Raising a Complex Number to a Decimal Value So for my class i have to make a java program that deals with complex numbers. I finished getting the root and power and i was wondering how to do a method that deals with powers such as 2.56. Now im not exactly sure how to take a complex number to the power of 2.56 and whether i would use the same method of De Moivres Formula.
I was thinking that taking something to the power of 2.56 is the same as taking it to the power of 256 and then taking the 100th root of it, but im not sure if that will work for complex numbers or if it will work in java.

Anyways any help would be appreciated...i barely remember anything about complex numbers...

There is not an universally agreed-upon way to do this (raise an arbitrary complex number to the power 2.56).  The usual way is to write the complex number $z$ in the form $r\exp(i\theta) = r(\cos\theta + i\sin\theta)$ where $r = |z| \geq 0$, $\theta$ is real (the equation is true due to Euler's Formula), then define

$$z^{2.256} = r^{2.56}\exp(2.56i\theta) =   r^{2.56}(\cos(2.56\theta) + i\sin(2.56\theta))$$

(using Euler's formula for the second equation). The problem with this is that for any $z$, many choices of $\theta$ satisfy $z=r\exp(i\theta)$.  If $z=0$, $\theta$ can be anything (but $0^{2.56}=0$, so there is no problem).  If $z \neq 0$, the most popular choice is to use $\theta = \operatorname{Arg}(z)$, where $\operatorname{Arg}(z)$ is the unique angle in $(-\pi,\pi]$ satisfying $\cos\theta= Re(z)/|z|$ and $\sin\theta= Im(z)/|z|$.  Note that it is a bad idea to say $\tan(\theta) = Im(z)/Re(z)$ because $Re(z)$ might be zero.  It is even worse to define $\theta = \tan^{-1}(Im(z)/Re(z))$ because not only can $Re(z)$ be zero, but this way, $\theta$ must be in a right-hand quadrant, which might not be correct.

While using $\operatorname{Arg}(z)$ is the most popular way to choose $\theta$, I don't know if there is any mathematical reason why it is the best way.

To compute $\operatorname{Arg}(z)$ using Java, if you are lucky, there will be a two-argument "Arg"-type function.  If this does not exist, you should use the regular inverse tangent function, but you will have to do several cases, depending on the signs of $x$ and $y$, and in some cases you will have to add or subtract $\pi$ from the output of the inverse tangent function.

EDIT: dfeuer makes a good observation in his/her answer, and I hope s/he doesn't mind if I steal it and repeat it here: if $n$ is an integer and $z$ a complex number, then $z^n$ is unambiguously defined and can be computed in the obvious way (repeated multiplication and/or repeated squaring) or by using the scheme above (you will get the same answer regardless of which sensible $\theta$ you use).  Of course, $0^n$ is undefined if $n<0$, and it is debatable whether $0^0$ should be given a value (in many applications, people seem to treat it as $1$).  If the exponent $\alpha$ is rational, then as s/he writes, then $z^\alpha$ has finitely many values that might make sense.  If your exponent is 2.56, which simplfies to $64/25$, there are "only" 25 possible possible values for $z^{2.56}$ if $z \neq 0$.

Any complex number $z = x + iy$ can be written in the form $z = r e^{i \theta}$, where $r = \sqrt{x^2 + y^2}$ and $\tan \theta = \frac{y}{x}$. ($0 \leq \theta < 2 \pi$)  Thus, you may use $z^{\alpha} = r^{\alpha} e^{i \alpha \theta}$ for any $\alpha \in \mathbb{R}$.


addition with a variable (mod) Given $2+x \equiv 7 \pmod 3$.

$2 + 0 = 2$

$2 + 1 = 3$

$2 + 2 = 4$

.
.
.

$2 + 5 = 7$

so, the answer will be $x = 5, 8, 11, 14, 17,\dots$

Is this correct? Because somebody told me the answer should be $x = 2, 5, 8, 11, 14, 17,\dots$

We are given that:

$(2+x) = 7 (\mod 3)$

Thus, $2+x-7 = 3k$ where $k$ is an integer.

Simplifying, yields:

$x-5 = 3k$

Therefore, $x\in (5,8,..)$


Monotone non-increasing functions (sequence) relations Not sure what is the topic of this question, but it seems like real analysis.

The question itself: Given $a(q) : \mathbb{N} \rightarrow \mathbb{R}$ monotone non-increasing positive sequence, s.t. $\sum_{q=1}^{\infty}a(q) = \infty$, there exists a monotone non-increasing sequence $a'(q)$, s.t. also - $\sum_{q=1}^{\infty}a'(q) = \infty$ and for all $t \geq 1$, $$\frac{a'(q)}{a(tq)} \xrightarrow{q \rightarrow \infty} 0$$

Any clues?

First, notice that, for each fixed $t$, the series $\sum_{q=1}^\infty a(tq)$ diverges.  The reason is that, if this series converged, then so would the series $\sum_{q=1}^\infty a(tq+1)$, $\sum_{q=1}^\infty a(tq+2)$, $\dots,\sum_{q=1}^\infty a(tq+t-1)$ because, by the "non-increasing" assumption, they're dominated by $\sum_{q=1}^\infty a(tq)$.  But the sum of all these convergent series would be all of $\sum_{q=1}^\infty a(q)$ except for the first $t-1$ terms, and this is assumed to diverge.

Second, notice that, still for fixed $t$, $\sum_{q=1}^\infty a(tq)/t$ diverges; we've just multiplied the divergent series from the preceding paragraph by a constant $1/t$.

With these observations in mind, we can define the required $a'(q)$ sequence as follows.

First, since $\sum_{q=1}^\infty a(q)$ diverges, find an $N(1)$ so large that $\sum_{q=1}^{N(1)}a(q)\geq1$, and set $a'(q)=a(q)$ for $1\leq q\leq N(1)$.  This ensures that, no matter how we define $a'(q)$ for larger $q$ (as long as it's positive), $\sum_{q=1}^\infty a'(q)$ will be at least $1$.  Call this stage $1$ of the definition.

Next, find $N(2)$ so large that $\sum_{q=N(1)+1}^{N(2)}a(2q)/2\geq1$.  Such an $N(2)$ exists because, as shown above, $\sum_{q=1}^\infty a(2q)/2$ diverges.  Set $a'(q)=a(2q)/2$ for $N(1)<q\leq N(2)$.  You thereby ensure that $\sum_{q=1}^\infty a'(q)$ will be at least $2$.  Call this stage $2$.

Continue similarly, setting $a'(q)=a(3q)/3$ at stage $3$ for $N(2)<q\leq N(3)$, where $N(3)$ is chosen so large that these $a'(q)$'s add up to at least $1$ and therefore $\sum_{q=1}^\infty a'(q)$ will be at least $3$.  And so forth for all later stages.

Since you keep making $\sum_{q=1}^\infty a'(q)$ at least $1$ bigger at each step, you clearly get this series to diverge.  Furthermore, if you fix some $t$ and then look at what happens at stages $s$ from $t$ on, you're setting $a'(q)=a(sq)/s\leq a(tq)/s$, so you have $a'(q)/a(tq)\leq 1/s$.  Since this holds for all stages $s\geq t$, and since $1/s\to0$ as $s\to\infty$, you get $a'(q)/a(tq)\to 0$ as required.

If $a(q)$ is a constant sequence, put $a'(q)=q.a(q)$ and you're done.

If $a(q)$ is decreasing, the subsequence $a(tq)$ is also decreasing.

If $\lim\limits_{q\to\infty}a(q)=-\infty$, take $a'(q)=1$ and you're done ($\sum\limits_{q=1}^{+\infty}a'(q)=\infty$).

If $\lim\limits_{q\to\infty}a(q)=\ell\in\mathbb{R}^*$, let $a'(q)=\frac{1}{q}$ so $\lim\limits_{q\to \infty}a'(q)=0$ and $\sum\limits_{q=1}^{+\infty}a'(q)=\infty$

If $\lim\limits_{q\to\infty}a(q)=0$, take $a'(q)=0$


You see a route 14 bus on the moon. What is the most likely number of bus routes on the moon? This question was asked on a forum and while many argued that the answer is 14 (since the probability of you seeing bus 14 is maximum in this case), I argued against it that they were working backwards.
My claim is that this question is invalid as there is no method to determine the probability of number of bus routes.

I'm looking for clarification as to the right answer (with proof obviously)

Your question touches on fundamental issues of the interpretation of probabilities. You won't get a proof, at least not in the mathematical sense, since this is not a mathematical question but an interpretational question.

Basically, there are two popular interpretations of probability theory, the frequentist one and the Bayesian one. In the frequentist interpretation, probabilities specify the relative frequencies of an event if you perform the "same" experiment many times. Obviously you can't create a large number of moons and count the number of bus routes on each, or, as you put it, "there is no method to determine the probability of number of bus routes", and thus in this interpretation there is no such thing as that probability. As William Feller put it:

  There is no place in our system for speculations concerning the probability that the sun will rise tomorrow.

In the Bayesian interpretation, probability theory allows us to reason about uncertain events, and more specifically to rationally update our assessments of how likely events are when new information comes in. In this framework, you always need some prior assessment of likelihoods, and then the theory tells you how to adjust that using the data you observe.

For some sorts of events, such as rolling a die, there are rational grounds for choosing prior probabilities (the same probability for each number). In other cases, such as bus routes on the moon, there isn't one obvious set of prior probabilities, but still reasoning about how prior assessments of likelihood should rationally be modified by incoming data can be useful.

In the present case, any prior assessment of the likelihood of various numbers of bus routes on the moon would presumably have exhibited a very dominant spike at $0$ and then a very low and rather flat tail for all other numbers. The Bayesian probability update requires us to multiply the a priori probability for each possible number of bus routes by the conditional probability that you would have observed a route $14$ bus if there were that many bus routes, and then normalize the resulting probabilities to $1$ to obtain the a posteriori probabilities. If we follow the assumption you seem to be making in the question, that the bus routes are numbered sequentially beginning with $1$ and we have an equal probability of encountering a bus from any one of the existing bus routes, then the conditional probability of observing a route $14$ bus given $n$ bus routes is zero for $0$ to $13$ bus routes and $1/n$ for $n$ bus routes if $n\ge14$.

Now although there is no way to agree on any particular prior in the present case, it seems rather plausible that apart from the dominant spike at $0$, the prior would have been relatively flat. That is, there was no strong a priori reason to favour, say, the number $15$ over the number $14$, and so the ratio of the prior probability for $15$ bus routes to the prior probability for $14$ bus routes would not have exceeded $15/14$. That's the ratio by which the observation of a route $14$ bus raises the probability for $14$ bus routes relative to the probability for $15$ bus routes. So although there's no proof and no one right answer, we can nevertheless plausibly argue that most sensible priors would not favour any number $n>14$ over the number $14$ by a factor of $n/14$, and thus, in a Bayesian framework, the number of bus routes with the highest a posteriori probability would be $14$.

I would have thought that on average your best guess is 28. I think the question does have an answer. Maybe it depends on the spread of which amounts of buses could be used (probably exponential - so the same likelihood in between 10 and 20 as between 100 and 200). But actually, I don't think that matters.

I think it's easy to overcomplicate the question. You don't question how someone flips a coin or how many sides a coin has or what surface it's landing on. Just because they mentioned the Moon doesn't mean the whole thing is completely hypothetical and ridiculous. Like is an even number of buses more likely?!

The game is this: I pick a thousand billion secret numbers. For each number N, I have to give you (the guesser) a random number between 1 and N. Then you have to guess N. What's the best strategy for you to use? We can measure your success based on number of correct guesses or based on average closeness.

An interesting question would be whether it matters what secret numbers I pick (what spread, etc)? If you use my strategy of doubling, I don't think it does. Anyone want to play?


Which integration formula for $\frac{1}{a^2-x^2}$ is correct? In my book integration formula for $\frac{1}{x^2-a^2}$ is given as $\frac{1}{2a}\ln(\frac{x-a}{x+a})$.

From the above formula we can write the formula for integration of $\frac{1}{a^2-x^2}$ as   $-\frac{1}{2a}\ln(\frac{x-a}{x+a}) = \frac{1}{2a}\ln(\frac{x+a}{x-a}) $ [as we know $\ln(\frac{1}{x})=-\ln(x)$]

But, in my book the integration of  $\frac{1}{a^2-x^2}$ is written as $\frac{1}{2a}\ln(\frac{x+a}{{a-x}}) $.

We know that the integration in the second case cannot be $\frac{1}{2a}\ln(\frac{x+a}{x-a}) $ and $\frac{1}{2a}\ln(\frac{x+a}{{a-x}}) $ at the same time. Which formula is the correct one and why?

Neither is correct.

It's not restrictive to assume $a>0$ (for $a=0$ the antiderivative has a different form; for $a<0$ change the intervals below accordingly).

The function $\frac{1}{x^2-a^2}$ is defined for $x\ne\pm a$ and is negative over $(-a,a)$, positive over $(-\infty,-a)$ and $(a,\infty)$.

The sign of $\frac{x-a}{x+a}$ is the same as the sign of $x^2-a^2$ (for $x\ne\pm a$, of course), so it's clear that
$$
\frac{1}{2a}\ln\frac{x-a}{x+a}
$$
cannot be an antiderivative over $(-a,a)$, because it is undefined there.

There's a simple way out. An antiderivative of $1/x$ is $\ln|x|$ (up to an arbitrary constant over $(-\infty,0)$ and an arbitrary constant on $(0,\infty)$, with no connection between each other).

The partial fraction decomposition is
$$
\frac{1}{x^2-a^2}=\frac{1}{2a}\frac{1}{x-a}-\frac{1}{2a}\frac{1}{x+a}
$$
so we can write
$$
\int\frac{1}{x^2-a^2}\,dx=
\frac{1}{2a}(\ln|x-a|-\ln|x+a|)=
\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|
$$
up to arbitrary constants in the intervals $(-\infty,-a)$, $(-a,a)$ and $(a,\infty)$.

Since $1/(a^2-x^2)=-1/(x^2-a^2)$, we have
$$
\int\frac{1}{a^2-x^2}\,dx=
-\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|=
\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|^{-1}=
\frac{1}{2a}\ln\left|\frac{x+a}{x-a}\right|
$$
again up to constants as before.

Let $f(x)=\frac{1}{x^2-a^2}$.
$f$ is continuous at $(-\infty,-|a|),(-|a|,|a|)$ and $(|a|,+\infty)$.

in each interval
$$2af(x)=\frac{1}{x-a}-\frac{1}{x+a}$$

and $$\int f(x)dx=\frac{1}{2a}\ln(\frac{|x-a|}{|x+a|}).$$

the final expression depends on which interval $J$  we want the antiderivative.
for example, if $a>0$ and $J=(-a,a)$, we have

$$\int f(x)dx=\frac{1}{2a}\ln( \frac{a-x}{x+a} )+C$$


A variant of the Knight's tour problem The knight's tour problem is a famous problem in chess and computer science which asks the following question: can we move the knight on an $n \ \times \ n$  chessboard such that it visits every square exactly once? The answer is yes iff $n\geq5$. Additionally, there are algorithms which can solve it in $O(n^2)$ time.

I have two variations of it discussed below.

Fix an $n \ \times \ n$ chessboard. In this variant, instead of one knight we have $m$ knights. These knights take turns moving (ie., one knight moves, after that another one, once all of them have moved, the first one moves again and the cycle repeats itself).

What is the largest $m$ such that for some initial starting position, each knight can tour the board (as described in the first paragraph) without threatening any other knight? In other words, there is no "instance" of the chessboard in which one knight may capture another knight (e.g., knight $A$ cannot be, for example, two cells right and one cell up from another knight $B$ when it is his turn). Note that there is obviously an upper bound on $m$ (e.g., $n^2$).

In essence, I'm looking for a function $f:\mathbb{N}_{\geq 5} \to \mathbb{N}$ which assigns to each $n$ the largest possible $m$.

My second variation is exactly the same, except now there are no restrictions on the turns. After one knight moves, any knight can move, including the one that has moved prior.  I do not know how these variations relate to each other; perhaps they are equivalent.

Also, as a function of $n$ what would the time complexity of an optimal algorithm be? Can either of these variants be solved in polynomial time?

For even $n$, $n \ge 6$, the answer is $m = n^2 - 1$ for both questions and the time is $O(n^2)$.

Obviously $m \le n^2 - 1$ is an upper bound, because at least one cell should be initially free. Let us see why it is reachable.

It is possible to construct a closed (cyclic) tour for one knight in the time of $O(n^2)$. The tour has additional property of symmetry, and the algo has property of good parallelization (i. e. it takes $O(1)$ on $n^2$ processors).

Let $C = v_1 v_2 \ldots v_n v_1$ be sequence of cells in such closed tour. Then we place the first knight into $v_n,$ the second one into $v_{n - 1}$, and so on, placing the last one, $m$th knight, into $v_2$. At the first turn the first knight moves to $v_1$, the second one moves to $v_n$, and so on, the last one moves to $v_3$. Then the first one moves to $v_2$ and so on and so on. Every knight walks through the entire tour.

For odd $n$ there is also a more tight upper bound: $m \le \frac{n^2 + 1}{2}$. It follows from a note that our graph is bipartite and has an odd order, so we should start every tour in the bigger part of graph (of $\frac{n^2 + 1}{2}$ cells). For the first question the upper bound is even less by $1$, because after the first turn all knights should be in the smaller part of graph (of $\frac{n^2 - 1}{2}$ cells).

May be some time later I'll investigate odd case more precisely.


Matrix Representation of Octonions Since quaternions $\mathbb{H}$ have a matrix representation as elements of $\text{SU}(2,\mathbb{C})$ as the following

$$   1 \mapsto \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\quad \mathrm i \mapsto \begin{pmatrix} \mathrm i_{\mathbb C} & 0 \\ 0 & -\mathrm i_{\mathbb C} \end{pmatrix},\quad \mathrm j \mapsto \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},\quad \mathrm k \mapsto \begin{pmatrix} 0 & \mathrm i_{\mathbb C} \\ \mathrm i_{\mathbb C} & 0 \end{pmatrix},
$$
I always wondered if there is also matrix representation of the octonions?

How is the non-associativity realised with matrices?

A $\mathbb R$-linear function $\phi:\mathbb O\to A$ to any real associative algebra which is multiplicative and unitary (so that $\phi(xy)=\phi(x)\phi(y)$ for all $x$, $y\in \mathbb O$, and $\phi(1)=1$) has to vanish on the bilateral ideal $I$ generated by the elements of the form $$(x\cdot y)\cdot z-x\cdot(y\cdot z)$$ with $x$, $y$, $z\in\mathbb O$. Now, this ideal is not zero because $\mathbb O$ is not associative, and therefore $I=\mathbb O$, because $\mathbb O$ has no non-trivial bilateral ideals (it is a division thing).

It follows that the map $\phi$ is in fact zero.

I tried to convert different dimensional base units to matrices. The reals are trivial, the complexes are easy, the quaternions are interesting, but solvable and the octonions are so tricky, that it does not work. My matrix looks like this:
$ M_{n m} = i_x^{o-1} o (i_n i_m)$
So e. g.
M = $ \begin{pmatrix} 0 & -1 & -2 & -3 \\ 1 & 0 & -3 & 2 \\ 2 & 3 & 0 & -1 \\ 3 & -2 & 1 & 0\end{pmatrix}$
for the quaternions, it can be for the octonions, too.
But to make sense of all these matrices, we should convert it to another (in this case 4) matrices with full of 0, 1 and -1 numbers. And if we multiplicate them, we get each other, right? I tested it, and it works at quaternions, but so does not at octonions. I am really sorry.
Maybe the multiplication table of octonions is wrong or the previous person was right talking about its problematics of the associativity.


Average Distance Between Random Points on a Line Segment Suppose I have a line segment of length $L$.  I now select two points at random along the segment.  What is the expected value of the distance between the two points, and why?

Byron has already answered your question, but I will attempt to provide a detailed solution...

Let $X$ be a random variable uniformly distributed over $[0,L]$, i.e., the probability density function of $X$ is the following

$$f_X (x) = \begin{cases} \frac{1}{L} & \textrm{if} \quad{} x \in [0,L]\\ 0 & \textrm{otherwise}\end{cases}$$

Let us randomly pick two points in $[0,L]$ independently. Let us denote those by $X_1$ and $X_2$, which are random variables distributed according to $f_X$. The distance between the two points is a new random variable

$$Y = |X_1 - X_2|$$

Hence, we would like to find the expected value $\mathbb{E}(Y) = \mathbb{E}( |X_1 - X_2| )$. Let us introduce function $g$

$$g (x_1,x_2) = |x_1 - x_2| = \begin{cases} x_1 - x_2 & \textrm{if} \quad{} x_1 \geq x_2\\ x_2 - x_1 & \textrm{if} \quad{} x_2 \geq x_1\end{cases}$$

Since the two points are picked independently, the joint probability density function is the product of the pdf's of $X_1$ and $X_2$, i.e., $f_{X_1 X_2} (x_1, x_2) = f_{X_1} (x_1) f_{X_2} (x_2) = 1 / L^2$ in $[0,L] \times [0,L]$. Therefore, the expected value $\mathbb{E}(Y) = \mathbb{E}(g(X_1,X_2))$ is given by

$$\begin{align} \mathbb{E}(Y) &= \displaystyle\int_{0}^L\int_{0}^L g(x_1,x_2) \, f_{X_1 X_2} (x_1, x_2) \,d x_1 \, d x_2\\[6pt]
 &= \frac{1}{L^2} \int_0^L\int_0^L |x_1 - x_2|  \,d x_1 \, d x_2\\[6pt]
 &= \frac{1}{L^2} \int_0^L\int_0^{x_1} (x_1 - x_2) \,d x_2 \, d x_1 + \frac{1}{L^2} \int_0^L\int_{x_1}^L (x_2 - x_1) \,d x_2 \, d x_1\\[6pt]
 &= \frac{L^3}{6 L^2} + \frac{L^3}{6 L^2} = \frac{L}{3}\end{align}$$

The pdf of the distance is highest closer to zero. This means that distance is more likely to be close to zero than any other positive value. This seems to be puzzling. Put another way, let us divide the segment into equal sized bucket, say 100 buckets .01 length each and number 0 to 99 with bucket i from i*.01 to (i+1)*.01. If you do the experiment for a large number of pairs of points and divide them into buckets, then bucket 0 will contain the largest number of samples. This seems counter intuitive at first. But it makes sense, since after selecting a point closer to the edge makes it twice more likely to select a second point closer to the first point than one farther from it. The closer point can be on either side of the original point but the farther point can only be on one side.


How does Hilbert's axiomatization relate to set theory? I'm studying Hilbert's axiomatization of Euclidean geometry, and I'm trying to combine my current understanding into my knowledge on mathematical logic (not very much).

At the beginning of this axiomatic system, it states the undefined terms are 'point', 'line', 'plane', 'lie', 'between', and 'congruence'. Therefore it seems that all later definitions should be based on these primitive notions. However, in the definition of 'segment', it states that a segment $AB$ is a set of two points on the same line. (or another variation where the segment contains also all points between two endpoints.)

My question is: how does 'set' come into play here? There is no notion of 'set' defined in this system, how is it valid to use the notion of set (or notions about sets like equivalent class) here? More generally, how does set relate to a specific mathematical theory?

Being a beginner on foundation related topics, I would appreciate if you can explain different concepts in a slow pace. In my understanding, Hilbert's system is a purely syntactic system with some undefined terms. It might be helpful if you can show me how the definitions (like segment) can be produced by FOL language.

The resolution is that Hilbert's axioms are not first-order axioms. They refer to sets (or "systems") as well as to individual objects.

This is most visible in the axiom of continuity, which directly refers to sets. The purpose of this axiom is to prevent issues which affect Euclid's axioms. One such issue is that the rational plane $\mathbb{Q} \times \mathbb{Q}$ satisfies many of the axioms of geometry (including the natural FOL versions of Euclid's axioms). Some axiom is needed to prove, for example, that the line $y = x$ intersects the circle of radius $1$ centered at the origin.

Hilbert's axioms are most naturally viewed in second-order logic. In that framework, "set" is a logical concept (like "equality" and "or") rather than a term of the theory being studied.

At the time that Hilbert was working on geometry, the distinction between first-order and second-order logic was not yet understood. So it is not at all surprising that Hilbert would not make any distinction, and would not view "set" as an undefined term of geometry. From the 19th century point of view there was simply "logic", which had not yet been formalized. It took until the 1940s or 1950s for our contemporary understanding of logic to be completely developed.

There are first-order axiom systems for geometry, such as Tarski's axioms. These are different from Hilbert's axioms in various ways, but in particular they do not try to include a "completeness" or "continuity" axiom analogous to Hilbert's axiom.

One of the primitive notions in Hilbert's approach is the relation of *betweenness" which is a relation among three variables $A,B,C$ written as $A*B*C$. This reflects our intuitive notion of point $B$ belonging to interval $[A,C]$. Thus Hilbert is not relying on naive Euclidean notions when he talks of segments. A book that details Hilbert's approach and is more accessible than Hilbert's is Greenberg's Euclidean and non-Euclidean geometries.


Why is the Expected Value different from the number needed for 50% chance of success? An event with probability $p$ of being success is executed $\frac{1}{p}$ times. For example, if $p=5\%$, the event would then be executed $20$ times.

The Expected Value for the total number of trials needed to get one success is $\frac{1}{p}$. In this case, it's $20$.

What I'm confused is, as p approaches zero, the chance of having a success in the first $\frac{1}{p}$ trials always approaches to $1-\frac{1}{e}$, or about $63\%$. This means: $P$(at least $1$ success in all $\frac{1}{p}$ trials) is about $63\%$.

This $63\%$ is higher than $50\%$. It seems to suggest that, if I take all $\frac{1}{p}$ trials and consider them as one big event, and do this big event multiple times, I'd get more successes than failures. But on the other hand, since $\frac{1}{p}$ times is the EV mentioned earlier, shouldn't the big event have an equal chance of being a success or a failure?

Let's take your example of an event that has a p=5% chance of success, and repeat it until it succeeds.  Let's call this one experiment, and let's call the number of times you had to repeat the event in one experiment the number of runs.  And let's repeat the experiment many, many times.

What can we say about the average number of runs you have to do for one experiment?  Well, that depends on what you mean by average:

Say we want the mean number of runs.  This would be the weighted infinite sum $(.05*1) + (.95*.05*2) + (.95^2*.05*3) + ... = 20$.  This is called the expected value, or the "expected number of runs."
Say we want the median number of runs.  This would be the number of runs at which you have exactly 50% chance of succeeding at or before that point.  In other words, the number of runs $x$ for which $(1-.95^x)=0.5$, which is $x = \log_{0.95}{0.5} \approx 13.5$.

Thus, if you were to repeat this experiment many times, you'd find that you'd have to do an average (mean) of 20 trials before getting a success.  However, in any given instance of the experiment, you'd find that you'd most likely finish by the 14th trial.  The rare cases that you do way, way more are what bump up the expected value (eg. approximately 0.6% of the time, you'll need to do more than 100 runs before succeeding!).

You appear to be making two intuitive mistakes:

Confusing the mean (expected value) with the median (value with $1\over 2$ smaller and $1\over 2$ bigger)
Creating an experiment which superficially looks like a geometric distribution, but isn't.

The mean of a geometric series is, as you say, $1\over p$. The median, however is $\lceil{-1\over{\log_2(1-p)}}\rceil$.

As @clementc points out, the experiment you construct is not a geometric distribution. Using the c.d.f. Of the geometric distribution, your random variable is:

$$Y=\begin{cases}
1 &:1-(1-p)^{1\over p}\\
0 &:(1-p)^{1\over p}
\end{cases}$$

When expressed this way, it is clear that this is not a geometric distribution. Further, each of those probabilities is clearly not equal to $1\over 2$

Now, if you use the median rather than the mean then they are (except for the effect of the ceiling function).


Why eigenvalues of an orthogonal matrix made with QR decomposition include -1? I want to make a real orthogonal matrix whose eigenvalues don't include -1.
However, eigenvalues of a matrix $Q \in \mathbb{R}^{n\times n}$ ($n$ is even number) made with QR decomposition of a random matrix always include -1.

e.g.

n = 10
Q,R = qr(rand(n,n))

Question.

Why eigenvalues of $Q$ include -1? Does this proceed from the algorithm of QR decomposition?
Is there any idea for get an orthogonal matrix whose eigenvalues doesn't include -1.

Edit 1

I noticed that "if a block diagonal matrix consists of $2\times 2$ appropriate rotation matrices, this matrix satisfies the condition".

i.e.

$$Q := \begin{bmatrix}
\cos(\theta_1)&\sin(-\theta_1)&&&\\
\sin(\theta_1)&\cos(\theta_1)&&&\\
&&\cos(\theta_2)&\sin(-\theta_2)&\\
&&\sin(\theta_2)&\cos(\theta_2)&\\
&&&&\ddots
\end{bmatrix}$$

But I want to make more complicated $Q$ as a test matrix.

Edit 2

An answer for the question 1

Let $Q\in \mathbb{R}^{2m \times 2m}$ be an orthogonal matrix.

(i) Determinant of a householder matrix is $-1$ ($\because$ Wikipedia)

(ii) If $Q$ is made with QR decomposition, $$
\begin{align}
\det(Q) &= \det(H_1)\det(H_2)\cdots\det(H_{2m-1})\\
& = -1
\end{align}
$$

(iii) $\det{Q}=-1 \Rightarrow -1 \text{ is an eigenvalue of } Q.$

I assume you mean that $A$ is a randomly generated matrix and $A=QR$ is its QR-decomposition.

The (real) eigenvalues of an orthogonal matrix can only be $\pm 1$. If you are randomly generating $A$, then you are effectively randomly generating $Q$ as well. Since $Q$ is $n\times n$, it will have $n$ eigenvalues (not counting multiplicity), and effectively, each eigenvalue has a 50-50 chance of being 1 or -1. Thus the chance of $Q$ having at least one negative eigenvalue is $1-\frac{1}{2^n}$, which for $n$ large (i.e. n=10) is pretty small.
If you replace any orthogonal matrix $Q$ with $Q^2$ then all the eigenvalues will be $+1$ (since if $v$ is an eigenvector of $Q$ corresponding to $-1$, then $Q^2v = (-1)(-1) v = v$ so all the eigenvalues of $Q^2$ are 1). Of course, this will change the matrix $A$, but it might be suitable for your purposes.


Pass a logic function using only NAND So I had a truth table and using a Karnaugh map I simplified a function. I obtained.

$ f = \overline{A_3}A_2\overline{A_1} + \overline{A_2}\overline{A_0} + A_3\overline{A_0} $

Then using the distributive property of boolean algebra:

$ f = \overline{A_3}A_2\overline{A_1} + \overline{A_0} (\overline{A_2} + A_3) $

Ok, with this we have the minimum of logic gates to use.

Now I need to convert this to NAND.
What seemed easier to me was to take the logigram (or electrical scheme) and directly change the gates to their equivalents with NAND. I obtained:

$ f = \overline { \overline{\overline{A_3}A_2\overline{A_1}} + \overline{\overline{A_0} \overline{A_2 \overline{A_3}}} }$

Now I have two questions about this:

How should I proceed algebraically to pass to one expression to another (I know I need to apply DeMorgan laws or use the complement of the function... I would thank if someone can give like a tip or something on how to start doing it.
I'm thinking if there is any way to simplify the NAND expression... I see that I have \$ A_2 \$ twice and also \$ \overline{A_3} \$ twice... I'm thinking if there is any way to go even further in the simplification... If someone could also give a tip on how to proceed to not ruin the rule of using NAND only. Thanks very much!

You can apply DeMorgan's Law and not distribute your NOTs.

for example:

$ab+c \longleftrightarrow \overline {\overline{a b} \overline c} $


Line integral around intersection of sphere and plane The unit sphere is intersected by the plane $x + y = 14$. Find the line integral of $F = \langle yz + y, xz+5x,xy+2y\rangle$ around the intersection.

$$\iint\nabla\times F\cdot\textbf{n}\ dA$$

the unit normal vector is easily found by looking at the plane equation: $\frac{1}{\sqrt2}\langle1,1,0\rangle$

Curl of $F = \langle 2,-1,4\rangle$. So the curl of $F$ dotted with that = $1/\sqrt2$, so we have

$$\iint\ \frac{1}{\sqrt{2}}\ dA$$

It's finding an expression for the skewed circle I'm having trouble with. By some algebra, I was able to find the circles projection in the $xz$-plane:

$$2\left(\frac{x-1}{2}\right)^2 + z^2 = \frac 32$$

But I'm not sure what I'm supposed to do now...

The unit circle $x^2 + y^2 = 1$ is a great circle of the unit sphere. This unit circle also happens to intersect the circle of intersection between the unit sphere and the plane, hence the chord connecting those two points is a diameter of the skewed circle. This diameter is also of course the diagonal of the unit square in the xy-plane, so its length is $\sqrt2$.

Then the area of the skewed disk is $A = \pi (\frac{\sqrt2}{2})^2 = \frac{\pi}{2}.$

\begin{align}
1 & =
x^{2} + y^{2} + z^{2}
=
x^{2} + \left(1 - x\right)^{2} + z^{2}
=
2\left(x^{2} -x\right) + 1  + z^{2}
\\[5mm] & =
2\left(x - {1 \over 2}\right)^{2} + {1 \over 2}  + z^{2}
\end{align}

$$
2\left(x - {1 \over 2}\right)^{2} + z^{2} = {1 \over 2}
\quad\Longrightarrow\quad
{\left(x - 1/2\right)^{2} \over \left(\color{red}{1/2}\right)^{2}}
+
{z^{2} \over \left(\color{red}{\sqrt{2\,}/2}\right)^{2}} = 1\,,
\quad
\mbox{( Ellipse )}
$$

$S_{\rm A}$: Surface area.
$\displaystyle
S_{\rm A} = \pi\times\left(\color{red}{1 \over 2}\right)\times\left(\color{red}{\sqrt{2\,} \over 2}\right)
= {\sqrt{2\,} \over 4}\,\pi$.

Integral de linea: $\left(1/\sqrt{2\,}\right)\left(\sqrt{2\,}\,\pi/4\right)
=
\color{#ff0000}{\displaystyle{\large{\pi \over 4}}}$


Solve the equation $x^{x^5} = 5$ over $\Bbb R.$ Find real numbers $x$ (if any) such that $x^{x^5} = 5.$

I have shown that $x \notin \Bbb Z.$ Does there exist any $x \in \Bbb R \setminus \Bbb Z$ which satisfies the above equation? Any help will be highly appreciated.

\begin{align}
x^{x^5}&=5
\tag{1}\label{1}
\end{align}

Equation \eqref{1} can be solved formally
with the help of the Lambert W function:

\begin{align}
x^5\ln x&=\ln 5
,\\
5x^5\ln x&=5\ln 5
,\\
x^5\ln (x^5)&=5\ln 5
,\\
\ln (x^5)\,\exp(\ln (x^5))&=\ln 5\exp(\ln5)
.
\end{align}

At this point we can apply the Lambert W function:
\begin{align}
\operatorname{W}(\ln (x^5)\,\exp(\ln (x^5)))
&=
\operatorname{W}(\ln 5\exp(\ln5))
,
\end{align}
and since the argument
$\ln 5\exp(\ln5)>0$,
there is just one solution

\begin{align}
\ln (x^5)&=\ln5
,\\
x^5&=5
,\\
x&=5^{\tfrac15}\approx 1.379729661
.
\end{align}

Edit

Or, less formally, without Lambert W function:

\begin{align}
(x^{x^5})^5&=5^5
,\\
x^{5x^5}&=5^5
,\\
(x^5)^{(x^5)}&=(5)^{(5)}
,\\
x^5&=5
.
\end{align}

Here are graphs that show the solution, $x = \sqrt[5]{5}$:


Show that $\prod _{i<j}(x_i-x_j)$ can be divided wihout remainder in $\prod_{i<j}(i-j)$ Let $x_1,...,x_n$ be a natural numbers,

  show that $\prod _{i<j}(x_i-x_j)$ can be divided wihout remainder in $\prod_{i<j}(i-j)$

I know $\prod \left(x_i-x_j\right)$ is the result of Vandermonde Determinant, so is related with Det?

Lemma. For every m-coloring vertices of complete graph $K_n$ ($m \leq n$) there are at least $n-m$ edges which have  two vertices of the same color.

proof: Inducion by $n$ - number of vertices. For $n=1,2,3$ this fact is obvious.

$n-1 \rightsquigarrow n$:
Let $V$ is the set of vertices, of $K_n$ ($\left\vert{V}\right\vert= n$).
$m \leq n$ hecnce from pigeonhole principle we have at least two vertices of the same color, for example $a$ and $b$. Consider graph $K_{n-1}$ with set of vertices $V \setminus {a}$. From induction, we have at least $n-1-m$ edges with 2 vetrices of te same color. These edges plus edge $\{a,b\}$ gives us the desirable at least $(n-m)$-element edge set.

Consider now $A := \prod_{i<j} (i-j)$. Note that given difference: $i-j=m$ occurs exactly $n-m$ times in this product ($A$). So $m^{n-m} | A$.

We ask why $m^{n-m} | \prod_{i<j} (x_i-x_j)$ ?
Consider graph $K_n$ of $V=\{1,2,\ldots,n\}$ veritices. Each vertex $v$ has color from $\{0,1,\ldots,m-1\}$ $m$-element set and this color equals $\chi(v) := x_v \bmod m$.
From above lemma we have at least $n-m$ egdes with both vertices of the same color, so there are at least $n-m$ pairs: $x_{i_1}-x_{j_1}, \ldots, x_{i_{n-m}}-x_{j_{n-m}}$, such that $m | x_{i_k}-x_{j_k}$, hence $m^{n-m} | \prod_{i<j} (x_i-x_j)$. This is true for all $m = i-j$, so we have $\prod_{i<j} (i-j) | \prod_{i<j} (x_i-x_j)$.


Set of Points where X Fails to be Locally Connected I am stuck on a problem!  Suppose $X$ is a compact, connected metric space.  Let $L(X)$ be the set of points at which $X$ is not locally connected (here, locally connected means the point has a local basis of connected, open neighborhoods).  We call the weaker notion that $x$ has a neighborhood basis of connected - but not necessarily open - sets connected im kleinen, or cik.

Proposition: $L(X)$ has no isolated points.

I can't prove it.  It's false for compact metric spaces: Just take a sequence of points converging to $0$.  So the connectedness is necessary.  It is known that the set of points where $X$ fails to be cik has no isolated points.

Anyone have an idea, or a reference?

Got it.  If anyone sees a problem in the proof, let me know!  Let $N(X)$ be the set of points where $X$ is not cik.

Since the set of points where $X$ is cik has no isolated points and $N(X) \subset L(X)$, if $x \in L(X)$ is an isolated point then $X$ must be cik at $x$.  Let $V_n$ be a neighborhood basis (not necessarily open) of connected sets for $x$.  Since $x$ is isolated, it has a neighborhood $U$ such that $X$ is locally connected at every other point in $U$; pick an index $n$ so that $V_n \subset \overline{V_n} \subset U$.  For fixed $V_n$ each point of $\partial(V_n)$ has an open connected neighborhood contained in $W_\alpha \subset U$.

Then the union $Z_n$ of $V_n$ and these neighborhoods is connected, and it's open since $V_n = \text{int}(V_n) \cup \partial{V_n}$, i.e. it can be written as the union of the open $W_\alpha$ and $\text{int}(V_n)$.  Thus $Z_n \subset U$ is an open, connected neighborhood of $x$.  But $U$ was an arbitrary neighborhood of $x$, proving the proposition.

I think the first sentence of your proof is flawed. The rest is likely flawed as well, since you don't really use the assumption that $X$ is a continuum.

Suppose $p$ is an isolated point of $L(X)$.

Then there is an open set $U$ such that $p\in U$, $X$ is locally connected at each point in $U\setminus \{p\}$, and $p$ has no connected open neighborhood in $U$.

Since the component of $p$ in $U$ is not open, $p$ has no connected neighborhood in $U$ (so $X$ is not cik at $p$).

Now you want to just follow the proof of Theorem 3.4 in this paper.  The author uses weakly locally connected instead of cik.  The theorem shows that there is a continuum $C$ such that $p\in C\subseteq L(X)$ and $|C|>1$. Clearly $p$ is not an isolated point of $C$, so $p$ is not an isolated point of $L(X)$.

The basic idea is that in $\overline U$ there must be a sequence components $C_n$ such that $p\in \overline{\bigcup_{n<\omega} C_n}$ and $C_n\cap \partial U\neq\varnothing$ for each $n<\omega$. The limit of this sequence of continua is a continuum $C$ that contains $p$ and intersects $\partial U$. If $x\in C\cap U\setminus \{p\}$, then $x$ has no connected neighborhood in $U$ (the neighborhood would meet some $C_n$ and contradict its maximality).


Measure with uncountable union of sets I have a question about one of the properties included in the measure definition:

$\mu(\cup_{k=1}^\infty E_k)=\sum_{k=1}^\infty \mu(E_k)$, with $\{E_k\}_{k=1}^{\infty}$ being pairwise disjoint sets.

If we replace this condition with:

$\mu(\cup_{\alpha \in A} E_{\alpha})=\sum_{\alpha \in A} \mu(E_{\alpha})$, where $A$ is some uncountable set of indicators.

Then is it possible that the whole power set of $\mathbb{R}$: $P(\mathbb{R})$ could be measurable (we should also change the condition in the $\sigma$-algebra definition)? Is it even possible to define the right side of the new condition (uncountable sum)?

It's not clear to me whether you're talking about Lebesgue measure or an arbitrary measure.
Lebesgue Measure on $\Bbb R.$
Lebesgue measure on $\Bbb R$ is not uncountably additive. Say $S_x=\{x\}$. Then $\bigcup_xS_x=\Bbb R$, but $$\sum_xm(S_x)=0\ne\infty=m(\Bbb R).$$
On the Other Hand,
let $\mu$ be counting measure on $\Bbb R$. Then $\mu$ is defined on the entire power set, and it's easy to see that it's additive for disjoint unions of any cardinality.

Let us define $\mu:\mathscr{E} \to [0,+\infty]$ where $\mathscr{E} \subset \mathcal{P}(X)$.

Recalling the following:

Lemma: If $(a_i)_{i \in I} \subset [0,+\infty[$ and $\sum_{i \in I} a_i < +\infty$ then only a many countable $a_i$ are strictly positive.

We observe that there is something that does not work, in general, if we assume $\mu(\cup_{\alpha \in A} E_{\alpha})=\sum_{\alpha \in A} \mu(E_{\alpha})$ for any $A$ with arbitrary cardinality.

Whenever $\mu(\cup_{\alpha \in A} E_{\alpha}) < +\infty$ then only many countable of $E_{\alpha}$ could have positive measure. This can in general produce contradictions.


What's the diameter of a circle suppose to be if What's the diameter of a circle suppose to be, if the area is suppose to be same as equilateral of triangle with the circumference of 3meters?

Answer: 74cm

No idea how I am suppose to start.

The area of a triangle is $\frac12 \cdot b \cdot h$ We know that each side of the triangle is one meter because it is an equilateral triangle, and each angle is 60°. A height must connect a vertex to the base so that it is perpendicular to the base. That means that the height bisects the vertex angle, so you have a 30-60-90 triangle. The sides of a 30-60-90 triangle are in a ratio of $x:x\sqrt3:2x$ with x opposite the 30° angle, $x\sqrt3$ opposite the 60° angle and 2x opposite the right angle, so the height is $\frac{2}{\sqrt3}$ times the hypotenuse, or 1.15 meters. You can then find the area of the triangle by plugging the height and base into the area formula. Next, we must find the radius of the circle using the area. The area of a circle is expressed as $A=\pi \cdot r^2$ so we can substitute the area of the triangle for A and solve for r. It should be $\sqrt{\frac{A}{\pi}}$. Multiply that by 2 to get the diameter in meters and convert it to centimeters by dividing be 100.

Equate the area of the circle and the area of the equilateral triangle. You can easily obtain the radius of the circle from there and hence, the diameter.


If $F$ is closed and $x \not\in F$, show that there are disjoint open sets $U,V$ such that $x \in U$ and $F \subset V$ If $F$ is closed and $x \not\in F$,

(a) show that there are disjoint open sets $U,V$ such that $x \in U$ and $F \subset V$.

(b) Can $U$ and $V$ be chosen such that their closures $\bar{U} , \bar{V}$are disjoint as well ?

Attempt:
Assume $d(x,F) = t$

Consider the function $d(y,F)$ which is continuous, so the set $A = \{y | d(x,F) > t/2\}$ is open and contains $x$, similarly the set $B = \{y | d(x,F) < t/2\}$ is also open and contains F, both A and B are disjoint hence the required sets.

Can I consider the closure of the A and B for the second part (b) of the question?

One missing piece in this proof: You need to know that $t>0$ in order to prove that $A,B$ are open. So, for that purpose, you need to prove that $t>0$.

Regarding part (b), the closure of $A$ and $B$ might, by bad luck, contain the same point $z$, such that $d(z,F)=t/2$. However, you should be able to rewrite your proof to avoid that bad situation and obtain disjoint closures.

Let r =d(x,F)/100
Now we define U=B(x,r) and V=UB(a,r)
a€F
Clearly, x€U and F is contained in V.
Then we claim that U and V are disjoint.
If not, then let there exists z€UnV.
=> d(x,z)<r and d(z,y) < r
⇒ d(x,y) < 2r= d(x,F)/50,which is a contradiction .
Hence our claim "U and V are disjoint " is true. Q.E.D


Is the dual of a finitely generated module finitely generated? I recently thought of this and have no idea whether over a general commutative unitary ring the dual of a finitely generated module is finitely generated. This must be known.

If $I$ is an ideal of $R$, the dual of $R/I$ is isomorphic to $\mathrm{Ann}(I) = \{r \in R : rI = 0\}$, and this doesn't have to be finitely generated. Take for instance $R = k[y,x_1,x_2,\dotsc]/(y x_i : i \geq 1)$ and $I=(y)$.

A more natural question would be: How can we characterize commutative rings with the property that duals of f.g. modules over that ring are f.g.? Even more natural: How can we characterize commutative rings with the property that hom modules between f.g. modules are f.g.? For example, noetherian commutative rings satisfy this property (see the comment by Keenan Kidwell). But I think that there are more examples (perhaps coherent rings?).

Given a finitely generated module $B$, then $B^* = \text{Hom}_R(B,R)$ is a finitely generated module.

Indeed, take a module $B'$, we have a $n$ and $\varphi:B^n\rightarrow B'$ a surjective morphism.

Now for $B^*$. Note first that

$$(B^*)^n = \text{Hom}_R(B,R)^n = \text{Hom}_R(B^n,R) = (B^n)^*$$

To construct our surjective morphism, we take $\{e_i\}_{i\in I}$ a basis of $B^n$. This gives us $\{e_i^*\}_{i\in I}$ where $e_i^*(e_j) = \delta_{ij}$. This set can be completed into a basis denoted $\{e_j^*\}_{j\in J}$ since it is a linearily independent set. We then define our surjective morhism as follows
            \begin{align*}
             \varphi' &:(B^*)^n = (B^n)^*\rightarrow B' \\ &:\sum_{j\in J} e_j^*r_j\mapsto \sum_{i\in I} \varphi(e_i^*)r_i
         \end{align*}

i.e. $\varphi'$ is a composition of a projection on the submodule generated by the dual elements of the original base of $B^n$ and of $\varphi$

NB: I think this is correct but not sure


Nullity of linear transformation I'm struggling to find the nullity $N(T)$ of the following linear transformation (in the canonical basis of $\mathbb{R^{2\times2}}$

$
M = \begin{bmatrix}
       0 & 0 & 0 & 0           \\[0.3em]
       -1 & -3 & 3 & 0  \\[0.3em]
       0 & 0 & 1 & 0  \\[0.3em]
       0 & 0 & 0 & 1
     \end{bmatrix}
$

What's making me feel confused is that we are considering a basis of matrices. I knew how proceed in the case of for example the canonical basis of $\mathbb{R^2}$ (just multiply by a column vector (a,b,c,d) and the I would have a system of 4 linear equations (equal to zero)). But the presence of matrices 2x2 is driving me confused because I can't even try to multiply my matrix of the linear transformation to a matrix 2x2

Am I making myself clear? If you could help me, please...

Thanks

Hint:

A matrix
$$
A=
\begin{bmatrix}
a&b\\c&d
\end{bmatrix}
$$
is represented in the canonical basis as:
$$
A=a\begin{bmatrix}
1&0\\0&0
\end{bmatrix}
+b\begin{bmatrix}
0&1\\0&0
\end{bmatrix}
+c\begin{bmatrix}
0&0\\1&0
\end{bmatrix}
+d\begin{bmatrix}
0&0\\0&1
\end{bmatrix}
$$
so,  in this basis, you can represent it as a vector:
$$
\vec A=
\begin{bmatrix}
a\\b\\c\\d
\end{bmatrix}
$$
So your matrix $M$ acts as:
$$
M\vec A=
\begin{bmatrix}
       0 & 0 & 0 & 0   \\
       -1 & -3 & 3 & 0  \\
       0 & 0 & 1 & 0  \\
       0 & 0 & 0 & 1
     \end{bmatrix}\begin{bmatrix}
a\\b\\c\\d
\end{bmatrix}
$$
and you have to solve :
$$
\begin{bmatrix}
       0 & 0 & 0 & 0   \\
       -1 & -3 & 3 & 0  \\
       0 & 0 & 1 & 0  \\
       0 & 0 & 0 & 1
     \end{bmatrix}\begin{bmatrix}
a\\b\\c\\d
\end{bmatrix}=0
$$

If $\mathbf{x}=(x,y,z,w)$, then right away you should see that $z=w=0$ and $x=-3y$. So the nullity has basis: $(-3,1,0,0)$ and is of dimension $1$.

If $\mathbf{x}$ is really a $2\times 2$ matrix, then you can think of it as $\left[\begin{matrix} x & y \\ z & w\end{matrix}\right]$. Converting between a matrix and its representation as a vector should be clear.


A question about probability I have met a interesting question:

  If today rains, the probability that tomorrow rains is $0.6.$ If today doesn't rain, the probability that tomorrow rains is $0.2.$ Given Tuesday rained, what's the probability that Monday rained?

I have no idea how to solve this.

If I make the question a bit more complicated:
Given Tuesday rain, what's the probability that the Sunday just before rained?

If there is some constant probability $p$ of rain on any given day, in the absence of information about any other day's rainfall, then we have $p = 0.6p + 0.2(1-p)$, which can be solved to give $p=1/3$.

So, probability of rain on Monday and Tuesday (if we don't already know that it rained on Tuesday) is $p \cdot 0.6 = 1/5$.  Probability of rain on Tuesday but not Monday (if we don't aleady know that it rained on Tuesday) is $(1-p) \cdot 0.2 = 2/15$.

But these are the cases when we KNOW that it rained on Tuesday; so the probability that there's rain on Monday given that it rained on Tuesday is $\dfrac{1/5}{1/5+2/15} = \dfrac{3}{5}$.

However, it's possible that there is no such constant probability $p$.  In this case, as time goes on, the probability of rain from day to day will approach 1/3 in the limit.  This is only an issue if the world has existed for a finite number of days.  In this case, though, there is not enough information to solve the problem.

Let, $P(M)$ be the probability that it rains on monday, $P(T)$ be the probability that it rains on the next day which is tuesday

Let, $P(T/M)$ be the probability that it rains on tuesday given that it rained the day before, which is monday, $P(M \cap T)$ be the probability that it rains on two consecutive days, monday and tuesday

$P(T/M)=0.6$

$P(T/\overline{M})=0.2$

$P(T/M)$ and $P(T/\overline{M})$ are mutually exclusive events

i.e. $P((T/M) \cup (T/\overline{M})) = P(T/M) + P(T/\overline{M}) = 0.8$

i.e. $P(T) = 0.8$

we have from conditional probability,

$P(M \cap T) = {P(M)}{P(T/M)} = P(M)(0.6)$ -------(1)

we also have,

$P(M \cup T) = P(M) + P(T) - P(M \cap T)$ -------(2)

using equations (1) & (2), we have,

$P(M \cup T) = P(M) + 0.8 - P(M)(0.6)$

$P(M) = \frac{5}{2}P(M \cup T) - 2$ -------(3)

we, know that the probability of any event is $\le 1$

set, $P(M \cup T) \le 1$ in equation (3) to get $P(M) \le \frac{1}{2}$

i.e. $$0 \le P(M) \le \frac{1}{2}$$


How to solve $(a-x)^{1.4}-bx=0$? I'm trying to solve the following equation algebraically :
$$(1.722-x)^{1.4}-0.565x=0$$

I can find the solution using Matlab using symbolic function solve or an approximation using Taylor series-
the answer is $1.039.$

I have thought of starting from this one $$(1.722-x)^7-(.565x)^5=0$$ but it turns out that there is no algebraic solutions of polynomial equations of degree $\ge5$ by the Abel–Ruffini Theorem.

Is there a way of doing this ? How ? If not, how do I know ?

Another sub-question that came to me while doing my searches is:

Why is there only one solution provided by Matlab whereas $$(1.722-x)^7-(.565x)^5=0$$ should have $7$ solutions as the equation is of degree $7$?

What about the Abel–Ruffini theorem with an equation of the form $ax^6+bx^3+c$ ? I can solve it using $y=x^3$ and therefore contradicts the Theorem. I suppose I have misinterpreted it.

In integers, your equation is:

$$(1722 - 1000x)^7 - 10^6 (565x)^5 = 0$$

Or:

$$2\cdot(861-500x)^7 - 5^7\cdot 113\cdot x^5 = 0$$

The way to determine if this has a solution in radicals is to calculate the Galois group, which is a subgroup of $S_7$, and determine if it is solvable.  If so, then there is a solution in radicals.

You can do this calculation with the online calculator for Magma:

P<x>:=PolynomialRing(Rationals());

GaloisGroup(2*(861-500*x)^7 - 5^7 * 113 * x^5);

This returns a group of order $5040$.  In other words, the Galois group is exactly $S_7$, and so there is no solution in radicals.  There may be solutions in terms of other functions, but this is a question of much broader scope.

Incidentally, there are indeed seven distinct roots of this equation, but six of them are complex.


On the complete rational solutions of $x^3+y^3+z^3=1$ I have to check that the rational solutions of $x^3+y^3+z^3=1$ are given by giving rationals values to $(s,t)$ at the formulas:

$$x(s,t)=\frac{3t-\frac{1}{3}(s^2+st+t^2)^2}{t(s^2+st+t^2)-3}$$

$$y(s,t)=\frac{3s+3t+\frac{1}{3}(s^2+st+t^2)^2}{t(s^2+st+t^2)-3}$$

$$z(s,t)=\frac{-3-(s^2+st+t^2)(s+t)}{t(s^2+st+t^2)-3}$$

After some dirty work, I can see that these expressions for $x,y,z$ satisfy the equation $x^3+y^3+z^3=1$, but I still have to see that all the rational solutions are of that form, and I don't know how to approach this (it is necessary, because there could be other solutions that can't be expressed as above, and then these formulas wouldn't be a complete solution).

Previously to this, I have found explicit expressions for the rational solutions of $x^2+y^2+z^2=1$ and the more general equation $x_1^2+...+x_n^2=1$, both through the stereographic projection. I have tried to do something similar with the cubic equation but it doesn't seem to work. I tried to transform it to the form $x^3+y^3+z^3=t^3$ and find the integer solutions to this, but I couldn't and I hardly found information about it on internet, and the little I've found doesn't match with the family of solutions I am working with (which is the one I'm interested in)

I genuinely don't know how to approach this, so I would appreciate any idea. Thanks.

Above equation shown below:

$x^3+y^3+z^3=1$  -----$(1)$

Solution shown by (OP) is not a general solution. Hence he can refer to paper by
Mr. Ajai Choudhry who arrived at a general solution in his paper shown at the link below,

https://projecteuclid.org/euclid.rmjm/1181071714#export

Above eqn. (1) has solution in the said paper & is shown below:

$dp = c(-a^3- b^3+ c^3 )$

$dq = (-a^4+ 2a^3 b- 3a^2 b^2+ 2ab^3- b^4 )+ (a+ b) c^3$

$dr = (a^4- 2a^3 b + 3a^2 b^2- 2ab^3+ b^4 )+ (2a- b) c^3$

$dt = c(a^3+(a-b)^3+ c^3 )$

Condition :   $ c^3 > (a^3+ b^3) $  and 'd' is a constant of multiplication

And $(x,y,z)=(p/t,q/t,r/t)$

For $(a,b,c,d)=(2,1,3,1)$ we get:$(x,y,z)=(-11/93,30/93,92/93)$

Re: Adam Bailey & Ai Yagami. I have corrected the typos that you'll have mentioned.

Regarding query by Ai Yagami about eqn.

$x^3+y^3+z^3=1$ -----(1)

Since $(x,y,z)=(3/6,4/6,5/6)$ is a solution to equation (1), than substituting the numerical solution in equation given by (OP) & solving for (s & t) we get the below mentioned equation:

$3=((s^2+st+t^2)(s+3t))/(s+2t+1)$ ----(2)

Above eqn. (2) has solution $(s, t)=(1,1)$

But, equation given by (OP) is not satisfied at (s,t)=(1,1). Hence it is a contradiction.


Calculating $\arg(-1+\sqrt 3 \cdot i)$ $$\arg(-1+\sqrt 3 \cdot i)$$

My try:

$$\arg=\arctan\bigg(\frac{\sqrt 3}{-1} \bigg)=-\frac{\pi}{3}$$

But the answer should be $\Large \frac{2 \pi}{3}$

Notice that the point $(-1, \sqrt 3\  i)$ lies in the second quadrant of complex plane hence its argument is given as $$\arg=\pi-\tan^{-1}\left|\frac{\sqrt 3}{-1}\right|$$
$$=\pi-\tan^{-1}(\sqrt 3)$$$$=\pi-\frac{\pi}{3}=\color{red}{\frac{2\pi}{3}}$$

  Edited:
  $$\color{red}{\arg(x+iy)}=\begin{cases} \color{blue}{\tan^{-1}\left|\frac{y}{x}\right|}\ \ \forall\ \  x>0, y>0\ \text{(point (x, y) lies in first quadrant)}\\
 \color{blue}{\pi-\tan^{-1}\left|\frac{y}{x}\right|}\ \ \forall\ \  x<0, y>0\ \text{(point (x, y) lies in second quadrant)}\\
\color{blue}{-\pi+\tan^{-1}\left|\frac{y}{x}\right|}\ \ \forall\ \  x<0, y<0\ \text{(point (x, y) lies in third quadrant)}\\
\color{blue}{-\tan^{-1}\left|\frac{y}{x}\right|}\ \ \forall\ \  x>0, y<0\ \text{(point (x, y) lies in fourth quadrant)}
\end{cases}$$

Your attempt is right the change in answer is because the argument of a complex number is defined as the angle the complex number makes with positive x axis in anticlockwise direction so u should manipulate the answer and find the principle value of arctan of the $\mod(y/x)$  in the question . I hope my answer would be of some help


Regular functions and morphisms of varieties for field not algebraically closed Let us work over a field that is NOT algebraically closed. Specifically, $K=\mathbb R$.
We can think of the affine line $\mathbb A_K^1$ as a ringed space $(\mathbb A_K^1,\mathcal O_{\mathbb A^1_K})$, where for an open set $U\subseteq\mathbb A^1_K$, we have $\mathcal O_{\mathbb A^1_K}(U)=(\text{the ring of regular functions over $U$})=\{\frac fg\mid f,g\in K[x] \,\&\,(\forall P\in\mathbb A^1_K)(g(P)\ne 0)\}$.
I think that according to this, $\frac1{1+x^2}$ should be a regular function over the entire $\mathbb A^1_K$. Moreover, it seems to me that $f:\mathbb A_K^1\to\mathbb A_K^1$ defined as $f(x)=\frac1{1+x^2}$ is a morphism from the ringed space $(\mathbb A^1_K,\mathcal O_{\mathbb A^1_K})$ to itself (in the sense that it pulls back regular functions to regular functions). Am I right?
I am probably confused by either different sources using different definitions or me seeing statements that only hold for $K$ algebraically closed. E.g. the Wikipedia article Morphism of algebraic varieties states

The regular functions on $\boldsymbol A^n$ are exactly the polynomials in $n$ variables [...]

,
which I only agree with for $K$ algebraically closed and not in the case I've shown above.

The definition of a regular map needs to be modified substantially when you work over a field that isn't algebraically closed. There are different ways to set things up at different levels of sophistication; one way to say it is that regular maps should have the property that they remain regular maps after extension of scalars, and your proposed map does not remain regular after extension of scalars to $\mathbb{C}$ because of the poles at $x = \pm i$.
You are implicitly thinking only in terms of what are called the $K$-points of a variety; this is only safe to do when $K$ is algebraically closed. For example if $K = \mathbb{F}_q$ is a finite field then the polynomial $f(x) = x^q - x$ vanishes on $K$-points but is not considered to be identically zero as a regular map, because it does not remain identically zero after extension of scalars, e.g. to the algebraic closure $\overline{\mathbb{F}_q}$. We also have issues such as varieties, for example $\text{Spec } \mathbb{R}[x, y]/(x^2 + y^2 + 1)$, which have no $K$-points (here $K = \mathbb{R}$) but which it would be incorrect to consider empty; here again the issue is that points exist after extension of scalars.
Here I am implicitly taking a functor of points approach which I personally find conceptually cleanest. The standard approach you will find in most textbooks is based on the Zariski spectrum of all prime ideals (which capture points over extensions of scalars but somewhat indirectly), not just maximal ideals corresponding to $K$-points.
On the other hand, there is a somewhat separate field of real algebraic geometry in which we only consider $\mathbb{R}$-points, and which is mostly not studied by algebraic geometers but by model theorists or something like that.

If you define $\mathcal{O}(\mathbb{A}^1_{\mathbb{R}})$ as the set of functions with the following property (P1):
P1. $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ with the property that for any point $p\in \mathbb{R}^2$ there is an open subset $U(p)\subseteq \mathbb{R}^2$ and two polynomials $p(x),q(x)\in \mathbb{R}[x]$ such that the restriction $f_{U(p)}$ of $f$ to $U(p)$ satisfies $f_{U(p)}=\frac{p(x)}{q(x)}$,
then it follows (as you noted) the function $\frac{1}{x^2+1}$ is a regular function on $\mathbb{R}^2$. Hence $\mathcal{O}(\mathbb{A}^1_{\mathbb{R}})\neq \mathbb{R}[x]$.  If $z=a+ib \in \mathbb{C}$ with $b\neq 0$, it follows
the polynomial $p_z(x):=(x-z)(x-\overline{z})=x^2-2ax+a^2+b^2\in \mathbb{R}[x]$ is an irreducible polynomial - it has no real roots.
Hence if you use definition P1 to define the ring of regular functions, it follows the function $f(x):=\frac{1}{p_z(x)}$ is a regular function. If $S\subseteq A:=\mathbb{R}[x]$ is the set of polynomials on the form $p(x):=\prod_{z_i\in \mathbb{C}-\mathbb{R}} (x-z_i)(x-\overline{z_i})$ it follows $S$ is a multiplicatively closed subset, and the ring of fractions $S^{-1}A$ is a sub-ring of $\mathcal{O}(\mathbb{A}^1_{\mathbb{R}})$. Hence with definition P1 it follows the ring $\mathcal{O}(\mathbb{A}^1_{\mathbb{R}})$ is much larger than $A$.
Question: "The regular functions on $\mathbb{A}^n$ are exactly the polynomials in n variables [...], which I only agree with for K
algebraically closed and not in the case I've shown above."
Answer: You want (for many reasons) to recover the polynomial ring $\mathbb{R}[x]$ as the ring of global sections of the structure sheaf. This problem is "solved" by the affine spectrum $Y:=Spec(\mathbb{R}[x])$. What you define above is a "sheaf of rings" on the real algebraic variety $\mathbb{A}^1_{\mathbb{R}}$ in the sense of "real algebraic geometry", and this sheaf differ from the sheaf defined using the affine spectrum in the sense of Hartshornes book "Algebraic geometry". The topological space of the variety $X:=\mathbb{A}^1_K$ you consider above is the set of real numbers $r\in \mathbb{R}$. The topological space of $Y$ consists of all real numbers plus the set of all maximal ideals in $A:=\mathbb{R}[x]$ with residue field the complex numbers. Hence the two topological spaces differ. Your space $X$ is by definition the set of rational points $Y(K)$ with the induced topology.
Note: There is nothing "wrong" with the ringed space $(X, \mathcal{O}_X)$ - it has properties different from $Y$. In real algebraic geometry it follows projective space $\mathbb{P}^n_K$ may be realised as a real affine algebraic variety - there is a closed embedding into real affine space. Moreover any real projective variety is affine. The reason for this is that the structure sheaf has more local sections, as you have observed in your comment above. Hence there are "more maps". This does not happen with the Hartshorne definition: You cannot construct a closed embedding of $Proj(k[x_0,..,x_n])$ into
affine space $Spec(k[y_1,..,y_d])$.
As an example: The global section $s(x):=\frac{1}{p_z(x)}$ for $z\in\mathbb{C}-\mathbb{R}$ is a unit - the polynomial $p_z(x)$ is non-zero on $\mathbb{R}$, and with your definition of "regular map" it follows $s(x)$ is a global section - it lives in $\mathcal{O}_{\mathbb{A}^1_K}(\mathbb{A}^1_K)$. The function $s(x)$ is not a global section of $\mathcal{O}_Y(Y)\cong \mathbb{R}[x]$.
Real algebraic varieties are relevant for people studying complex algebraic varieties and their topology. You may for any complex projective manifold $X \subseteq \mathbb{P}^n_{\mathbb{C}}$ construct the Weil restriction $W_{\mathbb{C}/\mathbb{R}}(X)$. If $dim(X)=d$ it follows  $dim(W_{\mathbb{C}/\mathbb{R}}(X))=2d$ and $W_{\mathbb{C}/\mathbb{R}}(X)$ is a smooth scheme over $\mathbb{R}$ - the "underlying" real algebraic variety of $X$. The variety
$W_{\mathbb{C}/\mathbb{R}}(X)$ has the same topology as $X$ and you may realize $W_{\mathbb{C}/\mathbb{R}}(X)$ as a real affine algebraic variety. Hence the "underlying" real algebraic variety of $X$ is affine. Moreover: Any real affine algebraic variety may (topologically) be realized as a hypersurface. Hence if you study the topology of $X$ it follows there is a real affine hypersurface $Z(f)$ with the same topology as $X$ (here I am extremely "vague").
You find this discussed here:
https://math.stackexchange.com/posts/3957121/edit
and here:
Different definitions of regular map (between affine varieties)?


Inverse function theorem for partial derivatives of a vector function I have a simple vector function:

$$\mathbf{y} = a\mathbf{x}\qquad a\in\mathbb{R},\;\mathbf{x},\mathbf{y}\in\mathbb{R}^n$$

The inverse is obviously:

$$\mathbf{y}^{-1} = {1\over a}\mathbf{x}$$

The inverse function theorem (at least for scalars) states that:

$$\left[f^{-1}(y)\right]'=\frac{1}{f'(f^{-1}(y))}$$

But the partial derivatives of $\mathbf{y}$ and $\mathbf{y}^{-1}$ with respect to $a$ are not related that way:

$$\begin{align}(\mathbf{y})'_a &= \mathbf{x} \\ (\mathbf{y}^{-1})'_a &= -{1\over a^2}\mathbf{x} \end{align}$$

What is the relation between the above derivatives, then?

The derivative of a function $F:\mathbb{R}^n\to \mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix

$$J_F(x) = \begin{pmatrix}
    \dfrac{\partial F_1}{\partial x_1}(x) & \cdots & \dfrac{\partial F_1}{\partial x_n}(x)\\
    \vdots & \ddots & \vdots\\
    \dfrac{\partial F_n}{\partial x_1}(x) & \cdots & \dfrac{\partial F_n}{\partial x_n}(x) \end{pmatrix}$$

And the Jacobian of $F^{-1}$ at point x is the inverse of the Jacobian of $F$ at point x :

$$J_{F^{-1}}(x) = \left(J_F(x)\right)^{-1}$$

You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem

The derivative w.r.t. $a$ is

$${\mathbb{d}\mathbf{y}\over \mathbb{d}a} = \mathbf{x}$$

and the derivative of inverse is the reciproval of above, i.e.

$${\mathbb{d}a\over \mathbb{d}\mathbf{y}} = {1\over \mathbf{x}}$$

The notation caused confusion about what is the inverse function. The inverse is function of $a$.


Whether any even number can be written as sum of odd no of primes? Whether any even number can be written as sum of odd number of primes?(3,5,7.. primes) I know that Goldbach's conjecture state that a even number can be written as sum of two primes.

D=A+B+C+...+n such that where no of elements in the equation is a odd number and where A,B,C are prime numbers and D is even number

whether  the above one is true?

All positive integers except $1$ and $4$ can be written as the sum of an odd number of primes.

$6=2+2+2$ can be written as the sum of an odd number of primes.

$7=2+2+3$ can be written as the sum of an odd number of primes.

$8=2+3+3$ can be written as the sum of an odd number of primes.

$9=3+3+3$ can be written as the sum of an odd number of primes.

By adding an even number of $2$s to these, any larger number can be written as the sum of an odd number of primes.

Looking at smaller numbers, $2$, $3$ and $5$ are equal to themselves and so are equal to the sum of an odd number of primes, since one is odd, while $1$ and $4$ cannot be written as the sum of an odd number of primes.

If Goldbach's conjecture is true, then you can do this by letting one of the primes is 2.D should be larger than 5.


Long summation question, including sets I have a really long question I'm absolutely stuck on, I don't even know where to begin:

Given:

$n \in \mathbb{Z}, \geq 2$

let $S$ be the set of all nonempty subsets of {2,3,...,n}. For each $S_i \in S$ let $P_i$ be thje product of the elements of $S_i$ prove or disprove:

$\sum\limits_{i=1}^{2^{n-1}-1} P_i = \frac{(n+1)!}{2} -1$

HINT: You can prove it quite straightforwardly by induction on $n$ if you notice that the non-empty subsets of $\{2,\ldots,n,n+1\}$ are of three kinds:

non-empty subsets of $\{2,\ldots,n\}$;
$S\cup\{n+1\}$, where $S$ is a non-empty subset of $\{2,\ldots,n\}$; and
$\{n+1\}$.

For a set $S$, consider the product $\prod_{x \in S}(1+x)$. Each term in this expansion is the product of elements of one proper subset of $S$, except for the term $1$. Applying this to your set and simplifying gives the answer.


How did Newton find derivative of basic functions before formulating systematic Calculus? I think I read somewhere that Newton tried to find derivatives of basic functions like $x^2$ before formulating systematic calculus; how did/would he do it?

The earliest formulations of something that looks like calculus were not in terms of functions but of curves given by algebraic relations. So if we have
$$ y=x^2 $$
and $(y+p,x+q)$ is another point on the curve, we would also have
$$ (y+p) = (x+q)^2 $$
and if we multiply these equations out and subtract we get
$$ p = 2qx + q^2 $$

Now what happens in the earliest sources is that it is simply postulated, without any backing by theoretical definitions or limits that when $p$ and $q$ are small, we can remove their higher powers, so ignoring the $q^2$ term yields us
$$ p = 2qx \qquad\text{or, in other words,}\qquad p:q = 2x:1 $$
which can be used to draw a tangent.

The justification for this procedure was initially just that it worked in practice, but the unexplained ignoring of terms like $q^2$ drew quite a lot of contemporary criticism.

This was in the generation before Newton (chiefly Fermat and Descartes). They could produce painstaking geometric proofs in the Euclidean tradition (which was the gold standard for proofs in those days) that for each of the actual curves they considered what came out of this procedure was the right result -- but they didn't have the definitions and machinery to explain or prove rigorously why the procedure always works.

Newton brought in the new idea of a quantity $x$ that varied with time and its time rate of change $\dot x$, but he was still building on the "ignore higher powers of the increments" method. But he did it masterfully and was able to get more out of it than his predecessors did.

Surely any child with an interest in numbers will notice that in a quadratic sequence like 1, 4, 9, 16, 25, ..., the differences between successive terms form a linear sequence like 3, 5, 7, 9, ...? That doesn't need calculus; but someone with a sophisticated mathematical brain will start to recognise that there's a relationship between the function that generates the first sequence and the function that generates the second, and will develop such thinking to discover the principles of calculus.


Prove composition is a measurable function Let $(X, \Sigma )$ be a measurable space and $f:X \rightarrow \mathbb R$ a measurable function. If $h: \mathbb R \rightarrow \mathbb R $ is continuous, then prove that $h \circ f$ is a measurable function.

I unfortunately lost my lecture notes so I really need some help with this module (linear analysis).

Please can someone guide me on how to do this question.

Thanks

Since $h$ is continuous, $h^{-1}((-\infty ,a))$ is open and thus borelien. Since $f$ is mesurable and $h^{-1}((-\infty ,a))$ is borelien, $f^{-1}(h^{-1}((-\infty ,a)))$ is measurable. Therefore $h\circ f$ is measurable.

On an abstract level, the composition of two measurable maps $f:X\to Y$ and $g:Y\to Z$ is measurable; this is very easy to verify from the definition of a measurable map.

The only remaining step is to prove, as Rick did, that continuous real maps are measurable with respect to the Borel sigma-algebra.


How can I show that $ \cos^{-1}(x)=-\int\frac{dx}{\sqrt{1-x^2}}+c? $ How can I show that
  $$
\cos^{-1}(x)=-\int\frac{dx}{\sqrt{1-x^2}}+c?
$$

One has: $$\forall x\in]-1,1[,(\cos\circ\arccos)(x)=x.$$
Therefore, one gets: $$\forall x\in ]-1,1[,-\arccos'(x)\sin(\arccos(x))=1.$$
For all $y\in\mathbb{R}$, $\sin^2(y)=1-\cos^2(y)$, using this identity for $y=\arccos(x)$ with $x\in ]-1,1[$ and noticing that $\sin\geqslant 0$ on $]0,\pi[\ni\arccos(x)$, one has: $$\sin(\arccos(x))=\sqrt{1-x^2}.$$
Finally, one has: $$\forall x\in]-1,1[,\arccos'(x)=-\frac{1}{\sqrt{1-x^2}}.$$

Just differentiate the function $x\mapsto \cos(\cos^{-1}(x)) = x$

Use the chain rule $(f\circ g)' = g'.f'\circ g.$


How to find the area of the rectangular region 2 ≤ x ≤ 5, -1 ≤ y ≤ 3 How to find the area of the rectangular region $2 ≤ x ≤ 5$,  $-1  ≤ y ≤ 3$.

I tried to plot the graph in $xy$-plane, but I'm not sure how to find the area.

Area of a rectangle is given by base$\times$height. The base of this rectangle is the change in $x=|5-2|=3$ and the height of the rectangle is the change in $y=|3-(-1)|=4$. The area is then
$$3\times4=12$$

The area of a rectangle of length $L \times l$ is $L\cdot l$.

One of the lengh $L_x$ is equal to the interval of variation of x: $L_x = 5-2 = 3$.

Same goes for $L_y = 3 - (-1) = 3 + 1 = 4$

The area is $3\cdot 4 = 12$.


Finding Second Derivative using implicit differentiation Find $y'' of$

$9x^2 + y^2 = 7$

Work

I've been working on this one for a while now, I'm not sure where I'm going wrong. I posted an image of my work, I'm hoping someone can take a look at it and give me some feedback

Notice, We have $$9x^2+y^2=7$$
differentiating both the sides w.r.t. $x$, we get
$$\frac{d}{dx}(9x^2+y^2)=\dfrac{d}{dx}(7)$$
$$18x+2y\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=\frac{-9x}{y}\tag 1$$
Differentiating (1) w.r.t. $x$, we get
$$\frac{d^2y}{dx^2}=\frac{y(-9)+9x\frac{dy}{dx}}{y^2}$$
setting the value of $\frac{dy}{dx}$ from (1), we get
$$\frac{d^2y}{dx^2}=\frac{y(-9)+9x\frac{(-9x)}{y}}{y^2}$$
$$\color{red}{y''=\frac{d^2y}{dx^2}=-\frac{9(9x^2+y^2)}{y^3}}$$


Proof about Countability and Induction Let n be a positive integer. Using the fact that the union of two countable sets is countable, use induction to prove that the union of n countable sets is countable.

I did the following.

**We begin by induction on n.
Suppose n = 1.

Then we have 1 set and by above, it is countable.

Suppose the union of k sets is countable, for some integer k.

We must show that the union of k+1 sets is also countable.**

From here I am not sure how to "Prove" this without just flat out saying that these would also be countable  by the theorem.

You're right in that it's pretty straightforward. Nevertheless, you have to be careful that you don't make illogical jumps.

You have the base case right. You have 1 countable set and so it is trivially countable.

Now consider the union of $k$ countable sets $S_1, S_2, \ldots S_k$ and assume this union is countable. Let us call this union $A$ i.e.

$$A = \bigcup_{i=1}^{k}{S_i}$$.

Now consider another countable set $S_{k+1}$. We want to prove that $\bigcup_{i=1}^{k+1}S_i$ is also countable. Well we have:

$$
\bigcup_{i=1}^{k+1}S_i = \left(\bigcup_{i=1}^{k}S_i \right) \cup S_{k+1} \\
= A \cup S_{k+1}
$$

From the induction hypothesis, we know $A$ is countable and $S_{k+1}$ is countable by assumption, and by our rule, we know the union of two countable sets if countable. Since the union of $S_1, S_2, \dots S_{k+1}$ is equal to the union of the countable set $A$ and the countable set $S_{k+1}$, we can conclude that $\bigcup_{i=1}^{k+1}S_i$ is countable.

Note we had to be careful in that we cant just say, the $k+1$ sets are countable because the union of the first $k$ are. Your rule says only the union of two countable sets is countable so you need to make that step of calling the union of the first $k$ a set $A$ and then taking its union with $S_{K+1}$

By definition, a set, $A$, is countable if there is a function $A\rightarrow\mathbb{N}$ such that every $n\in\mathbb{N}$ has finitely many preimages. (This means that for every $n\in\mathbb{N}$ there are only finitely many $a\in A$ with $f(a)=n$.) This is because $A_n$ can be defined to be $f(a)=n$, and then we have expressed A as a countable union of finite sets.

To prove that the set of all finite subsets of $\mathbb{N}$ is countable, we may consider $f(A)=\max A$. The number of sets with maximal element $n$ is finite ($2^{n-1}$), so we may conclude that $A+a$ is indeed countable, given both $A$ and $a$ have countable maximums.

If we let $k=A$ and $1=a$, then we have proven that $k+1$ is countable, because $A+a$ is countable.


Prove that a DE solution does not exist after a certain time I am unable to begin/know what do in order to show the following:

  Consider the one dimensional system
  $$
\dot x(t)=-\text{sgn}(x(t))=\begin{cases}
-1 & \text{if } x(t)\ge 0 \\
1 & \text{if } x(t)<0
\end{cases}
$$
  For the initial condition $(t_0,x_0)=(0,c)$ with $c>0$, we can define the solution $\varphi(t)=c-t$ for $t\le c$. Show that there does not exist a piecewise continuous solution, however, for $t>c$.

Any help/proof would be appreciated!

Assume $\varphi$ exists for $t > c$. Put $\psi := - \varphi$, then $\psi$ is a solution of $x' = \text{sgn}$. Now consider the sum of $\varphi$ and $\psi$ and their derivative in $t = c$...


$2^{49}$ ways to choose a set of integers $\leq 50$ with odd sum Show that the number of ways one can choose a set of distinct positive integers, each smaller than or equal to $50$, such that their sum is odd, is $2^{49}$.

My attempt: Suppose set $A=\{1,2,3,...,50\}$. I need to find the number of subsets $S\subset A$ where sum of elements of $S$ is odd. There are a total $2^{50}$ subsets of $A$, including empty subset $\phi$ (with sum of elements $=0$). How can I prove that exactly half of these subsets have sum of their elements odd?

Sum of the elements of $A \text{ is }1+2+...+50 = 1275$, an odd number. Thus if I split $A$ into two subsets, one of them will always have odd sum, and the other one even!

[Sorry for answering my own question, I just came up with this after posting the question]

$A=(1,2,・・・,50)=(1,2,1,2,・・・,1,2)$ $mod 2$

First 1 or 0 changes even or odd.Next 2 or 0 doesn't changes.・・・ so problem is even if all are odd numbers$(1,3,5,・・・49),$ their sum is $2^{n-1}$


Puzzle About Cubes (from the book thinking mathematically) I want to confirm my solution to the given problem (solutions were not available in the book)

  I have eight cubes. Two of them are painted red, two white, two blue
  and two yellow, but otherwise they are indistinguishable. I wish to
  assemble them into one large cube with each color appearing on each
  face. In how many different ways can I assemble the cube?

The answer i got was 96 is this correct?
I also tried to generalize
    the question such that given $n$ cubes and $\sqrt n$ number of colors, I came up with

  $$n!\cdot2(n-\sqrt{n})!\cdot(n-2\sqrt{n})!\cdot2(n-(4\sqrt{n}-4))!$$

is the above generalization correct?

Thanks

EDIT

well to further explain my question, the rational behind the answer to the 1st question was we have 6 faces and if we take one of the faces we have 4! possibilities to arrange the 4 colors. that gives us 24 possibilities. and if move to the other faces, we have 2 faces of 2! possibilities and 3 faces of 1! possibilities thus total arrangements are 4!x2!x2!x1!x1!x1! = 96

for the second question yes the colors should be n^1/3 but for the generalization n should be  no of colors.

for example if we take the current question

n=4

if we take the front face as the base face we have n! = 24
if we take the left face because of step 1 we have $(n-\sqrt n)! =2$
if we take the back face because of step 2 we again have $(n-\sqrt
    n)! =2$
if we take the right face because of step 1 and step 3 we have
$(n-2\sqrt n)! =1$
if we take the top face because of step 1,2,3,4  we have $(n-(4\sqrt
    n -4))! =1$
if we take the bottom face again because of step 1,2,3,4  we have
$(n-(4\sqrt n -4))! =1$

thus the result 4!x2!x2!x1!x1!x1! = 96

If you are allowed to rotate the assembled cube freely there is exactly one way to assemble it.

Proof. Let $\{0,1,2,3\}$ be the set of colors. Color each vertex of the assembled $2$-cube with the color of the $1$-cube it belongs to.  Vertices of the assembled cube having the same color are necessarily space-diagonally opposite. Assume that a vertex having color $0$ is facing you. Then  its three adjacent vertices have colors $1$, $2$, $3$, arranged either counterclockwise or clockwise. Now convince yourself of the following: If $1$, $2$, $3$ are arranged counterclockwise, then $1$, $2$, $3$ will be arranged clockwise if you look at the assembled cube with the opposite $0$-vertex facing you.$\qquad\square$

It is so far unclear how this problem can be meaningfully formulated for an $n\times n\times n$ assembled cube.

Yes think like this to make a cube max side is $2$ as $2^3=8$ qnd we have $8$ eight cubes. Now take tge base . The four cubes can be arranged in $4!$ ways in the base. Now as we want each colour on each face   $2$ cubes with colour different than base colour cubes can be arranged on top of two base cubes maybe row or column. So now they can be arranged in $2$ ways and same can be done with the cubes in other row or column .so these are $2$ ways. Hence total ways are $4!.2.2=96$ and may be number of colours in generalized formula should be $n^{1/3}$


Find function $ f(x) $ to ensure the limit has certain value If $\lim_{x\to1}$  $\frac{f(x)}{(x-1)(x-2)} = -3$ , then provide a possible function $y = f(x)$
*I don't understand what the question is asking me and how I should solve it. Can a possible function be $y = f(1)$? Or must I do something else to figure out the answer? I'd appreciate if anyone can help me out.

HINT
Let
$$f(x)=(x-1)g(x)$$
such that
$$\lim_{x\to 1}\frac{f(x)}{(x-1)(x-2)} = \lim_{x\to 1}\frac{(x-1)g(x)}{(x-1)(x-2)} =\lim_{x\to 1}\frac{g(x)}{x-2}=-3$$
Refer also to

Why are we allowed to cancel fractions in limits?

For making the limit to exist we gotta take the function as f(x)=(x-1)g(x).
So let's assume g(x) to be a polynomial function of degreer 1 i.e. g(x)=(x-n) where n is any real number.Therefore f(x)=(x-1)(x-n).
So just simply apply the limits and you will be left with (n-1).
Put it equal to -3 , you will get n = -2.
There you have your f(x)=(x-1)(x+2)


Dilogarithm inversion formula: $ \text{Li}_2(z) + \text{Li}_2(1/z) = -\zeta(2) - \log^2(-z)/2$ I have been chugging through some proofs regarding the dilogarithm, also known as
Spencer's function.

\begin{alignat}{2}
  &  \operatorname{Li}_2(z) + \operatorname{Li}_2(-z) = \frac{1}{2} \operatorname{Li}_2(z^2)
  && \text{(Double Identity)} \tag{1} \\
  &  \operatorname{Li}_2(z) + \operatorname{Li}_2(1-z) = \frac{\pi^2}{6} - \log z \log (1 - z) \
  && \text{(Eulers reflection formula)} \tag{2} \\
  &  \operatorname{Li}_2(-z) + \operatorname{Li}_2\left( \frac{z}{1+z} \right) = -\frac{1}{2} \log^2(z+1)
  && \text{(Landen's Identity)} \tag{3} \\
  &  \operatorname{Li}_2(z) + \operatorname{Li}_2\left( \frac{1}{z} \right) = - \frac{\pi^2}{6} - \frac{1}{2}\log^2(-z) \ \
  && \text{(Inversjons formula)} \tag{4}
\end{alignat}
With the basis of $\text{Abel's Identity}$ (see Proving Abel's identity for the Dilogarithm.). I have been able to prove
all the the identities except the last. (My proof for $(3)$, was somewhat convoluted, so hints there would be appreciated..). I have yet to prove $(4)$, although I have given it two attempts below

Attempt 1

In the Abel Identity let $x=y=1-z$ and divide by $2$ to obtain
$$
\frac{1}{2}\log(-z)^2 = \operatorname{Li}_2\left( -\frac{1+z}{z}\right) - \frac{1}{2} \operatorname{Li}_2\left( \left[ \frac{1+z}{z} \right]^2 \right) -   \operatorname{Li}_2(1+z)
$$
By now using $(1)$ with $z = [1+1/z]^2$ and inserting it into the equation above I obtain
$$
\frac{1}{2}\log(-z)^2 = - \left[ \operatorname{Li}_2\left( \frac{1}{z} + 1\right)- \operatorname{Li}_2(1 + z)\right]
$$
and from here I am stuck. It is close to what I want but I can not find
any way to transform the right hand side.

Attempt 2

From chat the suggestion was to instead look at the integral definition, this gives
\begin{align*}
\operatorname{Li}_2(z) + \operatorname{Li}\left( \frac{1}{z}\right)
   & = - \int_0^z \frac{\log(1-t)}{t}\,\mathrm{d}t
       - \int_0^{1/z} \frac{\log(1-t)}{t}\,\mathrm{d}t \\
   &  =  - \int\limits_0^1 {\frac{{\log \left( {z - t} \right) + \log \left( {1 - zt} \right) - \log z}}{t}dt} \\
     & = -\zeta(2) + \int_0^1 \frac{\log z - \log(1-zt)}{t}\,\mathrm{d}t
\end{align*}
The last step used that
$$
\int_0^1 \frac{\log(1-t)}{t}\,\mathrm{d}t
= \int_0^1 -\frac{1}{t} \sum_{n=1}^\infty \frac{t^n}{n} \,\mathrm{d}t
= \sum_{n=1}^\infty \frac{1}{n^2}
= \zeta(2)
$$
And this is where I stopped.
I think this argument can be finished by series expansion, but I got lost in the algebra.
If possible I would very much like to prove this identity from Abel's Identity and the three equations stated above. Any hint or solutions is much appreciated as always =)

The classical proof (Lewin) of the 'inversion formula' $(4)$ is simply to use :
$$\operatorname{Li}_2(z)=-\int\frac{\log(1-z)}z\,dz$$
and set $z:=-\frac 1x\,$ to get the derivative :
$$\frac d{dx}\operatorname{Li}_2\left(-\frac 1x\right)=-\frac{\log\left(1+\frac 1x\right)}{-\frac 1x}\frac 1{x^2}=\frac{\log\left(1+\frac 1x\right)}x=\frac{\log(1+x)-\log(x)}x$$
Integrating this again gives (with the constant determined at $x=1$) :
$$\operatorname{Li}_2\left(-\frac 1x\right)+\operatorname{Li}_2(-x)=2\,\operatorname{Li}_2(-1)-\frac12\log^2(x)$$

($\,x:=-z\,$ gives your relation of course)

Concerning the 'Landen identity' $(3)$ the same method applies :
\begin{align}
\frac d{dx}\operatorname{Li}_2\left(\frac x{1+x}\right)&=-\frac{\log\left(1-\frac x{1+x}\right)}{\frac x{1+x}}\frac 1{(x+1)^2}=-\frac{\log\left(\frac 1{1+x}\right)}{x(x+1)}\\
&=\log(1+x)\left(\frac 1x-\frac 1{1+x}\right)\\
\end{align}

so that integrating again gives (with constant determined by $x=0$) :
$$\operatorname{Li}_2\left(\frac x{1+x}\right)=-\operatorname{Li}_2(-x)-\frac 12\log^2(1+x)$$

For the proof of li_2(x) +li_2(-x)  =½li_2(x²)
rewrite the dilogs at the right hand side as sums
  We have €(x^n/n²) +((-x) ^n/n²) from 1 to infinity.
Then notice the odd terms keep cancelling out.
Leaving only the even terms so rewrite sum
   €(x^2n/4n² +(-x) ^2n/4n²)
   2/4€(x^2n/n²)
Which is recognized as
½li_2 (x²)


$\mathbb{R}^3$ spanned by skew lines Let $a,b,c,d$ be vectors in $\mathbb{R}^3$ such that $a+\lambda b$ ($\lambda\in \mathbb{R}$) , and $c+\mu d$ ($\mu\in \mathbb{R}$) are skew lines, i.e $b$ and $d$ are not proportional and  $a+\lambda b\neq c+\mu d$ for any $\lambda, \mu\in \mathbb{R}$, in other words both the lines are niether parallel nor intersect each other in $\mathbb{R}^3$.
Can any one tell how these skew lines span the whole of $\mathbb{R}^3$.

Short answer:
Two different lines will at the very least span a 2-dimensional space. This however, can only be the case if these lines lie in the same plane. So the lines must span a 3-dimensional space.
Longer answer:
Let $\langle.,.\rangle:\mathbb{R^3}\times\mathbb{R^3}\to\mathbb{R}$ denote the inner product on $\mathbb{R^3}$. Define $u:=a+\lambda b$, where $\lambda=-\frac{\langle a,a\rangle}{\langle a,b\rangle}$. Then $\langle a,u\rangle=\langle a,a\rangle-\frac{\langle a,a\rangle}{\langle a,b\rangle}\langle a,b\rangle=0$ (by properties of the inner product). This means that $a$ and $u$ are orthogonal i.e. they are linearly independent. Now if both $c$ and $d$ lie in the span of $a$ and $u$, then we have two points on some plane, which means the line between the points $c$ and $d$ must also lie on this plane. But then the line $c+\mu d$ must lie in the same plane as the span of $a$ and $u$ (which contains the line $a+\lambda b$), which means the lines would intersect. As a result, c and/or $d$ isn't in the span of $a,u$. Without loss of generality, assume $d$. Then $a,u,d$ are 3 linearly independent vectors, which obviously must span $\mathbb{R}^3$. $a$ and $u$ come from the line $a+\lambda b$ and $d$ comes from the line $c+\mu d$, which would also imply that these lines span $\mathbb{R}^3$.

Denote $L_1 =\{x\in\mathbb{R}^2 : \exists \lambda  :x=a+\lambda  b \} ,$ $L_2 =\{x\in\mathbb{R}^2 : \exists \mu:  x=c+\mu d \} ,$ then
$$L_1 +L_2 =\mathbb{R}^2$$


Number Theory: $x^2+y^2=a^2$ Is there a coprime triple $(x,y,z)$ such that $x^2+y^2=a^2, x^2+z^2=b^2, y^2+z^2=c^2$, where $a,b,c$ are integers

P.S. such solution doesn't exist for $a,b,c<1000$, as the computer says

P.P.S. the statement above is WRONG

It's called Euler brick,the smallest one is $x=44,y=117,z=240$

http://en.wikipedia.org/wiki/Euler_brick


Decidability of the Riemann Hypothesis vs. the Goldbach Conjecture In the most recent numberphile video, Marcus du Sautoy claims that a proof for the Riemann hypothesis must exist (starts at the 12 minute mark).  His reasoning goes as follows:

If the hypothesis is undecidable, there is no proof it is false.
If we find a non-trivial zero, that is a proof that it is false.
Thus if it is undecidable there are no non-trivial zeros.
This constitutes a proof the hypothesis is true, thus it is decidable.

This makes perfect sense, however only seconds earlier he states that the Goldbach conjecture might be undecidable.  It seems to me that the exact same reasoning we applied to the Riemann hypothesis could be applied to the Goldbach conjecture.  Just switch out the words "non-trivial zero" for "even number that cannot be represented by the sum of two primes" and to me this reasoning looks fine.

In fact as far as I can tell any statement that can be verified or falsified by example can be plugged into this to generate a proof it is decidable.

Since Marcus du Sautoy is considerably more qualified than I to speak about this I trust that there is something more complex going on behind the scenes here.  Does this apply to the Goldbach conjecture?  If not why not?  What am I not understanding here?

The issue here is how complicated is each statement, when formulated as a claim about the natural numbers (the Riemann hypothesis can be made into such statement).

For the purpose of this discussion we work in the natural numbers, with $+,\cdot$ and the successor function, and Peano Axioms will be our base theory; so by "true" and "false" we mean in the natural numbers and "provable" means from Peano Arithmetic.

We will say that a statement is "simple" if in order to verify it, you absolutely know that you do not have to check all the natural numbers. (The technical term here is "bounded" or "$\Delta_0$".)

For example, "There is a prime number small than $x$" is a simple statement, since verifying whether or not some $n$ is prime requires only checking its divisibility by numbers less than $n$. So we only need to check numbers below $x$ in order to verify this.

On the other hand, "There is a Fermat prime larger than $x$" is not a simple statement, since possibly this is false but only checking all the numbers above $x$ will tell us the truth of this statement.

The trick is that a simple statement is true if and only if it is provable. This requires some work, but it is not incredibly difficult to show. Alas, most interesting questions cannot be formulated as simple statements. Luckily for us, this "provable if and only if true" can be pushed a bit more.

We say that a statement is "relatively simple" if it has the form "There exists some value for $x$ such that plugging it in will turn the statement simple". (The technical term is $\Sigma_1$ statement.)

Looking back, the statement about the existence of a Fermat prime above $x$ is such statement. Because if $n>x$ is a Fermat prime, then the statement "$n$ is larger than $x$ and a Fermat prime" is now simple.

Using a neat little trick we can show that a relatively simple statement is also true if and only if it is provable.

And now comes the nice part. The Riemann hypothesis can be formulated as the negation of a relatively simple statement. So if the Riemann hypothesis was false, its negation was provable, so Riemann hypothesis would be refutable. This means that if you cannot disprove the Riemann hypothesis, it has to be true. The same can also be said on the Goldbach conjecture.

So both of these might turn to be independent, in the sense that they cannot be proved from Peano Arithmetic, but if you show that they are at least consistent, then you immediately obtain that they are true. And this would give us a proof of these statements from a stronger theory (e.g. set theory).

You could also ask the same about the twin prime conjecture. But this conjecture is no longer a relatively simple statement nor the negation of one. So the above does not hold for, and it is feasible that the conjecture is consistent, but false in the natural numbers.

To me Marcus du Sautoy's claim seems to be inconsistent especially in light of Asaf's answer. If you can construct the same two logical statements about the RH and GC then it cannot be that one is decidable and the other is not in the same system.

Since the GC is about a basic arithmetic truth you may not need a stronger axiomatic system than the Peano Axioms (PA). So if anything it would make more sense to propose the RH (which is a claim in complex analysis and probably requires ZF) is undecidable and the GC is decidable in ZF. But as Asaf mentioned the decidability of RH can be formalised as a relatively simple statement.


Proofs that the median and a 50th percentile aren't always the same thing. Is my reasoning correct? For our purposes we need to find at the least one case when a 50th percentile and the median aren't the same thing.
I will present three such cases.

Here the first case (proof) goes as this:
Suppose we have set of numbers {1,2,3}. Its median is 2. But 2 is NOT a 50th percentile because 2 isn't a percentile at all, it's a tertile. It has 1/3 of datapoints below it and 2/3 of datapoints below or equal to it.

The second proof:
Suppose we have set of numbers {1,2,3,4,5}. Its median is 3. But 3 is NOT a 50th percentile. For the inclusive definition of a percentile it's a 60th percentile, while for the exclusive definition it's a 40th percentile.

The third proof:
Suppose we have a set of numbers {0,10}. Its median is 5. And it's also a 50th percentile. But there are other 50th percentiles that aren't equal to 5 (and thus, aren't medians), namely any number in interval [0;10) for the inclusive defintion of a percentile and any number in interval (0;10] for exclusive definition of a percentile.

For a correct definition of the quantiles, you must use the (empirical) $\text{cdf}$ of the distribution, i.e.

$$\text{cdf}_X(x)=\frac{\#\{k:x_k\le x\}}n.$$

When $X=\{1,2,3\}$, the $50\%$ percentile is $2$, because $\dfrac{\#\{1\}\le50\%}3$, while $\dfrac{\#\{1,2\}}3>50\%$, the "jump" occurs at $2$ and this is where the $\text{cdf}$ meets the $50\%$ horizontal.

If you don't adopt such a definition, neither the centiles nor the median would exist !

The median, the central quartile and the $50\%$ centile are exact synonyms.

Turns out, that there are at the least 8 ways to calculate a percentile (probably even more)

I found this calculator that calculates percentiles in 8 different ways for the same dataset: https://www.wessa.net/rwasp_percentiles.wasp

I was unable to make calculations for {1,2,3} and {0,10}, probably they were too small. But I was able to make calculations for similar sets, {1,2,3,4,5,6,7,8,9} and {0,1,2,3,4,5,6,7,8,9} respectively.

Turns out that I'm both right and wrong in my examples. For some methods the median is indeed different from the 50th percentile, while for some it is equal to the median.

Let's review the results. For an analogue of the first case, set {1,2,3,4,5,6,7,8,9}, I got following results (the median is 5):

It seems like the methods used to calculate the 50th percentile are based on definitions of percentiles that allow to percentiles to exist even for such set as {1,2,3,4,5,6,7,8,9}. I have no idea what these definitions are though.

Now the second case, set {1,2,3,4,5} and the median is 3:

Only one method gave 50th percentile different from the median.

Now the third case, set {0,1,2,3,4,5,6,7,8,9} and the median is 4.5

3 methods gave result different from the median.

So, to summarize it all: Some methods can sometimes give the 50th percentile that is different from the median. Some methods never give 50th percentile that is different from the median. Some methods (I'm looking at you, "Weighted Average at Xnp"!) always give 50th percentile result that is different from the median. But even when 50th percentile is different from the median, the difference is relatively small. Probably the difference can be ignored altogether.


Can there be a single game of Chess which includes all the possible situations that may arise during Chess? First of all, I am sorry as there were some completely inappropriate posts posted by my account earlier. This happened because I forgot to log my account out from the computer in Net Cafe I last enetered and a devil used my account. Anyways, come to the question.

I guess that first of all, I have to calculate in numbers that how many different situations may arise in chess using permutation/combination/Probability and then check that how many different situations arises in the biggest possible game of chess (not w.r.t time). If the number of situations in the biggest game is equal to the number calculated earlier, then I can say that answer is Yes otherwise No. However this all stays a guess. Permutations/combinations/Probability: I am an amateur w.r.t to these theories. So, I would be grateful if anybody helps by giving the answer to my title in yes or no and then explain why.

By considering the position of pawns it is easy to see that the answer to this question is no. Consider the following two positions:

Since once the A pawn has moved forward it will always be in a rank higher than 4, any game that has contained the first position cannot in the future contain the second and vice versa.

No, for the simple fact that the number of pieces on the board can only diminish or stay the same as the game goes on, and never go back up again (promotion still takes a piece of the board, namely a pawn), so once you reach a position where something has been captured, you have excluded the possibility of reaching a position where there are more pieces on the board; and since pawns can't move backwards, it is impossible to reach all positions prior to the capture.


How to find period of this periodic function? How can I find a period of this function?

$$2\sin{3x} + 3\sin{2x}$$

Is here any way how to sum both sinuses?

You can always go the hard way analyzing, for arbitrary $T$:
$$
    2 \sin(3 (x+T) ) + 3 \sin(2(x+T)) = \\ 2 \sin(3x) \cos(3T) + 2 \cos(3x) \sin(3T) + 3 \sin(2x) \cos(2T) + 2 \cos(2x) \sin(2T)
$$
When $T$ equals a period you should have
$$
   \sin(3T) = 0, \quad \sin(2T) = 0, \quad \cos(3T) = 1, \quad \cos(2T) = 1
$$
Since $\sin(3T) = \sin(T+2T) = \sin(T) \underbrace{\cos(2T)}_1 + \cos(T) \underbrace{\sin(2T)}_0 = \sin(T)$, and $\cos(3T) = \cos(T) \cos(2T) - \sin(T) \sin(2T) = \cos(T)$. We have:
$$
    \sin(T) = 0, \quad \sin(2T) = 0, \quad \cos(T) = 1, \quad \cos(2T) = 1
$$
Repeating the exercise we see that $\sin(2T) = 0, cos(2T)=1$ is implied by $\sin(T)=0$ and $\cos(T) = 1$. Solving $\sin(T)=0$ and $\cos(T)=1$ is easy. There are infinitely many solutions:
$$
    T = 2 \pi n, \quad n \in \mathbb{Z}
$$
The minimal solution is $T = 2\pi$.

the correct ans will be $2\pi$, as the period of first is $2\pi$ by $3$ and second is $2\pi$ by $4$.