Geometric proof of number theory result: If the geometric and quadratic means of integers $a$ and $b$ are themselves integers, then $a=b$ We start with the following problem:

  Let $a$ and $b$ be positive integers such that their geometric and quadratic means are integers. Show that $a=b$.

One possible approach is to write down the corresponding diophantine equations and to do infinite descent on $u^4-v^4=w^2$ where $w=a^2-b^2$ which then has to be zero.

However, the means also have simple geometric representations. So my question is if there is a way to do the infinite descent geometrically.

Even if the two numbers are non-integer positive real numbers

$$ \sqrt{\frac{a^2+b^2}{2}} = \sqrt{ab} $$

requires that ( by squaring and simplfying )

$$  a^2+b^2-2ab =0 \rightarrow (a=b) $$

The RMS of $(a,b)$  is in general greater than GM. They are equal only if the numbers $(a,b)$ are themselves equal.


Limit of real function, theoretical exercise I'm a freshman in mathematics and this is my exercise:

  Prove that for function $f:\langle -a,a\rangle\setminus\{0\}\longrightarrow\langle0,+\infty\rangle$ so that is
  $$\lim_{x\to 0} \left(f(x)+\frac{1}{f(x)}\right)=2$$
  is also
  $$\lim_{x\to 0} f(x)=1.$$

I've tried to solve this problem from the Cauchy's definition of limit
$$(\forall \varepsilon>0) (\exists\delta>0) (\forall x \in \langle -a,a\rangle\setminus\{0\})\qquad (0< \lvert x-0\rvert  < \delta) \implies \Big( \Big| f(x)+\frac{1}{f(x)} \Big| < \varepsilon \Big)$$

by trying to get $\Big|f(x)-1\Big|<\varepsilon $ but I got that just in case $f(x)\in\langle0,1]$.

Any thoughts on solving the problem this way or I should use another approach?

Notice that to prove $\lim_{x \to 0} f(x) = 1$, it is equivalent to show that for any sequence $\{x_n\}$ converging to $0$, we have
$$\lim_{n \to \infty} f(x_n) = 1.$$

Let $\{x_n\}$ be any sequence converging to zero (for $\{x_n\}$ being contained in the domain of $f$, $\{x_n\} \subset (-a, a) \backslash\{0\}$). By condition, we have
$$\lim_{n \to \infty} \left(f(x_n) + \frac{1}{f(x_n)}\right) = 2,$$
which implies that the sequence $\left\{f(x_n) + \frac{1}{f(x_n)}\right\}$ is bounded above. Since $f(x_n) > 0$ for each $n$, it follows that $0 < f(x_n) < f(x_n) + \frac{1}{f(x_n)}$, hence the sequence $\{f(x_n)\}$ is also bounded. By Bolzano-Weierstrass theorem, $\{f(x_n)\}$ contains a convergent sequence, say, $\{f(x_{n_k})\}$.

Suppose that $\lim_{k \to \infty} f(x_{n_k}) = c \geq 0$. Clearly $c > 0$, otherwise
$$\lim_{k \to \infty} \left(f(x_{n_k}) + \frac{1}{f(x_{n_k})}\right) = 2 \tag{1}$$ would be violated. Therefore $\{1/f(x_{n_k})\}$ also converges with its limit $\frac{1}{c}$. Hence, $(1)$ gives that
$$c + \frac{1}{c} = 2.$$
Solving this for $c$, we obtain $c = 1$.

Now for any convergent subsequence $\{f(x_{n_m})\}$ of $\{f(x_n)\}$, similar argument as above gives that its limit must be $1$. Therefore $\lim_{n \to \infty} f(x_n) = 1$, and the proof is complete.


Measurable function of a transformation still measurable I'm looking for a maybe simpler or more elemental proof of the following statement:

Let $f:\mathbb{R^n}\to \mathbb{R}$ be (Lebesgue-)measurable.
Then $F:\mathbb{R^{2n}}\to \mathbb{R}$ such that $F(x,y)=f(x-y)$ is (Lebesgue-)measurable.

The proof I know works in the following way:
First show that $f$ is measurable on $\mathbb{R^{2n}}$.
Then $T(x,y)=(x-y)$ is Lipschitz and hence $f\circ T$ is measurable (by a theorem ).

Can one somehow argue without the Lipschitz continuity in the last step or with a different ansatz? Several books I've seen introduce the convolution using that $f(x-y)$ is measurable but proof that Lipschitz $T$ maps measurable sets onto measurable sets later in the text.

Alright, so this is as elementary and intuitive as I could think of. We assume only that if $U \subset R^n$ is measurable and $T$ is a rotation on $R^n$, then $T(U)$ is measurable. This is a standard fact about the Lebesgue measure and should be covered quite early in most books. For simplicity, consider $n = 1$.

Define $G:R^2 \to R$ by $G(x,y) = y-x$. Since $F^{-1}(U) = G^{-1}(f^{-1}(U))$ for each $U \in R$ and $f$ is assumed measurable, we only have to show that $G^{-1}(U) \subset R^2$ is measurable for each measurable $U \subset R$ in order to prove that $F$ is measurable.

If $U = \{c\}$ consists of a single point, then
$$G^{-1}(\{c\}) = \{(x,y) \in \mathbb{R}^2: y-x = c\} = \{(x,x+c)\in \mathbb{R}^2: x \in \mathbb{R}\}$$

which is the graph of the function $y(x) = x + c$. The slope of this function is $1$ which corresponds to an angle with the $x$-axis of $45$ degrees.

If we let $T: R^2 \to R^2$ be the linear transformation that rotates clockwise by $45$ degrees, then

$$T(G^{-1}(\{c\})) = T(\{(x,x+c)\in \mathbb{R}^2: x \in \mathbb{R}\}) = \{(x,c) \in R^2: x \in R\}$$

Now if $U \subset R$ is measurable, then
$$U = \bigcup_{c \in U} \{c\}$$
which means that
$$T(G^{-1}(U)) = \bigcup_{c \in U} T(G^{-1}(\{c\})) = \bigcup_{c \in U}T(\{(x,x+c)\in \mathbb{R}^2: x \in \mathbb{R}\})$$
$$ = \bigcup_{c \in U}\{(x,c) \in R^2: x \in R\} = R \times U$$

which is the product of two measurable sets and hence measurable (by definition).
Thus
$$G^{-1}(U) = T^{-1}T(G^{-1}(U))$$
is measurable since $T^{-1}$ is also a rotation.

The rotation technique needs to be preceded by a translation to the origin, and then shifted back. If you rotate the line y=x+1 by 45 degrees, it doesn't become parallel to the x axis unless you do this shifting stuff...


Limits - It's always possible to escape an indeterminate form? In general, when we are solving limits and get to an indeterminate form($\frac{0}{0}$, $\frac{\infty}{\infty}$, $0^0$,etc.), we use some technique to get rid of the indeterminate form(factorization, conjugate, variable substition, etc.) and then finish solving the limit.

My question is, it's always possible to get rid of indeterminate forms? If it's not, what conclusion should i take about the limit I'm trying to solve?

If you are studying a limit of the form
$$\lim_{x\to a}\frac{f(x)}{g(x)},$$
where $f$ and $g$ are nonzero smooth functions with $f(a)=g(a)=0$, then you can get rid of the indeterminate form $\frac00$ if $f$ and $g$ are analytic. In this case you have that $f$ and $g$ can be written as their Taylor's series at $a$ and so, since they are nonzero functions there must be a derivative which is different from zero, say $f^{(n)}(a)$ is the first non-zero derivative for $f$ and $g^{(m)}(a)$ is the first nonzero derivative for $g$. Hence,
$$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{\frac1{n!}f^{(n)}(a)(x-a)^n}{\frac1{m!}g^{(m)}(a)(x-a)^m}$$ and this limit can be computed.
The good news is that most elementary functions in calculus are analytic (polynomias, $\sin x$, $\cos x$, $e^x$, $\log x$) so most of the times indeterminate forms can be solved.
However, there are $C^\infty$ functions which are not analytic. As Renart wrote, the typical example is $f(x)=e^{-1/x^2}$. Its Taylor series at $0$ is zero and so if both $f$ and $g$ have this pathology (see Renart's example), then one would would not be able to resolve the indeterminate form with simple tools (Hopital, Taylor's formulas, and the such).

Edit
I will try to reply to TMM's criticism. Consider two functions $f$ and $g$ continuous at $a$ with $f(a)=g(a)=0$ and $f(x)\ne 0$, $g(x)\ne 0$ for $x$ near $a$. The formally correct answer to the question "it's always possible to get rid of indeterminate forms" is yes, but only in the sense that either the limit $\lim_{x\to a}\frac{f(x)}{g(x)}$ exists or it does not.
Now if the question "is there a general technique to always get rid of indeterminate forms" then the answer is yes for analytic function and no for $C^\infty$ functions. A general technique means a theorem. Hopital's, Taylor's formula, the squeeze theorems are theorems which can be used to compute limits but they are theorems.
Now if $f$ and $g$ are $C^\infty$ and not analytic at $a$, with all derivatives
$f^{(n)}(a)=g^{(n)}(a)=0$ for all $n$, then there is no general theorem which can determine if the limit exists or not.
So yes, if you want, my answer is incomplete but only because there is no complete answer in this case.

The wikipedia page on indeterminate forms defines them as being restricted to the following seven forms:
$$0/0,~ \infty/\infty,~ 0\times\infty,~ \infty-\infty,~ 0^0,~ 1^\infty,~ \infty^0 $$
and then goes on to provide a handy table for converting all of these into the form $0/0$ or $\infty/\infty \text{,}$ whose limit can then be found using L'Hôpital's rule


On The Parametric Equation Of A Parabola Let's look at a parabola with an equation $(y-k)^2=4a(x-h)$. I'm struggling to understand why its parametric equation would be $x=h+at^2$ and $y=k+2at$. Since it is being subtracted by $h$ and $k$ respectively, why would its $x$ and $y$ values increase/have $h$ and $k$ added to them instead?

I understand that since each $x$-value is being subtracted by $h$, each $x$-value needs to be $h$ bigger to "achieve" the same y-value. But that also means that for each $y$-value, each $x$-value will be h smaller right? So how does subtracting by $h$ increase the value of $x$ by $h$, as shown in the parametric equation $x=h+at^2$? I ask the same question for $y=k+2at$.

Anyways, I have no idea what the "x" output the parametric equation gives out even means? Does the parametric equation give out the x-value which will be inputted into the equation? But then no matter where the graph is, that doesn't change what x-value would be inputted into the equation, right? So why would subtracting x by h change the parametric equation of x either, if x will remain unchanged?

Or does the parametric equation give out the x-value as in the x-value to be plotted on the graph? But then if so, I still don't get why the sign would be +h, instead of -h. Yes, in the graph, for every y-value, the x-value of a graph y^2=4a(xx-h) would be left of the original graph. But how do we know the parametric equation refers to the x-value for every given y-value. And if that is not what is referred to by the parametric equation, why are we subtracting?

Can someone explain this to me as simply as possible, since I'm still a beginner. I'm just putting this as a precaution, and try not to use Calculus in the response, since I haven't learnt it yet.

Let's start with the simplest case, in which $h$ and $k$ are both $0$.  Then the parabola is
$$y^2 = 4ax$$
and the parametric equation is
$$y=2at, \hskip{0.5in} x=at^2$$
and you can directly verify that
$$y^2 = (2at)^2 = 4a^2t^2 = 4a(at^2) = 4ax$$
so everything works nicely.

Now let's consider the general case.  We have
$$(y-k)^2 = 4a(x-h)$$
If we introduce (temporarily) the new variables $u = x-h, v = y-k$ then the equation can be written as
$$v^2 = 4au$$
which has exactly the form of the simple case we already considered.  So we know the parametric equations for $u$ and $v$ are
$$v=2at, \hskip{0.5in} u=at^2$$
But now we remember that $u = x-h$ and $v = y-k$.  That means the parametric solutions are
$$y-k=2at,\hskip{0.5in}  x-h=at^2$$
which leads directly to the solution
$$y=k+2at,\hskip{0.5in}  x=h+at^2$$

Now let's look back and try to understand what's happening.  The main idea is that the "shift" induced by replacing $x$ and $y$ with $x-h$ and $y-k$, respectively, corresponds to a shift in the solution that also replaces $x$ and $y$ with $x-h$ and $y-k$.  The "plus" sign just comes from moving the negative terms to the other side of the equals sign.

Let us try
$$(y-k)^2=4a(x-h)=4a^2t^2,$$
then
$$y=k\color{red}\pm 2at,\quad x=h+at^2.\tag1$$
Formulas $(1)$ take in account that the parabola has two branches.

Therefore, the issue parametric function presents only one branch of parabola.


Would this be a legal move in a Matrix? Here I have a matrix:

\begin{pmatrix}
    b^2c^2 & bc & b+c \\
    c^2a^2 & ca & c+a \\
    a^2b^2 & ab & a+b \\
    \end{pmatrix}

I was wondering can I do the following operation on this:

$C2 \to C2 \times C2$

So it would give me the following:

\begin{pmatrix}
    b^2c^2 & b^2c^2 & b+c \\
    c^2a^2 & c^2a^2 & c+a \\
    a^2b^2 & a^2b^2 & a+b \\
    \end{pmatrix}

This is needed in order to prove the determinant is equal to zero.

You need to think in terms of homogeneous symmetric polynomials' degrees. Subtract $ab+bc+ca$ times the second column from the first, thus changing $b^2c^2$ to $-abc(b+c)$. Now add $abc$ times the third column to the first; it should vanish then. Explicitly$$\left(\begin{array}{c}
b^{2}c^{2}\\
c^{2}a^{2}\\
a^{2}b^{2}
\end{array}\right)-\left(ab+bc+ca\right)\left(\begin{array}{c}
bc\\
ca\\
ab
\end{array}\right)+abc\left(\begin{array}{c}
b+c\\
c+a\\
a+b
\end{array}\right)=\left(\begin{array}{c}
0\\
0\\
0
\end{array}\right).$$

Since this will (for most $a$ and $b$) change the value of the determinant, lets use co-factor multiplication by your matrix $M$ to prove this, which shouldn't be too bad given it is a $3x3$ matrix. I will denote each $2x2$ co factor as $A_i$ for simplicity.
$$det(M) = (b+c)det(A_1) - (c+a)det(A_2) + (a+b)det(A_3)$$
$$det(M) = (b+c)(a^3c^2b-a^3b^2c)-(c+a)(c^3b^2a-c^3a^2b)+(a+b)(b^3c^2a -b^3a^2c)$$
Hmm... maybe this wasn't the right approach but whatever too late now
$$det(M) = (a^3c^2b^2-a^3b^3c+a^3c^3b-a^3b^2c^2)-(b^3c^3a-b^3a^2c^2+b^3c^2a^2-b^3a^3c)+(c^3b^2a^2-c^3a^3b+c^3b^3a-c^3a^2b^2)$$
Jesus christ... notice that we can take out a $a^3, b^3$ or $c^3$
$$det(M) = a^3(c^2b^2-b^3c+c^3b-b^2c^2)-b^3(c^3a-a^2c^2+c^2a^2-a^3c)+c^3(b^2a^2-a^3b+b^3a-a^2b^2)$$
And then do even better by noticing we can take out a $ab, bc$ or $ac$
$$det(M) = a^3bc(cb-b^2+c^2-bc)-b^3ca(c^2-ac+ca-a^2)+c^3ba(ba-a^2+b^2-ab)$$
Now some terms cancel... thank god....
$$det(M) = a^3bc(-b^2+c^2)-b^3ca(c^2-a^2)+c^3ba(-a^2+b^2)$$
Rearrange some more..
$$det(M) = -a^3bc(b^2-c^2)-b^3ca(c^2-a^2)-c^3ba(a^2-b^2)$$
Factor out a $-abc$
$$det(M) = -abc(a^2(b^2-c^2)+b^2(c^2-a^2)+c^2(a^2-b^2))$$
Distribute
$$det(M) = -abc((a^2b^2-a^2c^2)+(b^2c^2-b^2a^2)+(c^2a^2-c^2b^2))$$
Reorder terms
$$det(M) = -abc(a^2b^2-b^2a^2-a^2c^2+c^2a^2+b^2c^2-c^2b^2)$$
Thus
$$det(M) = -abc(0)$$
$$det(M) = 0$$
I retract my former statement that this wouldn't be too bad, but luckily I denoted the co factors as $A_i$, saving me loads of time.


$\lim_{(x,y)\to(0,0)} (y-\sin(y))/(x^2+y^2)$ $$
f(x,y) = \begin{cases} 0 & \text{if } (x,y)=(0,0), \\[6pt]
\dfrac{y-\sin y}{x^2+y^2} & \text{otherwise.} \end{cases}
$$

is $\lim\limits_{(x,y)\to(0,0)} \dfrac{y-\sin y}{x^2+y^2} = 0 \text{ ?}$

according to wolfarm no but can someone show me why?

I'm not familiar with functional analysis, but this question seems to be just a typical limit.

We will show the limit is $0$.

Observe that
$$\left|\frac{y-\sin y}{x^2+y^2}\right|=\frac{\left|y-\sin y\right|}{x^2+y^2}\le\frac{\left|y-\sin y\right|}{y^2}$$ so it is enough to prove that $\lim_{y\to 0}\frac{\left|y-\sin y\right|}{y^2}=0$. This can be done with the well known inequality $\frac{t^3}{3!}\ge t-\sin t$ for $t\ge 0$, or $\frac{|t|^3}{3!}\ge |t-\sin t|$ for all $t\in\mathbb{R}$.

Now, One of the clasical method for obtaining the limit of function with two variables in $(0,0)$ is by this fact that you can
assume $x=r \, \cos(\theta)$ and $y=r \, \sin(\theta)$, then calculate the limit with new values for $x$ and $y$ when $r \to 0$, if after calculation limit, the final values of limit depend on to $\theta$, we conclude that, the function dose not have limit in $(0,0)$ but if the final values of limit be independent of $\theta$, we say the function has limit in $(0,0)$ with value that we obtained. so we have

$$\lim_{(x,y) \to (0,0)}\,\dfrac { y-\sin(y) } {x^{2}+y^{2} }
=\lim_{(r) \to (0)}\frac{r\,\sin(\theta)-\sin(r\,\sin(\theta))}{r^2}
$$
$$
\lim_{(r) \to (0)}\frac{r\,\sin(\theta)-\sin(r\,\sin(\theta))}{r^2}
=\lim_{(r) \to (0)}\frac{\frac{1}{3!}r^3\sin^3(\theta)}{r^2}=
\lim_{(r) \to (0)}\frac{1}{6}r\sin^3(\theta)=0
$$
that means your limit exist in $(0,0)$.


3D software like GeoGebra Does it exist a free interactive geometry software, like GeoGebra, which works for 3D geometry? I would be able to draw spheres, great circles, and so on.

GeoGebra 5.0 (with 3D) has been released:

https://www.geogebra.org/

If you're on a mac, theres a plotting software build in.
Go to Applications / Utilities / Grapher.
It's easy and powerful, can plot 2D and 3D graphs.


Finding Prime Number I have to show that for any prime p , there are either infinitely many or no positive integer a , or no positive integer a so that 6p divides $ a^p+1 $ . I have to find all those primes for which there exists no solution . How can I solve this problem ?

Firstly, note that by unique factorization, for any given $p\gt 3$, saying that $6p|a^p+1$ is the same as saying that $6|a^p+1$ and $p|a^p+1$ (the cases $p=2$ and $p=3$ should be considered separately - they're easy to do by hand).

Next, investigate the constraint that $p|a^p+1$: using F$\ell$T, we get $a^p\equiv a\bmod p$ and so $a^p+1\equiv a+1$.  It must be the case that $a+1\equiv 0\bmod p$, or in other words $a=kp-1$.  Contrariwise, if $a\equiv -1\bmod p$, then $a^p\equiv -1\bmod p$ (remember that this assuming $p$ is odd; we handled the case $p=2$ separately!) and so $p|a^p+1$.  Therefore, for any $p$ there are infinitely many $a$ with $p|a^p+1$ : all those with $a\equiv -1\bmod p$.

The other half is similarly straightforward: for $6$ to divide $a^p+1$ then clearly $a$ must be odd (i.e. $a\equiv 1\bmod 2$).  Furthermore, it must be the case that $a^p\equiv -1\bmod 3$; given that $p$ is odd, this is equivalent to saying that $a\equiv -1\bmod 3$.

Putting this all together, there are three conditions: $a\equiv -1\bmod p$, $a\equiv 1\bmod 2$, and $a\equiv -1\bmod 3$.  Now, you can use the Chinese Remainder Theorem to show that such an $a$ must exist for any $p\gt 3$, and in fact that infinitely many such $a$ exist: given some $a$ that satisfies the conditions, then $a+6kp$ does as well for any $k$.

Use Fermat Little Theorem ...

$ \frac{ a^p+1 }{6p}
=(\frac{a^p}{6p} +\frac{1}{6p}) $

Now since $ 6p $ will never divide $ 1 $ and as per Fermat Little Theorem $ a^p=a(mod p) $
where p is prime . Therefore the first term is also not divisible , So no positive integer a exist for which 6p divides $ a^p+1 $.


Verifying differential equations (without substitution or integration) I am aware that several similar threads exist on this forum, however, my particular query is different from any previous question I've seen here. All  other answers have a 'y'term on both sides, and involve substitution as a means of verification. My question is a bit different.

Basically, I want to know how I can verify this answer (without re-integrating it).

Let's say a question was:

  $
\text{Find } \frac{dy}{dx} \text{ for } y = 4x^2 + 3x -6
$

And my answer was: $\frac{dy}{dx} = 8x + 3$.

Is there any way to verify that my answer is correct (without using a calculator or otherwise integrating it)? I.e. any way to check that the derivative of $4x^2 + 3x -6$ is indeed $8x + 3$? If so, I would be most grateful if you could kindly explain the process of verification.

Thanks in advance for your answers!
Vignesh

[Edit] I did state that I needed a way to prove this without using a calculator, or any other similar electronic device. Essentially like in a non-calculator exam. Is there any way to prove this on paper?

You can always resort to definitions and take the limit of the difference quotient

$$f'(x) = \lim_{h \to 0} \frac{4(x + h)^2 + 3(x + h) - 6 - (4x^2 + 3x - 6)}{h}$$

and check that it does indeed give $f'(x) = 8x + 3$, although you're far more likely to make an error here (and spend a nontrivial amount of time), compared to using the basic differentiation rules for polynomials carefully.

Without integration, I don't really know of any other methods to verify that your derivative is without a doubt correct.

For something like this quadratic, you can get evidence by solving $f'(x) = 0$ and verifying that this happens at the $x$-coordinate of the vertex $\frac{-b}{2a}$, but this doesn't extend very well to other kinds of functions. It also doesn't guarantee a correct answer, it will just (potentially) show you if you're wrong.

If you just need to quickly check your work, you can plug it into
Wolfram Alpha.

If you want, you can do this in either Mathematica

  D[4*x^2+3*x+6,x]

or Python's SymPy package

from sympy import *
x = Symbol('x')
diff(4*x**2 + 3*x + 6, x)


If $a^b\equiv c^b\mod p$, can we conclude that $a\equiv c\mod p$? If $a^b\equiv c^b\mod p$, is is true that $a\equiv c\mod p$, where $b$ is odd and $p$ is prime?

We know that if $a\equiv c\mod p$, then $a^b\equiv c^b\mod p$. Is the reverse true?

As the other answers pointed out, the reverse is generally false.

However, if $b$ and $p-1$ are coprime, then $a^b \equiv c^b \bmod p$ implies $a \equiv c \bmod p$.

Proof. If $a \equiv 0 \bmod p$ or $c \equiv 0 \bmod p$ the result is clear (because $p$ is prime), so we will assume that $a \not\equiv 0 \bmod p$ and $c \not\equiv 0 \bmod p$. In this case, Fermat's little theorem shows that $a^{p-1} \equiv 1 \bmod p$ and $c^{p-1} \equiv 1 \bmod p$.

Therefore, for all integers $u,v$, the relation $a^b \equiv c^b \bmod p$ implies
$$
(a^b)^u(a^{p-1})^v \equiv (c^b)^u(c^{p-1})^v \bmod p
$$
By Bezout's theorem we can find integers $u,v$ such that $bu + (p-1)v = 1$, so what we have actually proven is $a^1 \equiv c^1 \bmod p$.

The condition that $b$ and $p-1$ are coprime is actually necessary and sufficient if $p$ is a prime number.

Indeed, if $b$ and $p-1$ are not coprime, we can find integers $b',k$ and $d \geq 2$ such that $b=b'd$ and $p-1 = kd$. Then for all $x$ which is a generator of the cyclic group $(\Bbb Z/p\Bbb Z)^\times$, the integers $a = x^k$ and $c = 1$ will satisfy
$$
a^b \equiv x^{kb'd} \equiv (x^{p-1})^{b'} \equiv 1 \equiv c^b \bmod p
$$
but $a \not\equiv c \bmod p$ because $x^k \equiv 1 \bmod p$ would imply $(p-1) \mid k$, which contradicts $0 < k < p-1$.

$2^3=1^3 \mod 7$ and  $2 \neq 1\mod 7$ but $(2)=(1)$ as ideals since we are in a field. So if you mean $(a), (b)$ as ideals the assertion is true since we have only one nonzero ideal in a field. But if you mean $(a)=(b)$ as elements there are counter examples.


How do you show one way equivalences in mathematics? In the real world, it seems you can often take a set of things and create new things with them. Let's say you want to make bread, for example. Normally, you can create bread by putting together flour, milk, eggs, and a few other ingredients, and cooking.

It would seem, that this could be written as:

  $flour+milk+eggs+(other ingredients)=bread$

I believe there is a problem with this though, in that you can't create milk or flour (in any practical way) from bread. It just wouldn't work.

So, in specific to that problem, I'm wondering of there is a way to identify one way equivalences in mathematics. That is $x$ and $y$, such that $x$ can turn into $y$, but $y$ cannot turn into $x$. I guess that can lead to the more complicated issue of causation, which I'm also wondering about.

I'd like to expand upon my comment because this is a very interesting question.

First, I think the word "causation" is really throwing people off. Causation does have a specific meaning that's been modeled in various ways (I'm thinking statistics and mathematical logic), but none of the ways I know about capture what (I think) you're after.

The link I posted in my comment is to a post about "Resource Convertibility," which really perfectly captures what you've been thinking about, or at least the example you gave. I don't know much about it, but it's fun to think about, so I'll try to say a little bit.

You've got a very interesting idea that surprisingly intersects with some current research!

Unfortunately, it may turn out that equality has connotations that are inappropriate, as you've already seen:

  $$\rm flour\ + milk\ +eggs\ + (other\ ingredients) = bread$$

is indeed something of an unfortunate notation, because equality is symmetric and as you've pointed out, our conversion here is really one way.

Tobias Fritz has evidently been thinking about this, and decided that an inequality was really the way to go: It makes more sense to write

$$\rm flour\ + milk\ +eggs\ + (other\ ingredients) \ge bread$$

and think of this statement as something like "having flour, milk, eggs, and other stuff is at least as good as having bread" (highly paraphrased from link above). It's also probably best to avoid causation and speak strictly about the ability to convert the things on the left to the things on the right (as the act of bringing milk and eggs together certainly doesn't cause bread to form of its own volition!).

The key features of his formulation are that

You have some means of "comparing" objects, using $\ge$.

Everything is comparable to itself (i.e., $\rm bread \ge bread$).
Comparisons are transitive. For example if $\rm bread\ ingredients \ge bread$, and $\rm bread \ge toast$, then $\rm bread\ ingredients \ge toast$ (just bake the ingredients into bread, then slice and toast!).
Finally, antisymmetry (think $x \ge 3$ and $3 \ge x$ means $x = 3$) is used as a means to decide what resources are equivalent. We have, for example (by visiting one's favorite financial institution and trading cash for different cash),

$$\rm five\ \$1\ bills \ge one\ $5\ bill \qquad and \qquad one\ $5\ bill \ge five\ \$1\ bills.$$

We also have the ability to add objects, as in the $\rm flour + milk + \ldots$ example (but the $+$ really just lets us build a shopping cart from what I can tell; adding doesn't convert anything, that's all encapsulated in $\ge$).

With these abilities, you get to say you're studying fancy things called ordered commutative monoids, and you can read much much more in the series of posts linked above, as well as the paper that ensued (there's a lot of fancy notation and theorems, but there's some value to skimming at least portions of the paper, if you find the blog series interesting enough).

Has any of this turned out to be useful, from a practical standpoint? I have no idea! But John Baez (the person running the blog) as been involved in, and popularized, efforts to build a framework to talk and think about network theory: Chemical reactions (think $H + O \ge H_2O$), birth-death processes, resource conversion, etc. It turns out that classical mathematics can only say so much about these subjects.

This perspective on Resource Convertibility is just one of many efforts to find a good framework.

I think the mathematical concept you want is implication.

For example, think about mathematical equations that only work sometimes. How about $x^2=x$. If you plug in a random value of $x$, that equation isn't true. But it happens to work when $x=1$. Written mathematically,

$$ x = 1 \quad \Longrightarrow \quad x^2 = x$$

The symbol $\Longrightarrow$ is read as implies, or as the then part of an if-then statement. You can also think about it as meaning mathematical causation -- e.g. the fact that $x=1$ causes the equation $x^2=x$ to be true. This arrow doesn't always go both ways. If it's true that $x^2=x$, that doesn't imply that $x=1$. (It could instead be true that $x=0$.)

Warning: there are some problems with this perspective, namely that in mathematical logic, any false statement implies any statement you want. This is worth thinking about some more!


Quasilinear PDE definition? Quasilinear PDE definition?

Here it's written that:

  Definition 3: A partial differential equation is said to be
  quasilinear if it is linear with respect to all the highest order
  derivatives of the unknown function.

What does mean to be linear with respect to all the highest order derivatives?

Does it mean that the terms containing the highest order derivatives are linear, but there might be other terms that are nonlinear?

Yes, that's it. The term
$$\frac{\partial u}{\partial y}\frac{\partial ^2u}{\partial y^2}$$
is linear with respect to $\frac{\partial ^2u}{\partial y^2}$, but the term
$$\left(\frac{\partial ^2u}{\partial y^2}\right)^2$$
is not linear.

If you have a second order derivative, it does not matter if the equation has a term like $u^2$ to be quasilinear. You just need to check the highest order derivatives (second order ones in the examples of the link).

The simplest definition of a quasi-linear PDE says:

  A PDE in which at least one coefficient of the partial derivatives is really a function of the dependent variable(say u).

For example,

$\mathrm{\frac{\partial^2u}{\partial x_1^2} +u\frac{\partial^2u}{\partial x_2^2} =0}$


How to prove that $ \sin \angle{GAB}+\sin \angle{GBC}+\sin \angle{GCA} \le \frac{3}{2} $ for a triangle $ABC$ with centroid $G$? Let $ G $ be the centroid of $ \triangle ABC $ , such that  $ \measuredangle{GAB}=x,\measuredangle{GBC}=y,\measuredangle{GCA}=z $.

How do I prove  that :
$$  \sin x +\sin y +\sin z\le \frac{3}{2} $$

With little effort, one can prove that $$\sin x = \frac{1}{\sqrt{1+(2\cot\alpha+\cot\beta)^2}}$$, where $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$. It's also easy to check that the cotangents of angles of any triangle satisfies the following: $$\cot\alpha\cot\beta+\cot\beta\cot\gamma+\cot\gamma\cot\alpha = 1$$. Therefore, the problem is equivalent to the following inequality:
$$x,y,z\in\mathbb{R} : xy+yz+zx=1\rightarrow\sum_{x,y,z}\sqrt\frac{1}{1+(2x+y)^2}\leq\frac{3}{2}$$
Now, while I believe that an avid contest-type inequality solver could find an elegant solution to above, it can also be solved by Lagrange multiplier method.
To begin with, let's make the following substituion $$(2x+y, 2y+z, 2z+x)= (a,b,c)\leftrightarrow (x,y,z) = (\frac{4a-2b+c}{9},\frac{4b-2c+a}{9},\frac{4c-2a+b}{9})$$
Now our problem is equivalent to $$-2(a^2+b^2+c^2)+5(ab+bc+ca) = 27\rightarrow \sum_{a,b,c}\frac{1}{\sqrt {1+a^2}}\leq\frac{3}{2}$$.
Now it's a routine exercise of Lagrange multiplier method that yields the inequality is attained where $a=b=c=3x=3y=3z$, which means our triangle must be equilateral for this to hold.

There is a theorem in geometry telling the point $G$ is inside the triangle.

So , Join it to $A , B$ and $C.$

It is obvious that $x + y + z < 90$

By using lagrange extreme theorem for multi variable functions, Find the maximum

of $\sin x + \sin y + \sin z$ with the condition $x + y + z < 90$

Its maximum is $\frac{3}{2}$


Algorithm for determining whether two imaginary quadratic numbers are equivalent under a modular transformation Let $\Gamma = SL_2(\mathbb{Z})$.
Let $\mathcal{H} = \{z \in \mathbb{C}\ |\ \mathcal{Im}(z) > 0\}$ be the upper half complex plane.
Let $\sigma = \left( \begin{array}{ccc}
p & q \\
r & s \end{array} \right)$ be an element of $\Gamma$.
Let $z \in \mathcal{H}$.
We write $$\sigma z = \frac{pz + q}{rz + s}$$
It is easy to see that $\sigma z \in \mathcal{H}$ and $\Gamma$ acts on $\mathcal{H}$ from left.

Let $\alpha \in \mathbb{C}$ be an algebraic number.
If the minimal plynomial of $\alpha$ over $\mathbb{Q}$ has degree $2$, we say $\alpha$ is a quadratic number.
There exists the unique polynomial $ax^2 + bx + c \in \mathbb{Z}[x]$ such that $a > 0$ and gcd$(a, b, c) = 1$.
$D = b^2 - 4ac$ is called the discriminant of $\alpha$.
Since $D \equiv b^2$ (mod $4$), $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).
Conversly suppose $D$ is a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). Then there exists a quadratic number $\alpha$ whose discriminant is $D$.

Let $D$ be a negative non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).
We denote by $\mathcal{H}(D)$ the set of quadratic numbers of discriminant $D$ in $\mathcal{H}$.
By this question, $\mathcal{H}(D)$ is $\Gamma$-invariant.

My question
Let $\alpha, \beta$ be explicitly given elements of $\mathcal{H}(D)$.
Is there algorithm for solving the following problems?
If yes, what is it?

Determine whether there exists $\sigma \in \Gamma$ such that $\alpha = \sigma \beta$.
If there exists such $\sigma$, determine the components of the matrix $\sigma$.

Remark
My motivation for the above question came from this and this.

We do not need the assumption that $\alpha$ and $\beta$ are quadratic. We can do the test for any two points in the upper half plane.

The crucial concept is the fundamental domain: the region $F$ in which a point $z$ satisfies $-1/2\leq\text{Re}(z)<1/2$ and $|z|>1$ or $|z|=1$ and $\text{Re}(z) \leq 0$. Points in $F$ cannot be mapped onto each other by modular transformations, apart from $z=i$ which can be mapped onto itself. However, the images of $F$ under modular transformations cover the upper half plane. This is a standard construction: http://en.m.wikipedia.org/wiki/Modular_group

A strategy for testing whether two numbers are equal modulo a modular transformation is to map them both into $F$. In the remainder of this answer I will define such a function.

The transformations $z \mapsto z+1$ and $z \mapsto -1/z$ are modular transformations and in fact generate the group. I will show that iterating $z \mapsto g(f(z))$ maps any point into $F$ in a finite number of steps, where:

$$f(z) = z - \lfloor \text{Re}(z)+1/2 \rfloor$$
$$g(z) = z \text{ if } z\in F \text{ else } -1/z$$

Suppose $\text{Im}(z)<1/3$. Let $w=f(z)$. Then $-1/2\leq\text{Re}(w)<1/2$, and the imaginary part is unchanged, so $|w| < \sqrt{13/36}$. Let $z'=g(w)$. Then $\text{Im}(z')>(36/13)\text{Im}(z)$. Therefore, we need to iterate the mapping $z \mapsto z'$ at most $-\log_{36/13}(\text{Im}(z))$ times to ensure $\text{Im}(z) \geq 1/3$.

From this point two further iterations of $z \mapsto g(f(z))$ suffice to map $z$ into $F$, as I will now show.

After $z \mapsto f(z)$, we also know $|\text{Re}(z)| \leq 1/2$. It's then possible that $z \in F$ and we're finished. If not we also know $|z| \leq 1$. Those four bounds describe a shape with four sides: two circles and two straight lines.

After $z \mapsto -1/z$ (the action of $g$ in this case) we have a different shape. The first bound ($\text{Im}(z) \geq 1/3$) maps to $|z-3i/2| \leq 3/2$. That's a circle of radius $3/2$ centred on a point with zero real part. We can weaken it to $|\text{Re}(z)| \leq 3/2$. The second and third bound ($|\text{Re}(z)| \leq 1/2$) map to $|z-1| \geq 1$ and $|z+1| \geq 1$. The fourth bound ($|z| \leq 1$) maps to $|z| \geq 1$.

So now we have a shape with five sides. If you draw it you will see it looks like three copies of $F$ side by side. Applying $f$ maps them all onto the central copy and ensures that $\text{Im}(z) \neq +1/2$. Applying $g$ one last time ensures that if $|z| = 1$ then $\text{Re}(z) \leq 0$.

So for an arbitrary starting point in the upper half plane, we have constructed a modular transformation that maps it into $F$. If we do this for two points, we can see if they map to the same point in $F$. If so, we can map them onto each other, otherwise it is impossible.

Let $F = \{z \in \mathcal{H}\ |\ -1/2 \le Re(z) \lt 1/2, |z| \gt 1$ or $|z| = 1$ and $Re(z) \le 0\}$. It is well-known that any two distincts points of $F$ are not equivalent under the action of $\Gamma$.
Let $S = \left( \begin{array}{ccc}
0 & -1 \\
1 & 0 \end{array} \right)$,
$T = \left( \begin{array}{ccc}
1 & 1 \\
0 & 1 \end{array} \right)$.
Let $z \in \mathcal{H}$.
We will show that applying $S, T$ or $T^{-1}$ consecutively on $z$, we can map $z$ into $F$.
Hence we can find $\sigma \in \Gamma$ such that $\sigma z \in F$.
We consider the following algorithm.

Algorithm 1

$z = T^{-\lfloor \text{Re}(z)+1/2 \rfloor}(z)$, where $\lfloor a \rfloor$ denotes the greatest integer $n$ such that $n \le a$.
If $|z| \gt 1$, then stop.
If $|z| \lt 0$, then $z = S(z)$ and go to 1.
If $|z| = 1$, then go to 3.
If $-1/2 \le Re(z) \le 0$, then stop.
If $0 \lt Re(z) \lt 1/2$, then $z = S(z)$ and stop.

If the algorithm terminates, then $z \in F$.
We will prove that it terminates in finite steps.

Lemma 1
$S(\{z \in \mathcal{H}\ |\ Im(z) \ge 1/3\}) = \{z \in \mathcal{H}\ |\ |z - 3i/2| \le 3/2\}$.

Proof:
Let $w = -1/z$.
Then $z = -1/w$.
Since $Im(z) = (z - \bar z)/2i$, $Im(z) = (-1/w + 1/\bar w)/2i$.
Hence $(-\bar w + w)/2i \ge w\bar w/3 = |w|^2/3$.
Let $w = x + yi$.
Then $y \ge (x^2 + y^2)/3$.
Hence $x^2 + y^2 - 3y \le 0$.
Hence $x^2 + (y - 3/2)^2 \le (3/2)^2$.
QED

Lemma 2
$S(\{z \in \mathcal{H}\ |\ Re(z) \lt 1/2\}) = \{z \in \mathcal{H}\ |\ |z + 1| \gt 1\}$.

Proof:
Let $w = -1/z$.
Then $z = -1/w$.
Since $Re(z) = (z + \bar z)/2$, $Re(z) \lt 1/2$ implies $-1/w - 1/\bar w \lt 1$.
Hence $-\bar w - w \lt |w|^2$.
Let $w = x + yi$.
Then $-2x \lt x^2 + y^2$.
Hence $(x + 1)^2 + y^2 \gt 1$.
QED

Lemma 3
$S(\{z \in \mathcal{H}\ |\ Re(z) \ge -1/2\}) = \{z \in \mathcal{H}\ |\ |z - 1| \ge 1\}$.

Proof:
Let $w = -1/z$.
Then $z = -1/w$.
Since $Re(z) = (z + \bar z)/2$, $Re(z) \ge -1/2$ implies $-1/w - 1/\bar w \ge -1$.
Hence $\bar w + w \le |w|^2$.
Let $w = x + yi$.
Then $2x \le x^2 + y^2$.
Hence $(x - 1)^2 + y^2 \ge 1$.
QED

Proposition 1
The algorithm terminates in finite steps.

Proof(based on the idea of apt1002):

Suppose $Im(z) \lt 1/3$ in the first place.
After executing step 1, $-1/2 \le Re(z) \lt 1/2$.
Hence $|z|^2 \lt 1/4 + 1/9 = 13/36$.
Then $Im(S(z)) = Im(z)/|z|^2 \gt (36/13)Im(z)$.
Since $\text{lim}_{n\rightarrow \infty} (36/13)^nIm(z) = \infty$,
after executing finite steps, the algorithm terminates or $Im(z) \ge 1/3$.
Hence we may suppose $Im(z) \ge 1/3$ in the first place.

After executing step 1, if $|z| \ge 1$, the algorithm terminates.
Hence we may suppose $|z| \lt 1$.
Let $w = -1/z$.
Then $|w| \gt 1$.
By Lemma 1, Lemma 2 and Lemma 3, $|Re(w)| \le 3/2, |w + 1| \gt 1, |w - 1| \ge 1$.

Case 1: $-3/2 \le Re(w) \lt -1/2$

Let $z = T^{-\lfloor \text{Re}(w)+1/2 \rfloor}(w) = w + 1$.
Then $-1/2 \le Re(z) \lt 1/2$.
Since $|z| \gt 1$, the algorithm terminates.

Case 2: $-1/2 \le Re(w) \lt 1/2$

Since $|w| \gt 1$, the algorithm terminates.

Case 3: $1/2 \le Re(w) \lt 3/2$

Let $z = T^{-\lfloor \text{Re}(w)+1/2 \rfloor}(w) = w - 1$.
Then $-1/2 \le Re(z) \lt 1/2$.
Since $|z| \ge 1$, the algorithm terminates.

Case 4: $Re(w) = 3/2$

Let $z = T^{-\lfloor \text{Re}(w)+1/2 \rfloor}(w) = w - 2$.
Then $Re(z) = -1/2$.
Since $Im(z) = Im(w) \ge \sqrt 3/2$, $|z| \ge 1$.
Hence the algorithm terminates.
QED


Two variable function on an unbounded set I have the following function

$$f(x,y) = 3x^2 +5y^2 +2y -4x -2xy + 1$$

I know that $f$ has a critical point in $x=9/14$ and $y=-1/14$. Also, I know that it will be a local minimum. How do I know (or check) that this value is a global minimum, that is, the lowest value of this function?

The domain is unbounded therefore I cannot apply Weierstrass Theorem. Do I count the limits? If yes, which of them? Is there a way to check that this function has a global minimum without even finding a local one?

$f_x = 6x-4 - 2y = 0 = f_y=10y+2-2x \implies 3(10y+2)-4-2y=0\implies28y=-2\implies y = -\dfrac{1}{14}\implies x = \dfrac{9}{14}$ . Thus $\left(\dfrac{9}{14}, -\dfrac{1}{14}\right)$ is the only critical point that is a local minima as confirmed by you. Hence this is also the global minima since the domain is an open unbounded set.


How to solve differential equation with one differential term I know how to solve a second order differential equation of the form
$y''P(x)+y'Q(x)+yR(x)=G(x)$. So when I come across equations of the form
$\frac{d^2y}{dx^2}=y$  I use the same method of substituting for $y$ with $e^{rx}$ however that does not get me the answer. I was wondering if there is another way to do it.

Edit:
I also would like to know how an equation of the form $\frac{d^2y}{dx^2}=x$ can be solved as well.

Assume $y = e^{rx}$ then this yields
$$r^2e^x = e^x$$$$r^2 = 1$$
$$r= \pm1$$

Since both solutions are possible then the general solution for $y$ is any linear combination of the $2$ solutions
$$y = c_1e^{-x} + c_2e^{x}$$

As requested by OP I'll solve the differential equation

$$\frac{d^2y}{dx^2} = x$$
Method $1$

We can use the same approach above. We first begin by solving for the homogeneous solution
assume $y = e^{rx}$ and then
$$r^2 = 0 \implies r = 0$$
Which is a repeated root.
Thus the homogeneous solution is
$$y_h = c_1x + c_2$$ We then find a particular solution by undetermined coefficients or whatever method you would like and you'll find that
$$y_p = \frac{x^3}{6}$$
So
$$y = \frac{x^3}{6} + c_1x + c_2$$

Method $2$

Use a substitution Let $w = \frac{dy}{dx}$ then
$$\frac{dw}{dx} = x$$
This can be solved by separation of variables and you would find that
$$w = \frac{x^2}{2} + c_1 \implies \frac{dy}{dx} = \frac{x^2}{2} + c_1$$
$$\implies y = \frac{x^3}{6} + c_1x + c_2$$

Let dy/dx=D

Then your problem $d^2y/dx^2-y=0$ becomes $(D^2-1)$$y$ = $0$
and hence $D^2-1=0$ implies $D= +-1$
hence the solution will be $y=c_1e^{-x} + c_2e^{x}$.
Hope you can solve it after that for $c_1$ and $c_2$.


Need Critique on my solution. Convergence in probability of a product of sequences of random variables Came across this problem in my self education. Found 4 solution here, but none looked simple enough to me... So I cooked up one of my own. Now it looks too simple :) Am I missing something?
I would really appreciate any constructive critique on my solution

Let me summarize the issues we already discussed (in the comments):

You cannot apply  the theorem of boundedness of convergent sequences since this would require that $X_n(\omega)$ converges for each $\omega \in \Omega$ (i.e. almost sure convergence) - but here only convergence in probability is assumed. In fact, it is not difficult to construct a sequence $(X_n)_{n \in \mathbb{N}}$ which converges in probability but is unbounded.
At the end of the proof you use $$\mathbb{P} \left( |Y_n-Y| > \frac{\epsilon}{2M} \right) \to 0 \quad \text{as} \, \, n \to \infty$$ for a random variable $M$. Mind that convergence in probability does only imply $$\mathbb{P} \left( |Y_n-Y| > \frac{2\epsilon}{c} \right) \to 0$$ for some constant $c>0$ (by "constant" I mean that it does not depend on $\omega$).

Here is a sketch how to fix your proof:

Using a very similar argumentation as in your question, we can show that

$$\mathbb{P}(|X_n Y_n-XY| \geq \epsilon) \leq \mathbb{P}\left(|Y_n| \cdot |X_n-X| > \frac{\epsilon}{2} \right) + \mathbb{P} \left( |X| \cdot |Y_n-Y| > \frac{\epsilon}{2} \right) =: I_1+I_2.$$

We estimate the terms separately. For $I_1$ note that

$$\begin{align*} I_1 &\leq \mathbb{P}\left(|Y_n| <R, |Y_n| \cdot |X_n-X| > \frac{\epsilon}{2} \right) + \mathbb{P} \left( |Y_n| \geq R, |Y_n| \cdot |X_n-X| > \frac{\epsilon}{2} \right) \\ &\leq \mathbb{P} \left( |X_n-X| > \frac{\epsilon}{2R} \right) + \mathbb{P}(|Y_n| \geq R) \tag{1}. \end{align*}$$

Now

$$\mathbb{P}(|Y_n| \geq R) \leq \mathbb{P}\left(|Y_n-Y| \geq \frac{R}{2} \right) + \mathbb{P} \left( |Y| \geq \frac{R}{2} \right). \tag{2}$$

If we plug $(2)$ into $(1)$ and let $n \to \infty$, we get

$$\limsup_{n \to \infty} \mathbb{P}\left(|Y_n| \cdot |X_n-X| > \frac{\epsilon}{2} \right)\leq \mathbb{P} \left( |Y| \geq \frac{R}{2} \right).$$

Finally, we can let $R \to \infty$ and conclude

$$\lim_{n \to \infty} \mathbb{P}\left(|Y_n| \cdot |X_n-X| > \frac{\epsilon}{2} \right) = 0.$$

A similar (even easier) argumentation applies to $I_2$. Consequently,

$$\lim_{n \to \infty} \mathbb{P}(|X_n Y_n-XY|  \geq \epsilon)=0.$$

I don't like these sorts of proofs that unnecessarily dive into a sea of $\epsilon$'s. Consider the following alternative characterization of convergence in probability.

Theorem. $X_n \to X$ in probability if and only if for every subsequence of the $X_n$ there is a further subsequence that converges to $X$ almost surely.

See, e.g. Williams: Probability with Martingales exercise EA13.1.

Using this theorem, let's prove your result. Suppose $X_n \to X, Y_n \to Y$ in probability. Let a subsequence $X_{n_k}Y_{n_k}$ of $X_nY_n$ be given. By the theorem take a subsequence and then another subsequence so that we have a subsequence on which $X_{n_{k_\ell}},Y_{n_{k_\ell}}$ both converge almost surely. Then of course $X_{n_{k_\ell}}Y_{n_{k_\ell}}$ converges almost surely to $XY$, so the theorem implies $X_nY_n \to XY$ in probability.

I encourage you to try and prove this theorem, it is not very difficult (though it will require some $\epsilon$'s which are, in my opinion, better spent on the theorem than your result.)


Are there any special rules when making a substitution in an integral? Please consider the integral:

$$\int_{-a}^{a}x^2dx=\frac{2a^3}{3}$$

I would like to know why I can't make the substitution:

$$u=x^2$$

When I make the substitution, the limits of the integral will be the same, and the integral itself will be zero, which is the wrong answer. So why does this simple change of variables not work as I have expected?

Please note that I do not want help solving the integral, I know how to solve it several ways. My question is why does this specific attempt at a solution not work?

This is a great question, and I find that variants are often poorly understood. Let's carry out the substitution completely. Since $a$ clearly doesn't matter, I'll suppose that $a = 1$ for simplicity.

$$ \int_{-1}^1 x^2 dx = \frac{1}{2} \int_{-1}^1 x \cdot 2x dx.$$

In this form, we substitute $u = x^2$, so that $du = 2x dx$. Then in the integral, we have

$$ \frac{1}{2} \int_{-1}^1 \underbrace{x}_{\pm\sqrt u} \cdot \underbrace{2x dx}_{du}.$$

To complete the substitution, we need to substitute $x = \pm \sqrt u$. I write $\pm$ to indicate that sometimes we have $x = \sqrt u$ and sometimes we have $x = - \sqrt u$. In particular, when $x$ is negative, we choose the negative square root, $x = - \sqrt u$. This ambiguity is very important.

Namely, for $x$ in $[-1, 0]$, we have that $x = -\sqrt u$ and for $x$ in $[0, 1]$ we have that $x = \sqrt u$. Then to perform the substitution, we split the integral into these two intervals and handle each separately.

$$\begin{align}
\frac{1}{2} \int_{-1}^1 x \cdot 2x dx &= \frac{1}{2} \int_{-1}^0 x \cdot 2x dx + \frac{1}{2} \int_{0}^1 x \cdot 2x dx \\
&= \frac{1}{2} \int_1^0 (- \sqrt u) du + \frac{1}{2} \int_0^1 \sqrt u du \\
&= \frac{1}{2} \int_0^1 \sqrt u du + \frac{1}{2} \int_0^1 \sqrt u du \\
&= \int_0^1 \sqrt u du = \frac{2}{3} u^{3/2} \bigg|_0^1 = \frac{2}{3}.
\end{align}$$

The exact nature of the substitution is very important. This causes some introductory texts to state that substitutions must be injective. But a careful analysis shows otherwise. $\spadesuit$

I like Alfred's point but I would say it slightly differently.  The $dx$ at the end is not just the “other end” of the integral delimited on the left by $\int$.  It's also not just the indicator of which variable is being integrated.  It's part of the integral itself.  In other words, we don't integrate functions $f(x)$ so much as forms $f(x)\,dx$.

When we make a substitution, we have to substitute fully the entire form from one involving $x$ to one involving $u$.  When integrating $x^2\,dx$, if we substitute $u = x^2$, we must substitute for $dx$ in terms of $u$ and $du$.  Since $\frac{du}{dx} = 2x$, we write $du = 2x\,dx$.  Thus $dx = \frac{1}{2x}\,du = \frac{1}{2\sqrt{u}}\,du$.  So
$$
 \int x^2 \,dx = \int u \frac{1}{2\sqrt{u}}\,du = \frac{1}{2}\int \sqrt{u}\,du
$$
That's different from $\int u\,du$.

I think you were just switching the differential from $dx$ to $du$ without substituting.  You can avoid this mistake if you remember that $dx$ and $du$ are part of the integral.


Equation with expected value of inverse of a binomial variable I need to solve an equation $\Bbb E\frac{1}{a + X} = b$ where $X\sim \mathrm{Bin}(p; n)$ in terms of $p$, where $a,b> 0$ and integer $n$ are given. As far as I understood from this question, the LHS does not admit a particularly elegant closed form, so no big luck inverting it. On the other hand, I rather know an approximate value of $p$ for $n$ large enough. Perhaps, in that case it may make sense to assume that $X$ is almost normal: I have not found any results on how hard is to compute the LHS in that case. Any ideas?

If you don't want to use the LHS closed-form from the answer in the your link, you can in that closed form substitute $Q=(1-p)^n \, _2F_1\left(a,-n;a+1;p/(p-1)\right)$ by some approximation. To make one note that

$$Q^{(m)}(0)=\frac{(m!)^2 \binom{n}{m}}{\underset{i=0}{\overset{m}{\prod }}(a+i)}\quad,\quad Q^{(m)}(1)=\frac{(m!)^2 \binom{n}{m}}{\underset{i=n-m}{\overset{n}{\prod}}(a+i)}$$

It doesn't make sense to consider just $n$ large, because the LHS converge to 0 pointwise except for $p=0$. (I suppose you want $0\leq p\leq1$)

When $p$ is constant, we have that $E\left[\frac{1}{1+X}\right]=O\left(\frac{1}{n\cdot p}\right)$
as $n\rightarrow\infty$ as shown by Cribari-Neto et al. (2000). See also the paper by Chao, M. T., and W. E. Strawderman. “Negative Moments of Positive Random Variables.” Journal of the American Statistical Association, vol. 67, no. 338, 1972, pp. 429–431. JSTOR, www.jstor.org/stable/2284399.


Paradox: Is $1 \in (0,1)$? Consider the set of numbers such that $x \in (0,1)$.

Their decimal expansion is $0.b_0b_1b_2\ldots$, with $b_n \in \{0,1,2,3,4,5,6,7,8,9\}$, and they are not all zero (or else $x = 0$).

Then choose all $b_n = 9$, we have $0.999\ldots = 1$.

But $0.999\ldots = x \in (0,1)$, so $1 \in (0,1)$.

Where did we go wrong?

EDIT:

Right, the answer is that not all $0.b_0b_1b_2 \ldots$ are in $(0,1)$. Here's a follow-up:

So define $x_m = 0.b_0b_1b_2 \ldots$ where all of the $b_n$ are $9$ except $b_m = 8$. Then the limit of $x_m$ an be thought of as the largest element of $(0,1)$. But hey, wait a second... $(0,1)$ is open in the usual topology for $\mathbb{R}$! What have we done now..?

You declared that $x$ had a decimal expansion consisting of the digits $1$ through $9$ without restriction, this is false.

The correct thing to say would be that  their decimal expansion is

$$0.b_{0}b_{1}b_{2}..$$ with $b_{n} \in \{0, 1, 2 ,3,4,5,6,7,8,9 \}$,but not all $0$ and not all $9$.

If the assumption you're making is that 0.999... $\neq$ 1, then this question is pretty much the same as Is it true that $0.999999999\dots=1$?. The number a decimal expansion represents is the infinite sum of the part before the decimal and each digit multiplied by 10 to the power of the negative of its position after the decimal. In the case of the decimal notation 0.999..., 0.9 is $\frac{1}{10}$ away from 1; 0.99 is $\frac{1}{100}$ away from 1. 0.999... is larger than any of the numbers in that sequence so it is exactly 1.

Here, I'm using the assumption that $\mathbb{R}$ is a complete ordered field. How do we know that? How do we define $\mathbb{R}$? To avoid using the unproven assumption that $\mathbb{R}$ is a complete ordered field, we could define a real number as a decimal expansion where 0.999... $\neq$ 1 and say that's how we defined it and define addition, multiplication, and inequality in the intuitive way. Then it wouldn't be a complete ordered field because $(1 - 0.999...) + 0.999... = 0 + 0.999... = 0.999... \neq 1$. I think most mathematicians agree to define $(\mathbb{R}, 1, 0, +, \times, \leq)$ in such a way that $(\mathbb{R}, 1, 0, +, \times, \leq)$ is a complete ordered field and agree that it's unimportant which complete ordered field it's defined to be because $(\mathbb{R}, 1, 0, +, \times, \leq)$ can be explicitly constructed in ZF and shown to be a complete ordered field which is unique up to isomorphism. One way to construct it is to construct the nonnegative real numbers as a decimal expansion and define a notation will a string of trailing 9's to represent the same number as the one that increases the last digit before the trailing 9's by 1 and changes all the 9's after it to a 0 and then stick the - operation onto the beginning of any notation you already defined the meaning of to represent its additive inverse.


Show that $\sum_{k=1}^\infty \frac{\ln{k}}{k^2}$ converges or diverges Show that $\sum_{k=1}^\infty \frac{\ln{k}}{k^2}$ converges or diverges

I know that I must use the Limit Comparison test and my instinct tells me that this series will converge. I cannot, however, think of another series $b_k$, to compare my given series to in order to show that it converges.

Any hints?

Thank you!

Equals the constant in http://oeis.org/A073002 , which gives further links.


Number of pairs of points whose distance is one Let $S$ be a set of $n$ points in the plane, the distance between any two of which is at most one. Show that there are at most $n$ pairs of points of $S$ at distance exactly one.

My attempt on proof:

Construct a graph $G$ in which $2$ vertices are adjacent if and only if their distance is exactly one.
Then graph contains $3$ vertices in the form of equilateral triangle (so, $3$ pairs). I don't know how to generalize this.

Hint 1:

given a point, where is located a point at distance 1 from this point ?

Hint 2:

considering the graph whose vertices are the points, and edges are the pairs of points at distance $1$, if a vertex has degree 3, show that one of its neighbours has degree one.

Hint 3:

once you have a cycle in this graph, can you have an other ?

[Edit, since a full proof was provided and mine was asked for]

I. intersection between unit circles.

If two distinct unit circles $\mathcal{C}_1$ and $\mathcal{C}_2$ intersect, their intersection is made of two points, located on the bisector of the segment joining the centres of the circles. This intersection splits $\mathcal{C}_1$ into two arcs.The arc contained on the side of the bisector which does not contain the center of $\mathcal{C}_1$ is shorter than a semi-circle, and is contained in $\mathcal{C}_2$. The intersection of $k$ unit discs is a convex shape whose boundary is made of arcs from the $k$ circles. Each of these arcs is smaller than a semi-circle, and we will refer to such a shape as a polyarc. Since the discs we will intersect are distinct, there cannot be tangent intersections between their boundaries.

II. If a circle intersects an arc of a polyarc, a corner of the polyarc is outside the circle.

By convexity, the polyarc $\mathcal{P}$ and the circle $\mathcal{C}$ intersect in two points. If $\mathbf{p}_1$ is our intersection, by convexity as well, since all the corners of $\mathcal{P}$ are inside $\mathcal{C}$ the second intersection $\mathbf{p}_2$ is located on the same arc as $\mathbf{p}_1$. The portion of the arc between $\mathbf{p}_1$ and $\mathbf{p}_2$ is shorter than a semi circle, and therefore has to be the portion of the circle defining the arc which lies inside $\mathcal{C}$. Therefore, the two corners of the arc are outside $\mathcal{C}$, which is a contradiction.

III. If a vertex of the graph has valence 3, one of its neighbours has valence 1.

Let us consider $\mathbf{p}_0$ of valence 3, and $\mathbf{p}_1$, $\mathbf{p}_2$ and $\mathbf{p}_3$ its neighbours. Wlog, $\mathbf{p}_2$ is between $\mathbf{p}_1$ and $\mathbf{p}_3$ on the circle centered at $\mathbf{p}_0$. The polyarc resulting from the intersection of the discs of $\mathbf{p}_0$, $\mathbf{p}_1$ and $\mathbf{p}_3$ has 3 corners, and $\mathbf{p}_0$ is one of them. If $\mathbf{p}_2$ has a neighbour $\mathbf{p}_4$, the circle $\mathcal{C}$ centred at $\mathbf{p}_4$ passes through $\mathbf{p}_2$ and cuts an arc of the polyarc at this point. From II, one of the corners of the polyarc lies outside the circle, and this cannot be $\mathbf{p}_0$. Therefore half of the arc containing $\mathbf{p}_2$ is outside $\mathcal{C}$, and either does $\mathbf{p}_1$ or $\mathbf{p}_3$, which is a contradiction.

IV. There can be only one cycle

All the vertices of a cycle have valence at least 2, and are therefore corners of the polyarc. In addition, from II, adding a new vertex can only increase the number of corners of the polyarc by 1. Therefore the polyarc has no other corners than the vertices of the cycle. If a new vertex $\mathbf{p}_1$ is added strictly inside the polyarc and has a neighbour $\mathbf{p}_2$, the circle centered at $\mathbf{p}_2$ passing through $\mathbf{p}_1$ intersects the polyarc and a corner of the polyarc is outside this circle, which is a contradiction. $\mathbf{p}_1$ is therefore located on the boundary of the polyarc, and one of the corners is among its neighbours. Since the corners all have valence at least 2, $\mathbf{p}_1$ has valence 1 and cannot be part of a cycle. There is therefore only one cycle. If we have a cycle, cutting an edge of the cycle transforms the graph into a tree which has at $n-1$ vertices. Otherwise, the graph is a forest and has less than $n$ vertices.

Hint: To think about it another way, you're trying to prove that if there are $\gt n$ pairs of points of $S$ at distance exactly one, then there are at least two points that are greater than one unit away from each other.


Can I shift this integral along the complex plane? Does
$$\int_{-\infty}^\infty \text{e}^{\ a\ (x+b)^2}\ \text dx=\int_{-\infty}^\infty \text{e}^{\ a\ x^2}\ \text dx\ \ \ \ \ ?$$
hold, even if the imaginary part of $b$ is nonzero?

What I really want to understand is what the phrase "By analogy with the previous integrals" means in that link. There, the expression $\frac{J}{a}$ is complex but they seem to imply the integral can be solved like above anyway.
The reusult tells us that the integral is really independend of $J$, which is assumed to be real here. I wonder if we can also generalize this integral to include complex $J$. In case that the shift above is possible, this should work out.
But even if the idea is here to perform that substitution, how to get rid of the complex $a$ to obtain the result. If everything is purely real or imaginary, then this solves the rest of the problem.

Let us write $b= r+it$. The real part of $b$ does not matter as you have already proven yourself. So wlog $r=0$.

For shifting along the imaginary axis, we have to employ the residue theorem. We have
$$
\begin{align}
\int_{-\infty}^\infty f(x+i t) \,dx&- \int_{-\infty}^\infty f(x)\, dx\\
&=\int_{-\infty-it}^{\infty-it} f(x) \,dx- \int_{-\infty}^\infty f(x)\, dx \\
&= 2\pi i \sum \text{Res}(f)+ \int_{\infty-it}^{\infty} f(x) \,dx
- \int_{-\infty-it}^{-\infty} f(x) \,dx
\end{align},$$
where $\sum \text{Res}(f)$ is the sum over the residues of $f$ in the area $z\in \mathbb{C}$ with $-t<\text{Im}\, z<0$.

So the two integrals are the same if there are no residues and if the two integral at $\pm \infty$ vanish (both of which is the case for your example as long as $\text{Re}\,a <0$).

Put
$$(x+b)^2=u^2\Longrightarrow 2(x+b)dx=2udu\Longrightarrow dx= du$$ so we get

$$\int_{-\infty}^\infty e^{a(x+b)^2}dx=\int_{-\infty}^\infty e^{au^2}du$$
and your question's answered in the affirmative.

Added: You may want to divide the integral in two rays $\,(-\infty,0)\,,\,(0,\infty)\,$ and then do the above, to avoid problems with the signs after taking the square root.


Inverse of the sum of matrices I have two square matrices - $A$ and $B$. $A^{-1}$ is known and I want to calculate $(A+B)^{-1}$. Are there theorems that help with calculating the inverse of the sum of matrices? In general case $B^{-1}$ is not known, but if it is necessary then it can be assumed that $B^{-1}$ is also known.

In general, $A+B$ need not be invertible, even when $A$ and $B$ are. But one might ask whether you can have a formula under the additional assumption that $A+B$ is invertible.

As noted by Adrián Barquero, there is a paper by Ken Miller published in the Mathematics Magazine in 1981 that addresses this.

He proves the following:

Lemma. If $A$ and $A+B$ are invertible, and $B$ has rank $1$, then let $g=\mathrm{trace}(BA^{-1})$. Then $g\neq -1$ and
$$(A+B)^{-1} = A^{-1} - \frac{1}{1+g}A^{-1}BA^{-1}.$$

From this lemma, we can take a general $A+B$ that is invertible and write it as $A+B = A + B_1+B_2+\cdots+B_r$, where $B_i$ each have rank $1$ and such that each $A+B_1+\cdots+B_k$ is invertible (such a decomposition always exists if $A+B$ is invertible and $\mathrm{rank}(B)=r$). Then you get:

Theorem. Let $A$ and $A+B$ be nonsingular matrices, and let $B$ have rank $r\gt 0$. Let $B=B_1+\cdots+B_r$, where each $B_i$ has rank $1$, and each $C_{k+1} = A+B_1+\cdots+B_k$ is nonsingular. Setting $C_1 = A$, then
$$C_{k+1}^{-1} = C_{k}^{-1} - g_kC_k^{-1}B_kC_k^{-1}$$
where $g_k = \frac{1}{1 + \mathrm{trace}(C_k^{-1}B_k)}$. In particular,
$$(A+B)^{-1} = C_r^{-1} - g_rC_r^{-1}B_rC_r^{-1}.$$

(If the rank of $B$ is $0$, then $B=0$, so $(A+B)^{-1}=A^{-1}$).

I know the question has been answered multiple times with great answers, but with my answer you don't need to memorize any lemmas or formulas.

Suppose $(A+B)x=y$, then $x=(A+B)^{-1}y$. This is all we need to get. The steps are:

(1) Start with $(A+B)x=y$.

(2) Then $Ax=y-Bx$, so $x=A^{-1}y -A^{-1}Bx$.

(3) Multiply $x$ in step (2) by $B$ to get
 $$Bx=BA^{-1}y -BA^{-1}Bx$$
 which is equivalent to
 $$(I+BA^{-1})Bx=BA^{-1}y $$
 or,
 $$Bx=(I+BA^{-1})^{-1}BA^{-1}y $$

(3) Substitute this $Bx$ into the $x$ in step (2) to get
 $$x=A^{-1}y -A^{-1}(I+BA^{-1})^{-1}BA^{-1}y $$

(4) Now factorizing the $y$ gives you the required result.
 $$x=(A^{-1} -A^{-1}(I+BA^{-1})^{-1}BA^{-1})y $$

(5)The assumptions we have used are $A$ and $I+BA^{-1}$ are nonsingular.

(6) We can factorize the $A^{-1}$ to get:
$$(A+B)^{-1}=A^{-1}(I -(I+BA^{-1})^{-1}BA^{-1})$$


Why aren't logarithms defined for negative $x$? Given a logarithm is true, if and only if, $y = \log_b{x}$ and $b^y = x$ (and $x$ and $b$ are positive, and $b$ is not equal to $1$)[1], are true, why aren't logarithms defined for negative numbers?

Why can't $b$ be negative? Take $(-2)^3 = -8$ for example. Turning that into a logarithm, we get $3 = \log_{(-2)}{(-8)}$ which is an invalid equation as $b$ and $x$ are not positive! Why is this?

For the real, continuous exponentiation operator -- the used in the definition of the real, continuous logarithm -- $(-2)^3$ is undefined, because it has a negative base.

The motivation stems from continuity: If $(-2)^3$ is defined, then $(-2)^{\pi}$ and $(-2)^{22/7}$ should both be defined as well, and be "close" in value to $(-2)^3$, because $\pi$ and $22/7$ are "close" to 3. But the same line of reasoning says that $(-2)^{22/7}$ should be  $8.8327...$, which isn't very close to $-8$ at all, and what could $(-2)^{\pi}$ possibly mean?!?!

You can define a "discrete logarithm" operator that corresponds to the discrete exponentiation operator (i.e. the one that defines $(-2)^3$ to be repeated multiplication, and thus $8$), but this has less general utility -- my expectation is that one would be much better off learning just the continuous logarithm and dealing with the signs "manually" in those odd cases where you want to solve questions involving discrete exponentiation with negative bases. (and eventually learn the complex logarithm)

In $\log_{−2}(8) = 1/2$, the way to solve it normally is to write: $(-2)^{1/2} = 8$.
But in this case, $(-2)^{1/2}$ is undefined. Hence, negative bases for log cannot be used.


Subspace of a topological space with cofinite topology I'm thinking about topological spaces with a cofinite topology $S$ and I proved a subspace ($M$) of $S$ with a relative topology has a discrete topology if $M$ is finite. Then what if $M$ is infinite?

I tried $\mathbb{Z}$ with a cofinite topology and its subset $\mathbb{N}$ and I think $\mathbb{N}$ has a cofinite topology as its relative topology too. Is this correct?

You are correct.  The open sets of $\mathbb{Z}$ under the cofinite topology are the $U$ such that $\mathbb{Z} \setminus U$ is finite.  The open sets of $\mathbb{N}$ under the subspace topology are of the form $\mathbb{N} \cap U$ for some open subset $U$ of $\mathbb{Z}$; so will $\mathbb{N} \setminus (\mathbb{N} \cap U)$ be finite?  Of course.  If it were infinite, then $U$ would not have finite complement in $\mathbb{Z}$, a contradiction!

This means every open set in $\mathbb{N}$ w.r.t. the subspace topology is cofinite in $\mathbb{N}$.  Now we need to establish that every cofinite subset of $\mathbb{N}$ appears as an open set in the subspace topology:

Suppose a subset $V$ is cofinite in $\mathbb{N}$.  This corresponds to a cofinite subset of $\mathbb{Z}$ that intersects with $\mathbb{N}$ to give $V$, namely $V \cup \{x \in \mathbb{Z} \ | \ x \leq 0 \}$.  So $V$ is open in $\mathbb{N}$ under the subspace topology.

We have thus established equality between the sets of open sets under the two topologies.

A subspace of cofinite space is cofinite. Let $(X,T)$ be cofinite and $(Y,S)$ be subspace. $S$ is cofinite if $Y \cap G$ has finite complement in $Y$. Complement of ($Y \cap G$) in $Y = Y \cap (\text{complement of }G)$, which is finite. Thus $(Y,S)$ is a cofinite space. [Note that complement of $G$ is finite].


General Topology, Showing that if $A$ is open then its supremum is not in $A$ unless $A$ is closed Prove: Show that if $A$ is open the sup($A$) is not in $A$ and it is if $A$ is closed.

Definitions: A set $A\in \mathbb{R}$ is open if and only each of its points is an interior point. A point $p\in A$ is a interior point of $A$ if and only if $p$ belongs to some open interval $S_{p}$ which is contained in $A$. A set $A$ is closed if its compliment is open or we could say $A$ is closed if and only if $A$ contains  each of its points of accumulation. A point $p\in \mathbb{R}$ is an accumulation point if and only if every open set $G$ containing $p$ contains a point of $A$ different from $p$

proof: Let $A$ be the set of the real numbers. Let $p\in A$ be an interior point of $A$. Then, there exists an open interval $N(p,\epsilon)$ such that $$p\in N(p,\epsilon)\subset A$$ Now, since $A$ is open, then each points of $A$ are an interior point. Now, let $A$ be a partially ordered set $X$ and suppose $a = $ sup($A$)$\in A$, then there exists an $m\in x$ such that $x\leq m$ $\forall x\in A$. Then, since each element of $A$ is an interior point then $a$ also an interior point of $A$. So, there exists a $\delta$ neighborhood of $a$ contained in $A$, so $a + \delta/2\in A$ which contradicts the $a = $ sup($A$). Therefore, $a\notin A$.

Let's now suppose $A$ is closed, then $p\in \mathbb{R}$ is an accumulation point of $A$ which implies every open set $G$ containing $p$ contains a point of $A$ we will call $q$, where $p\neq q$.

I am pretty lost with this problem, and I think my reasoning is crap can anyone help me?

I dont think in general it is true that if $A$ is closed then $sup A \in A$ ... for an example set of $\mathbb{N} \subset \mathbb{R}$ but $sup \mathbb{N}$ is not in $\mathbb{N}$ ...if it is given that  $A$ is bounded then that claim for closed set is true...

Sorry for the confusion I have made previously. I have re-written the whole post. Hopefully, it is now much more helpful.

Proposition $1$:
If $E$ is a nonempty set of $\mathbb R$ which is bounded above,
then
(a) sup$E\in \bar E$. In particular, if $E$ is closed in $\mathbb R$, then sup$E\in E$.

(b) sup$E\notin$ Int $E$. In particular, if $E$ is open in $\mathbb R$, then sup $E\notin E$.

Proof: Let $\alpha$ = sup$E$ which exits in $\mathbb R$ by the least upper bound property of $\mathbb R$. Let $\varepsilon > 0$ be given. There exists $x\in E$ such that $\alpha-\varepsilon < x \leq \alpha$. Hence, every open neighborhood of $\alpha$ intersects with $E$. It follows that $\alpha \in \bar E$ and $\alpha \notin$ Int $E$.

Proposition $2$:
If $E$ is a nonempty set of $\mathbb R$, then sup$E\in \bar E$ (The closure is relative to the set of extended real numbers). In particular, if $E$ is closed in the set of extended real numbers,
then sup$E\in E$.

Proof:
Case 1: If $E$ is bounded above, then the proof is just the same as Proposition $1$.
Case 2: If $E$ is not bounded above, then sup$E$ = $\infty$. Let $M\in \mathbb R$. Then there exists $x\in E$ such that $M \lt x \lt \infty$. Thus, every open neighborhood of $\infty$ intersects with $E$. It follows that $\infty \in \bar E$.

Comment of the "counterexample" given by the previous reply:

  $\mathbb N \subset \mathbb R$. sup $\mathbb N$ is not in $\mathbb N$ ...

We must be careful about which metric space underlying we refer to. Openness/closeness relative to $\mathbb R$ is not the same as that relative to the set of extended real line.

$\mathbb N$ is closed relative to $\mathbb R$. However, in the context of allowing the use of extended real numbers as supremum, it makes little sense to consider the induced topology on $\mathbb R$ instead of the topology of the set of extended real numbers.

$\mathbb N$ is not closed relative to the set of extended real numbers, in which case the extended real number $\infty$ is a limit point of $\mathbb N$. So, $\mathbb N$ is not a counterexample to Proposition 2.


Nice sets of 100-1000 3D points For a combinatorics problem, I need point sets of 100 to 1000 3D points, with the restriction that generation is 1 line of code.  For example:

Tuples[Range[-3, 3], {3}]
Select[Tuples[Range[30], {3}], IntegerQ[Norm[#]] &]
Select[Tuples[Range[-7, 7], {3}], BitXor @@ Abs[#] == 0 &]

You can use the code of your choice, but no fetching data from elsewhere is allowed.  What are some really nice 3D point sets?  Here's a nice set of points where the code is too long.

base = {{0, 0, 0}, {180, 180, 180}, {252, 252, -252}, {420, 420, 420}, {1260, 1260, -1260}, {0, 0, 420}, {0, 0, 1260}, {0, 180, 360}, {0, 315, 315}, {0, 360, 180}, {0, 420, 840}, {0, 630, 630}, {0, 840, 420}, {140, 140, 420}, {180, 180, -540}, {252, 252,756}, {420, 420, -1260}};
tetrahedral[{a_, b_, c_}] := Union[{{a, b, c}, {a, -b, -c}, {b, c, a}, {b, -c, -a}, {c, a, b}, {c, -a, -b}, {-c, a, -b}, {-c, -a, b}, {-b, c, -a}, {-b, -c, a}, {-a, b, -c}, {-a, -b, c}}];
points = Union[Flatten[tetrahedral[#] & /@ base, 1]];

The 149 points make 241 lines of 5 points, that's kinda interesting. Is it possible to get more interesting with less code?

If you are interested in thinking of points as vertices of a graph embedding then you have a lot of options. I personally like the symmetry of circulant graphs. For some other interesting graph families, that are pretty simple to generate, you could look at random spanning trees, regular graphs, graphs of certain images, etc. below is some Mathematica code for generating a random circulant graph with edge density parameterized by k


Show that a scalar integral over a parametrized curve is unchanged if you reparametrize it. Show that a scalar integral over the parametrized curve $(\ln (t), t, t)$ for $t \in [2, 4]$ is unchanged if the curve is reparametrized.

I have no idea how to show this.

The parameterization is not unique. So can take any other parameters and solve the question .it will show the same.


Why does Totally bounded need Complete in order to imply Compact? Why does "Totally bounded" need "Complete" in order to imply "Compact"?
Shouldn't the definition of totally bounded imply the existence of a convergent subsequence of every sequence?

No, total boundedness of $\langle X,d\rangle$ implies that every sequence in $X$ has a Cauchy subsequence. Completeness of $\langle X,d\rangle$ implies that every Cauchy sequence in $X$ actually has a limit point in $X$ and therefore converges. The two together therefore imply that every sequence in $X$ has a convergent subsequence, i.e., that $X$ is sequentially compact. Finally, there is a theorem that a metric space is sequentially compact if and only if it’s compact, so total boundedness plus completeness imply compactness.

The confusion is understandable. We restate the two definitions:

Total boundedness says that every uniform open cover has a finite subcover.
Covering compactness says that every open cover, including non-uniform covers, have a finite subcover.

(In metric spaces, open balls can be used in place of open sets/covers.)
A more illustrative counter-example would be $(0, 1)$. It is not compact because I can find an open cover that has no finite subcover, namely
$$B(1, {1 \over 2}) \cup B({1 \over 2}, {1 \over 3}) \cup \cdots = \bigcup_{n = 1}^{+ \infty} B({1 \over n}, {1 \over n + 1})$$
If we stop the construction of the union at any finite $n$, the interval $(0, {1 \over n} - {1 \over n + 1}]$ won't be covered.
It is totally bounded, though. There exists a finite, positive real number, namely $1$, such that, $\forall x, y \in (0, 1)$, $\ d(x, y) < 1$, making $(0, 1)$ bounded. On the real line, bounded is equivalent to totally bounded.


Making a 3x3 magic square that adds up to 1 I got this problem from my sister in 6th grade. I spent 3 hours on this and couldn't figure it out for the life of me.

You're given an empty 3x3 square. You have to fill it with numbers from -9 to 9. Columns and rows and diagonals add up to 1. You cannot repeat any number.

I can't seem to find any combination that would give sums of 1 for all of them. Is there some algorithm for this?

The teacher claimed that none of her students had been able to solve this problem the past 14 years she's given it to them.

Suppose the sum of
each row, column, and
diagonal is s.
Then the sum of
all the digits is 3s.

If z is the center value,
then,
adding all the lines
through the center,
since the center is
counted 4 times
and all the other cells once,
4s-3z = 3s,
or
s = 3z.

Therefore
the sum of all the digits
is 3s or 9z.
Therefore
the sum of all the digits
is divisible by 9.

But if s = 1,
then the sum of all the digits
is 3,
which is not divisible by 9.
This is a contradiction,
so no such square can exist.

I don't think it can be done. I made sort of a program to see all the possible answers and the closest you can get is rows and columns to add 1, but only one of the diagonals. The other is going to be a multiple of 3, ranging from -27 to 27.


Maximum and minimum modulus of a polynomial function $f(z)=z^{10}+1000i$ Let $f(z)=z^{10}+1000i$. Then on the annulus $1 \leq \mid z \mid \leq 2$, what are the maximum modulus and minimum modulus of $f(z)$?

From geometric interpretation, I think the maximum modulus and minimum modulus are all achieved on the circle $\mid z \mid =2$. Also, each value is $2024$ and $24$ respectively (when $z^{10}=1024i$ and $ -1024i$). Is my result correct? Could anyone please give me a feedback?

Hint
$$|z_1|-|z_2|\leq |z_1+z_2| \leq |z_1|+|z_2|$$

Equalities can be achieved only when $z_1=\lambda z_2$ or $z_2=\lambda z_1$ for some $\lambda \in \mathbb R$, with positive $\lambda$ achieving the sum and the negative achieving the difference.

You can convert it into a easy trigonometric problem.
Write $z$=$re^{i\theta}$ . Now substituting z in required expression we have
 $f(z)$ = $r^{10}e^{10i\theta}+1000i$ = $r^{10}\cos{10\theta} + ir^{10}\sin{10\theta}+1000i$ = $r^{10}\cos{10\theta}$ + i($r^{10}\sin{10\theta}$+$1000$)

Modulus of $a+ib$ is equal to $\sqrt{a^2+b^2}$ . Using this we have,
$\mid f(z) \mid = \sqrt{r^{100}\cos^2{10\theta}+r^{100}\sin^2{10\theta}+1000^2+2.1000.r^{10}\sin{10\theta}}$
$\mid f(z) \mid = \sqrt{r^{100}+1000^2+2.1000.r^{10}\sin{10\theta}}$

For maximum modulus take r=2 and $10\theta = \pi/2$
For minimum modulus $10\theta = -\pi/2$ and take r such that it minimises $r^{100}-2.1000.r^{10}$


Complex equations (find z that satisfy certain conditions)... I'd like to find all z in C (complex set) that satisfy :

2*Im(z³) = 3*z*Im(z²)

where Im(z) = imaginary part of z.

You are trying to solve $$2\operatorname{Im}(z^3)=3z\operatorname{Im}(z^2).$$
The left-hand side is real. So $3z\operatorname{Im}(z^2)$ must be real, which implies $z$ is real. Also, if $z$ is real then $\operatorname{Im}(z^3)=\operatorname{Im}(z^2) = 0$. So the equation is satisfied for all real $z$. As such, the solution is the set of real numbers, $\mathbb{R}$.

z^3= (a+ib)^3 = (a^3 - b^2 - 3ab^2) + i(3a^2b)

Im(z^3)= 3a^2b

2Im(z^3)=2(3a^2b)=6a^2b

then z^2 = (a+ib)^2 = (a^2 -b^2) +i2ab

Im(z^2)= 2ab

3zIm(z^2)= 3(a+ib)2ab = (3a+i3b)(2ab)= 6a^2b+iab^2

Hence, 2Im(z^3)= 3zIm(z^2)

= 6a^2b = 6a^2b+iab^2

which means that iab^2 = 0  so ab^2 =0

the condition for the real and imaginary part of z =
Re(z) + 0Im(z) = Re(z)


Lines formed mid point of sides and cevians from opposite point are concurrent $AD, BE, CF$ are three concurrent lines in $\triangle ABC$ meeting opposite sides in $D, E, F$ respectively. Show that the joins of the midpoints of $BC, CA, AB$ to the midpoints of $AD, BE, CF$ are concurrent.

(Should be done by Ceva's theorem, Menelaus theorem, Stewart's theorem)

I tried by using trig form of Ceva's theorem and tried to do something similar to Cevian nests proof by connecting A'B'C' triangle but I failed. So please consider giving a hint or something and post the answer later on if I need it.

Source:CTPCM

Hints:

(1)
Pick 2 vectors e.g.

$\overrightarrow{AB}$ and $\overrightarrow{AC}$

as a basis in the plane, and express all other vectors $\overrightarrow{AX}$ as linear combination of them (where X is any point on this drawing or whichever of the points you need to involve in your solution).

This means, for any point X you should be able to find 2 numbers $k_1, k_2$ such that

$\overrightarrow{AX} = k_1 \overrightarrow{AB} + k_2 \overrightarrow{AC}$

Then it's all a matter of picking good start (number) parameters and playing with the equations.

Good parameters maybe are:

AF:FB = a -> number

AE:EC = b -> number

Then BD:DC you can express in terms of a,b using Ceva.

(2) Denote:

$point\ S_1 = A'D' \cap B'E'$

$point\ S_2 = A'D' \cap C'F'$

Prove that $\overrightarrow{AS_1} = \overrightarrow{AS_2}$

If you do this, it would mean the points $S_1$ and $S_2$ coincide.

My high school math is rusty but that (or similar) should be the general idea.

(3)
Also, for arbitrary points in the plane ABCD, prove and use this lemma:

$\overrightarrow{FE} = 1/2 (\overrightarrow{CA} + \overrightarrow{DB})$

(on the drawing below F and E are midpoints).

Seems it might be useful for this problem.

I would do this using Cartesian coordinates.  Given any triangle, we can set up a coordinate system with origin at one vertex and x-axis along one side.  In that coordinate system, the vertices of the triangle are at (0, 0), (a, 0), and (b, c) for some numbers, a, b, and c.  The midpoints of the three sides are (a/2, 0), (b/2, c/2), and ((a+b)/2, c/2).

The medians are given by y= ((a+b)/c)x, y= (2c/(2b-a))x+ 2c/(a-2b), and y= (ac/(4b-2a))x+ ac/(2a-4b).   Show that the solutions to any two of those equation are the same.


The Modulus of all the roots of a Polynomial are equal to $1$ Suppose the real number $\lambda \in (0,1)$, and let $n$ be a positive integer. Prove that all roots of the polynomial $$f\left ( x \right )=\sum_{k=0}^{n}\binom{n}{k}\lambda^{k\left ( n-k \right )}x^{k}$$ have modulus equal to $1.$

The Putnam problem 2014 B4 is similar:  Show that for each positive integer $n,$ all the roots of the polynomial $\sum_{k=0}^n 2^{k(n-k)}x^k$ are real numbers.

The following solution is taken from The Modulus of a Polynomial are the Same on AoPS. For other solutions see A family of polynomials whose zeros all lie on the unit circle on Math Overflow.
The idea is to derive a recursion formula for
$$
f_n(z) = \sum_{k=0}^{n} \binom{n}{k}\lambda^{k(n-k)} z^k
$$
and prove the statement by induction. $\lambda$ is a fixed real number in the interval $(0, 1)$.
$f_0(z) = 1$ and $f_1(z) = 1+z$ surely have only roots of modulus one.
Now assume that $n \ge 1$ and all roots of $f_n$ have modulus one. Using the recurrence formula for binomial coefficients
$$
 \binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}
$$
one gets
$$
 f_{n+1}(z) = f_n(\lambda z) +  \lambda^n z f_n(\frac{z}{\lambda}) \, .
$$
Now let $z^*$ be a zero of $f_{n+1}$. We have to show that $|z^*| = 1$.
From the above recurrence formula we get that
$$
 f_n(\lambda z^*) = -  \lambda^n z^* f_n(\frac{z^*}{\lambda})
$$
Note that $f_n(\frac{z^*}{\lambda})$ cannot be zero: Otherwise both $\lambda z^*$ and $\frac{z^*}{\lambda}$ would have modulus one, which is not possible.
Denoting the roots of $f_n$ with $z_1, \ldots z_n$ we have $f_n(z) = (z-z_1) \cdots (z-z_n)$ so that
$$
 z^* = - \frac{f_n(\lambda z^*)}{\lambda^n f_n(\frac{z^*}{\lambda})}
= - \prod_{k=1}^n \frac{\lambda z^* - z_k}{z^* - \lambda z_k}
$$
and therefore
$$ \tag{*}
 |z^*| = \prod_{k=1}^n \left |\frac{\lambda z^* - z_k}{z^* - \lambda z_k} \right | \, .
$$
An elementary calculation (using $|z_k|=1$ for all $k$) shows that
$$
|\lambda z^* - z_k|^2 - |z^* - \lambda z_k|^2 = (1 - \lambda^2)(1 - |z^*|^2) \, .
$$
It follows that if $|z^*| < 1$ then all factors on the right-hand side of $(*)$ are larger than one, and if $|z^*| > 1$ then all factors on the right-hand side of $(*)$ are less than one. So both cases lead to a contradiction, and the only possibility is that $|z^*| = 1$. This concludes the proof.

Write the above polynomial in the form:     $a+bx+cx^2$... and observe that if we put $f(x)=0$ it is same as putting $x^k f(1/x)=0$ {property of binomial coefficients and the power of lambda} conclude that if $x$ is a root  then $1/x$ is also has the same modulus. Hence $|x|=1$


Prove that $1+\frac{1}{8n}\sqrt{\pi_1n}<\frac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}<(1+\frac{1}{8n}+\frac{1}{128n^{2}})\sqrt{\pi_2n}$ Within the confines of the O-level syllabus, prove that:

$$1+\dfrac{1}{8n}\sqrt{\pi_1n}<\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}$$

for all positive integer $n$, where $\pi_1=3.141$ and $\pi_2=3.142$

Using A-level mathematics, show that the result remains true with $\pi_1=\pi_2=\pi$ where $\pi$ is the familiar constant associated with a circle.

This is a question found in Hammersley's "On the enfeeblement of mathematical skills by "Modern Mathematics" and by similar soft intellectual trash in schools and universities"

When I first look at this problem, I wonder if I can prove it via Wallis'product?
$$\prod_{i=1}^{\infty}\dfrac{2n}{2n-1}.\dfrac{2n}{2n+1}=\dfrac{\pi}{2}$$

$$
\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)} = \dfrac{2n!}{(2n-1)!} = 2n \\ \therefore \;\; 1+\dfrac{1}{8n}\sqrt{\pi_1n}<\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}
\\ = 1+\dfrac{1}{8n}\sqrt{\pi_1n}<2n<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}
$$

Considering the case of $$\pi_1 = \pi_2 = \pi$$ since this generality can be used for the other case too.

Splitting the inequalities into two,

\begin{align}
I_1:
1+\dfrac{1}{8n}\sqrt{\pi n}<2n \\
1+\dfrac{1}{8\sqrt n}\sqrt{\pi}<2n \\
\dfrac{1}{8\sqrt n}\sqrt{\pi}<2n - 1 \\
\dfrac{1}{8}\sqrt{\pi}<(2n - 1)\cdot\sqrt n \tag{1}\label{1}
\end{align}
Since n is a positive integer, minimum value of n is 1, which when substituted in \eqref{1} gives $$ 0.2215 < 1 $$ which is true.

\begin{align}
I_2:
2n<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi n}\\
\dfrac{2n}{\sqrt{\pi n}}<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right) \\
\dfrac{2n}{\sqrt{\pi n}} - 1<\left(\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right) \\
\dfrac{2n}{\sqrt{\pi n}} - 1<\dfrac{1}{8n}\left(1+\dfrac{1}{16n}\right)\\
\dfrac{2n}{\sqrt{\pi n}} - 1<\dfrac{1}{8n}\left(\dfrac{16n+1}{16n}\right)\\
(128n^{2})\cdot\dfrac{2\sqrt n}{\sqrt{\pi}} - 1< 16n+1\\
\dfrac{256n^{\frac{5}{2}}}{\sqrt \pi} - 128n^{2}<16n+1\\
\dfrac{256}{\sqrt \pi} < ( 128n^2 + 16n + 1 )\cdot n^{\frac{2}{5}} \tag{2}\label{2}
\end{align}

Since n is a positive integer, minimum value of n is 1, which when substituted in \eqref{2} gives $$ 144.43 < 145 $$ which is true.


Prove that the eigenvalues of skew-Hermitian matrices are purely imaginary I would like some help on proving that the eigenvalues of skew-Hermitian matrices are all pure imaginary. I have gotten started on it, but am getting stuck.

Attempt at proof: $Av=\lambda v \implies A \bar{v}=\bar{\lambda}\bar{v}.$

Also, $v^TA^T=\lambda v^T \implies v^TA^T\bar{v}=\lambda v^T \bar{v} \implies ?$

Should I conjugate both sides next?

Let $x$ an eigenvector of $A$ associated to the eigenvalue $\lambda$ then

$$\langle Ax,x\rangle=\overline \lambda||x||^2=\langle x,A^*x\rangle=-\langle x,Ax\rangle=-\lambda\langle x,x\rangle$$
so
$$\overline\lambda=-\lambda$$
hence $\lambda$ is pure imaginary complex.

$AX=λX$
so $X^H AX=X^H λX$

$-A^H X=λX$, so $X^H A^H X=-X^H λX$

$X^H AX=-X^H λ^H X$

Hence, $λ=-λ^H$


Proving $C_0^1(\Omega)$ is not dense in $H^1(\Omega)$. Hi I am trying to work on the following problem:

(a) $C_0^1(\Omega)$ is dense in $L_2(\Omega)$

(b) $C_0^1(\Omega)$ is dense in $H_0^1(\Omega)$.

(c) Explain why $C_0^1(\Omega)$ is not dense in $H^1(\Omega)$.

I know how to do (a) and (b) but I couldn't find how to solve (c). Any help would greatly appreciated. Thanks in advance.

So By following the comments given below I got

$$f_n(x)=\begin{cases}n^2x^2,\,\,\,\,\,\,\,0\le x\le\frac{1}{n}\\1,\,\,\,\,\,\,\,\frac{1}{n}\le x\le 1-\frac{1}{n}\\n^2x^2,\,\,\,\,\,\,\,1-\frac{1}{n}\le x\le 1\end{cases}$$

Clearly $f_n(x)\to 1$ but $f(x)=1\not\in C_0^1(\Omega)$ where as $f_n(x)\in C_0^1$, therefore $C_0^1(\Omega)$ is not dense in $H^1(\Omega)$.

I still have a doubt about the fact that is $f_n(x)$ in $C_0^1(\Omega)$, since the derivative is not continuous anymore.

You can check that via looking at the trace. Let $D$ denote the closure of $C_0^1(0,1)$ in $H^1(0,1)$. By definition, the trace of $u \in C_0^1(0,1)$ is zero at both end points. Moreover, it is continuous. Hence, the trace of any function in $D$ is $0$. However, there are functions in $H^1(0,1)$ with non-zero trace, e.g., $x \mapsto 1$.

Note: f(x) = 1 has compact support, so this entire example is invalid.


Coffee machine breaking with probability 0,01 - check if two random variables are independent and find bound for $P(X > n)$ There is a coffee machine that makes a cup of coffee with probability $0.9$, makes a cup of hot water with probability $0.09$ and breaks forever with probability $0.01$.

Let $X$ be the number of cups of coffee the machine makes before it breaks and $Y$ be the number of cups of hot water before it breaks.

Find the least $a > 0$ such that $P(X > n) = O(a^n)$.
Are random variables $X$ and $Y$ independent?

I tried something like this:
$$P(X > n) = \sum_{i = n+1}^{+\infty} P(X = i)$$
$$P(X = i) = \sum_{k=i+1}^{+\infty} P(X=i|Z=k) \cdot P(Z=k)= \sum_{k=i+1}^{+\infty} 0.99^{k-1} \cdot 0.01 \cdot {k - 1 \choose i} \cdot 0.9^i \cdot 0.09^{k-1-i}= 0.01 \cdot 0.9^i \sum_{k=i+1}^{+\infty} 0.99^{k-1} \cdot 0.09^{k-1-i} {k - 1 \choose i} $$

I can't go any further with this and I think it is important to count this to be able to check if $X$ and $Y$ are independent.

The joint distribution of X,Y

$$f_{XY}(X=n,Y=m) = {(n+m)\choose(n)} (0.9)^{n}(0.09)^{m}(0.01)$$

$$f_X(X=n) = \sum_{m=0}^{\infty} {(n+m)\choose(n)}(.9)^{n}(.09)^{m}(0.01)$$

$$f_X(X=n) = \frac{1}{91} \left(\frac{90}{91}\right)^n\tag 1$$

$$P(X\gt n) = \sum_{i=n+1}^{\infty} \frac{1}{91}\left(\frac{90}{91}\right)^i$$

$$P(X\gt n) = \frac{90}{91}.\left(\frac{90}{91}\right)^n$$

Thus the least $$a = \frac{90}{91}$$

and X and Y are not independent from the joint distribution as $$f_{XY}(X=x,Y=y) \ne f_X(X=x)f_Y(Y=y)$$

Remark:
I got the result of (1) from wolfram


Differentials and implicit differentiation Consider this example. Suppose $x$ is a function of two variables $s$ and $t$,
$$x = \sin(s+t)$$
Taking the differential as in doing implicit differentiation [1],
$$dx = \cos(s+t)(ds+dt) = \cos(s+t)dt + \cos(s+t)ds$$

I know the right way of taking the differential of $x$ is by
$dx = \dfrac{\partial x}{\partial s}ds + \dfrac{\partial x}{\partial t}dt$ [2]
But why do the above method [1], which I cannot make any sense of mathematically, give the same result as [2]?

I do not understand method [1] because $s$ and $t$ are supposed to be independent variables not functions to be differentiated. i.e. $ds$ just means $s-s_0$, same with $dt$.

Briefly, in implicit differentiation, I find it very helpful to treat every variable as a function.  Because in
[1], the traditional method, chain rule the chain rule should apply to every variable or
[2], the differential method, one of ds, dx, dt, etc. should apply to every variable.

The exception is in the traditional, non differential method.  If you're taking derivative with respect to x, then x doesn't need to be a function.  Simply take derivative of x like normal.

This is a special case: you may consider a simple substitution $y = s + t$

$x = \sin(y)$

$dx = \cos(y)dy$

$dx = \cos(s+y)(ds + dt)$

if you do this with $x=s^t$

$y = s^t$

$x = y$

$dx = dy$

$dx = ts^{t-1}ds + s^t\ln(s)dt$

which isn't useful at all...


Integral of product of two error functions (erf) In the course of my research I came across the following integral:

$$\int\nolimits_{-\infty}^{\infty}\operatorname{erf}(a+x)\operatorname{erf}(a-x)dx$$

where $$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^2}dt$$ is the familiar error function.

Does anyone know if this is solvable?  If so, a suggestion on how to do this would be very much appreciated.  I can tolerate a solution in terms of $\operatorname{erf}$ functions of $a$.  I have tried my usual sources in these matters (G&R, the tables on the Wolfram website, Abramowitz's Handbook of Mathematical Formulas, the web) but couldn't find anything of use.

Thank you!

$\operatorname{erf}(x)$ is an odd function, therefore,
$$
\begin{align}
\int_{-\infty}^\infty(\operatorname{erf}(a+x)+\operatorname{erf}(a-x))\;\mathrm{d}x
&=\lim_{L\to\infty}\;\int_{-L}^L(\operatorname{erf}(x+a)-\operatorname{erf}(x-a))\;\mathrm{d}x\\
&=\lim_{L\to\infty}\;\int_{-L+a}^{L+a}\operatorname{erf}(x)\;\mathrm{d}x-\lim_{L\to\infty}\;\int_{-L-a}^{L-a}\operatorname{erf}(x)\;\mathrm{d}x\\
&=\lim_{L\to\infty}\;\int_{L-a}^{L+a}\operatorname{erf}(x)\;\mathrm{d}x-\lim_{L\to\infty}\;\int_{-L-a}^{-L+a}\operatorname{erf}(x)\;\mathrm{d}x\\
&=4a\tag{1}
\end{align}
$$
since $\lim\limits_{x\to\infty}\operatorname{erf}(x)=1$ and $\lim\limits_{x\to-\infty}\operatorname{erf}(x)=-1$.

Furthermore,
$$
\begin{align}
\int_{-\infty}^\infty\left(\operatorname{erf}(a+x)\operatorname{erf}(a-x)+1 \right)\;\mathrm{d}x
&=\int_{-\infty}^\infty(\operatorname{erf}(a+x)+1)(\operatorname{erf}(a-x)+1)\;\mathrm{d}x\\
&-\int_{-\infty}^\infty(\operatorname{erf}(a+x)+\operatorname{erf}(a-x))\;\mathrm{d}x\tag{2}
\end{align}
$$
To evaluate
$$
\begin{align}
\int_{-\infty}^\infty(\operatorname{erf}(a+x)+1)(\operatorname{erf}(a-x)+1)\;\mathrm{d}x
&=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^{a+x}\int_{-\infty}^{a-x}e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t\;\mathrm{d}x
\end{align}
$$
note that $s\le a+x$ and $t\le a-x$; i.e. $s-a\le x\le a-t$ and $s+t\le2a$. Thus,
$$
\begin{align}
\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^{a+x}\int_{-\infty}^{a-x}e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t\;\mathrm{d}x
&=\frac{4}{\pi}\int\int_{s+t\le2a}\int_{s-a}^{a-t}e^{-s^2-t^2}\;\mathrm{d}x\;\mathrm{d}s\;\mathrm{d}t\\
&=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-s-t)_+\;e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t
\end{align}
$$
Change variables: $u=(s+t)/\sqrt{2}$ and $v=(s-t)/\sqrt{2}$ so that $s=(u+v)/\sqrt{2}$ and $t=(u-v)/\sqrt{2}$:
$$
\begin{align}
\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-s-t)_+\;e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t
&=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-\sqrt{2}u)_+\;e^{-u^2-v^2}\;\mathrm{d}u\;\mathrm{d}v\\
&=\frac{4}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}(2a-\sqrt{2}u)\;e^{-u^2}\;\mathrm{d}u\\
&=4a(\operatorname{erf}(\sqrt{2}a)+1)-\frac{4}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}\sqrt{2}u\;e^{-u^2}\;\mathrm{d}u\\
&=4a(\operatorname{erf}(\sqrt{2}a)+1)-\frac{2\sqrt{2}}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}\;e^{-u^2}\;\mathrm{d}u^2\\
&=4a(\operatorname{erf}(\sqrt{2}a)+1)+\frac{2\sqrt{2}}{\sqrt{\pi}}\;e^{-2a^2}
\end{align}
$$
Therefore,
$$
\int_{-\infty}^\infty\left(\operatorname{erf}(a+x)+1\right)\left(\operatorname{erf}(a-x)+1\right)\;\mathrm{d}x
=4a\left(\operatorname{erf}(\sqrt{2}a)+1\right)+\frac{2\sqrt{2}}{\sqrt{\pi}}\;e^{-2a^2}\tag{3}
$$
Thus, the convolution of $\operatorname{erf}(x)+1$ with itself is $2x(\operatorname{erf}(x/\sqrt{2})+1)+\frac{2\sqrt{2}}{\sqrt{\pi}}e^{-x^2/2}$.

Subtract $4a$ from $(3)$ using $(1)$ and $(2)$ to get
$$
\int_{-\infty}^\infty\left(\operatorname{erf}(a+x)\operatorname{erf}(a-x)+1 \right)\;\mathrm{d}x
=4a\operatorname{erf}(\sqrt{2}a)+\frac{2\sqrt{2}}{\sqrt{\pi}}\;e^{-2a^2}\tag{4}
$$
My guess is you want either $(3)$ or $(4)$.


What is the probability that a point chosen randomly from inside an equilateral triangle is closer to the center than to any of the edges? My friend gave me this puzzle:

  What is the probability that a point chosen at random from the interior of an equilateral triangle is closer to the center than any of its edges?

I tried to draw the picture and I drew a smaller (concentric) equilateral triangle with half the side length. Since area is proportional to the square of side length, this would mean that the smaller triangle had $1/4$ the area of the bigger one. My friend tells me this is wrong. He says I am allowed to use calculus but I don't understand how geometry would need calculus. Thanks for help.

You are right to think of the probabilities as areas, but the set of points closer to the center is not a triangle. It's actually a weird shape with three curved edges, and the curves are parabolas.

The set of points equidistant from a line $D$ and a fixed point $F$ is a parabola. The point $F$ is called the focus of the parabola, and the line $D$ is called the directrix. You can read more about that here.

In your problem, if we think of the center of the triangle $T$ as the focus, then we can extend each of the three edges to give three lines that correspond to the directrices of three parabolas.

Any point inside the area enclosed by the three parabolas will be closer to the center of $T$ than to any of the edges of $T$. The answer to your question is therefore the area enclosed by the three parabolas, divided by the area of the triangle.

Let's call $F$ the center of $T$. Let $A$, $B$, $C$, $D$, $G$, and $H$ be points as labeled in this diagram:

The probability you're looking for is the same as the probability that a point chosen at random from $\triangle CFD$ is closer to $F$ than to edge $CD$. The green parabola is the set of points that are the same distance to $F$ as to edge $CD$.

Without loss of generality, we may assume that point $C$ is the origin $(0,0)$ and that the triangle has side length $1$. Let $f(x)$ be equation describing the parabola in green.

By similarity, we see that $$\overline{CG}=\overline{GH}=\overline{HD}=1/3$$

An equilateral triangle with side length $1$ has area $\sqrt{3}/4$, so that means $\triangle CFD$ has area $\sqrt{3}/12$. The sum of the areas of $\triangle CAG$ and $\triangle DBH$ must be four ninths of that, or $\sqrt{3}/27$.

$$P\left(\text{point is closer to center}\right) = \displaystyle\frac{\frac{\sqrt{3}}{12} - \frac{\sqrt{3}}{27} - \displaystyle\int_{1/3}^{2/3} f(x) \,\mathrm{d}x}{\sqrt{3}/12}$$

We know three points that the parabola $f(x)$ passes through. This lets us create a system of equations with three variables (the coefficients of $f(x)$) and three equations. This gives

$$f(x) = \sqrt{3}x^2 - \sqrt{3}x + \frac{\sqrt{3}}{3}$$

The integral of this function from $1/3$ to $2/3$ is $$\int_{1/3}^{2/3} \left(\sqrt{3}x^2 - \sqrt{3}x + \frac{\sqrt{3}}{3}\right) \,\mathrm{d}x = \frac{5}{54\sqrt{3}}$$

This gives our final answer of $$P\left(\text{point is closer to center}\right) = \boxed{\frac{5}{27}}$$

I am just passing by here. Perhaps the idea is to have something elegant, but this is just straightforward: an integral in polar coordinates with the help of Mathematica. I reuse the picture of the solution of Zubin. Let J be the midpoint of CD. By symmetry, we can restrict the analysis to CFJ. Without loss of generality, let the length of FJ be $1$. The total area of CFJ is $\sqrt{3}/2$. We need the area above the curve in CFJ. Pick an arbitrary point L on the curve inside CFJ.  Let $\theta$ be the angle between FJ and FL. Some trigonometry gives us that the length of FL is $x = (1+ \cos\theta)^{-1}$. (It helps to draw the straight line that extends FL and meets CJ at some point K and note that the length of FK is $1/\cos \theta$.) So, the area above the curve is the integral of $(1+\cos \theta)^{-2}/2$ from $0$ to $\pi/3$. Using mathematica, the primitive is $((2 + \cos\theta)\sin\theta)/(6(1 + \cos\theta)^2)$. So, the area above the curve is $((2 + \cos(\pi/3))\sin(\pi/3))/(6(1 + \cos(\pi/3))^2) = 5 \sqrt 3/ 54$ . So, the probability to be above the curve is $(5 \sqrt 3/ 54)/(\sqrt 3/2) = 5/27$.

This solution generalizes to any regular polygone. We integrate from $0$ to $\pi/n$ instead of from $0$ to $\pi/3$ and the total area of CFJ becomes $\tan(\pi/n)/2$ instead of $\sqrt 3/2$, every thing else stays the same.


Calculate trigonometric integral $ \int_{-\infty}^{\infty}{\sin(x^2)}\,dx$ Recently, I came across the following integral:
$$
\int_{-\infty}^{\infty}{\sin(x^2)}\,dx=\int_{-\infty}^{\infty}{\cos(x^2)}\,dx=\sqrt{\frac{\pi}{2}}
$$
What are the different ways to calculate such an integral?

In this answer is given a real method for showing this.

Using contour integration we can show that
$$
\begin{align}
\int_{-\infty}^\infty\left[\cos(x^2)+i\sin(x^2)\right]\,\mathrm{d}x
&=\int_{-\infty}^\infty e^{ix^2}\,\mathrm{d}x\\
&=\frac{1+i}{\sqrt2}\int_{-\infty}^\infty e^{-x^2}\,\mathrm{d}x\\
&=(1+i)\sqrt{\frac\pi2}
\end{align}
$$
To show the equality of these integrals, integrate $e^{iz^2}$ along the contour
$$
[-R,R]\cup Re^{i\pi[0,1/4]}\cup[R,-R]e^{i\pi/4}\cup-Re^{i\pi[1/4,0]}
$$
which must be $0$ since $e^{iz^2}$ has no singularities. Note that the integral along the two curved contours vanishes as $R\to\infty$. Each is bounded by
$$
\begin{align}
&\left|\,\int_0^{\pi/4}e^{iR^2(\cos(t)+i\sin(t))^2}R(-\sin(t)+i\cos(t))\,\mathrm{d}t\,\right|\\
&\le\int_0^{\pi/4}e^{-R^2\sin(2t)}R\,\mathrm{d}t\\
&\le\int_0^\infty e^{-R^24t/\pi}R\,\mathrm{d}t\\
&=\frac\pi{4R}
\end{align}
$$
Thus, the integral along $[-\infty,\infty]$ equals the integral along $[-\infty,\infty]e^{i\pi/4}$.

Step 1 is to substitue $u=x^2$ giving
$$
I = \int_{-\infty}^\infty \sin(x^2) dx = 2 \int_0^\infty \sin(x^2) dx =
2 \int_0^\infty  \sin u \frac{du}{2\sqrt{u}}= \int_0^\infty   \frac{\sin u}{\sqrt{u}} du
$$
The next step is to recognize this related to Fresnel's sine integral
$$
S(x) = \frac{1}{\sqrt{2\pi}}\int_0^x   \frac{\sin t}{\sqrt{t}} dt
$$
The third step is to notice that an asymptotic form for $S(x)$ is
$$
S(x) = \frac{1}{2} - \frac{1}{\sqrt{2\pi}x}\cos^2 x + O\left( \frac{1}{x^2} \right)
$$
(I suppose that anybody who does not know this from early childhood should go back to reading Gradshteyn and Ryzhik each night until it puts you to sleep. ROTFL)

Taken to infinity, this gives $S(x) \rightarrow \frac{1}{2}$ so
$$ I = \sqrt{2\pi} \frac{1}{2} = \sqrt{\frac{\pi}{2}}$$


Expected Number of Coin Tosses to Get Five Consecutive Heads A fair coin is tossed repeatedly until 5 consecutive heads occurs.

What is the expected number of coin tosses?

Let $e$ be the expected number of tosses. It is clear that $e$ is finite.

Start tossing. If we get a tail immediately (probability $\frac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\frac{1}{4}$), then the expected number is $e+2$. Continue $\dots$. If we get $4$ heads then a tail, the expected number is $e+5$. Finally, if our first $5$ tosses are heads, then the expected number is $5$. Thus
$$e=\frac{1}{2}(e+1)+\frac{1}{4}(e+2)+\frac{1}{8}(e+3)+\frac{1}{16}(e+4)+\frac{1}{32}(e+5)+\frac{1}{32}(5).$$
Solve this linear equation for $e$. We get $e=62$.

By conclusion and some other mathmatical proof. To get consecutive n heads or tails expected number is $2^n+1 -2$.

So, here answer is $2^6- 2 =62$.