinality) sets, both compatible with Peano axioms; rather they define a class of sets, each containing a subset, isomorphic to $\{0, 1, 2, 3, \ldots\}$. For example, let us consider the set $\mathbb N$ in its usual sense, and the set $\mathbb C \setminus {\mathbb N}^+$. Both sets fulfill the Peano axioms (except for the fifth one), but they are quite different in itself.

b) Of course, the second set from the previous paragraph has nothing in common with a real=world object called "the natural numbers".

Way of proving some theories by indcution is very importand technique without it mathematics would not be the same so it was added to formalize our intuition. From what I know suprise was that this axiom was not consequence of rest of the axioms but is needed to be stated explicite.


How to use xor properly I need to know how to use XOR properly on more than two variables.
I have following example.
a xor b xor c
Now, the way I understand it is that:
$a$ xor $b$ = $a$ * not $b$ + not $a$ * $b$
That part is fairly simple for me to understand but how does the next part work? I imagine it goes like this.
($a$ * not $b$ + not $a$ * $b$) xor $c$
So, how can I use XOR on more than one variable?

I will use $\oplus$ for exclusive OR and $\bar x$ for NOT $x$. Just expand the expression:

$$\begin{align*}
(x\oplus y)\oplus z&=(x\bar y+\bar xy)\oplus z\\
&=(x\bar y+\bar xy)\bar z+\overline{(x\bar y+\bar xy)}z\\
&=x\bar y\bar z+\bar xy\bar z+\overline{(x\bar y)}\;\overline{(\bar xy)}z\\
&=x\bar y\bar z+\bar xy\bar z+(\bar x+y)(x+\bar y)z\\
&=x\bar y\bar z+\bar xy\bar z+(\bar xx+\bar x\bar y+xy+y\bar y)z\\
&=x\bar y\bar z+\bar xy\bar z+(\bar x\bar y+xy)z\\
&=x\bar y\bar z+\bar xy\bar z+\bar x\bar yz+xyz
\end{align*}$$

If you examine that final result, you’ll see that it has every product (conjunction) with an odd number of non-negated variables. This is true in general: $x_1\oplus x_2\oplus\ldots\oplus x_n$ is the sum (disjunction) of all expressions of the form $y_1y_2\ldots y_n$, where each $y_k$ is either $x_k$ or $\overline{x_k}$, and the number of $k\in\{1,\ldots,n\}$ such that $y_k=x_k$ is odd. Thus, for instance,

$$\begin{align*}
w\oplus x\oplus y\oplus z&=wxy\bar z+wx\bar yz+w\bar xyz+\bar wxyz\\
&\quad+w\bar x\bar y\bar z+\bar wx\bar y\bar z+\bar w\bar xy\bar z+\bar w\bar x\bar yz\;.
\end{align*}$$

This more general result can be proved by induction on the number of terms.

Ensure that niether (b+a or b+c or a+c) are larger than your max datatype allows for. In other words in 16bit you can have XOR 255 and XOR (2^16 -256). This would be bit masking. Note the ^ is sometimes used for shifting instead of "to the power of".

Use XOR when trying to find the 2's Complement "the opposite of which bits have been set."

An example is 7 XOR 255 and may get you a byte of data for the number (248) 1111 1000 which is (15*16) + (8*1). However this is false if you dealing with 16 bit numbers 7 XOR 255 will be 0000 0000 1111 1000. Unless the byte order is swapped for 16 bit machines the same number might be stored as 1111 1000 0000 0000 which is very confusing. Like in 16 bit PCM data sounds. If you start multiplying these you might get unpredictable results especially if you start going beyond your MAX Integer value and mess around with "Casting" of datatypes. NEGATIVE 1 could be stored like 11111110

Search for "bit masking" and "two's complement" and karnaugh map.

What do you mean by not b? That it doesnt exist or that If B is true !B is false but do you mean the 2's complement? In order to find the twos complement you need all the bits to be set or 1s. You can do that with OR X where X has 1's for every bit. to set bits and AND0 to clear them.

Usually you need to know the number of bits used in your data type. And how negative numbers are stored in computers specifically your computer or raspberry pi or arduino or arm or motorolla processor or .... foobar named machine.

This technique is more common in low level programming "Assembly Languages".

Note: In decmial numbers we can multiply by 10 to append a zero in binary thats like multiplying by 2 and called a shift. If your left hand is in binary consider it times 16. So if you multiply a by b you could be shifting and so I disagree with the other answers on this page.

Back to 7 XOR 255 if you multiply both by 2 you would get problems 1110 XOR 1 1111 1110 this might set an overflow bit. I dont think XOR is reliable look on here for XOR Encryption and see problems especially in 64 bits and xxd command versus hexdump.


Computational Algorithm To Find Basis of Linear Equations Solution Set Suppose I have a set of linear equations and I am trying to find out the basis of solution space. For example Check This One. My question is that if there is an algorithm or process which is computationally implementable in a computer program.

You may want
to have a look at Gaussian algorithm or similar pages in the Internet.

It seems that the following is a much better source:
Maxima, search for Function: linsolve ([expr_1, …, expr_m], [x_1, …, x_n])


Stitching two analytic functions? Let $f$ be an analytic function on the open unit disc and let $g$ be an analytic function on the complement of its closure.
Further assume that the two functions have a the same continuous limit on the common boundary of their domain.

Is it possible to build an entire function that settles with $f$ and $g$.

The question is a it vague I'm afraid...

By the assumption

  Further assume that the two functions have a the same continuous limit on the common boundary of their domain

the function

$$h(z) = \begin{cases} f(z) &, \lvert z\rvert \leqslant 1 \\ g(z) &, \lvert z\rvert \geqslant 1\end{cases}$$

is well-defined, and continuous on all of $\mathbb{C}$. Furthermore, it is holomorphic on $\mathbb{C}\setminus \partial \mathbb{D}$.

For a point $z_0 \in \partial\mathbb{D}$, consider the disk $U := D_{1/2}(z_0)$ and use Morera's theorem to find that $h$ is holomorphic on $U$: Let $\Delta \subset U$ be a closed triangle, and $A = \Delta \cap \overline{\mathbb{D}}$, $B = \Delta \setminus \mathbb{D}$. Then

$$\int_{\partial\Delta} h(z)\,dz = \int_{\partial A} h(z)\,dz + \int_{\partial B} h(z)\,dz$$

since the integrals over $\Delta\cap \partial\mathbb{D}$ cancel due to the opposite orientation this arc has in $\partial A$ and $\partial B$ [or $\Delta \cap \partial\mathbb{D}$ consists of one or two isolated points or is empty, in which cases that part of the integrals is $0$ trivially]. For $\varepsilon > 0$ let $A_\varepsilon = \Delta \cap \{ z : \lvert z\rvert \leqslant 1-\varepsilon\}$ and $B_\varepsilon = \Delta \cap \{ z : \lvert z\rvert \geqslant 1+\varepsilon\}$. By Cauchy's integral theorem, we have

$$\int_{\partial A_\varepsilon} h(z)\,dz = 0 = \int_{\partial B_\varepsilon} h(z)\,dz.$$

By the uniform continuity of $h$ in $U$, we have

$$\int_{\partial A} h(z)\,dz = \lim_{\varepsilon \searrow 0} \int_{\partial A_\varepsilon} h(z)\,dz = 0$$

and

$$\int_{\partial B} h(z)\,dz = \lim_{\varepsilon \searrow 0} \int_{\partial B_\varepsilon} h(z)\,dz = 0,$$

whence Morera's theorem tells us that $h$ is holomorphic on $U$.

Hence $h$ is entire.

Since both functions are continuous along the boundary (in particular, having no poles), they can be analytically continued. For example, let $p$ be a point on the unit circle, and let $U$ be a neighborhood of $p$ small enough so that both $f(z), g(z)$ can be analytically continued.

Then $f(z) - g(z) = 0$ along the arc of the unit circle contained in $U$, so $f(z) - g(z) = 0$ everywhere on $U$. In particular, the analytic continuation of $g(z)$ into $U$ just agrees with $f(z)$. This holds for arbitrary $p$.

So, yes -- the "join" of the two functions will be entire.

EDIT: This answer is not correct (see comments).


Topological Vector Space: $\dim V=n\implies V\cong\mathbb{K}^n$ Let $V$ be a finite dimensional Hausdorff topological vector space.

Prove that it is is isomorphic to the Euclidean vector space of the same dimension:
$$\dim V=n\implies V\cong\mathbb{K}^n$$

I think this works: Assume without loss of generality that $V = \mathbb{F}^n$. Let $\tau$ be the Euclidean topology and $\tau_1$ be any other (Hausdorff) linear topology on $V$.

a) Suppose $x_{\alpha} \to 0$ in $(V,\tau)$, then each component $x_{\alpha}^i \to 0$ in $\mathbb{F}$. Since addition and scalar multiplication are continuous in $(V,\tau_1)$, it follows that
$$
x_{\alpha} = \sum_{i=1}^n x_{\alpha}^i e_i \to \sum_{i=1}^n 0e_i = 0
$$
in $(V,\tau_1)$. Hence, the identity map
$$
I : (V,\tau) \to (V,\tau_1)
$$
is continuous. So,
$$
\tau_1\subset \tau
$$

b) Now consider $B =\{x \in V : \|x\|_2 < 1\}$ and $S = \{x\in V : \|x\|_2 = 1\}$. $S$ is compact in $(V,\tau)$ and since $\tau_1 \subset \tau$, $S$ is compact in $(V,\tau_1)$ as well. Since $(V,\tau_1)$ is Hausdorff, $S$ is closed in $(V,\tau_1)$. Since $0\notin S$, there exists a circled $\tau_1$-neighbourhood $V$ of $0$ such that
$$
V\cap S = \emptyset
$$
I claim that $V\subset B$: If not, then $\exists x \in V\setminus B$. So $\|x\|_2 \geq 1$, whence
$$
\frac{x}{\|x\|_2} \in V\cap S
$$
This is a contradiction, so $V\subset B$. Hence, $B$ is a $\tau_1$-neighbourhood of $0$. Since scalar multiples of $B$ form a $\tau$-base at $0$, it follows that
$$
\tau \subset \tau_1
$$

Proof:

Let $V$ be a finite dimensional Hausdorff topological vector space and $E$ an Euclidean vector space of the same dimension.

Since both dimensions agree there is a linear bijection $L:E\to V$.

Now since the Euclidean space carries the product topology the projections $\pi_k:E\to\mathbb{K}$ are continuous. Similar vector addition and scalar multiplication in $V$ are continuous. But $L(x)=\sum_{k=1}^n\pi_k(x)L(e_k)$. Thus $L$ is continuous as composition of continuous functions.

Next, consider the unit sphere $S$. It is compact in $E$. Thus by continuity $LS$ is compact in $V$. But $V$ is Hausdorff so $LS$ is closed. So its complement is an (open) neighborhood of zero and so is also the balanced core. Lets call it $U$. The crucial observation will be $U\subseteq LB$.

Lemma: $U\subseteq LB$
By construction $U$ is disjoint from $LS$. Since $L$ is injective $L^{-1}U$ is disjoint from $S$. But $L^{-1}U$ is also balanced by linearity of $L$. Thus $L^{-1}U\subseteq B$ and so $U\subseteq LB$ since $L$ was also surjective.

Finally if $N\in\mathcal{N}_0$ then there exists $\delta>0$ such that $\delta B\subseteq N$ and therefore $\delta U\subseteq \delta LB\subseteq N$. But $U$ was a neighborhood of zero so also any of its dilates $\delta U$ and therefore $LN$ is a neighborhood of zero too. That proves that the inverse of $L$ is continuous at zero. But it is linear so it was continuous everywhere.

Concluding that $L$ is an isomorphism. c.q.f.d.


Empty set in a simplicial complex Should the empty set be considered a simplex in a simplicial complex?
Which justifications exist for the answer?

I guess it is somewhat comparable to $1$ not being a prime number.

This is a useful construction in homotopy theory and is probably easiest explained as the augmented simplex category.

I think your answer is yes.  If the intersection of two simplices in a simplicial complex is a face (sub-simplex) of each, then an empty set must be a simplex.


Stabiliser of a curve under Affine Transformations? Let $\gamma$ be a curve in the plane, and let $\text{Im}(\gamma) \subset \mathbb{R}^{2}$ be its image in the plane. Is it possible to completely specify the affine transformations of $\mathbb{R}^{2}$ which send $\text{Im}(\gamma)$ to itself?

For example, if $\gamma(t)=(t,t^{2})$, then the linear maps $(x,y) \to (\lambda x,\lambda^{2} y)$ are (I think) the whole collection of such maps. If $\gamma$ is a circle, then the orthogonal group does the trick. I would expect that "usually" the group of stabilising actions would be trivial, but I'm not sure how to tell.

These linear maps $(x,y)→(λx,λ^2y)$ is called parabolic rescaling and is crucial in proving decoupling theorem and Vindogronov main theorem.

Now assume we have a curve defined by $f(x,y)=0$. the CORRECT ANSWER OF THIS PROBLEM is those curve that do not have "error term". I will explain what I said in the following.

1.If $f(x,y)=0$ is a plane $V$. then we know the affine map send $image(\gamma)$ to itself is just a affine map make $V$ to be a invariance subspace.

2.if $f(x,y)=0$ is a quadratic curve. Then it is easy to see $f(x,y)=0$ could not to be a elliptic. now change a basis we can assume the new basis is the eigenvalue of affine map $A$. So $Ax=\lambda_1x, Ay=\lambda_2 y$.Now I claim $f(x,y)$ should be have only two terms, every unknown quantity a term. for example $x^2-y^2,x^2-y,x^2-y$. These is just derive by $\frac{f(\lambda_1^kx,\lambda_2^ky)}{\lambda_1^{w_1k}\lambda_2^{w_2k}}= f(x,y),\forall k\in N^*$.

3.For the cubic and higher polynomial the result is similar.


Nontrivial closed ideal of $\mathbb{B(H)}$, $\mathbb{H}$ is a non-separable Hilbert space. $\mathbb{H}$ is a non-separable Hilbert space. Give an example of nontrivial closed ideal $I$of  $\mathbb{B(H)}$, that is different from $\mathbb{B_0(H)}$ which is the ideal of compact operators.

Any disscussion would me welcomed!    Any hints would be appreciated!

You can see the ideal of compact operators as the ideal generated by the finite-rank operators. You can play the same game and consider the ideal generated by those operators which have separable range (or any other cardinality less than the dimension of $\mathbb H$).

Take any $0\neq x\in \mathbb{H} $ and cosider ideal $\mathcal{M} =\{ T\in \mathcal{B} (\mathbb{H} ) : T(x) =0\}.$


Dates with 8 consecutive digits In many places, dates are written as DD/MM/YYYY. For example, the 25th of April 1736 is written as 25/04/1736. Dates such as this one that use 8 consecutive digits (not necessarily in order) will be called illions.

What is the first illion after 2015?
Why must there be a 0 in every illion in the years 2000 to 2999?
Why must every illion in the years 2000 to 2999 have 0 as the first digit of the month?
How many illions are in the years 2000 to 2999?

(You cannot use the same digit twice in an illion)

I have tried to list all of the possibilities however, I have found this highly time consuming. I was wondering what method I should use to solve these questions without listing all the possibilities.

Thank you :)

(1) As argued in (2) below, the month contains a $0$ or is $12$, hence either the year is $>2999$ or does not contain a $0$. This makes $2134$ the earliest possible year. However, the leading digit of the day is $0$, $1$, $2$, or $3$, so that we must up the year to $2145$ at least, which allows day $30$ but then conflicts with the $0$ in the month. The next option is that the year does not use $1$ (nor $0$), which happens for the first time in $2345$. We can find an illion in that  year: We know we use $0$ in the month (preferably as leading digit). Then the leading digit of the day must be $1$. Now the lowest valid month is $06$ and we can take $17/06/2345$

(2) If the two-digit month does not contain a zero and has two distinct digits, it must be December. But then the use of $2$ conflicts with the leading $2$ of the year. Actually, we see that the same argument applies to the year range $1000$ to $2999$.

As in (2), we know that the month cannot be $12$ or $11$. In cannot be $10$ either, because that makes $0,1,2$ used up by month and year, hence the day must have leading digit $3$, but neither $30$ nor $31$ are allowed

For (4), you may want to first enumerate valid day/month combinations (without $2$!) and look for years that fit the gaps.

0 to 7 are the lowest consecutive numbers you can use.
Months only consist of 0,1,2.
Use the 2,3,4,5 for the year. That is the next closest. Now your trying to find the shortest month so the 0 is used. Now you cant use the 1 here otherwise you wont be able to make a date up for the day. Eg 67/01/2345 is impossible. Therefore you use the lowest number as June is before July and the 7 goes to the day. 17/06/2345


Lucas and Fibonacci Numbers Problem:

  Let
  \begin{align*}
A_0 &= 6 \\
A_1 &= 5 \\
A_n &= A_{n - 1} + A_{n - 2} \; \textrm{for} \; n \geq 2.
\end{align*}
  There is a unique ordered pair $(c,d)$ such that $c\phi^n + d\widehat{\phi}^n$ is the closed form for sequence $A_n$.

  Find $c$ using the Fibonacci and Lucas number sequences.

My Solution:

  $A_n$ is just $3L_n$+$2F_n,$ where $L_n$ is the $n^{th}$ Lucas number and $F_n$ is the $n^{th}$ Fibonacci number. Since the closed form of the Fibonacci sequence is $$F_n = \frac{1}{\sqrt{5}} \left( \phi^n - \widehat{\phi}^n \right),$$ and the closed form of the Lucas sequence is $$L_n = \phi^n + \widehat{\phi}^n,$$
  we get the closed form of $A_n$ as $\dfrac{2}{\sqrt{5}}+3.$

What is wrong with my solution?

EDIT: My answer is actually right.

Any sequence of the Fibonacci-type, i.e., $f_n=f_{n-1}+f_{n-2}$, can be expressed as a sum of a Fibonacci-term and a Lucas-term, as follows

$$f_n=\left(f_1-\frac{f_0}{2}\right) \frac{\phi^n-\psi^n}{\phi-\psi}+\frac{f_0}{2} (\phi^n+\psi^n) $$

where $\psi=-1/\phi$. For a more detailed description and generalization, see my post at Decimal Fibonacci Number?

Although the problem is old there are two additional methods that can be considered.

First Method

Given
\begin{align*}
A_0 &= 6 \\
A_1 &= 5 \\
A_n &= A_{n - 1} + A_{n - 2} \; \textrm{for} \; n \geq 2.
\end{align*}
then, since Fibonacci and Lucas numbers satisfy the difference equation, one can take
$$A_{n} = b \, F_{n} + c \, L_{n}.$$
First note that $F_{n} = \{ 0, 1, 1, ...\}_{n\geq 0}$ and $L_{n} = \{2, 1, 3, ...\}_{n \geq 0}$. For $A_{0} = 6$ then $6 = 2 c$ or $c = 3$ and leads to $A_{n} = b F_{n} + 3 L_{n}$. For $A_{1} = 5 = b + 3$ then $b = 2$ and leads to
$$A_{n} = 2 \, F_{n} + 3 \, L_{n}.$$
In terms of Fibonacci numbers only the solution is $A_{n} = 5 \, F_{n} + 6 \, F_{n-1}$.

Second Method

The generating function for $A_{n+2} = A_{n+1} + A_{n}$ takes the form
$$\sum_{n=0}^{\infty} A_{n} \, t^{n} = \frac{A_{0} + (A_{1} - A_{0}) \, t}{1 - t - t^2}$$
and comparing to the generating function of the Fibonacci numbers,
$$\sum_{n=0}^{\infty} F_{n} \, t^{n} = \frac{t}{1 - t - t^2},$$
and yields
$$\sum_{n=0}^{\infty} A_{n} \, t^{n} = \sum_{n=0}^{\infty} (A_{0} \, F_{n+1} + (A_{1} - A_{0}) \, F_{n}) \, t^n$$
and yields the general form
$$A_{n} = A_{0} \, F_{n+1} + (A_{1} - A_{0}) \, F_{n} = A_{1} \, F_{n} + A_{0} \, F_{n-1}.$$
This result agrees with the first.


We can always find a projective line $L\subset\mathbb{CP^2}$ intersecting degree $d$ projective curve at $d$ distinct points Let $X$ be a smooth projective plane curve$\{ [Z_0, Z_1, Z_2]\in \mathbb{CP^2}\ \vert \ p(Z_0, Z_1, Z_2) = 0 \}$ defined by non-singular homogeneous polynomial $p$ of degree $d$.

Then there exists a projective line $L\subset\mathbb{CP^2}$ intersecting $X$ at $d$ distinct points.

Is there a way to prove the above statement using only elementary knowledge of Riemann surfaces? I do not know much of algebraic geometry.

This is straightforward and you do not require any theory of Riemann surfaces. Let
$$
C := Z(p) = \{ x=[x_0:x_1:x_2]\in\Bbb C\Bbb P^2 ~\mid~ p(x) = 0 \}.
$$
Observe that a line $L\subseteq\Bbb C\Bbb P^2$ can be described as
$$
L = \{ [x_0+ty_0:x_1+ty_1:x_2+ty_2] ~\mid~ t\in\Bbb C\}
$$
for certain points $x=[x_0:x_1:x_2]$ and $y=[y_0:y_1:y_2]$. The choice of the projective coordinates here does not matter since any potential scaling can be moved in to the parameter $t$. Let
$$
p_L(T) := p( x_0+Ty_0, x_1+Ty_1, x_2+Ty_2) \in \Bbb C[T],
$$
it is a polynomial of degree $d$ in one variable.

Now the points of $L\cap C$ are given by the $t\in\Bbb C$ such that $p_L(t)=0$, i.e. they are the zeros of $p_L$. You have to argue that $x$ and $y$ can be chosen such that $p_L$ has $d$ distinct roots.

The polynomial $p_L$ has a multiple root if and only if the discriminant of $p_L$ vanishes. The discriminant of $p_L$ is a polynomial in the coefficients of $p_L$ and the coefficients of $p_L$ are polynomials in the entries of $x$ and $y$ - therefore, the discriminant is a polynomial
$$\Delta\in \Bbb C[X_0,X_1,X_2,Y_0,Y_1,Y_2].$$
Since there is at least one point $(x_0,x_1,x_2,y_0,y_1,y_2)$ where $\Delta$ is nonzero, you can pick these coordinates to define the line $L$ and it follows that this particular $L$ will intersect $C$ in $d$ distinct points.

This construction also shows that in fact, most of the lines you can choose will intersect $C$ in $d$ distinct points, because $\Delta$ vanishes only on a set of zero measure inside $\Bbb C^6$.

Here's an algebraic proof. Choose any point $P\in X$ that is simple (there are infinitely many such points), you can prove that all but a finite number of lines passing through $P$ intersect $X$ in $d$ points.

Here's a couple of hints:

You can suppose $P=[0:1:0]$ (applying a change of coordinates on $\mathbb{P}^2$ doesn't change the number of intersection points).

Consider the lines passing through $P$ of type $L_\lambda=\{[\lambda:t:1]\: |\: t\in \mathbb{C}\}\cup \{P\}$ for $\lambda\in \mathbb{C}$.
Write $X$ as a polynomial in $\mathbb{C}[Z_0,Z_2][Z_1]$.
Prove that $X(\lambda,t,1)$ for some $\lambda$ doesn't have double roots in $t$.
Deduce that $X$ intersects $L_\lambda$ in $n$ different points.

This is a more general result, and more algebraic, but I think it's an easy method to prove the property.

Note that this is not true in characteristics $p$, for example all lines passing through $P$ intersect $Z_0^{p+1}-Z_1^pZ_2$ in strictly less than $p-1$ points. Edit: Still, I believe that even in characteristic $p$ there is always a line intersecting a curve of degree $d$ in $d$ different points (over the algebraic closure)


Bivariate continuous density function: How is it possible to define conditional density? Maybe this is a very trivial question for an experienced mathematician, but it is something mysterious for someone from an engineering background, like me; this is bugging me for a long time:

Let's define a bivariate, continuous distribution $p_{x,y}(x,y)$ which is $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}p_{x,y}(x,y)dxdy = 1$. We can define a conditional density which for example is $p_{x|y}(x|y=Y)$ in this setting. The definition of conditional probability is: $p_{x|y}(x|y=Y) = \frac{p_{x,y}(x,y=Y)}{p_{y}(y=Y)}$. Now, something which confuses me is that: $p_y$ is a continuous density function,too and by definition it must be $p_{y}(y=Y) = 0$ since $\int_{Y}^{Y}p_{y}(y)dy = 0$. So this must make the conditional density function $p_{x|y}$ undefined. But this is clearly not the case. How is this contradiction resolved?

Thanks is advance.

We want to define $p_{X|Y=b}$ such that
$$ X|Y=b \sim p_{X|Y=b} $$
in the same way that is understood
$$ Z \sim p_Z $$
when $Z$ is random variable.

Let $ x\in \mathbb{R}$, you can think that the probability of $X|Y=b$ belongs to a small interval centered on $x$ is directly proportional to $p(x,b)$. This suggests defining
$$ p_{X|Y=b}(x) = \frac{p_{X,Y}(x,b)}{p_Y(b)} $$
when $p_Y(b) \neq 0$.

In fact, this is enough to give a good definition of $ p_{X|Y=b}$ (since that the set of $b$'s where $p_Y(b)=0$ has Lebesgue measure null). In addition, it is easy to verify that
$$ \mathbb{P}(a_1<X\le a_2|Y=b) = \int_{a_1}^{a_2} p_{X|Y=b}dx $$
and
$$ \mathbb{E}[X|Y=b] = \int_{-\infty}^\infty xp_{X|Y=b}(x)dx $$
how we wanted.

$$p_Y(y_0) \neq \int_{y_0}^{y_0} p_Y(y)\text{d}y = P(y_0 \leq Y \leq y_0)= P(Y \leq y_0)-P(Y \leq y_0)= 0 \,\,,$$
for $y_0$ in the domain of $Y$ (necessarily).
In other words, it might be the case that $p_Y(y_0)= P(Y=y_0)=0$, but this is not necessarily true. As a matter of fact, it must be the case that $$p_{X,Y}(x,y_0) = P(X=x,Y=y_0) \leq p_Y(y_0) =P(Y=y_0) \,\,\,\text{for any $x \in D_X$}\,,$$
where $D_X$ is the domain of $X$.


Hypergeometric function with negative $b$ and $a>c>0$ Recall the definition of the hypergeometric function
$$_2F_1(a,b,c;z)=\sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{n!(c )_n}x^n$$
where $(a)_n$ is defined to be $a(a+1)\cdots(a+n-1)$.

We suppose that none of the $a,b,c$ or the difference between any two of them is an integer.

Here, I am interested in the case where $a>c>0,b<0$.

What's the behaviour of this function on $[0,1]$?
I used Mathmatica to plot it and it suggests that it always goes negative somewhere on $[0,1]$. As $_2F_1(a,b,c;0)=1$, this suggests that the function has at least a zero on $[0,1]$.

Is there anyway to prove it?

$_2F_1(a,b;c;z)$

$=(1-z)^{-a}{}_2F_1\left(a,c-b;c;\dfrac{z}{z-1}\right)$ (according to http://en.wikipedia.org/wiki/Hypergeometric_function#Transformation_formulas)

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(a)_n(c-b)_nz^n}{(c)_nn!(1-z)^{n+a}}$


5 friends are sitting together, what is the probability that 2 are NOT sitting together five friends including Bilyana and Bojana are sitting in a row in a theatre determine the probability that they are not sitting together.
This is part of a homework assignment. I don't even know where to start with this.

We will put on a couple of the seats a sign saying "Only to be used by Bilyana or by Bojana. Anyone else, Bulgarian or not, keep off."

There are $\binom{5}{2}$ equally likely ways to select the two seats. Exactly $4$ of these choices leave us with the two B's sitting together. Thus our probability is $1-\frac{4}{\binom{5}{2}}$.

First you need to figure out the sample space.
Since there are five friends and 5 seats in a row, you need to line them up.
5 x 4 x 3 x 2 x 1 = 120
Next you need to find how many ways they can sit together.
2 x 1 x 3 x 2 x 1 = 12
Since there are 4 other cases of them sitting together multiply them.
12 x 4 cases= 48
The probability of them sitting together is 48/120
Since this is the probability of them sitting together, subtract 120-48= 72
This is assuming that the other cases are ones where they aren’t sitting together.
So it’s 72/120 or 60%


Prove that function is continuous. Possible Duplicate:
  Prove that function is homeomorphism.

Let
$$
f: \prod\limits_{1}^{\infty} ( \{0,2 \}, \mathcal{T} _{\delta}) \to ([0,1], \mathcal{T}_{e}):\{n_i \} \mapsto \sum_{i=1}^{\infty} \frac{n_i}{3^i}
$$
Prove that $f$ and $f^{-1}$ are continuous.

Let $X$ be the product that is the domain of $f$, and let $C=f[X]$.

You already know that $f$ is a bijection. To show that $f^{-1}$ is continuous, it suffices to show that $f$ is an open map, one that takes open sets to open sets.

For $n\in\Bbb Z^+$ let $\Phi_n$ be the set of functions from $\{1,\dots,n\}$ to $\{0,2\}$. For each $\varphi\in\Phi_n$ let $$B(\varphi)=\{x\in X:x_k=\varphi(k)\text{ for }k=0,\dots,n\}\;.$$ Let $\Phi=\bigcup_{n\in\Bbb Z^+}\Phi_n$, and let $\mathscr{B}=\{B(\varphi):\varphi\in\Phi\}$; then $\mathscr{B}$ is a base for the product topology on $X$.

The proof of the following lemma is very straightforward, and I’ll leave it to you.

  Lemma. If $f[B]$ is open in $C$ for all $B\in\mathscr{B}$, then $f$ is an open map, and $f^{-1}$ is continuous.

To show that each $f[B]$ is open, you need to figure out what $f[B]$ is. Suppose that $\varphi\in\Phi_n$. By definition

$$\begin{align*}
f[B(\varphi)]&=\left\{\sum_{k\ge 1}\frac{x_k}{3^k}:x\in B(\varphi)\right\}\\
&=\left\{\sum_{k=1}^n\frac{\varphi(k)}{3^k}+\sum_{k\ge n+1}\frac{x_k}{3^k}:x\in B(\varphi)\right\}\;;
\end{align*}$$

show that this set is equal to one of the $2^n$ ‘blocks’ of the set $C_n$ in this construction of the middle-thirds Cantor set, and explain why each of those blocks is an open subset of $C$.

To show that $f$ is continuous, show that those blocks are a base for the topology of $C$, and use what I’ve done above to show that their inverse images under $f$ are open inn $X$.

It is a basic problem in fractal ,actually, the cantor set is satified the strong open set condition that directly leads to your result.


Prove that $f^g$ is differentiable using limits Let $A \subseteq \mathbb{R}$ and $f:A \to (0, \infty), \: g:A \to \mathbb{R}$ be two differentiable functions. Prove that $f^g:A \to \mathbb{R}$ is also differentiable, using limits.

Let $x_0 \in A$ be arbitrary. In other words, we have to prove that there is a finite number $$l=\lim_{x\to x_0}\frac{f(x)^{g(x)}-f(x_0)^{g(x_0)}}{x-x_0}$$
I tried to get a common factor ouside the limit:
$$l=f(x_0)^{g(x_0)}\lim_{x \to x_0}\frac{\frac{f(x)^{g(x)}}{f(x_0)^{g(x_0)}}-1}{x-x_0}$$
but I'm not quite sure how to continue.

You are proceeding in right direction. The next step is to write $f^{g} /f_{0}^{g_{0}}$ in terms of exponential and logarithmic functions as $$\exp\log\frac{f(x) ^{g(x)}} {f(x_{0})^{g(x_{0})}} =\exp(g(x)\log f(x) - g(x_{0})\log f(x_{0}))$$ and note that the argument of $\exp$ function tends to $0$ as $x\to x_{0}$. Denoting this complicated argument by $t$ we can see that the derivative is given by $$\lim_{x\to x_{0}}f(x_{0})^{g(x_{0})}\cdot\frac{\exp(t)-1}{t}\cdot\frac{t}{x-x_{0}}$$ and this is same as $$f(x_{0})^{g(x_{0})}\lim_{x\to x_{0}}\frac{g(x)\log f(x) - g(x_{0})\log f(x_{0})}{x-x_{0}}$$ I hope you can take it from here.

Unfortunately, I'm also not sure how to go about this using limit. Let $F=f^g$.

\begin{align*}
\log F(x) &= g(x) \log f(x)\\
\frac{F'(x)}{F(x)} &= g'(x)\log f(x) + \frac{g(x)f'(x)}{f(x)}\\
F'(x) &= f(x)^{g(x)} \left\{g'(x)\log f(x) + \frac{g(x)f'(x)}{f(x)}\right\}
\end{align*}


Prove that $6$ divides $n^3+11n$? How can i show that
$$6\mid (n^3+11n)$$

My thoughts:
I show that
$$2\mid (n^3+11n)$$
$$3\mid (n^3+11n)$$

And
$$n^3+11n=n\cdot (n^2+11)$$
And if $n=x\cdot 3$ for all $x \in \mathbb{N}$ then:
$$3\mid (n^3+11n)$$
And if not:

The cross sum of$$n^2+11$$
is multiple of 3.

Can this be right or is there a simple trick?

$$n^3+11n=\underbrace{(n-1)n(n+1)}_{\text{Product of three consecutive integers}}+12n$$

See The product of n consecutive integers is divisible by n factorial

OR The product of n consecutive integers is divisible by n! (without using the properties of binomial coefficients)

Here is a proof by induction,

setting $n=1$, one should get
$$n^3+11n=1^3+11\cdot 1=12$$
obviously, the above number $12$ is divisible by $6$ hence statement is true for $n=1$.
Assume that the number $n^3+11n$ is divisible by $6$ for $n=k$ then
$$k^3+11k=6\lambda$$or $$k^3=6\lambda-11k\tag 1$$
Now, setting $n=k+1$, $$(k+1)^3+11(k+1)$$$$=k^3+3k^2+14k+12$$
$$=6\lambda-11k+3k^2+14k+12$$
$$=6\lambda+3(k^2+k+4)$$
since, $k^2+k+4$ is even for all integer $k$ hence setting $k^2+k+4=2m$,
$$=6\lambda+3(2m)$$
$$=6(\lambda+m)$$
since, $(\lambda +m)$ is an integer hence, the above number $6(\lambda+m)$ is divisible by $6$

Hence, $n^3+11n$ is divisible by $6$ for all integers $n\ge 1$


Coefficient of $x^2$ in the polynomial How to find the coefficient of $x^2$ in the polynomial
$$(1-x)(1+2x)(1-3x)....(1+14x)(1-15x)$$

Is there any particular way to look at such problems?

The terms $x^2$ appear as the product of $x$'s from two factors, while the $1$'s are used from the remaining factors.

Hence taking all pairs of factors,

$$C_2=\sum_{i=1}^{14}\sum_{j=i+1}^{15}(-1)^ii(-1)^jj.$$

You can ease this computation by taking all pairs twice, including the diagonal ones and

$$2C_2=\sum_{i=1}^{15}\sum_{j=1}^{15}(-1)^ii(-1)^jj-\sum_{i=1}^{15}((-1)^ii)^2\\
=\left(\sum_{i=1}^{15}(-1)^ii\right)^2-\sum_{i=1}^{15}i^2\\
=8^2-1240.$$

Since the constant term in each binomial factor is $1$, then we can express the coefficient of the $x^2$ term as the sum of the following series:

$-2+3-4+5-6+7...+15$

$-2-6+8-10+12-14...-30$

$...$

$15-30+45-60+75-90...-210$

The sum of all these sums is the answer, because for each factor, you're multiplying the $x$ term with the $x$ term from another factor and then just multiplying by the $1$ for the rest of terms to yield a $x^2$ term.


Can a negation have an inverse? I have started a middle/highschool level geometry text. The text is warming me up for proofs by going through basic logic, which was introduced alongside Euler diagrams.

I am given the following problem:

"Each of the lettered statements below is followed by some other statements. Identify the relation of each of them to the lettered statement if possible. Write "converse," "inverse," "contrapositive," or "original statement," as appropriate.

[Below:]

"Statement D: Lady kangaroos do not need handbags."

"35) If a kangaroo is not a lady, it needs a handbag."

My question:

I believe that conditional statement D is in the form 'if not a, then b'. So: $ \neg a \Rightarrow b$.

That interpretation has a ...negation or negative at $a$ ? If that is so, I think that problem 35) is presenting the inverse which I thinking as: $\neg (\neg a \Rightarrow b)$, which resolves to be $a \Rightarrow \neg b$.

The only notation I have been exposed to so far is the conditional notation  $\Rightarrow$, used so far with $a \Rightarrow b$. I have been shown "not" and assume "not" means "$\neg$". (I am sorry for any symbol confusion. I'm new to logic, stackexchange, and latex.)

Translation from English to symbols is not trivial. You could first deal with the anaphoric pronoun by using predicate symbols and a variable, and then analyze the intended meaning of both statements as follows.

D: $\text{kangaroo}(x)\implies
    (\text{lady}(x) \implies
     \neg \text{needhandbag}(x))$
35: $\text{kangaroo}(x)\implies
     (\neg \text{lady}(x) \implies
      \text{needhandbag}(x))$

Then you could use shorthand for the atomic statements:

D: $K\implies (L\implies \neg H)$
35: $K\implies (\neg L\implies H)$

Now if you focus on the consequents, you can look at the four possible truth assignments (eight with $K$) and see that there is at least one case where $L\implies \neg H$ comes out true but $\neg L\implies H$ as false: the former is compatible with lord kangaroos not needing handbags. Since D and 35 are not equivalent, "contrapositive" and "original statement" are ruled out. I'm not sure what "converse" and "inverse" are, but given the way the question is worded, one should apply, and you probably know which.

(Having "contrapositive" as one of the alternative answers suggests that the framers of the question may intend a translation to a single implication. I think that would be awkward.)

This question is phrased badly, I think (in the book, not your own.) I interpret it to mean:

P="If a kangaroo is a lady"

Q="it does not need a handbag"

$P \implies Q$.

Being statement D, whereas (35) is:

$\neg  P \implies \neg Q$, which is a logical fallacy called denial of the antecedent.

To make (35) a contrapositive, you could interpret statement D as follows:

P="if it does not need a handbag"

Q="it is a lady kangaroo"

$P \implies Q$.

In which case:

"If a kangaroo is not a lady, it needs a handbag" which translates symbollically to:

$\neg Q \implies \neg P$ i.e the contrapositive.

About your question, "negation" makes no claim about the nature of the statement itself. If the first statement is "I am not a man," then the negation of this is "I am a man." In other words, just because a sentence includes the word "not" doesn't make it a negation, it just means it's a statement, either true or false.


Why can't the interval construction argument used to show $\mathbb{R}$ is uncountable be used for other infinite sets? I read the following proof as to why the set of real numbers is uncountable.

Assume that $\mathbb{R}$ is countable. Then we can enumerate $\mathbb{R} = \{x_1, x_2, x_3, \cdots\}$ and be sure that every real number appears somewhere on the list. Let $I_1$ be a closed interval that does not contain $x_1$. Next, given a closed interval $I_n$, construct $I_{n+1}$ to satisfy both of the following:

(i) $I_{n+1} \subseteq I_n$

(ii) $x_{n+1} \not\in I_{n+1}$

Now consider the intersection $\cap_{n=1}^{\infty} I_n$. If $x_{n_0}$ is some real number from the enumerated list, then $x_{n_0} \not \in I_{n_0}$, so $x_{n_0} \not \in \cap_{n=1}^{\infty} I_n$. But we assumed the list contained every single real number, so this implies $\cap_{n=1}^{\infty} I_n = \emptyset$. However, the Nested Interval Property asserts that $\cap_{n=1}^{\infty} I_n \neq \emptyset$, hence the contradiction.

Question: I don't see why this same argument can't be applied onto $\mathbb{Q}$ and show that the set of rationals is uncountable (which is of course nonsense), or even more generally, show that any infinite set $S = \{s_1, s_2, s_3, \cdots \}$ is uncountable, by using the same construction of intervals used above.

A closed interval in $\mathbb R$ is an example of a compact set. Let $C$ be a compact set and let $\mathcal F$ be a set of closed sets in $C$. Suppose that for every finite subset $F\subseteq \mathcal F$, $\bigcap F\ne\varnothing$. Then in fact $\bigcap \mathcal F\ne \varnothing$. A non-degenerate closed interval in $\mathbb Q$ is not compact.

It has to do with topological properties of $\mathbb{R}$. In particular, we take adventure of the fact that $\mathbb{R}$ is totally ordered.

If we let $X$ be any topological space, that doesn't have a total ordering on it, then $\forall x,y$ we don't have a well defined notion of $x<y$, so the idea of an interval doesn't make sense. A simple example is $\mathbb{C}$ which cannot be made to be totally ordered.


Help with integral $\int_{-\infty}^\infty e^{iαx-α^2 t} dα$ Solve by integration

$$\int_{-\infty}^\infty e^{-iαx-α^2 t}   dα$$

solve by integration , this integral is - infinity to infinity and exponent value

This question has a short answer and a long one.

completing the square you have:
$$
e^{−iαx−α^2t}=e^{-t(\alpha+\frac{ix}{2t})} \,e^{\frac{-x^2}{4t}} \\
\Rightarrow \int_{-\infty}^{\infty} e^{−iαx−α^2t}\, d \alpha=e^{\frac{-x^2}{4t}} \, \int_{-\infty}^{\infty}e^{-t(\alpha+\frac{ix}{2t})^2} \,d\alpha \\=e^{\frac{-x^2}{4t}} \, \int_{-\infty}^{\infty}e^{-t\alpha^2} \,d\alpha \\=e^{\frac{-x^2}{4t}} \sqrt{\frac{\pi}{t}}
$$

But in the complete answer, we have to prove that

$$
\int_{-\infty}^{\infty}e^{-t(\alpha+b)^2} \,d\alpha=\int_{-\infty}^{\infty}e^{-t\alpha^2} \,d\alpha
$$

For that we can just use a change of variable, but $b$ is imaginary and that means that the limits of integration would change to a line in the complex plane.

So to prove that we need another approach and what is commonly done is to use the fact $e^{z^2}$ is an entire function, so any closed line integral in the complex plane will be $0$. So you choose a rectangle whose segments parallel to the real line go from $-\infty$ to $\infty$ , and the other segments go from $0$ to some $ic$. Then you solve the 4 integrals for each segment and using that the sum of the 4 of them is $0$, you should get the desired result, that is, it is fine to complete the squares even if you have to deal with an imaginary term.

Hint: Complete the square in the argument of the exponential in terms of $\alpha$ and then use Gaussian integrals.

Added: Note that, you need to make a change of variables after completing the square. See here.


Sum of roots of $p-3 = 5p^{2/3} - 6p^{1/3}=0$? In the middle of a solution of a question, the equation $$p-3 = 5p^{2/3} - 6p^{1/3}=0$$ is given. I am required to find out the value of the sum of roots for the next step, which for an equation $ax^3+bx^2+cx+d$ is equal to $-b/a$. I tried to put the equation in this format by cubing both sides, however I get $$p^3-27-9p^2+27p = 125p^2 - 216p - 90p(5p^{2/3} - 6p^{1/3})$$ which is in terms of $p^{2/3}$, therefore I think it cannot be right. What should I do? Is there any other method, or have I missed something in this method?

If
$$p^3-27-9p^2+27p = 125p^2 - 216p - 90p(5p^{2/3} - 6p^{1/3})$$
(I haven't checked that your algebra is correct) you can substitute
$$5p^{2/3} - 6p^{1/3}=p-3$$
to get rid of the fractional powers.

Note.  The pair of equations $p-3 = 5p^{2/3} - 6p^{1/3}=0$ has no solution, I am assuming you meant the single equation $p-3 = 5p^{2/3} - 6p^{1/3}$.


Interesting inequality $\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$ Prove that $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[k]{\frac{1+x^k}{2}}$$ for all real $x \ge 1$ and for all positive integers $m$  and $k \le 2m+1$.

My work. If $k \le 2m+1$ then $$\sqrt[k]{\frac{1+x^k}{2}}\le \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}.$$ I need prove that for all real $x \ge 1$ and for all positive integers $m$ the inequality $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$$ holds.