A question about algebra or number theory? Suppose $r\in \mathbb{Q}$, and $\cos^2(r \pi)\in \mathbb Q$, can we proof $$4\cos^2(r \pi)\in \mathbb Z ?$$

Let $\alpha=\cos(\frac mn\pi)$ with $\gcd(n,m)=1$, $\beta=\sqrt{\alpha^2-1}=i\sqrt{1-\alpha^2}=i\sin(\frac mn\pi)$, $\zeta=\alpha+\beta$. Then by assumption, $$\Bbb \alpha^2\in \Bbb Q,\quad \beta^2\in\Bbb Q,\quad \zeta^n=1.$$
However, instead of $X^n-1$, the minimal polynomial of $\zeta$ is a cyclotomic polynomial of degree $\varphi(n)$. Nevertheless, all roots of this polynomial are roots of unity and hence the constant term is $\pm1$.
As
$$ \zeta^2=(\alpha+\beta)^2=\underbrace{\alpha^2+\beta^2}_{\in\Bbb Q}+2\alpha\beta$$
we have
$$ (\zeta^2-(\alpha^2+\beta^2))^2=4\alpha^2\beta^2\in\Bbb Q$$
i.e., $\zeta$ is also a root of
$$\tag1X^4-2(\alpha^2+\beta^2)X^2+(\alpha^2-\beta^2)^2 $$
and this must be a multiple of the minimal polynomial of $\zeta$. In particular, $\varphi(n)\le 4$.
This leaves us with the following possibilities:

$\varphi(n)=1$: Then $n=1$ or $n=2$ and we deal with $\cos 0$, $\cos \frac\pi2$, $\cos\pi$, $\cos\frac{3\pi}2$, i.e., $\alpha\in \{-1,0,1\}$
$\varphi(n)=2$: Then $n=3$ or $n=4$ or $n=6$ and we deal with $\cos \frac\pi6$, $\cos \frac\pi4$, $\cos \frac\pi3$, $\cos\frac{2\pi}3$, $\cos\frac{3\pi}4$, $\cos\frac{5\pi}6$, i.e., $\alpha\in\{\pm\frac12,\pm\frac{\sqrt2}2,\pm\frac{\sqrt 3}2\}$
$\varphi(n)=3$: This is not possible
$\varphi(n)=4$: Then $n=5$ or $n=8$ or $n=12$. The corresponding cyclotomic polynomials are $X^4+X^3+X^2+X+1$, $X^4+1$, $X^4-X^2+1$. The first does not match $(1)$ and from the others we get $\alpha^2+\beta^2\in\frac12\Bbb Z$ and $\alpha^2-\beta^2=\pm1$, hence indeed
$$\alpha^2=\frac{(\alpha^2+\beta^2)+(\alpha^2 -\beta^2)}{2}\in\frac14\Bbb Z $$

For $r=k\in\mathbb{Z}$ we have $\cos(r\pi)=\pm1\in\mathbb{Z}$. For $r=\frac13$ we get $\cos(r\pi)=\frac12$. So $4\cos^2(r\pi)=1\in\mathbb{Z}$.

Answer to modified question:  https://en.m.wikipedia.org/wiki/Niven%27s_theorem


Stuck with proof techniques I have to prove the following question,

  Let A and B are subsets of a universal set U. Prove that A is a subset of
  B iff B' is a subset of  A'

Now I don't understand how do I prove this using proof techniques. Please guide me.

Thanks.

We want to prove the following:

$$A\subseteq B\iff B^c\subseteq A^c$$

We need, therefore, to prove two implications.

Assume $A\subseteq B$. We want to show that $B^c\subseteq A^c$, that is $x\in B^c$ implies $x\in A^c$. Suppose $x\in B^c$, then $x\notin B$. Since $A\subseteq B$ is to say $x\in A\rightarrow x\in B$, we have that if $x\notin B$ then $x\notin A$. In particular, $x\in A^c$. We therefore proved $\Rightarrow$.
Suppose now that $B^c\subseteq A^c$. Note that $(B^c)^c = B$. By the previous proof we have that $B^c\subseteq A^c\Rightarrow (A^c)^c\subseteq (B^c)^c$. However that means exactly that $A\subseteq B$.

Another approach is to use logical equivalence of $\alpha\rightarrow\beta\iff\lnot\beta\rightarrow\lnot\alpha$.

Since $A\subseteq B$ is $x\in A\rightarrow x\in B$, this as the above equivalence tells us, is the same as $x\notin B\rightarrow x\notin A$, which is precisely to say $B^c\subseteq A^c$.

See my formal proof at http://www.dcproof.com/MathSE-2011-09-30.htm

This proof was generated with the aid of my DC Proof software available at

http://www.dcproof.com

Dan


Approximating the error function erf by analytical functions The Error function

$\mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}\,dt$

shows up in many contexts, but can't be represented using elementary functions.

I compared it with another function $f$ which also starts linearly, has $f(0)=0$ and converges against the constant value 1 fast, namely

$\tanh{(x)} = \frac {e^x - e^{-x}} {e^x + e^{-x}}$.

Astoningishly to me, I found that they never differ by more than $|\Delta f|=0.0812$ and converge against each other exponentially fast!

I consider $\tanh{(x)}$ to be the somewhat prettyier function, and so I wanted to find an approximation to $\text{erf}$ with "nice functions" by a short expression. I "naturally" tried

$f(x)=A\cdot\tanh(k\cdot x^a-d)$

Changing $A=1$ or $d=0$ on it's own makes the approximation go bad and the exponent $a$ is a bit difficult to deal with. However, I found that for $k=\sqrt{\pi}\log{(2)}$ the situation gets "better". I obtained that $k$ value by the requirement that "norm" given by

$\int_0^\infty\text{erf}(x)-f(x)dx,$

i.e. the difference of the functions areas, should valish. With this value, the maximal value difference even falls under $|\Delta f| = 0.03$. And however you choose the integration bounds for an interval, the area difference is no more than $0.017$.

Numerically speaking and relative to a unit scale, the functions $\text{erf}$ and $\tanh{(\sqrt{\pi}\log{(2)}x)}$ are essentially the same.

My question is if I can find, or if there are known, substitutions for this non-elementary function in terms of elementary ones. In the sense above, i.e. the approximation is compact/rememberable while the values are even better, from a numerical point of view.

The purpose being for example, that if I see somewhere that for a computation I have to integrate erf, that I can think to myself "oh, yeah that's maybe complicated, but withing the bounds of $10^{-3}$ usign e.g. $\tanh(k\cdot x)$ is an incredible accurate approximation."

It depends on how much accuracy you need and over what interval.  It seems that you are happy with a few percent.  There is an approximation in Abromowitz & Stegun that gives $\text{erf}$ in terms of a rational polynomial times a Gaussian over $[0,\infty)$ out to $\sim 10^{-5}$ accuracy.

In case you care, in the next column, there is a series for erf of a complex number that is accurate to $10^{-16}$ relative error!  I have used this in my work and got incredible accuracy with just one term in the sum.

And another one not so nice but reversible by $\ln(x)$  to depressed quartic equation solvable by Ferrari's formulas and more accurate $\operatorname{erf}(x)= \operatorname{sgn}(x) \tanh[1.152|x| + 0.064|x|^4]$ (error below 0.5%).


Finding $\phi$ so that we have $3$ equal sets Find all angles $\phi$, if any, for which

  the set $S=\left \{ \sin(\phi),\sin(2\phi),\sin(4\phi) \right \}$,

  the set $C=\left \{ \cos(\phi),\cos(3\phi),\cos(9\phi) \right \}$,

  and the set $T=\left \{ \tan(\phi),\tan(4\phi),\tan(16\phi) \right \}$
  to be three equal sets.

What I have noticed; in $S$, the angles can be arranged to be in geometric progression. Similarly in $C$ and $T$, but I do not know how is that useful.

I could not even start solving this problem. Any help will be appreciated.

Just, consider three cases:
$$\sin\theta=\cos\theta,$$

$$\sin\theta=\cos3\theta$$ and
$$\sin\theta=\cos9\theta.$$
I got that we have no such value of $\theta$.


Properties of Nilpotent Matrices I have $2$ questions concerning nilpotent matrices and commutativity

  $\bullet$ If $M$ is an invertible matrix and $N$ is nilpotent, is then $M-N$ invertible ?

When $M$ and $N$ are commute yes, the inverse is then $(M^{k-1}+M^{k-2}N+\dots N^{k-1})M^{-k}$ if $N$ has nilpotent index $k$, but if they do not commute this is not the inverse, but there might exist one, or not ?

  $\bullet$ Is the sum of $2$ nilpotent matrices again nilpotent ?

Of course under the condition that they don't commute, otherwise one can use binomial formula. I only know that the product of nilpotent matrices can be non-nilpotent

can you help ?

No the the first question
$$
\begin{bmatrix}0&1\\1&0\end{bmatrix}
-
\begin{bmatrix}0&1\\0&0\end{bmatrix}
=
\begin{bmatrix}0&0\\1&0\end{bmatrix}.$$

No to the second question:
$$
\begin{bmatrix}0&1\\0&0\end{bmatrix}
+
\begin{bmatrix}0&0\\1&0\end{bmatrix}
=
\begin{bmatrix}0&1\\1&0\end{bmatrix}
$$

As you see, it is really the same example, with
$$
M = \begin{bmatrix}0&1\\1&0\end{bmatrix}
$$
invertible, and
$$
N = \begin{bmatrix}0&1\\0&0\end{bmatrix},
\qquad
M - N = \begin{bmatrix}0&0\\1&0\end{bmatrix}
$$
nilpotent.

If M is invertible matrix of order n then the characteristic polynomial of M is $$C_M(x)=(x-a_1)^{m_1}(x-a_2)^{m_2}(x-a_3)^{m_3}... (x-a_n)^m_n$$ where $a_{i's}$ are eigenvalues and are non zero since $M$ is invertible and $\sum_{i=1}^n m_i=n$ and the characteristic polynomial of N is $$C_N(x)=x^n$$
Now the characteristic polynomial of $M-N$ is
$$C_{M-N}=(x-(a_1+0))^{m_1}(x-(a_2+0))^{m_2}......(x-(a_n+0))^{m_n}$$
Thus we can see that none of the eigenvalue is zero hence $M-N$ is invertible.
For the case of M and N both nilpotent the characteristic polynomial of M+N is
$$C_{M+N}(x)=x^n $$
Thus all the eigenvalues are zero and thus $M+N$ is also nilpotent if both $M$ and $N$ are nilpotent


A math contest problem $\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x \ \mathrm dx$ A friend of mine sent me a math contest problem that I am not able to solve (he does not know a solution either). So, I thought I might ask you for help.

  Prove:
  $$\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x dx=-\pi^2\left(4\,\zeta'(-1)+\frac23\right).$$

If we let $x\mapsto e^{-x}$ and then use the series of $\log(1-e^{-x})$, we obtain that

$$\underbrace{\int_0^{\infty} \sum_{n=1}^{\infty}\log(4 \pi^2) \frac{e^{-y n}}{n} \ dy}_{\displaystyle \zeta(2) \log(4 \pi^2)} -\underbrace{\sum_{n=1}^{\infty}\int_0^{\infty} \log(4 \pi^2+y^2) \frac{e^{-y n}}{n} \ dy}_{\displaystyle -2\sum_{n=1}^{\infty}\frac{\operatorname{Ci}(2\pi n)}{n^2}+2\log(2)\zeta(2)+2\log(\pi)\zeta(2)}=\pi^{2} ( 4 \log A - 1 )$$

where in the last integral I used the exponential integral and $1.22a$ from this paper.

Q.E.D.


How to characterize the non-trivial solutions of this non-linear differential equation? Are they all periodic? Nota Bene:  These remarks are provided by Robert Lewis on Wednesday 15 July 2015 6:50 PM PST  in effort to provide context and motivation for this engaging question.  Our OP originally asked:

"How to prove that if $f$ is a non-trivial solution of the non-linear differential equation $x^{''}+\sin x=0$, then $f$ is periodic?"

Which cannot be done, since the equation has non-trivial, non-periodic solutions (see below), although most of the solutions are in fact periodic.  Therefore we need to broaden the scope of inquiry as indicated in the (now edited) title.  Essentially what we now seek is an answer to something more along the lines of

How may we rigorously describe the non-trivial solutions to

$$x'' + \sin x = 0;$$

are any or all of them periodic?

This, I think, captures much of the spirit of the original question whilst expanding the scope so that an answer is possible.

I have probed, though not to excessive depth, the question data base here and did not find this exact question; hopefully, it's not a duplicate.  It is, to my mind, a good question since it illustrates, in a fundamental way, some essential basic principles of analyzing non-linear ODEs.

These things being said, let me add that attempts based upon linearization or power series analysis might prove problematic, since the questions asked are global in nature.

With these modifications and caveats on the table, I bid you have at it, Ladies and Gentlemen of Math Stack Exchange.

Multiplying by $x' $ we get $$x'' x' +x' \sin x =0 $$
hence $$\left(\frac{1}{2} x'^2 \right)' -(\cos x )' =0$$ thus $$x'^2 -2\cos x =c$$ hence $$\pm\int \frac{dx}{\sqrt{2\cos x +c} }=t+c_1 $$
substituting $u =\cos x , $ $dx =\frac{- du}{\sqrt{1-u^2}}$ we get $$\pm \int \frac{du}{\sqrt{(1-u^2)(2u +c)}} =t+c_1 $$


How to calculate recursive series? $r_1 = r_0 (1 + e)$

$r_2 = r_1(1 + 2e)$

$r_3 = r_2(1 + 3e)$

$r_n = r_{n-1}(1 + ne)$

Can I get $r_n$ in closed form (a single formula) in terms of $r_0$ and $e$ ?

Also, as n tens to infinity what will $r_{n}$ converge to ? I guess it will also go to infinity. (given e is smaller than 1).

I have to equate $r_n$ to a term to solve for $e$.

I can't think of any formula for (1+e)(1+2e)(1+3e)...(1+ne). Is it too difficult to analyse. I guess the higher powers will be minimal as e < 1.

Thanks a lot.

You can divide everything by $r_0$, then multiply it back in at the end, so you are looking for $\prod_{i=1}^n(1+ie)$.  If $e \gt 0$ the infinite sequence does not converge as all the terms are greater than $1+e$.  If $e \lt 0$ the terms in the infinite product become large and negative, so the product gets very large in absolute value and oscillates in sign.  We need to consider finite products only.

You can collect the terms, saying $$\prod_{i=1}^n(1+ie)=1+\sum_{i=1}S_i(n)e^n$$  $S_i(n)$ is the sum of products of all combinations of $i$ different numbers taken from $\{1,2,3,\dots n\}$  $S_i=\frac 12n(n+1)$, the usual triangular number.  To evaluate $S_2$, imagine adding in the terms where the numbers are the same, in which case we have $S_1^2$, so $S_2(n)=S_1(n)^2-\sum_{i=1}^ni^2=\frac 14n^2(+1)^2-\frac 16n(n+1)(2n+1)$  One can keep going, but it gets messier. They are known as Newton's identities

The limit:

Remark:$e$ seems not equal to 2.718... so we have to look at the case $1+n_0 e =0$ !

if $e$ not in $\{ -\frac1{k}, k \in \mathbb{N}^*\}$, and $r_0 \neq 0$, then
 $r_n \to \pm\infty$ depending of $e$ and $r_0$.

if $e= -\frac1{k}$, then $r_n =0$ for all $n \ge k$.

if $r_0=0$, $r_n =0$ for all $n \in \mathbb{N}$.

The closer form:

(suppress.Thanks to Did). The coefficients are easily evaluated.

Let $a_{n,k}=0$ for $k \gt n$ and $a_{0,0}=1$ and $r_n = r_0 \sum_{k=0}^n a_{n,k} e^k$, then we have $r_{n+1} = r_0 ((n+1)e+1)\sum_{k=0}^n a_{n,k} e^k$ hence $a_{n+1,k}=a_{n,k-1}(n+1)+ 2*a_{n,k}$.

$a_{n+1,k}=a_{n,k-1}(n+1) +2 a_{n,k}$.


What does one sample average tell me about the average of other samples? I'm sure this has been asked before but I can't find the exact variant I'm looking for.

I have an infinite population of elements. I take a sample of those, say 10000, and make a measurement for each of them. The result could be "pass" or "fail". I get 3 "fail"s. Based on this information now I would like to know, what is the probability that another sample of 10000 elements, will give me the value of 10 "fail"s. How do i calculate this?

There are two parts to what you want.

First, you need to use your sample to estimate the proportion of fails.

On the basis of the fact that you have 3 fails out of 10000, the estimate of the proportion is given by $\hat p= \frac 3 {10000}$. There are ways to give bounds to this estimate. Have a look at:

http://www.ics.uci.edu/~jutts/8/Lecture20Compact.pdf

and

http://onlinestatbook.com/2/estimation/proportion_ci.html

for further details.

Second, you want to find the probability of achieving 10 fails in the next sample of 10000. For this you can use the Binomial distribution. If an event has probability of occurring $p$ and you perform $n$ trials, then we say that the number of times that it will occur has a Binomial distribution $X ~ B(n,p)$. You will use $n=10000$ and $p=\hat p$. Trying different values of $\hat p$ - remember you will have a set of possible values - will give slightly different results.

The probability of 10 fails is given by $P(X=10)$, for which there is a standard formula. Look at:

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

If you want to use Excel, then have a look at this:

http://homepages.wmich.edu/~bwagner/StatReview/Binomial/binomia%20in%20excel.htm

If you don't assume independence or even uniform distribution you can still use Chebyshev's inequality, assuming that is that you know the variance and mean value. And if you don't know even that then there's not much that you can say.


Does $y = f(x) = ax+b$ actually have two mappings inside it? I’m just a high school student, so I may be somewhat logically flawed in understanding this.
According to wikipedia, the definition of function requires an input $x$ with its domain $X$ and an output $y$ with its domain $Y$, and the function $f$ maps $x$ to $y$.
But how about $f(x)$? I often see syntaxes such as $f(1) = 0$  in my text book. Doesn’t that mean it is $f(x)$ being first assigned a value and then transfer the value into $y$? So, there must be two transitions/mappings between the input $x$ and the output $y$ right?
My conceptual model of function is like this: A definition of function requires an input $x$ with its domain $X$, a forwarder $f(x)$ with its domain $F$ and an output $y$ with its domain $Y$. The function $f$ first maps $x$ to $f(x)$ then maps $f(x)$ to $y$.
These two definitions are not quite the same.

Equality is not just a mapping. It means two things are one and the same.
We wrote $f(x) = y$ not to indicate that we are mapping $f(x)$ to $y$;
it means that $f(x)$ is $y$.
Consider a function $f$ that maps any value $x$ in the domain $\mathbb Z$ (the integers) to a value $y$ in the codomain $\mathbb Z$.
(Codomain rather than domain is the correct word here.)
In particular, $f$ is the function that maps each integer to the previous integer, so $f$ maps $3$ to $2$, $f$ maps $2$ to $1,$ and $f$ maps $1$ to $0.$
Then the notation $f(1)$ is simply a way of saying
"the number that $f$ maps $1$ to."
So if you ask, "What is $f(1)$," the answer is $0$,
because $f$ maps $1$ to $0$.
Again: $f$ maps $1$ to $0$, therefore the number $f$ maps $1$ to is $0$;
that is, $f(1)$ is $0$. We express this in the equation, $f(1) = 0$.

It is certainly possible to set up a situation in which a member of one set is mapped to something which is then mapped to something else.
But that is not how a single function is defined.
A function defines exactly one transition from one set to another,
and the notation $f(x)$ is one way to express the final result of that transition.
If $f$ maps from a domain $X$ to a codomain $Y$ then $f(x)$ is already a member of $Y$ and needs no further "transitioning" to get to $Y$.
Notice that in the previous paragraph I never mentioned any variable named $y$. Indeed I too take issue with the part of the Wikipedia article that says, "this relation is denoted by $y = f (x)$ ... where the element $x$ is the argument or input of the function, and $y$ is the value of the function ... ." The symbol $y$ is completely superfluous here; if we use $x$ to represent the input of the function, the usual denotation for the value of the function is $f(x)$, and a denotation that actually shows the relation of an element $x$ in the domain to an element in the codomain is
$x \mapsto f(x).$
Here $f(x)$ is the element of the codomain and $\mapsto$ is the relation.

I see what you are saying, but there are ways to think about this without using more than one mapping, as the other answers have pointed out.
But here is what I think you are getting at (which I agree with):
$x \mapsto f(x) \mapsto ax+b$. where we have a general function of $x$ and we send it to some expression in terms of the variable $x$.
Then $x$ maps to $f(x)$ as normal by using the domain and codomain.


How do I find this limit: $\lim_{n\to\infty} \left[ n-{n\over e}\left(1+{1\over n}\right)^n \right]$? I'm working on a practice exam for my masters quals and I am having difficulties with the following limit. According to wolfram alpha, it's value is 1/2. Does anyone know how to find this limit?

$$\lim_{n\to\infty} \left[ n-{n\over e}\left(1+{1\over n}\right)^n \right].$$

Just use the fact that
$$\left(1+\frac1n\right)^n=e^{n\ln(1+\frac1n)}=e^{n\left(\frac1n-\frac{1}{2n^2}+o(\frac{1}{n^2})\right)}=e\cdot e^{-\frac{1}{2n}+o(\frac1n)}=e\left(1-\frac{1}{2n}+o\Bigl(\frac{1}{n^2}\Bigr)\right),$$and the result follows.

Hint: try logarithms. $$f(n) - g(n) = \ln\frac{f(n)}{g(n)}$$


Find last 5 significant digits of 2017! Since there are less powers of $5$ than of $2$ and since $10 = 2 \cdot 5$, I counted the number of zeros in $2017!$:
$\left \lfloor{ \frac{2017}{5^1}}\right \rfloor + \left \lfloor{ \frac{2017}{5^2}}\right \rfloor +\left \lfloor{ \frac{2017}{5^3}}\right \rfloor +\left \lfloor{ \frac{2017}{5^4}}\right \rfloor = 502$

Now I need to find $\frac{2017!}{10^{502}} \pmod{10^5}$

Using Chinese remainder theorem I split the $\pmod{10^5}$ into $\pmod{2^5}$ and $\pmod{5^5}$. Since there are a lot more powers of $2$ left, $\frac{2017!}{10^{502}}\equiv 0 \pmod{2^5}$

I am having trouble with getting $\frac{2017!}{10^{502}} \pmod{5^5}$. What would be the best way to compute it?

Clarification: significant digits - the ones that follow up before the zeros at the right side of the number.

OK, for what it's worth, the answer can be derived by casting out all factors of $5$ and multiplying the resulting numbers incrementally $\bmod 100000$, which also allows us to also divide out from the running product the same power of $2$ as the power of $5$ that we just cast out.

As a way of checking any more elegant mathematical approach, therefore, the result is

   $15968$

As an aside, I'll share the process I used to calculate this and justify a shortcut that I used.

The basic idea here was to multiply successive numbers into a running product, excluding powers of $5$ and adjusting powers of $2$. The process for each number was:

find the highest power of $5$ that divides the new number
divide that out of the number
multiply the adjusted number into the running product and divide by an equivalent power of $2$
find the residue $\bmod 100000$

Note that this gives identical running $\bmod 100000$ values to @skyking's approach.

There is a intermittent problem with this approach, in that the running product will potentially be wrong for a couple of numbers after a large power of $5$ is divided out. Effectively, the value should be saturated with factors of $2$ from $8!$ onwards, but dividing out a suitable power of two may disturb this briefly. For example, the value for $15!$ is out in this process - $24368$ instead of the correct value, $74368$. Nevertheless, once a suitable number of factors of two in subsequent numbers are multiplied back into the product, the $\bmod 100000$ value gets back on track.

$2017$, occurring as it does directly after $2016$ = $2^5\cdot 63$, has an accurate running product by this method.

An continuously accurate running value can be produced, if necessary, by building up a "reserve" of a few powers of $2$ separately from a running product calculation, by stripping out powers of two from the successive values until the reserve is full. This is then used to compensate for powers of $5$ encountered, and multiplying these back in to the reduced running product. This correct gives the value for $15$ etc.


Company A has 40% of German cars and company B has 10% of German cars. All other cars are Korean.Help needed solving Bayesian Probability There are 2 taxi companies. Company A has 40% of German cars and company B has 10% of German cars. All other cars are Korean. Taxi aggregator assigns orders to A or B with probabilities 1/3 and 2/3 respectively. If German car arrives, what's the probability that it's from company B?
So my solution for this problem would be using Bayes Theorem
According to me the question is asking to find P(car being from company B|Being German)
So using Bayes theorem it would be equal to
$P(B|G)=(P(B) * P(G|B))/P(G)$  and that would be
$P(B|G)=((2/3)*0.1)/0.25= 4/15$ which is wrong. I would like to know where am i making mistake
Here P(B) is probability of car being from company B
P(G) is probability of car being German
I found P(G) considering there are 100 cars in  both company A and B so $50/200=1/4=0.25$

Law of total probability:
$P(G) = P(G|A)P(A) + P(G|B) P(B) = 0.4*1/3 + 0.1*2/3 = 0.2$.
By the way, the cars should be Mercedes Benz please.


Probability numbers from 1-10 I have a problem containing probablility and don't know how to solve it ?

Numbers from 1 to 10 are ordered randomly.
What is the probability that 1 2 3 appear in the sequence in that particular order - allow other numbers to be between them.

thanks in advance

Hint:

The relevant numbers $1,2,3$ can appear in $3!=6$ different orders that have equal probability to occur.

do not consider 1,2,3 and focus instead on numbers from 4 to 10.
They can occur in 6! ways (ordered randomly).
you have to multiply 6! for the number of ways in which (1,2,3) can present themselves in the sequence( for example, 123 can be at the beginning of the number, in the middle(allow other numbers to be between them)or at the end.
The number of positions of (1,2,3) inside a sequence of ten numbers are the combinations without repetition ten, three.
So favourable cases 6! per C(10,3)/10!(number of possibilities).


Is $i^i$ mathematically valid? WARNING: SLIGHT NSFW

http://www.smbc-comics.com/index.php?db=comics&id=2934#comic

Uhh...guys, mathematically speaking, how accurate is this comic. From what I remember in High School

$$a^b= \underbrace{\,a\cdot a\cdot  \ldots \cdot a \,}_{b\ \text{ times}} $$
So does that mean that
$$i^i= \underbrace{\,i\cdot i\cdot  \ldots \cdot i \,}_{i\ \text{ times}} $$

How can you multiply a number by itself an imaginary number of times?

To piggy back off of @Crostul and @dbanet, the answer to the question is that the comic is mathematically accurate. One could even ask does real exponentiation make sense? If you have something like $3^{\pi}$, is that really $$\underbrace{3\cdot 3\cdot\dots \cdot 3}_{\pi \text{ times}}$$How do you multiply something $\pi$ times? Even simpler, how do you even multiply something a rational number of times? The answer is actually not too hard: Rational exponentiation makes some intuitive sense, in that $3^{2/5}=\sqrt[5]{3^2}$, or it is the number such that its 5th power is $3^2$. Then, real exponentiation can be defined to be a limit or supremum of powers of all rational numbers less than the real number. Complex exponentiation requires Euler's formula to make any sort of sense.

For a good reference which, if you've at least seen calculus before should be readable (though maybe difficult), would be Needham's Visual Complex Analysis. It has a great intuitive explanation of Euler's formula.

No, you have to use Euler's formula for imaginary exponents.
In fact, as stated here,
$\,i^i = e^{-\pi/2}.\,$

Indeed, on complex plane point ${\left(0,i\right)}{}$ corresponds to the angle $\,\theta = \dfrac{\pi}{2}.\,$ Therefore $\,i = e^{\,i\pi/2},\,$ and
$$
i^{\,i} = \left(e^{\,i\pi/2}\right)^i = e^{\,i^2\pi/2} = e^{-\pi/2}
$$


Seating $2n$ people around a table with each person has at most $n-1$ friends I am trying to show that with the setting in the title, that it is always possible to arrange the seats so that no person sits beside his/her friend.

I am not good at this kinds of problems at all, the only method I could think of is to use induction, but with the induction hypothesis (statement holds for $2(n-1)$ people) the situation seems complicated when I have $2n$ people to arrange as I could not easily single out $2$ person, any suggestions?

This question can be answered using Ore's Theorem, which states that if $G$ is a (finite, simple) graph with $n\ge3$ vertices, and if $$\deg v+\deg w\ge n$$ for every pair of non-adjacent vertices $v$ and $w$, then $G$ is Hamiltonian, that is, there is a cycle in the graph that contains every vertex in the graph. The Wikipedia essay contains a short proof of Ore's Theorem.

The Wikipedia essay also notes that Ore's Theorem is a strengthening of Dirac's Theorem, which states that if each vertex of an $n$-vertex graph $G$ has degree at least $n/2$, then $G$ is Hamiltonian. Dirac's Theorem will suffice for solving the question here.

Construct the graph as follows:
Assume that the people are vertices and there is an edge between two vertices if and only if they are friends. Then at most $n-1$ enemies is the same as having at least $n$ friends, i.e degree of each vertex is at least $n$. Then by Dirac's theorem there exists a Hamiltonian cycle when $n \ge 3$ (In this case it will be $2n\ge 3$).
The trivial case ($n=1$) can be constructed by inspection.


Formula for $\sum_{k=1}^n k\binom nk^2$. I know that
$$\binom n1^2+\binom n2^2+\binom n3^2+\binom n4^2+\dots+\binom nn^2=\binom{2n}n.$$

Is there a similar formula
$$\binom n1^2+2\binom n2^2+3\binom n3^2+4\binom n4^2+\dots+n\binom nn^2=\cdots?$$

Thanks in advance.

$$r\cdot\left(\binom nr\right)^2=\binom nr\cdot r\cdot\binom nr$$

Now, $$r\binom nr=r\frac{n!}{(n-r)! r!}=r\frac{(n-1)!\cdot n}{\{(n-1)-(r-1)\}!\ (r-1)!\cdot r}=n\binom{n-1}{r-1}$$

Again, $$\frac{(1+x)^{2n-1}}{x^{n-1}}=(1+x)^n\left(1+\frac1x\right)^{n-1}$$

Now, observe that the coefficient of $x$  in the Right Hand Side is $\displaystyle \sum_{0\le r\le n}\binom nr\binom{n-1}{r-1} $

What about the Left Hand Side?


Finding first three terms of a GP whose sum and sum of squares is given? We have to find three numbers in G.P. such that their sum is $\frac{13}{3}$ and the sum of their squares is $\frac{91}{9}$.

Here's what I have tried to do:-

$$a+ar+ar^2=\frac{13}{3}$$
$$a^2+a^2r^2+a^2r^4=\frac{91}{9}$$

$$\frac{a^2+a^2r^2+a^2r^4}{a+ar+ar^2}=\frac{91}{9}*\frac{3}{13}$$

$$\frac{a+ar^2+ar^4}{1+r+r^2}=\frac{7}{3}$$

We have not been taught how to solve a biquadratic equation in school, so how should I proceed from here on?

$a+ar+ar^2=\frac{13}{3}$ [equation 1]

$a^2+a^2r^2+a^2r^4=\frac{91}{9}$ [equation 2]

Square the first equation:

$a^2(1+r+r^2)^2 = \frac{169}{9}$

$a^2(1 + r^2 + r^4 + 2r + 2r^2 + 2r^3) = \frac{169}{9}$ [equation 3]

Now take equation 3 minus equation 2:

$2a^2r(1 + r + r^2) = \frac{26}{3}$

Using equation 1 again,

$2ar \frac{13}{3} = \frac{26}{3}$

So $ar = 1 \implies r = \frac{1}{a}$

Now put that back into equation 1 to get:

$a + 1 + \frac 1a = \frac{13}{3}$

Multiply throughout by $a$, rearrange:

$a^2 - \frac{10}{3}a + 1 = 0$

Solving that will give you the nice solutions $a = 3$ or $a = \frac 13$, which are both admissible, and the respective values of $r$ are the reciprocals of $a$, $\frac 13$ and $3$.

So the possible sequences are $3,1,\frac 13$ and $\frac 13, 1, 3$. Both possibilities are admissible because they haven't told you if the GP is increasing or decreasing.

Take 3 numbers $a/r$ ,a,ar $a/r$+a+ar=$13/3$ and $a^2/r^2$+$a^2$+$a^2r^2$=$91/9$ from this take $a^2$ common so on sim plifying we get $a^2/r^2$(1+$r^2+r^4$)=$(91/9$ Then square 1 st condition by using idw
Identity $(a+b+c)^2$=$a^2+b^2+c^2+2ab+2bc+2ac=$169/9$ from this we get $91/9$+$2a(a/r+a+ar)$=$169/9$ so $91/9$+2a$(13/3)$=$169/9$ from here a=1   plug  it in any of the condition you get r as 1/3 or 3 thus terms are 1,3,1/3.thanks!


Using polar coordinates to find the area of an ellipse Considering an ellipse with the $x$ radius equal to $a$ and the $y$ radius equal to b$:$

I figured that some kind of parameterization might be:

$x=a\cos\theta$

$y=b\sin\theta$

and then polar $r^2$ is just $x^2 + y^2$

But then I tried to come up with some unit of infinitesimal area using triangles $\left(\dfrac{d\theta r^2}{2}\right)$ which does not give the correct answer. I read somewhere that my polar coordinates are wrong and that they are actually

$x=ar\cos\theta$

$y=br\sin\theta$

But this does not make sense to me as an engineer because that seems like it would have the dimension of area equal to the dimension of a distance. The integral also takes $r$ from $0$ to $1$ which I thought was eliminated because the equation for $r$ should be in terms of $\theta$ and the constants $a$ and $b.$

I would like some explanation of what I am doing wrong that would make some "physical" sense (or why physical intuition might fail for this problem)

Notice, since the ellipse: $x=a\cos\theta$ & $y=b\sin\theta$ is equally divided into four symmetrical regions hence,  the area of ellipse in Cartesian coordinates is given as $$=4\int_0^ay\ dx$$
Now changing in Polar-coordinates by setting $y=b\sin\theta$ & $x=a\cos\theta$ or $dx=-a\sin\theta\ d\theta$, one should get area of ellipse $$=4\int_{\pi/2}^0(b\sin\theta)(-a\sin\theta\ d\theta)$$
$$=4ab\int_0^{\pi/2}\sin^2\theta\ d\theta$$
$$=4ab\int_0^{\pi/2}\frac{1-\cos2\theta}{2}\ d\theta$$
$$=4ab\left(\frac{1}{2}\int_0^{\pi/2}\ d\theta-\frac 12\int_0^{\pi/2}\cos2\theta\ d\theta\right)$$

$$=2ab\int_0^{\pi/2}\ d\theta$$$$=2ab\frac{\pi}{2}=\color{red}{\pi ab}$$

I am quite surprised that no one pointed out the following fact: the parametrisation in question does not contain the usual plane polar angle; what is present there is the eccentric anomaly. That is why you can't use the normal techniques for finding areas in plane polar coordinates, and you have to resort to (different forms of) Green's theorem in a plane.


Origin of the words arithmetic and geometric progression Why are arithmetic progression and geometric progression called arithmetic and geometric respectively?

Without denying the truthfulness of Michael Hardy's answer, I was always under the impression that the reason for connecting products to geometry was due to the fact that the height of a straight edge triangle is the geometric mean of the projections of its other two sides onto the hypotenuse. This and the fact that geometric shapes such as rectangles provide a very intuitive understanding for why multiplication is commutative.

Imagine a line of length $1$, and the next of length $2$, and the next of length $4$, and then $8$, etc.  The pair of lines of lengths $1$ and $2$ has the same geometric shape as the pair of lengths $2$ and $4$, and similarly $4$ and $8$, etc.


Does there exist a nicer form for $\beta(x + a, y + b) / \beta(a, b)$? I have the expression

$$\displaystyle\frac{\beta(x + a, y + b)}{\beta(a, b)}$$

where $\beta(a_1,a_2) = \displaystyle\frac{\Gamma(a_1)\Gamma(a_2)}{\Gamma(a_1+a_2)}$.

I have a feeling this should have a closed-form which is intuitive and makes less heavy use of the Beta function. Can someone describe to me whether this is true?

Here, $x$ and $y$ are integers larger than $0.$

$$
\beta(1+a,b) = \frac{\Gamma(1+a)\Gamma(b)}{\Gamma(1+a+b)} = \frac{a\Gamma(a)\Gamma(b)}{(a+b)\Gamma(a+b)} = \frac{a}{a+b} \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} = \frac{a}{a+b} \beta(a,b).
$$
If you have, for example $\beta(5+a,8+b)$, just repeat this five times for the first argument and eight for the second:
$$
\frac{(4+a)(3+a)(2+a)(1+a)\cdot(7+b)(6+b)\cdots (1+b)b}{(12+a+b)(11+a+b)\cdots (1+a+b)(a+b)}\beta(a,b).
$$

You can also write this as a ratio of rising factorials.   The rising factorial is defined as $(a)_k = a(a+1)\cdots(a+k-1)$, with $(a)_0 = 1$.  Then use the recurrence relationship for the Gamma function to reduce the ratio of the beta functions to
$$\frac{\beta(a+x,b+y)}{\beta(a,b)} = \frac{(a)_x(b)_y}{(a+b)_{(x+y)}}.$$
I think that's as concise as it gets.  Of course, some folks aren't too fond of rising factorial notation...


How to find an angle in range(0, 360) between 2 vectors? I know that the common approach in order to find an angle is to calculate the dot product between 2 vectors and then calculate arcus cos of it. But in this solution I can get an angle only in the range(0, 180) degrees. What would be the proper way to get an angle in range of (0, 360)?

I'm adapting my answer on Stack Overflow.

2D case

Just like the dot product is proportional to the cosine of the angle, the determinant is proprortional to its sine. And if you know the cosine and the sine, then you can compute the angle. Many programming languages provide a function atan2 for this purpose, e.g.:

dot = x1*x2 + y1*y2      # dot product
det = x1*y2 - y1*x2      # determinant
angle = atan2(det, dot)  # atan2(y, x) or atan2(sin, cos)

3D case

In 3D, two arbitrarily placed vectors define their own axis of rotation, perpendicular to both. That axis of rotation does not come with a fixed orientation, which means that you cannot uniquely fix the direction of the angle of rotation either. One common convention is to let angles be always positive, and to orient the axis in such a way that it fits a positive angle. In this case, the dot product of the normalized vectors is enough to compute angles.

Plane embedded in 3D

One special case is the case where your vectors are not placed arbitrarily, but lie within a plane with a known normal vector $n$. Then the axis of rotation will be in direction $n$ as well, and the orientation of $n$ will fix an orientation for that axis. In this case, you can adapt the 2D computation above, including $n$ into the determinant to make its size $3\times3$.
One condition for this to work is that the normal vector $n$ has unit length. If not, you'll have to normalize it.
The determinant could also be expressed as the triple product:

$$\det(v_1,v_2,n) = n \cdot (v_1 \times v_2)$$

This might be easier to implement in some APIs, and gives a different perspective on what's going on here: The cross product is proportional to the sine of the angle, and will lie perpendicular to the plane, hence be a multiple of $n$. The dot product will therefore basically measure the length of that vector, but with the correct sign attached to it.

I write the formula like I wrote in excel . (xa,ya,xb,yb put in the cells a2,b2,c2,d2).
angle(vector.a,vector.b)
=(180/pi())* abs(pi()/2*((1+sign(a2))* (1-sign(b2^2))-(1+sign(c2))* (1-sign(d2^2)))
+pi()/4*((2+sign(a2))*sign(b2)-(2+sign(c2))*sign(d2))
+sign(a2*b2)*atan((abs(a2)-abs(b2))/(abs(a2)+abs(b2)))
-sign(c2*d2)*atan((abs(c2)-abs(d2))/(abs(c2)+abs(d2))))
The formula is giving the angle of two vectors a and b from 0 to 360 degrees,
in left wise direction for any value of the vectors coordinates.
For xa=ya=0  and  or xb=yb=0 the result is undefined.


Meaning of horizontal bar in old formula (paranthesis?) When reading an old paper from 1921* I find formulas like:

$\rho + \frac{\rho(1- \rho^2)}{2\overline{n - 1}} \big( 1+  \frac{9 - 14\rho^2}{6\overline{n-1}} \big)$

which is said to be the median of the estimated sample correlation, $r$.

What is the meaning of the horizontal bar above (n - 1)?

My best guess is that it is an old alternative notation for paranthesis but I am not able to find any real justification for this assumption. The paper also include expressions such as:

$2 \overline{s-1}r = \overline{s-2}+s \tanh(z - \phi)$
where the need of paranthesis around $s-1$ can be questioned.

If there where complex numbers involved I guess it could also be the complex conjugate but that does not really make sense here.

*R A Fisher "On the probable error of a coefficient of correlation deduced from a small samlpe", Metron 1921,

The horizontal bar is called a vinculum and is an alternative to brackets. See Wikipedia and Jeff Miller.


How do I find the matrix X that multiplied with another one A equals Y (matrices are not square) I have the following matrices:

$A$: shape 7,9 constant values not to be changed
$X$: shape 402,7 with values that multiplied with A should result in the desired solution
$Y$: shape 402,9

Equation: $Y = A  X$

Constraints: Values in X must be >= 0

Since they aren't square, algorithms solving the equations don't work (as in $A^{-1} Y = X$)

Does a good algorithm exist for getting matrix $X$ in this problem?

Edit:

I'm currently using scipy.optimize.minimize() (python) to minimize the error by optimizing X with L-BFGS-B. MeanError between $Y$ and $AX$ is currently around 34% on average.

As $X$ has $402 \times 7$ unknown entries, and $Y$ has $402 \times 9$ entries, these unknown entries will have to be solutions of an overdetermined linear system, i.e., with more equations than there are unknowns.

In general, in such a case, we will have no solutions.

But it is not excluded that solutions exist in particular cases, for example if there is a "physical" reason for such an existence.

A more reasonable attitude in this case (too many constraints vs. number of unknowns) is to consider your issue as an optimization problem, i.e., find $X$ such that

$$\|AX-Y\|$$

is minimum (where $\| ... \|$) is a matrix norm, e.g., Frobenius norm.

Edit :

Probably the simpler approach is to multiply left and right hand side of matrix equation $AX=Y$ by $A^T$, giving :

$$AX=Y \ \implies \ A^TA=A^TY \ \iff \ X=(A^TA)^{-1}A^T Y \tag{1}$$

Three remarks :

1) Why are we authorized to take the inverse of $A^TA$ ? Because it is square and moreover semi-definite positive, therefore has $\geq 0$ eigenvalues. It can happen that some of its eigenvalues are very close to zero generating a numerical error (we say that it is an unstable method, but maybe in your application you will not suffer from this phenomenon).

2) Why do we find a solution $X=...$ whereas we have said that no solution exists ? Because in (1) you may have noticed that there is an implication symbol : we have enlarged the set of solutions. This is not an equivalence sign : with this $X$ we cannot go the other way and say that $AX=Y$ ! Nevertheless it is not difficult to show that this $X$ is the solution of the minimization problem mentionned above (see http://www.sci.utah.edu/~gerig/CS6640-F2012/Materials/pseudoinverse-cis61009sl10.pdf).

3) Matrix $A^+:=(A^TA)^{-1}A^T$ is called the pseudo-inverse of $A$ ; in this way we can reduce (1) to

$$AX=Y \ \implies \ X=A^+ Y \tag{2}$$

More elaborate methods using SVD (Singular Value Decomposition) are less prone to instability...

As I'm sure you've noticed, what you need is $YA^{-1}$ , which is an issue since A is not square. This is where the concept of generalized inverses comes in.

First realize that if any such X exists, then the rows of Y will be linear combinations of the rows of A. If this is true, now find a g-inverse of A.

All non 0 matrices have one and they suffice your needs. The overall process you'll need to go through is this:

Find a rank-factorization for A, say (P,Q).
Find a left inverse of P, say C.
Find a right inverse of Q, say D.
Define $A^{-1}$ = DC

Use this to get $X$.


Can we have an uncountable number of isolated points? Is this possible? I've been trying to think of an example or defend why not, and I'm struggling in both directions.

Since you use the real-analysis tag, I will assume that you’re talking about $\Bbb R$ with the usual topology. If so, the answer is no: every discrete subset of $\Bbb R$ is countable.

To see this, note first that the set $\mathscr{B}$ of open intervals with rational endpoints is a countable base for the topology. Now suppose that $D\subseteq\Bbb R$ is discrete. Then for each $x\in D$ there is a $B_x\in\mathscr{B}$ such that $B_x\cap D=\{x\}$. $\mathscr{B}$ is countable, so if $D$ were uncountable, there would have to be distinct points $x,y\in D$ such that $B_x=B_y$, which is absurd: that would mean that

$$\{x\}=B_x\cap D=B_y\cap D=\{y\}\;,$$

yet $x\ne y$.

If you are asking the question in general, however, rather than about $\Bbb R$ with the usual topology, then the answer is yes; Ross Millikan has given a simple example.

No, each neighborhood around each point is disjoint and contains a rational not contained in any other point's neighborhood.


prove that $n(n+1)$ is even using induction Prove that $n(n+1)$ is even using induction

The base case of $n=1$ gives us $2$ which is even.

Assuming $n=k$ is true,

$n=(k+1)$ gives us $ k^2 +2k +k +2$

while $k(k+1) + (k+1)$ gives us $k^2+2k+1.$

whats is the next step to prove this by induction? I can't seem to show

$ k^2 +2k +k +2$ = $k^2+2k+1$

$n(n+1)$ is an even number.
Take any $n\in\mathbb N$, then $n$ is either even or odd.

Suppose $n$ is even $\implies n= 2m$ for some $m\implies n(n+ 1) = 2m(n+ 1)\implies n(n+ 1)$ is even.
Suppose $n$ is odd $\implies n+1$ is even $\implies n+1 = 2m$ for some $m \implies n(n+1) = 2nm\implies n(n+1)$ is even.


How to investigate the $\limsup$, the $\liminf$, the $\sup$, and especially the $\inf$ of the sequence $(\sqrt[n]{|\sin{n}|})_{n=1}^{\infty}$? How to investigate the $\limsup$, the $\liminf$, the $\sup$, and especially the $\inf$ of the sequence $(\sqrt[n]{|\sin{n}|})_{n=1}^{\infty}$?

Edit: The limit of this sequence is already investigated years ago in this post: Calculate $\lim_{n \to \infty} \sqrt[n]{|\sin n|}$. So the $\limsup$, the $\liminf$, and the $\sup$ are clearly 1. Sorry for did not search wisely.

The problem is far from being trivial.

The sequence given by $a_n=e^{in}$ is dense in the unit circle since $\pi$ is an irrational number. By considering $\text{Im}(a_n)$, we get that the sequence given by $b_n=\sin(n)$ is dense in the interval $[-1,1]$.

However, that is not enough to prove that the wanted $\sup,\inf,\liminf,\limsup$ are $1,0,0,1$.

By Lagrange's theorem there is an infinite number of rational numbers $\frac{p_n}{q_n}$ such that:

$$ \left| \pi-\frac{p_n}{q_n}\right| \leq \frac{1}{q_n^2} $$
holds, and in such a case:
$$ \left|\sin(p_n)\right|\leq \frac{1}{q_n}, $$
but that is not enough to ensure that $\left|\sin(p_n)\right|^{\frac{1}{p_n}}$ is arbitrarily close to zero.

The problem strongly depends on the irrationality measure of $\pi$. Since $\pi$ has a finite irrationality measure (it is conjectured to be $2$, but until today no one has proved something better than $7.6$) our $\inf$ is indeed a minimum and it is strictly greater than zero.

Well first of all try to calculate the absolute value of |sinn|. Since the sine function is defined on (-1;1) the absolute value of |sinn|<1. Then the order of the root, as it tends to infinity the root will get smaller and smaller so basically it will tend to 0. Try to figure out yourself the interval of definition of the function and from there you'll find the sup, inf etc. My first post here on the math section, don't kill me Hope I helped.


wedge product of $m$ vectors in $\mathbb{R}^n$ I came across the symbol $|v_1 \wedge \dots \wedge v_m|^{-1}$ in a paper -  this is the norm of the wedge product of vectors $v_k \in \mathbb{R}^n$ .
I thought it's meaning was self-evident until I tried to compute it.

First of all $|v_1 \wedge \dots \wedge v_m|$ is a number since we are taking the reciprocal.

If there were $m = n$ vectors then $|v_1 \wedge \dots \wedge v_n| = \det (v_i \cdot v_j) $ and it is the volume of the paralleliped generated by the vectors $v_1, \dots, v_n$.

If we have $m < n$ vectors, given in coordinates, we have the volume of the $m$-dimensional parallelipiped inside of $n$-dimensional space.  This volume is 0 unless we use the $m$-dimensional volume measure.

In exterior algebra, if I have the coordinates of the vectors, $v_1, \dots, v_m$,  how do I compute this volume?

In clifford algebra, this could be easily computed.  A wedge product of $m$ linearly independent vectors can be rewritten as a geometric product of $m$ orthogonal vectors $u_1 u_2 \ldots u_m$.  The product of this $m$-vector with its reverse $u_m u_{m-1} \ldots u_2 u_1$ is guaranteed to be a scalar; call this $|v_1 \wedge v_2 \wedge \ldots \wedge v_m|^2$.

If you would rather not use clifford algebra, then you can think of the metric as having a natural extension to $k$-vectors.  For instance, can you identify what $g(e_1 \wedge e_2 , e_1 \wedge e_2)$ should be?  How about $g(e_1 \wedge e_2, e_1 \wedge e_3)$?

You could then expand $V = v_1 \wedge v_2 \wedge \ldots \wedge v_m$ in terms of the basis $m$-vectors, and since you know how the metric acts on these basis $m$-vectors (it's very simple for any orthonormal basis), you can compute the norm.  Realistically, if this is Euclidean space with an orthonormal basis, then the metric will just tell you to take the sum of the squares of the components.

Write each $v_k$ as a column vector. The $m$-volume is the square root of the sum of the squares of the $m \times m$ minor determinants of the matrix $[v_1 | v_2 | \cdots | v_m]$.


What is the single most influential book every mathematician should read? If you could go back in time and tell yourself to read a specific book at the beginning of your career as a mathematician, which book would it be?

A Mathematician's Apology by G H Hardy. I did in fact read this in high school, and it raised my view of mathematics from a thing of utility to a thing of beauty and wonder. It inspired me to go on to study mathematics at Cambridge myself.

It's a pity that the "introduction" by C P Snow is longer than the original and contains a rather depressing view of Hardy's later life. I would recommend readers to skip the introduction altogether and concentrate on Hardy's own words.

I would read a book about Perelman's proof of the Poincaré conjecture (or even the papers themselves). Oh, you mean the book had to be written when I was starting?


Bounds for volume integral I'm working through some practice problems on volume integrals while preparing for an exam, and I don't understand the bounds in one of them. Here's the prompt in the textbook:

Find the volume of the region bounded by $z=x+y, 0\leq x\leq 7, 0 \leq y\leq 7$, and the planes $x=0, y=0$, and $z=0$.

I know (from the solutions manual and my professor) that the answer is given by  $\int_0^7\int_0^{7-x}\int_0^{x+y} dzdydx$ (and they don't explain why), but why is $7-x$ the upper bound for $y$? I thought it would be just $7$.
Edit: I am aware that the bounds for x and y on this integral give us a triangle in the first quadrant of the xy-plane with the axes and the line y=7-x as its sides, so maybe I should have been more specific with my question:

How could my professor and the solutions manual be interpreting the instructions (especially the inequalities) to get the triangle and not the square on the xy-plane?

For $[0,7]\times [0,7]$ in $OXY$ plane with roof $z=x+y$, floor $z=0$ volume is
$$\int\limits_0^7\int\limits_0^{7}\int\limits_0^{x+y} dzdydx$$
Addition 1.
this integral corresponds exactly to
$$\left\{\begin{array}{}
0 \leqslant x \leqslant 7 \\
0 \leqslant y \leqslant 7 \\
0 \leqslant z \leqslant x+y
\end{array}\right\}$$
which is in OP and is exactly answer to OP and is volume over whole rectangle, not only it's lower triangle .
Addition 2.
If we want to have volume over triangle, then it is set:
$$\left\{\begin{array}{}
0 \leqslant x \leqslant 7 \\
0 \leqslant y \leqslant 7-x \\
0 \leqslant z \leqslant x+y
\end{array}\right\}$$
so integral will be
$$\int\limits_0^7\int\limits_0^{7-x}\int\limits_0^{x+y} dzdydx$$
excuse me for repeating, that it is not initial OP.


What does it mean to "prove $1+1=2$"? It is a famous bit of trivia that it took Russell and Whitehead about 300 pages to prove that $1+1 = 2$. However, this seems more like a definition rather than theorem. As far as I know, $2$ is just the symbol we use as shorthand for $1+1$, where $1$ is the multiplicative identity.

You have to define what system you are working in.  In Peano Arithmetic, PA, $1$ and $2$ are not part of the language.  They are abbreviations for $S(0)$ and $S(S(0))$, where $S$ is intended as the successor function, so you are being asked to prove $S(0)+S(0)=S(S(0))$.  You can follow the Wikipedia proof, which you may need to update a bit depending on how your version of the axioms is written.  Basically this should look a lot like your axioms that define addition.  PA does not specify a multiplicative identity, you have to prove that $S(0)$ is one by induction.

Axioms for the natural numbers usually define only a single number: either $0$ or $1$. The existence of any other natural number must be inferred from these axioms using the axioms and rules of logic or set theory (if applicable). If addition on $N$ is given in your axioms, it would be a trivial exercise to prove that 1+1=2.

Suppose, for example, you are given the axioms:

$1\in N$
$\forall x \in N: \exists y\in N: y=x+1$

Applying both of these axioms, we can obtain:

$\exists y\in N: y=1+1$

Applying the rule of existential specification (for $y = 2$), we could infer that $2 \in N$ and $2=1+1$. This inference can be thought of as the "definition" of $2$.

If addition on $N$ is not given in your axioms, you would have to construct (i.e. prove the existence of such a function) using the axioms of your set theory before you could "define" or infer that $2=1+1$ as above.


If we take 5 shots with a basketball, what's the probability that you make 2 given you make at least 1? Probability of making a shot is $\displaystyle \frac{3}{5}$.

So $2$ shots would be $\displaystyle \dbinom{5}{2} \left(\frac{3}{5}\right)^2 \left(\frac{2}{5}\right)^3$ right?

To find the probability in question, you have to first check what the probability of making at least $1$ shot is, and also the probability of making exactly $2$ shots, and then divide the latter by the former. That is, the probability is $$\frac{\text{(probability of landing 2 shots)}}{\text{(probability of landing at least 1 shot)}}$$

Probability of landing at least $1$ shot:

This is the same as $1$ minus the probability of missing all $5$ shots. The answer, therefore, is $$1 - \left(\frac{2}{5}\right)^5$$

Probability of landing exactly $2$ shots:

This is the answer that you had, i.e. $$\dbinom{5}{2} \left(\frac{3}{5}\right)^2 \left(\frac{2}{5}\right)^3$$

Final answer:

$$\frac{\dbinom{5}{2} \left(\frac{3}{5}\right)^2 \left(\frac{2}{5}\right)^3}{1-\left(\frac{2}{5}\right)^5} = \boxed{\frac{240}{1031}}$$

The other answers are OK, but you should keep in mind that it is supposed the probability of making two different shots are independent of each other. In real life, for most of the players, if you managed to make one shot, it influences the probability of making the following shots. For example, if the probability of making a single shot is 0.6, the probability of making 5 independent shots would be 0.07777, but the probability of making 5 consecutive shots might be much higher, depending on the player.


Is a logic system with predicate symbols, individual constants, negation, and conjunction (no variables and no quantification) decidable? Suppose I have a logical system with predicate/relation symbols and individual constants symbols, negation and conjunction, but no variables or quantification. For instance, suppose I have the individual constants a and b, and a 2-place predicate P, in the language. I can formulate atomic sentences like P(a, b) or P(b, a). Suppose also that negation and conjunction are available, so I can formulate complex sentences like P(a, b) & P(b, a) or -P(a, b) (where & is the conjunction symbol and - is the negation symbol).

Is this decidable, in the way that propositional logic is? Or do I need to add variables and quantification before it becomes undecidable?

(How would this be different from propositional logic, anyway?)

You need quantification to make your system essentially different from propositional logic. Given a formula $\phi$ with no quantifiers, pick a distinct propositional variable $v_{xy}$ for each atomic predicate $P(x,y)$ occurring in $\phi$. Let $\phi'$ be the formula of propositional logic obtained by replacing $P(x, y)$ by $v_{xy}$ throughout $\phi$. Then $\phi$ is valid iff $\phi'$ is valid.

In order for the structure to be valid, the initial sentence (if A then B), and the inverse, converse, and contrapositive would have to bear their correct truths.
If you observe an event, and it is true, but its contrapositive is false, then your original event was either the inverse or converse of a different truth table.
In short, your whole axiomatic structure has to be consistent.
As for your abbreviated syntax, what I see is a computer application, in that:
The statement -P(a,b) can be read "If -P(a,b) exists (as input or output), then TRUE." But in that case, your statement is the Predicate in another structure.

So I'd say that if you can't pull out of your structure any theorems that are amalgams of your atomic statements which would negate the axioms/atomic statements, AND there are no other theorems which can be assembled within your structure from your statements and operators that would afford 'floating' answers ( like in buffers), then it would be decidable by definition, though I would look for reducibility.
Karnaugh maps are nice because they do it ( reducing a logic structure to the fewest gates {or least number of embedded theorems}) in a third-grade chart.
Consider the T/F status for 2=2 and x=2: the first is an atomic statement that is always true. The second is true IFF x has the value of 2.


A question which is to be proved assuming Bertrand Postulate I am trying  exercise problem of David Burton Elementary Number Theory and I am unable to think about this problem.

  Problem is->Prove that for every n$\geq$ 2 there exists a prime p with p$\leq$n< 2p .

Can someone please help how to think about this problem.
I shall be really thankful.

According to Bertrand’s postulate, there is a prime $p$ with $k\lt p\lt2k$ for all integers $k\gt1$. Apply this for $k=\left\lceil\frac n2\right\rceil$ to obtain a prime $p$ with $\left\lceil\frac n2\right\rceil\lt p\lt2\left\lceil\frac n2\right\rceil$ and thus $\frac n2\lt p\le n$. Now multiply the left inequality by $2$ and move it to the right.

(This proof doesn’t apply for $n=2$, since in that case $\left\lceil\frac n2\right\rceil=1$, so we have to check separately that there is a prime $p=2$ such that $p\le2\lt2p$.)

Assuming $n>2$ is non prime, now take the nearest prime (smaller than) $p$ of $n$, we can't have $p<2p\leq n$ because that would imply there is a prime $q$ such that $p<q<2p\leq n$ .(bertrand) contradicting the assumption that $p$ is the nearest prime. so the exact inequality is : $p<n<2p$.(why $n\neq 2p ?$) for $n$ is prime, thats an easy task.


What am I doing when I separate the variables of a differential equation? I see an equation like this:

$$y\frac{\textrm{d}y}{\textrm{d}x} = e^x$$

and solve it by "separating variables" like this:

$$y\textrm{d}y = e^x\textrm{d}x$$
$$\int y\textrm{d}y = \int e^x\textrm{d}x$$
$$y^2/2  = e^x + c$$

What am I doing when I solve an equation this way?  Because $\textrm{d}y/\textrm{d}x$ actually means

$$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$

they are not really separate entities I can multiply around algebraically.

I can check the solution when I'm done this procedure, and I've never run into problems with it.  Nonetheless, what is the justification behind it?

What I thought of to do in this particular case is write

$$\int y \frac{\textrm{d}y}{\textrm{d}x}\textrm{d}x = \int e^x\textrm{d}x$$
$$\int \frac{\textrm{d}}{\textrm{d}x}(y^2/2)\textrm{d}x = e^x + c$$

then by the fundamental theorem of calculus

$$y^2/2 = e^x + c$$

Is this correct?  Will such a procedure work every time I can find a way to separate variables?

The basic justification is that integration by substitution works, which in turn is justified by the chain rule and the fundamental theorem of calculus.

More specifically, suppose you have: $$\frac{dy}{dx} = g(x) h(y)$$
Rewrite as:
$$\frac{1}{h(y)} \frac{dy}{dx} = g(x)$$  Add the implicit dependency of $y$ on $x$ to obtain
$$\frac{1}{h(y(x))} \frac{dy}{dx} = g(x)$$

Now, integrate both sides with respect to $x$:
$$\int \frac{1}{h(y(x))} \frac{dy}{dx} \, dx = \int g(x) \, dx$$ If we do a variable substitution of $y$ for $x$ on the left-hand side (i.e., use the integration by substitution technique), we replace $\frac{dy}{dx} dx$ with $dy$. Thus we have $$\int \frac{1}{h(y)}\, dy = \int g(x) \, dx,$$
which is the separation of variables formula.

So if you believe integration by substitution, then separation of variables is valid.

Seperation of variables involves manipulating the differentials (the $dx$'s and $dy$'s in your equation). A differential is the infinitesimal change in a variable, and can be treated as a variable in its own right in many applications. With this perspective, $dy$ is a function of $x$ and $dx$, and the derivative $dy/dx$ is the ratio of these two differentials, which is a function of $x$. What you are doing is simply performing an algebraic manipulation of these variables and then using calculus to remove the differential terms.


normal test to student test, logic behind it Let's say we have :
$$ X_i  \sim \ N( \mu, \sigma^2 ) $$
iid

I'm constructing this test function, in order to test two hypothesis on $\mu$ : $$ \mathbb { 1} {\{ \sum^n X_i < q_a \} } $$
where $q_a$ is the $a$-quantile of $$ P_{ \mu_0 } ( \sum^n X_i < q_a  ) $$

If I know $\sigma$, this test is just fine. Because then I can rewrite the test function as :
$$ \mathbb { 1} \{ \frac{  \sum X_i - n \mu_0 }{ \sigma \sqrt{n} } < \phi^{-1} (a) \}  $$

But if I don't know $\sigma$, how can I handle things ? I have been told to replace it by $S$, such that : $$ S^2 = \frac 1 {n-1} \sum^n (X_i - \overline X)^2 $$

But how do you conclude that the test is now :

$$ \mathbb { 1} \{ \frac{  \sum X_i - n \mu_0 }{ S \sqrt{n} } < t_{n-1, a} \}  $$
where $t_{n-1, a} $ is the $a$-quantile of the student law with n-1 degrees of freedom?

The idea is that we are testing for two different things. The first one is a test for a normal where we know $ \sigma $, the second for an unknown $\sigma$.  We aren't replacing anything in the first case to get to the second. The thing is it is two totally different ideas. The only link I found is that :

$ \bar{X} $ is independant to $ S^2$. Hence :

$$ \frac {\bar{X}  - \mu } { \frac {S}{ \sqrt n }  } = \frac {        \frac { \sqrt n (\bar{X} - \mu )  }{ \sigma }       }{\sqrt {\frac { (n-1) S^2 }{(n-1) \sigma^2 } }     } \sim t_{n-1} $$

In both case we want a test with $\alpha$ level of risk, so take the $1-\alpha$ quantile. Which leads us to the result mentionned in the question.

So the conclusion is, we have two different tests, asssociated to two different random variables, depending on whether we know the variance or not.

It is well known that $S_{XX}/\sigma^2= \frac {1} {n\sigma^2} \sum (X_i - \overline X)^2$ has the $\chi_{n-1}^2$ distribution.

It is also known that if $Z$ is a standard normal, and $X$ is $\chi_k^2$ then $T=\frac{Z}{\sqrt{X/n}}$ has the $t_k$ distribution.

I'll let you see how to apply that here


Does Brownian motion visit every point uncountably many times? Let $B_t$ be a one-dimensional standard Brownian motion.

  Is it true that, almost surely, for every $x \in \mathbb{R}$ the set $\{t : B_t = x\}$ is uncountable?

Let $A_x$ be the event that $\{t : B_t = x\}$ is uncountable.  It is well known that $\mathbb{P}(A_0) = 1$.  By recurrence and the strong Markov property, we also have $\mathbb{P}(A_x) = 1$ for every $x$.  I am asking about the probability of $A = \bigcap_{x \in \mathbb{R}} A_x$.  Because of the uncountable intersection, it is not even obvious that this set is measurable.

George Lowther in this comment says the answer is yes, but I don't see how to prove it.

Actually there is an almost sure lower bound on the Hausdorff dimension of the level sets of Brownian motion that follows from the Ray-Knight theorem.
$$ a.s \quad \forall a\in \mathbb{R}, \quad \text{dim}\{t \geq 0 | B(t) =a\} \geq \frac{1}{2}.  $$

This is a theorem (6.48, p. 170) in the book, "Brownian Motion" by Peres and Morters.

Consider $C_x$ be the event that the set ${t: B_t = x}$ is countable.

For any $x$, the measure of $C_x$ is zero. Countable zero-measure sets union is also a zero-measure set. $A = \bigcap A_x = -\bigcup -A_x = -\bigcup C_x$ and hence has a measure of $1$.


Let $f: D_4 \rightarrow C_{24}$ be a homomorphism. Show that for all $a \in D_4$, the following is true $f(a)^2 = e$. Let $f: D_4 \rightarrow C_{24}$ be a homomorphism. Show that for all $a \in D_4$, the following is true $f(a)^2 = e$.

What I thought we could do, was write out $D_4$, as it has just 8 elements.
$D_4 = \{1,\rho,\rho^2,\rho^3,\sigma,\sigma\rho,\sigma\rho^2,\sigma\rho^3 \}$ with $\sigma$ being the reflection, having order 2 and $\rho$ being the rotation, having order 4. Then we could just show for every element here the condition $f(a)^2 = e$ holds. As we know that $f(a)^2 = f(a^2)$, we know that the elements with an order that divides 2 out of $D_4$, work. That are $1$, $\rho^2$, $\sigma$ and $\sigma\rho^2$. We still need to show that the condition holds for the other elements of $D_4$.

That is where I don't know what to do anymore. Is what I have done correct? If so, how should I continue? Are there better ways of tackling such problems?

Thanks for reading,

K.

Let $\varphi:D_4\to C_{24}$ be an arbitrary homomorphism. Let $a,b\in C_{24}$ be such that $\rho\mapsto a$ and $\sigma\mapsto b$. Then we have two relations: $bab=a^{-1}$ and $b^2=0$. Since $C_{24}$ is commutative, these relations implies that $a=a^{-1}$, or that $a^2=0$. Hence, we have that $a^2=b^2=0$, so $\varphi(x)^2=0$ for any possible $x\in D_4$.

My original, incorrect answer. Thanks to Zoe H for pointing out where I went wrong.

  I don't believe this is correct. Define a homomorphism $\phi: D_4\to\mathbb{Z}/24\mathbb{Z}$ given by $\rho\mapsto 6$ and $\sigma\mapsto 0$. Then $\phi(\rho)^2 = 12\ne 0$.

Then we could just show for every element here the condition $f(a)^2 = e$ holds. As we know that $f(a)^2 = f(a^2)$, we know that the elements with an order that divides 2 out of $D_4$, work. That are $1$, $\rho^2$, $\sigma$ and $\sigma\rho^2$.

That's true, and a nice idea! But, it turns out that the only elements whose orders don't divide $2$ are $\rho$ and $\rho^3$: products of a single rotation and a single reflection (like  $\sigma \rho$) are reflections, and have order $2$.

Thus, to continue with this approach, you can focus on showing that $a^2 \in \ker f$ for $a \in \{\rho, \rho^3\}$ and any homomorphism $f\colon D_4 \to C_{24}$. You can see that $(\rho)^2 = \rho^2 = (\rho^3)^2$, so you just need to figure out why $\rho^2 \in \ker f$ for any $f$.

One approach to do that: It turns out that the commutator subgroup of $D_4$ is $\langle \rho^2 \rangle$ (although it suffices here to know only that $\rho^2$ is in the commutator subgroup). Assuming you've learned the theorem that "$G/\ker f$ is abelian if and only if the commutator subgroup of $G$ is contained in $\ker f$", look at the possibilities for $G/\ker f$. The only possible sizes for $G/\ker f$ here are $1, 2$, and $4$ (why not $8$?), and all groups of these orders are abelian.

An alternative approach is simply to look at the possible sizes for normal subgroups of $D_4$ (i.e., possibilities for $\ker f$) and use the first isomorphism theorem to think about $f(D_4) \cong D_4 / \ker f$.

As above, the only possibilities for $|f(D_4)| = |D_4 / \ker f|$ are $1, 2,$ or $4$.

Certainly there's nothing to worry about if $|f(D_4)| \in \{1,2\}$.

The only normal subgroup of $D_4$ having order $2$ is $\langle \rho^2 \rangle$ (the only other subgroups of order $2$ are those generated by a single reflection, and these are never normal), which happens to be the center $\Bbb Z(D_4)$ of $D_4$. Probably you've learned that $G / \Bbb Z(G)$ can never be cyclic, implying that $D_4 / \langle \rho^2 \rangle$, having order $4$ and not being cyclic, must be isomorphic to $C_2 \times C_2$, where every element has order dividing $2$.


Proving by definition that $\lim_{(x,y) \to (1,2)}\frac{3x-4y}{x+y}=-\frac{5}{3}$ Proving by definition that $\lim_{(x,y) \to (1,2)}\frac{3x-4y}{x+y}=-\frac{5}{3}$

Take $\epsilon>0$, I want to find $\delta>0$ such that:

$$\lVert (x-1,y-2)\rVert <\delta \Rightarrow \left\lvert \frac{3x-4y}{x+y}+\frac{5}{3}\right\rvert<\epsilon$$

So I started by adding both fractions and obtained:

$$\left\lvert \frac{3x-4y}{x+y}+\frac{5}{3}\right\rvert=\frac{7}{3}\left\lvert\frac{2x-y}{x+y}\right\rvert=\frac{7}{3}\left\lvert\frac{2x-y-2+2}{x+y}\right\rvert=\frac{7}{3}\left\lvert\frac{2(x-1)-(y-2)}{x+y}\right\rvert$$

Now, I have $\lvert x-1\rvert\leq\sqrt{(x-1)^2+(y-2)^2}<\delta$ and $\lvert y-2\rvert\leq\sqrt{(x-1)^2+(y-2)^2}<\delta$

However, im not being able to bound $\frac{1}{\lvert x+y\rvert}$

Am I on the correct track? Any suggestions? Thank you very much.

$\lim_{(x,y) \to (1,2)}\dfrac{3x-4y}{x+y}
=-\dfrac{5}{3}
$

I like to let
variables go to zero,
so let
$x = 1+u, y=2+v$.

Then

$\begin{array}\\
\dfrac{3x-4y}{x+y}+\dfrac{5}{3}
&=\dfrac{3(1+u)-4(2+v)}{1+u+2+v}+\dfrac{5}{3}\\
&=\dfrac{3u-4v-5}{3+u+v}+\dfrac{5}{3}
\qquad\text{You can see here that the limit will be -5/3}\\
&=\dfrac{3(3u-4v-5)+5(3+u+v)}{3(3+u+v)}\\
&=\dfrac{14u-7v}{3(3+u+v)}
\qquad\text{and the constant term cancels out}\\
&=\dfrac{14u-7v}{9+3(u+v)}\\
\end{array}
$

From this we see that
if $u$ and $v$ are small
the the numerator is small
and the denominator
is not small,
being around $9$.

To make this rigorous,
if $|u|, |v| < c$
where
(to get a lower bound
on the denominator)
$0 < c < \frac12$,
then
$|14u-7v|
\lt 21c
$ and
$|9+3(u+v)|
\ge 9-3|u+v|
\ge 9-3(2c)
\gt 6
$
so
$|\dfrac{14u-7v}{9+3(u+v)}|
\le \dfrac{21c}{6}
=\dfrac{7c}{2}
$.

Therefore,
to make
$|\dfrac{14u-7v}{9+3(u+v)}|
\lt d$,
choose
$\dfrac{7c}{2}
\lt d$
or
$c \lt \dfrac{2d}{7}
$.

Finally,
to go back to
the original problem,
if
$|x-1| < \dfrac{2d}{7}$
and
$|y-2| < \dfrac{2d}{7}$
where
$\dfrac{2d}{7} < \frac12$
then
$|\dfrac{3x-4y}{x+y}+\dfrac{5}{3}|
\lt d
$.

On bounding $\frac{1}{|x+y|}$. If you take $\delta<1/2$, then
$$
|x-1|<\delta
$$
and
$$
|y-2|<\delta
$$
yields
$$
1/2<x
$$
and
$$
3/2<y
$$
and so
$$
|x+y|=x+y>2
$$
which is the part you are worried about.

The intuition behind this: Note that you are worried about $|x+y|$ getting very small, but of course it wont since in the limit, $x\approx 1$ and $y\approx 2$.


If a commutator has an eigenvalue $0$, do the two operators share an eigenvector? Let $A,B$ be two diagonalizable linear operators, such that their commutator $\left[A,B\right] = AB-BA$ has an eigenvalue $\lambda = 0$. Does this mean $A$ and $B$ share an eigenvector?

Of course, the vice-versa is correct. If they do share an eigenvector, the commutator necessarily has an eigenvalue of $0$. I'm not sure though if this is a sufficient condition, or only a necessary one.

Clearly we can pick a basis so that one of the operators is diagonal already. Here's a small counterexample of this sort:
$$ A = \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \mu \end{pmatrix}  \qquad B = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}. $$
The eigenvalues of $B$ are $\pm \sqrt{2}$ and $0$, and all eigenvectors are of the form $(1,a,\pm 1)$, none of which live in one of the eigenspaces of $A$. But $AB-BA$ has first row and first column all zero, so $(1,0,0)$ is an eigenvector with eigenvalue zero.

Here is another counterexample. Let $n>2$, the underlying field is real and

$A=\operatorname{diag}(n,\,n-1,\,\ldots,\,1)$,
$T$ be the skew-symmetric Toeplitz matrix whose first row is $(0,\,1,\,\frac12,\,\ldots,\,\frac1{n-1})$,
$C$ be any skew-symmetric matrix with a nonzero off-diagonal entry on every column when $n$ is odd, or any hollow, singular symmetric matrix with a nonzero off-diagonal entry on every column when $n$ is even. E.g. when $n$ is even, one may take $C=\pmatrix{S&S\\ S&S}$ where $S$ is any hollow symmetric matrix with a nonzero off-diagonal entry on every column,
$B=T\circ C$, the Hadamard product of $T$ and $C$.

Note that $C$ is singular and $B$ is is symmetric when $n$ is odd or skew-symmetric when $n$ is even. Moreover, $(AB-BA)_{ij}=(a_i-a_j)t_{ij}c_{ij}=c_{ij}$. Hence $A,B$ are diagonalisable and $AB-BA=C$ is singular. However, by construction, every column of $B$ possesses a nonzero off-diagonal entry. Hence $A$ and $B$ do not share any eigenvector.


Integrate $\int \sin (2x) \cos (2x)\,dx$ I have

$$\int \sin(2x) \cos (2x)\,dx = \frac12 \int \sin(4x)\,dx = -\frac18 \cos(4x),$$

but I also have

$$\int \sin(2x) \cos (2x)\,dx = \frac12 \int \sin 2x \cdot 2 \cos 2x \, dx = \frac14 \sin^2(2x).$$

Which one is correct, and why is the other method wrong?

They are differing by an integration constant, because of $\cos(2 y) = \cos^2(y) - \sin^2(y) = 1 - 2 \sin^2(y)$, and hence are the same as indefinite integration produces an anti-derivative up to a constant

In second method lost step it is not

$$\int \sin(2x) \cos (2x)\,dx = \frac12 \int \sin 2x \cdot 2 \cos 2x \, dx = \frac14 \sin^2(2x).$$

It has to be
(since 2sin(2x). cos(2x)=sin(4x))

$$\int \sin(2x) \cos (2x)\,dx = \frac12 \int \sin 2x \cdot 2 \cos 2x \,dx = \frac12 \int \sin(4x) dx=-\frac18\cos(4x)$$
(Since $$\int \sin(4x)dx=-\frac14 \cos(4x).$$)


Canonical probability distribution associated with the "harmonic mean" Is there a canonical continuous probability distribution, the center of which is best characterized with the harmonic mean, given by
  $$
\mathrm{HM}(X) = n \cdot \left( \sum\limits_{k=1}^{n} x_k^{-1} \right)^{-1}?
$$

With "canonical" I mean: similar to the way we typically associate the normal distribution with the arithmetic mean,
$$
\mathrm{AM}(X) = \sum\limits_{k=1}^{n} \dfrac{x_k}{n},
$$
or the log-normal distribution with the geometric mean,
$$
\mathrm{GM}(X) = \prod\limits_{k=1}^{n} x_k^{\frac{1}{n}}.
$$

Update: Because zero is not invertible, I guess that the support of such a distribution would have to be a subset of $\mathbb{R} \setminus \{ 0 \}$, maybe $\mathbb{R}_{>0}$.

The inverse normal distribution does not furnish a suitable example, because if $Y = Z^{-1}$ where $Z \sim \operatorname{Normal}(0,1)$, then $\operatorname{E}[Y]$ is indeterminate.  We can, however, consider a double-inverse gamma distribution:  define $$f_X(x) = \frac{|x|}{2}e^{-|x|}, \quad -\infty < x < \infty.$$  It is trivial to see that this function indeed defines a density.  Now let $Y = X^{-1}$, from which we find that the density of $Y$ is $$f_Y(y) = f_X(y^{-1})y^{-2} = \frac{1}{2y^2|y|} e^{-1/|y|}, \quad y \ne 0.$$  This function does have a well-defined expectation since $$\int_{y=0}^\infty yf_Y(y) \, dy = \frac{1}{2}.$$  Then, due to $f_Y$ being an even function, we trivially find $\operatorname{E}[Y] = 0$.

Now, whether the harmonic mean of an IID sample drawn from $Y$ is in some sense the "best" estimator of the population mean because $\bar x$ is the "best" estimator of the mean of $X$ and $Y = 1/X$, I am not so sure.  This is because we can say that the estimator $\tilde y = n (\sum_{i=1}^n y_i^{-1})^{-1}$ has expectation $$\operatorname{E}[\tilde y] = n \operatorname{E}\left[\left(\sum_{i=1}^n y_i^{-1}\right)^{-1}\right],$$ but it cannot be said that the RHS is in general equal to $$n \left(\operatorname{E}\left[\sum_{i=1}^n y_i^{-1}\right]\right)^{-1},$$ in as much as we cannot generally write $$\operatorname{E}[g(X)] = g(\operatorname{E}[X]):$$  that is, the expectation of a function of a random variable does not generally equal the function evaluated at the variable's expected value.  If you could say that, then the expectation passes through the sum via linearity and you'd get $$n \left(\sum_{i=1}^n \operatorname{E}[y_i^{-1}]\right)^{-1} = n \left(n \operatorname{E}[X]\right)^{-1} = \operatorname{E}[X]^{-1}.$$  And again, you run into the same problem/fallacy:  you can't claim that this last expression equals $\operatorname{E}[Y]$.  Thus, the idea to consider inverse distributions seems dubious to me.

It's quite unclear what you mean by "canonical" and "associated with" since

$AM(F_n)\overset{a.s.}\longrightarrow E(F)$ for any $F$ (if $E(F)$ exists) - not just normal $F$.
$GM(F_n)\overset{a.s.}\longrightarrow median(F)$ for any $F=\exp(S)$ if $E(S)=median(S)$ - not just normal $S$.
Pending that clarification, $HM(F_n)\overset{a.s.}\longrightarrow mode(F)$ for $$F\sim\Gamma(\alpha,\beta)$$
for $\alpha>1$, since $mode(\Gamma(\alpha,\beta))=\frac{\alpha-1}{\beta}=(E(Inv\Gamma(\alpha,\beta))^{-1}$.
In fact, the above convergence holds for any distribution $F$ s.t.
$$\int_R \frac {dF(u)}{u}=\frac 1 {mode(F)}$$
These include all distributions with $[0,\infty)$ support that have a mode at $0$. If not, since HM is skewed towards small values of the distribution, it picks up the mode if $f(0)=0$, $f$ rises to its peak and then falls off slower than it rose up.
A discrete example on $\{a,b,c\}$: Pick any positive values $a$, $c$ and probabilities $p_a$, $p_c$ s.t. $p_b=1-p_a-p_c>\max\{p_a,p_c\}$ and set $b^{-1}=\frac{p_a}{p_a+p_c}a^{-1}+\frac {p_c}{p_a+p_c}c^{-1}$ to get $HM(F_n)\to b$. Set $p_a=p_c<1/3$, to get $b=HM(a,c)$.


Surprisingly elementary and direct proofs What are some examples of theorems, whose first proof was quite hard and sophisticated, perhaps using some other deep theorems of some theory, before years later surprisingly a quite elementary, direct, perhaps even short proof has been found?

A related question is MO/24913, which deals with hard theorems whose proofs were simplified by the development of more sophisticated theories. But I would like to see examples where this wasn't necessary, but rather the theory turned out to be superfluous as for the proof of the theorem. I expect that this didn't happen so often. [Ok after reading all the answers, it obviously happened all the time!]

Goedel's original completeness/compactness proofs in logic are very hard and technical. Modern versions of the proofs are considerably simpler and do not use any sophisticated new theory.

Most proofs of the fundamental theorem of algebra use some topological/homotopical deep(ish) results or some deep(ish) results of complex analysis. In fact, the theorem can be proved using only the definition of complex numbers and absolute value, using very elementary properties of complex numbers, and the entire proof is about half a page long (such a proof is actually a minimum modulus argument but applied to a polynomial so that the computation can be done directly without appeal to the general minimum modulus principle). The proof is extremely elementary.

The proof that Euclide's fifth postulate is independent of the other axioms of Euclidean geometry by exhibiting models (i.e., the sphere and the hyperbolic plane) where the postulate fails (in different ways) is completely elementary. Certainly all those who tried to prove/disprove the fifth postulate did so while walking on a counterexample. However, the numerous attacks on the problem prior to its settlement in the 20th century were quite sophisticated and laborious. The barrier was conceptual - the understand of the importance of models - not technical.

Brouwer's topological proof of his fixed point theorem used (I believe) homology and/or the fundamental group. A perfectly elementary proof using Sperner's Lemma was later discovered (I'm not entirely sure about the chronological order here, so I may be mistaken).

Initial proofs that the higher homotopy groups are all abelian were greatly simplified by the Eckman-Hilton argument (which is completely elementary).

The Mac Lane coherence theorem in category theory is rather technical and is enormously simplified by considering the Yoneda embedding in the 2-categorical setting. All of the machinery was already present, but it required some re-assembly to figure out a rather elementary proof.

One might suppose that there needs to be some profound insight or conception needed.  That is, one either has to deconstruct the notation, or see something in a different light.

For example, a recent question on whether other sets of weights, other than $1,3,9,27,81$ might yield all numbers from $1$ to $121$.  One might be tempted to solve this by seeking the largest of the weights, as one does filling a backpack.  The actual proof involves hunting down the smallest weight.

When one devises notation, such as my Polygloss, the notation gives rise to insights that were not apparent without it.

One thing, for example, is that it's not apparently obvious that a dot-product of vectors in an oblique coordinate system could be done relatively easily.  But this can indeed be done by the use of a matrix, which normally does not appear in the process.  One multiplies one of the vectors by a matrix, and takes the dot product of this and the second vector.

I have had many profound insights into many different things, and the chief stumbling block is that one has to deconstruct the current notions and objections.  For example, the notion that a base consists of digits $0$ to $b-1$ goes, when one deals with alternating arithmetic.


How to solve $x^3 = x^{x^2-2x}$? Well i guess this is somehow pretty easy but there is something i don't understand.

I know that if $\;x>0\;$ then I can compare the exponents: $\; 3=x^2-2x$ , and from here I get that $\;x=3\; or \;x=-1\;$, but because $\;x>0\;$ that leaves me only with $\;x=3$ .

Second thing is that if both bases from both sides are equal to $1$ then its another solution, and therefore $\;x=1\;$ is another solution.

For conclusion we get that the solutions are: $\;x=3\; or \;x=1$ .

Now my question is why $\;x=-1\;$ is also a solution? How do i get to this solution? am I suppose to just try place it and check because I somehow got it for $\;x>0$ ? are there any steps I can follow for solving this kind of equations?

Another question is how do you call this kind of equation? when I was looking for "exponential equations" I could only find ones with numbers in the bases .

Thanks!

$x^a=x^b\implies a=b$ for all $x$,whether $x>0$ or $x<0$. You got $3=x^2-2x$, and solved this quadratic equation to get $x\in\{3,-1\}$. So $-1$ is indeed a solution. It depends on the equation as to which methods will you follow, there is no general algorithm. But yes, keep using exponent rules, and you will eventually lead to a solution.

look on rh(n)
and get x^2-2*x=3 (mod n)
Because rh(2)=-1 then we get
(-1)^2+2=3=3 (mod 2)
and thats why it true


On the definition of a Turing Machine as stated in Soare I am currently reading R. Soare's book Turing Computability. At page 7 the author gives the definition of a Turing Machine for the first time he writes,

  Definition. A Turing machine (automatic machine, a-
  machine) $M$ includes a two-way infinite tape divided into cells, a reading
  head which scans one cell of the tape at a time, and a finite set of internal
  states $Q = \{q_0,q_1,\ldots,q_n\}$, $n \ge 1$. Each cell is either blank ($B$) or has written on it the symbol $1$. In a single step the machine may simultaneously: (1) change from one state to another; (2) change the scanned symbol $s$ to another symbol $s' \in S =\{1,B\}$; and (3) move the reading head one cell to the right ($R$) or left ($L$). The operation of $M$ is controlled by a partial map $\delta: Q\times S \to Q\times S\times \{R,L\}$ (which may be undefined for some arguments).

Now in this definition a lot of things are not made precise enough. They are as follows,

What is the precise mathematical definition of "a two-way infinite tape"?
What is the precise mathematical definition of a "cell"?
What exactly is a "reading head"?
What exactly is a "scanned symbol"? What are the differences between a scanned symbol and an ordinary symbol?

Next in the book it is written that,

  We begin with $M$ in the starting state $q_1$ scanning the leftmost cell
  containing a $1$, called the starting cell.

My questions are,

What is meant by "scanning" here?
How can we precisely define "the leftmost cell"?

The notion of a "tape" and all these things is probably pretty intuitive to the computer science students but for a student coming from pure mathematics background, they are not at all intuitive to me and I don't understand the definition of a Turing Machine at all unless it is expressed in a precise form.

The missing bit of this definition is that it doesn't tell you about the state of a Turing machine - with that bit of information, the definition looks a lot more rigorous. In essence, the definition tells you how to define a Turing machine (which is not that useful alone), but not how to run it.

A Turing machine can be said to consist of the following information:

A finite set $Q$ of states.
A finite set $S$ of symbols.
A partial map $\delta:Q\times S \rightarrow Q\times S \times \{R,L\}.$

The state of the Turing machine can be thought of in a number of ways; most conveniently, you can consider a tape as a map $t:\mathbb Z\rightarrow S$ and the reading head as some $n\in \mathbb Z$ and the machine's internal state to be an element $q\in Q$. The "scanned symbol" is the term under the reading head - i.e. $t(n)$. Then, a single step of the Turing machine transforms a state $(t,n,q)$ to $(t',n',q')$ as follows:

Compute the tuple $(q',s,d)=\delta(q,t(n))$.
Define $t'(m)$ as being equal to $t(m)$ if $m\neq n$ and equal to the computed $s$ at $n$ - i.e. $t'$ is just $t$ with the term under the reading head changed.
Define $n'$ as being $n+1$ if $d=R$ and $n-1$ if $d=L$ - this is how the reading head moves.
The value $q'$ is already read as part of the tuple.

So, essentially, the Turing machine always has this "tape" which is really just a function and some position on that tape, which is really just an integer, and some state, which is some element of a given finite set.

The rest refers to which state one begins a Turing machine in; in particular, to represent algorithms, we imagine that the tape $t:\mathbb Z\rightarrow S$ is constant outside of some interval $[0,n]$; in that interval, we write some arbitrary string of symbols representing the input (and, apparently, this book wants the first symbol to be $1$, but this is not really necessary) - this is a detail that matters less how it is handled. Then, we start updating the Turing machine's state according to the above rules.

One can also simplify the definition by taking the state to just be a map $t:\mathbb Z\rightarrow S$ and imagining that the reading head is always at $0$, but the output element of $\{R,L\}$ shifts the whole function (by precomposing $t$ with $x\mapsto x\pm 1$) either left or right after writing the new value at $0$.

Two-way infinite tape and cells:

Think of an unrolled roll of toilet paper that is infinitely long, and extends to the left and to the right infinitely far.  There is a one-to-one map between cells and the integers, $\mathbb{Z}$.  Each cell sits in one-to-one relation with an integer.

Cells are the square pieces of toilet paper.  You can write a $1$ or a $B$ on each one.  The reading head is just a device that can determine if a given cell has a $1$ or a $B$ written on it.

A scanned symbol is simply the $1$ or $B$ written in a cell.  A cell might exist somewhere on the tape, but if it hasn't been read (scanned) it cannot be used to change the state of the Turing machine.

Scanning means reading successive squares of the toilet paper.


Any good books for mathematicians in order to better understand the "physics way of treating things"? Any good books for mathematicians in order to better understand the "physics way of treating things"?

In order to understand physics calculations better from a math, but non-physics background.

I have speculated that part of physics is not that much about calculation or proofs, but more about forming proper intuitions about the problems ("what happens physically", "what are important parameters", "how do they interact").

The "physics way of doing things" often involves creating a mathematical model of a physical situation that is (a) complex enough to capture the significant parts of the physical situation and at the same time (b) simple enough to be solvable.

For example, a simple model of ballistic motion will neglect air resistance, wind effects and Coriolis force.

Of course, making the correct choice of which factors to include and which to leave out depends on some degree of physical "intuition".

The Feynman Lectures on Physics are a good place to start to understand this style of thinking.

Some theoretical physicists now believe that if you create an enormously complex mathematical model, with many more spatial dimensions than currently observed by experimental physicists and engineers, and incorporating an uncountable number of differing candidate model universes with vastly differing and mostly unrealistic physical properties; that you can then take make an extra a priori leap of faith - that the right answer to any given physical question is to be found in realm of the highly complex mathematical model somewhere and now all you need to do is spend your entire research career searching this "mathematical model" for the right candidate "mathematical model" that describes what actually happens in our "real universe".

However I personally doubt that theoretical physics has now basically morphed into engineering: that is into a field of study dominated by building bigger computers and finding more efficient search algorithms.


Do prime numbers satisfy this? Is this true that $n\log\left(\frac{p_n}{p_{n+1}}\right)$ is bounded, where $p_n$ is the $n$-th prime number?

Seems unbounded:

Let  $g_n = p_{n+1} - p_n$ be the prime gap, then Westzynthius's result (see link below) states that $\lim\sup \left[ g_n/(\log p_n) \right] = \infty$, hence

$$\lim \sup n \log(p_{n+1}/p_n) = \lim \sup n \log (1 + g_n/p_n) = \lim \sup n g_n/ p
 _n =  \lim \sup  g_n/\log n = \infty$$

http://en.wikipedia.org/wiki/Cram%C3%A9r%27s_conjecture

$\frac{p_n}{p_{n+1}}<1$ for all $n$, so when $n\to\infty$, $(\frac{p_n}{p_{n+1}})^n\to0$.
so
$$n\log(\frac{p_n}{p_{n+1}})=\log(\frac{p_n}{p_{n+1}})^n\to-\infty$$
so this is not bounded


Given $n \in N$, prove that $f: [0, \infty) \rightarrow [0, \infty) f(x) = x^n$ is increasing without calculus (derivates) I have to prove this using $x_1 < x_2 \rightarrow f(x_1) < f(x_2)$.
I have no idea on how to do this, but what I have tried so far is:
Consider $x_1, x_2  \in [0, \infty)$ such that $x_1 < x_2 \rightarrow f(x_1) < f(x_2)$
$\rightarrow x_1^n < x_2^n$
I do not know what to do next

Say $x,y \in [0,\infty) $ and $x\lt y$. Then $$f(x) \lt f(y) \\ \iff x^n \lt y^n \\ $$
Since $\ln x$ is an increasing function, $$\iff n\ln x \lt n \ln y \\ \iff \ln x \lt \ln y \\ \iff x\lt y $$
But that’s of course true.


Show that $\lim_{n\rightarrow\infty} \frac{\binom{n}{k}}{2^n} =0$ Show that the following limit holds
$$
\lim_{n\rightarrow\infty} \frac{\binom{n}{k}}{2^n} =0
$$
for a fixed value of $k$
I really am just stuck at the first step here. Normally I would consider tackling this using L'Hopitals rule, however $\binom{n}{k}$ is not differentiable. I was considering using the binomial theorem, but that is for sums, not just the single scenario. Any help appreciated!

If you accept as known $\binom{n}{k} \leqslant \frac{n^k}{k!}$, then you obtain it by estimation.
Addition: added second part, as it can be helpful for somebody.
$$\frac{n^k}{k^k} \leqslant \binom{n}{k} \leqslant \frac{n^k}{k!}$$

Set k = n since it is the greatest it will ever be. The numerator will never surpass 1. So you simply have the ratio of 1 to infinity.


Integral vanishes on all intervals implies the function is a.e. zero I am having trouble with the following problem:

  $f:\mathbb{R}\to \mathbb{R}$ is a measurable function such that for all $a$:
  $$\int_{[0,a]}f\,dm=0.$$
  Prove that $f=0$ for $m$ almost every $x$ (here $m$ is the Lebesgue measure).

I have no problem proving this for $f$ non-negative, or under the assumption that $f$ is integrable. But the question only assumes that $f$ is measurable and no more.

My idea was the usual thing; we look at the set of points where $f$ is positive and negative and assume one of these has measure greater than zero. Then I wanted to estimate one of these by an open set, look at the integral on the open set and show that it had to be greater than zero, a contradiction. But a key part of this attack is the assumption of the absolute continuity of the integral, which only holds in the case where $f$ is integrable.

Alternatively, if it were integrable one could simply estimate $f$ by a continuous function, where the result is quite obvious.

Ultimately we are going to show that $f$ is integrable, but it is not clear to me how to show this before showing it is zero a.e. So there must be a simpler way. Does anyone have suggestions?

The function $f$ must be integrable (one of $\int f^+$ or $\int f^-$ is finite) in order for the symbol $\int f$ to be defined. So, I'll assume this is the case. In fact, then, since $\int_0^a f$ exists and is finite for any $a$, it follows that $\int_c^d |f|<\infty$ for any numbers $c$, $d$.

We show $f$ is almost everywhere
 $0$ on any interval $[c,d]$; this will imply the desired result.

Suppose $f>0$ on the set of positive measure $E\subset[c,d]$. Choose a closed subset $F$ of $E$ with positive measure. We then have $\int_F f>0$. Now let $U=[c,d]\setminus F$. As $U$ is open, we may write $U$ as a disjoint union of open intervals: $U=\bigcup_{k=1}^\infty (a_k,b_k)$.

Now, since $\int_c^d |f|<\infty$
$$
0=\int_{[c,d]}f=\sum_{k=1}^\infty\int_{a_k}^{b_k}f+\int_F f.
$$
Since $\int_F f>0$, it follows that $\sum\limits_{k=1}^\infty\int_{a_k}^{b_k}f$ is negative. But then $\int_{a_n}^{b_n} f$ must be negative for some $n$.  However, this proves untenable upon observing that
$$
\int_{a_n}^{b_n} f =\int_0^{b_n} f - \int_0^{a_n} f =0.
$$

Similarly, one can show $f$ cannot be negative on a set of positive measure.

Any measurable function is uniformly approximated by continuous function on $(a, b)$. given statement is true for any continuous function and hence it is true for uniform limit of continuous functions, that is for measurable function


Closed form for sine graphic rotated by 45 degrees? Is there a non-parametric closed form for a function looking like a sine rotated 45 degrees?

I have encountered also a similar question but it asks for a function resembling the rotated sine, but not necessarily exact.

Let $(X_0,Y_0)$ be a point on the rotated graph. Then, converting to polar coordinates, the point is $(\sqrt{X_0^2+Y_0^2},arctan \frac{Y_0}{X_0})$. Rotating 45 deg. gives  ($\sqrt{X_0^2+Y_0^2},arctan \frac{Y_0}{X_0}$+45 deg.$)$. That is $(\sqrt{X_0^2+Y_0^2}cos(arctan \frac{Y_0}{X_0}+45$ deg.$),\sqrt{X_0^2+Y_0^2}sin(arctan \frac{Y_0}{X_0}+45$ deg.$))$ in Cartesian coordinates. If this point is $(x,y)$, then $y=sin$ $x$, so the closed form $\sqrt{x^2+y^2}sin(arctan \frac{y}{x}+45$ deg.$)=sin(\sqrt{x^2+y^2}cos(arctan \frac{y}{x}+45$ deg.$)$ works.
For $45$ deg. rotation clockwise instead, change $+45$ deg. to $+45$ deg.


Does an internal set with countably many infinitesimals contain hyperfinitely many infinitesimals? Let $\mathscr{A}$ be an internal subset of $^*\mathbb{R}$ that contains a countable family, say $\{a_n\}_{n\in\mathbb{N}}$, of infinitesimal members, that is $a_n\approx 0$ for all $n\in\mathbb{N}$.

We can extend the sequence $a\colon\mathbb{N}\to\mathscr{A}$ to an internal hypersequence with the same name $a\colon{^*\mathbb{N}}\to\mathscr{A}$. It follows that the set
$$ \{n\in{^*\mathbb{N}}\mid |a_n|<\frac{1}{n}\} $$
is internal by the Internal Definition Principle and includes $\mathbb{N}$ by our assumption. Hence, by the Overflow Principle, there exists $H\in{^*\mathbb{N}}\setminus\mathbb{N}$ such that $a_N\approx 0$ for all $N\in{^*\mathbb{N}}$ with $N\leq H$.

Since the extended hypersequence $a$ is internal, $\{a_0,\dots,a_H\}$ is an infinite hyperfinite set of infinitesimal elements of $\mathscr{A}$.

Is this procedure correct? Is there a less sophisticated way to prove the statement?

Perhaps a slighly shorter argument exploits saturation.  By countable saturation you can find an infinitesimal $\alpha>0$ such that each of the countably many infinitesimals is smaller than $\alpha$ in absolute value.  Then the subset $\{x\in\mathcal{A}\colon |x|<\alpha\}$ is internal, infinite, and therefore is infinite hyperfinite.

The existence of such an upper bound $\alpha$ can be established as follows.  Suppose not; then wlog we can assume that the sequence $\langle\epsilon_n\colon n\in \mathbb N\rangle$ of positive infinitesimals is nondecreasing. Consider the hyperreal interval $S_n=[\epsilon_n,\frac{1}{n}]$. Then $\langle S_n\colon n\in\mathbb N\rangle$ is a nested sequence and by saturation it must have a common element $c$. But $c$ can be neither infinitesimal nor appreciable. The contradiction proves the existence of $\alpha$.

"Hyperfinitely many" in Robinsonian NSA translates to "finitely many" in IST, so you are asking whether there is a finite subset of an infinite set.


Circular definition in proof I am confused by the following exercise in Velleman's How To Prove It:

'Suppose $m$ and $n$ are integers. If $mn$ is even, then either $m$ is even or $n$ is even.

Proof: Suppose $mn$ is even. Then we can choose an integer $k$ such that $mn=2k$. If $m$ is even then there is nothing more to prove, so suppose $m$ is odd. Then $m=2j+1$ for some integer $j$. Substituting this into the equation $mn=2k$, we get $(2j+1)n=2k$, so $2jn+n=2k$, and therefore $n=2k-2jn=2(k-jn)$. To prove $n$ is even it suffices to find an integer $c$ such that $n=2c$, from the above $c=k-jn$ works. Since $k-jn$ is an integer, it follows that $n$ is even.'

There are two things I don't understand:

1) How could $n=2(k-jn)$ be legitimate? Isn't that circular? Surely if he is to solve for n all n must be on one side of the equation?

2) I am not quite sure what he did. It seems that he split them into two cases to exhaust the possibilities of m. If one can prove exhaustively that the conclusion necessarily happens, the proof is correct. But in terms of the entire equation that doesn't seem exhaustive to me, i.e. what about $n$?

Also, I understand that by existential instantiation the definition of an even number $$\exists x(m=2x) $$ leads to m=2k, and the same goes with m=2j+1.But why is it suffice to find an integer c such that n=2c? Is he going to use existential generalization to make n=2c into $$ \exists x(n=2x) $$ i.e. the definition of even number, to complete a direct proof? But this is not explicit in the proof.

I am trying to self-study this and my maths is, needless to say, dire. So I really appreciate any help on this, thank you so much!

The assumption of the exercise is that $m,n$ are given to us as integers, and that $mn$ is an even integer and therefore $mn=2k$ where $k$ is an integer.

The proof goes by cases. The case that $m$ is even has been handled, and now we are in the opposite case where $m$ is odd, so $m=2j+1$ where $j$ is given to us as an integer. (Note: proofs in different cases are independent of each other; the proof in one of the cases does not have to be "exhaustive", i.e. it does not have to apply to the other case).

The equation $n=2(k-jn)$ has been derived from those assumptions using the laws of algebra, it is perfectly valid, and $j,k,n$ are all integers, by assumption.

Perhaps you are confused because it looks like this equation is being used to "solve for $n$", but that's not what is happening in this proof. The quantity $n$ is not being defined by this equation $n=2(k-jn)$. Instead, $n$ has already been given to us as an integer, at the very beginning of the proof.

All that's happening with the equation $n=2(k-jn)$ is that $n$ is being expressed as a product of $2$ times another integer, that integer being $k-jn$. Since $2$ times any integer is an even integer, we may conclude (in the case that $m$ is odd) that $n$ is an even integer.

Let me attempt to emphasize the point by an extreme example. If you did not already believe that $4$ is an even integer, I could perhaps convince you by writing the equation
$$4 = 2(34 - 8 \cdot 4)
$$
which expresses $4$ as a product of $2$ times another integer, that integer being $34 - 8 \cdot 4$. I do not ask that you do the arithmetic to discover exactly which integer $34 - 8 \cdot 4$ happens to equal, instead I ask you to understand that the result of multiplying two integers and subtracting them from another integer is always an integer.

Part of your confusion is due to the fact that Velleman is glossing over the '$m$ is even' case.  The explicit proof for even '$m$' - which he is expecting you to see for yourself directly - is this:

If $m$ is even, then $m$=2$j$ for some $j$.  Then $mn$ is (2$j$)$n$.  By associativity, this is equal to 2($jn$).  So to show $mn$ = 2$k$ for some $k$, take $k$=$jn$.


Your last inequality is wrong.

Try $u=v=w=1$.

Thus, it's not correct.