Limit of product of sequences is the product of the limits of the sequences I want to prove that if:

$$\lim_{n \to \infty}s_n = L_1, \lim_{n \to \infty}t_n = L_2$$

then $$\lim_{n \to \infty}(s_n t_n) = L_1L_2$$

where $s_n, t_n$ are complex sequences (I'm working through Rudin's baby rudin, and I did this proof slightly different than him, so I would like it if someone took the time to look through my proof and verify whether it's correct)

My attempt (scratch proof: finding the appropriate epsilons etc):

By definition of limit, there are positive integers $N_1,N_2$ such that:

$$n >N_1 \Rightarrow |s_n - L_1| <\epsilon'$$
$$n >N_2 \Rightarrow |t_n - L_2| <\epsilon''$$

where we can freely choose $\epsilon', \epsilon'' >0$

Let $\epsilon > 0$

Now, if $n > \max\{N_1,N_2\}$, then:

$$|s_nt_n - L_1L_2| = |s_n(t_n - L_2) + s_nL_2 - L_1L_2|$$
$$\leq |s_n||t_n - L_2| + |L_2||s_n - L_1|$$
$$\leq |s_n|\epsilon'' + |L_2|\epsilon'$$

We want this expression to be smaller than $\epsilon$, but we can't make our $\epsilon$ depend on $n$, therefore we have to find an upper bound for $|s_n|$

There is a positive integer $N_3$ such that:

$$n > N_3 \Rightarrow |s_n - L_1| < 1$$
$$\Rightarrow |s_n| < 1 + |L_1|$$

so for $n > \max\{N_1,N_2,N_3\}$, we have:

$$ |s_n|\epsilon'' + |L_2|\epsilon' < (1 + |L_1|)\epsilon'' + |L_2|\epsilon'$$

and by chosing $\epsilon'' = \epsilon' = \frac{\epsilon}{1+|L_1|+|L_2|}$, the expression becomes smaller than $\epsilon$

Rigorous proof

Let $\epsilon > 0$

By definition of limit, there are positive integers $N_1,N_2, N_3$ such that:

$$n >N_1 \Rightarrow |s_n - L_1| <\frac{\epsilon}{1+|L_1|+|L_2|}$$
$$n >N_2 \Rightarrow |t_n - L_2| <\frac{\epsilon}{1+|L_1|+|L_2|}$$
$$n > N_3 \Rightarrow |s_n - L_1| < 1 \Rightarrow |s_n| < 1 + |L_1|$$

and for $n > \max\{N_1,N_2,N_3\}$, we have:

$$|s_nt_n - L_1L_2| = |s_n(t_n - L_2) + s_nL_2 - L_1L_2|$$
$$\leq |s_n||t_n - L_2| + |L_2||s_n - L_1|$$
$$< (1 + |L_1|)\frac{\epsilon}{1+|L_1|+|L_2|} + |L_2|\frac{\epsilon}{1+|L_1|+|L_2|} = \frac{\epsilon}{1+|L_1|+|L_2|}(1+ |L_1| + |L_2|) = \epsilon$$

Alternative proof:

Let $\epsilon > 0$

$(s_n)$ converges, so $(|s_n|)$ converges and it is bounded, meaning that there is a positive real number $S$ such that $|s_n| \leq S$ for every positive $n$.

By definition of limit, there are positive integers $N_1,N_2$ such that:

$$n >N_1 \Rightarrow |s_n - L_1| <\frac{\epsilon}{S + |L_2|}$$
$$n >N_2 \Rightarrow |t_n - L_2| <\frac{\epsilon}{S + |L_2|}$$

Now, for $n > \max\{N_1,N_2\}$, we have:

$$|s_nt_n - L_1L_2| = |s_n(t_n - L_2) + s_nL_2 - L_1L_2|$$
$$\leq |s_n||t_n - L_2| + |L_2||s_n - L_1|$$
$$< (S+|L_2|)\frac{\epsilon}{S + |L_2|} = \epsilon $$

Note that this wouldn't work whenever $S + |L_2| = 0$. However, if $S=0$, then $s_n$ is the null-sequence and the theorem is trivial, so we can safely assume $S \neq 0$ such that the denominator never becomes $0$.

QED

Is this correct?

Your “Rigorous proof” is not correct. Look at the way you chose $N_1$, $N_2$, and $N_3$. Only the sequence $(s_n)_{n\in\mathbb N}$ is mentioned, not the sequence $(t_n)_{n\in\mathbb N}$. You can't get a correct proof without using both of them.

Here is how I would prove the same statement. Please keep in mind that I always try to avoid to introduce asymmetries when the hypothesis are symmetric. In this case, what is supposed about both sequences is the same and therefore I will avoid to treat them in two different ways.

For each $n\in\mathbb N$, you know that$$s_nt_n-L_1L_2=(s_n-L_1)L_2+(t_n-L_2)L_1+(s_n-L_1)(t_n-L_2)$$and therefore that$$|s_nt_n-L_1L_2|\leqslant|s_n-L_1|.|L_2|+|t_n-L_2|.|L_1|+|s_n-L_1|.|t_n-L_2|.$$Let $\varepsilon>0$ and choose $N_1\in\mathbb N$ such that$$n\geqslant N_1\Longrightarrow|s_n-L_1|<\min\left\{\frac\varepsilon{3\bigl(|L_2|+1\bigr)},\sqrt{\frac\varepsilon3}\right\}.$$Also, choose $N_2\in\mathbb N$ such that$$n\geqslant N_2\Longrightarrow|t_n-L_2|<\min\left\{\frac\varepsilon{3\bigl(|L_1|+1\bigr)},\sqrt{\frac\varepsilon3}\right\}.$$Let $N=\max\{N_1,N_2\}$. Then, if $n\geqslant N$,\begin{multline*}|s_n-L_1|.|L_2|+|t_n-L_2|.|L_1|+|s_n-L_1|.|t_n-L_2|<\\<\frac{\varepsilon|L_2|}{3\bigl(|L_2|+1\bigr)}+\frac{\varepsilon|L_1|}{3\bigl(|L_1|+1\bigr)}+\frac\varepsilon3<\varepsilon.\end{multline*}

hint

there is an error in line 37.

the easiest way is to use the fact that a convergent sequence is bounded.

$$|s_n|\le S $$

and
$$|s_nt_n-L_1L_2|\le $$
$$S|t_n-L_2|+|L_2||s_n-L_1|\le$$
$$S\frac {\epsilon}{2S}+|L_2|\frac {\epsilon}{2|L_2|} $$

if $$SL_2\ne 0$$

If $L_2=0$ and $S\ne 0$

$$|s_nt_n|\le S\frac {\epsilon}{S} $$
for enough large $n $.


How to interpret material conditional and explain it to freshmen? After studying mathematics for some time, I am still confused.

The material conditional “$\rightarrow$” is a logical connective in classical logic. In mathematical texts one often encounters the symbol “$\Rightarrow$”, which is read as “implies” or “if … then ….” It is customary and reasonable to treat “$\Rightarrow$” as the material conditional, i. e. as equivalent to saying “the antecedent is false or the consequence is true (or both)”.

Now there is some controversy about whether the material conditional really captures conditional statements because it doesn't really say anything about a causal connection between the antecedent and the consequence. This is quite often illustrated by the means of statements in natural languages such as “the moon is made out of cheese $\Rightarrow$ all hamsters are green” – since the moon isn't made out of cheese, is this statement true? This remained problematic to me.

While I came to accept the material conditional as a good way of describing implications and conditionals, I'm having a hard time to explain this usage to freshmen whenever I get asked.

My questions are: How can we best justify the interpretation of “$\Rightarrow$” as a material conditional? Why is it so well-suited for mathematics? How can we interpret or read it to understand it better? Can my confusion about it be led back to some kind of misunderstandig or misinterpretation of something?

I have yet very poor background in mathematical logic (I sometimes browse wikipedia articles about it), but I'd have no problem with a technical answer to this question if it clarifies the situation.

Perhaps this will help to capture the truth-functional character of material implication:

The truth-value of an inclusion (subset) relation between sets corresponds to the truth-value of an implication relation, where $\subseteq$ corresponds to the $\rightarrow$ relation.

E.g., suppose $A\subseteq B$. Then if it is true that $x\in A$, then it must be true that $x\in B$, since $B$ contains $A$. However, if $x\notin A$ (if it is false that $x \in A$), it does not mean that then $x\notin B$, since if $A\subseteq B$, then $B$ may very well contain elements that $A$ does not contain.

Similarly, suppose we have that  $p\rightarrow q$. If $p$ is true, then it must be the case that $q$ is true.  But if $p$ is false, that does not necessarily mean then that $q$ is necessarily false. (For all we know, perhaps $q$ is true regardless of whether or not $p$ is true.) So $q$ can be true, while $p$ is false.

I don't know if this analogy helps or not. But it was the above analogy (correspondence) that helped me to firmly grasp the logic of material implication.

Here's a more down-to-earth example you may have already stumbled upon:

CLAIM: "If (it rains), then (I'll take an umbrella)":

I'd be lying (my assertion would be false) if (it rains = true), and I do not (take an umbrella).

But perhaps it's cloudy out, and I decide I'll take an umbrella , just in case it rains. In this case:

If it doesn't rain (it rains = false), but I took my umbrella (true), my claim above would not be a lie (it would not be false).

Sure you can have a valid "implication relationship" between two errant pieces of nonsense.

If we have M=>H between moon and hamster populations then the truth table is
M=>H == not(M and not H).  In modal logic (many-world) terms this means we allow all worlds to be possible EXCEPT the one that has a cheesy moon and at least one non-green hamster.

This is the main effect of the implication relationship, it acts as a prohibition on one case, indeed this is the essence of an asymmetric dependency. M=>H means M depends on H in the sense that we cannot have M without H. If H is false then M cannot be true. Asymmetric dependency relationships are very common, hence the utility of this connector.

In the other direction it matters not, there are at least two worlds that have a normal moon, and in only one of them the hamsters are all green. You are NOT forced to live in that world!

The Wason selection task is a good illustration of utility of material conditional support for deductive reasoning:

https://en.wikipedia.org/wiki/Wason_selection_task


Particle starts at $(0,-3)$ and moves clockwise around origin on graph $x^2+y^2=9$, find parametric equation Question

  particle starts at $(0,-3)$ and moves clockwise around origin on graph $x^2+y^2=9$, revolve in $9$ seconds find parametric equation in term of $t$.

What I've done so far:

I first thought that the graph ought to be $x^2+y^2=9$

so then I say that $x=\frac{9}{2}\cos{t}=x$ and $y=\frac{9}{2}\sin{t}$

but then this particle in this graph travels CCW so then I change it to :

$x=\frac{9}{2}\cos{t}=x$ and $y=\frac{9}{2}\sin{-t}$

but then I found out that when I plug in $t=0$, I do not get -3

How do I phase shift this parametric equation so that it satisfies the fact that the particle starts at $(0.-3)$?

Let's start with the standard clockwise parametrization,

$$x=3\cos (t)$$

$$y=-3\sin (t)$$

With $t \in [0,2\pi]$. This starts at $(3,0)$ and then moves clockwise.

We want it to start at $(0,-3)$. If we rotate $(3,0)$ clockwise $90$ degrees we get $(0,-3)$. So shift $\frac{\pi}{2}$ radians clockwise. Let,

$$x=3\cos (t+\frac{\pi}{2})$$

$$y=-3\sin (t+\frac{\pi}{2})$$

Now we need to deal with the fact that it should take $9$ seconds to do a full revolution. Right now it takes $2\pi$ seconds because the period of both is $2\pi$. If the new period is $9$ then $2\pi$ over the horizontal shift $a$ should be $9$.

$$\frac{2\pi}{a}=9$$

$$a=\frac{2\pi}{9}$$

So,

$$x=3\cos (\frac{2\pi}{9}t+\frac{\pi}{2})$$

$$y=-3\sin (\frac{2\pi}{9}t+\frac{\pi}{2})$$

$$
x=-3\sin\left(\dfrac{2\pi}{9}t\right), \quad y=-3\cos\left(\dfrac{2\pi}{9}t\right)
$$


How to take every third element in a series? $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = 1.644934$  or  $\frac{\pi^2}{6}$

What if we take every 3rd term and add them up?

A = $ \frac{1}{3^2} + \frac{1}{6^2} + \frac{1}{9^2} + \cdots = ??$

How to take every 3rd-1 term and add them up?

B = $ \frac{1}{2^2} + \frac{1}{5^2} + \frac{1}{8^2} + \cdots = ??$

How to take every 3rd-2 term and add them up?

C = $ \frac{1}{1^2} + \frac{1}{4^2} + \frac{1}{7^2} + \cdots = ??$

I am not sure how to adapt Eulers methods as he used the power series of sin for his arguments: https://en.wikipedia.org/wiki/Basel_problem

Note that we have

$$\psi'(z)=\sum_{n=0}^\infty \frac{1}{(n+z)^2}$$

where $\psi'(z)$ is the derivative of the digamma function.  Hence, we can write

$$\sum_{n=0}^\infty \frac{1}{(3n+1)^2}=\frac19 \psi'(1/3)$$

and

$$\sum_{n=0}^\infty \frac{1}{(3n+2)^2}=\frac19 \psi'(2/3)$$

Interestingly, since we have

$$\sum_{n=0}^\infty \left(\frac1{(3n+3)^2}+\frac1{(3n+2)^2}+\frac1{(3n+1)^2}\right)=\frac{\pi^2}{6}$$

we find that

$$\psi'(1/3)+\psi'(2/3) = 4\pi^2/3$$


Is $x=2,y=13$ the unique solution? Problem:
Find all positive integers $x$ and $y$ satisfying:
$$12x^4-6x^2+1=y^2.$$
If $x=1, 12x^4-6x^2+1=12-6+1=7,$ which is not a perfect square.
If $x=2, 12x^4-6x^2+1=192-24+1=169=13^2$, which is a perfect square. Thus, $x=2,y=13$ is a solution to the given Diophantine equation.
However, after testing a few more small cases, it seems as if $12x^4-6x^2+1$ can never be a perfect square if $x>2$. I have tried to prove this, but to no avail. Here is the gist of what I considered:
$12x^4-6x^2+1=y^2 \iff 12x^4-6x^2 = (y+1)(y-1)$. Since the L.H.S. is a multiple of $2$, it follows that $2 \mid (y+1)(y-1) \Rightarrow y$ is odd $\Rightarrow y+1$ and $y-1$ are both even.
Hence, $4 \mid (y+1)(y-1) \Rightarrow 4 \mid  12x^4-6x^2 \Rightarrow 4 \mid 6x^2 \Rightarrow x $ is even. Let $x=2m$ and $y+1=2k$, so $12x^4-6x^2=192m^4-24m^2=4k(k-1) \Rightarrow 48m^4-6m^2=k(k-1)$. At this point, I am at a loss of how to continue. We could continue with divisibility arguments, but it seems to be a never-ending process?
Another method I tried was to let $y=x+k$, thus $12x^4-6x^2+1 = x^2 + 2xk+k^2 \iff k^2+2xk-(12x^4-7x^2+1)=0$, which is a quadratic in terms of $k$. However, stuff like the discriminant or sum and product of roots did not seem to yield any important information.
Any hints provided to point me in the right direction will be much appreciated.

This is an addition to my previous answer, which by pointed out by Mike doesn't sufficiently solve the problem.
We use the result, $$3(2x-1)^2(2x+1)^2 = (2y-1)(2y+1).$$
We have that $\gcd(2a-1,2a+1)=1$ for positive integer $a$ (by Euclidean Algorithm if you need convincing, however this is trivial enough to state).
It follows that $2y-1$ cannot be divisor to $2x-1$ or $2x+1$ without being divisor to the full square term (in other words, if $2y-1 \vert 2x-1,$ then $2y-1 \vert (2x-1)^2$ and similar). Since it also applies the other way around, it follows that one of the following are true:

$2y-1 = 1$ or $2y-1 = 3$
$2y-1 = (2x-1)^2$ or $2y-1 = 3(2x-1)^2$
$2y-1 = (2x+1)^2$ (cannot be $3(2x+1)^2$ as we must have $2y+1 > 2y-1$)

For the first, we have $y = 1$ or $y=2.$ No $x$ makes this true, so we can simply ignore this case.
The second case (after making $y$ in terms of $x$ and rearranging)
$2y-1 = (2x-1)^2$ gives us the polynomial $4x^4 + 4x^3 - 7x^2 +2x = 0,$ which has roots $-2, 0, 1/2, 1/2.$ None of these are positive integers.
$2y-1 = 3(2x-1)^2$ gives a polynomial with rational roots $1/2,0$ and the rest are irrational.
$2y -1 = (2x+1)^2$ gives us rational roots (-1/2, 0, 2). 2 is thus an answer!
I may have some cases I overlooked but it seems that this can solve it, albeit in a depressingly bashy way.

We try to get a perfect square term on the LHS.
Note that multiplying both sides by 4 keeps the right side as a square.
$$48x^2 - 24x^2 + 4 = (2y)^2$$
Now, completing the square yields $$3(4x^2 -1)^2 +1 = (2y)^2.$$
Try to now find the solutions to $3(2x-1)^2(2x+1)^2 = (2y-1)(2y+1).$


$X$, $Y$, and $Z$ are continuous IID, $\text{P}(X<Y<Z)$? So we have 3 continuous IID random variables $X$, $Y$ and $Z$. I want to find $p = \text{P}(X<Y<Z)$.

The easy way to do it is to use symmetry and thus we have $3!$ equally likely orderings and thus $p = 1/3!$

I was thinking about doing it the hard way and go for the integral of the joint pdf over a specific region. This is where I'm stuck. I'm not sure about the limits of the integral for each one. Any help or guidance is appreciated.

Let $F$ be the common distribution function. Then
\begin{align*}
P(X < Y < Z) &= E\big( E\left(\mathbb{I}_{X < Y} \mathbb{I}_{Y < Z}\mid Y, Z\right)\big)\\
&= E\big(F(Y)\mathbb{I}_{Y < Z} \big)\\
&=E\Big(E\big(F(Y) \mathbb{I}_{Y < Z} \big) \mid Y\Big)\\
&=E\left(F(Y)-F(Y)^2\right)\\
&=\int_{-\infty}^{\infty}F(y)dF(y) - \int_{-\infty}^{\infty}F(y)^2dF(y)\\
&=\frac{1}{6}.
\end{align*}

Assuming they have pdfs,

$$\int_{-\infty}^{\infty}\int_{-\infty}^{z}\int_{-\infty}^{y} f_{X,Y,Z}(x,t_y,t_z)dxdt_ydt_z$$

$$ = \int_{-\infty}^{\infty}\int_{-\infty}^{z}\int_{-\infty}^{y} f_X(x)f_Y(t_y)f_Z(t_z)dxdt_ydt_z$$

$$ = \int_{-\infty}^{\infty}f_Z(t_z)\int_{-\infty}^{z}f_Y(t_y)\int_{-\infty}^{y} f_X(x)dxdt_ydt_z$$

To begin let

$$u=\int_{-\infty}^{z}f_Y(t_y)\int_{-\infty}^{y} f_X(x)dxdt_y$$

$$dv = f_Z(t_z)dt_z$$

Can you take it from here?

Otherwise, what you said.


In a village, $90\%$ of people drink Tea, $80\%$ Coffee, $70\%$ Whiskey, $60\%$ Gin. Nobody drinks all four. What percentage of people drinks alcohol? In a small village $90\%$ of the people drink Tea, $80\%$ Coffee, $70\%$ Whiskey and $60\%$ Gin. Nobody drinks all four beverages. What percentage of people of this village drinks alcohol?

I got this riddle from a relative and first thought it can be solved with the inclusion-, exclusion principle. That the percentage of people who drink alcohol has to be in the range from $70\%$ to $100\%$ is obvious to me

When $T$, $C$, $W$, and $G$ are sets, and I assume a village with $100$ people, then what I am looking for is
$$\lvert W\cup G\rvert = \lvert W\rvert+\lvert G\rvert-\lvert W\cap G\rvert$$
I know that
$$\lvert T \cap C \cap W \cap G \rvert = 0$$
and also the absolute values of the singletons.

But I do not see how this brings me any closer, since I still need to figure out what $\lvert W\cap G\rvert$ is and that looks similar hard at this point

On the way there I also noticed that $\lvert T\cap C\rvert \ge 70$ and similar $\lvert W\cap G\rvert \ge 30$

By now I think there is too little information to solve it precisely.

If you add up the percentages, they come out to $300\%$. This means that the average number of beverages per person is $3$. No one drinks more than that, so no one can drink less than that, either. Since everyone drinks exactly three beverages, everyone has exactly one beverage that they don't drink. So no one doesn't drink both whiskey and gin, i.e. everyone drinks alcohol.

Rephrase the question equivalently:

10% don't drink tea, 20% don't drink coffee, 30% don't drink whiskey, and 40% don't drink gin. This must sum to at least 100% (since nobody doesn't not drink all 4).

Clearly they sum to at most exactly 100%, so the "not-X-drinkers" are disjoint sets, and so everybody drinks at least one of whisky or gin.


Find Matrix $P$ so $B=PA$ I have this problem.

$$A =  \left(\begin{array}{ccc}
1 & 2 & 3 \\
2 & 4 & 6 \\
1 & 2 & 3 \end{array}\right)$$

$$B =  \left(\begin{array}{ccc}
1 & 2 & 3 \\
0 & 0 & 0 \\
0 & 0 & 0 \end{array}\right)$$

Find matrix $P$ invertible so $B=PA$.

It's pretty clear in this case.

But I wonder is there a way to find such a matrix in general case, when it's not obvious.

Any help will be appreciated.

By row reducing, you can find matrices $P_1,P_2$ such that $P_1A$ and $P_2B$ are in the same row-echelon form.  From there, we can say that
$$
P_1A = P_2B \implies B = (P_2^{-1}P_1)A
$$
This will work whenever such a $P$ exists.

For your example: note that $B$ is in row-echelon form.  So, in this case, we have $P_2 = I$.  For $A$, we have
$$
\pmatrix{
1&2&3&&1&0&0\\
2&4&6&&0&1&0\\
1&2&3&&0&0&1}
\mathop{\leadsto}^{R_3\to R_3-R_1}\cdots
\mathop{\leadsto}^{R_2\to R_2-2R_1}\\
\pmatrix{
1&2&3&&1&0&0\\
0&0&0&&-2&1&0\\
0&0&0&&-1&0&1}
$$
So,
$$
P_1 = \pmatrix{
1&0&0\\
-2&1&0\\
-1&0&1
}
$$

when you row reducing a matrix $A$ by hand, it is much easier to keep track of the row operations, if we augment the matrix $A$ with a column like this:
$ \left( \begin{array}{lll|l} 1 & 2 & 3 & a\cr 2 & 4 & 6 & b\cr 1 & 2 & 3 & c\end{array} \right) \to
\left( \begin{array}{lll|l} 1 & 2 & 3 & a\cr 0 & 0 & 0 & -2a + b\cr
0 & 0 & 0 & -a+c\end{array} \right)
$
you decode this as multiplying by the matrix $E$ on the left where
$ E \pmatrix{a \cr b\cr c}=\pmatrix{1 & 0 & 0\cr-2 & 1 & 0\cr -1 & 0 & 1}\pmatrix{a\cr b\cr c\cr d} =  \pmatrix{a &\cr-2a+b\cr -a+c}.$

that is $EA = U.$ of course, if you are calculator or computer matrix software, you augment by the identity matrix as suggested by others.


Did I take the derivative correctly? $x^y=y^x$ Need to differentiate following equation:
$$x^y=y^x$$
My attempt:
$$x^y\log(x) \cdot y' = y^x\log(y)\cdot1;  $$
$$y'=\frac{y^x\log(y)}{x^y\log(x)}$$
Please tell me if I've made a mistake.

Not quite. The idea is to take the $\log$ of both sides and implicitly differentiate. $$ \begin {eqnarray*} x^y &=& y^x \\ y \cdot \log (x) &=& x \cdot \log (y) \\ y' \cdot \log(x) + \dfrac {y}{x} &=& \dfrac {x}{y} \cdot y' + \log (y) \\ y' \cdot \left( \log(x) - \dfrac {x}{y} \right) &=& \log(y) - \dfrac {y}{x} \\ y' &=& \dfrac {\log (y) - \dfrac {y}{x}}{\log (x) - \dfrac {x}{y}}, \end {eqnarray*} $$so our answer is $$ y' = \boxed {\dfrac {x \cdot \log (y) - y}{y \cdot \log (x) - x} \cdot \dfrac {y}{x}}. $$


To what extent were mathematicians in previous centuries aware of the lack of rigour in their methods? By modern standards, much of pre-modern mathematics isn't rigorous. Famous examples include Euler's solution to the Basel problem or literally anything involving sets before Cantor and Russel came along, when a "set" was simply a handwaving notion of "all things that have some property", later found to be quite problematic.

To what extent were the mathematicians of the past aware of these shortcomings? Did they even feel mathematics needed to be a rigorous subject, or was there an idea that anything goes as long as it works?

I feel this question is not just of historical interest. Modern mathematics is usually judged (by modern mathematics) to be rigorous, but we have no reason to believe we are able to assess our methods correctly unless our predecessors were capable of assessing theirs.

Lagrange launched a contest at the end of the 18th century whose goal was to clarify the notions of infinity and infinitesimal. There was no clear winner, but Lazare Carnot submitted an entry which eventually became a popular book. This is only one episodes illustrating the fact that rigor has indeed been a concern historically.

A century earlier, there was a debate at the French academy focusing on similar issues, opposing Rolle and Varignon.

In the 19th century, Cauchy lists rigor as one of the objectives of his approach to analysis.

So to answer your question, mathematicians in previous centuries were aware of lack of rigor, but in most cases attributed it to their predecessors only. This seems to be the case today, as well.

It is a little known fact that much of Bernhardt Riemann's work had remained un-rigorous until Hilbert's arrival on the mathematical scene some fifty years after Riemann's death.  Riemann was aware of his shortcomings, but many before him were not.  Did you know that complex numbers were not properly accepted until the nineteenth century?  The historian D.T. Whiteside considered Newton's first proposition in the Principia to be problemmatic.  According to Michael Atiyah, it took hundreds of years after Newton's death for calculus to be made properly rigorous.  It is worth pointing out that in Newton's day, Euclidean geometry was considered to be more rigorous than the new-fangled algebraic approach introduced by Descartes (according to historian of mathematics S. Hollingdale).  Also, Euler was ignorant of the fact that he had made some completely incorrect deductions concerning alternating series.  Of course, we know from the correspondence of G.H. Hardy that the great Ramanujan did not possess properly rigorous proofs for several of his 4000+ theorems.  In the nineteenth century there was at least one mathematician who had erroneously believed during his lifetime he had correctly proved a special case of Fermat's last theorem.  According to Felix Klein, even the great Gauss did not appear to be to concerned about being rigorous when he was involved with more practically inclined investigations.  But Gauss was well aware of this.


Is this valid when deriving quadratic equation? When deriving the quadratic formula, isn't the square root of $(x+\frac{b}{2a})^2$ the absolute value of $(x+\frac{b}{2a})$? It's usually just represented as $(x+\frac{b}{2a})$ without absolute value and then $\frac{b}{2a}$ is subtracted from left side and boom, theres the quadratic formula. I just don't understand why its not absolute value of $(x+\frac{b}{2a})$. For example the square root of $x^2$ is the absolute value of $x$, which is equal to $\pm{x}$. Sorry if this is confusing its more of a conceptual thing. Thank you in advance.

philalethesnew

You seem to be asking why after the step $\;\left(x+\cfrac b{2a}\right)^2=\cfrac{b^2-4ac}{4a^2}\;$ , they don't go to

$$\left|x+\cfrac b{2a}\right|=\sqrt{\frac{b^2-4ac}{4a^2}}$$

But we actually do! To write the above is exactly the same as to write

$$x+\cfrac b{2a}=\pm\sqrt\frac{b^2-4ac}{4a^2}$$

just as $\;x^2=a\implies |x|=a\;\; (\,a\ge 0)\;$ is exactly the same as $\;x^2=a\implies x=\pm a\;\;(a\ge0)\;$ , under the assumption, of course, that once we take the + sign and the second time we take the - sign.

When they get to point of having
$$\bigl{(}x+\frac{b}{2a}\bigr{)}^2 =-\frac{c}{a}+\bigl{(}\frac{b}{2a}\bigr{)}$$ the square root of both sides has no absolute value sign on the left because the right side can be $\pm$.
$$\biggl{|}x+\frac{b}{2a}\biggr{|}\ne\sqrt{-\frac{c}{a}+\biggl{(}\frac{b}{2a}\biggr{)}}$$

One complete derivation explained simply is shown here.


Combinatorics problem: form a team of three people with at least one medic, having 5 nurses and 4 medics. At an ambulance stations there are 5 nurses and 4 medics. Find in how many ways can we create a team with three people, where at least one of them is a medic.

I solved the problem this way: there are $4\cdot C_5^2$ ways to form a team with one medic and two nurses and $C_4^2 \cdot 5$ ways to form a team with two medics and one nurse and $C_4^3$ ways to form a team with three medics.

So the answer will be $4\cdot C_5^2 + C_4^2 \cdot 5 + C_4^3 = 74$.

I am not sure if I am right with this answer, so this is the reason I am asking this question.

The quicker way to look at it is that there are $\binom{9}{3}$ ways of selecting a team of 3 people, of which $\binom{5}{3}$ don't have a medic. So the answer is $\binom{9}{3} - \binom{5}{3} = 84 - 10 = 74$, which agrees with your calculation.

Best way

(Count of those with at least 1 medic) = (Count of all combinations) - (Count those
                                         with no medics)

$${}^9\mathrm C_3-{}^5\mathrm C_3 = 74$$


How to convert a decimal to a fraction easily? Given a decimal like $1.5$, I want to find an easy way to write it as $3/2$ in this case is easy but given a decimal such as $0.3760683761$, how could I figure out it can be written as $572/1521$? Thx.

Edit

In case anyone is interested in more about this, I found the following links (more could exist, but what I have is enough - Thx to your contributions):

Link 1, Link 2

Here's a different way to proceed, using as little as possible of the theory of continued fractions.  First write $$\frac01 < 0.3760683761 < \frac11$$ so right away we have a pair of fractions that approximate 0.3760683761.  They are  not very good approximations though!

Now if we have $\frac ab$ on the left and $\frac cd$ on the right, we should check $\frac{a+c}{b+d}$ to see where it lies.  In this case $\frac ab = \frac01$ and $\frac cd = \frac 11$ so $\frac{a+c}{b+d} = \frac{0+1}{1+1} = \frac12 = 0.5$:

$$\frac01 < 0.3760683761 < \frac12< \frac 11$$

so forget about $\frac11$, which is too big.  Now repeat:  Next is $\frac{0+1}{1+2} = \frac13$, which is too small:

$$\frac01< \frac13 < 0.3760683761 < \frac12$$

Next is $\frac{1+1}{2+3} = \frac25$, which is too big:

$$\frac13 < 0.3760683761 < \frac25 < \frac12$$

Next is $\frac{1+2}{3+5} = \frac38 = 0.375$, which is an excellent approximation:

$$\frac13 < \frac38 < 0.3760683761 < \frac25 $$

Unfortunately for this example, there is no better approximation than $\frac38$ until we get to $\frac{44}{117}$.  usually the convergence is quicker, but this happens to be an unusually badly-behaved example, and so takes several more steps:

$$\frac13 < \frac38 < \frac{44}{117} <  0.3760683761< \frac{41}{109} < \cdots < \frac{17}{45} < \frac{14}{37} < \frac{11}{29} < \frac8{21} < \frac 5{13} < \frac25 < \frac12$$

But it does find $\frac{44}{117} \approx .37606837606837606837$, which is extremely close, and perhaps is what you were looking for.  On the other hand perhaps $\frac 38$ was close enough and you could stop there.

If you want $0.3760683761$ exactly, you can continue the process, which will eventually terminate with exactly $\frac{3760683761}{10000000000}$.

One pleasant feature of this algorithm is that it always produces fractions in lowest terms, so you never have to worry about reducing them afterwards.

The theory of continued fractions provides a way to perform this calculation much more quickly, producing the continued fractional result $$0.3760683761 \approx \cfrac1{2+\cfrac 1{1+\cfrac1{1+\cfrac1{1+\cfrac1{14}}}}}$$

The theory of primes is that any number can be written as a product of prime factors, so 572/1521 can be written as 2*2*11*13 / 3*3*13*13 which reduces down to 44/117.

The theory of decimals is that any decimal number can be written as a product and ratio using three whole numbers, e.g. xxx.xxxx = A * B/C.


recurrence relation for proportional division Consider the following recurrence relation, for a function $D(x,n)$, where x is a positive real number and n is a positive integer:

$$ D(x,1) = x $$

$$ D(x,n) = \min_{k=1..n-1}{D(xk/n,k)} \ \ \ \  [n>1] $$

This formula can be interpreted as describing a process of dividing a value of x to n people: if there is a single person ($n=1$), then he gets all of the value x, and if there are more people, they divide x in a proportional way.

By induction on n, it is possible to prove that:

$$ D(x,n) = x/n $$

PROOF: For $n=1$, this is given. Assume it is true for $1, 2, ... n-1$. Then:

$$ D(x,n) = \min_{k=1..n-1}{D(xk/n,k)} = \min_{k=1..n-1}{x/n} = x/n $$

Now consider a slight modification of the formula:

$$ E(x,1) = x $$

$$ E(x,2) = x/2 $$

$$ E(x,n) = \min_{k=2..n-1}{E(x(k-1)/n,k)} \ \ \ \ [n>2] $$

The modified version models division with loss - in each division process, we lose some value: we give to $k$ people, only $(k-1)/n$ of the original value.  What is the solution to this formula?

By induction on n, it is possible to prove that, for $n \geq 2$:

$$ E(x,n) = x/n(n-1) $$

PROOF: For $n=2$, this is given. Assume it is true for $2, ... n-1$. Then:

$$ E(x,n) = \min_{k=2..n-1}{E(x(k-1)/n,k)} = \min_{k=2..n-1}{x/nk} = x/n(n-1) $$

This is interesting - even a small loss in the division process, leads to a large reduction of the value each person gets, from $O(1/n)$ to $O(1/n^2)$.


Isotropic vectors of quadratic form How can i find all isotropic vectors of quadratic form
$x_{1}^2+4x_{2}^2+8x_{3}^2-4x_{1}x_{2}+8x_{1}x_{3}-14x_{2}x_{3}$

I don't understand how to approach.

Use Lagrange's Reduction (complete the square). I'll use $\;x,y,z\;$ for ease:

$$x^2+4y^2+8z^2-4xy+8xz-14yz=(x^2-2x(2y+4z))+4y^2+8z^2-14yz=$$

$$=\left(x-2y-4z\right)^2-\color{red}{4y^2}-16yz-16z^2+\color{red}{4y^2}+8z^2-14yz=$$

$$=\left(x-2y-4z\right)^2-8\left(z^2+\frac{15}2yz\right)=(x-2y-4z)^2-8\left(z+\frac{15}4y\right)^2+\frac{225}2y^2$$

Thus, your quadratic is equivalent to $\;\tilde x^2+\tilde y^2-\tilde z^2\;$ (signature $\,(2,1)\,$) , so it actually is a cone.

$$$$


Let $G$ be a group with $|G|=102 ( = 2 \cdot 3 \cdot 17)$, with $|Z(G)| = 2$. Prove the following: Let $G$ be a group with $|G|=102 ( = 2 \cdot 3 \cdot 17)$, with $|Z(G)| = 2$. Prove the following:

$|Z(G/Z(G))| = 1$
$G/Z(G)$ has a subgroup of order 17
$G$ has a subgroup of order 34

I am preparing for my Algebra exam and I have been stuck on this question for a bit.
1.
So the only thing I know is that $|Z(G/Z(G))| = 1$, which means that $Z(G) \in G/Z(G)$ is the only element for which $A \cdot Z(G) = Z(G) \cdot A$ for every $A \in G/Z(G)$. But I don't know how to prove that there are no other subgroups for which that holds. At first I thought it might be a general property, but apparently it isn't (according to this example: $Z(G/ Z(G)) = 1$ Prove or disprove.).
So I know that it has to do something with the order of the groups, but I don't know how to use the information I have to get the conclusion. So far what I know is that with Lagrange I get $|G/Z(G)| = 102/2 = 51 = 3 \cdot 17$. I see that both are prime but I don't know how this is supposed to help me any further.
2.
I thought of maybe using the fact that $|Cl(A)|$, for every $A \in G/Z(G)$, has to be a divisor of $|G/Z(G)|$, but then I have that $|Cl(A)|$ for every $A$ has to be either 3 or 17, but how to prove that there exists one of 17 I don't know. Or maybe I can use the Class equation but I'm not sure.
I unfortunately don't know how to get any further with this question.
3.
I suppose that this will follow from (1) and (2), since if $G/Z(G)$ has a group of order 17, since $|Z(G)| = 2$, it follows with Lagrange that $G$ has a subgroup of order 34.

Can someone please help me by giving hints that will help me solve this problem? This is a rather introductory course, I've seen other people solve this with 'Sylow theory' but we haven't seen that yet in this course.

A proof for 1) without Sylow: Note that for any Group $H$ if $H/Z(H)$ is cyclic then $H$ is abelian.
So if $|Z(G/Z(G))|>1$ we get $[G/Z(G):Z(G/Z(G))]=1,3$ or $17$. By the above remark $G/Z(G)$ will be abelian. Since $|G/Z(G)|=3\cdot 17$ it follows that $G/Z(G)$ is cyclic and hence $G$ is abelian, a contradiction.
Since $|Z(G)|=2$ claim 3) will follow from 2).
For 2) notice that by Cauchy's theorem there is an element and therefore also a subgroup of order $17$ in $G/Z(G)$.
Edit: A self contained proof that $G/Z(G)=:H$ contains a subgroup of order $17$ (similar to the proof of Cauchy's theorem): Let $x_1,\dots,x_r$ be a system of representatives for the non-trivial conjugacy classes of $H$ (i.e. those with more than one element). Then we have $$H=\underbrace{Z(H)}_{\{e\}}\cup\bigcup_{i=1}^r[x_i]$$ where $[x_i]$ denotes the conjugacy class of $x_i$ and the union is disjoint. This gives us the class equation: $$3\cdot 17=1+\sum_{i=1}^r |[x_i]|$$ Since $|[x_i]|$ divides $|H|$ and $|[x_i]|>1$ there has to be some $i$ with $|[x_i]|=3$. It follows that the centralizer $C_H(x_i)$ has order $17$ which is our subgroup of order $17$.

Hints:

We have that $\;\left|G/Z(G)\right|=\frac{102}2=51\;$. Now, how many groups of order $\;51\;$ are there? This question is pretty symple if you know semidirect product, but if you don't it is also pretty simple knowing only Sylow Theorems

2)+ 3) are immediate consequences of (1) and Sylow Thms. and the given info (what happens when you take the product of two subgroups $\;HK\;$ ,with $\;K\lhd G\;$ ?)


number of element in a principal ideal domain can be $25/36/35/15$? Could any one tell me number of element in a principal ideal domain can be $25/36/35/15$ ?

I just know a principal ideal domain is generated by a single element. what the knowledge I need to find this result?

Thank you

First, you need to know that any finite integral domain is a field. Then you should know that there is a finite field of cardinality $n$ if and only if $n$ is a prime power.

Note that what you said is probably not what you meant:

  a principal ideal domain is generated by a single element.

In any ring $R$, we have that $R$ is equal to the ideal generated by $1_R$, so this is true,  but the statement you probably meant is

  every ideal of a principal ideal domain is generated by a single element.

25 elements can have a principal ideal  domain because for if n is p^2 then there exist a field of order P^2 moreover since a field is pid then every ideal is principal.


Irreducibility in $\mathbb{Z}[\sqrt{-2}][X]$ I'm working on the following question:

(a) Prove that the ring $R = \mathbb{Z}[\sqrt{-2}]$ is Euclidean.
(b) Show that $R/(3+2\sqrt{-2})$ is a field. What is the characteristic of this field?
(c) Show that the polynomial $X^4+3$ is irreducible over the field $\mathbb{F}_{17}$ of $17$ elements; and deduce that $f(X) = X^4 − 170X^3 + 9 + 4\sqrt{−2} \in R[X]$ is irreducible.
(d) Is the polynomial $Y^4 −f(X)\in R[X,Y]$ irreducible? (Why?)

I have done part (a). For (b) I said that by (a), $R$ is a PID, hence prime and irreducible are equivalent. Thus $(3+2\sqrt{-2})$ is maximal if and only if $3+2\sqrt{-2}$ is prime. Now using the norm function defined by
$$\varphi\colon R\setminus\{0\}\to\mathbb{N}_{>0}$$ where $\varphi(a+b\sqrt{-2}) = a^2+2b^2,$ we have that $\varphi(3+2\sqrt{-2}) = 9 + 8 = 17,$ which is prime in $\mathbb{Z}$, hence $3+2\sqrt{-2}$ is prime in $R$. Therefore $R/(3+2\sqrt{-2})$ is a field. I'm not sure how to determine the characteristic. I had hoped to get some relation using $$3+2\sqrt{-2}=0,$$ but I wasn't getting anywhere.
For (c), I don't know the best way to do this. One way would be to check that for $n\in\{0,1,\dots,16\}$ we have that $n^4+3\not\equiv 0\bmod{17}$ to show that there is no linear factor. Then we could test that
$$X^4+3 = (X^2+aX+b)(X^2+cX+d)$$ has no solutions for $a,b,c,d\in\mathbb{F}_{17}$. Is there a better approach?
For (d), I'm thinking $Y^4-f(X)$ is irreducible by Eisenstein?

for part (b) to find the characteristic of the field $J=R/(3+2\sqrt{-2})$ find the order of $1$ in the additive subgroup of $J$ .


Geodesics that paint curves on the projective space bundle Let $M$ be a geodesically complete connected riemannian manifold.

Let $p \in M$ be a point and $c: \mathbb{R} \to M$ an arbitrary geodesic that doesn't intersect p. Our aim is to find a "nice" map sending points on the curve $c$ to tangent lines at $p$.

Pick a point $q$ on the curve $c$ and connect it with $p$ via a geodesic curve (this is possible by completeness of $M$). Here is the relavent picture:

Every such geodesic corresponds to a unique tangent line through $p$. Ideally we want to have a cannonical choice of such geodesic for every point of $c$. Assuming there is such a choice we could obtain a function $f: c(\mathbb{R}) \to \mathbb{P}T_pM$. where $\mathbb{P}T_pM$ is the tangent projective space or equivalently the fibre at $p$ of the $(1,dim(M))-$grassman bundle over $M$.

Now my question has two parts:

1) Given a initial point $q$ and an initial choice of geodesic connecting $p$ and $q$, is there then a cannonical choice for every other point on $c$? What i have in mind is a choice that "minimizes" in some suitable sense the variation of the geodesics with respect to variation of the initial point $q$. (I'm not so familiar with jacobi fields but i have a strong sense that this is what's missing here).

2) Under what conditions on $M$ is the map $f$ (defined above) smooth (resp. continuous) as a curve in $\mathbb{RP}^{n-1}$ where $n=dim(M)$. By which i mean the composition $f \circ c : \mathbb{R} \to \mathbb{P}T_pM \cong \mathbb{R P}^{n-1}$ is smooth (resp. continuous).

Here's the picture I'd like to have in mind in this context. It might be wrong though, so you better take it with a pint of salt.

Note: The picture is far from being literal. $M$ is drawn as a surface so the fiber should be a projective line while i drew a projective plane.

I do not know what you mean by "nice", but here is a construction. Let  $q=c(0)$. Then $c$ lifts to a parameterized straight line $\tilde {c}$ in the tangent space $T_qM$, such that $c(t)=\exp_q(\tilde{c}(t))$. The point $p$ lifts (noncanonically, unless $\exp_q$ is a diffeomorphism) to a point $\tilde{p}\in T_q(M)$. Now, connect $\tilde{p}$ to $\tilde{c}(t)$ by straight line segments in $T_qM$. These segments project to the required curves $a(s,t)$ in $M$ connecting $p$ to $c(t)$. The velocity vectors
$$
\frac{\partial }{\partial s} a(s,t)|_{s=0}
$$
define the lines in $T_pM$ that you are asking for and, hence the map to $PT_pM$.

Note that the curves $a(s,t)$ above are not geodesics (in general). In order to use geodesics you have to assume that for the geodesics $\gamma(s,t)$ connecting $p$ to $c(t)$, the points $p, c(t)$ are not conjugate (otherwise you cannot make a smooth choice of $\gamma(s,t)$). For instance, if you assume that $M$ has nonpositive curvature, you can use geodesics $\gamma(s,t)$ for the construction.  The clean way to do so is to lift $c$ to the tangent space $T_pM$ via the exponential map $\exp_p$ (since it is a covering map by Cartan-Hadamard theorem) and then use straight line segments to connect via the same procedure I described above.

Not sure if i understood all of your requirements well.To the extent it is clear to me, a geodesic flow field of the type you suggest is certainly possible in reply to first part of your question. The following from Wolfram Alpha demonstration needs only a minor modification to be suitable to your description.

From the link suggested you can pick one out of three geodesic regimes for returning geodesics behavior at hyperbolic points (saddle points, Gauss curvature negative) with  low $ \alpha $ and high initial height.

A minimum of two geodesics possible, infinitely many also by rotation.

In this option geodesics go to a minimum radius as per Clairaut's Law on surfaces of revolution and cannot proceed beyond the tangential  circle of no return by virtue of invariant minimum radius in-built.They are tangential  to a circle of radius $ r \sin \alpha$ after boundary ring is touched before returning.

The pattern can be made ( by design) to return, or go to infinity on the other side of the surface which can be open/closed. CNC filament winding is performed on vessels of light weight design.

FilamentShell

The other two options ( non-returning and asymptotic minimum radius ) are not  suitable.

The start point can be converted to line ( as a line to line map ) or its points can be picked/isolated.

I can prepare to upload another image to required dimensions if requirement of second part of your question  is more clear with suitable examples.

HyperboloidGeodesics


How to calculate 3D arc between two lines I have three known points P1,P2,P3 in X,Y,Z coordinates and known radius. How do I find the center point(Pc), start point(Pa) and end point(Pb) of the arc between two lines in 3D?

I could solve it for 2D, but it is confusing to calculate for 3D case.

In fact, it's a 2D problem, you just have to choose the right coordinate system. Let $$\vec{P_2P_1}=P_1-P_2$$ the vector pointing from $P_2$ to $P_1,$ and $$n_a=\frac{\vec{P_2P_1}}{|\vec{P_2P_1}|}$$ the normalized (length $1$) vector, $$\vec{P_2P_3}=P_3-P_2$$ the vector pointing from $P_2$ to $P_3,$ and $$n_b=\frac{\vec{P_2P_3}}{|\vec{P_2P_3}|}$$ the normalized vector. The angle $\alpha$ between these two directions is defined by the scalar product: $$\cos\alpha=n_a\cdot n_b.$$ Clearly, $P_c$ lies on the bisector of this angle. The distance from $P_2$ to $P_a$ and $P_b$ must be the same, $r/\tan\frac{\alpha}{2}$, so
$$P_a=P_2+\frac{r}{\tan\frac{\alpha}{2}}\,n_a$$ and $$P_b=P_2+\frac{r}{\tan\frac{\alpha}{2}}\,n_b,$$ and you get $P_c$ from
$$P_c=P_2+\frac{r}{\sin\alpha}\,(n_a+n_b).$$
While these formulas show the simple geometry more clearly, @Nominal Animal pointed out that it might be better to avoid calculating trigonometric functions. Indeed, we can use the cross product to express $\sin\alpha$:
$$\sin\alpha=|n_a\times n_b|.$$ For calculating $P_a, P_b$, we can use
$$\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1+\cos\alpha}=\frac{|n_a\times n_b|}{1+n_a\cdot n_b}.$$ So everything can be determined by vector operations, alone.

The equation of a sphere is $(x-x_0) ^2 + (y - y_0)^2 + (z-z_0)^2 = r^2$ where the center of the sphere is $(x_0, y_0, z_0)$. This should give you three equations (one for each point) with three unknowns. You can then solve for the unknowns. Can you take it from here?

Bob


Prove that the below set is closed in X! Let $(X,μ)$ be a topological space,
$g, f:X\longrightarrow\Bbb R$ continuous functions.
Prove that $\{ x\in X: g(x)=f(x)\}$ is closed in $X$.
I tried to prove that with the definitions of a continuous and a closed function, and with theories about a continuous function (between two spaces), but i got nothing.
I said that since f and g are both continuous so for every closed set ${K}$ in ${R}$ , $f^{-1}{K}$ is closed in X , same about g.
I also looked at $g\circ f^{-1}:\Bbb R\longrightarrow R$.
$g\circ f^{-1}(K) = g(f^{-1}(K)$ is closed in $X$.
Then $g\circ f^{-1}(K)=\operatorname{id}(K)=K$ is closed in ${X}$, but that did not help me to prove that the given set is closed.

Let your set of interest be $I = \{x \in X \mid f(x) = g(x)\}$ ($I$ for "intersection").  We claim that $I$ is closed by showing its complement is open.

Take a point $x \notin I$ so that $x \in X$ and $f(x) \neq g(x)$.  Then we can find open sets $U$ and $V$ in $\mathbb{R}$ with $U$ surrounding $f(x)$, $V$ surrounding $g(x)$, and $U \cap V = \varnothing$.  Now pull back $U$ and $V$ under $f$ and $g$ respectively to get the open set $W=f^{-1}(U) \cap g^{-1}(V)$.  Clearly $x \in W$ and no point of $W$ can belong to your set $I$.
For instance, if $x \in I$ and $x \in W$ then $f(x) = g(x) \in U \cap V$, which is impossible as $U$ and $V$ are disjoint.  This $W$ gives a neighborhood around $x$ that stays within $X-I$, so $X-I$ must be open.  Hence $I$ is closed, as desired.

Note: this argument generalizes to functions $f, g: X \to Y$, where $Y$ can be any Hausdorff space.

Put $h=f-g$.
Your set is
$$h^{-1}(\{0\})$$

which is colsed  as reciprocal image of the closed $\{0\}$ by a continuous function.


Remainder of $35! $ when divided by $41$ I use $p=41 $ is a prime, so

$\begin{align}
(41-1)! &\equiv -1 \mod 41\\
(40)! &\equiv -1 \mod 41\\
40\times39\times38\times37\times36\times(35)! &\equiv -1 \mod 41\\
(-1)\times(-2)\times(-3)\times(-4)\times(-5)\times(35)! &\equiv -1 \mod 41\\
120(35)! &\equiv +1 \mod 41\\
(-3)(35)! &\equiv +1 \mod 41\\
(3)(35)! &\equiv  -1 \mod 41 \\
(3)(35)! &\equiv  -42 \mod 41 \ \to \div 3
\\(35)! &\equiv  -14\equiv 27 \mod 41 \\
\end{align}$

IS my work true ?
   can anyone says another idea ?

with a simple division, $$35!\equiv 27(mod41)$$


Solving the limit: $\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]$ Find the value of the limit $$\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]$$ where $[\cdot]$ denotes the greatest integer function or the box function.

My attempt: I am aware of the standard limits $$\lim_{x\to0}\left(\frac{\sin^{-1}(x)}{x} \right) = 1 $$ and $$\lim_{x\to0}\left(\frac{\tan^{-1}(x)}{x} \right) = 1 $$ but am not sure how will I apply the box function on this limit.

Any detailed explanation to help me understand this concept will be appreciated.

From

$$\arcsin|x|>|x|$$ and $$\arctan|x|<|x|$$ for small $|x|$, you draw

$$100\frac{\arcsin|x|}{|x|}>100$$ and

$$100\frac{\arctan|x|}{|x|}<100.$$

Then as the limits of these expressions are both $100$ you can find a neighborhood of $0$ where

$$\left\lfloor100\frac{\arcsin|x|}{|x|}\right\rfloor=100$$ and

$$\left\lfloor100\frac{\arctan|x|}{|x|}\right\rfloor=99.$$


Puzzle - reordering people in a line Eighty players numbered I through $80$ are standing in a row, one behind the other, in the
increasing order of their numbers. The physical director of the players performed eight successive
inspections of the players and in each of the respective inspections he sent the first $10$, $20$, $30$, $40$,
$50$, $60$, $70$ and $80$ players, from the front of the row to the end of the row. Each time the players
being sent back one after another. After these eight rounds of inspections what is the position of
the player numbered $5$?

Every time I solve it I gets confused, but still is there any general solution to this problem

If I understand this correctly...

Sending one player to the back moves player $i$ to position $i - 1$ (wrapping around by adding $80$ as needed).  The director does it $10 + ... + 80 = 360$ times, so this sends $5$ to position $5 - 360$, which wraps around to $5 - 360 + 5 * 80 = 5 + 40 = 45$.

from logical point of view as i am seeing it and if i get your question right , it will be like

Original Position : 10 20 30 40 50 60 70 80

first iteration : 80 70 60 50 40 30 20 10

2nd iteration : 10 20 30 40 50 60 70 80

3rd iteration : 80 70 60 50 40 30 20 10

4th iteration : 10 20 30 40 50 60 70 80

5th iteration : 80 70 60 50 40 30 20 10

6th iteration : 10 20 30 40 50 60 70 80

7th iteration : 80 70 60 50 40 30 20 10

8th iteration : 10 20 30 40 50 60 70 80

so the last one is the same , the player position will be same.comment if this is correct


if $\int_\Bbb{R}\vert v\vert<\infty \rightarrow \int_\Bbb{R}\vert uv\vert<\infty$ then $u$ is bounded. Let $u\in\mathcal{C}^0(\Bbb{R};\Bbb{R}).$
Assume that for all $v\in\mathcal{C}^0(\Bbb{R};\Bbb{R})$ such that $\int_\Bbb{R}\vert v\vert$ converges then $\int_\Bbb{R}\vert uv\vert$ converges as well.

How can I prove that $u$ is bounded ?
I tried by contradiction, if I can construct a function such that $\int_\Bbb{R}\vert uv\vert$ diverges but such that $\int_\Bbb{R}\vert v\vert$ converges.
Any idea how can I do that ?

Suppose $u$ is unbounded. Take $\{x_n:n\geq 0\}\subset \Bbb R$ where $x_0=0$ (or anything in $\Bbb R$) and  $|u(x_n)|>\max (1+|u(x_{n-1})|, 1+4^n)$ for  $n\in \Bbb N.$

For each $n\in \Bbb N$ take $r_n>0$ such that

(i). The interval $J_n=[-r_n+x_n,r_n+x_n]$ satisfies $\forall x\in J_n \;(\; |u(x)-u(x_n)|<1).$

(ii).  $\;J_m\cap J_n=\emptyset$ when $ 0<m< n.$

Condition (ii) is possible. Otherwise for some  $m\in \Bbb N$  the point $x_m$ would be a limit point of the set $\{x_{n'}:n'>n\}$ for every $n>m,$ which would make $u$ unbounded on every nbhd of $x_m,$ contradicting the continuity of $u.$

Let $S=\Bbb R\setminus \cup_{n\in \Bbb N}J_n.$ There exists a continuous $v$ with $\int_S|v|<1$ such that for $x\in J_n$  we have $$v(x)=\frac {2^{-n}\cdot sgn (u(x_n))}{2r_n}$$       (Note: $sgn(y)=y/|y|$ for $y\ne 0.$)

Then $\int_{\Bbb R}|v|<2$ and $\int_{\Bbb R}|uv|\geq \sum_{n\in \Bbb N} |J_n| \cdot |4^nv(x_n)|=\sum_{n\in \Bbb N}2^n=\infty.$

Take the contrapositive: if u is unbounded, then |uv| diverges. For simplicity, I'll take u and v to be positive.

Since u is unbounded:
for all M there exists x$_M$ such that u(x$_M$)>M.

Since u is continuous:
for all $\epsilon$ there exists $\delta _{\epsilon}$ such that if |x-x$_M$|<$\delta _{\epsilon}$ then |u(x)-u(x$_M$)|<$\epsilon$.

So now take $\epsilon$ = M/2, and it follows that if |u(x)-u(x$_M$)|<$\epsilon$ then u(x)>M/2.

So choose x$_n$ such that u(x$_n$) > 2$^n$ and x$_{n} > x_{n-1}+10$
Note that the existence of x$_{n}$ such that  x$_{n} > x_{n-1}+10$ follows from continuity (continuous functions are bounded on finite intervals, so if u is unbounded, then u must be unbounded outside of the finite interval [0,$ x_{n-1}+10$]

Choose $\delta _n$ such that

if |x$_n$-x| <  $\delta _n$ then u > (2$^n$)/2
and
$\delta _n$ < 1

For each x$_n$, we have an interval of length 2$\delta _n$ for which u > 2$^{n-1}$.  Let v be equal to (2$^{-n}$)/$\delta _n $ on that interval. Then the integral of uv on that interval is at least 2$\delta _n $2$^{n-1}$ * 2$^{-n}$/$\delta _n $ = 1.

If we set v to be zero outside of these intervals, the total integral of uv is  the sum over all these intervals, and that sum clearly diverges. (Note that strictly speaking, this violates the continuity of v, but for each n, we can take v as going to zero in an arbitrarily small distance.)

The integral of v is its value over these intervals. For each interval, v is 2$^{-n}$ for an interval of at most 1, so the integral for each interval is at most 2$^{-n}$, which converges.


Does $\pi(\partial M)\subset \partial(M/\sim)$? Let $M$ a set and $\sim$ an equivalence relation. Let $\pi: M\longrightarrow M/_\sim$ the projection. Do we have that $\pi(\partial M)\supset \partial (M/_\sim)$ ? (where $\partial A$ denote the boundary). I think yes but I have problem to prove it.

My question behind is : if we know the boundary of $M$ can we easily find the boundary of $M/_\sim$ ?

I made a slight typo in one of my comments. Let me summarize the discussion here and try to answer your question. Let's compare $\pi(\partial M)$ and $\partial(M/\sim)$.

If $M=[0,1]$ and we identify the endpoints, the quotient is $S^1$ and so the image of the boundary is not always contained in boundary of the quotient.

If $M=[0,1]$ again and we identify $x \sim 1-x$, call the quotient $N$, which is again a copy of $[0,1]$. But $1/2 \in M$ is in the preimage of the boundary of $N$, and clearly not a boundary point of $M$. So the image of the boundary does not always contain the boundary of the quotient.

Now, you gave an example of the Mobius band. Let $M=[0,1] \times [0,1]$. Set $(0,y) \sim (1,1-y)$ and let $N$ be the quotient. Then the boundary of $M$ is $\partial M = \{ (0,s) \}\cup\{ (1,s) \}\cup\{ (t,0) \}\cup\{ (t,1) \}$ and the boundary of $N$ is $\partial N = \{ [t,0]\} \cup \{[t,1]\}$, and you notice that $\pi(\partial M)$ contains $\partial N$. Why is that so? Well all of the points you identified lay in the boundary of $M$. It's almost tautological: either a boundary point of $N$ was unchanged by the identification, in which case it was a boundary point of $M$, or it was identified in some way to "become" a boundary point: but then it was in the boundary of $M$ in the first place. This failed in the above example because we identified lots of non-boundary points.

We could also compare the interiors. Pick any point in the interior of $M$, and you can find a little open set that is mapped identically by a restriction of the quotient map. So the interior of $M$ is mapped into the interior of $N$ when all the identifications are on boundary points.

Let me know if you have questions (or if I made any errors).

The answer about the conclusion is no
For example, take $M=\mathbb{R}^2$ and the relation is $x ~ y \Leftrightarrow x-y \in \mathbb{Z}^2$. The quotient space is $\mathbb{R}^2 / \mathbb{Z}^2 \cong \mathbb{S}^2$. Obviously that $\partial M= \emptyset$ because $M$ is open, and $\partial (M/\sim)=\mathbb{S}^2$. There aren't anything to conclude about $\partial (M/\sim)$ from $\partial M$


"Exponential Madness" (Gauss's challenge) From Euler's identity, we see that

$e^{i\pi}=-1$

$\Rightarrow e^{2ik\pi}=1$ [squaring both sides].

This equation surely holds for all integers $k$.

EDIT: From the second equation we get $e^{1+i2k\pi}=e$

So, $e^{1+i2k\pi}=\left (e^{1+i2k\pi}\right )^{(1+i2k\pi)}$

and $e=e^{(1+i2k\pi)^2}=e^{1+i4k\pi}e^{-4k^2\pi^2}$  $ ...(*)$

But $e^{i4k\pi}=1$ from Euler's identity and $e^{1+i4k\pi}=e$

So $e=ee^{-4k^2\pi^2}$ and hence $e^{-4k^2\pi^2}=1$   $[$from $(*)$$]$

But this is true only for $k=0$ whereas we started out with a statement true for all $k\in \mathbb{Z}$. What is the catch here?

NOTE:

$$(x^k)^2 \not = x^{k^2}, \text{ instead} \quad (x^k)^2 = x^{2k}$$

For the second update note that the summation/ product of  exponents is valid only for real exponent. In other words $(e^{1+2ki\pi})^{1+2ki\pi} \not = e^{(1+2ki\pi)^2}$

Thank you for an entertaining and instructive paradox. Most of your humourless commenters have missed the point, because the point involves the use of reason rather than faith in scriptural authority.

The catch is that the only values of $z$ for which $e^z$ has one value are those for which the real part of $z$ is an integer. Thus $$e^{(1-4k^2\pi^2)+4k\pi{i}}$$ is single-valued only when $1-4k^2\pi^2$ is an integer: that is, when $k=0$.

I am not going to quote formulae to prove this, because formula don't provide truth, only reflect it. But here is some motivation.

$1^{1/1}$ has just one value: $1$.
$1^{1/2}$ has two values: $1, -1$.
$1^{1/4}$ has four values: $1, -1, i, i$.
$1^{1/n}$ has $n$ values.

For irrational exponents $x$, $1^x$ has infinitely many values, one of which is $1$.

As long as a family of values all behave in the same way, this multi-valuedness doesn't much matter. What your paradox has done is point out a sequence of operations where the multiplicity of the values causes trouble if you ignore it. Well done for pointing it out! To be brutal about it, not many people would think that (for example) $1^\pi$ has infinitely many values.  (In general they all have the form $e^{2k\pi^2{i}}$ for some integer $k$.


Holomorphic branch of $\log f$ and $f^{1/n}$ I am doing old complex analysis questions to prepare me for the final exam. I came upon this question, from which I could not solve the last part.
Let $0<r<R$ and assume $f$ has no zeros in a neighbourhood of $A(0,r,R)$.
Let $\rho\in(r,R)$. Prove that there is a holomorphic branch of $\log f$ on $A(0,r,R)\setminus \{z\in\mathbf{C}\mid r<\operatorname{Re} z<R,\operatorname{Im} z=0\}$.
This I proved with a theorem in my book that says that if $D$ is a simply connected domain in $\mathbf{C}$ and $f$ has no zeros on $G$, then there exists a holomorphic branch of $\log f$ on $D$.

(a) Take such a holomorphic branch $\log f$. Calculate $\lim_{y\to 0^+}\left[\log f(\rho+iy)-\log f(\rho-iy)\right]$.
(b) Prove that there exist holomorphic branches of $f^{1/n}$ on $A(0,r,R)$.

I have no idea how to do (a). For (b) I thought of doing something like $g(z)=e^{\frac1n\log(f(z))}$. But then the theorem does not work anymore, since the annulus is not simply connected.
Can someone provide some help?

First, (b) is false;  $r=1$, $R=2$, $f(z)=z$ is a counterexample.

It's not clear exactly what counts as an answer to (a). Say $g=\log f$ in $D=A(0,r,R)\setminus \{z\in\mathbf{C}\mid r<\operatorname{Re} z<R,\operatorname{Im} z=0\}$. Then $g'=f'/f$. So, writing $g(p)-g(q)$ as the integral of $g'$ over a path from $q$ to $p$, it follows that $$\lim_{y\to0}(g(\rho+iy)-g(\rho-iy))=-\int_{|z|=\rho}\frac{f'(z)}{f(z)}\,dz,$$which may well be the expected answer.

Here's a generalization of a correct version:

    Suppose $V$ is an open set and $f\in H(V)$ has no zero. The following are equivalent:

    (a') There exists $\log f\in H(V)$

    (b') For every $n\in\Bbb N$ there exists $f^{1/n}\in H(V)$.

Of course it's trivial that (a') implies (b'). Assume (b').

Let $g_n=f^{1/n}$. Say $\gamma$ is a closed curve in $V$, and define $$I=\frac1{2\pi i}\int_\gamma\frac{f'}f,$$
$$I_n=\frac1{2\pi i}\int_\gamma\frac{g_n'}{g_n}.$$

Since $f'/f=ng_n'/g_n$ it follows that $$I=nI_n\quad(n\in\Bbb N).$$

Since $I$ and $I_n$ are integers this implies that $$I=0.$$
Since $I=0$ for every closed path  $\gamma$ there exists $L\in H(V)$ with $$L'
=f'/f.$$Hence $fe^{-L}$ is constant (wlog $V$ is connected), and so there exists $c$ with $$fe^{-(L+c)}=1.$$


Infiniteness of non-twin primes. Well, we all know the twin prime conjecture.
There are infinitely many primes $p$, such that $p+2$ is also prime.
Well, I actually got asked in a discrete mathematics course, to prove that there are infinitely many primes $p$ such that $p + 2$ is NOT prime.

Let $p\gt 3$ be prime. If $p+2$ is not prime, we are happy. If $p+2$ is prime, then $(p+2)+2$ is not, since one of $x,x+2,x+4$ is divisible by $3$.

Added: Dolda2000 noted that a more interesting question is whether there are infinitely many primes that are not members of a twin pair. For this we can use the fact that there are infinitely many primes of the form $15k\pm 7$. If $p$ is such a prime, then one of $p-2$ or $p+2$ is divisible by $3$, and the other is divisible by $5$, so if $p\gt 7$ then neither $p-2$ nor $p+2$ is prime.

"I actually got asked in a discrete mathematics course, to prove that there are infinitely many primes p such that p+2 is NOT prime". (I insert the question here to prevent confusing if it was modified.)
I use only that there are infinitely many prime numbers. (An elementary proof of Euclid.)
If $p>3$ ($p$ is prime number) then denote
$$
P_{-1}:=\{p|p=3k-1,\text{for some }k\}
$$
and
$$
P_{1}:=\{p|p=3k+1,\text{for some }k\}.
$$
(Every prime number, greater than $3$, has remainder $\pm 1$ dividing by $3$.)

So at least one of them contains infinitely many prime numbers.

(a) If $P_{1}$ contains finitely many prime numbers then the statement is true, because $P_{-1}$ contains infinitely many prime numbers so if $k$ is large enough then $p+2=(3k-1)+2\in P_{1}$ not a prime number.

(b) If $P_{1}$ contains infinitely many prime numbers then the statement is true, because taking them $3|(3k+1)+2$.


Confidence interval for sample I have a sample of size $n=19593$ of count data

  1      2     3    4    5   6   7   8   9   10  11  12  13  14  15  16  17  18  19 20 21 22 23 24 25 26 27
18890  3032  1542  727  599 530 409 363 298 266 261 273 231 268 129 131 145 120 119 29 62 48 0  18 0  38 10

I want to find a confidence interval for the center of the data, specifically in the form $n \mu$. Do I have enough data to estimate $\sigma$ and use the central limit theorem. The negative binomial seems to be the best fit among known distributions, but I  don't think it is the right fit because of the long tail in the data, so would bootstrapping be a better approach and using the median instead of the mean to measure the center of the data.

Confidence intervals for mean of a highly skewed population using a large sample.

This seems to be a followup from a previous post in which you started
asking about CIs for the population mean in addition to your original question (about retrieving data from confidence intervals for the frequencies of
various values). Here the values $y$ and their respective frequencies $f$ are provided.

Your explicit question in the current post is about a comparison of a traditional
confidence interval for the population mean $\mu$ based on the CLT and a nonparametric bootstrap.
As @Bey speculates, there is good agreement.

Traditional confidence interval. First, the traditional CI.
We begin by inputting your data and checking for formatting errors.

 y = c(1, 2,  3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15,
       16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27)
 f = c(18890, 3032, 1542, 727, 599, 530, 409, 363, 298, 266, 261,
       273, 231, 268, 129, 131, 145, 120, 119, 29, 62, 48, 0, 18,
       0, 38, 10)
 n = sum(f);  x = rep(y, times=f);  n
 ## 28538  # n
 obs.mean = sum(y * f)/n;  obs.mean;  mean(x) #check
 ## 2.740872  # x.bar
 ## 2.740872  # agreement: data correctly parsed
 obs.sd = sd(x);  pm = c(-1,1);  z.star = qnorm(.975)
 obs.mean + pm*z.star*obs.sd/sqrt(n)  # trad'l CI
 ## 2.696125 2.785619

So the standard CI of the population mean $\mu,$ assuming normality is $(2.70, 2.78).$ The distribution is indeed very seriously skewed,
but the sample size is huge and the traditional interval should be OK.

Nonparametric bootstrap CI. Now to compare with results of a nonparametric bootstrap. I am
bootstrapping the distribution of the variability $V = \bar X - \mu,$
of the sample mean about the population mean for (very large) samples of size  $n = 28,538.$

If we knew the distribution of $V,$ we could find $v_L$ and $v_U$
cutting 2.5% of the probability from the respective tails of its
distribution so that
$$P(v_L \le V = \bar X - \mu \le v_U) = P(\bar X - v_U \le \mu \le \bar X - v_L) = 0.95,$$
and $(\bar X - v_U, \,\bar X - v_L)$ is a 95% CI for $\mu.$

The resulting CI turns out to be $(2.69, 2.79),$
essentially identical to the result based on the CLT. The bootstrap
interval is a little longer because here we are making no assumptions
about the distribution of the population (except that it has a mean).
So it does not matter whether the data are Poisson, negative binomial, or from some other severely skewed distribution. (The code below assumes that the code above has been run
just previously. In typical bootstrap notation, $*$'s are used for resampled quantities.)

 # Bootstrap World--Estimated Distribution of V
 B = 1000;  re.x = sample(x, B*n, repl=T)
 RDTA = matrix(re.x, nrow=B)          # B x n matrix of resamples
 re.mean = rowMeans(RDTA)             # vector of B `x-bar-star's
 re.v = re.mean - obs.mean            # vector of v-star's
 v.UL = quantile(re.v, c(.975,.025))  # estimated quantiles of V

 # Real World--Bootstrap Confidence Interval
 obs.mean - v.UL
 ##    97.5%     2.5%
 ## 2.692651 2.785346

Notice that no estimate of the variance is used in the bootstrap procedure. The variability of $V$ is 'deduced' from repeated sampling.

You are right - your distribution is definitely not normal.

There seems to be a misconception in some of the answers - large $N$ does not imply the distribution is close to normal! The empirical distribution will converge to the actual distribution for a large $N$ - and not normal if the actual distribution isn't. (Only means converge to normal according to CLT.)

For your distribution, however, you can find confidence intervals by taking percentiles - for example the range between 2.5th percentile and 97.5th percentile is a 95% confidence interval.

The picture below illustrates this on your distribution.

Or in general, take $1-\alpha$ confidence interval by choosing $\hat F_{\alpha/2}$.


Computing $C_0^2+C_1^2+C_2^2+C_3^2+ \cdots +C_n^2$ If $C_k$ denotes binomial coefficient of choosing $k$ objects from  a set of $n$ objects how to calculate this:

$$C_0^2+C_1^2+C_2^2+C_3^2+\cdots +C_n^2$$

Consider a group of $2n$ people of which $n$ are men and $n$ are women. In how many ways can we choose $n$ people from this group? In $\binom{2n}{n}$ ways.  Now we count the same number in a different way. Each such chosen group consists of $k$ men and $n-k$ women for some $k=0,1,2,...,n$. First we choose $k$ men from $n$ men; this can be done in $\binom{n}{k}$ ways. Then select $n-k$ women from $n$ women; this can be done in $\binom{n}{n-k}$ ways. So, number of ways of choosing $n$ people of which $k$ are men and $n-k$ are women is $\binom{n}{k} \binom{n}{n-k}$. And $k$ ranges from $0,1,2,...,n$. Therefore, the total number of ways of choosing $n$ people is $\sum_{k=0}^n{\binom{n}{k} \binom{n}{n-k}}$. Thus we get, $$ \binom{2n}{n} = \sum_{k=0}^n{\binom{n}{k} \binom{n}{n-k}} $$
Now use the identity $\binom{n}{k} = \binom{n}{n-k}$. [To see this, observe that number of ways of choosing $k$ objects from $n$ objects is the same as discarding $n-k$ objects from $n$ objects.] We get, $$ \binom{2n}{n} = \sum_{k=0}^n{\binom{n}{k} \binom{n}{k}} $$
$$ \binom{2n}{n} = \sum_{k=0}^n{{\binom{n}{k}^2 }} $$

From Wikipedia
$$\sum_{k=0}^n {n \choose k}^2 = {2n\choose n}.$$


A detail in the proof of Stone representation Theorem Let $(\mathcal{B},\sqcap,\sqcup,\leq)$ be a Boolean algebra. Let $x,y\in\mathcal{B}$. I want to prove the following implication:
$$x\sqcap y'\leq 0\Rightarrow  x\leq y$$
where $y'$ is the complement of $y$.

I have checked that this works in the case $\mathcal{B}=\mathcal{P}(X)$, with $\sqcap$=intersection, $\sqcup$=union, $\leq=\subseteq$, $0=\varnothing$.....but I can't prove it for a generic Boolean algebra.

Hint: $x=x\sqcap1=x\sqcap(y\sqcup y')=(x\sqcap y)\sqcup(x\sqcap y')=\ldots$

Since $$x⊓y′≤0$$
$$(x⊓y′)⊔y ≤0 ⊔ y = y$$
$$(x⊔y)⊓(y'⊔y)≤ y$$
$$(x⊔y)⊓1 ≤ y$$
$$(x⊔y)≤ y$$
$$x ≤ y$$


The bound of valuation Let $V_p$ be the $p$-adic valuation. We know that $(p - 1)! + 1\equiv0\mod p$ for the prime $p$ by Wilson's theorem. I wonder if there is an upper bound for $V_p((p - 1)! + 1)$.

Also I do not know how to prove the following statement: Let $p\equiv 7\mod8$ be a prime. Then
$\sum\limits_{r = 1}^{\frac{p - 1}{2}}r(\frac{r}{p}) = 0$, where $(\frac{\cdot}{\cdot})$ is the Legendre symbol.

Could anybody help me to answer these questions? Thanks.

$$\lfloor \frac{n-1}{p-1} \rfloor - k \leq v_p(n !) \leq \lfloor \frac{n-1}{p-1} \rfloor$$

with $k$ defined by $ p^k \leq n < p^{k+1} $


Prove that the bilinear form can be presented as a product of two linear forms Let $ f:\Bbb R^3 \times\Bbb R^3 \to \Bbb R $ be a bilinear form such the the rank of $f$ is 1.
Prove that $f$ can be presented as a product of the linear forms, such that:

$$f(x,y)=(b_{1}x_1+b_2x_2+b_3x_3)(c_{1}y_1+c_{2}y_2+c_3y_3)$$

For every $$x=(x_1,x_2,x_3), y=(y_1,y_2,y_3)$$

Actually I don't know how to approach this question. I have to use the fact that the rank=1. But how?
Thanks,
Alan

The general shape of a bilinear form is
$$(v,w)\longmapsto v^{\top}Qw,$$
where $v$ and $w$ are vectors and $Q$ is a square matrix.

In the other hand for two linear forms (on $\Bbb R^3$) we have
$$\alpha(x_1,x_2,x_3)=[b_1,b_2,b_3]
\left(\begin{array}{c}
x_1\\
x_2\\
x_3
\end{array}\right)=b_1x_1+b_2x_2+b_3x_3,$$
and
$$\beta(y_1,y_2,y_3)=[c_1,c_2,c_3]
\left(\begin{array}{c}
y_1\\
y_2\\
y_3
\end{array}\right)=c_1y_1+c_2y_2+c_3y_3,$$

The two matrices for $\alpha$ and $\beta$ respectively are used to
construct their  tensor product  (or Kronecker product) which is
$$\alpha\otimes \beta=\left(\begin{array}{c}
b_1\\
b_2\\
b_3
\end{array}\right)(c_1,c_2,c_3)
=
\left(\begin{array}{ccc}
b_1c_1 & b_1c_2 & b_1c_3\\
b_2c_1 & b_2c_2 & b_2c_3\\
b_3c_1 & b_3c_2 & b_3c_3
\end{array}\right).
$$

Then your bilinear form would be
$$f=
(x_1,x_2,x_3)
\left(\begin{array}{ccc}
b_1c_1 & b_1c_2 & b_1c_3\\
b_2c_1 & b_2c_2 & b_2c_3\\
b_3c_1 & b_3c_2 & b_3c_3
\end{array}\right)
\left(\begin{array}{c}
y_1\\
y_2\\
y_3
\end{array}\right)
.$$


Not able to understand this graph for rational function . I am not getting why there are 3 lines drawn in that way .

If $x=2$ or $x = -2$ you get $\frac {x^2-3x-2}{x^2 - 4}$ will have $x^2 - 4 = (\pm 2)^2 -4 = 0$ and the whole thing is undefined and can not be plotted.  For values where $x$ is close to $0$ then $x^2 - 4$ will be very small in magnitude and so $\frac 1{x^2- 4}$ will be very large in magnitude and the function will stretch to very large positive or negative values and as $x\to 0$ the $|\frac {x^2-3x-2}{x^2 - 4}| \to \infty$.  The vertical lines $x =2$ and $x=-2$ are the vertical asymptotes and as $x$ is very close to $\pm 2$ the graph will get very close to those vertical lines.
$y = 1$ is the horizontal asymptote.  As $x$ gets very very large then the the proportional difference between the numerator and the denominate will get insignificant and the value of $\frac {x^2 - 3x-2}{x^2 -4}$ will get closer and closer to $1$. (for example if $x=10$ you have $\frac {x^2-3x-2}{x^2-4} =\frac {100-30 -2}{100 -4} = \frac {68}{96}$ but and $x =100$ you have $\frac {x^2-3x -2}{x^2 - 4}=\frac {10000-300 -2}{10000 - 2}=\frac {9698}{9992}$.)
More formally $\frac {x^2 -3x -2}{x^2 - 4} = \frac {x^2 -4 -3x +2}{x^2 -4}=1 -\frac {3x-2}{x^2-4}$ and as $x \to \pm \infty$ we have $\frac {3x-2}{x^2 -4} \to 0$.
so the graph gets close to $y=1$ for large positive and negative values of $x$.
This is the nature of assymptotes.  It's pretty important you understand them.  Keep studying.

As the final sentence above "Example 2" says, those are the asymptotes of the function, lines which the graph approaches as $x$ approaches $\pm\infty$, or as $x\to\pm2$ (in which case $y$ approaches $\pm\infty$).


How many diagonals does a decagon have? How many diagonals does a decagon have?

I have just learnt permutations, dispositions, combinations. How can I solve it with these concepts? I drew it and it was $35$ diagonals. How can I prove it with this method?

A diagonal joins a vertex to one of the vertices that do not include that vertex itself and the immediately adjacent vertices. So: for each vertex there are seven diagonals. Times 10 equals 70; each diagonal is counted twice, so the final answer is 35.

Now, using combinations and such: There are $\binom{10}{2}\;$("10 choose 2") pairs of vertices, which equals 45. So there are 45 line segments joining pairs of vertices. Exactly 10 of those are sides of the decagon, the others are diagonals. Answer: 35.  (Corrected; original had "10 choose 9" for no reason other than my lack of concentration.)

(N-1 choose 2 ) -1

Example,
(10-1 choose 2) -1
= (9 choose 2) -1
= 36-1
= 35 diagonal lines

This will work for any regular shape

Example 2/ 20 sides would be
C19,2 -1
=170
Check: C20,2 - 20 =170

Very easy formula developed by Shawn Covrigaru


show this inequality to $\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $ Let $a,b$ and $c$ be positive real numbers. Prove that
$$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$

This problem is from Iran 3rd round-2017-Algebra final exam-P3,Now I can't find this inequality have solve it,maybe  it seem can use  integral to solve it?

my attempts:

I took $p=3a+2b,$ $2a+2b+c$. $k=3$  and I wanted to use this integral:
$$\dfrac{1}{p^k}=\dfrac{1}{\Gamma(k)}\int_{0}^{+\infty}t^{k-1}e^{-pt}dt$$
but I don't see how it  helps.

Using binomial inequality for
$$|c-a|<2a+2b+c,$$
one can get
$$\dfrac1{(3a+2b)^3}=\dfrac1{(2a+2b+c)^3}\left(1-\dfrac{c-a}{2a+2b+c}\right)^{-3}
\ge \dfrac1{(2a+2b+c)^3}\left(1+3\dfrac{c-a}{2a+2b+c}\right),$$
$$\dfrac{a^3b}{(3a+2b)^3}-\dfrac{a^2bc}{(2a+2b+c)^3} \ge \dfrac{a^2b}{(2a+2b+c)^3}\left(a-c-3a\dfrac{c-a}{2a+2b+c}\right),\tag1$$
The issue inequality can be presented in the form of
$$S\ge 0\tag2,$$
where
$$S = \sum\limits_\bigcirc \left(\dfrac{a^3b}{(3a+2b)^3}-\dfrac{a^2bc}{(2a+2b+c)^3}\right) \ge  \sum\limits_\bigcirc\dfrac{a^2b(a-c)}{(2a+2b+c)^3}
+3\sum\limits_\bigcirc\dfrac{a^3b(a-c)}{(2a+2b+c)^4}.$$

In according with the rearrangement inequality for the productions of
$$a\cdot a\cdot a\cdot \dfrac b{2a+2b+c}
+ b\cdot b\cdot b\cdot \dfrac c{2b+2c+a}
+ c\cdot c\cdot c\cdot \dfrac a{2c+2a+b}$$
and
$$a\cdot a\cdot c\cdot \dfrac b{2a+2b+c}
+ b\cdot b\cdot a\cdot \dfrac c{2b+2c+a}
+ c\cdot c\cdot b\cdot \dfrac a{2c+2a+b}$$
between $a$ and $c$ (where the ratio can be placed arbitrary), one can write
$$\sum\limits_\bigcirc\dfrac{a^3b}{(2a+2b+c)^3} \ge
\sum\limits_\bigcirc\dfrac{a^2bc}{(2a+2b+c)^3}$$
and similarly
$$\sum\limits_\bigcirc\dfrac{a^4b}{(2a+2b+c)^4} \ge
\sum\limits_\bigcirc\dfrac{a^3bc}{(2a+2b+c)^4},$$
so $(1)$ is valid.

$\textbf{Proved.}$


FInd a set of $n$ numbers to minimize number of distinct products I would liked to find a set of $n$ natural numbers none of which are perfect squares that minimizes the number of distinct products of any two numbers in the set.

Stated more formally: find a generating set $S$ of $n$ non-perfect square natural numbers such that the number of unique products of any pair of numbers in the set $S$ is minimized. We only care about products of distinct elements in $S$.

According to Cauchy–Davenport, you always have at least $2n-1$ distinct products, and this is achieved by geometric series: $a,ab,\ldots,ab^{n-1}$. Now see if you can arrange that none of these numbers is a perfect square.


Solve the initial value problem: $\frac{dx}{dt}=2t \sin x$; $x(0)=\frac{\pi}{2}$ Solve the initial value problem: $\frac{dx}{dt}=2t \sin x$; $x(0)=\frac{\pi}{2}$
I am almost done. But I cannot seem to solve the equation for x(t) in terms of t.
This is what I got:
$$
\frac{dx}{\sin x}=2tdt\\
\csc x =2tdt\\
-\ln|\csc x+\cot x|=t^2+C\\
\ln \frac{1}{|\csc x+\cot x|}=t^2+C\\
\frac{1}{\csc x+\cot x}=\pm e^Ce^{t^2}
$$

So now I need to know what x(t) is. How do get the expression for x(t)?

\begin{align*}
\frac{dx}{dt}&=2t \sin x\\
\frac{dx}{\sin x}&=2t \ dt\\
\csc x \ dx&= 2t \ dt\\
-\log(\csc x + \cot x)&=t^2+C_1\\
\log((\frac{1+\cos x}{\sin x})^{-1})&=t^2+C_1\\
\log (\tan\frac{x}{2})&=t^2+C_1\\
\tan{\frac{x}{2}}&=Ce^{t^2}\\
x(t)&=2\arctan(Ce^{t^2})
\end{align*}

For the initial condition:
\begin{align*}
\frac{\pi}{2}&= 2\arctan(Ce^{0})\\
\tan( \frac{\pi}{4}) &=C\\
C&=1
\end{align*}


Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution. Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.

My attempt is as follows:

$$\left(x-\dfrac{2k-\sqrt{4k^2-20}}{10}\right)\left(x-\dfrac{2k+\sqrt{4k^2-20}}{10}\right)<0$$
$$\left(x-\dfrac{k-\sqrt{k^2-5}}{5}\right)\left(x-\dfrac{k+\sqrt{k^2-5}}{5}\right)<0$$
$$x\in\left(\dfrac{k-\sqrt{k^2-5}}{5},\dfrac{k+\sqrt{k^2-5}}{5}\right)$$

As it is given that it has got only one integral solution, so there must be exactly one integer between $\dfrac{k-\sqrt{k^2-5}}{5}$ and $\dfrac{k+\sqrt{k^2-5}}{5}$

Let $x_1=\dfrac{k-\sqrt{k^2-5}}{5}$ and $x_2=\dfrac{k+\sqrt{k^2-5}}{5}$ , then $[x_2]-[x_1]=1$ where [] is a greater integer function.

But from here, how to proceed? Please help me in this.

Your idea is good; you want to find all positive integers $k$ for which there is precisely on integer between the roots of
$$5x^2-2kx+1=0.$$
Then the distance between the roots can be at most $2$, where the distance between the roots is precisely
$$\frac{1}{5}\sqrt{(-2k)^2-4\cdot1\cdot5}=\frac25\sqrt{k^2-5},$$
as you already found. This is at most $2$ if and only if $\sqrt{k^2-5}\leq5$, or equivalently $k\leq5$. This leaves only $5$ values of $k$ to check.

hint

Observe that

$$x_2=\frac{k+\sqrt{k^2-5}}{5}=\frac{1}{k-\sqrt{k^2-5}}$$

$ x_2 $ is integral if  $ (k -\sqrt{k^2-5})=\pm 1$ ,

By the same, $ x_1 $ is integral if
$ (k+\sqrt{k^2-5}) =\pm 1$.

This gives the possible values for $ k$ :
$-4, -3, 3, 4$.
the sum of positives values is $7$.


How to prove this inequality given the equation Given $ a>0 $ and $ a^5-a^3+a=3$, how can I prove the inequality: $ a^6 \ge 5 $ ?
I have tried factorizing the equation, solving for $ x, x^2, x^3 $ and then adding equations made together. Last but not least I tried, to solve for $x$ and then raise it to the power of $ x^6 $, but none of these seem to work

Given $a^5 - a^3 + a = 3$, we can factorise the left hand side to write $a(a^4-a^2+1)=3$.

The second factor here is reminiscent of cyclotomic polynomials/finite geometric series - in any case, we can write it as $\frac{a^6+1}{a^2+1}$, and clear fractions to get to $a(a^6+1)=3(a^2+1)$.

Now, we spot the $a^6$ we're looking for on the left hand side, and rearrange to see that $a^6 = 3\frac{a^2+1}{a}-1=3(a+\frac{1}{a})-1$.

There are a number of ways of showing $a+\frac{1}{a}\geq2$ for $a>0$ - the easiest is probably differentiating with respect to $a$ to find a minimum at $a=1$, but you can equally use the arithmetic-geometric inequality or a number of other methods. Substituting $a+\frac{1}{a}\geq2$ back into the previous inequality gets us to $a^6\geq3(2)-1=5$ and we're done.

We also could have, upon spotting the $a^6$, subtracted 5 and looked for a reason for it to be positive - in this case, we'd see that $a^6 - 5 = 3(a+\frac{1}{a})-6=3(a+\frac{1}{a}-2)$, and pushing through this a bit takes us to $\frac{3(a^2-2a+1)}{a}=\frac{3(a-1)^2}{a}$.

It's then clear that $a^6=5+\frac{3(a-1)^2}{a}$, which displays fairly immediately that the inequality holds.

Define $f: \mathbf{R} \to \mathbf{R}: x \mapsto x^5-x^3+x-3$. Then $$
f^\prime(x)=5x^4-3x^2+1=5\left(x^2-\frac{3}{10}\right)^2+\left(1-\frac{9}{20}\right)>0.
$$
Therefore $f$ is strictly increasing, so it is sufficient to check that $f(5^{1/6})<0$.


Finding a closed form for $\sum\limits_{\substack{0\le n\le N\\0\le m\le M}}\left|nM-Nm\right|$ I am trying to figure out if there is a closed form for the following sum:
$$
\sum_{\substack{0\le n\le N \\ 0\le m\le M}}\left|(N-n)(M-m)-nm\right|=\sum_{\substack{0\le n\le N \\ 0\le m\le M}}\left|nM-Nm\right|.
$$
Clearly, from symmetry, if we remove the absolute values, the sum evaulates to $0$.

Using the symmetry, I tried to evaluate the sum as
$$
2\cdot\sum_{\substack{0\le n\le N \\ 0\le m\le M \\ (N-n)(M-m)\gt nm}}\left((N-n)(M-m)-nm\right).
$$
However, that expression ended up containing sums of floor functions, for which I did not know a closed form.

Is there another way to obtain a closed form for the above sum?

Edit: If no closed form can be provided an efficient algorithm to compute the above sum given $N,M$ would also be good.

Edit 2: Since the time I have placed the bounty, I was able to find a closed form for the sum:

  $$\frac{1}{6}\left[MN(2 M N + 3 (M + N + 1)) + M^2 + N^2-\gcd(M,N)^2\right].$$

So now I change the question to a challange: derive the above form from the sum. The prettiest derivation (if there are any) will get the bounty.

$\def\peq{\mathrel{\phantom{=}}{}}$Suppose $M, N > 0$ and $d = (M, N)$, $M = md$, $N = nd$.\begin{align*}
\sum_{\substack{0\leqslant k\leqslant M\\0\leqslant j\leqslant N}} |kN - jM| &= \sum_{\substack{0\leqslant k\leqslant M-1\\0\leqslant j\leqslant N-1}} |kN - jM| + \sum_{k = 0}^{M - 1} (NM - kN) + \sum_{j = 0}^{N - 1} (MN - jM)\\
&= \sum_{\substack{0\leqslant k\leqslant M-1\\0\leqslant j\leqslant N-1}} |kN - jM| + \frac{1}{2} NM(M + 1) + \frac{1}{2} MN(N + 1). \tag{1}
\end{align*}
Note that$$
\sum_{\substack{0\leqslant k\leqslant M-1\\0\leqslant j\leqslant N-1}} (kN - jM) = \frac{1}{2} NM(M - 1) - \frac{1}{2} MN(N - 1),
$$
then\begin{align*}
\sum_{\substack{0\leqslant k\leqslant M-1\\0\leqslant j\leqslant N-1}} |kN - jM| &= \sum_{\substack{0\leqslant k\leqslant M-1\\0\leqslant j\leqslant N-1\\kN\geqslant jM}} (kN - jM) - \sum_{\substack{0\leqslant k\leqslant M-1\\0\leqslant j\leqslant N-1\\kN<jM}} (kN - jM)\\
&= 2\sum_{\substack{0\leqslant k\leqslant M-1\\0\leqslant j\leqslant N-1\\kN\geqslant jM}} (kN - jM) - \sum_{\substack{0\leqslant k\leqslant M-1\\0\leqslant j\leqslant N-1}} (kN - jM)
\end{align*}
which implies$$
(1) = 2\sum_{\substack{0\leqslant k\leqslant M-1\\0\leqslant j\leqslant N-1\\kN\geqslant jM}} (kN - jM) + MN(N + 1). \tag{2}
$$

Now for each $0 \leqslant k \leqslant M - 1$, define $$r_k = kn - m\left[ \dfrac{kn}{m} \right],$$ i.e. $r_k$ is the remainder of $kn$ modulo $m$.
\begin{align*}
&\peq \sum_{\substack{0\leqslant k\leqslant M-1\\0\leqslant j\leqslant N-1\\kN\geqslant jM}} (kN - jM) = \sum_{\substack{0\leqslant k\leqslant dm-1\\0\leqslant j\leqslant dn-1\\kdn\geqslant jdm}} (kdn - jdm) = d \sum_{\substack{0\leqslant k\leqslant dm-1\\0\leqslant j\leqslant dn-1\\kn\geqslant jm}} (kn - jm)\\
&= d\sum_{k = 0}^{dm - 1} \sum_{j = 0}^{\left[ \frac{kn}{m} \right]} (kn - jm) = d\sum_{k = 0}^{dm - 1} \left( kn \left( \left[ \frac{kn}{m} \right] + 1 \right) - m·\frac{1}{2} \left[ \frac{kn}{m} \right] \left( \left[ \frac{kn}{m} \right] + 1 \right) \right)\\
&= d\sum_{k = 0}^{dm - 1} \left( kn - \frac{m}{2} \left[ \frac{kn}{m} \right] \right) \left( \left[ \frac{kn}{m} \right] + 1 \right) = d\sum_{k = 0}^{dm - 1} \left( kn - \frac{m}{2}·\frac{kn - r_k}{m} \right) \left( \frac{kn - r_k}{m} + 1 \right)\\
&= d\sum_{k = 0}^{dm - 1} \left( \frac{1}{2} kn + \frac{1}{2} r_k \right) \left( \frac{kn}{m} + 1 - \frac{r_k}{m} \right) = \frac{d}{2m} \sum_{k = 0}^{dm - 1} (kn(kn + m) + m r_k - r_k^2). \tag{3}
\end{align*}

Note that $(m, n) = 1$ implies for each $0 \leqslant s \leqslant d - 1$,$$\{kn \mid sm \leqslant k \leqslant (s + 1)m - 1\}$$ is a complete residue system modulo $m$, therefore$$
\{r_{sm}, \cdots, r_{(s + 1)m - 1}\} = \{ 0, \cdots, m - 1\}.
$$
Thus$$
\sum_{k = 0}^{dm - 1} r_k = \sum_{s = 0}^{d - 1} \sum_{k = sm}^{(s + 1)m - 1} r_k = \sum_{s = 0}^{d - 1} \sum_{k = 0}^{m - 1} k = \frac{1}{2} m(m - 1)·d,\\
\sum_{k = 0}^{dm - 1} r_k^2 = \sum_{s = 0}^{d - 1} \sum_{k = 0}^{m - 1} r_k^2 = \sum_{s = 0}^{d - 1} \sum_{k = 0}^{m - 1} k^2 = \frac{1}{6} (2m - 1)(m - 1)m·d,
$$
then\begin{align*}
(3) &= \frac{d}{2m} \sum_{k = 0}^{dm - 1} kn(kn + m) + \frac{d}{2} \sum_{k = 0}^{dm - 1} r_k - \frac{d}{2m} \sum_{k = 0}^{dm - 1} r_k^2\\
&= \frac{dn^2}{2m} \sum_{k = 0}^{dm - 1} k^2 + \frac{1}{2} dn \sum_{k = 0}^{dm - 1} k + \frac{d}{2} × \frac{1}{2} m(m - 1)·d - \frac{d}{2m} × \frac{1}{6} (2m - 1)(m - 1)m·d\\
&= \frac{dn^2}{2m} × \frac{1}{6} (2dm - 1)(dm - 1)dm + \frac{1}{2} dn × \frac{1}{2} (dm - 1)dm\\
&\peq + \frac{d}{2} × \frac{1}{2} m(m - 1)·d - \frac{d}{2m} × \frac{1}{6} (2m - 1)(m - 1)m·d\\
&= \frac{1}{6} d^4 m^2 n^2 + \frac{1}{4} d^3 mn(m - n) + \frac{1}{12} d^2 (m^2 + n^2 - 3mn) - \frac{1}{12} d^2.
\end{align*}

Combining with (2),\begin{align*}
\sum_{\substack{0\leqslant k\leqslant M\\0\leqslant j\leqslant N}} |kN - jM| &= 2\sum_{\substack{0\leqslant k\leqslant M-1\\0\leqslant j\leqslant N-1\\kN\geqslant jM}} (kN - jM) + MN(N + 1)\\
&= 2\left( \frac{1}{6} d^4 m^2 n^2 + \frac{1}{4} d^3 mn(m - n) + \frac{1}{12} d^2 (m^2 + n^2 - 3mn) - \frac{1}{12} d^2 \right)\\
&\peq + d^2 mn(dn + 1)\\
&= \frac{1}{3} d^4 m^2 n^2 + \frac{1}{2} d^3 mn(m + n) + \frac{1}{6} d^2 (m^2 + n^2 + 3mn) - \frac{1}{6} d^2\\
&=\frac{1}{6} \left( MN(2MN + 3(M + N + 1)) + M^2 + N^2-(M,N)^2\right).
\end{align*}

Partial solution:

We can break the expression into four parts $E_1 + E_2 + E_3 + E_4$

where $$E_1 = \sum_{\substack{0\le n\le N/2 \\ 0\le m\le M/2}}(MN -mN-nM) = $$
$$E_2 = \sum_{\substack{0\le n\le N/2 \\ M/2\le m\le M}}|MN -mN-nM|$$
$$E_3 = \sum_{\substack{N/2\le n\le N \\ 0\le m\le M/2}}|MN -mN-nM|$$
$$E_4 = \sum_{\substack{N/2\le n\le N \\ M/2\le m\le M}}(mN + nM - MN) = \sum_{\substack{0\le n\le N/2 \\ 0\le m\le M/2}}(mN + nM)$$

$E_1 + E_4 = M^2N^2/4$

Both $E_2$ and $E_3$ can be rewritten as

$$ E_2 = E_3 = \sum_{\substack{0\le n\le N/2 \\ 0\le m\le M/2}}|MN/2 -mN-nM|$$

so $$E_2 + E_3 = \sum_{\substack{0\le n\le N/2 \\ 0\le m\le M/2}}|MN -2mN-2nM|$$

So $$LHS = M^2N^2/4 + \sum_{\substack{0\le n\le N/2 \\ 0\le m\le M/2}}|MN -2mN-2nM|$$
and may be we can recursively move forward?


About supporting hyperplanes of convex sets Let $K \subset R^n$ a convex set, and $x \in \partial K$ such that that there exists  a closed ball $B(x_0,R) \subset K$ of positive radius  with $x \in B(x_0,R) $.

My intuition tells me that there exist exactly one supporting hyperplane for $K$ at $x$. How can I prove this?
I know that exist a hyperplane of $K$ at  $x$ by the convexity of $K$ . I don't know how prove that it's unique.

Thanks in advance!

If $H$ is a supporting hyperplane to $K$ at $x$, that means $x\in H$, and $K$ lies entirely on one side of $H$ (including $H$ itself). Describing the hyperplane by a linear equation, $H = \{ y\in \mathbb{R}^n : \lambda(y) = 1\}$ (every Hyperplane has such a representation), we have either $\lambda(k) \leqslant 1$ for all $k\in K$ or $\lambda(k) \geqslant 1$ for all $k\in K$. Let us suppose we have $\lambda(k) \leqslant 1$ for all $k\in K$.

Since $B(x_0,R) \subset K$, we have $\lambda(y) \leqslant 1$ [in fact $\lambda(y) < 1$, since the ball is open] for all $y\in B(x_0,R)$. That means the ball $B(x_0,R)$ also lies entirely on one side of $H$, and since $x\in \partial B(x_0,R)$, the hyperplane $H$ is a supporting hyperplane for $B(x_0,R)$ at $x$.

But we know (hopefully, otherwise prove it, it's not difficult using the normal vector to the hyperplane) that supporting hyperplanes to balls are unique. Thus there is also only one supporting hyperplane to $K$ at $x$.

HINT:

A supporting hyperplane of $K$ at $x$ is also a supporting hyperplane of $B$ at $x$.


How does one show that $\int_{0}^{1}{1\over 1+\phi x^4}\cdot{\mathrm dx\over \sqrt{1-x^2}}={\pi\over 2\sqrt{2}}?$ Consider

  $$\int_{0}^{1}{1\over 1+\phi x^4}\cdot{\mathrm dx\over \sqrt{1-x^2}}={\pi\over 2\sqrt{2}}\tag1$$
  $\phi$;Golden ratio

An attempt:

$x=\sin{y}$ then $\mathrm dx=\cos{y}\mathrm dy$

$(1)$ becomes

$$\int_{0}^{\pi/2}{\mathrm dy\over 1+\phi \sin^4{y}}\tag2$$

Apply Geometric series to $(2)$,

$$\int_{0}^{\pi/2}(1-\phi\sin^4{y}+\phi^2\sin^8{y}-\phi^3\sin^{12}{y}+\cdots)\mathrm dy\tag3$$

$${\pi\over 2}-\int_{0}^{\pi/2}(\phi\sin^4{y}-\phi^2\sin^8{y}+\phi^3\sin^{12}{y}-\cdots)\mathrm dy\tag4$$

Power of sine seem difficult to deal with

How else can we tackle $(1)?$

On the path of Aditya Narayan Sharma.

Define for $a\geq 0$,

$\displaystyle F(a)=\int_0^1 \dfrac{1}{(1+ax^4)\sqrt{1-x^2}}dx$

Perform the change of variable $y=\sin x$,

$\displaystyle F(a)=\int_0^{\tfrac{\pi}{2}} \dfrac{1}{1+a(\sin x)^4}dx$

$(\sin x)^2=\dfrac{1}{1+\dfrac{1}{(\tan x)^2}}$

Perform the change of variable $y=\tan x$,

$\begin{align}\displaystyle F(a)&=\int_0^{+\infty} \frac{1}{(1+x^2)\left(1+a\left(\tfrac{1}{1+\frac{1}{x^2}}\right)^2\right)}dx\\
&=\int_0^{+\infty} \dfrac{1+x^2}{(1+a)x^4+2x^2+1}dx\\
\end{align}$

Perform the change of variable $y=x\sqrt[4]{1+a}$,

$\displaystyle F(a)=\dfrac{1}{\sqrt[4]{1+a}}\int_0^{+\infty} \dfrac{1+\tfrac{x^2}{\sqrt{1+a}}}{x^4+\tfrac{2}{\sqrt{1+a}}x^2+1}dx$

Perform the change of variable $y=\dfrac{1}{x}$,

$\displaystyle F(a)=\dfrac{1}{\sqrt[4]{1+a}}\int_0^{+\infty} \dfrac{x^2+\tfrac{1}{\sqrt{1+a}}}{x^4+\tfrac{2}{\sqrt{1+a}}x^2+1}dx$

Therefore,

$\begin{align}
\displaystyle F(a)&=\frac{1+\frac{1}{\sqrt{1+a}}}{2\sqrt[4]{1+a}}\int_0^{+\infty} \dfrac{x^2+1}{x^4+\tfrac{2}{\sqrt{1+a}}x^2+1}dx\\
&=\frac{1+\frac{1}{\sqrt{1+a}}}{2\sqrt[4]{1+a}}\int_0^{+\infty} \dfrac{\tfrac{1}{x^2}+1}{x^2+\tfrac{1}{x^2}+\tfrac{2}{\sqrt{1+a}}}dx\\
\end{align}$

Perform the change of variable $y=x-\dfrac{1}{x}$,

$\begin{align}
\displaystyle F(a)&=\frac{1+\frac{1}{\sqrt{1+a}}}{2\sqrt[4]{1+a}}\int_{-\infty}^{+\infty} \dfrac{1}{x^2+2+\tfrac{2}{\sqrt{1+a}}}dx\\
&=\frac{1+\frac{1}{\sqrt{1+a}}}{2\sqrt[4]{1+a}}\left[\dfrac{\sqrt{1+a}}{\sqrt{2}\sqrt{1+a+\sqrt{1+a}}}\arctan\left(\dfrac{x\sqrt{1+a}}{\sqrt{2}\sqrt{1+a+\sqrt{1+a}}}\right)\right]_{-\infty}^{+\infty}\\
&=\frac{\sqrt{1+\frac{1}{\sqrt{1+a}}}}{2\sqrt{2}\sqrt[4]{1+a}}\pi\\
&=\boxed{\frac{\sqrt{1+\sqrt{1+a}}}{2\sqrt{2}\sqrt{1+a}}\pi}
\end{align}$

Since $\sqrt{1+\phi}=\phi$ then,

$\begin{align}\displaystyle F(\phi)&=\frac{\sqrt{1+\sqrt{1+\phi}}}{2\sqrt{2}\sqrt{1+\phi}}\pi\\
&=\frac{\sqrt{1+\phi}}{2\sqrt{2}\phi}\pi\\
&=\dfrac{\phi}{2\sqrt{2}\phi}\pi\\
&=\boxed{\dfrac{1}{2\sqrt{2}}\pi}\\
\end{align}$

For $x>0$,

Let,

$\displaystyle F(x)=\int_0^{\tfrac{\pi}{2}} \dfrac{1}{1+x(\sin t)^4}dt$

$\displaystyle H(x)=\int_0^{\tfrac{\pi}{2}} \ln\left(1+x(\sin t)^4\right)dt$

Therefore,

$xH^{\prime} (x)+F(x)=\dfrac{\pi}{2}$

That is,

$\boxed{F(x)=\dfrac{\pi}{2}-xH^{\prime}(x)}$

Using Olivier Oloa's formula, https://math.stackexchange.com/q/873905,

$\displaystyle \boxed{F(x)=\frac{\sqrt{1+\sqrt{1+x}}}{2\sqrt{2}\sqrt{1+x}}\pi}$


Is the Commutative Property of Addition on the Reals a Postulate or Proven? My Calculus book from back in the day (Calculus Second Edition Michael Spivak) starts out by stating 12 basic properties of numbers which he labels P1-P12.  He states:

  "Most of this chapter has been an attempt to present convincing
  evidence that P1-P12 are indeed basic properties which we should
  assume in order to deduce other familiar properties of numbers."

I had always taken this to mean that those properties were axioms upon which (with the help of a few other properties) he was going to build up the whole of Calculus.

However, a google search for such a proof pops up the following link:
https://proofwiki.org/wiki/Real_Addition_is_Commutative

Which purports to be be a proof of the Commutative Property of Addition of the Reals.

I then begin to question whether I even know what an Axiom is and look it up in Wikipedia arriving at:
https://en.wikipedia.org/wiki/Axiom

Which immediately provides $a+b=b+a$ as an example of a non-logical axiom for which the terms axiom, postulate or assumption are interchangeable.

In a sense, a postulate means here is where I want to start. If you start at a more elementary level, then your postulates may become theorems. The field postulates for real numbers are a good example of that. In (usually) Foundation of Mathematics, the basic properties of the integers, rationals, and reals are all derived from basic set theory.

I think what you are asking about is not only unclear but illogical, sorry if that is a little harsh. It is not sensible to question things we should already inherently understand, and which everyone inherently sees by instinct.

There exists a philosophy about the whole of mathematics that it is only based on axioms, postulates, etc. We don't really KNOW anything, we just assume a few axioms to arrive at the conclusion of the matter, and that serves all practical purposes. This is a constructivist argument and sort of assumes that understanding is inherently relative.

I believe what the author is saying, and what is true of axiomatic systems, is that one has to check the system by making sure it doesn't contradict itself, become too specific to contain all of the subject matter at hand (too many axioms), or too general to be cohesive, or maybe express certain arguments (too few axioms).

So he is trying to "axiomatise" the system of numbers the book will use, so that it will have a logical base and not be inherently flawed, or otherwise unintelligable. So although he may not be trying to deal with the most basic level, he is trying to convey the information as logically as possible.

There will always be people trying to provide the most technical answer, but if a basic assumption makes sense to you, just stick with that and move on.


Two relations defined on same sets be equal? We say two relations $R_1$ and $R_2$ are equal if:
$$R_1\subseteq \prod_{i=1}^n A_i$$
$$R_2\subseteq \prod_{i=1}^m B_i$$
then,
$$n = m ,$$
$$A_i=B_i\quad \forall i,\,1\le i \le n,$$
$$let,s_j\,\,and\,\,t_j\,\,be\,\,tuples\,\,of\,\,R_1\,\,and\,\,R_2\,\,respectively$$
$$s.t.,\,\,s_j=<a_1,a_2,a_3,\ldots,a_j>\epsilon\,R_1\land$$
$$t_j=<b_1,b_2,b_3,\ldots,b_j>\epsilon\,R_2$$
$$\Rightarrow\,(a_1=b_1\,\land a_2=b_2\,\land\ldots\land a_n=b_n)$$
$$\forall \,j\,\epsilon \mathbb N)$$
where $A_i$ and $B_i$ are sets.
So, will the below relations be equal?
$$Name = \{x,y\}\,\,and\,\,City = \{a,b\}$$
let, $$R_1=\{<p,q>|\quad p\,\epsilon\,Name\land q\,\epsilon\,City,\quad p\,\,was\,\,born\,\,in\,\,city\,\,q\}$$
$$R_2=\{<p,q>|\quad p\,\epsilon\,Name\land q\,\epsilon\,City,\quad p\,\,lives\,\,in\,\,city\,\,q\}$$
suppose we get,
$$R_1=\{<x,a>,<y,b>\}$$
$$R_2=\{<x,a>,<y,b>\}$$
That is, x was born in the same city which he/she lives in.
Can we say,$$R_1=R_2$$

The relations $R_1$ and $R_2$ are equal if and only if both $x$ and $y$ live in the same city as they were born in. That is, for all $p\in\{x,y\}$ and all $q\in\{a,b\}$ we have
$$
\text{$p$ was born in city $q$} \quad\Longleftrightarrow\quad \text{$p$ lives in city $q$.}
$$
You wrote down the same set of pairs to define $R_1$ and $R_2$, so indeed they are equal.
Equality of relations is a special case of equality of sets, and the sets you wrote down are equal.


Is the undefined value in Isabelle formalized in mathematics? I just encountered the concept of "undefined" values for any type from this blog article that " every Isabelle function is total". In addition

... This design has an important consequence: It allows the existence
of a polymorphic expression that inhabits any type, namely
undefined :: 'a
The naming of this term alone has caused a great deal of confusion for Isabelle beginners, or in communication with users of different systems, so I implore you to not read too much into the name.In fact, you will have a better time if you think of it as arbitrary or, even better, unknown.
we can observe an important fact: undefined is not an extra value
besides the “usual ones”. It is simply some value of that type ...

So for example, if you define a function (e.g. division of natural numbers) without a specified value at certain input values (e.g. 0):
fun division :: "nat ⇒ nat ⇒ nat"  where
   "division n (Suc m) = n div (Suc m)"

then in Isabelle/HOL, at this input (0), the value of the function does exist, but is undefined/arbitrary/unknown.
⋀a. division a 0 = undefined

If I understand it right, in ordinary math, division at zero does not have a value. But in Isabelle/HOL logic, the notion is replaced by one that the value is undefined/unknown.
My question is:
Is there any rigorous formalization of the undefined/unknown value in mathematics elsewhere, and are there axioms to govern how "undefined" values behave?
For example, from this Math.SE question, we have  $0 \cdot \infty = 0$ etc. Are there similar formal rules about the "undefined" value in Isabelle/HOL?
In Isabelle/HOL, there is:
lemma "undefined = undefined" by auto

By comparison, I am not so sure that $\infty = \infty$.

Disclaimer: there are several proof assistants for HOL and Isabelle/HOL is the one that I am least familiar with. The different proof assistants all take slightly different lines on the design of their mathematical libraries, but have the same basic mechanisms for defining the objects in those libraries. The principle here is common to all of them.
As discussed in the blog article that you reference, a feature of HOL is that the mechanism for introducing a new constant is not restricted to simple equational definitions, but allows you to give implicit and possibly loose specifications of the constant. So, for example, you could introduce $\pi$ with the defining property:
$$ 0 < \pi \land \forall x\cdot \sin x = 0 \Leftrightarrow \exists m:\Bbb{Z}\cdot x = m \times \pi$$
(There is a small price to pay: a proof obligation is imposed that ensures that the the defining property is consistent.)
The above is an implicit specification, but can be shown not to be loose: i.e., you can prove that there is exactly one real number $\pi$ that satisfies the above property. As a conceivable example of a loose specification, after defining the complex numbers as pairs of reals with the usual definition of the arithmetic operators, you could define $\mathbf{i}$ by:
$$ \mathbf{i}^2 = -1 $$
(The above example is not very likely in practice as it is more convenient to be less abstract and to require $\mathbf{i} = (0, 1)$. The blog article contains other relevant examples, e.g., see its discussion of $\mathtt{fromSome}$.)
$\mathtt{undefined}$ in Isabelle/HOL is an extreme example of loose specification: its defining property is just $\mathbf{true}$. It has a polymorphic type such that it can be instantiated to any type. Its instance in any type has all the properties that are common to all elements of the type. So you have properties like::
$$
(\mathtt{undefined} : \tau) = (\mathtt{undefined} : \tau) \\
\lnot (\mathtt{undefined} : \Bbb{N}) < 0 \\
(\mathtt{undefined} : \Bbb{R}) < 0.0 \lor (\mathtt{undefined} : \Bbb{R}) = 0.0 \lor (\mathtt{undefined} : \Bbb{R}) > 0.0
$$
where $\tau$ can be any type. $\mathtt{undefined}$ has no other special properties: In particular, $f(x) = \mathtt{undefined}$ does not imply that $x$ is not in the domain of $f$. The definition of $\mathtt{division}$ is just saying that $\mathtt{division} \, a \, 0$ is equal to $\mathtt{undefined}$, some unspecified, but perfectly ordinary natural number. As the blog article suggests, it would perhaps have been more appropriate to call it $\mathtt{unspecified}$, rather than $\mathtt{undefined}$.
There are logics that do explicitly model undefined terms. Tennant's natural logic is one example. See also this article on free logics. HOL is not one of these logics.
As regards the conventions for the use of $\infty$ in ordinary mathematics, good texts will make it quite clear what conventions are being used. I don't think equations like $\infty = \infty$ involving $\infty$ are considered as problematic. E.g., Rudin's Principles of Mathematical Analysis states a convention that $x + \infty = \infty$ when $x$ is finite.

In mathematics founded on the ZFC set theory the value of a function applied at $x$ that is not in the domain of the function is the empty set. In Isabelle/ZF this is lemma apply_0 in the func theory, and in Metamath theorem ndmfv.


What is the metric of a surface defined by $0=\phi^2 t^4-x^2-y^2-z^2$? This is a follow up to this post.  We've worked out the formula for a surface (still haven't got a name for it), and now I want to know if there's a metric that can be made from:$$0=\phi^2 t^4+x^2+y^2+z^2$$Unlike the Minkowski metric, this one has a dependence on the absolute value of $t$, so it's almost like the metric needs to be some function of $t$ (e.g. $ds^2(t)= <some formula >$), but that's where I get stuck.  Ideally, I'd like to find an expression for the curvature of the surface we've defined.

OK, I worked this out on a spreadsheet and it seems to give the desired result (a formula for the distance between two points on this surface).  Let me know if there's something I got wrong:$$ds^2(t)=dt^2a_0^2(dt-2t)^2-dx^2-dy^2-dz^2$$I don't know if the notiation is right, but I know the metric depends on the time.  If there's a way to express this independent of the time, I'd love to see it.