"Which answer in this list is the correct answer to this question?" I received this question from my mathematics professor as a leisure-time logic quiz, and although I thought I answered it right, he denied. Can someone explain the reasoning behind the correct solution?

  Which answer in this list is the correct answer to this question?

  All of the below.
  None of the below.
  All of the above.
  One of the above.
  None of the above.
  None of the above.

I thought:

$2$ and $3$ contradict so $1$ cannot be true.
$2$ denies $3$ but $3$ affirms $2,$ so $3$ cannot be true
$2$ denies $4,$ but as $1$ and $3$ are proven to be false, $4$ cannot be true.
$6$ denies $5$ but not vice versa, so $5$ cannot be true.

at this point only $2$ and $6$ are left to be considered. I thought choosing $2$ would not deny $1$ (and it can't be all of the below and none of the below) hence I thought the answer is $6.$

I don't know the correct answer to the question. Thanks!

// gcc ImpredictivePropositionalLogic1.c -o ImpredictivePropositionalLogic1.exe -std=c99 -Wall -O3

/*
Which answer in this list is the correct answer to this question?

(a) All of the below.
(b) None of the below.
(c) All of the above.
(d) One of the above.
(e) None of the above.
(f) None of the above.
*/

#include <stdio.h>
#define iff(x, y) ((x)==(y))

int main() {
  printf("a b c d e f\n");
  for (int a = 0; a <= 1; a++)
  for (int b = 0; b <= 1; b++)
  for (int c = 0; c <= 1; c++)
  for (int d = 0; d <= 1; d++)
  for (int e = 0; e <= 1; e++)
  for (int f = 0; f <= 1; f++) {
    int Ra = iff(a, b && c && d && e && f);
    int Rb = iff(b, !c && !d && !e && !f);
    int Rc = iff(c, a && b);
    int Rd = iff(d, (a && !b && !c) || (!a && b && !c) || (!a && !b && c));
    int Re = iff(e, !a && !b && !c && !d);
    int Rf = iff(f, !a && !b && !c && !d && !e);

    int R = Ra && Rb && Rc && Rd && Re && Rf;
    if (R) printf("%d %d %d %d %d %d\n", a, b, c, d, e, f);
  }
  return 0;
}

This outputs:

a b c d e f
0 0 0 0 1 0

The main point I'd like to get across is that you cannot assume at the outset that there is only 1 satisfying assignment.  For example consider the question:

Which of the following is true?
    (a) both of these
    (b) both of these

You might be tempted to say that both (a) and (b) are true.  But it is also consistent that both (a) and (b) are false.  The tendency to assume singularity from definitions isn't correct when the definitions are impredictive.

I see this question a little differently:

If I take the #5 statement as a denial of the set of items containing 1-4, then I am judging the truth or falsity of the statement on the basis of set 1-4.

If I take the statement of #6 however, I am actually judging a different set than I am with #5. I am including in that set all items with the denial of #5 included. What I would be saying then is that "none of the above items in set 1-5 are true because the set itself is not structured correctly". Since 6 is outside the previous set, it becomes the "best answer". Notice I don't bother evaluating the truth of number 5 against number six, because it generates a contradiction. I consider number 5 to be simply a "liar" and move forward. Why would I not do that with #5? Because #6 is not in the set of items 1-5 and before six arrives, there is no contradiction in the set of items 1-4 and so 5 has a claim to truth. Once 5 becomes part of the set, you cannot deny a denial. He has no real existence or purpose at that point.

An analogy might be to have a group of black cats (1-4). A tall stranger, #5 shows up and says  "These black cats all suck". Okay. So we take number five under advisement and say that is a possibility for all black cats 1-4. Another tall stranger, #6 shows up and says all black cats suck, and so does tall stranger #5 because he is trying to generate problems with me by contradiction. Since my statement includes him, and his statement does not include me, give me priority. He isn't following the rules of set construction. How do you know? Because I am here, last in this set. He, on the other hand, is not last in the set. His position betrays the lie.


Wave equation with initial and boundary conditions - is this function right? If $y(x,t)$ satisfies the 1-dimensional wave equation

$$\frac{\partial^2y}{\partial t^2}=c^2\frac{\partial^2y}{\partial x^2}\quad\text{for }0\leq x \leq l$$

with boundary conditions

\begin{align}
y(0,t)&=0,\\
\frac{\partial y}{\partial x}(l,t)&=0
\end{align}

and initial conditions

\begin{align}
y(x,0)&=x(2l-x),\\
\frac{\partial y}{\partial t}(x,0)&=0,
\end{align}

is it true that

$$y(l,t)=l^2-c^2t^2\quad\text{for }0\leq ct\leq l\text{ ?}$$

Given that the Fourier series of $x(2l-x)$ is

$$\frac{32l^2}{\pi^3}\sum_{m=0}^{\infty}\frac{\sin((2m+1)\pi x/2l)}{(2m+1)^3},$$

I've got the full solution to be

$$y(x,t)=\sum_{m=0}^{\infty}\frac{32l^2}{(2m+1)^3\pi^3}\cos\left(\frac{(2m+1)\pi ct}{2l}\right)\sin\left(\frac{(2m+1)\pi x}{2l}\right)$$

(which seems to satisfy the equation and all 4 conditions), but when I set $x=l$, the resulting series doesn't come to $l^2-c^2t^2$.

Is there a mistake in the question asking me to show $y(l,t)=l^2-c^2t^2$, or have I made a slip in the calculations?

Many thanks for any help with this!

Your approach is that first handle these three B.C.s and I.C.s $y(0,t)=0$ , $\dfrac{\partial y}{\partial x}(l,t)=0$ and $\dfrac{\partial y}{\partial t}(x,0)=0$ . You find that the separation constant is best to take $-\dfrac{(2m+1)^2\pi^2c^2}{4l^2}$ and you get $y(x,t)=\sum\limits_{m=0}^\infty C(m)\sin\dfrac{(2m+1)\pi x}{2l}\cos\dfrac{(2m+1)\pi ct}{2l}$ .

Now you handle the I.C. $y(x,0)=x(2l-x)$ , so you have to face $\sum\limits_{m=0}^\infty C(m)\sin\dfrac{(2m+1)\pi x}{2l}=x(2l-x)$ . But the following should be extremely carefull!

You are facing to find an unusual kernel inversion, so all the calculations should be start from first principle!

Luckily the method in http://tutorial.math.lamar.edu/Classes/DE/FourierSineSeries.aspx still hold in this case.

$\sum\limits_{m=0}^\infty C(m)\sin\dfrac{(2m+1)\pi x}{2l}=x(2l-x)$

$\sum\limits_{m=0}^\infty C(m)\sin\dfrac{(2m+1)\pi x}{2l}\sin\dfrac{(2n+1)\pi x}{2l}=x(2l-x)\sin\dfrac{(2n+1)\pi x}{2l}$

$\int_0^l\sum\limits_{m=0}^\infty C(m)\sin\dfrac{(2m+1)\pi x}{2l}\sin\dfrac{(2n+1)\pi x}{2l}dx=\int_0^lx(2l-x)\sin\dfrac{(2n+1)\pi x}{2l}dx$

$\sum\limits_{m=0}^\infty C(m)\int_0^l\sin\dfrac{(2m+1)\pi x}{2l}\sin\dfrac{(2n+1)\pi x}{2l}dx=\int_0^lx(2l-x)\sin\dfrac{(2n+1)\pi x}{2l}dx$

$\because\int_0^l\sin\dfrac{(2m+1)\pi x}{2l}\sin\dfrac{(2n+1)\pi x}{2l}dx$

$=\int_0^l\dfrac{1}{2}\biggl(\cos\dfrac{(m-n)\pi x}{l}-\cos\dfrac{(m+n+1)\pi x}{l}\biggr)dx$

$=\begin{cases}\biggl[\dfrac{l}{2(m-n)\pi}\sin\dfrac{(m-n)\pi x}{l}-\dfrac{l}{2(m+n+1)\pi}\sin\dfrac{(m+n+1)\pi x}{l}\biggr]_0^l&\text{when}~m\neq n~\text{and}~m+n\neq-1\\\biggl[\dfrac{x}{2}-\dfrac{l}{2(m+n+1)\pi}\sin\dfrac{(m+n+1)\pi x}{l}\biggr]_0^l&\text{when}~m=n\\\biggl[\dfrac{l}{2(m-n)\pi}\sin\dfrac{(m-n)\pi x}{l}-\dfrac{x}{2}\biggr]_0^l&\text{when}~m+n=-1\\\left[0\right]_0^l&\text{when}~m=n~\text{and}~m+n=-1\end{cases}$

$=\begin{cases}0&\text{when}~m~\text{and}~n~\text{are integers and}~m\neq n~\text{and}~m+n\neq-1\\\dfrac{l}{2}&\text{when}~m~\text{and}~n~\text{are integers and}~m=n\\-\dfrac{l}{2}&\text{when}~m~\text{and}~n~\text{are integers and}~m+n=-1\end{cases}$

$\therefore C(n)\dfrac{l}{2}=\int_0^lx(2l-x)\sin\dfrac{(2n+1)\pi x}{2l}dx$

$C(n)=\dfrac{2}{l}\int_0^lx(2l-x)\sin\dfrac{(2n+1)\pi x}{2l}dx$

$C(m)=\dfrac{2}{l}\int_0^lx(2l-x)\sin\dfrac{(2m+1)\pi x}{2l}dx$

From the result of http://integrals.wolfram.com/index.jsp?expr=x%282l-x%29sin%28%28%282m%2B1%29pi+x%29%2F%282l%29%29&random=false,

$\int_0^lx(2l-x)\sin\dfrac{(2m+1)\pi x}{2l}dx=\dfrac{16l^3}{(2m+1)^3\pi^3}$ when $m$ is integer

$\therefore y(x,t)=\sum\limits_{m=0}^\infty\dfrac{32l^2}{(2m+1)^3\pi^3}\sin\dfrac{(2m+1)\pi x}{2l}\cos\dfrac{(2m+1)\pi ct}{2l}$ for $0\leq x\leq l$

To check whether $y(l,t)=l^2-c^2t^2$ for $0\leq ct\leq l$ correct or not, note that you are difficult to simplify $\sum\limits_{m=0}^\infty\dfrac{32l^2}{(2m+1)^3\pi^3}\cos\dfrac{(2m+1)\pi ct}{2l}$ , so you should take the inverse process that consider $\sum\limits_{m=0}^\infty f(m)\cos\dfrac{(2m+1)\pi ct}{2l}=l^2-c^2t^2$ , determine whether $f(m)$ equals to $\dfrac{32l^2}{(2m+1)^3\pi^3}$ or not.

Note that you are facing to find an unusual kernel inversion again, so all the calculations should be again to start from first principle!

Luckily the method in http://tutorial.math.lamar.edu/Classes/DE/FourierCosineSeries.aspx still hold in this case.

$\sum\limits_{m=0}^\infty f(m)\cos\dfrac{(2m+1)\pi ct}{2l}=l^2-c^2t^2$

$\sum\limits_{m=0}^\infty f(m)\cos\dfrac{(2m+1)\pi ct}{2l}\cos\dfrac{(2n+1)\pi ct}{2l}=(l^2-c^2t^2)\cos\dfrac{(2n+1)\pi ct}{2l}$

$\int_0^\frac{l}{c}\sum\limits_{m=0}^\infty f(m)\cos\dfrac{(2m+1)\pi ct}{2l}\cos\dfrac{(2n+1)\pi ct}{2l}dt=\int_0^\frac{l}{c}(l^2-c^2t^2)\cos\dfrac{(2n+1)\pi ct}{2l}dt$

$\sum\limits_{m=0}^\infty f(m)\int_0^\frac{l}{c}\cos\dfrac{(2m+1)\pi ct}{2l}\cos\dfrac{(2n+1)\pi ct}{2l}dt=\int_0^\frac{l}{c}(l^2-c^2t^2)\cos\dfrac{(2n+1)\pi ct}{2l}dt$

$\because\int_0^\frac{l}{c}\cos\dfrac{(2m+1)\pi ct}{2l}\cos\dfrac{(2n+1)\pi ct}{2l}dt$

$=\int_0^\frac{l}{c}\dfrac{1}{2}\biggl(\cos\dfrac{(m-n)\pi ct}{l}+\cos\dfrac{(m+n+1)\pi ct}{l}\biggr)dt$

$=\begin{cases}\biggl[\dfrac{l}{2(m-n)\pi c}\sin\dfrac{(m-n)\pi ct}{l}+\dfrac{l}{2(m+n+1)\pi c}\sin\dfrac{(m+n+1)\pi ct}{l}\biggr]_0^\frac{l}{c}&\text{when}~m\neq n~\text{and}~m+n\neq-1\\\biggl[\dfrac{t}{2}+\dfrac{l}{2(m+n+1)\pi c}\sin\dfrac{(m+n+1)\pi ct}{l}\biggr]_0^\frac{l}{c}&\text{when}~m=n\\\biggl[\dfrac{l}{2(m-n)\pi c}\sin\dfrac{(m-n)\pi ct}{l}+\dfrac{t}{2}\biggr]_0^\frac{l}{c}&\text{when}~m+n=-1\\\left[t\right]_0^\frac{l}{c}&\text{when}~m=n~\text{and}~m+n=-1\end{cases}$

$=\begin{cases}0&\text{when}~m~\text{and}~n~\text{are integers and}~m\neq n~\text{and}~m+n\neq-1\\\dfrac{l}{2c}&\text{when}~m~\text{and}~n~\text{are integers and}~m=n\\\dfrac{l}{2c}&\text{when}~m~\text{and}~n~\text{are integers and}~m+n=-1\end{cases}$

$\therefore f(n)\dfrac{l}{2c}=\int_0^\frac{l}{c}~(l^2-c^2t^2)\cos\dfrac{(2n+1)\pi ct}{2l}dt$

$f(n)=\dfrac{2c}{l}\int_0^\frac{l}{c}~(l^2-c^2t^2)\cos\dfrac{(2n+1)\pi ct}{2l}dt$

$f(m)=\dfrac{2c}{l}\int_0^\frac{l}{c}~(l^2-c^2t^2)\cos\dfrac{(2m+1)\pi ct}{2l}dt$

From the result of http://integrals.wolfram.com/index.jsp?expr=%28l%5E2-c%5E2x%5E2%29cos%28%28%282m%2B1%29pi+cx%29%2F%282l%29%29&random=false,

$\int_0^\frac{l}{c}~(l^2-c^2t^2)\cos\dfrac{(2m+1)\pi ct}{2l}dt=\dfrac{16l^3}{(2m+1)^3\pi^3c}$ when $m$ is integer

$\therefore f(m)=\dfrac{32l^2}{(2m+1)^3\pi^3}=C(m)$

Hence $y(l,t)=l^2-c^2t^2$ for $0\leq ct\leq l$ is correct

You just have to find the Fourier series of $l^2 - c^2t^2$ for $t \in [0, \frac lc)$, then compare the series with the solution as $x = l$. The actual period is $\frac{4l}c$, but you can do two extensions of $l^2 - c^2t^2$ to cover the period. Equivalently, you can just use the cosine terms that are present in the solution as the basis.


Divisibility by prime numbers Suppose prime number $p$ and an arbitrary number $r$ is given.
What methods do you know for detection that $r$ is divisible by $p$?

The best way that's known for doing this is also the most straightforward way: You have to divide and check. How do you do that? I'll give you two answers: One for if you're a human, and one for if you're a computer.

If you're a human:

Chances are, the numbers are represented in base 10. In this base, for certain values of $p$, there are certain "tricks" you can use to speed up the division. For example, dividing a number by 2 or 5 reduces to checking its least significant digit (and it is no coincidence that these are the factors of 10). Dividing by 3 or 9 can be done by summing the digits of $r$ and checking the divisibility of the sum. And there are more rules. Unfortunately, this cannot really (easily) be generalized for larger values of $p$, and you'll end up just having to do long division (or another division algorithm that humans can do - I don't personally know one).

If you're a computer:

We are given two integers, $a$ and $b$ (it turns out we don't care whether $b$ is prime - that's why I changed the letters...) and we want to check whether $a$ is divisible by $b$.

The "naïve" algorithm is simple: Try dividing $a$ by $b$, and see whether you get an integer.

How does a computer divide two numbers? A simple way to do this would be to use "long division" (or "pen-and-paper division"), which is exactly how you learned to divide in elementary school. (The difference is that since you're a computer, you'll probably work in base 2, base 16, base 256 or some other base which is a power of 2, instead of the "human" base 10. Any computer would agree this is much easier and faster!)

Let's try to analyze how fast this algorithm is, at least in terms of complexity. If this term is new to you, it means something like this: If we graph the speed of the long division algorithm as a function of how many digits there are in the input, what sort of graph do we get? Linear? Logarithmic? Quadratic?

Denote by $d_1$ the number of digits in $a$, and by $d_2$ the number of digits in $b$ (these are actually the logarithms of $a$ and $b$, rounded, taken in the base that the numbers are represented in). Now, recall how long division works:

At each step, we look at $d_2$ digits (give or take 1) of $a$, and divide the number made up of those digits by $b$. This "smaller" division step can be done by subtracting $b$ repeatedly until we get a value smaller than $b$, and that's our remainder from this "small" step. The amount of subtractions we need to do is no more than a constant (for example no more than 10 if we work in base 10), so this step has time complexity $O(d_2)$ (the time each subtraction takes). If you're unfamiliar with this notation, it means "no more than $d_2$ multiplied by some constant inherent in the algorithm which we don't care to specify".
The amount of such "small divisions" we need to perform is equal to $d_1$ (give or take up to $d_2$), as we move "further down" the number $a$.

So, this simple algorithm's running time is $O(d_1 d_2)$. This is certainly not the fastest or most suitable way to do this (take a look at this Wikipedia article for a survey of some faster methods), but it's not too shabby! For instance, you can check a number $a \approx 2^{1024}$ for divisibility by a number $b \approx 2^{512}$ in about half a million steps, which is very fast in modern standards.

How much this can be improved turns out to be a pretty difficult question. If this interests you, I suggest you look at this Wikipedia article: There are 6 different non-trivial multiplication algorithms listed there, and the "Newton-Raphson division" method allows you to "convert" any multiplication algorithm to a division algorithm running at the same speed.

One thing's for sure: No division algorithm will be strictly faster than $O(d_1 + d_2)$. The reason is that this is the time complexity of just reading the digits, and you can't divide correctly without knowing all the digits. So the "game" of optimizing division is between $O(d_1 + d_2)$ and $O(d_1 d_2)$.

I'll show you with an example:

Check that 311 is prime

The square root of 311 is roughly 17 (prime).

Check if the primes starting at 2 and leading up to 17 divide evenly into 311. None do. Therefore, 311 is a prime number.

Yay!


Prove: if $px^2 - (p+q)x + p = 0$ has an integer root, $px^2 + qx + p^2 - q = 0$ has an integer root too, for coprime $p,q \in \mathbb{N}$ For $\mathbf{coprime}$ $p,q \in \mathbb{N}$, how can I prove that:

if equation

$$px^2 - (p+q)x + p = 0$$

has a root $x \in \mathbb{Z}$,

$$px^2 + qx + p^2 - q = 0$$

has an integer root as well?

Couldn't get a working idea.

In this answer, $p$ and $q$ are coprime integers such that $q\geq 0$ (and $p$ can be zero or negative).  We want to show that if $px^2-(p+q)x+p=0$ has an integer solution, then so does $px^2+qx+p^2-q=0$.

If $x=t$ is an integer root of $px^2-(p+q)x+p=0$, then
$$t(p+q-pt)=p.$$
If $t=0$, then $p=0$.  Because $\gcd(p,q)=1$, this means $q=1$.  Then, the equation $px^2+qx+p^2-q=0$, which is equivalent to $x-1=0$, has an integer solution $x=1$.

From now on, suppose that $t\neq 0$.  That is, $p=st$ for some integer $s$.  Hence,
$$p+q-pt=\frac{p}{t}=s.$$
Therefore,
$$q=s+p(t-1)=s+st(t-1)=s(t^2-t+1).$$
Therefore, $s$ divides $p$ and $q$, which implies $s=\pm1$ as $p$ and $q$ are coprime.  Since $q\geq 0$, we get $s=1$.

Therefore, $p=t$ and $q=t^2-t+1$.  Hence, $$px^2+qx+p^2-q=tx^2+(t^2-t+1)x+\big(t^2-(t^2-t+1)\big)$$ so that
$$px^2+qx+p^2-q=tx^2+(t^2-t+1)x+(t-1)=(tx+1)(x+t-1).$$
That is, $x=1-t$ is an integer solution to the quation $px^2+qx+p^2-q=0$.

The condition $q\geq 0$ is necessary.  Note that $p=-2$ and $q=-3$ satisfies the condition that $px^2-(p+q)x+p=0$ has an integer solution, but $px^2+qx+p^2-q=0$ does not have an integer solution.

From the first equation, the product of the roots is 1. So they are both 1 or both -1.        If they are both 1, substituting x=1 into the first equation gives q=p.                        Since p and q are co-prime, p=1,q=1 or p=-1,q=-1                                              If they are both -1, substituting x=-1 into the first equatio gives q=-3p.                  Since p and q are co-prime, p=1, q=-3 or p=-1, q=3                                           In all four cases for p and q, substituting the values for p and q into the second equation gives solutions for x that are integers.


What is the difference between weak and strong convergence? What is the difference between strong and weak convergence?

I am reading "Introductory functional analysis" by Kreyszig and I dont appreciate the differences between the two.

Definition of strong convergence:

A sequence $(x_n)$ in a normed space $X$ is said to be strongly convergent if there is an $x \in X$ such that $$\lim_{n \to \infty}||x_n-x||=0$$

Definition of weak convergence:

A sequence $(x_n)$ in a normed space $X$ is said to be weakly  convergent if there is an $x \in X$ such that $$\lim_{n \to \infty}f(x_n)=f(x)$$

I do not appreciate the differences between the two, does anyone have an example to highlight the differences?

  How does the proof differ in showing if a sequences converges weakly
  or strongly?

It all boils down to realizing that it is natural (and interesting!) to consider different topologies on the same set $X$, each of which comes with a notion of convergence.

In this scenario we have on one hand the strong topology, i.e. the topology induced by norm on $X$, and the so called weak topology on the other, i.e. the coarsest topology for which every element in $X^*$ is continuous.

The first thing to realize is that $\tau_{\text{weak}} \subset \tau_{\text{strong}}$. Indeed,

  strong implies weak: if $x_n \to x$ and $f \in X^*$, then by definition of $X^*$ we have that $f(x_n) \to f(x)$. In other words, strong convergence implies weak convergence, weakly closed implies (strongly) closed etc.

The other implication does not hold as shown by the following example:

  Weak does not imply strong: consider the Banach space $\ell^p$, $1<p<\infty$ and let $e_i$ be the sequence $(0,\dots,1,0\dots)$, with $1$ in the $i$-th position. Then it is easy to show that $\{e_i\}$ converges weakly to $0$ in $\ell^p$ but not strongly.

Here you can find a proof that for finite dimensional spaces the two topologies coincide.

It is worth mentioning it is not true that the topologies are different if and only if $X$ is infinite dimensional. For a counterexample you can take a look at this question ($\ell^1$ is a weird space).

A weak convergence is defined in an inner product while a strong convergence is defined in a norm.

A strong convergence is also a weak convergence but not vice versa.


Inverse of a function containing the ceiling function over the natural numbers I am wondering if there exists an inverse function for $\lceil{e^{x}}\rceil$ over the natural numbers. I don't think it is a trivial task to derive an inverse function for a function containing a ceiling or a floor even if you know there is a bijection, and that an inverse function does exist. Am I wrong?

Thanks for any insight...

Yes, if we regard $x \mapsto \lceil e^x \rceil$ as a function with codomain its image, it is bijective, and this doesn't depend on whether we include $0$ in $\mathbb{N}$ or not. For $x \in [0, \infty)$, $\frac{d}{dx} e^x = e^x \geq 1$, and so
$$e^{x + 1} = e^x + \int_x^{x + 1} e^x \, dt \geq e^x + \int_x^{x + 1} \,dt = e^x + 1.$$
In particular, $$\lceil e^{x + 1} \rceil > \lceil e^x \rceil,$$
so the function is injective. In particular it is bijective onto its image and so admits an inverse.

While that function is injective, it is not surjective, so it can't has an inverse in the common sense.

What it does have is a right inverse, that is, a function $g$ such that $f \circ g = id$, and you can simply define it as $g(x) = y$ such that $f(y) =x$, which is guaranteed existing a unique.

Finding a closed form for that is another thing, and in general it's not possible, in this case I don't know.


How do I show that the following set is a vector space? Question
Determine if the following set is a vector space:
$$ \left\{ \begin{bmatrix} a \\ b \\ c \\ d\end{bmatrix} : \begin{matrix} a - 2b & = & 4c \\ 2a & = & c + 3d\end{matrix}\right\}$$
I have a clue that I'm suppose to use the axioms to prove that it is a vector space, like that there exists the zero vector, its closed under addition and multiplication but I have no clue how to go about it

I am going to show a part of it just to give you the idea. You can do the rest. If $V_{1}$ and $V_{2}$ belongs to the vector space (S), we want to show that $V = V_{1}+V_{2}$ also belongs to that vector space. In other words, we want to show $V$ has the properties defined in the set.
$$ V_{1} = \begin{bmatrix} a_{1} \\ b_{1} \\ c_{1} \\ d_{1}\end{bmatrix}, \space  V_{2} = \begin{bmatrix} a_{2} \\ b_{2} \\ c_{2} \\ d_{2}\end{bmatrix} \longrightarrow V = \begin{bmatrix} a=a_{1}+a_{2} \\ b=b_{1}+b_{2} \\ c=c_{1}+c_{2} \\ d=d_{1}+d_{2}\end{bmatrix} $$
Since the vectors $V_{1}$ and $V_{2}$ belong to S, we can say:
$$ \text{Because $V_{1} \in S$}: \space a_{1} - 2b_{1} = 4c_{1} $$
$$ \text{Because $V_{2} \in S$}: \space a_{2} - 2b_{2} = 4c_{2} $$
Now, let's add Left-hand side together, and Right-hand side together:
$$ (a_{1}+a_{2}) - 2(b_{1}+b_{2}) = 4(c_{1}+c_{2}) $$
$$ a - 2b = 4c $$
Similarly, you can show $2a=c+3d$ can be satisfied as well.
Good Luck.

Presumably you are using the usual operations of addition in the vector space and multiplication by real scalars.  You now need to verify the axioms of a vector space with this subset of $\Bbb R^4$ and these operations.  The same $0$ vector works.  You need to show that it is in your subset.  Once you do that, the fact that it is the additive identity is inherited from $\Bbb R^4$.  Now you need to show that it is closed under multiplication by scalars, so show that any vector in your set can be multiplied by a real number and stay in the set.  Addition works the same way:  you are to take two vectors in your set and show their sum is still in the set.  Just write it out and see where it takes you.


Domain of $(f \circ g)(x)$ when $f(x) = x^2 + 9$ and $g(x) = \sqrt{x + 3}$ I am in need of some help. Right now I am doing some test corrections for my Honors High School Pre-calc class. This class is Algebra-based calculus. So the question is:

  Find $(f\circ g)(x)$ and its domain when $f(x)=x^2+9$ and $g(x)=\sqrt{x+3}$.

I solved $(f\circ g)(x)$ to get $x+12$. So I understand how to get $(f\circ g)(x)$, but I do not understand how to get the domain. According to my test, it says that answer for the domain is $[-3,\infty)$. When I graph it on Desmos.com and my TI-84 Plus I get a straight line, so I really don't know how they got that answer. I would really appreciate it if someone could help me! Thanks in advance!

The domain of this function is indeed $[-3, \infty)$, assuming $\sqrt[]{x}$ is undefined when $x<0$.

This is because $$(f\circ g)(x) = \left[\sqrt[]{x+3}\right]^2+3.$$ Notice that when $x < -3$, we are taking the square root of a negative number, which is undefined. Therefore, the only values of $x$ for which $(f\circ g)(x)$ is defined are those satisfying $x \geq -3$.

Let $D(f)$ be a domain of the function $f$.

Thus, by definition we have:
$$D(f\circ g)=D(g)\cap\{x|g(x)\in D(f)\}$$
In our case
$$f(g(x))=\left(\sqrt{x+3}\right)^2+9,$$
$$D(f)=\mathbb R$$ and $$D(g)=[-3,+\infty).$$

Thus, the domain is $[-3,+\infty)$.


Magnitude in complex sum of vectors If $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b}=2\hat{i}+3\hat{j}+2\hat{k}$ and $\vec{c}$ be a vector non collinear with $\vec{b}.$ Find magnitude of
$$
\frac{\vec{a} \cdot (\vec{b}\times \vec{c})}{|\vec{b}\times \vec{c}|^2}(\vec{b}\times \vec{c})+\frac{(\vec{a}\cdot \vec{b})\vec{b}}{|\vec{b}|^2}+\frac{(\vec{a}\cdot \vec{c})\vec{c}}{|\vec{c}|^2}
$$

What I tried:
Vector component of $\vec{a}$ along $\displaystyle \vec{c}=(|\vec{a}|\cos \alpha)\hat{c}=\frac{(\vec{a}\cdot \vec{c})}{|\vec{c}|^2}\vec{c}$
Vector component of $\vec{a}$ along $\displaystyle \vec{b}=(|\vec{a}|\cos \beta)\hat{b}=\frac{(\vec{a}\cdot \vec{b})}{|\vec{b}|^2}\vec{b}$
Vector component of $\vec{a}$ along $$\displaystyle \vec{b }\times\vec{c}=(|\vec{a}|\cos \gamma)(\hat{b}\times \hat{c})=\frac{(\vec{a}\cdot (\vec{b}\times\vec{c}))}{|\vec{b}\times\vec{c}|^2}{(\vec{b}\times \vec{c})}$$
How do I solve it?
Help me please.

As @MMM says, the result is not pretty in general. To illustrate this, I wrote a Mathematica script which assumes $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$ where $x,y,z$ are real numbers.

a = {1, 2, 3}; b = {2, 3, 2}; c = {x, y, z};
d = a.Cross[b,c]/Norm[Cross[b,c]]^2 Cross[b,c] + a.b/Norm[b]^2 b + a.c/Norm[c]^2 c;

And then

Assuming[x\[Element]Reals && y\[Element]Reals && z\[Element]Reals, Simplify[d]]

which yields
$$\vec{d}=\left(\frac{x (x+2 y+3 z)}{x^2+y^2+z^2}+\frac{(2 y-3 z) (5 x-4 y+z)}{13 x^2-4 x (3 y+2 z)+8 y^2-12 y z+13 z^2}+\frac{28}{17}\right)\hat{i}\\+\left(\frac{y (x+2 y+3 z)}{x^2+y^2+z^2}-\frac{2 (x-z) (5 x-4 y+z)}{(3 x-2 y)^2+4 (x-z)^2+(2 y-3 z)^2}+\frac{42}{17}\right)\hat{j}\\+\left(\frac{z (x+2 y+3 z)}{x^2+y^2+z^2}+\frac{(3 x-2 y) (5 x-4 y+z)}{(3 x-2 y)^2+4 (x-z)^2+(2 y-3 z)^2}+\frac{28}{17}\right)\hat{k}$$
It is evident that the magnitude of such vector is far from pretty. But just to be sure,

Assuming[x\[Element]Reals && y\[Element]Reals && z\[Element]Reals, Simplify[Norm[d]]]

results in
$$|\vec{d}|=\sqrt{\frac{P(x,y,z)}{Q(x,y,z)}}$$
where
$$P=3922 x^4+3728 y^4+6738 z^4+x^3 (2256 z-476 y)+x^2 \left(5297 y^2-2004 y z+8052 z^2\right)-2 x \left(1876 y^3+5331 y^2 z+5954 y z^2-136 z^3\right)-776 y^3 z+3225 y^2 z^2+1044 y z^3,\text{ and}\\
Q=17 \left(x^2+y^2+z^2\right) \left(13 x^2-4 x (3 y+2 z)+8 y^2-12 y z+13 z^2\right)$$

HINT

Let $\vec{z} = \vec{b} \times \vec{c}$. Note that $\vec{b}, \vec{c}$ are not collinear, so $\vec{z} \perp \vec{b}$ and $\vec{z} \perp \vec{c}$ and so $\mathcal{B} = \{\vec{z}, \vec{b}, \vec{c}\}$ is a basis for $\mathbb{R}^3$.

Each component of your sum is basically the projection of $\vec{a}$ on each of the elements of that basis $\mathcal{B}$, so the end result should be very related to $a$. Can you finish the problem now?

UPDATE

If $a,b$ are some vectors, then the projection of $a$ onto $b$ is given by
$$
p_b(a) = \frac{a \cdot b}{|b|^2} b.
$$
Note that $p_b(a)$ is always in the direction of $b$.

Now if you have a basis for some vector space, you can write any element of that vector space as a linear combination of the elements of the basis. For example, consider the standard basis in $\mathbb{R}^2$ and note that I can write any vector $(u,v)^T$ as
$$
\begin{bmatrix} u \\ v \end{bmatrix}
 = \begin{bmatrix} u \\ 0 \end{bmatrix}
 + \begin{bmatrix} 0 \\ v \end{bmatrix}
 = u \begin{bmatrix} 1 \\ 0 \end{bmatrix}
 + v \begin{bmatrix} 0 \\ 1 \end{bmatrix}
$$
so each element $(u,0)^T$ and $(0,v)^T$ is the projection of the original vector $(u,v)^T$ onto one of the elements of the standard basis.

Finally, in a feature unique to $\mathbb{R}^3$, if $x,y \in \mathbb{R}^3$ are non-collinear, then their cross product $x \times y$ is perpendicular to them both. Thus, the vectors $\{x,y, x \times y\}$ form a basis of $\mathbb{R}^3$.

Your problem sums three different terms, each of which is a projection of one vector onto the 3 elements of a basis of $\mathbb{R}^3$, constructed exactly through such a cross product.

When you sum the sizes of the projections, what should you get?


Newtons Method to approximate inflection point Here's a question from my tutorial which I'm having difficulties with.

Consider the function $$f(x) =\frac{e^x}{( 1+ x^2)}$$

a) Show that $f$ has an inflection point at $x = 1$

My answer:

$$f'' (x) = \frac{e^x (x^4 -4x^3 +8x^2 -4x-1)}{(x^2 +1)^3}$$

For inflection point $f''(x)=0$, so
$$f''(1)=\frac{e^1 (1^4 -4(1)^3 +8(1)^2 -4(1) -1}{(1^2 +1)^3}=\frac{e(0)}{8}=0$$
so $f(x)$ has an inflection point at $x=1$.

b) $f(x)$ has another inflection point near $x=-0.25$. Use Newton's method to approximate the x-coordinate of this inflection point, correct to $3$ decimal places.

My unfinished answer

I set up Newton's formula as error, $e = 0.5 \times 10^{-3}$

$$X_{n+1}=X_n - f'(x)/f''(x)$$

$$X_{n+1} = X_n - \frac{(X_n^2 - 1)^2 (X_{n}+1)}{(X_n^4-4X_n^3+8X_n^2-4X_n-1)}$$

Now my problem is I can't figure out where to start my approximation, as looking at the graph of $f(x)$ is not so helpful either.

Can someone please help me out.

the second derivative can be simplified to $$f''(x)={\frac {{{\rm e}^{x}} \left( x-1 \right)  \left( {x}^{3}-3\,{x}^{2}+5
\,x+1 \right) }{ \left( {x}^{2}+1 \right) ^{3}}}
$$


How to derive Lobachevsky's formula for the angle of parallelism I'm lightly studying some non-Euclidean Geometry and in the book I am reading there is no proof or derivation from where the Lobachevsky formula for angle of parallelism comes from:

$$\Pi(x)=2\tan^{-1}\left(e^{-x}\right)$$

Any help? Thanks

P.S. I couldn't find anything by google search either.

Hint An interesting relation between arcus tangent and exponential function are made clear in one complex variable

$$\arctan(z) = \frac{1}{2}i\left[\ln(1-iz) -\ln(1+iz)\right]$$

Maybe it can help you somehow.

Lemma 1: Small scale Lobachevsky is Euclidean.
Lemma 2: The area of a polygon is $$\rho^2\times defect$$
First, let us derive a helpful small scale formula. [Picture 1][1]
This is small scale so $d\approx\tan\alpha \Delta x$.
The trapezoid's area is $$-\rho^2\Delta\alpha\approx xd\approx x\tan\alpha \Delta x $$
Let's solve this differential equation:$$\int\cot\alpha\ d\alpha=-\int\frac{x}{\rho^2}dx+C\Rightarrow\ln\sin\alpha=-\frac{x^2}{2\rho^2}\Rightarrow\sin\alpha=e^{-\frac{x^2}{2\rho^2}}\Rightarrow\underline{\alpha\approx\frac{\pi}{2}-\frac{x}{\rho}} $$
Now we can find the general formula. [Picture 2][2]
The smallest triangle's area is $\rho^2(\phi-\frac\pi2-\Delta\alpha)$. It is negligible compared to $\Delta x$ and $\Delta\alpha$ so we can write that it is zero. Using our first, small-scale formula we get the following differential equation:$$0=\phi-\frac\pi2-\Delta\alpha\approx-\frac{\Delta x\sin\alpha}{\rho}-\Delta\alpha$$ It only remains to solve it:$$\int{d\alpha\over\sin\alpha}=-\int{dx\over\rho}+C\Rightarrow\ln\tan\frac{\alpha}{2}=-{x\over\rho}+C\Rightarrow\underline{\alpha=2\arctan(\cot{\beta\over 2}e^{-{x\over\rho}})}$$
This can be used to derive everything in Lobachevskian Geometry. It's the first and the most important formula and it can be written in many ways:$$\cot{\alpha\over 2}\cot{\beta\over 2}=e^\frac{x}{\rho}$$ $$\tanh{x\over\rho}=\frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta}$$ $$\tanh{x\over 2\rho}=\frac{\cos\alpha+\cos\beta}{1+\cos(\alpha-\beta)}=\frac{1+\cos(\alpha+\beta)}{\cos\alpha+\cos\beta}$$ $$\sinh{x\over\rho}=\frac{\cos\alpha+\cos\beta}{\sin\alpha\sin\beta}$$ $$\cosh{x\over\rho}=\frac{1+\cos\alpha\cos\beta}{\sin\alpha\sin\beta}$$
[1]: https://i.stack.imgur.com/SkWSa.jpg
[2]: https://i.stack.imgur.com/szDHH.jpg


Permutation question: Pick out 2 items from 4 items which 2 duplicated, is there any formula? If given ABBC, find the permutations if 2 letters are picked.
If I calculate manually, they are:
AB, AC, BA, BB, BC, CA, CB, 7 permutations.
Is there any formula to solve this type of question?
(Meaning, picking r number of letters from n number of letters with duplicated letters)

Suppose you had $n$ unique letters ${X_k}$, and letter $X_k$ were present in $r_k$ copies.

Then the number of unique pairs of letters, can be computed as $n (n-1) + \sum_{k} \mathrm{sgn} (r_k-1)$.

The term $n(n-1)$ count the number of pairs where letters are distinct, and remaining sum counts same letter pairs.

Permutation of n things taken r at a time out of which k are repeated is

$\frac{n!}{(n-r)!k!}$

In your question , n=4 r=2, k=2

so total permutations is  $\frac{4!}{2!2!}$=6


Is the following subset connected? Is the following subset connected?
$A=\{a,d,e\}$ subset of a topological space $(X,\tau)$, $X=\{a,b,c,d,e\}$ and $\tau = \{X,\phi,\{a,b,c\},\{c,d,e\},\{c\}\}$.

My ans:

The closed sets of $\tau$ are given by $\{\phi,X,\{d,e\},\{a,b\},\{a,b,d,e\}\}$. Clearly the set $A \neq U \cup V$ for some U and V, both open (closed) in $\tau$.
Hence, the set $A$ is connected.

But my solution key says this set is disconnected. what am I doing wrong here and also, is my idea to solve this question correct? Thanks.

You have that
$$A =\left(\lbrace a,b,c\rbrace \cap A\right) \cup \left(\lbrace c,d,e \rbrace) \cap A\right)$$
so $A$ can be written as the union of two open disjoints subsets of itself, for the topology on $A$ induced by the given topology on $X$. Therefore $A$ is not connected.

$A\subseteq\{a,b,c\}\cup\{d,e,c\}= U\cup V$ provides a disconnection of $A$. Although $U\cap V\ne\emptyset$,  this is not required for a disconnection. What is required is that $U\cap\{d,e\}=\emptyset=V\cap\{a\}$.


The space of sequences which are eventually zero in $l^2$ is not a Hilbert space. Define $V$ to be the space of sequences which are eventually zero, i.e.

$$V=\bigcup_{N=1}^{\infty}\{(x_n)_{n=1}^{\infty}\in l^2: x_n=0 \; \text{for}\; n\ge N\}.$$

Is $V$ a Hilbert space with respect to the $l^2$ inner product?

I don't think this is a Hilbert space. The example I came up with is $x^n=(1,1/2,\dots,1/n,0,\dots)$. Then $x^n\in V$, and $||x^n-x^m||_2\to 0$ as $m,n \to \infty$, by the tail convergence as $\sum 1/n^2$ converges. Also, this sequence clearly converges to $(1,1/2,1/3,\dots, 1/n, 1/(n+1),\dots)$, by the same reasoning as above, yet this element is not in $V$.

This is the simplest example I could come up with, but what other examples aside from the harmonic series can we think of? I would appreciate any suggestions.

This was recently discussed in my Real Analysis class, so this is not my idea.

First, we need to state the following result.

If $H$ is a Hilbert space (complete, inner product space), and $M$ is a closed proper subspace of $H$, then there exists a $z \in H, z \neq \vec{0} $ such that $z\bot M$. This result will fail to hold if $H$ is not complete (not Hilbert).

We shall define $V$ as in the question and construct a closed proper subspace $M$ of $V$ and show that if $z \bot M$ then $z = \vec{0}$.

$$V:=\{x:\mathbb{N}\to\mathbb{R}: \exists n_{0} \in \mathbb{N}: x_{n} = 0 \ \forall\ n\geq n_{0} \}$$
Let us endow $V$ with the inner product $$<x,y> = \sum_{n=1}^{\infty}x_{n}y_{n}$$
Let $f:V\to \mathbb{R}$ with
$$ f(x) = \sum_{n=1}^{\infty}\frac{x_{n}}{n}$$
We can check that $f$ is linear and $f$ is bounded.
$$ |f(x)| = |\sum_{n=1}^{\infty}\frac{x_{n}}{n}| \leq \bigg( \sum_{n=1}^{\infty}\frac{1}{n^{2}} \bigg)^{1/2}\bigg( \sum_{n=1}^{\infty}x_n^{2} \bigg)^{1/2}=\sqrt{\frac{\pi^{2}}{4}}||x|| $$

Let us now define the kernel of $f$ as $M:=\{x \in V:f(x)=0\}.$ Then $M$ is a closed linear subspace of $V$ and also a proper subset of $V$.

We will now show that $z = \vec{0}$ is the only element of $V$ such that $z\bot M$, implying that $V$ is not complete, hence not Hilbert.

Consider $z = (z_{n})_{n} \in V$, and $y_{n} = e_{1} - n e_{n}$, where the $i^{th}$ entry of $e_{i}$ is 1, and rest all are $0$.We can observe that $y_{n} \in M$.
$$ f(y_{n}) = 1 + \bigg(\frac{-n}{n}\bigg)=0 $$
If $z\bot M$, then $z \bot y_{n}, i.e\ <z,y_{n}>=0\ \forall n \in \mathbb{N}$.
$$ <z,y_{n}> = z_{1} - n z_{n} = 0 \implies z_{n} = \frac{z_{1}}{n}.$$

Since $z \in V, \exists\ n_{0}: z_{n_{0}}=0 \implies z_{1}=0$.

If $z \bot M$, we showed that $z = \vec{0}$, hence $V$ is not complete. So $V$ is not Hilbert.

If an = 2-n/2 for n >= 1, define a  =  (an), a point in el2, N0, (N0 = {0, ..., infinity}), belongs to the unit sphere S.

If we like, we can also divide the finite-dimensional approximations, say bn,k = ak if 0 <= k <=n and 0 if k > n, by their own lengths to also get a sequence of points, say cn, on the unit sphere S of el2(N). Of course these points approach the point a in the el2 limit.

The only advantage being: Now, the points of the eventually-zero ones whose el2 limit is not another eventually-zero point aren't just anywhere, they're on the unit sphere.


Is there more to this relationship with the Fibonacci numbers? So I recently thought of a cool way to represent the Fibonacci sequence, which provides many identities really interestingly. The key is to define

$$x^2=x+1$$

And consider the integer sequences given by

$$x^n=a_nx+b_n$$

These sequences satisfy $F_{n+2}=F_{n+1}+F_n$ and $a_1=a_2=b_2=b_3=1$, thus producing the Fibonacci sequence. This is easily verifiable:

\begin{align}\color{blue}{a_{n+2}}x+\color{green}{b_{n+2}}&=x^{n+2}\\&=x^nx^2\\&=x^n(x+1)\\&=x^{n+1}+x^n\\&=a_{n+1}x+b_{n+1}+a_nx+b_n\\&=(\color{blue}{a_{n+1}+a_n})x+(\color{green}{b_{n+1}+b_n})\end{align}

One can also come up with a simple $\mathcal O(\log(n))$ algorithm to compute $F_n$ using exponentiation by squaring:

\begin{align}\color{blue}{a_{2n}}x+\color{green}{b_{2n}}&=x^{2n}\\&=(x^n)^2\\&=(a_nx+b_n)^2\\&=a_n^2x^2+2a_nb_nx+b_n^2\\&=a_n^2(x+1)+2a_nb_nx+b_n^2\\&=\color{blue}{a_n(a_n+2b_n)}x+\color{green}{a_n^2+b_n^2}\end{align}

And similarly for $x^{2n+1}$. This also quickly gives some other cool identities using the fact that $x^{n+k}=x^nx^k$, for example.

I was thinking, however, that this is just way too convenient. I'm unsure how to generalize this. For example, what if I wanted to start at different integers?

I'm also curious to know if there's something more to this. Some deeper math behind the scenes that can explain why I'm able to write the Fibonacci sequence like this aside from just brute force showing it satisfies the definition.

As far as starting at different integers goes, we can consider the sequences given by

$$x^n(a_0x+a_1-a_0)=a_nx+b_n$$

which preserves the recurrence relation and lets us choose what we want $a_0$ and $a_1$ to be.

As far as I can tell, it is possible to do this to any recurrence relation of the form

$$x_{n+k}=y_1x_{n+k-1}+\dots+y_kx_n$$

by considering the corresponding

$$x^k=y_1x^{k-1}+\dots+y_k$$

and considering the sequences given by

$$x^nP(x)=a_nx+b_n$$

for some polynomial $P$ which corresponds to the initial conditions, provided $x$ is irrational to ensure uniqueness.

So this leaves me the question of whether or not there's more math relevant here aside from my just stumbling upon this. I ask this because I think this kind of identity is just "too good to be true", especially for me to not have noticed this kind of thing despite having seen plenty of recurrence relations before.

You have builе a Fibonacci (?) field $Q[x]/(x^2-x-1)$. Every element of this field can be written as binomial, so it has representation in 2x2 matrices:
$$
ax+b\qquad\Leftrightarrow\quad \begin{pmatrix}a+b & a\\ a & b\end{pmatrix} = aF+bI
$$
You can check that $F^2=\begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}^2=F+I $

You $O(\log n)$ algorithm is a well known the Doubling Method, which is essentially fast multiplication algorithm for matrix $F$.

To generalize, you should just select right first element. So in ordinary fibonacci, you start with (1,0), which is $X_1=1x+0$, next element is $X_2=x X_1$, then $X_3=x^2 X_1$ and so on. If you want to start with (-1, 2), then $Y_1=-1x+2$, $Y_2=x Y_1$, $y_3=x^2 Y_1$ and so on. You can use the same algorithm of fast multiplication to find $Y_n$.

Edit. As an example, I want to show how to build a tribonacci field $Q[x]/(x^3-x^2-x-1)$.

If we have $y=\alpha x^2+\beta x+\gamma$, then $$xy = \alpha x^3+\beta x^2+\gamma x = (\beta+\alpha)x^2 + (\gamma+\alpha)x+\alpha$$
So element $x$ is equivalent to matrix:
$$
T = \begin{pmatrix}1 & 1& 1\\ 1 & 0&0\\0&1&0\end{pmatrix}.
$$
So our field can be represented with $Q[I, T]$

Good for you on discovering these.

As you suspected, these are already known as
Fibonacci polynomials.

Here is a reasonable place to start:

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


Name when two functions are equal under integration (expectation)? What is it called when $E[X] = E[Y]$? That is,

$$\int x f(x)\,dx = \int y g(y)\,dy.$$

What I want to say is not that the expectation of $X$ is equal to that of $Y$ but rather (the equivalent statement) that $xf(x)$ integrates to the same value as $xg(x)$. I would like the statement to say something along the lines of:

"$f$ is [definition] with respect to $g$"

Assume that the expectations, $\int f$, and $\int g$ are finite.

I don't think equality of integrals over a single set ($\mathbb{R}$ in this case) endows the integrands with special names.  In probability theory, we would say just that $X$ and $Y$ have the same first moments, but I think you're aware of that.

OTOH since $X,Y$ are $\mathbb{R}$-valued, if you can show that $\int_a^b x\,f(x)\, dx = \int_a^b y\,g(y)\,dy$ for all open intervals $(a,b)$, then the integrands would be said to be equal almost everywhere.


Typesetting the mathematical expression "const." How do I write in the right way, that something is constant?

I think, usually one does write = const., which should be an abbreviation for the Latin constat.

Should I write the expression italic ($\textit{const.}$) or upright ($\textrm{const.}$)? As far as I know, fixed mathematical operators should be written upright, but this isn't an operator...

Formally speaking, the right way to doing it would be to write something like
$$\int 2x\, \mathrm{d}x = x^2 + k, \ \text{where } k \text{ is constant}$$
If you did insist on doing it that way, to me it would look weird if it was italicised: italicised letters inside equations or formulae are usually interpreted as variables, so you technically end up with some ambiguity if any of the letters $c,o,n,s,t$ are used as variables. So I'd certainly prefer
$$\int 2x\, \mathrm{d}x = x^2 + \text{const.}$$
...but, of course, there is no Mathematical Institute of the Universe to govern this kind of thing, so as they say on Jersey Shore, you do you.

I believe your question has a definite answer if you fix a particular formal language, e.g. the language of first-order logic, in which you express a well-formed formula.

You could specify a constant symbol from the signature of the language to address a particular term in the domain of discourse. You could use quantifiers (if those exist in your language) to express a more complex association of an expression with a fixed member of the domain.


Set closed under multiplication I'm struggling with some very basic logic, I guess it is in the simplest question that will probably exist on stack exchange.

Define a set $N_{2,3} = \left\{ n \in \mathbb{Z} : 2 \nmid n \text{ and } 3 \nmid n \right\}$. Then prove it is closed under multiplication.

I was trying to use the contrapositive, and had the following solution;

By considering the contrapositive, we seek elements $n \in \mathbb{Z} : 2\mid n \text{ or } 3\mid n$. So given two elements $a,b \in N_{2,3}$ we seek to prove that $ab\in N_{2,3}$. Suppose then w.l.o.g. that $2\mid a$, then we can say $a=2c$ for some $c\in\mathbb{Z}$. Now $ab = (2c)b = 2(cb)$, and clearly $2\mid ab$, hence $ab \in N_{2,3}$.

Something just doesn't feel right with all of this, but I can't put my finger on it. I was also thinking about just saying that taking two elements $a,b \in N_{2,3}$, neither $3$ nor $2$ divide $n$ or $m$, therefore their product $ab$ is also not divisible by $3$ or $2$, hence the set is closed under multiplication.

Thanks for any help!!

By the set construction: "If $2\nmid ab$ and $3\nmid ab$, then $ab\in N_{2,3}$ ".

The contraposition is that: "If $ab\notin N_{2,3}$, then $2\mid ab$ or $3\mid ab$".

However, as you have shown, for any $a,b$ that are both $\in N_{2,3}$ it will always be so that $2\nmid ab$ and $3\nmid ab$.

So every such $ab$ must be in $N_{2,3}$.

tl;dr  When considering the contrapositive you do not want to find a case were the consequent holds.   You seek to deny that it ever can (for any product of elements from the set).

This is actually so stupid of a logic problem youll kick yourself a little bit.

This is the list of all numbers that you can write as prime factorizations without using a 2 or 3.  All numbers can be written as a prime factorial.  Thus any number that is in the set written without 2 or 3 is in the set of multiples.

Lets to 35 as an easy example:
not divisible by 2, nor 3... Check
(1^1)(5^1)(7^1)=35.  35 not divisible by 3, nor is 1,5, or 7 check.

Now lets say we want to know if 385 is in our set.

well 385 cannot be divided by 3 nor 2 check
its also 35*11 =  (1^1)(5^1)(7^1)*(11^1)

do you see whats going on yet?

no matter what numbers come after you cannot write the numbers in the set without 2 or 3 as a prime factorization.  Thus the resulting manipulation is inconsequential.  The set is closed under multiplication because it must be for that prime factorization to exist


The Poisson Distribution Customers arrive at a casino as a Poisson process of rate 100 customers per hour. Upon
arriving, each customer must flip a coin, and only those customers who flips heads actually enter the casino.

So how can I find

the PMF of N? Let's say for the
number of customers who arrive
between 5PM to 7PM.
the pdf of T? The    time between
successive customers    arriving.

The key observation is that the new process, denote it by $N = \lbrace N(t) : t \geq 0 \rbrace$, has rate $\lambda$ which is half the original, that is $\lambda=50$.
Then the answers to your questions follow immediately. Denote the original process by $N_0$, and set $\lambda_0 = 100$.
Then, the distribution of $N(1)$ (that is, the number of customers who actually enter the casino in one hour) is obtained (using the law of total probability) as follows.
$$
{\rm P}(N(1) = k) = \sum\limits_{j = k}^\infty  {{\rm P}(N(1) = k|N_0 (1) = j) {\rm P}(N_0 (1) = j)}.
$$
Given $N_0 (1) = j$, $N(1)$ is binomial$(j,1/2)$. Hence,
$$
{\rm P}(N (1) = k) = \sum\limits_{j = k}^\infty  {\frac{{e^{ - \lambda _0 } \lambda _0^j }}{{j!}}{j \choose k}\frac{1}{{2^j }}} = \cdots = \frac{{e^{ - \lambda _0 /2} (\lambda _0 /2)^k }}{{k!}}.
$$

EDIT: For a proof that the new process $N$ indeed has stationary independent increments, see, for example, p. 211 here (where $\lambda$ and $p$ correspond to $100$ and $1/2$, respectively).

Note that the pdf of $T$ is an exponential distribution.


Problem with $y'=x^2+y^2$ when $y(0)=0$ Referring to my question "Simple non-linear differential equation" and the associated answers and comments, the provided solutions in terms of fractional order Bessel functions or $\frac12$ order parabolic cylinder functions all seem to have a problem with requiring $y(0) = 0$. Never the less the series $$y = \frac{x^3}{3} + \frac{x^7}{63} + \frac{2x^{11}}{2079} + ..... $$ is a solution with $y(0) = 0$ so can anybody please explain why the Wolfram-solutions fail at $x =0$?

If in the solution given by Wolfram you let $C[1]\to\infty$, you obtain the solution wth $y(0)=0$:
$$
y(x)=-\dfrac{x^2\,J_{-\frac{5}{4}}\Bigl(\dfrac{x^2}{2}\Bigr)-x^2\,
   J_{\frac{3}{4}}\Bigl(\dfrac{x^2}{2}\Bigr)+J_{-\frac{1}{4}}\Bigl(\dfrac{x^2}{2}\Bigr)}{2\,x\,J_{-\frac{1}{4}}\Bigl(\dfrac{x^2}{2}\Bigr)}.
$$

I have tried to use the Bessel functions recurrence relations with respect to order and obtained this simplyfied expression

which does approach x^3/3 for small x.
Further, I have veryfied by hand that it is actually the solution.
PROBLEM SOLVED!


Limit of $x\log x\cdot f(x)$ when $f$ is integrable function Suppose that $f\in L^1(0,+\infty)$ is a monotone decreasing, positive function. Prove that $$\lim_{x \to +\infty}x(\log x)\cdot f(x)=0.$$

The assertion is not true. Consider
$$
f_0=\sum\limits_{n\geqslant1}(n^2\mathrm e^{n^2})^{-1}\mathbf 1_{[0,\mathrm e^{n^2}]},
$$
then $f_0$ is nonincreasing, $g_0(x)=f_0(x)x\log(x)$ does not converge to zero when $x\to\infty$ since $g_0(\mathrm e^{n^2})\geqslant1$ for every $n\geqslant1$, and the integral of $f_0$ is $\sum\limits_{n\geqslant1}n^{-2}$, which is finite, hence $f_0$ is integrable.

One can modify this example slightly to get a decreasing function $f$, for example
$$
f(x)=\sum\limits_{n\geqslant1}(n^2\mathrm e^{n^2})^{-1}\exp(-x\mathrm e^{-n^2})\mathbf 1_{[0,\mathrm e^{n^2}]}(x).
$$
Then $f$ is decreasing, $g(x)=f(x)x\log(x)$ does not converge to zero when $x\to\infty$ since $g(\mathrm e^{n^2})\geqslant\mathrm e^{-1}$ for every $n\geqslant1$, and $f\leqslant f_0$ hence $f$ is integrable.

I think it's not true, for example $f(x)=1/x$

and we have

$\lim_{x\rightarrow 0}xln(x)(1/x)=+\infty$


Derivative of the $f(x,y)=\min(x,y)$ I just encountered this function $f(x,y)=\min(x,y)$. I wonder what  the partial derivatives of it look like.

$$
f(x, y) = \min(x,y) = \begin{cases}
x & \text{if } x \le y \\
y & \text{if } x \gt y
\end{cases}
$$

The function isn't differentiable along $y = x$, but the partial derivatives are straightforward otherwise.

$$
\frac{\partial f(x, y)}{\partial x} = \begin{cases}
1 & \text{if } x \lt y \\
0 & \text{if } x \gt y
\end{cases}
$$

$$
\frac{\partial f(x, y)}{\partial y} = \begin{cases}
0 & \text{if } x \lt y \\
1 & \text{if } x \gt y
\end{cases}
$$

Here is a plot of the function to help you see the derivatives and why it's not differentiable along $y = x$:

It refers to the partial derivatives of functions of two variables or more/simply multiple variables,with their notation $f\{x,y,z,.....\}$, depending on how many variables are given.this method can be done by differentiating with respect to individual variable (with respect to $x$,you fix $y$ and $z$ and treat them as constants, with respect to $y$, you fix $x$ and $z$ and treat them as constants and the same goes for $z$ $zd_n$ other given variables. We can also use the formula.


Estimation with complex exponent Context (not necessary to understand the task). I'm working through so called Eigenvalue Estimation Problem from the field of Quantum Computing. More precisely, I try to measure the inaccuracy of the quantum circuit when an eigenvalue measured is irrational (the circuit is designed for rational phases).

The task can be reduced to the estimation with the lower bound of an (modulus of) expression:

$$
\frac 1 {2^m} \frac {1 - e^{2\pi i \delta 2^m}} {1 - e^{2\pi i \delta }}
$$

Here $m$ is a natural number, $\delta$ is a real from $[0,1)$, $i$ is the imaginary unit.

It is known that the desired lower bound is $2 / \pi$. It is suggested to get upper bound for denominator (which is clearly $2\pi\delta$) and lower bound for numerator (which is problematic for me) and then divide them. It is known that numerator $N$ can be estimated like this: $ N > 2\pi\delta2^m(2/\pi) > 4\delta 2^m$. Can anyone explain, how to obtain the bound in the left part of the inequality?

It is another approach which finds more exact lower bound (you can see it in my program for example)

It gives lower bound 1 (as limit)

Note, it will works only if b converges to +0 (for very small b and fixed m)

The solution turns to be direct consequence of Jordan's inequality:

$$
    \sin x \geqslant x \frac 2 {\pi}
$$

We would like to obtain lower bound for $|1 - e^{2\pi i \delta 2^m}|$, that is $|AC|$ on the following graphics with unit circle:

$OB$ is altitude and, at the same time, bisector and median, so

$$|AC| = 2 |AB| = 2 \sin {\frac a 2} \geqslant 2 \frac a 2 \frac 2 {\pi} = \frac {2a} \pi .$$

Given that $a = 2\pi \delta 2^m$ we have desired $|AC| = |1 - e^{2\pi i \delta 2^m}| \geqslant 4 \delta 2^m$. The whole fraction from the post turns to be estimated as follows:

$$
\frac 1 {2^m} \frac {1 - e^{2\pi i \delta 2^m}} {1 - e^{2\pi i \delta }}
\geqslant
\frac 1 {2^m} \frac {4 \delta 2^m} {2 \pi \delta} = \frac 2 {\pi}.
$$


simple probability identity I got a little stuck on a simple proof of the following probability identity.

Given

$\mathbb{P}(A^c \cap B^c)=1-\mathbb{P}(A)-\mathbb{P}(B)+\mathbb{P}(A\cap B)$

how to prove for any set $X$,

$\mathbb{P}(X \cap A^c \cap B^c)=\mathbb{P}(X)-\mathbb{P}(X\cap A)-\mathbb{P}(X\cap B)+\mathbb{P}(X\cap A\cap B)$

Looks very intuitive; just replace the whole space by $X$. But how to prove it simply and rigorously? Thanks.

Write $X$ as a disjoint union:
$$
X=(X\cap A\cap B)\sqcup (X\cap A^c\cap B)\sqcup (X\cap A\cap B^c)\sqcup (X\cap A^c\cap B^c).
$$
This gives:
$$
P(X)=P(X\cap A\cap B)+P(X\cap A^c\cap B)+P(X\cap A\cap B^c)+P(X\cap A^c\cap B^c).
$$
Hence
$$
P(X\cap A^c\cap B^c)=P(X)-P(X\cap A\cap B)-P(X\cap A^c\cap B)-P(X\cap A\cap B^c).
$$
Now the result follows from $P(X\cap A\cap B)+P(X\cap A^c\cap B)=P(X\cap B)$ and $P(X\cap A\cap B)+P(X\cap A\cap B^c)=P(X\cap A)$.

so.. i am using the latter to derive the former..

write P(X)  as $\mathbb{P}(X\cap (A U A^c))$ , equal to $P(X \cap A)+P(X \cap A^c)$, cancel some terms
then expand $P(X \cap A^c)$ to $P(X \cap A^c \cap (B U B^c))$, which is equal to $P(X \cap A^c \cap B)+ P(X\cap A^c \cap B^c)$, cancel the terms to get the answer.


Isomorphism between projective varieties $\mathbf{P}^{1}$ and a conic in $\mathbf{P}^{2}$ I'm trying to establish an isomorphism between the projective line $\mathbf{P}^{1}$ and the conic in $\mathbf{P}^{2}$ defined by $Y=Z(g)$, where $g=x^2+y^2-z^2$. This is part of exercise I.3.1 in Hartshorne's Algebraic Geometry.

I defined $\varphi : \mathbf{P}^{1} \rightarrow Y$ by $\varphi (a,b) = (a^2 -b^2,2ab,a^2+b^2)$ and showed that it's a morphism of varieties. I'm having trouble defining the inverse map. First, I tried using $\psi : Y \rightarrow \mathbf{P}^{1}$ defined by $\psi (a,b,c) = (b,c-a)$ to get $$\psi\varphi(a,b)=\psi(a^2 -b^2,2ab,a^2+b^2)=(2ab,2b^2)=(a,b) \text{ if }b\neq0$$ $$\varphi\psi(a,b,c)=\varphi(b,c-a)=(2a(c-a),2b(c-a),2c(c-a))=(a,b,c)\text{ if }a\neq c$$
This is problematic because of the restrictions on $a,b,c$, and also because $(1,0,1)\in Y$ but $\psi(1,0,1)=(0,0)\notin \mathbf{P}^{1}$. Then I tried using $\psi : Y \rightarrow \mathbf{P}^{1}$ defined by $$\psi (a,b,c) = \begin{cases}(b,c-a) \text{ if } a\neq c \\ (1,0) \text{ if } a=c \end{cases}$$
This is troublesome because now I can't show continuity: if $Z(T)$ is a closed set in $\mathbf{P}^{1}$, then $\psi^{-1}(Z(T))$ is a closed set in $Y$. (I need $\psi^{-1}(Z(T))$ to be the zero set of some homogeneous polynomials in $k[x,y,z]$ of positive degree, where $k$ is a field of characteristic zero.)

Does anyone know how to establish the definition and continuity of the inverse map to $\varphi$? I will only be able to understand answers within the scope and context of Hartshorne up to chapter I section 3. Thanks.

1) The isomorphism $\phi$ is between $\mathbb P^1$ and a conic  $Y\subset\mathbb P^2$, not a cone.
(This conic is related to the cone in $\mathbb A^3$ with the same equation, but you should not confuse these two algebraic varieties).
2) As in all categories there are morphisms between varieties but you must keep in mind a most important point: those morphisms are defined locally and there is no hope that a morphism can be given by a single "formula": see the "Morality" below.
3) Your question beautifully illustrates the difficulty  in 2) and allows for a pedagogical illustration  of how to correctly define the inverse morphism $\psi=\phi^{-1}:Y\to \mathbb P^1$, namely:

a) The conic $Y$ has an open (in the Zariski topology!) covering $\{U,V\}$, where $U=Y\setminus \{[-1:0:1]\}$ consists of the points $p=[x:y:z]\in Y$ satisfying $y\neq 0$ or $x+z\neq0$ and $V=Y\setminus \{[1:0:1]\}$ consists of the points $q=[x:y:z]\in Y$ satisfying $z-x\neq 0$ or $y\neq 0$.
b) We have a morphism $\psi_ U:U\to \mathbb P^1:[x:y:z]\mapsto [y:x+z]$
and a morphism $\psi_ V:V\to \mathbb P^1:[x:y:z]\mapsto [z-x:y]$
c) The morphisms $\psi_ U$ and $\psi_ V$ coincide in the (huge!) open intersection $U\cap V\subset Y$and thus together define the required isomorphism $\psi=\phi^{-1}:Y\to \mathbb P^1$.
d) Notice that our maps $\psi_ U$ and $\psi_ V$ are morphisms (and thus continuous) but to state merely that they are continuous in the Zariski topology is worthless: for example any set-theoretic bijection between curves is a homeomorphism but is absurdly far from being an isomorphism in general.

Morality
The morphism $\psi$ is a perfectly defined inverse of $\phi$, but it is impossible to write it in the form $\psi([x:y:z])=[f(x,y,z):g(x,y,z):h(x,y,z)]$ with homogeneous polynomials of the same degree $f(x,y,z),g(x,y,z), h(x,y,z)\in k[x,y,z]$.

You must define $\psi(1,0,1)=(1,0)$. This allows to remove the condition $b\ne 0$ in the first composition. This allows also to remove the condition $a\ne c$ in the second composition, because, on $Y$, $a=c\implies (a,b,c)=(1,0,1)$.

I remains to prove that this extended definition of $\psi$ is continuous on $Y$, which is not difficult. In fact, if one want the continuity for the Zariski topology, the closed sets are the finite subsets, and continuity results from the fact that we have bijections. Slightly more difficult is to show that the extended definition of $\psi$ is a morphism. The proof depends on your definition of a morphism. Generally the definition relies on homomorphisms of quotients of polynomial rings. This means that $a, b, c$ have to be viewed as indeterminates, and thus one can divide by $b$ and $a-c$ (which are nonzero polynomials) without taking care of any specific value.

For the continuity for the usual topology, one can use the geometric interpretation that follows.

Geometric interpretation: Let $L(a,b)$ be the line of equation $a(x-z)+by=0$. This line passes through $(1,0,1)$ and, if $b\ne 0$, intersects $Y$ in exactly another point, which is $\varphi(a, b)$. Conversely, any point of $Y$, different from $(1,0,1)$, defines a unique line passing through $(1,0,1)$; this defines $(a, b)$, and this is the definition of $\psi$. If $b=0$, the line $L(a,b)$ is tangent to $Y$ at $(1,0,1)$. That is, it is the limit of the secant $L(a,b)$ when $b\to 0$. The implies that the limit of the point $\varphi(a, b)$ is $(1,0,1)$. This explains why the extended definition of $\psi$ is continuous. Again, a complete algebraic proof depends on the algebraic background that you use.


An inequality for metric spaces: $|d(x, z) − d(y, z)| \le d(x,y)$ Question : Prove $|d(x, z) − d(y, z)|$ is less than or equal to $d(x, y)$.

I know I have to use the triangle inequality but I'm just not sure how to apply it with a negative $d(y,x)$.

The claim is invariant under exchange of $x$ and $y$. Thus without loss of generality we can assume $d(x,z)\ge d(y,z)$. Then $|d(x,z)-d(y,z)|=d(x,z)-d(y,z)$, which is $\le d(x,y)$ by the triangle inequality.

First note that $|a-b|\leq |a|+|b|$\
Now, $|d(x,z)-d(y,z)|\leq |d(x,z)|+|d(y,z)|=d(x,z)+d(y,z)\leq d(x,y) (\because \text{triangle inequality}).$


Prove that if $B\subseteq A$, then $A\backslash (A\backslash B) = B$. So I tried this problem and I just wanted to make sure I did it right. Here it is:
Proof:
Given: $B \subseteq A$
I.) Let $x\in A\backslash(A\backslash B)$.
It follows that $x\in A$ AND $x\notin A\backslash B$, but $x\notin A\backslash B$ is logically equivalent to $x\notin A$ OR $x\in B$.
By the distributive law for logic that gives ($x\in A$ AND $x\notin A$) OR ($x\in A$ AND $x\in B$), which is just False OR ($x\in A$ AND $x\in B$).
That leaves $x\in A$ AND $x\in B$ which implies that $x\in B$, and thus $A\backslash(A\backslash B) \subseteq B.$
II.) Let $x\in B$.
Then its clear that $x\in A$ (because $B \subseteq A$) and $x\notin A\backslash B$ (because $x\in B$). However that's the same thing as saying that $x\in A\backslash(A\backslash B)$, so $B\subseteq A\backslash(A\backslash B)$.
Therefore $A\backslash (A\backslash B) = B$ if $B\subseteq A$.
Is my proof correct?

Your proof looks ok, but it can be much simpler if you just usebasic properties
$$
A \setminus (A \setminus B) = A \cap \overline{(A \cap \overline B)}=A\cap (\overline A \cup B) = (A \cap \overline A)\cup (A \cap B)=A \cap B = B
$$

let $$A=B\cup C$$ such that $B$ and $C$ are disjoint. Then, $$A-(A-B)=A-C=(B\cup C)-C=B\cup C-C=B$$ (Because $B$ and $C$ are disjoint)


Why is this particular case often used to introduce the property $P(A \cup B) = P(A) + P(B) - P(A \cap B)$? I'm taking up a probability course and my teacher, when explaining the fondamental properties of probability, listed this one:
$$\text{If } A \text{ and }B \text{ are mutually exclusive events, then }P(A \cup B)  = P(A) + P(B)$$
Then, later, under an "other properties" list, you had:
$$P(A \cup B)  = P(A) + P(B) - P(A \cap B)$$
which, per my understanding, is the general case of the first statement.
I have seen this approach taken in some books and by other teachers as well, where the particular case where the probability of the intersection is $0$ is treated as the "main" case, and the general case is listed separately, as if the two had no correlation, whereas the former is really just the latter with the additional hypothesis that $A \cap B = \emptyset$.
Is there any reason why this is? Are there any cases in which, mathematically, it makes more sense to treat a particular case as the main instance of a property or theorem?

We have $A = (A \smallsetminus B) \cup (A \cap B)$ and $B = (B \smallsetminus A) \cup (A \cap B)$. Notice that $A \smallsetminus B$ and $A \cap B$ are disjoint. Similarly $B \smallsetminus A$ and $A \cap B$ are disjoint. Now
$$ A \cup B = (A \smallsetminus B) \cup (A \cap B) \cup (B \smallsetminus A) \text{.}$$ The sets in the previous equation are pairwise disjoint. Use the equations for the probabilities of disjoint sets and some algebra to calculate the probability of $A \cup B$.

This is best viewed by drawing some venn diagrams.
In the first case, if it is said that they are mutually exclusive (never touching or intersecting) then we can simply add up the elements in A and B. The venn diagram for this would look like two circles labeled A and B separated from each other.
Otherwise, if nothing is said about being mutually exclusive, we assume that they are touching or intersecting and so you need to subtract or take away the overlapping part as to not double count the middle area (because we already include all of A and all of B). The venn diagram for this would be two circles labeled A and B with some overlap in the middle.
The two instances mean two different things and they usually start by teaching the nonoverlap property before learning about the other one.


What does it mean to say a language is context-free? What does it mean to say a language is context-free?

A context-free grammar is defined as a grammar in which every production rule is of the form $A \rightarrow \alpha$, where $A$ is a variable and $\alpha$ is a sequence of variables and terminals.

Formally, a context-free grammar can be defined as a 4-tuple $(V, \Sigma, R, S)$, where $V$ is a finite set consisting of the variables, $\Sigma$ is a finite set consisting of the terminals, $R$ is a set of production rules (in the form mentioned above), and $S \in V$ is the starting variable.

The language of a context-free grammar is the set of strings that can be derived from its start variable.  A context-free language is any language that is generated by a context-free grammar.

For example, $\{ 0^n1^n : n \ge 0 \}$ is context-free because it is generated by the context-free grammar $(\{S\}, \{0, 1\}, R, S)$, where the set of rules, $R$, is $$S \rightarrow 0S1 \mid \varepsilon.$$

(Note: I am using $\varepsilon$ to denote the empty or null string.)

As seen in this example, the set of context-free languages contains languages that are not regular. Also, since it is easy to mimic a DFA with a context-free grammar, the set of regular languages is a proper subset of the set of context-free languages.  Pushdown automata are the automata cousins of context-free grammars; they accept context-free languages and there exist algorithms to convert between the two models.

Note that the language $\{ a^nb^nc^n : n \ge 0 \}$ is not context-free (try to write a context-free grammar to generate this language and you will get a feeling for why).  It can be proven that the language is not context-free with the pumping-lemma for context-free languages, which I will leave as an exercise for the reader.

There are more powerful grammars, such as context-sensitive grammars, which allow the production rules to have the form $\beta A \gamma \rightarrow \beta\alpha\gamma$, where $\alpha$, $\beta$, and $\gamma$ are sequences of variables and terminals.  Context-sensitive grammars are as powerful as linear bounded automata (LBA).

My understanding is that a language is context-free if all statements can be understood without requiring external context, which is true for no natural language and those aspects of programming languages that do not rely on data or APIs from external sources.


Color the edges of $K_6$ red or blue. Prove that there is a cycle of length 4 with monochromatic edges. Color the edges of $K_6$ red or blue. Prove that there is a cycle of length 4 with monochromatic edges.

Attempt: I know that i have to...

prove that there must be TWO vertices with “red-degree” at least 3, or two with "blue-degree" at least 3. Call these U and V. (We'll assume we're in the red case; the blue case is similar.) Now, ask if they’re connected by a red edge, and how many red-adjacent neighbors they share. If you analyze this, there should be only a few cases to consider – and each should be relatively straightforward. For each case, you’ll show either that that case can’t actually happen in $K_6$, or that it gives rise to a monochromatic 4-cycle.

I’ll do most of it, leaving a few details for you to check. It’s very helpful to make diagrams for the various cases.

Since each vertex of $K_6$ has degree $5$, each vertex must have at least $3$ edges of the same color. Call a vertex red if it has at least $3$ red edges and blue if it has at least $3$ blue edges. Since there are $6$ vertices altogether, there must be at least $3$ vertices of the same color; suppose that there are at least $3$ red vertices, say $u,v$ and $w$. There are several cases.

First suppose that all three of the edges $uv,vw$, and $uw$ are blue. Let the other $3$ vertices be $x,y$, and $z$. The vertices $u,v$, and $w$ are red, so every edge from one of $u,v$, and $w$ to $x,y$, or $z$ must be red, and $u,x,v,y,u$ is a red $4$-cycle.

Thus, we may assume that two of these red vertices, say $u$ and $v$, are connected by a red edge. Each of them is connected by red edges to (at least) $2$ other vertices. Say $u$ is connected by red edges to $u_1$ and $u_2$, and $v$ is connected by red edges to $v_1$ and $v_2$.

If $\{u_1,u_2\}=\{v_1,v_2\}$, we may assume that $u_1=v_1$ and $u_2=v_2$. In this case $u,u_1,v,u_2,u$ is a red $4$-cycle.
If $\{u_1,u_2\}\cap\{v_1,v_2\}=\varnothing$, then the $6$ vertices $u,v,u_1,u_2,v_1,v_2$ are distinct, and one of them, say $u_1$, must be $w$. There are at least $3$ red edges at $w$, one of which is $uw$. If $wv_1$ is an edge, then $u,w,v_1,v,u$ is a red $4$-cycle, and similarly if $wv_2$ is a red edge. If neither $wv_1$ nor $wv_2$ is a red edge, then $wu_2$ and $wv$ must both be red, and $u,v,w,u_2,u$ is a red $4$-cycle.

The remaining possibility is that the sets $\{u_1,u_2\}$ and $\{v_1,v_2\}$ have exactly one element in common; say $u_1=v_1$. I’ll leave it to you to check that if any of the edges $u_1u_2,u_1v_2,u_2v_2,uv_2$, or $vu_2$ is red, then there is a red $4$-cycle, so we may assume that all of these edges are blue. This means that $u_2$ and $v_2$ are blue vertices, so either $w=u_1$, or $w$ is the sixth vertex, different from all of $u,v,u_1,u_2$, and $v_2$.

If $w=u_1$, let $x$ be the sixth vertex; the edge $wx$ must be red, since $w$ is a red vertex, and $wu_2$ and $wv_2$ are blue. If any of the edges from $x$ to $u,v,u_2$, or $v_2$ is red, it’s easy to find a red $4$-cycle, so assume that all are blue; then $x,u_2,v_2,u,x$ is a blue $4$-cycle.
If $w$ is the sixth vertex, it has red edges to at least $3$ of the vertices $u,v,u_1,u_2$, and $v_2$. In particular, it must have a red edge to at least one of the vertices $u_1,u_2$, and $v_2$. Suppose that $wu_1$ is red; you can check that no matter which of the edges $wu,wv,wu_2$, or $wv_2$ is red, there is a red $4$-cycle. (E.g., if $wu$ is red, we can use $w,u,v,u_1,w$, and if $wu_2$ is red, we can use $w,u_2,u,u_1,w$.) Thus, we may assume that $wu_1$ is blue and $wu_2$, say, is red. But then either $wv$ is red, and we have a red $4$-cycle $w,u_2,u,v,w$, or $wv$ is blue, and we have a blue $4$-cycle $w,u_1,u_2,v,w$.

I'm not sure if you're trying to prove the right thing, because there's not necessarily a monochromatic cycle of length 4 (see the picture). Actually, the Ramsey number R(3,3)=6, which means we can always find a monochromatic cycle of length 3.

You can refer to this link if you want to see the monochromatic 3-cycle proof.

https://plus.maths.org/content/friends-and-strangers


Isn't reflex angle between $180^\circ$ and $360^\circ$? I was solving this question.

Given that $\sin\theta = 5/13 $, find $\theta$, $\cos\theta$, and $\tan\theta$ where $\theta$
(a) is an acute angle, and
(b) is a reflex angle.

I could get the acute angle which is $22.62^\circ$. So, I did $360^\circ - 22.62^\circ$, to get the reflex angle, but the answer is  $157.38^\circ$ which is $180^\circ - 22.62^\circ$. Why is that? Do I have a wrong understanding of a reflex angle? Please help.

It sounds like what you are trying to ask is the following:

  Given that $\sin\theta=\frac{5}{13}$ find $\theta$, $\cos\theta$, and $\tan\theta$, where (a) $\theta$ is an acute angle, and (b) $\theta$ is a reflex angle.

I've just looked up reflex angle, and the definition says that $\theta$ is a reflex angle iff $180^{\circ}<\theta<360^{\circ}$. It follows that if $\theta$ is a reflex angle, then $\sin\theta<0$, so there is no solution if $\theta$ is a reflex angle.

If $\theta$ is acute than $\theta=\sin^{-1}\left(\frac{5}{13}\right)\approx22.6^{\circ}$.

Also, if $\theta$ is acute then $\cos\theta>0$ and $\tan\theta>0$. From the fact that $\sin^2\theta+\cos^2\theta=1$, you get that $\cos\theta=\frac{12}{13}$, which gives you that $\tan\theta=\frac{5}{12}$.

But $\sin (360 - \theta) \ne \sin\theta$.

so you can not figure that if $\sin \theta=\frac 5{13};0\le \theta < 360^{\circ}$, that that would mean the reflexive angle $\psi = 360-\theta$ will have $\sin \psi = \sin \theta$.  In fact just the opposite;  $\sin \psi = -\sin \theta$.

So:

$\sin \theta = \frac 5{13}$.

Find $\theta$:  $\frac 5{13}>$ so $0< \theta < 180$.  So if a) $\theta$ is accute then $\theta = \arcsin \frac 5{13}$.  If b) $\theta$ is reflexive this is not possible and there is no solution.

$\cos \theta = \pm \sqrt{1-\sin^2 \theta} = \pm\frac {12}{13}$.  So if a) $\theta$ is accute  then $\cos \theta \ge 0$ and $\cos \theta = \frac {12}{13}$.  If b) then $\theta$ is impossible so there is no solution.

$\tan \theta = \frac {\sin \theta}{\cos \theta}$.  So if a) $\theta$ is accute then $\tan \theta = \frac {\frac 5{13}}{\frac {12}{13}} = \frac 5{13}$ and, again, for (b) there is no solution.

Now it's not sure if I'm supposed to leave it as $\theta = \arcsin \frac {5}{13}$ or if I'm supposed to do some fancy geometry and trigonometry to find out what the angle of right triangle with sides of $5,12,13$ is.  But I'm not going to grub in the mud unless I know there's a potato down there.

I could punch keys into a calculator to find $\theta = \arcsin \frac {5}{13}$ but nobody wants some stupid decimal when $\arcsin \frac {5}{13}$ is much more descriptive than $\theta = 22.6whogivesapileoffetiddingokidneys1087668.....$

So $\theta = \arcsin \frac 5{13}=\arctan \frac 5{12}$.  Final answer.

.....

I suspect the person asking the question meant obtuse.

In which case b)  $90 < \theta < 180$ and as $\sin{180 - \theta}=\sin \theta$ we have $\theta = 180 -\arcsin \theta$.  And $\cos \theta < 0$ so $\cos \theta =-\frac 5{12}$ and $\tan \theta =-\frac 5{12}$.  By the way we can express $\theta = \arctan -\frac 5{12}$.


How do you work out the angle in this square? I have labelled all the angles that I can work out.

But I can't think of any other way to find the other angles without being 100% sure.

Thank you!

P.S. I have attached the official question - I have scribbled all over it with lines and therefore have drew out again by hand to make things clearer.

$\triangle PTS$ is isosceles because $PT =PS (= PQ) $. Hence $\angle PST = \frac{(180 - 30)}{2}=75^{\circ}$ and $\angle TSR = 90 - 75 = 15^{\circ}$. By symmetry $\triangle STR$ is isosceles so the required angle is $(180 - 30)=150^{\circ}$

Ok, This is the way I found.

Construct a segment parallel to SR, with midpoint T. This creates 2 right triangles.
Extend ST or TR, creating an Isosceles triangle inside one of the new right triangles.
Bisect the isosceles triangle into 2 congruent right triangles. You will note that both of them will be 30˚/60˚/90˚ triangles
This shows you that angle STR has two compliments of 30˚
Angle STR must be 120˚

Hope that helps!


rational angles with sines expressible with radicals An angle x is rational when measured in degrees. sin(x) is can be written using radicals. What are the conditions on x? If nested square roots are allowed?

What I know so far:
If sin(x) can be expressed in radicals then so can $\cos{x}=\sqrt{1-\sin^{2}{x}}$, and $\tan{x}=\sin{x}/\cos{x}$. As can sin{x/2} and sin{x/3}, because I can use trig identities to get them in terms of sin{x} and the resultant quadratics and cubics can be solved.

We prove a standard constructibility result for angles in radians, then make the adaptation to degrees.

Theorem: Let $x=\frac{m}{n}$, where $m$ and $n$ are positive relatively prime integers.  Then the $\frac{2\pi m}{n}$-radian angle is Euclidean constructible iff $n$ is a power of $2$ times a possibly empty product of distinct Fermat primes.

Note that the $\frac{2\pi m}{n}$-radian angle is constructible iff the $\frac{2\pi}{n}$-radian angle is constructible. One direction is obvious. For the other direction, since $m$ and $n$ are relatively prime, there exist integers $x$ and $y$ such that $xm+ny=1$.  Multiply both sides by $\frac{2\pi}{n}$. We obtain
$$x\frac{2\pi m}{n} +y(2\pi)=\frac{2\pi}{n}.$$
By assumption, the $\frac{2\pi m}{n}$-radian angle is constructible, and therefore so are $x$ copies of it. Clearly the $y(2\pi)$-radian angle is constructible, and therefore the $\frac{2\pi}{n}$-radian angle is constructible.

It is a standard fact about constructible regular polygons that the regular $n$-gon is constructible iff $n \ge 3$ is of the shape a power of $2$ times a product of distinct Fermat primes. This result takes care of everything but $n=1$ and $n=2$, which are obvious.

Adapting to Degrees: Let $a$ and $b$ be relatively prime positive integers.  We ask for the possible values of $a$ and $b$ such that the $\frac{a}{b}$-degree angle is constructible.  This is the case iff the $\frac{2\pi a}{360b}$-radian angle is constructible. Let $d=\gcd(a,360)$. So we are interested in the constructibility of the $\frac{2\pi m}{n}$-radian angle, where
$$m=\frac{a}{d} \text{ and } n=\frac{360b}{d}.$$
By the result for radians, we have constructibility precisely if $\frac{360b}{d}$ is a power of $2$ times a product of distinct Fermat primes.

Suppose $d$ is not divisible by $3$.  Then $\frac{360b}{d}$ is not of the right shape, since it is divisible by $3^2$.  So for constructibility we need $3|a$.  In addition, since we assumed that $a$ and $b$ are relatively prime, $b$ cannot be divisible by $3$.

The other problematic prime is $5$.  If $a$ is not divisible by $5$, then $b$ cannot be divisible by $5$, else the Fermat prime $5$ would occur more than once in the factorization of $\frac{360b}{d}$.  And if $a$ is divisible by $5$, again $b$ cannot be, since $a$ and $b$ are relatively prime.  Thus in either case $5$ cannot divide $b$.  We have proved:

Theorem: Let $x=\frac{a}{b}$, where $a$ and $b$ are positive relatively prime integers.  Then the $\frac{a}{b}$-degree angle is Euclidean constructible iff
(i) $a$ is a multiple of $3$ and (ii) $b$ is a power of $2$ times a possibly empty product of distinct Fermat primes greater than $5$.

Comment: Presumably the second result (about degrees) has been proved many times.  The part about avoiding the primes $3$ and $5$ in the denominator is an "accident" caused by the choice of degree as the unit.  If the Babylonians had decided to have a $340$-unit circle, $3$ would no longer be special, but $5$ and $17$ would be.

Algebraically exact value of Sine for all integer angles is possible.

Please visit https://archive.org/details/ExactTrigonometryTableForAllAnglesFinal for the list of exact values for Sine of integer angles in degrees.


$f(x) = \sqrt{ |x|} - \log (1+|x|)$ has no Upper Bound on $\mathbb R$ Is the following  function bounded on  $\mathbb R$ ?

$f(x) = \sqrt {|x|} - \log (1+|x|)$

My Attempt:  I can prove that $f(x) $ is an increasing function on $(0 , \infty)$ and $f(x) $ is decreasing on $(-\infty ,0)$ and $f(x) $ is continuous on $\mathbb R$ and $f(x) \geq 0$  for all $x \in  \mathbb R$. But I can not proceed further.

Can anyone please give me any hint or some thing?

Let $ x = e^{k}-1$ where $ k >> 2$.

Then $ \sqrt{x} > e^{k/2} - 1$ and $ \log (1 + x) = k$, and clearly $e^{k/2} - k  - 1 $ tends to infinity.

Hint: use (prove) the fact that for any $M > 0$ there is $X > 0$ that for all $x > X$

$$ x^\frac{1}{2} > M + log(1 + x)$$


What is the best/most effective way to gain experince in using set theory to understand math, calculus in particular? Note: While this question as phrased in the title is somewhat subjective, what I'm looking for as an answer should be specific enough to still have a clear/valid answer(s) to it. Also, apologies in advance if this is somewhat long winded but I feel the full context of my question is needed for clarity.

First let me provide some context. When it comes to learning new areas of math (or in this case, arguably, more fundamental/abstract areas), one problem all students face is 'vision'. Taking new concepts and adding them to old ones is often difficult since to do that requires seeing how things interrelate. However your typical education in math usually leaves you vastly unprepared to do this.

In a sense, most modern math education teaches students backwards. Most of the maths that students learn is gutted of the more important (and abstract) details. Details such as the 'real' proofs of certain theorems being replaced with a shorthand version that, while proves the part of the concept being focused on, ignores the parts that give it its foundation. "Proofs in a vacuum" as I call them since they usually don't bother proving (sometimes not even mentioning) other concepts that the main idea depends on.

This leads me to my problem. I have a rather strong foundation in multiple areas of math (particularly those often relevant to physics, such as calculus), but since my education has focused on results/applications rather then math as a whole I'm left without the 'glue' that ties it all together.

The language of set theory (as well as basic forms of logic) is still new to me. While I have a strong (enough) foundation in the basics after recently completing a class in discrete math, taking the concepts (and building onto them where needed) and applying them so as to add to my understanding of calculus for example is still difficult.

My question is this: What are some methods/pet projects I should try an implement to 'break out' of this novice phase as far set theory is concerned? To bridge the gap between knowing basic ideas concerning sets and relations between them and concepts from calculus for example? I don't want to look at an equation and see an equation. I wanna be able to see it as a more abstract entity.

Keep in mind what I'm looking for in an answer is:

Specific topics to research that show how sets and relations are used to construct other others of math
Go to examples for me to play around with so as to see how set theory is applied in practice (for example, how relations can be used to construct basic operations like addition, or differential operators).
Examples of a common (but not obvious) mistakes people tend to make applying set theory to describe common concepts in other area of math.

If I understand your question correctly, although you have embedded a critique of how abstract math is taught, your question is really about how you can shore up your own understanding of abstract math via an enhanced understanding of set theory.

I would suggest that you learn about how the real number system is built up in steps from Peano's axioms for the natural numbers. There are three steps:

From Peano's axioms for the natural numbers to the integers.
From the integers to the rational numbers.
From the rational numbers to the real numbers.

There is one still deeper step:

From the ZFC axioms of set theory to Peano's axioms for the natural numbers (via the construction of the ordinal numbers).

I cannot recommend a single source to you for all four of these steps, although steps 2, 3, 4 at least are in many books at the advanced undergraduate or early graduate level, with steps 2 and 3 more likely to be in number theory books and steps 3 and 4 more likely to be in advanced calculus books. For Step 2, I like the closing two chapters of the book "Theory of Numbers" by B. M. Stewart. For Step 1, I like the early sections of Cohen's slim little atom bomb of a book "Set theory and the continuum hypothesis".

Ultimately, I believe you will find your task to be difficult.

The problem lies with Zermelo's emphasis of Fregean principles over Cantorian principles and the subsequent criticism of Zermelo by Skolem.

Applied mathematics relies fundamentally on Liebniz' principle of identity of indiscernibles as expressed within the axioms for metric spaces.  Logicians and analytical philosophers have rejected this as a logical principle despite the fact that this rejection leads to contradiction when the sign of equality is interpreted as mere substitutivity.  For them, the priniciple of identity of indiscernibles constitutes an epistemic warrant for uses of the sign of equality.  But, they are concerned with extensional semantics.  So, in first-order logic, the principle of identity of indiscernibles is taken to constitute "real definition".  This is because the logic carries what is called "existential import".  The objection is intended to admit the possibility of two distinct objects having all of the same properties.  Thus,
they cannot be discerned from one another by their properties.  The claim is that a definition cannot determine an individual.

If you look carefully at the axiom of extension in standard texts on set theory, you will find that it uses a conditional rather than a biconditional.  The conditional defers to the treatment of identity in first-order logic.  And, the identity of indiscernibles is not part of the
standard account of identity for that logic.  One simply has that $x = x$ means "x is substitutable for x" and that $x = y$ means "y is substitutable for x".  What an applied mathematician usually understands as an equivalence
relation can be derived from these rules with substitutivity tricks.

This interpretation for identity in set theory follows from Skolem's criticism.  In Zermelo's original axioms, the sign of equality is interpreted with respect to the singleton sets in the theory.  As for Zermelo emphasizing Fregean views over Cantorian views, there is a difference between "the extension of a concept" and "a collection taken as an object".  There is much in the historical development of set theory that has given it the form that it has.  The fact that Cantor's views had been set aside is one of the points which Lawvere has emphasized in his development of sets using category theory.  In his work, the identity of indiscernibles is taken for granted as a necessary logical law.  And, given that Cantor's notion of set had been closely bound to his topological ideas, Lawvere's program should not be dismissed merely because analytical philosophers have different priorities from a significant number -- maybe even a majority -- of mathematicians.

I strongly suggest that you look at Max Black's discussion of the principle of identity of indiscernibles in the paper,

http://home.sandiego.edu/~baber/analytic/blacksballs.pdf

It may help you to decide which conception of set will be more amenable with your objectives.

And, there is one other point to make.  You speak of sets in the constructive manner with which they had been introduced historically.  The notion of formalism in mathematics has deprecated this relationship to other parts of mathematics.  This is largely why Skolem's criticisms had been accepted.  One may speak of how mathematical objects may be represented within set theory, but, set theory is not as widely considered to be "foundational" as it had been in the past.  This is largely because of how arithmetic may be used to represent statements about what can and what cannot be proven.  But, that is different from what is meant by formalism, and, formalism divorces set theory from the role of being "glue".

Modern set theory is simply not what most applied mathematicians think it is.  One may "believe" in sets, but the contentiousness of the positions in Professor Black's paper may very well have divorced set theory from applied
mathematics.  And, this insistence on a logical aesthetic at the expense of mathematical forms may be traced back to passages from Aristotle.

What you seem to be looking for is a single progressive development leading to elements used by applied mathematics.  I have found "Vectors and Tensors" by Bowen and Wang,

https://www.amazon.com/Introduction-Vectors-Tensors-Two-Mathematics/dp/048646914X

to be useful in this regard.  There is an online version if you look for it.  It comes in two volumes, and, the transition from algebraic development to metric spaces is in the first chapter of the second volume.  It is this transition which should motivate the study of Willard's self-verifying systems of arithmetic rather than Peano arithmetic.  But, unless one is particularly interested in how foundational studies relate to applied mathematics, one would not even recognize this.

While I hope to have given you alternative ways to think about your situation, I suspect that you will still pursue set theory out of your own curiosity.  Good luck.


Homomorphisms between fields are injective. How would I prove this?

I know that I must show $f(a)=f(b) \Rightarrow a = b$

I also know I must use the definition of homomorphism, ie:

$f(a+b)=f(a)+f(b)$

$f(ab)=f(a)f(b)$

$f(1)=1$

I am assuming that a contradiction would be a good approach to solve this, but not quite sure on specifics.

Suppose $f(a) = f(b)$, then $f(a-b) = 0 = f(0)$. If $u = (a-b) \ne 0$, then $f(u)f(u^{-1}) = f(1) = 1$, but that means that $0 f(u^{-1}) = 1$, which is impossible. Hence $a - b = 0$ and $a = b$.

To prove it is an injective we must show that if $f(a) = f(b)$ then $a = b$.
Now suppose $f(a) = f(b)$ but $a \ne b$, then we have that $f(a) - f(b) = 0$ and $u = a - b \ne 0$.
So $f(u u^{-1}) = f(u)f(u^{-1}) = 1$, but $f(u) = f(a - b) = f(a) - f(b) = 0$. We have that $f(u)f(u^{-1}) = 1 = 0 f(u^{-1})$ in which for field is impossible. Thus $a = b$.


Surjectivity of the induced map of affine algebraic sets For a morphism $f: X\rightarrow Y$ of affine algebraic sets, I want to show that if the induced map $f^*:k[Y]\rightarrow k[X]$ is surjective then $f(X)$ is closed.

I am trying to prove that $f(X)=Z(\ker f^*)$. The direction $f(X)\subset Z(\ker f^*)$ is easy but I am stuck for the other one.

I know it was asked before (Why does surjectivity of the induced map show that a morphism of affine varieties has closed image?) but the solution offered showed that $f$ surjective implied $f^*$ has a closed image.

Thanks!

Let  $X \subset k^m$.  For every $1 \le i \le m$ there exist  $g_i$ in $k[Y]$ so that for the coordinate function $x_i$ in $k[X]$ we have

$$x_i = g_i(f(x))$$
We get a map $g \colon Y \to k^m$ such that
$$x = g(f(x))$$ for all $x \in X$. ($\tiny{ \text{ We got the inverse map, just a bit more work needed.} }$)

Let $Z$ the Zariski closure of $\phi(X)$ in $Y$. Let's show first that
$$g(Z) \subset X$$
Indeed, let $p_{\alpha}(x)=0$ equations giving $X$. Want to show that $p_{\alpha}(g(z))= 0$ for all $z \in Z$. Now $p_{\alpha}(g(f(x))= p_{\alpha}(x) = 0$ for all $x \in X$. Since $Z$ is the Zariski closure of $f(X)$ we conclude that $p_{\alpha}(g(z)) = 0$ for all $z \in Z$.

We have maps
$$f \colon X \to Z\\
g \colon Z \to X$$

and $$\mathbb{1}_X = g \circ f$$ Let's show that $$f\circ g = \mathbb{1}_Z$$ It's enough to show the equality on the dense subset $f(X)\subset Z$. We have

$$f\circ g( f(x) = f(g(f(x)) = f(x)$$

So $g$ is the inverse map of $f$, and therefore $f(X) = Z$ is closed. Moreover, the map $f\colon X \to f(X)$ is an isomorphism.

$\tiny{ \text{The proof works in other categories (analytic, smooth).}}$

Here is a proof of the other inclusion wich does not take the detour via affine schemes:

Because of the inclusion already shown, we can switch the codomain from $Y$ to $Z(\operatorname{ker}(f^*))$. This corresponds to taking the quotient of $k[Y]$ by $\operatorname{ker}(f^*)$. This quotient is isomorphic to $k[X]$ by the homomorphism theorem because $f^*$ is surjective. But then this implies that $X$ and $Z(\operatorname{ker}(f^*))$ are isomorphic via $f$. In particular $f(X)=Z(\operatorname{ker}(f^*))$.

what we are using here is the equivalence between algebraic sets and reduced finitely generated k-algebras. I hope I didn't mix up the order in which things are usually shown here.


Finding the area of triangle if length of medians are given My question is:

  In triangle ABC ,length of median from vertex A is $13$ , length of
  median from vertex B is $14$ , length of median from vertex C is $15$.
  Compute the area of triangle ABC.

One strategy, probably not optimal, is to find the lengths of the sides, and then use Heron's Formula.

Take a triangle $XYZ$, with sides $x$, $y$, $z$ as usual, and let $m$ be the length of the median from $Z$ to the side $XY$, which has length $z$.

Let $P$ be the midpoint of $XY$.  We have divided our triangle into two triangles, $ZPX$ and $ZPY$.  Let $\theta=\angle ZPX$ and $\phi=\angle ZPY$. Note that $\theta+\phi=\pi$.  By the Cosine Law, we have
$$y^2=m^2+\left(\frac{z}{2}\right)^2-mz\cos\theta.$$
Also by the Cosine Law, we have
$$x^2=m^2+\left(\frac{z}{2}\right)^2-mz\cos\phi.$$
But $\cos\phi=-\cos\theta$. Add. The awkward cosine terms cancel, and we get
$$x^2+y^2=2m^2+2\left(\frac{z}{2}\right)^2,$$
which yields the result
$$m^2=\frac{1}{4}\left(2x^2+2y^2-z^2\right).$$

The rest is downhill. We go back to the triangle of the question. The three lengths of the medians give us three linear equations in the squares of the side lengths. If the sides of our triangle are $a$, $b$, and $c$, we get
$$4\cdot 13^2=2a^2+2b^2-c^2;\quad 4\cdot 14^2=2b^2+2c^2-a^2;\quad 4\cdot 15^2=2c^2+2a^2-b^2.$$
Solve. A simple  way is to use the symmetry of the equations by first of all "adding" the above three equations to first find $a^2+b^2+c^2$.

Now that we  have the sides, we can use Heron's Formula.

Area: $$\Delta=\dfrac 43\sqrt {s(s-a)(s-b)(s-c)}$$
Where $a,b$ & $c$ are median lengths & $s$ is the semi-perimeter.


Prove $ \sqrt{\arctan(x)} = (1/2) \arccos((1-x)/(1+x))$ $$ \sqrt{\arctan(x)} = \dfrac{1}{2} \arccos\left(\dfrac{1-x}{1+x}\right)$$

I have been trying to solve this problem for the past hour, but I'm not able to solve it as I have just started solving difficult trigonometric problems. I'm not able to get any logic to solve this problem. I'm not able to put any trigonometric formula to solve this please help.

Here is 2-steps plan:

Salvage your accept rate from its current appalling value.
Prove that
$$ \arctan(\sqrt{x}) = \frac{1}{2} \arccos\left(\dfrac{1-x}{1+x}\right).
$$

Arctan . Jpeg

Click on that blue thingy above - and no, I swear this isn't a spam. The math formula programming was simply too tiring to learn in 3min. flat, so I've provided a written vers instead. Don't judge my handwriting, perhaps?
Hope it helped! :)


Solving Second Order Linear Non-homogeneous Differential Equation I am trying to solve the following:
$$
y''+4y=\tan(t).
$$

I have used the method of variation of parameters. Currently I am at a point in the equation where I have this: $$u_1= \int \frac{\tan t \cos2t}{2}$$

I am stuck here

$\cos 2t = 2\cos^2 t - 1$ hence

$$\int \frac{\tan t \cos2t}{2} dt =  \int \sin t \cos t \  dt - \frac 12 \int \tan t \ dt \\ = \frac 12 \sin^2 t - \frac 12 \ln(\sec t) + C$$
$$ = \frac 12 \left( \sin^2 t + \ln(\cos t) \right) + C$$


Number of normals to a parabola from a given point I know that from any point a maximum of three normals could be drawn to a parabola because the equation of normal is cubic.
But I want to know the condition on the point for the number of normals

Claude's technique can be extended slightly to find the the points and normal lines given a particular point off of the parabola. Using this, I obtained an animation of some of the normals.

Only two perpendicular tangents can be drawn from a point P on a parabola directrix ( standard result, need not be again proved). Taking this as given we complete the rectangle by drawing parallels to arrive at a unique opposite point inside point Q ... from which we conclude accordingly only 2 unique normals can be drawn from an arbitrary but inside point like Q.


calculating the dihedral angle of two planes I have an ordered pair of planes, meeting at a hinge at their intersection line. The planes and the line are oriented. I want a formula for the dihedral angle $\theta$, calculated as follows:

Orient everything so that the hinge is oriented along the positive $z$ axis. The two planes are now lines in the $xy$ plane. Measure the angle between them, starting at the first plane and rotating in the direction of the first plane's orientation until you hit the second plane.

There is the obvious formula $n_1\cdot n_2 = \cos \theta$, where $n_1$ and $n_2$ are the plane normals, but to determine the correct angle in $[0,2\pi]$ I then need to also look at $\sin \theta$. Is it possible to write down a single formula for $\theta$ in terms of the positions and orientations of the planes/line, differentiable whenever the angle lies in $(0,2\pi)$?

Here's a cross section of the situation, with two oriented planes. If you're rotating the red/green plane in the direction of its green side, does that mean clockwise or counterclockwise?

So I don't think you can use the orientation of the planes to fix an interpretation of the angle. However, you write that you have a preferred orientation of the line of intersection, and that sounds more promising.

Namely, you can decide to turn the first plane counterclockwise when looking in the preferred direction along the intersection line.

Somewhat tedious formula-churning implementation of this: Fix some point on the intersection line and consider the plane that is normal to it at that point. Equip that plane with an orthonormal coordinate system that creates a right-handed system together with the preferred direction of the intersection. The orthonormal coordinates create a parametric representation of the normal plane; compose that with the equations for the two original planes; that gives 2D equations within the normal plane for its intersection with the original plane. Now find the angle in the normal plane using standard 2D techniques (e.g as the difference of atan2 values).

Computationally slicker: You can find $|\cos\theta|$ by the dot product or normals in 3D without needing to project into a specific cross section; the only remaining information you need is whether to use $\theta_0$, $\pi-\theta_0$, $\theta_0-\pi$ or $-\theta_0$, where $\theta_0$ is the arccosine of the dot product. But you can actually distinguish between those four cases simply by the combination of (a) the sign of the dot product and (b) whether the cross product of the normals is parallel or antiparallel to the chosen direction of the intersection line. I'd probably make some embarassing mistake if I tried to work our exactly what the connection should be, but there are only 4! possible ways to do it anyway, so just figure it out with trial and error...

$a_1 x + b_1 y + c_1 z + d_1 = 0$ is the equation for the first plane.
The normal to the plane is $(a_1,b_1,c_1)$.
$a_2 x + b_2 y + c_2 z + d_2 = 0$ is the equation for the second plane.
The normal to the second plane is $(a_2,b_2,c_2)$
Let $\alpha$ be the angle betweeen the two planes.
Then, $\cos \alpha = |n_1 \cdot n_2| / \text{len}(n_1) \cdot \text{len}(n_2)$.
The top has the absolute value of the cross product, and the bottom has the lengths of $n_1$ and $n_2$ normal vectors multiplied together.  Take the $\arccos$ of both sides of the equation to remove the $cos$ on the left side.

Below is the link where I am getting my information from:

https://math.tutorvista.com/geometry/angle-between-two-planes.html