Finding $r$ in quadratic integer equation $$M=f(n)=\frac{n(n+1)}{2}$$
$$M=g(k)=40k+30+r$$
I have found that $r=2,6$ but $r$ can only be one. Since $n, k \in \mathbb{N}$, I wanted to see which vaue for r would lead to integer solutions for k and n.
$$n^2 +n -2(40k+30+r)=0$$
$$n=\frac{-1+\sqrt{1-4(1)(-1)(80k+60+2r)}}{2}$$
$$n= \frac{-1+\sqrt{320k +241+8r}}{2}$$
$$n=\frac{-1+\sqrt{320k +257}}{2},\frac{-1+\sqrt{320k+289}}{2}$$
now $n$ is only an integer if $320k +257$ or $320k + 289$ are perfect squares.
I don't want to add any more context.

$\color{blue}{\mbox{Sufficiency}}$
So we must look at when $320k+257$ and $320k+289$ , separately, are perfect squares. Indeed, note that if these are perfect squares then they are odd hence perfect squares of odd numbers . In particular, we have that the numerator of the expression for $n$ is an even number so $n$ is an integer provided they are perfect squares.
$\color{green}{\mathrm{320k+257}}$
The idea behind showing that no number of the form $320k+257$ is a square, is to show that no number of a simpler form is a square. Let me show you.

Every number of the form $320k+257$ is of the form $5l+2$ for some $l$.

This is easy, since $320k+257 = 5(64k) + 55 + 2 = 5(64k+11)+2$ so we can take $l = 64k+11$.

No number of the form $5k+2$ can be a perfect square.

Suppose that $x^2 = 5k+2$. Let $5r$ be the smallest multiple of $5$ such that $5r<x$. Then $0<x-5r < 5$. We have $$(x-5r)^2 = x^2-10xr+25r^2 = 5(5r^2-2xr +k) + 2$$
So $(x-5r)$ has the same property as $x$ in terms of having a square $2$ more than a multiple of $5$, which means that some number out of $0,1,2,3,4$ must have a square of the form $5k+2$. But this never happens, as can be easily checked!
This shows that no number of the form $5k+2$ is a perfect square, and hence that no number of the form $320k+257$ is a perfect square.
$\color{fuchsia}{\mathrm{320k+289}}$
Using the logic we used previously works somewhat : if $x^2 = 320k+289$ then $(x-320r)$ also has the same property for any $r$. In particular, having all the solutions from $0,1,2,...,319$ whose squares leave a remainder of $289$, will give us the set of all $x$ for which $x^2$ leaves a remainder of $289$ upon division by $320$.
Why? For example, take $k=0$, so $320k+289 = 289 = 17^2$. Then $x = 17$. Now take any number of the form $17+320l$, and square it, it will also leave a remainder $289$ upon division by $320$. For example, $320+17, 17-320,17-10*320$ would all have this property.
Which means, all we need are the numbers from $0,1,...,319$ whose squares leave a remainder $289$ upon division by $320$. I won't explain how one can simplify this (which one can) but the solutions are $17,47,113,143,177$.
That is, let $S = \{17,47,113,143,177\}$ then every $k$ such that $320k+289$ is a perfect square, is of the form $\left\{\frac{(s+320l)^2 + 289}{320} : s \in S, l \in \mathbb Z\right\}$.
$\color{red}{Further \ reading}$
You may look up the notion of quadratic congruences, which has a large theory dedicated to it. In particular, the Jacobi symbol may be used in some instances to decide if squares of the type $ak+b$ exist, and we can use prime factorizations of $a$ to obtain solutions to the problem or provide bounds on the number of solutions.
Quadratic congruences, incidentally, are motivated by the kind of problem you consider i.e. when is a certain kind of quadratic form the square of another number.


Is the sum of two prime ideals a prime ideal? Is the sum of two prime ideals (of a ring) again a prime ideal?

Please can someone give me a hint? I know that the intersection of two prime ideals is again a prime ideal iff one is contained in the other, but I'm not sure about the sum.

Th answer is no, consider the ring $\mathbb{Z}$ and the prime ideals $(2)$ and $(5)$. Since two and five are relative primes then every number can be written as a linear combination over $\mathbb{Z}$ of 2 and 5. To see this. note that if if $k$ is any interger, since we know that
$$5-(2)2=1$$
by multiplying by $k$ we obtain the expression
$$k5-(2k)2=k$$
which is a $\mathbb{Z}$-linear combination of $k$ in therms of $2$ and $5$. Note that $k5\in(5)$ and $-2k(2)=-4k\in(2)$. This means that $(2)+(5)=\mathbb{Z}$, which is not a prime ideal.

Note that the sum of ideals is an ideal when one is contained in the other. This is not possible in $\mathbb{Z}$ with the exemption when of the ideals is the prime ideal $(0)$, but other rings can give more interesting examples. Have a look at polynomials rings for example.

I think (correct me if I'm wrong) that if you are not so restrictive in your definition of prime ideals (allowing 1 to belong to a prime ideal) the answer is positive.

For instance, let $p_{1}$ and $p_{2}$ be two prime ideals of $R$, and take the natural projection map $\pi: R \longrightarrow R/p_{1}$, then $\pi(p_{2})$ is a prime ideal because $\pi$ is a surjective map and $p_{2}$ is a prime ideal. But then, $\pi^{-1}(\pi(p_{2}))$ is a prime ideal in R, and $\pi^{-1}(\pi(p_{2}))=p_{1}+p_{2}$.


Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.

Not a full solution but if you use the equality: $(x+y)^3 = x^3 + y^3 + 3x^2y + 3y^2x$, you can reach the equation $z^3 + z = 4$ where $z$ is the value of the above expression.  This leads to $z = 1$ as a solution.  I cubed your formula and simplified.  I do not, however, know how to solve $z^3 + z = 4$.  Cubing both sides is safe only if $z$ is positive.


I'm trying to prove $ M \subset N \Rightarrow M \cup(N \setminus M) = N $ I'm trying to prove:
$$
M \subset N \Rightarrow M \cup(N \setminus M) = N
$$
My trial:
$
\{x|x\in M \cup(N \setminus M)\} \Leftrightarrow \{x|x\notin \overline{M \cup(N \setminus M)}\}
$
using de Morgan
$ \Leftrightarrow \{x|x\notin \overline{M} \cap \overline{(N \setminus M)}\}$
$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \notin \overline{(N \setminus M)}\}$
$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \in (N \setminus M)\}$
$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \in N \wedge x\notin M\}$
aand we got a problem here. $x$ cannot be both $x\notin \overline{M}$ and  $x\notin M$.
My steps seems logical to me. Can someone please spot my mistake. Thanks a lot!

Your error: $x\notin A\cap B$ is the negation of $x\in A\cap B$. $$x\in A\cap B \text{ iff }x\in A \land x\in B$$
So,$$x\notin A\cap B \text{ iff }x\notin A \lor x\notin B$$
Please notice how the $\land$ becomes $\lor$ on negation! In general, $\lnot(p\land q) = \lnot p\lor \lnot q$.

Alternative proof strategy: Whenever I want to show $A = B$, where $A,B$ are two sets, I aim to establish $A\subset B$ and $B\subset A$. Works on most occasions!

Suppose $M\subset N$. Then, for every $x\in M$, $x\in N$. Note that
$$N\setminus M = \{x\in N: x\notin M\}$$
Suppose $x\in M \cup N\setminus M$. Then, $x\in M$ or $x\in N\setminus M$. If $x\in M$, then $x\in N$ by $M\subset N$. If $x\in N\setminus M$, then $x\in N$ by definition of $N\setminus M$. So, $M\cup N\setminus M \subset N$.
Next, suppose $x\in N$. Then, either $x\in M$ or $x\notin M$. If $x\in M$, $x\in M\cup N\setminus M$ by definition of the union. If $x\notin M$, then clearly $x\in N\setminus M$, since we assumed $x\in N$, to begin with (refer to the definition of $N\setminus M$ again). As a result, $x\in M\cup N\setminus M$ by definition of the union, once again. So, we have $N\subset M\cup N\setminus M$.
$M\cup N\setminus M \subset N$ and $N\subset M\cup N\setminus M$ together, yield $M\cup N\setminus M = N$. So we have shown $$M\subset N\implies M\cup N\setminus M = N$$ Done!

Lastly, the following image says more than words ever can:

So clearly if $x\in N\setminus M$ then $x\in N$ and if $x\in M$ then $x\in N$ from the hypothesis $M\subseteq N$ so that
$$M\cup N\setminus M\subseteq N$$
and thus we have only to show the other inclusion. So clearly if $x\in N$ then either $x\in M\cap N=M$ or $x\notin M\cap N=M$ (any event is true or false!) and thus $x\in M$ or $x\in N\setminus M$ so that $x\in M\cup N\setminus M$  and thus the statement follows immediately.


General misconception about $\sqrt x$ I noticed a large portion of general public (who knows what square root is) has a different concept regarding the surd of a positive number, $\sqrt\cdot$, or the principal square root function.

It seems to me a lot of people would say, for example, $\sqrt 4 = \pm 2$, instead of $\sqrt 4 = 2$. People even would correct a statement of the latter form to one with a $\pm$ sign. Some also claim that, since $2^2 = 4$ and $(-2)^2 = 4$, $\sqrt 4 = \pm 2$. Some people continue to quote other "evidences" like the $y=x^2$ graph. While most people understand there are two square roots for a positive number, some seem to have confused this with the surd notation.

From an educational viewpoint, what might be lacking when teaching students about surd forms? Is a lack of understanding to functions a reason for this misconception?

Now I have noticed another recent question that hinted that poster was confused. Following @AndréNicolas's comment below, might these confusion really come from two different communities using the same symbol?

They probably just think that $\sqrt{x}$ means any number whose square is $x$, and don't know that the definition is just the positive root. I don't really think it's anything more than that.

In my opinion it's lack of logical thinking that generates such confusion.

Also mixing natural language with mathematics gives problems:

  $Q_1$: What are the squares roots of $4$?
  $A_1$: They are $2$ and $-2$.

Proper interpretation of $Q_1$: find the extension of the set $\{x\in \Bbb R\colon x^2=4\}$.

  $Q_2$: What is the square root of $4$, i.e., $\sqrt 4$?
  $A_2$: The square root of $4$ is $2$.

Proper interpretation of $Q_2$, once it has been established the truth of statement $(\forall x\in \Bbb R^+)(\exists !y\in \Bbb R^+)(y^2=x)$, what is the only positive number such that its square is $4$?

I find that most students don't have enough logic in them to make the above translations and since mathematicians insist on abusing natural language, confusion is bound to rise.

Once students known that $(\forall x\in \Bbb R^+)(\exists !y\in \Bbb R^+)(y^2=x)$ and that given a positive real number $x$, $\sqrt x=y$, there is absolutely no danger of mistakes. It comes directly from the definition.


Bayesian Network vs Markov Decision Process I am wondering if somebody can tell me anything about the practical differences between using Markov Decision Processes and and Bayesian Networks in reasoning about probabilistic processes?

Read chapter 7 "Making Decisions" of David Barber's book on Bayesian ML. http://web4.cs.ucl.ac.uk/staff/D.Barber/textbook/090310.pdf

From what little I know about BN, they are just representations of Markov Chains (MC). That is, you know the current state $x$ and that gives you the distribution of the successor state given by $P(x)$. I guess, in some cases you will also have $P$ dependent on the current time, or maybe values of some hidden states. Nevertheless, the bottom line is that the dynamics is given to you, and you can only talk about "what is the probability of this or that".

In MDP the setting is similar with a crucial difference that $P(x,u)$ depends on the input $u$ which is provided externally. In that case, we are talking about optimization - how to choose $u$ optimally to maximize/minimize probability of this or that.

Note that once you've fixed the way you choose $u$, that will give you a fixed dynamics which can possibly be represented as a BN.


Does $\sum_{k=1}^{n} \left(\frac{1}{k \cdot \log(n)}\right) $ converge to 1? I'm trying to show convergence of the following:

$$
\lim\limits_{n \to \infty} \sum_{k=1}^{n} \left(\frac{1}{k \cdot \log(n)}\right)
$$

How shall I proceed?

Hint:  Draw a graph of $f(x) = \dfrac{1}{x}$ over $[1,n]$,and see that: $\dfrac{1}{2}+\dfrac{1}{3}+\cdots +\dfrac{1}{n}<\displaystyle \int_{1}^n \dfrac{1}{x}dx = \ln n < 1+\dfrac{1}{2}+\cdots +\dfrac{1}{n-1}$

Since $\log(n)$ can be viewed as a number in the sum:

$$ \lim\limits_{n \to \infty} \sum_{k=1}^{n} \left(\frac{1}{k \cdot \log(n)}\right) = \lim\limits_{n \to \infty} \frac{1}{\log(n)} \sum_{k=1}^{n} \left(\frac{1}{k }\right) = \lim\limits_{n \to \infty} \frac{\sum_{k=1}^{n} \left(\frac{1}{k }\right)}{\log(n)} \cdot  $$ since $$\lim\limits_{n \to \infty} \sum_{k=1}^{n} \left(\frac{1}{k }\right)$$ diverges so look at this, you can apply L'Hopital Rule (that will be your work).


Why does $\lim_{n \to\infty}a_{n+1} = \lim_{n\to\infty}a_n$? Assume that $\{a_n\}$ is a convergent sequence. How to use the definition of a limit of a sequence to prove that $\lim\limits_{n \to\infty}a_{n+1} = \lim\limits_{n\to\infty}a_n$?

Write $L=\lim_{n\to\infty}a_n$. This means that for every $\epsilon>0$ there exists a whole number $N$ such that for all integers $n\ge N$, $$|L-a_n|<\varepsilon$$ Now, to prove $\lim_{n\to\infty}a_{n+1}=L$ as well, consider any $n\ge N$ and note that $n+1\ge N$. So by our initial statement, $|L-a_{n+1}|<\varepsilon$.

I think this is just a matter of notation.

For $n \ge 1$,
considering $a_{n+1}$ means the sequence $a_2, a_3, a_4, a_5, \dots$, and
considering $a_n$ means the sequence $a_1, a_2, a_3, a_4, a_5, \dots$.

Limits are a matter of the eventual behaviour of the sequence, and it is not surprising that the limit (if it exists) is the same for "both" — it's the same sequence whose eventual behaviour you're considering.

Put differently, $$\lim_{n\to\infty} a_{n+1} = \lim_{m-1\to\infty} a_m = \lim_{m\to\infty} a_m = \lim_{n\to\infty} a_n$$
as $m-1\to\infty$ and $m\to\infty$ are the same thing.


Why does the lower limit of Integration change to a zero and where does the 8 come from? I'm working through an example from Stewart's book, and I can not work out why the lower bound of integration changes to a $0$. Also I am confused as to where the $8$ comes from?

I am assuming there is a rule being applied here that I am unaware of.
I can't find any similar examples.

  $$4\int_{-\pi/2}^{\pi/2}\cos^4\theta\ d\theta=8\int_0^{\pi/2}\cos^4\theta\ d\theta$$

Imagine trying to find the area of a circle.  It suffices to find the area of a semicircle, then doubling it, right?

For any function that satisfies $$f(x)=f(-x),$$ we have $$\int_{-a}^af(x)\,dx=2\int_0^af(x)\,dx.$$  We call these functions even functions.

The integral can be though of as the purple shaded area in the figure below.

Because the integrand is even (it is symmetric across the $y$-axis), the area on the left of the $y$-axis (the blue area) is equal to the area on the right side of the $y$-axis (the blue area).  The area of the red region is the integral over the interval $[0,\pi/2]$.

More generally, we have the following:

  Theorem: Suppose that $f : [-L,L] \to\mathbb{R}$ be an even, continuous function.  Then
  $$ \int_{-L}^{L} f(x)\,\mathrm{d}x = 2 \int_{0}^{L} f(x)\,\mathrm{d}x. $$

Proof:  By the linearity of integration,
$$ \int_{-L}^{L} f(x) \,\mathrm{d}x
= \int_{-L}^{0} f(x)\,\mathrm{d}x + \int_{0}^{L} f(x)\,\mathrm{d}x. \tag{1}$$
Making the change of variables $u = -x$, the first term becomes
$$ \int_{-L}^{0} f(x)\,\mathrm{d}x
= -\int_{L}^{0} f(-u)\,\mathrm{d}u
= \int_{0}^{L} f(-u)\,\mathrm{d}u. $$
Since $f$ is even, $f(-u) = f(u)$ for any $u \in [-L,L]$.  Therefore
$$ \int_{0}^{L} f(-u)\,\mathrm{d}u
= \int_{0}^{L} f(u)\,\mathrm{d}u. $$
Substituting this back into (1), we obtain
\begin{align}\int_{-L}^{L} f(x) \,\mathrm{d}x
&= \int_{-L}^{0} f(x)\,\mathrm{d}x + \int_{0}^{L} f(x)\,\mathrm{d}x \\
&= \int_{0}^{L} f(u)\,\mathrm{d}u + \int_{0}^{L} f(x)\,\mathrm{d}x \tag{these terms are equal}\\
&= 2\int_{0}^{L} f(x)\,\mathrm{d}x.
\end{align}
The last identity holds, as $u$ and $x$ are just dummy variables.  After integration, the result is a constant which does not depend on either of these variables.


Injectivity and surjectivity of $f\colon P(X) \times P(Y) \to P(X \times Y)$ Let $X$ and $Y$ be nonempty sets and a function such as:

$$f: P(X) \times P(Y) \to P(X \times Y)$$
$$(A,B)\to A \times B$$

The question is if this function is injective or surjective, but I do not even understand the function. If $X={1,2,3}$ and $Y={1,2}$, $P(X)=\{\{1\},\{2\},\{3\},\{1,2\},\{2,3\},\{1,3\},\{1,2,3\},0\}$ and $P(Y)=\{\{1\},\{2\},\{1,2\},0\}$, who is $f(\{1,2\},\{1,2,3\})$?

First, to understand the function itself:

$f:\mathcal{P}(X)\times\mathcal{P}(Y)\to \mathcal{P}(X\times Y)~~~~~(A,B)\mapsto A\times B$

The function takes an ordered pair of sets, $(A,B)$ where $A$ is a subset of $X$ and $B$ is a subset of $Y$, and maps it to the set of all ordered pairs of the elements where the first element is in $A$ and the second is in $B$.

Your specific example, letting $X=\{1,2\}$ and $Y=\{1,2,3\}$ we have

$f(\color{red}{\{1,2\}},\color{blue}{\{1,2,3\}}) = \{(\color{red}{1},\color{blue}{1}),(\color{red}{1},\color{blue}{2}),(\color{red}{1},\color{blue}{3}),(\color{red}{2},\color{blue}{1}),(\color{red}{2},\color{blue}{2}),(\color{red}{2},\color{blue}{3})\}$

Similarly:

$f(\color{red}{\{2\}},\color{blue}{\{1,3\}})=\{(\color{red}{2},\color{blue}{1}),(\color{red}{2},\color{blue}{3})\}$

As for the question of injectivity, assuming that $X$ is nonempty and contains at least one element $x$:

$f(\emptyset,\emptyset)=\emptyset\times \emptyset=\emptyset$

also

$f(\{x\},\emptyset)=\{x\}\times\emptyset = \emptyset$

So, we have shown that there exist two inputs, $(\emptyset,\emptyset)$ and $(\{x\},\emptyset)$ which both map to the same output, therefore so long as $X$ is nonempty the function will not be injective.  Reversing the roles of $X$ and $Y$ will show the same result for if $Y$ is nonempty.  We get as a result, $f$ is injective if and only if $X$ and $Y$ are simultaneously empty and is not injective otherwise.

As for the question of surjectivity.  You should know

$|\mathcal{P}(X)\times \mathcal{P}(Y)|=|\mathcal{P}(X)|\times|\mathcal{P}(Y)|=2^{|X|}\times 2^{|Y|}=2^{|X|+|Y|}$

and

$|\mathcal{P}(X\times Y)|=2^{|X\times Y|} = 2^{|X|\times |Y|}$

For large enough sets $X$ and $Y$, one has $2^{|X|+|Y|}<2^{|X|\times|Y|}$ which proves that $f$ cannot be surjective.  If $X$ and/or $Y$ are small enough however this isn't true and isn't sufficient to prove or disprove surjectivity.  For example if $X$ or $Y$ are empty, then the function will indeed be surjective.

For an explicit example, note that that with $X=\{1,2\}$ and $Y=\{1,2,3\}$ there is no input which will map to $\{(1,1),(2,1),(2,2)\}$ despite this being an element of the codomain.  All of the outputs of the function will be "balanced" so to speak.


how to write a recursive definition so the question asks to define s(n) as the number of strings of a's b's and c's of length n that do not contains "aa". write a recursive definition for s(n). what is s(0),s(1),s(2),s(3).
i had to write out every combination from head but i am pretty sure i didnt have to do that.
i got s(0)=1,s(1)=3,s(2)=8,s(3)=22

from what i tried to do i got s(n)=2(n-1)+2(n-2) but not sure if i am right.

can someone explain how to do this kind of problems.

A more complete answer may be in order.  From my comment,

$$s(n) = s_a(n) + s_b(n) + s_c(n)$$

Now look at what happens when you add a letter to the end.  Adding a c to the end of any string with no "aa"s is still a string with no "aa"s.  So...

$$s_c(n+1) = s(n)$$

Similarly for adding a b to the end.

$$s_b(n+1) = s(n)$$

Now adding an a to the end only counts if the previous string did not end in a.

$$s_a(n+1) = s_b(n) + s_c(n)$$

And so we have, by adding all the pieces:

$$s(n+1) = s_a(n+1) + s_b(n+1) + s_c(n+1) = 2 s(n) + s_b(n) + s_c(n) = 2 s(n) + 2 s(n-1)$$

where that last equality uses the formulas for $s_b$ and $s_c$ derived earlier.

Show that the algorithms $(2n+1)$ are recursive.                                                                                Can someone help to get the result of this problem? Thanks


Ito Isometry and quadratic variation Here is a confusion regarding stochastic integrals. Let
$Y_t=\int_0^tW_sds$ where $W_t$ is a Brownian Motion. Now $dY_t=W_tdt$. So from this expression one can conclude that $dY_t \cdot dY_t=d[Y_t,Y_t]$ where $[\ ]$ is the quadratic variation process and it must be zero as $dt \cdot dt =0$. But people apply Fubini-stochastic integral exchange trick and see $Y_t=\int_0^t(t-u)dW_u$. So by applying Ito isometry one can see that $E(Y_t^2)=\int_0^t(t-u)^2du \neq 0$. Can someone tell me where I'm going wrong? Why this discrepancy?

Every step is correct, including the identities $Y_t=\displaystyle\int_0^tW_s\mathrm ds=\int_0^t(t-s)\mathrm dW_s$ and $[Y]_t=0$, except the underlying assertion that the process $Y^2-[Y]$ should be a martingale and consequently that $\mathbb E(Y_t^2)$ and $\mathbb E([Y]_t)$ should coincide for every $t$.

Recall that the quadratic variation process $[Y]$ is defined as
$[Y]_t=Y_t^2-2\displaystyle\int_0^tY_s\mathrm dY_s$. In particular, $Y^2-[Y]=2\displaystyle\int_0^\cdot Y\mathrm dY$ is a martingale when $Y$ is, but not otherwise.

Here, $Y$ is not a martingale since, for every $s\lt t$, $\mathbb E(Y_t\mid\mathcal F_s)=Y_s+(t-s)W_s\ne Y_s$.

Edit: A general method (which is not a trick) to compute $\mathbb E(Y_t^2)$ is indeed to use Itô's formula for $Y^2$. One gets
$$
\mathrm d(Y_t^2)=2Y_t\mathrm dY_t+\mathrm d[Y]_t=2Y_tW_t\mathrm dt.
$$
Fubini on $\Omega\times[0,t]$ with the measure $\mathbb P\otimes\mathrm{Leb}$ yields
$$
\mathbb E(Y_t^2)=\mathbb E\left(\int_0^t2Y_sW_s\mathrm ds\right)=\int_0^t2\mathbb E(Y_sW_s)\mathrm ds.
$$
Fubini again yields
$$
2\mathbb E(Y_sW_s)=2\int_0^s\mathbb E(W_uW_s)\mathrm du=\int_0^s2u\mathrm du=s^2,
$$
hence
$$
\mathbb E(Y_t^2)=\int_0^ts^2\mathrm ds=\frac{t^3}3.
$$

Your second representation is correct, but you cannot treat it as a stochastic integral (so will not be a martingale) as the integrand depends on the end point of the integral, $t$. Even if it were a martingale, you cannot apply the Itou isometry to it for the same reason. Basically, the problem is that $Y_t$ are integrands against completely different processes in the second representation.


Solving Cubic Equations with Lagrange Resolvent? I'm having difficulties understanding my textbook's decription of solving cubic equations using Lagrange Resolvents and symmetric polynomials.

Here's what I understand:
$$ x^3 + px - q = (x-r)(x-s)(x-t)$$
We can also write:
$$\lambda = r+ws+w^2t$$
$$\mu =wr+s+w^2t$$
where $1, w, w^2$ are the cubic roots of 1. I then understand that $\lambda^2 + \mu^3$ and $\lambda^3\mu^3$ are symmetric polynomials in r, s, and t. It is also solvable that the elemntary symmetric functions in r, s, t are $0, p, q$ where
$$r+s+t=0$$ $$rs+rt+st=p$$ $$rst=q$$
The part where I get confused is that the book claims that $\lambda^3$ and $\mu^3$ are the  roots of the quadratic polynomial $q(x)=x^2-(\lambda^3+\mu^3)x+\lambda^3\mu^3$, which seems obvious to me. Then they claim you can use the quadratic formula to solve for $\lambda^3$ and $\mu^3$ in terms of $p$ and $q$, thus allowing you to solve a system of equations to acquire $r,s,t$.

How can you use the quadratic formula to "explicitly solve for $\lambda^3$ and $\mu^3$ in terms of $p$ and $q$"?

A symmetric polynomial in $r$, $s$, and $t$ will be a polynomial in the elementary symmetric functions $0$, $p$, and $q$, i.e. a polynomial in p and q. You need to to know those polynomials explicitly to solve the quadratic equation explicitly. This involves getting your hands dirty.

You can see this worked out carefully in this paper by Svante Janson (pp. 1-6). What he calls $u$ and $v$ are, up to scalar factors, what you call $\lambda$ and $\mu$. In his notation, $u+v$ is one of the roots of your polynomial, $uv=-p/3$ (so $u^3v^3=-p^3/27$), and $u^3+v^3$ is the product of the roots, i.e. $-q$. This will give you your explicit quadratic polynomial with roots $u^3$ and $v^3$. Janson takes a slightly different approach, showing that $u^3-v^3=\sqrt{-\Delta/27}$ where the discriminant $\Delta=-4p^3-27q^2$. You can get $u^3$ and $v^3$ directly from those two equations.

To convert the notation, I believe that $v=\lambda/3$ and $u=w^2\mu/3$ if you take $r=\beta_1$,$s=\beta_2$,and $t=\beta_3$.

A good historical reference is Chapter 6 or William Dunham's "Journey through Genius," which covers the discovery of "Cardano's formula," which is what this is... without the Galois theory, because, well he did it about 250 years before Galois. It was one of the first significant results in mathematics that went beyond the legacy of the Greeks.

I use this formula all the time. If you have found the roots of the resolvent quadratic. Then all you have to do is find the cube roots of x1^3 and x2^3. Your answer should be $(-a+x1^(1/3)+x2^(1/3))/3.$ The other two roots are $(-a+w*x1^(1/3)+w^2*x2^(1/3))/3$ and $(-a+w^2*x1^(1/3)+w*x2^(1/3))/3.$ Where w=(-1+sqrt(-3))/2 and w^2=(-1-sqrt(-3))/2.


Solve the quartic equation With $x \in R$. Solve the equation $$x^4+(x-1)(x^2-2x+2)=0$$

Idea 1: $(x^2-x+1)(x^2+2x-2)=0$

Idea 2:$(2x^2+x-1)^2=(3x-3)^2$

Idea 3:$<=>x^4(x-1)x^2-2(x-1)^2=0$

Let $y=x-1$.....

I need another way because I am collecting methods of solving with quartic equations

Some methods of solving the quatric equations are described Here


Noetherianess of a locally noetherian affine scheme without axiom of choice I use the definition of a noetherian ring given by Qiaochu in this:
A commutative ring is noetherian if, for any nonempty collection of ideals $\mathcal{I}$, there is some $I \in \mathcal{I}$ which is not properly contained in any $J \in \mathcal{I}$.

Can we prove the following theorem without using Axiom of Choice?

Theorem
Let $A$ be a commutative ring with unity.
Let $f_1,\cdots, f_n$ be elements of $A$ such that $A = (f_1,\cdots, f_n)$.
Suppose that each $A_{f_i}$ is noetherian.
Then $A$ is noetherian.

As for why I think this question is interesting, please see(particularly Pete Clark's answer):
Why worry about the axiom of choice?

Remark
Under the condition of the theorem, we can prove without the axiom of choice
that $A$ satisfies the ascending chain condition(see for example Hartshorne).
However, we need the axiom of dependent choice to prove that $A$ is noetherian in our sense using this fact.

Notation
Let $A$ be a commutative ring with unity.
Let $I$ be an ideal of $A$.
Let $f$ be an element of $A$.
We denote by $I_{(f)}$ the inverse image of $IA_f$ by the canonical homomorphism $A \rightarrow A_f$.
Namely $I_{(f)} = \{x \in A | \ $ There exists an integer $n \gt 0$ such that $f^n x \in I  \}$.

Let $\mathcal{I}$ be a nonempty collection of ideals of $A$.
We denote by $\mathcal{I}_{(f)}$ the set $\{I_{(f)}\ | \ I \in \mathcal{I} \}$.

Lemma 1
Let $A$ be a commutative ring with unity.
Let $f$ be an element of $A$.
Let $\mathcal{I}$ be a nonempty collection of ideals of $A$.
Let $J$ be an element of $\mathcal{I}$.
Suppose $JA_f$ is maximal in the set $\{IA_f \ |\ I \in \mathcal{I} \}$.
Then $J_{(f)}$ is maximal in $\mathcal{I}_{(f)}$.

Proof:
Let $I$ be an element of $\mathcal{I}$ such that $J_{(f)} \subset I_{(f)}$.
Then $J_{(f)}A_f \subset I_{(f)}A_f$.
Hence $JA_f \subset IA_f$.
By the assumption, $JA_f = IA_f$.
Hence $J_{(f)} = I_{(f)}$.
QED

Lemma 2
Let $A$ be a commutative ring with unity.
Let $f_1,\cdots, f_n$ be elements of $A$ such that $A = (f_1,\cdots, f_n)$.
Let $I, J$ be ideals of $A$ such that $I \subset J$.
Suppose $I_{(f_i)} = J_{(f_i)}$ for all $i$.
Then $I = J$.

Proof:
Let $x \in J$.
It suffices to prove that $x \in I$.
Since $J \subset J_{(f_i)}$ for all $i$, $J \subset I_{(f_i)}$ for all $i$,
Hence there exists an integer $m \gt 0$ such that $f_i^m x \in I$ for all $i$.
By this question, we can prove without the axiom of choice that $A = (f_1^m,\cdots, f_n^m)$.
Hence there exists $g_1,\cdots, g_n \in A$ such that $1 = g_1f_1^m + \cdots + g_n f_n^m$.
Then $x = g_1f_1^m x + \cdots + g_n f_n^m x \in I$ as desired.
QED

Proof of the theorem
We define inductively sets of nonempty ideals $\mathcal{I}_0, \mathcal{I}_1, \cdots, \mathcal{I}_n$
and ideals $J_1, \cdots, J_n$ as follows.
We define $\mathcal{I}_0 = \mathcal{I}$.
Let $k$ be an integer such that $1 \le k \le n$.
Suppose $\mathcal{I}_{k-1}$ is defined.
Let $J_k$ be an element of $\mathcal{I}_{k-1}$ such that
${J_k}_{(f_k)}$ is a maximal element of $\mathcal{I_{k-1}}_{(f_k)}$ which exists by Lemma 1.
We define $\mathcal{I}_k = \{I \in \mathcal{I}_{k-1} \ |\ I_{(f_k)} = {J_k}_{(f_k)}\}$.
We claim that $J_n$ is a maximal element of $\mathcal{I}$.
Let $I \in \mathcal{I}$ be such that $J_n \subset I$.

We first show by induction that $I \in \mathcal{I}_k$ for $k = 0, 1, \cdots, n$.
Since $\mathcal{I}_0 = \mathcal{I}$, $I \in \mathcal{I}_0$.
Let $k$ be an integer such that $1 \le k \le n$.
Suppose $I \in \mathcal{I}_{k-1}$.
Since $J_n \in \mathcal{I}_k$, ${J_n}_{(f_k)} = {J_k}_{(f_k)}$.
Since ${J_n}_{(f_k)} \subset I_{(f_k)}$, ${J_k}_{(f_k)} \subset I_{(f_k)}$.
Since $I \in \mathcal{I}_{k-1}$, ${J_k}_{(f_k)} = I_{(f_k)}$.
Hence $I \in \mathcal{I}_k$ as desired.

Next we prove that ${J_n}_{(f_k)} = I_{(f_k)}$ for all $k$.
Since $J_n \in \mathcal{I}_k$, ${J_n}_{(f_k)} = {J_k}_{(f_k)}$.
On the other hand, since $I \in \mathcal{I}_k$, ${J_k}_{(f_k)} = I_{(f_k)}$.
Hence ${J_n}_{(f_k)} = I_{(f_k)}$ as desired.

By Lemma 2, $J_n = I$.
Hence $J_n$ is a maximal element of $\mathcal{I}$.
QED


Simplification of: AB + A'C + BC in boolean algebra I am trying to understand the simplification of the boolean expression:

AB + A'C + BC

I know it simplifies to

A'C + BC

And I understand why, but I cannot figure out how to perform the simplification through the expression using the boolean algebra identities. I was wondering if someone could show me the steps needed to do this. Thank you in advance.

\begin{align*}
&\mathrel{\phantom{=}}AB+A'C+BC\\
&=AB+A'C+BC(A+A')          \quad \text{($A+A'=1$, Complementarity law)}\\
&=AB+A'C+ABC+A'BC\\
&=AB+ABC+A'C+ABC           \quad \text{(Associative law)}\\
&=AB+A'C                 \quad   \text{(Absorption law)}
\end{align*}


Limit of Sequence Whose Elements Depend on Previous Elements I was solving some exercises on limits, when I came across this problem:

  The point $C_1$ divides a segment $AB = l$ in half; the point $C_2$ divides a segment $AC_1$ in half; the point $C_3$ divides a segment $C_2C_1$ in half; the point $C_4$ divides $C_2C_3$ in half, and so on. Determine the limiting position of the point $C_n$ when $n \to \infty$.

I did successfully answer this particular exercise. However, it led me to a more general question.

How do you evaluate the limit of a sequence whose elements depend on prior elements? Sources on this topic are also welcome.

As another example, the Fibonacci sequence is one whose elements rely on prior elements (however, unlike the exercise above, there is no limit).

Note: I'd appreciate if nothing more advanced than limits are used. This exercise was found at the beginning of a calculus and analysis book, and all that I've covered so far are limits.


Multiplication partitioning into k distinct elements Let's say I have a list with the prime factors of a number $n$ and their corresponding exponents. Is there a formula to calculate how many multiplications with $k$ distinct factors of $n$ are possible? For example, $12 = 2^2 \times 3$, so $F(12,2)$ would be:

$[2,6][3,4][1,12] = 3$

Note that the order of each set of factors does not matter. I have a feeling I only need the exponents of each prime factor, but I'm at a loss on how to actually proceed.

This problem is a straightforward application of the Polya Enumeration
Theorem and very similar to the problem discussed at this
MSE link.

Remark. What follows below does permit repeated factors,
the formula for distinct factors is at the end.

Recall  the recurrence by Lovasz  for the cycle  index $Z(S_n)$ of
the multiset operator $\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{MSET}_{=n}$ on $n$ slots, which is
$$Z(S_n) = \frac{1}{n} \sum_{l=1}^n a_l Z(S_{n-l})
\quad\text{where}\quad
Z(S_0) = 1.$$

We have for example,
$$Z(S_3) = 1/6\,{a_{{1}}}^{3}+1/2\,a_{{2}}a_{{1}}+1/3\,a_{{3}}$$
and
$$Z(S_4) = 1/24\,{a_{{1}}}^{4}+1/4\,a_{{2}}{a_{{1}}}^{2}
+1/3\,a_{{3}}a_{{1}}+1/8\,{a_{{2}}}^{2}+1/4\,a_{{4}}.$$

Suppose the prime factorization of $n$ is given by
$$n=\prod_p p^v.$$

Applying PET it now follows almost by inspection that the
desired count is given by the term
$$F(n, k) = \left[\prod_p X_p^v\right]
Z(S_k)\left(\prod_p \frac{1}{1-X_p}\right)$$
where the square bracket denotes coefficient extraction
of formal power series.

As an  example of what  these substituted cycle indices  look like
consider
$$Z(S_3)\left(\frac{1}{1-X_1}\frac{1}{1-X_2}\frac{1}{1-X_3}\right)
\\ = 1/6\,{\frac {1}{ \left( 1-X_{{1}} \right) ^{3}
\left( 1-X_{{2}} \right) ^{3}\left( 1-X_{{3}} \right) ^{3}}}
\\+1/2\,{\frac {1}{ \left( -{X_{{1}}}^{2}+1
 \right)  \left( -{X_{{2}}}^{2}+1 \right)  \left( -{X_{{3}}}^{2}+1 \right)
 \left( 1-X_{{1}} \right)  \left( 1-X_{{2}} \right)
\left( 1-X_{{3}} \right) }}
\\+1/3\,{\frac {1}{ \left( -{X_{{1}}}^{3}+1 \right)  \left( -{X_{{2
}}}^{3}+1 \right)  \left( -{X_{{3}}}^{3}+1 \right) }}.$$

It  should be  clear  that  coefficient extraction  here  is fast  and
efficient using the Newton binomial series which says that
$$[Q^{\mu k}] \left(\frac{1}{1-Q^\mu}\right)^\nu
= {k+\nu-1\choose \nu-1}.$$

We get for $F(n,1)$ the sequence
$$1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,\ldots$$
which is as it ought to be.

For $F(n,2)$ we get the sequence
$$1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 3, 1, 3, 1, 3,\ldots$$
which points us to OEIS A038548.

As another example consider $F(n,4)$ which yields
$$1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 4, 1, 2, 2, 5, 1, 4, 1, 4,\ldots$$
which points us to OEIS A21320
where the above calculation is confirmed.

 As a last example consider $F(n,5)$ which yields starting at $n=20$
$$4, 2, 2, 1, 7, 2, 2, 3, 4, 1, 5, 1, 7, 2, 2, 2, 9, 1, 2, 2,\ldots$$
which points us to OEIS A252665.

The  following   Maple  code  implements this cycle index computation.

with(combinat);
with(numtheory);

pet_cycleind_symm :=
proc(n)
option remember;

    if n=0 then return 1; fi;

    expand(1/n*add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;

pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;

    res := ind;

    polyvars := indets(poly);
    indvars := indets(ind);

    for v in indvars do
        pot := op(1, v);

        subs1 :=
        [seq(polyvars[k]=polyvars[k]^pot,
             k=1..nops(polyvars))];

        subs2 := [v=subs(subs1, poly)];

        res := subs(subs2, res);
    od;

    res;
end;

F :=
proc(n, k)
    option remember;
    local v, q, sind, res;

    v := op(2, ifactors(n));

    sind :=
    pet_varinto_cind(mul(1/(1-cat(`X`, q)), q=1..nops(v)),
                     pet_cycleind_symm(k));

    res := sind;
    for q to nops(v) do
        res := coeftayl(res, cat(`X`, q)=0, v[q][2]);
    od;

    res;
end;

Addendum. For distinct factors the formula is
$$G(n, k) = \left[\prod_p X_p^v\right]
Z(P_k)\left(\prod_p \frac{1}{1-X_p}\right)$$
where the square bracket denotes coefficient extraction
of formal power series and $Z(P_n)$ is the cycle index of the
set operator $\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{SET}_{=n}$ which was also used in the linked-to
computation from above.

Starting at $n=20$ we get for $G(n,3)$ the sequence
$$2, 1, 1, 0, 4, 0, 1, 1, 2, 0, 4, 0, 2, 1, 1, 1, 4, 0, 1, 1,\ldots$$
which points us to OEIS A088434.

Starting at $n=120$ we get for $G(n,4)$ the sequence
$$8, 0, 0, 0, 0, 0, 3, 0, 1, 0, 1, 0, 3, 0, 0, 1, 1, 0, 1, 0,\ldots$$

The Maple code for $G(n,k)$ is as follows.

with(combinat);
with(numtheory);

pet_cycleind_set :=
proc(n)
option remember;

    if n=0 then return 1; fi;

    expand(1/n*add((-1)^(l-1)*a[l]*pet_cycleind_set(n-l), l=1..n));
end;

# duplicate code omitted

G :=
proc(n, k)
    option remember;
    local v, q, sind, res;

    v := op(2, ifactors(n));

    sind :=
    pet_varinto_cind(mul(1/(1-cat(`X`, q)), q=1..nops(v)),
                     pet_cycleind_set(k));

    res := sind;
    for q to nops(v) do
        res := coeftayl(res, cat(`X`, q)=0, v[q][2]);
    od;

    res;
end;

Addendum.  Observe that  we  can easily  compute $$G(n,5)$$  where
$$n=2^{50}\times 3^{20} \times 5^{10}$$  which has $11781$ divisors so
that    the    brute    force    method   would    have    to    check
$$1889560697901637761$$  five-tuples,  which  is  not  feasible.   The
answer is $$27879304592.$$

Yes! From a number theoretical perspective, this is known as the divisor function. Or, even more generally, $\sigma_k (n)$ where the sigma function denotes the sum of all divisors of an integer to the power $k$. If we set $k=0$ we see that this is exactly the same as counting all possible divisors of this integer.

The general form of the divisor function (sometimes denoted $d(n)$) is as follows:

$$\prod_{p^\alpha || n} (\alpha +1)$$ where $p^\alpha || n$ means that $p^\alpha |n$ but $p^{\alpha +1}$ does not. Also, p denotes a prime divisor. Also, if the notation is getting to you, just thinking of it as factorizing an integer, and then taking the product of 1 more than all the exponents of each divisor. (Do an example!)


Is it wrong to deduce the premise from the conclusion? My professor gave me the following argument to prove:

  Let $x$ be an even number. Then $x+x^2$ must be an even number.

I wrote down the following:

  Hypothesis: $x = 2n$ for some $n\in \mathbb{Z}$
  Thesis: $x + x^2 = 2m$ for some $m \in \mathbb{Z}$

  Let us consider a direct proof.

  $x+x^2=2m$ (replacing $x=2n$)
  $2n + 4n^2=2m$
  $2(n + 2n^2) = 2m$ (since $n\in\mathbb{N}$, we can replace $m=n+2n^2$)
  2m = 2m

My professor says I only proved the converse (that if $x+x^2$ is even, then $x$ must be even) I claimed that showing that I can reach an identity from any equation using the act that $x=2n$ suffices as a formal proof, but he wasn't convinced. How can I prove I'm right?

Consider the invalid statement: "Let $x$ be a bird. Then $x$ is a duck."

Now, your approach is to assume the consequent, that $x$ is a duck, then demostrate that the duck is a bird, and decide that the statement was valid.  Opps.

So, no, affirming the consequent is a falacy.

sorry it's not clear if the universe of discourse is natural numbers or integers since you reference both.

Assuming integers:

if $ x $ even -> $ (x + x^2) $ even.

because you're given that x is even, pretend $ \exists i \in \Bbb Z (x=2i) $

you need to show that $ \exists j \in \Bbb Z  (x^2 + x = 2j) $

$$ x = 2i $$
$$ x^2 = 4i^2 $$
$$ (x^2 + x) = 4i^2 + 2i = 2j $$
$$ j = 2i^2 + i $$


Show $\mathfrak{gl}(n,\mathbb{R})_{\mathbb{C}}\cong \mathfrak{u}(n)_{\mathbb{C}}$ I would appreciate help showing the complexification of these Lie algebras are isomorphic:

$$\mathfrak{gl}(n,\mathbb{R})_{\mathbb{C}}\cong \mathfrak{u}(n)_{\mathbb{C}}$$

where $\mathfrak{gl}(n,\mathbb{R})= M_n(\mathbb{R})$ and $\mathfrak{u}(n)=\lbrace X\in M(\mathbb{C}):X+X^{*}=0\rbrace$

Revealing my ignorance, I especially am confused regarding the complexification of $\mathfrak{u}(n)$ since it already is comprised of complex matrices. And how can $\mathfrak{gl}(n,\mathbb{R})_{\mathbb{C}}$ be isomorphic to $\mathfrak{u}(n)_{\mathbb{C}}$ in view of the stipulation regarding elements of $\mathfrak{u}(n) $that $X+X^{*}=0$

Thanks

Let $H_i$ denote the $n$-by-$n$ diagonal matrix with $1$ at the $(i,i)$-entry and $0$ elsewhere.  For $i,j=1,2,\ldots,n$ with $i<j$, let $X_{i,j}$ be the matrix with $1$ at the $(i,j)$-entry and $0$ elsewhere, whilst $Y_{i,j}$ denotes $X_{i,j}^\top$.  Note that the matrices $H_i$'s, $X_{i,j}$'s, and $Y_{i,j}$'s form a Chevalley basis of the Lie algebra $\mathfrak{gl}(n,\mathbb{C})$, noting that $$\mathfrak{gl}(n,\mathbb{R})_\mathbb{C}=\mathbb{C}\,\underset{\mathbb{R}}{\otimes}\,\mathfrak{gl}(n,\mathbb{R})\cong\mathfrak{gl}(n,\mathbb{C})\,.$$
(An isomorphism $\mathbb{C}\,\underset{\mathbb{R}}{\otimes}\mathfrak{gl}(n,\mathbb{R})\cong\mathfrak{gl}(n,\mathbb{C})$ should be easy to find).

A basis for $\mathfrak{u}(n)\subseteq\mathfrak{gl}(n,\mathbb{C})$ is given by $\tilde{h}_i$, $\tilde{x}_{i,j}$, and $\tilde{y}_{i,j}$ where $i,j=1,2,\ldots,n$ and $i<j$, where
$$\tilde{h}_i:=\sqrt{-1}\,H_i\,,\,\,\tilde{x}_{i,j}:=\frac{1}{\sqrt{2}}\,\left(X_{i,j}-Y_{i,j}\right)\,,\text{ and }\tilde{y}_{i,j}:=\frac{\sqrt{-1}}{\sqrt{2}}\,\left(X_{i,j}+Y_{i,j}\right)\,.$$
Now, set
$$h_i:=1\,\underset{\mathbb{R}}{\otimes}\,\tilde{h}_i\,,\,\,x_{i,j}:=1\,\underset{\mathbb{R}}{\otimes}\,\tilde{x}_{i,j}\,,\text{ and }y_{i,j}:=1\,\underset{\mathbb{R}}{\otimes}\,\tilde{y}_{i,j}\,.$$
Then, we see that the elements $h_i$'s, $x_{i,j}$'s, and $y_{i,j}$'s form a Chevalley basis of $\mathfrak{u}(n)_\mathbb{C}=\mathbb{C}\,\underset{\mathbb{R}}{\otimes}\,\mathfrak{u}(n)$.  Consequently, an isomorphism $\varphi:\mathfrak{gl}(n,\mathbb{C})\to\mathfrak{u}(n)_\mathbb{C}$ is the linear extension of the assignments
$$\varphi\left(H_i\right)=h_i\,,\,\,\varphi\left(X_{i,j}\right)=x_{i,j}\,,\text{ and }\varphi\left(Y_{i,j}\right)=y_{i,j}$$
for $i,j=1,2,\ldots,n$ with $i<j$.

P.S.:  I said something wrong.  I called the bases I provided Chevalley bases, but they are actually not.

This answer is incorrect - See comment below

In an attempt to answer my own question, can I not say that two finite dimensional vector spaces with the same dimension are isomorphic?

Thus $\dim \mathfrak{gl}(n,\mathbb{R})= \dim \mathfrak{u}(n)= n^2$ And their respective complexifications have dimension $2n^2$.


if A = {a,b,c} . Prove OR disprove that P(A) "power set of A" is a group under the union operation So my thinking is that

$$P(A) = \{\{\emptyset\},\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}\} $$

although it's clearly closed under union , assiociatve and the identity exists which is the empty set but still it's not a group since there are no inverses exists for any of the elements except for that of the empty set itself.

But is that valid to say , I mean how can I do it more rigorously?

The empty set is the identity but $B\cup C$ contains $B$ so if $B$ is non empty $B\cup C$ is non empty.  Hence a nonempty element is not invertible.

Let $~ X= \{1,2,3\}~$ and $~P(X)~$ be the power set of $~X~$. Examine if $~P(X)~$ is a group under the composition $~*~$ defined by $$~A*B= A\cup B ,~~~~~~~ \forall A,B \in P(X)~.$$


The system of Diophantine equations. Often seen similar systems of equations. Usually consider such systems in which decisions no. Such as there.  Is there $a,b,c,d\in \mathbb N$ so that $a^2+b^2=c^2$, $b^2+c^2=d^2$?

I think it would be more interesting to solve the system in which there are solutions.  For example to find out whether such a system solution?

$$\left\{\begin{aligned}&a^2+b^2+c^2=q^2\\&c^2+q^2=k^2\end{aligned}\right.$$

What is the right approach? And how to solve it?

The $2.n.m$ Diophantine equations have well-known complete solutions (a.k.a. parameterizations) for all $n,m \ge 1$ (for example, see Bradley's paper or Barnett's paper).
In particular, the solutions for the $2.1.2$ and $2.1.3$ equations you are asking about have been known for centuries. If $q$ is odd (which is essentially equivalent to “all primitive integer solutions”), you’re looking for
\begin{align}
  k &= r^2+s^2  \\
  c &= 2rs  \\
  q &= r^2-s^2
\end{align}
where
$$q = t^2+u^2+v^2+z^2$$
and either
$$c = t^2+u^2-v^2-z^2  \tag{$\star$}$$
or
$$c = 2(tz-uv)\tag{$\star\star$}$$
With respect to $q$, you are are ultimately finding $r,s,t,u,v,z$ such that
\begin{align}
  r^2 &= s^2+t^2+u^2+v^2+z^2;
\end{align}
this is the $2.1.5$ Diophantine equation, for which the complete solution (a.k.a. parameterization) is known. You then simply need to intersect that result with the solutions for ($\star$) and ($\star\star$) to obtain the complete solution to your original system of equations.

If you solve the system of equations:

$$\left\{\begin{aligned}&a^2+b^2+c^2=q^2\\&c^2+q^2=w^2\end{aligned}\right.$$

When the standard approach solution and using a replacement.

$$p=9t^2-10tk+5k^2$$

$$s=5t^2-10tk+9k^2$$

$$x=7t^2-10tk+7k^2$$

$$y=4(t^2-k^2)$$

$$z=5t^2-14tk+5k^2$$

Then the solution can be written as :

$$a=2zx$$

$$b=z^2+y^2-x^2$$

$$c=2yx$$

$$q=z^2+y^2+x^2$$

$$w=p^2+s^2$$

$t,k$ - integers which we ask.


CAR- & Weyl-Algebra: Uniqueness? Given a Hilbert space: $\mathcal{h}$

Consider representations of the CAR-algebra: $\mathcal{A}_\text{CAR}^{(\prime)}(\mathcal{h})$

In Bratelli & Robinson it is stated the uniqueness:
$$\Phi:\mathcal{A}_\text{CAR}(\mathcal{h})\to\mathcal{A}_\text{CAR}'(\mathcal{h})$$

My guess would be: $\Phi[a(f)]:=a'(f)$

Extended to the algebra via:
$$\Phi[A+B]:=\Phi[A]+\Phi[B],\,\Phi[\lambda A]:=\lambda\Phi[A]$$
$$\Phi[AB]:=\Phi[A]\Phi[B],\,\Phi[A^*]:=\Phi[A]^*,\Phi[1]:=1$$
Especially, it is an isometry then: $\|\Phi[A]\|=\|A\|$

Similarly for the Weyl algebra: $\mathcal{W}^{(\prime)}(\mathcal{h})$

Why do these construction work?

Ok, I think I got it now...

Both work perfectly fine as they are always nondegenerate:
$$\mathcal{A}_\text{CAR}:\quad a(f\neq0)\neq0$$
$$\mathcal{W}:\quad W(f)\neq0$$

A counterexample is provided by the angular momentum algebra:
$$\mathcal{J}:\quad [J_i,J_j]=\imath\varepsilon_{ijk}J_k$$
There one has one trivial representation: $J_x=J_y=J_z=0$


Fundamental group of the complement of $3$ pairwise linked circles in $\mathbb R^{3}$ I'm reading through the Hatcher book for a course on algebraic topology. Here it is explained how to find the fundamental group for the complement of two linked circles.
I'm trying to proof the case with three pairwise linked circles, but I can't seem to reproduce any results.
I tried moving to $S^3$, and moving the point on infinity. This way you would get two linked circles and a line passing through them. This could be (I think) deformation retracted to a torus with two linked circles cut out, the circles going around the middle 'hole'.
At that point I'm stuck on how to either:

a) Moving on from here to another deformation which gives me a space or wedge sum of spaces I know the fundamental group of.

b) Finding suitable open sets for using Van Kampen's theorem.

I'd be very interested in hearing anyone's thoughts on this, and also in seeing if there is a way to extend the answer to the complement of $n \in \mathbb N$ pairwise linked circles. (I expect the answer to be $\mathbb Z^n$, based on the case of 1 and 2 linked circles)

Not a complete answer

Hatcher shows two unlinked circles have a fundamental group $\mathbb{F}_2$ which is humungous - it's the free group of all possible wors in two letters.  The fundamental group of two linked circles is simply $\mathbb{Z} \oplus \mathbb{Z}$ - we can exploit the linking of the rings in order to express any entanglement as a special type.

In order to find the homotopy group of the Borromean rings you may need to draw as many as 4 rings.

3 pairwise linked circles will have three generators, looping ones around each of the possible three circles.  Do these generators commute?  Sure we can imagine one the circles disappearing and getting two circles.  We have shown:

$$ \pi_1(3\text{ linked circles}) \subseteq \mathbb{Z}\oplus \mathbb{Z} \oplus \mathbb{Z}$$


Does $\sum_{n=1}^{\infty}\ln(n\sin(\frac{1}{n}))$ converge? I must determine whether the following series converges: $$\sum_{n=1}^{\infty}\ln\left(n\sin\left(\frac{1}{n}\right)\right)$$

I know that in general, I must use the limit comparison test, but I cannot find an expression to which I can compare it. For instance, I have tried the usual process:

For $n$ large, we have that $\lim_{n\to \infty}n\sin\frac1n=1$, and so, $\ln(1)=0$. This fails the divergence test, but it cannot be concluded automatically that the series is convergent either. How may I proceed here? Any help would be appreciated.

Taylor's theorem tells us that $\sin x = x - \dfrac1{3!} x^3 + \dfrac1{5!} \cos(\xi) x^5$ for some $\zeta \in [0,x]$ whenever $x > 0$. Since $0 \le \cos \xi \le 1$ when $x = \dfrac1n$:

$$x - \frac1{3!} x^3 \le \sin x \le x - \frac1{3!} x^3 + \frac1{5!}x^5$$

Substituting $x = \dfrac1n$:

$$\frac1n - \frac1{3!} \frac1{n^3} \le \sin \frac1n \le \frac1n - \frac1{3!} \frac1{n^3} + \frac1{5!} \frac1{n^5}$$

Multiplying all sides by $n$:

$$1 - \frac1{3!} \frac1{n^2} \le n \sin \frac1n \le 1 - \frac1{3!} \frac1{n^2} + \frac1{5!} \frac1{n^4}$$

Applying Taylor's theorem on $\ln(1-x)$ when $x>0$:
$$\ln(1-x) = - \frac 1 {1 - \xi} x$$
for some $\xi \in [0,x]$

Therefore:
$$-\frac{x}{1-x} \le \ln(1-x) \le -x$$

That is:
$$1-\frac{1}{1-x} \le \ln(1-x) \le -x$$

Combining the above two sections:

$$\ln \left(1 - \frac1{3!} \frac1{n^2}\right) \le \ln \left( n \sin \frac1n \right) \le \ln \left(1 - \frac1{3!} \frac1{n^2} + \frac1{5!} \frac1{n^4} \right)$$

Then:

$$\frac{-1}{6n^2-1} \le \ln \left( n \sin \frac1n \right) \le - \frac1{3!} \frac1{n^2} + \frac1{5!} \frac1{n^4}$$

Applying sandwich theorem:

$$\sum_{n=1}^\infty \frac{-1}{6n^2-1} \le \sum_{n=1}^\infty \ln \left( n \sin \frac1n \right) \le \sum_{n=1}^\infty - \frac1{3!} \frac1{n^2} + \frac1{5!} \frac1{n^4}$$

and both sums can be easily shown convergent.


The complement of a finite set in a countable set is countable Let $X$ be a countable set. Let  $A  \subset  X$.

How can I prove that if $A$ is finite then $X  \setminus  A$ is countable?

I started off by supposing that $X  \setminus  A$ is finite and then need to show that this leads to a contradition.

The result that $X  \setminus  A$ is countable will then follow because all subsets of $X$ are countable or finite

Suppose for the sake of contradiction that $X- A$ is finite. What does that imply about $(X-A) \cup A$?

Let $X$ be a countable set.

Therefore, there exists an injective function $f:X\to \mathbb{N}$

By definition of injective, we have $\forall a,b\in X:[f(a)=f(b)\implies a=b] $

  Let  $A  \subset  X$.

Let $g$ be a function $g:X \setminus A \to \mathbb{N}$ such that $\forall a\in X\setminus A: g(a)=f(a)$

Suppose $x,y\in X\setminus A$
By definition of $\setminus$, we have $x,y \in X$
Suppose $g(x)=g(y)$
By definition of $g$ ,we have $g(x)=f(x)$
By definition of $g$ ,we have $g(y)=f(y)$
$f(x)=f(y)$ (substituting 4 and 5 into 3)
$x=y$ (from 2 and since $f$ is injective)
$g(x)=g(y)\implies x=y$ (conclusion from 3 and 7)
$\forall a,b \in X\setminus A: [g(a)=g(b) \implies a=b]$ (conclusion from 1 and 8)
$g$ is an injective function $g:X\setminus A\to \mathbb{N}$ (from 9 and definition of $g$)
By definition, $X\setminus A$ is countable  (regardless of whether $A$ is finite)


limit of Wiener process divided by time Why $ \lim_{t\to\infty} \frac{W_t}{t} = 0 $

I tried to write it as Wiener process is defined which is:
$ \lim_{t\to\infty} \frac{W_t}{t} = \lim_{t\to\infty}\lim_{N\to\infty} \frac{\sum_{k=1}^{\lfloor Nt \rfloor} \varepsilon_{k}}{\sqrt{N}*t}*\frac{\lfloor Nt \rfloor}{\lfloor Nt \rfloor} $
and then use a Law of large numbers but I got expression $ 0*\infty $

any hints or solution would be appreciated.

Ok, I tried different approach, we know that, $$ Z_t=t*W_\frac{1}{t}  $$
is also a wiener process, then wanted limit is equiwalent to
$$ \lim_{t\to 0 } \frac{Z_\frac{1}{t}}{\frac{1}{t}}=\lim_{t\to 0} t*Z_\frac{1}{t} $$
which is 0 , is it true ?


$a,b,c\in\mathbb{N}\:\land\:\frac ab+\frac bc+\frac ca\in\mathbb{N}\Rightarrow abc=n^3,\:n\in\mathbb{N}$ If $a$, $b$ and $c$ are natural numbers such that $$\frac ab+\frac bc+\frac ca\in\mathbb{N},$$ prove that $abc$ is a perfect cube.

All I could get was $ab^2+bc^2+ca^2\equiv_{abc}0$.

Theorem $\rm\displaystyle\ \ For\,\ a,b,c,n\in\Bbb Z\!:\ \frac{a}b+\frac{b}c+\frac{c}a =n\:\Rightarrow\: abc= m^3,\ $ for some $\rm\,m\in \Bbb Z$

Proof $\ $ Scaling by $\rm\,abc\,$ yields $\rm\quad  a^2 c + b^2 a + \color{#C00}{c^2 b}\, =\, nabc\qquad\qquad \color{brown}{(E)}$

We prove  $\rm\,\ a = kj^2/d^2,\, b = ik^2/d^2,\, c = ji^2/d^2,\,\ $ so $\rm\,\ {\bf abc = m^3}\:$ for $\rm\: m = ijk/d^2\in\Bbb Z,\: $

for $\rm\ i = (b,c),\ j = (a,c),\  k = (a,b),\  d = (a,b,c).\ $ Note $\rm\:d\mid i,j,k\: $ so $\rm\,d^2\!\mid jk,i^2,j^2,k^2.$

$$\begin{eqnarray}\rm j^2 k\,
&=&\ \ \ \rm (a,c)^2 (a,b)\\
&=&\ \ \ \rm (a^2,ac,c^2)(a,b)\\
&=&\,\rm a(a^2,ac,c^2,ab,cb,\ \ \ \color{#C00}{c^2b/a)}\ \ \ so,\ by\ \color{brown}{(E)}/a\\
&=&\,\rm a(a^2,\color{#e70}{ac},c^2, ab,\color{#0A0}{bc},\overbrace{n\color{#0A0}{bc}\!-\!\color{#e70}{ac}\!-\!b^2)}\\
&=&\,\rm a(a^2,ac,c^2,ab,bc,b^2)\\
&=&\,\rm a(a,b,c)^2 =\, ad^2\\
\end{eqnarray}$$

Finally, by symmetry, also  $\rm\,\ ik^2 = bd^2,\,\ ji^2 = cd^2.\quad$ QED

Remark $ $ There is much literature on this equation, e.g. see Dave Rusin's page and OEIS A085705, where you will find mention of the connection with elliptic curves.  Since the above proof uses only gcds, it holds true (mod units) over any gcd-domain, i.e. any domain where every pair of nonzero elements has a gcd. Hence this proof is more general than proofs employing prime factorization. For example, it works in the ring $\,\Bbb I\,$ of all agebraic integers, where no primes exist (since $\rm\, a = \sqrt{a}\sqrt{a}),\:$ but which, nevertheless, is a Bezout domain, so a gcd domain.

Consider polynom $(X-\frac a b)(X- \frac b c)(X - \frac c a)$.
It is equal to $X^3 - u*X^2 + v*X - 1$, where $u = \frac a b + \frac b c + \frac c a$.
and $v = \frac b a + \frac c b + \frac a c$.

for $X = 1$, you get $1 - u + v - 1 = v - u$. So you have $(1-\frac a b)(1- \frac b c)(1- \frac c a) = v-u$.

You can rewrite that: $(b-a)(c-b)(a-c)=c(b^2-a^2)+a(c^2-b^2)+b(a^2-c^2)$

but not sure it leads anywhere.


Good math books to get on with my education? im looking for a set of good math books.

Since i live in norway im going through the norwegian school system, where i choose to go whats called a electrical line. Im not gonna go to deep into the education system here, but the problem is that this "line" is primarily applied to by teenagers that dosnt do it to good in school. However i did do it quite good in school at junior high, and the math we have now is the same as we had during the first and second year of junior high. And i want to do some more advanced math, and after taking some tests i got an offer, but that offer had forced me to loose at least 3 hours of computer class, and maybe a couple of houres of automation, and in addition to that i would need to take the child math i already took. So i said no. However now that im trying to learn some fuctional programing(already have a solid fundation i imperative style) it would be nice to improve my math skills.

So does anyone have some good recomadations ?

https://www.regjeringen.no/en/topics/education/school/the-norwegian-education-system/id445118/
http://en.wikipedia.org/wiki/Education_in_Norway

I loved this series, as a child, especially How to count without counting, Graphs and their uses, and First concepts in topology.  Of course, you will have your own taste in different topics, but these books are just wonderful.

If you are just in grade 9, 10, 11 or 12, you really are not ready for things like combinatorics and topology.  I assume you know basic algebra and the easy parts of trigonometry. In that case, I would suggest you just use one book to work on your own, and the book I would try is Thomas and Finney's book Calculus and Analytic Geometry.  You can get it used for about \$20, and the skills you learn will surely be of use even if later you decide not to go further in math.

BTW, I love David's suggested books, I just think you may not be ready for them.


How do I solve the following equality-constrained quadratic program? I am trying to minimize:
$$(x_1-k_1)^2 + (x_2-k_2)^2 + (x_3-k_3)^2 +\ldots+ (x_n-k_n)^2$$

subject to following equality:
$$B = 1 + x_1 + x_2 + x_3 + x_4+\ldots+x_n.$$

Is there  a closed form solution for this?   I found the  case where $n=2$ using Lagrange multipliers and could do the same for case $n=3$, but does anyone see a simple solution?

Wikipedia gives this solution, but I'm not sure if the matrix has an inverse so I'm weary of going down that route.  Also inverting matrix takes some computer time and may not be feasible for solving different instances of the problem thousands of times.

Also, I put $B$ above as a constant, which it is, but I'm interested in the case where $B = (1+k_1)\times(1+k_2)\times\ldots\times(1+k_n)$, but I'm not sure if that makes the problem easier.  My intuition is that it does not and just complicates things.

We have a quadratic program

$$\begin{array}{ll} \text{minimize} & \|\mathrm{x} - \mathrm{k}\|_2^2\\ \text{subject to} & 1_n^T\mathrm{x} = c\end{array}$$

where $c = -1 + \displaystyle\prod_{i=1}^n (k_i+1)$. Using a Lagrange multiplier,

$$\mathcal{L} (x,\lambda) := \frac{1}{2}\|\mathrm{x} - \mathrm{k}\|_2^2 + \lambda (1_n^T \mathrm{x} - c)$$

Taking the partial derivatives and finding where they vanish,

$$\nabla_x \mathcal{L} = \mathrm{x} - \mathrm{k} + \lambda 1_n = 0_n \qquad \qquad 1_n^T \mathrm{x} = c$$

Hence,

$$1_n^T \mathrm{x} - 1_n^T \mathrm{k} + \lambda 1_n^T 1_n = 0_n$$

and

$$\lambda = \dfrac{1_n^T \mathrm{k} - c}{n}$$

Thus, the minimum is attained at

$$\mathrm{x} = \mathrm{k} + \left(\dfrac{c - 1_n^T \mathrm{k}}{n}\right) 1_n$$

and the minimum is

$$\|\mathrm{x} - \mathrm{k}\|_2^2 = \left\|\left(\frac{c - 1_n^T \mathrm{k}}{n}\right) 1_n \right\|_2^2 = \frac{\left(c - 1_n^T \mathrm{k}\right)^2}{n}$$

You can rewrite the constraint as:

$$0= 1-B +\sum_{i=1}^nx_i$$

and then turn it into a cost function:

$$\lambda\left\|1-B+\sum_{i=1}^{n}x_i\right\|_2$$

If we want to we can rewrite this with matrices, $\bf 1$ being the column vector full of ones:

$$\min_{\bf x} \left\{ \|{\bf x}-{\bf k}\|_2 + \lambda\|1-B+{\bf 1}^T{\bf x}\|_2\right\}$$

You will probably get better faithfulness to the constraint if you solve a sequence of linear problems using  for example iterated reweighting approach replacing the 2-norm with another norm for the constraint term:

$$\min_{\bf x} \left\{ \|{\bf x}-{\bf k}\|_2 + \lambda\|1-B+{\bf 1}^T{\bf x}\|_k\right\}$$

you could aim for different $k$, but I think $k=1$ would be a good choice.

EDIT I erroneously claimed that 1-norm term in this case would be the Lasso method, but as pointed out below in comments, in that case the 1-norm is of course applied on the $\bf x$ vector alone. Even as the constraint is in fact a scalar the $k$ norm is still relevant as the deviation for the constraint would be punished differently from the terms in the least-squares term.


Trying to understand why circle area is not $2 \pi r^2$ I understand the reasoning behind $\pi r^2$ for a circle area however I'd like to know what is wrong with the reasoning below:

The area of a square is like a line, the height (one dimension, length) placed several times next to each other up to the square all the way until the square length thus we have height x length for the area.

The area of a circle could be thought of a line (The radius) placed next to each other several times enough to make up a circle. Given that circumference of a circle is $2 \pi r$ we would, by the same reasoning as above, have $2 \pi r^2$. Where is the problem with this reasoning?

Lines placed next to each other would only go straight like a rectangle so you'd have to spread them apart in one of the ends to be able to make up a circle so I believe the problem is there somewhere. Could anybody explain the issue in the reasoning above?

The main issue is that you don't form an area by placing lines next to each other--you need to place strips next to each other.  As you say, to form an $a\times a$ square, you can place $n$ strips of dimension $a\times w$ next to each other, where $w=a/n$, giving total area $naw=a^2$.

Your suggestion amounts to forming the area of a circle of radius $r$ by placing $n$ strips of dimension $w\times r$ next to each other radially, where $w=2\pi r/n$, giving area $nrw=2\pi r^2$.  But look what happens if you do this:

The problem is that the strips overlap, so the total area of the $n$ strips is greater than the area of the circle.  If you run the animation (by reloading the page, if necessary) you can convince yourself that in the $n\rightarrow\infty$ limit, half of each strip contributes to the final area.  (Observe that as strips are added in the counterclockwise direction, roughly half of each strip gets covered by subsequent strips.)

We can fix the problem of overlapping strips more easily by using triangles of base $w$ and height $r$ instead of rectangles:

This gives area $\frac{1}{2}nrw=\pi r^2$.

Actually, you can use your method, but you have to be a bit careful.  Start with a circle of radius $r$ centered at the origin.  We can think of the total area of the circle as adding the "lengths" of all the circumferences of all the circles centered at the origin which have a radius $x \leq r$:

$$\text{Area of Circle} = \int_0^{r} 2 \pi x~\textrm{d}x = \pi r^2$$

In reality, we are approximating the circumference of each circle by an annulus of small width, and then letting that width tend to zero.

This method works with finding the area of the square too.  Suppose you have a square centered at the origin.  Then, the area of the square is the perimeter of each square centered at the origin which ha a side length $x < r$.

$$
\text{Area of Square} = \int_{-r/2}^{r/2} 4x dx = r^2.
$$

The reason this doesn't work the way you originally thought was in order to add "fattened" radii all around the circle, the fattened radii overlap, and so you are over counting.


How to prove $f(z) = \sum_{n=-\infty}^{\infty}e^{2\pi inz}e^{-\pi n^2}$ has a unique zero inside a unit square in the 1st quadrant. There are three parts of the question, the first two which I proved.

$a.$ Proving $f(z)$ is entire analytic.

$b.$ $f(z+1) = f(z)$ and $f(z+i) = e^{\pi}e^{-2\pi iz}f(z)$

$c.$ Inside the unit square in the 1st quadrant, prove that $f(z)$ has a unique $z_0$ such that $f(z_0)=0$

I need some help regarding the $c.$ part of the question. I don't understand how to apply the argument principle here (which is a hint). It is also given that $f(z)$ does not vanish on the boundary of the square.

What I could do:

$$f(0) = f(1) = 1+2\sum_{n=1}^{\infty}e^{-\pi n^2}$$

and

$$f(i) = f(1+i) = e^{\pi}\big(1+2\sum_{n=1}^{\infty}e^{-\pi n^2}\big)$$

Using the hint given below by a user: I end up with $$\dfrac{1}{2\pi i}log f(z)$$ and the integral evaluates to -1/2 and 1/2 on the two vertical sides and 0 on the horizontal lines. Resulting in the closed integral to 0. Any ideas where I might be going wrong?

Any idea how to proceed?

Just an idea but I'm getting zero. Perhaps you can spot something I missed.

Let $\;C_1,C_2,C_3,C_4\;$ the unit square's side begining with the one on the real axis and counterclockwise. Then, since the function is entire and non-zero on the square, we get

$$\left.\int_{C_1}\frac{f'(z)}{f(z)}dz=\log f(z)\right|_0^1=\log f(1)-\log f(0)=0$$

and the same on the other horizontal side, yet on the rightmost vertical side:

$$\int_{C_2}\frac{f'(z)}{f(z)}=\log f(1+i)-\log f(1)=\log\left(e^\pi e^{-2\pi i}f(1)\right)-\log f(1)=\pi$$

and on the other vertical side:

$$\int_{C_4}\frac{f'(z)}{f(z)}=\log f(0)-\log f(i)=\log f(0)-\log\left(e^\pi e^{0}f(0)\right)=-\pi$$


What's the difference between stochastic and random? What's the difference between stochastic and random?

A variable is random. A process is stochastic. Apart from this difference, the two words are synonyms.

There is absolutely a difference between stochastic process and randomness. For example, if I take one step then let's suppose my friend takes two steps. Now my friend's steps are not random, those are dependent on my steps. That means my friend's steps has a process which is the number of steps I take. But it is random because my friend doesn't know how many steps I will take. So the steps I take is a random walk.


Find the area of the region enclosed by the inner loop of the curve $r = 4 + 8 \sin \theta$ The loop is generated for $\theta \in \left[\frac76\pi, \frac{11}{6}\pi\right]$
(this is from setting $r = 0$).

So, $$A = \int_{\frac76\pi}^{\frac{11}{6}\pi} \frac12(4 + 8 \sin \theta)^2 \,d\theta=\\
= \int_{\frac76\pi}^{\frac{11}{6}\pi} 8 + 32 \sin \theta + 32 \sin^2\theta\, d\theta =\\
= \int_{\frac76\pi}^{\frac{11}{6}\pi} 8 + 32 \sin \theta + 16(1 - \cos(2\theta))\, d\theta=\\
= \int_{7\pi/6}^{11\pi/6} 24 + 32 \sin \theta - 16 \cos(2\theta)\, d\theta =\\
= \bigg[24\theta - 32 \cos \theta - 8 \sin(2\theta)\bigg]_{\frac76\pi}^{\frac{11}{6}\pi} =\\
= (44\pi - 16\sqrt{3} + 4\sqrt{3}) - (28\pi + 16\sqrt{3} - 4\sqrt{3}) =\\
= 16\pi - 24\sqrt{3}$$

Is my answer right?

As much as Wolfram|Alpha is impressive, here is the answer again using the SageMath free/libre software:

sage: integral((1/2)*(4 + 8 * sin(x))^2, x, 7*pi/6, 11*pi/6).simplify()
16*pi - 24*sqrt(3)

Yes.

Another way of doing this would be:
$$\int_{7\pi/6}^{11\pi/6} \int_{0}^{4 + 8\sin \theta} rdrd\theta$$
$$ \ldots = 16\pi - 24\sqrt{3}$$
If you fancy multiple integrals.


integrating using student t distribution Evaluate the integral

$\int_0^\infty\frac{1}{1+x^2}dx$

using the Student t distribution.

I don't know where to start. I am assuming that I can't just do regular integration.  I don't know how I am supposed to use the Student T distribution.  Does it have to do with the pdf of  the Student T distribution?

Can anyone help me?

The $t$ density is
\begin{eqnarray*}
  f \left( x \right) & = & \frac{\Gamma \left( \frac{\nu + 1}{2}
  \right)}{\sqrt{\nu \pi} \Gamma \left( \frac{\nu}{2} \right)} \left( 1 +
  \frac{x^2}{\nu} \right)^{- \frac{\nu + 1}{2}}
\end{eqnarray*}
If you write your integrand (for $\nu = 1$)
\begin{eqnarray*}
  \frac{1}{1 + x^2} & = & \left( 1 + \frac{x^2}{\nu} \right)^{- \frac{\nu +
  1}{2}}
\end{eqnarray*}
then you notice that
\begin{eqnarray*}
  \int_0^{\infty} \frac{1}{1 + x^2} \mathrm{d} x & = & \frac{\sqrt{\pi} \Gamma
  \left( \frac{1}{2} \right)}{\Gamma \left( 1 \right)}\\
  & = & \frac{\pi}{2}
\end{eqnarray*}

What is the density of this distribution? If you look that up, you might be able to specialize it, and use the fact that it must integrate to 1, to get what you seek.


on sequences of Lebesgue measurable subsets of a compact set In preparation for the real analysis qualifying exam at my grad school, I've been working through the recommended textbook Modern Real Analysis, by Zimmer (it's free online here).

For the last few days, I've been trying to figure out problem 4.27 (on p124 of the text, which is p132 of the pdf), which asks us to prove the following:

Theorem. If $\{E_k\}_{k=1}^\infty$ is a sequence of Lebesgue measurable subsets of a compact set $K\subseteq \mathbb{R}^n$ such that $\inf_{k\geq 1} \lambda(E_k)>0$, then there's some point which belongs to infinitely many $E_k$'s (i.e. there's some point which belongs to $E_k$ for infinitely many $k$).

Definitions, Etc.

Lebesgue measure is denoted by $\lambda$.
The Lebesgue outer measure $\lambda^*$ is defined to be
$$\lambda^*(A):= \inf\left\{\sum_{I\in\mathcal{S}} \mathrm{vol}(I) \right\},$$
where the infimum is taken over countable all covers $\mathcal{S}$ of $A$ by sets of the form $[a_1,b_1]\times \cdots \times [a_n,b_n]$ (which Zimmer calls closed $n$-dimensional intervals, or just closed intervals if the dimension is clear).
A set $A\subseteq\mathbb{R}^n$ is defined to be Lebesgue measurable if
$$\lambda^*(A)=\lambda^*(A-S)+\lambda^*(A\cap S)$$
for any set $S\subseteq \mathbb{R}^n$.

My Attempt

If there are finitely many distinct $E_k$'s, this is trivial. So, assume there are infinitely many distinct $E_k$'s (I don't know if this assumption will be useful, but since we're working with compactness, I figured it might). From each $E_k$, choose a point $x_k$. Then because $K$ is compact, there's a subsequence $\{x_{k_j}\}_{j=1}^\infty$ of $\{x_k\}_{k=1}^\infty$ which converges to some point $x\in K$. At this point, I'm stuck. My gut tells me that this point $x$ (or some similarly constructed point) should be in infinitely many $E_k$'s, but I can't for the life of me figure out how to prove this.

Could someone point me in the right direction?

Suppose every point is contained in at most finitely many $E_k$, so the indicator functions $\chi_{E_k}:K\to \mathbb R$ converge to $0$ pointwise. Since $K$ has finite measure, we can apply Egorov's theorem, so for any $\epsilon>0$ we have a set $S\subseteq K$ such that $\lambda(K\setminus S)<\epsilon$ and $\chi_{E_k}|_S$ converges uniformly to $0$. But since $\chi_{E_k}|_S$ takes only the values $0$ and $1$, we must have that this sequence is eventually identically $0$ hence $E_k\cap S=\emptyset$ for sufficiently large $k$. But then $$\lambda(E_k)\leq \lambda(E_k\cap S)+\lambda(K\setminus S)=\epsilon$$
and since $\epsilon$ is arbitrary this violates the fact that $\inf_{k\geq 1} \lambda(E_k)>0$. Thus some point is contained in infinitely many $E_k$.

We can use the following: Given inf E_k is larger.
Make use of limsup of the sets.
Let A_1 be union from 1 to infinity of E_k's this is of course geq E_1 hence infimum > 0
Let A_2 be union from 2 to infinity of E_k's this is of course geq E_2 hence infimum > 0
and so on

Take the intersection of those. Those are the set of points that are to be found in
infinitely many sets E_k. This intersection being larger than or equal to infimum has positive measure meaning it contains uncountably many points.


Find the equation of the diameter of the circle having equation $x^2+y^2-6x+2y-8=0$ Find the equation of the diameter of the circle having equation $x^2+y^2-6x+2y-8=0$.

My Work:
$$x^2+y^2-6x+2y-8=0$$
Comparing the above equation with $x^2+y^2+2gx+2fy+c=0$, we get:
$$g=-3$$,
$$f=1$$
Now,
$$\textrm {centre}=(-g,-f)$$
$$=(3,-1)$$

Search for "Completing the Square".

Solving your system $x^{2}+y^{2}−6x+2y−8=0$, you can add and substract some numbers to get the classical circle equation $(x-h)^{2}+(y-k)^{2}=r^{2}$.

Well, you can check that your problem is equivalent to $$(x-3)^2+(y+1)^2=18$$ So your radius is $\sqrt{18}=3\sqrt{2}$, and your center $(3,-1)$

I dont know what you mean about "the equation of the diamater" but if you have a point $(a,b)$ the equation that passes through the point and the diamenter is:
$$y=\frac{b-3}{a+1}(x-a)+b$$


many value of delta corresponding to single epsilon? I have to prove that limiting value of $f(x,y)=xy^2/(x^2+y^2)$ is zero when $(x,y) \to (0,0)$ using epsilon delta relation.

My solution is
$|xy^2/(x^2+y^2)|< |x|< \delta$  as $\sqrt{x^2+y^2}<\delta$
so here I got $\varepsilon=\delta$.
Now I can also take $\varepsilon= n \times \delta$ where $n>1$

In this way I have many value of $\delta$ corresponding to one value of $\varepsilon$. Is that correct ?

The number $\varepsilon$ is fixed from the start. Therefore, you can't take $\varepsilon=n\times\delta$.

On the other hand, if $\delta=\varepsilon$ works, then $\delta=\frac\varepsilon2,\delta=\frac\varepsilon3,\ldots$ will also work. There's no problem there.

Right, $0\leq\left|\frac{xy^2}{x^2+y^2}\right|\leq|x|.$

The limite 0 follows immediately from the squeeze theorem. (are you familiar with?)


Isomorphism between (even, odd) and (positive, negative) This is a problem in Conceptual Mathematics: A First Introduction to Categories

Find an isomorphism
$$
(\{{\rm odd}, {\rm even}\}, +) \stackrel{f}{\to} (\{{\rm positive}, {\rm negative} \}, \times)
$$
HINT: There are only two invertible maps of sets from $\{{\rm odd}, {\rm even}\}$ to $\{{\rm positive}, {\rm negative} \}$. One of them 'respects the combining rules', but the other doesn't

I've a little stumped with this question. This is one of the maps I've come up with
$$
f(x): \begin{cases}
-k & {\rm if} & x = 2k - 1, ~~k=1,2, ...\\
k & {\rm if} & x = 2k, ~~ k = 0, 1, 2,
\end{cases}
$$
This will map odd to positive numbers, and even to negative numbers, and is invertible. The issue is that it does not preserve the combining rules, e.g. if $x = 2m - 1$ is odd, and $y = 2n$ even
$$
f(x + y) = f(2(m + n) - 1) = -(m + n) \tag{1}
$$
but on the other hand
$$
f(x) \times f(y) = f(2m + 1) \times f(2n) = (-m) \times (n) = -mn \tag{2}
$$
Clearly
$$
f(x + y) \not= f(x)\times f(y)
$$
Any other suggestion?

Let g be the map of sets from {odd,even} to {pos, neg} that sends odd to pos and even to neg.
g is an isomorphism for obvious reason (existence of an inverse)
Now, suppose you want to "add structure" to g and make it an isomorphism between $(\{{\rm odd}, {\rm even}\}, +)$ and $(\{{\rm pos}, {\rm neg} \}, \times)
$. Let's prove that the following equation doesn't not hold:
$g(odd + even) \stackrel{?}{=}  g(odd) * g(even)$
$g(odd + even)    = g(odd)$  (odd + even = odd, combining rule for elements in {odd,even})
             = pos             g(odd) = pos by construction of g

g(odd) * g(even) = pos * neg       by construction of g
             = neg             (pos * neg = neg, combining rule in ({pos, neg}, *))

g(odd+even) is different from g(odd) * g(even) that's why the combining rule is not followed by g.


Isolated singularities of this function? I have a function $\sin (\frac 1 z) \sin(wz)$ where $w=\exp (\frac {i\pi} {4} )$. I need to find out the singularities of this function.

My progress :-

This function will have singularities on points where it isn't defined i.e. $z=0$.

This singularity may be removable, pole, or essential.

I'm unable to see how to go further. Computations on WolframAlpha returned limit as 0 along the real axis. It returned an infinity when $z$ was assumed purely imaginary. Hence I guessed it is a essential singularity.
Could someone tell me a non WolframAlpha solution/insight? Thank you :)

You can characterize the different kinds of singularities by limits of the function. If it has a pole in $z_0 \in \mathbb{C}$, then we have that $\lim_{z \to z_0}|f(z)| = \infty$. If it has a removable singularity, then it has a finite limit (the value which you can give the function to expand it to a holomorphic function at that point if your initial function was already holomorphic in its domain). An essential singularity is "any other kind of singularity", which happens to have specific characteristics. Well, if it is just a singularity which is not removable and not a pole, then it is enough to find sequences $\{z_n\}_{n \in \mathbb{N}},\{z'_n \}_{n \in \mathbb{N}} \subset D$ (where $D$ is the domain of the given function) such that $z_n \to z_0$, $z'_n \to z_0$ with $|f(z_n)|\to \infty$ and $f(z'_n) \to a \in \mathbb{C}$.
(Actually, having an essential singularity at $z_0$ is actually equivalent to having a sequence $z_n \to z_0$ such that $lim_n f(z_n)$ does not exist.)
In this case, you can check that $z_n := \frac{i}{n}$ is such that $|f(z_n)|\to \infty$ and that $z'_n:= \frac{1}{\pi n}$ is such that $f(z'_n) \to 0$.
Hope this helps :)

Looking at the limits along the real and imaginary axes is a fine idea. That was your idea, so good for you. But why did you go to WolframAlpha to find these limits? Any student of complex analysis needs to be able to do such things on his own. OK, you "guessed" the correct answer, but it would be better to be able to prove/explain how these limits imply an essential singularity at $0.$


The function given by the Rao-Blackwell theorem is a statistic $(X_1,\ldots,X_n)$ is a random sample, $V_n$ is an unbiased estimator of the population parameter $\theta$ and $T_n$ is a sufficient statistic for $\theta$. Then by Rao-Blackwell theorem the rv
$$\varphi(T_n):=\mathbb E_\theta[V_n\mid T_n]$$
is an unbiased estimator of $\theta$ whic is uniformly better than $V_n$.
I know the definitions of conditional expectation and distribution, but I can't get how $\varphi(T_n)$ is an estimator (a function of the random sample that does't depend on $\theta$) using sufficiency

For simplicity, consider the discrete case:
$$\mathbb E_\theta[V_n\mid T_n=t]=\sum_{\lbrace x:T_n(x)=t\rbrace}\mathbb P_\theta(X=x\mid T_n=t)V_n(x)$$
By sufficiency of $T_n$, $\mathbb P_\theta(X=x\mid T_n=t):=g(x,t)$ is independent of $\theta$. So,
$$\mathbb E_\theta[V_n\mid T_n=t]=\sum_{\lbrace x:T_n(x)=t\rbrace}g(x,t)V_n(x)$$
is independent of $\theta$ and is a statistic.

The definition of sufficiency is that the conditional distribution of all of the data given the value of the sufficient statistic does not depend on $\theta,$ i.e. it is the same for all distributions in the family.
If the conditional distribution does not depend on $\theta,$ then the conditional expectation does not depend on $\theta.$


Number in power of $ log x$ stuck Find $x$ if $$x^{\log 26} - x^{\log 24} = x$$

I can’t fiqure this out , I read all law of log but unfurtunately I can’t solve this. Any help about this?

Sorry for my bad english.


Identify heavier ball in group of $14$ We have $14$ identical looking balls, one of them is heavier than the rest of $13$ (all the others weigh exactly the same), how many of weighings (minimum) to find the heavier one?
And I want to know if there is mathematical formula for this problem?

(Sorry for my English)

You need at most 3 weightings.

One possible solution algorithm is:

  Divide the balls in 2 groups of 7 balls.
  Compare the 2 groups.
  Divide the heavier group of 7 balls into 3 groups (3-3-1)
  Compare the 2 groups containing 3 balls. If they are equal, then the 1 remaining is the heaviest.
  Else: Divide the heavier group of 3 balls into 3 groups containing 1 ball.
  Compare 2 of those balls. If they are equal, the remaining ball is the heaviest.

One can also start with 3 groups (5-5-4) and terminate in 3 weighting steps (and I bet there are more different solutions). It is obvious that you can't terminate with 2 weighting steps at most, but I can't prove this formally.

You need just three weighings.

Number the balls in odd numbers, and the remaining number is 20, except 13.  Then use numbers 0 and 26

So the balls are numbered, 0,1,3,5,7,9,11,15,17,19,20,21,23,25,26 or in base 3,

000, 001, 010, 012, 021, 100, 102, 120, 122, 201, 210, 212, 221,222

454 v 454 v 454  for the odd numbers, and 5,4,5 for this set.

Label the pans 0 and 2, and leave 1 off.  Weigh the three digits, and record which pan goes down, or 1 for equal.  The number written is the base 3 value of the heavy ball.

This is actually a three-state system, and the trinary numbers give the weighings,  The solution is good for 26 balls, numbered in trinary numbers 0 to 27 (222) except 13 (111).

0 = put ball in left pan   heavy 0 makes this pan go down

1 = leave ball off  heavy 1 makes pans equal

2 = put ball in right pan.  heavy 2 makes right pan go down.

Since the heavy ball is comprised of three trinary numbers, the weighings will give up to 26 unique numbers.

The solution as given is a slightly modified version of the problem where one coin is known good, and one of thirteen coins is defective, can be either heavy or light.  This has a solution in

M = left pan;   0 = left off;  1 = right pan.

The 13 coins are numbered from 1 to 13, in trinary, with the even numbers in reverse sign, viz 1, -2, 3, -4, ..., -13.  In trinary, they look like

001, 0m1, 010, 0mm, 1mm, m10, 1m1, m01, 100, m0m, 11m, mm0, 111.

The known good coin is numbered -7, ie m1m.

You weigh the coins according to first, second, third digit, and get the numbers as m = left pan down, 0 = balance, 1 = right pan down.

If the number is even (ie an odd number of 0's), swap the m and 1's, eg m10 = 1m0,

You have then determined the number of the coin (as the weightings are 9,3,1), and whether it's heavy or light.  So 1m0 gives +9-3.1 = +6, coin 6 is defective, and heavy.

But you can't do it with 14 coins in three weighings unless you know the defect is +.  In that case, you can do as many as 26 weighings.


Why is this a smooth manifold? I read that the space of quantum states, i.e. the space of density operators
$\mathcal{S}_n = \{ \rho \in H_n : \rho \geq 0, \, \, Tr[\rho]=1\}$
is a smooth manifold of dimension $n^2-1$, without further explanation. Could anyone help me understand why this is the case?
Here $H_n$ denotes the space of hermitian matrices: $H_n = \{ A \in \mathbb{C}^{n \times n} : \, A^* =A  \}$, and $\rho \geq 0$ means that the matrix $\rho$ is positive semi-definite.

This is false, since the set has a boundary, but is not a even a manifold with boundary, since the boundary (the set of singular matrices with trace equal to one) is very singular. However, the set of positive definite hermitian matrices with trace equal to one is, indeed, an open manifold.

Preimage theorem:

Theorem. Let $f:X\to Y$ be a smooth map, and let $y\in Y$ be a regular value of $f$. Then $f^{-1}(y)$ is a submanifold of $X$. If $y\in {\text{im}}(f)$, then the codimension of $f^{-1}(y)$ is equal to the dimension of $Y$. Also, the tangent space of $f^{-1}(y)$ at $x$ is equal to $\ker(df_{x})$.


Uniqueness Proof, Discrete Math Help While reading I came across Uniqueness Proofs. Where a theorem asserts the existence of a unique element with a particular property. In order to prove this two steps are needed, Prove existence and Prove Uniqueness. The example given is Show that if $a$ and $b$ are real numbers and $a ≠ 0$, then there is a unique real number $r$ such that $ar + b = 0$.

I could do the existence portion. $r = -\frac{b}{a}$

I would like some clarification on proving the uniqueness part.

Suppose that $s$ is a real number such that $as + b = 0$. Then $ar + b = as + b$, where $r = -\frac{b}{a}$. You subtract $b$ from both sides and divide both sides by $a$ to get $r = s$. Then it says that this means if $s ≠ r$ then $as + b ≠ 0 $and that this establishes uniqueness.

I guess my issue is how exactly does this prove uniqueness? When would placing some random variable in the same spot as the previous not end with the two variables being the same? One example I thought of where it wouldn't be unique I wasn't able to follow the same steps to disprove.
Consider this
$n^2 = 4$

following the previous example suppose that $s$ is another real number such that
$s^2 = 4$

$\implies n^2 = s^2$

square root both sides
$n = s$

now in the other one they just jumped to the conclusion that this means if $s ≠ n$ then $s^2 ≠ 4$

however... -2 and 2 could fill meaning this example does not have a unique solution. Could someone please give some clarification so I can better understand uniqueness proofs.

To show an object is unique one approach (the one taken here) is to assume that there is a second object that satisfies the given conditions. Then if you can show that this second object is actually the first object then you've shown that all objects that satisfy the condition are identical. In particular, by supposing that $s$ and $r$ are both arbitrary solutions to $ax+b=0$ and showing that $s=r$, you've shown that every solution to $ax+b=0$ is identical, i.e. the solution is unique (if it exists). Note that you can show uniqueness without showing existence.

That fact (that if you divide equal real numbers by equal real nonzero numbers, you get equal real numbers) may have been earlier in your book, or they may have assumed you know it.  It's certainly true for the basic arithmetic operations (addition, subtraction, multiplication, and division), and comes from the fact that the operations are invertible (with zero a special case for multiplication/division).


Proving Euler's identity (fluid mechanics) by differentiating the determinant $J=\partial(x,y,z)/\partial(X,Y,Z)$ I am looking at some fluid mechanics lecture notes. In this context, $X,Y,Z$ are Lagrangian variables (at time $t = 0$) and $x,y,z$ are Eulerian variables (at time $t > 0$). Euler's identity says that $$\frac{\mathrm{D}J}{\mathrm{D}t}=J\mathbf{\nabla}\cdot \mathbf{u}$$ where $\mathbf{u}=\mathbf{u}(x(t),y(t),z(t),t)$ is the velocity of the fluid, $J=\frac{\partial (x,y,z)}{\partial(X,Y,Z)}$ is the Jacobian determinant and $\frac{\mathrm{D}}{\mathrm{D}t}=\frac{\partial }{\partial t}+\mathbf{u \cdot \nabla}$ is the convective differential operator.

The notes say this:

Whilst this seems to look quite straightforward, I am not sure what property has been used here or why it is permissible for us to be able to differentiate each row of the determinant separately and add up the resulting determinants. Could somebody please offer an input or key result that is being used?

Consider the case $n=2$. Observe
\begin{align}
\frac{d}{dt}
\begin{vmatrix}
a_{11}(t) & a_{12}(t)\\
a_{21}(t) & a_{22}(t)
\end{vmatrix}
=&\ \frac{d}{dt} \sum_{\sigma \in S_2}(-1)^{\operatorname{sgn}\sigma} a_{1\sigma(1)}(t) a_{2\sigma(2)}(t)\\
=&\ \sum_{\sigma \in S_2}(-1)^{\operatorname{sgn}\sigma} [a_{1\sigma(1)}'(t) a_{2\sigma(2)}(t)+a_{1\sigma(1)}(t) a_{2\sigma(2)}'(t)]\\
=&\ \sum_{\sigma \in S_2}(-1)^{\operatorname{sgn}\sigma} a_{1\sigma(1)}'(t) a_{2\sigma(2)}(t)+\sum_{\sigma \in S_2}(-1)^{\operatorname{sgn}\sigma} a_{1\sigma(1)}(t) a_{2\sigma(2)}'(t)\\
=&\
\begin{vmatrix}
a_{11}'(t) & a_{12}'(t)\\
a_{21}(t) & a_{22}(t)
\end{vmatrix}
+
\begin{vmatrix}
a_{11}(t) & a_{12}(t)\\
a_{21}'(t) & a_{22}'(t)
\end{vmatrix}.
\end{align}
Note I have used the definition: If $A=(a_{ij})$ is an $n\times n$ matrix then
\begin{align}
\det A = \sum_{\sigma \in S_n} (-1)^{\operatorname{sgn}\sigma} a_{1 \sigma(1)}\cdots a_{n\sigma(n)}
\end{align}
where $S_n$ is the permutation group of $n$ objects.

An additional resource which I found was useful for this problem are those slides:

http://www3.dicca.unige.it/rrepetto/linked-files/fluid-dynamics-lecture-notes.pdf

The appendix (Slide 154 - 156) it explains the topic as well.

The approach there is as follows:

$$J=\operatorname{det}\left(\begin{array}{ccc}\frac{\partial x_{1}}{\partial \xi_{1}} & \frac{\partial x_{1}}{\partial \xi_{2}} & \frac{\partial x_{1}}{\partial \xi_{3}} \\ \frac{\partial x_{2}}{\partial \xi_{1}} & \frac{\partial x_{2}}{\partial \xi_{2}} & \frac{\partial x_{2}}{\partial \xi_{3}} \\ \frac{\partial x_{3}}{\partial \xi_{1}} & \frac{\partial x_{3}}{\partial \xi_{2}} & \frac{\partial x_{3}}{\partial \xi_{3}}\end{array}\right)$$

Definition of the material derivative:

$$\frac{D}{D t}\left(\frac{\partial x_{i}}{\partial \xi_{j}}\right)=\frac{\partial}{\partial \xi_{j}} \frac{D x_{j}}{D t}=\frac{\partial u_{i}}{\partial \xi_{j}}$$

Apply the chain rule:

$$\frac{\partial u_{i}}{\partial \xi_{j}}=\frac{\partial u_{i}}{\partial x_{1}} \frac{\partial x_{1}}{\partial \xi_{j}}+\frac{\partial u_{i}}{\partial x_{2}} \frac{\partial x_{2}}{\partial \xi_{j}}+\frac{\partial u_{i}}{\partial x_{3}} \frac{\partial x_{3}}{\partial \xi_{j}}=\frac{\partial u_{i}}{\partial x_{k}} \frac{\partial x_{k}}{\partial \xi_{j}}$$

As an example for the first row, insert the previous results:

$$ \operatorname{det}\left(\begin{array}{ccc}
\frac{\partial u_{1}}{\partial x_{k}} \frac{\partial x_{k}}{\partial \xi_{1}} & \frac{\partial u_{1}}{\partial x_{k}} \frac{\partial x_{k}}{\partial \xi_{2}} & \frac{\partial u_{1}}{\partial x_{k}} \frac{\partial x_{k}}{\partial \xi_{3}} \\
\frac{\partial x_{2}}{\partial \xi_{1}} & \frac{\partial x_{2}}{\partial \xi_{2}} & \frac{\partial x_{2}}{\partial \xi_{3}} \\
\frac{\partial x_{3}}{\partial \xi_{1}} & \frac{\partial x_{3}}{\partial \xi_{2}} & \frac{\partial x_{3}}{\partial \xi_{3}}
\end{array}\right)
$$

Simplify the result:

$$ \operatorname{det} = \left(\begin{array}{ccc}\frac{\partial u_{1}}{\partial x_{1}} \frac{\partial x_{1}}{\partial \xi_{1}} & \frac{\partial u_{1}}{\partial x_{1}} \frac{\partial x_{1}}{\partial \xi_{2}} & \frac{\partial u_{1}}{\partial x_{1}} \frac{\partial x_{1}}{\partial \xi_{3}} \\ \frac{\partial x_{2}}{\partial \xi_{1}} & \frac{\partial x_{2}}{\partial \xi_{2}} & \frac{\partial x_{2}}{\partial \xi_{3}} \\ \frac{\partial x_{3}}{\partial \xi_{1}} & \frac{\partial x_{3}}{\partial \xi_{2}} & \frac{\partial x_{3}}{\partial \xi_{3}}\end{array}\right)=\frac{\partial u_{1}}{\partial x_{1}}J$$


The eternally sleeping beauty. (Thought experiment regarding uniform distribution on the natural numbers) A cousin of mine recently confronted me with a thought experiment that in essence contained an analogical situation to the following problem:

Assume you are a beauty with the following properties:

-You know there was a first day on which you woke up.

-You know each time you fall asleep, you lose your memories of the previous times you woke.

-You are immortal and live in an temporally infinite universe.

You are confronted with the question: What probability do you ascribe to the even "Today is the n-th time I woke up."?

It seems to me that there is no answer within Kolmogorov's probability theory, since any day seems equally likely and you cannot have an uniform distribution over the natural numbers. Is the question not well defined? I would love to read your thoughts.

Due to the finite lifetime of princesses (sorry little children but princesses are not immortal) there are a finite number of values this could take (even taking into account some Disney story this could be a large value) so there is a finite but small probability you could define to each day.

Now even with an immortal princess there is a finite age to the planet/universe so there are still limitations on this.

If we now allow the princess some planning she may very well leave notes to herself which could allow her to get the answer with high probability (or zero probability if we follow a storyline like Momento).

In short without greater bounds on the question anyone could mess it around however they like so I think I'll go with your answer than the question is not well defined.

My thoughts:

Since the princess didn't really know how many days she woke up,
So: Her guess is a purely random guess.
->numerator for the probability is 1(1 guess)
She is immortal and lives in a temporary infinite universe,
So: The limit of the numerator is $$\infty$$

$$1/\infty=0$$

So probability for P(Correctly answer"Today is the n-th time I woke up.")=0


Modules over algebras In his book Global Calculus, S. Ramanan defines modules over a sheaf of algebras. He demands the abelian group $\mathcal{M}(U)$ to be provided with the structure of $\mathcal{A}(U)$-module for every open $U$. Note that here $\mathcal{A}$ is a sheaf of algebras, not rings.

Strangely enough, I could not find the definition of ‘module over algebra’. Everything I came up with was the following:

  If algebra $A$ over $R$ is a homomorphism $R\longrightarrow A$ of commutative rings, and module $M$ over ring $A$ is a homomorphism $A \longrightarrow \mathop{\mathrm{End}} M$ in the same category, then module over algebra should be an arrow from $R\longrightarrow A$ to $R\longrightarrow \mathop{\mathrm{End}} M$ (an arrow between objects under $R$),

but the homomorphism $R\longrightarrow \mathop{\mathrm{End}} M$ is not given, unless $M$ is already an $R$-module (mostly a real vector space, for $\mathcal{A}$ is usually a sheaf of $\mathbb{R}$-algebras of differentiable functions).

How should I think of modules over algebras in this context? Is there a common definition?

A module over an $R$-algebra $A$ is just an $A$-module.

(When one is dealing with a bimodule $M$ over an $R$-algebra $A$, in general one only considers those such that the action of $R$ on the left and on the right on $M$ coincide, but there is no such condition for one-sided modules)

I'm not sure to understand your question: what do you mean by "the homomorphism $R \longrightarrow \mathrm{End} M$ is not given"? -If you have an $A$-module structure on $M$, that is, a morphism $A \longrightarrow \mathrm{End}M$, and $A$ is an $R$-algebra, that is, a morphism, $R \longrightarrow A$, of course you are given  a morphism $R \longrightarrow \mathrm{End} M$: it's just the composition of the previous ones.


Doubt about step of the proof that $\limsup\limits_{n\to\infty}\left|\sum\limits_{\ell=1}^mz^n_\ell\right|^{1/n}=\max\limits_{1\le\ell\le m}|z_\ell|$ I'm studying some questions of A problem book of complex analysis, Daniel Alpay. The Question is:

  Let $z_1$, $\ldots$, $z_m$ be complex different from $0$.
  Show that $$\limsup\limits_{n\to\infty}\left|\sum\limits_{\ell=1}^m{z_\ell}^n\right|^{1/n}=\max\limits_{1\le\ell\le m}|z_\ell|$$

My doubt:

Let $M :=\max\limits_{1\le\ell\le m}|z_\ell|$ and $S_n:=\sum\limits_{\ell=1}^mz_\ell^n$

I prove that $\limsup _{n \rightarrow \infty}|S_n|^{1/n}\leq M$. But I don't saw anything to do for prove that $\limsup |S_n|^{1/n}\geq M$.

I suppose that using contradiction that $\limsup |S_n|^{1/n} = \tilde{M}<M$ and take $z_l = M e^{i2\pi l}$, it'll be done.

Could Someone help me?

We import an elementary result in Number Theory. Lemma: If $(x_1,...,x_k)$ is a  finite sequence of real numbers and  $\epsilon >0$ then there are infinitely many $n\in \Bbb N$ such that  $\max_{1\leq j\leq k} d(nx_j,\Bbb Z)<\epsilon,$ where $d(y,\Bbb Z)=\min \{|y-z|:z\in \Bbb Z\}.$

WLOG, $\;M=1.$ Let $A=\{l:|z_l|=1\}.$ Let $B=\{l:|z_l|<1\}.$ Let $|A|$ be the number of members of $A$. Let $|B|$ be the number of members of $B.$

Note: $|A|\geq 1$ but $B$ could be empty. In what follows, we follow the convention that a sum such as  $\sum_{l\in B}f(l)$ is $0$ if $B$ is empty

Given $\epsilon >0,$ and given $n_0\in \Bbb N:$

(i).   Take $\epsilon'\in (0, \min (1,\epsilon)]$ and take $n_1\geq n_0$ such that $(1-\epsilon')^{1/n_1}>1-\epsilon.$ Then $\forall n \geq n_1\;(\;(1-\epsilon)^{1/n}>1-\epsilon\;).$

(ii). Take $\epsilon''\in (0,\epsilon']$  such that $\forall n\geq n_1\;\forall l\in B\;(\;|z_l|^n<\frac {\epsilon'}{2(1+|B|)}\;).$

(iii). Take $\epsilon'''\in (0,\epsilon'']$ such that $\forall t\in (-\epsilon''',\epsilon''')\;( |1-e^{it}|<\frac {\epsilon'}{2|A|}).$

(iv). By the Lemma, take $n_2\geq n_1$ such that $\forall l\in A\;(d(n_2t_l,\Bbb Z)<\epsilon'''.$

(v). Now for $l\in A$ we have $(z_l)^{n_2}=1+\delta_l$ with $|\delta_l|<\frac {\epsilon'}{2|A|}.$  And for $l\in B$ we have $|z_l|^{n_2}<\frac {\epsilon'}{2(1+|B|)}$.

Hence $$|\sum_{l=1}^m (z_l)^{n_2} |=|\sum_{l\in A}(1-\delta_l) +\sum_{l\in B}(z_l)^{n_2}|   \geq$$   $$\geq |A|-\sum_{l\in A}|\delta_l| - \sum_{l\in B}|z_l|^{n_2}   >$$ $$>|A|-\epsilon'/2 -\epsilon'/2.$$

So we have $n_2\geq n_0$  and  $$|\sum_{l=1}^m(z_l)^{n_2}|^{1/n_2}>$$ $$>(|A|-\epsilon')^{1/n_2}\geq$$ $$\geq (1-\epsilon')^{1/n_2} >1-\epsilon =M-\epsilon.$$

I tried to finish with another manner:

Cauchy Hadamard Theorem:

let be the power series $\sum_{n\geq 0} a_n z^n$. Then, the radius of convergence is $R = \frac{1}{\limsup |a_n|^{1/n}}$

consider the function

$g(z) = \sum_{l\in\{1,\ldots,m\}} \frac{1}{1-z_l z} = \sum_{n\geq 0}{(z_1^n +\ldots +z_l^n) z^n}$

The radius of convergence at origin is $ R = \min\frac{1}{|z_l|} = \frac{1}{\max |z_l|}$.

Using the C-H Theorem, it would be done, but It's incomplete.

So, I think about other:

WLOG suppose that

$$ |z_1|\leq |z_2| \leq \ldots \leq |z_k| < |z_{k+1}| =\ldots |z_m| = M$$

We have $m-k$ z's with maximum modulus. Applying Triangle Inequality, we have

$$ \Bigg|\sum_{l=0}^{m}z^n_l \Bigg|^{1/n} \leq \Bigg(\sum_{l=0}^{m}|z|^n_l\Bigg) \leq (m M^n)^{1/n}\Rightarrow \Bigg|\sum_{l=0}^{m}z^n_l \Bigg|^{1/n} \leq m^{1/n} M$$

So

$$ \lim_{n\rightarrow \infty} m^{1/n} M = M \Rightarrow \limsup_{n\rightarrow\infty} \Bigg|\sum_{l=0}^{m}z^n_l \Bigg|^{1/n} \leq M$$

Divide the expression by $z^n_m = M^n e^{in\alpha_m}$:

$$\Bigg(\sum_{l=0}^{m}z^n_l \Bigg) = z^n_m\Bigg(\sum_{l=0}^{m}\Big(\frac{z_l}{z_m} \Big)^n\Bigg) = z^n_m\Bigg(\sum_{l=0}^{k}\Big(\frac{z_l}{z_m} \Big)^n+ \sum_{l=k+1}^{m}\Big(\frac{z_l}{z_m} \Big)^n\Bigg) = z^n_m\Bigg(\sum_{l=0}^{k}\Big(\frac{z_l}{z_m} \Big)^n+ \sum_{j=k+1}^{m}e^{in\theta_j} \Bigg)\Rightarrow$$

$$\Rightarrow \Bigg|\sum_{l=0}^{m}z^n_l \Bigg|^{1/n} = |z_m|\Bigg|\sum_{l=0}^{k}\Big(\frac{z_l}{z_m} \Big)^n+ \sum_{j=k+1}^{m}e^{in\theta_j} \Bigg|^{1/n}$$

So, we have

$$\limsup_{n\rightarrow \infty}  \Bigg|\sum_{l=0}^{m}z^n_l \Bigg|^{1/n} = M \limsup_{n\rightarrow \infty} \Bigg|\sum_{l=0}^{k}\Big(\frac{z_l}{z_m} \Big)^n+ \sum_{j=k+1}^{m}e^{in\theta_j} \Bigg|^{1/n}$$

We need to prove that
$$\limsup_{n\rightarrow \infty} \Bigg|\sum_{l=0}^{k}\Big(\frac{z_l}{z_m} \Big)^n+ \sum_{j=k+1}^{m}e^{in\theta_j} \Bigg|^{1/n} = 1$$

  Fact: The orbit of irrational rotation on the circle are dense in the unit circle.

Consider this one

$$ \Bigg|\sum_{l=0}^{k}\Big(\frac{z_l}{z_m} \Big)^n+ \sum_{j=k+1}^{m}e^{in\theta_j} \Bigg|$$

We know that

$$\lim_{n\rightarrow \infty} \Big(\frac{z_l}{z_m} \Big)^n = 0$$

But, the second therm we could have rational angles $p_j/q_j$ and irrationals $\alpha_j$.

WLofG we need to prove that

$$\limsup_{n\rightarrow \infty}\Bigg|\sum_{l=k+1}^{k'}e^{in\alpha_j}+ \sum_{j=k'+1}^{m}e^{inp_i/q_i} \Bigg| = m-k$$

For rationals, take a subsequence of $N_n = n (2 \prod_j q_j)$. This sequence will land all angles at zero.

1 is a accumulation point. using the fact above(that irrational rotation is dense in unit circle), $e^{n\alpha_j}$ never repeat your values.

Using the Diophantine approximation for $\alpha_j \in \mathbb{R}\backslash\mathbb{Q}$ this numbers become near to rationals. with this, we can approximate $e^{i\alpha_j}$ for any $e^{ip/q}$. We have an accumulation point  for each $n\alpha_j$, that is, for any $\epsilon >0$ there are two iterates $m>n$ such that $|e^{im\alpha_j}-e^{in \alpha_j}|<\epsilon$, so $|e^{i(m-n)\alpha_j}-1|<\epsilon$.

So, this number is $\epsilon$ close to $1$. We are done.


Differential Equation $dy/dx=y^{1/3}$ and condition $y(x_0)=y_0$ has infinite solutions Prove that the differential equation
  $$\frac{dy}{dx}= y^{1/3}$$
  with the initial value of $y(x_0)=y_0$ has infinite solutions.

I don't really understand the problem if I have to show that there are infinite solutions depending on the initial conditions or if it is something like if I proposed $(x_0,y_0)=(0,0)$ and prove that for that case the equation has infinite solutions. if you think is the first one could you explain how to do it.

If $y_0 > 0$ then $f(y) = y^{1/3}$ is Lipschitz in a neighbourhood of $y_0$ (the derivative is bounded). So by Picard-Lindelöf there is a unique solution.

On the other hand, for $y_0 = 0$, $f$ is no longer Lipschitz and so we can no longer expect a unique solution. In fact two solutions are found readily:

$$ y(x) = 0 \text{ and } y(x) = \left[\frac{2}{3}(x - x_0)\right]^{3/2}. $$

But as Wikipedia alludes to in its article on the Peano existence theorem, the transition between the two solutions can happen at any point (not just at $x_0$). So the general solution is

$$ y_a(x) = \begin{cases}
0 & x \le a, \\
\left[\frac{2}{3}(x - a)\right]^{3/2} & x > a,
\end{cases} $$

for any parameter $a \ge x_0$.