Finding all differentiable $f: [0,+\infty) \rightarrow [0,+\infty)$ such that $f(x) = f'(x^2)$ and $f(0)=0$ After some investigation it seems fairly obvious to me that the only such function is the zero function, however I haven't been able to prove it. By considering $$\alpha  =\sup\{x\in[0,+\infty) :f(x) = 0\},$$ I was able to show that $\alpha$ can only be $1$ or $0$ but I could not weed out those two possibilities. Any hints/solutions welcome.

EDIT 1

Because of the continuity of $f$, we must have $f(\alpha) = 0$. Note that because of the relation given we have $$\int_0^{\sqrt \alpha}2xf'(x^2)\,\mathrm dx = f(\alpha),$$ but because of the relationship given this implies
$$\int_0^{\sqrt \alpha}2xf(x)\,\mathrm dx = f(\alpha).$$
If $\alpha$ is strictly between $0$ and $1$, then $\sqrt \alpha > \alpha$, but then splitting the integral we get
$$\int_{\alpha}^{\sqrt \alpha}2xf(x)\,\mathrm dx = f(\alpha) = 0.$$ But by our choice of $α$, this integral should be non-zero since our function is positive. Hence $\alpha$ cannot be between $0$ and $1$.

Now suppose it is greater than $1$, then we have $$f(\alpha^2) =\int_0^{\alpha}2xf(x)\,\mathrm dx = 0.$$
Since our function is $0$ on $[0,\alpha]$ (Note that it is increasing), this is again a contradiction because $\alpha^2 > \alpha$. Therefore $\alpha$ is $0$ or $1$.

EDIT 2

I forgot to mention the important condition that $f(0)=0$.

As Alex already noticed, a slightly more general statement holds:

  Let $f:[0, \infty) \to [0, \infty)$ be continuous,  differentiable
  on $(0, \infty)$, and $c \ge 1 $.

  If $f(0) = 0$ and $f(x) = f'(x^c)$ for all $x > 0$ then $f = 0$.

Proof: $f'(x) = f(x^{1/c}) \ge 0$, so that $f$ is increasing.

This in turn implies that $f'$ is increasing on $(0, \infty)$, so that $f$ is convex.

  Step 1: $f(x) = 0$ for $0 \le x \le 1$.

From the convexity and $f(0) = 0$ it follows that
$$
 f(t) \le t \cdot f(1) \quad \text{ for } 0 \le t \le 1 \, .
$$
On the other hand, the mean-value theorem gives
$$
 f(1) - f(0) = f'(\xi) (1 - 0)
$$ for some $\xi \in (0, 1)$, therefore
$$
f(1) =  f'(\xi) =  f(\xi^{1/c}) \le \xi^{1/c} \cdot f(1)  \, .
 $$
$\xi^{1/c}$ is strictly less than one, so that $f(1) \le 0$
follows.

Since $f$ is increasing, $f(x) = 0$ for $0 \le x \le 1$.

  Step 2: $f(x) = 0$ for $x \ge 1$.

For $x \ge 1$
$$
 f'(x) = f(x^{1/c}) \le f(x)
$$
so that we can use a standard (Grönwall's inequality type) argument:
$h(x) = e^{-x} f(x)$ satisfies
$$
 h'(x) =  e^{-x} (f'(x) - f(x)) \le 0
$$
so that $h$ is decreasing on $[1, \infty)$:
$$
 e^{-x} f(x) \le e^{-1} f(1) = 0 \\
 \implies f(x) \le 0 \implies f(x) = 0 \, .
$$

You can use chain rule to get:

$f'(x) = 2xf''(x^2)$

Now you have:

$f'(x^2) = 2 x^2 f''(x^4)$

Now substitute it on first equation:

$f(x) = 2 x^2 f''(x^4)$

And since $x \in [0,\infty)$ we have 3 cases:

(i) f(x) is constant $\Rightarrow$ $f(x) = f^{(n)}(x)=0$

(ii) f(x)  is non decreasing $\Rightarrow$ $f^{(n)}(x)\geq0$, but in order to equations to hold $f^{(n)}(x)=0$; $n=0,1,...$

(iii)f(x)  is non increasing $\Rightarrow$ $f^{(n)}(x)\leq0$, but in order to equations to hold $f^{(n)}(x)=0$; $n=0,1,...$

Concluding that $f(x) = 0$


Why is $\{(x,y) \mid x,y\in\mathbb{C}\}$ an infinite dimensional vector space over $\mathbb{Q}$? $V=\{(x,y)\mid x,y \in \mathbb{C}\}$ is a vector space over $\mathbb{C}$, $\mathbb{R}$, and $\mathbb{Q}$. Then $\operatorname{dim} (V)=2$ over $\mathbb{C}$, and $4$ over $\mathbb{R}$, but why it is infinite over $\mathbb{Q}$?

Suppose that it were a finite-dimensional space of dimension $n \in \Bbb N$.
Let $B = \{v_1, \ldots, v_n\} \subset \Bbb C^2$ be a basis.
Then, every $v \in \Bbb C^2$ could be written as $$v = q_1z_1 + \cdots + q_nz_n$$ for some unique choice of scalars $q_1, \ldots, q_n$.
This gives us an injection $\Bbb C^2 \to \Bbb Q^n$ defined by
$$v \mapsto (q_1, \ldots, q_n).$$
This would give us a contradiction since $\Bbb C^2$ is uncountable whereas $\Bbb Q^n$ is not.

With a bit more effort, one can show that no countable basis exists either. For this, you may have to use the fact that the set of finite subsets of $\Bbb Q$ is countable.

You can see the dimension of a vector space $V$ over a field $\Bbb F$ as the number of parameters $a,b,c,... \in \Bbb F$ you need to use in order to write a generic element of $V$. Examples:
You can write every $(x,y)\in V$ as $(a,b)$ with $a,b\in \Bbb C$. $a$ and $b$ are 2 complex parameters so the dimension of $V$ over $\Bbb C$ is 2.
You can also write every complex number $a+bi$ with $a,b\in \Bbb R$. But to write every element $(x,y)\in V$ you need two complex numbers, so you can write it as $(a+bi, c+di)$ with $a,b,c,d\in \Bbb R$. You need 4 real parameters so the dimension of $V$ over $\Bbb R$ is 4.
How many rational parameters do you need to write every element of $V$?
If we say two and write $(x,y)\in V$ as $(a,b)$ with $a,b\in \Bbb Q$ we can find at least one element of $V$ that cannot be written by using just two rational numbers: $(π, i)$. Is this an element of $V$? Yes. But π is irrational so you need to sum an infinite number of rational parameters to represent it. (EDIT)>> So you must add it to a basis of $V$, so the dimension is three, at least. You can always find an irrational number with this same property and you'll have to add it to a basis of $V$. Since the linearly independent irrational numbers are infinitely many, the dimension of $V$ over $\Bbb Q$ is infinite.


Why the term and the concept of quotient group? The basic concept of Quotient Group is often a confusing thing for me,I mean can any one tell the intuitive concept and the necessity of the Quotient group,
I thought that it would be nice to ask as any basic undergraduate can learn the intuition seeing the question.
My Question is :

Why is the name Quotient Group kept,normally in the case of division, let us take the example of $\large \frac{16}{4}$ the Quotient of the Division is '$4$' which means that there are four '$4$'s in $16$, I mean we can find only $4$ elements with value $4$

  So how can we apply the same logic in the case of Quotient Groups,like consider the Group $A$ and normal subgroup $B$ of $A$, So if $A/B$ refers to "Quotient group", then does it mean:

    Are we finding how many copies of $B$ are present in $A$??, like in the case of normal division, or is it something different ??

I understood the Notion of Cosets and Quotient Groups,b ut I want a different Perspective to add Color to the concept. Can anyone tell me the necessity and background for the invention of Quotient Groups?

Note: I tried my level best in formatting and typing with proper protocol,if in case, any errors still persist, I beg everyone to explain the reason of their downvote (if any), so that I can rectify myself, Thank you.

Often I think of a quotient group in terms of (loss of) information. When we move from a group to its quotient group we lose some information about the identity of the elements. For example, when we map an element of the additive group of integers $\mathbf{Z}$ to the quotient group $\mathbf{Z}/10\mathbf{Z}$ we lose the information of all the other digits save the least significant one. In other words after moving down to the quotient group we can no longer tell the difference between 9, 999, or 314159. In this sense we then equate 9 with 99 et cetera

Why would we want do this, as it amounts to loss of information? Well, there are several reasons. Sometimes we are only really interested in the residual information. For example, when we study the set of numbers of the form $a+b\root 3\of 2+c\root 3\of 4$, where $a,b,c$ are integers, and we want to start adding, subtracting and multiplying them, we quickly notice that those operations are very similar to the corresponding operations involving polynomials $a+bx+cx^2$. The difference is that we are only interested in the value of the polynomial at a single point $x=\root 3\of 2$. This shows in the multiplication rule, because the polynomial $x^3$ takes the value $2$. In order to make this correspondence between polynomials and numbers more accurate we are forced to equate the polynomial $x^3-2$ with the polynomial $0$. This time we get a quotient ring instead of a quotient group (see algebra textbooks for such details), but the idea is that some things we have learned about polynomial algebra will carry over to our set of numbers, and that gives us the benefit of economy of thinking. We don't need to relearn everything from scratch, if the next time we are interested in $\root 3\of 3$ instead.

Sometimes quotient groups are forced upon us. We are not in possession of all the information. A simple example is the following. Assume that somebody is counting coins, but the only counting aid available to him is a light switch. Every time he tallies one more coin he will toggle the light switch: lit, dark, lit, dark,... He may or may not be able to keep track of the actual tally, but if somebody else comes to the room, or the tallyman gets confused, the status of the light switch will only tell whether an odd or an even number of coins have been counted, i.e. we have moved from the group $\mathbf{Z}$ to the quotient group $\mathbf{Z}/2\mathbf{Z}$. Another very common quotient group in mathematics is used to decribe an angle of rotation. Let's say that we are studying a planar object spinning about its center of mass. It may have completed God knows how many full revolutions, but when we enter the room and observe its position, we have no way of knowing anything else but the current direction pointed at by, say, a small arrow somebody painted on the object for this purpose. A full revolution corresponds to an angle of rotation $2\pi$, so the total angle of rotation will have an uncertainty that can be any integer multiple of $2\pi$. In other words, we can only see an element of the quotient group $\mathbf{R}/2\pi\mathbf{Z}$, not an element of $\mathbf{R}$.

$$
\begin{array}{c|c|c}
A & B & C \\ D & E & F \\ G & H & I \\ J & K & L
\end{array}
$$

$12\div 4 = 3$ because $3$ is how many sets of $4$ it takes to make a set of $12.$

A normal subgroup of order $4$ in a group of order $12$ has $3$ cosets; thus the quotient group has order $3$.


Tricky question related to sequences and series I have the following question with me:

"In the sequence $$1,9,7,7,4,7,5,3,9,4,1,....$$ every term from the fifth one is the sum of the previous 4 modulo 10. Do the numbers $1977$(apart from beginning) and $0197$ ever occur in the sequence? If yes, does it occur a finite number of times or not?"

How do I solve this question?  The only advancement I have made is that when the sequence is written modulo 2, I get $$1,1,1,1,0,1,1,1,1,0....$$

How do I proceed from here?

Hint: you can prove results about sequences like this without explicitly computing anything using two facts:

The number of possible sequences of four digits is finite, while the sequence you are given is infinite.
Any four consecutive digits in the sequence (uniquely) determine the whole sequence both forwards and backwards

Obviously you have to show that these facts are true for the sequence in hand, and then work out how to use them to get the result you need.

The things about explicitly computing residues can help in determining the period of a recurring sequence.

First, you can consider the sequence with index in whole integers. For example, we can define $a_{0} = 0$ since $0 + 1 + 9 + 7\equiv 7$ mod 10 and $a_{-1} = 7$ since $7 + 0 + 1 + 9 \equiv 7$ mod 10. Now try to explain that there exists $m\neq n$ such that $(a_{m}, a_{m+1}, a_{m+2}, a_{m+3}) = (a_{n}, a_{n+1}, a_{n+2}, a_{n+3})$ by using the pigeonhole principle.


Help with a recurrence with even and odd terms I have the following recurrence that I've been pounding on:

$$
a(0)=1\\
a(1)=1\\
a(2)=2\\
a(2n)=a(n)+a(n+1)+n\ \ \ (\forall n>1)\\
a(2n+1)=a(n)+a(n-1)+1\ \ \ (\forall n\ge 1)
$$

I don't have much background in solving these things, so I've been googling around looking for different techniques. Unfortunately, I haven't been able to correctly apply them yet. So far I've tried:

Characteristic equations
Generating functions
Plugging it into Wolfram Alpha
Telescoping
Observation

I'm sure that one or more of these is the way to go, but my biggest problem right now is figuring out how to deal with the idea that there are different equations for odd and even values of $n$. I'm not necessarily looking for a solution (though it would be gratefully accepted), but any nudges in the right direction would be much appreciated.

To follow up, it turned out that the speed problem described in my comment, below was related to the Decimal implementation in Python 2.7. Switching to Python 3.3 (and Java) made everything many orders of magnitude better.

I have an almost closed form. It uses the functions $z_2(n)$, which is the number of zeros in the binary representation of $n$ (after the first non-zero digit, of course) and $Z_2(n)=\sum_{i=1}^n z_2(n)$. I am quite sure these can be calculated quickly, but I'll leave that to you.

Section 1: A Closed Form For S(n).

Running off of Michael Biro's answer, where they define $S(n)=a(n+2)-a(n)$ and calculate that $S(2n)=S(n)+1$ and $S(2n+1)=S(n-1)$, we can find a closed form for $S$ and then find a neat sum for it.

In particular, rewrite Michaels two properties as:
$$S(n)=S(2n+3)=S(2n)-1.$$
and consider that this measures the numbers of even numbers we pass through as we iterate the function $f(2n+3)=n$ and $f(2n)=n$. We also have the edge conditions
$$S(0)=1$$
$$S(1)=2$$
$$S(2)=5$$
Now, this relation is clearly dealing with something happening in binary - but even better than binary is to write $n$ in base $2$, leading with the digit $1$ or $3$ and otherwise using the digits $0$ and $3$ - meaning that the expansion of $2n$ is the expansion of $n$, with a $0$ appended, and the expansion of $2n+3$ is the expansion of $n$ with a $3$ appended. This means that the map $f$ just chops off the last digits of this representation! A table of such representations follows:
\begin{array}{ll}
n & n_2 \\
 1 & 1 \\
 2 & 10 \\
 3 & 3 \\
 4 & 100 \\
 5 & 13 \\
 6 & 30 \\
 7 & 103 \\
 8 & 1000 \\
 9 & 33 \\
 10 & 130 \\
 11 & 1003 \\
 12 & 300 \\
 13 & 133 \\
 14 & 1030 \\
 15 & 303 \\
 16 & 10000 \\
 17 & 1033 \\
\end{array}
What is notable is that if the pattern starts with "3", then it will iterate to $0$ under $f$ after some number of steps, meaning $S(n)$ would be the number of $0$s in the expansion (how many times the orbit of $n$ under $f$ passed through even numbers), plus $S(0)=1$. If the pattern starts with "10", then $n$ first passes through $2$ under $f$, so $S(n)$ would be the number of zeros in the expansion (excluding the first one used in the pattern "10") plus $S(2)=5$. If the pattern starts with "13", then $n$ passes through $1$ under $f$ and so $S(n)$ equals the number of zeros in the pattern, plus $S(1)=2$.

This looks kind of intimidating to start with, but it's actually not too bad. Notice that if we chop off the most significant digit of our representation, the resulting representation is divisible by $3$, since it only uses the digits $0$ and $3$ and, moreover, if we divide that by $3$ and write it in normal binary, it's the same as if we replaced every $3$ by a $1$. Therefore, if we let $z_2(n)$ be the number of zeros in the normal binary representation of $n$, then we can interpret the above as follows, letting $2^b$ be the place value of the $1$ in our representation of $n$ if it exists (which must happen if $3$ doesn't divide $n$):
$$S(n)=\begin{cases}z_2\left(\frac{n}3\right)+1&&\text{if }n\equiv 0\pmod 3 \\ z_2\left(\frac{n+2\cdot 2^b}3\right)+2&&\text{if }5\cdot 2^{b-1} \leq n  \\ z_2\left(\frac{n+2\cdot 2^b}3\right)+4&&\text{if }n < 5\cdot 2^{b-1}\end{cases}$$
where we add $2\cdot 2^b$ to $n$ to convert the first digit from a $1$ to a $3$ and the condition that $5\cdot 2^{b-1}\leq n$ is equivalent to the expansion of $n$ beginning with "13". We require knowing $b$ to calculate the above, but noticing that we must have that
$$n=2^b+3\cdot k$$
for some $0\leq k \leq 2^b-1$, implying $2^b\leq n \leq 4\cdot 2^b - 3$, it is obvious that $b$ must either be $\lfloor\log_2(n)\rfloor$ or $\lfloor\log_2(n)\rfloor-1$ and that $n-2^b$ is divisible by $3$. So, we could write:
$$b=\begin{cases}\lfloor\log_2(n)\rfloor && \text{if }2^{\lfloor\log_2(n)\rfloor}\equiv n\pmod 3\\\lfloor\log_2(n)\rfloor-1 && \text{if }2^{\lfloor\log_2(n)\rfloor-1}\equiv n\pmod 3\end{cases}.$$

We could quickly write the expansion of some large integer using this information. Say we wanted to know the representation of $124315$ which is $1$ mod $3$ and is between $2^{16}$ and $2^{17}-1$ so $b'=16$. We note that $124315-2^{16}$ is divisible by $3$, so we know our expansion will have a $1$ in the $2^{16}$ position followed by the binary representation of $\frac{124315-2^{16}}3=1110010110011011_2$ with $1$'s replaced by $3$'s. This gives:
$$124315=13330030330033033_2$$

Section 2: A Closed Form For $a(n)$

We are going to have to split into cases for this, which is annoying. In particular, from the definition of $S$, we obviously have:
$$a(2n)=1+\sum_{i=0}^{n-1}S(2i)$$
$$a(2n+1)=1+\sum_{i=0}^{n-1}S(2i+1)$$
Let's start with the even case because it works out more nicely. In particular, let's rewrite $S(2i)=S(i)+1$ and get:
$$a(2n)=1+\sum_{i=0}^{n-1}S(i)+1$$
which means that we are taking the sum of $S$ over all numbers less than $n$. To do this, we just need to find all representations in our system giving a number less than $n$. For convenience, we'll define $[n]$ to be the greatest integer strictly less than $n$ (so $[5]=4$ for instance). Next, we get to four cases

First, let's list every representation starting with "3". This is equivalent to listing all binary representations less than $\frac{n}3$. So, since $S(3n)=z_2(n)+1$, if we define $$Z_2(n)=\sum_{i=1}^{n}z_2(i)$$ then the contribution of these $2\left[\frac{n}3\right]$ terms to the sum is: $$Z_2\left(\left[\frac{n}3\right]\right)+2\left[\frac{n}3\right]$$
Next, consider representations starting with "13". Choose $b_1$ to be such that $5\cdot 2^{b_1-1}\leq n\leq 4\cdot 2^{b_1}-3$, which would be the place value of the $1$ in the largest such representation less than or equal to $n$. Notice that if the $1$ is in a lower place value, but the string still began "13", then the representation would still be lesser than $n$, since it would be bounded above by $2\cdot 2^{b_1}< 5\cdot 2^{b_1-1} \leq n$. Therefore, if we take an binary number less than $\frac{n-2^{b_1}}{3}$, converted all the 1s to 3s and prepended a "1", we would get a valid representation starting with "3". The contribution of these terms to the sum would be:
$$Z_2\left(\left[\frac{n-2^{b_1}}3\right]\right)+3\left[\frac{n-2^{b_1}}3\right]$$
It is possible that no suitable $b_1$ exists, in which case choose $b_1$ to be the largest value such that $4\cdot 2^{b_1}-3<n$, meaning that any number with the $1$ in the place with value $2^{b_1}$ works, giving a contribution of $2^{b_1-1}-1$ terms as:
$$Z_2\left(2^{b_1-1}-1\right)+3\cdot 2^{b_1-1}-3$$
Now, consider representations starting with "100". Choose $b_2$ to be such that $$2^{b_2}\leq n \leq 7\cdot 2^{b_2-2}-3.$$ The contribution of these terms will be
$$Z_2\left(\left[\frac{n-2^{b_2-1}}3\right]\right)+7\left[\frac{n-2^{b_2-1}}3\right]$$
or, if no such $b_2$ exists, letting $b_2$ instead be the greatest value such that $2^{b_2}\leq n$, giving a contribution of:
$$Z_2\left(2^{b_2-2}-1\right)+7\cdot 2^{b_2-2}-7$$
Finally, we consider representation starting with "103". Choose $b_3$ to be such that $7\cdot 2^{b_3-2}\leq n \leq 5\cdot 2^{b_3-1}-3$. The contribution of thse terms will be
$$Z_2\left(\left[\frac{n-2^{b_3}}3\right]\right)+6\left[\frac{n-2^{b_3}}3\right]$$
and if no such $b_3$ exists, choose the largest such that $7\cdot 2^{b_3-2}<n$. Then, the contribution will be:
$$Z_2\left(2^{b_3-2}-1\right)+6\cdot 2^{b_3-2}-1.$$
We have not yet counted the strings "10", and "1". They contribute $9$ to the sum (assuming $n>2$).

Now, if you just sum up all those contributions, making the big assumption that I didn't make a single computation error in this entire post, you should get $a(2n)$. It's not really worth compiling those into a single equation, since there are essentially $8$ cases. I'll leave the odd cases to you, but it should proceed analogously. You just need to compute $Z_2(n)$ quickly, but I think this is probably easy enough since it has the identity:
$$Z_2(2^{k}-1)=k\cdot (2^{k-1}-1)$$
and should have nice identities for $Z_2(2^{k_1}+2^{k_2}-1)$ and so on as we specify more digits. (There's still substantial computation to do here, but it should take linear time in the number of digits, rather than quadratic time, as the naive algorithm would take)

If you consider each element and $a_0$, $a_1$, $a_2$,... You eventually see $a_2$ is just y(0)$a_{0}$.  And $a_3$ is just y(1)$a_{1}$.  $a_4$ = y(2)*y(0)$a_0$. $a_5$ = y(3)*y(1)$a_1$. Assuming you get a different answer for $a_0$ and $a_1$.  You will always get different results for even and odd inputs.


(Physics) Finding velocity with angle and speed Hi all, I'm very sorry if this is the wrong place to post a physics problem, but considering it's all maths I was hoping someone could help me. It's been a long time since I did maths on this or a similar level (nearly a decade) but I am learning physics for an exam I am planning to take this year and I have a long way to go!

I am trying to solve the a) calculation in the image.

I managed to work out the vertial (y axis) calculation. By doing:
tan(30) on the calculator which gave 0.58
0.58*200m/s = 115m/s

Now I have tried to do the a) calculation and I know the answer should be 231m/s as it's in the back of the book, but I don't know how to get it. The closest I've got so far is 221ms. I have been using the Khan Academy video to help me but I can't get it right - I'm so tired from trying...
http://www.youtube.com/watch?v=ZZ39o1rAZWY&list=SPAD5B880806EBE0A4

Can someone explain to me clearly why it is 231m/s. I tried to figure out the time and distance like he does in the video but I wasn't getting the right answer.

Thanks!

From Pythagoras $\sqrt{115^2+200^2}\approx 231$  Similarly, $\frac {200}{\cos 30^\circ}\approx 231$

Hey I got the answer of actual velocity of missile.
Just use Pythagoras theorem in the triangle.
Let velocity of missile be $x$,so
$$X^2=(200)^2+(115)^2$$
$$X^2=40000+13225$$
$$X^2=53225$$
$$X  =230.7054$$
$$X  = 231 m/s$$


Zeroes of $z^4+e^z$ in the unit disk How many zeroes does $f(z)=z^4+e^z$ have in the unit disc?

ADDED: can you calculate them?

Here the same question is asked about the disk of radius $2$. It can be solved easily by Rouché's theorem since when $|z|=2$, if $z=x+iy$ then $|e^z|=e^x\leq e^2$ so
$$|f(z)-z^4|=|e^z|\leq e^2\leq 9<16=|z^4|$$ therefore there are $4$ roots in that disc (up to multiplicity).

Now when $|z|=1$ we don't have this inequality, nor can we use the same trick by subtracting $e^z$ instead, since when $|z|=1$, $1/e\leq |e^z|\leq e$, one side is less than $1$ and the other greater.

This can be used to show that there are no roots in the right half of the disc, where by going around a curve approximating the boundary of the right half of the disc, so always $x>0$,  we get $e^x>1\geq |z|$. A direct calculation also shows no roots are on the $y$ axis.

Any ideas on how to deal with the left half?

This is not a solution, just a graphical representation of the modulus $|f(z)|$ (with colors tending to red when values tend to zero). This shows that there are two roots inside the unit disk with conjugate (approximate) values $-0.67003127469869344 \ \pm 0.5161249765067623 i$ (the two others are outside the unit disk).

Another way to be convinced graphically of the fact that there are 2 zeros inside the unit circle is by plotting $\frac{\text{arg }f(e^{it})}{2 \pi}$ (unwrapped) (using the argument principle) .


Laplace transform and Differentiation help I need to know how to differentiate something like these

a) Differentiate with respect to s  $$ b/(s^2 +b^2)$$

b) Solve the given equation
                               $$f(t) = s(1/s^2 + 2/s^3)$$

c) Reverse Laplace transform of $$(3s - 15)/(2s^2 -4s + 10)$$

For c),
$$
\frac{{3s - 15}}{{2s^2  - 4s + 10}} = \frac{3}{2}\frac{{s - 5}}{{s^2  - 2s + 5}} = \frac{3}{2}\frac{{s - 1 - 4}}{{(s - 1)^2  + 4}} = \frac{3}{2}\frac{{s - 1}}{{(s - 1)^2  + 2^2}} - 3\frac{2}{{(s - 1)^2  + 2^2 }}.
$$
Now let $\alpha = -1$ and $\omega = 2$, and consider 8-9 of this table.

EDIT: By 9 and 8 of that table, the inverse transforms of $\frac{{s + \alpha }}{{(s + \alpha )^2  + \omega ^2 }}$ and  $\frac{\omega}{{(s + \alpha)^2  + \omega^2 }}$ are $e^{-\alpha t} \cos(\omega t)$ and $e^{-\alpha t} \sin (\omega t)$, respectively. Hence the inverse transform of $\frac{{3s - 15}}{{2s^2  - 4s + 10}}$ is
$$
\frac{3}{2}e^t \cos (2t) - 3e^t \sin (2t).
$$

I'm going to GUESS that in (c), you really meant $(3s-15)/(2s^2-4s-10)$, despite the fact that you wrote $3s-\Big(15/(2s^2-4s-10)\Big)$.

The quadratic polynomial $2s^2-4s-10$ is $0$ when $s=1\pm\sqrt{6}$, so it factors as $2(s-1-\sqrt{6})(s-1+\sqrt{6})$.  Then you can do a partial-fraction decomposition as
$$
\frac{3s-15}{2s^2-4s-10} = \frac{A}{s-1-\sqrt{6}} + \frac{B}{s-1+\sqrt{6}}.
$$
You need to figure out numbers what $A$ and $B$ are.

Then find the inverse Laplace transform of each of the two terms separately.

Later edit:  Oh.  I see you had $+10$, not $-10$.  So the quadratic is irreducible.  But you should still find it useful to bear this kind of thing in mind.  I see that two others went through partial fractions without mentioning the term "partial fractions".  And they completed the square without mentioning the term "completing the square".


Truth Table - implies false I'm work with a task where I am not exactly sure if I proceed right. The task is saying: "We define the operation $\oplus$  by $a \oplus b = (a \wedge \neg b) \vee (\neg a \wedge b)$. Give the truth table of:"

$$(a \oplus b) \to \text{False}$$

So my question is, do I answer this with a truth table like this:

Or do I answer like this:

Or is it something I am not aware of? Thanks for any answer! :D

"False" is not a variable, so you should not add it into the truth table. It makes no sense to say "what if false is true", which is exactly what the second row in your first table asks.

Your second table is correct.


how is the signum function neither continuous nor discontinuous at $x=0$. In my book, there is a sentence that says exactly this:

"The function $\mathrm{sgn}(x)= \dfrac{x}{|x|}$ is neither continuous nor discontinuous at $x=0$. How is this possible?"

It was easy for me to tell it is not continuous at $x=0$ as there is no limit existence due to left-right limit inequality, or simply because the graph is broken at $x=0$. But I can't understand the second part, which claims it is also not discontinuous.

I've always thought that "if it is not continuous, then discontinuous", but apparently it seems to be wrong. How is this function not discontinuous?

source: A complete course: calculus(8th edition)

A function cannot be continuous and discontinuous at the same point.

Yes the function is discontinuous which is right as per your argument.

I think the question wanted to convey this..

It has a jumped discontinuity which means if the function is assigned some value at the point of discontinuity it cannot be made continuous.

But the function is definitely discontinuous at x=0.

the sgn function is a discontinuous function (isolated/jump discontinuity). it will be continuous if domain of the function is changed to R- or R+.

You can get some info about the types of discontinuity from the following sources:
https://en.wikipedia.org/wiki/Continuous_functio
https://en.wikipedia.org/wiki/Classification_of_discontinuities#Removable_discontinuity


The shape of a room that's bigger on the inside I have an advanced facility with many floors containing experimental technology. One of these floors goes on forever. What shape is it?

More specifically: within the room, you can move as far as you like in any given direction. It is possible to travel an infinite distance in a straight line, in the sense that you'll need to travel that same distance in order to get back to your starting point; travelling costs time and energy; there is (eventually) a speed of light delay in communications, etc etc. There are no observable edges to the room, so upon entering you appear to be have emerged the centre of a vast featureless space. Objects do not visibly appear distorted when moving through the space, but everything in the room is small enough to have been brought in through the door (let's say vehicle-garage-size).

This room is however not particularly useful, because despite its apparent size, you can't put very much in it - the storage capacity of the room is still limited by the containing building:

the volume within the room is no greater than the volume of any other floor of the building with an equivalent external footprint. Trying to put more objects into the room than would fit into a conventional space will result in Bad Things™ happening (certainly not anything a protagonist would want to experience first-hand).
the mass that can be contained within the room is limited by the supporting structure of the building, the same as on any other floor. Overload it and the floor will collapse. This is likely to result in problems connected to the first point, too, unless whatever process is maintaining the spatial distortion is turned off quickly by the destruction.

Trying to flood the level with water or cement would certainly be a very bad idea.

The cause of this effect is obviously waaay beyond the scope of a SE question, and anyway the protagonist isn't interested. However, for us creators with weaker math-fu, what would be a rigorous mathematical description of the effect on space that's being observed within the room?

I'm leaning towards something like a four-dimensional version of a shape along the lines of a Menger sponge, which demonstrates a similar effect to the room, in that it has infinite surface area but zero volume (or perhaps its inverse, which - I assume - also has infinite surface area, but unit volume). Moving on the surface of a Menger sponge would be slightly similar to a 2D version of the above, in that you could move infinitely on certain paths, but remain within an enclosed space.

Is there a fractal shape that fits the results described above and can be used to describe the space within the room? i.e. infinite lengths possible between points in the space it encloses, finite 3D volume for the whole shape.

The space doesn't have to appear empty, if unpacking a 4D shape into 3D would necessarily result in singularities or mathematically correct terms creating areas that can't be moved through cleanly / spaces that don't line up (it would actually be a bonus to have "free" floor and support struts throughout, that aren't real objects with mass), similarly to how you can't travel in an endless straight line in any direction on the infinite surface of the Menger sponge (I think).

(solutions that allow an infinite volume to be contained within the space are acceptable, if this is an otherwise impossible request - I can always use the weight limit to prevent the space from being useful, but limiting volume seems more interesting for the story)

One possibility is a flat $3$-torus.  This is the shape you get by taking a cube (or more generally, a parallelopiped) and declaring that the opposite faces of the cube are actually equal to each other.  This means that when you walk far enough in one direction, you "wrap around" back to the other side.  In particular, this means you can keep walking infinitely long in any direction, and room has no "edges".  The total volume of the torus is the same as the volume of the cube you started with, so it is finite.  And locally, the geometry is the same as the cube, so it is just ordinary Euclidean geometry, so it looks just like normal space as long as you don't look far enough to see things wrapping around.

However, this might not be very satisfying for the property of being able to "walk infinitely far", because while you can walk for an infinitely long time, after you walk a little bit, you'll find yourself right back where you started (or, technically, arbitrarily close to where you started, unless the direction you're walking is a rational linear combination of the edges of the cube).  And there is a uniform bound on the length of the shortest path between any two points.

Alternatively, if you're willing to drop the assumption that you can move infinitely far in every direction and just want there to exist directions you can go infinitely far in, it is easy to get such a shape without having any sort of weird topology.  Just take any nice ordinary subset of $\mathbb{R}^3$ like a ball, and attach some "tendrils" to it going out to infinity.  If you make the tendrils get thinner and thinner fast enough as they go out, the volume will still be finite.  I'm not sure this really captures your intuition though, as the directions in which you can go out to infinity are very restricted, and if you try to actually travel along them for very long, they will quickly become too thin for you to fit down them (assuming you are an actual physical object rather than a point with no volume).  The latter problem seems unavoidable though: if you can move an object with positive volume down an infinitely long path that doesn't loop back around to itself (as happens in the torus), then by tracing its path, you can see that your space must have infinite volume.

By the way, I'm assuming here that when you say there are "no edges", you're including that there is no floor or ceiling to the room (so up and down are among the directions you can move in).  If you want to have a floor and ceiling, you should probably just take the two-dimensional versions of these shapes (a flat $2$-torus, or a planar shape with tendrils), and let the vertical directions have ordinary geometry (in more mathematical terms, this means you take a product of the 2-dimensional shape with a closed interval, with the interval being the vertical direction).

Since at the moment you enter the room, all of the walls are infinitely distance, then if any given infinite distance is equal to all other infinite distances, then the room must be circular.

If it has a floor and a ceiling which are not infinitely distant, then your room looks like a coin which is lying on either of its faces.

If either the floor or the ceiling is infinitely distant while the other is not, then the coin looks like a bowl or a dome.

And if both the floor and ceiling are infinitely distant, then your room is a sphere.


Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$. Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.

I think we'll have to use number theory to do it. Simply solving the equations won't do.
If we divide the second equation by the first, we get:
$$x^2 - xy + y^2 = 1 + z.$$
Also, since they are integers $z^2 \ge z \implies -z^2 \le -z$. This implies
$$x + y = 1 - z \ge 1 - z^2 = x^3 + y^3.$$
This shows that atleast one of $x$ and $y$ is negative with the additive inverse of the negative being larger than that of the positive.
I have tried but am not able to proceed further. Can you help me with this?
Thanks.

A general identity that is nice to know is
$$x^3+y^3=(x+y)(x^2-xy+y^2).$$
Given that $x+y=1-z$ and $x^3+y^3=1-z^2$ we have $z=1-x-y$ and hence
$$x^3+y^3=1-z^2=(1-z)(1+z)=(x+y)(2-x-y).$$
This means that either $x+y=0$, so $y=-x$ and $z=1$, or
$$x^2-xy+y^2=2-x-y.$$
This is a quadratic in $x$, and quadratics are easy. Applying the quadratic formula yields
$$x=\frac{y-1\pm\sqrt{(1-y)^2-4(y^2+y-2)}}{2}=\frac{1}{2}\left(y-1\pm\sqrt{-3(y+3)(y-1)}\right),$$
and for these $x$ to be integers we need $-3(y+3)(y-1)$ to be a square, so certainly this must be nonnegative. It follows that
$$y\in\{-3,-2,-1,0,1\},$$
and these few cases can be checked by hand. For $(x,y,z)$ we find the solutions
$$(-2,-3,6),\ (-3,-2,6),\ (0,-2,3),\ (-2,0,3),\ (1,0,3),\ (0,1,3),$$
and $(x,-x,1)$ for all $x\in\Bbb{Z}$ from before.


Does noncompact manifold or orbifold have the homotopy type,of CW complex? I forget for a while, we don't need the compactness condition here right?

According to The Topology of CW-Complexes by Lundell and Weingram
(Van Nostrand Reinhold, 1969) the answer is yes for (separable) manifolds.

For smooth manifolds the following holds.
By the existence of Morse functions one can deduce a handle-body decomposition of a manifold.
This decomposition then yields a CW structure on a space homotopy equivalent to the manifold.
I don't think that compactness is needed in any of the above arguments.


Darboux integral epsilon-delta proof in piecewise continuous function Using Darboux sums, if $f$ is a piecewise continuous function in $[a,b]$, then

It is integrable in $[a,b]$.

Given $\epsilon>0$, there is $\delta>0$ such that for every $P$ partition:
$$||P|| < \delta\implies S\left(f,P\right) - I\left(f,P\right) < \epsilon$$

I already know how to prove the first proposition, my question is for the second one. Please and thank you. This can be proven adding a point $c$ to $[a,b]$.

If $f$ is continuous then for any $\epsilon> 0$ there exists a $\delta$ such that for all $(x,y)$ if $|x-y| < \delta \implies |f(x) - f(y)|< \epsilon.$

$\|P\|< \delta \implies |x-y|<\delta$

$|f(x) - f(y)|< \epsilon$ for all $(x,y)$ in $P$ is an equivalent statement as $|\sup(f(x)) - \inf(f(x))|< \epsilon$ for all $x$ in $P.$


Prove solutions to 3rd order differential equation form 3-dimensional vector space Prove that the solutions to $ \dfrac{ \mathrm{d^{3}y}}{ \mathrm{dx^{3}}} -6 \dfrac{ \mathrm{d^{2}y}}{ \mathrm{dx^{2}}}+11 \dfrac{ \mathrm{dy}}{ \mathrm{dx}}-6y=0$ form a 3-dimensional vector space:

Its simple to verify the set of solutions, $V$, form a vector space by noting $ \dfrac{ \mathrm{d}}{ \mathrm{dx}} ( \alpha f + \beta g) = \alpha \dfrac{ \mathrm{df}}{ \mathrm{dx}} + \beta \dfrac{ \mathrm{dg}}{ \mathrm{dx}} $, however I am unsure how to show $\dim V= 3$.

I know that in general, an nth order linear differential equation has n linearly independent solutions, but that has not been used in the previous part of the book this question is from and seems lengthy to prove, so is there an obvious way?

(Its clear $e^x$ is a solution but not sure how this helps).

Hint: The characteristic polynomial has three solutions $\lambda=1,2,3$:

Hence the general solution is $y=c_1\exp(x)+c_2\exp(2x)+c_3\exp(3x)$. The Wronskian determinant is non-zero, hence these are three linearly independent solutions spanning a three dimensional vector space (show by checking definition of vector space). This can be directly guessed by looking at the integer factors of $6$.


The degree of Gauss map If $M$ is an $2m$-dimensional closed orientable hypersurface in $\mathbb R^{2m+1}$, then we have a Gauss map $G:M\rightarrow S^{2m}$.

I have known from my differential geometry book that deg$G=\frac{1}{2}\chi(M)$ where $\chi(M)$ is the Euler characteristics of $M$. So my question is: is there an easy way to prove this conclusion if we assume the Poincaré–Hopf theorem and why is this argument only applied to the even dimension?

The Euler characteristic is zero for any odd-dimensional compact orientable manifold (this follows from Poincaré duality).


Standard deviation of the weighted mean How do you find the standard deviation of the weighted mean?

The weighted mean is defined: $\bar{x}_w = \frac{\sum{wx}}{\sum{w}}$

The weighted standard deviation (since it is not specified, I take it as of the distribution) is defined:

$$s_w = \sqrt{\frac{N'\sum_{i=1}^N {w_i(x_i-\bar{x}_w)^2}}{(N'-1)\sum_{i=1}^N{w_i}}},$$

where $N'$ is the number of nonzero weights, and $\bar x_w$ is the weighted mean of the sample (source)

For an unweighted sample, calculating the standard deviation of the mean from the standard deviation of the distribution is described on Wikipedia.

How do I calculate it for the weighted mean, and how is the expression derived?

I think that your answer is

$$\frac{\sum w_i(\theta_i-\bar{\theta})(\theta_i-\bar{\theta})'}{\sum w_i}$$

TO understand what the weights do, assume that the weights are all integers (if they are rational you can multiply out the LCM of the denominators). To keep it simple assume you have 3 data points with weight 2, 3 and 5.

Your mean is then
$$
\frac{2 x_1 + 3 x_2 + 5 x_3}{2 + 3 + 5}
$$

Now suppose you go back to your $3$ data points and want to make $x_3$ important (i.e. bias the mean towards $x_3$). One way to do this is to make up additional data by writing $x_3$ 5times.  You do the same for $x_2$ except you write it $3$ times and $x_1$  two times.

To be specific, suppose that $x_1=1$, $x_2=0$, $x_3=8$.

$$
\text{Unweighted mean} = \frac{1+0+8}{3} = 3
$$
With weights, your data is now $\left[1,1, 0,0,0,8,8,8,8,8\right]$. So your weighted mean is
$$
\text{Weighted mean} = \frac{1+1+0+0+0+8+8+8+8+8}{10} = 4.2$$
See how the mean shifted from $3$ towards $8$ to become $4.2$.

Weights are a way to make some data more important than others.


Poker Probability Problem I can't even begin to grasp how to approach such a statistics problem.

Given 9 different cards: two cards for 2 players, and 5 community cards.
Find n, where n is the number of same suited cards out of 9; so that 50% of the time at least one player has a flush.

For example:
If all 9 cards are the same suit, a flush will be present 100% of the time. Conversely, if only 4 cards are the same it drops to 0%.
n = 5 will rarely have a flush, and finding such a percentage is where I'm having trouble.

Thanks in advance to whoever broadens my view.

If you think about it you can easily remove $n=8$ as a possibility.

For $n=7$, this is the case where two cards are not from the suit. Fix one of these cards and notice that in most hands ($5$ out of every $9$) this card will not be in any of the players' hands (sorry for using the word hands with different meanings), and this of course corresponds to the case where at least one of the players has a flush.


Should BODMAS not be BODMSA? Let me blindly follow for a second BODMAS.
$$
\begin{align*}
1-3+2
&=1-(3+2)\\
&=1-5\\
&=-4
\end{align*}
$$
(The brackets are just to make the error clear - I wouldn't write them in "real life".)
This is correct according to BODMAS, because BODMAS doesn't talk about positive or negative numbers. It just says that addition should come before subtraction.

What am I missing here? Is it like "i before e except after c", and then you realise their is a word and the whole rhyme is as follows:

  i before e,

  Except after c,

  Or when sounded as "a,"

  As in neighbour and weigh.

So, am I missing some subtlety? Is the rhyme incomplete? Is it not true that subtraction has greater precedence than addition, and so BODMSA is actually correct and BODMAS is incorrect?

NOTE: This issue is true for all other mnemonics I know of, such as BIDMAS, PEMDAS, PIDMAS...

In the Danish educational system we are taught the order of operations in the following hierarchy:

Brackets
Roots and powers
Multiplication and division
Addition and subtraction

along with the rule of carrying out operations of equal order in the reading direction. In effect we have a BO(D,M)(A,S) system where (D,M) and (A,S) are pairs of equal order. This yields the same results for any calculation as all international standards prescribe, so no need to worry. The funny thing is that in our system you are meant to calculate as follows

$$
a\cdot b/c=(a\cdot b)/c
$$
whereas the English BODM(A,S) prescribes
$$
a\cdot b/c=a\cdot(b/c)
$$
Of course those two expressions are equal to one another, so it produces no controversies about Danish and international order of operation standards.

My best guess is that BODMAS was ordered that way in order to form a pronouncable word rather than the tongue twisting BODMSA knowing that A and S are equal as long as you apply the rule of the reading direction too.

NOTE: Actually PEMDAS would cause problems too unless we interpreted it as PE(M,D)(A,S). For instance $a/b\cdot c$ would equal $a/(b\cdot c)$ if multiplication really came before division. But in reality what is meant is reading from the left to the right performing operations of equal order, namely $a/b\cdot c=(a/b)\cdot c$.

It must be BODMSA at all times. BODMAS is fake. Don't try to defend it and reject the impeccable BODMSA on the grounds of pronunciation. Can the 'nice' pronunciation right the wrong?
Someone suggests we teach our students to consider, for example: 12-5+4 as 12+(-5)+4 without regarding the minus sign. I ask my fellow teacher, "what about 12-(-5)+4? Is the minus sign still insignificant?
In fact, we must admit to give priority to subtraction over addition in the order since it is correct at all times, whether it be at the left or the right.
BODMAS says 12-9÷3+5=4; also 12-(-5)+4=13 while BODMSA gives correct answers of 14 and 21 respectively.
It's true doing the operations from left to right is correct doesn't mean we should embrace what is not true.


Proving Theorem on Sub-Group Consider the below theorem on Sub-Groups

  Theorem : Let $(G, \circ)$ be a group and $H$ be a non-empty finite subset of $G$. Then $(H, \circ)$ is a subgroup of $(G, \circ)$ if and only if $a \in H, b \in H \implies a \circ b \in H$

Proving Forward : If $(H, \circ)$ is a subgroup of $(G, \circ)$, then $a \in H, b \in H \implies a \circ b \in H$ using Closure Property.

Proving Backward : If $a \in H, b \in H \implies a \circ b \in H$, then $(H, \circ)$ is a subgroup of $(G, \circ)$.

In order to prove this we have to satisfy 4 properties of group i.e.

Closure
Associative
Existence of Identity
Existence of inverse

Property 1: Satisfied

Property 2: Satisfied. Since $H$ is a non-empty subset of $G$ and $\circ$   is associative on $G$, $\circ$ is assoicative on H.

Property 3: ?

Property 4: ?

How can I prove existence of identity and existence of inverse with just information of Closure Property.

Since $H$ is non-empty, pick any old $a \in H$. Consider the map; $H \to H$ given by:

$h \mapsto ah$ (left-multiplication by $a$).

This map is injective (since it's injective on all of $G$), and since $H$ is finite, it's also surjective on $H$ (can you think of an example where this fails for some group $G$ and an infinite subgroup $H$?).

So there must be some $h_0 \in H$, with $ah_0 = a$ (it might be $a$ itself).

Similarly, the map $h \to ha$ is also surjective. This means for any $b \in H$, we have some $x \in H$ with $xa = b$.

Thus $bh_0 = (xa)h_0 = x(ah_0) = xa = b$, for any $b \in H$.

A similar argument shows that there is an $h_1 \in H$ with $h_1a = a$, and for any $c \in H$, another element $y \in H$ such that $ay = c$.

Hence $h_1c = h_1(ay) = (h_1a)y = ay = c$, for any $c \in H$.

So we have a left-identity, and a right-identity. Thus:

$h_1 = h_1h_0 = h_0$.

Thus $h_1 = h_0$ is an identity for $H$. Now it is easy to see that any element $H$ has both a left-inverse, and a right-inverse (just consider the pre-image of the element $e$ under the maps $x \to hx$ and $x \to xh$). So if we suppose $x$ is a left inverse for $h$, and $y$ a right-inverse:

$x = xe = x(hy) (xh)y = ey = y$

so two-sided inverses exist for any $h \in H$, and being unique in $G$, must be unique in $H$.

For any $a\in H$, $\;\{a^n\mid n\in\mathbf N\}\subset H $. Hence this set is finite, so there are exponent $i,j,\enspace i<j$, such that $a^i=a^j$. As we're in a group, the cancellation law gives $a^{i-j}=e\in H$, whence $a^{-1}=a^{i-j-1}\in H\;$ (note $i-j-1\ge 0$).


Is $\{v_1,v_2,v_3\}$ a basis for $H$? I don't understand the answer to the exercise that I'm doing, can anyone explain it in a more clear manner?

  Let $ v_1 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, v_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, v_3 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $, and let $H$ be the set of vectors in $\Bbb R^3$
  whose second and third entries are equal. Then every vector in $H$ has
  a unique expansion as a linear combination of $v_1, v_2, v_3$, because

  $ \begin{bmatrix} s \\ t \\ t \end{bmatrix} =  s \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + (t-s) \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}  + s \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $
  for any $s$ and $t$. Is $\{v_1,v_2,v_3\}$ a basis for $H$?

  Why or why not?

The answer given was:

  In order for the set to be a basis for $H$, $\{v_1,v_2,v_3\}$ must be a spanning set for $H$; that is, $H=\operatorname{Span}\{v_1,v_2,v_3\}$. The exercise shows that $H$ is a subset of $\operatorname{Span}\{v_1,v_2,v_3\}$, but there are vectors in $\operatorname{Span}\{v_1,v_2,v_3\}$ which are not in $H$ ($v_1$ and $v_3$, for example). So $H \neq \operatorname{Span}\{v_1,v_2,v_3\}$, and $\{v_1,v_2,v_3\}$ is not a basis for $H$.

I see $v_1$ and $v_3$ making $H$, why does it say they are not in $H$?

You'll notice that $H$ is indeed a subset of $\Bbb R^3$, but to be more precise, it's a subspace of it resembling $\Bbb R^2$ (it's a plane). To see this, notice that the first variable is free, and the third dependent upon the third.

Now, you can use the vectors $v_1, v_2, v_3$ to express points in that plane, but you'll notice that you can also use those vectors to express points that are not in the plane, $v_1$ for example, so $Span(v_1,v_2,v_3)$ does not span $H$.

you can express the standard basis $\{e_1, e_2, e_3\}$ as linear combinations of $\{v_1, v_2, v_3\}$ in the following way: $e_3 = \pmatrix{1\\0\\0} = v_2 - v_3, e_1 = \pmatrix{1\\0\\0}= v_1 - v_2 + v_3, e_2 = v_3.$  therefore $\{v_1, v_2, v_3\}$ spans $R^3.$  to show independence, suppose $$c_1v_1 + c_2 v_2 + c_3v_3 = 0$$ then you have $$\pmatrix{c_1\\c_2 + c_3\\c_1 + c_2} = \pmatrix{0\\0\\0}\to c_1 = 0, c_2 = 0, c_3 = 0. $$


Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero? If $a_n$ is a sequence such that
  $$a_1 \leq a_2 \leq a_3 \leq \dotsb$$
  and has the property that $a_{n+1}-a_n \to 0$, then can we conclude that $a_n$ is convergent?

I know that without the condition that the sequence is increasing, this is not true, as we could consider the sequence given in this answer to a similar question that does not require the sequence to be increasing.

$$0, 1, \frac12, 0, \frac13, \frac23, 1, \frac34, \frac12, \frac14, 0, \frac15, \frac25, \frac35, \frac45, 1, \dotsc$$

This oscillates between $0$ and $1$, while the difference of consecutive terms approaches $0$ since the difference is always of the form $\pm\frac1m$ and $m$ increases the further we go in this sequence.

So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge?  Or is this still not sufficient?

An easy way to visualize why this can't be true is to try putting some points on a number line.

Start with 1 point in [0, 1):

2 points in [1, 2):

And so on:

Now you have a sequence that grows to infinity but keeps getting closer together.

The condition $a_{n+1}-a_n \to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $\lim\limits_{n\to\infty}(a_{n+m(n)}-a_n)=0$ for all $m(n)\in \mathbb{N}$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.

Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.


Is there an analogue of Lebesgue’s Dominated Convergence Theorem for a net $ (f_{\alpha})_{\alpha \in A} $ of measurable functions? Is there an analogue of Lebesgue’s Dominated Convergence Theorem for a net $ (f_{\alpha})_{\alpha \in A} $ of measurable functions defined on a measure space $ (\Omega,\Sigma,\mu) $, where the index set $ A $ (assumed to be directed) is not necessarily $ \mathbb{N} $? That is, if

$ (f_{\alpha})_{\alpha \in A} $ converges pointwise almost everywhere to a measurable $ f $,
there exists a measurable $ g $ such that $ |f_{\alpha}| \leq |g| $ almost everywhere for each $ \alpha \in A $, and
$ \displaystyle \int_{\Omega} |g| ~ \mathrm{d}{\mu} \leq \infty $,

then does $ \displaystyle \lim_{\alpha \in A} \int_{\Omega} f_{\alpha} ~ \mathrm{d}{\mu} = \int_{\Omega} f ~ \mathrm{d}{\mu} $?

Here is an interesting case study involving a directed and well-ordered index set.

Assume that the Continuum Hypothesis holds, i.e., $ 2^{\aleph_{0}} = \aleph_{1} $. As the cardinality of the closed interval $ [0,1] $ is $ 2^{\aleph_{0}} $, we can find a bijection $ \phi: \omega_{1} \to [0,1] $. Define a net $ (f_{\alpha}: [0,1] \to [0,1])_{\alpha < \omega_{1}} $ of Lebesgue-measurable functions by
$$
\forall \alpha < \omega_{1}, ~ \forall x \in [0,1]: \quad
{f_{\alpha}}(x) \stackrel{\text{def}}{=}
\begin{cases}
1 & \text{if $ x = \phi(\beta) $ for some $ \beta \leq_{\mathbf{On}} \alpha $}; \\
0 & \text{elsewhere}.
\end{cases}
$$
Consider an ordinal $ \alpha < \omega_{1} $. As $ \alpha $ is countable, it follows that $ f_{\alpha} $ assumes the value $ 1 $ for at most countably many arguments and the value $ 0 $ everywhere else. Hence, $ f_{\alpha} $ is Lebesgue measurable and is zero almost everywhere. We thus have
$$
\forall \alpha < \omega_{1}: \quad
\int_{[0,1]} f_{\alpha} ~ \mathrm{d}{\mu} = 0.
$$
However, we have $ \displaystyle \lim_{\alpha \to \omega_{1}} f_{\alpha} = 1_{[0,1]} $, which yields
\begin{align}
       \lim_{\alpha \to \omega_{1}} \int_{[0,1]} f_{\alpha} ~ \mathrm{d}{\mu}
& =    \lim_{\alpha \to \omega_{1}} 0 \\
& =    0 \\
& \neq 1 \\
& =    \int_{[0,1]} 1_{[0,1]} ~ \mathrm{d}{\mu} \\
& =    \int_{[0,1]}
       \left( \lim_{\alpha \to \omega_{1}} f_{\alpha} \right)
       \mathrm{d}{\mu}.
\end{align}

Most likely wrong, see comment below (leaving it as the mistake may be instructive)

The same theorem does apply, from what I recall. Take any sequence $(a_n)$ such that $a_n \xrightarrow[n\to\infty]{} a$, and set $g_n\stackrel{\rm def}{=} f_{a_n}$. Can't you apply your version of Lebesgue's Dominated Convergence Theorem to $(g_n)_{n\in\mathbb{N}}$, which implies what you want as $(a_n)_{n\in\mathbb{N}}$ is arbitrary?


Differential of square $d(w^2)$or square of differential$(dw)^2$!? in Curved space
it seems $d(w^2)=(dw)^2$ how is it possible!?

$$x^2+y^2+z^2+w^2=\kappa^{-1}R^2,$$
first:
$$dw=-w^{-1}(xdx+ydy+zdz),$$

$$\kappa^{-1}R^2-(x^2+y^2+z^2)=w^2,$$
$$dl^2 = dx^2 + dy^2 + dz^2+dw^2,$$

$$dl^2 = dx^2 + dy^2 + dz^2
+\frac{(xdx+ydy+zdz)^2}{\kappa^{-1}R^2-x^2-y^2-z^2}$$

Remark:
in general as i know $dx^2=2xdx$ so $dw^2$ should be:

$$d(w^2)=(2w)w^{-1}(xdx+ydy+zdz)=2(xdx+ydy+zdz),$$
but here

$$d(w^2)=(dw)^2=w^{-2}(xdx+ydy+zdz)^2,$$
how is it possible?

In general, $dx^2=2dx$, but you've been confused by notation enlightening, where the parenthesis are omitted because there should be no possible confusion. Imagine that page (or any work on general relativity) with every parenthesis explicitly written. My eyes'd start to bleed.


Equivalent definition of uniform continuity at infinity I am trying to make an equivalent definition of uniform cointiniity for functions that converge at infinity. Thank you in advance for your time.

Given $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x)\rightarrow l$ when $x\rightarrow \infty$

*The traditional definition of uniform coninuity that I know:

$f$ is uniformly continuous iff for every $\epsilon>0$ there is a $d>0$ such that for every x,y (pertaining to $\mathbb{R}$)  $|x - y|< d$ implies $|f(x) - f(y)|< \epsilon.$

*This are two definitions that I would like to prove to be true (given same $f$) :

1) $f$ is uniformly continuous iff for every $\epsilon>0$ there is an $M>0$ such that for every
$x,y$(pertaining to $\mathbb{R}$) $ x,y>M$  implies $|f(x) - f(y)| < \epsilon$

2) $f$ is uniformly continuous iff for every $\epsilon>0$ there is an
$N>0$ such that for every
two sequences $\{X_n\}$,$\{Y_n\}$ that diverge to $(\infty$) $Ñ>N$ implies
$|f(X_ñ) - F(Y_ñ)|< \epsilon$

Are 1) or 2) equivalent to the traditional definition of uniform continuity?

Since you only consider asymptotic behavior of the function as $x\to +\infty$, I assume you actually mean to study functions $f: R^+\to R^+$, since for functions on $R$ itself one can't conclude anything without knowing something about the asymptotic behavior at $-\infty$.

If so, the functions $f$ you consider, namely those that satisfy $\lim_{x\to\infty}f(x)=1$, are necessarily uniformly continuous.  Your conditions (1) also guarantees that there will be a limit value at infinity, and therefore the function is uniformly continuous.

Your condition (2) is a bit problematic because $N$ should be dependent on the sequences chosen (you must change the order of the quantifiers).

no, there could be an N such that after some well-described first upper limit, the functional behavior approaches abruptly a differing limit; the upper bound could be a spike jump (upwards or downwards) from the previous lower/upper limit due to accumulated built-up potential.  Thus, the continuity constraint plays a key-role.  Does it infer to monotonic or non-monotonic functions, that's the main question to consider.  For ratioanl functions that could be an evasive question.


$x^q < y^q \implies x < y$ proof $x^q < y^q \implies x < y$ , $q $ is a rational number. For a natural number it is quite easy. We can show it by induction. As $x<y \leftrightarrow x*x < y*x < y*y$ and so on..But how could it be done for any rational number ?
I guess it is only possible for $x,y > 0$

Let $x>0$, $y>0$, $q>0$. Suppose that $q=m/n$ with $m,n\in\mathbb N$. From the fact that $x^q<y^q$ we get

\begin{align*}
0<y^q-x^q&=y^{m/n}-x^{m/n}=\left(y^{1/n}\right)^m-\left(x^{1/n}\right)^m\\
&=\left(y^{1/n}-x^{1/n}\right)\cdot\left(y^{(m-1)/n}+y^{(m-2)/n}\cdot x^{1/n}+\cdots+y^{1/n}\cdot x^{(m-2)/n}+x^{(m-1)/n}\right)
\end{align*}

The expression in the last brackets is positive therefore

$$
y^{1/n}-x^{1/n}>0\quad\Rightarrow\quad y^{1/n}>x^{1/n}\quad\Rightarrow\quad \underbrace{y^{1/n}\cdot y^{1/n}\cdots y^{1/n}}_{\text{$n$-times}}>\underbrace{x^{1/n}\cdot x^{1/n}\cdots x^{1/n}}_{\text{$n$-times}}\quad\Rightarrow\quad y>x.
$$

$q >0, z>0$.

$f(z):=z^{1/q}$ is strictly increasing, since

$f'(z)=(1/q)z^{(1/q-1)} >0 $.

For $x^q<y^q$ :

$f(x^q)=x<f(y^q)=y$.

What happens if $q <0$ ?


Probability of car starting first time The Question

One a given day, Claude's car has an 80% chance to starting first time
and Andre's car has a 70% chance of the same, Given that one of the
cars has started first time, what is the chance that Andre's car
started the first time?

My Understanding

Event A: Claude's car starting first time
Event B: Andre's car starting first time

We want at least one car started first time, also event B occurs, so there are only these $2$ cases: AB and A'B. I calculated the probability of these $2$ separately and multiply them together, which is $0.8\times 0.7\ {+}\ 0.2 \times 0.7=0.7$, which makes sense since event A and B are independent events, the occurrence of A should not affect the probability of B. I have to try something else because this is not the answer.
Then I tried to calculate the probability of B', which is when Andre's car is not starting first time.
there's only one case for this to be happening while at least one car is starting first time. A B'.
the probability for this is $0.8 \times 0.3 =0.24$. then I subtracted this from $1$ to get $76\%$.
the answer for this is $\frac{35}{47}$
It's close but not really. Can someone help me with this? I can't think of more ways of solving it.

Let me rename the events, because I keep thinking event A refers to Andre.
Let A be the event that Andre's car starts the first time and let C be the event that Claude's car starts the first time.  We are asked for $$\Pr(A|A\cup C)=\frac{\Pr(A\cap(A\cup C))}{\Pr(A\cup C)}=\frac{\Pr(A)}{\Pr(A\cup C)}$$
Now, $\Pr(A\cup C)=\Pr(A)+\Pr(C)-\Pr(A\cap C) = .7+.8-.7\cdot.8=.94$, so the required probability is $$\frac{.7}{.94}=\frac{35}{47}$$

I think simplest way is using a tree diagram
enter image description here
P(At least one of them starts)=0.8X0.7+0.2X0.7+0.3X0.8=0.94
P(Andre starts first)=0.8X0.7+0.2X0.7=0.7
P(Andre given started)=0.7/0.94=35/47
Hope this helps


Right triangle inscribed in a square. Find the square area? I hope it's valid to ask for "a more neat solution" of a problem on this network, despite the fact that I don't have a strict definition of the word "neat".
Here is the square and the right triangle inscribed in it.

I did the following:
$$AC = Ah + hB$$
$$4\sin\theta = 4\cos\theta + 3\sin\theta$$
So
$$\tan\theta = 4$$
But
$$\sin\theta = \frac{\tan\theta}{\sqrt{1+\tan^2\theta}}$$
Therefore
$$AC = 4\sin\theta = \frac{16}{\sqrt {17}}$$
$$\text{Area} = \left(\frac{16}{\sqrt {17}}\right)^2$$

Without using trigonometry:

enjoy...

By coloring triangles the same I am trying to emphasize the similar triangles, which are not necessarily equal (although red and blue triangles are). I don't mean that the triangles with the same color has the same area, be careful.

Noting CAh and hBf are similar triangles in the linear ratio 4:3, redraw the original figure in a 4x4 square. The square of its hypotenuse Ch² is therefore 17 (see figure) - which is too large, it should be 4² or 16. Since the area of any similar figure is proportional to the square of any of its linear dimensions, simply scale the 4x4 square by 16/17 to find the area of the original figure: 16²/17

redrawn figure


Is this series conditionally convergent or absolutely convergent? $ \sum_{k=1}^{\infty}\left(-1\right)^{k+1}\frac{k^{k}}{\left(k+1\right)^{k+1}} $ Is this series conditionally convergent or absolutely convergent? How to prove it ?

I tried ratio test and root test, but did not work I think.

$
\sum_{k=1}^{\infty}\left(-1\right)^{k+1}\frac{k^{k}}{\left(k+1\right)^{k+1}}
$

To show that $\sum_{k=1}^{\infty}\left(-1\right)^{k+1}\frac{k^{k}}{\left(k+1\right)^{k+1}}$ converges, it suffices to show that $k \mapsto \frac{k^{k}}{\left(k+1\right)^{k+1}}$ is monotonically decreasing and then we can apply the alternating series test.

Consider $$f(x) = \ln\frac{x^x}{(x+1)^{x+1}} = x \ln x - (x+1)\ln (x+1)$$ We have $f'(x) = \ln x - \ln (x+1) < 0$ so $f$ is monotonically decreasing.

Then $e^{f(x)} = \frac{x^x}{(x+1)^{x+1}}$ is also monotonically decreasing so the claim follows.

Hints

Use the alternating series test to determine whether the series is convergent. For absolute convergence use the limit comparison test, noting that
$$
\lim_{k\to \infty}\frac{k^k}{(k+1)^{k+1}}\biggr/\frac{1}{k}=\lim_{k\to \infty}
\left(1-\frac{1}{k+1}\right)^{k+1}=\frac{1}{e}
$$
and compare with the harmonic series.


First order logic. Describe that a set has more than 2 elements. I would like to describe that a set has at least 3 elements using first order logic, would this be a valid way to do that?

$\forall x\exists y\exists z(\neg(x=y)\wedge\neg(x=z)\wedge\neg(y=z))$

I can't find this example anywhere in my book, but it seems to do what I want, comments?

You need to assert the existence of three elements $x, y, z$ (so $\forall x$ is incorrect), and
you need to assert something about those existent elements, otherwise you are only saying "there exists at least three elements $x, y, z$". Presuming you want to assert that they all belong to some set we'll call $A$, then
you need a proposition, e.g., $A(x)$ to denote $x \in A$, where $A$ is some set.

Without the proposition, you are claiming only the existence of at least three elements, but nothing about their being contained in some set.

$$\exists x \,\exists y\, \exists z\,\Big(A(x) \land A(y) \land A(z) \land \;\lnot(x= y )\;\land \;\lnot (y= z) \;\land\; \lnot(x= z)\Big)$$

Or, simply, $$\exists x \,\exists y \,\exists z\,\Big(x\in A \;\land\; y\in A\; \land \;z\in A\;\land\; \lnot(x= y )\;\land \;\lnot (y= z) \;\land\; \lnot(x= z)\Big)$$

If we may use universal and existential quantifiers in FOL, then here are some fun axiomatizations.

#import: written-on(word,paper).

my-minus-one(N,M) <->  (N > 1)  &  (M = N-1)

forall x. same(x,x)

% confirms non empty domain.
legal-pad(paper) <-> min-elements(1, paper)

exactly-n-elements(1,paper) <->
(exists paper word.
    (   written-on(word,paper)
      & ~(exists other-word.
           (  written-on(other-word,paper)
             & ~same(other-word,word)))))

exactly-n-elements(2,paper) <->
(exists paper word1 word2.
  (  written-on(word1,paper)
   & written-on(word2,paper)
   & ~same(word1,word2)
   & ~ (exists other-word.
         (   written-on(other-word,paper)
          & ~same(other-word,word1)
          & ~same(other-word,word2)))))

exactly-n-elements(N,paper) <->
   range-elements(N,N,paper)

range-elements(N,M,paper) <->
 (exists paper.
  ( min-elements(N,paper)
    & max-elements(M,paper)))

min-elements(1,paper) <->
 (exists paper word1.
     (written-on(word1,paper)))

min-elements(2,paper) <->
 (exists paper word1 word2.
   (  written-on(word1,paper)
    & written-on(word2,paper)
    & ~same(word1,word2)))

max-elements(2,paper) <->
(exists paper word1 word2.
   (  written-on(word1,paper)
    & written-on(word2,paper)
    &  ~(exists other-word.
          (  written-on(other-word,paper)
            & ~same(other-word,word1)
            & ~same(other-word,word2)))))

% exists a paper with no elements
exactly-n-elements(0,paper) <->
  (exists paper
     ~(exists word1. written-on(word1,paper)))

max-elements(1,paper) <->
 exactly-n-elements(0,paper) v exactly-n-elements(1,paper)

containsAtLeastOneUnique(paper1,paper2) <->
  (exists word. (written-on(word,paper1) -> ~written-on(word,paper2)))

disjoint(paper1,paper2) <->
  ~(exists word. (written-on(word,paper1) & written-on(word,paper2)))

subset(paper1,paper2) <->
    (forall word.
      written-on(word,paper1) -> written-on(word,paper2) )

union(paper1,paper2,paper) <->
 (forall word.
   (written-on(word,paper) <->
       (written-on(word,paper1) v written-on(word,paper2))))

union(paper1,paper2,paper) <->
  (exists scratchpad.
      union(paper1,paper2,scratchpad)
     & ~containsAtLeastOneUnique(paper,scratchpad))

union-disjoint(paper1,paper2,paper) <->
   (   union(paper1,paper2,paper)
     & disjoint(paper1,paper2))

min-elements(4,paper) <->
  (exists paper1 paper2.
     min-elements(2,paper1)
   & min-elements(2,paper2)
   & union-disjoint(paper1,paper2))

min-elements(N,paper) <->
  (exists paper1 paper2.
     min-elements(1,paper1)
   & min-elements(M,paper2)
   & legal-pad(paper2)
   & union-disjoint(paper1,paper2)
   & my-minus-one(N,M))

max-elements(N,paper) <->
  (exists paper1 paper2.
     max-elements(1,paper1)
   & max-elements(M,paper2)
   & legal-pad(paper1)
   & legal-pad(paper2)
   & union-disjoint(paper1,paper2)
   & my-minus-one(N,M))

equal_papers_v2(paper1,paper2) <->
   ( ~containsAtLeastOneUnique(paper1,paper2)
     & ~containsAtLeastOneUnique(paper2,paper1)
     & legal-pad(paper1)
     & legal-pad(paper2))

equal_papers_v1(paper1,paper2) <->
 ( legal-pad(paper1)
   & (forall word.
      (written-on(word,paper1) <-> written-on(word,paper2))))


Minimum of sum of squares over sums I am trying to minimize $\phi(\alpha)$, where $\alpha \in \mathbb{R}^K$.

$\phi(\alpha) = \frac{R^2 + G^2 \gamma \sum_{i=0}^{K} A_i \alpha_i^2}{\sum_{i=0}^{K} A_i \alpha_i} $

Where, $A_i = \gamma \beta^i (K+1-i)$. $K, \beta, \gamma$  are constants. $K$ is a positive integer, $\beta \in (0,1)$, $\gamma = 1 - \beta$.

I am trying to obtain a sequence $\alpha_i$ which will minimize $\phi(\alpha)$. I am proceeding the standard way of differentiating $\phi(\alpha)$ with respect to each $\alpha_i$ and setting it to zero. Further, I know $\phi(\alpha)$ is a convex function. My final objective is to obtain a solution in closed form, however it is turning out to be a nightmare.

I am actually solving a generalization of a problem. The special case being when $A_i$ are constants. In which case it is easy to argue that $\phi(\alpha)$ is a symmetric convex function and the solution should the one in which $\alpha_i$ are constants.

Am I proceeding in the correct direction? Please suggest.

I seeked a numerical solution to this problem. This problem being a {convex}./{linear} cannot be directly solved with CVX for example. Epi-graph trick and bisection give a way to solve this problem numerically.

Thus this problem can be transformed to,

$\underset{\alpha_0,\cdots,\alpha_K}{ minimize }\ t \\
 subject\ to\ \ \ numerator \leq t \times denominator $

Which in turn can be solved by bi-section. Here is the code with matlab/CVX. Repeat this code with various t, which can be initially chosen as average of an upper bound and an lower bound and updating the upper bound or lower bound based on if the problem is feasible or not.

cvx_begin
variable alphaS(K)
minimize 1
subject to
( R^2 + G^2 * sum_square( diag(b) * alphaS ) ) - t * (2*sum(alphaS) ) <= 0
alphaS >= 0
cvx_end

Entire matlab script,

b = zeros(K,1);
for p=1:K
    b(p) = 1-beta^p;
end

maxSubDivisions = 20;

% epigraph trick / bisection
upper = 2;
lower = 0;

loopCount=0;
while( 1 )
    loopCount = loopCount + 1;

    t=(upper+lower)/2;
    cvx_begin quiet
    variable alphaS(K)
    minimize 1
    subject to
    ( R^2 + G^2 * sum_square( diag(b) * alphaS ) ) - t * (2*sum(alphaS) ) <= 0
    alphaS >= 0
    cvx_end

    display( sprintf( '#%d : %s', loopCount, cvx_status ) );
    if( strcmp( cvx_status, 'Infeasible' ) == 1 )
        lower = t;
    end

    if( strcmp( cvx_status, 'Failed' ) == 1 )
        break;
    end

    len = length(strfind( cvx_status, 'Solved' ));
    if( ( len == 1) )
        upper = t;
        optArg = alphaS;
        optFVal = t;

        if( length(strfind( cvx_status, 'Inaccurate' )) == 1 )
            break;
        end
        if( loopCount > maxSubDivisions )
            break;
        end

    end

end


Does the following define a Mahlo cardinal? Let M be a cardinal with the following properties:
  - M is regular
- $\kappa < M \implies 2^\kappa < M$
- $\kappa < M \implies s(\kappa) < M$ where $s(\kappa)$ is the smallest strongly inaccessible cardinal strictly greater than $\kappa$

My question is: Is M a Mahlo cardinal ? If so, how does the definition above connect to the usual definition in terms of stationary sets ?

Motivation: My intuition about a Mahlo cardinal is that it you cannot reach it by taking unions, power sets or "the next inaccessible cardinal", which is my definition above.
My worry is, I might have arrived at something much smaller than Mahlo.

No, your condition doesn't imply Mahloness.

First, note that your first two conditions simply state that $M$ is inaccessible, and the third one gives that $M$ is limit of inaccessibles. Now consider the first cardinal $M_0$ which is inaccessible limit of inaccessibles. That is, every inaccessible below $M_0$ is not a limit of inaccessibles. Now consider $C$ consisting of limits of the set $I$ of inaccessible cardinals smaller than $M_0$. By choice of $M_0$, $C$ contains no inaccessibles below $M_0$, so if we show $C$ is a club below $M_0$, we will get that $M_0$ is not Mahlo.

$C$ is closed: this is true because the set of limit points of any set is closed. For the sake of completion, I will sketch the argument: suppose that each of $\alpha_i$ is a limit point of $I$, where $(\alpha_i)_{i<\kappa}$ is an increasing sequence of cardinals with limit $A$. Let $\beta<A$. This means that there is some $\alpha_i$ between $\beta$ and $A$. Since $\beta<\alpha_i$, there is some element of $I$ which is between $\beta$ and $\alpha_i$, hence between $\beta$ and $A$. So $A$ is a limit point of $I$, so $A\in C$. Hence $C$ is closed.

$C$ is unbounded below $M_0$: Let $\alpha<M_0$. Then $s(\alpha),s(s(\alpha)),s(s(s(\alpha))),...$ is an increasing sequence in $I$ of length $\omega$. Since cofinality of $M_0$ is (clearly) greater than $\omega$, limit of this sequence, which is an element of $C$, is between $\alpha$ and $M_0$. Hence $C$ is unbounded below $M_0$.

Indeed, the condition "inaccessible limit of inaccessibles" (or simply "regular limit of inaccessibles") is a condition known as 1-inaccessibility. It can be generalized: we can speak of 2-inaccessibles, which are regular limits of 1-inaccessibles, 3-inaccessibles which are regular limits of 2-inaccessibles and so on. This can be obviously extended to a successor ordinal, and for limit ordinal $\alpha$ (e.g. $\alpha=\omega$) we say that cardinal is $\alpha$-inaccessible if it's $\beta$-inacessible for $\beta<\alpha$ (this isn't circular definition, but rather it proceeds by transfinite induction).

For every ordinal $\alpha$, notion of $\alpha+1$-inaccessibility is strictly stronger than that of $\alpha$-inaccessibility. As can be shown, every Mahlo cardinal $\kappa$ is not only 1-inaccessible, 2-inaccessible, $\omega$-inaccessible, but even $\kappa$-inaccessible. Notion of Mahloness is even stronger than that, but for that, I'd recommend reading this section on Wikipedia article.

Thanks a lot for your answers, Wojowu and Andreas.
I find I am not able to simply reply with a comment while being a guest, so I am writing it this way.

The mental picture I am getting of Mahlo versus inaccessibles is somewhat like $\aleph_1$ versus the Veblen hierarchy of countable ordinals - you keep recursively strengthening definitions and taking fixed points, but just never get there. Don't know how strong the analogy is, though.


Parametric solution of the Diophantine equation $\frac{3}{n}=\sum\frac{1}{a}$ Assmue  $n>3$ is a odd number,Prove that there exists distinct odd numbers $a,b,c$ such
$$\dfrac{3}{n}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\tag{1}$$
I'm reading a lot about the Erdös-Straus Conjecture (ESC), a conjecture that states that for every natural number p≥2, there exists a set of natural numbers $a,b,c$, such that the following equation is satisfied:
$$\dfrac{4}{p}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$$
 see Egyptian fraction
But still I have no idea about how to attack the  one $(1)$

This provides only a solution for $n$ not both prime and $-1\mod 12$, as well as how I derived it.

Let's try to find a couple such triples first.
\begin{align}
\frac35&=\frac13+\frac15+\frac1{15}\\
\frac37&=\frac13+\frac1{11}+\frac1{231}\\
\frac39&=\frac15+\frac19+\frac1{45}
\end{align}
We see that in the first and third case, we used a fraction $\frac1n$ so we can try to solve

$$\frac2n=\frac1a+\frac1b$$

instead. The above equation reduces to $n=\frac{2ab}{a+b}$ so we'll solve $a+b\mid 2ab$ for odd $a,b$. First let $\gcd(a,b)=d$ so that $d\mid a+b$. Now note that $a+b\mid 2ab-2b(a+b)$ so $a+b\mid 2b^2$ and so also $a+b\mid 2a^2$. From those two follows:

$$a+b\mid \gcd(2a^2,2b^2)=2\gcd(a^2,b^2)=2\gcd(a,b)^2=2d^2$$

Now let $x=a+b$. We have $d\mid x$ and $x\mid 2d^2$. We write $dk=x$ and $x\gamma=2d^2$ and solving for $x$ and $d$ yields $d=\frac{k\gamma}2$ and $x=\frac{\gamma k^2}2$. We can write $a=\alpha d$ and $b=\beta d$, and using the above expressions we see $k=\alpha+\beta$, which we can substitute into the expression for $d$, after which we substite that expression into $a=d\alpha$ and $b=d\beta$ to get that
\begin{align}
a=\tfrac12\gamma\alpha(\alpha+\beta)\\
b=\tfrac12\gamma\beta(\alpha+\beta)
\end{align}
where $\alpha,\beta,\gamma$ are all odd. Now we see that

$$\frac2n=\frac1a+\frac1b=\frac{\alpha+\beta}{\tfrac12\gamma\alpha\beta(\alpha+\beta)}=\frac2{\alpha\beta\gamma}$$

so that we need to write $n=\alpha\beta\gamma$. Note that for $a$ and $b$ to be distinct, $\alpha$ and $\beta$ need to be distinct - and we wanted that, so let's choose $\alpha=1$. Also, since we don't want $b$ to be equal to $n$ (since we actuall chose $c=n$, the variable that we chose at the very start), we need $n$ to be composite, since then we can choose $b$ a divisor of $n$ not equal to $1$ or $n$ itself (and since $n$ should be odd, those divisors will also be odd automatically). Now let $n=rs$, where $r,s$ are both positive integers, and $r\equiv1\mod 4$, but also $r\neq1$, then we have a solution

$$\frac3{rs}=\frac1{rs}+\frac{1}{\tfrac12s(r+1)}+\frac1{\tfrac12rs(r+1)}$$

and since $r\equiv 1\mod4$, $r+1$ is only divisible by $2$ once, so that $\tfrac12(r+1)$ is odd, and so $\tfrac12s(r+1)$ and $\tfrac12rs(r+1)$ are odd, so this is a solution.

So we handled all $n$ that have at least one divisor $1\mod 4$ (that divisor not being equal to $1$) and so we're left with all $n$ that are prime and $3\mod 4$.

Let's try some examples where $n\equiv3\mod4$ is prime.
\begin{align}
\frac37=\frac13+\frac1{11}+\frac1{231}\\
\frac3{11}=\frac15+\frac1{15}+\frac1{165}\\
\frac3{19}=\frac17+\frac1{67}+\frac1{8911}
\end{align}
What we see here is that $\frac37-\frac13=\frac2{21}$, and $\frac3{19}-\frac17=\frac2{133}$. Since we already solved $\frac2x=\frac1a+\frac1b$, we can use this; let $n=12p+7$ be prime. Now we can choose $c=4p+3$ and see
\begin{align}
\frac{3}{12p+7}-\frac1{4p+3}&=\frac{12p+9}{(12p+7)(4p+3)}-\frac{12p+7}{(12p+7)(4p+3)}\\
&=\frac{2}{(12p+7)(4p+3)}
\end{align}
Which we know is solvable, since we can just look at our solution above to that kind of equation, and find $a=\tfrac12(12p+7)(4p+3)((12p+7)(4p+3)+1)$ and $b=\tfrac12((12p+7)(4p+3)+1)$.

Now we've narrowed it down to primes $-1\mod12$.

It was necessary to write the solution in a more General form:

$$\frac{t}{q}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

$t,q$ - integers.

Decomposing on the factors as follows: $p^2-s^2=(p-s)(p+s)=2qL$

The solutions have the form:

$$x=\frac{p(p-s)}{tL-q}$$

$$y=\frac{p(p+s)}{tL-q}$$

$$z=L$$

Decomposing on the factors as follows: $p^2-s^2=(p-s)(p+s)=qL$

The solutions have the form:

$$x=\frac{2p(p-s)}{tL-q}$$

$$y=\frac{2p(p+s)}{tL-q}$$

$$z=L$$

To find solutions, you can always write.  $3L-q=1 ; 2$

Although formally, this problem reduces to the factorization.  To such a formula.

$$\frac{t}{q}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

$$k=\frac{qL(p+s)}{ps(tL-q)}$$

$$x=pk$$

$$y=sk$$

$$z=L$$

The calculation procedure is simple. To try all possible options ranging from $1$.

$$tL-q=1;2;3;4;5....$$

A large bust is not needed. It suces to restrict our if the number obtained is less than $2$.


Proof Involving Imaginary Number: Where's the wrong one? Here are the propositions:

$$i=\sqrt{-1}$$
$$i^2=-1$$
$$(i)(i)=-1$$
$$\sqrt{-1}\sqrt{-1}=-1$$
$$\sqrt{(-1)(-1)}=-1$$
$$\sqrt{1}=-1$$

There's an error in the propositions above. I think it's in the fifth line where $\sqrt{(-1)(-1)}=-1$.

Are $\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $a=(\sqrt{a})(\sqrt{a})$ different from each other?

1) The rule "$\sqrt{a}\sqrt{b}= \sqrt{ab}$" is not true for a and b non-real.

2) As Elio Joseph said, defining "i" as "$\sqrt{-1}$ is not rigorous.  Better is to define the complex numbers to b pairs of real numbers with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication by (a, b)x(c, d)= (ac- bd, ad+ bc).  We think of the real numbers as "embedded" in the complex numbers by a<==> (a, 0).  The "imaginary numbers" are the set of pairs (0, b).  And, of course, "i" is (0, 1):  (0, 1)x(0, 1)= (0x0- 1x1, 0x1+ 1x0)= (-1, 0) which is what we associate with "-1".

The error is that you can't say that $$i=\sqrt{-1}$$

because the function $\sqrt.$ is not defined on $\mathbb C$ properly.

You can't assigned a unique value for $\sqrt{z}$ if $z\in \mathbb C$ like you do for real positive numbers.

Therefore, writing $\sqrt{-1}$ instead of $i$ may and will lead to wrong results.


Proving very basic statements. I'm just talking about (b), (c) and (d) in this question.

The way I see it, (b) is asking to prove that:

$$n \mod m = n \mod m$$which is like asking to prove that $1 = 1$.

(c) is also asking to prove that 1=1 essentially.

So how am I supposed to go about "proving" these statements b c and d. I don't know what to say other than "it just is".

The relation says that $n\sim m$ iff $n\ mod\ 4\ =\ m\ mod\ 4$.
So b) is trivially true. Similar observations hold for c) and d) as well. for d), if $n$ and $m$ on dividing by 4 leave remainder say $r$ and $m$ and $k$ leave remainder say $q$ then $r\ =\ q$ and hence $n\sim k$.

The best thing to understand your relation is to workout some examples, e.g., part a) will give you some clear picture as to what the relation is trying to convey.


3D Vectors, why aren't my angles adding up to 180 for my triangle? I have 3 vectors
$$A\langle0,-1,-1\rangle$$
$$B\langle-1,0,-1\rangle$$
$$C \langle-1,-1,-2\rangle$$
Where C is $A+B$

Simply put I wanted to find the angle between $A$ and $B$

Using dot product I found that $A\cdot B = 1$ and then dividing by the magnitudes $||A||=\sqrt{2}$, $||B|| = \sqrt{2}$, and $||C||=\sqrt{6}$

Giving me an angle of $60^\circ$

However when I find the angle between $A$ and $C$ I find an angle of $30^\circ$ and this is the same answer I get for the angle between $B$ and $C$.

$A\cdot C=3$ then $cos\theta=\frac{3}{\sqrt{2}\sqrt{6}}$ where $\theta=30^\circ$
This is the same for $B\cdot C$

Adding all of these angles together I only get a grand total of $120^\circ$, how is this possible as I have to have $180^\circ$

Can anyone help me figure out what I am missing or where I went wrong?

The problem is that there are two possible angles for any dot product, $\theta$ and $\pi - \theta$  (Draw the intersecting lines and you will see the adjacent supplementary angles.)

Since 60 degrees is too small, try 180 degrees - 60 degrees = 120 degrees.
That will work.

You could also try 180 degrees - 30 degrees = 150 degrees but that would be too large.

Your vectors $A$ , $B$ and $C$ form a triangle if and only if $A + B + C = 0$


Convergence of a complex power series Consider the series:
$$\sum_{n=1}^{\infty} \frac{n!}{n^n}z^n$$

I found radius of convergence using almbert $R=e$

I know that this series converges for $|z|<e$ , does it converge for |z|=e

Letting $z = ee^{i\theta}$, your sum becomes

$$\sum_{n=1}^{\infty} \left(\frac{e}{n}\right)^nn!\,e^{in\theta}.$$

In order for the sum to converge, we need that the summand goes to zero as $n$ tends to infinity. Does that happen here?

Let's use Stirling's formula: $n! \sim \sqrt{2\pi n} (n/e)^n$, so you have $$\sum_{n=1}^\infty \frac{n!}{n^n} e^n e^{int} \sim \sum_{n=1}^\infty \sqrt{2\pi n}  e^{int}$$ if $|z|=e$. But $\lim\limits_{n\to\infty}|\sqrt{2\pi n}  e^{int}| = \infty$, and hence еhis series diverges.


Why are there no impossible modular residues of cubes for some moduli? I was working on a problem about cubes but I kept on getting stuck on some cases because for some moduli, for example 10, there are no impossible residues. For modulo 10:

\begin{array}
.N  &  N^3\pmod{10} \\\hline
1  & \hfill 1  \hfill\\
2  & \hfill 8  \hfill\\
3  & \hfill 7  \hfill\\
4  & \hfill 4  \hfill\\
5  & \hfill 5  \hfill\\
6  & \hfill 6  \hfill\\
7  & \hfill 3  \hfill\\
8  & \hfill 2  \hfill\\
9  & \hfill 9  \hfill\\
0  & \hfill 0 \hfill
\end{array}

All the residues are used in the $N^3\pmod{10}$ table.
Also for modulo $7$ all cubes are congruent to either $1$ or $-$1 modulo $7$. My question is whether there is a mathematical reason for why this is the case. Thanks

Yes, $\bmod 10\,$ the cube map $\,c(n) = n^3$ is a bijection by $\, a \equiv a^9 = (a^3)^3 = c(c(a))\,$ therefore $\,c\,$ is self-inverse. Indeed $\,a^9\equiv a\pmod{\!10}\,$ follows by a slight generalization of little Fermat / Euler.

You can observe this in your table: $\,2\mapsto 8\mapsto 2,\,$ $\, 3\mapsto 7\mapsto 3\,$ and others are fixed points so $\,c(c(a)) = a,\,$ i.e. $\,c^2 = 1,\,$ so $\,c^{-1} = c,\,$ i.e. $\,c\,$ is the permutation $\,(2\ 8)\,(3\ 7)\,$ in cycle notation.

More generally if $\,n\,$ is a product of distinct primes $\equiv 2\pmod{\!3}\,$ then   $\,\phi = \phi(n)\equiv 1\pmod{\!3}.\,$ Hence $\bmod 3\!:\ \phi\equiv 1\,\Rightarrow\,2\phi+1\equiv 0\,$ so $\,\color{#c00}{2\phi+1 = 3k}\,$ so the above linked theorem yields

$$\bmod n\!:\,\ 0\equiv a(a^{2\phi}-1)\,\Rightarrow\,a\equiv (a^{\large\color{#c00}k})^{\large 3}\equiv a^{\large \color{#c00}{2\phi+1}} \qquad\qquad\!\!\!$$

e.g. $\, \begin{align} n &= 2\cdot 5\cdot 11 = 110\\ \phi &= 1\cdot 4\cdot 10 = 40\end{align}\ $ so $\,2\phi+1 = 27\cdot 3\ $ so $\ a\equiv (a^{\large 27})^{\large 3}\pmod{\!110}$

When $\,n\,$ is a prime $\equiv 2\pmod{\!3}\,$ the above specializes to a simple case of cubic reciprocity.

For the 2nd question, by Fermat, $\!\bmod 7\!:\ n\not\equiv 0\,\Rightarrow 0\equiv n^6-1 = (n^3-1)(n^3+1)$ so $7$ divides one of the factors by Euclid's Lemma, so $\,n^3\equiv \pm1.\,$  More generally see Euler's Criterion.

Yes, there is a mathematical reason. It's easiest to understand for prime moduli. Primes other than $3$ leave a remainder of $1$ or $2$ when you divide by $3$. Work out your tables for $p=5,7,11,13,17,19$ and you should see the pattern.

Fermat's Little Theorem will help you with a proof.

You can read about this at the beginning of the  wikipedia page on cubic residues.


How to prove this limit does not exist? I have the following limit:

$$\lim_{(x,y) \to (0,0)}  \frac{\sin(xy)}{\sqrt{x^2 + y^2 + |y|} \times \|(x,y)\|}$$

Where I'm using the usual euclidean norm. I've been trying to find a curve that approximates (0,0) in a way such that the limit is not 0, but I'm really struggling to do so.

We have that

$$\lim_{(x,y) \to (0,0)}  \frac{\sin(xy)}{\sqrt{x^2 + y^2 + |y|} \times ||(x,y)||}=\lim_{(x,y) \to (0,0)} \frac{\sin(xy)}{xy} \frac{xy}{\sqrt{x^2 + y^2 + |y|} \times ||(x,y)||}$$

and

$$\lim_{(x,y) \to (0,0)}  \frac{\sin(xy)}{xy}=1$$

and by polar coordinates

$$\frac{xy}{\sqrt{x^2 + y^2 + |y|} \times ||(x,y)||}=\frac{r\cos \theta \sin \theta}{\sqrt{r^2+r|\sin \theta|}}\to 0$$

indeed

$$0\le \frac{r|\cos \theta| |\sin \theta|}{\sqrt{r^2+r|\sin \theta|}}\le \frac{r|\cos \theta| |\sin \theta|}{\sqrt{r|\sin \theta|}} \frac{\sqrt{r|\sin \theta|}}{\sqrt{r|\sin \theta|}}=|\cos \theta|\sqrt{r|\sin \theta}| \to 0$$

Let $x=rc,y=rs$ with $c^2+s^2=1$ and $r$ small. Then the numerator is on the order of $r^2cs$ and the denominator on the order of $\sqrt{r|s|}$ (because $r^2$ is neglectible, unless $s=0$) times $r$. Hence the limit converges to zero like $\sqrt r$.

If we can't neglect $r^2$ because $s$ tends to zero,
$$\frac{r^2cs}{r\sqrt{r^2+r|s|}}=\frac{cs}{\sqrt{1+\dfrac{|s|}r}}$$ tends to zero anyway.


Integers $d$ for which the Negative Pell equation is soluble for both $d$ and $2d$? Let $\text{NPE}_d$ denote the negative Pell equation:
$$ x^2-dy^2=-1$$
Where $d$ is a given positive nonsquare integer and integer solutions are sought for x and y.

we know that (in this paper):

  Theorem : The equation $\text{NPE}_d$ has integer solutions if and only if there exist two integers $a(d)=a$ and $b(d)=b$ such that $d=a^2+b^2$ and there exists a Pythagorean triplet $(A,B,C)$
  such that $|aA-bB|=1$ and in this case $(Aa+Bb,C)$ is a solution.

Obviously if $\text{NPE}_d$ has integer solutions then $d$ cannot be divisible by any prime $p$ such that $p=3\mod 4$.

  My question: Is there any characterization for the integers $d$ for which $\text{NPE}_d$ and $\text{NPE}_{2d}$ have both integer solutions.

I used the characterization above, but I can't link the couple $(a(d),b(d))$
to $(a(2d),b(2d))$ because the theorem doesn't give us much information

The sequence  of the elements $d$ for which $\text{NPE}_d$  is soluble is OEIS A031396.

Thank you for your help.

It is necessary to use the formula.    http://www.artofproblemsolving.com/blog/101140

$$x^2-dy^2=-1$$

$$z^2-2dg^2=-1$$

It will give us a series of solutions using the Pell equation:  $$p^2-2s^2=k$$

Let us use the fact that the following solution can be found knowing the previous, formula.

$$p_2=3p_1+4s_1$$

$$s_2=2p_1+3s_1$$

Using certain series of solutions you can find ratio formula.

$p^2-2s^2=1$     $;$     $(p_1;s_1) - (3;2);$  $d=s^2+1$

$p^2-2s^2=-1$   $;$   $(p_1;s_1) - (1;1) ;$   $d=s^2+4$

$p^2-2s^2=-7$    $;$   $(p_1;s_1)  - (1;2) ;$    $d=s^2+1$

$p^2-2s^2=7$    $;$   $(p_1;s_1) - (3;1) ;$   $d=s^2+4$

$p^2-2s^2=17$   $;$   $(p_1 ; s_1)  - (5;2) ;$  $d=s^2+9$

While it is difficult to say. Limited to any solution of these series or not.  Clearly what is involved in this equation Pell.


sequence of square root of positive operators is convergent I'm trying to prove the next:

a) If $A_{n}\geq 0,$ $A_{n}\rightarrow A$ in norm, then $\sqrt{A_{n}}\rightarrow\sqrt{A}$ in norm,

b) Suppose that $A_{n}\rightarrow A$ strongly for a sequence $\{A_{n}\}.$ Then $\sqrt{A_{n}}\rightarrow\sqrt{A}$ strongly.

I'm stuck prove this. I've seen the proof of a) using spectral theorem, but I don't familiar with this. I was thinking in a proof more elementary.

I was thinking in something of the form
$$||\sqrt{A_{n}}-\sqrt{A}||=||\frac{A_{n}-A}{\sqrt{A_{n}}+\sqrt{A}}||,$$ and then bounding denominator an use the hypotesis of convegence in norm, but I guess this is not correct and useless.

Any kind of help is thanked in advanced.

Start with the binomial series
$$
 (1-z)^{\alpha}=1-\sum_{n=1}^{\infty}c_n z^n,\;\;\;\; |z| < 1, \;\; 0 < \alpha < 1.
$$
where $c_n \ge 0$ for all $n \ge 1$. It follows that $\sum_{n=1}^{\infty}c_n=1$,
by considering the limit as $z\rightarrow 1$ on the real line. This provides one way to construct the positive square root of an operator $0 \le A \le I$:
$$
        A^{1/2} = (I-(I-A))^{1/2}=1-\sum_{n=1}^{\infty}c_n(I-A)^{n}
$$
By scaling, you can assume your sequence $\{ A_n \}$ satisfies $0 \le A_n \le I$, with $0 \le A \le I$. With that you can show that $A_n^{1/2}\rightarrow A^{1/2}$ in operator norm.


Took me 20 seconds to figure out: 85228

Here is the python code:

    from sympy.abc import x
    from sympy import expand
    from math import factorial
    d1 = expand((x+x**2+x**3+x**4+x**5+x**6)**10).as_coefficients_dict()
    print d1[x**20]


Couples around a table Find the number of ways in which $n$ men and $n$ women can sit around a round table such that every man can pair off with a woman sitting next to him (to form $n$ couples).

Anyone can help with this problem? Thanks :)

EDIT: Ignore rotations of the table. For example, when $n=1$ there should be only 1 seating arrangement.

First of all, we can assign seats in two stages. First, decide which $n$ seats will be for men and which for women. Second, arrange men on male seats and women on female seats.

For the second step, there are always $(n!)^2$ possibilities. So the answer to the problem is $k \cdot (n!)^2$, where k is the number of ways to decide which $n$ seats will be for men. So what is left is to find $k$.

For convenience, let us assign numbers to seats (say, clockwise): $1, 2, \ldots, 2n$. When every male seat is paired with a nearby female seat, there are two possibilities: either the paired seats are $(1, 2), (3, 4), \ldots, (2n-1, 2n)$, or they are $(2, 3), (4, 5), \ldots, (2n-2, 2n-1), (2n, 1)$.

Let $A$ be the set of all male-female seat assignments that allow pairing of the first kind, and $B$ the set of all assignments that allow pairing of the second kind. Now $k$ is the cardinality of their union $A \cup B$.

It is easy to find $|A|$. In each pair we pick arbitrarily which seat will be male and which female. There are $2^n$ choices, so $|A| = 2^n$. In a similar fashion, $|B|=2^n$.

Now, to find $|A \cup B|$, we note that $|A \cup B| = |A| + |B| - |A \cap B|$. What is $A \cap B$? It is the set of all male-female assignments that can be split into pairs in both ways: "$(1, 2), (3, 4), \ldots$" and "$(2, 3), (4, 5), \ldots$". There are only 2 assignments like this: one is when we assign all odd seats to men and even seats to women, the other is when we do the opposite. So $|A \cap B| = 2$.

Then we see that $k = 2^n + 2^n -2$, and the answer is $(2^{n+1}-2) \cdot (n!)^2$.

UPDATE: Ok, I assumed this was clear, but maybe not. I assume that all seats are distinguishable, and so are all the $2n$ people. So, $2n$ people enter a room, and you know each of them personally. And $2n$ seats are standing at fixed places around the table, and each seat is familiar to you, and each is precious in its own way, like an old friend. Now people take their seats at the table, and for you it is important who sits where exactly. If two men swap positions - you notice, because you distinguish them. If everyone stands up and moves to the seat to the right, thus rotating the whole arrangement - you also notice, because you know and distinguish the seats as well as people.

If you decide you don't want to distinguish between different men and different women, you drop the $(n!)^2$ factor. If you decide in addition that you don't even distinguish between sexes, you divide the answer by 2.

If your understanding is that people are all distinguishable, but rotations of the table don't really change the arrangement, then simply divide my answer by $2n$. The justification is simple: given one allowed arrangement, you can get $2n$ distinct arrangements from it by rotations. I have counted each one as a separate arrangement, but for you they are all the same, so you divide the answer by $2n$ to get $(2^n - 1) \cdot n! \cdot (n-1)!$.

This is what I think:

Imagine that there are $n$ seats where the men can sit, and $n$ seats in between these seats where the women can sit.
The number of ways the men can sit is $n!$, and the number of ways where the women can sit is also $n!$.
But here are 2 ways of seating these people:
man,woman,man,woman... or woman,man,woman,man...

Therefore the total number of ways is $2(n!)^2$.

Proof:

If $n=2$ then $2(n!)^2=8$

Assume that odd numbers are female and even numbers are male.
Combinations: $1234,1432,2143,2341,3214,3421,4123,4321$


Dual cone and sum of closed cones Picture below is from the 35 page of  Schneider R.-Convex Bodies_ The Brunn-Minkowski Theory-Cambridge University Press (2013) , I think $C^o$ is always closed no matter $C$ is closed or not. Because the inner product is continuous.  Besides the  I think the sum of closed convex cones must be closed, because the sum is continuous . Where is my mistake ?

Yes, $C^\circ$ is always closed.

However, the sum of the closed convex cones
$$A = (-\infty,0] \times \{(0,0)\} \\ B = \operatorname{cl-cone}(B_{1}(1,1,0))$$
is not closed. Here, $B_{1}(1,1,0))$ is the closed ball centered at $(1,1,0)$ with radius $1$, touching all three coordinate planes.

I just think the answer of gerw is a little complex. So I according his way make the below picture . Green and Black are two closed cones , but the sum of them is open ( the sum is $R^2$).


Uniform convergence of $n\left(\sqrt{x+\frac{1}{n}}-\sqrt{x}\right)$ I need to say whether or not $f_n(x)=n\left(\sqrt{x+\frac{1}{n}}-\sqrt{x}\right)$ is uniformly convergent on $(0,\infty)$.

I've found that the function is locally convergent to $f(x)=\frac{1}{2\sqrt{x}}$ and was trying to find $\sup{|f_n(x)-f(x)|}$.

I got the derivative $f_n'(x)= \frac{2nx\left(x-\sqrt x\sqrt{x+\frac{1}{n}}\right)+\sqrt{x}\sqrt{x+\frac{1}{n}}}{...}$ and could not find $x$ so that $f_n'(x)=0$

Any ideas?

Note that each of the functions $\displaystyle f_n(x) = n\biggl(\sqrt{x+\frac{1}{n}}-\sqrt{x}\biggr)$ is bounded on $(0,\infty)$, with $f_n(x) \leq \sqrt{n}$.  Since $\displaystyle f(x) = \frac{1}{2\sqrt{x}}$ is unbounded on $(0,\infty)$, the sequence $\{f_n\}_{n=1}^\infty$ does not converge uniformly to $f$.

As already pointed out by several others, the convergence is not uniform on $(0,\infty)$ but it may be worth noting that it is uniform on every subinterval $[c,\infty)$ for $c > 0$.

This follows from the following computation:

$$
\begin{split}
|f_n(x) - f  (x)| &= \left|\frac{1}{\sqrt{x+\dfrac1n}+\sqrt{x}} - \frac1{2\sqrt{x}}\right| \\
&= \left|\frac{\sqrt{x}-\sqrt{x+\dfrac1n}}{2\sqrt{x}\left( \sqrt{x+\dfrac1n}+\sqrt{x} \right)} \right|\\
&= \left|\frac{\dfrac1n}{2\sqrt{x}\left( \sqrt{x+\dfrac1n}+\sqrt{x} \right)^2} \right|
\le \frac1n \cdot \frac{1}{8c\sqrt{c}}.
\end{split}
$$


Solve $ \lim_{x\to ∞}\frac{x+\sin(x)}{2x+\cos(x)} $ I'm having trouble finding the limit
$$ \lim_{x\to ∞}\frac{x+\sin(x)}{2x+\cos(x)} $$

I've started with
$$ \lim_{x\to ∞}\frac{\frac{x}{x}+\sin(x)}{\frac{2x}{x}+\cos(x)}= \lim_{x\to ∞}\frac{1+\sin(x)}{2+\cos(x)}=\frac{1}{2}\lim_{x\to ∞}\frac{\sin(x)}{\cos(x)} $$

but I don't know how to go on from here or if I should've started with a different method.

Thank you in advance!

It should be $$ \lim_{x\to ∞}\frac{x+\sin{x}}{2x+\cos{x}}=\lim_{x\rightarrow\infty}\frac{1+\frac{\sin{x}}{x}}{2+\frac{\cos{x}}{x}}=\frac{1}{2}$$

$$\frac{x+\sin x}{2x+\cos x}=\frac12+\frac{2\sin x-\cos x}{2(2x+\cos x)}\to \frac12+0$$


How to find the transformation matrix given two vectors and their particular transformations? I know this is probably a pretty simple thing to do, but I can't really wrap my head around it: I have two vectors in R^2, (1,2) and (1,0), whose transformations are (1,2,0) and (3,0,1) respectively. How do I find out the transformation matrix from this information?

I know I could manually write down a new system of equations using the elements of the matrix as my unknowns, but I supposed that's too tedious to be the right solution.

Also, do basis transformation matrices have anything to do with this? I thought I could use a new matrix consisting of my two column vectors in R2 (representing a change of basis from the standard basis vectors), but I'm not sure if doing the same thing to my transformed vectors in R^3 makes any sense (can I group them up too? Would that be a change of basis in the range?). I'm sorry if I'm mixing up two unrelated things, but I had a hunch the problem might be related to that.

Thanks for the help!

You could try the following. First map the two vectors in R^2 to the standard basis vectors in R^2. Then find a mapping that maps the standard basis vectors in R^2 to the ones in R^3.

In particular, in your example, this would yield something like this: The matrix $A$ that maps the standard basis vectors to (1,2) and (1,0) is:

$$A =
        \begin{bmatrix}
        1 & 0 \\ 2 & 1
        \end{bmatrix}
$$

The matrix $B$ that maps (1,0) and (0,1) to (1,2,0) and (3,0,1) is the following one:

$$B =
\begin{bmatrix}
1 & 3 \\ 2 & 0 \\ 0 & 1
\end{bmatrix}
$$

So, you first want to do the inverse of $A$ to map (1,2) and (0,1) to the standard basis vectors, and then apply $B$ to get to (1,2,0) and (3,0,1). So, the resulting matrix becomes:

$$BA^{-1} = \begin{bmatrix}
-5 & 3 \\ 2 & 0 \\ -2 & 1
\end{bmatrix}$$

HINT....You can write this as a matrix equation. Let the transformation matrix be $M$

Then $$M\left(\begin {matrix}1&1\\2&0\end{matrix}\right)=\left(\begin{matrix}1.2&3\\0&0.1\end{matrix}\right)$$

Now post multiply by the inverse and calculate $$M=\left(\begin{matrix}1.2&3\\0&0.1\end{matrix}\right)\left(\begin {matrix}1&1\\2&0\end{matrix}\right)^{-1}$$

Now you can finish this.


If $\int f^*(x)g(x) dx$ is the scalar product in the vector space of $L^2$, what is $f(x)g(x)$? For two functions $v=f(x)$, $w=g(x)$ from the vector space $L^2$, the scalar product $v^Tw$ is defined to be

$\int f^*(x)g(x) dx$.

However, I can also just multiply one function with another, $f(x)\cdot g(x)$. What is the corresponding algebraic operation?

It's not a vector space operation, because in a vector space $V$ the only $V \times V \to V$ operation is the addition.