Show that at some time the hour hand of the first clock points to the tip of the hour hand of the second clock Consider two round clocks of different sizes lying on a table:

As shown on the picture, the clocks can be oriented differently but they are both set to the same time.

The problem is to show that at some time of the day, the hour hand of the first clock points to the tip of the hour hand of the second clock.

Implicit assumptions:

The hands rotate continuously
The clocks don't overlap

My try: This seems like an application of the intermediate value theorem, but I find it very hard to make my argument rigourous.

Intuitively, if we extend the hour hand of the first clock infinitely, it swaps the entire second clock in some portion of the day that last less than 6 hours. During that time, surely it must meet the tip of the hour hand of the second clock. But how to prove this rigourously.

This assume that the clocks don't overlap.
Let $O$ denote the cneter of the first  clock
If the clocks don't overlap, than by convexity of the second clock, there exists a line $\ell$ through $O$ such that the second clock is entirely in one of the half planes defined by it.
Fix a system of coordinates with origin $O$ at the centre of the first clock, and with $\ell$ as $y$-axis and with the second clock in the negative $x$ half plane.
Let $a(t)$ denote the position vector of the tip of the first clock's hour hand at time $t$, similarly $b(t)$ for the second clock.
Let $n(t)$ denote the normal vector to $a(t)$ in clockwise direction, or more formally just let $n(t)=a(t+3\,\mathrm h)$.
Now define the function $f(t)=\langle n(t),b(t)\rangle$.
We know that $a(t)$ and $n(t)$ describe a circle around $O$, especially that there are times $t_+$ and $t_-$ where $n(t_+)$ is on the positive $x$-axis and $n(t_-)$ is on the negative $x$-axis, and these times are uniquely determined up to multiples of twelve hours. Then clearly $f(t_+)<0$ and $f(t_-)>0$. By the 12 hour periodicity, we may assume wlog. that $t_-<t_+<t_-+12\,\mathrm h$.
As $f$ is continuous ($a,n,b$ are continuous in $t$ by assumption and the scalar product is continuous), we conclude that $f$ has a zero $\tau_1$ in $(t_-,t_+)$ and a zero $\tau_2$ in $(t_+,t_-+12\,\mathrm h)$. At a zero of $f$, $a(t)$ must either point to $b(t)$ or away from it. As $\tau_1,\tau_2$ are separated by $t_+$, $a(\tau_1$ and $a(\tau_2)$ are in different half planes, at the one belonging to negative $x$ values we must have that $a(\tau)$ points to $b(\tau)$.

Note that the claim indeed becomes false if we allow the clocks to overlap. Foe example, put the in the same position but slightly rotated against each other.

I think a more general result (with Dali-like molten clocks) holds, bu there are a number of subtleties involved:

Conjecture. Let $a,b\colon S^1\to\mathbb C$ be continuous with nonintersecting images, $a[S^1]\cap b[S^1]=\emptyset$. Assume that $\mathbb C\setminus (a[S^1]\cup b[S^2])$ has among others two components $U_\infty$ and $U_0$ where $U_\infty$ is not bounded and $\partial U_\infty$ intersects both $a[S^1]$ and $b[S^1]$, and $U_0$ is bounded and $\partial U_0$ intersects $a[S^1]$.
Assume  $0\in U_0$. Then there exists $t\in S^1$ such that $b(t)=\lambda\cdot a(t)$ for some $\lambda>1$.

Let the tip of the hour-hand on the big click simply be a point on the plane.

Let the small hour-hand represent a line originating from another point on the plane (the center of the small clock). That line originates at an angle.

Angles are continuous, so it follows that for every single point on the plane, you can reach it with a straight line from the center of the small clock.

(The mental picture you should have is a single point with an arbitrarily long line extending from it, representing the small clock and hand. Over to the top-right, you have a circle, representing the possible positions of the tip of the hour hand of the big clock.)

Then you want to show that you can sweep that line in one clockwise revolution around the center of the small clock, and will certainly hit the hour-hand point of the big clock.

And that is clear: the point of the hour-hand on the big clock is continuously moving on a circle. We can use the line to bisect the circle into two parts: the part it has gone over (A) and the part it has not gone over (B). At every time, the point will be either in A or in B.

We start out with A being empty and B being all the points. Later, the line will have swept the entire circle, so B will be an empty set and A will contain all the points. That means that the point of the hour hand will move from set B to set A at some point. And the only way this move is possible is if it is 'swept' by the line, i.e. if it intersects the line, i.e. if it is pointed to by the small hour hand.


100 blue-eyed islanders puzzle: 3 questions I read the Blue Eyes puzzle here, and the solution which I find quite interesting. My questions:

What is the quantified piece of information that the Guru provides that each person did not already have?
Each person knows, from the beginning, no fewer than 99 blue-eyed people to be on the island. Then how is considering the 1 and 2-person cases relevant, if each person can dismiss these 2 cases immediately as possibilities?
Why must they wait 99 nights if, on the first 98 or so of these nights, they're simply verifying something that they already know?

EDIT: Most answers seem to concentrate on question 1 which I understand partly: but I remain confused because of different answers.
Can someone answer questions 2 and 3?

Here's the story of one blue-eyed islander. The Guru said she saw someone with blue eyes. He looked around and thought "Hey, I don't see anyone with blue eyes. I guess she means me." And so he left right away.

Here's the story of two blue-eyed islanders. The Guru said she saw someone with blue eyes. They looked around and thought "OK, I see someone with blue eyes. I guess she means him," and they stayed. But the next day came, and they thought "Hey, that blue-eyed guy didn't figure it out. I guess he must have seen someone else with blue eyes, but I don't see anyone else with blue eyes. I guess that means me." And so they left together on the second day.

Here's the story of three blue-eyed islanders. The Guru said she saw someone with blue eyes. They looked around and thought "OK, I see two people with blue eyes. I guess she means one of them," and they stayed. A day passed, and nobody left, and they thought to themselves "OK, this is the day those two guys figure it out." But another day passed, and nobody left. The blue-eyed people thought "Wait; those two guys didn't figure it out yet. I guess they must have seen another person with blue eyes, but I don't see anyone else with blue eyes. I guess that means me." And so they left together on the third day.

Here's the story of four blue-eyed islanders. The Guru said she saw someone with blue eyes. They looked around and thought "OK, I see three people with blue eyes. I guess she means one of them," and they stayed. A day passed, and nobody left, but they were not worried; they knew it would take a couple of days. A second day passed, and nobody left, and they all thought to themselves "OK, this is the day those three guys figure it out." But another day passed, and nobody left. The blue-eyed people thought "Wait; those three guys didn't figure it out yet. I guess they all must have seen another person with blue eyes, but I don't see anyone else with blue eyes. I guess that means me." And so they left together on the fourth day.

...and this is why they have to wait the full 99 days. It's not important that the Guru can see someone with blue eyes, unless there's only one islander. What's really important is that, given that the Guru can see someone with blue eyes, "those blue-eyed guys" should be able to figure it out among themselves, and that takes a specific amount of time for a given number of blue-eyed islanders. It's only when they can't do this for a number of islanders that doesn't include you that it becomes clear you must have blue eyes too.

The quantifiable piece of information provided is that at least one person on the island has to have blue eyes.  Since they are perfect, logical machines, they will count how many individuals have blue eyes, and come to the conclusion that, if no other person has blue eyes, then they must, and they must leave the island.
If, however, they see at least one person with blue eyes, they have only the evidence that one person has blue eyes, and thus, shall remain.  If, however, they see TWO blue-eyed people, they must assume that these two blue-eyed people are assuming that the other is the only one remaining, and thus shall wait for the two to leave.  If, however, they see THREE blue eye people, they must assuem that each blue-eyed person is assuming there are two blue-eyed people, and that each of those two blue-eyed people will leave. Because they are perfectly logical beings, they can make no presumption of their own eye color, but cannot be aware of it either.  But they are aware that other individuals can observe eye colors, and that they will base their own presumptions on those eye colors that they see, so they will presume that their own eye color is excluded, and that each blue-eyed individual also excludes their own eye color.
Is actually a short-form explanation of 2.  In much, much shorter form, it is because they must exclude themselves from the count, even if they have blue eyes.  Ironically, not a single person on the island will know their own eye color until the last day becsause of this exlusion requirement (They MUST stay if they do not have blue eyes), because they assume every blue-eyed person is exluding themselves as well.


If $\tan A+\sec A=e^x$, find $\cos A$ If $\tan A+\sec A=e^x$, find $\cos A$

Here is what I've tried:

\begin{align}&\frac{\sin A}{\cos A}+\frac{1}{\cos A}&=e^x\\
\implies&\frac{1+\sin A}{\cos A}&=e^x\end{align}

Now, I've squared both sides, but, in vain. How should I continue? Am I proceeding wrong? Any help is much appreciated.

HINT:

As $\sec A+\tan A=e^x$

Now  $\sec A-\tan A=\dfrac1{\sec A+\tan A}=?$

Can you solve for $\sec A?$

Finally $\cos A\cdot\sec A=?$

Given
tanA+secA=e^x.   ------------(I)
Calculate.   ( secA-tanA) using tan^2A+1=sec^2A
secA-tanA=1/e^x.  ------------(ii)
Add.  (i)&(ii)
2secA=e^x+1/e^x
And then calculate cosA


Lie bracket of vector fields on $R^2$ Compute the Lie bracket$$\Big[-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y},\frac{\partial}{\partial x}\Big]$$

on $R^2$

Can you help me please?

I am not 100% sure that this is what you want, but I think that
$$\begin{align}
\Big[-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y},\frac{\partial}{\partial x}\Big] &= \left(-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\right)\frac{\partial}{\partial x} - \frac{\partial}{\partial x}\left(-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\right) \\
&= -y\frac{\partial^2}{\partial x^2}+x\frac{\partial}{\partial y}\frac{\partial}{\partial x}  +y\frac{\partial^2}{\partial x^2} - \frac{\partial}{\partial x}\left(x\frac{\partial}{\partial y}\right) \\
&= \dots
\end{align}
$$

Use the product rule on the last part:
$$\begin{align}
\frac{\partial}{\partial x}\left(x\frac{\partial}{\partial y}\right) = \frac{\partial}{\partial y} + x \frac{\partial^2}{\partial x\partial y}
\end{align}
$$

[−y∂/∂x+x∂/∂y,∂/∂x]=(−y∂/∂x+x∂/∂y)(∂/∂x)−(∂/∂x)(−y∂/∂x+x∂/∂y)= - ∂/∂y


Proving an Integral inequality from a given integral inequality Problem: Let $f$ and $g$ be continuous, non-negative function on $[0, 1]$, with
$$\int_{0}^{1}e^{-f(x)}dx \geq \int_{0}^{1}e^{-g(x)}dx. $$
Prove that,
$$\int_{0}^{1}g(x)e^{-f(x)}dx \geq \int_{0}^{1}f(x)e^{-g(x)}dx. $$
This problem is an equivalent form of a problem from the book "More Calculus of a variable" by Peter Mercer. In the original problem there was $xf(x)$ and $xg(x)$ in place of $f(x)$ and $g(x)$.
I couldn't make much progress. I tried contradiction and method which led to conlclude that its enough to show,
$$e^{g(x)}-e^{f(x)}+f(x)e^{f(x)}-g(x)e^{g(x)} < 0.$$ But sadly this is not true in general.
I also tried integration by parts to seperate $\displaystyle\int^{1}_{0}e^{-g(x)}\,dx$ from $\displaystyle\int^{1}_{0}f(x)e^{-g(x)}\,dx$ to use the given inequality. But I didn't happen.
Please help me. I am in  high school so please don't use any very advanced inequality.

From the original inequality:
$$\int_0^1 e^{-f(x)}dx\ge\int_0^1e^{-g(x)}dx \rightarrow f(x)\le g(x)$$
Normally you would want to prove this specifically but if you are in highschool we will just take it as intuitive. Also, we have the statement:
$$e^{-f(x)} \ge e^{-g(x)}$$
Using this:
$$g(x)e^{-f(x)}  \ge f(x)e^{-f(x)}$$
But also:
$$f(x)e^{-f(x)}  \ge f(x)e^{-g(x)}$$
Therefore:
$$g(x)e^{-f(x)}  \ge f(x)e^{-f(x)}\ge f(x)e^{-g(x)}$$
Drop the middle term then integrate:
$$g(x)e^{-f(x)}  \ge f(x)e^{-g(x)} \rightarrow \int_0^1 g(x)e^{-f(x)} dx  \ge \int_0^1f(x)e^{-g(x)}dx$$


Finding $\lim_{x,y \rightarrow 0,0} \frac{x \ln(1+y)}{2x^2+y^2}$ Find bounds for $\lim_{x,y \rightarrow 0,0} \frac{x \ln(1+y)}{2x^2+y^2}$

I am finding maximum and minimum for function and one of critical case is to find possible minimal and maximal value of given function in $0,0$. But how can I do this due to this limit doesn't exists (for example we can take $x,y = {1\over n},{1\over n}$ and $x,y = {2\over n},{1 \over n}$

Every level curve

$$ f(x,y)= \frac{x \ln(1+y)}{2x^2+y^2}=c $$

for $c\le0.353$ passes through $(0,0)$ but not so for $c>0.354$, so the maximal point $(0,0,z_{\text{MAX}})$ should have $z_{\text{MAX}}$ lie somewhere between those two quantities.

For each level curve, the point nearest $(0,0)$ appears to lie along the line $y=\sqrt{2}x$. Replacing $y$ with $\sqrt{2}x$ gives

$$ y=\frac{\ln(1+\sqrt{2}x)}{4x} $$

and

$$ \lim_{x\to0}\frac{\ln(1+\sqrt{2}x)}{4x}=\frac{\sqrt{2}}{4}\approx0.3535533906 $$

This would be the correct maximum if indeed the point on each level curve closest to $(0,0)$ actually lies along the line $y=\sqrt{2}x$.

Let's have $f(x,y)=\dfrac{x\ln(1+y)}{2x^2+y^2}$

For $x\neq 0$ we have $f(x,0)=0\to 0$

On the other hand $f(x,x)=\dfrac{x\ln(1+x)}{3x^2}=\dfrac{\ln(1+x)}{3x}\sim\dfrac x{3x}\to \dfrac 13$

Since $f$ has at least two different limits along different paths, then $f$ has no limit in $(0,0)$


Prove that the infinite series is divergent I would like to know how I proof that the following infinite series is divergent:
$$\sum_{i=1}^n (-1)^n*(\frac{n}{n+1})^n$$
I tried the following tests: ratio test, root test, Leibniz criterion, divergence test, but they lead to an inconculsive result. The other only tests I am allowed to use is the comparison test and Cauchy's convergence test , but I don't know how to proof that the infinite series is divergent with these.

The series has the $n^{th}$ term $a_n=\dfrac{1}{\left(-1-\dfrac{1}{n}\right)^n}$ does not converge to $0$. Thus the series diverges.

Observe
\begin{align}
\limsup_{n\rightarrow \infty}\ (-1)^n \left(\frac{n}{n+1}\right)^n = e^{-1} \neq 0.
\end{align}

Edit: By the request of Ted Shifrin, I have updated my answer.

Heuristically speaking, an infinite series converges provided the terms that you are adding eventually becomes negligible. Otherwise, if you keep on adding ''big" terms to your partial sum
\begin{align}
\sum^N_{n=1} a_n + "\text{infinitely many big terms}"
\end{align}
then there's no hope that the series will converge. For instance consider the example
\begin{align}
-1 + 1 + -1 + 1 + -1+\ldots
\end{align}
We see that if we add an even number of terms then the partial sum  will be $0$, but if we add an odd number of terms the partial sum will be $-1$. Hence our partial sum will fluctuate between $0$ and $-1$, which means the sequence of partial sums will not converge. The reason why the partial sum does not converge is mainly due to that fact that each term we add to the  partial sum is not negligible. In fact the new term that we added to the partial sum greatly affects the value of the partial sum.

Back to the problem. In the given series, we see that for large enough $N$ the series looks more or less like the following
\begin{align}
\ldots + e^{-1}-e^{-1} + e^{-1}-e^{-1} + \ldots
\end{align}
which is almost exactly the same as our above example. Hence there's no way the given series converges.


Compact operators on Hilbert Space I m working on the following problem:

Let $K:H\rightarrow H$ be a compact operator on a Hilbert space. Show that if there exists a sequence $(u_n)_n\in H$ such that $K(u_n)$ is orthonormal, then $|u_n|\rightarrow \infty$.

Here is my argument: It is suffice to show for all $u\in H$, there exists $M>0$ such that $|u_n-u|\geq M$, that is $(u_n)$ has no limit in $H$. Suppose $u_n\rightarrow u$ (all the convergence are strong in my argument), then $Ku_n\rightarrow Ku$. I claim that we can extract a subsequence $(Ku_{n_k})_k$ that is divergent. Denote $E_k=\overline{span\{Ku_{n_k}\}}$,  and we may choose $Ku_{n_k}$ such that $||Ku_{n_k}||=1$ (trivial) and $dist(Ku_{n_{k+1}}, E_k)\geq 1/2$. So $|Ku_{n_k}-Ku_{n_{k+1}}|\geq 1/2$ for all $k$, which violates the convergence of the sequence $Ku_n$.

Since I did not use the fact that $K$ is a compact operator, there must be something wrong with the proof. Can someone let me know which part is problematic? Any hints for the right approach?

I'd proceed as follows:

An orthonormal set has no limit point.
Therefore, any set containing an infinite orthonormal set is not compact.
If $|u_n|$ is bounded, then $K$ maps a bounded set to a set whose closure is not compact.

Your proof got confused in the first step

Remember that a compact operator is continuous, You use continuity when said
$Ku_{n}\rightarrow Ku$. Thus the proof is fine.


If $f:[0,\infty)\to\mathbb R$ is continuous and $\tau:=\inf\left\{t\ge0:|f|\ge\varepsilon\right\}\in(0,\infty)$, is $|f(\tau)|=\varepsilon$? Let $f:[0,\infty)\to\mathbb R$ be continuous and $$\tau:=\inf\left\{t\ge0:|f|\ge\varepsilon\right\}$$ for some $\varepsilon\ge0$. Assume $\tau\in(0,\infty)$. Is it possible to conclude $|f(\tau)|=\varepsilon$?

Clearly, since $\tau\in(0,\infty)$, there are $t_{<\varepsilon},t_{\ge\varepsilon}\ge0$ with $$|f(t_{<\varepsilon})|<\varepsilon$$ and $$|f(t_{\ge\varepsilon})|\ge\varepsilon$$ and hence $\tau\in(t_{<\varepsilon},t_{\ge\varepsilon}]$. If $|f(\tau)|>\varepsilon$, then there would be a $t_\varepsilon\in(t_{<\varepsilon},\tau)$ with $$|f(t_\varepsilon)|=\varepsilon;$$ in constrast to the definiton of $\tau$. So, $$|f(\tau)|\le\varepsilon.\tag0$$

If $|f(\tau)|>\epsilon$ then there is a neighborhood $U\subset (0,\infty)$ of $\tau$ such that $|f(x)|>\epsilon$ for $x\in U$. Therefore there are points $<\tau$ in which $|f(x)|>\epsilon$.

Likewise, if $|f(\tau)|<\epsilon$ there is a neighborhood $U\subset (0,\infty)$ of $\tau$ in which $|f(\tau)|<\epsilon$ holds, contradicting the definition of $\tau$ because $\{t\geq0:\ |f(t)|\geq \epsilon\}\subset U^c$.

Let $$I:=\left\{t\ge0:|f(t)|\ge C\right\}.$$ Assume $$|f(\tau)|<C.\tag1$$ Let $$\varepsilon:=C-|f(\tau)|>0.$$ $f$ is continuous at $\tau$ and $\tau>0$ $\Rightarrow$ $\exists\delta\in(0,\tau)$ with $$f(B_{\delta}(\tau))\subseteq B_{\varepsilon}(f(\tau))\tag2$$ and hence $$|f(t)|\le|f(\tau)|+|f(t)-f(\tau)|<C\;\;\;\text{for all }t\in B_\delta(\tau)\tag3.$$ Thus, $$I\subseteq[0,\infty)\setminus B_\delta(\tau)$$ and hence $$\forall t\in I:|t-\tau|\ge\delta\tag4.$$ $\tau$ is a lower bound for $I$ $\Rightarrow$ $$\forall t\in I:\tau\le t\tag5$$ and hence (since $t\le\tau-\delta<\tau$ is in contradiction to $(5)$ for all $t\in I$) $$\forall t\in I:t\ge\tau+\delta;\tag6$$ in contradiction to the definition of the infimum. By $(0)$, we obtain $$|f(\tau)|=C.$$


How to find the reminder when $982^{40167}$ is divided by 15? I probably have to use Euler's function but I'm not sure how.

$$\phi(15) = 8$$ which means $$a^8 \equiv 1\pmod {15}$$

$$ 40167 = 7 + 5020 \times 8$$

And $$982 = 65 \times 15 + 7$$

$$982^{40167} \equiv 7^{7} \equiv 13 \pmod {15}$$

EDIT: one may figure out how to calculate $7^{7}$

The story is really simple:

$$7^2 \equiv 49 \equiv 4 \pmod{15} $$

Therefore,

$$7^4 \equiv 4\times4 \equiv 1 \pmod{15} $$

$$7^7 \equiv 7^4 7^2 7 \equiv 1\times 4 \times 7 \equiv 13 \pmod {15}$$


Conditional expectation with multiple conditioning For any r.v.s $X$ and $Y$:
$$E(Y|E(Y|X)) = E(Y|X)$$
But I cannot seem to be able to prove this. I tried using Adam's Law with extra conditioning ($E(Y|X) = E(E(Y|X,Z)|Z)$) but I don't seem to get anywhere with it.
What I tried is the following:
$$g(X) = E(Y|X)$$
$$E(Y|g(X)) = E(E(Y|X,g(X))|g(X))$$
Since the event $X$ happened and $g(X)$ happened are equivalent, conditioning on both $X$ and $g(X)$ is the same as conditioning on only one of them.
Is there any intuitive interpretation of this ?
Does this also mean that conditioning on $X$ or any function $g$ of $X$ is the same ?

The argument you already have is a pretty good non-measure theory argument.  I will just formalize that below, it may help to give confidence about some details.
Using your argument structure: Let $g(X)=E[Y|X]$.  Then
\begin{align}
E[Y|g(X)] &\overset{(a)}{=} E[E[Y|g(X),X]|g(X)]\\
&\overset{(b)}{=} E[E[Y|X]|g(X)]\\
&=E[g(X)|g(X)]\\
&\overset{(c)}{=}g(X)
\end{align}
where (a) uses the law of iterated expectations; (b) uses $E[Y|g(X),X]=E[Y|X]$; (c) uses $E[Z|Z]=Z$ for any random variable $Z$. $\Box$

The step (b) more closely examined is:
$$E[Y|g(X),X]=E[Y|X]$$
and this intuitively means that if we already know $X$, then the additional information $g(X)$ adds nothing new.

Notes:

Conditioning on $X$ is generally not the same as conditioning on $g(X)$, but it works in this particular problem.

A measure-theory derivation could be given along the lines of my first comment on your answer.  You can also justify $E[Y|g(X),X]=E[Y|X]$ more formally by measure theory ("the sigma algebra generated by $(g(X),X)$ is the same as the sigma algebra generated by $X$").

A formal measure theory definition talks about "versions of" a conditional expectation, and I do not go into such detail in this answer (some people may want to replace my equalities with equalities that hold "with probability 1").

Given the level of formality of the book, I think what the exercise is going for is primarily conceptual. I.e., what does the below conditioning mean:
E(Y|E(Y|X), X) | X = x
This represents the expectation of Y, if I know what the expectation of Y given X would be... and now I also know X! So I'm exactly in the situation where I know X, and I know what the expectation of Y given X is, which is what the expectation of Y given that conditional information is. I.e., E(Y|X). And then you Adam's Law it up via taking out what's known given X, which E(Y|X) is as that's a function of X.
(So it's more specific than just any g(X) - its essential to the argument that g(X) = E(Y|X) in particular.)


Is there any expression to calculate the sum of (at least) 3 cosines? I'm envolved in a waves problem and I have to calculate $\cos(A)+\cos(B)+\cos(C)$, where $A$, $B$ and $C$ are independent angles. I want to find a expression similar to the sum-product identity: $$\cos(A)+\cos(B)=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) $$

I have tried the same path that leads you to the mentioned identity, but it is different. In the two cosines case it is easy because you loose two terms as they have opposite signs, however, in the three cosines case, you can't loose them as they are odd.

Any help is welcome!

Thank you for your time :)

EDIT:
Someone told me that you can use:
$$
cos(A)+(cos(B)+cos(C))+(cos(A)+cos(B))+cos(C)+cos(B)+(cos(A)+cos(C))
$$
And use the sum-product identity in those sums in brackets in order to obtain the triple of the sum we want to obtain, so we obtain:
$$
cos(A)+cos(B)+cos(C)=\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)+\cos\left(\frac{B+C}{2}\right)\cos\left(\frac{B-C}{2}\right)+\cos\left(\frac{A+C}{2}\right)\cos\left(\frac{A-C}{2}\right)
$$
While it is not a product, it is an interesting thing.

A classmate found this.

If you write:
$$
[cosA+cosB]+cosC=[2cos(\frac{A+B}{2})cos(\frac{A-B}{2})]+cosC
$$
$$
cosA+[cosB+cosC]=...
$$
$$
cosB+[cosA+cosC]=...
$$
And you do the same as done in the first equation, but in the following ones, and them sum up, you obtain the following expression:
$$
cosA+cosB+cosC=cos(\frac{A+B}{2})cos(\frac{A-B}{2})+cos(\frac{B+C}{2})cos(\frac{B-C}{2})+cos(\frac{A+C}{2})cos(\frac{A-C}{2})
$$

It is not a unique product, but is the most near solution I got.
Thank you all. I like this place, it is a nest of creativity.

Well, I was trying to sleep but an idea came to my mind xD
If you use the sum-product identity twice:

$$
cos(A)+cos(B)+cos(C)+cos(D)=2cos(\frac{A+B}{2})cos(\frac{A-B}{2})+2cos(\frac{C+D}{2})cos(\frac{C-D}{2})
$$
Now, if we want to have the sum of the first three cosines, we just have to do $ D= \frac \pi 2$, so the cosine of D will equal zero. Then:
$$
2cos(\frac{C+D}{2})cos(\frac{C-D}{2})=2\left(cos(\frac C 2)cos(\frac \pi 4)-sen(\frac C 2)sen(\frac \pi 4)\right)\left(cos(\frac C 2)cos(\frac \pi 4)+sen(\frac C 2)sen(\frac \pi 4)\right)=\frac{2}{\sqrt{2}}\left(cos(\frac C 2)-sen(\frac C 2)\right)\left(cos(\frac C 2)+sen(\frac C 2))\right)
$$
Where we used the cosine of a sum of angles. And if we use the equation for the cosine of a double angle:
$$
{\sqrt{2}}\left(cos^2(\frac C 2)-sen^2(\frac C 2)\right)={\sqrt{2}}cos(C)
$$

So the final result should be:
$$
cos(A)+cos(B)+cos(C)=2cos(\frac{A+B}{2})cos(\frac{A-B}{2})+{\sqrt{2}}cos(C)
$$

However, there is something very strange in this equation as we can use the sum-product identity again to remove cos(A)+cos(B) and we will obtain:
$$
cos(C)={\sqrt{2}}cos(C)
$$
What can only be true if cos(C)=0 and then $C=\frac \pi 2$

Why does this happen?


Solve for $y$ in the given equation Solve:
$$y^{y\sqrt {y}}=(y\sqrt {y})^{y}$$

My Attempt:
$$y^{y\sqrt {y}}=(y\sqrt {y})^{y}$$
$$y^{y^{\dfrac {3}{2}}}=(y^{\dfrac {3}{2}})^y$$

How do I solve further?

The domain is $y>0$.

We need to solve
$$y^{y\sqrt{y}}=y^{\frac{3}{2}y},$$
which gives $y=1$ or
$$y\sqrt{y}=\frac{3}{2}y,$$
which is $y=\frac{9}{4}$ and we get the answer: $\{1,\frac{9}{4}\}$.


Calculate limit with floor function or prove it doesnt exist Please help calculating the following limit:

$$  \lim_{x\to\frac{\pi}{2}} \frac{\lfloor{}\sin(x)\rfloor}{\lfloor x\rfloor}  $$

I used $$ t = x - \frac{\pi}{2} $$ and got:

$$  \lim_{t\to 0} \frac{\lfloor{}\sin(t+\frac{\pi}{2})\rfloor}{\lfloor t+\frac{\pi}{2}\rfloor}  $$

for t close to 0 we get from arithmetic of limits that the denominator is 1 but not sure how to go from here..

Thanks

The floor function is continuous on $\Bbb R\setminus \Bbb Z$ so
$$\lim_{x\to\frac\pi2}\lfloor x\rfloor=\left\lfloor\frac\pi2\right\rfloor=1$$
and since

$$0\le \sin x<1\,\forall x\in I:=(0,\pi)\setminus\{\frac\pi2\}$$
hence $\lfloor \sin x\rfloor=0\;\forall x\in I$ so the desired limit is $0$.

In a punctured neighborhood of $\pi/2$ we have $\lfloor\sin x\rfloor=0$ and $\lfloor x\rfloor=\color{red}1$, so the limit is $0$.


Measure 11 liters using bottles of 16, 6, and 3 liters This question has been bugging me for a day and finally I gave up and decided to ask the community for it so here's how it goes:

Suppose we have 3 bottles with capacities of $16,6$ and $3$ liters, all of which are full of water currently.

In what way can we move the water from a bottle to another one (spilling is also an option) that only $11$ liters of water remain in the $16$-liter bottle?

EDIT:Thanks for your comments and now I believe you may have misinterpreted my question and therefore I think I'll unmark the answer for now(which I realized was not true after a while).

EDIT2:I forgot to mention the following conditions but they pretty much could be interpreted from the question iteself:

We cannot measure a bottle half-way.
The only scales we have are only these 3 bottles (This one is more like an emphasize on the first condition but anyway I needed to add it)
There's no reservoir where we can keep the spilled water.
There are no other water resources from which we can add water.

and finally I need to point out the fact that the question may or may NOT have an answer so it's more like a "prove or disprove" kind of question.

EDIT3:Multiple transfers are allowed (Thanks to @David for pointing it out)

Thanks for your help in advance.

It is trivially possible. Just spill water from the $16$ litre bottle until it is empty. By the intermediate value theorem, there is an instant when the there is exactly $11$ litre left.

Nothing in the question indicates that this is about a construction, so an abstract existence proof will do.

Discard all the water from $3$ liter bottle
Fill it full with water from $16$ liter bottle
Now, $16$ liter bottle has $13$ liter water, mark the level of the water
Discard all the water from $3$ liter bottle
Fill it full with water from $16$ liter bottle
Now, $16$ liter bottle has $10$ liter water, mark the level of the water
Divide the portion of the bottle between $10$ and $13$ mark into $3$ equal parts
Fill one part with water from any one of the other two bottles


How to write n number of loops mathematically I have the following code. I want to write it mathematically. Any help would be appreciated.

int in1=1, in2=-1;
double totalDistance=0.0;
double min_dis = Double.MAX_VALUE;
for(int i=0;i<N1;i++){
    for(int j=0;j<N2;j++){
        dis = calculateDistance(Solutions(i),Solutions(j));
                totalDistance = dis;
                if(totalDistance< min_dis){
                    min_dis=totalDistance;
                    in1=i;
                    in2=j;
            }
        }
    }
print(in1, in2)

Mathematically, I write it as follows:
\begin{equation}
argmin_{i,j \in \{1,2, \dots, n_1\} \times \{1,2, \dots, n_2\}} dis(S_i, S_j)
\end{equation}
$\times$ refers to Cartesian product between sets and ${dis(S_i, S_j)}$ is the distance.

I want to know is it correct mathematical representation. If it is correct, I want to extend it to n loops. Following is an example of 3 loops.

int in1=1, in2=-1, in3=-1;
    double totalDistance=0.0;
    double min_dis = Double.MAX_VALUE;
    for(int i=0;i<N1;i++){
        for(int j=0;j<N2;j++){
            double dis1 = calculateDistance(**S1**(i), **S2**(j));
            for(int z=0;z<N3;z++){
                double dis2 = calculateDistance(**S2**(j), **S3**(z));
                totalDistance = dis1+ dis2;
                if(totalDistance< min_dis){
                    min_dis=totalDistance;
                    in1=i;
                    in2=j;
                    in3=z;
                }
            }
        }
    }
    print(in1 in2 in3);

How can I represent the generalized version in mathematical notation?

Thanks in advance.

The mathematical expression is not correct. Here we focus on the first code snippet. The consideration for the second part and the general case is similarly.

  Some aspects need to be considered in more detail:

  There may be more than one pair with minimum distance, while the code returns always a single element.
  The order of the for-loops: This is important in case there are more than one element with minimum distance.
  Technical detail: The index range is $0\leq i < N$ instead of  $1\leq i\leq N$.

In the following we use the notation
\begin{align*}
\operatorname{calculateDistance(Solutions(i),Solutions(j))}\quad&\longrightarrow\quad d(S_i,S_j)\\
\operatorname{N1}\quad&\longrightarrow\quad N_1\\
\operatorname{N2}\quad&\longrightarrow\quad N_2\\
\operatorname{in1}\quad&\longrightarrow\quad i_1\\
\operatorname{in2}\quad&\longrightarrow\quad i_2\\
\end{align*}

  First step: $2$ loops

  Let $N_1,N_2>0$. We obtain,    assuming symmetry of the distance function

  \begin{align*}
A_2&:={\operatorname{argmin}}_{0\leq i_2<i_1<\max\{N_{i_1},N_{i_{2}}\}}\{d(S_{i_1},S_{i_2})\}\\
&=\{(i_1,i_2):d(S_{i_1},S_{i_2})\leq d(S_{j_1},S_{j_2}),0\leq i_1,j_1<N_1, 0\leq i_2,j_2<N_2\}\\
\\
B_2&:=\min_{0\leq i_2<N_2}\min_{0\leq i_1<N_1}\{(i_1,i_2):(i_1,i_2)\in A_2\}
\end{align*}
  with $B_2$ consisting of the one element which is returned by the code.

There may be more than one pair with the same minimum distance, so that $A_2$ will contain usually more than one element. Assume
the set $A_2$ contains the four elements
\begin{align*}
A_2=\{(4,3),(4,5),(8,2),(10,6)\}
\end{align*}
In this case the code will return $(4,3)$. Since $4$ is the smallest value of the first coordinates, the candidates from $A_2$ are $(4,3)$ and $(4,5)$. Out of these two values the code selects the one with the lowest second coordinate giving $(4,3)$.

This implies that the order of the minimum settings in $B_2$ is essential. Note that
\begin{align*}
B_2=&\min_{0\leq k_2<N_2}\min_{0\leq k_1<N_1}\{(k_1,k_2):(k_1,k_2)\in A_2\}=\{(4,3)\}\quad\text{while}\\
&\min_{0\leq k_1<N_1}\min_{0\leq k_2<N_2}\{(k_1,k_2):(k_1,k_2)\in A_2\}=\{(8,2)\}\\
\end{align*}

  General step: $k$ loops

  Let $N_1,N_2,\ldots,N_k>0,k\geq 2$. We obtain

  \begin{align*}
A_k&:={\displaystyle{{\text{argmin}}_{{0\leq i_l<i_{l+1}<\max\{N_{i_l},N_{i_{l+1}}\}}\atop{1\leq l\leq k-1}}}
\left\{\sum_{j=1}^{k-1}d(S_{i_j},S_{i_{j+1}})\right\}}\\
\\
B_k&:=\min_{0\leq i_k<N_k}\min_{0\leq i_{k-1}<N_{k-1}}
\cdots\min_{0\leq i_1<N_1}\{(i_1,i_2,\ldots,i_k):(i_1,i_2,\ldots,i_k)\in A_k\}
\end{align*}
  with $B_k$ consisting of the one element which is returned by the code.

Code review:

$\operatorname{in1}$ should be initialised with $-1$
Since both indices $i$ an $j$ are the argument for the call $\operatorname{calculateDistance(S(i),S(j))}$ and the call is symmetrical with respect to the arguments it is not sound, that $i$ and $j$ have different range. We would rather assume
\begin{align*}
&0\leq i_1,i_2< N_1\\
\end{align*}
We expect a distance function $d$ to act symmetrically: $d(x,y)=d(y,x)$ for all $x,y$ and also $d(x,x)=0$ for all $x$. So, we would rather code
\begin{align*}
&0\leq i_2 < i_1<N_1\\
\end{align*}

Define the set of minimizers:

$S = \arg\min_{ x \in \mathbb{N}^n} \{ \sum_{i=0}^{n-2} dis(S_{x_i},S_{x_{i+1}}) : x_i < n_i \; i = 0,1,\ldots,n-2 \}$

The solution is the lexicographic first element of $S$:

$\min\{x \in S \}$

where $\min$ performs lexicographic minimization. This returns the same element of $S$ that your code outputs.


How to solve this absolute value inequality? $ |x| + |x - 2| \gt 5 $ I'm not sure how to solve this inequality. Can someone please explain step-by-step? Thanks!

$ |x| + |x - 2| \gt 5 $

Since $\mid x\mid$ and $\mid{x-2}\mid$ will always be positive, you can change the equation to $x+x-2 \gt 5$. This gives you $2x-2\gt 5$
then  $2x \gt 7$ and the final answer is $x \gt \frac72$
Also since x can be negative you must set another equation as $|x|+|x-2| \lt -5$ and solve this as $x+x-2 \lt -5$ and get $2x-2 \lt -5$ then $2x \lt -3$ then $x \lt -\frac{3}{2}$ so $x \gt \frac72$ or $x \lt -\frac{3}{2}$


Determine whether a set is a vector space, problem with axioms 4 & 5 Vector spaces have been confusing me for a while, I haven't really grasped a clear image of the idea yet.

The question is simple though

  Determine whether the set of real numbers (x,y) with operations
  $$(x_1,y_1)\oplus(x_2,y_2) = (x_1+x_2+1, y_1+y_2+3)$$  $$ k  \cdot(x,y) =  (kx,y) $$ is a vector space or not. Check all axioms.

The axioms that passed were 1,6,2,3,9 & 10.
The axioms that failed were 7,8

The axioms that confused me were 4 & 5.

In axiom 4 the rule is that there should exist $\vec 0$ such that $$ \vec0 + \vec u =  \vec u$$

Here's where it confused me, is $\vec 0$ always $= (0,0)$?

Because in this case it wouldn't work, since
$$\vec0 + \vec u = (0,0) + (x_1,y_1)$$
$$= (x_1+1, y_1+3)$$
Unless $\vec 0$ = (-1,-3) then this axiom would be true.

But then when I checked axiom 5, where the rule is
$$-\vec u + \vec u = \vec 0$$
If we apply it we get
$$-\vec u + \vec u = (-x_1,-y_1) + (x_1,y_1)$$
$$= (1,3)$$

So first I don't know if the zero vector could be something different and in this case it doesn't help. Second, in axiom 5, is the negative in $-\vec u$ considered as a scalar and should be treated it as such? So is $-\vec u$ actually $(-x_1, y_1)$ ? Or is it correct as I previously did?

Axiom 4. For each element $x$ in $V$ there is an element $y$ in $V$ s.t.
$$x+y=0_V.$$

In your case $0_V = (-1, -3),$ so if $u=(a,b), v=(-a-2,-b-6)$, clearly
$$(a,b)\oplus(-a-2,-b-6)=(-1,-3)=0_V.$$
Your error is that you directly said $\large-\vec{u}=-1\color{blue}{\cdot_R}\vec{u},$ then you're using an operation not defined! And even if you correctly use the defined operation,
$$\large-1\cdot \vec{u}=(-x_1,y_1)\color{red}{=_?}-\vec{u},$$
you haven't prove it!
In short, $\Large-\vec{u}$ means nothing more than it's the inverse of $\Large\vec{u}.$

Well first, if you get one axiom to fail then it is not a vector space . Second ,O^-->  does not have to be (0,0) ,it can be any vector (pair) with U satisfying U circled addition V=V for every vector (pair) ( .Third given U=(a,b) ,you must find a vector  W =(c,d) which ,when added to U (add means using your circled addition).
(p,q)+(r,s)=(p+q,r+s) is NOT the addition used for this problem .
  In the definition and list of axioms for a vector space the addition is always denoted by '+' .but you must change this to circled '+'.


Total number of injective functions The number of injective functions possible from A to B such that  p'th element of A cannot map with p'th element of B where |A|=3 and |B|=5 is ?

My attempt:-

Total number of injective functions possible from A to B = 5!/2! = 60.

1) Number of ways in which one element from set A maps to same element in set B is
   (3C1)*(4*3) = 36.

2) Number of ways in which two elements from set A maps to same elements in set B is
  (3C2)*(3) = 9.

3)Number of ways in which three elements from set A maps to same elements in set B is 1

So, answer should be 60-(36+9+1) = 14.

But it seems that my answer is wrong. Can someone point out the mistake in my approach ?

For clarity, let $A = \{1, 2, 3\}$ and let $B = \{1, 2, 3, 4, 5\}$, as @drhab suggested.

You did not apply the Inclusion-Exclusion Principle correctly.

The correct answer is $60 - 36 + 9 - 1 = 32$.

When we subtract those cases in which one element of $A$ is mapped to the corresponding element of $B$, we have subtracted those cases in which two elements of $A$ are mapped to corresponding elements of $B$ twice, once for each way we could designate one of those elements as the element of $A$ that is mapped to the corresponding element of $B$.

Since we only want to exclude those cases in which two elements of $A$ are mapped to corresponding elements of $B$ once, we must add those cases back.

However, we have not excluded the case in which all three elements of $A$ are mapped to the corresponding elements of $B$ since we subtracted them three times, then added them three times.  We subtracted them three times when we counted those cases in which one element of $A$ is mapped to the corresponding element of $B$, once for each way we could designate one of the three elements as the one that is mapped to the corresponding element of $B$.  We added them three times when we counted those cases in which two elements of $A$ are mapped to the corresponding elements of $B$, once for each of the $\binom{3}{2}$ ways we could designate two of the three elements as the elements of $A$ that map to the corresponding elements of $B$.  Therefore, we must subtract the case in which all three elements of $A$ are mapped to the corresponding  elements of $B$.

Since you have 5 different choices for 3 different numbers.
The first element in A has 5 choices from B.
Now, as the first element has chosen one element in B, you will only have 4 choices left in B.
So, the second element only has 4 choices from b.
And, the final element will have 3 choices.
By the principle of multiplication,
There are 5*4*3 = 60 total injective functions.

This can be summed up as |A||B|.

Hope that helped.


Injective and surjective proof. Let  $f:\mathbb{R}\to\mathbb{R}$ be defined by $f(x)=-2x^2+7x+15$.

a) Is $f$ injection or surjection?

b) Determine $f(\mathbb{R})$.

c) How much results has the Equation $f (x) = b$ as a function of $b\in\mathbb{R}$?

d) Enter two largest possible domains $A\subset\mathbb{R}$ and matching image regions $B \subset\mathbb{R}$ so that $f : A \to B$ is bijective and determine the respective inverse function.

Can someone please explain me how to see if function is injection or surjection when we have x^2. What means to determine f(R) and how to solve that?

Well here is how I would go about it.

Since every parabola is symmetric about some line then it can not be injective,since it takes on every value twice,namely once before intersection with axis,and once after.Thus not injective.

Also from negative sign next to largest degree it is obvious that this function has a maxima.Thus any number larger then its maxima is not value of the function.Thus for some real number large enough there is no x such that it is equal to f(x).

To determine the image of the function it is only necessary to determine the maxima of the function(let it be called t from now on),then since function is not bound below then its image would be set off all numbers less than or equal to its maxima.

For D the largest injective domain would be either all x up to maximum point,or all x after maximum point,and the range would be still the image of f under R.This comes due to symmetry.

Find a counterexample to prove that the function $f(x)$ is not an injective nor surjective function.

Say we have $-1.5,5 \in \mathbb{R}$, then $f(-1.5)= f(5) \in \mathbb{R}$, but $-1.5 \neq 5$, hence it is not injective.
Say we have $22 \in R$. There is no $x \in \mathbb{R}$ such that $-2x^2+7x+15 = 22$, hence it is not surjective.

Let the range be $Y:=\{y \in \mathbb{R},\;  y=-2x^2 + 7x +15\}$. Intuitively, what are all possible values of $f(x)$ for all $x \in \mathbb{R}$?


Stabilising chain of submodules Take $S$ to be a commutative ring with unit, and $N \subset M$ to be $S$-modules. I am interested in the necessary and sufficient condition for it to be the case that for every collection of submodules $\{M_i\}_{i=1,\dots,n}$, such that $M_i \cap N = 0$, they form a stabilising chain,
$$M_1 \subset M_2 \subset \cdots \subset M_{n}.$$
I believe a necessary and sufficient condition is that every $M_i$ such that $M_i\cap N=0$ is finitely generated. I know that if every submodule was finitely generated, then $M$ would be Noetherian, and thus I am guaranteed this stabilising chain.
However, if I am only requiring that $M_i$'s such that $M_i \cap N = 0$ be finitely generated, this does not exclude the possibility of submodules which are not finitely generated to exist in $M$ so $M$ is not being required to be Noetherian, and so I am not guaranteed this stabilising chain. How can I approach this?

EDIT: This answer is completely wrong, so don't bother reading it. I'll leave it up for the record; details are in the comments, but I amateurishly confused the set $M \setminus N \cup \{ 0 \}$ with the submodule that it generates.
Let $M_N$ be the smallest submodule of $M$ containing $\left( M \setminus N \right) \cup \{ 0 \}$. The submodules of $M_N$ are exactly the submodules $M_i \leq M$ s.t. $M_i \cap N = \{ 0 \}$. The property in question is thus equivalent to whether $M_N$ is Noetherian or not. There are two cases.

Trivially, $M_N$ could simply be $M$ itself, i.e. $M$ has a generating set in $M \setminus N$.
$M_N$ could be a proper submodule of $M$. Since $M \setminus M_N = N \setminus \{ 0 \}$, this works out to say that at least one non-identity of element $N$ cannot be written as a linear combination of elements of $M_N$ over $S$.


Constructing Infinite countable bounded subset of irrationals without accumulation point I was solving some problems in countabilty,That's when this question arises.

$\mathbb{R}-\mathbb{Q}$ is uncountable, I think of a countable subset of irrationals, obviously finite set is countable, I think of a infinite subset, i.e $\{n\sqrt2:n\in\mathbb{N}\}$.I thought If we can add condition , that is that set should be bounded. The set that comes to my mind is $\{x\in(0,1):x\in \mathbb{Q}^c\} $ but that is not countable. After thinking longtime I found the set $\{\sqrt2+\frac{1}{n}:n\in \mathbb{N}\}.$ I thought to add more conditions. In this set I found that lot of elements are very close to $\sqrt2$ i.e $\sqrt2$ is a accumulation point. That's Why I want a set without accumulation point.

I am so curious to know whether to construct such set! Can someone help with this?

Define a relation $R$ on $[0,1]$ as $xRy$ if $x-y\in \mathbb{Q}$

Then the equivalance class of $[1/\sqrt2]$ is bounded, infinite$(\{1/\sqrt2+\epsilon/2^n\}\subset [1/\sqrt2] $ ), countable( since we can map every element to the distance from $1/\sqrt2$, i.e every element to maps to a rational number), no accumulation point(Since every class is dense in $[0,1]$


If two topological spaces have the same topological properties, are they homeomorphic? Topological properties are investigated because we can show that two spaces are not homeomorphic by finding one property that holds in one space but not the other. But what if no topological property can distinguish two topological spaces? So I ask:

  If two topological spaces have the same topological properties, must they be homeomorphic?

Edit: I don't really have any particular class of topological properties in mind, because what I am thinking is really every single topological property, whenever it is well-defined for a topological space. I just didn't know that the class of topological properties is so large, that even "homeomorphic to a space $X$" is itself a topological property, making my question trivial.

Well, it usually goes the other way. A property $P$ of a topological space $X$ is deserved to be called topological if $P(X)$ holds if and only if $P(Y)$ holds whenever $X$ and $Y$ are homeomorphic. An example of a topological property is "$X$ is connected" while an example of a non-topological property is "$X$ is a subset of $\mathbb{R}^n$. The latter can be upgraded to a topological property by requiring instead "$X$ can be embedded in $\mathbb{R}^n$".

With this convention, given a topological space $X$ there is a smart-ass topological property one can define using $X$: "$P(Z)$ holds iff $Z$ is homeomorphic to $X$". Since homeomorphism is an equivalence relation, this is indeed a topological property and clearly $X$ satisfies $P$. If $Y$ is another space that satisfies the same topological properties as $X$, then $P(Y)$ must hold and so $X$ and $Y$ are homeomorphic.

The claim is the following:

If two spaces $X$ and $Y$ are not homeomorphic, then one can find a property that holds in one of the spaces and not the other. The contropositive of this is:

If one cannot find a property that holds in once space but not the other then $X$ and $Y$ are homeomorphic. So, yes, thie claim is true.


Calculus: Improper integrals vs series What is the difference between improper integrals and the a series?
For example, if you solve a type one improper integral from 1 to infinity, the answer is different than if you solve the same function using a geometric series.

In your case, taking $1 \to \infty$:

Series sum over integers $n = 1 \to \infty$: $\displaystyle \sum_{n=1}^\infty f(n)$.

Integrals "sum" over all reals $x = 1 \to \infty$: $\int_{n=1}^\infty f(x) dx$.

Added: this might help in understanding the relationship between partial summation (series) and integration, from Wikipedia's entry on "Series":

  Partial summation takes as input a sequence, $\{ a_n \}$, and gives as output another sequence, $\{ S_N \}$. It is thus a unary operation on sequences. Further, this function is linear, and thus is a linear operator on the vector space of sequences, denoted $\Sigma$. The inverse operator is the finite difference operator, $\Delta$. These behave as discrete analogs of integration and differentiation, only for series (functions of a natural number) instead of functions of a real variable. For example, the sequence $\{1, 1, 1, ...\}$ has series $\{1, 2, 3, 4, ...\}$ as its partial summation, which is analogous to the fact that $\int_0^x 1\,dt = x.$

More clearly, for series  we use $\sum$ which are discrete values summed over integers and for integrals we use $\int_a^b$ which are continuous interval


$T^3=\frac{1}{2}(T+T^*) \rightarrow$ T is self adjoint Let $T$ be a normal transformation on a finite-dimensional Hilbert space; that is, $TT^*=T^*T$, where $T^*$ is the adjoint of $T$.

Prove that if $T^3=\frac{1}{2}(T+T^*)$, then $T$ is self adjoint.

I have tried to do some math on $(Tv,u)$ but I was not successful in proving the following: $(Tu,v)=(u,Tv)$ which is what I need for self-adjoint transformation.

Edit:
$(T^3,v)=\frac{1}{2}\left(\left(T+T^*\right),u \right)= \left( Tu,v\right)+  \left( T^*u,v\right)=\left( u,T^*v\right)+\left( u,Tv\right)=\left( u,T^3v\right)$
Therefore, $T^3$ is self adjoint. Does it mean that $T$ is self adjoint?

Thanks!

It is not true that if $T$ is normal and $T^3$ is selfadjoint, then $T$ is selfadjoint. Example:
$$
T=\begin{bmatrix}e^{i\pi/3}&0\\0&e^{i\pi/3}\end{bmatrix}.
$$
Then $T^3=I$ (or, even when $n=1$ and $T=e^{i\pi/3}$).

But here we know more. Since $T$ is normal, it is unitarily diagonalizable. So we may assume that $T$ is diagonal. We can then see that the equality $T^3=\text{Re}\,T$ will be satisfied by the eigenvalues of $T$. If
$$\tag1
\lambda^3=\text{Re}\,\lambda
$$
and we write $\lambda=re^{it}$, we immediately get
$$\tag2
r^3e^{3it}=r\cos t.
$$
There is the obvious solution $r=0$, i.e. $\lambda=0$. For $r\ne0$, the equality $(2)$ becomes
$$\tag3
r^2\cos 3t=\cos t$$ and $$\tag4 r^2\sin 3t=0.
$$
The equality $(4)$ gives us $t=k\pi/3$. And now we have
$$\tag5
r^2=\frac{\cos k\pi/3}{\cos k\pi}=(-1)^k\cos k\pi/3.
$$For $k=0,1,2,3,4,5$, the right-hand-side in $(5)$ is
$$
0,-\frac12,-\frac12,1,-\frac12,-\frac12.
$$
As $r^2>0$, the only acceptable values of $k$ are $0$ and $3$, which give $\lambda=1$ and $\lambda=-1$ respectively.

In summary, the only possible eigenvalues of $T$ are $0,1,-1$, and thus $T$ (being normal) is selfadjoint.

We have $2A^3 =A+A^*$ hence $A+A^* =(2A^3)^*=2(AAA)^* =2(A^* )^3$ thus the operator $A^3 $ is self-adjont. And therefore the operator $A=f(A^3 ) $ where $f(x) =\sqrt[3]{x} $ is self - adjont.


$\int_\Omega f d\mu = 0 $ if and only if $f(x)=0$ almost everywhere can someone give me a hint on what kind of theorem/definition I should make use of to solve this?

Let $(\Omega,\mathfrak A, \mu)$ be a measure space and $f:\Omega \to \mathbb R$ a non-negative measurable function.
 Show that
$$\int_\Omega f d\mu = 0 $$ if and only if $$f(x)=0$$ a.e.

Sketch:

(1) Suppose not, then there is a positive measure subset $A$ where $f\not=0$.

(2) Let $A_n$ be the set of points where $f>\frac{1}{n}$.

(3) Observe that $A_n\subseteq A_{n+1}$ and $\bigcup A_n=A$.

(4) Conclude that some $A_n$ has positive measure using the properties of measures.

(5) Conclude that $\int f d\mu\geq\frac{1}{n}\mu(A_n)>0$.

If @$f(x) = 0$ a.e. is clear that $\int_{\Omega} fd\mu = 0$.

Now, suppose that $f$ is non-negative, $\int_{\Omega} fd\mu = 0$, but $f$ is not $0$ a.e. It means that there is $A$ such that $\mu(A) >0$ and $f(x) \neq 0, \forall x \in A$.

Then,

$\int_{\Omega} f d\mu \geq \int_{A} f d\mu > 0$. Contradiction!


Assign n professors to 2n courses in two semesters and in spring no professor teaches the same pair of course So the problem is

  A math department has n professors and 2n courses, two assigned to each professor each
  semester. How many ways are there to assign the courses in the fall semester? How many
  ways are there to assign them in the spring so that no professor teaches the same two
  courses in the spring as in the fall? If all the assignments are equally likely, what is the probablity of this event.

For the first part, I believe it is 2n!/(2)**n. But I keep having trouble wrapping my head around the second and third one. I was thinking if there are no professor teaches the same course, then it is a lot like getting them assigned first, and ask each professor to switch course with another professor. So the first one has (n-1) choice, second one has (n-2) choice, and it will end up being (n-1)!, and by rule of product, there are [2n!/(2)**n]*(n-1!). But I feel that this number might be too big, and I might be double counting something. And for the third question, I think the denominator would be [2n!/(2)**n]**2 by rule of product. But I don't know what to do with nominator.

Thanks in advance!

Part II:

Using the principle of Inclusion and Exclusion,

you get then number of ways to be $ = \sum_{0}^{n} (-1)^n {n\choose k} {2n-2k\choose 2,2,2..}$ k number of 2's.

ofcourse for n= 1, the number of ways$=0$ as @Bram28 has asked you to confirm.

The answer for part III is $\frac{Part II Answer}{Part I answer}$ that is numerator is the number of ways the same courses are not taught by the professor in Spring and the denominator is the total number of ways courses could be assigned in a semester.


Help with this differential equation $(3x^2y+2xy+y^3)dx + (x^2+y^2)dy = 0$ We have:
$(3x^2y+2xy+y^3)dx + (x^2+y^2)dy = 0$
Say
$M(x, y) = 3x^2y+2xy+y^3$
and
$N(x, y) = x^2+y^2$
We want $M_y = N_x$, but they are not. So we suppose there is a $π(x, y)$ so that
$π(x, y)M(x, y)dx + π(x, y)N(x, y)dy = 0$.
The method I know is that $π(x, y) = μ(φ(x, y))$ with $φ(x, y)$ given, but in this equation there is no such function given.
How can I work this out?

Suppose $(\pi M)_y  = (\pi N)_x$, i.e.
$$
\pi_yM + \pi M_y = \pi N_x + \pi_x N
$$
If we make the simplifying assumption that $\pi_y = 0$ (i.e. $\pi$ is only a function of $x$), we get
$$
\frac{M_y-N_x}{N} = \frac{\pi_x}{\pi}
$$
from which we get
$\pi(x) = e^{3x}$.

Given a differential equation of the form $Mdx+Ndy=0$ if we had $M_y=N_x$ then the equation is called exact and exact equations are easy to solve as one would try to find a function $f$ such that $\frac{\partial f}{\partial x}=M$ and $\frac{\partial f}{\partial y}=N$ (so that  $Mdx+Ndy=0$ becomes same as $\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}$).
Now if $Mdx+Ndy=0$ isn't exact we'd like to multiply a factor $\pi(x,y)$ with it so that it becomes exact ( this $\pi$ is called the integrating factor ).
If such a $\pi$ exists then we must have that $\frac{\partial (\pi M)}{\partial y} =\frac{\partial (\pi N)}{\partial x}$ . If you assume that $\pi$ is a function of x alone or a function of y alone and try to solve in terms of $M$ and $N$  (in one of the two cases you'd be able to do that) and multiply the $\pi$ hence found with the given equation it'd become exact and hence you'd be able to solve it easily.


use mathematical induction to show that $n^3 + 5n$ is divisible by $3$ for all $n\ge1$ What I have so far
Base: $n^3 + 5n$
Let $n=1$
$$
1^3 + 5(1) = 6
$$
$6$ is divisible by $3$
Induction step: $(k+1)^3 + 5(k+1)$
$(k^3 + 3k^2 + 8k + 6)$ is divisible by $3$
I kind of get lost after this point. For starters, how do I prove that this isn't applicable for any number less than $1$? Also, where do I go after this? Thank you!

In the induction step, we assume that it holds for $n = k$.
That means that $\exists a \in \mathbb{Z} : k^3 + 5k = 3a$.
Then, for $(k+1)$, we get
$$
\begin{split}
(k+1)^3 + 5(k+1)
 &= k^3 + 3k^2 + 8k + 6 \\
 &= (k^3 + 5k) + 3k^2 + 3k + 6 \\
 &= 3a + 3k^2 + 3k + 6 \\
 &= 3(a + k^2 + k + 2).
\end{split}
$$
So, if $k^3 + 5k$ is a multiple of $3$, then $(k+1)^3 + 5(k+1)$ must also be a multiple of $3$.

You correctly expanded $(k+1)^3+5(k+1)$ into $k^3+3k^2+8k+6$. What you now need to do is to show that $k^3+3k^2+8k+6$ is divisible by $3$, using the inductive hypothesis that $k^3+5k$ is divisible by $3$. To do this, simply write
$$k^3+3k^2+8k+6=(k^3+5k)+3k^2+3k+6$$
and note that each of the four terms on the right hand side is divisible by $3$.
As for your question,  how do I prove that this isn't applicable for any number less than $1$?, you don't need to prove anything about numbers less than $1$ (and in this case the assertion does still hold: $n^3+5n$ is divisible by $3$ for all integers $n$, positive or negative). Induction only needs to go in one direction.


Solution of a specific integral using the Cauchy Integral formula I have a question regarding a integral I just can't find the right solution to. It's this beauty:
$$\int_\gamma \frac{\cos(\pi z)}{z^2+1} dz$$ where $\gamma = 3 \cdot e^{it}$.
I used the Cauchy Integral Formula (see here) with $n=0$, $a=1$ and $f(z) = \frac{\cos(\pi z)}{z+1}$ and it got me a result of $-\pi i$. However, this integral should be zero according to my professor. Any help is appreciated!

It is not clear to me as to how you applied CIF for you function $f$. To apply CIF you need an analytic function.
Poles are at $\pm i$ and the  residues are $\cos (\pi i)=\frac {e^{-1}+e^{1}}{2 i}$ and $\cos (-\pi i)=\frac {e^{-1}+e^{1}}{ -2i }$. So the integral is $0$ by Residue Theorem.

You don't need any advanced Complex Analysis theorems to compute that integral. Let $f(z)=\frac{\cos(\pi z)}{z^2+1}$. In order to compute $\oint_\gamma f(z)\,\mathrm dz$, all that I will need is that $f$ is integrable and even. In fact, $\oint_\gamma f(z)\,\mathrm dz$ is equal to $\int_0^{2\pi}f(3e^{it})3ie^{it}\,\mathrm dt$, which, in turn, is equal to $3i$ times$$\int_0^\pi f(3e^{it})e^{it}\,\mathrm dt+\int_\pi^{2\pi}f(3e^{it})e^{it}\,\mathrm dt.\tag1$$But\begin{align}\int_\pi^{2\pi}f(3e^{it})e^{it}\,\mathrm dt&=\int_0^\pi f\left(3i^{i(t+\pi)}\right)e^{i(t+\pi)}\,\mathrm dt\\&=\int_0^\pi f\left(-3e^{it}\right)\left(-e^{it}\right)\,\mathrm dt\\&=-\int_0^\pi f\left(3e^{it}\right)e^{it}\,\mathrm dt,\end{align}since $f$ is an even function. So, $(1)$ is equal to $0$, and therefore $\oint_\gamma f(z)\,\mathrm dz=0$.


Notation explanation for $\text{div} \ \omega = \nabla^i \omega_i $ and $\Delta = -\nabla^i \nabla_i $. Given a connection $\nabla$ on the tangent bundle, the divergence of a vector field $X\in C^{\infty}(TM)$ is defined as $div(X) = C_1^1(\nabla X)$ where $C$ is the contraction.

If $\{e_i\}$ is a local reference frame and $\{\eta^i\}$ is the induced coframe, then locally
$$(1)\quad \quad  div(X) = \nabla X(e_i,\eta^i)=\eta^i(\nabla_i X) = e_i(X^i) + X^r\Gamma_{i r}^i $$
where $\Gamma_{i r}^k$ are the Christoffel's symbols.

Given a pseudo-Riemannian metric,  if  instead of a vector field we have a 1-form $\omega\in C^{\infty}(T^*M)$, we can define
$$(2) \quad \quad div(\omega) := div(\omega^\#).$$

  Sometimes I have encountered this notation for the divergence of a 1-form :
  $$ (3)\quad \quad   \text{div} \ \omega = \nabla^i \omega_i$$
  What does the symbol $\nabla^i$ mean?
  Is $\omega_i$  defined by $\omega = \omega_i dx^i$? Or it is  $\text{div} \ \omega = (\nabla^i \omega)_i$ the i-component of the covariant tensor $\nabla^i \omega$?
  I also need to explain why , in the same paper, the author states that $\Delta = -\nabla^i\nabla_i$ as if $\Delta f = -\nabla^i\nabla_if$ since $\nabla_i f $ is a function it seems that $\nabla^i$ act on functions.

Locally $(2)$  is written as $div(\omega) = e_i( g^{ij}\omega_j)+ g^{r j}\omega_j\Gamma_{i r}^i $ I can't see clearly how is defined $\nabla^i$.

This is index notation as traditionally used in the Ricci calculus, these days typically interpreted by mathematicians as abstract index notation.

When we write $\mathrm{div}\,X = \nabla_i X^i$, what is meant is that we first take the covariant derivative of $X$ and then perform the contraction indicated by the repeated indices; i.e. $\nabla_i X^i = (\nabla X)^i_i$, where the RHS can be interpreted either as abstract index notation denoting the contraction $C(\nabla X)$ or as a coordinate formula using Einstein summation notation. Importantly, this is not the same thing as first taking the coordinate components $X^i$ (which are scalars) and then differentiating, which would yield instead $\partial_i X^i$, a coordinate-dependent quantity. Thus we usually don't think of the expression $\nabla_i X^i$ as being an operator $\nabla_i$ acting on the component functions $X^i$ - instead, it's the operator $\nabla$ acting on the vector field $X$, with the indices just telling us how to contract. (Hopefully this answers your last question regarding $\nabla^i \nabla_i f.$)

In the case of a one-form $\omega$, the covariant derivative is a $(0,2)$-tensor $\nabla\omega = (\nabla_j \omega_i) \eta^j \otimes \eta^i$. (Note again that in this notation, $\nabla_j \omega_i$ is not the same as $\partial_j \omega_i$ - we are taking the covariant derivative before plugging in the reference frame.) To take the trace of this to obtain a divergence, we first raise an index with the metric as you described, yielding $\mathrm{div}(\omega^\sharp) = \nabla_i \omega^i := \nabla_i (\omega^\sharp)^i$ where of course $\omega^i =
 (\omega^\sharp)^i = g^{ij} \omega_j.$ Since the covariant derivative is $g$-compatible and the metric is symmetric, we can think of this as $\mathrm{tr}_g \nabla \omega = g^{ij} (\nabla \omega)_{ij};$ so we could just as fairly raise the first index instead of the second, yielding $\nabla^i \omega_i.$

For a function, $\omega$, of n variables, $x_1$, $x_2$, ..., $x_n$, "grad $\omega$", or $\nabla \omega$, is defined as $\frac{\partial \omega}{\partial x_1}+ \frac{\partial \omega}{\partial x_2}+ \cdot\cdot\cdot+ \frac{\partial \omega}{\partial x_n}$.  That can be thought of as the "scalar product" of the "vector operator" $\nabla= \frac{\partial}{\partial x_1}+ \frac{\partial}{\partial x_2}+ \cdot\cdot\cdot+ \frac{\partial}{\partial x_n}$ with $\omega$ which is denoted, in Einstein summation notation, as "$\nabla_i \omega$".

Similarly, divergence of a vector valued function (one-form) $\vec{\omega}= f_1(x_1,x_2,...,x_n)\vec{i_1}+ f_2(x_1,x_2,...,x_n)\vec{i_2}+ \cdot\cdot\cdot+ f_n(x_1,x_2,...,x_n)\vec{i_n}$, where "$i_m$" is the mth basis vector, is the scalar vector $\frac{\partial f_1}{\partial x_1}+ \frac{\partial f_2}{\partial x_2}+ \cdot\cdot\cdot+ \frac{\partial f_n}{\partial x_n}$ which can be thought of as the dot product of the vector operator $\nabla$ and the vector $\vec{\omega}$.  That is denoted, in Einstein summation notation, as $\nabla_i\omega^i$.


*-Groups or "Group with Involution"? Wikipedia has a basic reference on *-rings in the *-algebra article, which defines a *-ring as a ring equipped with a * operator which is an antiautomorphism and an involution.

However, I am not interested in rings equipped with such an operator, but Groups equipped with one.  I tried searching for "*-group" to find references, but wasn't able to find much (I also tried the groupified versions of the alternative names Wikipedia suggested for *-rings: "involutive ring, involutory ring, and ring with involution", with no luck).

I am wondering if such a structure as a "*-group" exists in the literature, and if so, what name does it go by?

Is there some reference that could give a basic introduction to the subject? (Or maybe the structure is trivial, which might explain so little has been written about it?)

(For the context of why I am interested, I am thinking about generalizations of a Counted B-Tree for non-commutative sums, where segments can be added together, subtracted off from each other, and "flipped" to reverse the order).

All groups are "*-groups", with the involution being inversion. Inversion is an involution as $(a^{-1})^{-1}=a$, and is an antiautomorphism as $(ab)^{-1}=b^{-1}a^{-1}$.


Expected value of a guessing game I'm trying to solve project euler 527, I don't understand how the expected value of B(6) is taken.

  A secret integer t is selected at random within the range 1 ≤ t ≤ n.

  The goal is to guess the value of t by making repeated guesses, via
  integer g. After a guess is made, there are three possible outcomes,
  in which it will be revealed that either g < t, g = t, or g > t. Then
  the process can repeat as necessary.

  Normally, the number of guesses required on average can be minimized
  with a binary search: Given a lower bound L and upper bound H
  (initialized to L = 1 and H = n), let g = ⌊(L+H)/2⌋ where ⌊⋅⌋ is the
  integer floor function. If g = t, the process ends. Otherwise, if g <
  t, set L = g+1, but if g > t instead, set H = g−1. After setting the
  new bounds, the search process repeats, and ultimately ends once t is
  found. Even if t can be deduced without searching, assume that a
  search will be required anyway to confirm the value.

  Your friend Bob believes that the standard binary search is not that
  much better than his randomized variant: Instead of setting g =
  ⌊(L+H)/2⌋, simply let g be a random integer between L and H,
  inclusive. The rest of the algorithm is the same as the standard
  binary search. This new search routine will be referred to as a random
  binary search.

  Given that 1 ≤ t ≤ n for random t, let B(n) be the expected number of
  guesses needed to find t using the standard binary search, and let
  R(n) be the expected number of guesses needed to find t using the
  random binary search. For example, B(6) = 2.33333333 and R(6) =
  2.71666667 when rounded to 8 decimal places.

  Find R(1010) − B(1010) rounded to 8 decimal places.

Im trying to get the expected value B(6) as follows:

$$B(6) = \frac16 + 2\cdot\frac13 + 3\cdot\frac23 = 2.8333333$$

I got this from an answer on another thread, the probability to get a correct guess at the n-th time is $\frac{2^{n-1}}{6}$.

As for R(6) I don't understand how to get it.

Could someone explain how to get the B(6) and R(6) exactly?

First of all, the expected value for B (6) is not 2.83333 but rather 2.333333 as is stated in an example in the problem.  Second of all, it's kind of against the spirit of project euler to just look up the solution, but here are some recursive formulas to help you along (you'll need a better approach for the full scale problem though.

For B (n), let $ i = \left \lfloor {n + 1 \over 2} \right \rfloor $.  Then
$$
B (n) = 1 + {i - 1 \over n} B (i - 1) + {n - i \over n} B (n - i)
$$
This is one guess plus the number of guesses afterwords if the secret number is lower than the guess times the probability of the secret number being less than the guess plus the number of guesses afterwords if the secret number is higher than the guess times the probability of the secret number being higher than the guess.

For R (n), the randomization makes it less simple, but for any choice of the guess i the expected value for secret values below and above is easily defined recursively as the probability of the secret number being in that range times the recursively calculated expected value.  We can't really choose i so we sum over all i and multiply by 1/n, the probability of each one being chosen.  Then the above and below recursive terms combine and we get
$$
R (n) = 1 + {2 \over n^2} \sum_{i = 1}^{n - 1}{i R (i)}
$$

You can easily expand these by hand for n = 6 and I think you might see a pattern.

Here's a simple way to understand B(6)

when t=1, then the g progression would be 3 then 1

t=2, g progression 3,1,2

t=3, => 3

t=4, => 3,5,4

t=5, => 3,5

t=6, => 3,5,6

In total there were 14 guesses and there were 6 possible t. 14/6=2.3333333, which is the expected number of guesses.

That of course, is not the way to do B(1000000000)!

Cheers.


Proof: $\arg(z\cdot w)= arg(z) + arg(w)$ and $\arg\left( \frac{z}{w} \right)=\arg(z) − \arg(w) $ Note: I've seen similar questions but I'd like to ask you for review the way of proving shown below.

$\arg(z\cdot w)= arg(z) + arg(w)$

Let's say that $arg(z) = \alpha$ and $arg(w) = \beta$.
Then: $z = |z|(cos\alpha + isin\alpha)$
$z\cdot w = |z|\cdot |w|[(cos \alpha\cdot \cos\beta −\sin \alpha \cdot \sin β)+i(\cos \alpha \cdot sin \beta +\cos \beta\cdot
\sin \alpha)]$
$z \cdot w = |z|\cdot|w|[\cos(\alpha + \beta) + i\sin(\alpha + \beta)]$

Point 4. based on:

$\cos(\alpha + \beta) = \cos\alpha\cdot\cos\beta -\sin\alpha\sin\beta$

and

$\sin(\alpha + \beta) = \sin\alpha\cdot\cos\beta + \sin\beta\cos\alpha$

$\arg\left( \frac{z}{w} \right)=\arg(z) − \arg(w) $

$z = w \cdot \frac{z}{w}$

Using the previous equation: $\arg(z\cdot w)= arg(z) + arg(w)$

$\arg(z) = \arg(w) + \arg\left( \frac{z}{w} \right)$

$\arg\left( \frac{z}{w} \right) = \arg(z) - \arg(w)$

Your work [above] is based on trigonometry, but trig is based on complex arithmetic. Sometimes it helps to cut out the trig and use Euler's identity
$$exp(a+bi)=cos(a)+I*sin(b)$$ which of course is obvious in the series definitions of $exp, cos, sin$. Let $z=exp(a1+i*b1)$ and $w=exp(a2+i*b2)$ now by third-grade algebra $$z*w=exp^{a1+i*b1}*exp^{a2+i*b2}=exp^{(a1+i*b1)+(a2+i*b2)}=exp^{(a1+a2)+i*(b1+b2)}.$$ Since $b1, b2, (b1+b2)$ are your arg's you immediately get your first equation. The others are obvious using the same approach.


Multivariable limit $\lim_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1}$ $$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} $$

According to my textbook the limit equals $2$.

What I have tried:

Using the squeeze theorem:

$$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 2} $$
$$ 0 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le 0 $$

I have also tried to use the squeeze theorem with two other equations and obtained different values:

$$ \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 - 1}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 + 1}{\sqrt{x^2 +y^2 + 1} - 2} $$
$$ -1 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le -1 $$

Remember the difference of squares algebraic identity.

$$ A^2 - B^2 = (A - B)(A + B) $$

Why is that useful?  With $A = \sqrt{x^2 + y^2 + 1}$ and $B = 1$, the denominator of your expression is $A - B$.  With that in mind,

$$ \begin{align}
\frac{x^2 + y^2}{\sqrt{x^2 + y^2 + 1} - 1} &= \frac{x^2 + y^2}{\sqrt{x^2 + y^2 + 1} - 1} \cdot \frac{\sqrt{x^2 + y^2 + 1} + 1}{\sqrt{x^2 + y^2 + 1} + 1} \\
&= \frac{(x^2 + y^2)\left(\sqrt{x^2 + y^2 + 1} + 1 \right)}{x^2 + y^2} \\
&= \sqrt{x^2 + y^2 + 1} + 1
\end{align} $$

Now, you can evaluate the limit as $(x, y) \to (0, 0)$ simply by evaluation, since this expression is continuous at the origin.

$$ \lim_{(x, y) \to (0, 0)} \sqrt{x^2 + y^2 + 1} + 1 = \sqrt{0^2 + 0^2 + 1} + 1 = 2. $$


Disquisitiones Arithmeticae Chapter 8? Does anyone know where I can find the posthumously published (I think) chapter 8 of Gauss's Disquisitiones Arithmaticae?

According to the German edition of Wikipedia you can find the eighth chapter in here.

I have read some information that it may have been published as 'General Investigations of Curved Surfaces' however I am having some difficulty confirming.


Is there a number $\mathscr{D}_2 \neq \mathscr{D} = {{3003}^2}\cdot{22021}$ satisfying a certain condition? (Note:  This question is tangentially related to this earlier one.)

Let $$\sigma(x) = \sum_{d \mid x}{d}$$ denote the sum of divisors of $x \in \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers or positive integers.

Recall that a Descartes number is an odd number $n = km$, with $1 < k$, $1 < m$, satisfying $$\sigma(k)(m+1)=2km.$$  ($m$ is called the quasi-Euler prime of $n$.)  Note that we define $\sigma(m) := m + 1$ even when $m$ is composite (that is, we pretend that $m$ is prime).

Notice that the lone Descartes number $\mathscr{D}$ that is known is
$$\mathscr{D} = k'm' = {{3003}^2}\cdot{22021}.$$

  In particular, note that:
  $$\frac{\sigma(k')}{m'} + \frac{\sigma(m')}{k'} = \frac{670763}{819} \approx 819.002,$$
  where
  $$\lfloor{\frac{670763}{819}}\rfloor = 819 = 2\cdot{{3003}^2} - \sigma({{3003}^2}) = D({{3003}^2}),$$
  and $D(x)=2x-\sigma(x)$ is the deficiency of the positive integer $x$.

Here are my questions:

  (1) Is there a number $k_2 \times m_2 = \mathscr{D}_2 \neq \mathscr{D} = {{3003}^2}\cdot{22021}$ satisfying
  $$\lfloor{\frac{\sigma(k_2)}{m_2} + \frac{\sigma(m_2)}{k_2}}\rfloor = D(k_2)?$$

Note that I have no restrictions on $k_2$, $m_2$, and therefore, $\mathscr{D}_2$.

  (2) When does the relationship
  $$\lfloor{\frac{\sigma(k_3)}{m_3} + \frac{\sigma(m_3)}{k_3}}\rfloor = D(k_3)$$
  hold, for some integers $1 < k_3$, $1 < m_3$?

After trying to search (in Sage Cell Server) for examples satisfying the equation
$$\lfloor{\frac{\sigma(k_3)}{m_3} + \frac{\sigma(m_3)}{k_3}}\rfloor = D(k_3)$$
for some integers $1 \leq k_3 \leq 1000$, $1 \leq m_3 \leq 1000$, I believe that it holds for most $k_3$ and $m_3$:

Pari/GP script #1

for(x=1, 1000, for(y=1, 1000, if(floor((sigma(x)/y)+(sigma(y)/x))==2*x-sigma(x),print(x,"   ",y))))

Output #1

I have suppressed the output - there are several lines returned.

Pari/GP script #2

for(x=1, 1000, for(y=1, 1000, if(floor((sigma(x)/y)+(sigma(y)/x))==2*x-sigma(x) && gcd(x,y)==1,print(x,"   ",y))))

Output #2

I have suppressed the output - there are several lines returned.

Pari/GP script #3

for(x=1, 1000, for(y=1, 1000, if(floor((sigma(x)/y)+(sigma(y)/x))==2*x-sigma(x) && gcd(x,y)==1 && issquare(x),print(x,"   ",y))))

Output #3

9   20
9   26
9   35
9   43
9   47
25   168
25   248
25   266
25   268
25   297
25   314
25   326
25   333
25   413
25   427
25   437
25   479
25   487
25   491
49   660
49   768
49   852
49   858
49   940
49   992
484   25


Cardinality of infinite sets - Help with intuition Yesterday I studied the cardinality of infinite sets and, I must say, it was the first time I have ever distrusted math. That is, it was the first time I have not bought - hook, line and sinker - something about math that was proven to me by a mathematics professor.

This is not to say that I think it is untrue. I am saying that the truth of the cardinality of infinite sets is so at odds with my intuition that I am having trouble believing it.  And I could use a bit more explanation, I think, to clear up a few things.

The distinction between the rules governing finite and infinite sets. For example, we can say that the odd numbers are a proper subset of the integers. We know this absolutely, correct? Yet, we can then prove using diagonalization that the set containing all odds and the set containing all integers have the same cardinality, which is Aleph-null.

How is that we can allow such discontinuity in our understanding of these sets? How can both of those two statements be simultaneously true? How, if A is a proper subset of B, can |A|=|B|? I know this doesn't hold when we limit the conversation to finite sets. However, I don't understand why we can't make the same assertion about these two infinite sets - or at least not definitively say they both have cardinality Aleph-null.

Proving that the reals don't form a bijective map to Aleph-null. This was proven by trying to map the reals onto the naturals. More specifically, the professor considered only the continuum and showed that the continuum can't map to the naturals and therefore neither can the reals.

The professor did this by showing that any list of natural numbers could never contain all the reals because we can always create at least one new number not contained in the list.

HOWEVER, it felt like a bit of flashy wording to me. If we can do this and then say "look, we didn't account for this number" and furthermore not be allowed to say "okay, but I still have this natural number to which I can map that real number" in this instance, why do the same arguments not apply to other instances? That is, why can't I do the exact same thing when dealing with the mapping between odds/integers?

I know this isn't the typical question for stackexchange, but I really want to nail this down. Don't get me wrong, I know the information. I can apply the information. However, I don't necessarily believe the information and that bothers me a bit.

Your (understandable) disbelief seems to hinge mostly on one idea: The assertion that if $A\subsetneq B$ then $\#A<\#B$.

This is obviously true for finite sets, but as you have learned not for infinite ones, at least given the standard definition of cardinality (not necessarily the only possible one), which asserts that two sets are of equal cardinality if they can be put in bijective correspondence.

A very simple example using this definition is the following: Take the natural numbers $\mathbb N$. Now remove the number $1$. You have a new set that is s strict subset of $\mathbb N$, but if you relabel (i.e. put in correspondence) every element $n$ of that new set with the "label" $n-1$, you suddenly have every single natural number showing up again, despite having removed one at the beginning. Therefore, by the definition of cardinality given above, the cardinalities of the two sets are indeed equal.

The whole apparent paradox disappears the moment you understand that it is not the nature of infinite sets that violates your intuition, but the (re)definitions of common concepts such as "size" when applied to those sets. You are welcome to define cardinality in a way that is compatible with the intuition of inclusion and relative size. The reason why mathematicians don't do the same is that it leads to inconsistencies in set theory.

To put it succinctly: Of course there are more integers than even integers – except if you take "more" to mean set-theoretic cardinality. No conflict with intuition at all.

I answer just this question.

How, if A is a proper subset of B, can |A|=|B|? I know this doesn't hold when we limit the conversation to finite sets. However, I don't understand why we can't make the same assertion about these two infinite sets - or at least not definitively say they both have cardinality Aleph-null.

u/BobbyP27 answers this on
r/explainlikeimfive excellently!

I think the problem is you are thinking of "infinite" to be "a very big number". It is not a very big number, it's a different kind of thing. A similar problem exists with zero, in that it's not just "a really small number", it's actually zero. For example if I take a really small number like 0.0000001 and double it, I get 0.0000002. If I take 0 and double it, I still get zero. $2 \times 0$ is not bigger than $1 \times 0$. If I have an infinite number of numbers between 0 and 1, then they are separated by 0. If I double all of those numbers, then they are separated by  $2 \times 0$, so they are still separated by 0.

Post Script
Unlike Stack Exchange, Reddit answers can be irretrievably deleted, therefore important to record them! Like parent comment of this daughter comment.


Doubly exponential sequence behaviour from inequality I am investigating a strictly decreasing sequence $(a_i)_{i=0}^\infty$ in $(0, 1)$, with $\lim_{i\to\infty}a_i=0$, such that there exist constants $K>1$ and $m\in\mathbb{N}$ such that $$\frac{a_{i-1}^m}{K} \leq a_i \leq K a_{i-1}^m$$ for all $i$. Even though $K>1$, is it of the right lines to conclude that $a_i \sim \alpha^{m^i}$ for some constant $0<\alpha<1$?

Thanks,
DW

[Edit: Now that the question has been changed a day later, I removed analysis of the old version of the first inequality.  Perhaps sometime I will update to fully answer the new question.  The following still applies to the second inequality.]

On the other hand, f For the second inequality, $$a_i\leq Ka_{i-1}^m\implies a_i\leq K^{1+m+m^2+\cdots+m^{i-1}}a_0^{m^i}\leq \left(K^{2/m}a_0\right)^{m^i}.$$  You could apply a similar inequality with each $N\in\mathbb N$ in place of $0$ to get $$a_i\leq \left(K^{2/(m^{N+1})}a_N^{1/m^N}\right)^{m^i}.$$  By choosing $N$ such that $K^{2/m}a_N<1$, you at least get $\displaystyle{a_i=O\left(\alpha^{m^{i}}\right)}$ with $\alpha=\left(K^{2/m}a_N\right)^{1/m^N}\in(0,1)$.