absolute value of 2 complex exponentials Sorry if this is a dumb question.  Trying to understand what this simplifies to:

$$f(t) = |e^{iwt} - e^{iwv}|$$

I don't understand how to get a nice numerical answer to this as the exponentials use different variables to one another...

Its related to the proof of the Fourier Condition of Lipschitz regularity in Stephane Mallat's "Wavelet tour of signal processing book".   (Theorem 6.1, page 165)

Many thanks for any answers!

EDIT: Looking at folks answers it seems pretty clear to me you can get a few different solutions to this...for extra clarification, the "solution" in Mallat's book is:

For (condition 1) $$\frac{1}{|t - v|}<= |w|$$
$$\frac{|exp(iwt) - exp(iwv)|}{|t - v|^\alpha} <= \frac{2}{|t - v|^\alpha}  <=  2|w|^\alpha$$

For (condition 2)  $$\frac{1}{|t - v|}>= |w|$$

$$\frac{|exp(iwt) - exp(iwv)|}{|t - v|^\alpha} <= \frac{|w||t - v|}{|t - v|^\alpha}  <=  |w|^\alpha$$

So far only one person has given a solution to the problem at the top which could give the solution for condition 1...I still have no idea how Mallat's achieved it for condition 2!

We have that

$$|e^{i\omega t} - e^{i\omega v}|\le |e^{i\omega t}|+| e^{i\omega v}|\le 1+1=2.$$

To get a better bound we note that (assume $\omega>0,$ $t<v$, without lost of generality)

$$|e^{i\omega t} - e^{i\omega v}| = \left|\int_{\omega t}^{\omega v} ie^{it}\,dt\right|\le \int_{\omega t}^{\omega v} |ie^{it}|\,dt  = \omega(v-t).$$

(see figure below)

  $$\tag{1} |exp(iwt) - exp(iwv)|=2 \left|\sin \left( \dfrac{w(v-t)}{2} \right)\right|$$

on a geometrical basis that can be explained in this way :

Let $A=(cos(wt),sin(wt))$ and  $B=(cos(wv),sin(wv))$.

Thus angle $AOB=|wv-wt|=|w(v-t)|$. Let us denote this angle by $2\alpha$.

Thus, we have $\alpha = \dfrac{|w(v-t)|}{2}.$

We have to express $AB=2OH$ where $H$ is the midpoint of $AB$.

The essential point is that $AOB$ is an isosceles triangle ; consequences:

$AH$ is orthogonal to $AB$ and
angle $\alpha$ = angle $HOA$ = angle $HOB = w|t-v|/2. $

Thus  $AH = \sin(\alpha)$ ; then $AB=2AH=2\sin(\alpha)$ yielding relationship (1).

Remark: As a consequence of (1), this expression is bounded by

$$ 2 \left| \dfrac{w(v-t)}{2} \right| =\left| w(v-t) \right|$$


Why is the tangent of 22.5 degrees not 1/2? Sorry for the stupid question, but why is the tangent of 22.5 degrees not 1/2?
(Okay... I get
 that that the tangent of 45 degrees is 1 ("opposite" =1, "adjacent" =1, 1/1 = 1.  Cool.  I am good with that.) Along those same lines, if the "opposite" drops to 1/2 relative to the "adjacent" i.e., "opposite" = 1, "adjacent" = 2 therefore 1/2. What am I missing?  Thanks in advance for your help.

It helps to realize that the tangent of an angle can be viewed as the length of a segment tangent to the unit circle (hence the name):

As the above helps illustrate, as an angle sweeps from 0 to 90 degrees, the corresponding value of tangent ranges from 0 to infinity. Behavior that exotic cannot reasonably be expected to follow a simple rule like "halving the angle halves the tangent"; just look at the reverse: Doubling a $45^\circ$ angle does a lot more than double its tangent value!

I'll leave you to ponder the deeper implications.

The basic error being made is the assumption that you are converting one unit into another unit such as when you convert meters into yards.  Degrees are units but tangent represents a ratio.


Show that $2003$ divides the numerator of $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{1335}$. I came up with this problem on a book in number theory :

  Let $p$ and $q$ be natural numbers such that
  $$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{1335} = \frac{p}{q} \,.$$
  Show that $2003 \mid p$.

Any ideas of how to start solving it?
Suggestions and helps would be greatly appreciated!

Suppose that $k>1$ is a positive integer such that $3k-1$ is a prime number.  Then the numerator of the alternating series
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{2k-1}$$
is divisible by $3k-1$.  For example when $k=2$, we have $2k-1=3$ and $$1-\frac12+\frac13=\frac{5}{6}$$ with numerator divisible by $5=3k-1$.  For $k=4$, we have $2k-1=7$ and $3k-1=11$, and $11$ divides the numerator of
$$1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17=\frac{319}{420}.$$  The claim may not hold if $3k-1$ is not prime.  For example if $k=3$, we have $3k-1=8$, and $8$ doesn't divide the numerator of
$$1-\frac12+\frac13-\frac14+\frac15=\frac{47}{60}.$$

The OP's problem comes from $k=668$, where $2k-1=1335$ and $3k-1=2003$.  Here is a proof of this claim.

Let $S$ denote the series to be evaluated.  Then,
$$S=\left(1+\frac12+\frac13+\frac14+\ldots+\frac{1}{2k-1}\right)-2\left(\frac12+\frac14+\ldots+\frac{1}{2k-2}\right).$$
Hence
$$S=\left(1+\frac12+\frac13+\frac14+\ldots+\frac{1}{2k-1}\right)-\left(1+\frac12+\ldots+\frac{1}{k-1}\right).$$
This makes
$$S=\frac1k+\frac1{k+1}+\ldots+\frac1{2k-2}+\frac{1}{2k-1}.$$
Note that $3k-1$ must be odd, so $k$ is even.  That is,
$$S=\left(\frac1k+\frac1{2k-1}\right)+\left(\frac{1}{k+1}+\frac{1}{2k-2}\right)+\ldots+\left(\frac{1}{\frac{3}{2}k-1}+\frac{1}{\frac{3}{2}k}\right).$$
That is
$$S=\frac{3k-1}{k(2k-1)}+\frac{3k-1}{(k+1)(2k-2)}+\ldots+\frac{3k-1}{\left(\frac{3}{2}k-1\right)\frac32k}.$$
Thus
$$S=(3k-1)\left(\frac{1}{k(2k-1)}+\frac{1}{(k+1)(2k-2)}+\ldots+\frac{1}{\left(\frac{3}{2}k-1\right)\frac32k}\right).$$
Because $3k-1$ is a prime greater than $k,k+1,\ldots,2k-2,2k-1$, it is not cancelled out from the numerator of $S$.

I think it is an interesting problem to see whether there exists an integer $k>1$ such that $3k-1$ is not prime but $3k-1$ divides the numerator of $S$.  I conjecture that such $k$ don't exist.


Bus arrival poisson paradox I have a question about the waiting time paradox for poisson processes(in this case in terms of bus arrivals).

Suppose I know that buses arrive with poisson distribution(lambda). I arrive at fixed time t.

I want to figure out three things -

The mean length of time I will wait until the next bus.
The mean length of time since the last bus.
Mean length of time between bus arrivals.

I think 1 should be 1/lambda - the poisson process is memoryless, so it shouldn't matter when I arrrive.

I think 3 should also be 1/lambda, because that is the mean of the exponential distribution, which determines the distribution of times between buses.

I'm not sure about 2.

Can anyone help/tell me if my reasoning is wrong/right?

This type of problem is always tricky, as the discussion here has already shown.

As an alternative to clever thinking we can solve the problem of items 1. and 2. by just observing what happens to the passenger and the busses. Instead of doing this in the real world (thus possibly catching a cold or standing in the pouring rain) we make a simulation. The result is, of course, $1/\lambda$ for items 1. and 2. For item 3. the same result follows simply from the definition of the Possion process.

We assume the passenger to arrive at some fixes instant of time $t_p = 20$. For each trial we generate an array of a sufficiently large number of bus arrival times ($n = 50$) based on a Poisson process with $\lambda = 1$. Then we pick the bus arrival times $t_1$ and $t_2$ just before and just after $t_p$ respectively, calculate the time differences $dt_1 = t_p - t_1$ and $dt_2 = t_2 - t_p$ and study their statistics.

The time difference $\tau$ of two neighbouring events of a Poisson process is distributed according to an exponential law

$$f(\tau,\lambda) = \lambda \;exp (-\lambda \tau)$$

A random variable $r$ following this distribution is generated from a basic random number $R$ equally distributed between $0$ and $1$ is defined by

$$r=\log \left(\frac{1}{R}\right)$$

The array of bus arrival times $t_k$ is then generated by cumulating the differences $r_i$, i.e.

$$t_k=\sum _{i=1}^k r_i$$

We have done 1000 trials. The results are presented here as histograms of the $dt_1$ and $dt_2$ compared to the exponential law.

The agreement with the exponential law is reasonable.
The central moments are for $dt_1$ and $dt_2$ resp.

means = {1.04494,0.966218}
standard deviations = {1.06179,0.949569}

We can also have a look on the results of all 1000 trials

Appendix: the Mathematica code

Scenario $t_1$ < $t_p$ < $t_2$

$t_1$      arrival time of bus just missed
$t_p$  arrival time of passenger
$t_2$  arrival time of bus to be taken by passenger

$dt_1 = t_p - t_1$  length of time by which the previous bus was missed
$dt_2 = t_2 - t_p$  length of time the passenger has to wait for the bus to be taken

The code

r := Log[1/RandomReal[]];
(* random time difference between two busses \
according to Poisson process exponentially distributed with lambda = 1 *)

tp = 20; (* time of arrival of the passenger *)
nn = 10^3; (* number of trials *)

m1 = {};(* store times to bus just missed *)
m2 = {};(* store times to wait for bus  *)

Do[
 tdiff = Array[r &, 50]; (* array of time differences between consecutive busses *)
 tsbus = FoldList[Plus, 0, tdiff]; (* array of arrival times of busses *)
 p = Position[tsbus, Select[tsbus, # < tp &][[-1]]][[1,1]]; (* position of arrival of passenger between the busses *)
 t1 = tsbus[[p]]; (* arrival time of bus just missed *)
 t2 = tsbus[[p + 1]]; (* arrival time of bus to be taken *)
 dt1 = tp - t1;(* length of time by which the passenger missed the previous bus *)
 dt2 = t2 - tp;  (* length of time the passenger has to wait for the bus *)
 AppendTo[m1, dt1]; (* collect times differences *)
 AppendTo[m2, dt2],(* collect times differences *)
 {nn}] (* number of trials *)
Print["means = ", Mean /@ {m1, m2}];
Print["standard deviations = ", StandardDeviation /@ {m1, m2}];

means = {1.04494,0.966218}

standard deviations = {1.06179,0.949569}

The values are close to unity as it should be for $\lambda = 1$

First two are 1/$\lambda$ for sure. We just discussed the question today with Prof.

Third one is tricky. From your point of view, you reach to the bus stop during interval with length 2/$\lambda$ . Because expected values of the time since the last bus passed and next bus will come are same; which is 1/$\lambda$. This may look counter-intuitive but actually not. In exponential distribution next bus may come very very late (actually there is no probability), so you have an higher probability to go to bus stop during these long intervals.

But for a somebody already sitting in the bus, if he takes off the bus and waits for the next one at the stop, his initial expected time is 1/$\lambda$. Now is the tricky part. Let's say you go to bus stop and ask this guy how long he has been waiting. The answer does not change anything because you are expected to wait for 1/$\lambda$ time again.

I hope it helped, did not confuse you further.


$L_p$ space and convergence Let $f_i\rightarrow f$ $m$-a.e on $[0,1]$, $m$ is a measure and $\int|f_i(x)|^4dm$$\le1$ for all $i$.Then $\int|f_i(x)|^2dm\rightarrow \int|f(x)|^2dm$. how to prove it?

in my solution i prove that $|f_i(x)|^2\in L_2$ for all $i$ and $|f(x)|^2\in L_2$.
then use some statement (attached picture) where $r = 1,p = 2,F_i(x) =|f_i(x)|^2,F(x) =|f(x)|^2,C = 1$ and $f_n(x)\Rightarrow f(x)$ mean convergence in measure, but when measure is finite from a.e.convergence $\Rightarrow$ convergence in measure. It's right solution?

Yes, you can use the quoted theorem.

We can also argue only with the dominated convergence theorem. Indeed, for a fixed $R$, we have
$$\left|\int f_i^2-\int f^2\right|\leqslant \left|\int f_i^2\mathbf 1\{|f_i|\leqslant R\}-\int f^2\mathbf 1\{|f|\leqslant R\}\right|\\
+\left|\int f_i^2\mathbf 1\{|f_i|\gt R\}-\int f^2\mathbf 1\{|f|\gt R\}\right|\\
\leqslant \left|\int f_i^2\mathbf 1\{|f_i|\leqslant R\}-\int f^2\mathbf 1\{|f|\leqslant R\}\right|+\frac 2{R^2},$$
where we used $$\int f_i^2\mathbf 1\{|f_i|\gt R\}\leqslant \frac 1{R^2}\int f_i^4\leqslant \frac 1{R^2}$$
and the same inequality holds for $f$ by Fatou's lemma.
By dominated  convergence, we have for each $R$ such that
$m\{x\mid \left|f(x)\right|=R\}=0$,
$$\lim_{n\to +\infty}\int f_i^2\mathbf 1\{|f_i|\leqslant R\}-\int f^2\mathbf 1\{|f|\leqslant R\}=0.$$
This is due to the fact that $\mathbf 1\{|f_i|\leqslant R\}\to\mathbf 1\{|f|\leqslant R\}$ almost everywhere.

It seems like a simple application of Dominated Convergence Theorem. From assumption, we have that $|f_i|^2\longrightarrow |f|^2$ a.e on $[0,1]$. But
\begin{equation*}
\int|f_i|^2\, dm\le \left(\int|f_i|^4\, dm\right)^{\frac{1}{2}}\le 1
\end{equation*}
where the first inequality follows from Holder inequality. Thus, DCT implies that
\begin{equation*}
\lim_{i\to\infty} \int |f_i|^2\, dm = \int \lim_{i\to\infty} |f_i|^2\, dm = \int |f|^2\, dm
\end{equation*}


Change of variables in triple integral Let $D$ be the region in $xyz-$space defined by inequalities $1 \le x \le 2, 0 \le xy \le 2 $ and $0\le z \le 1$. I want to evaluate $\displaystyle \int\int\int_D (x^2y + 3xyz) \text{dxdydz}$ by applying the following transformation in $uvw-$space:

$$u = x, v = xy, w =3z$$
over an appropriate region $G$.

So,  I sketch the region $D$ and write $x,y,z$ as
$$x=u, \enspace y = \frac{v}{x} = \frac{v}{u} \enspace z = \frac{w}{3}$$ and found the Jacobian $J(u,v,w)$ as $\frac{1}{3u}$.

Now, I have two questions:

1.
 I know that I have to write the boundaries of $D$ in terms of $x,y,z$ which comes from the $6$-sides of it. However, since we have $0\le xy \le 2$ the bottom of $D$ is some kind of right triangle with $y=1$ line at the bottom, $x = 1$ line on the left side and some part of $y = \frac{1}{x}$ curve. So I have some kind of prism which has $5$-sides. In that case, how do we write $6$-conditions for the boundaries of $u,v$ and $w$?

2.For boundaries of $G$, I found $1\le u \le 2$ and $0 \le w \le 3$.  I found $u \le v \le u^2$ by using $0 \le y \le \frac{2}{x} $ and considering $1 \le u \le 2$. However, it is clear that these cannot be the boundaries.

I think the first question solves the second one. Thank you.

This is not a good example to apply change of variables.  As the limits of z do not depend on x or y and is constant between 0 and 1.  Only y depends on x.  Hence the integral can be written as

$$\int_{z=0}^{1}\int_{x=1}^{2}\int_{y=0}^{\frac{2}{x}}(x^2y + 3xyz)dydxdz = 2 + 3ln 2$$


Finding an expression for the diameter of a circle. The figure shows two circles of radius $1$ that touch at $P$. 1]1 $T$ is a common tangent line; $C_1$ is the circle that touches $C$, $D$, and $T$; $C_2$ is the circle that touches $C$, $D$, and $C_1$; $C_3$ is the circle that touches $C$, $D$, and $C_2$. This procedure can be continued indefinitely and produces an infinite sequence of circles $\{C_n\}$. Find an expression for the diameter of $C_n$.

I'll work with the radii instead of the diameters. Obviously, any diameter is twice the corresponding radius. Now, note from the figure below that the triangle $OAB$ is rectangular at $A$.

Thus,
$$
\overline{OA}^{\,2} + \overline{AB}^{\,2} = \overline{OB}^{\,2}
$$

So,
$$
(\underbrace{1}_{\overline{OA}})^2 + (\underbrace{1 - R_1}_{\overline{AB}})^2 = (\underbrace{1 + R_1}_{\overline{OB}})^2
$$
which gives you

  $$
R_1 = \frac{1}{4}
$$

Similarly, by making $B$ be the centre of the ever-smaller circles, you should be able to see that
$$
(\underbrace{1}_{\overline{OA}})^2 + (\underbrace{1 - 2R_1 - 2R_2 - \cdots - 2R_{n-1} - R_n}_{\overline{AB}})^2 = (\underbrace{1 + R_n}_{\overline{OB}})^2
\qquad(1)
$$
for $n > 1$. Now define

  $$
S_{n} \equiv \sum_{k\,=\,1}^{n}R_k
\qquad (n \ge 1)
\qquad(2)
$$

so that $(1)$ can be written as
$$
1 + (1 - 2S_{n-1} - R_n)^2 = (1 + R_n)^2
\qquad(3)
$$
for $n > 1$. However,
$$
(1 - 2S_{n-1} - R_n)^2 =
1 + 4S_{n-1}^2 + R_n^2 - 4S_{n-1} - 2R_n + 4S_{n-1}\,R_n
$$
and $(3)$ gives
$$
1 + 1 + 4S_{n-1}^2 + R_n^2 - 4S_{n-1} - 2R_n + 4S_{n-1}\,R_n =
1 + 2R_n + R_n^2
$$
which can be simplified to
$$
(1 - 2S_{n-1})^2 = 4(1 - S_{n-1})R_n
$$
so

  $$
R_n = \frac{1}{4}\frac{(1 - 2S_{n-1})^2}{(1 - S_{n-1})}
\qquad (n > 1)
\qquad (4)
$$

This gives $R_n$ in terms of the radii of only the circles with indices smaller than $n$ so it's very amenable to a recursive calculation. For example, we already have $R_1 = 1/4$. Then $S_1 = 1/4$ and $R_2 = 1/12$. Then $S_2 = 1/3$ and $R_3 = 1/24$. Then $S_3 = 3/8$ and $R_4 = 1/40$, and so on.

Computing a few more values in succession, it's easy to discern the pattern:

  $$
S_n = \frac{n}{2(n+1)}
\qquad\mbox{and}\qquad
R_n = \frac{1}{2n(n+1)}
\qquad (n \ge 1)
$$

That these satisfy $(2)$ and $(4)$ can be easily proven by induction on $n$.

Notice, the radius say $r_1$of circle $C_1$ is given by the formula (see derivation) $$\color{red}{r_1=\frac{ab}{(\sqrt a+\sqrt b)^2}}$$ setting radii $a=b=1$ of larger identical circles, we get  $$r_1=\frac{1\cdot1}{(1+1)^2}=\frac{1}{4}$$

Now, the radius $r_2$ of circle $C_2$ externally touching the circles $C$, $D$ & $C_1$ is given by the standard formula (see derivation)
$$\color{red}{r_2=\frac{abc}{2\sqrt{abc(a+b+c)}+(ab+bc+ca)}}$$ Setting the values of $a=1, b=1, c=r_1=\frac{1}{4}$, we get radius of circle $C_2$ as follows $$r_2=\frac{1\cdot1\cdot\frac{1}{4}}{2\sqrt{1\cdot1\cdot\frac{1}{4}\left(1+1+\frac{1}{4}\right)}+1\cdot1+1\cdot \frac{1}{4}+\frac{1}{4}\cdot 1}=\frac{1}{12}$$

Similarly,
the radius $r_3$ of circle $C_3$ externally touching the circles $C$, $D$ & $C_2$ is calculated by setting the values of $a=1, b=1, c=r_2=\frac{1}{12}$ in the above standard formula as follows $$r_3=\frac{1\cdot1\cdot\frac{1}{12}}{2\sqrt{1\cdot1\cdot\frac{1}{12}\left(1+1+\frac{1}{12}\right)}+1\cdot1+1\cdot \frac{1}{12}+\frac{1}{12}\cdot 1}=\frac{1}{24}$$ Thus, continuing the same procedure, we can calculate radius of any circle $C_n$ using above standard formula.


How to calculate $p_i$ in Blau's Index of Heterogeneity I came across this:

"Therefore, board gender diversity and board racial diversity are calculated using Blau's index of heterogeneity $(1 - \sum p_{i}^2)$, where $p_i$ is the proportion of group members in each of the $i$ categories."

I'd like to know, if the category just includes $2$ (male and female) then how can I calculate Blau's Index? How can I calculate $p_i$?

For example, if ten men and ten women in total, what is the index?

Sorry, English is not my first language.

Blau's Index calls for adding the squared proportion of individuals in each category, summing them up, then subtracting from 1.  In the example of two individuals, half (.5) are male and half (.5) are female.  .52 is .25.  Add .25 and .25 to obtain .5.  Subtract this from 1 and you have a diversity index of .5.  The proportions are the same when there are ten females and ten males so the answer would be the same.  However, different proportions such as 3 females and 7 males do result in a different answer.  .32 is .09 and .72 is .49.  Added together we have .58 to subtract from 1 and a diversity index of .42. Pi represents each category's proportion with each proportion (.5 and .5 or .3 and .7 in the respective examples).


prove $\lim\limits_{x \to \infty}(1+ 1/n)^n = e$ without using l'Hospitals rule I did $y=(1+1/n)^n)$ so $ln(y)=n*ln(1+1/n)$ so then $ln(y)=Ln(1+1/n)/(1/n)$ which sith l'hospitals rule becomes $ln(y)=(-1/(n^2 + n))/(-1/n^2)$ which simplifies to $n/(n+1)$ which using l'hospitals rule is $1$ so the answer is e. how do you solve this without l'hospitals rule

Let $f(x) = \ln(x)$ and then go through the process of evaluating $f'(1)$ using the formal definition of the derivative, noting ahead of time that the actual value of $f'(1)$ is $1$:

$$ 1 = f'(1) = \lim_{h \to 0}\frac{f(1+h)-f(1)}{h} = \lim_{x \to 0}\frac{f(1+x)-f(1)}{x} = \lim_{x \to 0}\ln(1 + x)^{1/x}; $$

and so,

$$ e^{1} = e^{\lim_{x \to 0}\ln(1+x)^{1/x}} = \lim_{x \to 0}(1+x)^{1/x}. \tag{$*$}$$

Now, let $n = \frac{1}{x}$.

Then substituting $n$ into $(*)$ and noting that as $x \to 0^{+}$ we have $n \to \infty$, it follows that

$$\lim_{n \to \infty}\left( 1 + \frac{1}{n} \right)^{n} = e.$$

This can be done straight from the definition of $\ln$ as $\ln(x) \equiv \int_{1}^x dt/t$ and $\exp$ as its inverse function. Then we have
\begin{multline}
\ln((1+1/n)^n) = \int_1^{(1+1/n)^n} \frac{dt}{t} = \int_1^{(1+1/n)^{n-1}} \frac{dt}{t}  + \int_{(1+1/n)^{n-1}}^{(1+1/n)^n} \frac{dt}{t} \\ = \int_1^{(1+1/n)^{n-1}} \frac{dt}{t} + \int_1^{1+1/n} \frac{du}{u} = \ln((1+1/n)^{n-1}) +  \int_1^{1+1/n} \frac{du}{u}.
\end{multline}
Repeating this $n$ times gives
$$
\ln((1+1/n)^n) = n\int_1^{1+1/n} \frac{du}{u}.
$$
It's not hard to show from the properties of the integral that
$$
\lim_{n\rightarrow\infty} n\int_1^{1+1/n} \frac{du}{u} = 1,
$$
and that $\ln$ is continuous, so it is valid to say
$$
\lim_{n\rightarrow\infty}\ln((1+1/n)^n) = \ln\left(\lim_{n\rightarrow\infty}(1+1/n)^n.\right)
$$
Thus,
$$
\ln\left(\lim_{n\rightarrow\infty}(1+1/n)^n\right) = 1
$$
and so we have
$$
\lim_{n\rightarrow\infty}(1+1/n)^n = \exp(1) \equiv e
$$


Showing two matrices are similar I have to show that each of the following matrices

$$ \frac{1}{\sqrt{2}}
\begin{pmatrix}
0&1&0\\
1&0&1\\
0&1&0
\end{pmatrix}\quad ,
\frac{1}{\sqrt{2}}
\begin{pmatrix}
0&-i&0\\
i&0&-i\\
0&i&0
\end{pmatrix}
 ,
\begin{pmatrix}
1&0&0\\
0&0&0\\
0&0&-1
\end{pmatrix}$$

are equivalent to one of the following

$$\begin{pmatrix}0&0&0\\0&0&1\\0&-1&0\end{pmatrix},\begin{pmatrix}0&0&-1\\0&0&0\\1&0&0\end{pmatrix},\begin{pmatrix}0&1&0\\-1&0&\\0&0&0\end{pmatrix}$$

It is a relatively simple but time consuming problem to construct a similarity transformation. I have nine real independent variables and 9 equations to check for each matrix. I want to know if there is any process or software that could help me do this calculation?

Since you mentioned a "similarity transform", I presume your "equivalent" means "similar"; also I presume this is over $\mathbb C$.  Well, the first thing I would do is find the eigenvalues of each matrix; the second (if necessary) is the Jordan canonical form.  As for software, Maple will handle this quite easily; I imagine
most other CAS's will also.

EDIT: Oops, all of your first three matrices are similar to each other, and none is similar to any of the last three.  Are you sure you quoted the question right?

Two matrices are similar if their traces are equal.


Uniform convergence of $\sin(x^n)/x^n$ Let $f_n(x)=\sin(x^n)/x^n$.
Show that $f_n$ converges uniformly to $1$ in the interval $(-\delta,\delta)- \{0\}$, with $0<\delta<1$.

So far what I've tried is to take the infinite norm of the function in the interval minus $1$ and see if its limit when $n$ tends to $\inf$ is equal to 0.

It looks like the function is strictly positive and less than 1 in the interval, plus it looks crecient in (-1,0) and decrecient in (0,1), though I haven't managed yet to formalize this.

This is false. Let $f_n(x)=\frac{\sin x^n}{x^n}$ and  take $x_n=1-\frac{1}{n}$. Then  $|f_n(x_n)-1|\rightarrow 1-\frac{\sin (e^{-1})}{e^{-1}}\neq 0$ Therefore  $\lim\limits_{n\rightarrow \infty} \sup\limits_{x\in (-1,1)\setminus \{0\}} |f_n(x)-1|>0$

Since $$\left|\frac{\sin u}{u} -1\right|=\left|\frac{\sin u -u}{u}\right|\leqslant 1- \cos u$$ we have $$\left|\frac{\sin (x^n )}{x^n} -1\right|=\left|\frac{\sin (x^n ) -x^n}{x^n}\right|\leqslant 1- \cos (x^n )$$

and since the sequence $(x^n ) $ tends to $0$ almost uniformly on $(-1,1 )$ thus the sequence $\left( \frac{\sin (x^n )}{x^n} \right)$ tends to $0$ almost uniformly on $(-1,1 ).$


Why does the derivative of sine only work for radians? I'm still struggling to understand why the derivative of sine only works for radians. I had always thought that radians and degrees were both arbitrary units of measurement, and just now I'm discovering that I've been wrong all along! I'm guessing that when you differentiate sine, the step that only works for radians is when you replace $\sin(dx)$ with just $dx$, because as $dx$ approaches $0$ then $\sin(dx)$ equals $dx$ because $\sin(\theta)$ equals $\theta$. But isn't the same true for degrees? As $dx$ approaches $\theta$ degrees then $\sin(dx \,\text{degrees})$ still approaches $0$. But I've come to the understanding that $\sin(dx \,\text{degrees})$ approaches $0$ almost $60$ times slower, so if $\sin(dx \,\text{radians})$ can be replaced with $dx$ then $\sin(dx \,\text{degrees})$ would have to be replaced with $(\pi/180)$ times $dx$ degrees.

But the question remains of why it works perfectly for radians. How do we know that we can replace $\sin(dx)$ with just $dx$ without any kind of conversion applied like we need for degrees? It's not good enough to just say that we can see that $\sin(dx)$ approaches $dx$ as $dx$ gets very small.  Mathematically we can see that $\sin(.00001)$ is pretty darn close to $0.00001$ when we're using radians. But let's say we had a unit of measurement "sixths" where there are $6$ of them in a full circle, pretty close to radians. It would also look like $\sin(dx \,\text{sixths})$ approaches $dx$ when it gets very small, but we know we'd have to replace $\sin(dx \,\text{sixths})$  with $(\pi/3) \,dx$ sixths when differentiating. So how do we know that radians work out so magically, and why do they?

I've read the answers to this question and followed the links, and no, they don't answer my question.

Radians, unlike degrees, are not arbitrary in an important sense.

The circumference of a unit circle is $2\pi$; an arc of the unit circle subtended by an angle of $\theta$ radians has arc length of $\theta$.

With these 'natural' units, the trigonometric functions behave in a certain way. Particularly important is
$$\lim_{x\to 0} \frac{\sin x}{x} = 1 \quad\quad - (*)$$

Now study the derivative of $\sin$ at $x = a$:

$$\lim_{x \to a} \frac{\sin x - \sin a}{x-a} = \lim_{x \to a}\left( \frac{\sin\left(\frac{x-a}{2}\right)}{(x-a)/2}\cdot \cos\left(\frac{x+a}{2}\right)\right)$$

This limit is equal to $$\cos a$$ precisely because of the limit $(*)$. And $(*)$ is quite different in degrees.

Simply put: Because a radian is defined as the unit of measurement that makes $\sin(dx)\approx dx$.

As you have realized, for any unit of measurement you define as the basis of $\sin$, you'll have $\sin(dx)\approx \alpha\ dx$ for some $\alpha$. There is a specific unit of measurement for which $\alpha=1$. Call this unit a radian, and you're done.

Well, not really done because there are a lot of connections you still need to prove. But that's a basic idea that underlies much of math: You can define things however you like, but you can only define each once. If you want several different things to be true about a concept, you have to take one as the definition and prove the rest.


Finding the minimum of the maximum value of $|P(x)|$ In the book "Problem Solving Strategies", I found the following problem

  Find the polynomial $p(x) = x^2 + px + q$ for which $\max\limits_{x\in[-1,1]}|P(x)|$ is minimal.

I have no idea how to deal with this problem. Any help will be highly appreciated.

$max |P(x)| =\max \{1+|p|+q,|q-\frac {p^{2}} 4|\}$ since the maximum value is attaiined at one of the points $-1,1,-\frac p 2$. Claim: either $1+|p|+q \geq \frac 1 2$ or $|q-\frac {p^{2}} 4| \geq \frac 1 2$. I leave it to you to prove this by contradiction. (It is quite easy: if both inequalities fail you will get the contradiction $\frac {p^{2}} 4 <-|p|$). ). It follows that $max |P(x)| \geq \frac 1 2$ always. This value is attained when $p=0$ and $q =-\frac 1 2$ so the required polynomial is $x^{2} -\frac 1 2$.

I presume that you mean "$P(x)= x^2+ px+ q$", not "p(x)".  That is a parabola opening upward.  It takes on its maximum value on any interval at one of the endpoints of the interval.  So the maximum of P(x) on [-1, 1] is either 1-p+ q or 1+ p+ q.
The smallest value of either of those occurs when p= q= 0 or $P(x)= x^2$.


Regular/Right hyperbolas through three points Find all equations of regular hyperbolas passing through the points $A(\alpha,0),B(\beta,0),$ and $C(0,\gamma)$.

Attempt: I am assuming here that regular means right. We can then write the equation of such as $(x-x_0)^2-(y-y_0)^2 =a^2$. Thus by the given we have $(\alpha-x_0)^2 -y_0^2 = a^2, (\beta-x_0)^2 -y_0^2 = a^2,$ and $x_0^2 - (\gamma-y_0)^2 = a^2.$ Therefore $\alpha-x_0 = \pm(\beta-x_0) \implies \alpha = \beta$ or $\alpha - 2x_0 = -\beta$.

I am not sure what to do next because we have a lot of variables and should I break this up into cases then with $x = \beta$ or $\alpha -2x_0 = -\beta$?

A conic section is characterized by five real degrees of freedom. Knowing three distinct points on it will account for three degrees of freedom (not six since each point might still move along the conic without changing it). Assuming that it's a right hyperbola should account for another degree of freedom, so we should end up with a one-parameter family of conics. But how to find them?

One useful tool is projective geometry, where you can use an homogeneous coordinate vector like $[x:y:0]$ to describe a point at infinity. This is also the direction of the asymptotes, since the asymptotes are just tangents at these points at infinity. If $[x_1:y_1:0]$ describes one tangent direction, $[y_1:-x_1:0]$ is the perpendicular one. So together with the given three points, that's five points defining a conic section. You just have to compute its equation.

This $x_1$ and $y_1$ may look like two parameters, but since these are homogeneous coordinates, the only thing that matters is their ratio. You could for example write $x_1=\cos(t),y_1=\sin(t)$ and characterize things in terms of an angle $t$. Adding a multiple of $90°$ to that angle would result in the same conic, since it would be defined by the same set of points. So you could also w.l.o.g. take $y_1=1$ and $x_1\in[-1,1)$ to obtain a unique description of each hyperbola.

I'm using Sage to work out the formula since I'm lazy:

PR.<x1,y1,alpha,beta,gamma> = QQ[]
A = vector((alpha, 0, 1))
B = vector((beta, 0, 1))
C = vector((0, gamma, 1))
D = vector((x1, y1, 0))
E = vector((y1, -x1, 0))
M1 = A.cross_product(B).column()*C.cross_product(D).row()
M2 = A.cross_product(C).column()*B.cross_product(D).row()
M = (E*M1*E)*M2 - (E*M2*E)*M1
M += M.transpose()
M = M.parent()(M/gcd(M.list()))
M

$$\begin{pmatrix}
2 x_{1} y_{1} \gamma &
- x_{1}^{2} \gamma + y_{1}^{2} \gamma &
- x_{1} y_{1} \alpha \gamma -  x_{1} y_{1} \beta \gamma \\
- x_{1}^{2} \gamma + y_{1}^{2} \gamma &
-2 x_{1} y_{1} \gamma &
- x_{1} y_{1} \alpha \beta + x_{1} y_{1} \gamma^{2} \\
- x_{1} y_{1} \alpha \gamma -  x_{1} y_{1} \beta \gamma &
- x_{1} y_{1} \alpha \beta + x_{1} y_{1} \gamma^{2} &
2 x_{1} y_{1} \alpha \beta \gamma
\end{pmatrix}$$

If you wanted to, you could turn this matrix representation into a polynomial equation, and also plug in $y_1=1$ as discussed above:

$$x_1\gamma\,(x^2-y^2)
-(x_1^2-1)\gamma\,xy
-x_1(\alpha+\beta)\gamma\,x
-x_1(\alpha\beta-\gamma^2)\,y
+x_1\alpha\beta\gamma = 0
\quad\text{with }x_1\in[-1,1)
$$

  I am assuming here that regular means right.

So am I. Googling for “regular hyperbola” I found one match which claims

  For a regular hyperbola the eccentricity is $\sqrt{2}$.

According to Wikipedia, that's just the eccentricity of a rectangular i.e. right hyperbola. So I think your interpretation of the term is likely correct.

Use the equation $$ A x^2 + B y^2 + D x + E  y = 1 $$

With the points given you have three equations

$$ \begin{align} A \alpha^2 + D \alpha & = 1 \\ A \beta^2 + D \beta & = 1 \\
B \gamma^2 + E \gamma 7 & = 1 \end{align} $$

I include a skew parameter $\zeta = -B/A$ also to parametrize all the curves and together I get the solution

$$\begin{align}
  A & = -\frac{1}{\alpha \beta} \\
  B & = \frac{\zeta}{\alpha \beta} \\
  D & = \frac{\alpha+\beta}{\alpha \beta} \\
  E & = \frac{\alpha \beta - \zeta \gamma^2}{\alpha \beta \gamma}
\end{align} $$

These coefficients give all the curves passing through the points, for every $\zeta \neq 0$.

Note that when $\zeta<0$ the result is an ellipse with a circle at $\zeta =- 1$.


Cardinality of the set of all infinite monotonically decreasing sequences of naturals Find the cardinality of the set of all infinite monotonically decreasing sequences of naturals.

I think it's $\aleph_0$. I marked this set in $A$, and said that $\forall n\in\Bbb N \ (n,n,n,...)\in A$, Therefore, $\aleph_0\le|A|$. But I'm not sure how to show that $|A|\le\aleph_o$. Thanks

A monotonically non-increasing sequence of natural numbers is eventually constant, so it is completely identified by specifying the finite initial subsequence up to the point at which it becomes constant. There are only countably many finite sequences of natural numbers.

Find the cardinality of the set of all infinite monotonically decreasing sequences of naturals.

There are no infinite monotonically decreasing sequences of naturals, since the natural numbers are well-ordered. Thus, the set of all such sequences is empty, and has cardinality $0$.

You seem to have gotten decreasing mixed up with non-increasing.


diophantine equation $x(x^2-x)+y(y^2-y)=2xy$ OK so I accidentally posted a wrong equation in my previous question and I didn't realize it after it was solved. Hope it helped someone and sorry. This is the more challenging one I wanted to solve.

HINT

$x^3+y^3-(x+y)^2=(x+y)(x^2+y^2-xy-x-y)$

Where the second part has already been discussed in your last question(here).

If $x=y$,
this becomes
$x(x^2-x) = x^2$.
A solution is $x = 0$.
If $x \ne 0$,
$x^2-x = x$
or $x=2$.

If $x > y > 0$,
$x(x^2-x) < 2x^2$
or $x^2-x < 2x$
or $x < 3$.

This then reduces to
trying $x = 2$, $y=1$
which works.


is this formula semantically entailed from the empty set of premises? I did the truth table of the below logic:

((p ∨ q) → r) → ((p → r) ∨ (q → r))

However I didn't quite understand what semantically entailed form the empty set of premises? What that mean exactly?

As far as i understand, Whatever P I pick, the conclusion should always be true. So in this case, is it semantically entailed form the empty set of premises?

I think it is not because

((p ∨ q) → r) <> p

in case p is T q is T and r is false

In this setting, semantic entailment $S \models Q$ simply means that if you write down the truth table and throw away the rows where any of the statements in S are false (i.e. you keep only the rows where every statement in $S$ is true), then $Q$ is identically true in the remaining rows.

Thus, your truth table does indeed prove

$$ \models ((p \vee q) \to r) \to ((p \to r) \vee (q \to r)) $$

However I didn't quite understand what semantically entailed form the empty set of premises? What that mean exactly?

Is it entailed by only the rules of inference? That is: is it a tautology?

$$\models((p\lor q)\to r)\to((p\to r)\lor (q\to r))$$


Projections: Orthogonality Given a unital C*-algebra $1\in\mathcal{A}$.

Consider projections:
$$P^2=P=P^*\quad P'^2=P'=P'^*$$

Order them by:
$$P\perp P':\iff\sigma(\Sigma P)\leq1\quad(\Sigma P:=P+P')$$

  Then equivalently:
  $$P\perp P'\iff 0=PP'=P'P\iff\Sigma P^2=\Sigma P=\Sigma P^*$$

How can I check this?

(Operator algebraic proof?)

I'm assuming that by $\sigma(\Sigma P)\leq1$ you mean that $\|\Sigma P\|\leq1$.

Suppose that $\|P+Q\|\leq1$. So $0\leq P+Q\leq 1$. Then $(P+Q)^2\leq P+Q$ (just conjugate with $(P+Q)^{1/2}$). That is,
$$
P+Q+QP+PQ\leq P+Q,
$$
or $QP+PQ\leq0$. If we conjugate this inequality with $Q$, we get $QPQ+QPQ\leq0$. But $QPQ\geq0$, so $QPQ=0$. Then
$$
0=QPQ=(PQ)^*PQ,
$$
and then $PQ=0$. By taking adjoints, $QP=0$.

$$
\
$$

If $PQ=0$, it follows by taking adjoints that $QP=0$. And
$$
(P+Q)^2=P^2+Q^2+QP+PQ=P+Q.
$$

$$
\
$$

If $(P+Q)^2=P+Q$, then by the C$^*$-identity
$$
\|P+Q\|^2=\|(P+Q)^2\|=\|P+Q\|,
$$
so either $\|P+Q\|=0$ (which by positivity would force $P=Q=0$) or $\|P+Q\|=1$.

Let's enumerate the conditions:
(1) $P\perp P'$;
(2) $0=PP'=P'P$;
(3) $\Sigma P$ is a projection.

We can assume that $A\subset B(H)$ for some Hilbert space $H$ (this basically follows from the GNS construction); also, remember that for positive operators $T\in B(H)$ we have $\langle T\xi,\xi\rangle\geq 0$ for all $\xi\in H$.

For (1)$\Rightarrow$(2), assume that $\xi\in P(H)$ and note that $\langle (P+P')\xi,\xi\rangle\leq\langle\xi,\xi\rangle=\|\xi\|^2$ by assumption (since $0\leq P+P'\leq 1$). Since $\langle P\xi,\xi\rangle=\|\xi\|^2$, we have $ \|P'\xi\|^2=\langle P'\xi,\xi\rangle=0$, so that $\xi=(1-P')\xi+P'\xi=(1-P')\xi$. Thus for all $\eta\in H$, note that $P\eta\in P(H)$, so that we have $P\eta=(1-P')P\eta$ by what we found above. This means that $P'P\eta=0$ for all $\eta\in H$, so $P'P=0$. Taking the adjoint yields (2). (3) follows from (2) by just computing, and any projection $Q\in A$ satisfies $0\leq Q\leq 1$, so if (3) holds, then $\sigma(\Sigma P)\subset[0,1]$ immediately.


Evaluate $\lim_{n\to \infty}\frac{1}{\sqrt[4]{{n^4}+n+2}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}}$ I am new to analysis and I have no clue how to solve this limit. This is an exam problem from my analysis 1 course, there are one or two similar ones on the exam.

$$\lim_{n\to \infty}\frac{1}{\sqrt[4]{{n^4}+n+2}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}}$$

The only thing I tryed was this silly idea to rewrite it as one single fraction
and apply Stolz-Cesaro theorem, but it got way too messy so I doubt that is the way.

I can't find explanations generally on these limits of sequences of the type $\frac{1}{f(x_n)}+\cdots+\frac{1}{f(x_{n+k})}$ (I hope this is a good representation). Should series be involved in solving these kinds of limits ?

EDIT: The limit is supposed to be solved only with the knowledge prior to derivatives and integrals.

Thanks in advance

Also note that $\frac{1}{n+1}<\frac{1}{n^4+n+k} <\frac{1}{n}$
so $$ \sum_{k=1}^{4n-1} \frac{1}{n+1}<S_n=\sum_{k=1}^{4n-1} \frac{1}{(n^4+n+k)^{1/4}} <\sum_{k=1}^{4n-1} \frac{1}{n}.$$  So, we note that
$$ \lim_{n \rightarrow \infty} \frac{4n-1}{n+1}=4= \lim_{n \rightarrow \infty} \frac{4n-1}{n}= \lim_{n \rightarrow \infty} S_n$$

When $n \rightarrow \infty$ the first term of the sequence $1/(n^4+n+2)^{1/4} \rightarrow 0$ it can well be neglected.

Note that $$\frac{1}{(n^4+n+k)^{1/4}} > \frac{1}{n+1}$$ can be checked to be true for all $1<k\le 4n-1$


Find all elements $x$ in $Z_{99}$ such that ⟨$x$⟩ $=$ ⟨$11$⟩ What I know is that ⟨$11$⟩ $=$ $(0, 11, 22, 33, 44, 55, 66, 77, 88$), but I can't understand how any elements in $Z_{99}$ can produce this result other than $11$. Any pointers?

Hint: $|x|=99/\gcd(x,99)$. Look for elements $x\in\langle 11 \rangle$ such that $|x|=9$. Then $\langle x \rangle = n\langle 11 \rangle$.


Algebraic equation (find $b$ and $c$) The goal is, given the field extension $\,\mathbb{Q}\subset\mathbb{C}$, to find the minimal polynomial for the element
$$\eta=\cos\left(\frac{2\pi}{5}\right)$$

I define the element
$$\xi=\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)$$
and due to $\,\left|\xi\right|=1$, I rewrite $\,\eta\,$ as
$$\eta=\frac{\xi+\overline{\xi}}{2}=\frac{1}{2}\left(\xi+\frac{1}{\xi}\right)$$
The next step is to find $\,b,c\in\mathbb{Q}\,$ such that $\,f\left(\eta\right)=0$, where $\,f(x)=x^2+bx+c\in\mathbb{Q}\left[X\right].$ So, I am trying to solve (unsuccessfully) the following equation
$$\frac{1}{4}\left(\xi+\frac{1}{\xi}\right)^2+\frac{b}{2}\left(\xi+\frac{1}{\xi}\right)+c=0$$

I know the solution is $\,b=\frac{1}{2},\,c=-\frac{1}{4}\,$ but when it comes to obtaining those values myself, I fail again and again. HELP!

Note $\xi = e^{2i\pi/5}$. Since $\xi$ is a complex fifth root of unity, we know $\xi^4 + \xi^3 + \xi^2 + \xi + 1 = 0$. Now divide through by $\xi^2$ to get $$\xi^2 + \xi + 1 + \frac{1}{\xi} + \frac{1}{\xi^2} = 0$$ Hence $$\left(\xi+ \frac{1}{\xi}\right)^2 - 2 + \left(\xi + \frac{1}{\xi}\right) + 1 = 0$$ and so $$4\eta^2 + 2\eta - 1 = 0$$ and since $\eta\not\in \mathbb{Q}$, the minimal polynomial of $\eta$ is $4x^2 +2x - 1$, or making it monic $x^2 + \frac{1}{2} x - \frac{1}{4}$, as per your solution.


how calculate integral $\int \dot{x} \; dx$ Let $x$ depend on $t$. $\dot{x}$ is derivative $x$ over $t$. I want to calculate the integral $\int \dot{x} \; dx$. I asked similar question about differentiation here. Any thoughts and ideas are appreciated. Thank you!

The integral $\int \dot x dx$ cannot be evaluated explicitely unless the form of the function $x(t)$ is also given. This can be easily understood in the following way

$$\int \dot x dx=\int (\dot x)^2 dt$$

that cannot be furtherly explicited. This kind of computations generally come out from studies on mechanics with dissipative systems. If you have a differential equation like

$$\ddot x=-\dot x+F(x)$$

you can multiply both sides by $\dot x$ and integrating obtain

$$\int dt \frac{d}{dt}\left(\frac{{\dot x}^2}{2}\right)-\int dx F(x)=-\int \dot x dx$$

that cannot be reduced anymore even if lhs can be expressed through an energy integral.

You can write it as $\int \frac{dx}{dt}dx$ which, assuming appropriate smoothness conditions on $\dot{x}$ is the same as $\frac{d}{dt}\int x dx = \frac{d}{dt} (\frac{x^2}{2} + C) = x\dot{x}$


How do I find $\liminf$ and $\limsup$ if $a_{2n}=\frac {a_{2n-1}}2$ and $a_{2n+1}=\frac12+\frac {a_{2n}}2$? Its given that $a_1=a>0$ and that for any $n>1$ two things happen:
$$a_{2n}=\frac {a_{2n-1}}2$$
$$a_{2n+1}=\frac12+\frac {a_{2n}}2$$

How do I find $\lim\inf$ and $\lim\sup$
I am trying to look at

$a_{2n+1}$ and $a_{2n-1}$
$a_{2n}$ and $a_{2n-2}$

But I am unable to prove that they are bounded.

NOTE: Look at joeys answer for correct solution.

EDIT: This response was written for the recurrence equations
$$ a_{2n} = \frac{a_{2n-1}}{2}, a_{2n+1} = \frac{1}{2}+\frac{a_{2n}}{2}. $$
The exact answer may differ depending on the exact edit to the question, but the techniques remain valid, assuming the coefficients are not drastically changed.

Since there are different rules for the even and odd terms, it makes sense to suspect that the even terms will behave differently from the odd terms. So let's investigate the two sets of terms separately.

For $n\in\mathbb{N}$, let $b_n = a_{2n-1}$ and $c_n = a_{2n}$. From the relations provided, we have
$$ b_{n+1} = a_{2n+1} = \frac{1}{2}+\frac{a_{2n}}{2} = \frac{1}{2} + \frac{1}{2}\left(\frac{a_{2n-1}}{2}\right) = \frac{1}{2} + \frac{b_{n}}{4}$$
Similarly, we have
$$ c_{n+1} = a_{2n+2} = \frac{a_{2n+1}}{2} = \frac{1}{2}\left(\frac{1}{2} + \frac{a_{2n}}{2}\right) = \frac{1}{4} + \frac{c_n}{4}.$$

We now look at the behavior of $b_n$ as $n\rightarrow\infty$. Suppose it converged to some value, say $b$. What would $b$ have to be? If we take our equation
$$ b_{n+1} = \frac{1}{2} + \frac{b_n}{4} $$
and let $n\rightarrow\infty$, then both $b_{n+1}$ and $b_n$ would converge to $b$, and hence
$$ b = \frac{1}{2} + \frac{b}{4}\implies b=\frac{2}{3}.$$
So we have a hunch that $b_n$ converges to $\frac{2}{3}$. How do we prove it? Well, let $d_n = b_n - \frac{2}{3}$, and let's try to show that $d_n\rightarrow 0$. Now,
$$ b_{n+1} = \frac{1}{2} + \frac{b_n}{4}\implies d_n + \frac{2}{3} = \frac{1}{2} + \frac{d_n + \frac{2}{3}}{4} = \frac{2}{3} + \frac{d_n}{4}\implies d_{n+1} = \frac{d_n}{4}$$
which clearly shows that $d_n\rightarrow 0$. Hence, $b_n\rightarrow\frac{2}{3}$. Similar reasoning yields that $c_n\rightarrow c$, where
$$ c = \frac{1}{4} +\frac{c}{4}\implies c = \frac{1}{3}.$$

Hence, the odd terms tend to $\frac{2}{3}$, and the even terms tend to $\frac{1}{3}$. This shows that
$$ \limsup\limits_{n\rightarrow\infty}{a_n} = \frac{2}{3} $$
and
$$ \liminf\limits_{n\rightarrow\infty}{a_n} = \frac{1}{3}. $$

It could of course be that I made a mistake in my calculations so proceed with caution!

We'll show by induction that for any $n \geq 1$:
$$
a_{2n} =  \frac{a}{2\cdot 4^n} + \sum_{i=1}^{n-1} \frac{1}{4^i}
$$
where a sum $\sum_{i=1}^0\frac{1}{4^n}$ is assumed to be zero.

The first step: $n=1$ is trivial. Now let $n$ be arbitrary we'll first prove the form of $a_{2(n+1)}$: by the induction hypethesis we may assume that:
$
a_{2n} =\frac{a}{2\cdot 4^n} + \sum_{i=1}^{n-1} \frac{1}{4^i}
$ thus we get:
$$
a_{2(n+1)} = a_{2n+2} = \frac{a_{2n+1}}{2} = \frac{\frac{1}{2} + \frac{a_{2n}}{2}}{2} = \frac{1}{4} + \frac{a_{2n}}{4} \overset{IH}{=}  \frac{1}{4} + \frac{\frac{a}{2\cdot 4^n} + \sum_{i=1}^{n-1} \frac{1}{4^i}}{4} \\= \frac{1}{4} + \frac{a}{2\cdot 4^{n+1}} + \sum_{i=1}^{n-1} \frac{1}{4^{i+1}}= \frac{a}{2\cdot 4^{n+1}} + \sum_{i=1}^{n} \frac{1}{4^{i}}
$$
which proves the claimed equality for $a_{2n}$ we now have for arbitrary $n$:
$$
a_{2n+1} = \frac{1}{2} + \frac{a_{2n}}{2} = \frac{1}{2} + \frac{\frac{a}{2\cdot 4^n} + \sum_{i=1}^{n-1} \frac{1}{4^n}}{2} = \frac{a}{4^{n+1}} + \frac{1}{2} \sum_{i=0}^n \frac{1}{4^i}.$$
Can you deduce the result from this? I can work it out further if you want but I think it's clear how you should continue now (just calculating the sums yields that the liminf is 1/3 and the limsup is 2/3).

Now without liminf and limsup, first we show that $a_n \geq 1/3 $ always holds by induction. For $a_{2n+1}$ this is immediately clear since we have $a_{2n+1} \geq 1/2$. For $a_{2n}$ we proceed by induction so suppose we have $a_{2n} \geq 1/3$ and we show that then also $a_{2(n+1)} \geq 1/3$, we have:
$$
a_{2(n+1)} = a_{2n + 2} = \frac{a_{2n-1}}{2} = \frac{1}{4} + \frac{a_{2n}}{4} \geq \frac{1}{4} + \frac{\frac{1}{3}}{4} = 1/3,
$$
in the same way one shows that $a_n \leq 2/3$ for all $n$. Now it remains to show that $\lim_{n \rightarrow \infty} a_{2n} \leq 1/3$ and $\lim_{n\rightarrow \infty} a_{2n+1} \geq 2/3$ if you show this you are done.


Integral (square root function multiplied by exponential function) did i do it right? I'm trying to determine $\int x^3\sqrt{x^2 +1}\, dx$

I said that $u(x) = x^2 + 1$

and then that $dx = 2x\,dx$
so I rewrote the integral as

$$\int x^3\sqrt{x^2 +1}\,2x\,dx$$

which is also

$$\int2x^4\sqrt{x^2 +1} \,dx$$

and then it is easy to integrate, is that all legal to do?

I'm afraid not. You should not have an equation with a "differential factor" on one side and not on the other (that is, $dx=2x$ is nonsense). For more on how to deal with differential factors, you might find the second part of this answer (from "Now, if I wrote..." through "...let's get back to your problem.") useful.

What you can say is that $$\frac{du}{dx}=2x,$$ so that $$du=2x\,dx,$$ so that $$x\,dx=\frac12\,du.$$ Then your substitution gives you $$\int x^3\sqrt{x^2+1}\,dx=\frac12\int x^2\sqrt{u}\,du.$$ We're not quite there, yet, though. Can you rewrite $x^2$ in terms of $u$?

HINT:  when completing a $u$ substitution, the typical flow is as follows:

$$\int x^3\sqrt{x^2+1}dx$$

Substitute $u=x^2+1$, then $du=2xdx$ and we have $x^2=u-1$:

$$\int {(u-1)\sqrt u du\over 2}$$

From there, the remaining steps are to integrate, then reverse-substitute to obtain the function relative to $x$, and to add the unknown constant.


Prove that there exist linear functionals $L_1, L_2$ on $X$ Let $X$ be a linear space, $p, q$ sublinear functionals on $X$, and $L$ a linear functional on $X$ such that $|L(x)| ≤ p(x) + q(x),$ for all $x ∈ X$. Prove that there exist linear functionals $L_1, L_2$ on $X$ such that $L(x) = L_1(x) + L_2(x),$ and $|L_1(x)| ≤ p(x), |L_2(x)| ≤ q(x),$ for all $x ∈ X.$

My Work:

First I thought to use Hahn Banach Theorem. But since there is no known subspace it was useless. Then I tried to make $L(x)$ as $L(x)=\frac{L(x+\lambda)+L(x-\lambda)}{2}$ for some scalar $\lambda$ but failed to find suitable $L_1$ and $L_2$. I think this problem is little bit tricky. I want to try it myself and I only need a hint to start. Can somebody please give me a hint?

Note that specifying a pair of linear functionals on $X$ is the same as specifying a single linear functional $X\oplus X$, by writing $M(x\oplus y)= L_1(x)\oplus L_2(y)$. Also, if $p,q$ are sublinear functionals on $X$ then $r(x\oplus y)=p(x)+q(y)$ defines a sublinear functional on $X\oplus X$.

Consider now applying Hahn-Banach to extend the functional $M$ on the diagonal of $X\oplus X$ given by $M(x\oplus x)=L(x)$ to all of $X\oplus X$.

Update: Using the Hahn-Banach theorem as suggested above is only enough to obtain linear functionals $L_1,L_2$ satisfying $L_1(x)\le p(x)$ and $L_2(x)\le q(x)$. Note the absence of the absolute value on the left hand side of these inequalities. Once you put the absolute values in, then the result no longer holds in general, as shown by Mizar's counterexample. If, however, we restrict to the case where $p(-x)=p(x)$ and $q(-x)=q(x)$ then the inequalities remain true when the absolute values are introduced, as can be seen by applying them to both $x$ and $-x$.

As you wish only a hint, here it is. Let $X_0=\ker L$ and $x_0\in X\setminus X_0$ be an arbitrary element. Then each element $x\in X$ has a unique representation $x=x’+\lambda_x x_0$, where $x’\in X_0$ and $\lambda_x\in\Bbb R$. Then $L(x)=L(\lambda_x x_0)\le p(\lambda_x x_0)+q(\lambda_x x_0)$. So the problem can be reduced to one-dimensional case and a space $\langle x_0\rangle$.


Evaluate $\int_0^\infty e^{-x^2}\cos\frac{t}{x^2}\,\mathrm dx$ How to integrate

$$\int_0^\infty e^{-x^2}\cos\frac{t}{x^2}\,\mathrm dx$$
where $t$ is a constant/parameter?

Maybe:
$$\int_0^\infty e^{-x^2}\cos\frac{t}{x^2}dx=\int_0^\infty \frac{1}{2}e^{-x^2}\left(e^\frac{it}{x^2}+e^\frac{-it}{x^2}\right) dx\\
=\int_0^\infty \frac{1}{2}\left(e^{-x^2+\frac{it}{x^2}}+e^{-x^2-\frac{it}{x^2}}\right) dx$$

Complete the square for the second term $\left(x+\sqrt t\frac{1-i}{\sqrt2}x^{-1}\right)^2-\sqrt {2t}i=x^2+itx^{-2}$

And similar for the first term...

Is it possible to integrate this out with the substitutions $$u=x+(1-i)\sqrt\frac{t}{2}x^{-1}$$ and
$$s=x+(1+i)\sqrt\frac{t}{2}x^{-1}$$?

I can give you only the results for different $t$s.

Let $t=1$ be, then

$$\int _0^{\infty }e^{-x^2}\cos \left(\frac{1}{x^2}\right)dx = \frac{e^{-\sqrt{2}}\sqrt{\pi}\cos(\sqrt{2})}{2}$$

If $t=2$ is, then
$$\int _0^{\infty }e^{-x^2}\cos \left(\frac{2}{x^2}\right)dx = \frac{e^{-2}\sqrt{\pi}\cos(\sqrt{4})}{2}$$
If $t=3$, then
$$\int _0^{\infty }e^{-x^2}\cos \left(\frac{3}{x^2}\right)dx = \frac{e^{-\sqrt{6}}\sqrt{\pi}\cos(\sqrt{6})}{2}$$
So:
Is $t$ a constant, then we have:
$$\int _0^{\infty }e^{-x^2}\cos \left(\frac{t}{x^2}\right)dx = \frac{e^{-\sqrt{2t}}\sqrt{\pi}\cos(\sqrt{2t})}{2}$$


Is $\aleph_0 = \mathbb{N}$? Some very wise people here have just told me that  $\aleph_0 = \mathbb{N}$, i.e. that the cardinality of the set of natural numbers is just the set of natural numbers itself. Is this now the general consensus in mathematics, real analysis in particular? Or have I got completely the wrong end of the stick as usual?

There's nothing deep going on here. Its just that:

Its often convenient to identify $\mathbb{N}$ with the least infinite ordinal $\omega$.
Its often convenient to identify each well-orderable cardinal number $\kappa$ with the least ordinal $\alpha$ such that $|\alpha| = \kappa$. This is called the von Neumann cardinal assigment.

Under these identifications, we find that $\mathbb{N} = \omega = \aleph_0 = |\mathbb{N}|.$ I wouldn't read too much into it though.

You have N≠ℵ0, Infact one is the set (N), while ℵ0 is a transfinite number.
Asking  ℵ0=N, is somewhat.. you are asking.. Is 2={1,2} ? which is obviously not.


Inequation: quadratic difference equations Given:

$$\frac{(x - 3)}{(x-4)} > \frac{(x + 4)}{(x + 3)}$$

Step 1:

$$(x + 3)(x - 3) > (x + 4)(x - 4)$$

Step2 : Solving step 1:

$$x^2 - 3^2 > x^2 - 4^2$$

*Step 3:

$ 0 > -16 + 9$ ???

As you see, I can delete the $x^2$, but there is no point in doing that.

What should be the next step?

One way to do this more carefully is to take $$\frac {x-3}{x-4}-\frac {x+4}{x+3}=\frac 7{(x-4)(x+3)}\gt 0$$ and this is clearly true iff $(x-4)(x+3)\gt 0$

To do this more formally, note that from $a\gt 0$ we have $a^2\gt 0$ and we can deduce $\frac {a}{a^2}=\frac 1a\gt 0$

The thing is that the sign in the inequality is not an equal sign but a less than sign which requires that the left side be greater than the right side. If you add 9 to -16,  you will get -7. The answer to the inequality is 0>-7 because a negative number is less than 0.


$A \in M_n$ is reducible if and only if there is a permutation $i_1, ... , i_n$ of $1,... , n$ Let $A \in M_n$. $A$ is called reducible if $A$ is permutation similar to a matrix of the form

$$\begin{pmatrix}
B & 0\\
C &D
\end{pmatrix}
$$
where $B$ and $D$ are square matrices.

Reading a book I got stuck in the following lines:

Obviously, $A \in M_n$ is reducible if and only if there is a permutation $i_1, ... , i_n$ of $1,... , n$ and an integer $s$ with $1 \leq s \leq n-1$ such that
$A [i_1,. . . , i_s | i_{s+1} , . .. , i_n] = 0$.

I am confused with... $A [i_1,. . . , i_s | i_{s+1} , . .. , i_n] = 0$...what does this means?

It means that the submatrix of $A$ given by
$$
\begin{bmatrix}
a_{i_1,i_{s+1}} & \cdots & a_{i_1,i_n}\\
\vdots & & \vdots\\
a_{i_s,i_{s+1}} & \cdots & a_{i_s,i_n}
\end{bmatrix}
$$
is equal to the zero matrix.

It means that there are $s$ rows with indexes $i_1$,\dots, $i_s$ and $n-s$ columns with indexes $i_{s+1}$,\dots, $i_n$ such that the submatrix of these rows and columns is a zero matrix.


Expressing P(XY > K) in terms of P(X > a, Y > b)? I'm trying to express the value $P(XY > K)$ in terms of $P(X > a, Y > b)$ for some values of $a$ and $b$ that are probably related to $K$.  I'm generally looking for a formula where $P(X > a, Y > b)$ is present in some form or other - it's fine if it's inside an integral or whatnot.

I've tried $\int_{0}^{\infty} P(X > z, Y > K/z)dz$, but this seems to overcount by quite a lot.  Any advice on how I can express this value in the desired format?

Remember that $$f_{X,Y}(x,y)=\frac{\partial}{\partial x}\frac {\partial}{\partial y} F_{X,Y}(x,y)=\frac{\partial}{\partial x}\frac {\partial}{\partial y}\mathbb{P}(X<x,Y<y)$$
Then $$\begin{align} \mathbb{P}(XY>k)&=\iint_{xy>k}f_{X,Y}(x,y)\mathrm{d}x\mathrm{d}y \\ &=\iint_{xy>k}\frac{\partial}{\partial x}\frac {\partial}{\partial y}\mathbb{P}(X<x,Y<y)\mathrm{d}x\mathrm{d}y\end{align}$$
The integral bounds will depend on the supports for both $x$ and $y$.

Since X and Y are independent events, $$ P(XY>K)=P(X>z \text{ and } Y>K/z)=P(X>z)\,P(Y>K/z). $$
So summing over all possible $z$ gives as a limit the integral
$$P(XY>K) = \lim_{t\to0^+}\int_t^{1/t}{P(X>z)\,P(Y>K/z)\,dz}$$
(we need the limit with $t$ because we cannot divide by $0$ in $K/z$).


Please compute $\int_0^{\infty}e^{-(x+ai)^2}dx$, where, $i^2=-1$. I'm trying to do the following Fourier (Hankel?) transform for a cylindrincally symmetric function:
$$
\int_0^{\infty} \!\int_0^{\infty} \! \left( Are^{-(r^2+z^2)/\delta^2}\right)J_1(k_r r) e^{-ik_z z} r \, \mathrm{d}r\, \mathrm{d}z
$$
Trying to compute the integral in $z$ I got
$$\int_0^{\infty}e^{-\left(\zeta+i k_z \delta/2\right)^2}\,\mathrm{d}\zeta.$$
Any ideas on how to attempt this?

Also, any hints that might be helpful in computing the rest of the integral in $r$ would be really helpful. I have little experience with Bessel functions.

Consider a closed contour $\gamma$ in the complex plane consisting of four straight line segments:

From $z = 0$ to $z = R \in \mathbb{R}$;
From $z = R$ to $z = R + i a$;
From $z = R + i a$ to $z = i a$;
From $z = ia$ to $z = 0$.

Let $f(z) = e^{-z^2}$.  Since this function has no poles in the complex plane, its integral around $\gamma$ is zero:
$$
\oint f(z) \, dz = \int_{\gamma_1} f(z) \, dz + \int_{\gamma_2} f(z) \, dz +\int_{\gamma_3} f(z) \, dz +\int_{\gamma_4} f(z) \, dz = 0.
$$
Taking each of these four integrals in turn, and looking at their limit as $R \to \infty$:

We have $$
\int_{\gamma_1} f(z) \, dz = \int_0^R e^{-x^2} \, dx,
$$
and in the limit $R \to \infty$, this approaches $\sqrt{\pi}/2$.
We have
$$
\int_{\gamma_2} f(z) \, dz = i \int_0^a e^{-(R + iy)^2}\, dy = i e^{-R^2} \int_0^a e^{2 i R y} e^{y^2} \, dy.
$$
The absolute value of the integrand in the last step is bounded above by $e^{a^2}$, and so the integrand will always be smaller in magnitude than $a e^{a^2}$ by the estimation lemma.  Thus, as $R \to \infty$, the integrand over $\gamma_2$ will vanish.
We have
$$
\int_{\gamma_3} f(z) \, dz = \int_R^0 e^{-(x + ia) ^2} \, dx = - \int_0^R e^{-(x + ia) ^2} \, dx,
$$
and so as $R \to \infty$ this approaches the negative of the integral we're looking for.
We have
$$
\int_{\gamma_4} f(z) \, dz = i \int_a^0 e^{-(iy)^2}\, dy = - i \int_0^a e^{y^2} \, dy.
$$

Thus, putting all four integrals together and taking the $R \to \infty$ limit, we conclude that
$$
\frac{\sqrt{\pi}}{2} - \int_0^\infty e^{-(x + ia) ^2} \, dx - i \int_0^a e^{y^2} \, dy = 0 \\
\boxed{  \int_0^\infty e^{-(x + ia) ^2} = \frac{\sqrt{\pi}}{2} - i \int_0^a e^{y^2} dy.}
$$

The remaining integral above is related to Dawson's integral.  It can also be related to the imaginary error function obtained by defining the real error function $\mathrm{erf}(x)$ in the customary way and then extending it analytically to the complex plane $\mathrm{erf}(z)$.  In this case, we have
$$
\int_0^a e^{y^2} dy = - \frac{i \sqrt{\pi}}{2} \mathrm{erf}(i a).
$$
Like the "real" error function, its values cannot be calculated exactly for arbitrary values of its argument.  The NIST Digital Library of Mathematical Functions lists several interesting identities obeyed by these functions, as well as tables of values in various sources and software packages that can calculate it.  Most reasonably sophisticated mathematical software (Mathematica, Maple, Octave, MATLAB, etc.) should also be able to calculate the values of the imaginary error function.

Let u= x+ ai.  Then du= dx.  When x= 0, u= ai, and as x goes to infinity so does u.  That integral becomes $\int_{ai}^\infty e^{-u^2}du$ which can be written in terms of the "error function".


Simple discontinuity of real valued function Consider a real valued function $f$ defined on $[a,b]$. Say it has a simple discontinuity at a point $x \in (a,b)$, with $f(x-) \neq f(x+)$ (LHL not equal to RHL).

Is it necessary that $f(x)$ is equal to either $f(x-)$ or $f(x+)$, or can it be different from both?

Kindly help!


Graphing the binomial coefficients: appearances? Is it appropriate to describe the function
$$f(x) = \binom{x}{2}$$ as parabolic?
How does it vary over $c$:
$$f(x) = \binom{x}{c}?$$
And, is there any intuition behind these graphs? For example, their resemblance to the polynomials of degree $c?$

I tried inputting one into wolfram alpha and it spat out a continuous function, which confused me. This is what prompted me to ask the question.

Yes, you can generalise the factorial, choose function, etc. to not whole numbers too.
$$\binom{x}{k}=\frac{x!}{k!(x-k)!}$$
$$=\frac{x(x-1)\dots(x-k)!}{k!(x-k)!}$$
$$=\frac{x(x-1)\dots(x-k+1)}{k!}$$
So for $k=2$ you will get:
$$\binom x2=\frac{x(x-1)}{2!}=\frac{x^2-x}{2}$$
And for $k=3$:
$$\binom x3=\frac{x(x-1)(x-2)}{3!}=\frac{x^3-3x^2+2x}{6}$$
And so on.

I have usually used the binomial formula with discrete whole numbers. If you want to plot this, you get points on a grid.

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$


Proving $\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}$ (Dixon's identity) I found the following formula in a book without any proof:

$$\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}.$$

I don't know how to prove this at all. Could you show me how to prove this? Or If you have any helpful information, please teach me. I need your help.

Update : I crossposted to MO.

// I thought it would be nice to have an actual solution on this page (especially if that is not one of solutions posted @ MO).

OK, so we want to compute the coefficient
$$
[z^{2m}w^{2m}]\left\{(1-z)^{2m}(1-w)^{2m}(z-w)^{2m}\right\}
$$
i.e. the residue
$$
\operatorname*{res}_{z,w}\frac{(1-z)^{2m}(1-w)^{2m}(z-w)^{2m}}{z^{2m}w^{2m}}\frac{dz}z\frac{dw}w;
$$
after the substitution $Z=z/(1-w)$, $W=w/(1-z)$ (using that $z-w=\frac{Z-W}{1-ZW}$ and $\frac{dz}z\frac{dw}w=\frac{dZ\,dW}{ZW(1-ZW)}$) we see that it's equal to
$$
\operatorname*{res}_{Z,W}\frac{(Z-W)^{2m}}{(1-ZW)^{2m}Z^{2m}W^{2m}}\frac{dZ\,dW}{ZW(1-ZW)}=
[(ZW)^{2m}]\left\{\frac{(Z-W)^{2m}}{(1-ZW)^{2m+1}}\right\}
$$
which is equal to $(-1)^m\binom{2m}{m}\binom{3m}m$.

// The substitution looks somewhat magical. I've learned it from Gessel–Stanton paper mentioned by Todd Trimble @ MO.

Try this: use Newton's binomial on both sides of $$(1-x)^{2n}(1+y)^{2n}(1+z)^{2n}=[(1-x)(1+y)(1+z)]^{2n}$$

then track down and equate the coefficients at terms of the form $(xyz)^k$ on both sides. This should yield the required sum.


Real-valued function and extended-real-valued function, what's the difference? This might be a simple question, but  it is really confusing. I have not found any explicit definition or example of extended real-valued function.

My guess is that f defined by f: x -> 1/x on (0,1) is REAl-valued, while g defined by g: x -> 1/x on [0,1) is EXTENDED REAL-valued. Is this the case?

Could someone please give some examples?

In general, an $E$-valued function is a function that takes values in $E$.

The extended reals are $\mathbb{R}\cup\{\pm \infty\}$. So an extended-reals-valued function takes values that are either in $\mathbb{R}$, or possibly $\pm \infty$.

The only difference between Real valued function and Extended Real valued function is interval of domain . . .
In case of Real Valued Function Domain is (-∞,+∞) and in case of Extended Real Valued Function Domain is [-∞,+∞].
Now another Question arises that ∞ is a real number?
absolutely no . . .
But in mathematical analysis and calculus ∞ play an important role that's why it is named as extended real numbers , moreover it also satisfies some laws of binary operation (i.e addition and multiplication)
just as follows
5 + ∞ = ∞ + 5 = ∞
5 * ∞ = ∞ * ∞ = ∞


can you simplify values in a monoid? Assume I have a monoid $M$  and I am not guaranteed that all elements have an inverse.

Say I have the property that:

$a^m = a^{m+n}$

Can I claim than it must be that $i=a^n$?

Why or why not?

In general you cannot simplify expressions in monoids this way. Consider a monoid $M = \{i,x\}$, where $i$ is the neutral element and $x$ satisfies $x^2=x$. Then, $x^m=x^{n+m}$ for $n,m \geq 1$, but $i \neq x^n$.

However, in some monoids you can do this (for example, in groups and submonoids of groups).

The answer is yes, if the monoid is a cancellative monoid.

Here's an example of a non-cancellative monoid under multiplication: $$\{[1],[4]\}\subseteq \mathbb{Z}/6\mathbb{Z}$$

$[4][4]=[4]$, but $[4]\neq [1]$


A circulant matrix and its transpose It is well-known that a circulant matrix $A$ of size $n \times n$ is isomorphic to a polynomial $$p(x) \bmod x^n - 1.$$  If we consider the transpose $A^T$, what is the corresponding polynomial called? Is there something such as transpose of a polynomial?

The answer is yes, there is indeed something such as the transpose of a polynomial. Being somewhat esoteric, it rarely [to my knowledge] occurs outside coding theory. The transpose is $$A^T \sim a(x) ^T  = \sum^{n-1}_{i=0} a_{n-i} x^i.$$

Easily verified, it also holds that
$$ a(x)^T = a(x^{n-1}) \bmod (x^n-1). $$

Using the equipment provided by @CarlLöndahl, another way of presenting that polynomial's transpose is

$a(x)^T = a(x^{-1}) \mod (x^n - 1)$

Furthermore, the product

$a(x)^T a(x) \sim A^TA$

is also an $n \times n$ circulant, one which is of some interest in autocorrelation modelling, and even musical Pitch Class Theory.

It allows you to evaluate the difference distribution of a set. For example where $a(x) = 1 + x + x^3 + x^4$ represents a set of integers, $\{0, 1, 3, 4 \}$ from $\mathbb{Z}_6$.

The multiset of differences is

$\{ 1-0, 3-0, 4-0, 3-1, 4-1, 4-3 \} \cup \{ 0-1, 0-3, 0-4, 1-3, 1-4, 3-4 \}$

$= \{ 1, 3, 4, 2, 3, 1 \} \cup \{ -1, -3, -4, -2, -3, -1 \}$

$\equiv_6 \{ 1, 3, 4, 2, 3, 1 \} \cup \{ 5, 3, 2, 4, 3, 5 \}$

$= \{ 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5 \}$

From which we may account 2 of 1, 2 of 2, 4 of 3, 2 of 4 and 2 of 5. Present also is 4 of 0 if you include differences between identical elements, to take the total to the expected 16.

But instead of counting them 'by hand' we can multiply $a(x)$ by its transpose, $a(x^5)$ or, more conveniently in this case by $a(x^{-1})$ - it saves a little time with the exponents:

$(1 + x + x^3 + x^4)(1 + x^{-1} + x^{-3} + x^{-4})$

$= (1 + x^{-1} + x^{-3} + x^{-4}) + (x + 1 + x^{-2} + x^{-3}) + (x^3 + x^2 + 1 + x^{-4}) + (x^4 + x^3 + x + 1)$

$= 4 + 2x + x^2 + x^{-4} + 2x^3 + 2x^{-3} + x^4 + x^{-2} + 2x^{-1}$

$\equiv_6 4 + 2x + 2x^2 + 4x^3 + 2x^4 + 2x^5$

whereupon we recover the frequency of differences of size $d$ directly by reading off the coefficients of $x^d$ in $a^T(x)a(x)$.

One may of course recover the same result by multiplication of the two $6 \times 6$ circulant matrices (in either order - it's commutative) derived from row shifts of (1, 1, 0, 1, 1, 0) with its transpose - the circulant matrix derived similarly from (1, 0, 1, 1, 1, 0). The result will be the (perforce symmetric) circulant matrix (4, 2, 2, 4, 2, 2).


Is $(\mathbb R^2,+,\cdot )$ a field where $(a,b)\cdot (c,d)=(ac-bd, ad+bc)$? Is $(\mathbb R^2,+,\cdot )$ a field where $(a,b)\cdot (c,d)=(ac-bd, ad+bc)$ ?
I think it is because $\varphi :\mathbb R^2\to \mathbb C$ defined by $\varphi (a,b)=a+ib$ is a ring isomorphism, but my teacher says it's not, and I don't understand why. Any idea ?

To check if it is a field, let's check the requirements for a field: closure with respect to addition, closure with respect to multiplication, additive identity, multiplicative identity, additive commutativity, additive associativity, multiplicative commutativity, multiplicative associativity, additive inverse, and multiplicative inverse.
Our potential field is defined as $(a,b)$, where $a,b\in\mathbb{R}$.  Our addition operation is defined as $(a,b)+(c,d)=(a+c,b+d)$.  Our multiplication operation is defined as $(a,b)\cdot(c,d)=(ac-bd,ad+bc)$.
The sum and product of real numbers are real numbers, so we have closure with respect to both addition and multiplication.
The additive identity is $(0,0)$.
The multiplicative identity is $(1,0)$.
$$(a,b)+(c,d)=(a+c,b+d)=(c,d)+(a,b)$$
$$((a,b)+(c,d))+(e,f)=(a+c,b+d)+(e,f)=(a+c+e,b+d+f)=(a,b)+(c+e,d+f)=(a,b)+((c,d)+(e,f))$$
$$(a,b)\cdot(c,d)=(ac-bd,ad+bc)=(c,d)\cdot(a,b)$$
$$((a,b)\cdot(c,d))\cdot(e,f)=(ac-bd,ad+bc)\cdot(e,f)=(ace-bde-adf-bcf,acf-bdf+ade+bce)=(a,b)\cdot(ce-df,cf+de)=(a,b)\cdot((c,d)\cdot(e,f))$$
Additive inverse of $(a,b)$ is $(-a,-b)$.
Multiplicative inverse of $(a,b)\neq(0,0)$ is $(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2})$.
This meets all the criteria for a field, so it is a field


Proving $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ function is bijective. Given $f: \mathbb{R}\rightarrow\mathbb{R}$ is a $C^1$ function and $\forall t.|f'(t)| \leq c < 1$,$g: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is defined as $g(x,y) = (x + f(y), y+f(x)).$ I am trying to show $g$ is bijective.

It is sufficient to show $g$ is 1-to-1 and onto, but how exactly is this approached with such an $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ function?

You can construct the inverse to show the function is bijective: Assuming f bijective:

take $h:= (x-f(y), y-f(x) )$ gives you the inverse, so $g$ is bijective.


Proof that there exists an integer $n \geqslant2$ such that $n^2$ divides $2^n + 3^n$ Proof that there exists an integer $n \geqslant 2$ such that $n^2$ divides $2^n + 3^n$. I came up with this problem and I don't have a clue how to start, or even if it is not trivial at all.

Suppose there is an $n\in\mathbb{N}$ such that $2^n+3^n$ is divisible by $n^2$. Then, $2^n+3^n\equiv 0 \bmod n^2$. In particular, $2^n+3^n\equiv 0 \bmod n$. Suppose for convenience that $n=p\neq 2,3$ is a prime number. Then, by Fermat's little theorem, $2^p\equiv 2 \bmod p$ and $3^p\equiv 3 \bmod p$, so
$$2^p+3^p\equiv 2+3\equiv 5 \bmod p.$$
Hence, if $2^p+3^p\equiv 0 \bmod p$, we must have $p=5$. And indeed, $p=5$ works,
$$2^5+3^5 = 275 = 11\cdot 25.$$
Hence, we have shown that the only prime $p$ such that $p^2$ divides $2^p+3^p$ is $p=5$.


Constructing invertible matrix with entries in a principal ideal domain Am trying to solve this exercise in S. Bosch's book Algebra. From the Viewpoint of Galois Theory (Exercise 5, page 81).

Assume that $A$ is a principal ideal domain, and assume that $({a_{11}},\dots,{a_{1n}})$ is a set elements in $A$ for which $\gcd{({a_{11}},\dots,{a_{1n}})}=1.$ Find an $n\times n$ square matrix with entries in $A$ having $({a_{11}},\dots,{a_{1n}})$ as its first row, and which is invertible in $A$.

I may be missing something completely obvious here (sorry!) but my approach is as follows. Since $\gcd{({a_{11}},\dots,{a_{1n}})}=1,$ we have an $n$-tuple of elements of $A$, ${({x_{1}},\dots,{x_{n}})}$ such that $${\sum^n_{i=1}{x_{i}{a_{1i}}}}=1.$$
So I want to construct an $n\times n$ square matrix whose first row is $({a_{11}},\dots,{a_{1n}})$, and whose determinant is ${\sum^n_{i=1}{x_{i}{a_{1i}}}}$. In the case $n = 2$ this seems straight forward. If ${a_{11}}x+{a_{12}}y=1$, then consider the $2\times 2$ matrix whose first row is $({a_{11}},{a_{12}}),$ and the second row is $(-y,x)$. But I have no way of treating the case $n>2$. Is it in general possible to construct a matrix having determinant ${\sum^n_{i=1}{x_{i}{a_{1i}}}}$ having $({a_{11}},\dots,{a_{1n}})$ as its first row (even forgetting the condition ${\sum^n_{i=1}{x_{i}{a_{1i}}}}=1$)?

Because we are working over a PID, all matrices have a Smith normal form. Let's apply that to the $1\times n$ matrix $R=(a_{11},\ldots,a_{1n})$.

The existence of a Smith normal form implies that there exists a $1\times 1$ matrix $P$ and and an $n\times n$ $Q$, both with determinant $1$, such that
$$
PRQ=\left(d_1,0,0,\ldots,0\right),\tag{1}
$$
where $d_1$ is the first invariant factor of $R$ (and this the only invariant factor given that $R$ has a single row only). Furthermore, $d_1$ is the gcd of the $1\times1$-minors of $R$.

In our special case we can immediately deduce that $d_1=1$ and that the factor $P$ really isn't there. Recalling that $\det Q=1$ we know that $Q^{-1}$ exists and has entries in our PID $A$. Therefore we can rewrite $(1)$ in the form
$$
R=\left(\begin{array}{cccc}1&0&\cdots&0\end{array}\right)Q^{-1}.\tag{2}
$$

But, expanding the right hand side of $(2)$, we see immediately that the first row $Q^{-1}$ is equal to $R$.

  Therefore $Q^{-1}$ is the sought after matrix and we are done.

The argument generalizes to the following:

  A $k\times n$ matrix $R$ with entries in a PID $A$ and all its invariant factors $d_1=d_2=\cdots=d_k=1$ can be completed to an $n\times n$ matrix $B$ with entries from $A$ and $\det B=1$

For the purpose of the task at hand invoking the full power of Smith normal forms is a bit overkill. We can build the required matrix $Q$ in stages using the following observation as an inductive step.

Assume that at least one of $a,b\in A$ is non-zero. Because $A$ is a PID, $d=\gcd(a,b)$ exists. Furthermore, by Bezout's identity we can find elements $u,v\in A$ such that $d=ua+vb$. As $d$ is a factor of both $a$ and $b$, we can find $x,y\in A$ such that $a=dx$, $b=dy$. To this data we associate the $2\times2$ matrix
$$
Q(a,b)=\left(\begin{array}{rr}u&-y\\ v&x\end{array}\right).
$$
This matrix has the following key properties:

As $d\cdot\det Q(a,b)=d(ux+vy)=ua+vb=d$ and the cancellation law holds in $A$, we see that $\det Q(a,b)=1$.
We have the matrix product identity
$$
(a\quad b) Q(a,b)=(d\quad 0).\tag{3}
$$
We then prove by the following

Claim. For any natural number $n$ and any vector $(a_1,a_2,\ldots,a_n)\in A^n$ there exists an $n\times n$ matrix $Q\in M_n(A)$ such that $\det Q=1$ and
$$
(a_1\quad a_2\quad\cdots\quad a_n)Q=(d\quad 0\quad\cdots\quad0),
$$
where $d=\gcd(a_1,a_2,\ldots,a_n)$.

Proof. An induction on $n$. The base case $n=1$ is trivial as we can use $Q=I_1$. Assume then that $n\ge2$. Let's write $d_2=\gcd(a_2,a_3,\ldots,a_n)$.
The induction hypothesis implies that we can find a matrix $Q_2\in M_{n-1}(A)$ with determinant $1$ such that
$$
(a_2\quad a_3\quad\cdots\quad a_n)Q_2=(d_2\quad 0\quad\cdots\quad0).\tag{4}
$$
Let
$$
\tilde{Q}_2=\left(\begin{array}{c|c}
1&\mathbf{0}\\
\hline
\mathbf{0}&Q_2\end{array}\right)
$$
be the $n\times n$ matrix gotten by adding an extra row and a column to $Q_2$ with all the other new entries equal to zero save for a single $1$ at position $(1,1)$. Obviously $\det\tilde{Q}_2=1$ also. In view of identity $(4)$ we have
$$
(a_1\quad a_2\quad\cdots\quad a_n)\tilde{Q}_2=(a_1\quad d_2\quad 0\quad\cdots\quad0).\tag{5}
$$
By basic properties of gcds we have $d=\gcd(a_1,d_2)$. The claim thus follows by multiplying $(5)$ from the right with the matrix
$$
Q_1=\left(\begin{array}{c|c}
Q(a_1,d_2)&\mathbf{0}\\
\hline
\mathbf{0}&I_{n-2}\end{array}\right),
$$
where $Q(a_1,d_2)$ is the $2\times2$ matrix from our earlier considerations, and the zero blocks have sizes $2\times (n-2)$ and $(n-2)\times 2$ respectively. Clearly $\det Q_1=1$, so $\det Q=1$ also. QED.

Expanding for the determinant along the first row (consisting of $a_{1,i}$'s) gives you $n$ inter-related conditions on $n$ $(n-1 \times n-1)$-subdeterminants where the $i^\text{th}$ subdeterminant must be  $(-1)^{i + 1} \cdot x_i$.Well, this gets arbitrarily complicated for $n > 2$, but it is exactly what you (implicitly) did for the basic $2$-by-$2$ case. It is enough to get $n\times n-1$ independent equations which shall impose conditions on our  $n\times n-1$ unknowns ( since we already know the first row or equivalently, $n$ entries). We thus have less equations than unknowns and we can thus strategically impose some values on some entries to reduce the number of unknowns.This strategy might be the core of the solution.


Probability of engine failure A 2-engine airplane has 0.23 probability of an engine failure in general, 0.1 is probability of a failure in the right engine and 0.15 is the probability of a failure in the left engine.

What is the probability of a failure in both engine?
What is the probability of a failure in only one engine?

I solved it as follows:

$P=0.1\cdot 0.15$
$P=0.1\cdot(1-0.15)+0.15 \cdot(1-0.1)$

I didn’t develop intuition for probability yet and the fact that I didn’t use the probability of failure in general makes me think I got it all wrong.

Your solution would be OK if the events involved were independent. But they aren't.

Call $A=$ failure in 1st engine and $B=$failure in second engine.

You must use this:
$$P(A)+P(B)=P(A\cup B)+P(A\cap B)$$

That is
$$0.1+0.15=0.23+P(A\cap B)$$
so the probability of a double failure is $0.02$.

In general, the probability of a single failure of an engine is p.  The probability that one will fail on a twin-engine aircraft is 2p.  The probability that both will fail is p^2.  There are similar relationships for more engines.  2p^3, p^4, etc.

In this specific case, you have 0.1 in the right side and 0.15 on the left side, so you have 0.1 + 0.15 = 0.25 as the probability of either failing.  Not sure where your 0.23 came from.  For both to fail, you would get 0.1 x 0.15 = 0.015.


Calculate the height of a distant object using estimated angles from two different points. I knew how to do this long ago, found the exact problem in my old trig book, but I can't seem to work it out.

Say I'm at an unknown distance from a mountain, called point P, and I estimate the angle of elevation to the top of the mountain is 13.5 degrees. Then I move to point N, which is 100 meters closer to the mountain, and I estimate the angle of elevation to be 14.8 degrees. What is the height of the mountain?

I remember this being enough information to solve both triangles, but without the distance to the mountain, or the height of the mountain, I'm at a loss.  Hint's would be appreciated.

Or...

Use the Law of Sines to find the longest side in the triangle with the 100 m side (you know all the angles).

This longest side is also the hypotenuse of a another, right angle triangle where you know the angle opposite the height you want...

Let $O$ and $M$ the base point and the tip of the mountain, respectively. Then
$$NM=100\cdot\frac{\sin13.5}{\sin1.3},$$
and $OM$ is the height of the mountain, where
$$OM=NM\cdot\sin14.8\approx262.854.$$
Note that the input of the sine is in degrees.


Father and daughter river crossing puzzle. There are 3 men and 3 girls. lets name them ABC, A's daughter X, B's Daughter Y, C's daughter Z.

Rule to cross the river:

At once Only two Can go on boat.
Only Men know how to drive the boat (So there has to be atleast one man on boat)
No girl crosses river with strange man, (that means, only father can take a girl on boat)
No girl stays with strange men on shore, (that means, father should not leave his daughter with other men, either she should be alone, or with other girls or with her father)

How do they cross river?

I assume that the boat can only carry two persons, otherwise there is not much of a puzzle.

Martini has given the natural and obvious solution. The weakness of this solution is that in the second step, A can do nothing to prevent B from stepping ashore and ravish X.

The original version of this problem, posed by Alcuin of York (teacher of Charlemagne) more than 1200 years ago, is slightly different. There three men are travelling with their sisters, and each man desires the other girls. The question is how to cross the river without any girl being defiled. Martini's solution is not permitted, but the girls are able to row.

A mathematical method to solve problems of this kind is to look for paths in graphs. Each permissible distribution of people among the various places is a node in the graph, and two nodes are connected by an edge if it is possible to get from one node to the other with one boat trip. A solution of the problem is a path from a start node to an end node. There are simple and effective algorithms for finding paths in graphs, which makes it feasible to solve problems of this kind with hundreds of constraints.

1- one father nd his daughter crosses the river nd only father came back
2- then two girls go to other side nd a girl come back
3- then two other father go nd one girl with her father came back
4- then both father go nd the single girl came back
5- two girl go nd one girl came back
6- then two last girls goes.


Prove closed disc $D^n$ is homeomorphic to the cone $CS^{n-1}$ I need to find a continuous surjective map from $D^n$ to $CS^{n-1}$.

For 2 dimensions, we can use $$f: S^1 \times I /S^{1} \times \{1\} \rightarrow D^2$$ with $f(\theta,t) = (1-t)e^{i \theta}$ which is continuous and surjective.

But how can I generalise this to n dimensions.

Let's fix some notations:
$$D^n = \{ (x_1, \dots, x_n) \in \mathbb{R}^n \mid x_1^2 + \dots + x_n^2 \le 1 \}$$
$$S^{n-1} = \partial D^n = \{ (x_1, \dots, x_n) \in \mathbb{R}^n \mid x_1^2 + \dots + x_n^2 = 1 \}$$
and the cone $CX$ of a space $X$ is given by $(X \times I) / (X \times \{1\})$.

So first let's define $f : S^{n-1} \times I \to D^n$; it will be exactly like your map $(1-t)e^{i\theta}$, except it will work for all dimensions. Let
\begin{align}
f : S^{n-1} \times I & \to D^n \\
(x_1, \dots, x_n, t) & \mapsto ((1-t)x_1, \dots, (1-t)x_n)
\end{align}

Now clearly if $\sum_i x_i^2 = 1$, then $\sum_i ((1-t)x_i)^2 = (1-t)^2 \sum_i x_i \le 1$ (because $0 \le (1-t)^2 \le 1$). Polynomials are continuous, so $f$ is continuous. And if $(y_1, \dots, y_n) \in D^n$, let $t = 1 - \sqrt{y_1^2 + \dots + y_n^2}$, and $x_i = y_i / \sqrt{(y_1^2 + \dots + y_n^2)}$. Then $f(x_1, \dots, x_n, t) = (y_1, \dots, y_n)$, so $f$ is surjective.

It is rather easy to check that
$$f(x_1, \dots, x_n, t) = f(y_1, \dots, y_n, s) \iff t = s = 1 \text{ or }(t = s \text{ and } (x_1, \dots, x_n) = (y_1, \dots, y_n))$$
So we have first that $f$ factors through the quotient $CS^{n-1} = (S^{n-1} \times I) / (S^{n-1} \times \{1\})$, because $f(x_1, \dots, x_n, 1) = f(y_1, \dots, y_n, 1)$ for all $x_i, y_i$. And secondly, the induced map $g : CS^{n-1} \to D^n$ is injective, because there were no other possibilities for two different elements of the domain to map to the same point in $D^n$.

So finally we get a bijective, continuous map $g : CS^{n-1} \to D^n$. Since $CS^{n-1}$ is compact and $D^n$ is Hausdorff, this is a homeomorphism.

If you use polar coordinates for $D^n$ you can define $f(r,\theta_1,\dotsc,\theta_{n-1})=((\theta_1,\dotsc,\theta_{n-1}),r)\subset S^{n-1}\times  I$, now this is surjective and continuous, if you compose with the projection map you get what you want.


Change of Basis - Homework Question Please help me understand what is being asked, I feel I am missing something.

Compute the change of basis matrix for each of the bases, and use it to find the coordinate vector v with respect to B

a) B={(1,2), (1,-2)}, v=(-1,3), (n=2)

b) Same thing just in $R^3$. I Imagine if I understand for R$^2$ I can get this one by myself.

I see a Basis, but only one. All examples in my text and online say change from Basis B to Basis C, and give both bases. So I go and compute the transition matrix $P_B \rightarrow c $ By creating an augmented matrix $ \left[
      \begin{array}{c|c}
        C&B\\
      \end{array}
    \right]$ And reduce it to get the inverse of B. Then I use $P_B \rightarrow c $ times [v]$_B$ to to change the coordiantes to [v]$_C$ But in this question, where C?? Or am I trying to answer a question that hasn't been asked?

Thanks.

They are probably assuming the other basis to be the canonical, $\{(1,0),(0,1)\}$.


How do you find the automorphism? How exactly would you find all the automorphism of something like $Z_8$ or $U(8)$?

I read that there are $4$ automorphisms of $Z_8$, but how did they come about it?

Please explain this as if talking to a beginner. Thanks

In the case of $\Bbb Z/8\Bbb Z$, a homomorphism is determined by where it sends a generator, and an automorphism must send generators to generators. What are the generators? Work from there.

In the case of $U(8)=(\Bbb Z/8\Bbb Z)^\times$, automorphisms permute elements of the same order, and this groups has three elements of order three. Find out which permutations are possible.

This entry is full of hints...maybe a fuller answer would be useful?
$U(8)$ has the operation of multiplication modulo 8.  $1$ is the identity.  To be a group, each element must have an inverse.  If we imagine starting with the set of numbers $0,1,2,3,4,5,6,7$,  then we see that $0, 2, 4$, and $6$ do not have inverses: you cannot multiply them by anything in the set which will give an answer of $1 \bmod 8$.  So we are left with $3,5,7$.  Each is its own inverse.  (So each is cyclic with order two - meaning, you multiply it by itself two times and get back to the identity.)  For example, $3*3=9\equiv 1 \bmod 8$.
So the elements of the group $U(8)$ are $1,3,5,7$.  It is of order four, meaning it has four elements.
To be an automorphism it must also be an isomorphism.  That means:
$$\phi(a * b) = \phi(a) * \phi(b)$$
that is, if we take two elements a and b, we can perform the group operation before mapping or after mapping and get the same result.
For example, if our automorphism map is the $3\to 3,5\to 7$ and $7\to 5$ then, for example we can perform the operation $3*5$ in the original group and get the result $7$, which maps to $5$ in the automorphic group.
Or, we can map $3\to 3$ and $5\to 7$, and perform the operation in the automorphic group, i.e. $3*7$, with the result being $21 \bmod 8 = 5$. We show that this works for all cases, which shows that it is an automorphism.
For the case of $\mathbb{Z}/8\mathbb{Z}$, the set is $\{0,1,2,3,4,5,6,7\}$ and the group operation $\bmod 8$ addition.  $1,3,5$ and $7$ are generators, meaning that I can take any one of them, add it to itself a number of times, and get the whole group.  For the isomorphism formula above to work, this must also be true in the automorphic group.  Because $1$ is a generator, once I map it to a different generator, I determine the rest of the automorphism.  For example, if I map $1\to 3$, then since $1+1=2$ (original group), and $3+3=6$ (mapped group), I must map $2\to 6$.  $1+1+1=3$ (original group), $3+3+3=9\equiv 1 \bmod 8$ (mapped group), so I must map $3\to 1$, etc..  So I really have only four choices, $1\to1, 1\to3,1\to 5,1\to 7$, and any one of these gives an automorphism.


N different normal distributions - probability A is largest Following on from this question, I now have a harder version I need to solve. If we have N numbers selected from N normal distributions, each with its own mean and variance, what is the probability that a particular number is the largest. So for example we choose $A,B,C$ and $D$ from distributions with means $M_a,M_b,M_c,$ and $M_d$ and variances $V_a,V_b,V_c$ and $V_d$, then what is the probability that:

$$A > \max\{B,C,D\}$$

Thank you Ian for pointing out my mistake (independence issues). I've got ANOTHER answer and some code for you. I tested it out properly with R via Monte Carlo simulation. I'll provide the R code and the Mathematica code. I used Mathematica to numerically evaluate the integral I derive below.

So let's say we have $N$ distributions, and each $X_i$ is distributed according to the $i^{th}$ distribution. Let's assume without loss of generality that we are interested in $P(X_1 \ge \max_i X_i)$. I'm going to be a little loose with the notation, but I think it'll be clear.

$$\begin{align}
P(X_1 \ge \max_i X_i) &= P(X_1 \ge X_2, X_1 \ge X_3, \ldots, X_1 \ge X_N)\\
&= \int_{-\infty}^\infty P(X_1 \ge X_2, X_1 \ge X_3, \ldots, X_1 \ge X_N, X_1 = x) dx\\
&= \int_{-\infty}^\infty P(X_1 \ge X_2, X_1 \ge X_3, \ldots, X_1 \ge X_N| X_1 = x) f_{X_1}(x) dx\\
&= \int_{-\infty}^\infty P(X_1 \ge X_2, X_1 \ge X_3, \ldots, X_1 \ge X_N| X_1 = x) f_{X_1}(x) dx\\
&= \int_{-\infty}^\infty P(X_1 \ge X_2|X_1)P(X_1 \ge X_3|X_1)\ldots P(X_1 \ge X_N| X_1 = x) f_{X_1}(x) dx\\
&= \int_{-\infty}^\infty f_{X_1}(x) \prod_{i=2}^N F_{X_i}(x) dx\\
\end{align}$$
where $f$ denotes the PDF and $F$ is the CDF. At this point, it's reasonable to evaluate the integral numerically.

Here is some Mathematica code to show an example of how to evaluate the above expression. Say

$X_1$ ~ N(1,2)
$X_2$ ~ N(0,1)
$X_3$ ~ N(2,1)
$X_4$ ~ N(1.5,3)

Then to get $P(X_1 \ge \max_i X_i)$ we just run the following in Mathematica

NIntegrate[
 CDF[NormalDistribution[0, 1], x]*CDF[NormalDistribution[2, 1], x]*
  CDF[NormalDistribution[1.5, 3], x]*
  PDF[NormalDistribution[1, 2], x], {x, -Infinity, Infinity}]

We can get answer of 0.216147.

Now let's check our answer with R. Here's code for running a Monte Carlo simulation. Note that it took me around 2 minutes to run.

x1 = rnorm(10^8, 1, 2)
x2 = rnorm(10^8, 0, 1)
x3 = rnorm(10^8, 2, 1)
x4 = rnorm(10^8, 1.5, 3)
indicators = ifelse(x1 >= pmax(x2, x3, x4), 1, 0)
mean(indicators)

I just ran it and I got 0.2161365, which is pretty close to what we expect. I tried different numbers of variables and different distributions, and it always worked. Hopefully I understood your problem correctly and this helps you.

Let us say the $A$ we are looking for is $A_k$ and we want to see if it is largest in the set $\{A_1,\cdots,A_n\}$. My initial thought is to fix $A_k$ and then calculate the probability that none of the other $A_i$ is pairwise larger and then use complementary probability:

$$P\left(\max_i A_i\leq A_k| A_k\right) =1-\prod_{\forall i \neq k} P\left(A_i\leq A_k\right)$$

Then what would be needed is to sum or integrate over all $A_k$:

$$P\left(\max_i A_i \leq A_k\right) = \sum_{\forall A_k} P\left(\max_i A_i \leq A_k|A_k\right)\cdot P(A_k)$$

If you have a table for normal distribution function, this becomes easy, just the product of a series of table lookups ( maybe you want interpolation in between table entries ). You will need to calculate which entry of the table to check for each $A_i$ but it will be a a really nice function of its mean and variance. This will be nice to express with vectors too, in case you want to try it in a computational language like R or Matlab or Python or Octave.


Are 400 and 45 correct for CD and VL in roman,respectively? can VL and CD be roman numerals for 45 and 400 respectively? By the way,I already tried CCCC for 400.

$400$ would be $CD$ ($500-100$), the same way $4$ is $IV$ ($5-1$)

$45$ is $XLV$ ($(50-10)+5$).

$CCCC$ would probably be understandable (it has been shown that some less educated romans used $VIIII$ for nine instead of $IX$), but it is not the appropriate way to write it.

Roman numbers from 1 to 500

Yes CD is the Roman numeral for 400 but for 45 it's XLV


Difference in definition of a cover Let $X$ be a topological space,

Ive seen some define a cover, as a collection, $A$, of sets such that $X=\bigcup A$, while others use the condition $X\subseteq \bigcup A$.

Why the difference? My lecturer uses the first definition but to me the second seems more natural. After all, to cover something doesn't necessarily mean to cover it exactly.

The difference is : is $X$ a subspace of a larger space $Y$ ?

If it isn't, you only have one natural definition : a cover is a collection of subspaces $A_i$ such that $X = \bigcup_i A_i$.

But if $X\subset Y$, then you have two natural definitions : a cover of $X$ can either be a collection of subspaces of $X$ such that $X = \bigcup_i A_i$ (def $1$), or a collection of subspaces of $Y$ such that $X\subset \bigcup_i A_i$ (def $2$).

Of course a cover for def $1$ is in particular a cover for def $2$. And if you have a cover $(A_i)$ for def $2$, you get a cover for def $1$ by using $B_i = A_i\cap X$.

This behaves well because for instance $B_i$ is open/closed in $X$ iff $A_i$ is open/closed in $Y$. On the other hand, if $(A_i)$ is an open cover for def $1$, it may not be an open cover for def $2$, because an open subset of $X$ is not necessarily open in $Y$ if $X$ is not itself open if $Y$. So in that case you have to replace $A_i$ with $A'_i\subset Y$ such that $A'_i$ is open in $Y$ and $A_i = X\cap A'_i$ (which exists by definition).

In the end you can see that the two definitions give the same thing if you're ready to modify a little your covers when you switch definitions.

$X\subset\bigcup A\implies X=\bigcup \left(X\cap A\right)$


The denominator of a fraction is 4 more than twice the numerator. Determine the fraction. The denominator of a fraction is $4$ more than twice the numerator. When both the numerator and denominator are decreased by $6$, the denominator becomes $12$ times the numerator. Determine the fraction.

I tried the following,

Let the numerator be $x$ and the denominator be $y$

Therefore, Fraction$=$ $x$/$4$$+$$2$$x$

Because, both the numerator and denominator are decreased by 6

Therefore, the new fraction becomes $x$$-$6/$($$4$$+$$2$$x$$)$$-$$6$$=$$12$$x$

I do not know how to proceed further.

Using your notation you have:

$$y=2x+4$$

And the next condition reads:

$$12(x-6)=y-6$$

Because if the denominator is 12 times the numerator that means that your fraction is equal to $\frac{1}{12}$.

So from this system of equations you can get the values of $x$ and $y$. I will give them to you but I'll let you go through the last steps of the process to get them:

$$x=7 \qquad y=18$$

let the numerator be x and let the denominator be y...
in the first part of the question it is said that,x/2x*4(*is plus).in second part of the question it is said that :
x-6/(2x*4)-6.now finally at the last part of the question it is said that: 2x-2=12x[x-6]
2x-2=12x-72
2x-12x= -72+2
-10x=-70
x=7...
in first part of the question we saw that...y=2x+4...now substitute the value of x in that....you get
2.7+4=18...therefore the required fraction is 7/18....hope u understand...[• indicates multiplication]


How to show that if two Platonic solids have the same number of edges, vertices, and faces, then they are similar in $\mathbb{R}^{3}$? Note: It appears that some of the terms here do not have standardized definitions, so some sources may give conflicting info.
I was looking into the proof that there are only five Platonic solids in Basic Concepts of Algebraic Topology by F.H. Croom at page 29, Theorem 2.7. To clarify,

We define a Platonic solid as a simple, regular polyhedron homeomorphic to $S^{2}$.
We define a simple polyhedron to be a polyhedron that does not self-intersect itself.
We define a regular polyhedron to be a polyhedron whose faces are regular polygons all congruent to each other and whose local regions near the vertices are all congruent to each other.

Using homology theory, one can prove that the Euler formula $V-E+F=2$ must hold for Platonic solids. Then by using Euler's formula and invoking a counting argument, we find that there are five possible tuples $(V, E, F)$.
This is a beautiful proof, but I am unsatisfied with a question: How do we know there can't be two non-similar Platonic solids that have the same $(V, E, F)$-tuple?
Almost all sources I've looked at seem to assume it is obvious that two Platonic solids with the same $(V, E, F)$-tuple are similar, and it is not obvious to me.
Does anyone have any suggestions for how to prove this? Alternatively, does anyone know of a reference where this is proved rigorously?

Edit 1: It's not completely clear, but it seems like the definition I used for "regular polyhedra" is different than the one commonly used. Note that I am not assuming any global symmetry, so if any global symmetry is to be invoked, it needs to be proven.
Edit 2: I've been made aware of Cauchy's rigidity theorem, which is proven in, e.g., Proofs From the BOOK by Aigner & Zeigler. One can show that any two Platonic solids that have the same $(V, E, F)$-tuple must be combinatorically equivalent. However, in order for the theorem to apply, we need to show that our Platonic solids are convex. I can't seem to think of any rigorous argument for why the Platonic solids have to be convex.
And actually, you don't need to show that the entire polyhedron is convex. If I'm not mistaken, the proof for Cauchy's rigidity theorem only relies on the vertices of the polyhedron being locally convex. So really it suffices to show the vertices are convex.

Edit: This answer invokes rotational symmetry about the vertices, which isn't required in the OP, so this is not a complete answer without a proof that such symmetry ought to hold.

Given $V,E,F$, you can compute the number of sides $n$ of each face ($2E/F$) and the degree $k$ of each vertex ($2E/V$).
By regularity, a Platonic solid with degree $k$ should be invariant under a rotation by $360/k$ degrees about a vertex. This forces the arrangement of regular $n$-gons at a vertex to be "rigid", because this rotational symmetry along with a specific internal angle of each polygonal face forces a unique relative position of the edges from a vertex, up to rotation of the whole configuration in space. See Eric Wofsey's comment for more elaboration. (Also note that the uniqueness of such a configuration is reliant on there being no self-intersections between the polygons around the vertex; if you allow this, the Kepler-Poinsot polyhedra result.)
Furthermore, this rigid configuration about a vertex is fully determined by the position of three of the edges in the configuration; two edges are enough to determine the configuration up to a reflection, and the third tells us whether the vertex is convex or not (whether the corner points "out" or "in").
So, if we start with a vertex and an orientation of its local neighborhood, we get a rigid structure of the polygons meeting at a single vertex. This gives us a ring of "boundary" vertices, where the neighboring polygons have not yet been fully specified. We repeatedly choose a boundary vertex where two boundary polygons meet, which has at least 3 fixed edges attached to it and therefore forces a unique attachment of polygons from that vertex by the above paragraph. By adding in these polygons to our rigid structure, this ring of boundary vertices grows outward, and we can repeat the process. So long as we perform this extension on the boundary vertices closest (in the graph-theoretic sense) to our starting vertex, we will eventually specify the local neighborhood of every vertex in the graph (possibly after infinitely many operations; we aren't assuming that things will naturally close up here.)
The result of this procedure is a surface without boundary, so if the polyhedron is homeomorphic to the $2$-sphere, this process must have terminated after finitely many points and closed up into a unique finite solid.
(Of course, this does not show that any such polyhedron does exist - see this question for an algebraic topology proof of existence. Essentially, the vertex-extending process described above forms a covering space of $S^2$, and so must actually yield the sphere.)

OK so we can imagine two solids, each with the same triple of $(V,E,F)$ each has the same number of faces. And you calculate the number of edges that each face has as $2E/F$ (each edge can be in two faces). So they have the same number of faces and each face has the same number of edges. How many edges meet at each vertex? This is $2E/V$ (each edge hits two vertices). So these two polyhedra have the same number of faces, each with the same number of sides and the same number of edges meeting at each vertex so they must be the same polyhedron. We could build each to the precise formula of faces, number of edges per face and number of edges per vertex and we have no room to make choices. The recipe precisely defines the polyhedron.


Cauchy Sequence that Does Not Converge What are some good examples of sequences which are Cauchy, but do not converge?

I want an example of such a sequence in the metric space $X = \mathbb{Q}$, with $d(x, y) = |x - y|$.  And preferably, no use of series.

Another one, same idea:
$$
a_n = \left(1+\frac{1}{n}\right)^n
$$
a sequence of rationals, but its limit $e$ is not rational.

This agood example for acauhy seqeance
but is not convergent.Take Xn€Q suhc that
        Xn^2<2 this imples Xn<2^2
    then Xn convergent to 2^2 which dont belong  to Q.
      So is not convergent.


which order do we perform the exponents $x^{2*1/2}$? For $x^{2*1/2}$, do we do $\sqrt x^{2} $ or $ \sqrt{x^{2}}$ ? the reason is because the second one is defined for the entire real line and is equal to |x| whereas the first one is only defined for positive numbers. But at the same time, the rules of exponents say the order is commutative so we should not have two different answers. How do we reconcile this?

Neither, to calculate $x^{2*1/2}$ you do the multiplication first to get $x^1=x$.
And
$\sqrt{x}^2=(x^{1/2})^2$
$\sqrt{x^2}=(x^2)^{1/2}$
but in general, $x^{ab}\ne (x^a)^b$

First see the comments of jjagmath and myself that follow this answer.

The rule of exponents does say that order is commutative, so you never have two different answers.  However, the domain of the two functions is different.  Therefore, an element $x$ can be in the domain of one of the functions, and out of the domain of the other function.
Personally, I think the expression $x^{2 \times (1/2)}$ is ambiguous and might reasonably be construed to represent either

$(x^2)^{(1/2)}$ or
$x^{2 \times (1/2)} = x^{1}$, since $[2 \times (1/2)] = 1.$

It depends on the order of evaluation.
Personally, I would not evaluate it as $\left(\sqrt{x}\right)^2$, because I regard that as being represented by
$x^{(1/2) \times 2}$, which I regard as different from
$x^{2 \times (1/2)}$.

Note that $\sqrt{x^2} = |x|$
and that (in Real Analysis)
$\left(\sqrt{x}~\right)^2$ is undefined when $x < 0.$
When $x \geq 0$, then $\left(\sqrt{x}~\right)^2$ evaluates the same way that $\sqrt{x^2}$ does.