Correct order of the growth function $5 \log( \log n)  $

$n (\log n)^2$

$\sqrt{n} \log n$

$n^{\frac{4}{3}}$

$n \log (\log n)$

$7 \sqrt{n}$

What is the ascending order of the growth function? Please give the explanation as well.

If you take sufficiently large value of n,then ascending order according to the value will be,
$5 log( log n)  < 7 \sqrt{n}<\sqrt{n} log n<n log (log n)<n (log n)^2<n^{\frac{4}{3}}$

So,if we sort them from Most efficient to least efficient,it will be

$5 log( log n)  > 7 \sqrt{n}>\sqrt{n} log n>n log (log n)>n (log n)^2>n^{\frac{4}{3}}$

If we want for sufficient large n, then if we name the equation from top to down as from 1 to 6 then ordering looks like 1<5<2 , 7<3<2 and 6<4.


Determine if equation will generate perfect squares Given the following quadratic equations:

$4n^2 + 128n - 131$
$4n^2 + 16n - 11$
$4n^2 + 24n - 3$

Is it possible to determine how many values of n will generate a perfect square?  Or better yet, is it possible to determine which values of n will generate a perfect square?

You can complete the square.  For the third example, $4n^2+24n-3=(2n+6)^2-39.$  As $(n+1)^2-n^2=2n+1$, this cannot be a perfect square for any $2n+6$ greater than $20$, leaving only $7$ possibilities.

We can do better.  You want $(2n+6)^2-39=p^2.$  This means $39=(2n+6-p)(2n+6+p)$.  There will be one solution for each way to factor $39$ into two numbers.  Because of the $2n,$ you need the sum of the numbers to be even, which will be true for odd numbers like $39$ and will be the case for your other three examples.

HINT: Let's rearrange the first equation:

$4n^2-4n+132n+1-132=(2n-1)^2+132(n-1)=m^2$ now we can easily see that $n-1$ has to be $0$ so $n=1$

For the second equation we may write:

$4n^2+4n+12n+1-12=(2n+1)^2+12(n-1)=m^2 \Rightarrow n=1$

For the third equation:

$4n^2+20n+4n+25-28=(2n+5)^2+4(n-7)=m^2 \Rightarrow n=7$


The number of positive integral solutions of $abc =$ $30$ is The number of positive integral solutions of $abc =$ $30$ is

My attempt
Factors of $30$ are $3 \cdot 5 \cdot 2$.
Therefore $a$ will have three choices, similarly $b$ & $c$ will have $2$ choices & $1$ choice.
So,it's answer must be $3 \cdot 2 \cdot 1$

But answer in my textbook is given is $3 \cdot 3 \cdot 3$.
.It states that"$2$ can be assigned to either $a$ or $b$ or $c$".  Similarly, each of $3$ and $5$ can be assigned in $3$ ways.  Thus, the number of solutions is $3 \cdot 3 \cdot 3$

According to me, if $a$ is already assigned to $2$ than how can other letters ($b$ or $c$) have same value as that of $a$.

Please explain me where am I wrong?

Hint: (Slightly different point of view than the book; but will lead to same answer).

There are many ways to write $30$ as a product of three factors, such as $1\cdot 1\cdot 30$ or $1\cdot 5\cdot 6$.

I think you have neglected some of these cases.


Using fundamental theorem of calculus and differnetiability Say we have know that $$f(x)=\begin{cases} 1/n &\text{if  $x=\frac{1}{n}$} \\ 0 &\text{else} \\ \end{cases}$$ is $\in R[0,1]$ with a value of $0$.

and say we define $g(x)=xf(x)$, and $G(x)=\int_{0}^{x}g$

Then how could I prove that G is differentiable on $[0,1]$ but that $G'(c) \neq g(c)$ whenever g is not continuous at c, moreover if $F(x)= \int_{0}^{x}f$ then how could I prove it is differentiable on $[0,1]$ and that there exists $a,b$ in the interval with f being discontinuous at a and b and $F'(c)=f(c)$ but $F'(a) \neq f(a)$

My thoughts;

I am pretty confused. I know that g is R-int as it is the product of two R-int functions. I also know that f has a value of zero, so I think g will evaluate to zero. So G would be zero everywhere and hence differentiable, but I am not sure,
I cannot use results of Lebesque , ie about countable sets etc

Does anyone at all have any advice please? I really dont know how to proceed and need help
I am still confused. Can anyone please help?

You're right.  $G(x) = \int_0^x g(y)\,dy = 0$ for all $x$.  Thus $G$ is differentiable but $G'(x) \ne g(x)$ for some $x$.  That is because $g(x)\ne 0$ only for $x=1/n$ with integral $n$, and that's just a countable set, and the integral of a function which is zero outside a countable set is zero.

I don't get what you mean by "with a value of $1$" at the end of your description of $f$.

Do you need a proof that a Riemann integrable function differing from $0$ on a countable set has Riemann integral zero?


$m$ square-free integer such that $4\vert m-1\implies 2$ irreducible in $\mathbb{Z}[\sqrt{m}]$ Let $m$ square-free integer such that $4\vert m-1$. I need to prove that
$2$ is irreducible in $\mathbb{Z}[\sqrt{m}]$.

What I taught is to use the following theorem:

LEMMA: Let $R$ a Euclidian domain with respect to $\phi$, and $\alpha=\min\{\phi(a):a\in R(\{0\}\bigcup R^{*})\}$, where $R^{*}$ denote the invertible elements of $R$. So the element $b\in R(\{0\}\bigcup R^{*})$ such that $\phi(b)=a$ is irreducible in $R$.

Well, IF $\mathbb{Z}[\sqrt{m}]$ is a Euclidian domain with respect to $\phi(a+b\sqrt{m})=|a^2-mb^2|$, so I can use the Lemma, since $\alpha=4\implies b=2$ is irreducible in $\mathbb{Z}[\sqrt{m}]$.

I just need to prove that $\mathbb{Z}[\sqrt{m}]$ is a Euclidian domain.

Is easy to see that $\phi$ is multiplicative, but the Euclidian Algorithm part is confuse to me.

Is $\mathbb{Z}[\sqrt{m}] $ actually a Euclidian domain (supposing $m$ square free and $4\vert m-1)?$

If $4\mid m-1$, then $\mathbb{Z}[\sqrt{m}]$ is not integrally closed and so cannot be an Euclidean domain.

Indeed, if $4\mid m-1$, then the integral closure of $\mathbb{Z}[\sqrt{m}]$ is $\mathbb{Z}[\frac{1+\sqrt{m}}{2}]$.


What is the probability of randomly chosen real number is even? I read in the book A First Course in Probability by Sheldon Ross the following statement:

  Technical Remark. We have supposed that $P(E)$ is defined for all the events $E$ of the sample space. Actually, when the sample space is an uncountably infinite set, $P(E)$ is defined only for a class of events called measurable. However, this restriction need not concern us as all events of any practical interest are measurable.

The set of Real numbers is an infinitely uncountable set then how can we calculate the probability of even real number?

The answer is that there is $0$ chance that a randomly chosen number is even, because the set of real numbers is much larger than the set of even numbers (a bigger type of infinity, to be precise).

This gets into a branch of math called measure theory. The idea is, given a set of real numbers, can we give certain measure to it? Probably the most common measure (as far as I know), is Lebesgue measure. It extends the idea of lengths of intervals. For example, any interval $[a,b]$, has measure $b-a$. This is simple enough, but more complicated sets are more complicated to measure. Some sets can not be measured. In a Probability space $S$,  we can then say the probability of an event $E$ is $\frac{\mu(E)}{\mu(S}$. Here $\mu$ is the measure function. It turns out in this case $\mu(E) = 0$ and $\mu(S) = \infty$, so the probability is 0.

If you want to see better why this works, the basic requirements of a measure are

Non-negativity: For all E $\mu(E) \ge 0$.
Null Empty Set: $\mu(\emptyset) =0$.
Countable Additivity: For a countable collection of pairwise disjoint sets $\{E_i\}$, $\mu(\bigcup\limits_{i=1}^{\infty} E_{i}) = \sum_{i=1}^{\infty} \mu(E_i)$.

The measure of any point is $0$, as as the integers are a countable collection of points, their measure is the sum of a bunch of zeros, and so is zero.

We can write the reals as an infinite union of intervals of length one, and thus the reals have measure $\infty$.

If you want a more detailed explanation of measure the wikepedia page does a pretty good job explaining it, and it is short. https://en.wikipedia.org/wiki/Measure_(mathematics).


Solving a differential equation, substitution has been recommended I have the differential equation, $x$ is a function of $t$:

$xx'=-a(t+b)+cx$ where $a,b$ and $c$ are constants

We have been told that using the homogeneous substitution method might be useful, but when I sub in $v=\frac xt$ I get:

$vt(v+v't)=-a(t+b)+cvt $

this seems more complicated.

This is part of a larger question which was solve the following system of equations:

$x'=1$  which implies $x=t+x_0$
$y'=0$ which implies $y=y_0$
$z'=3y -\frac{2xy^2}{z}$ so subbing in the values above gives $z'=3y_0-\frac{2y_0^2(t+x_0)}{z}$

As you can see the equation that I generalised at the start refers to the third equation. I'm not sure if this context helps in answering the question.

$xx'=-a(t+b)+cx$

$x'=-\dfrac{a(t+b)}{x}+c$

This is an ODE of the form $x'=f\left(\dfrac{x}{t+b}\right)$ .

In fact the substitution $v=\dfrac{x}{t+b}$ can leads the ODE to becomes a separable ODE.

Let $v=\dfrac{x}{t+b}$ ,

Then $x=(t+b)v$

$x'=(t+b)v'+v$

$\therefore(t+b)v'+v=-\dfrac{a}{v}+c$

$(t+b)v'=-\dfrac{a}{v}+c-v$

$(t+b)\dfrac{dv}{dt}=-\dfrac{v^2-cv+a}{v}$

$\dfrac{dt}{t+b}=-\dfrac{v}{v^2-cv+a}dv$

$\int\dfrac{dt}{t+b}=-\int\dfrac{v}{v^2-cv+a}dv$

If you think $\int\dfrac{v}{v^2-cv+a}dv$ is too tedious to be simplified, you may leave it as for the solution, however in fact $\int\dfrac{v}{v^2-cv+a}dv$ is just a standard integral of the form http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions#Integrands_of_the_form_xm_.2F_.28a_x2_.2B_b_x_.2B_c.29n, means that leaving it as for the solution tends to be unreasonable.

This by no means gets all solutions but here are a whole heap of them:

Guessing $x(t) = et + f$ allowing for coefficients in $\mathbb{C}$, we see that this equation reads:

$$ e^2 t + ef + a(t + b) - cet - cf = 0 $$

$$ (e^2 + a - ce)t + (ef - ab - cf) = 0$$

The first is a quadratic in $e$ so you can solve that, the second is then a linear equation in $f$ which you can solve provided $e - c \neq 0$.

If $e = c$, then $c + \sqrt{c^2 - 4a} = 2c$ meaning $a=0$. But then we just have $x' = c$ in which case $x = ct + x_0$ is a solution


Integer partitions of $N$ into three distinct parts How many ways are there to partition 10 into 3 distinct parts?

For example, $\{7,2,1 \}$, $\{6,3,1 \}$ and $\{5,4,1\}$ are valid partitions, but $\{4,5,1\}$ would not be an additional partition because it has the same elements as $\{5,4,1\}$. $\{8,1,1\}$ is not admissible because it has two 1's.

The Wikipedia article on partitions indicates how to see that the number of partitions into at most three parts is the same as the number of partitions with greatest part at most three.  It gives the number of partitions of $n$ into at most three parts as the nearest integer to $\frac {(n+3)^2}{12}$, which for $n=10$ is $14$.  For exactly three parts, we subtract the $5$ partitions into $2$ parts and the $1$ partition into $1$ part, leaving $8$.  They are:
$$1,1,8\\1,2,7\\1,3,6\\1,4,5\\2,2,6\\2,3,5\\2,4,4\\3,3,4$$
Of these, four are into distinct parts.  There is also equality between all partitions into distinct parts and all partitions into odd parts, but I don't know how to equate that to your problem with both distinct parts and a given number.

enter image description hereThe formula <C(n+2,2)/6-0.75*(n+1)+89/72> is the answer of the number of solutions of a+b+c=n, where a,b,c>0 and <x> is the integer closest to x


Find all pairs $(a,b)\in \mathbb{R^2}$ such that $af_n^2+bf_{n+1}^2 $ is a member of the Fibonacci sequence for all $n\in\mathbb{N}$. Question : Let$\{f_n\}$ be the Fibonacci sequence $\{1,1,2,3,...\}$. Find all pairs $(a,b)\in \mathbb{R^2}$ such that $af_n^2+bf_{n+1}^2 $ is a meber of the sequence for all $n\in\mathbb{N}$.

I used  $f_n=\frac {1}{\sqrt 5}[\phi^n-(-1)^n\phi^{-n}]$ with $\phi=\frac {1+\sqrt 5}{2}$. To reduce the given expression but then was not getting any helpful ideas to  proceed. So any help/hint will be appreciated. Can we find such $a,b$ so that $af_n^k+bf_{n+1}^k$ for some $k>2$ in positive integers?

Thank You.


$\int_{\Omega} |f| d\lambda = 0 \Rightarrow \{x \in \Omega: f(x) \neq 0\}$ is a null set Let $(\Omega, \mathscr A, \lambda)$ be a measure space.
For an $\mathscr A$-measurable numerical function $f: \Omega \rightarrow \Bbb R$, it holds that
$\int_{\Omega} |f| d\lambda = 0 \Rightarrow \{x \in \Omega: f(x) \neq 0\}$ is a null set.

The argument works the following way:
Define
$S := \{x \in \Omega: f(x) \neq 0\}$. Since $f$ is measurable, it follows that $S \in \mathscr A$. For every natural number $k \ge 1$, we define
$\phi_k := \inf\{k|f|, X_s\}$ with $X_s$ being the Indicator function.
Hence,
$\phi_k \uparrow X_s$, and since $\phi_k \le k|f|$, we get $\int \phi_k d\lambda = 0$.
While I understand the very last step here, I don't see why $\phi_k \uparrow X_s$. $X_s = 0$ for every $x \notin S$, and since $S$ is defined the way it is, it's only possible that $|f| = 0$, and therefore, $k|f| = 0$ for every $k$. So how does $\phi_k$ converge against $X_s$ here?

From construction it is clear that $\phi_k \leq \chi_S$ for all $k$. To see that $\phi_k$ converges pointwise to $\chi_S$ consider first if $x\in S$. Then $f(x)\neq 0$ and so $k |f(x)|\geq 1$ for sufficiently large $k$. Thus, eventually $k|f(x)|\geq \chi_S(x)$ and $\phi_k(x) = \chi_S(x)$. On the other hand, if $x\not\in S$ then $f(x)=0$ and $\chi_S(x)=0$, so it is clear that $\phi_k(x)  = \chi_S(x)$ for all $k$.

Together, these two cases imply that $\phi_k$ converges to $\chi_S$ pointwise.

By Markov's inequality, $\forall \varepsilon > 0$,

$$P(f \ne 0) = P(|f| \ge \varepsilon) \le \frac{E[|f|]}{\varepsilon} = 0$$

I know yours is a measure but not probability space but the same holds true for measure space.


Prove that the sum of the squares of two odd integers cannot be the square of an integer. Prove that the sum of the squares of two odd integers cannot be the square of an integer.

My method:

Assume to the contrary that the sum of the squares of two odd integers can be the square of an integer. Suppose that $x, y, z \in \mathbb{Z}$ such that $x^2 + y^2 = z^2$, and $x$ and $y$ are odd. Let $x = 2m + 1$ and $y = 2n + 1$. Hence, $x^2 + y^2$ = $(2m + 1)^2 + (2n + 1)^2$
$$= 4m^2 + 4m + 1 + 4n^2 + 4n + 1$$
$$= 4(m^2 + n^2) + 4(m + n) + 2$$
$$= 2[2(m^2 + n^2) + 2(m + n) + 1]$$
Since $2(m^2 + n^2) + 2(m + n) + 1$ is odd it shows that the sum of the squares of two odd integers cannot be the square of an integer.

This is what I have so far but I think it needs some work.

Let $a=2n+1$, $b=2m+1$. Then $a^2 + b^2=4n^2 + 4n +4m^2 +4m+2$. This is divisible by $2$, a prime number, but not by $4=2^2$. Hence it cannot be the square of an integer.

just think base2:
a and b, being odd numbers will look like $$ xxxx1_2$$
square of a and b will be odd and also look  like $$ xxxx1_2 $$
sum of $$xxxx1_2 + xxxx1_2 = xxx10_2 $$

square of even number $$xxxx0_2$$ will look like $$xxx00_2$$

so digits on position '2' will be always 0 for square of even number but for sum of squares of odd numbers will be 1:

$$ xxxx10_2 \neq xxx00_2 $$


What is the coefficient of x^2 in the expansion of (x+2)^4(x+3)^5 I'm missing something here.

I've calculated the $x^2$ coefficient of $(x+2)^4$ as 24 with constant term 16. And $x^2$ term coefficient of $(x+3)^5$ as 270 with constant term 243.

if I'm correct here then the answer should be $(16*270)+(24*243)$? but this does not appear to be the case.

any assistance appreciated.

Hint:

$$\color{blue}{(x+2)^4} \cdot \color{green}{(x+3)^5} = \color{blue}{(x^4+\square x^3 + \square x^2+ \square x + 16)}\cdot (\color{green}{x^5+\square x^4+\square x^3+\square x^2+\square x + 243)}$$

The term for $x^2$ in the resulting final expansion could have been made by combining a constant blue term with an $x^2$ green term, an $x$ blue term with an $x$ green term, or an $x^2$ blue term with a constant green term.

Wouldn't the $x$ term of one and the $x$ term of the other combine in the product to produce an $x^2$ term in the result?


Equation of the line, vertical I need to find the equation of the line, so I want to use y-y1=m(x-x1)  but for a vertical line the slope is undefined.  So I am unsure how to proceed.

A vertical line in the plane is defined by the equation $x = c$, where $c$ is some real. This makes sense because along the vertical line, the $x$ value doesn't change.

How about below parameterization.

$ \rho = x \cos{\theta} + y \sin{\theta}$

where

$ \rho $ = perpendicular distance to the line from the origin (0,0)

$ \theta $ = angle of a line perpendicular to our line with the x axis

you can take a look at http://en.wikipedia.org/wiki/Hough_transform.

when line slope is $90^o$, $\theta = 0^o$, then $ \rho = x $.


How to prove that a complex function is differentiable at a point and where is it analytic in its domain I know the definition of complex differentiable functions and I know the Cauchy-Reimann equations that are used to prove that a function is differentiable in general, but my question is how to prove that a complex function is differentiable at a point?

Suppose $z_0 \in \mathbb{C}$.  We say that a function $f$ is complex differentiable at $z_0$ if $\lim_{z \to z_0} \frac{f(z) - f(z_0}{z - z_0}$ exists and is finite.  Implicit in this definition is the fact that complex differentiation is not defined at a single point $z_0$ but for all points in a open set $B(z_0, r)$ about $z_0$.  What I am saying is that a function cannot be complex differentiable at just one point $z_0$ without being complex differentiable in an open set (neighbourhood) about that point.  You stumbled upon a very good question that many students in complex analysis never think about.


Given a radius and velocity calculate position of an aircraft banking to make a turn I have a radius, R, for an aircraft traveling at velocity, V. If we start at point, (X,Y), what is the position of the point at time, t.

For example:
The aircraft is at point (0,0) and traveling at 250 knots and initiates a turn with a bank angle, phi, of 5 degrees. Assume that the aircraft can instantaneously rotate to the five degree bank. The equation for the turn radius, R where g is the acceleration due to gravity (9.81) is:
\begin{equation}
\text{R} = \frac{V^2}{\text{g} \tan{\phi}}
\end{equation}

For this example, R = 10.4 nautical miles. Where is the aircraft at t = 2 if the aircraft is traveling at a heading of 90 degrees (straight along the y axis)?

See banked turn in  aeronautics

In a motion along a circular path centered at $(0,0)$ (see figure below) with constant angular velocity $\omega$ the position of a point $P(x,y)$ (in Cartesian coordinates and $P(r,\varphi)$ in cylindrical coordinates) starting at $(r,0)$ at $t=0$ is given by the law of the uniform circular motion:

$$x=r\cos(\omega t)$$

$$y=r\sin(\omega t),$$

where $r$ is the radius of the circumference and $\omega=\dfrac{d\varphi}{dt}$. The magnitude of the tangential velocity is $v=\omega r$ and the radial velocity is zero. Thus $\omega=\dfrac{v}{r}$.

$$x=r\cos\left(\frac{vt}{r}\right)$$

$$y=r\sin\left(\frac{vt}{r}\right)$$

In your case $r=R$, $v=V$ (it is assumed that the aircraft mantains the speed $V$) and the centre of the circumference is $(X,Y)=(-R,0)$. If you make the change of coordinates $X=x-R$ and $Y=y$, you get the following $X,Y$ coordinates as a function of time $t$:

$$X=R\left(\cos\left(\frac{V}{R}t\right)-1\right)$$

$$Y=R\sin\left(\frac{V}{R}t\right)$$

In this $XY$-coordinate system the motion starts at $(X,Y)=(0,0)$ ($t=0$).

I assumed that the direction of the turn is to the left. If it is to the right, then one has a symmetric motion with respect to the $y-$axis:

$$X=R\left(-\cos\left(\frac{V}{R}t\right)+1\right)$$

$$Y=R\sin\left(\frac{V}{R}t\right).$$

The numerical result is obtained for $t=2$ and the other given data.

Added: For a different starting position $\left( X_{0},Y_{0}\right) $ at $t=0$ we have to incorporate these values in the motion equations. Integrating $%
\omega =\dfrac{d\varphi }{dt}$, we obtain $\varphi =\omega t+C$, where $C$ is
a constant. Then

$$x=r\cos \left( \omega t+C\right) ,y=r\sin \left( \omega t+C\right) $$

$$x_{0}=r\cos \left( C\right) ,y_{0}=r\sin \left( C\right) $$

and

$$\dfrac{y_{0}}{x_{0}}=\dfrac{\sin \left( C\right) }{\cos \left( C\right) }%
=\tan \left( C\right) ,C=\arctan \left( \dfrac{y_{0}}{x_{0}}\right) $$

$$x=r\cos \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right) \right)
,y=r\sin \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right) \right) .$$

Since

$$X=x-r=r\cos \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right)
\right) -r,Y=y=r\sin \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}%
\right) \right) $$

we get

$$X_{0}=x_{0}-r,Y_{0}=y_{0},\dfrac{y_{0}}{x_{0}}=\dfrac{Y_{0}}{X_{0}+r}$$

$$r=R,\omega =\dfrac{v}{r}=\dfrac{V}{R},\dfrac{y_{0}}{x_{0}}=\dfrac{Y_{0}}{X_{0}+R}.$$

Finally, we obtain the equations of the motion of the aircraft in the $X,Y$-plane:

$$X=x-R=R\left( \cos \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R} \right) \right) -1\right) $$

$$Y=y=R\sin \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R}\right) \right),$$

where $\left( X_{0},Y_{0}\right) $ is the position of the aircraft at $t=0$.

If the direction of the turn is to the right, then the motion is symmetric with respect to the $y-$axis:

$$X=-x+R=R\left( -\cos \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R} \right) \right) +1\right) $$

$$Y=y=R\sin \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R}\right) \right).$$

Note: I changed the notation; in the question the position of the aircraft
at $t=0$ is $\left( X,Y\right) $.

The velocity vector is $\overrightarrow{v}=v\overrightarrow{e}_{\varphi }$ and the acceleration vector $\overrightarrow{a}=-\dfrac{v^{2}}{r}\overrightarrow{e}_{r}$

Use the following:

$L$ = lift
$R$ = radius
$\theta$ = bank angle
$M$ = mass
$C$ = centrifugal force
$G$ = gravity

Then

$$L\cos(\theta)=MG\\L\sin(\theta)=C=\frac{MV^2}{R}$$

and so

$$\tan(\theta)=\cfrac{\frac{MV^2}{R}}{MG}=\frac{V^2}{RG}$$

Ergo

$$R=\frac{V^2}{\tan(\theta)G}$$


How can I prove that $\sqrt{2}$ is not inside Q($\sqrt{5 + \sqrt{5}}$)? I want to prove it, but although it is quite intuitive I don't know how to prove it mathematically. I have tried using the tower rule but I haven't got anything crear. Thanks in advance.

Another approach to the answer: first show that the Galois group of $\mathbb{Q}(\sqrt{5+\sqrt{5}})$ is isomorphic to $\mathbb{Z}/4\mathbb{Z}$. In particular $\mathbb{Q}(\sqrt{5})$ is the unique degree-2 subfield, by the Galois correspondence, since $\mathbb{Z}/4\mathbb{Z}$ only has one subgroup of order $2$.
So that reduces us to showing that $\mathbb{Q}(\sqrt{2}) \ne \mathbb{Q}(\sqrt{5})$. And this is true since $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{5})$ each have a unique (up to rational multiple) element of trace $0$, and $\sqrt{2}, \sqrt{5}$ can't be rational multiples of each other since $2/5$ isn't a square.
Still, it takes a computation to show that the Galois group is $\mathbb{Z}/4\mathbb{Z}$ (and that $\mathbb{Q}(\sqrt{5+\sqrt{5}})$ is a splitting field). One proof that the Galois group is cyclic is as follows: there must be an automorphism $\phi$ taking $\sqrt{5+\sqrt{5}}$ to its Galois conjugate $\sqrt{5-\sqrt{5}}$ (both are roots of $x^4 - 10x^2 + 20$, which is irreducible by Eisenstein with $p=5$). Squaring both sides of the equation $\phi(\sqrt{5+\sqrt{5}}) = \sqrt{5-\sqrt{5}}$ shows $\phi(\sqrt{5}) = -\sqrt{5}$. Next,
$$\sqrt{5+\sqrt{5}} \sqrt{5-\sqrt{5}} = \sqrt{20} = 2\sqrt{5},$$
so we see that $$\phi(\sqrt{5-\sqrt{5}}) = \phi(\frac{2\sqrt{5}}{\sqrt{5+\sqrt{5}}})=-2\sqrt{\frac{5}{5-\sqrt{5}}}=-\sqrt{5+\sqrt{5}}.$$
From this it follows that $\phi^2 = -\mathrm{id}$ (on roots of $f$), so $\phi$ has order $4$.

To expand on Michael's answer:
$$\sqrt 2 + \sqrt{5+\sqrt 5} = \sqrt{2+5+\sqrt 5 + \sqrt{10+2\sqrt 5}} =\sqrt{7+\sqrt 5 + \sqrt{10+2\sqrt 5}} $$
Which is pretty clearly degree 8(note that $\sqrt{10+\sqrt {20}}$ is undenestable because $10^2-20=80$ isn't a perfect square. see https://en.wikipedia.org/wiki/Nested_radical)


Let $A=(a_{ij})$ be a $n \times n$ matrix where $a_{ij} = \max(i,j)$. Find the determinant of $A$. Let $A=(a_{ij})$ be a $n \times n$ matrix where $a_{ij} = \max(i,j)$. Find the determinant of $A$. How to find the solution of this kind of problem?

Hint: For $2\le k\le n$ substract the $k$-th row from the $k-1$-row.

Subtract the 1st row from the 2nd row we have the new 2nd row to be :

1 0 0 ... 0.
Expand the determinant on this 2nd row we have it equal to:

-1*/A/ where A is the new minor of the form:

2 3 4 5 ... n

3 3 4 5.... n

4 4 4 5 ... n

Repeat step 1 by subtracting 1st row of A from 2nd row of A and expand along the 2nd row we get:

1*/B/ with B is the minor of A by deleting the 2nd row and 2nd column of A. Repeating this step on the way to the last step that we have: the final minor to be:

(n - 1)   n

n      n

this one has determinant = n(n - 1) - n*n  = -n. So the final answer is: -n if n = even, and n if n = odd.


List of interesting math podcasts? mathfactor is one I listen to.  Does anyone else have a recommendation?

I would like to draw attention to the beautifully produced podcast Relatively Prime hosted by Samuel Hansen, which was funded through a successful kickstarter campaign. All eight episodes are well worth a listen and include interviews with some very high profile and influential mathematicians.

There's a new maths podcast that you may be interested in that reviews "Maths at the Movies: Interstellar plus SPECIAL GUEST" in its first series. It's a slightly anarchic look at movies, mathematics, and life: www.mathsat.co.uk (direct link to podcast MP3 audio). There are some additional podcasts included as other blog entries.


Find the area of the circle with extern point Points $A, B, C, D$ are on a circle such that $AB = 10$ and $CD = 7$. If $AB$ and $CD$ are extended past $B$ and $C$, respectively, they meet at $P$ outside the circle. Given that $BP = 8$ and $\angle APD = 60^\circ$, ﬁnd the area of the circle.

I have no idea.. Could someone help? I can find the length of $PC$ by power of a point, and maybe use Ptomely's Theorem, but how can I find the area of the circle?

The first step is to add a point $H$ to the figure, located half way between the points $A$ and $B$. Note that the line segment $MH$ is perpendicular to the line through the points $A$, $H$, $B$ and $P$. So we can apply Pythagoras to the triangle $MHP$, yielding:

$$13^2 + (MH)^2 = (MP)^2$$

Similarly we apply Pythagoras to the smaller triangle $MHB$:

$$5^2 + (MH)^2 = R^2$$

Subtracting the second equation from the first gives an expression for the distance $(MP)$ between the centre $M$ and the point $P$ in terms of the radius $R$:

$$(MP)^2 = R^2 + 144$$

The second step is to apply the same procedure to the other side of the picture. We first define the point $I$, half way between the points $D$ and $C$. The angle between the line segment $MI$ and the line through $D$, $I$, $C$, $P$ is $90$ degrees. So we can apply Pythagoras to the triangles $MIP$ and $MIC$. In combination with the result already obtained for $(MP)$, we find that $[(PC) + 3.5]^2 - 3.5^2 = 144$ from which it follows that the length $(PC)$ is equal to $9$.

The third step is to derive a formula for the radius $R$ in terms of the other lengths. To do this we use the final bit of information, namely that the angle $APD$ is equal to $60$ degrees. Take the cosine of this angle, use the fact that angle $APD$ = angle $APM$ + angle $DPM$, and apply the well-known rule for the cosine of a sum of two angles. Substituting all our results so far, we get a somewhat messy equation. After tidying up we obtain:

$$R^4 + 71R^2 - 10512 = 0$$

This is a quadratic equation for $R^2$ which has the solution $R^2 = 73$. Therefore the surface area of the circle equals $\pi * 73$.

The clue given is: Circle and its area depend  only upon the power and included angle between any two secants or tangents that you are free to choose.

So we can decide to settle for tangents squared or power conveniently.

If power is $T^2$, radius R, half included angle $ \beta$, then

Area = $\pi R^2 = \pi (T \tan \beta)^2 $

In this case $ \beta = 30^{0}, T =\sqrt {PB\cdot PA} = \sqrt {80} = 4 \sqrt 5, $ Area = $ \pi \ 80/3. $

EDIT1:

There is an error about pair of secants' included angle considered. My main attempt however is to find Circle area in terms of its power the included angle of  secants only.. that I believe would be more instructive and elegant.


Can non-linear transformations be represented as Transformation Matrices? I just came back from an intense linear algebra lecture which showed that linear transformations could be represented by transformation matrices; with more generalization, it was later shown that affine transformations (linear + translation) could be represented by matrix multiplication as well.

This got me to thinking about all those other transformations I've picked up over the past years I've been studying mathematics.  For example, polar transformations -- transforming $x$ and $y$ to two new variables $r$ and $\theta$.

If you mapped $r$ to the $r$ axis and $\theta$ to the $y$ axis, you'd basically have a coordinate transformation.  A rather warped one, at that.

Is there a way to represent this using a transformation matrix?  I've tried fiddling around with the numbers but everything I've tried to work with has fallen apart quite embarrassingly.

More importantly, is there a way to, given a specific non-linear transformation, construct a transformation matrix from it?

As Harry says, you can't (the example of affine transformations can be tweaked to work because they're just linear ones with the origin translated). However, approximating a nonlinear function by a linear one is something we do all the time in calculus through the derivative, and is what we often have to do to make a mathematical model of some real-world phenomenon tractable.

No.  Everything is determined by a choice of basis.  For a more in-depth answer, I would need to explain the first two weeks of linear algebra and draw some commutative diagrams.

If you'd like a better explanation, see pages 12-14 of Emil Artin's monograph Geometric Algebra.


proving that this map is not an isometry between the surfaces Consider the parametrized surfaces
  $$\phi(u,v)= (u\cos(v),u\sin(v),\log(u))$$
  $$\psi(u,v)= (u\cos(v),u\sin(v),v)$$
  Prove that the Gaussian curvature equals in both surfaces in each $(u,v)$. And then prove that $\psi \circ \phi^{-1}$ is not an isometry (i.e  the differential is not an isomorphism).

I proved that the Gaussian curvature are the same for both surfaces, but I'm struggling with the computation of the differential (or maybe there is another way to do it).

Compute the corresponding first fundamental forms.


Is $\lim_{x \to a} \sqrt[K(x)]{L(x)},$ indeterminate, where $\lim_{x \to a} K(x)=\infty, \lim_{x \to a} L(x)=\infty$? Is $\lim_{x \to a} \sqrt[K(x)]{L(x)},$ indeterminate, where $\lim_{x \to a} K(x)=\infty, \lim_{x \to a} L(x)=\infty$?

I do not know how one would show that this is true or otherwise.

It depends.  Let K(x)=L(x)=x, with a infinite. The expression converges to 1. I presume one could make up example where it is indeterminate.


$i^i$ is real number. But $\ln(i^i)=i\cdot \ln(i)=\frac{i}{2}\ln(-1)$. But $\ln(-1)$ is not defined. $i^i$ is a real number. But,
$\ln(i^i)=i\cdot\ln(i)=\frac{i}{2}\ln(-1)$. But $\ln(-1)$ is not defined. So how can $i^i$ be a real number?

$$i^i=\left(\left|i\right|e^{\arg\left(i\right)i}\right)^i=\left(1e^{\arg\left(i\right)i}\right)^i=\left(1e^{\frac{\pi}{2}i}\right)^i=\left(e^{\frac{\pi}{2}i}\right)^i=e^{\frac{\pi}{2}i\cdot i}=e^{-\frac{\pi}{2}}$$

$$\ln\left(i^i\right)=\ln\left(e^{-\frac{\pi}{2}}\right)=-\frac{\pi}{2}$$

Because we know that $\ln(-1)=\pi i$


Are there positive integers $a$ and $b$ such that $a(a^2-1)=2b^2$? If $a$ and $b$ are positive integers, does a solution to the follwing equation exist?
$$a(a^2-1)=2b^2$$

I have tried graphing to no avail, can anybody help?

$a(a^2-1)=a(a-1)(a+1)$. The three factors are consecutive integers and hence all coprime with the possible exception of a single factor of $2$ if $(a-1)$ and $(a+1)$ are both even. In order for three coprime factors to multiply to $2b^2$, two of the factors must be squares (the third must be twice a square). But the factors differ from each other by either $1$ or $2$, and there are no integer squares other than $0,1$ that differ from each other by either $1$ or $2$. So one solution is: $a-1=0$, $a=1$, $a+1=2$ yielding $a(a^2-1)=a(a-1)(a+1)=0$ corresponding to $b=0$. There are no other integer solutions.

$b=±\frac{\sqrt{a(a^2 -1)}}{\sqrt 2}$

You just put the value of $a$ to get the value of $b$.
And you can put any value of $a$ such that $$a(a^2-1) \ge 0$$

That is when $a \in [-1,0] \cup [1,+\infty[$


Not understanding this division problem $\frac{52}{x}=13$

It says to next step

$\frac{52}{13}=x$

Ok, I can do future problems like this, but is there a rule that explains this? What just happened to both sides of the equal sign?

There are two steps here:
starting with
$$\frac{52}{x}=13$$
we multiply both sides by $x$ to get
$$\frac{52}{x}\cdot x=13\cdot x$$
$$52=13\cdot x$$
and then dividing both sides of that by $13$ to get
$$\frac{52}{13}=\frac{13\cdot x}{13}$$
$$\frac{52}{13}=x$$

I'd count several steps here.  Anyways...

We start with $$\left(\frac{52}{x}\right)=13$$.  Multiplying both sides by x we have $$\left(\left(\frac{52}{x}\right)*{x}\right)=(13*{x}).$$  Since $$\left(\frac{x}{1}\right)=x$$ we have  $$\left(\left(\frac{52}{x}\right)*\left(\frac{x}{1}\right)\right)=(13*{x}).$$  Applying the product rule for quotients, we have $$\left(\frac{(52*x)}{(x*1)}\right)=(13*{x}).$$  Since multiplication commutes, we have $$\left(\frac{(x*52)}{(x*1)}\right)=(13*{x}).$$  Applying the product rule for quotients again, we have  $$\left(\left(\frac{x}{x}\right)*\left(\frac{52}{1}\right)\right)=(13*{x}).$$  Since $$\left(\frac{x}{x}\right) =\left(\frac{1}{1}\right)$$ we obtain $$\left(\left(\frac{1}{1}\right)*\left(\frac{52}{1}\right)\right)=(13*{x}).$$  Applying the product rule for quotients once again we have  $$\left(\frac{(1*52)}{(1*1)}\right)=(13*{x}).$$  Since $(1*1)=1$, and $(1*52)=52$ we obtain  $$\left(\frac{52}{1}\right)=(13*{x}).$$  Now we multiply both sides by $$\left(\frac{1}{13}\right)$$ and obtain $$\left(\left(\frac{1}{13}\right)*\left(\frac{52}{1}\right)\right)=\left(\left(\frac{1}{13}\right)*\left(13*{x}\right)\right).$$  Since 13 equals $$\left(\frac{13}{1}\right)$$ we obtain $$\left(\left(\frac{1}{13}\right)*\left(\frac{52}{1}\right)\right)=\left(\frac{1}{13}\right)*\left(\left(\frac{13}{1})*{x}\right)\right).$$  Since * associates we have $$\left(\left(\frac{1}{13}\right)*\left(\frac{52}{1}\right)\right)=\left(\left(\left(\frac{1}{13}\right)*\left(\frac{13}{1}\right)\right)*{x}\right).$$  Applying the product rule for quotients on both sides we obtain $$\left(\frac{(1*52)}{(13*1)}\right)=\left(\left(\frac{(1*13)}{(13*1)}\right)*{x}\right).$$  Since $(1*52)=52$, $(13*1)=13$, and $(1*13)=13$, we obtain $$\left(\frac{52}{13}\right)=\left(\left(\frac{13}{13}\right)*{x}\right).$$  Since $$\left(\frac{13}{13}\right)=1$$ we obtain $$\left(\frac{52}{13}\right)=(1*x).$$  Since $(1*x)=x$, we obtain $$\left(\frac{52}{13}\right)=x$$

Even reasoning in this way steps have gotten left out, such as how we can "multiply both sides by x."


Do we have $(f^{-1}(\mathcal O_{\operatorname{Spec}A}))(\operatorname{Spec}B)=A$? Given a morphism of schemes $f:\operatorname{Spec}B\to\operatorname{Spec}A$, do we have $(f^{-1}(\mathcal O_{\operatorname{Spec}A}))(\operatorname{Spec}B)=A$?

No, suppose $A = k[x]$ and $B = k[x] / (x)$ for some field $k$, and $f$ corresponds to the quotient map $A \to B$.  Then $\Gamma(\operatorname{Spec} B, f^{-1}(\mathcal{O}_{\operatorname{Spec} A})) \simeq k[x]_{(x)}$.

I believe the following holds: If $f: X \rightarrow S$ is a map of schemes and if $p \neq q\in X$ are points with $f(p)=f(q):=s \in S$ you get a canonical isomorphism
$$f^{-1}(\mathcal{O}_S)_p \cong \mathcal{O}_{S,s}\cong f^{-1}(\mathcal{O}_S)_q.$$
Since $p \neq q$ we do not have an isomorphism
$$ \mathcal{O}_{X,p}\neq \mathcal{O}_{X,q}$$
in general. Hence $f^{-1}(\mathcal{O}_S) \neq \mathcal{O}_X$ in general.
Question: "Do we have $f^{-1}(Spec(A))(Spec(B))=B$?"
Answer: We have a canonical map
$$f(Spec(B)): f^{-1}(\mathcal{O}_{Spec(A)})(Spec(B)\rightarrow \mathcal{O}_{Spec(B)}(Spec(B))\cong B,$$
but since the sheaves $f^{-1}(\mathcal{O}_{Spec(A)})$ and $\mathcal{O}_{Spec(B)}$ differ the map $f(Spec(B))$ is not an isomorphism in general.
Note: If there is a spelling mistake in the question, and we replace $A$ with $B$
the answer is positive when the map is an open immersion:
There seems to be some confusion on an exercise on adjoint functors.
If you look in Hartshorne Chapter II.1 you will find for any map of topological space $f: X\rightarrow Y$ (exercise 1.18) an isomorphism
$$M1.\text{   }Hom_X(f^{-1}G, F)\cong Hom_Y(G, f_*F).$$
Hence if $F:=(\tilde{f}, f^{\#}): X \rightarrow S$ is a map of schemes there is canonically a map of sheaves of rings on $X$:
$$f: \tilde{f}^{-1}(\mathcal{O}_S)\rightarrow \mathcal{O}_X.$$
The map $f$ corresponds to the map
$$f^{\#}: \mathcal{O}_S \rightarrow \tilde{f}_*\mathcal{O}_X$$
via the isomorphism $M1$. If $X:=Spec(B), S:=Spec(A)$ it follows the map $f$ is a map of sheaves of rings on $X$, not on $S$. Hence it does not make any sense to evaulate $f$ at $S$, since $S$ is not an open set in $X$. It only makes sense to evaluate $f$ at $X:=Spec(B)$. Moreover $A$ is not a subring of $B$ hence you never recover $A$ from $\mathcal{O}_X(U)$ or $\mathcal{O}_X(X)$.
Example 1. Let $S:=Spec(A), X:=D(f)$ for $f\in A$. We get an inclusion of the basic open set $D(f)$ into $S$:
$$\tilde{i}: X:=D(f) \rightarrow S$$
and a canonical map
$$i: \tilde{i}^{-1}(\mathcal{O}_S) \rightarrow \mathcal{O}_{D(f)},$$
and the map $i$ is an isomorphism: You may check that $\tilde{i}^{-1}(\mathcal{O}_S)\cong \mathcal{O}_{D(f)}$. In this special case you get an isomorphism
$$ i(D(f)): \tilde{i}^{-1}(\mathcal{O}_S)(D(f)) \cong A_f \cong\mathcal{O}_{D(f)}(D(f)),$$
but you do not get an isomorphism in general.
The natural question to ask is: for a general map of rings $\phi: A \rightarrow B,$
Question: Is the canonical map
$$f(X):\tilde{f}^{-1}(\mathcal{O}_S)(X) \rightarrow \mathcal{O}_X(X) \cong B$$
an isomorphism?"
Answer: The answer is "no", by the argument given in the previous post.
Note: if you take the "pull back"
$$F^*(\mathcal{O}_S)$$
you get
$$F^*(\mathcal{O}_S):= \mathcal{O}_X \otimes_{\tilde{f}^{-1}(\mathcal{O}_S)} \tilde{f}^{-1}(\mathcal{O}_S) \cong \mathcal{O}_X$$
hence
$$F^*(\mathcal{O}_X)(X)\cong \mathcal{O}_X(X) \cong B.$$
To people reacting negatively to this, denouncing it as "vandalism": Do exercise II.1.18 in HH and consult the definition of the notion "morphism of schemes" in Chapter II in HH.


Find the area of the shaded region in the figure Find the area of the shaded region in the figure

What steps should I do? I tried following the steps listed here https://answers.yahoo.com/question/index?qid=20100305030526AAef8nZ

But I got 150.7 which is wrong.

Hint: Break the shaded region up into two shapes. One is a portion of the circle (you know the portion because of the given angle), and the other is a triangle (which is equilateral). Find the area of each shape and then add them.

Notice, the area of the equilateral triangle (indicated by red) $$=\frac{1}{2}(12)(12)\sin\frac{\pi}{3}$$ $$=72\frac{\sqrt{3}}{2}= 36\sqrt{3}$$  The area of the major circular sector subtending at the center an angle $2\pi-\frac{\pi}{3}=\frac{5\pi}{3}$ $$=\frac{1}{2}\left(\frac{5\pi}{3}\right)(12)^2=120\pi$$ Hence, the area of the shaded portion $$=\text{area of (red) triangle}+\text{area of major sector}$$ $$=\color{blue}{36\sqrt{3}+120\pi\approx 439.34494 \ }$$


Find the equations of the two circles which pass through the point $(2,0)$ and have both the $y$-axis and the line $y-1=0$ as tangents I have this question:

  Find the equations of the two circles which pass through the point $(2,0)$ and have both the $y$-axis and the line $y-1=0$ as tangents.

By plotting $y=1$ and $(2,0)$, it's obvious that the smaller circle has a radius of 1 and therefore an equation of $(x-1)^2 + y^2 = 1$. When you look at the larger one, it seems to touch these 3 co-ordinates: $(r,1)$, $(2,0)$, $(0, 1-r)$ where $r$ is its radius.

The larger circle's equation is apparently $(x-5)^2+(y+4)^2=25$. How do I go about finding this algebraically?

Let the equation of the circle be $(x-a)^2+(y-b)^2=r^2$  where $r\ge0$

Now the distance of any tangent from the centre of the circle = radius

For $y$ axis $x=0,r=\dfrac{|a|}{\sqrt{1^2+0^2}}=|a|\implies a=\pm r$

For $y-1=0, r=\dfrac{|b-1|}{\sqrt{1^2+0^2}}=|b-1|\implies r^2=(b-1)^2$

Case $\#1:$ If $b-1\ge0, r=b-1\iff b=r+1$

So, we have $(x\pm r)^2+\{y-(r+1)\}^2=r^2$

Now find $r(\ge0)$ using the fact : the circle passes through $(2,0)$

Case $\#2:$

If $b-1<0, r=-(b-1)=1-b\iff b=1-r$

Like $\#1$

Consider center (a,b)

Step 1) The circle touches y axis, giving the radius r= ± a . So the equation
         will be :    ......(X-a)² +(Y-b)²=a².......

Step 2) The circle touches line y=1, so the resultant equation after
         substituting y=1 will have one solution. (Think in a way that we want
         to find the intersection points between the circle and the line y=1).
          (x-a)²+ (1-b)²=a² will have one solution if and only if......
          (1-b)²=a².....

Step 3) Substitute point (2,0):  (2-a)²+(0-b)²=a²

Step 4) Rearrange:  (1-b)²=a² and (2-a)²+b²=a² . Solve the system. (a,b,r)=(1,0,1) or (5,-4, 5)


Sets of Prime Numbers Generated By an Irreducible Monic Polynomial Given a non-constant integral irreducible monic polynomial $f(x)$, the prime factors of its value at integers $x\in\mathbb{N}$ forms a set $\mathcal{P}(f)$.

Is it possible that $\mathcal{P}(f)\cap\mathcal{P}(g)=\emptyset$ for two such polynomials $f$ and $g$.

The answer is no. See Theorem 3 in Prime Numbers in Certain Arithmetic Progressions:

  If $f,g \in \mathbb{Z}[x]$ are non-constant, then $P(f) \cap P(g)$ is infinite.

The proof there is quite short so I might as well include it:

  Suppose $\alpha$ is a root of $f$ and $\beta$ is a root of $g$. By Dedekind 's theorem, except for a finite number of exceptions, $p$ is a prime divisor of $f$ exactly when $p$ has a first degree prime ideal factor in the field $\mathbb{Q}(\alpha)$. A similar statement holds for $g$. Consider the field, $\mathbb{Q}(\alpha, \beta)=\mathbb{Q}(\theta)$ for some $\theta \in \mathcal{O}_K$. If $h$ is the minimal polynomial of $\theta$ then Schur's Theorem assures us that there are infinitely many primes which have a first degree prime ideal factor in $\mathbb{Q}(\alpha, \beta)$. These primes thus have a first degree prime ideal factor in both $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$. Thus, except for a finite number of exceptions, these primes are in both $P(f)$ and $P(g)$.

Let $f(y)=y^2-2$, and let $g(x)=x^3+5$, then this works, because if they had a solution in common you'd have a solution for $y^2=x^3+7$, but this is known not to have any integer solutions. See Silverman's Arithmetic of Elliptic curves page 296.


Prove that $\forall a\in \mathbb{R\smallsetminus Q}$, there exist infinitely many $n\in \mathbb N$ such that $\lfloor{an^2}\rfloor$ is even. Question: Prove that  $\forall a\in \mathbb{R\smallsetminus Q}$, there exist infinitely many $n\in \mathbb N$ such that $\lfloor{an^2}\rfloor$ is even.

If $a=\sqrt2$ and $x^2-2y^2=1(x,y\in\mathbb N)$ then $x-\sqrt2y=\dfrac{1}{x+\sqrt2y},$
$$9xy-9\sqrt{2}y^2=\dfrac{9y}{x+\sqrt2y}\in(3,4)$$
so $\lfloor{(3y)^2\sqrt2}\rfloor=9xy-4$ is even since $y$ is even.

If $k\in\mathbb N,\sqrt k \notin \mathbb N$ then we can prove it for $a=u+v\sqrt k,u,v\in\mathbb Q$ as above.

I know that if $a$ is an counterexample and the continued fraction representations of$a=[a_0; a_1,a_2,....]$ then $a_{k},a_{k+2},a_{k+4},\cdots$ must be all even for some $k\in \mathbb N$.

We use the following lemma, which can be proved by using van der Corput's theorem:

  If $\alpha$ is an irrational number, then the sequence $\alpha n^2$ is equidistributed modulo 1.

Assume $\alpha > 0$. If there is only finitely many $n$ such that $\lfloor \alpha n^2 \rfloor$ is even. Then for sufficiently large $n$, we can write $$\alpha n^2 = 2N_n+1+r_n$$
where $0<r_n <1$ and $N_n$ is an integer.
Hence
$$\frac{1}{2} < \left\{\frac{\alpha}{2}n^2 \right\} = \frac{1+r_n}{2} < 1 $$

$\{x \}$ means the fractional part of $x$. Thus, the sequence $\dfrac{\alpha}{2}n^2$ is not equidistributed modulo 1, a contradiction.

In the same way, we can prove there are infinitely many $n$ such that $\lfloor \alpha n^2 \rfloor$ is odd.

Too long for a comment:

Assume w.l.o.g. that $a<1$ and write $a=0.a_1 a_2a_3\ldots$. If there are infinitely many $i$ such that $a_{2i}=0$, then we are done, because we can take the corresponding even powers of 2 as the squares. Otherwise, there is a $k$ such that $2^{2k}a$ has only ones in its binary rep after the point, which means that the fractional part of $2^{2k}a$ is at least $1/3$.

Thus, it is probably sufficient to show that there are infinitely many $n$ such that the fractional part of $n^2a$ is less than 1/3.


Solution Form of a Ridge Regression Problem The linear ridge regression loss function:
$$
J(\beta)=\Sigma_{i=1}^n(x_i^T\beta-y_i)^2+\lambda\Sigma_{j=1}^p\beta_j^2= \Vert X\beta-Y \Vert^2 + \lambda\beta^TI\beta \text{ (matrix form)}
$$
where $x_i$'s are the input vectors, $y_i$'s are the outputs (observations), $\beta$ is the vector of coefficients, and $\beta_j$'s are the elemenents of $\beta$, has the solution:
$$
\hat{\beta}=(X^TX+\lambda I)^{-1}X^TY
$$

On the other hand, in my textbook, it is said that by setting the derivative of $J(\beta)$ to $0$, we can obtain the solution $\hat{\beta}$ of the form:
$$
\hat{\beta}=\Sigma_{i=1}^n \alpha_ix_i \tag{*}
$$
where:
$$
\alpha_i=\frac{-1}{\lambda}(x_i^T\beta-y_i)
$$

How do we obtain (*)?

Since $J(\beta) = \frac{1}{2}\sum_{i=1}^n(x_i^T\beta-y_i)^2 + \frac{1}{2}\lambda\beta^T\beta$, we have $\nabla J(\beta)=\sum_{i=1}^n(x_i^T\beta-y)x_i + \lambda \beta$. Thus setting $\nabla J(\beta)$ to zero gives
$\sum_{i=1}^n(x_i^T\beta-y)x_i + \lambda \beta = 0$, that is $\beta = -\frac{1}{\lambda}\sum_{i=1}^n(x_i^T\beta-y)x_i$, a fixed-point equation for the coefficients $\beta$.


Particular solution for $y'' + y' - 2y = 3xe^x$ For the equation $y'' + y' - 2y = 3xe^x.$

the auxiliary equation $\lambda^2+\lambda-2 = 0$ and $\lambda_1= 1$ and $\lambda_2 =-2$

so generic solution is $$C_1e^x+C_2e^{-2x}$$

am I right that particular solution will be:

$$x(Ax+B)e^x \rightarrow (Ax^2+Bx)e^x$$

After opening the brackets I got $Ax^2$ but my book has another answer, so I am a bit confused about how to act in this cases

First your generic solution (which you should call the homogeneous solution) is correct.To find the particular solution you should use the method of undetermined coefficients next.The complete solution is $y=y_h+y_p$. Where $y_h$ is the homogeneous solution and $y_p$ is the particular solution.

Your guess for the particular solution should be of the form $$A x e^x + Bx^2 e^x  $$ where $A$ and $B$ are constants to be determined (That is your guess is correct). You now need to find
\begin{align}
\frac{d}{d x}A x e^x + Bx^2 e^x \\
\frac{d}{d x^2}A x e^x + Bx^2 e^x \\
\end{align}
and make the appropriate substitutions. After simplifying you should get $$(3 A+2B) e^x+6Be^x x=3 e^x x. $$
From which you can determine A and B.

make for the particular solution the ansatz $$y_P=(Ax^2+Bx+C)\exp(x)$$
i think your idea is good
for your Control the solution is given by $$y(x) ={{\rm e}^{x}}{\it \_C2}+{{\rm e}^{-2\,x}}{\it \_C1
}+1/6\, \left( 3\,x-2 \right) x{{\rm e}^{x}}
$$


Do commutative matrices share the same eigenvectors? Let two square matrices $A$ and $B$ represent linear operators on a vector space $V$ over $\mathbb{C}$. Suppose they are commutative. Then

$$ABx=BAx,\,\,\forall x\in V$$

Then let $\tilde{x}$ be an eigenvector of $B$. Setting, $x=\tilde{x}$, we see that

$$\begin{align}
AB\tilde{x}&=BA\tilde{x}\\
A\lambda \tilde{x} &= BA\tilde{x},\,\,\exists \lambda\in\mathbb{C}\\
\lambda (A\tilde{x}) &= B(A\tilde{x})\\
&\\
\implies A\tilde{x}&=\alpha\tilde{x},\,\,\exists \alpha\in\mathbb{C}
\end{align}$$

So every eigenvector of $B$ is also an eigenvector of $A$. $\blacksquare$

MY QUESTIONS: Is this valid? If it isn't, what have I done wrong? If it is, does this mean that the two matrices are similar by the matrix formed by making each column an eigenvector of the two matrices?

It's obviously false. The identity matrix commutes with every other matrix, but any nonzero vector is an eigenvector for the identity, which can be not true for a non identity matrix.

Yes, it is true. May be $P$ the Matrix of all eigenvectors and $diag([\lambda])$ the diagonal Matrix of all Eigenvalues $\lambda$. Then $A=P^Tdiag([\alpha])P$ and $B=P^Tdiag([\lambda])P$. These matrices are similar; only the Eigenvalue spectrum can differ.


You can add more coefficients in the Taylor expansion
$$\log(1-x)\leq-x-x^2/2+...$$
in order to obtain the sharper bound
$$b_{n}+(1-b_{n})\ln{(1-b_{n})}<b^2_{n}/2+b_n^3/2$$
and so on, if it is necessary.


Total derivative notation help consider the function

$$f = f(x(t),y(t))$$

I know that its total derivative wrt t is

$$\frac {df}{dt} = \frac {\partial f} {\partial x} \frac {dx}{dt} + \frac {\partial f}{\partial y} \frac {dy}{dt}$$

and that the total derivative wrt x is
$$ \frac {df} {dx} = \frac {\partial f} {\partial x} \frac {dx}{dx} + \frac {\partial f} {\partial y} \frac {dy} {dx}$$

However I am not fully familiar with the notation and the forms in which it takes during more extreme conditions, such as the following, could anyone fill the blanks in for me?

1)$$f = f(x(t,w),y(t,w))$$
$$\frac {df}{dt} = ?$$
2)
$$f = f(x(g(t,w)),y(g(t,w)))$$
$$\frac {df}{dt} = ?$$
3)
$$f = f(x(g(t,w),z),y(g(t,w),z))$$
$$\frac {df}{dt} = ?$$
4)
$$f = f(x(g(t)),y(g(t)))$$
$$\frac {df}{dt} = ?$$

how would I go about writing these properly? Any answer will help illustrate the semantics in much greater detail

I shall write for the most extreme case (3)

$$\frac{df}{dt} =\frac{\partial f}{\partial x}\frac{\partial x}{\partial g}\frac{dg}{dt}+\frac{\partial f}{\partial x}\frac{\partial x}{\partial g}\frac{\partial g}{\partial w} \frac{dw}{dt}+\frac{\partial f}{\partial x}\frac{\partial x}{\partial z}\frac{dz}{dt}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial g}\frac{dg}{dt}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial g}\frac{\partial g}{\partial w} \frac{dw}{dt}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial z}\frac{dz}{dt}$$

A good rule of thumb could be imagining this as an expanding tree.

In general if you have the function $H(t)= F(G(t))$ and you want to find the derivative of $h$ with respect to $t$. You have the general rule:
$$\frac{d H}{dt} = DF(G(t))(\frac{d G}{dt}).$$ It seems to be strange notation with respect to your level, but in this case you can take,for example, $$G(t)= (x(t),y(t))$$ and use $$Df((x,y))((z_1,z_2))= \frac{\partial f}{\partial x}z_1+ \frac{\partial f}{\partial y}z_2.$$


Is there a Math symbol that means "associated" I am looking for a Math symbol that means "associated" and I don't mean "associated" as something as complicated as isomorphism or anything super fancy.

I am looking for a symbol that means something like "$\triangle ABC$ [insert symbol] $A_{1}$" (as in triangle ABC "associated" with area_{1}) Or want to say something like "The eigenvector associated with the eigenvalue"

You get the idea.

If you want to use such a symbol for association in some sense, you would have to define its meaning precisely first in your exposition. I suggest using $\sim$ for your purpose.

You can always make up a symbol (I've done so on several occasions), but, like user4594 said, whenever you first use it in a work, you need to always precisely define it. If you find yourself needing to use the word association or correlation a lot, and you don't want to use existing notation (e.g. corr(A,B)), then you can insert a symbol and either make a footnote that defines the symbol or include a legend/key at the beginning or end of what you are creating.
Using a symbol for every relationship between arguments often unnecessarily burdens your reader, but sometimes, I find that it helps me remember things better if I can assign symbols to certain words, especially if I end up using the words repeatedly in a single work.
Here is an example of a group of symbols I have created (not designed, but assigned):
"⌽A, ⦵B, ⍉C, ⦵D"
...meaning:
"If A, then B, else if C, then D."
I know that "A⟶B" ("If A, then B") is more commonly used, but in my particular document, a ⟶ meant something else entirely.
If you want to make a novel symbol but don't want to personally design one (font editors can get really complicated), there are some fonts you can download that consist mostly or entirely of symbols (many of which you will probably have never seen before). Some examples, at least in Windows (I don't own a Mac or Linux machine), are Cambria Math, Wingdings (including 2 and 3), Segoe UI Symbol is quite extensive (it even has symbols of playing cards and dominoes among other things), as is Segoe UI Emoji, and Symbol. The font Segoe UI Historic has a lot of random characters that were probably part of some ancient languages, like    (click here if you can't see them). There are plenty of options to choose from. If you use an obscure symbol, and you want to send the document to someone else, either send them the font file you used (impractical) or save the document as a *.pdf or image file, because if the other person doesn't have the font installed on their device, they will probably just see boxes.
Usually, though, I only do things like that when I'm making notes for myself (I use plenty of custom shorthands when I take notes). Neuropsychological research has suggested that the human mind has a rough limit, reported as anywhere from 3-7 distinct items, to what can be held in working memory at a time (Cowan, 2010), so it may be counterproductive to use too many symbols, as your ideas will be lost to your readers while they try to keep all the symbols straight. It's probably a good idea to apply the adage: when in doubt, just write it out.


Determining n in sigma ($\Sigma_{x=0}^n$) Reffered here by https://mathoverflow.net/questions/41750/determining-n-in-sigma-x0n

I'm not entirely sure if this question falls under MathOverflow but neither of my Calculus AP  teachers in high school could help me with this:

Given $\Sigma_{x=0}^n {f(x)\over2}$ and the output of the summation, how would you find $n$? I've learned how to determine the $n$ given an arithmetic or geometric sequence, but not for an arbitrary function.

Specifically, when $f(x) = 40 + 6\sqrt{x}$.

12 Oct 2010. Edit:
It seems like I need to explain the entire situation for finding $n$, the number of trapezoids, for trapezoidal rule. It started on a simple review question for Calc AP and a TI-83 program that my calc teacher gave to me to solve the definite integral with trapezoidal rules. Aiming to major in Computer Science, I took it a bit further and completely took apart the program resulting in my original question on StackOverflow: https://stackoverflow.com/questions/3886899/determining-the-input-of-a-function-given-an-output-calculus-involved

Since there were tumbleweeds for a response, I took it as a personal challenge to reverse engineer the trapezoidal program into an algebraic form with my notes found on my forum: http://www.zerozaku.com/viewtopic.php?f=19&t=6041

After reverse engineering the code into some algebra, I derived the formula:
$$TrapRule(A, B, N) = {(B-A)\over N}({F(A)\over2}+\sum_{k=1}^NF(k)+{F(B)\over2})$$

Given the values of A and B are constant for the definite integral, I should be able to isolate and solve for $N$. The problem, however, was determining $N$ in $sum_{k=1}^N$ and I came to the conclusion that it was an issue that I called recursive complexity because it was impossible to determine without recursively adding for the summation.

Eventually, I found MathOverflow and they referred me here. I was hoping only to get help on the issue for a summation because its beyond my skill as a high school student. Now that others have proposed other solutions for my dilemma, I guess I can throw out my thesis Dx

Thanks for the help though, I'll definitely be returning for more.

You are trying to calculate the inverse of the function
$$S(n)=\sum_{x=0}^n f(x)/2$$

In general the inverse would not exist (think $f(x)=0$).

If the inverse exists, a trivial way to solve your problem would be to start calculating the sum for $n=0,1,2,\ldots$ and stop when the result matches the number provided.

If you are looking for a formula for $S^{-1}(n)$, there is no reason for there to be such a formula for every $f(x)$. The arithmetic and geometric progression examples you gave were lucky exceptions.

If you don't require trapezoids of the same width, you can get a closed form by taking a partition of $[0,n]$ into intervals with endpoints  $n,nq,nq^2,\dots nq^k$ for $q<1$ and $k$ large (or infinite), or by using $(i/n)^2, i=0 \dots n$ as the endpoints.  The first approach leads to the sum of a geometric series and the second is the sum of values of a polynomial.

Better answers are possible if you specify what is the problem you are ultimately trying to solve.  Is it finding the number of equal-width trapezoids needed to achieve a given accuracy in the integral $\int_0^n f(x) dx$?


Derivative of squared norm of a complex function Assuming $f(x)$ is a complex function, its squared norm is defined as $$|f(x)|^2 = f(x) . f^*(x)$$

What is derivate of $|f(x)| ^ 2$?

For a complex variable a, |a|^2 derivative exist only at zero. Try evaluating at x_0+i*y_0 not zero. Approach from y=y_0 and you get 2*x_0 but approach from x=x_0 and you get -2*y_0*i. And at zero you get limit r^2 over re^(thetai) that go to zero.


Functions as relations I´ve got just a short question. Is this relation a function?

$R = \{(a,b) \in N \times N | 3b \leq a < 3b+3\}$

In my opinion it is, because every a in N has got exactly one b in N. But I´m really not sure. Can an inequality be a function?

Thank you for your help =)

Yes, this relation is indeed a function. That inequality can be written in the following way:
$$3b \leq a <3 (b+1) $$

And it is easy to see that each $a \in \mathbb{N}$ is always between a certain pair of consecutive multiples of $3$. So, for each $a $ there is exactly one $b $ that "matches" with it.

You could write it as:
$$f: \mathbb{N} \rightarrow \mathbb{N}$$
$$b=f(a)=\lfloor \frac{a}{3} \rfloor $$

(Where $\lfloor x \rfloor $ is the floor function, that gives the biggest integer less or equal to $x $.)

By the way, you cannot write it in the form $a=g(b) $ because there is more than one $a$ for each $b $; for example, $f(1)=0$ and $f(2)=0$.

Note: I assumed that $0 \in \mathbb{N}$. If you don't use that convention, you have to exclude ${0,1,2} $ from the domain of the function.

No it is not a function. A function is of the form f(x), where f is a function and x is a parameter. What you presented is not a function. It would have to be presented differently and stated in terms of a function with parameters.


Impossibility of proving a foundation to be consistent An argument came to my mind that seems really mind-blowing and I haven't found it anywhere. Here's how it goes:

We call a formal system F embodied in classical logic a foundation of mathematics when each and every statement of mathematics is translatable into its language and interpretable accordingly.

By Gödel's second incompleteness theorem, we know that any foundation of mathematics cannot prove its own consistency because it necessarily should include arithmetic.

If we came up with a mathematical proof of the consistency of F, then by definition of F we would be able to mimic that proof inside F, contradicting Gödel's second theorem, proving F to be inconsistent, so F is useless.

In other words, if F is consistent, not only are we not able to prove its own consistency inside F, we are not able to prove its consistency using tools beyond it.

Therefore, if a foundation F really is absolutely consistent, there is no way to know it is.

Is this a good argument?

I think the argument leaves out some other methods of proving the consistency of F, methods that are not at all mathematical. But knowing that F is itself a mathematical object, what else can we use to study it?

This kind of question is dicsussed and (for me) well answered into:

Torkel Franzén, Gödel's theorem: An incomplete guide to its use and abuse (2005), Ch.5.3 Consistency Proofs, page 107-on:

  it is a simple matter to prove $\mathsf {PA}$ consistent using only ordinary mathematics. [...] We say that a theory $T$ is arithmetically sound if every arithmetical theorem of $T$ is true. An arithmetically sound theory is consistent, since an inconsistent theory proves the false arithmetical statement $0=1$. [...] to prove that  $\mathsf {PA}$ is arithmetically sound  [...]
  we need to define "true arithmetical sentence," then we need to show that
  the axioms of $\mathsf {PA}$ are all true arithmetical sentences and that the rules of reasoning of $\mathsf {PA}$ lead from true premises to true conclusions. Inevitably, this involves some logical and mathematical formalities.

  the proof that the axioms of $\mathsf {PA}$ are true and the rules of
  reasoning of $\mathsf {PA}$ lead from true statements to true statements uses just the same axioms and rules of reasoning as those embodied in $\mathsf {PA}$, plus a little bit of set theory or some mathematics at a comparable degree of
  abstraction. The proof is sometimes said to be carried out in $\mathsf {ZFC}$, but logically speaking this is enormous overkill. Only a very much weaker set
  theory is needed to carry out the proof, specifically a fragment of $\mathsf {ZFC}$ known as $\mathsf {ACA}$.

  So what we have here is a mathematical proof, formalizable in the weak
  set theory $\mathsf {ACA}$ [...], using ordinary mathematical principles, that $\mathsf {PA}$ is consistent. A common objection to this description of what has been achieved is that the proof is really no proof of consistency, for $\mathsf {ACA}$ is logically stronger. So all we can say that the proof shows is the consistency assuming the consistency of $\mathsf {ACA}$. In other words, the proof shows that if $\mathsf {ACA}$ is consistent, then $\mathsf {PA}$ is consistent. This whole line of thought is predicated on the assumption that we
  have doubts about the consistency of $\mathsf {PA}$ and are trying to allay those doubts by means of a consistency proof. But when we regard the axioms
  and principles formalized in $\mathsf {PA}$ and $\mathsf {ACA}$ as straightforwardly part of our mathematical knowledge, the soundness proof for $\mathsf {PA}$ is not intended to allay any doubts at all. It is quite
  simply an essentially trivial proof of a basic result in logic.

  There is, therefore, no basis in Gödel's theorem for the idea that a consistency proof is not a proof in exactly the same sense as any other mathematical proof is a proof. Every mathematical proof is based on certain basic axioms and rules of reasoning. A consistency proof such as the one sketched by no means yields a justification of the axioms and rules of reasoning formalized in $\mathsf {PA}$. It is just a proof of an arithmetical statement, a proof which establishes the statement as true in the same way and in the same sense of "establish" as do other proofs of arithmetical statements using those same axioms and rules of reasoning.

In a nutshell, every "absolute" foundational system F will be obviously open to some sort of "skeptical challenge": any conceivable system will be based on some assumptions, and it is always possible to ask for the justification of these assumptions.

Regarding an "absolute" consistency proof of $\mathsf {PA}$, the natural numbers sequence $\mathbb N$ is a model of it and is a "standard" result of logic that a theory having a model is consistent.

I'd say it's not really true that we can "never know" the system is consistent. For a first example, Gentzen gave a consistency proof for PA by using a principle which PA can't formalize but which many/most accept as true (transfinite induction up to the countable ordinal $\varepsilon_0$).

Of course, PA isn't a foundation for all, or even most, of mathematics, so it's not surprising that it would fail to capture some or many aspects of mathematical practice. ZFC is a better example. Is it impossible to really know it's consistent? Assuming ZFC is consistent, a proof of its consistency can't be formalized (mimicked) within ZFC, so the proof would have to rely on a new principle or axiom. One such axiom is the existence of an inaccessible cardinal: if $\kappa$ is such a cardinal then $V_{\kappa}$ is a (set) model of ZFC, so ZFC is consistent. However, one may ask how certain we are that inaccessibles exist — surely that's less certain than the consistency of ZFC. Even simpler possible axioms can be cited, but they beg the question: $\operatorname{Con}(ZFC)$, for example.

I have to conclude that the jury remains "out" on the answer to your question. "Certainty" concerning the consistency of a formal system that provides a foundation for mathematics would require some principle not captured or proven by the system, and in which the mathematical community would have great confidence. I'm not sure that such a principle has presented itself so far.


Why does the Dominated Convergence Theorem fail for evaluating $\lim_{n \to \infty} \ n^2 \int_{0}^{1} xe^{-n^2x^2}dx$? I read in a book that to try to use the dominated convergence theorem for:

$$
\lim_{n \to \infty} \  n^2 \int_{0}^{1} xe^{-n^2x^2}dx
$$

fails. The reasoning was written to be that $nxe^{-n^2x^2} \to 0$ as $n \to \infty$ for all $x\in [0,1]$, so that we cannot interchange integral and limit. However, I fail to see why the Dominated Convergence theorem fails exactly in this case. Would anyone have any ideas? Thanks!

Because $\sup_{n\ge 1} n^2 x e^{-n^2x^2}={1\over x\cdot e}$ for $x\in(0,1]$.

To show that the conclusion of the dominated convergence theorem does not hold, assume it does i.e. assume there exists some dominating function. Then show that there is a contradiction.

Related exercise in Probability w/ Martingales:

Apply the same logic to $g:= \sup_n |n^2xe^{-n^2x^2}|$.

If we suppose g is integrable, then by the dominated convergence theorem, we have

$$0 = \int_{0}^{1} \lim_{n \to \infty} f_n dx  = \lim_{n \to \infty} \int_{0}^{1} f_n dx = 1/2$$

The reason why we get the supremum of the integrand ie $g: =\sup_n |n^2xe^{-n^2x^2}|$ is because DCT assumes an integrable g exists s.t. $|f_n| \le g$. The smallest possible g is  $\sup_n |f_n|$ so if $\sup_n |f_n|$ doesn't work, by monotonicity of integration, any $g \ge \sup_n |f_n|$ won't work and hence DCT does not apply


Can we know the region in the xy-plane given this information? Consider a closed smooth bounded curve enclosing a region S in the XY-plane. Suppose that we know the length of all lines which is the intersection of S with an infinite line parallell to the x-axis. And we know the same for the y-axis. Can we reconstruct S using only this data?
If not, suppose we know the same for some other axes, is there a finite set of axis enough to reconstruct S?

No it is not possible to reconstruct such a shape. For instance, if you take a rectangle $A$ for which the sides form a $45°$ angle with the $XY$-axis, and rotate it by $90°$ to obtain a rectangle $B$, then the intersection of any line parallel to the x or y axis with $A$ will have the same length than the intersection with $B$.

For the generalization, I have no idea.

We know the function $f(x)$, where $x$ is a point on the $x$ axis, defined as the length of the intersection of, the line paralell to the $y$-axis which goes through $x$, with the set $S$, so it can have multiple components and we know the total length. I am owner of question


Find a differentiable $f$ such that $\mathrm{Zeros}(f)=\mathbb{any\; closed \;set}$ Let $B\subset \mathbb{R}^2$ be a closed set.

How to prove that there is a differentiable function $f:\mathbb{R}^2\longrightarrow\mathbb{R}$ such that $$Z(f)=B$$

where $$Z(f)=\{x\in\mathbb{R}^2:f(x)=0\}$$

Any hints would be appreciated.

I will adapt the  MO answer by Harald Hanche-Olsen, filling in some details, and taking into account that you don't ask for $C^\infty$, but only for a differentiable function.

Let
$$E_0=\{x\colon\operatorname{dist}(x,B)\ge 1 \}$$
and for $k=1,2,\ldots$ let
$$E_k=\{x\colon 2^{-k}\le \operatorname{dist}(x,B)\le  2^{1-k}\}$$
These sets cover the complement of $B$. Also let
$$F_k=\{x\colon  \operatorname{dist}(x,E_k)\le  2^{-k-2}\}$$
be an "enlargement" of $E_k$ which still stays away from $B$.

Let $\omega$ be a smooth function on $\mathbb R^2$ such that $\omega\ge 0$, $\omega(x)=0$ when $|x|\ge 1/2$, and $\omega(x)>0$ when $|x|\le 1/4$. Let $\omega_k(x) = \omega( 2^{ k}x)$.

The convolution of $\chi_{F_k}$ with $\omega_k$ has the following properties:

it is as smooth as $\omega$ is
it is nonnegative
it is zero on the set $ \{x\colon  \operatorname{dist}(x,F_k) >  2^{-k-1}\}$, which contains the set $ \{x\colon  \operatorname{dist}(x,B ) < 2^{-k-2}\}$
it is strictly positive on $E_k$
it does not exceed $4^{-k}\int_{\mathbb R^n} \omega$.

Define
$$f=\sum_{k=0}^\infty  (\chi_{F_k}*\omega_k) \tag{1}$$
and observe that

$f$ is strictly positive on the complement of $B$, and vanishes   on $B$.
$f$ satisfies an estimate of the form $f(x)\le C(\operatorname{dist}(x,B ))^2$, because at distance about $2^{-k}$ from $B$
it takes values about $4^{-k}$
By the above, $f$ is differentiable on $B$.
$f$ is also differentiable on the complement of $B$, because every point of this complement has a neighborhood in
which only finitely many terms of (1) are nonzero.

As Harald Hanche-Olsen notes, introducing a rapidly decaying weight one can make the sum $C^\infty$ smooth, e.g.,
$$f=\sum_{k=0}^\infty  2^{-k^2} (\chi_{F_k}*\omega_k )$$

Would something like this work?

Let $B \subset \mathbb{R}^2$ be a closed set. Define a function $f$
$$ f(x) = \text{dist}(x,B)$$
where
$$ \text{dist}(x,B) = \inf_{b\in B} d(x,b)  $$


Find the missing number Find the missing numbers in the bottom middle circle. (Clue: left halves of the
  central circles relate to the left circles and the right halves to the right circles)

circles are attached in the image

I am not able to find the numbers with the clue also. Please help.

I hope this helps. This might be giving more than a 'hint' but still.

Sorry for the messiness.

My answer is $6,2$
By using clue we are multiply by the left circles ( like $1\cdot 2$ and $3\cdot 4$ and $5\cdot 6$ we get left central circles values $2$ and $12$ and $30$ respectively)... similar way we use bottom of the left half central circles....then we get the value $6$
Similarly , we divide by the right circles values (like $4÷1$ and $6÷3$ and $9÷3$ we get right half central circles values $4$ and $3$ and $3$ respectively) similarly way we use bottom of the right half of the central circles value is $2$


Why 6 races are not sufficient in the 25 horses, 5 tracks problem The horse-racing puzzle has been asked on mathSE several times (1, 2, 3, 4); there is also a generalization.
I restate the puzzle below:

  25 horses all run at different speeds. You can race 5 horses at a time to rank them in order of speed, but you do not actually learn their numerical speeds by racing them. How many races are required to determine the third-fastest horse?

The correct strategy for doing it in 7 races has been described in detail here and elsewhere on the internet.
But why isn't 6 races enough?

The only attempt at showing 6 races is not enough (in this answer) is pretty hand-wavy:

  Now let's prove that $7$ races is optimal.

  Using the same idea, we can draw a directed graphs to represent these races and relations. The circles or nodes are horses, and the directed paths represent one horse being faster than another. Here would be what one race would look like, with the fastest horse on the left:

  $$
\circ\rightarrow\circ\rightarrow\circ\rightarrow\circ\rightarrow\circ
$$

  So basically, what we want to end up with to find the three fastest horses is a parent node and a total of at most $5$ children with a depth of $2$ from this parent, with all of the other nodes being under $5$ children. We need a race to compare these $5$ children, and the other races to set up the graph. Each race will place $4$ horses, as the fastest horse needs to be a horse that has already raced in order to be compared. The only exception is the first race, which can place $5$. $\lceil 24/4 \rceil = 6$, and we add that to our check race to show that the best possible solution is $7$ races, and it can in fact be attained.

I really don't follow this reasoning, and I don't think it provides rigorous justification.

Note as well, that 6 races IS enough if you get lucky. So a correct argument will have to appeal to some sort of worst-case-scenario.

Does anyone have a rigorous version of this argument, or a different rigorous argument, to show 6 races is not enough? I'd like to have a solution that is as elegant as possible.

Assume that you have an algorithm to find the first three places in 6 races. Assume that after the fifth race $n$ horses have raced.
If $n=25$, then the last race can detemine only the winner, but not the second nor the third place: it has to race one horse from each previous race, so if the first three places were all in the same heat or 1,2 in the same and 3 in a different, it couldn't tell the difference.

So $n<25$, and in the last race you have to race the new horses against the first three places A1, A2, A3 (possibly without knowing the order) of the $n$ horses that already have raced, otherwise you cannot determine the overall three fastest horses with the last race.

Hence $n=23$ or $n=24$.

If you assume $n=23$, then it is impossible to determine the three fastest horses out of 23 with 5 races:

You cannot have less than three different winners, since only two repetitions are allowed.

If you have three different winners that have not been compared, then any of the 5 second places could be A2 or A3.

If you have four or more different winners that have not been compared, all of them could be A1,A2, or A3.

If $n=24$, then you have to know a set of four horses out of the first 24 which contains A1,A2 and A3. But this is impossible in 5 races, since you have at least 4 different winners that have not been compared, and so these four horses and the second places could be all A1,A2 or A3.

Hence all cases lead to a contradiction, and so the assumption that 6 is enough, is untenable.

$A_1$ $A_2$ $A_3$ $A_4$ $A_5$
$B_1$ $B_2$ $B_3$ $B_4$ $B_5$
$C_1$ $C_2$ $C_3$ $C_4$ $C_5$
$D_1$ $D_2$ $D_3$ $D_4$ $D_5$
$E_1$ $E_2$ $E_3$ $E_4$ $E_5$
We race $A,B,C,D,E$ groups.-$5$ races.
Now you say that we have to determine the $3$rd fastest horse. So if we take a look at the situation here, $A_1$ runs faster than everyone. So we have got the first one. Now we have to compare $A_2$ and $A_3$ with $B_1$ and $B_2$ and $C_1$.
I know you would think that why am I writing all this, this is the classic solution. But the rigorous solution is also this: In the starting, we had to compare $35$ horses. And after what seemed convenient, (the $5$ races), $10$ horses were left to compare.
These types of questions only demand a creative  solution, there cannot be a rigorous proof for these types of question. Its like asking a unique solution for $x+y+z=10$. (Just an example).
Because even in the $7$ race method, we first assume the order of finishing positions of horses. So there are many, many solutions, but there cannot be a solution lesser than this.
Edit: In these types of questions, we only take the worst case scenario, not that the person may get lucky. If this was true, the answer would obviously be $0$. And I assumed that you know all this.


When a symmetric densely defined operator is an adjoint operator? I am wondering if it is possible to say that if a symmetric differential operator is densely defined then the operator is self-adjoint indeed?

More Precisely, Let $A:D(A)(\subset H)\to H$ a densely defined operator on the Hilber space $H$ and we have $<Ax,y>=<Ay,x>$; is $A$ a self-adjoint operator?

What more condition does it need to become self adjoint?

Please help me with this, I think in an infinite dimensional spaces every assertion would has some subtleties as in this case?

If $A : \mathcal{D}(A)\subseteq H\rightarrow H$ is symmetric on its domain, then $A$ is selfadjoint iff $A\pm iI$ are surjective. If these operators are surjective, then the domain is automatically dense, which saves some checking.

no of course not. en effet, you must require $A=A^\ast$, you are missing $D(A)=D(A^\ast)$.


Integral domain $\implies$ $x\to ax$ is injective. $$R \mbox{ integral domain then in } ax=bx \implies ax-bx = 0 \implies (a-b)x = 0 \implies a=b \mbox{ or } x = 0$$

But what is this $x=0$? I'm supposed to prove only that $a=b$. What can I do to eliminate the hippotesis that $x=0$?

This map is in general not injective unless $a$ is non-zero.  Certainly if you take $f: \mathbb{Z} \rightarrow \mathbb{Z}$ by $f(n)=0n$, every integer is mapped to $0$, so it fails to be injective.

I suppose it is worth noting that if $R$ is the zero-ring, $R=\{0\}$, then $f(x)=0x$ is actually injective.  But if there are at least two elements in $R$, it is not injective since everything gets mapped to zero.

I think your main confusion is that it is the $a$ which is fixed, not the $x$. So let $a$ be a non-zero element and let $f: R \rightarrow R$ be a map defined by $f(x)=ax$.  Then if $f(x)=f(y)$, we have $ax=ay$.  Then since $R$ is an integral domain, $a(x-y)=0$, so $a=0$ or $x-y=0$.  But $a \neq 0$, so $x-y=0$ which implies $x=y$ so $f$ is injective.

The function  $f:R\to R$ is defined by $f(x)=ax$ forall $x\in R$.

$f(x)=f(y)\implies ax=ay\implies a(x-y)=0\implies$  either $a=0$ or $x=y$

If $a=0\implies f(x)=0\forall x$ in that case $f$ is not injective.

I think you should have $a\neq 0$ in your hypothesis


How to show that $\left|\int_0^x \frac{\sin u} u \, du\right|$ is bounded? How to show that $\left|\int_0^x \frac{\sin u} u \, du\right|$ is bounded?

Is it because $\left|\int_0^x \frac{\sin u} u \, du\right|\leq \left|\cos1-\int_0^x \frac{\cos u}{u^2} \, du\right|$, which is finite?

As is well known, $\lim_{x \to \infty}\int_0^x \frac{\sin u }{u } \, du  = \frac{\pi}{2}$  which proves boundedness, but without using this fact we can argue as follows.

Fixing $a > 0$, by Bonnet's mean value theorem for integrals
 there exists $\xi_x \in (a,x)$ such that for all $x > a$,

$$\left|\int_a^x \frac{\sin u}{u}\, du\right| = \left|\frac{1}{a}\int_a^{\xi_x} \sin u\, du\right| = \frac{|\cos \xi_x - \cos  a|}{a} \leqslant \frac{2}{a},$$

and $\int_0^a \frac{\sin u }{u} \, du$ is finite.

Prove that $\sin(x) \leq x$ for $x \geq 0$
For $x < 0$, note that the integrand is even.


Find the equation of the parabola in which the ends of the latus rectum Find the equation of the parabola in which the ends of the latus rectum have the coordinates $(-1,5)$ and $(-1,-11)$ and the vertex is $(-5,-3)$.

I could think of assuming the equation of parabola as $(y-k)^2=4a(x-h)$ and plug in those three points to get three linear equations and solve for the unknowns. But, I wonder if there's any other approach?

$h$ and $k$ are the coordinates of the vertex, hence $h=-5$ and $k=-3$. As focus is $F=(-1,-3)$ (midpoint of latus rectum) and vertex $V=(-5,-3)$, then $a=VF=4$.


Express $\frac{\sin 7\theta}{\sin \theta}$ in powers of $\sin \theta$ only By using DeMoivre's theorm express
$$\frac{\sin 7\theta}{\sin \theta}$$
in the powers of Sine only

answer given in the book is
$$7-56\sin ^2\theta+112\sin ^4 \theta-64\sin^6 \theta$$
can any one help to solve the question

Steps To Carry Out

1) First take a look at this link which is a guide for DeMoivre's formula.

2) Using step 1 show that

$$\sin (7x) = 64\sin \left( x \right)\cos {\left( x \right)^6} - 80\sin \left( x \right)\cos {\left( x \right)^4} + 24\sin \left( x \right)\cos {\left( x \right)^2} - \sin \left( x \right)$$

3) Replace $\cos^2(x)=1-\sin^2(x)$ and obtain

$$\sin (7x) = 7\sin \left( x \right) - 56\sin {\left( x \right)^3} + 112\sin {\left( x \right)^5} - 64\sin {\left( x \right)^7}$$

4) Divide by $\sin(x)$

$${{\sin (7x)} \over {\sin (x)}} = 7 - 56\sin {\left( x \right)^2} + 112\sin {\left( x \right)^4} - 64\sin {\left( x \right)^6}$$

$sin(7\theta)=sin(\theta+6\theta)=sin(\theta).cos(6\theta)+sin(6\theta)cos(\theta)$ then we have $cos(6\theta)=cos(3\theta+3\theta)=cos(3\theta).cos(\theta)-[sin(3\theta).sin(3\theta)]$ now we know $cos(3\theta)=4cos^3(\theta)-3cos(\theta),sin(\theta)=3sin(\theta)+4sin^3(\theta)$. Now you can just back substitute and get the job done. Hope its clear.


If $g(X)\in \Bbb Z[X]$ irreducible polynomial, then $\langle g(X) \rangle \trianglelefteq \Bbb Z[X]$ is not maximal I would like to prove that given an irreducible polynomial $g(X)\in \Bbb Z[X]$, then the ideal $\langle g(X) \rangle \trianglelefteq \Bbb Z[X]$ is not maximal.

One can think to prove
$$\langle g(X) \rangle \subsetneq \langle p,g(X)\rangle \subsetneq \Bbb Z[X]$$
and maybe to use the ring isomorphism
$$\frac{\Bbb Z[X]}{\langle p,g(X) \rangle}\cong\frac{\Bbb Z_p[X]}{\langle \overline g(X) \rangle }$$
where $p$ is a prime number and $\overline g(X)$ is $g(X)$ with coefficients in $\Bbb Z_p$. But how could we proceed?

I face difficulty to prove $\langle p,g(X)\rangle \subsetneq \Bbb Z[X]$.

I know that this may be an easy question, but I have stuck. Also, any other ideas are welcome!

Hint: $\langle g\rangle$ cannot contain any elements with degree lower than the degree of $g$ (apart from the zero polynomial).

Taking into consideraction Arthur's and Wuestenfux's answers, I ll try to write down the full answer.

At first, if $\deg g(X)=0\implies g(X):=m\in \Bbb Z$, then $\Bbb Z[X]/\langle m \rangle \cong \Bbb Z_m[X]$, which is not a field, so $\langle g(X)\rangle $ is not maximal.

Now, we will assume that $\deg g(X)\geq 1$.

$\langle g(X)\rangle \subsetneq \langle p ,g(X) \rangle$, because if so, $p\in \langle g(X)\rangle \implies p=q(X)g(X),\ q(X) \in \Bbb Z[X \implies \deg (p)=\deg q(X)+\deg g(X) \implies 0=d_1+d_2,$ where $d_1,d_2\in \Bbb N$ and $d_2\geq 1$, contradiction.
$\langle p ,g(X) \rangle \subsetneq \Bbb Z[X]$: We will show that
$$1\notin \langle p ,g(X) \rangle  \iff \langle p ,g(X) \rangle  \neq \Bbb Z[X].$$
We suppose, contrary, that  $1\in \langle p ,g(X) \rangle$. Then,
\begin{alignat*}{2}
1\in \langle p ,g(X) \rangle \iff 1=a(X)p+b(X)g(X), \tag{*}
\end{alignat*}
for some $a(X),b(X)\in \Bbb Z[X]$. We consider the following cases:

(1) If $a(X),b(X)=0$, we have immediately contradiction, through (*).

(2) If $a(X)\neq 0,\ b(X)=0$, then $1=b(X)g(X) \implies \deg g(X)=0$, contradiction.

(3) If $b(X)\neq 0,\ a(X)=0$, then $1=a(X)p \implies p|1$, contradiction.

(4) If $a(X)\neq 0,\ b(X)\neq 0$, then if we take degrees in (*), we will have
$$0=\max\{ \deg(a(X)p),\underbrace{\deg (b(X)g(X))}_{\geq 1} \},$$
so again we have a contradiction.


How can we prove $(P \to \neg P) \to \neg P$ in this system? It's been days that I'm stuck in a simple proof of $(P \to \neg P) \to \neg P$ using an axiomatic system, and, whenever I think I'm closer to it, I just found I'm walking in circles.

The system goes as follow:

Ax1: $(\varphi_1 \to (\varphi_2 \to \varphi_3)) \to ((\varphi_1 \to \varphi_2)\to(\varphi_1 \to \varphi_3))$
Ax2: $\varphi_1 \to (\varphi_2 \to \varphi_1)$
Ax3: $\varphi_1 \to ((\neg \varphi_1) \to \varphi_2)$
Ax4: $((\neg \varphi_1) \to \varphi_1) \to \varphi_1$
Ax5: $(\neg \varphi_1) \to (\varphi_1 \to \varphi_2)$
Ax6: $\varphi_1 \to ((\neg \varphi_2) \to (\neg (\varphi_1 \to \varphi_2)))$
Rules: Modus Ponens (Deduction theorem is also acceptable).
Language: $\neg$ and $\to$ as primitives.

Is it provable at all?

Any help will be highly appreciated.

We wish to prove the following:
$$\vdash (P \to \neg P) \to \neg P$$

Notice how similar this is to your Axiom 4 - we just need to replace all instances of $P$ with $\neg P$ and then $\neg \neg P$ with $P$. This gives us a clue about where to start. If your language doesn't define $\neg P$ as $P \to \bot$ (or if it doesn't even have a symbol for $\bot$) then you can just skip Lemma 1.

Lemma 1: $P \to \neg \neg P$.
Proof: Axiom 3 states that $P \to ((\neg P) \to \bot)$, which is longhand for $P \to \neg \neg P$.

Lemma 2: $\neg \neg P \to P$.
Proof: by deduction theorem, we may assume $\neg \neg P$ and try to prove $P$. Axiom 5 gives $\neg \neg P \to (\neg P \to Q)$; modus ponens gives $\neg P \to Q$. Axiom 4 gives $(\neg P \to P) \to P$, so letting $Q = P$ and modus ponens gives $P$, as required.

That is, we have shown that $\neg \neg P$ is equivalent to $P$.

Therefore, a proof of $(P \to \neg P) \vdash \neg P$ (which is equivalent by the deduction theorem) is as follows:

$(\neg \neg P \to \neg P) \to \neg P$ (Axiom 4)
$P \to \neg P$ (hypothesis)
$\neg \neg P \to P$ (lemma 2)
$(P \to \neg P) \to (\neg \neg P \to (P \to \neg P)) $ (Axiom 2)
$\neg \neg P \to (P \to \neg P)$ (modus ponens, lines 4 and 2)
$(\neg \neg P \to (P \to \neg P)) \to ((\neg \neg P \to P) \to (\neg \neg P \to \neg P))$ (Axiom 1)
$(\neg \neg P \to P) \to (\neg \neg P \to \neg P)$ (modus ponens, lines 5 and 6)
$\neg \neg P \to \neg P$ (modus ponens, lines 3 and 7)
$\neg P$ (modus ponens, lines 1 and 8)

I think I showed that it can be provable in your formal system.
Let me explain.

We rely on the formal proofs in the site
"Metamath Proof Explorer".
The site possesses the proof of $\left(\phi \to \neg\phi\right) \to \neg\phi$ (Theorem pm2.01 (158)). We assume that the proofs of this formula and relevant lemmas are all correct, for they are computer-checked :)

However, there is a big gap remaining. The proof linked above is not based on your formal system. In fact, the formal system used in the above site has the following three axioms:

your Ax1 (Axiom ax-2 (6)):

$\left(\phi \to \left(\psi \to \chi\right)\right)\to\left(\left(\phi\to\psi\right)\to\left(\phi\to\chi\right)\right)$
your Ax2 (Axiom ax-1 (5)):

$\phi\to\left(\psi\to\phi\right)$
the axiom of contraposition (Axiom ax-3 (7)):

$\left(\neg\phi\to\neg\psi\right)\to\left(\psi\to\phi\right)$

In addition, it has exactly the same inference rule as yours, i.e. the rule of Modus Ponens (Axiom ax-mp (8)).

So, The only axiom lacking in your formal system is ax-3 in the list above. The problem has been reduced to proving ax-3 in your system. Let's meta-prove that the system has such proof.

Under the assumptions $\neg\phi\to\neg\psi$, $\psi$ and $\neg\phi$, you have $\psi$ and $\neg\psi$, and therefore $\phi$ by Ax3 and the inference rule. By Deduction Theorem, you have $\neg\phi\to\phi$ assuming $\neg\phi\to\neg\psi$ and $\psi$; thus by Ax4 there is a proof of $\phi$ with the same assumption. Again by Deduction theorem, there is a formal proof of ax-3 in your formal system, without any assumption.

I did not directly give the proof of $\left(\phi \to \neg\phi\right) \to \neg\phi$; however from the above we showed that there is some proof of $\left(\phi \to \neg\phi\right) \to \neg\phi$, in your system. That is, firstly retrieve the proof of $\left(\phi \to \neg\phi\right) \to \neg\phi$ in the Metamath, in which the used axioms are ax-1, 2, and 3. It is not a proof in your system, but you can substitute every single occurrence of ax-3 by the proof of ax-3 in your system, obtained above.
Then the resulting sequence of well-formed formulas is a proof in your system, for it has no longer has an occurrence of ax-3 as an axiom; instead ax-3 is only an intermediate result proved using ax1, 2, 3 and 4. Of course, ax-1  and 2 are no problem at all, as they are your ax2 and 1.


Simplifying $\cos(x) \times \cos(y)$ So I got a problem wrong on my homework because I simplified $\cos (x) \times \cos (y)$ to be $\cos (xy)$, and my teacher wrote on my paper that it can't be simplified. I was wondering if anyone could give me an explanation as to why you can't simplify those two.

It is depends what you mean by saying simplify...
Sometimes it can be useful to use this identity:
$\cos(x) \cos(y) = \frac{1}{2} \left[ \cos(x+y) + \cos(x-y)\right]$
Usually you can change trigonometric function multiplication to sum or backwards.

Multiplication of any two functions with different variables cannot be simplified into one function of one variable.
This kind of simplification is extremely rare:

(1) $f(x)f(y)=g(xy)$

Examples of function that fulfill (1) is $f(x)=ax$ or $f(x)=ax^n$.
Other function that fulfill (1) probably esoteric...


Why is the matrix-defined Cross Product of two 3D vectors always orthogonal? By matrix-defined, I mean

$$\left<a,b,c\right>\times\left<d,e,f\right> = \left|

\begin{array}{ccc}
i & j & k\\
a & b & c\\
d & e & f
\end{array}

\right|$$

...instead of the definition of the product of the magnitudes multiplied by the sign of their angle, in the direction orthogonal)

If I try cross producting two vectors with no $k$ component, I get one with only $k$, which is expected. But why?

As has been pointed out, I am asking why the algebraic definition lines up with the geometric definition.

Here's an explanation in terms of the Hodge dual and the exterior (wedge) product.

Let ${e_1, e_2, e_3}$ be the standard orthonormal basis for $\mathbb{R}^3$. Consider the two vectors $a = a_1 e_1 + a_2 e_2 + a_3 e_3$ and $b = b_1 e_1 + b_2 e_2 + b_3 e_3$. From the matrix computation we obtain the familiar formula

$a\times b = (a_2 b_3 - a_3 b_2) e_1 + (a_3 b_1 - a_1 b_3) e_2 + (a_1 b_2 - a_2 b_1) e_3$.

But (see note at the bottom)

$a \wedge b = (a_1 b_2 - a_2 b_1) e_1 \wedge e_2 + (a_2 b_3 - a_3 b_2) e_2 \wedge e_3 + (a_3 b_1 - a_1 b_3) e_3 \wedge e_1$,

where the wedge $\wedge$ represents the exterior product. One can now compute the dual of this latter expression using that the left contraction of $(e_1 \wedge e_2)$ onto $(e_3 \wedge e_2 \wedge e_1)$ is $e_3$ (and similar relations). The result is that

$a \times b = (a \wedge b)^*$,

that is, the cross product of $a$ and $b$ is the dual of their exterior product.

Geometrically, this is an incredible picture. The exterior product is the plane element spanned by both $a$ and $b$, and the dual is the vector orthogonal to that plane.

This is my favorite interpretation of the cross product, but it's only helpful, of course, if you're familiar with exterior algebra and the Hodge dual.

Note: The wedge product can be found by formally computing

$(a_1 e_1 + a_2 e_2 + a_3 e_3) \wedge (b_1 e_1 + b_2 e_2 + b_3 e_3)$

using the distributivity and anticommutation relations of the exterior product.

The obvious but slightly trite answer is "because that's just how the cross-product works as an operation".

If you're looking for an intuitive reason, the cross-product by definition produces a vector that is orthogonal to the two operand (input) vectors. You know that the base vectors $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ are all orthogonal, thus if your two input vectors lie on the $(x, y)$ plane (i.e. only $\mathbf{i}$ and $\mathbf{j}$ components), you know that any orthogonal vector must have only a component in the $z$ direction (multiple of $\mathbf{k}$).


Why Not Define $0/0$ To Be $0$? For every number $x$, $x\times 0=0$, hence $\dfrac{0}{0}$ can be any number!  So $\dfrac{0}{0}$ "is knows as indeterminate" [1].
But what if we define it to be $0$? I already have an answer, but don't know how convincing it is:

$1=\dfrac{1}{1}=\dfrac{1}{1}+0=\dfrac{1}{1}+\dfrac{0}{0}=\dfrac{1\times 0}{1\times 0}+\dfrac{0\times 1}{0\times 1}=\dfrac{0}{0}+\dfrac{0}{0}=0+0=0$, a contradiction.

Is there any better explanation why not to define $\dfrac{0}{0}$ to be $0$ (or any other number)?

Thanks.

[1] http://en.wikipedia.org/wiki/Division_by_zero#In_algebra

You can define $0/0$ if you want, but that would be missing the point. One considers quotients as an operation. So what you want to be able is to have the multiplicative inverse of $0$, which can be easily seen to not make sense.

The idea of considering the fraction $p/q$ is to be able to think about it as $p\times1/q$. But, what arithmetic can you do with $0/0$?

Informally, $0/0$ is indeterminate, so it would be wrong to give it a value. It would be like o items out of 0 things. Is it none of the divisor or all of it, or maybe halfway? Since there's nothing to compare, all of these could be seen as valid viewpoints, so you cannot assign it a specific value.


Constructing a polynomial with certain zeroes. I want to construct a polynomial $f(x)$ that has zeroes at $-9,\,-5,\,0,\, 5,\, 9$.

Can somebody provide a method (or perhaps some hints) for solving this?

And nowhere else, I guess, otherwise the zero polynomial will do. And I assume your are working over the rationals, say.

Just use the fact that $f(x)$ has a zero at $a$ if and only if $x - a$ divides $f(x)$.


Coin probability problem I tag these questions as homework, as they are older exam questions and every year

Can you try to solve/explain how to solve with a method some of these?

If something is answered in an old post please post the link.

1. We toss a fair coin until we get either two heads either two tails. If the probability getting head is $p$, find the expected number or tosses

Important update: We assume that we stop after $2$ consecutive rolls that bring the same result(e.g. HTHTHTHTHTHTHTHH,THTHTHTHTT). The given probability is for getting ONE head.

To community: I'm familiar with many terms of probability theory so with a good analysis I think I might understand your answers.

Reminder:Their diffictulty may seem easy to medium but not everyone is great at this.

\begin{align}
X&:= \# \text{ tosses required} \\
\Pr(X\geq 1) &= 1 \\
(\forall n\geq 1) \hspace{20pt}\Pr(X\geq n+1) &= \Pr(\underbrace{HTHT\cdots HT}_{n \text{ times}}) + \Pr(THTH\cdots TH) \\
&= \begin{cases}
2[p(1-p)]^{n/2}, &n \text{ even}\\
[p(1-p)]^{(n-1)/2}, &n \text{ odd}
\end{cases}\\\\
\implies E[X] &= \sum_{i=1}^\infty \Pr(X\geq i) \\
&= 1 + \sum_{n=1}^\infty \Pr(X\geq 2n) + \sum_{n=1}^\infty \Pr(X\geq 2n+1) \\
&= 1 + \sum_{n=1}^\infty \Pr(X\geq (2n-1)+1) + \sum_{n=1}^\infty \Pr(X\geq (2n)+1) \\
&= 1 + \sum_{n=1}^\infty [p(1-p)]^{n-1} + \sum_{n=1}^\infty 2[p(1-p)]^{n} \\
&= 1 + 1+ \sum_{n=1}^\infty [p(1-p)]^{n} + 2\sum_{n=1}^\infty [p(1-p)]^{n} \\
&= 2+ 3\sum_{n=1}^\infty [p(1-p)]^{n}\\
&= 2+ \frac {3p(1-p)} {1-p(1-p)} \\
&= \frac {3} {1-p(1-p)} -1 \\
\end{align}

I've assumed independence. I might have made an algebraic error somewhere, but a sanity check is to notice that $\displaystyle{ \lim_{p\to 0}E[X] = \lim_{p\to 1}E[X] = 2}$, which is to be expected (mind the pun). Also, the result is symmetrical around $p=1/2$.

$$P(T=2k)=p^{k+1}(1-p)^{k-1}+p^{k-1}(1-p)^{k+1}=\{p(1-p)\}^{k-1}$$
for events $HTHTHT\cdots HTHH~\text{and}~~THTHTH\cdots THTT$
$$P(T=2k+1)=(1-p)^{k}p^{k+1}+(1-p)^{k+1}p^k=\{p(1-p)\}^k$$
for events $THTHTH\cdots THTHH$ and $HTHTHT\cdots HTHTT$

Given the probabilities calculate the expectation of $T$.


What will be the value of $\frac{1-x^x}{x \log x}$ as $x \rightarrow 0$? We were taught L'hopital's rule and this is one of the questions from the assignment. I can't wrap my head around the indeterminate form that we get when we put in the limit.


From measure $P_X$ and knowing that $P_X=\int f_X d\mu$, how to find $f_X$? If we know that $P_X=\int f_X \ d\mu$, where $\mu$ is another measure(could be lebesgue) and I know what I get from $P_X(B), \ \forall B\in \mathcal{B}(\mathbb{R})$, how can I deduce the function $f_X$?
Any help would be appreciated.

Edit: Some comments have suggested $\frac{d P_X(]-\infty,x])}{dx}=f(x)$, using the fundamental theorem of calculus. However, the only version I know is the following:

If $f:[a,b]\rightarrow \mathbb{R}$ is continuous, then $f$ is integrable, and the function $F$ given by $F(x)=\int^x_a f \ dm$ is differentiable for $x \in (a,b)$ with derivative $F'=f$. The measure $m$ is Lebesgue.

We usually define $F(x)=P_X(]-\infty,x])$. So, what changes do we need to do to the text of the above theorem to be able to apply it to the case  $F(x)=P_X(]-\infty,x])$?

Formally you can use

$$
f(x) = \mathsf{E}[ \delta(X-x)]
$$

where $X$ is the random variable, $x$ the values it can attain, and $\mathsf{E}$ the expectation value. Alternatively you can compute the cdf:

$$
cdf(x) = \mathrm{Prob}(X < x) = \mathsf{E}[ \theta(x-X)]
$$

where $\theta$ is the Heaviside function. You can then obtain $f(x)$ by differentiating the $cdf$.


Finding the angles of a parallelogram. In a parallelogram, one angle is $2/5th$ of the adjacent angles. Determine the angles of the parallelogram.

I tried the following,

Let the adjacent angles be $2x$

Let the other angle be $y$

Accordingly, $y=6/5*2x$

What should I do next? Please help.

In a parallelogram, addition of 2 adjacent angles always produces $\pi$. Can you take it from here?

Also, I think you might have misunderstood the question. If any of the adjacent angles(the angles to the clockwise and the anti-clockwise direction to any angle  $\angle A$ are mutually equal) is x, then the angle in question is 2x/5.

What you assumed would probably be true if the question read, "In a parallelogram, one angle is 2/5th of the SUM OF adjacent angles. "

Sum of two adjacent angles in a parallelogram is $180^{\circ}$, so you have $$x + \frac{2x}{5} = 180^{\circ}.$$