Two men, each placing one ball in five out of ten boxes There are 10 boxes and each box can hold any number of balls. A man having 5 balls randomly puts one ball in each of the arbitrary chosen five boxes. Then another man having five balls again puts one ball in each of the arbitrary chosen five boxes. The probability that there are ball(s) in at least 8 boxes. Assuming all boxes and balls are identical. Both men can choose any arbitrary box.

My attempt:

I was really not able to pull through the solution. What I thought was to let the first man choose five boxes by $^{10}C_5$, and put the balls in those 5 boxes each. There is $1$ way to do that.

Now the second man comes and chooses his 5 boxes and put each ball in those 5 boxes. It can be done in $^{10}C_5 * 1$

Now these are the following ways the boxes can be filled:
$$1 1 1 1 1 1 2200$$
$$1111111120$$
$$1111111111$$

Now I am not able to understand how to apply combinatorics and find the number of ways these conditions can happen. Please correct me if I was wrong in my attempt at the solution.

I will approach the problem $\require{enclose}\enclose{horizontalstrike}{\text{two}}$ three different ways, and show that the answer arrived at is the same in both ways.  In doing so, I hope to highlight what is good and what is incorrect about the other posted answers.

Method 1:

We allow ourselves to imagine each box as having a label, yet the balls as continuing to be unlabelled.  The people themselves can't see the labels, so this in no way affects the probabilities involved but it will be convenient for us in defining our sample space.

We take as our sample space the ways in which the balls are distributed after each person places their five balls into the boxes.  Each way is indeed equiprobable by assumption.  There are $\binom{10}{5}\binom{10}{5}$ number of ways of accomplishing this.

Each of these ways can be seen as a string of ten characters taken from $a$'s, $b$'s, $c$'s, and $x$'s such that the number of $a$'s plus the number of $c$'s is equal to five and the number of $b$'s is equal to the number of $a$'s.  The locations of the $a$'s represent those boxes in which only the first man placed a ball, $b$'s represent the boxes in which only the second man placed a ball, $c$'s represent the boxes in which both men placed a ball, and $x$'s represent empty boxes.  Again, the set of strings satisfying these conditions are in direct bijection with the outcomes in our sample space.

To count the number of "good" outcomes, i.e. those in which at least eight boxes have balls in them (i.e. at most two are empty), we look to count the strings as described above such that there are two, one, or zero $x$'s.

Two $x$'s will imply that there are $3$ $a$'s, $3$ $b$'s, two $c$'s and two $x$'s.  There are $\binom{10}{3,3,2,2}=\binom{10}{3}\binom{7}{3}\binom{4}{2}\binom{2}{2}=\frac{10!}{3!3!2!2!}$ number of ways in which this can occur.  (note: this is in contradiction with both zzcnick's and anonymous's answers, they seem to have not taken into account that the person who placed each ball is relevant to the probability using this sample space)
One $x$ will imply that there are $4$ $a$'s, $4$ $b$'s, one $c$ and one $x$.  There are $\binom{10}{4,4,1,1}=\binom{10}{4}\binom{6}{4}\binom{2}{1}\binom{1}{1}=\frac{10!}{4!4!1!1!}$ number of ways this can occur.
Zero $x$'s will imply that there are $5$ $a$'s, $5$ $b$'s, and zero $c$'s and $x$'s.  There are $\binom{10}{5,5}=\binom{10}{5}$ number of ways in which this can occur.

Adding these favorable cases together and taking the ratio with the number of outcomes in our sample space we get the probability:

$$\dfrac{\binom{10}{3,3,2,2}+\binom{10}{4,4,1,1}+\binom{10}{5,5}}{\binom{10}{5}\binom{10}{5}}=\frac{31752}{63504}=\frac{1}{2}$$

Method 2:

Continuing from Morgan's observation, we may approach this via conditional probability.  After the first person has picked his five boxes to place his balls, we may arbitrarily call those the first five boxes in our minds from left to right.  The remaining empty boxes from left to right will be labeled as being the second five boxes.  Again, the people with the balls can't see the labels on the boxes, so this in no way affects the probabilities, it is just a convenient way for us to organize our thoughts.  With those as the first five boxes, the question is simply how many of the second person's balls go into the first five boxes versus the empty five boxes.

Our sample space then is described as the number of ways in which the people place the balls according to these labels.  We can see that there are only $\binom{10}{5}$ ways, a much smaller sample space than before, but despite that this will still be an equiprobable sample space so we may use counting techniques.

We count how many favorable outcomes there are by again breaking into cases based on how many of the second person's balls were placed into already filled boxes.

Only two were put into the first five boxes while three in the second five.  There are $\binom{5}{2}\binom{5}{3}$ ways in which this occurs.
Only one was put into the first five boxes while four in the second five.  There are $\binom{5}{1}\binom{5}{4}$ ways in which this occurs.
None were placed in the first five boxes.  There are $\binom{5}{0}\binom{5}{5}$ ways in which this occurs.

Adding these favorable cases together and taking the ratio with the number of outcomes in our sample space we get the probability:

$$\dfrac{\binom{5}{2}\binom{5}{3}+\binom{5}{1}\binom{5}{4}+\binom{5}{0}\binom{5}{5}}{\binom{10}{5}} = \frac{100+25+1}{252}=\frac{126}{252}=\frac{1}{2}$$

Method 3: Very similar to method two, except we appeal to symmetry.

Recognize that the probability that at least eight boxes are filled is equivalent to the probability that at least three of the second person's balls are placed into empty boxes after the first person has placed his balls.

Recognize that this is equivalent to the probability that given five red boxes and five blue boxes that a person picking five boxes picks mostly red boxes.  Recognize that this is then equivalent to the probability that he picks mostly blue boxes.

Recognize now that the probability of success is equal to the probability of failure, and thus the probability of success is equal to $\frac{1}{2}$ by the law of total probability.

Each of the three conditions you present represents a different way to fill the ten boxes with a different amount of balls so that at least 8 boxes have one ball.

In the first case, you have two boxes with two balls, two boxes with zero balls, and six boxes with one ball. If you labelled the ten boxes, you are then looking for the amount of ways to choose 2 boxes to have two balls, 2 boxes to have zero balls, and 6 boxes to have one ball. Solving for this case gives:

\begin{align*}
{10\choose2}{10 - 2 \choose 2}{10 - (2 + 2) \choose 6}
= {10\choose2}{8\choose2}{6\choose6} =
\frac{10!}{2! * 2! * 6!}
\end{align*}

Repeat this for your remaining two cases, and you will have all the ways to arrange the balls in the ten boxes so that you have at least 8 boxes with 1 ball.


Method of dominant balance and perturbation Approximate the solutions of $$\epsilon x^4 + (x-1)^3=0$$

I can't perform a singular perturbation because if I let $\epsilon=0$ then I lose a root.  My professor suggests The Method of Dominant Balance for problems like this, but I don't really know how to proceed with that method.

If someone could just get me started, then I could do the rest in maple.

Thanks

If we naively just send $\epsilon \to 0$ then we get the equation

$$
(x-1)^3 = 0,
$$

so we can deduce that we have three roots tending to $x=1$ as $\epsilon \to 0$.  We'll suppose they have asymptotic series of the form

$$
x \approx 1 + \sum_{k=1}^{\infty} a_k \delta_k(\epsilon),
$$

where

$$
1 \gg \delta_1(\epsilon) \gg \delta_2(\epsilon) \gg \delta_3(\epsilon) \gg \cdots
$$

as $\epsilon \to 0$.

If we substitute just the first two terms $x \approx 1 + a_1 \delta_1(\epsilon)$ into the original equation

$$
\epsilon x^4 + (x-1)^3 = 0 \tag{$*$}
$$

and expand it we get

$$
a_1^4\epsilon\delta_1(\epsilon)^4 + 4a_1^3\epsilon\delta_1(\epsilon)^3 + a_1^3\delta_1(\epsilon)^3 + 6a_1^2\epsilon\delta_1(\epsilon)^2 + 4a_1\epsilon\delta_1(\epsilon) + \epsilon \approx 0.
$$

Now we'll apply the method of dominant balance.  First, certain terms are, by assumption, smaller than others, so they could not possibly be part of a dominant balance and may be ignored.  Specifically we know that

$$
a_1^4\epsilon\delta_1(\epsilon)^4 \ll 4a_1^3\epsilon\delta_1(\epsilon)^3 \ll 6a_1^2\epsilon\delta_1(\epsilon)^2 \ll 4a_1\epsilon\delta_1(\epsilon) \ll \epsilon,
$$

so if we ignore all but the largest of these then our equation becomes

$$
a_1^3\delta_1(\epsilon)^3 + \epsilon \approx 0.
$$

From this we see that

$$
\delta_1(\epsilon) = \epsilon^{1/3} \quad \text{and} \quad a_1^3 = -1,
$$

and with three choices for $a_1$---namely the three cube roots of $-1$---we obtain approximations for each of the three roots of the original equation with tend to $x=1$:

  $x \approx 1 - \epsilon^{1/3}$,
  $x \approx 1 + e^{i\pi/3} \epsilon^{1/3}$, and
  $x \approx 1 + e^{i5\pi/3} \epsilon^{1/3}$.

This suggests that we might be able to take

$$
\delta_k(\epsilon) = \epsilon^{k/3}, \quad k \geq 1,
$$

and indeed if we substitute

$$
x \approx 1 + \sum_{k=1}^{\infty} a_k \epsilon^{k/3}
$$

into $(*)$ and collect like powers of $\epsilon$ then we obtain equations for the coefficients $a_k$,

$a_1^3 + 1 = 0$,
$3 a_2 a_1^2+4 a_1 = 0$,
$3 a_3 a_1^2+6 a_1^2+3 a_2^2 a_1+4 a_2 = 0$, etc.

It just remains to find the last root of $(*)$.  If we simply expand it out we get

$$
\epsilon x^4 + x^3 - 3x^2 + 3x - 1 = 0. \tag{$**$}
$$

We will again use the method of dominant balance.  By ignoring the $\epsilon x^4$ term in the very beginning we essentially assumed it would not be part of a dominant balance, so to find the last root we must assume the opposite.  As such we'll be looking for balances between $\epsilon x^4$ and the remaining four terms, so we'll consider the cases

$\epsilon x^4 \asymp 1$,
$\epsilon x^4 \asymp 3x$,
$\epsilon x^4 \asymp 3x^2$, and
$\epsilon x^4 \asymp x^3$.

In case 1 we have $x \asymp \epsilon^{-1/4}$, but then the term $3x$, being $\asymp \epsilon^{-1/4}$, dominates the terms $\epsilon x^4$ and $1$ in the balance, so the balance is not dominant.

In case $2$ we have $x \asymp \epsilon^{-1/3}$, but then the term $-3x^2$, being $\asymp \epsilon^{-2/3}$, dominates both terms in the balance, so the balance is again not dominant.

In case $3$ we have $x \asymp \epsilon^{-1/2}$, but then the term $x^3$, being $\asymp \epsilon^{-3/2}$, dominates both terms in the balance, namely $\epsilon x^4$ and $x^2$, which are $\asymp \epsilon^{-1}$.

In case $4$ we have $x \asymp \epsilon^{-1}$, and this balance is dominant since

$$
\epsilon x^4 \asymp x^3 \asymp \epsilon^{-3} \gg 3x^2 \gg 3x \gg 1.
$$

Ignoring the terms outside of the balance we then have

$$
\epsilon x^4 + x^3 \approx 0,
$$

so that

  $$
x \approx - \epsilon^{-1}.
$$

We might then suspect that this root has an asymptotic series of the form

$$
x \approx -\epsilon^{-1} + \sum_{k=0}^{\infty} a_k \epsilon^k,
$$

and if we substitute this into $(*)$ or $(**)$ and collect like powers of $\epsilon$ we obtain equations for the coefficients $a_k$,

$a_0+3 = 0$,
$3 a_0^2+6 a_0-a_1-3 = 0$,
$-3 a_0^3-3 a_0^2+6 a_1 a_0+3 a_0+6 a_1-a_2-1 = 0$, etc.

We've thus accounted for all four roots of the polynomial equation and can calculate as many terms of their asymptotic series as we wish.

Let $\epsilon=\tau^3$, then we can solve the equation

$$0=\tau^3 x^4 + (x-1)^3$$

There are 4 roots.  When $\tau \to 0$, they are given by:

$$x_1=-3-\frac{1}{\tau^{3}}$$
$$x_2=1-\tau$$
$$x_3=1+\left(\frac{1}{2}-i\frac{1}{\sqrt{3}}\right)\tau$$
$$x_4=1+\left(\frac{1}{2}+i\frac{1}{\sqrt{3}}\right)\tau$$


Rational numbers - countability I have to show that the set of all finite sequences
$$
(q_1,q_2,\dotsc,q_k),\quad k \in \mathbb{N}
$$
 of rational numbers is countable.
To prove that the set $\mathbb{Q}$ of all rational numbers is countable, I used that the set $\mathbb{Z}\times\mathbb{N}$ is countable and can be listed as
$$
(a_1,b_1),(a_2,b_2),(a_3,b_3),\dotsc
$$
and then making a list of all elements in $\mathbb{Q}$:
$$
\frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3}.
$$
But how can I show for all finite sequences?

As $\mathbb Q$ is countable, you can solve the question for the $k$-uples of naturals instead (every rational can be traded for an integer).

Then for every $n$, you can enumerate the natural solutions of

$$\sum_{j=1}^k q_j=n$$ (by choosing all possible values of $q_k\in[1,n]$ and recursively solving $\sum_{j=1}^{k-1} q_j=n-q_k$), so by increasing $n$ you can enumerate all sequences. (No sequence will be missed, because every sum is enumerated, and for a given sum, no solution is missed. Also note that no sequence will be duplicated, but this inessential.)

The result also holds for sequences of arbitrary lengths (all $k$), by enumerating the pairs $(n,k)$ with the same method.

The general result you need is that the set of ordered pairs (x,y), where each of x, y is a member of a countable set, is itself a countable set.  This is the essence of the proof that $\mathbb{Q}$ is countable.  I leave it to you to figure out how to apply this to your problem (hint: induction on number of indices).


If $\lambda$ is a characteristic root of a non-singular matrix $A$, then prove that $det(A)/\lambda$ is a characteristic root of $Adj A$. Let $A$ be a $n\times n$ matrix. Then the characteristic equation is $\mathrm{det}(A-\lambda×I)=0$, with possibly characteristic roots of $A$ $\lambda_1,\dots,\lambda_n$. Let $A$ be regular, i.e. $\mathrm{det}(A)\neq0$. Now $\mathrm{Adj}(A)=\mathrm{det}(A)\mathrm{Inv}(A)$. I don't know how to find its characteristic roots.

Let $A$ be invertible. As you rightly remarked, then $\mathrm{adj}(A)=\mathrm{det}(A)A^{-1}$. We can even prove the following stronger proposition:

Proposition: For $A$ invertible, $\lambda\neq 0$ is an eigenvalue of $A$ iff $\mathrm{det}(A)\lambda^{-1}$ is an eigenvalue of $\mathrm{adj}(A)$:

Proof:
$$
Av=\lambda v\Leftrightarrow \mathrm{adj}(A)(Av)=\mathrm{adj}(A)(\lambda v)\Leftrightarrow \mathrm{det}(A)A^{-1}(Av)=\mathrm{det}(A)A^{-1}(\lambda v)\Leftrightarrow \mathrm{det}(A)v=\mathrm{det}(A)\lambda A^{-1}(v)\Leftrightarrow\mathrm{det}(A)\lambda^{-1}v=\mathrm{det}(A)A^{-1}(v)\Leftrightarrow\mathrm{det}(A)\lambda^{-1}v=\mathrm{adj}(A)v
$$
$\Box$

You may rephrase this result in terms of roots of the characteristic polynomial, as you naturally have for $\lambda\neq 0$ that $\mathrm{det}(A-\lambda E_n)=0$ iff $\lambda$ is an eigenvalue of $A$ iff $\mathrm{det}(A)\lambda^{-1}$ is an eigenvalue of $\mathrm{adj}(A)$ iff $\mathrm{det}(\mathrm{adj}(A)-\mathrm{det}(A)\lambda^{-1}E_n)=0$.

Obviously, the condition $\lambda\neq 0$ is w.l.o.g. as $0$ is a eigenvalue of $A$ iff $\det(A-0E_n)=\det(A)=0$ iff $A$ is singular.

AX=aX where a is the eigen value of non singular matrix A
(adj A)AX=adj A (aX)
|A|A^- (AX)= a (adj A)X
|A|X=a(adj A) X
|A|X/a=(adj A)X
(adj A)X=|A|X/a
=》|A|/a is the characteristic root of matrix adjA


$-(1 + m)^2 - t - (-1 + r - w) (1 + r + w)$, solve for $w,m$ with known $t,r$? Consider some known fixed numbers $t,r$, then  can we solve for $m,w$ in $-(1 + m)^2 - t - (-1 + r - w) (1 + r + w) = 0$ ?
Is there any way to figure out what are the integer solutions to this? I have tried looking at this every which way and nothing. I can manufacture many of this equations by choosing $t,r$  with known $w,m$ integers, but given $t,r$ only can we find $m,w$?
Update: $t,r,m,w \in \mathbb{N}$

You have
$$-(1+m)^2-t-(-1+r-w)(1+r+w) = 0 $$
$$\Longleftrightarrow -(1+m)^2-t-(r-(w+1))(r+w+1) = 0 $$
$$\Longleftrightarrow -(1+m)^2-t-r^2+(w+1)^2 = 0 $$
$$\Longleftrightarrow (w-m)(w+m+2)=t+r^2 $$
Now if $t$ and $r$ are given, you can check the dividers of $t+r^2$ and deduce the different possibilities for $w-m$ and $w+m+2$, and then deduce the conrresponding values of $m$ and $w$.


Miscellaneous Solids. How do I solve this problem? Can anybody help me with this

Find the volume in the first octant inside the cylinder $x^2/a^2 +y^2/b^2 =1$ under the plane $z=3x$. Use the given slice in the figure to compute the volume.

Using ellipse coordenates for the double integral:

$$\int\int_D\left(\int_0^{3x}1\,dz\right)\,dx\,dy=\int\int_D3x\,dx\,dy=3\,a^2\,b\int_0^{\pi/2}\int_0^1\,r^2\,\cos \theta\,dr\,d\theta=\boxed{a^2\,b}$$


Integer solutions to $ x^2-y^2=33$ I'm currently trying to solve a programming question that requires me to calculate all the integer solutions of the following equation:

$x^2-y^2 = 33$

I've been looking for a solution on the internet already but I couldn't find anything for this kind of equation. Is there any way to calculate and list the integer solutions to this equation?

Thanks in advance!

Suppose that $x=y+n$; then $x^2-y^2=y^2+2ny+n^2-y^2=2ny+n^2=n(2y+n)$. Thus, $n$ and $2y+n$ must be complementary factors of $33$: $1$ and $33$, or $3$ and $11$. The first pair gives you $2y+1=33$, so $y=16$ and $x=y+1=17$. The second gives you $2y+3=11$, so $y=4$ and $x=y+3=7$. As a check, $17^2-16^2=289-256=33=49-16=7^2-4^2$.

If you want negative integer solutions as well, you have also the pairs $-1$ and $-33$, and $-3$ and $-11$.

This is all fine, but you all take the initial knowledge of the prime composition of 33 as 11*3, but imagine you did not know that. You would not know it without a lot/ some trial and error , if I instead of 33 took 1771 instead and stated x^2-y^2=1771, then only the few, would know the prime factors of 1771

There are no known ways of solving this problem other than trial and error. Likewise with 33, if you had no knowledge that 11*3= 33

You will have to run through this calculation:  sqrt(1771)=42.08 and mod(1171,4)=3
Mod 3 means X^2 is even and y^2 is odd. Had the result been 1, it was opposite.

You can rewrite the equation to y^2= x^2-1771, so both y and x must be integers, x will have an even root and y will have an odd root.
Nearest even integer after 42.08 is 44, so that is where I start. (x^2 has to be bigger than 42.08)

We start our trial and error:
44^2-1771= y^2
46^2-1771= y^2
48^2-1771= y^2
If you calculate the first 3, they will generate a y^2, which is not a perfect square, that is, y will not be integer.

But 50^2= 2500-1771= 729 and that gives the roots sqrt(2500)= 50 and sqrt(729)=27, so one solution is (50+27)=77 and the other is (50-27)=23
and 77*23= 1771
It follows the format (x+y)(x-y)= X^2-Y^2 and (50+27)(50-27)= 1771

If we could find a way of getting directly to either 2500 or 729, we would at the same time get a lead to solve some of the Millennium math. problems. I am trying, but have not succeeded.

When you go deeper into this, you might find some short cuts, which can reduce the number of trial and error, but currently there are no ways to reduce it down to just one single cycle. If you find it, let me know.

In this case there are other solutions as well, because the prime composition of 1771 is 7*11*23 and we just solved for (7*11)*23
The product is not changing for other combinations, but the sum of (x+y) and the difference (x-y) is changing, which leads to different perfect squares, but with the same feature, that their difference is 1771.

If you only find ONE solution, your difference is a prime number. A prime number is always the difference of 2 consecutive squares and have no other solutions, than this combination, whereas composite numbers will at least have 2 solutions, the prime composition of 2 consecutive squares and at least one more.

It is amazing and surprising, that there is no way but brute force, to factor an odd number.... however as long as that goes on, we have our financial transactions fairly safe, they rely on the excessive time, it takes to factor a product of 2 large prime numbers.


Mistake in solving an equation involving a square root I want to solve $2x = \sqrt{x+3}$, which I have tried as below:

$$\begin{equation}
4x^2 - x -3 = 0 \\
x^2 - \frac14 x - \frac34 = 0 \\
x^2 - \frac14x = \frac34 \\
\left(x - \frac12 \right)^2 = 1 \\
x = \frac32 , -\frac12
\end{equation}$$

This, however, is incorrect.

What is wrong with my solution?

You made a mistake when completing the square.

$$x^2-\frac{1}{4}x = \frac{3}{4} \color{red}{\implies\left(x-\frac{1}{2}\right)^2 = 1}$$

This is easy to spot since $(a\pm b)^2 = a^2\pm2ab+b^2$, which means the coefficient of the linear term becomes $-2\left(\frac{1}{2}\right) = -1 \color{red}{\neq -\frac{1}{4}}$. This means something isn’t correct...

Note that the equation is rewritten such that $a = 1$, so you need to add $\left(\frac{b}{2}\right)^2$ to both sides and factor. (In other words, divide the coefficient of the linear term $x$ by $2$ and square the result, which will then be added to both sides.)

$$b = -\frac{1}{4} \implies \left(\frac{b}{2}\right)^2 \implies \frac{1}{64}$$

Which gets

$$x^2-\frac{1}{4}x+\color{blue}{\frac{1}{64}} = \frac{3}{4}+\color{blue}{\frac{1}{64}}$$

Factoring the perfect square trinomial yields

$$\left(x-\frac{1}{8}\right)^2 = \frac{49}{64}$$

And you can probably take it on from here.

Edit: As it has been noted in the other answers (should have clarified this as well), squaring introduces the possibility of extraneous solutions, so always check your solutions by plugging in the values obtained in the original equation. By squaring, you’re solving

$$4x^2 = x+3$$

which is actually

$$2x = \color{blue}{\pm}\sqrt{x+3}$$

so your negative solution will satisfy this new equation but not the original one, since that one is

$$2x = \sqrt{x+3}$$

with no $\pm$.

$\require{cancel}$Additional details in $\color{blue}{blue}$.  Important detail in $\color{green}{green}$.  Mistakes in $\color{red}{\cancel{\text{canceled red}}}$. Corrections in $\color{purple}{purple}$.

$\begin{equation}
\color{blue}{2x = \sqrt{x+3}}\color{green}{\ge 0}\\
\color{blue}{4x^2 = x+3}\color{green}{\text{!AND!} x\ge 0}\\
4x^2 - x -3 = 0 \\
x^2 - \frac14 x - \frac34 = 0 \\
x^2 - \frac14x = \frac34 \\
\color{blue}{x^2 -2\cdot\frac 18=\frac 34}\\
\color{blue}{x^2 -2\cdot\frac 18 +(\frac 18)^2=\frac 34+\frac 1{64}}\\
\color{red}{\cancel{(x - \frac12 )^2 = 1}}\color{purple}{(x - \frac18)^2 = \frac {49}{64}} \\
\color{blue}{x-\frac 18=\pm \frac 78}\\
\color{red}{\cancel{x = \frac32 , -\frac12}}\color{purple}{x = 1 , -\frac34}\color{green}{\text{!AND!} x\ge 0}\\
\color{purple}{x = 1}\\
\end{equation}$


Why are rotational matrices not commutative? Is there any intuition why rotational matrices are not commutative? I assume the final rotation is the combination of all rotations. Then how does it matter in which order the rotations are applied?

Here is a picture of a die:

Now let's spin it $90^\circ$ clockwise. The die now shows

After that, if we flip the left face up, the die lands at

Now, let's do it the other way around: We start with the die in the same position:

Flip the left face up:

and then $90^\circ$ clockwise

If we do it one way, we end up with $3$ on the top and $5, 6$ facing us, while if we do it the other way we end up with $2$ on the top and $1, 3$ facing us. This demonstrates that the two rotations do not commute.

Since so many in the comments have come to the conclusion that this is not a complete answer, here are a few more thoughts:

Note what happens to the top number of the die: In the first case we change what number is on the left face, then flip the new left face to the top. In the second case we first flip the old left face to the top, and then change what is on the left face. This makes two different numbers face up.
As leftaroundabout said in a comment to the question itself, rotations not commuting is not really anything noteworthy. The fact that they do commute in two dimensions is notable, but asking why they do not commute in general is not very fruitful apart from a concrete demonstration.

I'm discussing here non-commutativity of rotations made about current axes..

Rotations in 3d are non commutative because  rotation changes direction of every potential other axis except itself (unlike in 2d where it is nothing to change because it is only one "axis" of rotation - it can be reduced in 3D to rotation about Z axis ).

Denote succession of rotations about Z-axis as $Z_1,  Z_2,... Z_n$ (it's convenient to imagine that there is multitude of them by a small angular increment let's say $1^\circ$ or $2^\circ$ ) and one rotation about different axis $X$ ( it can be also not very big).  When we have only rotations $..Z_j...$ we can change order of rotations in any way - the result will be the same.

But if  we  insert  rotation $\mathbf{X}$ somewhere into middle of rotations succession   $Z_1,{Z_2}...{Z_n}$ the final result $Z_1,{z_2}...{Z_j}{\mathbf{X}}  {Z_k}...{Z_n}$ now depends on which place we inserted it because inserting it means that we are changing current axis of rotations for all successive rotations Z - the current Z axis, after this operation, is changed in the global frame.

It could be seen the other effect of inserting X rotation into different places of succession of $ Z_i $ if we would track the end of X versor (vector $i$) :

If this X rotation is made at the beginning the plane XY constant in
current frame for Z rotations is changed and the endpoint of X axis
is distancing itself from original XY plane with every successive
incremental rotation about changed now Z-axis.
If it would be made at the end, the distance from original XY plane
still would be equal $0$ !

As the versors $i,j,k$  after whole linear operation represents the resulting matrix we clearly see that these two operations are quite different.
Graphically one can imagine this track of the endpoint of $X$ versor on the surface of the unit sphere.

The simplest alternative presentation of non-commutativity in 3D (approach here is more algebraic than geometric as previous one) can be made using axis unit vector $v_1$ of   rotation, say, $R_1$.

($R_2$, we assume, has different axis with vector $v_2 \ne v_1$ and moreover we assume $R_2(v_1) \ne -v_1$ which together with $R_2(v_1) \ne  v_1$ states that transformed by $R_2$ vector $v_1$ is not lying in the axis of $R_1$).

Apply it to $R_2R_1(v_1)=R_2(v_1) = v_{21}$. (Obviously $v_{21} \ne v_1$).

In reverse order $R_1R_2(v_1)=R_1(v_{21} )$.
But  $v_{21} $ is not lying in the axis of $R_1$ !
Hence the result must be different than $v_{21}$ and $R_1R_2 \ne R_2R_1$.

They commute only if they share common axis or in the case of different axes they preserve each other axes with result vector  changing sign i.e. $R_2(v_1)=-v_1$ and $R_1(v_2)=-v_2$ what is possible when $v_1$ is orthogonal to $v_2$ and rotations are by $\pi$ angle.


Extremum, directional derivatives If a function $f: \mathbb{R}^n \to \mathbb{R}$ has all the directional derivatives zero at a point $c \in \mathbb{R}^n$, does it imply that $f$ has an extremum at $c \in \mathbb{R}$? If so, how do I prove it?

No, it doesn't because it is only a necessary condition for a local extremum. You should check your point $c$ for the fulfillment of a sufficient condition for a local extremum.


Multiplication of real and complex radicals If I have, for example, the product

$\sqrt{7+\sqrt{22}}\sqrt[3]{38+i\sqrt{6}} $

Can I perform the multiplication or this cannot be done and only remains to leave the product in this form?

Your question is ambiguous, because there is no such thing as a radical for complex numbers. It's not well defined, because you can't choose a specific complex number for your function (on $\mathbb R$ we choose the positive one).

So if you define $\sqrt.$ properly, may be you could perform your product, but it seems a lot of trouble.


Showing that the sequence of functions is not Cauchy I need to show that $ g_n(x)=x^{1/(2n-1)}  $ is not a Cauchy sequence in $C[-1,1] $ w.r.t. supremum norm. I tried to find the maximum of the difference of $g_n$ and $g_m$ by just differentiating but that did not really lead me anywhere.

Any suggestions would be greatly appreciated.

If you look at the graphs of $x^{1/n}$ for some large values of $n$, you should see a clear pattern in their behavior on $[0, 1]$: it looks as though they converge pointwise to:

$$f(x)=\begin{cases} 0\ \text{ if $x=0$} \\ 1\ \text{ else} \end{cases}$$

Of course with odd $n$ we get symmetrical behavior on $[-1, 0]$. Note that this convergence is pointwise, and not with respect to the supremum norm. Now, as pointed out in the comments, if you can show that $g_n$ does not converge, you will show that it is not Cauchy, so this suggests a plan of attack. If you remember the following fact, you can use what I've said to prove that $g_n$ does not converge w.r.t. the supremum norm:

  If $f_n$ converges to $f$ w.r.t. the supremum norm, then $f_n$ converges pointwise to $f$.


Total Order Relation Given $X = \{ a , b , c \}$ and $R = \{ (a,a) , (b,b) , (c,c) , (a,b) , (a,c)\}$

Is $R$ a total order on $X$?

I know that total order requires the relation to be comparable on all elements, anti-symmetric, and transitive.

What I am confused about is that I was told this relation is NOT total order, but I do not see how the elements cannot be compared, because they surely are anti-symmetric and transitive.

Can't $a$ be compared to itself? Surely $(a,b)$ and $(a,c)$ pass the comparability test, so all that remains must be the reflexive examples.

Clarification is greatly appreciated, thank you for your time, patience and assistance!

A total order relation requires 4 things:
1)refxivity(it is reflexive in this case)
2)anti-symmetricity(it is anti-symmetric)
3)transitivity(it is transitive)
4)comparibility
Now compatibility means that if you choose any two elements say a,b then either aRb or bRa
But in this case if we take for example b and c, neither bRc nor cRb so the relation is not compatible and hence is not a total order.

Is this relation transitive?. I don't think so.
We have (a,b) and (a,c).
To be transitive, (b,c) must be a part of relation too.
So, this is not a total order relation because is not transitive.


Prove: $(1-\frac{1}{d+1})^d>\frac{1}{e}$ I need to prove that $\left(1-\frac{1}{d+1}\right)^d>\frac{1}{e}$.

I guess that I have to use that $\left(1+\frac{1}{n+1}\right)^n\rightarrow e$ for $n\rightarrow\infty$ or better $<e$ or $\left(1+\frac{1}{n-1}\right)^n>e$ but I don't know exactly how it works.

Thank you for your help!

Approach 1

Using Bernoulli's Inequality, which is strict for $d\ge2$,
$$
\begin{align}
\frac{\left(1-\frac1{d\vphantom{+1}}\right)^{d-1}}{\left(1-\frac1{d+1}\right)^d}
&=\frac{\left(\frac{d-1}{d\vphantom{+1}}\right)^{d-1}}{\left(\frac{d}{d+1}\right)^d}\\
&=\frac{d}{d-1}\left(\frac{d^2-1}{d^2}\right)^d\\[9pt]
&\gt\frac{d}{d-1}\left(1-\frac1d\right)\\[15pt]
&=1
\end{align}
$$
Thus, $\left(1-\frac1{d+1}\right)^d$ is strictly decreasing and its limit is $\frac1e$. Therefore, we have
$$
\bbox[5px,border:2px solid #C0A000]{\left(1-\frac1{d+1}\right)^d\gt\frac1e}
$$

Approach 2

If $x\ge-n$, Bernoulli's Inequality says that for $x\ne0$ and $n\ge2$,
$$
1+x\lt\left(1+\frac x2\right)^2\le\left(1+\frac xn\right)^n
$$
Therefore, taking the limit as $n\to\infty$, we get that for $x\ne0$,
$$
1+x\lt e^x
$$
Thus,
$$
1+\frac1d\lt e^{1/d}
$$
Raise both sides to the $-d$ power
$$
\bbox[5px,border:2px solid #C0A000]{\left(1-\frac1{d+1}\right)^d\gt\frac1e}
$$

Well if you know l'hospitals rule you can show that $$(1-\frac{1}{x})^{x}$$ is equal to $e^{-1}$ as $x \to \infty$
and use this fact to prove your equation. If you don't, e can be defined as $$lim_{x->\infty}(1+\frac{1}{x})^{x}$$ and the function is less than that of e
Use this facts plus  $$lim_{x->\infty}(1+\frac{a}{x})^{x}=e^{a}$$ (try to prove this yourself as exercise)
Substitute the $x$ for $d+1$ and the $a$ for -1 all this guides you to the answer


The Price is Right optimal play The following situation happened on the Price is Right and I was curious about the optimal response.
The rules are:
A contestant rolls a wheel with 5 cent increments from 5 - 100 (20 numbers total). A contestant can choose to spin the wheel once and accept the number the wheel landed on (stay) or spin again and have this new number added to the previous number. If the contestant has a number that is over 100 cents they automatically lose. The object of the game is to roll the highest number out of a set of contestants.

GirlA rolled once and rolled 60 cents. GirlA knows GirlB will play afterwards. Should GirlA roll again? At what number is the expected value neutral / what range of numbers should GirlA stay on? If there is more than one player behind, what range of numbers is best to stay on?

I'll look at the two player version, although IIRC the TV show was for three players.  Also, instead of the values being 5 to 100 in steps of 5, WLOG just consider the wheel to be 1 to 20.

Well I believe the first thing to figure out is, if player one retires with value $x$, what is probably that player two wins?  He can get either more than $x$ on his first roll, or with two rolls:

hit $y$ on the first roll
hit more than $x - y$ but no more than $20 - y$ on the second roll

The chance of second player Winning in one roll is then

$$\sum_{y=x+1}^{20}\frac{1}{20} = \frac{20 - x}{20}$$

and winning in two rolls is

$$\sum_{y=1}^x\overbrace{\frac{1}{20}}^\text{Chance of hitting y} \times \underbrace{\sum_{z=x - y + 1}^{20-y}\frac{1}{20}}_\text{Chance of hitting a winning second roll}$$

$$=\frac{1}{20\times 20}\sum_{y=1}^x \sum_{z=x - y + 1}^{20-y} 1$$

$$=\frac{1}{20\times 20}\sum_{y=1}^x 20 - x$$

$$=\frac{20x - x^2}{20\times 20}$$

So if player 1 retires with $x$, then the chance that player two wins is $p_2(x) = \frac{20 - x}{20} + \frac{20x - x^2}{20\times 20} = \boxed{1 - \frac{x^2}{400}}$

Ok so now we have to answer the second part:  if you are player 1, and you roll $x$, should you reroll?  For this you calculate the chance of winning if you reroll, and the chance of winning if you won't reroll, and choose which one is larger.

If you reroll, in order to win:

roll value $z$ between $1$ and $20 - x$ inclusive
player 2 must lose with player 1 retiring on $x + z$

$$\begin{align} \text{reroll win chance} &= \sum_{z = 1}^{20 - x} \frac{1}{20} (1 - p_2(x + z)) \\
&= \frac{1}{20} \sum_{z = 1}^{20 - x} \frac{(x+z)^2}{400} \\
&= \frac{1}{20^3} \sum_{z = 1}^{20 - x} x^2+2xz +z^2 \\
&= \frac{1}{20^3} \left((20-x)x^2+2x\frac{(20 - x)^2 + (20 - x)}{2} +\frac{(20-x)((20-x)+1)(2(20-x)+1)}{6} \right)\\
&= \frac{17220 - x - 3x^2 -2x^3}{48000}
\end{align}$$

If you don't reroll, the chance of your winning is $1 - p_2(x) =\frac{x^2}{400}$.

So summarized:

$$
\newcommand{\cent}{{\mathrm{c}\mkern-6.5mu{\mid}}}
\begin{array} {|c|c|c|}
\text{Player 1 first roll} & \text{Winning chance if reroll} & \text{Winning chance if stay} \\
5 \cent & 0.358625 & 0.0025 \\
10 \cent   & 0.358125& 0.01 \\
15 \cent   & 0.357   & 0.0225 \\
20 \cent & 0.355   & 0.04 \\
25 \cent  & 0.351875& 0.0625 \\
30 \cent  & 0.347375& 0.09 \\
35 \cent  & 0.34125 & 0.1225 \\
40 \cent  & 0.33325 & 0.16 \\
45 \cent & 0.323125& 0.2025 \\
50 \cent & 0.310625& 0.25 \\ \hline
55 \cent & 0.2955  & 0.3025 \\
60 \cent & 0.2775  & 0.36 \\
65 \cent & 0.256375& 0.4225 \\
70 \cent & 0.231875& 0.49 \\ \hline
75 \cent & 0.20375 & 0.5625 \\
80 \cent & 0.17175 & 0.64 \\
85 \cent & 0.135625& 0.7225 \\
90 \cent & 0.095125& 0.81 \\
95 \cent & 0.05    & 0.9025 \\
100 \cent & 0       & 1
\end{array}$$

So player 1 should reroll if he gets anything less than $x=11$ or 55 cents.  But unless he gets at least 75 cents, he's probably going to lose anyway.  That's a pretty unfair game, player 1 only has a ${6 \over 20}$ chance of winning overall.

First, we have to lay out some assumptions. 1) The players have no "skill" in that they can get a "feel" for the wheel and improve their odds of hitting a specific number (or range) on subsequent spins. So, each spin is independent and random. 2) The goal for the players is only to beat the other two contestants, and they will not try to get to $1.00 if it jeopardizes their odds of beating the other contestants. The $1.00 bonus spin is just that, a separate bonus not calculated into your spin strategy. 3) All the players will spin optimally, as we don't have any prior information about their computational abilities. I suppose if you wanted to solve this more for "real world" application you could allow for some degree of sub-optimal play (such as in blackjack when people hit on 13 vs. 5 because "they have a gut feeling"). So, for the sake of this solution, we are only looking at the pure math, and aren't factoring in skill, greed, and other's play ability.

To solve the three person problem, it's easiest to work backwards. The last person to spin (Player C) doesn't have any choice other than if they tie the best score. If their first spin is too low, they must spin again. If higher, they win unless they spin > $1.00. If they tie, they can choose to go to a one spin spin-off, or spin again. Since the spin-off has a 50% chance, we can assume Player C will choose to spin again if they tie the best score with 0.50 or less (since they would have a better chance of beating the tie with a second spin). So, we essentially know the "rules" for the Player C.

We can then build out the conditional probabilities for each scenario of Player B's first spin to determine their optimal strategy vs. Player C.

1st Roll    Stay      Spin       Max
 0.05    0.00250     0.33988     0.33988
 0.10    0.00875     0.33944     0.33944
 0.15    0.02000     0.33844     0.33844
 0.20    0.03625     0.33663     0.33663
 0.25    0.05750     0.33375     0.33375
 0.30    0.08375     0.32956     0.32956
 0.35    0.11500     0.32381     0.32381
 0.40    0.15125     0.31625     0.31625
 0.45    0.19250     0.30663     0.30663
 0.50    0.23875     0.29469     0.29469
 0.55    0.28750     0.28031     0.28750
 0.60    0.34125     0.26325     0.34125
 0.65    0.40000     0.24325     0.40000
 0.70    0.46375     0.22006     0.46375
 0.75    0.53250     0.19344     0.53250
 0.80    0.60625     0.16313     0.60625
 0.85    0.68500     0.12888     0.68500
 0.90    0.76875     0.09044     0.76875
 0.95    0.85750     0.04756     0.85750
 1.00    0.95125     0.00000     0.95125

Assuming Player B beat Player A's score on their first spin, and had a choice whether or not to spin again, they should spin again on 0.50 or lower, and stay on 0.55 or higher. (Note that 0.55 gives the worst odds to win, similar to getting a 16 vs. 2 in blackjack. It's probably not enough to hold up, but if you take another card, you'll still lose.) So, here we have essentially solved the two-person problem.

NOTE: Player B has a disadvantage over Player C (45.76% chance of winning), because they are forced to choose first. This is very similar to blackjack, where the house has the edge by being the last player and just follows a strict set of rules.

Now that we know what Player B will do on their first spin (spin again if less than Player A or 0.55), we can calculate a similar matrix of odds for Player A's choices:

1st Roll    Stay      Spin       Max
 0.05    0.13750     0.37438     0.37438
 0.10    0.13875     0.36744     0.36744
 0.15    0.14250     0.36031     0.36031
 0.20    0.14875     0.35288     0.35288
 0.25    0.15750     0.34500     0.34500
 0.30    0.16875     0.33656     0.33656
 0.35    0.18250     0.32744     0.32744
 0.40    0.19875     0.31750     0.31750
 0.45    0.21750     0.30663     0.30663
 0.50    0.23875     0.29469     0.29469
 0.55    0.28750     0.28031     0.28750
 0.60    0.34125     0.26325     0.34125
 0.65    0.40000     0.24325     0.40000
 0.70    0.46375     0.22006     0.46375
 0.75    0.53250     0.19344     0.53250
 0.80    0.60625     0.16313     0.60625
 0.85    0.68500     0.12888     0.68500
 0.90    0.76875     0.09044     0.76875
 0.95    0.85750     0.04756     0.85750
 1.00    0.95125     0.00000     0.95125

This comes out very similar to the previous matrix, and 0.55 or higher is still the optimal choice for Player A to stop.

NOTE: Player A has a 46.38% chance to beat Player B, so again A is at the worst disadvantage since he has to go first.

Based on these given decisions, it is easier to now build out a 3-way probability matrix to determine each player's overall odds:

Player A   30.26%
Player B   33.01%
Player C   36.74%


Showing that $ab=0$ implies $ba=0$ in a ring Question:
Let $n$ be an integer greater than $1$. In a ring in which $x^{n}=x$ for all $x$, show that $ab=0$ implies $ba=0$.

$(ab)^{n}={ab\cdot\cdot\cdot ab}=a(ba\cdot\cdot\cdot ba)b$
there are $(n-1)$ factor of $b\cdot a$
$(ba)^{n-1}=b^{n-1}a^{n-1}=b^{n}b^{-1}\cdot a^{n}a^{-1}=bb^{-1}aa^{-1}$
I have looked at a number of solutions which begins with

"note that $(ba)^{n}$"

which I dislike.
Is there an alternative solution that follows from my attempt?
Thanks in advance.

$0 = ab = (ab)^n = b(ab)^n a = (ba)(ba)^n = (ba)^n+1$ implies $ba = 0$.
Using the result that in any ring with $x^n = x$ for all $x$ in the ring and if $a^m = 0$ for some $a$ in the ring then $a = 0$.


Accidental divisibility sequence? It seems (verified for $n \leq 3 \cdot 10^5$) that when the equation

$$p = n!+1 \bmod{\frac{n^2(n+1)^2}{4}}$$

is satisfied with some $n>1\in\mathbb N$ and $p\in\mathbb P$, it implies $n\in\mathbb P$ as well. The overall feel reminds me of a divisibility sequence like Fibonacci or $2^n-1$, which is the only time I've seen prime values locked to prime indices like that. However, I don't see many other factors being passed along with index factors, which is what I would normally expect.

Note that this applies only to the specific $p$ calculated as above using mod as an operation, and not to any arbitrary $p$ satisfying a congruence.

Can anyone explain why my sequence is behaving this way?

Upon reflection I think it has to do with my using Wilson's Theorem in a mangled way, and consequently I'm only getting non-$1$ values near primes and $2p$ semiprimes.

Some Mathematica code to show a table if you like:

upperBound = 100;
pretty[n_] :=
  If[Length@# > 1, CenterDot @@ #, First@#] &@
    FactorInteger[n] /. {{b_, 1} :> b, {b_, e_} :> Superscript[b, e]};
boldPrime[n_] := If[IntegerQ[n] && PrimeQ[n], Style[n, Bold, Blue], n];
Grid[Table[{n, boldPrime@pretty@n,
   boldPrime@pretty@Mod[n! + 1, Plus @@ (n^2 (n + 1)^2/4)]}, {n, 2,
   upperBound}], Alignment -> Right]

My updated thoughts:

For all $n>9$, it seems like $p=1$ except in one of three conditions:

$n+1$ is prime, in which case $n+1$ divides $p$
$n+1$ is an even semiprime, i.e. $2q$ with prime $q$, and in which case $2$ divides $p$
$n$ is prime, in which case $p$ may or may not be prime, presumably randomly.

If these are indeed the conditions (and I've verified them through $10^5$, so it seems very likely), that would be sufficient to explain the primality correlation. However, I'm still looking for a cogent explanation for why these particular conditions arise from my equation, and will happily accept an answer that can give me some insight into that.

You have $$41617=8!+1+\frac{8^2(8+1)^2}{4}$$
Yet 41617 is a prime and 8 isn't.

There are no smaller counterexamples as $4!+1+k\frac{4^2(4+1)^2}{4}$ will always be a multiple of $5$ (for any $k$), and $6!+1+k\frac{6^2(6+1)^2}{4}$ will always be a multiple of $7$.


Implicit differentiation to find property of a function Im asked to show that if $h(1)=0$ and $h'(x)={1\over{x}}$ then $a,b>0$ show $h(ab)=h(a)+h(b)$. Im expected to use implicit differentiation to show this property.

The question in its current form isn't correct. If we define $h(x)=\ln(x)+7$, then $h'(x)=\frac{1}{x}$, but
$$h(ab)=\ln(ab)+7,$$
$$h(a)+h(b)=\ln(a)+7+\ln(b)+7=\ln(ab)+14.$$

EDIT: With the extra information you have given, the solution follows readily from the fundamental theorem of calculus
$$f(x)=f(x_0)+\int_{x_0}^xf'(y)dy,$$
where in this case we set $x_0=1$. I can't however think of any way to do this using only 'implicit differentiation'.


Natural Deduction Proof: {A v B, ¬A v C} ⊢ B v C Prove using natural deduction: $(A \lor B), (\lnot A \lor C) ⊢ (B \lor C)$.
My work so far:
1     A v B
2     A v C
3      A      :AS
4       ¬A      :AS
5       _|_     :¬E 3,4
6    ¬¬A      :¬I 4-5
7  --
8  ¬A     :AS

This is the work that I have come up with, for the past 5 hours. Can you please help me?

You are setting up for disjunction elimination, so when you derive contradiction, use explosion to derive the target conclusion.$$\def\fitch#1#2{~~\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{~~1.~A\lor B\hspace{10ex}\textsf{Premise}\\~~2.~\lnot A\lor C\hspace{8.5ex}\textsf{Premise}}{\fitch{~~3.~A\hspace{12.5ex}\textsf{Assumption}}{\fitch{~~4.~\lnot A\hspace{8.5ex}\textsf{Assumption}}{~~5.~\bot\hspace{10ex}\textsf{Negation Elimination}\\~~6.~B\lor C\hspace{5.75ex}\textsf{Explosion (aka }\textit{Ex Falso Quodlibet}\textsf{)}}\\~~\vdots\\~~m.~B\lor C }\\~~\vdots\\~n.~B\lor C}$$
I am sure you can fill out the rest of this "proof by cases".


Draw a finite state machine which will accept the regular expression $(a^2)^* + (b^3)^*$ Draw a finite state machine which will accept the regular expression:

$(a^2)^* + (b^3)^*$

In particular, I am confused by the $+$ sign, what does it exactly mean? Most literature I could find about $+$ is $a^+$, which means 1 or more $a$; but here it is clearly not the same meaning.

Here is the drawing:

The idea is that the machine has two counters, one for as and one for bs, indicated by the orange boxes.  Since it accepts either a number of as that is a multiple of 2, or a number of bs that is a multiple of 3, it only needs to choose which of these counters to use.  Upon seeing the first symbol in the input, either a or b, it chooses which of the counters to use, and thereafter circulates around the counter.  One state of each counter is an accepting state.  If the wrong symbol appears while the machine is counting, it goes into ‘dead’ state $x$, from which it cannot escape; transitions to state $x$ are drawn in red.

Start at state $q_{0}$. On input of $a$, transition to state $q_{1}$. On input of $b$, go to $q_{2}$.

While on $q_{1}$, if there is an input of $a$, go to $q_{0}$.

While on $q_{2}$, if an input of $b$, go to $q_{3}$. On $q_{3}$, on input of $b$, go to $q_{4}$. While on $q_{4}$, go to $q_{2}$ on input of $b$.

The accepting states are $q_{0}$ and $q_{4}$. Of course, you could have $q_{0}$ and $q_{4}$ go to a separate accepting state on input $\epsilon$, the empty string.


A function $p \in C^2\big([0,R),\mathbb{R}\big)$ satisfies $p(x) ≥ \alpha$ and $(1 + \beta)[p'(x)]^2\leq p''(x)p(x).$ The following is exercise IV.3.1. from Analysis I by Amann and Escher.

  Suppose that $\alpha, \beta, R > 0$ and $p \in C^2\big([0,R),\mathbb{R}\big)$ satisfy $p(x) ≥ \alpha$ and $$(1 + \beta)[p'(x)]^2\leq p''(x)p(x),\quad x\geq0.$$ Show that $R < \infty$ and $p(x)\to\infty$ as $x\to R−$. (Hint: The function $p^{-\beta}$ is concave. Use a tangent line to $p^{-\beta}$ to provide a lower bound for $p$ (see Application 3.9(e)).

I think I found a counterexample. Let $\alpha:=\beta:=1$ and $p:[0,\infty)\to\mathbb{R}$ with $$p(x):=\frac{x+2}{x+1},\quad x\geq0.$$ Then $p'(x)=-1/(1+x)^2$, $p''(x)=2/(1+x)^3$, so $$(1 + \beta)[p'(x)]^2=\frac{2}{(1+x)^4}=\frac{2}{(1+x)^3}\frac{1}{1+x}\leq\frac{2}{(1+x)^3}\frac{2+x}{1+x}=p''(x)p(x)$$ for all $x\geq0$. Now $p(x)\geq1$, $R=\infty$ and $p(x)\to1$ as $x\to R-$.

Maybe I'm missing something. The question has been asked here but there are no answers yet.

I wrote an email to the author of this series and the following is the screenshot of the email I received:

Since in email system, it can not show Latex, I will type it down for convenience:
$0\leq (1+\beta)p'(x)\leq p''(x)p(x)$ for $0\leq x\leq R$.
In addition, you have to add “unless p is constant” at the end of the assertion.
Now you should be able to do the exercise.
Although the question was proposed three years ago, I hope this would help you somehow.


Proving a continuous function is a constant Let $f$ be a function that is continuous on an interval $[a,b]$ and differentiable  on $(a,b)$. Assume $f'(x)=0$ for all $x\in (a,b)$. Show f is a constant.

I feel like the obvious set up is by contradiction. If I assume $f$ is not a constant, then it must have a variable, but I'm unsure how to construct an $f$ (without loss of generality) which I can then differentiate to contradict $f'(x)=0$.

any suggestions to start this proof?

Let, $x,y\in [a,b]$ then by Mean value theorem we have,
$$f(x)-f(y)=(x-y)f'(\xi)$$ for some $x<\xi<y$ and as given $f'(t)=0,\forall t\in(a,b)$ so, $f'(\xi)=0\implies f(x)=f(y),\forall x,y\in[a,b]$

Hence, $f$- is constant.

Well, $f'(x)$ gives the rate of change of $f$ at a point $x$. If $f'(x)=0 \forall x \in (a,b)$, then $f(x)$ isn't changing, i.e. it is constant, on $(a,b)$.

That's the idea, you just need to show it mathematically.


Questions about absolutely continuous function It could be a silly question but the definition of absolutely continuous function says that
"A real valued function $f:[a,b]\rightarrow\mathbb{R}$ is absolutely continuous on $[a,b]$ if for all $\epsilon>0$, there is a $\delta>0$ such that for every finite disjoint collection $\{(a_k,b_k)\}_{k=1}^n$ of $(a,b)$, $\sum_{k=1}^n[b_k-a_k]<\delta$ implies $\sum_{k=1}^n|f(b_k)-f(a_k)|<\epsilon$."
As I know every absolutely continuous function on $[a,b]$ is continuous on $[a,b]$. But with this definition it seems not hold if we take strange values at $x=a,b$ because the definition only depends on open intervals in $(a,b)$. Is it right that this statement does not hold with above definition?

You're overthinking this. Let $\varepsilon>0$, and choose the corresponding $\delta$. Then for any $\eta<\delta$ such that $a+\eta\leqslant b$, we have
$$|f(a+\eta)-f(a)|<\varepsilon, $$
and hence $f$ is continuous at $a$. $f$ is continuous at $b$ by a similar argument.

There is no 'continuity' in a closed interval, mainly because of the +/-(δ>0) criteria at the upper/lower bounds.


Extremely lost and confused on how to apply law of total variance on problem I am a bit confused on the concept of conditioning and finding the new distribution of conditioned random variables.

An example,

Suppose that there is a new carnival in town and that citizens will begin to line up outside at $10:00$ am. Further suppose that the arrivals are independent, and that the number that arrive in any 1 minute interval is Poisson distributed with $\lambda=3$.

Now suppose the owners will check on the line $T$ minutes after people begin to show up, where $T$ is binomial distributed with $n=20$ and $p=0.8$, then find the expected number of citizens in the line when it is checked and the standard deviation.

What I tried:

I know the law of double expectation and variance

the first is that $E[P]=E[E(P\mid T)]$

and $Var(P)=E[Var(P\mid T)]+Var[E(P\mid T)]$ which I could find standard deviation from.

I know that $E(T)=np=16$ and $E(P)=\lambda=3 $

If I had to give an answer without being confident, I would say that the expected number of customers would be 48

So I know that $E(C|T)$  ( I now think it is $3T$) should be a random variable, how do I find its distribution? What about in general to find the distribution of conditional expectation?

I am having some trouble on the second part and this is why, to use the formula I listed for variance above I want to calculated $Var(P|T)$ , I thought this would be equal to $E(P^2|T)-(E(P|T))^{2}$ which would give me $9T-9T^{2}$, which has an expected value of $-2188.8$, adding this to the variance of the expected value $( 28.8)$ still leaves a very negative number which I know cannot be the case. so I must have a mistake somewhere. Here I also used that we know the variance of binomial to be npq , which in this case is =3.2 and used this to solve for $E(M^2)$ I have tried over and over now but cant figure out where and why I am not getting an answer that makes sense.

Still confused, is really no one willing to help clarify this at all?? I am also confused in one of the answers how he says that the r.v is poisson

Thanks

You should have $\operatorname{E}(P\mid T) = \lambda T.$

Remember that $\lambda$ is in persons-per-minute and $T$ is in minutes, so when you multiply them you get persons.

So $\operatorname{E}(\operatorname{E}(P\mid T)) = \operatorname{E}(\lambda T) = \lambda\operatorname{E}(T) = \cdots$ etc.

And $\operatorname{var}(P\mid T) = \lambda T$ (since the variance of a  Poisson distribution is the same as the expected value.


Need help with this limit $\lim \limits_{x\to \:1}\left(\frac{xe^{x^2}-e}{x-1}\right)$ I haven't been able to solve this limit, so help would be appreciated.
L'Hopital is not allowed.
Thanks in advance.

Hint:

Putting $f(x) = x e^{x^2}$, that limit is just $f'(1)$.


A mathematical function for counting forwards then backwards Is there a mathematical function for counting forwards then backwards though a number range. e.g.

Range of 0-10

The number 7 would equal 7

The number 11 would equal 9

The number 23 would equal 3

So the numbers are counting all the way to the upper limit then counting back down and then back up again and so on.

I come across this requirement a lot in programming and was wondering if there was a typical math formula/function for calculating this?

Yes, just use the cyclic group by first building a representation matrix of its generating element : a circulant matrix

$${\bf C} = \left[\begin{array}{cccccc}0&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\\1&
0&0&0&0&0\end{array}\right]$$

Now let first number be represented by ${\bf v} = [1,0,0,0,0,0]^T$. Then next number is $\bf Cv$, third number is ${\bf C}^2 \bf v$ and so on. Finally assign the value for each position to a column vector $w$. For your example you will want $w_1 = w_N$, $w_2 = w_{N-1}$ and so on, and the value of number at position $k$ of your sequence will be the scalar product:

$${\bf w}^T {\bf C}^k {\bf v}$$

It will be more general than sines and arcsines as you can prescribe any sequence of numbers in $\bf w$.

An example of ${\bf w} = [1,2,3,3,2,1]^T$ will give 1,2,3,3,2,1 in that order if you increase $k=0,1,2,3,4,5$, so you can explicitly enter the numbers you want to have in the sequence into the vector.

If you want to be even fancier/advanced you can construct ${\bf w}^T$ using the following:

$${\bf w}^T = [1,2,3]\left[\begin{array}{cccccc}1&0&0&0&0&1\\0&1&0&0&1&0\\0&0&1&1&0&0\end{array}\right]$$

In general the matrix will be an identity followed by a "opposite diagonal" matrix filled with ones.

The parent function of your function is a triangle wave. Indeed, we have

$$P(x) = \frac{2a}\pi \arcsin\big(\sin\big(\frac{2 \pi}p x\big) \big)$$

So let $a = 10$ and $p = 40$, then take the absolute value to get your function:

$$f(x) = \Big|\frac{20}\pi \arcsin\big(\sin\big(\frac{\pi}{20} x\big) \big)\Big|$$ where $|x|$ is the absolute value of $x$.


Is $\int_0^2 f(x) dx$ defined for $f(x)=x,x \ne 1$? Let $f(x)=x,x \ne 1$.

Is $$\int_0^2 f(x) dx$$ defined?

I'm currently a high school student, and we learn that the integral is the area under the graph. But in calculus textbooks and websites, I see some stuff about "continuity" and things being "integrable". So I was wondering if this simple case is considered "integrable" (because $\lim_{x\to1} f(x)$ is defined) or not (because it has a value "missing" altogether, rather than being discontinuous but still integrable like a step function).

Thanks!

Technically speaking, no, it isn't, since the function needs to be defined on the entire interval $[0,2]$ for the integral over $[0,2]$ to be well-defined. That said, you can define $f(1)$ however you want so that you have a function defined on $[0,2]$; if you do this, the integral is defined and its value is $2$.

A single bad point in the interval that is integrated has zero Lesbegue measure, so it has zero impact on the integral.   However, study of Lesbegue theory of measure and integration is normally done in a graduate level class on analysis of functions of real variables and you may not have been there yet.

If you limit yourself to the Riemann method of integration, you cannot define the upper and lower sums for that bad point, so the integral cannot be defined.


How to find $\lim_{x\to0}\frac{x-x\cos x}{\tan^3x}$ without using L'hopital rule or Taylor series? I found that exercise. However I can not find the answer. I know I can do it with L'hopital. However is there a way to solve it without using L'hopital nor Taylor series? If so show me please. I will appreciate it a lot. If there's no way, which solution seems easier? L'hopital?Taylor?

Hints: You know that $$\frac{\tan x}{x} \xrightarrow[x\to 0]{} 1$$ (as this is $\tan'(0)$) so $\frac{x}{\tan x} \xrightarrow[x\to 0]{} 1$.
You know that $\cos^2 x \xrightarrow[x\to 0]{} 1$.
And you know that
$$
\frac{1-\cos x}{\sin^2 x}=\frac{1-\cos x}{1-\cos^2 x} = \frac{1-\cos x}{(1-\cos x)(1+\cos x)} = \frac{1}{1+\cos x}\xrightarrow[x\to 0]{} \frac{1}{2}.
$$

Can you piece all of it together?

We know that
$\lim_{x\to 0}\frac{\sin x}{x} = 1$, which implies that $$\lim_{x\to 0}\frac{x}{\sin x} = 1$$

Now lets consider
 $$\begin{array}{lll}
\lim_{x\to 0} \displaystyle\frac{x-x\cos x}{\tan^3 x}&=&\lim_{x\to 0} \displaystyle\frac{x(1-\cos x)}{\tan x \tan^2 x}\\
&=&\lim_{x\to 0} \displaystyle\frac{x\cos x(1-\cos x)}{\sin x \tan^2 x}\\
&=&\bigg(\lim_{x\to 0} \displaystyle\frac{x}{\sin x }\bigg)\bigg(\lim_{x\to 0} \displaystyle\frac{\cos x(1-\cos x)}{ \tan^2 x}\bigg)\\
&=&1\cdot \lim_{x\to 0} \displaystyle\frac{\cos x(1-\cos x)}{ \tan^2 x}\\
&=&\lim_{x\to 0} \displaystyle\frac{\cos^3 x(1-\cos x)}{ \sin^2 x}\\
\end{array}$$

Can you take it from here?


How would I solve integrals of derivatives? Solving integrals of values like $\int f \, \mathrm{d} f $ is simple enough but how would I solve integrals of derivatives?

For example consider the integral:

$$ \Delta F = \int^b_a \int \frac{\mathrm{d} f}{\mathrm{d} t} \, \mathrm{d} f \, \mathrm{d} t $$

I guess this can be broken up into two parts.

$$ \frac{\mathrm{d} u}{\mathrm{d} t} = \frac{\mathrm{d} f}{\mathrm{d} t} \frac{\mathrm{d} f}{\mathrm{d} t} $$

and

$$ \Delta F = \int^b_a u \, \mathrm{d} t $$

I'm interested in this because its functional derivative is (I think):

$$ \frac{\delta F}{\delta f} = \frac{\partial u}{\partial f} - \frac{\mathrm{d}}{\mathrm{d} t}\frac{\partial u}{\partial f'} $$

$$ \frac{\delta F}{\delta f} = 0 - 2\frac{\mathrm{d} f}{\mathrm{d} t}$$

but that doesn't seem right to me and it should be something more like $\frac{\mathrm{d} f}{\mathrm{d} t}$.

Using a polynomial series I can represent the function as:

$$ \left[ Dt + \sum^{\infty}_{k=1} \sum^{\infty}_{l=1} \frac{c_k c_l k l}{(k + l -1)(k + l)} t^{k + l} \right]^b_a $$

but I don't know if I can deduce any useful properties from that.

$$I=\int\limits_a^b\int\frac{\mathrm df(t)}{\mathrm dt}\mathrm df(t)\mathrm dt$$

Notice that $$\mathrm df(t)=\frac{\mathrm df(t)}{\mathrm dt}\mathrm dt,$$
so $$I=\int\limits_a^b\int\frac{\mathrm df(t)}{\mathrm dt}\frac{\mathrm df(t)}{\mathrm dt}\mathrm dt\mathrm dt=\int\limits_a^b\int\left(\frac{\mathrm df(t)}{\mathrm dt}\right)^2\mathrm dt\mathrm dt=\int\limits_a^b\int f'\left(t\right)^2\mathrm dt\mathrm dt.$$

There most probably is no simpler form for the second antiderivative of the square of the first derivative than just $$\iint f'(t)^2\mathrm dt^2.$$


Accessible explanation of the link between the distribution of primes and the Riemann zeta function. I am looking for an accessible explanation of the link between the distribution of the primes and the Riemann zeta function.
I have read the related questions and answers here (eg this), and also via internet search, and also key recommended books (popular maths, like Prime Obsession, texts like Apostol, etc) - and they either gloss over the key points, or are aimed at readers with university level maths.
I would appreciate answers here, or pointers to explanations elsewhere.
My students understand the Euler product formula, complex functions, calculus.
Note - this is a repeat of a previous question (now deleted) as it was marked negative without explanation.

$\frac{-\zeta'(s)}{s\zeta(s)}$ is the Laplace transform of $\psi(e^u)$.
If $\psi(e^u)-e^u=O(e^{\sigma u})$ then $\frac{-\zeta'(s)}{s\zeta(s)}-\frac1{s-1}$ is analytic for $\Re(s)>\sigma$. The converse is the purpose of the Tauberian theorems, highly non-trivial, the topic of several textbooks.
The main results are

the PNT $\psi(e^u)-e^u=o(e^{u})$ (following from that $\frac{-\zeta'(s)}{s\zeta(s)}-\frac1{s-1}$ is analytic on $\Re(s)\ge 1$ and shifting the inverse Laplace transform integral to the left),

that under the RH $\psi(e^u)-e^u=O(e^{u/2} u^2)$ (shifting the inverse Laplace transform integral even more to the left after proving a lot of non-trivial facts on $\zeta(s)$)

and the Riemann explicit formula.


What is the expected distance between these two points? Suppose you have a straight line of length L, and a point is chosen at random along the line.

Now suppose a second point is chosen at random to the left of the first point.

What is the expected distance between these points?

Note:  I am aware that if two points are independently chosen at random, the expected distance is L/3, but is it the case in this scenario where the second point is dependent upon the first?  Please justify answers with mathematics.

We have the density function for uniform $x\in[0,L]$ and uniform $y\in[0,x]$
$$
f(x,y)=\frac 1L\cdot\frac 1x
$$
so it is
$$
\int_0^L\int_0^x (x-y)\frac 1L\cdot\frac 1x\ dy\ dx=\frac L4
$$
which can be confirmed by this Wolfram|Alpha-computation.

let [0, L] represent the line.

If X1 and X2 are the Random Variables representing the expected positions of the two points, then:

E(X1 - X2) = E(X1) - E(X2) = L/2 - L/4 = L/4


Show that in a field always $0\ne1$ Suppose that $F$ is a field and prove that $0\ne1$

According to the definition of a field I know that the zero element is different from the one element, but is there a scientific proof for that?

In a field, $0\neq 1$ by definition. This is to exclude the zero ring from being a field.

This leads to a natural question which is an easy but illuminating exercise: If we are in a ring with $0=1$, show that every element equals $0$.

There is a principle in proof theory that if you can find a set of variables such that axioms hold but a conclusion does not, then it is impossible to prove the conclusion from the axioms.
For example, suppose you know the following axioms:

$$x > y \tag{A1}$$
$$x > z \tag{A2}$$

Call $P$ a proposition to be proven as $y > z$.  Well, if we let $x = 3, y = 1, z = 2$.  Note that under this assignment, A1 and A2 are true, but $P$ is false.  Therefore you cannot prove $P$ from A1 and A2.

There are 10 or so axioms of field theory.  The one element field satisfies the axioms of field theory.  The one element field also satisfies $0_F = 1_F$, therefore it is impossible to prove $0_F \ne 1_F$ from just field axioms.


What is the relation between the linear combination and modular arithmetic? What is the relation between the linear combination and modular arithmetic? The linear combination is in a field and there must be some fundamental relation between them. What is it?

When first learning modular arithmetic, you learn how to add, subtract, and multiply congruence classes of integers mod $n$ --- just as you learned basic arithmetic on numbers when you were in elementary school. There is no deep relationship between linear combinations here.

A bit later, you begin to look at division. This takes a bit a care, since in general it is not true that $ac \equiv bc \pmod n \implies a \equiv b \pmod n$, or rather one cannot simply divide or cancel on both sides in a congruence. We ask: by which elements does it make sense to "cancel" from a congruence?

This boils down to finding modular inverses. The inverse of $a \pmod n$ is the integer $x$ such that $ax \equiv 1 \pmod n$. Multiplying by such an inverse is a lot like dividing each side by $a$, except that one can say that $a \equiv b \pmod n \implies ac \equiv bc \pmod n$.

So when does an element $a$ have an inverse mod $n$? This is deeply connected to linear combinations of $a$ and $n$. Namely, the congruence
$$ ax \equiv 1 \pmod n$$
is equivalent to the linear combination
$$ ax + ny = 1$$
for integers $x,y$.

Since division (or rather modular inverses) is so fundamental, one can think of this as a deep relationship between modular arithmetic and linear combinations.

Modular arithmetic uses the operator modulus to create a system which repeats or is cyclical. This is because the modulus or c in the example repeats it's self. Euclidean division is a = bx + c where b is the number of times x goes into a (output of division or input of multiplication) and c is the remainder (output of modulus) (there is no input because there are an infinite number of cases that satisfy this result but you could limit it to the first ones to make it). It is cyclical because if you consider x the base or basis which is trying to reach a then there exists a number which is not a linear combination of only x as the basis (b is just a scalar) so you have to choose  a c from another bases to make it work.


Divisibility Trick for 11 I am trying to prove the 11 alternating sum divisibility trick. I know that $10\equiv -1\pmod{11}$ so for every power of $10$ in a number, we should be able to substitute in $(-1)$ like so:

$$ a(10^n) + b(10^{(n-1)}+\cdots+ c(10) + d \equiv a(-1)^{n} + b(-1)^{n-1}+\cdots+c(-1)+d\pmod{11}$$

What I am having trouble understanding is the fact that starting from the left, we are taking an alternating sum of the digits $(a - b + \cdots$ ). Since each power of $10$ will always become a power of $(-1)$, how can we always use an alternating sum from the left side? That seems like it would change the underlying nature of what is happening with the mod depending on whether we have an even or odd number of digits in the number. For example, when we have a three digit number $a(-1)^3+b(-1)^2+c = -a+b-c$ but according to the rule we have $a-b+c$ verses when we have a four digit number $a(-1)^4+b(-1)^3+c(-1)^2+d = a-b+c-d$ which works with the rule.

Am I missing something obvious? Do I possibly just have the alternating sum rule written incorrectly in my notes?

Thanks!

You end with a $+$ for the final digit for the correct residue mod $11$. If you start from the left and pick the wrong sign for that but alternate anyway, it doesn't matter as the results will be each other's negative and if one is $0$ mod $11$ so is the other. So for divisibility it's equivalent.

Why are you shifting the variable names.  You can but its confusing.  It=n $a(-1)^2 + b(-1)^1+c=a-b+c$ you have $a,c$ in the odd positions and $b$ in the even position.

But with $a(-1)^3+b(-1)^2+c(-1)^1 + d= -a+b-c+d$ you have $a,c$ switching to the even position, while $b$ switches to the odd.  And $d$ is new in the odd.

For less confusing to list from the right.

$a = a$ ($a$ in odd, nothing in even [divisible by $11$ if and only if $a = 0$])

$b(-1)^1 + a = -b + a$  ($a$ in odd; $a$ will always be in odd.  $b$ in even.)

$c(-1)^2 + b(-1)^1 + a = c - b + a$ ($a,c$ in odd; $b$ in even.)

$d(-1)^3 +c(-1)^2 +b(-1)^1 + a = -d + c -b+a$ ($a,c$ in odd; $b$ in even.... and so on.)


Moebius Strip Parameterization Moebius strip $S$ is classically parameterized as follows:

$$
\phi(u,v)=\left( \left(2-v \sin \left(\frac{u}{2}\right)\right) \sin (u), \left(2-v \sin \left(\frac{u}{2}\right)\right) \cos (u), v\cos \left(\frac{u}{2}\right)\right)
$$
for $0 < u < 2 \pi$ and $-1<v<1$. I denote by $U$ the domain in $\mathbb{R}^2$ described so far. It is obvious that $\phi_u \wedge \phi_v \neq 0$ for any $(u,v) \in U$. How can I prove that Moebius strip is a regular surface with such a parameterization? I need to prove that $\phi$ is a homemomorphism onto its image and $\phi(U)$ is open in $S$. To this regard, it is simple after some computations to see that $\phi$ is injective. Clearly, $\phi$ is also continuous. Moreover, to prove that $\phi(U)$ is open in $S$, we observe that $\phi(U)$ is the Moebius strip minus the segment $v \to (0,2,v)$. At this point, I need to prove that $\phi^{-1}: \phi(U) \to U$ is continuous. Nevertheless, I cannot express in an easy way the relation of the coordinates $(u,v)$ in terms of $(x,y,z)$. In the comments, I tried to solve this point, but you will see that a cumbersome computation arises. How can I solve this exercise?

Your $\phi$ does not produce the segment $v\mapsto (0,2,v)\in S$. Therefore you need a second (similarly looking) chart $\psi$, valid for $-\pi<u<\pi$, that has this segment at $u=0$. The two charts $\phi$ and $\psi$ together then form an atlas of $S$.

That your $\phi$ is a homeomorphism of the open rectangular domain $R$ in the $(u,v)$-plane onto its image is easily checked by inspection: The formulas show that $\phi$ is continuous and injective, and one can also express $\phi^{-1}:\>\phi(R)\to R$ in terms of elementary functions whose continuity is understood.


If $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} $ find the ratio of $x$, $y$ and $ z$ I have this question from higher algebra by Hall and Knight:

if $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} $ then find the ratio of x,y and z?
There are two answers given for this question, the first is $\frac x4 =\frac y2 =\frac z3$ and the second is $\frac x1 =\frac y{-1} =\frac z0 $. Now I solve this question in the following manner:

Adding numerator and denominator gives $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y}= 2\frac{x+y}{x+y} =2$  (when x+y is not zero).This gives the first answer.

When (x+y ) is zero => $ y= -x $ that is  $\frac xy = -1$ and now there are two things

(i)If I put these values in the original expression I get $ \frac{y}{x-z} =\frac 0z= -1$ this implies 0= 1 where am I making the mistake?

(ii)Also from the original expression I have$  \frac{x+y}{z} = \frac{x}{y}$ multiplying by z I get $ x+y =z\frac xy$. This implies that z=0 and x:y:z =x:-x:0 = 1:-1:0 and this gives the second answer.But the problem is how z can be zero when it appears in the denominator in the expression

I think these things are very basic but still I am stuck.Could anyone please help me in knowing where the mistake is?

I too am concerned about an answer of $\frac{z}{0}$ being listed. That is a poor way of writing that $z$ is not related to $x$ and $y$ (if that is actually what it means).

What the second answer should say is "If both $x$ and $y$ are zero then $z$ can be any value." Asking this as a ratio question makes it very awkward.


Dot product from magnitudes and cross product Given $|{\vec{a}}|=2, |{\vec{b}}|=5$ and $|{\vec{a}}\times{\vec{b}}|=8$

Find ${\vec{a}}.{\vec{b}}$

Books says answer should be $6$, I think answer should $\pm6$ since $|sin \theta| = \frac{8}{10}$ and thus $ cos \theta =\pm \frac{3}{5}$

So Basically, why is $|{\vec{a}}\times{\vec{b}}|=|{\vec{a}}||{\vec{b}}|Sin \theta$ and mod isn't used around $Sin \theta?$

[Update]

I found an example for which ${\vec{a}}.{\vec{b}}=-6$

${\vec{a}}=2\hat{i}$

${\vec{b}}=-3\hat{i}+4\hat{j}$

Just exploit Lagrange's identity
$$|\vec a\times\vec b|^2=|\vec a|^2|\vec b|^2-\langle\vec a,\vec b\rangle^2,$$
hence $6$ and $-6$ are the solutions.

the equation $$|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin(\theta)$$ is the absolute value of the cross product and the length of the vector $\vec{a}\times \vec{b}$
see also here https://en.wikipedia.org/wiki/Cross_product


What is a formal definition of 'randomness'? What is a rigorous mathematical/logical definition of 'randomness'? Under what conditions can we truthfully apply the predicate 'is random'?

To avoid philosophical debates (I assume you are looking for the mathematical concept) one deals with random variables (they can be though as numerical characteristics of your experiment) which are functions defined on a probability space $(\Omega,\mathcal{B}, \mathbb{P} )$
$$X: \Omega \to \mathbb{R}$$

In order that this construction makes sense, one requires that you can ask some 'natural' questions about the result of your numerical characteristic such as: Is $X$  bigger than some $a$? what is the probability of this event?

You would like to consider $\{\omega \in \Omega : X(\omega) > a\} $ or briefly $[X>a]$. As $\mathcal{F}$ is the set of events, you require that $[X>a] \in \mathcal{F}$ then the probability of the event is given by $\mathbb{P}[X>a] \in [0,1]$.

that is, we require it to be a number between $0$ and $1$.

Lastly one demands that $\mathbb{P}[X\in \mathbb{R}] = 1$

An interesting point is the law of large numbers (a theorem) that states that as you repeat the experiment (with $X_1, \ldots, X_n, \ldots$ as the random variables representing the reproduction of the experiment (as independent random variables)) the mean value you observe converges to the probability on the event in question (almost surely ).

$$\lim_{n\to \infty} \frac{\sum_{j=1}^n 1_{[X_j >a](\omega)}}{n} = \mathbb{P}[A>a]\quad \omega\; a.s.$$

This is a remarkable result  you may find a better discussion on Durret's Book ( Probability: theory and examples). I started there.

Now, if there is randomness in the world or if this is nothing but a useful model is a deeper question that requires more than the formal construction we made above. you should check this quote of Einstein "God does not play with dice" and contrast it with James Clerk Maxwell:
“The true logic of this world is in the calculus of probabilities.”

a fuller discussion can be found on http://www.feynmanlectures.caltech.edu/I_06.html

Good luck

Food for thought:
Aperiodic infinite "pattern"
Since that is an oxymoron we can rather say,
The Logical Definition of Randomness:

Aperiodic infinite sequence of a set of values whose values must
appear in that infinite sequence to the exact proportion they appear
in the set of values

The predicate is said to be random if it is an outcome of an event that is a part of that aperiodic infinite sequence of a set of values given that the trials for the event were to be continued indefinitely
In the case of Pi, each new iteration of the calcuation to further the digits of Pi is a single trial, you cannot know the next digit until you calculate it. That is the equivalent of flipping an unbiased coin. There is a reason we only know Pi to the 2 quadrillionth digit. But even now, you cannot know the next digit until you run the next iteration - ergo the next "coin toss".


If a vector field is divergenceless and curless, is that vector field = 0? Just a simple question, from the title of the thread, is a vector field = 0 if the divergence is 0 and the curl is 0? I had trouble finding an answer anywhere online, proof of why or why not would be helpful, thanks!

$$\nabla \cdot \vec{F} = 0$$
$$\nabla \times \vec{F} = 0$$
$$\vec{F} = ?$$

No.  To get a feel for what's up, let $\phi$ be a non-constant harmonic function so $\nabla^2 \phi = 0$ but $\nabla \phi \ne 0$.  Set $\mathbf F = \nabla \phi \ne 0$; then   Then $\nabla \times \mathbf F = \nabla \times \nabla \phi = 0, \,$ since the curl of a gradient always vanishes.  Also, $\nabla \cdot \mathbf F = \nabla \cdot \nabla \phi = \nabla^2 \phi = 0$.  The divergence and curl of $\mathbf F$ both vanish, but not $\mathbf F$!

This line of reasoning can, like tape or film, be re-wound and run "backwards":  if $\mathbf F \ne 0$ and $\nabla \times \mathbf F = 0$, then (locally at least) there is a function $\phi$ with $\mathbf F = \nabla \phi \ne 0$; if now we also have $\nabla \cdot \mathbf F = 0$, then $\nabla^2 \phi = \nabla \cdot \nabla \phi = \nabla \cdot \mathbf F = 0$, and $\phi$ is harmonic.

The classic examples of such a field may be found in the elementary theory of electromagnetism:  in the absence of sources, that is, charges and currents, static (non -time varying) electric fields $\mathbf E$ and magnetic fields $\mathbf B$ have vanishing divergence and curl:  $\nabla \times \mathbf B = \nabla \times \mathbf E = 0$, and $\nabla \cdot \mathbf B = \nabla \cdot \mathbf E = 0$; the electrostatic potential function $\phi$ such that $\mathbf E = -\nabla \phi$ with $\nabla^2 \phi = 0$ exists by virtue of these facts; a similar assertion holds for the $\mathbf B$ field.

Hope this helps.  Cheerio,

and as always (as ol' James Clerk M. has taught us),

Fiat Lux!!!

No. For example, if we consider simple 2 dimension field,

$\vec{V} = a (x,-y) $     ,  for any scalar $a$ then

$ \vec{\nabla} . \vec{V} = 0 $ and
$ \vec{\nabla} \times  \vec{V} = 0 $


How to explain that division by $0$ yields infinity to a 2nd grader How do we explain that dividing a positive number by $0$ yields positive infinity to a 2nd grader? The way I intuitively understand this is $\lim_{x \to 0}{a/x}$ but that's asking too much of a child. There's got to be an easier way.

In response to the comments about it being undefined, granted, it is undefined, but it's undefined because of flipping around $0$ in positive or negative values and is in any case either positive or negative infinity.

Yet, $|\frac{2}{0}|$ equals positive infinity in my book. How do you convey this idea?

The first thing to point out is that division by zero is not defined! You cannot divide by zero. Consider the number $1/x$ where $x$ is a negative number. You will find that $1/x$ is negative for all negative $x$. As $x$ gets closer and closer to zero, $1/x$ gets bigger and bigger in the negative direction: $1/x \to -\infty$ as $x \to 0$ from the negative side. Next, consider the number $1/x$ where $x$ is a positive number. You will find that $1/x$ is positive for all positive $x$. As $x$ gets closer and closer to zero, $1/x$ gets bigger and bigger in the positive direction: $1/x \to +\infty$ as $x \to 0$ from the positive side.

$$\lim_{x \to 0^-} \frac{1}{x} \neq \lim_{x \to 0^+} \frac{1}{x}$$

Informally: what does $6 \div 3$ mean? It means, how many times do you add $3$ together to get $6$, and the answer is $2$. What does $7 \div 2$ mean? It means, how many times do you add $2$ together to get $7$, and the answer is $3\frac{1}{2}.$ What does $1 \div 0$ mean? It means, how many times do you add $0$ together to get $1$? Well: $0 = 0+0 = 0+0+0$, etc.
You have to keep adding zeros for all of eternity. In reality, you never get to $1$ and so there is no answer. It is not infinity: you can't have "infinitly many" zeros. But some people might say "You add $0$ together infinitely many times".

I say that division by $0$ is defined, that it can yield multiple results in different contexts. People need to stop treating $0$ as a number or an integer for it is not! It is a place holder, the empty or null set. First lets start with what the operation of division or a fraction is! It is the inverse or opposite operation of multiplication; simple enough!. What is multiplication? It is the repetitive operation of addition! What is addition? Addition is a 1 dimensional linear transformation specifically translation!

Consider this: we have an arbitrary point $p(n)$ where n is any value. A point itself is similar to but different than a vector. A point itself can have an infinite amount of dimensional coordinates but the point itself is $0$ dimensional where it has no association or awareness of any other point, direction (rotation), or dimensions. Once you introduce a second point and you associate the two, then you have a line segment that has distance, displacement or length. The addition or subtraction which is horizontal linear translation of two points that creates that line segment transforms those two points into a single 1 dimensional vector. A vector does not become $2D$ until it is introduce to another vector that has a separation by a angle of some degree. So at minimum you need 3 points at different locations to create at minimum two different vectors to where one of those points becomes a vertex or intersection between two line segments, lines or vectors.

So what does points, vectors, lines, rotations and different dimensional spaces have to do with division by $0$ and whether it is undefined or defined? Everything!

To explain this we need to understand some geometry, some algebra and some trigonometry and the relationships of vectors and angles.

Things need to know:

Slope of a line:
$$m = \frac{rise}{run} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\sin\theta}{\cos\theta} = \tan\theta$$
If a slope has the form $\frac{n}{d}$ then its perpendicular or the line orthogonal to it is its negative reciprocal as in $-\frac{d}{n}$
If $\tan\theta$ represents the slope of the line according to $\theta$ where $\theta$ is the angle above the horizontal in standard position then we can say that the perpendicular must be:
$$-\frac{\cos\theta}{\sin\theta} = -\cot\theta$$
Both the $\sin$ and $\cos$ have a domain of $\mathbb R$ and a range of $[-1,1]$, they are circular functions or are continuous and repeat forever with a period of $2\pi$ making them periodic wave functions.
Pythagorean Theorem & Equation of Circle Are The Same
$$A^2 + B^2 = C^2$$   $$X^2 + Y^2 = r^2$$
The relationship between the cosine of the angle between two vectors: $$\cos\theta = \frac{\vec a \cdot \vec b}{|\vec a||\vec b|}$$

When we think of slope and its definition that we are referring to it is the rate of the increase of height or elevation over distance. So when the ground that we walk on is perfectly flat and we have $0$ elevation we have $0$. To demonstrate this use the Unit Circle and the coordinate pairs on the circle $(x,y) = (\cos\theta, \sin\theta)$ and the vector or line has a length of $1$. So when their is $0$ rise the $\sin\theta$ is $0$ so the equation of the slope of the line or the tangent is as follows: $\frac{0}{\cos\theta} = 0$. This simply means we have no elevation but we have any amount of horizontal translation.

Now when we rotate the line $CCW$ around the unit circle and we take a pit stop at $45°$ or $\frac{pi}{4}$ we end up with a slope of $1$.

This takes the form of: $$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{ \frac{\sqrt{2}}{2} } { \frac{\sqrt{2}}{2} } = 1$$

So here we are going up 1 in height for every one in length. So far so go.

Let's continue to rotate this vector in the same direction and make a pit stop at $90°$ or $\frac{pi}{2}$.

In conventional math teachings and what many of us have been taught was oh-no we have a problem; vertical slope is undefined due to division by $0$.

I say we don't have a problem and that it is not undefined. Why?

Well it was okay to say we had $0$ rise with infinite horizontal translation resulting in $0$ slope. And now that we are at its perpendicular that slope isn't defined?

In the first case we had  $$\frac{\sin\theta}{\cos\theta} = \frac{0}{\cos\theta} = 0$$ but now that we have the opposite where we have no run but we do have rise we can not seem to defined this? $$\frac{\sin\theta}{\cos\theta} = \frac{\sin\theta}{0} =$$  $$? ... undefined$$

This is vertical slope and because of these assessments it also causes us to treat the tangent and cotangent to be undefined at specific values because of division by $0$.

However by definition of words and thought and sound intuition we know that our height is always increasing, but we are no longer moving horizontally. This is infinite slope or vertical slope. This is well defined if you ask me. Let's remove the $0$ for a moment and consider the following with some arbitrary variables.

If the slope = $\tan = \frac{\sin}{\cos}$ then its negative reciprocal must be $-\frac{\cos}{\sin} = -\cot$

So if the slope of a horizontal line is $0$ in the form of $\frac{0}{\cos}$ then to find the negative reciprocal to get its perpendicular we have to say that: Instead of using $0$ directly I will use the angle of the trig function that evaluates to $0$ since it has a valid range and domain to produce this value of $0$.

$$-1 = \frac{n}{d} * -\frac{d}{n}$$

Let $$\frac{n}{d} = \frac{\sin\left(0°\right)}{\cos}$$ and  $$-\frac{d}{n} = -\frac{\cos}{\sin\left(0°\right)}$$

then

$$-1 = \frac{\sin\left(0°\right)}{\cos} * -\frac{\cos}{\sin\left(0°\right)}$$

Initially this would appear to produce $-1 = 0 * 0$ if you multiplied the fractions out which wouldn't make any sense. However when you look closely the two $\sin$ functions without any known input values intuitively cancel each other out leaving you with $-1 = -\frac{\cos\theta}{\cos\theta}$. Let's prove this by looking at the dot product of two vectors and applying trig substitutions. Let's consider this: $p1 = (0,0), p2 = (1,0),  p3 = (0,1)$ which are the three points on the unit circle that would make a right triangle with the axis being the perpendicular. We need two vectors $\vec A \vec B$ and in this case we need the vectors to be $\vec A = p2 - p1$ and $\vec B = p3 - p1$ and this will give us the vectors as: $\vec A = (1,0)$ and $\vec B = (0,1)$.

If we take the dot product of these two vectors and divide them by their magnitudes it should look like this:

$$\cos\theta = \frac{\vec A(1,0) \cdot \vec B(0,1)}{|\vec A||\vec B|}$$

and this simplifies to: $$\cos\theta = \frac{0}{1} \implies \arccos(0) = 90°$$

Now that we have a relationship with $\arccos$ and we know that the these vectors do in fact intersect each other at a $90°$ angle making them perpendicular and orthogonal to each other, they are also both a normal to each other since their magnitudes or the length of these vectors both have an absolute value of $1$.

We know through trigonometry that there are relationships between the trig functions and since slope is defined as the $\tan$ of the angle $\theta$ above the horizontal in standard angle form where the vertical vector or leg that creates the right angle on the right side of the triangle with $\theta$ being the vertex on the left when the trig functions are used within the context of the unit circle with the unit circle being defined by the equation $X^2 + Y^2 = C^2 = X^2 + Y^2 = 1$. Since the radius vector of the unit circle has a magnitude of length $1$ we can apply some of these relationships between the $\sin$ and the $\cos$ to convert from one to the other based on the fact that the natural graphs of the two are a $\pm90°$ angle or a $\pm\frac{\pi}{2}$ radians horizontal translations of each other. We can use this to substitute back into the equation where we found the negative reciprocal of the horizontal normal vector with a magnitude of $1$.

NOTE: Before I do the substitution I will make the statement here that the sign of the vectors in this case only implies opposite direction or reflection of symmetry or the rotation of either $180°$ degrees or $\pi$ radians about a fixed point and in this case the fixed point is $(0,0)$. The reason I state this is because $\vec A$ that starts at the point $(0,0)$ and points to the point $(1,0)$ has the positive direction stemming out from $(0,0)$ while another vector $-\vec A$ starts at the point $(0,0)$ and points to the point $(-1,0)$ again stemming out from $(0,0)$. This relationship of vectors $\vec A$ and $-\vec A$ are the same exact vector when vector $\vec A$ is rotated $180°$ or $\pi$ radians the point at the end of the initial vector before the rotation located at $(1,0)$ is the same exact point that ends up at $(-1,0)$ along with all the points of the line or vector in between the fixed point $(0,0)$ and vector $\vec A$. This is evident by two supporting functions, the first was already used once but with different values and that is the relationship of $\cos\theta$ and the dot product between vectors $\vec A$ and $-\vec A$ which will yield $-1$; also if you take the $\arccos(-1)$ it will evaluate to $180°$ of rotation which is also the angle measure of any and all $1D$ Cartesian Linear Lines of the form $y = mx + b$. The second supporting function is the definition of the circumference of a circle $c = 2\pi r$ and since $r = 1$ this simplifies to $2\pi$ which also happens to be the period of the $sine$ and $cosine$ functions. Now that we have $2\pi$ which measures the arc length around the circle when the radius $r = 1$ when we divide that in half it gives us $\pi$ radians or $180°$.

Let's now apply the substitution of the $\sin$ and $\cos$ functions in order to remove the component of the fractional part of the $\tan$ that represents the axis directional component of vertical slope.

Given that when $-1$ is applied to the $\arccos()$ function and returns a value of $180°$ and that the first or left fraction represents the slope of a line whose slope is $0$ and that the second or the right negative fraction represents the slope of the line that is perpendicular to the first shown again here:
$$-1 = \frac{\sin\left(0°\right)}{\cos} * -\frac{cos}{\sin\left(0°\right)}$$
and knowing the trigonometric Cofunction identities with $$\sin\left(\frac{\pi}{2} - \theta \right) = \cos\theta$$
and
$$\cos\left(\frac{\pi}{2} - \theta \right) = \sin\theta$$
we can use these to make that substitution and since $\sin(0) = 0$ and it is found in the negative reciprocal we need to substitute all $\sin()$ functions with $\cos()$ functions

$$-1 = \frac{\cos(\frac{\pi}{2} - \theta)}{\cos\theta} * \frac{\cos\theta}{\cos(\frac{\pi}{2} - \theta)}$$

If we partially ignore the negative sign attached to the $-1$ on the left hand side of the equation since it only implies the facing direction of the vector as in right versus left we can see that $$|-1| = \frac{\cos(\frac{\pi}{2} - \theta)}{\cos\theta} * \frac{\cos\theta}{\cos(\frac{\pi}{2} - \theta)}$$ still holds true.

Conclusion To sum this up when a slope is $0$ it is the numerator or the rise component that is $0$ which means it has no slope for there is no rise in height and this is okay. When the slope is vertical slope we are normally taught that it is undefined because of division by $0$. However without using values and just using real life concepts since mathematics is supposed to model real life concepts when we do have vertical slope I say that it is very much defined. It is the opposite of $0$ slope from the rise being $0$ in the numerator and when slope is vertical we now have $0$ run or horizontal displacement but we are ever increasing in the up and down direction does imply that the slope is not $0$ but the slope is infinite or the $\lim$ approaches $\infty$ just like the horizontal component of $0$ slope approaches $\infty$. Imagine walking down an alley between two skyscrapers and that the alley represents the relationship of different translations between different inclines of slope. At first the ground is level so there is no slope then the alley has a hill and the slope is $1$ this means perfect rise over run as the increase in elevation is equivalent to the change in horizontal translation where the angle above the horizon is $45°$ and $\tan(45°) = 1$. Then the ground levels out again with $0$ slope and is flat and horizontal. Then all of a sudden you can go no farther because there is another skyscraper in front of you, there is no turning left or right, and there is no going back. The outside wall of the building has a ladder and you begin to climb up. Now you have $0$ horizontal translation and continuous vertical translation for each rung of the ladder you climb. There for vertical slope is defined as it is always 100% straight up, orthogonal, perpendicular and normal to $0$ or horizontal slope.

One thing to be aware of is if you have the unit circle with 2 fixed points one at $(0,0)$ and one at $(1,0)$ and your third point can move freely along the circumference of the unit circle we can make 3 vectors to form a triangle. We do this in the first quadrant so that the initial right triangle is in standard position. Two of the vectors will always equal 1 and initially the first vector will be composed of the first 2 fixed points. The 2nd vector will virtually start off as the same exact vector of the fixed vector but it will rotate about the origin around the unit circle. The 3rd vector initially doesn't exist or is the $\vec 0$ since this has a tail at $(1,0)$ that will remain fixed but will have its head tracing around the unit circle with the rotating vector radius. This vector will change its length at each interval of rotation. Now imagine the right triangle in standard position with a $45°$ angle and the rotating radius with a slope of 1 this will end up being the hypotenuse. This radius vector will remain the hypotenuse until it coincides with the $Y-axis$. Once it does and the angle becomes $90°$ and the once standard position of the right angle no longer exists on the right side of the triangle as the intersection of the axis now becomes your new right angle for only this measure of degree and when $\theta$ also equals $270°$. The radius vector that was the hypotenuse no longer is and the variable length vector now becomes the new hypotenuse with a length of $\sqrt 2$. When the two radius vectors are in this position or when the rotating vector is at $(0,-1)$ both of these right triangles yields the maximum area of any triangle that can be made by the unit circle while having two legs with a magnitude of its radius. Another peculiar thing to notice here is when we have this reversed right triangle such that the angle of rotation is at $90°$ degrees and we take the origin point $(0,0)$ and reflect it over this
new hypotenuse it will give us the point $(1,1)$ which is outside of the unit circle. This creates a perfect unit square with an area of $1$, this also shows the relationship of the Pythagorean Theorem that pertains to triangles and how it relates to the equation of the Unit Circle, and this is also one of its proofs.

So in this context of linear transformations, rotations, slope etc., division by $0$ is defined as it represents vertical or infinite slope. Now in other contexts division by $0$ may represent other things such as $0,1,\infty,D.N.E.(Does Not Exist)$ We have to remember that $0$ is not really a number nor an integer. It is a place holder and has no value as it represents the empty or null set, an approach to infinity or even infinity itself. $0$ in its purest form is only a character, digit or glyph.

How to explain this to a 2nd grader? Hmm, good luck as you seen what I had to do to prove all of the massed printed text books wrong!

Edit -Additional things to consider.

Since we already seen that $\tan(45°)$ or $\frac{\sin(45°)}{\cos(45°)} = 1$ let's consider the very first expression or equation that we are taught $1 + 1 = 2$. Since all integer values can be represented as fractions we can do this as well: Ant that is to substitute values of $1$ with the appropriate trig functions with the appropriate angles and this would give us...

$$1 + 1 = 2 \implies \tan(45°) + \tan(45°) = 2 * \tan(45°)$$
which can also mean:
$$1 + 1 = 2 \implies \frac{\sin(45°)}{\cos(45°)} + \frac{\sin(45°)}{\cos(45°)} = 2 * \frac{\sin(45°)}{\cos(45°)}$$

This holds true because the definition of slope is $\frac{\delta y}{\delta x}$
which is a fraction. So one can see that all numbers are in fact a form of slope including the approach to $\pm\infty$. So I tend to believe in this particular context that slope in the form of $\frac{\sin\theta}{\cos\theta}$ where the $\sin$ does not evaluate to $0$ and the $\cos$ does, the yielding result would be $\pm\infty$.

Added Note - Also some might say well it yields both $\pm\infty$ so it still has more than one answer so it still has to be undefined. This is a wrong assumption or assessment. It depends on the direction of the angle or rotation. If you angle $\theta$ is either $90°$ or $-270°$ then the answer is obviously $+\infty$ or vertical slope that is straight up and if your angle $\theta$ is either $-90°$ or $270°$ then it is $-\infty$ or vertical slope and it is straight down, and don't forget to count they periodicity of higher valued angles which would be $\theta + 180° * n$ where $n$ is some integer $>=1$ or $0$. This also works as a function because even though there are two inputs that give you a single output it is the nature of the $\sin()$ and $\cos()$ functions that gives this behavior of the slope of the line. This duality is due to the equation of a circle $x^2 + y^2 = r^2$ or the Pythagorean Theorem $A^2 + B^2 = C^2$.