What is an open set in a topological space? I recently started learning topology to help me understand limits and continuity better for calculus, and I am struggling with some of the definitions.
What I am getting confused with is why is every set in a topology considered to be open and when talking about sets in the topology we always say the set is open.
My intuitive notion of openness from  previous knowledge of mathematics is an interval that does not contain its endpoints, so there is an infinite sequence at the end points, e.g., $(0,1)$ is an open interval.
However, in topology, for example, the singleton $\{1\}$
is considered an open set—how is this so? Why are sets in a topology always open? And what is the definition of an open set in a topological space?
Thanks in advance.

There is no such thing as an "open set," only a open subset of a topological space. "The singleton $\{ 1 \}$ is an open set" is not a meaningful statement; the meaningful statement is whether the singleton $\{ 1 \}$ is an open subset of some topological space containing $1$. For example it is not an open subset of $\mathbb{R}$, or of the closed interval $[0, 1]$ (with their usual topologies). However, it is an open subset of itself, considered as a one-point space (which has a unique topology).
Sets in a topology are open by definition; that's what "open" means.
I suggest that when starting to learn topology you completely ignore the ordinary English meaning of the word "open" and just work with the axioms abstractly for awhile. It's not the most fun or intuitive way to do it but at least you aren't letting preconceptions get in the way. Also, I don't really recommend this as a way to better understand calculus; you'd be better off picking up a textbook on real analysis.

There are a number of things about topology that are idiosyncratic. It is typical to refer to singleton sets with the same symbol as used for the element of the singleton. This should act as an indication that one is actually working within the power set of the domain of definition.
A topology is given as a partition on the power set. Provided that a partition is compatible with the axioms, the partition will consist of open, closed, clopen (both open and closed) and none of the above. The axioms can be given in terms of either open sets or closed sets. So, these classifications reside in the power set of the power set of the domain of definition. Since one can have clopen sets, this classification does not produce disjoint collections.
As a set, the domain of definition has a sort of special relationship with the topology in which every singleton is open. This is called the discrete topology. Every set is open by arbitrary unions. So every set is closed as well by complementation.
Importantly, one can have topologies where not every set is open. The topology with only the empty set and the domain of definition as open sets satisfies the axioms. It is called the trivial topology.
From these remarks, you should surmise that "what is open" in a topology is a matter of stipulation.  For real analysis and calculus the stipulation comes from our naive conceptions of length and measurement. What becomes important are inequalities. The "in between" mathematical object is called a metric space. This governs how we understand distances between points.
Whatever topology book you are using, it could be highly useful to obtain a copy of "Counterexamples in Topology" by Steen and Seebach (reprints from Dover Books are inexpensive).  It is full of examples, including some simple ones accessible to beginners. It also summarizes what you will learn from main textbook. So, it will become a reference resource as you learn the subject better.


How to find out the probability? The question is:
A true or false test contains eight questions. If a student guesses the answer for each question, find the probability:
(a) 8 answers are correct. (b) 7 answers are correct. (c) 6 answers are correct. (d) at least 6 answers are correct.
I proceed as that total number of questions are eight, each question may be true or false, so $n(S)=16$, I don;t know whether am I correct to find the sample space or incorrect?
Also I don't know how to proceed and calculate the other parts of the questions?

You have 8 questions, each a choice from two answers.   The state space contains $2^8$ distinct outcomes selected without bias (ie having an uniform probability distribution).

Thus the probability for each individual outcome to occur is $1/256$.

  (a) 8 answers are correct.

Obviously only one outcome corresponds to this event.

  (a) 7 answers are correct.

We must count the ways to select seven from eight questions to give true answers.

So there are $\tbinom 87$ or simply $8$ outcomes in this event.

Thus the probability is $8/256$, or simply $1/32$.

  …

You can do the rest.

The answer for that question is given by the formula $$\Big(\frac{1}{n}\Big)^k$$ then multiply it to the number of combinations.

I will explain why. For the first question there will be 2 possiblities - a True or a False(let's just use T and F). For the 2nd question there will be 2 more for each possible event of the first. TT, TF, FT,FF. For the third there will be 2 more for each previously mentioned samples.

a. 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256 or $(\frac{1}{2})^8 = \frac{1}{256} * \frac{8!}{8!(8-8)!}$ = $\frac{1}{256}$

b. 2 x 2 x 2 x 2 x 2 x 2 x 2 = 128 or $(\frac{1}{2})^7 = \frac{1}{256}  * \frac{8!}{7!(8-7)!}$ = $\frac{1}{32}$

c. 2 x 2 x 2 x 2 x 2 x 2 = 128 or $(\frac{1}{2})^6 = \frac{1}{256} * \frac{8!}{6!(8-6)!}$ = $\frac{7}{64}$


Angle perpendicular lines and x-axis Let's say we have a line $L$ with positive slope that intersects $(0,0)$ and makes an angle of $\theta<90^\circ$ with the $x$-axis ("on top" of the x-axis, i.e. the angle is in quadrant 1). Given a line perpendicular to $L$ going through quadrant 1, what is the angle this line forms with the x-axis?

The answer, intuitively, seems to be $90-\theta$. But I don't know how to prove it.

You are right. To prove it just consider the right triangle formed by $x$ axis, line $L$ and the second perpendicular to $L$, then since the sum of the internal angles adds up to $180°$ the angle that the perpendicular line forms with the x-axis is $90°-\theta$.

The two lines and the x-axis form a triangle. One angle is $\theta$ and one is $90^{\circ}$.

The last angle can then only be ....


Set of limit points of topologist's sine curve $S$ Let $S=\{(x,\sin(1/x)):x \in (0,1]\}$ be the topologist's sine curve. Find the limit points $\lim S$ of $S$.

I claim $\lim S = S \cup \{(0,y):y \in [-1,1]\}$. But, how do you show that any of these points is such? Certainly there is some nice lemma for this? All I can think to do is revert to analysis by showing that, for $s=(x,\sin(1/x)) \in S$, and any open ball $B=B(z,\epsilon)$ which contains $s$, there is a point $s'=(x',\sin(1/x')) \in S$ with $s' \in B$, but I'm having trouble finding an $x'$...

Because $\sin(1/x)$ is continuous, any point in $S$ will be a limit point of $S$.

For a point $z=(0,y)$ where $y \in [-1,1]$, you need to show that every deleted neighborhood $B(z,r) \setminus \{z\}$ contains a point of $S$. This is true due to the rapid oscillation of $\sin(1/x)$ near $x=0$. To explicitly find an $x$ that fits in this ball, take some $n$ so that $\frac{1}{2\pi n} < r$. Then $\sin(1/x)$ takes on all values in $[-1,1]$ for $x \in\left(\frac{1}{2\pi(n+1)}, \frac{1}{2\pi n}\right)$, so it must cross inside the deleted neighborhood.

A picture may be better than many words. So $[-1,1]$ is also the limit points of  topologist's sine curve.


Find a formula for the sequence $(a_n)$ I would appreciate if somebody could help me with the following problem

Q : Find $a_n=?$

$$a_1=1, a_{n+1}=\frac{a_n+4}{a_n+1}(n=1,2,3,\cdots) $$


What is the recursive definition of $f(n)=1+(-1)^n$ for $n≥1$? I think it's: What is the recursive definition of $f(n)=1+(-1)^n$ for $n≥1$? I think it's:

1) $f(1)=0$

2) $f(n)=f(n-1)+2$ if $n$ is even

$f(n)=f(n-1)-2$ if $n$ is odd

Is it correct?

Your casewise approach works. Alternatively, you can say $f(1)=0,$ and $$f(n)=f(n-1)-2(-1)^{n-1}$$ for $n\ge 2.$

To come up with that, I started with the explicit formula, so that $$f(n)-f(n-1)=\left(1+(-1)^n\right)-\left(1+(-1)^{n-1}\right),$$ then worked to get it into like terms to simplify: $$f(n)-f(n-1)=1+(-1)^n-1-(-1)^{n-1}$$ $$f(n)-f(n-1)=(-1)^n-(-1)^{n-1}$$ $$f(n)-f(n-1)=-2(-1)^{n-1}.$$

Just as simply, we could have noticed that $-(-1)^{n-1}=(-1)^n,$ leading instead to the formula $$f(n)=f(n-1)+2(-1)^n.$$


near linearly dependent I'm considering a situation where vectors are nearly linearly dependent.

Suppose that $A$ is $2 \times 2$ matrix

$A= \left[
\begin{array}{cc}
1 & a \\
a & 1
\end{array}
\right] $

If $a=1$, $A$ is of course singular and linearly dependent. But if $a=0$ and $a=0.999$, $A$ is non-singular for both cases. But, apparently $a=0.999$ is more close to being singular or linearly dependent than $a=0$. In this $2 \times 2$ case, the determinant can be used to measure the closeness to being singular or linearly dependent.

Now consider $3 \times 2$ matrices $B_1$ and $B_2$

$B_1 = \left[
\begin{array}{cc}
1 & 0 \\
1 & 1 \\
1 & 2
\end{array}
\right] $

$B_1 = \left[
\begin{array}{cc}
1 & 0.999 \\
1 & 1 \\
1 & 1.001
\end{array}
\right] $

Mathematically, both $B_1$ and $B_2$ are linearly dependent. But, $B_2$ is more close to being linearly dependent than $B_1$.

My question is are there any ways to measure the closeness to being linearly dependent for such non-square matrix. In other words, how can I demonstrate that $B_2$ is close to being linearly dependent than $B_1$.

A square matrix of rank $r$ has $r$ non-zero eigenvalues (may be multiple). So it is singular iff one of its eigenvalues is zero. The determinant (i.e. the product of all eigenvalues) is not a good quantitative measure for singularity. For example the matrix
$\left[
\begin{array}{cc}
10^{10} & 0 \\
0 & 10^{-10}
\end{array}
\right] $
has determinant $1$ but clearly (for me) is more singular than the identity matrix.
The smallest eigenvalue by magnitude is a better measure for that (possibly normalized by the largest eigenvalue).

The generalization for non-square matrices you are looking for are singular values. The smallest singular value is zero for singular matrices and close to zero for almost singular matrices. In your example matrices they are 0.915272 and 0.001, respectively.

Definitely. For square matrices, just calculate the determinant: if the result is close to zero, then the columns of your matrix are close to be linearly dependent (though, strictly speaking, they are not exactly linearly dependent). For a non-square matrix $B$, the columns are nearly dependent if the determinant of $B^TB$ is nearly zero.


Invertible elements of $\mathbb{Z}/p^n\mathbb{Z}$ Let $p \ge 3$ be a prime and let us consider $U(p^n)$, namely the set of all invertible elements of $\mathbb{Z}/p^n\mathbb{Z}$. Is it true that $1+p \in U(p^n)$? If so, which is its order in $U(p^n)$?

HINT:

$$[k]_{p^n} \in U(p^n) \iff (k,p^n) = 1$$

where $[k]_{p_n}$ is an element of the quotient group with representant $k \in \mathbb{Z}$

The order of $U(p^n)$, $n\geq 1$, is given by Euler's totient function $\phi$ with $\phi(p^n)=p^n-p^{n-1}$.


understanding integration by change of variables I'm having trouble understanding integration by the method of change of variables. Can someone elaborate the method with an example?

For instance, $\int_0^{5\pi} \frac{\sin t}{2+\cos t } dt$ using $s=\cos t$ as a new variable.

Let $f: [c,d]\to \Bbb R$ be a continuous function and $\varphi: [a,b]\to [c,d]$ be a continuously differentiable, bijective function (sometimes bijectivity isn't strictly required, but it's easier to prove with this assumption).  Then

$$\int_a^b f(\varphi(t))\varphi'(t)\ dt = \int_{c}^{d} f(x)\ dx$$

Here's an image (from an old project of mine) that hopefully shows roughly what $\varphi$ does.  $\varphi$ is a function which maps one interval (on the real number line) to another interval.

This formula is easy to derive from the chain rule and the Fundamental Theorem of Calculus.

The thing that many calculus students don't understand immediately is that this formula can be run left-to-right or right-to-left.

Left-to-right
Sometimes the easiest thing to do is to manipulate our integrand until we get it into the form $f(\varphi(t))\varphi'(t)$.  Then we can simplify it by just setting $x = \varphi(t)$ and using the above formula.  For example, consider the integral $$\int_0^4 \frac{t}{(1+t^2)^2}\ dt$$  In this case if we realize that $$\frac{t}{(1+t^2)^2} = \frac{\frac 12(1+t^2)'}{(1+t^2)^2} = \frac 12 \left(\frac{1}{(1+t^2)^2}\right)\left(1+t^2\right)' = \frac 12f(\varphi(t))\varphi'(t)$$ then we can set $x=1+t^2 = \varphi(t)$.  First we note that $\varphi: [0,4]\to [1,17]$ is bijective so then we can use our formula to simplify this integral to $$\frac 12\int_{1}^{17} \frac 1{x^2}\ dx$$ which is much easier to evaluate.  This type of substitution is called a pushforward.

Right-to-left
Sometimes it'll be easier just to compose your function $f$ with a function that has some property that'll simplify your integral.  For instance consider this similar looking integral $$\int_0^{4} \frac{1}{1+x^2}\ dx$$ Here you can see that if you take $\varphi(t) = \tan(t)$ and compose it with the function $f(x) = \frac 1{1+x^2}$ then you'll be able to use the fact that $1+\tan^2(t) = \sec^2(t) = \cos^{-2}(t)$ to simplify this integral.  Then we just use the change of variables formula to get $$f(x) = \frac{1}{1+x^2} \\ \implies f(\varphi(t))\varphi'(t) = \left(\frac{1}{1+\tan^2(t)}\right)(\tan(t))' = \frac{\sec^2(t)}{\sec^2(t)} = 1$$ That definitely simplifies our integrand.

But before we can set up the new integral we'll need to find the inverse of $\tan$ and check for bijectivity.  The (restricted) inverse of $\tan$ is $\arctan$.  Using that we find that the endpoints of our interval in the domain of $\tan$ should be $\arctan(0) = 0$ and $\arctan(4)$.  And we know that $\tan: [0,\arctan(4)] \to [0,4]$ is a bijection.  Then our new integral becomes $$\int_{0}^{\arctan(4)} dt$$ which is much easier to evaluate.  This type of substitution is called a pullback.

Make sure that you understand the difference between pushforwards and pullbacks.  In elementary calculus courses they both go under the title of "$u$-substitution" but they actually require different mindsets to find the appropriate substitution.  For pushforwards we try to manipulate our integrand until we recognize it as a composition of the form $f(\varphi(t))\varphi'(t)$ (possibly with a constant factor that we can pull out of the integral).  For pullbacks we try to find a function $\varphi$ that has properties that'll help us simplify our integral.

For the integral in your question the best approach would be to use a pushforward.  That is, if you recognize that $$\frac{\sin(t)}{2+\cos(t)} = \frac{-(2+\cos(t))'}{2+\cos(t)} = -\left(\frac{1}{2+\cos(t)}\right)(2+\cos(t))'$$ then you can use the change of variables formula.

You may notice, however, that $\phi(t) = 2+\cos(t)$ is not bijective on $[0,5\pi]$.  However it is bijective on $[0,\pi]$, $[\pi, 2\pi]$, etc.  See if you can figure out how to handle this on your own.  Is there some condition you see (or can guess without proof) on when $\varphi$ can be non-bijective in pushforwards?

$\int \frac{\sin(t)}{2+cos(t)}dt$. Let $u=2+\cos(t) \to du=-\sin(t) dt$. Now we have $\int -\frac{1}{u} du = -\ln(u) +C$. Finally the integral is $-\ln(2+\cos(t)) +C$


Tricky Sum involving Binomial Coefficients and Sine I am stumped by the sum
$$\sum_{x=0}^n \binom{n}{x}\sin\big(\frac{\pi x}{n}\big)$$
but I can't figure it out. I tried expanding the taylor series of sine and using Euler's identity, but to no avail. Any hints?

PLEASE do not give me a full solution - I just need a hint.

Thanks!

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{x = 0}^{n}{n \choose x}\sin\pars{\pi x \over n} & =
\Im\sum_{x = 0}^{n}{n \choose x}\expo{\ic\pi x/n} =
\Im\sum_{x = 0}^{n}{n \choose x}\pars{\expo{\ic\pi/n}}^{x} =
\Im\pars{1 + \expo{\ic\pi/n}}^{n}
\\[5mm] & =
\Im\bracks{1 + \cos\pars{\pi \over n} + \ic\sin\pars{\pi \over n}}^{\,n}
\\[5mm] & =
\Im\bracks{\root{2 + 2\cos\pars{\pi \over n}}
\exp\pars{\ic\arctan\pars{\sin\pars{\pi/n} \over 1 + \cos\pars{\pi/n}}}}^{\,n}
\\[5mm] & =
2^{n/2}\,\bracks{2\cos^{2}\pars{\pi \over 2n}}^{\,n/2}\
\sin\pars{n\arctan\pars{2\sin\pars{\pi/\bracks{2n}}\cos\pars{\pi/\bracks{2n}} \over 2\cos^{2}\pars{\pi/\bracks{2n}}}}
\\[5mm] & =
2^{n}\,\verts{\cos\pars{\pi \over 2n}}^{\,n}
\sin\pars{n\arctan\pars{\tan\pars{\pi \over 2n}}}
\\[5mm] & =
2^{n}\,\verts{\cos\pars{\pi \over 2n}}^{\,n}\sin\pars{\pi \over 2} =
\bbx{2^{n}\,\verts{\cos\pars{\pi \over 2n}}^{\,n}} \\ &
\end{align}


Find the minimum of $(x(1+y)+y(1+z)+z(1+x))/\sqrt{xyz}$ over positive integers $x,y,z$ Let $x,y,z$ be positive integers.The least value
$$\frac{x(1+y)+y(1+z)+z(1+x)}{(xyz)^{\frac 12}}$$
I tried sum using arithmetic-geometric means inequality (seems promising as the denominator is similar to geometric mean of $x,y,z$). However, so far this didn't help my cause.

Take partial derivatives and then solve for them = $0$.


Cancelling out differentials I have been learning about differential equation on Khan Academy.

https://www.youtube.com/watch?v=DL-ozRGDlkY

And at 1:57 Sal multiplied the whole equation by dx and cancelled the dx in dy/dx. But isn't dy/dx something like an operator. Is it mathematically correct to do such cancellation?

So our good 'ol Sal is dealing with separable differential equations. The differential operator $\frac{dx}{dt}$ is indeed the same as $\frac{dx}{1} \cdot \frac{1}{dt}$, which means you could indeed cancel it out by multiplying both sides by one or the other. This is done because then each side may be integrated to reach a solution. As is $dx$ or $dt$ in the integral, $D=\frac{dx}{dt}$ is simply a value, and can be manipulated like any other variable.

I hope this helps!


absolute value inequality limit Could someone verify this proof? Prove $\lim\limits_{n\to\infty} C_n = \lim\limits_{n\to\infty} \dfrac{4n+3}{7n-5} = \dfrac{4}{7}$

Proof: Let $\epsilon > 0$ and take $N = \dfrac{41}{49\epsilon} + 1$ This implies that for all $n > \dfrac{41}{49\epsilon} + 1 \iff n-1 > \dfrac{41}{49\epsilon} \iff \epsilon(n-1) > \dfrac{41}{49} \iff \epsilon > \dfrac{41}{49n - 49} \iff$

$ \iff \epsilon > \dfrac{41}{49n-35} \iff \epsilon > \dfrac{28n+21-28n+20}{49n-35} \iff \dfrac{7(4n+3)-4(7n-5)}{7(7n-5)} \iff \epsilon > \dfrac{4n+3}{7n-5}-\dfrac{4}{7} \iff \epsilon >|\dfrac{4n+3}{7n-5}-\dfrac{4}{7}|$

End of proof

This is correct I guess, but extremely tedious. I would never actually prove a limit exists this way when we are dealing with quotients of polynomials. The standard thing to do is prove the following theorem:

  Theorem. If $\lim\limits_{n \to \infty} a_n = A$ and $\lim\limits_{n \to \infty} b_n = B$ and $c\in \mathbb{R}$ then
  $$
\lim\limits_{n \to \infty}(c a_n) = c A \text{ as long as $cA \neq 0 \cdot \infty$}
$$
  $$
\lim\limits_{n \to \infty}(a_n + b_n) = A + B \text{as long as $A + B \neq \infty - \infty$}
$$
  $$
\lim\limits_{n \to \infty}(a_n b_n) = AB , \text{ as long as $AB \neq 0\cdot \infty$}
$$
  $$
\lim\limits_{n \to \infty}\frac{a_n}{b_n} = \frac{A}{B}, \text{ as long as $B \neq 0$ and not $B=A=\infty$}.
$$

One you have this theorem we can write $$\frac{4n+3}{7n-5} = \frac{4 + \frac{3}{n}}{7 - \frac{5}{n}}$$ for all $n \neq 0$. Applying the quotient of limits part of the previous theorem we find
$$
\lim_{n \to \infty}\frac{4n+3}{7n-5} = \lim_{n \to \infty}\frac{4 + \frac{3}{n}}{7 - \frac{5}{n}} = \frac{\lim\limits_{n \to \infty}(4+\frac{3}{n})}{\lim\limits_{n \to \infty}(7 - \frac{5}{n})} = \frac{4+0}{7+0} = \frac{4}{7}.
$$

I have the same issue when it comes to proving limits with the inequality theorem.

I'm not sure the above is a proof by first principles though i.e. using the epsilon- delta method. Is there a way to explain how to use the ineuality theorem using first principles?


Use the sine rule to prove trig identity Using the sine rule:
$$ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$$

prove, for triangle ABC:

$$\sin\left(\frac{B-C}{2}\right) = \frac{b-c}{a} \cos\left(\frac A2\right)$$

Using the sine rule it's easy to translate the RHS into:

$$\sin\left(\frac{B-C}{2}\right) = \frac{\sin(B)-\sin(C)}{\sin(A)} \cos\left(\frac A2\right)$$

Yes, I can expand out the LHS, and use the difference of 2 sines in the RHS, but neither makes an obvious equality, especially with terms in a and A in the RHS.

A nudge in the right direction would really be appreciated. Thanks.

Just assume that $a/sin(A)=b/sin(B)=c/sin(C)=1/k$.

So $sin(A)=ka, sin(B)=kb,sin(C)=kc$.

Now substitute this and you will get the desired result.

Since $A+B+C=\pi$, the difference formula for $\sin$ gives \begin{align}\frac{\sin B-\sin C}{\sin A} \cos\frac A2&=\frac{2\cos\frac{B+C}2\sin\frac{B-C}2}{\sin(\pi-B-C)}\cos\frac{\pi-B-C}2\\&=\frac{2\sqrt{\frac{1+\cos(B+C)}2}\sin\frac{B-C}2}{\sin(B+C)}\sqrt{\frac{1+\cos(\pi-B-C)}2}\\&=\frac{2\sqrt{\frac{1+\cos(B+C)}2}\sin\frac{B-C}2}{\sin(B+C)}\sqrt{\frac{1-\cos(B+C)}2}\\&=\frac{\sin\frac{B-C}2}{\sin(B+C)}\sqrt{1-\cos^2(B+C)}\\&=\frac{\sin\frac{B-C}2}{\sin(B+C)}\sin(B+C)=\sin\frac{B-C}2\end{align} as required.


Finding the principal disjunctive normal form (PDNF) of a Boolean expression Find the principal disjunctive normal form (PDNF) of a Boolean expression
$$((p\wedge q) \rightarrow r)\vee((p\wedge q)\rightarrow \neg r).$$
I  tried by expanding it but I am stuck with the expression
$(\neg p \vee \neg q \vee r) \vee (\neg p \vee \neg q \vee \neg r)$. I don't know how to convert them into min terms. Please help me.

Expression mentioned is a Tautology.
$$((p.q) \rightarrow r)+((p.q)\rightarrow r')=((p'+q'+r)+(p'+q'+r'))=p'+q'+1 = 1$$
So PDNF corresponding to it is
$$(p'.q'.r')+(p'.q'.r)+(p'.q.r')+(p'.q.r)+(p.q'.r')+(p.q'.r)+(p.q.r')+(p.q.r)$$

The required normal form is ( if I am not wrong ):-

(p^q^r)v(~p^q^r)v(~p^~q^r)v(p^q^~r)

Once you get the form : (¬p∨¬q∨r)∨(¬p∨¬q∨¬r)
Treat this as (¬p)∨(¬q)∨(r)∨(¬p)∨(¬q)∨(¬r)
And convert the individual literals into minterms.

To convert any variable P to a minterm ,just add :  ^(Qv~Q)^(Rv~R)  to P
That is :P^(Qv~Q)^(Rv~R)And then use the distributive laws.


Why a linear numerator for fractions with irreducible denominators? For example: (2x^3+5x+1)/((x^2+4)(x^2+x+2)) breaks down to (ax+b/(x^2+4))+(cx+d/(x^2+x+2)). I have been told that since the denominators are irreducible, the numerators will be either linear or constant. Now my question is for something like (2x^3+5x+1/(x^2-4)) you would make it equal (A/(x+2))+(B/(x-2)), why do assume that the numerators are constant? Why couldn't the numerators be linear like the irreducible one?

With a fraction like $ax+b\over cx+d$ it is always possible to remove $a\over c$ copies of the denominator from the numerator, leaving a constant in the numerator:

$${ax+b\over cx+d}={ax+b-ax-\frac ac d\over cx+d}+\frac ac={b-\frac{ad}c\over cx+d}+\frac ac$$

In other words, we have the choice of leaving the fraction in a "linear over linear" form, or changing it to be in a "constant over linear plus constant" form.  Note that a function like $\dfrac a{bx+c}+d$ is easier to graph than one like $\dfrac{ax+b}{cx+d}$ because of the constant offsets compared with the affine transformation involved in the second case.

The answers to partial fraction reductions neglect the following....
(1) a function (numerator) such as ax + c is NOT lineAs previously noted  a polynomial of thaform is affine. Explicitlyan affine function is the composite of a linear function followed by a translation:
ax + b = (x + b) o ax. ax is the linear part, it satisfies the linearity conditis
f(a + b ) = f(a) + f(b) ; f(ka) = kf(a) ( for the experts: a linear function is a vector space homomorphism). A translation does not have those linearity properties, so the composite (affine function) is not linear. If the word "affine" is disliked, "1st. degree polynomial" is acceeptable.
(2) Partial fraction reduction is a consequence of the fact that polynomials  are integral domains, like Z, so possess a Division Algoritm prooperty.
For polnomials f and s, there are two polynomials: q and r such that
f  =  qs + r ; (degree r) < (degree s)....
if (degree f) < (degree s), there is no :q.

and since all polynomials can be factorised  with factors of degree 2 (fundamental theorem of algebra) r can be only of degree o or 1.
The decomposition of a Mobius function given in an earlier answer is treated in C.Sidne'ys Modern Basic Pure Mathematics.


A hausdorff topology that is not regular Consider, for $a$ and $b$ positive integers, the set $U(a,b)=\{an+b\mid n\in \mathbb{Z}\}\cap \mathbb{N}$. The set $\{U(a,b)\mid (a;b)=1\}$ is a basis for some topology on the naturals (I proved this).

Show that this space is Hausdorff but not regular.

Take $n, m\in \mathbb{N}$ such that $m\neq n$. I don't know hot to find disjoint open sets for each. Any hints?

To prove it's not regular, I tried taking the point $1$ and the set $C=\{2n\mid n\in \mathbb{N}\}$. Any open set $W$ such that $1\in W$ will have a basic set contained in it, but I don't see the problem.

Any help would be really appreciated.

For Hausdorff:

Consider $U(mn+1, m), U(mn+1,n)$. Since $\gcd(a\cdot k + 1, a) =1$, these are both basis elements, hence open. Further, they're disjoint, as if there exists $j,k$ such that
$$ (mn+1)k+m = (mn+1)j+n$$
Then $m = n $ modulo $mn+1$, but $m, n$ are distinct, both positive and less than $mn+1$, so this cannot happen. Thus there are no such $j,k$, and the sets are disjoint.

Now, for regular. Take $C,p$ as you've outlined above. Suppose for contradiction there are disjoint open sets $A, B$ containing $p, C$ respectively. Note that $A$ must contain a set of the form $U(k_1,1)$, where $k_1$ is even (otherwise $k_1+1 \in C$ and $k_1+1 \in U(k_1,1)$), as the only other basis sets containing $1$ also contain all of $\mathbb{N}$ - which obviously intersects with $B$.

Likewise, $B$ must contain contain $U(k_2,k_1)$ (with $gcd(k_2,k_1) = 1$) as it must contain $k_1$ (a positive even number, which is in $C$). Again if we have some other open set $U(a,b)$ containing $k_1$, then $a*n + b = k_1$ for some $n$, and hence this open set contains $U(a, a*n+b) = U(k_2, k_1)$.

However, $U(k_1,1) \cap U(k_2,k_1) \neq \emptyset$, because of Bezouts identity, since $gcd(k_2,k_1) = 1$, we have $i,j \in \mathbb{Z}$ such that $$i k_2 + j k_1 = 1 \implies i k_2 + k_1 = 1 - (j-1) k_1$$

To turn this element into a natural number (so we get our intersection) add a sufficiently large multiple of $k_1 \cdot k_2$ to each side. This is clearly an element of both $U(k_2, k_1)$ and $U(k_1, 1)$, so $A,B$ are disjoint. A contradiction.


Showing that there exists $c \in (a, b)$ with $f''(c) > 0$. Suppose we are given that a function, say $f$ is twice differentiable on $(a, b)$. Now $x_1, x_2, x_3 \in (a,b)$ and $f(x_2) < f(x_1)$ and $f(x_2) < f(x_3)$ (also $x_1 < x_2 < x_3$). I want to show that there is a $c \in (a, b)$ with $f''(c) > 0$. I've heard that we can apply the MVT three times here, but I do not see where that would help.

Graphically, this is intuitive and my belief is that there are many such $c$'s in $(a, b)$ such that $f''(c) > 0$. In fact, every $x  \in (a,b)$ has $f''(x) > 0.$

Anyways the mean value theorem states that: Suppose that $a , b \in \mathbb{R}$ with $a < b$. If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ then there exists a $c \in (a, b)$ such that $f(b) - f(a) = f'(c)(b-a)$.

Can you give me some hint?

First we apply the MVT twice

$$f(x_1) - f(x_2) = f'(c_1)(x_1-x_2),$$ so $f'(c_1) < 0$ for $c_1 \in (x_1,x_2)$.

$$f(x_3)-f(x_2) = f'(c_2)(x_3-x_2),$$ so $f'(c_2) > 0$ for $c_2 \in (x_2,x_3)$.

Now, $c_2 > c_1$, so $$f'(c_2)-f'(c_1) = f''(c)(c_2-c_1)$$ since the lhs is positive, and since $c_2 >c_1$ this $c$ gives $f''(c)>0$.

Notice that in all of this, we only look at the signs of things, and this is enough without worrying about any particular values.


Why is: $\lim_{n\to\infty}\frac1n\sum_{r=0}^{n}\left(\frac{r}{n}\right)=\int_0^1x\,dx$ Could somebody please explain how: $$\lim_{n\to\infty}\frac1n\sum_{r=0}^{n}\left(\frac{r}{n}\right)=\int_0^1x\,dx$$
I am just beginning integration and this has been highlighted as a very fundamental and important result/theorem in my senior's notes. Does it have any name ? Any books you could suggest would also be very helpful.

Firstly, notice that $x \mapsto x$ is a continuous function on $\mathbb{R}$, and hence uniformly continuous on the closed interval $[0,1]$, thus its Riemann integral on $[0,1]$ exists, and hence can be expressed as a limit of the left/right Riemann sum using partitions with strips of width $\frac{1}{n}$ for integer $n$. Secondly, the expression inside your limit on the left-hand side is just zero plus the right Riemann sum.

The $\text{LHS}$ is area under $y=x$ from $x=0$ to $x=1$.

It divides your triangle in infinitesimal thick rectangles with width $\frac1n$ and the term $\frac r n$ divides $x=0\to1$ as height of each rectangle.

$\text{RHS}$ is also area according to fundamental theorem of calculus.


A solution to a nonlinear recurrence equation I am interested in finding a solution to the following recurrence relationship:
$$
F_{n+1} = \sqrt{ A + B/F_n }
$$
OR equivalently, the equation:
$$
S_{n+1} = C + \frac{E}{\sqrt{S_n}}
$$
where $A,B, C, E$ are real positive numbers.
 I would appreciate to have a solution for any value of n, or at least to know if these sequences are bounded, or if they converge at all. The starting points are nonzero positive reals.

Many thanks in advance.

Thanks for those who help. I would like to add to this question, the possibility of having A and B being independently negative. Also, Does taking the absolute value inside the square root force convergence?
I mean: is the relation $$ F_{n+1} = \sqrt{ |A + B/F_n | } $$ better ( i.e. more guaranteed, i.e. wider range of possibilities ) to converge than the one above for any real value for A and B (positive or negative)?
Since F(0) is nonzero positive, we cannot be sure that F(n) is not zero now we are having negative values for A and/or B. So, any guess of a starting point to guarantee convergence at this case is also well appreciated.

Note:  I was reading the work of Rabinovic http://www.math.tamu.edu/~berko/papers/pdf/jmpRBH96.pdf ) , and I think the theorem reported therein can be a great basis for solving any nonlinear recursion if it could be written as a polynomial. A strategy that comes in my mind is: To expand the relation using Taylor series, and then write the general polynomial term as a function of F(n) . Then use the theorem reported in the that paper,
Is this idea sound?

Sorry for long blathering

In this case, a sketch is  very worthy.

It is clear how the recurrence is graphed, and how it is developing out.

We consider only the domain $0<x$.
In this function has an horizontal asymptote  $y_{\infty}= \sqrt{A}$, and remains above that.
The derivative
$$
y'(x) = {d \over {dx}}\sqrt {A + B/x}  =  - \;{B \over {2x^{\,2} \sqrt {A + B/x} }}
$$
is always negative if $0<B$.
The second derivative is always positive, for positive $A$ and $B$.

Then, since in the recursion remains $A<x$, if the absolute value of the
derivative at $x=\sqrt{A}$ is less than $1$
$$
\left| {y'(\sqrt A )} \right| = \;{B \over {2A\sqrt {A + B/\sqrt A } }} < 1
$$
the Priciple of Contraction guarantees that the recurrence
is convergent to $x=y(x)$.

HINT

Consider the sequence $(F_n)_{n \ge 0}$, given by the recurrence relation
$$
F_{n+1} = \sqrt{A+ \frac{B}{F_n}}, \quad F_0, A, B \in \mathbb{R}^+.
$$
If this sequence converges to some limit $F$, then $F$ must satisfy
$$
F = \sqrt{A + B/F}
$$
so you must solve $F^3 - AF - B = 0$, which has exact solutions using Cardano's formula. At least one of them is guaranteed to be real.

The second recurrence will similarly lead to $$(S-C)^2 = \frac{E^2}{S},$$ which is a cubic as well...


derivative of domain of integration Suppose $f(x,y)$ is a function $\mathbb{R}^2\times \mathbb{R}^2 \to\mathbb{R}$ and $\Omega(x)$ is a family of compact regions of the plane whose boundary curve $\gamma(s,x)$ varies smoothly in $x$. I need to take the derivative of
$$\int_{\Omega(x)} f(x,y)\,dy$$
with respect to $x$.

Intuitively I expect to get something like

$$\int_{\Omega(x)} \frac{\partial f}{\partial x}(x,y)\,dy + \int_{\gamma} f(x,\gamma(s,x))\frac{\partial \gamma}{\partial x}(s,x)\cdot n(s)\,ds$$
where $n(s)$ is the normal to $\gamma$... but how can I see this formally?

I've found the pertinent result -- my integral above is exactly the statement of the Reynolds transport theorem and a short proof can be found on page 22 of Advanced Transport Phenomena: Fluid Mechanics and Convective Transport Processes by Leal.

Combine three standard results:

If $x$ appears in an expression multiple times, we can differentiate the expression with respect to $x$ by differentiating with respect to each $x$ individually, holding the others constant, and adding up the results.  (This is just the ordinary multivariate chain rule, applied to a function $G(x_1, x_2, \ldots x_n)$ by taking $x_i(x) = x$.)
Differentiation under the integral (https://math.stackexchange.com/a/12911/3296)
Stokes's theorem


Using de Moivre's formula to find an expression for $\sin 3x$ in terms of $\sin x$ and $\cos x$ I was asked to use de Moivre's formula to find an expression for $\sin 3x$ in terms of $\sin x$ and $\cos x$.

De Moivre's formula is this: $$\cos nx+i\sin nx=(\cos x+i\sin x)^n$$
I plugged $3$ in for $n$ and got the following: $$\cos 3x+i\sin 3x = (\cos x+i\sin x)^3$$
At this point, I am stuck. I don't know how to take this and solve for $\sin 3x$. I could do: $\sin 3x=\frac{(\cos x+i\sin x)^3-\cos 3x}{i}$, but I do not think the answer is supposed to have an imaginary number in it.
If I expand my numerator I get:$$\frac{\cos^3x-3\sin^2x\cos x+i(3\cos^2x\sin x-\sin^3x)-\cos 3x}{i}$$ Simplifying this gets me: $$\frac{-3\sin^2x\cos x}i+3\cos^2x\sin x-\sin^3x$$ This still leaves me with an imaginary number. How do I get rid of it? And, are my other steps correct?

$$(\cos x+i\sin x)^3=\cos^3x+3i\cos^2x\sin x+3i^2\cos x\sin^2x+i^3\sin^3x$$
$$=\cos^3x+3i\cos^2x\sin x-3\cos x\sin^2x-i\sin^3x$$
$$=(\cos^3x-3\cos x\sin^2x)+i(3\cos^2x\sin x-\sin^3x)$$

Since this is $\cos 3x +i\sin3x$, we can conclude that $\cos3x=\cos^3x-3\cos x\sin^2x$ and $\sin3x=3\cos^2x\sin x-\sin^3x$.  You can simplify with the Pythagorean identities as you desire.

Expand the right hand side, i.e. the term $( ... )^3$;
Gather the imaginary part in the right hand side
Identify the imaginary parts.


Prove that a spanning tree of a connected multigraph contains at least one edge of every edge-cut Problem:

  Prove that a spanning tree of a connected multigraph contains at least one edge of every edge-cut, where an edge-cut is defined as the set of edges whose removal disconnects the graph.

Any hints/suggestions regarding this problem will be much appreciated.

Let $T$ be any spanning tree of a graph $G$ and $E$ an edge-cut. We need to show that $E\cap E(T)$ is non empty (then $T$ contains at least one edge of that arbitrary edge-cut). Let $v$ and $w$ be two vertices in different components of $G\setminus E$ (that is $G$ with edges $E$ removed). Since $T$ is a spanning tree, there is a path $P$ from $v$ to $w$ in $T$. If we had $E\cap E(T)=\emptyset$, then $T$ would be a subgraph of $G\setminus E$ and therefore $P$ would also connect $v$ and $w$ in $G\setminus E$, contradiction to the choice of $v$ and $w$. Hence $E\cap E(T)$ is non empty.

I think the definition of edge-cut you are using may not be quite correct. A cut of $G=(V,E)$ is defined to be a pair $(A,B)$ where $A,B\subset V$, $A\cup B=V$, and $A\cap B=\emptyset$ - in other words, we partition the vertex set $V$ into to sets $A$ and $B$. Now, based on this definition of cut, I would think that what is meant by edge-cut in this context is the set of edges that go between $A$ and $B$, explicitly $\{(v,w)\in E:v\in A, w\in B\}$. From these definitions, you should think about whether it is possible for a spanning subgraph to be connected if it misses any edges between $A$ and $B$ for any cut $(A,B)$.


Special Relativity-Book I would a good book to study the Special Relativity. In my course the professor has treated the following topics:

$(1)$ Lagrangian and hamiltonian dynamic of a charged particle;

$(2)$ Relaticistic cinematic;

$(3)$ Conservation principles in particle collisions.

In particoular I need a book full of exercises.

Thanks in advance.

I really like Gravity: An Introduction to Einstein's General Relativity by Hartle. It covers special relativity in the first few chapters. It also somehow manages to cover general relativity by holding your hand through differential geometry.

I really enjoyed reading S. Adams' book Relativity: an introduction to spacetime physics. It is easy to read, and it is written mathematically (for instance viewing Lorentz transformation as a rotation etc.).


Inequality $ \left(1+\frac{x}{n}\right)^n \leq \exp(x) $ How does one show this seemingly simple, but actually not that simple, inequality

$$
\left(1+\frac{x}{n}\right)^n \leq \exp(x)
$$

In this answer I define the natural logarithm as
$$
\log x =\int_{1}^{x}\frac{1}{t} \, dt \, .
$$
and the exponential function as the inverse of the logarithm.

We begin with the fact that
$$
\frac{x}{n} \geq \log\left(1+\frac{x}{n}\right) \, .
$$
which can be deduced geometrically:

$\log\left(1+\frac{x}{n}\right)$ is the area of the region bounded by the hyperbola $y=1/t$, the $t$-axis, and the vertical lines $t=1$ and $t=1+\frac{x}{n}$. The rectangle with a width of $x/n$ and a height of $1$ gives us an upper bound for this region. A little care is needed when $x/n$ is less than $0$, but the result still holds. Hence,
$$
n\log\left(1+\frac{x}{n}\right) \leq x \, ,
$$
and therefore
$$
\exp\left(n\log\left(1+\frac{x}{n}\right)\right)=\left(1+\frac{x}{n}\right)^n \leq \exp(x) \, .
$$

Assuming you can use the fact that the limit of that sequence is equal to $exp(x)$, then the sequence will be bounded by $exp(x)$, meaning that the inequality will be true.


Matrix Calculator Tools Does there exist any such software/online calculator where I can enter a matrix and am able to obtain the formula for an arbitrary power of the matrix?

It would be an added benefit if the entries of the matrix are allowed to be variables.

In case this question belongs somewhere else, I would be glad to be pointed in the direction.

Edit 1: I would like to add more detail as to what my aim is. I have a set of matrices consisting of two variables $a \in \mathbb{N}, b \in \mathbb{R}.$ I want to know whether for infinitely many of them the $a^{\text{th}}$ power is positive.

Right now I'm just handpicking some matrices and checking for them to see whether this is true. I would be happy to be suggested better methods for the same.

Edit 2: The matrices are as follows:

$$A(a,b)=\begin{pmatrix}\frac{a}{b}-1&-1&-1&-1\\
-1&\frac{a}{b}-1&-1&-1\\
1&-1&\frac{a}{b}-1&-1&\\
-1&0&-1&\frac{a}{b}-1\end{pmatrix};\ a\in \mathbb{N}, b \in \mathbb{R}$$

I want to check whether every entry of $A(a,b)^a$ is positive for infinitely many $a$ and $b.$

You can use Wolfram Alpha.

The general syntax is matrixpower[matrix in matrix form, power] and below are examples.

Here is one with only numbers.

Here is one with variables.

If you want to give a matrix $A$ and find $A^n$, you can also use this same approach and here is an example.

When a matrix is diagonalizable, you can use $A^n = P D^n P^{-1}$ from the Jordan Normal Form.

You should also consider using some of the free CAS tools like SAGE or Python.

Of course, professional CAS programs have things like Mathematica - MatrixPower.

You can also consider a free cloud account on the Wolfram Cloud, which gives you access to Mathematica commands, which are not all in Wolfram Alpha.

Update

$A^n(a, b) = \begin{pmatrix}
 \dfrac{-b n \left(\dfrac{a}{b}-1\right)^n+a \left(\dfrac{a}{b}\right)^n-b \left(\dfrac{a}{b}\right)^n}{a-b} & \dfrac{2 a \left(\dfrac{a}{b}-2\right)^n-2 b \left(\dfrac{a}{b}-2\right)^n-a \left(\dfrac{a}{b}-1\right)^n-a \left(\dfrac{a}{b}\right)^n+b \left(\dfrac{a}{b}-1\right)^n+b \left(\dfrac{a}{b}\right)^n+2 b n \left(\dfrac{a}{b}-1\right)^n}{a-b} & \dfrac{2 a \left(\dfrac{a}{b}-2\right)^n-2 b \left(\dfrac{a}{b}-2\right)^n-2 a \left(\dfrac{a}{b}-1\right)^n+2 b \left(\dfrac{a}{b}-1\right)^n+b n \left(\dfrac{a}{b}-1\right)^n}{a-b} & -\dfrac{b n \left(\dfrac{a-b}{b}\right)^n}{a-b} \\
 -\dfrac{b n \left(\dfrac{a-b}{b}\right)^n}{a-b} & \dfrac{2 a \left(\dfrac{a}{b}-2\right)^n-2 b \left(\dfrac{a}{b}-2\right)^n-a \left(\dfrac{a}{b}-1\right)^n+b \left(\dfrac{a}{b}-1\right)^n+2 b n \left(\dfrac{a}{b}-1\right)^n}{a-b} & \dfrac{2 a \left(\dfrac{a}{b}-2\right)^n-2 b \left(\dfrac{a}{b}-2\right)^n-2 a \left(\dfrac{a}{b}-1\right)^n+2 b \left(\dfrac{a}{b}-1\right)^n+b n \left(\dfrac{a}{b}-1\right)^n}{a-b} & -\dfrac{b n \left(\dfrac{a-b}{b}\right)^n}{a-b} \\
 \dfrac{b n \left(\dfrac{a-b}{b}\right)^n}{a-b} & \dfrac{-a \left(\dfrac{a}{b}-2\right)^n+b \left(\dfrac{a}{b}-2\right)^n+a \left(\dfrac{a}{b}-1\right)^n-b \left(\dfrac{a}{b}-1\right)^n-2 b n \left(\dfrac{a}{b}-1\right)^n}{a-b} & \dfrac{-a \left(\dfrac{a}{b}-2\right)^n+b \left(\dfrac{a}{b}-2\right)^n+2 a \left(\dfrac{a}{b}-1\right)^n-2 b \left(\dfrac{a}{b}-1\right)^n-b n \left(\dfrac{a}{b}-1\right)^n}{a-b} & \dfrac{b n \left(\dfrac{a-b}{b}\right)^n}{a-b} \\
 \left(\dfrac{a}{b}-1\right)^n-\left(\dfrac{a}{b}\right)^n & \left(\dfrac{a}{b}-2\right)^n-2 \left(\dfrac{a}{b}-1\right)^n+\left(\dfrac{a}{b}\right)^n & \left(\dfrac{a}{b}-2\right)^n-\left(\dfrac{a}{b}-1\right)^n & \left(-\dfrac{b-a}{b}\right)^n \\
\end{pmatrix}$

Now, you have to do cases based on $a, b$ to determine if things are positive. It may be easier to choose $a, b$ prior to taking the matrix to the $n-th$ power and seeing if you can determine it from that.

If you can diagonalize the matrix.  (Not all matrices can be diagonalized but most can)

$A = P^{-1}DP\\
A^2 = P^{-1}DPP^{-1}DP = P^{-1}D^2P\\
A^n = P^{-1}D^nP$

The diagonal matrix is easy to multiply by itself.


Writing Euclid's proposition "Ratios which are the same with the same ratio are also the same with one another" in contemporary style I need to write the Proposition 11 in Book 5 of Euclid's Elements according to contemporary style. It states that:

  Ratios which are the same with the same ratio are also the same with
  one another.

The clue in book says that it is about transitivity. However I could not indicate it mathematically.

May you help me to write it by using an if-then statement or another way? Thanks for your helps.

Let a, b and c be rational. If $a=b$ and $b=c,$ then $a=c.$


Checking the bound on an expectation inequality Let $X_1,X_2,\dots,X_n$ be independent random variables such that
  $\mathbf{E}(X_i)=0,\mathbf{E}(X_i^2)\leq 1$ for $1\le i\le n$ then
  show that $$\mathbf{E}\left(\max_{1\le i\le
> n}(X_1+X_2+\dots+X_i)^2\right)\leq \dfrac{n(n+1)}{2}$$

Now what I did was, the maximum would be some $(X_1+X_2+\dots+X_j)^2$ so we need to compute $$\mathbf{E}((X_1+\dots+X_j)^2)=\sum\mathbf{E}(X_i)^2+\sum\mathbf{E}(X_kX_\ell)=\sum_{i=1}^{j}\mathbf{E}(X_i^2)\le j$$
because $\mathbf{E}(X_kX_\ell)=\mathbf{E}(X_k)\mathbf{E}(X_\ell)=0$ is due to independence. But then $$\mathbf{E}\left(\max_{1\le i\le n}(X_1+X_2+\dots+X_i)^2\right)\le \max_{1\le j\le n}j=n$$
Which reduces the bound considerably. So, I am a bit worried this proof is flawed, can someone check it? Thanks a lot.

Your attempt is complete because you proved that
$$\max_{1\leqslant i\leqslant n}\mathbb E\left[\left(\sum_{j=1}^i X_j\right)^2 \right] \leqslant n,$$
but in general, for non-negative random variables $Y_i$, $\max_{1\leqslant i\leqslant n} \mathbb EY_i\lt \mathbb E\left[\max_{1\leqslant i\leqslant n} Y_i \right] $.

In this case, however, we can notice that
$$\max_{1\leqslant i\leqslant n}\left(\sum_{j=1}^i X_j\right)^2\leqslant
\sum_{i=1}^n\left(\sum_{j=1}^i X_j\right)^2,$$
then take the expectation and use the fact that the $X_j$'s are uncorrelated.


Maybe Poisson Distribution? The number of visitors I'm trying to solve the problem, can you help me?

To study the behavior of its costumers, a small shop registers the number of visitors that it receives per hour during a day. The results show that, on average, it has 1 visitors every hour. If we are told that in a given hour the shop has received 1 visitor at least, what is the probability that it will receive exactly 3 visitors within that hour?
HINT: Enter your answer with at least 3 decimal digits.

Let $N$ be the random variable giving the number of visitors in a given hour.

$$P(N=k)=e^{-\lambda}\frac{\lambda^k}{k!}$$

Here we know the expectation is $1$, so $\lambda=1$, since $E(N)=\lambda$.

The event $N\ge1$ the complement of $N<1$, which is $N=0$. So:

$$P(N\ge1)=1-P(N=0)=1-e^{-1}$$

Now, you want to know the probability that $N=3$ given that $N\ge1$, that is $P(N=3|N\ge1)$:

$$P(N=3|N\ge1)=\frac{P(N=3\wedge N\ge1)}{P(N\ge1)}$$

But of course, $N=3\wedge N\ge1$ is the same event as $N=3$, so

$$P(N=3|N\ge1)=\frac{P(N=3)}{P(N\ge1)}=\frac{\frac{e^{-1}}{3!}}{1-e^{-1}}=\frac{1}{6(e-1)}$$

Probability $P(3visitors)=e^{-\lambda}\frac{\lambda^k}{k!}=\frac{1}{6e}=0.06131$
where $\lambda=$no. of visitors per hour
$k=$ required no. of visitors.


Sequence convergence to zero Suppose that {$x_n$} is a sequence of positive numbers and

$\hspace{70mm}$ lim ($\frac {x_{n+1}}{x_n}$) = $L$

Show that  if $ L < 1$, then the lim $x_n = 0$

We know that if $L < 1$ then ${x_{n+1}}$ must be strictly less than ${x_n}$ and if ${x_{n+1}} < {x_{n}}$ for every $n$, then the sequence is monotone decreasing. Since the sequence is of all positive numbers, it is bounded below by zero, it's infimum. Is this sufficient?

Not sufficient. A sequence with $a_n=\frac{1}{2} + \frac{1}{n}$ is decreasing, bounded below by zero, monotone, and contains all positive numbers. Its limit is not zero.

Of course, its infimum is also not zero, but you haven't shown that $0$ is the infimum for $x_n$ either.

Hint: If $\frac{x_{n+1}}{x_n}$ would be $L$, then $x_{n+1}=L^nx_0$ which has a limit of $0$. Of course, you don't have $\frac{x_{n+1}}{x_n}=L$, but can you find a $L'<1$ such that you have $\frac{x_{n+1}}{x_n}<L'$ at least for $n>N$ for some $N$?

This was my first approach, but ultimately incorrect:

Consider the following:
$$ 0 < x_{n+1} < L x_n < L^2 x_{n-1} < ... < L^{n+1} x_0 $$

So, we ultimately have
$$ 0 < x_{n+1} < L^{n+1} x_0 $$

Now use squeeze theorem. :-)

(This is wrong because the ratio is only true in the limit; without carefully considering some $\epsilon$'s, this is no good.)

Try this approach instead

Try instead a contradiction -- suppose $L<1$ but $x_n \to c \neq 0$. Then, $x_{n+1}/x_n\to1$. Since $L\neq1$ (and we know we can't go off to infinity due to boundedness), we're done here.

** [Edit] That's wrong too, I think. **


What is an operator in mathematics? Could someone please explain the mathematical difference between an operator (not in the programming sense) and a function? Is an operator a function?

Based on your comment it sounds like you're actually asking about operations, not operators. A binary operation on a set $S$ is a special kind of function; namely, it is a function $S \times S \to S$. That is, it takes as input two elements of $S$ and returns another element of $S$. We can denote such an operation by a symbol such as $a \star b$ and then demand various additional properties of this operation, such as

associativity: $(a \star b) \star c = a \star (b \star c)$,
commutativity: $a \star b = b \star a$

and so forth. On the other hand, an arbitrary function $f : A \to B$ between two sets only takes a single input and returns an output which is not necessarily of the same type, so one can't speak of associativity or commutativity for such a thing. One might call a function $f : A \to A$ a unary operation but one still can't speak of associativity or commutativity for such a thing.

An operator is a mapping from one vector space or module to another. where a function is a map from any arbitrary set to another


When affine variety is complete? How to prove that an affine variety $X$is complete only if $\dim X=0$? It is clear that in this case $X$ must be a single point. But I don't known why its dimension should be zero. Could anyone help me? Thanks a lot!

If an affine variety $X\subset \mathbb A^n_k$ has more than one point (which is certainly the case if  $dim (X)\gt 0$) some coordinate function  $x_i$ is not constant on $X$ and we thus get a non constant regular function $x_i\in \mathcal O(X)\setminus k$.
On the other hand,  every regular function on a complete variety $Y$ is constant (in other words $\mathcal O(Y)= k$) and thus our positive-dimensional affine variety $X$ cannot be complete.

Edit
In the above answer, set in an elementary context not assuming schemes,  I Assumed $X$ irreducible and $k$ algebraically closed.
At a more advanced level one can only say that $\operatorname {dim} _k \Gamma(X, \mathcal O_X) \lt \infty $ for a scheme $X$ proper over a ( maybe not algebraically closed) arbitrary field $k$  . Thanks to @Alex for his comment which caused  this edit.

If I understand properly you are asking why an affine variety which is a single point has dimension zero?

It is very clear from the definition of dimension: A variety X has dimension n is there exists an chain of irreducible closed subsets of X $Z_0 \subsetneq Z_1\subsetneq...\subsetneq Z_n \subsetneq X$.

If X has to have dimension 1 then  u need at least  two points because you need two irreducible closed subsets of  X one containing the other properly.


Prove the cardinality of $\mathbb{Z}$ and $\mathbb{N}$ is the same. Prove the cardinality of $\mathbb{Z}$ and $\mathbb{N}$ is the same.

For the cardinality to be the same, there must exists a bijective function $f:\mathbb{N}\mapsto\mathbb{Z}$. If there exists a bijective function, then there exists an inverse $g:\mathbb{Z}\mapsto\mathbb{N}$ of it.

Suppose there exists such function. Then if $n\in\mathbb{N}$ and $z\in\mathbb{Z}$,

\begin{align}
(g\circ f)(n)&=n\\
g(f(n))&=n\\
g(z)&=n\\
f(g(z))&=f(n)\\
z&=f(n)\\
f(n)&=f(n)\\
f(n)-f(n)&=0\\
0&=0
\end{align}

Clearly this is true, so there exists such bijective function. Which means $\mathbb{Z}$ and $\mathbb{N}$ have the same cardinality.

But I feel this is incorrect, because if instead of $\mathbb{Z}$ and $\mathbb{N}$ I have to prove the equality of the cardinalities of $\mathbb{Z}$ and $\mathbb{R}$, the conclusion would be the same, but clearly this is not true.

This appears to be your argument, boiled down to the essentials:

"If a bijection exists, then it follows that 0=0.  Clearly 0=0 is true.  Therefore a bijection must exist."

The problem is that this is NOT a valid argument.  This is an instance of a logical fallacy known as 'Affirming the Consequent'.

Suppose such bijective function $f:\mathbb{N}\mapsto\mathbb{Z}$ doesn't exist.

Let $z\in\mathbb{Z}$ and $n\in\mathbb{N}$.

Then $\forall f:\mathbb{Z}\mapsto\mathbb{N}:(\text{$f$ is not a injection $\lor$ $f$ is not a surjection})$.

\begin{align}
\forall f:\mathbb{N}\mapsto\mathbb{Z}:(\text{$f$ is not a injection } \lor \text{$f$ is not a surjection})\\
\forall f:((\exists n_1,n_2:\lnot(f(n_1)=f(n_2)\implies n_1=n_2))\lor(\exists z:z\neq f(n))\\
\forall f:((\exists n_1,n_2:(f(n_1)=f(n_2)\land n_1\neq n_2))\lor(\exists z:z\neq f(n))\\
\end{align}

If this is false we are done.

Let $f:\mathbb{N}\mapsto\mathbb{Z}$ be arbitrary. We now need to show $(\exists n_1,n_2:(f(n_1)=f(n_2)\land n_1\neq n_2))\lor(\exists z:z\neq f(n))$.

If $\exists n_1,n_2:(f(n_1)=f(n_2)\land n_1\neq n_2)$ we have $n_1\neq n_2$ and applying $f$ to both sides:

$$f(n_1)\neq f(n_2)\tag{contradiction}$$

We concluded that $\exists n_1,n_2:(f(n_1)=f(n_2)\land n_1\neq n_2)$ must be false, so for the condition to be true $\exists z:z\neq f(n)$ must be true. So we need to find a function that takes a natural number as argument and maps it to the whole range of integers. Because we can construct a surjective function

$$
g(n)=\begin{cases}
n/2 & \text{$n$ is even}\\
-(n-1)/2 & \text{$n$ is odd}
\end{cases}
$$

that takes a natural number and can return every value of the integers the last statement is false, which means that supposing a bijective function $f$ does not exist has lead to a contradiction. Hence, a bijective function must exist. So the cardinality of $\mathbb{Z}$ and $\mathbb{N}$ must be the same.


Number of integral values of $x\in (\frac{\pi}{2},5\frac{\pi}{2})$ such that $\max(x^2+\cos x-1,1-x^2-\cos x)=\max(\cos x+x^2-1,x^2-1-\cos x)$ The number of integral values of $x\in(\frac{\pi}{2},5\frac{\pi}{2})$ such that

$$\max(x^2+\cos x-1,1-x^2-\cos x)=\max(\cos x+x^2-1,x^2-1-\cos x)$$ is satisfied is... ?

I understand that $\max(x,-x)=|x|$ but I could not simplify the right hand side of the equation.Please help me in this regard. Thanks.

For $x\in (\frac{\pi}{2},\frac{5}{2}\pi)$, we have that
$$\begin{align}(x^2+\cos x-1)-(1-x^2-\cos x)&=2(x^2+\cos x-1)\\\\&\ge 2(x^2-2)\\\\&\gt 2\left(\left(\frac{\pi}{2}\right)^2-2\right)\\\\&=\frac{1}{2}(\pi^2-8)\\\\&\gt 0\end{align}$$

and that
$$(\cos x+x^2-1)-(x^2-1-\cos x)=2\cos x\begin{cases}
\lt 0 & \text{if}\ x\in (\frac{\pi}{2},\frac{3}{2}\pi)\\[2ex]
\ge 0  & \text{if}\ x\in[\frac{3}{2}\pi,\frac{5}{2}\pi)
\end{cases}
$$

For $x\in (\frac{\pi}{2},\frac{3}{2}\pi)$, the equation can be written as
$$x^2+\cos x-1=x^2-1-\cos x,$$
i.e.
$$\cos x=0$$
which does not hold.

For $x\in[\frac{3}{2}\pi,\frac{5}{2}\pi)$, the equation can be written as
$$x^2+\cos x-1=\cos x+x^2-1$$
which always holds.

Therefore, the answer is $\color{red}{3}$. ($x=5,6,7$)

two possibilities. if the first part of LHS is greater then, the values of $x$ for which $cos(x)\,> -cos(x)$ are solutions. that is $5\,,\,6$.
 second possibility is second part of LHS and RHS are equal and greater than first part. there are two values for which second parts of max functions are equal and none of them are in the interval. hence there are only two values of x in the given interval which satisfy the equation namely, $5\,,\,6$. by  first part and second part I mean first  and second inputs to max function respectively.


$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}}$ How many solutions the following equation has in real numbers
?$$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}}$$
$1)1\quad\quad\quad\quad\quad\quad\quad2)2\quad\quad\quad\quad\quad\quad\quad3
 )3\quad\quad\quad\quad\quad\quad\quad4)4\quad\quad\quad\quad\quad5)\text{zero}$

Obviously we have $x\ge1390$. starting from $x=1390$ we can see LHS is very close to $1+1+1=3$ and RHS is close to zero. for higher values of $x$ we can see right side of the equation grows much faster than left side. so it seems the equation has not answer in real numbers.
I'm not sure how to proceed mathematically. I can rewrite it as follow:
$$\left(\sqrt{\frac{x-8}{1388}}-\sqrt{\frac{x-1388}{8}}\right)+\left(\sqrt{\frac{x-7}{1389}}-\sqrt{\frac{x-1389}{7}}\right)+\left(\sqrt{\frac{x-6}{1390}}-\sqrt{\frac{x-1390}{6}}\right)=0$$
I putted fractions with similar numbers in the same parentheses so it looks better now, but don't know how to continue.

Write $x=1396+t$. Then
$$\sqrt{1+\frac t{1388}}\:+\sqrt{1+\frac t{1389}}\:+\sqrt{1+\frac t{1390}}$$
$$=\sqrt{1+\frac t8}\:+\sqrt{1+\frac t7}\:+\sqrt{1+\frac t6}.$$
Clearly $t=0$ is one solution. Now, $t$ cannot be positive, because that would make the terms on the LHS each smaller than the terms on the RHS. Similarly, $t$ cannot be negative, for the opposite reason. Therefore $t=0$ or $x=1396$ is the unique solution.


Show $y^3+xy+x^2(x-1)^2\in\mathbb{R}[x,y]$ is irreducible I have tried applying Eisentein's Criterion to no avail, and I have also tried reducing modulo some proper ideal, with no luck either. Is there some standard way of showing polynomials in two variables are irreducible?

Suppose $f(x,y)=y^3+xy+x^2(x-1)^2$ and suppose we have the decomposition
$$f(x,y)=g(x,y)h(x,y).$$
Since degree of $f(x,y)$ with respect to $y$ is $3$, then we must have
$$g(x,y)=p(x)y+q(x)\quad \text{and}\quad h(x,y)=r(x)y^2+s(x)y+t(x)$$
where $p(x),q(x),r(x),s(x)\in\mathbb{R}[x]$ are polynomials with degree at most $4$.

Expanding $g(x,y)h(x,y)$ we reach
$$f(x,y)=p(x)r(x)y^3+(p(x)s(x)+q(x)r(x))y^2+(p(x)t(x)+q(x)s(x))y+q(x)t(x)$$
which implies $p(x)=r(x)=1$ are constant polynomials because in $\mathbb{R}[x]$ only constant polynomials are invertible.

Now by comparing coefficient, we see
$$s(x)=-q(x),\quad t(x)=-q(x)s(x)+x,\quad q(x)t(x)=x^2(x-1)^2$$
i.e. $$t(x)=q(x)^2+x$$ and therefore, $$q(x)(q(x)^2+x)=x^2(x-1)^2$$
which is impossible because LHS is a polynomial of degree multiple of $3$.

Perhaps consider mod 6, as then $y^3\equiv y$ and $y(x+1)+x^2(x-1)^2$ is irreducible


Middle School Level - Probability If you have 3 boxes and 7 coins fall in randomly, what is the probability that 3 coins fall into the first box?

My theory is that if you put one coin into each box until you have a remainder, by using modulus, you get box%coins = x. And then divide by the number of boxes to get the actual probability box%coins = x/boxes

This gives 1/3 as the probability in the case of 3 boxes and 7 coins. Is this correct?

Also, does it work for any number of boxes and coins?

There are $9 \choose 2$ ways seven coins can fall into three boxes, and $5 \choose 1$ ways for four coins to fall into two boxes, so the probability is
$$
{5\choose1}/{9\choose2}={5\over36}
$$


Multiplying by a $1\times 1$ matrix? For matrix multiplication to work, you have to multiply an $m \times n$ matrix by an $n \times p$ matrix, so we have $$\bigg(m \times n\bigg)\bigg(   n\times p \bigg).$$

But what about a $1 \times 1$ matrix? Is this just a scalar? But every matrix can be mulitplied by a scalar; so do $1 \times 1$ matrices break the rule?

A $1\times 1$ matrix is not the same thing as a scalar.  There is a one to one correspondence between them, but $[a]$ and $a$ are two different things, the first being a matrix and the second one being a scalar.  The confusion comes from the fact that the rings $R$ and $M_1(R)$ are isomorphic, and so $M_n(R)$ is isomorphic to $M_n(M_1(R))$.  When you talk about treating $[a]$ as a scalar, you are implicitly using this isomorphism.

Yes!  The simple answer is that a 1 by 1 matrix is a scalar and a scalar is a one by one matrix.  This makes sense because if you regard the dot product of two vectors (which always returns a scalar) as a row vector times a column vector then you always get a 1 by 1 matrix.  There is no distinction as they have all the same properties (commutativity, same inverse, associative, etc...), if you want to be "weird", you can leave the [brackets] around the 1 by 1 matrix and say that there is a isomorphism from the 1 by 1 group of matrices to the scalars of F (R or C, etc).   Define it as, g([A])=A.  If you want to be really neat you need to just check that if the matrix is [0] that it has no inverse, and therefore g([0])=0 and g^-1 does not exist for this particular "matrix".


How can I use this equation to solve for $t$? The problem given is, "Using the equation $R = R_{20} [1 + a_{20} (t-20)] $, find the temperature at which the resistance will increase to 1 Ohm from a level of 0.8 Ohm at 20C", where $a_{20} = 0.00393$ and $R_{20} = 0.8$ (Ohm).

I have the answer from the book (83.61C), but I've failed to rearrange the equation to make $t$ the subject and it seems to imply I should be using the equation as is. Am I missing something?

Note you have: $R=R_{20}\big(1+a_{20}(t-20)\big)$. First, use the distributive property on the inside, so you have $R=R_{20}(1+a_{20}t-20a_{20})$. Now use the distributive property again: $R=R_{20}+R_{20}a_{20}t-20R_{20}a_{20}$. Now you want to solve for $t$. So move everything that does not have a $t$ to the same side: $R-R_{20}+20R_{20}a_{20}=R_{20}a_{20}t$. Now to solve for $t$, divide by its coefficient:
$$
t=\frac{R-R_{20}+20 R_{20}a_{20}}{R_{20}a_{20}}
$$
Now all that is left to do is plug in $R=1, R_{20}=0.8,a_{20}=0.00393$ to get $t \approx 83.61323$, as you can see here.


Programming the Euclidean algorithm for arbitrary Euclidean Domains in Maple I have to implement the Euclidean algorithm for arbitrary domains in Maple.

I know how to do this for integers:

GCD:=proc(a,b) description "Computes the GCD of two integers";
    local r_0, r_1, aux;
    r_0 := a; r_1 := b;
    while r_1<>0 do;
        aux := irem(r_0, r_1);
        r_0 := r_1; r_1 := aux;
    od;
    return r_0;
    end proc:

But how do I represent elements from arbitrary domains in Maple? And how do I operate with them?

You will have to learn how to use Maple's object oriented programming interface. You can define a class whose instances will be Euclidean Domains and use methods from that class on objects of that class.

I don't know Maple. If there is a class for Rings already you will be able to build on that.

https://www.maplesoft.com/support/help/maple/view.aspx?path=object


Question about Angle-Preserving Operators This an exercise out of Spivak's "Calculus on Manifolds".

Edit:  There was a typo in the exercise as is noted below in the answers.  The statement has been edited to reflect this.

Given $x,y\in\mathbb{R}^{n}$, the angle between $x$ and $y$ is defined by

$$\angle(x,y) = \arccos\left(\frac{\langle x,y \rangle}{|x|\cdot |y|}\right),$$ where $\langle x,y \rangle$ denotes the standard Euclidean inner product.

A linear operator $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is said to be angle-preserving if
$\angle(T(x),T(y)) = \angle(x,y)$ for every $x,y\in\mathbb{R}^{n}$.

The exercise as stated:

Let $\{x_{1},\dots, x_{n}\}$ be a basis for $\mathbb{R}^{n}$.  Then suppose that $\lambda_{1}, \dots, \lambda_{n}\in \mathbb{R}$ are such that $Tx_{j} = \lambda_{j}x_{j}$ for each $j = 1,\dots, n$.

Then $T$ is angle-preserving only if (not if and only if!)$|\lambda_{i}| = |\lambda_{j}|$ for every $1\leq i\leq j\leq n$.

I'm having problems with the $(\Rightarrow)$ direction.

My best attempt (which seems to lead nowhere) is to suppose that $|\lambda_{j}|\neq |\lambda_{k}|$.  Then by assumption,
\begin{align*}
\angle(Tx_{j},Tx_{k}) & = \arccos\left(\frac{\langle Tx_{j},Tx_{k} \rangle}{|Tx_{j}|\cdot |Tx_{k}|}\right)\\
& = \arccos\left(\frac{\langle \lambda_{j}{x_{j}},\lambda_{k}{x_{k}} \rangle}{|\lambda_{j}{x_{j}}|\cdot |\lambda_{k}{x_{k}}|}\right)\\
& = \arccos\left(\frac{\lambda_{j}\lambda_{k}\langle {x_{j}},{x_{k}} \rangle}{|\lambda_{j}|\cdot|\lambda_{k}|\cdot|{x_{j}}|\cdot |{x_{k}}|}\right)\\
& = \arccos\left(\text{sign}(\lambda_{j})\text{sign}(\lambda_{k})\frac{\langle {x_{j}},{x_{k}} \rangle }{|{x_{j}}|\cdot |{x_{k}}|}\right)\\
\end{align*}
may also be calculated as
\begin{align*}
\angle(Tx_{j},Tx_{k}) & =  \angle(x_{j},x_{k})\\
& = \arccos\left(\frac{\langle x_{j},x_{k} \rangle}{|x_{j}|\cdot |x_{k}|}\right).
\end{align*}

Then since $\arccos$ is injective, I believe I can make the jump that $\text{sign}(\lambda_{j})\text{sign}(\lambda_{k}) = 1$, which does not resemble the conclusion that I should arrive at.

Note:  I wasn't sure what tag to put this under, so anyone who knows better please feel free to adjust.

Thanks for any help you can give.

If necessary, rescale each $x_i$ so they all have unit length and relabel the indices so that $|\lambda_1|\neq |\lambda_2|$.

Now, consider the vectors $v_1 = x_1 + x_2$ and $v_2 = x_1-x_2$.

First, I claim that $v_1$ and $v_2$ are orthogonal, for \begin{align*} \langle v_1, v_2\rangle &= \langle x_1+x_2,\; x_1-x_2\rangle \\\ &= \langle x_1, x_1\rangle + \langle x_1, x_2\rangle - \langle x_2,x_1\rangle - \langle x_2, x_2\rangle \\\ &= |x_1|^2-|x_2|^2 \\\ &=0\end{align*} where the last equality follows since both $x_1$ and $x_2$ are unit length.

Next, I claim that $Tv_1$ and $Tv_2$ are not orthogonal.  For, \begin{align*} \langle Tv_1,  Tv_2\rangle &= \langle \lambda_1 x_1 + \lambda_2 x_2,\; \lambda_1 x_1 - \lambda_2 x_2\rangle \\\ &= \lambda_1^2 \langle x_1, x_1\rangle +\lambda_1 \lambda_2 \langle x_1, x_2\rangle - \lambda_2\lambda_1 \langle x_2,x_1\rangle - \lambda_2^2 \langle x_2, x_2\rangle \\\ &= \lambda_1^2 |x_1|^2 - \lambda_2^2 |x_2|^2 \\\ &= \lambda_1^2 - \lambda_2^2\end{align*}

and this last line is $0$ iff $|\lambda_1| = |\lambda_2|$.

New bug in above arguments! If $AA^T=\sigma ^2I$ we cannot say $A=\sigma I$. Counterexample:  take a diagonal matrix with $+\sigma$ or $-\sigma$ on diagonal, and $0$ elsewhere.

The very final answer to these circle of ideas is
Which I have proved with ALL details, and optimal care to such issues as orientation.
For 1 => 2 see ALL Orthogonality preserving linear maps from $\mathbb R^n$ to $\mathbb R^n$?
Please challenge.


How can I show this statement. Show that there is no holomorphic fuction $f$ in the unit disc $D$ that  extends continuously to boundary of $D$ such that $f(z)=\frac{1}{z} ~for~ z\in \partial( D) $.

I tried to apply maximum principle but I couln't find the way to prove it.

Help me please.

I just update the full statement and I think it probably assume it is not constant fuction.

Thank you.

If such a function existed, then for all $r < 1$ you'd have $\int_{|z| = r} f(z)\,dz = 0$ by Cauchy's Theorem. Taking limits as $r \rightarrow 1$ would give $\int_{|z| = 1} f(z)\,dz = 0$ as well,
contradicting that  $\int_{|z| = 1} {1 \over z}\,dz = 2\pi i$.

By maximum principle we get that the function $g(z)=zf(z)-1$ can take the maximum value as $0$, but it can have the value less than $1$ just like in case if $z=0$, $g(0)=-1$, why it is a contradiction?


If $u_{n+2}-4u_{n+1}+4u_n=0$ then why $u_n=A2^n+Bn2^n$? Let $V=\{u\in\mathbb R^{\mathbb N}\mid\ u_{n+2}-4u_{n+1}+4u_n=0\}$, a vector space.

1) Show that $V$ has dimension $2$.

2) Show that $V=\text{Span}((2^n)_{n\in\Bbb{N}},(n2^n)_{n\in\Bbb{N}}).$

Let $u_n=r^n$. Then $$r^{n+2}-4r^{n+1}+4r^n=0\iff 0=r^2-4r+4=(r-2)^2\iff r=2.$$
Since I have a unique root, $u_n=A2^n+Bn2^n=D2^n$, so I can't get a sum of two linear independent solution. Is there a way as for ODE ? (i.e. for example if $f''-4f'+4=0$, then by setting $D(f)=f'$, then $D^2f-4Df+4=(D-2)^2f=0$ and thus $f$ is is a generalized eigenvector, and we can easily find a basis of $\{f\mid f''-4f+4=0\}$. Is there a similar way for sequences ?


An equation in two variables? Given 13x + 35y = 2000. How do I find positive integer solutions for this equation (without hit and trial).

My work :- I know I can use Bezout's Theorem to find integer solutions to this equation if I have first solution (x., y.). But I just want positive solutions. To find the first solution I tried using Euclid's algorithm but could not progress further. Please help. Or if you have a smaller way of doing this type of questions feel free to post them as answers. (for all integer solutions or only (+)ve integer solutions).

It's immediately obvious that $x$ has to be a multiple of $5$, if it has any hope of satisfying this. Hence write $x=5z$ and so we get:

$$35y+65z=2000$$

Now you can apply what Alfred suggested because $35+65=100$ which divides $2000$.


Probability of meeting This is a basic probability question.

Persons A and B decide to arrive and meet sometime between 7 and 8 pm. Whoever arrives first will wait for ten minutes for the other person. If the other person doesn't turn up inside ten minutes then the person waiting will leave. What is the probability that they will meet? I am assuming uniform distribution for arrival time between 7 pm and 8 pm for both of them.

My favourite way of solving uniform distribution related problems is to try and think of them in terms of relative volume/area/length. For instance, take a continuous uniform distribution on $[0, 1]$. The probability that $X \in (a, b)$, where $0 \le a \le b \le 1$, and $X$ is a random variable drawn from that distribution, is just the length of the interval $(a, b)$.

Let's do another example. Let $X$ be a uniform distribution over $(0, 5)$. What is the probability that $X \in (a, b)$, where $0 \le a \le b \le 5$?. It is just $\frac{\text{length of interval $(a,b)$}}{\text{length of interval (0, 5)}} = \frac{\text{length of interval $(a,b)$}}{5}$.

This generalizes as well. Let there be a uniform distribution over a square $S$. What is the probability that a randomnly drawn point will be in a particular region $H$?. The answer is $\frac{\text{area of $H$}}{\text{area of $S$}}$.

Let's apply this technique to the current problem.

The above picture represents the problem. It is easy to see that the distribution in the question is equivalent to placing a uniform distribution over the square in the diagram, where the times of arrival are the co-ordinates of a randomnly drawn point from the distribution. The shaded are is the favourable area. It represents points $(x, y)$ which are within $10$ of each other. Thus what we need is the area of that region divided by the area of the whole region.

I hope your are doing well, I tried a new method to resolve the meeting problem but I didn't find the same result.

We have to calculate $P(|X-Y| < 1/6)$ where X and Y are independents uniformly distributed between 0 and 1.

So : $P(|X-Y| < 1/6)  =  \int_{0}^{1} \int_{0}^{1} |x - y| * dx*dy$

Then : $P(|X-Y| < 1/6)  =  \int_{0}^{1} \int_{y - \frac{1}{6}}^{y + \frac{1}{6}}  dx*dy$

Finally : $P(|X-Y| < 1/6)  = \int_{0}^{1} \frac{1}{3} *dy = \frac{1}{3} $

Please help me to find my error.

Thank you


Solving $\sin(5x) = \sin(x)$ If I have an equation:

$$\sin(5x) = \sin(x)$$

In what case can I equate $$5x = x$$

Is it only when there is a multiply of $2\pi n$ on either side, where n is any integer so

$$ 5x = x+2\pi n$$

Also with this method can I get every possible solution or does that not work?

I know I can use the sine addition formula but I want to see others I way I can solve this.

You might find this useful:

$$
\sin(A)=\sin(B) \iff A=B +n2\pi\ \mbox{ or }\ A=\pi-B +n2\pi,
$$
with $n\in\mathbb{Z}$.
$$
\cos(A)=\cos(B) \iff A=B +n2\pi\ \mbox{ or }\ A=-B +n2\pi,
$$
with $n\in\mathbb{Z}$.

So you need to solve two equations separately:
$$5x=x+n2\pi$$
and
$$5x=\pi-x+n2\pi.$$
Good luck :)

Edit:
One way to prove
$$
\sin(x)=\sin(y) \iff x=y +n2\pi\ \mbox{ or }\ x=\pi-y +n2\pi,
$$
is to just verify the solutions work, using a consequence of Euler's formula:
$$\sin(x)=\frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}.$$
So, here we go:
$$
\sin(x)=\sin(y) \iff \frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}=\frac{\mathrm{e}^{iy}-\mathrm{e}^{-iy}}{2i} \iff \mathrm{e}^{ix}-\mathrm{e}^{-ix}=\mathrm{e}^{iy}-\mathrm{e}^{-iy}.
$$

Now note that $x=y+n2\pi$ is indeed a solution:
$$\mathrm{e}^{ix}-\mathrm{e}^{-ix}=\mathrm{e}^{i(y+n2\pi)}-\mathrm{e}^{-i(y+n2\pi)}=\mathrm{e}^{iy}\mathrm{e}^{in2\pi}-\mathrm{e}^{-iy}\mathrm{e}^{-in2\pi}=\mathrm{e}^{iy}-\mathrm{e}^{-iy}
$$
and that $x=\pi-y+n2\pi$ is indeed a solution:
$$\mathrm{e}^{ix}-\mathrm{e}^{-ix}=\mathrm{e}^{i(\pi-y+n2\pi)}-\mathrm{e}^{-i(\pi-y+n2\pi)}=\mathrm{e}^{-iy}\mathrm{e}^{i(2n+1)\pi}-\mathrm{e}^{iy}\mathrm{e}^{-i(2n+1)\pi}=\mathrm{e}^{iy}-\mathrm{e}^{-iy}.$$

The last equalities use $\mathrm{e}^{in\pi}$ is equal to $1$ if $n$ is even and equal to $-1$ if $n$ is odd.

Alternative solution: expand the LHS:
$$16\sin^5 x - 20\sin^3 x + 5\sin x = \sin(5x) = \sin x,$$
and $s = \sin x$ is a solution of
$$(16s^4 - 20s^2 + 4)s = 16s^5 - 20s^3 + 4s = 0.$$


An abstraction of matrices that can be added, inverted, transposed, multiplied $$x^TAx+x^Tb+c=(x-h)^TA(x-h)+k$$
where
$$h=-(A+A^T)^{-1}b$$
$$k=c-h^TAh$$

Determining $h$ and $k$ above is called "completing the square" and requires matrix addition, inversion, transposition, multiplication etc.

I am implementing algorithms (such as the above and others) in a generic way such that they can apply to any entity on which the requisite operations (addition, inversion, etc.) are defined.

I would call this library "matrix algebra" but I wondered whether there is a more abstract/generic name for "something that can be added, inverted, transposed, multiplied, etc.")?

Yes this would be an example of geometric algebra. A matrix is just a representation of general algebraic /geometric objects like vectors, multivectors, dual multivectors etc..


Is there an analytic function with zeroes only at $-2n$, and zeroes at $\frac12\pm it$, and further, symmetric zeroes within the critical strip? Is there an analytic function with zeroes only at:

every $-2n$,
$\frac12\pm it$, and
at least one at $\frac12\pm\epsilon\pm it$ where $0<\epsilon<\frac12, t\neq0$ (and these zeroes observing the known reflective symmetries within the critical strip)?

A answer assuming the Riemann hypothesis true would be fine.

To my mind the uniqueness of any analytic continuation of $\zeta(s)$ is suggestive of the existence of such a function being incompatible with the Riemann Hypothesis.

If not, uniqueness of $\zeta$ is of course a nice simple sufficiency for the Riemann hypothesis.

Take $\zeta(z)\left(z-\frac14-i\right)\left(z-\frac14+i\right)\left(z-\frac34-i\right)\left(z-\frac34+i\right)$.

Every condition that you mentioned holds.

I think this is an open question. As long as your function is automorphic, the Grand Riemann Hypothesis asks this exact question.


We all use mathematical induction to prove results, but is there a proof of mathematical induction itself? I just realized something interesting. At schools and universities you get taught mathematical induction. Usually you jump right into using it to prove something like

$$1+2+3+\cdots+n = \frac{n(n+1)}{2}$$

However.

I just realized that at no point is mathematical induction proved itself? What's the mathematical induction's proof? Is mathematical induction proved using mathematical induction itself? (That would be mind blowing.)

Suppose we want to show that all natural numbers have some property $P.$ One route forward, as you note, is to appeal to the  principle of arithmetical induction.

The principle is this: Suppose we can show that (i) $0$ has some property $P$, and also that (ii) if any given number has the property $P$ then so does the next; then we can infer that (iii) all numbers have  property $P$.

In symbols, we can use $\varphi$ for an expression attributing some property to numbers, and we can put the induction principle like this:

  Given (i) $\varphi(0)$ and   (ii) $\forall n(\varphi(n) \to \varphi(n + 1))$,  we can infer (iii) $\forall n\varphi(n),$

where the quantifiers run over natural numbers.

The question being asked is, in effect, how do we show that arguments which appeal to this principle are good arguments?

Just blessing the principle with the title "Axiom" doesn't yet tell us why it might be a good axiom to use in reasoning about the numbers. And producing a proof from an equivalent principle like the Least Number Principle may well not help either, as the question will just become why arguments which appeal to this equivalent principle are good arguments?

Is there nothing to be said then?

Well we can perhaps still hope for some conceptual clarification here. To that end, consider this informal argument.

  Suppose we establish both the base case (i) and the induction step
  (ii).

  By (i) we have $\varphi(0)$.

  By (ii), $\varphi(0)  \to \varphi(1)$. Hence we can infer
  $\varphi(1)$.

  By (ii) again, $\varphi(1)  \to \varphi(2)$. Hence we can now infer
  $\varphi(2)$.

  Likewise, we can use another instance of (ii) to infer $\varphi(3)$.

  And so on and so forth, running as far as we like through the
  successors of $0$ (i.e. through the numbers that can be reached by
  starting from zero and repeatedly adding one) until we get to the $n$'th successor of zero.

  But the successors of $0$ are the only natural numbers, so for every
  natural number $n$, $\varphi(n)$.

Set out like that, we see that the crucial point here is the last one. And what this reasoning reveals is how arithmetical induction principle is tantamount to a claim about the basic structure of the number sequence. The natural numbers are characterized as those which you can get to step-by-step from zero by repeatedly adding one, and that implies the absence of `stray' numbers that you can't get to step-by-step from zero by applying and reapplying the successor function. If a property gets passed down from zero, and from successor to successor, it gets passed on to every number because there are no 'stray' numbers outside the sequence of successors.

Now, how illuminating is that? We are in effect defining the natural numbers as those objects for which that informal line of argument is a good one, i.e. the numbers are those objects over which we can do arithmetical induction? Is this going round in circles?

Well, in a sense, yes! But note this isn't a vicious circle -- rather it is e.g. just how Frege and Russell famously defined the natural numbers. The point that is emerging from these reflections is that the principle of induction just cashes out the intuitive idea that the natural number are as those which you can get to step-by-step from zero by repeatedly adding one. To fix what we are talking about in talking of the natural numbers is to fix that we are talking about things for which the principle of arithmetical induction holds. That is why we can take the principle as axiomatic.

Note: Of course, you can do fancy things like model natural numbers in ZFC, and then prove in ZFC that a formal version of induction-over-these-number-representing-sets. But it is important to see that this does not help with the question that is being asked here. After all, what is the criterion for judging that a certain sequence of sets defined in ZFC is appropriate to be used as a model of the numbers? Not any old sequence will do. Some are far too long! An appropriate sequence is one where you can get step-by-step from the 'zero' by repeatedly 'adding one', and there are no 'stray' numbers in the sequence that you can't get to step-by-step from 'zero' by applying and reapplying the 'successor' function in the model. In other words, we choose the implementation of natural numbers in ZFC in order to ensure that the sequence in question has the right structure to make arithmetic induction work. Now, you can prove things in ZFC about that sequence of sets, including an analogue of arithmetical induction. But to take that as giving you some independent reason for thinking that induction holds over the numbers is to get things exactly upside down. Proving induction in ZFC is, to put it vividly, only taking out the inductive rabbit that you've already smuggled into the hat when you chose how to  implement the numbers!

Suppose that $ 1 \in S \text{ and that } n \in S  \Rightarrow n+1 \in S$, and suppose that there is a smallest number $N$ such that $N+1\notin S$...

Supposing that any subset of natural numbers has a minimal element, the above proves the principle of induction in set theory, whether some people like it or not.


Find the area of an equilateral triangle given the distances from an interior point to the vertices Given the distances from an interior point to the vertices of an equilateral triangle, find the area of that triangle.

I have already tried equating $\sqrt{3}\times a^2/4$ and sum of the area of three triangles interior to equilateral triangle  formed by given lines. But that approach is making a hard equation to solve.

Any solution using Computer programming language may also help.

The following relation holds:
$$3(p^4+q^4+t^4+a^4)=(p^2+q^2+t^2+a^2)^2$$
where
$p,q,t$ are the distances from the vertices and $a$ is the length of the side of the triangle

According to the link given by @Greg (above), the area of the equilateral triangle ABC with an interior point P, will be
((sqrt(3)(PA^2+PB^2+PC^2)/4) + 3(area formed by triangle with 3 lines as PA, PB and PC))/2]]


Proving $ \sum_{n=1}^{ \infty } f_n(1) $ converges when $ \sum_{n=1}^{ \infty } f_n(x) $ converges uniformly I have a question which I tried solving for a few hours.
My only solid direction so far is trying using Cauchy.
Let $(f_n(x))$ be a series of continuous functions in $[0,1]$.
Prove; If $ \sum_{n=1}^{ \infty }  f_n(x) $ converges uniformly at $[0,1)$, then $ \sum_{n=1}^{ \infty }  f_n(1) $ converges.
Thank you in advance, been trying a lot solving this. Much appreciated.

Maybe not the most elegant, but a detailed way to get it by "splitting $\varepsilon$."
Let $S_n\colon[0,1]\to\mathbb{R}$ be the function defined by
$S_n(x) = \sum_{k=1}^n f_k(x)$, for $n\geq 1$ and $x\in[0,1]$. Since the $f_n$'s are continuous on a compact set, they are uniformly continuous, so is $S_n$. Moreover, by assumption the sequence $(S_n)_n$ converges uniformly on $(0,1)$, to a limit which we will call $S\colon(0,1)\to\mathbb{R}$.
Note that $S$ is then uniformly continuous (can you see why?), and thus can be extended on $[0,1]$ to a continuous function. I.e., we get that $S$ is continuous on $[0,1]$.

Fix $\varepsilon > 0$. By uniform convergence, there exists $n_\varepsilon\geq 1$ such that, for every $n\geq n_\varepsilon$, $\sup_{x\in(0,1)} |S_n(x)-S(x)| \leq \varepsilon/3$.

Moreover, since $S_{n_\varepsilon}$ is continuous on $[0,1]$, there exists $\delta_\varepsilon>0$ such that, for all $x\in(1-\delta_\varepsilon,1)$, $|S_{n_\varepsilon}(1)-S_{n_\varepsilon}(x)| \leq \varepsilon/3$. (Using continuity at $1$).

Moreover, since $S$ is also continuous on $[0,1]$, there exists $\delta'_\varepsilon>0$ such that, for all $x\in(1-\delta'_\varepsilon,1)$, $|S(1)-S(x)| \leq \varepsilon/3$. (Using continuity at $1$).

By the triangle inequality, combining the two inequalities for $S_{n_\varepsilon}$, we get that for all $x\in(1-\min(\delta_\varepsilon,\delta'_\varepsilon),1)$,
$$
|S_{n_\varepsilon}(1)-S(1)| \leq \varepsilon
$$
Since $\varepsilon>0$ was arbitrary, this shows that the sequence $(S_n(1))_n$ converges, which is what we wanted.

If $f(x)=\sum_{n=1}^\infty f_n(x)$ converges uniformly on $(0,1)$, then $\forall \epsilon>0$ there exists $N$ s.t. $$\left|\sum_{n=N}^\infty f_n(x) \right| < \epsilon$$ for all $x\in (0,1)$. Furthermore $\sum_{n=N}^\infty f_n(x)$ is uniformly continuous on $(0,1)$ (and can therefore be continued to $[0,1]$) since (by uniform continuity of $f_n(x)$ on $(0,1)$ and uniform convergence of $f(x)$) $\forall \epsilon'>0$ there is $N'>N$ and $\delta(N')=\delta'>0$, s.t. $\forall x,x' \in (0,1)$ with $|x-x'|<\delta'$ $$\left|\sum_{n=N}^\infty f_n(x)-f_n(x') \right| \leq \sum_{n=N}^{N'-1} \underbrace{\left| f_n(x)-f_n(x') \right|}_{<\frac{\epsilon'}{N'-N}} + \underbrace{\left|\sum_{n=N'}^\infty f_n(x)\right|}_{<\epsilon'} + \underbrace{\left|\sum_{n=N'}^\infty f_n(x') \right|}_{<\epsilon'}<3\epsilon' \, .$$
For $x=1-\delta$ with $\delta >0$ this gives
$$\left|\sum_{n=N}^\infty f_n(1) \right| \leq \left|\sum_{n=N}^\infty f_n(1) - f_n(1-\delta) \right| + \left|\sum_{n=N}^\infty f_n(1-\delta) \right|<2\epsilon$$
The second term is less than $\epsilon$ by definition of uniform convergence, while the first term can be made less than $\epsilon$ for small enough $\delta$ due to the continuity of $\left|\sum_{n=N}^\infty f_n(x) \right|$ on $[0,1]$.


Integral $ \int \frac{\operatorname d\!x}{\sin^3 x} $ I need to calculate the following integral for my homework, but I dont know how. If someone show me step by step solution I would really appreciate it.
 $$\int \frac {1}{\sin^3(x)} dx$$

There is a standard first-year calculus response. Rewrite the integrand as $\dfrac{\sin x}{\sin^4 x}=\dfrac{\sin x}{(1-\cos^2 x)^2}$. The substitution $u=\cos x$ leaves us integrating $-\dfrac{1}{(1-u^2)^2}$. Now partial fractions.

There are in many cases more efficient procedures, but one can in principle handle in this way all
$$\int \sin^m x\cos^n x\,dx,$$
where $m$ and $n$ are integers and at least one of $m$ and $n$ is odd.

integral csc^3(x) dx

Use the reduction formula,

-cos(x) sin^2(x)^((m-1)/2) csc^(m-1)(x) 2F1(1/2, (m+1)/2, 3/2, cos^2(x)) = -(cos(x) csc^(m-1)(x))/(m-1) + (m-2)/(m-1)-cos(x) sin^2(x)^((m-3)/2) \csc^(m-3)(x) 2F1(1/2, (m-1)/2, 3/2, cos^2(x)), where m = 3:

= 1/2 integral csc(x) dx-1/2 cot(x) csc(x)

The integral of $\csc(x) \,\,\text{is} -\log(\cot(x)+\csc(x)):$
 = $-1/2 (\cot(x) \csc(x))-1/2 \log(\cot(x)+\csc(x))+$constant

Factor the answer a different way:

Answer:
   $= 1/2 (-\cot(x) \csc(x)-\log(\cot(x)+\csc(x)))+\text{constant}$

Sorry for the mess.


Why is $e^{(x+2x+4x+...+2^{k-1}x)}=e^{((2^k-1)x)}$? Could anyone explain me why we have the following equality, $e^{(x+2x+4x+...+2^{k-1}x)}=e^{((2^k-1)x)}$? I don´t see what manipulation we did to get it

it is known that $1+a+a^2+...+a^m=(a^{m+1}-1)/(a-1)$. To see this cross multiply and telescope. So $x+2x+...+2^{k-1}x=x(1+2+2^2+...+2^{k-1})=x(2^k-1)$. The result follows.


Multiplicity of a root of a polynomial :)

It's true that, if a polynomial has a root (let's say,  k, for example) with multiplicity n (n>1, for n integer), then it's true that the derivate polynomial have k as a root with multiplicity n-1.

The reciprocal is always true? I.e., it's true that if a derivative polynomial have k as a root with multiplicity n-1, then the original polynomial have k as a root with multiplicity n? If this is not always true, in which conditions this is true?

Thanks every one! :)

The complete story is this:

The following conditions are equivalent for a polynomial $f(x)$:

(i) $\;f(x)$ is divisible by $(x-a)^n$;

(ii) $\;f(a)=f'(a)=\dots=f^{(n-1)}(a)=0$.

It is a generalisation of the well-known result:

$f(x)$ has $a$ as a root  if and only if $f(x)$ is divisible by $x-a$.

suppose $f$ is a polynomial of degree $m$ and $f'$ has a root $k$ of multiplicity $k.$ then we can write $$f'(t) = (t-a)^{n-1}q(t), q(a) \neq 0.  \tag 1  $$
integrating $(1)$ from $a$ to $x,$ we get
$$\begin{align}f(x)-f(a) &= \int_a^x(t-a)^{n-1}q(t)\, dt \\
&=\int_0^{x-a}t^{n-1}q(t+a)\, dt\\
&=\int_0^{x-a} t^{n-1}\left(q(a)+a_1t+a_2t^2 +\cdots+a_{m-n+1}t^{m-n+1}\right)\, dt\\&=(x-a)^n\left(\frac{q(a)}n+\frac{a_1}{n+1}(x-a)+\cdots+a_{m-n+1}(x-a)^m\right)\end{align}$$ therefore if $f(a) = 0,$  then the zero $x=a$ of $f$ has multiplicity $n.$


How do I integrate (1/polynomial) without using partial fractions? I had a lecture earlier today where the use of partial fractions was introduced. He used partial fractions and a more 'brute force' method to $\int\frac{1}{(x^2 + 5x + 6)}\mathrm dx$. I could solve this using partial fractions but I need to be reminded of the more difficult method(which I've learned months ago) for my current maths subject's purposes. I've been trying to find a solution that yields to $\ln\left|\frac{(x + 2)}{(x + 3)}\right| + C$ to no avail. Can anyone help me how to solve the problem without using partial fractions?

By completing the square,

$$x^2+5x+6=\frac{(2x+5)^2-1}4.$$

Then use the change of variable $t=2x+5$ and get

$$I=\int\frac{dx}{x^2+5x+6}=2\int\frac{dt}{t^2-1}.$$

You should recognize the derivative of the inverse hyperbolic tangent

$$I=-2\,\text{artanh}(t)=\ln\left(\left|\frac{1-t}{1+t}\right|\right)=\ln\left(\left|\frac{x+2}{x+3}\right|\right).$$


Reparameterization Trick in VAE I was reading this web page on variational auto-encoders, and am unable to understand how the function below is generated. Based on my limited understanding, the sampling part of the VAE which uses a gaussian distribution cannot be backprop-ed. So we are forced to re-write the equation.
The part I do not understand is how we are able to write the gaussian equation
$$
\frac{1}{ \sqrt{2\pi\sigma^2}} e^{\frac{-(x-\mu)^2}{2\sigma^2}}
$$
into the things written below. If someone has a link to the proof or the derivation please post it here. Or if I have totally missed the point please kindly explain :)

This reparametrization is possible because of two properties of Gaussian random variables. I'll stick to the 1D case for simplicity:
1. If $\mathbf{X \sim N(\mu, \sigma^2)}$ and $\mathbf{\alpha \in \mathbb{R}}$ then $\mathbf{\alpha X \sim N(\alpha\mu, \alpha^2\sigma^2)}$.
Proof: If $\alpha = 0$ it is trivial. Suppose that $\alpha \neq 0$. We have
$$P(\alpha X \leq t) = P(X \leq t/\alpha) = \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{t/\alpha} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx$$
Setting $u = \alpha x$ we have
$$\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{t/\alpha} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx = \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{t} e^{-\frac{(u-\alpha\mu)^2}{2\alpha^2\sigma^2}} \frac{1}{\alpha}du = F_{N(\alpha\mu, \alpha^2\sigma^2)}(t)$$
That is, the cumulative distribution function of $\alpha X$ is that of a Gaussian $N(\alpha\mu, \alpha^2\sigma^2)$, so it is the case that $X \sim N(\alpha\mu, \alpha^2\sigma^2)\quad\blacksquare$
2. If $\mathbf{X \sim N(\mu, \sigma^2)}$ and $\mathbf{\beta \in \mathbb{R}}$ then $\mathbf{\beta + X \sim N(\beta + \mu, \sigma^2)}$.
Proof: As before,
$$P(\beta + X \leq t) = P(X \leq t - \beta) = \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{t-\beta} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx$$
Setting $u = x + \beta$:
$$\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{t-\beta} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx = \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{t} e^{-\frac{(u-\beta-\mu)^2}{2\sigma^2}} du = F_{N(\beta+\mu,\sigma^2)}(t) \quad\blacksquare$$
Consequence:
What (1) and (2) tell you is that for a Gaussian $X \sim N(\mu, \sigma^2)$ it holds $X - \mu \sim N(0, \sigma^2)$ and furthermore $\frac{X-\mu}{\sigma} \sim N(0, 1)$. Equivalently, if $X \sim N(0, 1)$ it holds $\sigma X \sim N(0, \sigma^2)$ and furthermore $\mu + \sigma X \sim N(\mu, \sigma^2)$. That is the reparametrization trick. Now, as others have stated, this is convenient in the context of VAEs because $\mu$ and $\sigma$ may depend on learnable parameters, and with this expression these are decoupled from the sampling process—you sample from the r.v. $N(0, 1)$.

In a VAE, we have an encoder distribution (approximate posterior) $q_\phi(z|x)$. It works by defining two functions $\mu_\phi(x)\in\mathbb{R}^n$, $\Sigma_\phi(x)\in\mathbb{R}^{n\times n}_\text{diag}$ and sampling from the latent space via: $$ z \sim \mathcal{N}(\mu_\phi(x),\Sigma_\phi(x)) $$
The problem is that this sampling operation is not differentiable. E.g. $\partial_\phi\mathcal{L}$ requires one to compute the dependence of $z$ on $\phi$, which is a stochastic function.

To get around this, we can instead write: $$ z = \mu_\phi(x) + \xi \sqrt{\Sigma_\phi(x)} $$ where $\xi\sim \mathcal{N}(0,I)$. Notice that now $z$ depends on $\phi$ deterministically, so we can compute the backprop derivatives, because the non-differentiable sampling operation has been moved "off to the side" of the computation graph.


Absolute convergence for all values except the values $z=\left( 1+\frac {a} {m}\right) e^{\frac {2k\pi i} {m}}$ I am trying to show that the series $$\sum _{n=1}^{\infty }\dfrac {nz^{n-1}\left( \left( 1+\dfrac {1} {n}\right) ^{n}-1\right) } {\left( z^{n}-1\right) \left( z^{n}-\left( 1+\dfrac {1} {n}\right) ^{n}\right) }$$ converges absolutely for all values of $z$, except the values $$z=\left( 1+\dfrac {a} {m}\right) e^{\dfrac {2k\pi i} {m}}$$
($a= 0, 1; k = 0,1,\ldots m-1;m=1,2,3,\ldots)$.

Since we are looking for absolute convergence D'Alembert's Ratio Test for absolute convergence and Gamma's convergence criterion come to mind.
So if we can show $$\lim _{n\rightarrow \infty }\left| \dfrac {U_{n+1}} {U_{n}}\right| =l < 1$$ or $$\lim _{n\rightarrow \infty }\left| \dfrac {U_{n+1}} {U_{n}}\right| = 1 +\dfrac {A_{1}} {n}+O\left( \dfrac {1} {n^{2}}\right) $$, where $A_{1}$ is independent of n and $A_{1} < -1$, then we'll establish the series is absolutely convergent.

I was hoping to first establish that the series is absolutely convergent and wishfully thinking that i might stumble across an expression while doing this to prove the exception values.

Although this may seem like a good plan while solving the limit i am coming up against undefined expression as $n\rightarrow \infty $

I am unsure if i am going down the right line here any suggestion, alternative approaches , help would be much appreciated.

Edit: Maybe Cauchy's test if $$\lim _{n\rightarrow \infty }\sup \left| U_{n}\right| ^{\dfrac {1} {n}} < 1$$, then the series converges absolutely might be what we need here.

First of all, the term
$$
\left(1+\frac1n\right)^n-1
$$
approaches $e-1$ eventually, so it doesn't play any role in the convergence. The same can be said of
$$
z^n-\left(1+\frac1n\right)^n
$$
when $|z|\leq1$, since $4\geq|z^n-(1+\frac1n)^n|\geq1$ eventually. So, for $|z|<1$, the general term in your series of absolute values is comparable with $n|z|^{n-1}$, and the series converges.

For $|z|>1$,
$$
\left|\dfrac {nz^{n-1} } {\left( z^{n}-1\right) \left( z^{n}-\left( 1+\dfrac {1} {n}\right) ^{n}\right) }\right|=\dfrac{n|z|^{n-1}}{|z|^{2n}|1-\frac1{z^n}|\,|1-\frac{(1+1/n)^n}{z^n}|}
$$
For $n$ big, the two differences in the denominator approach 1, so the convergence of the series is decided by
$$
\frac{n|z|^{n-1}}{|z|^{2n}}=\frac{n}{|z|^{n+1}},
$$
Which shows that the series converges absolutely for every $z$ with $|z|>1$.

The case $|z|=1$: here $z=e^{2\pi i t}$, with $t\in [0,1]$. When $t$ is rational, the series is not defined (as there are infinitely many values of $n$ where $z^n=1$. For $t$ irrational the terms of the series are defined, but they do not go to zero as $n\to\infty$, so the series is not convergent.

In all honesty, the general term of the series as written has no desire to tend to $0$ for $|z|=1$, so what absolute convergence can we talk about?


Given one primitive root, how do you find all the others? For example:  if $5$ is a primitive root of $p = 23$.

Since $p$ is a prime there are $\phi(p - 1)$ primitive roots. Is this correct?

If so, $\phi(p - 1) = \phi(22) = \phi(2) \phi(11) = 10$.  So $23$ should have $10$ primitive roots?

And, to find all the other primitive roots we need powers of $5$, say $k$, sucht that $gcd(k, p - 1) = d> 1$.  Again, please let me know If this true or not

So, the possible powers of $5$ are: $1, 2, 11, 22$.  But this only gives four other primitive roots.  So I don't think I'm on right path.

If $d=ord_ma,$ we know from here $ord_m(a^n)=\frac d{(n,d)}$ where $m,d$ are positive integers  and $a,n$  are integers

If $a$ is a primitive root $\pmod m, ord_ma=\phi(m)$

(Remember all integers may not have a primitive root )

So, $ord_m(a^n)=\frac {\phi(m)}{(n,\phi(m))}$

So,  $a^n$ is a primitive root $\pmod m,$ if $ord_m(a^n)=\phi(m)$ which needs $(n,\phi(m))=1$

If $m$ is prime, $\phi(m)=m-1$

The possible powers of 5 are all the $k$'s that $gcd(k,p−1)=1$, so $k$ is in the set {$1, 3, 5, 7, 9, 13, 15, 17, 19, 21$} and $5^k$ is in the set {$5, 10, 20, 17, 11, 21, 19, 15, 7, 14$} which is exactly of length 10.


Prove that f is differentiable at x=0 Suppose $0 < f(x) < x^2, \ \forall x$. Prove that $f$ is differentiable at $x=0$ and $f'(x)=0$.

I don't even know how to start with the first one. For the second question, I'm guessing we can use squeeze theorem to show that the limit must also go to $0$ as the limit of the bounds goes to 0.

Much appreciated!

I take it you mean $0 \leq f(x)$. It is easy to see that $f(0) = 0$.

For $h\neq 0$, we have $0 \leq f(h) \leq h^2$.

If $$h>0 \Longrightarrow  0 \leq \frac{f(h)}{h} \leq h$$ $$h<0 \Longrightarrow h \leq \frac{f(h)}{h} \leq 0$$

Observe that these are the difference quotients for the three functions at $x=0$. Now squeeze.

Once we have $ \ f: \mathbb{R} \to \mathbb{R} \ $ such that $ \ 0 \leq f(x) \leq x^2$, $\forall x \in \mathbb{R}$, is straightforward that $ \ f(0)=0$. Then, $\forall t \in \mathbb{R} \setminus \{ 0 \}$, we have that $$\displaystyle 0 \leq \frac{f(t)}{t} \leq \frac{t^2}{t} = t$$ By the Sandwich theorem, we conclude that $$\displaystyle f'(0) = \lim_{t \to 0} \frac{f(0+t) - f(0)}{t} = \lim_{t \to 0} \frac{f(t)}{t} = 0$$


What does $e^{i \pi \mathbf x} \in \mathbb C^2$ mean? I saw this in an book (coding theory) without explanation and I need help figuring out the meaning.

Consider $\mathbf x\in \{0,1\}^2$ and
$$
\mathbf s = e^{i \pi \mathbf x} \in \mathbb C^2 \tag 1
$$

I know $\{0,1\}^2$ and $\mathbb C^2$ are the cartesian products, i.e.
\begin{align}
\{0,1\}^2&=\{0,1\}\times \{0,1\}\\
&=\{(0,0),(0,1),(1,0),(1,1)\} \tag 2
\\
\mathbb C^2 &=\mathbb C \times \mathbb C = \{(z,y): z,y\in \mathbb C\} \tag 3
\end{align}
So $\mathbf x= (x_1,x_2)$ is a vector which take a value in the set $\{0,1\}^2$, for instance, $\mathbf x = (1,1)$.
So far so good, but what is $e^{i \pi \mathbf x} \in \mathbb C^2$?
Any examples?
Say $\mathbf x = (1,1)$, what does $e^{i \pi (1,1)} \in \mathbb C^2$ mean?
Any clarification would be appreciated.

In general you can without comment extend any function $f: A \to B$ to a function of the same name $f: A^n \to B^n$ by defining $f(a_1, a_2, ..., a_n) = (f(a_1), f(a_2), ... f(a_n)).$
In your particular case, $A = \left\{0, 1\right\}$, $B = \mathbb{C}$, $f(a) = e^{i \pi a}$, and $n = 2$.
So $e^{i \pi (x_1, x_2)} = (e^{i \pi x_1}, e^{i \pi x_2})$.
A simpler way to write exactly the same transformation is $f(x) = (-1)^x$. Basically the idea is to send all the $0$ bits to $1$ and send all the $1$ bits to $-1$.


How to prove that N(u) = 1 if and only if u is a unit in $Z[\sqrt-5]$ The norm of an element $u=a+b\sqrt-5$ in $Z[\sqrt-5]$ is defined as $N(u)= a^2 +5b^2$, now if $N(u) = 1$ then $a^2+5b^2 = 1$ but then how would i prove that it's a unit !?

Note that norm has the following property. $N(\alpha)=\alpha \bar{\alpha}$. Thus when $N(\alpha)=1$, this means $\alpha \bar{\alpha}=1$, hence $\alpha$ is a unit.

$N(u)=1$ implies $|a|=1$ and $b=0$, so $u=-1$ or $u=1$, hence $u$ must be a unit.