Network Flow Problem - How to solve with a matrix? Find the general flow pattern of the network. Assuming that the flows are all nonnegative, what is the smallest possible value for $x_4$?

Points of Intersection: Flow In = Flow Out

A: $x_1+x_4 = x_2$

B: $x_2 = x_3 + 100$

C: $x_3 + 80 = x_4$

The total flow in equals the total flow out, so you can set up other equations...

How do I turn this into a solvable matrix to get the answer?

Put it each column the corresponding variable, last column is used for constants. So rewrite each equation in the form $...=0$. Then, for $A$ we get $x_1-x_2+x_4=0$, $B$: $x_2-x_3-100=0$, $C$: $x_3-x_4+80=0$. Put that in a matrix:
$
 \begin{pmatrix}
1 & -1 & 0 & 1 & 0\\
0 & 1 & -1 & 0 & -100\\
0 & 0 & 1 & -1 & 80\\
 \end{pmatrix}
$

Solving that matrix results into $
 \begin{pmatrix}
1 & 0 & 0 & 0 & 20\\
0 & 1 & 0 & -1 & -20\\
0 & 0 & 1 & -1 & 80\\
 \end{pmatrix}
$

So, in other words, $x_1=20, x_2=-x_4-20, x_3=-x_4+80$. Since the flows are non-negative, it must hold that $x_4\le -20$ (for $x_2\ge 0$) and $x_4 \le 80$ for $x_3 \ge 0$. So, $x_4 \le -20$. But that is not possible, since $x_4 \ge 0$ so there are no solutions.


Using the Definition of Derivative to Derive the Derivative of a Function Assume $f$ is differentiable at point $a$, and $f(a)>0$. Determine the derivative of
$$g(x)=x\sqrt{f(x)}$$
in terms of $f'(a)$.

What I have done is substituting whatever is given and I don't really know how to continue already.

Really appreciate your help!

Okay using chain rule you would have:
$$\frac{d}{dx}g(x)=\frac{d}{dx}(x\cdot \sqrt{f(x)})$$
$$=1\cdot\sqrt{f(x)}+x \cdot \frac{f'(x)}{2\sqrt{f(x)}} $$
Using the actual definition of the derivative via limits, this is what we do.
$$\lim_{h\rightarrow 0} \frac{g(x+h)-g(x)}{(x+h)-(x)}=$$
$$\lim_{h\rightarrow 0} \frac{(x+h)\cdot \sqrt{f(x+h)}-x\cdot \sqrt{f(x)}}{(x+h)-(x)}=$$
$$\lim_{h\rightarrow 0} \frac{(x+h)\cdot \sqrt{f(x+h)}-x\cdot \sqrt{f(x)}}{(x+h)-(x)}=$$
Now we expand numerator to "split" the interval of the limit.
$$\lim_{h\rightarrow 0} \frac{(x+h)\cdot \sqrt{f(x+h)}-(x+\frac{h}{2})\sqrt{x+\frac{h}{2}}+(x+\frac{h}{2})\sqrt{x+\frac{h}{2}}-x\cdot \sqrt{f(x)}}{(x+h)-(x)}=$$
going to have to finish this later.


Prime conditional problem in modulo Any odd prime p can be written in the form of $x^2 + y^2$ when $p \equiv 1 \pmod{ 4}$. explain the proof.  Is there any such condition equations and satisfying the primes? if there, how to construct them? Also, is there any function, which will give a prime.

Regarding the proof: Here's an elementary argument. It can be modified to the following similar problems:

$x^2 + 2y^2 =p \leftrightarrow p \equiv 1,3 \mod 8$ or $p=2$.
$x^2+ 3y^2 = p \leftrightarrow p \equiv 1 \mod 3 $ or $p=3$.

Neccesity: Note that for any $x,y$, we have $x^2+y^2=0,1,2 \mod 4$, so for it to be a prime $p$ we need either $p=2$, or $p= 1 \mod 4$.

Sufficiencty: If $p=2$, $x=y=1$ will work. If $p \equiv 1 \mod 4$, then $t^2 + 1 = 0 \mod p$ is solvable, as it is equivalent to finding an element of order 4 in $(\mathbb{F}_{p})^{\times}$, which is a cyclic group of order $p-1$, i.e. divisible by $4$, so you can just take $t=g^{\frac{p-1}{4}}$ for any generator $g$.

By the pigeonhole principle, two of the elements $\{ ta-b \mod p| 0 \le a,b < \sqrt{p} \}$ are the same mod $p$, say: $ta-b=ta'-b' \mod p$, i.e.
$$t(a-a')=(b-b') \mod p$$
Square both sides of the last equation and move both terms to same side to get: $(b-b')^2 + (a-a')^2 \equiv 0 \mod p$. Note that $|a-a'|,|b-b'| < \sqrt {p}$, so $0< (b-b')^2 + (a-a')^2 < 2p$, implying:
$$(b-b')^2 + (a-a')^2 = p$$

Regarding more such conditions - I suggest the wonderful book "Primes of the form $x^2 + ny^2$" by David A. Cox. Another related book is "Introduction to the Construction of Class Fields" by Harvey Cohn.

Regarding a prime-generating function: consult this Wikipedia page. There are many formulas, most of them impractical. But there's a system of Diophantine equations "defining" the primes, which is maybe what you're looking for.

I do not know the prime function. It is my out of expectation. We can create your cited type problems. For example, primes can be written as primes $x^2$ + 2$y^2$ if and only if p = 1 or 3 (mod 8). Regarding your first question:

Let L = (1, 0, 1) and L has discriminant = -4. If (a, b, c) is reduced representative of some equivalence class with |b| < or = a < or = |c| => $b^2$ = 4(ac -1). Since |ac| > or = $b^2$, we should have ac -1 = 0. The only form of L is (1, 0, 1). Thus h (-4) = 1 and p can be expressible by (1, 0, 1) if and only if -4 = () (mod 4p). I hope you can do the necessary things to complete the solution of  $x^2$ + 2$y^2$ if and only if p = 1 or 3 (mod 8).
All the best!


Prove that the product of four consecutive positive integers plus one is a perfect square I need to prove the following, but I am not able to do it. This is not homework, nor something related to research, but rather something that came up in preparation for an exam.

  If $n = 1 + m$, where $m$ is the product of four consecutive positive
  integers, prove that $n$ is a perfect square.

Now since $m = p(p+1)(p+2)(p+3)$;

  $p = 0, n = 1$ - Perfect Square

  $p = 1, n = 25$ - Perfect Square

  $p = 2, n = 121$ - Perfect Square

Is there any way to prove the above without induction? My approach was to expand $m = p(p+1)(p+2)(p+3)$ into a 4th degree equation, and then try proving that $n = m + 1$ is a perfect square, but I wasn't able to do it. Any idea if it is possible?

Your technique should have worked, but if you don't know which expansions to do first you can get yourself in a tangle of algebra and make silly mistakes that bring the whole thing crashing down.

The way I reasoned was, well, I have four numbers multiplied together, and I want it to be two numbers of the same size multiplied together. So I'll try multiplying the big one with the small one, and the two middle ones.

$$p(p+1)(p+2)(p+3) + 1 = (p^2 + 3p)(p^2 + 3p + 2) + 1$$

Now those terms are nearly the same. How can we force them together? I'm going to use the basic but sometimes-overlooked fact that $xy = (x+1)y - y$, and likewise $x(y + 1) = xy + x$.

$$\begin{align*}
(p^2 + 3p)(p^2 + 3p + 2) + 1 &= (p^2 + 3p + 1)(p^2 + 3p + 2) - (p^2 + 3p + 2) + 1 \\
 &= (p^2 + 3p + 1)(p^2 + 3p + 1) + (p^2 + 3p + 1) - (p^2 + 3p + 2) + 1 \\
 &= (p^2 + 3p + 1)^2
\end{align*}$$
Tada.


What definition gives $\hat{x}\in X^{**}$? Definition: Let $X$ be a topological vector space and let $x\in X$. Then $x$ defines a linear functional $\hat{x}$ on $X^*$ via $\hat{x}(f)=f(x)$ $(f\in X^*)$.

let $X$ be a normed space and let $x\in X$. I am trying to show that $\hat{x}\in X^{**}$. And my attempts are:

Let $f\in X^*$ and let $(f_i)$ be a net in $X^*$ with $f_i\overset{\|\cdot\|}{\longrightarrow} f$ in $X^*$. Then $$|\hat{x}(f_i)-\hat{x}(f)|=|f_i(x)-f(x)|=|(f_i-f)(x)|\leqslant\|f_i-f\|\cdot\|x\|\rightarrow0.$$ Thus $\hat{x}(f_i)\rightarrow\hat{x}(f)$. Hence, $\hat{x}$ is a continuous linear functional on $X^*$; that is, $\hat{x}\in X^{**}$.
Using a corollary of the Hahn-Banach Theorem, $$||\hat{x}||=\sup\{|\hat{x}(x^*)|:\|x^*\|\leqslant1\}=\sup\{|x^*(x)|:\|x^*\|\leqslant1\}=\|x\|.$$ Thus $\hat{x}\in X^{**}$.

But my professor mentioned I didn't need any proof at all. It follows immediately by using definition in functional anaylsis. But I don't know what definition gives $\hat{x}\in X^{**}$? Any helps will be appreciated!!

You've correctly noticed that something needs to be shown, in particular that $\hat{x}$ is a continuous linear functional. So, you just need to show that $\hat{x}$ is a bounded linear functional, which is easy: take $\|f\|=1$, then
$$|\hat{x}(f)| = |f(x)| \leq \|x\|$$
hence $\|\hat{x}\|:=\sup_{\|f\|=1} |\hat{x}(f)|\leq\|x\|$, so is bounded.

Notice that with Hahn-Banach, you can additionally show that $\|\hat{x}\| = \|x\|$ (as you did), but that's not necessary here.

$V^*$ is the dual space of $V,$ and the double dual space $V^{**}$ is isomorphic to $V$. Can you take it from here?

Edit: I'm assuming $dim(V)<\infty$ here.


Is this Batman equation for real? HardOCP has an image with an equation which apparently draws the Batman logo.  Is this for real?

Batman Equation in text form:
\begin{align}
&\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\
&\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\
&\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\
&\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\
&\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0
\end{align}

As Willie Wong observed, including an expression of the form $\displaystyle \frac{|\alpha|}{\alpha}$ is a way of ensuring that $\alpha > 0$. (As $\sqrt{|\alpha|/\alpha}$ is $1$ if $\alpha > 0$ and non-real if $\alpha < 0$.)

The ellipse $\displaystyle \left( \frac{x}{7} \right)^{2} + \left( \frac{y}{3} \right)^{2} - 1 = 0$ looks like this:

So the curve $\left( \frac{x}{7} \right)^{2}\sqrt{\frac{\left| \left| x \right|-3 \right|}{\left| x \right|-3}} + \left( \frac{y}{3} \right)^{2}\sqrt{\frac{\left| y+3\frac{\sqrt{33}}{7} \right|}{y+3\frac{\sqrt{33}}{7}}} - 1 = 0$ is the above ellipse, in the region where $|x|>3$ and $y > -3\sqrt{33}/7$:

That's the first factor.

The second factor is quite ingeniously done. The curve $\left| \frac{x}{2} \right|\; -\; \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2}\; -\; 3\; +\; \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}-y=0$ looks like:

This is got by adding $y = \left| \frac{x}{2} \right| - \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2} - 3$, a parabola on the positive-x side, reflected:

and $y = \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}$, the upper halves of the four circles $\left( \left| \left| x \right|-2 \right|-1 \right)^2 + y^2 = 1$:

The third factor $9\sqrt{\frac{\left( \left| \left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right) \right| \right)}{\left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right)}}\; -\; 8\left| x \right|\; -\; y\; =\; 0$ is just the pair of lines y = 9 - 8|x|:

truncated to the region $0.75 < |x| < 1$.

Similarly, the fourth factor $3\left| x \right|\; +\; .75\sqrt{\left( \frac{\left| \left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right) \right|}{\left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right)} \right)}\; -\; y\; =\; 0$ is the pair of lines $y = 3|x| + 0.75$:

truncated to the region $0.5 < |x| < 0.75$.

The fifth factor $2.25\sqrt{\frac{\left| \left( .5-x \right)\left( x+.5 \right) \right|}{\left( .5-x \right)\left( x+.5 \right)}}\; -\; y\; =\; 0$ is the line $y = 2.25$ truncated to $-0.5 < x < 0.5$.

Finally, $\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like:

so the sixth factor $\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\sqrt{\frac{\left| \left| x \right|-1 \right|}{\left| x \right|-1}}\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like

As a product of factors is $0$ iff any one of them is $0$, multiplying these six factors puts the curves together, giving: (the software, Grapher.app, chokes a bit on the third factor, and entirely on the fourth)

Sorry but this is not the answer but too long for a comment:
Probably the easiest verification is to type the equation on Google you'l be surprised :
The easiest way is to Google :2 sqrt(-abs(abs(x)-1)abs(3-abs(x))/((abs(x)-1)(3-abs(x))))(1+abs(abs(x)-3)/(abs(x)-3))sqrt(1-(x/7)^2)+(5+0.97(abs(x-.5)+abs(x+.5))-3(abs(x-.75)+abs(x+.75)))(1+abs(1-abs(x))/(1-abs(x))),-3sqrt(1-(x/7)^2)sqrt(abs(abs(x)-4)/(abs(x)-4)),abs(x/2)-0.0913722(x^2)-3+sqrt(1-(abs(abs(x)-2)-1)^2),(2.71052+(1.5-.5abs(x))-1.35526sqrt(4-(abs(x)-1)^2))sqrt(abs(abs(x)-1)/(abs(x)-1))+0.9


Evaluating $\frac{(1+i)^{n+2}}{(1-i)^n}$ Evaluate $\dfrac{(1+i)^{n+2}}{(1-i)^n}$

I think that the meaning is that it need to be simplified.
Thanks


If squaring a number means multiplying that number with itself then shouldn't taking square root of a number mean to divide a number by itself? If squaring a number means multiplying that number with itself then shouldn't taking square root of a number mean to divide a number by itself?

For example the square of $2$  is $2^2=2 \cdot 2=4 $ .

But square root of $2$ is not $\frac{2}{2}=1$ .

taking square root means reversing the effect of squaring. Dividing a number by itself does not do that (but rather always returns 1 as you noted).

Compare your question to: if doubling a number means adding it to itself, shouldn't halving a number mean subtracting it from itself? Answer: obviously not.

If you take a number, $N$ and multiply it times $x$ and then again times $x$, you would end up with $Nx^2$. To get $N$ back, you would divide by $x$ and then divide again by $x$.

The word root is generally used to represent one of the zeros of a function; that is a solution to $f(x) = 0$. In particular, a square root of N  is a solution to the polynomial equation $x^2 - N = 0$. If $N$ is a non negative real number, then there is a positive square root and a negative square root of $N$. Usually we express the positive square root of $N$ as $\sqrt N$ and the negative square root of $N$ as $-\sqrt N$.

ALSO

I suppose that , if you want dividing by $x$ in the definition of square root, then you can say that the square root of $N$ means finding the number, $x$, such that $\dfrac Nx = x$.

A comparison

\begin{array}{ccc}
   \text{action} & \text{equation} & \text{inverse} \\
\hline
   \text{doubling} & 2x = x + x & x = \dfrac 12(2x) \\
   \text{squaring} & x^2 = x \cdot x & x = \sqrt{x^2} \\
   \hline
\end{array}

Notice that the inverse of doubling is not subtracting x and that the inverse of squaring is not dividing by x.


Is there no solution to the blue-eyed islander puzzle? Text below copied from here

  The Blue-Eyed Islander problem is one of my favorites. You can read
  about it here on Terry Tao's website, along with some discussion.
  I'll copy the problem here as well.

  There is an island upon which a tribe resides. The tribe consists of
  1000 people, with various eye colours. Yet, their religion forbids
  them to know their own eye color, or even to discuss the topic; thus,
  each resident can (and does) see the eye colors of all other
  residents, but has no way of discovering his or her own (there are no
  reflective surfaces). If a tribesperson does discover his or her own
  eye color, then their religion compels them to commit ritual suicide
  at noon the following day in the village square for all to witness.
  All the tribespeople are highly logical and devout, and they all know
  that each other is also highly logical and devout (and they all know
  that they all know that each other is highly logical and devout, and
  so forth).

  [For the purposes of this logic puzzle, "highly logical" means that
  any conclusion that can logically deduced from the information and
  observations available to an islander, will automatically be known to
  that islander.]

  Of the 1000 islanders, it turns out that 100 of them have blue eyes
  and 900 of them have brown eyes, although the islanders are not
  initially aware of these statistics (each of them can of course only
  see 999 of the 1000 tribespeople).

  One day, a blue-eyed foreigner visits to the island and wins the
  complete trust of the tribe.

  One evening, he addresses the entire tribe to thank them for their
  hospitality.

  However, not knowing the customs, the foreigner makes the mistake of
  mentioning eye color in his address, remarking “how unusual it is to
  see another blue-eyed person like myself in this region of the world”.

  What effect, if anything, does this faux pas have on the tribe?

The possible options are

Argument 1. The foreigner has no effect, because his comments do not tell the tribe anything that they do not already know (everyone in the tribe can already see that there are several blue-eyed people in their tribe).

Argument 2. 100 days after the address, all the blue eyed people commit suicide. This is proven as a special case of

Proposition. Suppose that the tribe had $n$ blue-eyed people for some positive integer $n$. Then $n$ days after the traveller’s address, all $n$ blue-eyed people commit suicide.

Proof: We induct on $n$. When $n=1$, the single blue-eyed person realizes that the traveler is referring to him or her, and thus commits suicide on the next day. Now suppose inductively that $n$ is larger than $1$. Each blue-eyed person will reason as follows: “If I am not blue-eyed, then there will only be $n-1$ blue-eyed people on this island, and so they will all commit suicide $n-1$ days after the traveler’s address”. But when $n-1$ days pass, none of the blue-eyed people do so (because at that stage they have no evidence that they themselves are blue-eyed). After nobody commits suicide on the $(n-1)^{st}$ day, each of the blue eyed people then realizes that they themselves must have blue eyes, and will then commit suicide on the $n^{th}$ day.

It seems like no-one has found a suitable answer to this puzzle, which seems to be, "which argument is valid?"

My question is...
Is there no solution to this puzzle?

Argument 1 is clearly wrong.

Consider the island with only two blue-eyed people.  The foreigner arrives and announces "how unusual it is to see another blue-eyed person like myself in this region of the world." The induction argument is now simple, and proceeds for only two steps; on the second day both islanders commit suicide. (I leave this as a crucial exercise for the reader.)

Now, what did the foreigner tell the islanders that they did not already know? Say that the blue-eyed islanders are $A$ and $B$. Each already knows that there are blue-eyed islanders, so this is not what they have learned from the foreigner. Each knows that there are blue-eyed islanders, but neither one knows that the other knows this.  But when $A$ hears the foreigner announce the existence of blue-eyed islanders, he gains new knowledge: he now knows that $B$ knows that there are blue-eyed islanders. This is new; $A$ did not know this before the announcement.  The information learned by $B$ is the same, but mutatis mutandis.

Analogously, in the case that there are three blue-eyed  islanders, none learns from the foreigner that there are blue-eyed islanders; all three already knew this.  And none learns from the foreigner that other islanders knew there were blue-eyed islanders; all three knew this as well.  But each of the three does learn something new, namely that all the islanders now know that (all the islanders know that there are blue-eyed islanders).  They did not know this before, and this new information makes the difference.

Apply this process 100 times and you will understand  what new knowledge was gained by the hundred blue-eyed islanders in the puzzle.

it looks like an administrator here deleted my answer. i just want to say that if you consider it important in any way to be any sort of ambassador to the layperson, then you need to seriously reconsider your attitude. theres a post on here that literally is written by a salesman, talking about the psychological reasons of why the 100 people would know. that one didnt get deleted. i wrote a clearly stated, reasoned response to the best of my ability, and it was snarkily ridiculed with no explanation, then deleted. terrible attitude dude. i am extremely disappointed.

if there is a problem with my logic that i am unaware of, then let me know. i am interested in discussion. i just dont think it is helpful to simply tell someone they havent answered the question without any explanation other than: its a math problem. it might be listed in the math section of this site, but as it is stated, it isnt a math problem. its a logic problem. the whole point of the problem is to determine what can and cannot be known with what information exists. if there is a flaw in my logic, please point it out.

i have re-answered the question, with a few clarifications. i expect you will delete it again mr. hardmath. but i just wanted you to know that if you do, it is very petty and uncool, especially considering you left up the answer from the salesman.

ANSWER:

the correct answer is option one: there would be no change. everyone already knows that everyone knows there are people with blue eyes. as long as there are at least 2 people amongst the islanders with blue eyes, everyone would see at least one other person with blue eyes, but not know if they themselves have blue eyes or not. in this case there are 100 people with blue eyes, and 900 without. but it doesnt matter how many people there are with blue eyes as long as there are more than 2 - they would have no way of knowing their own eye color, EVER. since there is no new information, why would anything change?

the problem that i see with the reasoning in the answers supporting option 2 is that it claims that the new knowledge is that the islanders now know that all of the other islanders know. this is not logical! all of the islanders must know that all the others know already! what one person sees, the others can see. they already understand that what they see the others can see, so therefore they already know that everyone already knows they already know. or whatever.

to take the above example of bob, cathy, and alice: all three must know that someone has blue eyes, and they all must know that they all know, because they all have blue eyes, therefore bob knows cathy knows and cathy knows alice knows bob knows, etc.

therefore, there is no new information. therefore, behavior does not change.

if someone disagrees with this, please explain. i have read through all of the answers and i really cannot understand why anyone thinks that the new information is that now all the islanders know that all the islanders know.


Is there an algorithm to generate primes iteratively and not test for primality? I apologise if I'm not clear. I know that you can find all prime numbers between $1$ and $n$ by iteratively testing all numbers between $1$ and $n$ and recording those that pass a primality test. Is there a way to algorithmically find all the primes between $1$ and $n$ without testing each number for primality?

If you want to find all of the primes from 1 to $n$, the best way by far is a sieve. The sieve of Eratosthenes is easy to write and fast. How fast? The basic version takes time $O(n\log\log n)$, which mean that it spends $\log n\log\log n$ time for each prime. This is really quick. In practice this means that it's faster to generate primes than to read them from disk (though not as fast as reading them from memory). Testing each number for primality would take longer, in theory, and much longer in practice ($100\times$ longer at best). Let me put it this way: sieving doesn't take that much longer than just listing the primes that you find.

You might want to look at a range of numbers large enough that you can't store all the primes in memory. That's OK -- there are sieves that solve that problem. 99% of the time the answer is the same: the sieve of Eratosthenes can be segmented to require only $\sqrt n$ or so memory, so finding the primes up to $10^{15}$ would take 4 MB instead of 125 TB.

You mentioned a function to find the next prime after a given prime $p$. There's no good way to do this; at present the best algorithm would be to sieve the small primes from a chunk somewhat bigger than $\log p$, then test the remaining numbers until you find one which is prime.* But sieves have an answer here, too. If you hit the end of the sieve and decide you want to go further, you can—without extra effort!—if you're using a class of sieves called incremental sieves. Sorenson [1] has an example; other ones I've seen ([2], [3]) are impractical.

If you like the idea of a sieve, but $\log n\log\log n$ is too slow for you, there are more advanced sieves which take only $\log n$ work per prime, some of which can be combined with a wheel to reduce this to $\log n/\log\log n.$ The Atkin-Bernstein sieve is the usual example but there are others. (There's nothing special about that sieve in terms of speed, the paper describes a linear sieve and gives a general trick, related to what Tao calls the W-trick, for shaving a factor of $\log\log n$.) This is the fastest that any existing sieves can go, and it's not that far from the best possible. I mean, think about it -- writing a number $n$ in decimal takes $\Theta(\log n)$ time because it has that many digits, and so methods like this produce primes faster than you can write them down.

In practice highly-optimized versions of the sieve of Eratosthenes are state of the art. The best use a variant of Oliveira e Silva; primesive may be the fastest freely available sieve. Bernstein's PrimeGen is innovative but not presently competitive.

Speed isn't the only problem, though. If you want to find the primes up to a very large $n$ then you could run out of memory, or you could run out of memory because you're running on a limited machine. Even if you have plenty of memory you might not want to use it all—possibly because other programs need it, but more likely because main memory is too slow and you want to run the sieve in the L3 or even L2 cache. In that case the dissected sieve [4] is excellent, and [5] promises to be an interesting alternative. Both run in $O(n^{1/3+\varepsilon})$ space which should be all you'll ever need if you're running from 1 to $n$ (rather than large isolated intervals like $10^{25}$ to $10^{25}+10^9$). But just in case there are sieves which use even less space like the pseudosquare prime sieve [6], though at that point you pay a steep performance penalty. And of course you could always do a partial sieve by small primes followed by a Miller-Rabin test and a primality-proving routine, for another performance hit.

* If you were really committed to doing this, theory says you would want to sieve up to something like $\exp(\log^2p/\log\log p)$ and the sieve range could be, say, $p+2$ to $p+5\log p$ (so around 1% of the time you'd need to sieve out a second interval), then use a Miller-Rabin test on all numbers not sieved out until one passed; at that point you can use the primality-proving test of your choice, maybe BPSW if less than $2^{64}$ or ECPP otherwise.

References

[1] Jonathan P. Sorenson, Two compact incremental prime sieves, LMS Jour. Comp. Math. 18 (2015), pp. 675–683. arXiv:1503.02592 [cs.DS]

[2] S. Bengelloun, An incremental primal sieve, Acta Informatica 23:2 (1986), pp. 119–125.

[3] Paul Pritchard, Improved incremental prime number sieves, ANTS-I (May 1994), pp. 280–288.

[4] William F. Galway, Dissecting a sieve to cut its need for space. In Algorithmic Number Theory (Leiden, 2000), volume 1838 of Lecture Notes in Comput. Sci. (2000), pp. 297–312.

[5] Harald A. Helfgott, unpublished conference slides (2016).

[6] Jonathan P. Sorenson, The pseudosquares prime sieve

You mean like this??

 Select[Range[100],PrimeQ]

Sorry if this isn't what your looking for


Series and Consistency Sometime ago I asked myself this question:

  Why manipulating $\sum_{n=1}^{\infty}\frac{1}{2^n}$ as $1$ leads to no contradiction in the arithmetic of $\mathbb{R}$ ?

More generally, why "treating" convergent series as numbers leads to no contradiction? I've found no satisfactory explanation to this question, maybe this is a meaningless question so nobody ever asked this.

I tried to come up with an answer, here it is:

We usually think of a convergent series as being a number, for example, $\sum_{n=1}^{\infty}\frac{1}{2^n}$ equals $1$.
But we should think the other way around, if $S$ is the set of all convergent series, we should think of "$=$" as being a function from $S$ to $\mathbb{R}$, that associates every convergent series to its limit. Of course, for every real number we can find a infinite series that converge to it, but there are many different series, so we need an equivalent relation on $S$, two series being equivalent if they have the same limit.

Is my answer ok?

Does my question even make sense?

Thanks.

I think you're misunderstanding what the notation "$\sum$" means. For an infinite summation, $$\sum_{i=1}^\infty a_i$$ is defined to be the unique real number $b$ such that the finite partial sums "approach $b$", or formally, such that$$\forall \epsilon>0\exists N\forall m>N\left(\left\vert b-\sum_{i=1}^ma_i\right\vert<\epsilon\right),$$ if such a $b$ exists (note that at most one such $b$ exists).

We can now prove that e.g. $\sum_{i=1}^\infty2^{-i}=1$, using the basic axioms of the real numbers (and this is a good exercise). We can also prove that some summations don't correspond to any real number, e.g. $\sum_{i=1}^\infty 7$.

Note that we're not allowed to just assume that an expression like "$\sum_{i=1}^\infty 2^{-i}$" corresponds to a real number (that is, is defined); we have to prove it. But once we've proved that it does, we've shown that "$\sum_{i=1}^\infty 2^{-i}$" is a name of some real number. Note that numbers have many names - e.g. $1=\sum_{i=1}^\infty 2^{-i}={7\over 7}=\sin{\pi\over 2}$. There's no problem with having multiple series converge to the same value, any more than there is with multiple fractions evaluating to the same number.

That said, the picture you've given of a function from [names] to [values] is a very important one, and will come up in many fields of mathematics (some keywords include free groups/rings/etc., groups/rings/etc. generated by some relations, and - in logic and universal algebra - term models). However, I'd argue that it's not really the right picture for this particular context.

One cannot always manipulate series is such a manner, even (conditionally) converging series.

  The Riemann series theorem says "if an infinite series is conditionally convergent, then its terms can be arranged in a permutation so that the new series converges to any given value, or diverges."

A famous one is the paradox shown on the Simpson's.  I will leave it to the reader to make conclusions about the manipulation of series.

You may find other examples here.


Factoring $x^3 - x^2 - x + 1$ I'm familiar with basic forms of polynomial factoring like completing the square or factoring (e.g. finding that $x^2+x-6=(x+3)(x-2)$, but I'm currently working on integration of rational functions by partial fractions and I find myself encountering the factorizations of polynomials unlike those I've worked with before.

Given: $x^3 - x^2 - x + 1$

When I first look at this, I'm unsure of where to start. It's very unlike the example above. It's degree $3$, for one thing, and it has $4$ terms.

When I look ahead to the solution, the book says $f(1) = 0$, so we know that $(x-1)$ is a factor. Okay, now that they mention that, I can see it fits, but what?! How did they come up with that? Is that a general rule? Does anyone have a link to where I can learn about it?

Further, once it's known that $(x-1)$ is a factor, how do I then factor an $(x-1)$ out of a longer polynomial like this?

In other words, how does $x^3 - x^2 - x + 1 = (x-1)(x^2-1)$?

Where can I learn more about this type of factorization?

There's some theorem (rational root theorem/ factor theorem?) that says that for all rational factors of the form p/q, p divides into the constant in your polynomial and q divides into the leading coefficient. That leaves to with 1 and -1. Testing 1, first, we find it's a root. -1 is also a root. Using polynomial division it becomes easier to test these numbers. Since it's a cubic it has three roots - it turns out that 1 is a double root.


Every open set $ G \subset \mathbb{R}^n$ can be approximated by finite union of non-degenerate compact intervals contained in $G$? For every open set $ G \subset \mathbb{R}^n$ and $ \epsilon > 0 $ there exist a finite collection of compact intervals $\{ I_j \quad|\quad j=1,2,...m\}$ in $ \mathbb{R}^n $ such that $\cup^{j=1}_{m}I_j \subset G $ and  $ m(G\backslash \cup^{j=1}_{m}I_j)< \epsilon $ where m denotes lebesgue measure.

NOTE : compact interval in $\mathbb{R}^n$ is $[a_1,b_1]\times[a_2,b_2]\times...\times[a_n, b_n]$ where $[a_i,b_i] $ are real closed intervals for each i.

EDIT: $m(G)$ is finite. sorry, I forget to write.

This follows from the fact that every open set in $\mathbb{R}^n$ is a countable disjoint union of compact intervals. Indeed, if $G = \cup_{k=1}^\infty I_k$, then since $\sum_k m(I_k) = m(G) < \infty$, we may take $K$ so that $\sum_{k > K} m(I_k) < \epsilon$. Then $\cup_{k=1}^K I_k$ does the trick.

See page 7 for the proof of the first fact I stated

http://assets.press.princeton.edu/chapters/s8008.pdf

This is false. For example take $G= \mathbb R^n$.


Expected value of repeated bets Suppose a game where a fair coin is flipped. Tails pays out $0.5$x the bet, and heads pays out $1.75$x. With a bet of \$1, the expected value of a single game is $0.5\cdot 0.5 + 0.5\cdot 1.75 = 1.125$. So if $1000$ games of this is played, \$125 is expected to be gained.

But what if instead of betting \$1, the entire bankroll is bet? So starting with \$100, after a single round there is a $50$% chance of \$50, and a $50$% chance of \$175. Regardless of outcome, the entire amount is bet again, so if it was heads, the next round \$175 would be bet and so on. What is the expected amount after $1000$ rounds?

My intuition is that you'd end up with \$0, since it would take more than one win to make up for a single loss as $0.5\cdot 1.75 = 0.875$, and when I run a few trials programmatically it would come out to $0$ (as in, $<\epsilon$) each time, but I don't know how to prove this

Computer simulations suggest that the bankroll is expected to get vanishingly small after the 1000 rounds. This is false, as will be shown below.

Without loss of generality let the initial bankroll be 1. If the number of heads obtained is $k$, the bankroll after playing 1000 rounds is
$$(1.75)^k(0.5)^{1000-k}$$
Thus the expected value of the bankroll after 1000 rounds is
$$\sum_{k=0}^{1000}\frac{\binom{1000}k}{2^{1000}}(1.75)^k(0.5)^{1000-k}$$
$$=\sum_{k=0}^{1000}\binom{1000}k(0.875)^k(0.25)^{1000-k}$$
$$=(1.125)^{1000}\sum_{k=0}^{1000}\binom{1000}k(0.875/1.125)^k(0.25/1.125)^{1000-k}$$
The sum in the last expression is over the probabilities of every possible result in a binomial $n=1000,p=0.875/1.125$ random variable, so is 1. The whole thing reduces to $1.125^{1000}=1.421\cdot10^{51}$.

This is counterintuitive, and the reason for this is that the distribution of ending values has a very long tail (positive-skewed) which is not very apparent from the simulation.

After 1000 rounds, we cannot say, what the expected emount is, as that is down to chance.

However, taking the worst possible outcome, every time (that is, losing), you will be left, with an extremely small number, which, however, is unequal to zero.

The amount will only me equal to zero, is if someone loses the game, an infinite amount of times


how many smaller circles(radius is equal) I can fit within a larger circle then the question is,the larger radius D,the small radius d,get the largest number of small circle put in the larger?

The answer can be closely approximated by this equation:

Number of Circles $= 0.83\frac{R_2^2}{r_1^2} - 1.9$  (rounded down to whole number)

where: $R_2$ = radius of larger circle
       $r_1$ = radius of smaller circle


How to calculate $\iint_D y\, dx\, dy$, $(D = \{(x, y) : x \le y \le 2x, 0 \le x \le 3\} )$? How to calculate $$\iint_D y\, dx\, dy,$$ where $$D = \{(x, y) \mid x \le y \le 2x, 0 \le x \le 3\} )?$$

This is the first time I have to solve such question, so I have definitely no idea how to do it.

If you can help me HOW to solve it, and not SOLVE IT for me,

I'll be very thankful.

What is the $D$ below the inner integral?

Thanks in advance!

Check that you can apply Fubini's theorem, and integrate first with respect to $y$.

$$\int \int_D ydxdy $$
= $$\int_0^3 \int_x^2xydxdy$$

= $$\int_0^3(x\int_x^2ydy)dx$$

= $$\int_0^3\frac{4x-x^3}{2}dx$$

Now you can do it


Prove that every maximal planar graph of order 4 or more is 3-connected I thought I might be able to use the fact that for a maximal planar graph the minimum degree of the  graph is at least 3, but I couldn't figure anything out. Am I headed in the right direction? Where do i go from there?

If any more detail is needed, my answer is based on this.

If an edge $e$ of a maximal planar graph is in exactly two triangles then $G/e$ is also maximal planar. (Since $G/e$ has exactly three fewer edges.)

If $G$ is a maximal planar graph with $n \geq 4$ vertices then there are at least $n$ such edges. (Induction on $n$.)

Let $G$ be a maximal planar graph with at least four vertices. Assume that there are vertices $u,v$ such that $G-\{u,v\}$ is disconnected. Let $X$ be one component of $G-\{u,v\}$ and let $Y$ be another.

Contract edges of $G$ until you're left with just $u$ and $v$ along with one vertex of $X$ and one vertex of $Y$.

Since there are now only four vertices, we must be left with $K_4$. But then removing $u$ and $v$ can't disconnect the graph, so we have a contradiction.

This is not true.  Consider $G=(V, E)$, where $V=\{1,2,3,4\}$ and $E=\{(1,2), (2,3), (3,4), (1,4), (1,3)\}$.  If we remove nodes 1 and 3, the graph is not connected, thus $G$ is not 3-connected.


Probability of independent not mutually exclusive events I am wondering if I am doing this right. If an event has a probability of happening of $0.9\%$, what is the probability it will happen if it is repeated $50$ times. I have tried to calculate this with this formula:

  $1-0.009^{50}$

Which gives $1$, so what I am wondering is it really true that if an event has a chance of only $0.9\%$ of happening once, if repeated $50$ times it will happen for sure, or have I used the wrong formula to calculate this?

$0.009^{50}$ is the probability the event happens $50$ times out of $50$
$1-0.009^{50}$ is the probability the event happens fewer than $50$ times out of $50$
$(1-0.009)^{50}$ is the probability the event happens $0$ times out of $50$
$1-(1-0.009)^{50}$ is the probability the event happens more than $0$ times out of $50$

Because your calculator can only calculate up to certain precision. For a value that is too small, say $0.009^5$ in your case, the calculator regards it as $0$.


Compact operator with closed image Let $K$ be a compact operator between two normed spaces. If $K(X)$ is closed, does this necessarily imply that $K(B)$ is closed?

where $B$ is the closed unit ball?

Consider $f:c_0\rightarrow R$ defined by $f(e_i)={1\over 2^i}$, it is compact and the image of the unit ball is $(-2,2)$. where $c_0$ is the set of sequences with finitely non zero terms endowed with $\|\|_{\infty}$.


if p then q unless r -- do i understand this? the statement is: if p, then q, unless r

I'm to convert it into a compound proposition.

I've reasoned that:

when p is true AND r is true, q is false
when p is true AND r is false, q is true
when p is false, we can't know if q is true

But, how to convert this into compound propositions?

$ (p \land \lnot r) \to q $
$ (p \land r) \to \lnot q $

How can we express both of these at the same time? or does each of them already contain the other?

Start with a truth table

 p | q | r || (p & ~r) -> q |(p & r) -> ~q  ||  p, then q, unless r
---|---|---||----------m----|--------m------||-----------------------
 T | T | T ||    F F   T    |   T    F  F   ||           ?
 T | T | F ||    T T   F    |   F    T  F   ||           ?
 T | F | T ||    F F   T    |   T    T  T   ||           ?
 T | F | F ||    T T   F    |   F    T  T   ||           ?
 F | T | T ||    F F   T    |   F    T  F   ||           ?
 F | T | F ||    F T   T    |   F    T  F   ||           ?
 F | F | T ||    F F   T    |   F    T  T   ||           ?
 F | F | F ||    F T   T    |   F    T  T   ||           ?

First conclusion $ (p \land \lnot r) \to q $ and $ (p \land r) \to \lnot q $ are not identical.

Then what is the truth table of "p, then q, unless r" ?

That question you have to do yourself, and you will need to argue for each line without reference to the formula's

GOOD LUCK

It's $\lnot r \rightarrow (p \rightarrow q) $ I think


Finite Measure Space: Integral Closure = Bochner Integral I can't sleep for so long time as the integral gives me headaches.
I was looking for so many approaches. Now another one. Hope this works...

Let $\Omega$ be a finite measure space and $E$ a Banach space.
Consider the function space $\mathcal{F}(\Omega,E)$ with the topology of $\mu$-almost everywhere convergence.
Define the integral of a a simple function $s=\sum_i a_i \chi_{A_i}$ as $I(s):=\sum_i a_i\mu(A_i)$.

Is the operator closable, i.e.:
$$(s_n,I(s_n))\to(0,c)\implies(c=0)$$
(Note that the measure space is finite $\mu(\Omega)<\infty$, so there can't be running blocks to infinity...)

Moreover does the closure coincide with the Bochner integral?

Ok to bad, the answer is again: No!

For $s_n:=n\chi_{[0,\frac{1}{n}]}$ it follows $s_n\to 0$ but $I(s_n)\to 1$.


Given $0 < x \le 1 \le y$. Calculate the minimum value of $\frac{x}{y + 1} + \frac{y}{x + 1} + \frac{x^2 + 3y^2 + 2}{6xy(xy + 1)}$. Given $0 < x \le 1 \le y$. Calculate the minimum value of $$\large \dfrac{x}{y + 1} + \dfrac{y}{x + 1} + \dfrac{x^2 + 3y^2 + 2}{6xy \cdot (xy + 1)}$$

We have that $$\frac{x^2 + 3y^2 + 2}{6xy \cdot (xy + 1)} \ge \left[ \begin{align} \frac{2 \cdot (x + 3y + 2)^2}{12xy \cdot (6xy + 6)}\\ \frac{2 \cdot (2xy + 4y)}{12xy \cdot (xy + 1)} \end{align} \right. = \frac{8 \cdot y(\sqrt{x} + 2)^2}{9 \cdot (3xy + 1)^2}$$

But that's all I have.

Show that $$\frac{x}{y+1}+\frac{y}{x+1}+\frac{x^2+3y^2+2}{6xy(xy+1)}\geq \frac{3}{2}$$
The equal sign holds if $x=y=1$


Find the spectrum of the operator $T: \ell^2(\mathbb{C}) \to \ell^2(\mathbb{C})$ defined by $(Tx)_n = \frac{x_n}{n}$ Consider the linear operator $T:\ell^2(\mathbb{C}) \to \ell^2(\mathbb{C})$ defined as
$$ (Tx)_n = \frac{x_n}{n}, \quad x \in \ell^2(\mathbb{C}). $$
I can show that it is bounded with norm $\|T\|=1$, which tells me that
$$ \sigma(T) \subseteq \{ \lambda \in \mathbb{C} : \,\, |\lambda| \le 1 \}. $$
I am also able to show that the point spectrum is
$$ \sigma_p(T) = \left\{ \frac{1}{n} \right\}_{n \in \mathbb{N} }.$$
Finally I know that the residual spectrum is empty because $T$ is self-adjoint.

How can I go on and find the continuous spectrum of $T$?
Is there a "more direct" way to show that the residual spectrum of $T$ is empty, without using the self-adjointness property?

One can show that $T$ is a compact operator: define $S_n:\ell^2(\mathbb{C}) \to \ell^2(\mathbb{C})$ by
$$(S_nx)_m = \left\{ \begin{matrix}
\frac{x_m}{m} & m \leq n \\
0 & m > n
\end{matrix}
\right.
$$
Note that the $S_n$ are finite-rank and that
$$((T-S_n)x)_m = \left\{ \begin{matrix}
0 & m \leq n \\
\frac{x_m}{m} & m > n
\end{matrix}
\right.
$$
so $\lVert T-S_n\rVert_2^2 = \frac{1}{n+1} \to 0$ as $n \to \infty$. Hence, $T$ is compact. By the Fredholm Alternative, the non-zero spectrum of $T$ consists purely of eigenvalues (i.e. the point spectrum). Since the spectrum must be closed, it must also contain $0$; as $T$ is injective and bounded, $0$ cannot belong to the point or residual spectra so it belongs to the continuous spectrum.

For $\lambda \notin \{0\} \cup \sigma_p(T)$, you can explicitly write down
$(\lambda I - T)^{-1}$.


How to show $\frac{d}{d x}\left(|x|^{1/2}\right)=\frac{x}{2|x|^{3/2}}$? $$\frac{d}{d x}\left(|x|^{\frac{1}{2}}\right)=\frac{x}{2|x|^{\frac{3}{2}}}$$
and also the second derivative $$\frac{d^{2}}{d x^{2}}\left(\sqrt{|x|}\right)=\frac{\delta(x)}{\sqrt{|x|}}-\frac{x^{2}}{4|x|^{\frac{7}{2}}}$$
I know this may seem basic. But I don't get how the Dirac delta (thanks for the comments for correcting me) comes in. Is it just a way someone has decided to write the result for $x >0$, $x<0$ and $x=0$ cleverly in one line. Is it arbitrary as long as the Kronecker term beats the other term?
If anyone has a clean derivation for these I'd be grateful.
EDIT: thanks for a lot of your answers, I've learnt a lot.
Still not entirely sure with the second derivative, the formula is here: https://www.wolframalpha.com/input/?i=second+derivative+of+%7Cx%7C%5E1%2F2, is this wrong? I still think it is probably correct. From taking something like. d/dx($\operatorname{sign}(x) \frac{1}{2|x|^{\frac{1}{2}}} )=1/2 ($$\frac{\delta(x)}{\sqrt{|x|}} -\frac{sign(x)*x}{2|x|^{5 / 2}})$ Don't know why first summand is missin 1/2. I know its distributions, but can't you still differentiate sign(x), you just have to be careful and when evaluating use test functions and integrals

Let $f(x) = |x|^{1/2}$. Of course, away from $x=0$, the answer of Robert Lee (without any Dirac delta) is correct, and you will just have
$$
\begin{align*}
f'(x) &= \frac{x}{2\,|x|^{3/2}}
\\
f''(x) &= \frac{-1}{4\,|x|^{3/2}}
\end{align*}
$$
for any $x\in\mathbb{R}\setminus\{0\}$ (since $x^2/|x|^{7/2} = 1/|x|^{3/2}$).

So, to know if there is a Dirac delta, you have to take the derivative in the sense of distributions. To get the first derivative, since $f'(x) = \frac{x}{2\,|x|^{3/2}}$
everywhere except at $x=0$, we see that $f'(x) = \frac{x}{2\,|x|^{3/2}}$ almost everywhere. Moreover, since $|f'| = \frac{1}{2\,|x|^{1/2}}$ is locally integrable, we obtain that the identity $f'(x) = \frac{x}{2\,|x|^{3/2}}$ is also valid locally in $L^1$ and so also in the sense of distributions.
This means that to compute $f''$ in the sense of distributions, we just need to compute the derivative of $f'$ in the sense of distributions. Let $\varphi$ be a smooth compactly supported (let say $\varphi = 0$ in $[-a,a]$ for some $a>0$) test function. Then by definition and the above computation
$$
\langle f'',\varphi\rangle = \langle(f')',\varphi\rangle = -\langle f',\varphi'\rangle = -\langle \tfrac{x}{2\,|x|^{3/2}},\varphi'\rangle
$$
and we can write this as an integral since both functions are locally integrable. Therefore
$$
\begin{align*}
\langle f'',\varphi\rangle &= -\int_{\mathbb{R}} \tfrac{x}{2\,|x|^{3/2}}\,\varphi'(x)\,\mathrm{d}x = -\int_{-a}^a \tfrac{x}{2\,|x|^{3/2}}\,\varphi'(x)\,\mathrm{d}x
\\
&= -\int_{-a}^a \tfrac{x}{2\,|x|^{3/2}}\,(\varphi(x)-\varphi(0))'\,\mathrm{d}x
\end{align*}
$$
Since $\varphi(x)-\varphi(0)$ is vanishing at $x=0$ and even better, $|\varphi(x)-\varphi(0)| ≤ C\,|x|$ (by the fundamental theorem of calculus, with $C= \sup |\varphi'|$ for instance), we can integrate by parts and use the value of $f''$ that was already computed for $x≠0$ to get
$$
\begin{align*}
\langle f'',\varphi\rangle &= \left[-\tfrac{x}{2\,|x|^{3/2}}\,(\varphi(x)-\varphi(0))\right]_{-a}^a - \int_{-a}^a \tfrac{1}{4\,|x|^{3/2}}\,(\varphi(x)-\varphi(0))\,\mathrm{d}x
\\
&= - \int_{-a}^a \tfrac{1}{4\,|x|^{3/2}}\,(\varphi(x)-\varphi(0))\,\mathrm{d}x
\end{align*}
$$
since $\varphi(a) = \varphi(-a) = 0$. The distribution $\mathrm{pf}(\frac{1}{|x|^a})$ for $a\in(1,2)$ is known as the finite part of $\frac{1}{|x|^a}$ and is defined exactly by
$$
\begin{align*}
\langle \mathrm{pf}(\tfrac{1}{|x|^a}),\varphi\rangle &= \int_{\mathbb{R}} \tfrac{1}{|x|^{a}}\,(\varphi(x)-\varphi(0))\,\mathrm{d}x
\end{align*}
$$
so in the sense of distributions
$$
f'' = \frac{-1}{4} \mathrm{pf}\left(\frac{1}{|x|^{3/2}}\right)
$$

Remarks: I do not know where you found the formula with a Dirac delta but it is obviously wrong, since there is no meaning to $\frac{\delta(x)}{\sqrt{x}}$.
Even if it is true that $\mathrm{sign}' = \delta_0$ in the sense of distributions, the following computation
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x}{|x|^{3/2}}\right) = \frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{sign}(x)\frac{1}{|x|^{1/2}}\right) = 2\delta_0(x)\left(\frac{1}{|x|^{1/2}}\right) + \mathrm{sign}(x)\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{|x|^{1/2}}\right)
$$
is not valid since one cannot multiply singular distributions in general.

Okay.... second go.....
If $1 =\int_{-\infty}^{\infty} \delta(x) dx$ then if we have a function $f(x)$ where $f(x)$ "equals $\frac 10$" when $x=0$  I guess can say:
$f(x) = 1*f(x) = [\int_{-\infty}^{\infty} \delta(x) dx]f(x)$ so
$f'(x)= [\int_{-\infty}^{\infty} \delta(x) dx]'f(x) + [\int_{-\infty}^{\infty} \delta(x) dx]f(x)= \delta(x)f(x) + f'(x)$.
When $x\ne 0$ then this is $f'(x)$.  If $x=0$ then... well I'm waving my hands wildly... but as $f(x)$ is assymptotic at $x=0$ then the derivative is infinite.
So.... $|x|^{\frac 12}$ is defined everywhere but has a "divot" at $x=0$.
the first derivative if $sign(x) \frac 1{2|x|^{\frac 12}} = \frac x{2|x|^{\frac 32}}$.
Not the first derivative "equals $\frac 10$ at $x = 0$ so if we take that the first derivative is $[\int_{-\infty}^{\infty} \delta(x) dx]\cdot\frac x{2|x|^{\frac 32}}$ then the second derivative will be $\delta(x)sign(x) \frac 1{2|x|^{\frac 12}} + [sign(x) \frac 1{2|x|^{\frac 12}}]'=$
$\pm \delta(x)sign(x) \frac 1{2|x|^{\frac 12}} -\frac{x^{2}}{4|x|^{\frac{7}{2}}}$
.....
Okay, somehow in my not really knowing what I was doing and handwaving I handwaved $\frac {sign(x)}2$ in whereas the author doing whatever s/he was doing and knowing what she was doing probably had a reason they "canceled" out.
Actually, I think I can almost see why.  $sign(x) \frac 1{2|x|^{\frac 12}} = \frac x{2|x|^{\frac 32}}$ is negative of $x < 0$ and positive otherwise so so we don't really have the first derivative of $[\int_{-\infty}^{\infty} \delta(x) dx]\cdot\frac x{2|x|^{\frac 32}}$ but the $sign(x)[\int_{-\infty}^{\infty} \delta(x) dx]\cdot\frac x{2|x|^{\frac 32}}$ and so the signs cancel out. And as we are doing it for negative and positive values of $x$ we ... are doing it twice????  Okay.... I'm handwaving again.
....
I don't know what the author was doing but I can see that as the first derivative is infinite at $x=0$ and as $\delta(x) =0$ whe $x \ne 0$ so the result is consistant and that weird chain rule for differentiating $f(x)_{x\ne 0}= 1 = \int \delta(x)dx$ I can squint and almost see what the author had in mind.


Sketching fuctions - standard meaning of terms? When sketching the graphs of polynomials, do certain terms always have the same meaning?

For example, in $f(x) = x^n + c$, the intersection with the y-axis is always $c$, no matter how high the degree of the polynomial (because $c$ is a constant of course). But I was wondering whether or not for example $x^1$ or $x^2$ and so forth also always have a certain 'effect' on a function, which would make sketching them easier.

The highest-order term of a polynomial always determines its general behavior very far from the origin. Low-order terms always are good estimates sufficiently close to the origin.


Example of $2$ non-isomorphic groups that have the same quotients I'm looking for an example of $2$ non isomorphic groups $G_1,G_2$  that are finitely generated and presented that have the same finite quotients (up to isomorphism) .

Thanks

In my answer to this question, I wrote down finite presentations from a paper by Baumslag and Solitar of two non-isomorphic groups that are each isomorphic to a quotient group of the other. So these two groups have exactly  the same quotient groups, not just the same finite quotient groups.

Let $Z_k$ be the cyclic group of order $k$.

Consider, $A = Z_2 \times Z_2$ and $B = Z_4$.

These are not isomorphic to each other and the only non-trivial quotient they have is $Z_2$.


Nth Term of Fractions How do I work out the Nth term of these fractions?
$$1,\frac{1}{4},\frac{1}{9},\frac{1}{16},\frac{1}{25},\dots$$
Would I need to change them all into decimals but that would be quite complicated and then go from there? Yes I can see a pattern, the denominators are squared numbers but what do I do after? I know that answer is going to have a $^2$ in it.

See if you can find any pattern in the fractions or their reciprocals.

$$T(n)= \frac{1}{n^2}$$ where $n$ is greater then equal to 0.


Inner dependence of Independent random vectors If $X = (X_1,X_2)$ and $Y = (Y_1,Y_2)$ , $X$ and $Y$ are stochastically independent can $X_1$ and $Y_1$ be dependent?

Hint:
$$
P(X_1\in A,Y_1\in B)=P(X\in A\times\mathbb{R},Y\in B\times\mathbb{R}).
$$

Let us consider that all variables are continuous and let us consider probability density functions $f$ that are distinguished by their arguments.

The independence of $X$ and $Y$ means
$$f(X,Y)=f(X)(Y)$$
The right hand side can be written using the chain rule as
$$f(X)f(Y) = f(X_2,X_1)f(Y_2,Y_1)=f(X_2|X_1)f(X_1)f(Y_2|Y_1)f(Y_1)=f(X_2|X_1)f(Y_2|Y_1)f(X_1)f(Y_1)=f(X_2,Y_2|X_1,Y_1)f(X_1)f(Y_1)$$
where the last equality stems from the fact that $X_2$ and $Y_2$ do not occur in $f(X_1)f(Y_1)$ and $f(X_1)f(Y_1)$ define a probability density on $X_2,X_1$.

The left hand side can be also decomposed by the chain rule
$$f(X,Y)=f(X_2,Y_2|X_1,Y_1)f(X_1,Y_1)$$

From the equality of left and right sides we can see the independence of $X_1$ and $X_2$.


Find the ages of the three Mabvuto is half as old again as Matthew, Matthew is half as old again as Ngilazi. Given that the sum of their ages is 114, find their ages. My work just messes up, I need someone to help me!

The key word is "again."  In English parlance, to say that Mabvuto is "half again as old" as Matthew means that Mabvuto is Matthew's age plus half Matthew's age.  That is, if we let $Mb$ denote Mabvuto's age and $Ma$ Matthew's, then $Mb = Ma + {1\over2}Ma= {3\over2}Ma$.  Likewise, the statement that Matthew is half again as old as Ngilazi implies $Ma = {3\over2}Ng$, where $Ng$ denotes Ngilazi's age.  Thus

$$114=Mb+Ma+Ng = {3\over2}Ma+{3\over2}Ng+Ng = {3\over2}\cdot{3\over2}Ng+{3\over2}Ng+Ng = {19\over4}Ng,$$

from which ones gets $Ng=24$, $Ma=36$, and $Mb=54$.

Denote the respective ages as $x,y,z$. Then you are given
$$ x=\frac12y$$
$$ y=\frac12z$$
$$x+y+z=114$$
You should be able to solve this.


markov chains and coin flips A coin that comes up heads with probability p is continually flipped until the pattern T T T H appears. Let X denote the number of flips, find EX.

If I use Markov chains is there a simpler way to set up the states than using each state as a different combo? Or should I proceed in that way so that I have 16 states?

Define the state of the  Markov chain to be the sequence of the previous 4 coin tosses. The invariant distribution $\pi$ puts mass $p(1-p)^3$ on the state $TTTH$.
Therefore the expected number of tosses needed to reach $TTTH$ again, starting at $TTTH$ is ${1\over\pi(TTTH)}={1\over p(1-p)^3}.$ For this pattern, that is the same as starting from scratch, so this answers the question.

There is a trick to solve this type of problems without any algebra.

In general you have two cases. The one where the pattern overlaps and the one without overlaps:

Overlap means that a part of the series at the beginning is repeated exactly at the end of the pattern. Example: A pattern HTHHTHTH. In this case the pattern 'HTH' is repeated at the beginning and EXACTLY at the end of the pattern. Also, the single 'H' is an overlap.

The above pattern has expected time $\frac{1}{p^5q^3} + \frac{1}{p^2q} + \frac{1}{p}$. Which means that you find the expected value of a series of tosses not taking into account the overlaps. And add the overlaps.

In your case there is no overlap. So the answer is simply: $\frac{1}{pq^3}$ as was pointed out.

Look at section 3.6.4 of Introduction to Probability models by Seldon Ross to see a thorough explanation of this formula.


How to show that $\sum_{k=1}^n k(n+1-k)=\binom{n+2}3$? While thinking about another question I found out that this equality might be useful there:
$$n\cdot 1 + (n-1)\cdot 2 + \dots + 2\cdot (n-1) + 1\cdot n = \frac{n(n+1)(n+2)}6$$
To rewrite it in a more compact way:
$$\sum_{k=1}^n k(n+1-k)=\frac{n(n+1)(n+2)}6.$$

This equality is relatively easy to prove:
$$\sum_{k=1}^n k(n+1-k)=
(n+1)\sum_{k=1}^n k - \sum_{k=1}^n k^2 =
(n+1) \frac{n(n+1)}2 - \frac{n(n+1)(2n+1)}6  = n(n+1) \left(\frac{n+1}2-\frac{2n+1}6\right) = n(n+1)\frac{3(n+1)-(2n+1)}6 =
\frac{n(n+1)(n+2)}6.$$
(We only used the known formulas for the
sum of the first $n$ squares and the
sum of the first $n$ numbers.)

Are there some other nice proofs of this equality? (Induction, combinatorial arguments, visual proofs, ...)

EDIT: Now I found another question which asks about the same identity: Combinatorial interpretation of a sum identity: $\sum_{k=1}^n(k-1)(n-k)=\binom{n}{3}$ (I have tried to search before posting. But the answers posted here so far gave me some new ideas for good keywords to search which lead me to finding that question.) The questions are, in my opinion, not exact duplicates since the other question asks specifically about combinatorial proofs and my question does not have that restriction. But I agree that this is a very minor distinction. In any case, if you think that one of them should be closed as a duplicate, then you can vote to close. I will refrain from voting to close/reopen on this question. (If one of the two questions is voted to be a duplicate of the other one, they probably cannot be merged, since the summation variables are off by one.)

Let us choose three numbers from $\{0,1,2,\ldots, n+1\}$, beginning with the middle one, which has to be some $k\in \{1,\ldots,n\}$. We then have $k$ choices for the smallest and $n+1-k$ choices for the largest of the three. It follows that
$${n+2\choose3}=\sum_{k=1}^n k(n+1-k)\ .$$

from an arithmetic point of view the sum of this series is just
$((n+1)^3-(n+1)/6)$
let's take an example
$1\cdot5+2\cdot4+3\cdot3+2\cdot4+1\cdot5$
here $n=5$
so the value will be
$\frac{(6^3-6)}{6}$
$=\frac{216-6}{6}$
$=\frac{210}{6}$
$=35$
this is another approach to get your answer.


Let $G$ be a simple group. Show that if $H$ is a subgroup $G$ such that $[G:H]= 3$ then $H=\{1\}$ and $|G|=3$. MY ANSWER: Since $G$ is simple the only normal subgroups of $G$ are $\{1 \} $ is $ G $ itself. As $ G / H $ is a quotient group it follows that $ |H|\neq 3 $ because otherwise the $ H $ index in $ G $ would be 1. Therefore, $ H = \{1 \} $ and therefore $ | G | = 3 $.

Can you check if my answer is right?

Your argument is thoroughly incorrect (and almost incoherent, to be honest). You do not know that $H$ is normal, you are only told it has index $3$ and is a subgroup. So you cannot assert that $G/H$ is even a group. Moreover, you cannot assert that $|H|\neq 3$, and your reasoning does not make sense; if $|H|$ were equal to $3$, that would not imply that the index is $1$... at least not immediately; what makes you think it does? It is certainly possible for a subgroup to have the same order as its index, and the order and the index of a subgroup need not be related at all.  All you know is that the subgroup has index $3$.

Instead, note that the action of $G$ on the left cosets of $H$ induces a homomorphism $G\to S_3$ whose kernel is a normal subgroup contained in $H$, and since $G$ is assumed to be simple, the kernel must be trivial. Can you take it from there?

On page 253 in this, we see that "Corollary 12.6. Let $G$ be a finite group, and let $H \leq G.$ Assume that $|G : H| = p,$
and let $p$ be the smallest prime divisor of $|G|$. Then $H\unlhd G.$ And your question deduces $p=3.$


How can I solve this nonlinear Partial differential equation? Can anyone help me, how to find the exact solution of this PDE:
$$
\begin{align}
u_{tt} &= u^2 u_{xx} \\
u(x,0)&=g(x) \\
u_t(0, t)&=0
\end{align}
$$
Give me roughly idea or method please.
Thank you.

Let $u(x,t) = X(x)T(t)$
$$\implies XT'' = X^2T^3X''$$
$$\implies XX'' = \frac{T''}{T^3} = \lambda$$
Now solving each equation in $X$ and $T$ separately, we have
$$(1) \space \space X'' = \frac{\lambda}{X} $$
$$(2) \space \space T'' = \lambda T^3$$
Can you solve each of these independently, apply the boundary conditions to get your solution?


The annihilator of an intersection is the sum of annihilators Given a subset $X$ of a vector space $V$, let $X^\circ$ be the annihilator of $X$, that is $X^\circ = \{y\in V^* | \; y(x)=0, \;\forall\; x\in X\}$, where $V^*$ is the dual space of $V$.

Question: If $M$ and $N$ are subspaces of an $n$-dimensional vector space $V$, then $(M\cap N)^\circ =M^\circ +N^\circ $.

It's easy to prove this result by using that $(M+N)^\circ =M^\circ \cap N^\circ $, which is easy to prove too.

My doubt is: Is there a "direct way" to prove that $(M\cap N)^\circ =M^\circ +N^\circ $ (without make use of other equality)?

Thanks!

There is a reason why the first relation
$$(M+N)^\circ =M^\circ \cap N^\circ \tag1$$
is easier to prove than
$$(M\cap N)^\circ =M^\circ +N^\circ\tag2$$
Indeed, (1) is true for arbitrary vector spaces. (The proof is short: a linear function annihilates $M+N$ if and only if it annihilates both $M$ and $N$.)

But (2) fails for general vector spaces (it holds for finite dimensional ones). More  specifically, the inclusion $(M\cap N)^\circ \supseteq M^\circ +N^\circ$ is true in general vector spaces, but not the reverse inclusion. So, any proof of (2) must use the finiteness of dimension somewhere. It cannot be a twin of the proof of (1).

(A gerw said, (2) can be made true in full generality by replacing $M^\circ +N^\circ$ with $\overline{M^\circ +N^\circ}$.)

You may work with orthonormal bases:
Let $\{u_i\} \cup \{v_i\} \cup \{w_i\} \cup \{x_i\}$ a basis in $X$, such that $\{u_i\}$ is a basis of $M\cap N$, $\{u_i\} \cup \{v_i\}$ of $M$ and  $\{u_i\} \cup \{w_i\}$ of $N$.

Then, $M^\circ$ is the linear hull of $\{w_i\} \cup \{x_i\}$, $N^\circ$ is the hull of $\{v_i\} \cup \{x_i\}$ and $(M \cap N)^\circ$ is the hull of $\{v_i\} \cup \{w_i\} \cup \{x_i\}$ (if I have done anything right).

Anyway, this is a mess. The easy proof you mentioned has also the advantage, that it works in infinite dimensions (in this case you have that the closure of $M^\circ + N^\circ$ is equal to $(M\cap N)^\circ$).


How to solve implicit solution of IVP The given question is:
$$
 dy/dt + 2y = 1\ ;\qquad    y(0)= 5/2
$$
when i solve this i get $\ln(-4)=c$
now the problem is how to solve $\ln(-4)$?

You can get rid of the "arbitrary constant" as follows.

Cauchy's problem:

$$\tag{1} \begin{cases} y^\prime (t) +2y(t)=1 \\ y(0)=5/2 \end{cases}$$

has unique global solution (because the ODE is linear with constant coefficients); this solution is continuous in $\mathbb{R}$, hence $y^\prime (t)=1-2y(t)$ is continuous in $\mathbb{R}$ and $y(t)$ is of class $C^1$ in $\mathbb{R}$... Bootstrapping one gets $y(t)$ of class $C^\infty$ in $\mathbb{R}$.

Moreover, the ODE in (1) has only one constant solution, namely $\bar{y}(t):=1/2$, and it forces its nonconstant global solutions to be either $>1/2$ or $<1/2$ everywhere in $\mathbb{R}$: in fact, if a solution $\phi(t)$ has $\phi(t_1)<1/2<\phi(t_2)$, then (by continuity) there exists $T$ s.t. $\phi(T)=1/2$ therefore $\phi(t)=\bar{y}(t)$ everywhere (by uniqueness), a contradiction. In the present case, since $y(0)=5/2>1/2$, one has $y(t)>1/2$ everywhere in $\mathbb{R}$; hence $y^\prime (t)=1-2y(t)<0$ in $\mathbb{R}$, so that $y(t)$ is strictly decreasing.

Now, let $t>0$; one can divide both sides of $y^\prime (t)=1-2y(t)$ by $1-2y(t)\neq 0$ and integrate over $[0,t]$ to get:

$$\int_0^t \frac{y^\prime (\tau)}{1-2y(\tau)}\ \text{d} \tau = \int_0^t \text{d} \tau \; ;$$

since $y(\tau)$ is of class $C^\infty$ and strictly decreasing, one can make the change of variable $\eta = y(\tau)$ in the LHside to get:

$$\int_{5/2}^{y(t)} \frac{1}{1-2\eta}\ \text{d} \eta =t\; ,$$

hence:

$$-\frac{1}{2}\log |1-2\eta|\Big|_{5/2}^{y(t)}=t \; ;$$

keeping in mind that $1-2y(t)$ is negative in $\mathbb{R}$, one finds:

$$\log \frac{4}{2y(t)-1} =2t$$

and finally:

$$\tag{2} y(t)=2e^{-2t}+\tfrac{1}{2} \; .$$

On the other hand, if $t<0$ one has to integrate over $[t,0]$ to recover the same elementary expression (2) for the solution. Therefore (2) gives the unique global solution to problem (1).

this is a linear differential equation and the solution of this equation is as follow:

IF= = e^∫▒〖2 dt〗
= e^2t
Now solution of equation is:
y e^2t = ∫▒〖1 .e^2t  dt〗
y e^2t = e^2t/2 + c  ……………………….eq(1)
Now put t=0 and y =  5/2
5/2 = 1/2+ c
C = 2
Putting this in eq(1), we get
y e^2t = e^2t/2+ 2
y=1/2+  1/2 e^(-2t)


Is $\frac{0}{\infty}$ indeterminate? I have been searching for an answer for this for half an hour and I can't seem to find one. I've lots of information about other combinations of $0$ and $\infty$ but I haven't seen anything that says whether $\frac{0}{\infty}$ is indeterminate?

This quotient is zero. This is true whether you mean projective infinity or extended positive infinity.

Since you say "indeterminate", I assume you're talking about limit forms. Well, division is continuous and defined at $0/\infty$, so this is, indeed, not an indeterminate form.

This is actually pretty easy to show that this limit form works out to $0$ by the $\epsilon-\delta$ definition of limit: you should try to do so on your own. (and ask a new question here on MSE about it if you get stuck)


Finding limit of a recursively defined sequence Let $(x_{n})_{n\geq1}$ be a sequence defined by:
$x_1=1$ and $x_n=n(x_{n+1}-\frac{n+1}{n^2})$. Calculate $\lim_{x\rightarrow \infty }nx_n$.
We can write $x_n=n(x_{n+1}-\frac{n+1}{n^2})$ as $(n+1) x_{n+1} = \frac{n + 1}{n^2}(n x_n + (n+1))$, and by the substitution $y_n=nx_n$ we obtain: $y_{n+1} =  \frac{n + 1}{n^2}y_n + \left(1 + \frac{1}{n}\right)^2$. How to go on?

It is clear that $ x_n> \frac{1}{n}.$

Using mathematical induction is easy to see that $ x_n < \frac{1}{n-4} $ for $n\geq6.$

Proof by mathematical induction:

$x_6=\frac{199}{450}<\frac{1}{2}$

$x_{n+1}= \frac{n+1}{n^2} + \frac{x_n}{n}< \frac{n+1}{n^2}+\frac{1}{n(n-4)}=\frac{n^2-2n-4}{n^2(n-4)}. $

Because $\frac{n^2-2n-4}{n^2(n-4)}<\frac{1}{n-3}<=>-n^2+2n+12<0$ assertion is proved.

It follows that for $n\geq6$  $$\frac{1}{n}<x_n < \frac{1}{n-4} $$

and consequently $\ lim_{n\to\infty} nx_n=1.$

If the sequence converges, then when $n\to \infty$, we can set $c=x_n=x_{n+1}$, and the limit $c$ satisfies
$$c=n\left(c-\frac{n+1}{n^2}\right)$$
from which $c$ is solved to be $1/n$.

The limit $\lim_{n\to \infty}nx_n$ is thus $1$.


How should one explain the conventional formula for relative variation of signed variables? The conventional formula for the relative variation between two signed values is $\displaystyle \frac{b-a}{|a|}$. Unfortunately, I could not find any clear justification for it, especially when considering examples like this one: according to the conventional formula, the relative variation between -10 and +20 is +300%, while the relative variation between -20 and +20 is +200%. This is puzzling, because the absolute variation for the former is +30 while the absolute variation for the latter is +40 and both datasets end at the same +20 value, yet the relative variation for the former is greater than the relative variation for the latter.
How can one justify this counter-intuitive scenario?

To better understand what is going on, it usually helps to think of the relative growth between two values of different signs as being composed of two separate parts: the relative growth from the first value to zero, plus the relative growth from zero to the second value. For example, the relative growth between $-10$ and $20$ is equal to the relative growth between $-10$ and $0$, plus the relative growth between $0$ and $20$.
The relative growth between $-10$ and $0$ is $\displaystyle \frac{0-(-10)}{|-10|} = \frac{10}{10} = 1 = 100\%$ according to our previous formula.
In fact, the relative growth between any negative value and $0$ is always equal to $100\%$, which actually makes sense when one comes to think about it. But things become a bit more tricky when we need to compute the relative growth between $0$ and a positive value, like $20$ in our previous example. There, we cannot simply use our formula, because it would lead to a division by zero. Instead, we have to compute this relative growth in relation to the previous one, by computing the ratio of two absolute growths.
For example, when going from $-10$ to $20$, we gain $10$ in absolute terms between $-10$ and $0$, then $20$ in absolute terms again between $0$ and $20$. Therefore, we gained twice as much going from $0$ to $20$ than we did when going from $-10$ to $0$. And as we have seen earlier, the relative growth between $-10$ and $0$ is $100\%$, therefore the relative growth between $0$ and $20$ should be twice that amount, or $200\%$, and the relative growth between $-10$ and $20$ should be the sum of our two relative growths, or $100\% + 200\% = 300\%%$, which matches what our adjusted formula gave us in the first place. In other words, we found two ways of getting to the same result, but this should not come as a surprise if you look at the original equation.
Indeed, if $a$ is negative, we can rewrite $\displaystyle \frac{b-a}{|a|}$ as $\displaystyle \frac{b-a}{-a}$. Then, we can split the fraction as $\displaystyle \frac{b}{-a} + \frac{-a}{-a}$ and further simplify it into $\displaystyle \frac{b}{-a} + 1$, or even $\displaystyle 1 + \frac{b}{-a}$. This addition has two addends, $1$ and $\displaystyle \frac{b}{-a}$. The former is the relative growth from the first value to zero (always equal to $1$), and the latter is the ratio of the absolute growth between the first value and $0$ and the absolute growth between $0$ and the second value, which we can write as $\displaystyle \frac{b - 0}{0 -a}$.
When we look at the formula through that angle, we realize that a relative growth between two values of different signs is only affected by the ratio between the two values $\displaystyle \frac{b}{-a}$, unlike the relative growth between two values of the same sign, which is the ratio between the difference of the two values and the first value $\displaystyle \frac{b-a}{a}$. But this is where the adjusted formula becomes misleading, because we should not look at the former as a ratio between two values. Instead, we should look at it as a ratio between two absolute growths $\displaystyle \frac{b - 0}{0 -a}$.
Make no mistake though: this is not a trivial arithmetic rewriting. Instead, by writing $\displaystyle \frac{b}{-a}$ as $\displaystyle \frac{b - 0}{0 -a}$, we clearly communicate the fact that our fraction is a ratio between two absolute growths, instead of a ratio between two values. And this is what explains that the relative growth between $-10$ and $20$ ($300\%$) is greater than the relative growth between $-20$ and $20$ ($200\%$).
As explained earlier, to grow from $-10$ to $20$, you first need to grow from $-10$ to $0$, then grow from $0$ to $20$, and the first growth component to $0$ is always equal to $100\%$. But when you grow from $0$ to $20$, while this absolute growth ($20$) is the same whether you started from $-10$ or $-20$, the relative growth is twice as much if you started from $-10$ than if you started from $-20$.
In other words, what the relative growth of values of different signs really measures is the relative growth after you have crossed the zero line, because the relative growth before you cross the zero line is always the same ($100\%$).

A quantity, $a$, is changed by $b-a$. What fraction of $a$ does that change constitute?
The answer is $$\dfrac{b-a}{a}$$
This is called the relative change because its value is computed "realtive to" $a$.
This is not a useful measure when $a$ is close to $0$.
Wikipedia: approximation error


Magnitude of velocity A car travels around a circular track with a radius of $r=250m$. When it is at point $A$ then $V_a=5m/s$ which increases at a rate of $\dot{v}=(0.06t)m/s$. Determine the magnitude of its velocity and its acceleration when it is $1/3$ around the track. My distance in this situation is obviously $524m$.

Asked this question earlier and while the answerer was extremely helpful the answer he gave me was incorrect or at least didn't give the answer my tutor is looking for. (I don't mean to appear ungrateful for the attempted help in my earlier question the gentleman was extremely helpful and nice.)

Help please

Hint: The acceleration would be the vector sum of the tangential and radial one, i.e $$|a|=\sqrt{(\frac{v^2}{r})^2+(\frac{dv}{dt})^2}$$
Now you have $a(t)=0.06t$, can you find $v(t)$? and subsequently $x(t)$? then you know $x=524m$, find t from this and then find v at that t and find |a| by the above relation.

Notice that the circular path of the car does not affect the velocity function. We use the radius only to find the distance traveled by the car so that we may use basic calculus to find the velocity and acceleration.

The distance is, as you have found, to be $\frac{1}{3} \times 2\pi(250),$ which is about $523.6.$

We know that
$$\int_{0}^{t} (0.06t + 5) dt = 523.6$$
$$\left[0.03t^{2} + 5t\right]_{0}^{t} = 523.6$$
$$0.03t^{3} + 5t^{2} = 523.6,$$
and we conclude that $t$ is about $9.94\text{s}.$

We can find the velocity simply by plugging in, so
$$v = 0.06 \times 9.94$$
$$v = \boxed{0.60\frac{\text{m}}{\text{s}}}.$$

Differentiating the velocity function, we have
$$a = \boxed{0.06\frac{\text{m}}{\text{s}^{2}}}.$$


Factorize $2a^3 - b^3 - c^3$ I need to factorize the expression $2a^3 - b^3 - c^3$. I see that one zero is achieved when $a=b=c$, but I can't find the factor(s).

I assume you're looking for factors that are polynomials with integer coefficients.

Your polynomial is not factorizable in this sense. To see why, look at what happens when $b=0,c=1$. You get $2a^3-1$, which you cannot factor over the integers.

$$2X^3-b^3-c^3=0$$

So $X=\sqrt[3]{\frac {b^3+c^3} 2}$ is a root, now do long division and use the quadratic formula.


Let $f: \mathbb{R}\setminus\{0\}\to\mathbb{R}$ be given by $f(x) = x + \frac{1}{x}$. Find $f((0,2))$. So for the question I know it's asking for the interval of the codomain that corresponds to the interval $(0,2)$ of the domain. I know that the upper part of the interval is $(y,2.5)$ but I don't know what $y$ would be. Since $0$ is not even in the domain of $f$, could $y$ be the empty set? This does not seem right to me though. Any help is appreciated.

First, find $f(x)$ at the boundaries.

At the lower bound $x = 0$
$$\lim_{x\to 0^+} \left(x + \frac{1}{x} \right)= \infty $$

At the upper bound $x = 2$
$$ \lim_{x\to 2^-} \left(x + \frac{1}{x} \right)= \infty = 2.5 $$

Second, check to see if the function has any local minima or maxima
$$ f'(x) = 1 - \frac{1}{x^2} = \frac{x^2-1}{x^2}$$

$ f'(x) = 0$ if $x = \pm 1$. One of these points, $x = 1$, lies within the interval of consideration. This is a local minimum as the function is very large and positive near $0^+$. This minimum value is $f(1) = 2$

To conclude, we found that $2 \le f(x) < \infty$ in $(0,2)$, so the codomain is $[2, \infty)$

Hint: $(x-1)^2=x^2-2x+1\geq 0$. What does this imply about
$$
x+\frac{1}{x}
$$
?


Find all group of order $20$ which is a semidirect product of a cyclic group of order $4$ by a cyclic group of order $5$ Problem: Find all group of order $20$ that are a semidirect product of a cyclic group of order $4$ by a cyclic group of order $5$.

My attempt: We knew that a cyclic group of order $n$ is isomorphic to $\mathbb{Z}_n$, so we have $C_4 \cong \mathbb{Z}_4$, $C_5 \cong \mathbb{Z}_5$ and $\mid C_4 \mid = \mid \mathbb{Z}_4 \mid = 4$, $\mid C_5 \mid = \mid \mathbb{Z}_5 \mid = 5$. Then we find a semidirect product of $\mathbb{Z}_4$ by $\mathbb{Z}_5$. To find this, we define an automorphism $\theta \colon \mathbb{Z}_5 \rightarrow \text{Aut}(\mathbb{Z}_4)$, while $\text{Aut}(\mathbb{Z}_n) = \{\sigma_k \mid k \in \mathbb{Z}, (k,n)=1\}$, so $\text{Aut}(\mathbb{Z}_4) = \{\sigma_1,\sigma_3\}$ and $\mid \text{Aut}(\mathbb{Z}_4) \mid = 2$.

My professor told that there is only the trivial homomorphism $\mathbb{Z}_5 \rightarrow \text{Aut}(\mathbb{Z}_4)$ since $(5,2)=1$.

So my question:

  Why from $(5,2)=1$ can we conclude that there is only one trivial homomorphism, and for the trivial homomorphism, how do I define it?

Why from $(5,2)$ we can conclude that there is only one trivial homomorphism

Let $f:\Bbb Z_5 \to \text{Aut}(\Bbb Z_4) \sim \Bbb Z_2$ be a homomorphism with $f(1)=a$. Now order of  $f(1)$ divides both $5$ and $2$. But their gcd is one, so $a$ must be identity and hence $f$ is trivial!

$$
\DeclareMathOperator{\Ker}{Ker}
\DeclareMathOperator{\Im}{Im}
\DeclareMathOperator{\Aut}{Aut}
$$
Also, there's another way to prove $f$ is trivial. Consider corollary from homomorphism theorem:
$$f: G → H, \quad\textrm{where $f$ is a homomorphism}$$
$$|G| = |\Im f|\cdot|\Ker f| $$ So in our case $5 = |\Im f| |\Ker f|$. We know that $|\Im f| <= |\Aut(Z_4)| = |\Aut(Z_2)| = 2$. Thus, the only possible way is that $|\Ker f| = 5 $ and $|\Im f| = 1$, which concludes $f$ is trivial homomorphism.


It's confusing to calculate Euler characteristic of this surface This pic below is an exploded view of a cone.

I'm trying to calculate the Euler characteristic of the surface made from the fragment $M$, i.e.,

At first I thought the Euler characteristic is 0, but the one who made this question says it is actually 1.

And this is actually a part of an exercise to find the total geodesic curvature of $\partial M$. I tried to figure it out using the Gauss-Bonnet theorem.

I'm sorry that the question is not clear.
$M$ is the region between top circle of cone and closed geodesic meeting the top circle of cone at one point.

If I'm reading that picture correctly, the Euler characteristic shouldn't exist.

The subset $M$ of the cone looks like it started as a topological disk facing left, and is curving to the right to connect back with itself, and will become a topological cylinder (a sphere with 2 disks removed).

A disk has $\chi=1$ , and a cylinder has $\chi=0$ . The surface $M$ is between those.

Nevertheless, Euler's formula can be applied to a polygonalization of $M$, formed as a square pyramid with 2 opposite triangles removed; the peak of the pyramid is the singular point.

The pyramid has $5$ vertices, $8$ edges, and $3$ faces. Thus $\chi=5-8+3=0$.

Gauss- Bonnet theorem  establishes the connection/identity between isometric (left hand side of equation) and Euler topological constant  (right hand side ). In short

$ \int K dA + \int \kappa_{g} \,ds= 2 \pi \chi$

where for compact surfaces the first term is total/integral curvature or solid angle in steridians, second term is  rotation in tangent plane measured in radians which together elegantly  sum up to Euler charactristic $2 \pi \chi.$ Sudden jumps with geodesics can be accommodated/interpreted as external angles sum $\Sigma \psi_{i}$ around the contour for the line integral.

For a Torus with cancelling geodesic sections:

$0+ 0=2 \pi \chi \rightarrow \, \chi=0 \tag1$

For a hemisphere bounded by an equator:

$2 \pi +0 = 2 \pi \chi \rightarrow \, \chi=1 \tag2 $

For a closed convex loop on a developable surface (Gauss curvature $K=0\,$ for cones, cylinders/developable helicoids) for either continuous ( like a circle in a flat plane or non-intersecting continuous loop on a curved surface) curves or discontinuous sloped curves (segment of circle $M$ like the one you sketched of developed cone patch):

$0+ 2 \pi=2 \pi \chi \rightarrow \, \chi=1 \tag3 $

The matter is thus established by isometry/topological considerations.

EDIT1:

I am in agreement with the question setter. I would suppose he wanted the student to recognize a group of such isometric/homeomorphic equivalents.


Semicircle Hanging Out of a Rectangle Let $ABCD$ be a rectangle so that $AB=CD=2$ and $BC=AD=x$, where $x<1$. A semicircle of radius $1$ and diameter $AB$ is drawn so that the semicircle intersects $CD$ at points $M$ and $N$. If the minor arc between $M$ and $N$ can be fit snugly in $ABCD$ with $MN$ on $AD$ (i.e. the minor arc can be fit tangent to the semicircle and has $MN$ on $AD$) then what is the value of $x$?

Apologies if this is difficult to decipher what I am exactly saying. I do not know how to draw a diagram, but if you have any clarifying questions I will be happy to answer it.

I have tried rotating the rectangle to form some sort of tangency argument, but to no avail...so like, draw the tangent minor arc fitting snugly in the rectangle as described. Then we "complete the semicircle" by drawing the rest of it underneath the original semicircle. Then we see we can make another copy of $ABCD$ which is a rotation of the first. Darn I wish I knew how to use geogebra lol...

Any ideas?

Not an answer, but I don't think we can put an image in a comment.  Here is an image.  The arc starting at $D$ is supposed to be the same size as the arc $MN$ and the two arcs are tangent.  Now I believe there is enough information to find $x$.

If I understand your question correctly, I think you're saying place the semicircle in a box and cut off the minor arc that doesn't fit, but try to fit it back in the box so that it's tangent to the original semicircle.  Now, in order to do this, you need to translate $M$ and $N$ to $M'$ and $N'$.  Let's say $M'=D$, and $N'$ lies on $\overline{AD}$ so that $\overline{M'N'}=\overline{MN}$.  Now, we essentially have two circles of equal radius that we are trying to move around so they are tangent at one point.  This means that we want their centers to be a distance of $r_1+r_2=2$ apart.

If we call $O_1$ the midpoint of $\overline{AB}$, let $O_1$ lie on the origin in the Euclidean plane.  Then $M=(-\sqrt{1-c^2},-c)$, $N=(\sqrt{1-c^2},-c)$, and the length of $\overline{MN}$ is $2\sqrt{1-c^2}$.  We then find that, since $M'=(-1,-c)$, $N'=(-1,2\sqrt{1-c^2}-c)$.  Since $M'$ and $N'$ lie on a circle with radius one, we can find the circle center on the perpendicular bisector of segment $\overline{M'N'}$.

Calling the second center $O_2$ and the midpoint of $\overline{M'N'}$ $X$, $\Delta M'XO_2$ is a right triangle with $\overline{M'O_2}=1$, $\overline{M'X}=\sqrt{1-c^2}$, so $\overline{O_2X}=c$.  Thus, $O_2=(-1-c,\sqrt{1-c^2}-c)$.

Now, we solve for $d(O_1,O_2)=2$. $(1+c)^2+(\sqrt{1-c^2}-c)^2=4$.  With some bash, this simplifies to $5c^4+4c^3-4c^2-8c+4=0$, and $c\approx 0.92464593203170915717\ldots$


Calculating $\int_0^{\pi/2} \frac{\ln(\sin x+\cos x)}{\sin x\cos x}\, dx$ How would you evaluate the following integral:
$$\int_0^{\pi/2} \frac{\ln(\sin x+\cos x)}{\sin x\cos x}\, dx$$

I can't see any helpful by parts or substitution ideas, e.g. $u=\sec x$ or $u=\csc x$ since the other integral that comes out of by parts don't evaluate nicely, so any help would be much appreciated. The integral should come out nicely as it’s part of a problem (at roughly first year undergraduate level) that says show a particular integral comes to something nice, which I’ve simplified to this.

Rewrite the integral as: $$\int_0^\frac{\pi}{2}\frac{2\ln(\sin x+\cos x)}{2\sin x\cos x}dx=\int_0^\frac{\pi}{2}\frac{\ln(1+\sin (2x))}{\sin (2x)}dx=\int_0^\frac{\pi}{2}\frac{\ln (1+\sin x)}{\sin x}dx =\frac{\pi^2}{8}$$ For the last integral look here and take $a=1$.

Note that if you use Weierstrass substitution you get:
$$I=2\int_0^1\frac{(1+t^2)(\ln(1+2t-t^2)-\ln(1+t^2))}{2t(1-t^2)}dt$$
This may work

EDIT

note:
$$\sin(x)+\cos(x)=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$$
also, we know the rule:
$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
but when applied to the original integral it does not help us.

Writing our integral as:
$$I=\int_0^{\pi/2}\frac{\ln\left(\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\right)}{\sin(x)\cos(x)}dx$$
we can now apply the substitution:
$$u=x+\frac{\pi}{4}$$
to obtain:
$$I=2\int\limits_{\pi/4}^{3\pi/4}\frac{\ln\left(\sqrt{2}\sin(u)\right)}{(\sin u+\cos u)(\sin u-\cos u)}du=2\int\limits_{\pi/4}^{3\pi/4}\frac{\ln\left(\sqrt{2}\sin(u)\right)}{\sin^2(u)-\cos^2(u)}du$$
now we would like to substitute with the $\sqrt{2}\sin(u)$ but this doesn't work with the limits. So we are going to try using the substitution $v=\sqrt{2}\sin(u)$ and split the integral up into two parts.
$$I_1=2\int\limits_{\pi/4}^{\pi/2}\frac{\ln(\sqrt{2}\sin(u)}{\sin^2(u)-\cos^2(u)}du=\sqrt{2}\int_1^\sqrt{2}\frac{\ln(v)}{\frac{v^2}{2}-\frac{2-v^2}{2}}dv=\sqrt{2}\int_1^\sqrt{2}\frac{\ln(v)}{v^2-1}dv$$$$
=\sqrt{2}\int_1^\sqrt{2}\frac{\ln(v)}{2(v-1)}-\frac{\ln(v)}{2(v+1)}dv$$
and the same can be done for the second part.


Convergence of a generalized Fourier series in different basis Using a generalized Fourier series, almost every arbitrary function can be approximated as a sum using a family of orthogonal functions which I call a basis, over a domain $D$.

For example, the traditional Fourier series utilizes a basis defined as $A = { e^{int} : n \in \mathbb{Z} }$ in the domain $[0,2\pi]$. Then the series for $f$ under $A$ is

$$S_{D} [f]_{A}(x) = \sum_{n=-\infty}^{\infty} a_{n} e^{inx}$$

However for a function such that $S_{D} [f]_{A}$ diverges almost everywhere, is it also true for all other basis or does there exist another basis $B$ with domain $H : H \cap D \neq \emptyset$ such that $S_{H} [f]_{B}$ is convergent almost everywhere in the intersection of the domains?

My personal guess is that yes there will always exist another basis such that the series for $f$ is convergent, however a proof or disproof is beyond my abilities.

Yes, there exists a basis, meaning an orthonormal basis for $L^2(D)$, such that the series for $f$ in that basis converges to $f$.

For example, take an orthonormal basis $\phi_1,\dots$ such that $\phi_1=cf$. Then the series representing $f$ is $$\sum_{n=1}^\infty c_n\phi_n=c_1\phi_1=f.$$


Kolmogorov's Truncation Lemma (iii) Probability with Martingales:

In the definition of $f$, is that really $z$ and not $\lceil |z| \rceil$, $\lfloor |z| \rfloor + 1$ or something?
How exactly do we have the part in the $\color{red}{\text{red}}$ box?

What I tried:

$$\sum_{n=1}^{\infty} \frac{E[Y_n^2]}{n^2}$$

$$ = \sum_{n=1}^{\infty} \frac{E[|X|^2 ; |X| \le n]}{n^2}$$

$$ = \sum_{n=1}^{\infty} \frac{E[|X|^2 1_{\{|X| \le n]\}}]}{n^2}$$

$$ = \sum_{n=1}^{\infty} \frac{E[|X|^2 1_{\{|X| \le n]\}}]}{n^2}$$

$$ = \sum_{n=1}^{\infty} E\left[\frac{|X|^2 1_{\{|X| \le n]\}}}{n^2}\right]$$

$$ = E\left[ \sum_{n=1}^{\infty} \frac{|X|^2 1_{\{|X| \le n]\}}}{n^2}\right] \ \text{by Fubini's theorem}$$

$$ = E\left[ \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{|X|^2 }{n^2}\right]$$

$$ = E\left[ |X|^2 \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{1}{n^2}\right]$$

Now observe that for integer $k$

$$\sum_{n=k}^{\infty} \frac{1}{n^2} \le \sum_{n=k}^{\infty} \frac{2}{n(n+1)} = 2 \sum_{n=k}^{\infty} \frac{1}{n(n+1)} = 2 \sum_{n=k}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) = 2 (1/k - 1/(k+1) + 1/(k+1) - 1/(k+2) + ...) = 2(1/k)$$

Hence

$$\sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{1}{n^2} \le \frac{2}{\max\{1,\lceil |X| \rceil\}}$$

$$\to \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{|X|^2 }{n^2} \le \frac{2|X|^2}{\max\{1,\lceil |X| \rceil\}}$$

$$\to E\left[ \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{|X|^2 }{n^2}\right] \le E\left[\frac{2|X|^2}{\max\{1,\lceil |X| \rceil\}}\right]$$

$$\to E\left[ \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{|X|^2 }{n^2}\right] \le E\left[\frac{2|X|^2}{\lceil |X| \rceil}\right] = E\left[2\lceil |X| \rceil\right]$$

$$\le E\left[2 (|X| + 1) \right] < \infty \ \because E[|X|] < \infty$$

Is that right?


When does something become a "mathematical object"? A mathematical object is an abstract object arising in philosophy of mathematics and mathematics.

Abstract object:

  Abstract and concrete are classifications that denote whether a term describes an object with a physical referent or one with no physical referents.

Denoting, specifying, describing an "object" (wasn't the abstract object the "object"?).

What is an "object" in this sense? There seems to be no canonical explanation of what makes a mathematical object an "object", whatever that implies.

An object has multiple meanings, and the mathematical one doesn't do the justice of explaining it.

"Math scholars" have told me that "mathematical objects" are used to represent numbers in general, matrices, etc. However, a matrice is considered an "array". Where does this premise reach a concrete definiton if arrays can be in fact numerical?

Why is an "array" considered different than a number, if they can both be the exact same thing?

It seems math is quite shaky in my introspection of varying angles of the subject.

Therefore, I ask this ... why should I take something serious beyond elementary logic participles if it has no clarity, distinction of properties, or common sense?

All of your criticisms are equally valid when applied to.. well, anything.  How does a football coach know what a "formation" is, and whether it really applies to football?  How does a software engineer know the difference between a "program" and the instructions executed by a computer?  How does a dog know that a "frisbee" is something that you can catch in your mouth?  How does a general use little flags to signify troop positions, when they are really just flags?

None of this is to say that these are not interesting questions—I personally find them quite fascinating.  But saying that they are reasons not to take something seriously is rather antisocial.  If a lover stares into your eyes on a moonlit night and professes his or her adoration, do you start measuring oxytocin concentrations?

I do think that many mathematicians are a bit too attached to the Cantorian or Platonist views, and have incorrectly made mathematics out to be about things which are more than what they are—and that starts many arguments unnecessarily (for example, when someone claims that a theorem is true "in all possible universes", as if that meant anything).  In my opinion, topos theory provides a better foundation for mathematics in this sense, because it is easier to understand the relationship between semantics, syntax, and the ever-elusive ontology.  One speaks of this topos or that topos (or "topic", if you prefer), and never needs to worry about whether something "is" this or "is" that.

One relatively recent paper which I think has helped advance this more enlightened way of thinking is the quantum mechanics paper (heavily inspired by the philosophical work of Heidegger) What is a Thing?.  There it is argued that set theory has not quite succeeded in providing the proper background for interpreting the world as it appears to us.  The "state space" of physics professes to arrange possible worlds into a set, and runs headfirst into various paradoxes as we realize that our experimental equipment itself changes what is being measured, blurring our picture of how things really work and necessitating the continual introduction of new concepts and interpretations.

In short: perhaps truth, in the pragmatic sense, is more sheaf-like than set-like.  But I digress.

If anybody tells you that you should take math seriously because it has figured out, once and for all, the correct way to divide the abstract from the concrete, and has firmly established the foundations for rational thought, then they are too caught up in their subject and you really shouldn't pay attention to them.  And, if you really want, you can simply walk away, shaking your head in disappointment that mathematicians have failed to live up to their promise.

But, however seriously you take it, mathematics remains a powerful force in the world.  While we're not particularly better than anybody else at explaining what we're talking about, what we are good at is bringing disparate things together under the same semantical umbrella—to a large extent, precisely because we are given the freedom not to explain ourselves.  Measure theory, for example, has allowed us to shuttle insights between discrete phenomena and continuous phenomena.  Algebra has, for hundreds of years, improved our speed of numerical reasoning by a billion-fold, by knowing when to compute and when to encode.  Algebraic geometry has provided a language that is equally at home with basic arithmetic, encryption, signal processing, causality, and phylogenetic trees.  And so people keep finding it useful, however many students will stand up angrily in our classes and insist that they don't think it could possibly be useful because something something.

In short, mathematics saves time for certain kinds of projects.  If you don't do any of those projects, then of course you don't need to take it seriously.  But it's under no obligation to explain itself, particularly not to somebody who thinks he is entitled to answers and "justice".  If you find the foundations lacking, then we would love for you to come make a career of improving those foundations.  If you are mostly complaining however, then pardon us while we focus on our other students.

Object = Thing.

There are two types of things, abstract things and concrete things. Read the wikipedea article of Type–token distinction carefully.

Abstract object = Type
Concrete object = Token

Example: The number $2$ is an abstract object. We can have two coins, two metre long rod, two kilogram apples etc. In all these examples two is a concept of being two. Two doesn't exist physically in contrast with apples, coins and rod, which exist physically and are all concrete objects.

I'm not sure about your other question of how is Matrix an object. Perhaps since Matrix is a kinda concept of imagination, so it's an abstract object. Matrix is not a concrete object. Perhaps we can say that every object, which is not concrete, is abstract--hence the Matrix is an abstract object.


Number of elements of order $7$ in a group If $G$ has at least $7$ elements of order $7$, then prove that $G$ has at least $12$ elements of order $7$.

I know that an element having order $7$ means that for any $a$ in $G$, $a^7=e$ where $e$ is the identity element. The inverses also have same order, so I thought maybe of the $7$ elements, $6$ had their inverse as well. However, then this would just mean there had to be $8$ elements, not $12$ for the $7$th element to have it's inverse.

If there is an element with order 7, we have a subgroup of the group which is isomorphic to the cyclic group with 7 elements, namely the subgroup generated by the element of order 7.

All elements in this subgroup also have order seven (by Lagrange's theorem) excepting the identity element. So there exists another element with order seven out of the subgroup. This element also generates a subgroup of order 7 which meets the other subgroup only in the identity element (because intersections of subgroups are subgroups and by Lagrange a group of order seven has only the trivial subgroups), so we get 5 more elements of order seven, namely the other elements in the second subgroup.

So all in all we have at least 12 elements of order 7.

If $a$ has order $7$, then $a^2,a^3,a^4,a^5,a^6$ have order $7$.
Now the last one, $b$, is such that it has a different cycle. So $b^2,b^3,b^4,b^5,b^6$ have order $7$.


Probability that a random bridge board does not contain a sequence? Two cards with adjacent values and the same suit produce a SEQUENCE. For example,
the heart-ten and the heart-jack form a sequence. The order of the values in
bridge is $$23456789TJQKA$$

-What is the probability that a random bridge board contains no sequence in any
 hand ? (That means, for example : If North holds the queen of spades, he neither
 holds the jack nor the king of spades )

A long time ago, I approached this problem by first determining all possible
combinations not containing a sequence (for example kt742) , and then calculate
the number of boards from that. I remember that the probability is very low.
But there should be a better method, perhaps inclusion-exclusion or something
like that.

if i get your question right,
an ideal sample set E will be a set of all subsets
E={S2,S3,S4,…,S12}

whereas is subset of three serial number s=[n-1,n,n+1]

your sample space for all possible options are equal 2^11
one of these event meet the required criteria which is  ‘E' in your case.
the probability of having Pr(E=1) = 1/(2^11)

hope it helps


Is there a formula for $\sqrt[3]{a}+\sqrt[3]{b}=\sqrt[3]{?}$ similar to $\sqrt{a}+\sqrt{b}=\sqrt{a+b+2\sqrt{ab}}\;$? I know this ancient formula:
$$\sqrt{a}+\sqrt{b}=\sqrt{a+b+2\sqrt{ab}}$$ And I would like to know if there is a similar formula for:
$$\sqrt[3]{a}+\sqrt[3]{b}=\sqrt[3]{?}$$
I can't find it. Thank you

Notice that your initial formula is nothing more than $(\sqrt{a}+\sqrt{b})^2 = a + b + 2\sqrt{ab}$.
Thus, starting from $(\sqrt[3]{a}+\sqrt[3]b)^3 = a + b + 3\sqrt[3]{ab} \ (\sqrt[3]{a}+\sqrt[3]{b})$, you can write (by taking cubic roots on both sides)
$$\sqrt[3]{a}+\sqrt[3]b = \sqrt[3]{a + b + 3\sqrt[3]{ab} \ (\sqrt[3]{a}+\sqrt[3]{b})}.$$

To @Babado (@Z Ahmed , @Mostafa Ayaz)
I already knew that formula, but what I want to find out is in the answer.I don’t understand how that works.
If I take cubic roots on both sides, then the answer brings me nothing. In the first formula I can know and operate, for example:
$\sqrt{ab}$
, but no:
$\sqrt{a}+\sqrt{b}$


Explanation of numeric experiment that approximates e? Recently I found this post on Reddit. It describes the following algorithm to find e:

  Here is an example of e turning up
  unexpectedly. Select a random number
  between 0 and 1. Now select another
  and add it to the first. Keep doing
  this, piling on random numbers. How
  many random numbers, on average, do
  you need to make the total greater
  than 1? The answer is e.

This means that you need on average ~2.7 random real numbers to make the sum greater than 1.

However, a random number between 0 and 1 would on average be equal to 0.5. So intuitively I would think that, on average, only 2 random numbers would be required to have a sum > 1.

So where did I go wrong in my thinking?

Update

I just figured it out myself: You need at least two numbers to have a sum > 1, but often you'll need three, sometimes you'll need four, sometimes five, etc... So it only natural that the average required numbers is above 2.

Thanks for the replies!

It is never possible to get greater than 1 on the first try, so the average must be strictly greater than 2.

Sorry, can't explain it either. It's obvious you need at least two random numbers to go greater than 1. And if the random numbers are on average .5 you'll need (edit) on average (/edit) maximum three numbers. So we end up in the range [2..3], but that's as close as I can get to e.

While the method sure looks curious, it must be about the least efficient way ever to calculate e digits. A quick-and-dirty test program I wrote only gave 4 correct significant digits after a billion iterations. Or maybe Delphi's PRNG isn't perfect.