Probablity of $k$ balls go into at most (including) $m$ boxes, with $n$ given boxes ($m\le n\le k$) Suppose I have $n$ boxes, and an infinity of balls. All balls are regarded the same.

I randomly throw balls into the boxes, each time the ball may be thrown into one of the box, with equal probability, i.e., $\frac{1}{n}$.

Now I throw $k$ times, each time one ball thrown into one box. Let $E_{k, m, n}$ denote the situation where the k balls go into at most (including) m boxes (with $m\leq n\leq k$). What is the probability for $E_{k,m,n}$ to happen?

We have to take .care of the fact that

(a) more than one ball can go into a box

(b) arrangements through stars and bars are not equi-probable.

(c) probabilities (including box n) must sum to 1.

I believe we shall have to apply PIE, as illustrated below for n = 6, k = 10

Let $m_k$ denote the # of ways k boxes can be filled. Applying PIE, we get

$m_1 = {6\choose 1}\cdot 1^{10} = 6$

$m_2 = {6\choose 2}\cdot [2^{10} - {2\choose 1}\cdot1^{10}] = 15,330$

$m_3 ={6\choose 3}\cdot [3^{10} - {3\choose 2}\cdot2^{10} + {3\choose 1}\cdot 1^{10}] = 1119600$

Proceeding similarly,

$m_4 = 12277800$

$m_5 = 30618000$

$m_6$ = 16435440

Divide by 6^10 for the probabilities, which will sum to 1, as they must.

You can now generalise for $P[E_{k,m,n}]$

Edit:

Oh, comment gives boxes as indistinguishable ! Shall have to rethink !

Edit 2:

In effect, there are n targets at which you are throwing k missiles which always hit some target or other. This makes the targets distinct, which is what I took in my answer

I think the answer could be as follows:

Using stars and bars, the number of ways k balls could get into m boxes is $(k-1)\choose(m-1)$ ways. the number of ways m boxes could be chosen amongst  n boxes is $n\choose m$.  The probability that each of the k balls could hit any of the boxes is $\frac{1}{n}$.If it is thrown k times, it is $(\frac{1}{n})^k$.  Thus

$E(n,m,k) = {(k-1)\choose(m-1)}.{n\choose m}.(\frac{1}{n})^k$


Determinant Equal to $0$ If Product of Two Matrices Equal to $0$ For two $n \times n$ matrices $M$, $N$ ($n$ is odd) s.t. $MN=0$, either $\det(M+M^T)=0$ or $\det(N+N^T)=0$.
I am trying to show this through two cases: 1) WLOG, if $M$=0, then we are done. 2) If the product of two non-zero square matrices is zero, then both factors must be singular. But for 2), it shows that $\det(M)=\det(N)=0$, which does not guarantee that $\det(M+M^T)=0$ or $\det(N+N^T)=0$, and I cannot seem to find a way to use the information that $n$ is odd here.

If $N$ has rank $r$ (that is, $r$ independent columns) then $M$ must send all those columns to $0$, so $M$ has nullity at least $r$, and rank at most $n-r$. In other words, their ranks add up to at most $n$.
Because $n$ is odd, $M$ and $N$ can't both have rank $\frac n2$: one of them has rank strictly less. Say that $M$ has rank less than $\frac n2$. Then so does $M^{\mathsf T}$; as a result, $M + M^{\mathsf T}$ has rank less than $n$, and is singular.

If $MN=0$ taking determinants you conclude either $det(M)=0$ or $det(N)=0$. Assume the first. Then also $det(M^t)=0$ and both $M$ and $M^t$ correspond to transformations with equal rank less than $n$. The maximum possible dimension of the range of $M+M^t$ is thus twice the rank of $M$ so even and consequently less than $n$. Therefore $M+M^t$ is singular. Similar argument if $det(N)=0$.


Basic problem on conditional probability. Need help in understanding the solution I am confused about below basic problem of conditional probability. Here is the problem and the given solution.

A student is taking a one-hour-time-limit makeup examination. Suppose the probability that the student will finish the exam in less than $x$ hours is $\frac{x}2, \forall 0 \leq x \leq 1$. Then, given that the student is still working after $.75$ hour, what is the conditional probability that the full hour is used?

Solution. Let $L_x$ denote the event that the student finishes the exam in less than $x$ hours, $0 \leq x \leq 1$, and let $F$ be the event that the student uses the full hour. Because $F$ is the event that the student is not finished in less than $1$ hour,

$P(F) = P(L_1^c) = 1 - P(L_1) = 0.5$

Now, the event that the student is still working at time .75 is the complement of the
event $L_.75$, so the desired probability is obtained from

$$ P(F|L_.75^c) = \frac {P(F\color{blue}{\cap}L_.75^c)} {P(L_.75^c)}$$

$$= \frac {P(F)} {1 - P(L_.75)} $$

$$= \frac {0.5}{0.625} = 0.8 $$

QUESTION:
I do not understand only the bottom expression. How did we get $P(F)$ from $P(FL_.75^c)$ ?
And how did we compute $P(L_.75^c)$ so we got 0.625?

Thank you!

Note that $FL_{.75}^c$ (a full hour is used and the student is not ready in less than $.75$ hours) is exactly the same event as $F$ (a full hour is used). You could say that $F\subset L_{.75}^c$ and consequently $F\cap L_{.75}^c=F$.

$P(L_x^c)=1-\frac{x}{2}$ leads directly to  $P(L_{0.75}^c)=1-\frac{0.75}{2}=0.625$

How about we use Venn Diagram to solve this problem

Area of Venn Diagram $= 1$

Area of Event $A$ i.e. $1$ hour to complete $=\frac12 \times 1=0.5$

Area of Event $B$ i.e. $\frac34$ hour to complete $=\frac12 \times \frac34= \frac38$

then Area $B'=1-\frac38=\frac58$

Area of $A' = 1-0.5=0.5$

Note: event $B$ is comprised/subset of Event $A$

So $P(A∩ \cap B')$ is area outside area of Event $A=1-0.5=0.5$

$P(A|B')=\frac{P(A \cap B')}{P(B')}=\frac{1/2}{5/8}=0.8$

Please correct me in case I am wrong. :)


Compound interest Differential Equation A college student starts a savings account with an initial balance of $\$0$. He plans to save money at a continuous rate of $\$200$ per week. Also, at every week he plans to increase this rate by $\$10$. (ex. At the 4th month he would be saving at a rate of $\$240$ per week). Additionally, the college student finds a bank account that pays continuously compounded interest at a rate of $4\%$ per year. Estimate the time it'll take for the college student to save $\$500,000$.
Hint: set up and solve a differential equation and plot the solution to make the final estimate.

My attempt:

The differential equation is hard to set up.

Let $S =$ amount saved.

Let $t =$ time.

$$\frac{dS}{dt} = \frac{0.04}{52}(200 + 10t)$$

I tried this differential equation but it doesn't satisfy the initial condition.

Can someone help me come up with the differential equation?

Thanks!

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$$
\mbox{Let}\quad
\left\{\begin{array}{rclcl}
b_{n} &:& \mbox{Balance after}\ n\ \mbox{weeks}.&& b_{0} = 0
\\[2mm]
s_{0} & : & \mbox{Initial week saving} & = & 200
\\[2mm]
\Delta s & : & \mbox{Amount added to the every week saving} & = & 10
\\[2mm]
r & : & \mbox{Bank interest} \pars{~\mbox{per one per week}~}
& = & {4/\pars{12\times 4} \over 100} = {1 \over 1200}
\end{array}\right.
$$
We assumed $4$ weeks per month.

  $$
\begin{array}{rclc}
b_{0} & = & 0
\\
b_{1} & = & s_{0}
\\
b_{2} & = & b_{1}\pars{1 + r} + \pars{s_{0}  + \Delta s}
\\
b_{3} & = & b_{2}\pars{1 + r} + \pars{s_{0}  + 2\Delta s}
\\
b_{4} & = & b_{3}\pars{1 + r} + \pars{s_{0}  + 3\Delta s}
\\
\vdots & = & \vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\vdots
\end{array}
$$

In general we have to solve:
$$
b_{n} = b_{n - 1}\pars{1 + r} + \bracks{s_{0} + \pars{n - 1}\Delta s}\,,\quad n=2,3,4,\ldots\,;\qquad b_{1} = s_{0}\tag{1}
$$

  Lets $\quad\ds{{\rm B}\pars{z} \equiv \sum_{n = 1}^{\infty}b_{n}z^{n}}\quad$ with
  $\quad\ds{\verts{z} < {1 \over 1 + r}}$:

\begin{align}
\sum_{n = 2}^{\infty}b_{n}z^{n} &= \pars{1 + r}
\sum_{n = 2}^{\infty}b_{n - 1}z^{n}
+s_{0}\sum_{n = 2}^{\infty}z^{n} + \Delta s\sum_{n = 2}\pars{n - 1}z^{n}
\\[3mm]{\rm B}\pars{z} - b_{1}z &= \pars{1 + r}\
\underbrace{\sum_{n = 1}^{\infty}b_{n}z^{n + 1}}_{\ds{=\ z\,{\rm B}\pars{z}}} +
s_{0}\,{z^{2} \over 1 - z} + \Delta s\,{z^{2} \over \pars{1 - z}^{2}}
\\[5mm]
\bracks{1 - \pars{r + 1}z}{\rm B}\pars{z}&
=s_{0}\,{z \over 1 - z} + \Delta s\,{z^{2} \over \pars{1 - z}^{2}}
\end{align}

  $$
{\rm B}\pars{z}
={s_{0}\ z/\pars{1 - z} + \Delta s\ z^{2}/\pars{1 - z}^{2} \over 1 - \pars{r + 1}z}
$$
  $$
b_{n}=
\frac{\left[\left(r + 1\right)^{n} - n r-1\right]\Delta s  +
\left[\left(r + 1\right)^{n} - 1\right] r\,s_{0}}
{r^{2}}
$$
  $$
\color{#66f}{\large b_{n}}
=\color{#66f}{\large 12000\braces{1220\bracks{\pars{1201 \over 1200}^{n} - 1} - n}}
$$

$$
b_{284} \approx 499,325.84\,,\qquad
b_{\color{#c00000}{\Large 285}} \approx 502,781.94\,,\qquad
b_{286} \approx 506,250.93
$$

  $$
\color{#c00000}{\Large 285} = \color{#66f}{\Large 5} \times 48
+ \color{#66f}{\Large 11} \times 4 + \color{#66f}{\Large 1}
$$

$$\color{#66f}{\large%
5\ \mbox{years}, 11\ \mbox{months and $1$ week}.  }
$$


Integrate bivariate normal distribution over circular region Context: Need to compute the probability that a 2D Gaussian random walk falls within distance $ d $ of some point $ p $ on the next step. (Assume the covariance  $ \Sigma $ is the identity matrix $ I $.)

Considering that there's no analytic expression for the CDF of a multivariate normal (according to Wikipedia), I expect that the same is true here... If that's the case, then how can I go about approximating such an integral?

The best thing I can think of is to approximate the circular area with a portion of an annulus, which would allow me to use the CDF of the univariate standard normal distribution ($F$). So if the RW is at $ S_t $ at time $ t $, the probability that it falls within the annulus which touches the circular region at the next time step is

$$ 2 [F(|S_t-p|+d) - F(|S_t-p|-d)] $$

if $ |S_t-p| > d $.

I may have made some sloppy mistakes, but I think this is the gist of it:

$\{(x,y) | x^2+y^2\leq d\} = \{(r,\theta) | r\leq d\}$,

So

$\int \int \frac{1}{2\pi} \exp\{-\frac{1}{2}(x^2+y^2)\} dy dx$

$=\intop\limits_0^{2\pi}\intop\limits_0^d \frac{1}{2\pi}\exp\{-\frac{1}{2}r^2\}r dr d\theta=1-e^{-\frac{1}{2}d^2}$


Is every vector in $\Bbb R$ a cross product? In learning about particle physics, I stumble upon an apparent paradox, so I must misunderstand something. I am reading about pseudovectors, which are vectors that do not change sign under an inversion. Cross products are examples of this. So in my mind, this distinction would in my mind imply that
$$\neg(\forall z ∈ ℝ³)(∃x,y ∈ ℝ³)(z =x × y).$$
This, on the other hand, seems ludicrous, since if any $z ∈ ℝ³$ would be expressible as a a cross product, all vectors should be, due to rotational invariance about the origin. Denying this seems to imply that some points in 3D-space are more ‘special’ than others.
Where is the fallacy here?
Example of a contradiction
We know that $\hat{z} = \hat{x} ×  \hat{y}$. Now for $P$ the inversion, i.e. $(x,y,z) ↦ (-x,-y,-z)$, we obtain
$$ -\hat{z} ≠ \hat{z} = \hat{x} ×  \hat{y} = P(\hat{x} ×  \hat{y}) = P(\hat{z}) = -\hat{z}.$$

To the question in the title; yes, every vector in $\Bbb{R}^3$ is a cross product of two vectors in $\Bbb{R}^3$. The argument you sketch for this is entirely correct, and there is no fallacy in your thinking.
The fallacy is in the way you (or your source) define pseudovectors. Of course the only vector in $\Bbb{R}^3$ that is invariant under inversion is the zero vector. However, for any pair of vectors $x,y\in\Bbb{R}^3$ you have
$$x\times y=(-x)\times(-y),$$
which shows that the cross product, as a map $\Bbb{R}^3\times\Bbb{R}^3\ \longrightarrow\ \Bbb{R}^3$, is invariant under precomposition by the inversion $x\ \longmapsto\ -x$. Perhaps this is what your source is getting at?
Of course this is just a very particular instance of the fact that the cross product 'plays nicely' with orthogonal transformation, which you already use in your sketch for the argument that every vector in $\Bbb{R}^3$ is a cross product. More precisely, the fact that for any orthogonal transformation $A$ you have
$$A(x)\times A(y)=(\det A)A(x\times y).$$

So I think I may have found a way to express this in the language of algebraic topology (which I am just learning). If we let $C$ be the cross product map, and $P$ the inversion again, then the following diagram commutes (I don't know how to draw an arrow from bottom left to upper right)
$\require{AMScd}$
\begin{CD}
ℝ^3 × ℝ^3 @>{P}>> ℝ^3 × ℝ^3\\
@A \text{id} A A @VVCV \\
ℝ^3 × ℝ^3 @>{C}>> ℝ^3
\end{CD}
If $C$ is a covering map here, which I think it should be, then $P$ is the unique lift of $C$ across $C$. Also, we could note that $P ∈ \text{Aut}( ℝ^3 × ℝ^3 / ℝ^3)$
Is this in any way useful..? XD


Behaviour of positive function on compact sets Let $v$ be any strictly positive bounded function on $\Omega \subset \mathbb{R}^n$ and zero on $\partial \Omega.$

Can we say that for each compact subsets (w.r.t usual topology defined on $\mathbb{R}^n $) of $\Omega$ there exists a constant $c$ such that $v \geq c >0$?

If not, what can be the example?

Let $n=1$, $\Omega =(-1,1)$, $u(\frac 1 n)=\frac 1 n$ for $n=1,2,...$ and $u(x)=1$ for all other $x \in \Omega$, $u(x)=0$ for $x =\pm 1$. Consider the compact set $[0,\frac  1 2]$.

Well let x(n)=p+ (1\n)p  with p in the compact set (say a say a closed ball about p lying in your open set and n sufficiently large. Let v(x(n)) =1/n with . v(q)=1 at other points in your open set ( and 0 on the boundary ).


Minimum of a function in $(0,1) \times (0,+\infty)$ I would like to minimize the function
$$ (\alpha,\theta) \mapsto F(\alpha,\theta) := -\theta x^\alpha + \sum_{k=1}^N \ln(1+p_k(e^{\theta \ell_k^\alpha}-1)) $$
where $\theta \in (0,+\infty)$, $\alpha \in (0,1)$ and the parameters are such that $p_k,x,\ell_k \in (0,1)$, with $\sum_{k=1}^N \ell_k = 1$.

I took the partial derivatives with respect to $\alpha$ and $\theta$ and I obtained
$$ \partial F/\partial \theta = -x^\alpha + \sum_{k=1}^N \ell_k^\alpha \frac{p_ke^{\theta \ell_k^\alpha}}{1+p_k(e^{\theta \ell_k^\alpha}-1)} $$
$$ \partial F/\partial \alpha = -\theta \ln(x) x^\alpha + \theta \sum_{k=1}^N \ell_k^\alpha \ln(\ell_k) \frac{p_k e^{\theta \ell_k^\alpha}}{1+p_k(e^{\theta \ell_k^\alpha}-1)} $$
but I'm stuck in trying to find a pair $(\alpha^*,\theta^*)$ that makes them null.

Can I say something about the minimum of $F$ (it if exists)?

You try to minimize a function on an open set, so there is no guarantee that such a minimum actually exist.

Looking for the zero of the gradient is a necessary condition, but not necessarily sufficient since your function F is not convex. Indeed, this points can also be a local maximum or a saddle point.

Now, looking for points that satisfy this necessary condition, you can simplify your expressions considering that the first equation gives
$$\partial F/\partial \theta=0 \rightarrow x^{\alpha}=\sum_{k=1}^N l_k^{\alpha} \frac{p_k e^{\theta l_k^{\alpha}}}{1+p_k(e^{\theta l_k^{\alpha}}-1)}$$
using this in the second equation leads to
$$
\partial F/\partial \theta=0 \rightarrow 0=-\theta \ln(x) \sum_{k=1}^N l_k^{\alpha} \frac{p_k e^{\theta l_k^{\alpha}}}{1+p_k(e^{\theta l_k^{\alpha}}-1)}+\theta \sum_{k=1}^N l_k^{\alpha} \ln(l_k) \frac{p_k e^{\theta l_k^{\alpha}}}{1+p_k(e^{\theta l_k^{\alpha}}-1)}
$$
simplifying and taking into account that $\theta>0$
$$
0= \sum_{k=1}^N l_k^{\alpha} (\ln(l_k)-\ln(x)) \frac{p_k e^{\theta l_k^{\alpha}}}{1+p_k(e^{\theta l_k^{\alpha}}-1)}
$$
and now the results highly depend on the $l_k$.


Find sequences such that $\lim_{n \to \infty} (\sqrt{a_n}-\sqrt{b_n})=\pi$, with $a_n,b_n \in Q$, increasing and defined by recursion For any real number $r$ we can find a pair of natural numbers $N$ and $M$, such that $\sqrt{N}-\sqrt{M}$ will approximate $r$ with any given precision (if we choose $N,M$ large enough).

That's why I tried to think of any increasing sequences of integer numbers which would allow to compute $\pi$ in this way. And I mean, without knowing the digits of $\pi$ beforehand. Thus, $a_n, b_n$ have to be defined by recursion themselves. However, I wasn't able to make any progress.

1) The easiest way is to use the known value of $\pi$ to compute various pairs of $N,p$, such that:

$$\sqrt{N+p}-\sqrt{N} \approx \pi$$

$$N=\left[\frac{(p-\pi^2)^2}{4 \pi^2} \right]$$

Here $[]$ is the floor function. For example:

$$\sqrt{30268+1103}-\sqrt{30268}=3.14159$$

2) If we let the numbers be rational, and not necessarily increasing, we can use the known series to make something resembling what I want:

$$\pi^2=6 \left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots \right)$$

$$\pi=\sqrt{6 \left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots \right)}-\sqrt{0}$$

3) Another way, is to use the Spiral of Theodorus (spiral of square roots):

$$\rho (n)=\sqrt{n}$$

$$\phi (n)=\sum_{k=1}^{n} \arcsin \frac{1}{\sqrt{n}}-\frac{\pi}{2}$$

The $\pi$ arises as a limit of radii difference at approximately the same angle:

$$\lim_{n \to \infty} \rho(\phi+2\pi)-\rho(\phi)=\pi$$

For example, the first such pair of numbers is:

$$\sqrt{21}-\sqrt{2}=3.17$$

However, I'm not sure if it's possible to properly define a sequence I want using this definition for a spiral - it's just too complicated (especially, finding pairs of rays with the closest angles).

  Can we find sequences $\{a_n\}, \{b_n\}$ with $a_n,b_n \in \mathbb{Q}$, increasing and defined by recursion, such that $$\lim_{n \to \infty} (\sqrt{a_n}-\sqrt{b_n})=\pi$$?

  Or can we prove that we can't?

Edit

I only consider rational sequences because I don't believe that any such integer sequences exist. However, I would be very happy if someone answers with integer sequences.

  The main thing though, I want something different from $$\lim_{n \to \infty} \sqrt{a_n} = \pi+c, ~~~~ \lim_{n \to \infty} \sqrt{b_n}=c$$

Let $\sum \limits _{k=0} ^\infty x_k$ be any series with positive and rational terms that converges to $\pi$. One such example may be

$$\sum _{k=0} ^\infty \frac 1 {16 ^k} \left( \frac 4 {8k+1} - \frac 2 {8k+4} - \frac 1 {8k+5} - \frac 1 {8k+6}\right)$$

but you may find others by considering MacLaurin series of various inverse trigonometric functions and evaluating them at special simple points (for instance the series of $\arcsin$ evaluated at $\frac 1 2$, multiplied by $6$).

Let

$$a_n = \left( \sum _{k=0} ^n x_k + 1 \right)^2, \quad b_n = \left( 1 - \frac 1 n \right) ^2 .$$

Notice that $\sqrt{a_n} - \sqrt{b_n} \to (\pi + 1) - 1 = \pi$. (The $1$ here is arbitrary, you may choose any other strictly positive number.)

To show that the sequences are recursive, notice that

$$a_{n+1} = a_n ^2 + 2 x_{n+1} a_n  + x_{n+1} ^2, \quad b_{n+1} = b_n ^2 - \frac 2 {n (n+1)} b_n + \frac 1 {n^2 (n+1)^2} .$$

(Notice, though, that the recursion is not linear - but you have not asked for such a property.)

Finally, the sequences are increasing being squares of other increasing sequences (for $a_n$ here it is where the positivity of $x_n$ matters). It is also obvious that the sequences are made only of rational numbers (here is where the rationality of $x_n$ matters.)


Probability of choosing positive 2-digit integer with 4 in either place GRE study exam guide has following

  If an integer is randomly selected from all positive 2-digit integers,
  what is the probability that the integer chosen has at least one 4 in
  the tens place or the units place?

I understand probability of being in 10s place is $1/9$ and the probability of being in the units place is $1/10$

When I add $1/9$ + $1/10$ the answer is $19/90$. However, answer says $1/5$.

Please explain

P(AorB)=P(A) +P(B) _P(both A and B)

P(both A and B)= P(A) * P(B) events are not mutually exclusive..
So P(A)= 1/9, P(B) = 9/90
               = 1/9 + 9/90 _ 1/9*9/90= 1/5 answer..


Monty Hall Problem Intuition I was thinking about the Monty Hall problem and I thought of a possible intuitive explanation:

You choose a door.
Monty gives you the option of sticking with your original choice or instead choosing both of the other two doors.
If you decided to switch (which now becomes an obvious choice), Monty first opens the door with the goat behind it (say, to add to the excitement), and then opens the other door.

My question then is, is this reasoning flawed? Is this even the same problem as before? Because now, choosing to switch from one door to two doors becomes quite obvious, and so does the $2/3^{rd}$ chance of winning the car on switching.

I cannot find a flaw in your reasoning.

My own reasoning (if you are interested).

Someone who sticks to his original choice will win if his original choice was correct.

Probability on that: $\frac13$.

Someone who switches will win if his original choice was wrong.

Probability on that: $\frac23$.

I think there's a slight difference from the original problem here.

In your version, neither the elimination nor the switching has any effect. In the original version, these processes matter. Here's how:

Let's say you choose Door 1. There's a 1/3rd chance you're right, and a 2/3rd chance you're wrong.

At this point, if you switch Doors (without Monty eliminating a choice for you) you're just moving to another door with the same 1/3rd chance of being right.

Now, Monty opens Door 2. You're then asked to switch or not.

At this point, if you switch, you will win if you were wrong at first (probability 2/3).

If you were right at first (probability 1/3) and you switch, you lose!

Essentially, when Monty removed that empty door, the question "What are the chances that you will win if you switch" became equivalent to "What are the chances that you were wrong at first", which is 2/3rd.


Statistically distinguishing two groups of data Setup: I have two groups of data points; call then $A$ and $B$. Each data point in $A$ and in $B$ has an $(x,y)$ position on a set of axes. Further, the elements in each group tend to be grouped together on these axes - for example, lets say that all elements of $A$ lie in a rectangle bounded by $(x_A, y_A)$ and $(x_A',y_A')$, while all elements of $B$ lie in a similar rectangle bounded by $(x_B, y_B)$ and $(x_B',y_B')$. Being arbitrary, the two regions may overlap; further I made them rectangular for simplicity, but they may as easily be ovoid or some other shape. Lastly, each data point has some characteristic uncertainty, $\sigma$.

Question: Suppose I have another data point; we'll call it $z$, with it's own uncertainty $\sigma_z$. I know $z$ belongs to either $A$ or $B$ but I do not know which. How can I quantify my confidence that $z$ is a member of one group or the other?

You can have $\sigma_z$ define some circle $C_z$ around $z$ with radius, dependent on $\sigma_z$ and weight of the circle as some decaying function of the distance from $z$.

Now compute the weights $w_A, w_B$ of the intersections of $C_z$ with regions of $A$ and $B$, respectively, and define a decision metric based on these two, picking the larger of the weights. Example confidence in $A$ could be
$$
\frac{w_A}{w_A+w_B}
$$


"Binomiable" numbers Is there a nice criterion to determine whether a given natural $m$ can be written as a binomial number $\binom{n}{k}$ with $1 < k < n-1$?

I've been thinking on this problem with a friend and all we concluded is that primes aren't "binomiables".

  CLAIM 1. Primes aren't "binomiables".

  Proof. Let $p$ a prime and suppose $p = \binom{n}{k}$.

  If $n$ is prime, then $n$ must be equals $p$, because $n\mid \binom{n}{k}$. But, in this case, $n\mid\mid \binom{n}{k}$, so $k=1$. (This would work to prime powers too.)
  If $n$ is composed, note that $\binom{n}{k} = p$ implies, a fortiori, that $p\mid \binom{n}{k}$ and consequently $p\leqslant n$. But under our conditions, $\binom{n}{k} > n$, contradiction. $\square$

Also, I've already saw in Proofs from THE BOOK the problem "Binomial coefficients are (almost) never powers", and although it is highly related, it doesn't provide much insight towards this problem (at least as far as I understood).

EDIT: Prime powers aren't binomiables too!

Let's state three results first:

Lemma. Let $p$ a prime. If $p\mid n$ and $0<k<p$, then $p\mid \binom{n}{k}$.

Proof. Let $n= pm$ and assume $0<k<p$. Consider the Vandermonde's identity:

$$\binom{pm}{k} = \sum_{j=0}^k \binom{p}{j}\binom{p(m-1)}{k-j}.$$

It follows that $\binom{pm}{k}\equiv \binom{p(m-1)}{k} \mod{p}$ (the
term $j=0$ in the sum). Repeating it $m$ times, we conclude
$\binom{pm}{k}\equiv \binom{p}{k} \mod{p}$, and from $p\mid
    \binom{p}{k}$ follows our statement. $\square$
Bertrand's postulate. If $n\geqslant 2k$, then the binomial $\binom{n}{k}$ has a prime factor $q > k$.
Theorem. [Erdös] The binomial $\binom{n}{k}$ is never a perfect power for $4\leqslant k\leqslant n-4$.

(This is the paragraph I mentioned in Proofs from THE BOOK.)

So there is only two cases to analyse, $k=2$ and $k=3$. In both cases we can assume $n \geqslant 6$.

  CLAIM 2. Prime powers aren't "binomiables".

  Proof. Let $p$ a prime, $m>1$, and suppose $p^m = \binom{n}{k}$.

  If $k=2$, then $p^m = \frac{n(n-1)}{2}$. By Bertrand's postulate, $p\neq 2$. Now observe that $\gcd(n, n-1) = 1$, so
  $$ p^m \mid n\quad \text{xor}\quad p^m\mid n-1. $$
  Then:
  $$2 =\frac{n(n-1)}{p^m} \geqslant n-1 \implies n \leqslant 3,$$
  that contradicts $n\geqslant 6$.
  If $k=3$, then $p^m = \frac{n(n-1)(n-2)}{6}$. Bertrand's postulate allows us to take $p\neq 2,3$. Hence, similarly to the previous case, we can deduce:
  $$p^m \mid n \quad \text{xor}\quad p^m\mid n-1 \quad \text{xor} \quad p^m\mid n-2.$$
  But then:
  $$ 6 = \frac{n(n-1)(n-2)}{p^m} \geqslant (n-1)(n-2) \implies n \leqslant 4,$$
  a contradiction. $\square$

EDIT 2: I've just realized that I didn't use the Lemma at all. Anyway, I think I'll let it here, it may be useful someday hahaha

Here is a partial answer for a special case: if $\sqrt{8n+1}$ is an odd integer, then $n$ is a binomial coefficient of the form $n \choose 2$.  Such a formula is possible because $n \choose 2$ is quadratic in $n$ and therefore has a straightforward solution.

I know they are composite.

To show this, suppose not, that is, $\binom mn=p$, a prime. Then $m\ge p$. Since we are excluding the trivial cases, $m\ge p+1$ and then

$$\binom mn\ge\binom{p+1}2=p\cdot\frac{p+1}2\ge\frac32p>p$$

a contradiction.

But not every composite number is a non-trivial binomial coefficient. For example, $9$ is not.


Bayesian Posterior Density Derivation Assume that $Y_{i} \sim N\left(\theta x_{i}^{2}, \sigma^{2}\right)$ for $i = 1,\ldots,n$ independently. Consider a Bayesian model with a known $\sigma = 1$ and with a prior distribution $\theta \sim N\left(0, \sigma_{0}^{2}\right)$ for some fixed value $\sigma_0 > 0$.
Show that the posterior density of $\theta$ is $$\theta \mid \mathbf{y} \sim N\left(\tilde{\theta},\left(\sum_{i=1}^{n} x_{i}^{4}+\sigma_{0}^{-2}\right)^{-1}\right)$$
and find $\tilde{\theta}$
By using the fact
$p(\theta\mid \textbf{y}) \propto {f (\textbf{y}|\theta)p(\theta)}$, how do i go about deriving the desired posterior density?
Thank you

Ok Let's start!
$$p(\theta|\mathbf{y}) \propto Exp\Bigg\{-\frac{\theta^2}{2\sigma_0^2}\Bigg\}Exp\Bigg\{-\frac{1}{2}\Sigma_i(y_i-\theta x_i^2)^2\Bigg\}$$
Now let's expand the exponent throwing away any element not depending on $\theta$ (it will be part of the normalization constant)
$$\propto Exp\Bigg\{-\frac{\theta^2}{2\sigma_0^2}+\frac{2\theta\Sigma_i y_ix_i^2}{2}-\frac{\theta^2\Sigma_i x_i^4}{2}\Bigg\}$$
$$\propto Exp\Bigg\{-\frac{1}{2}\Big[\theta^2\Big(\sigma_0^{-2}+\Sigma_i x_i^4\Big)-2\theta\cdot \Sigma_i y_ix_i^2\Big]\Bigg\}$$
$$\propto Exp\Bigg\{-\frac{1}{2\Big(\sigma_0^{-2}+\Sigma_i x_i^4\Big)^{-1}}\Bigg[\theta^2-2\theta\cdot\underbrace{\frac{\Sigma_iy_i x_i^2}{\sigma_0^{-2}+\Sigma_i x_i^4}}_{=\tilde{\theta}}+?\Big]\Bigg\}$$
As you can see in the exponent there is a missing square. To complete the gaussian is enough to sum and subtract the missing element, $\pm(\tilde{\theta})^2$, and multiply all by the normalization constant (but it is not necessary to answer to your question).
The mean you are requesting to shoe, in fact, is exactly the expression of $\tilde{\theta}$ I underbraced.
The variance they ask to show is evidently written...
I hope this help.


Is there a more elegant way of proving $\langle (1,2)(3,4), (1,2,3,4,5) \rangle = A_5$ I'm trying to show the following

  $\langle (1,2)(3,4), (1,2,3,4,5) \rangle = A_5$

I managed to prove this but I think my solution is very inelegant. Here's my argument

let $J = \langle (1,2)(3,4), (1,2,3,4,5) \rangle,$ then $\lvert J \rvert \geq 8$ and $\lvert J \rvert$ divides $\lvert A_5 \rvert = 60,$ so the possibilities are

$\lvert J \rvert = 10, 12, 15, 20, 30, 60$.

Then I sat down and calculated 13 more elements of $J$, so now the possibilities are $\lvert J \rvert = 30$ or $60$. But we can't have $\lvert J \rvert = 30$ because then $J$ would be normal (index 2 theorem) which would contradict $A_5$ being simple, so we must have $J = A_5$.

The calculation part make this proof quite long winded, is there a simpler way of getting the result?

As a more elementary approach $H=<(1 2)(3 4),(1 2 3 4 5)>$ contains an element of order 2, and element of order 5 and an element of order 3 (the product of the two generators).

A5 is generated by a 3 cycle. The product of the generators is 3 cycle so it generates A5


Evaluate the general infinite square root $$x = \sqrt{n\sqrt{n\sqrt{n}} \cdots}$$

I see that:

$$x = \sqrt{nx}$$

$$x^2 -nx = 0$$

Them:

$$x(x - n) = 0 \implies x \in \{0, n\}$$

How should I reject the $x = 0$ solution?  (any level proof is fine, analysis, calculus etc...)

Assuming that $x=0$ is also a solution with $n \neq 0$. Then you would have the relation $0 = \sqrt{n \sqrt{n \sqrt {n ...}}}$. Now one can take the square on both sides and it follows that either $n$ is zero or the iterated square root is zero. When $n=0$ you are done; else you continue proving the rejection of 0 by squaring again on both sides and compare solutions. Since this procedure is repeadet infinitely many, one concludes that $n=0$ which is a contradiction.


Proving that there is no ideal $I$ in $\mathbb{Z}[t]$ such that $\mathbb{Z}[t]/I \cong \mathbb{Q}$ I want to prove that there is no ideal $I$ in $\mathbb{Z}[t]$ such that $\mathbb{Z}[t]/I \cong \mathbb{Q}.$

The first part of the question asks us to show that if $\phi$ is a nonzero ring homomorphism from $\mathbb{Z}[t]$ to $\mathbb{Q},$ then $\phi(1)=1.$ Further, letting $\phi(t)=u/v$ in lowest terms, it asks me to show that $m/n\in \mathbb{Q}$ (where $m$ and $n$ are coprime) is in the image of $\phi$ if and only if every prime divisor of $n$ divides $v.$ I proved this, but I'm not sure how this allows me to prove that there is no ideal $I$ in $\mathbb{Z}[t]$ such that $\mathbb{Z}[t]/I \cong \mathbb{Q}.$ Any hints please?

The claim

  There exists an ideal $I$ such that $\Bbb Z[t]/I\cong \Bbb Q$

can be rephrased as

  There exists a surjective ring homomorphism $\Bbb Z[t]\to\Bbb Q$.

Indeed, if $f\colon\Bbb Z[t]/I\cong \Bbb Q$ is an isomorphism, then $\Bbb Z[t]\stackrel \pi\longrightarrow \Bbb Z[t]/I\stackrel f\longrightarrow \Bbb Q$ is a surjective homomorphism; and if $g\colon \Bbb Z[t]\to\Bbb Q$ is a surjective homomorphism, you can take $I=\ker g$.

Let $I$ be an ideal of $\mathbb Z[x]$ such that $\mathbb Z[x]/I=\mathbb Q$.  Thus, $I$ is a maximal ideal of $\mathbb Z[x]$, and so, by Miles Reid's book (under graduate commutative algebra) page 22, there exists a prime number $p$ such that $\mathbb Z[x]/I=\mathbb Q$ is a finite algebraic extension over $\mathbb Z_p$ (the field with $p$ elements), that is a contradiction. For more information about maximal ideals of $\mathbb Z[x]$ see Miles Reid's book (undergraduate commutative algebra).


Does $X ⊥ Y \leftrightarrow X ⊥ Y | Z$ implies $(X,Y) ⊥ Z$? Let $X, Y$ and $Z$ be random variables. Let

$p_1$ be the statement that $(X,Y) ⊥  Z$ (meaning $(X,Y)$ and $Z$  are independent),
$p_2$ be the statement that $X ⊥  Y$ (meaning $X$ and $Y$ are independent)
$p_3$ be the statement that $X ⊥  Y \mid Z$ (meaning $X$ and $Y$ are conditionally independent given $Z$)

What I have known is that if $p_1$ is true, then $p_2$ and $p_3$ imply each other.

I wonder if the reverse is true. That is, if $p_2$ and $p_3$ imply each other, will $p_1$ be true? To disprove it, is there a counterexample? Thanks.

No, the converse is not true.  Here are two counter-examples:

1) Consider the following triplets of values all with equal probability.  $p_2$ and $p_3$ are both true in that $X$ and $Y$ are unconditionally independent of each other and (given $Z$) are conditionally independent of each other, while $p_1$ is not true as the pair $(X,Y)$ is not independent of $Z$

 X    Y    Z
 1    1    1
 1    2    1
 2    1    1
 2    2    1
 1    3    2
 1    4    2
 2    3    2
 2    4    2

2) Consider the following triplets of values all with equal probability.  Neither $p_2$ and $p_3$ are true in that $X$ and $Y$ are neither unconditionally independent of each other nor (given $Z$) conditionally independent of each other, while $p_1$ is not true as the pair $(X,Y)$ is not independent of $Z$

 X    Y    Z
 1    1    1
 1    2    1
 2    1    1
 1    3    2
 1    4    2
 2    3    2

If $p_1$false, then $p_3$ does not imply $p_2$ because assuming $p_1$ and $p_3$, we can't decide whether $X$ and $Y$ are dependent or independent.

I <3 all Tim's.


How to solve this First Order PDE $xu_x - yu_y + yu = y$? The PDE is $xu_x-yu_y+yu=y$ .

The method of characteristics gives $\dfrac{dx}{x}=-\dfrac{dy}{y}=\dfrac{du}{y-yu}$

Then $x=-c_1y$ and thus $c_1=-\dfrac{x}{y}$ .

Then, I did $du=\dfrac{(y-yu)dy}{y}$ to try to solve for the second constant so that I can have that in terms of $f\left(-\dfrac{x}{y}\right)$ but I can't seem to do that.

$xu_x-yu_y+yu=y$

$yu_y-xu_x=y(u-1)$

$u_y-\dfrac{x}{y}u_x=u-1$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dy}{dt}=1$ , letting $y(1)=1$ , we have $y=t$

$\dfrac{dx}{dt}=-\dfrac{x}{y}=-\dfrac{x}{t}$ , letting $x(1)=x_0$ , we have $x=\dfrac{x_0}{t}=\dfrac{x_0}{y}$

$\dfrac{du}{dt}=u-1$ , we have $u(x,y)=f(x_0)e^t+1=f(xy)e^y+1$

$u \equiv w + 1\quad\Longrightarrow yw_{y} - xw_{x} + yw = 0$
$$
w = \phi\left(x\right)\varphi\left(y\right)
\quad\Longrightarrow\quad
y\varphi'\left(y\right) + y = x\phi'\left(x\right) = \mu = \mbox{constant}
$$
$$
y\varphi'\left(y\right) + y = \mu
\quad\Longrightarrow\quad
\varphi'\left(y\right) = {\mu \over y} - 1
\quad\Longrightarrow\quad
\varphi\left(y\right)
=
\mu\ln\left(y\right) - y + A\,,\ A: \mbox{constant}
$$

$$
\phi'\left(x\right) = {\mu \over x}
\quad\Longrightarrow\quad
\phi\left(x\right)
=
\mu\ln\left(x\right) + B\,,\quad B: \mbox{constant}
$$

$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
u
\color{#000000}{\ =\ }
1
+
\left[\vphantom{\LARGE A}\mu\ln\left(x\right) + B\right]
\left[\vphantom{\LARGE A}\mu\ln\left(y\right) - y + A\right]
\quad}
\\ \\ \hline
\end{array}
$$


Partial derivative in gradient descent for logistic regression I post this question here because I think this is a calculus problem

I'm a software engineer, and I have just started a Udacity's nanodegree of deep learning.

I have also worked my way through Stanford professor Andrew Ng's online course on machine learning and now I'm comparing.

I have a big doubt about Gradient Descent with sigmoid function because on Andrew Ng's course it is different from the one I see on Udacity's nanodegree.

From Andrew Ng's course, gradient descent is (First formula):

But, from Udacity's nanodegree is (Second formula):

Note: first picture is from this video, and second picture is for this other video.

But in this CS229 course notes from Andrew Ng's, on page 18, I have found the demonstration from Andrew Ng's gradient ascent formula. I only add it here as a demonstration because I haven't found this derivative process for gradient descent and I don't know how to do it:

Note: the formula above is for Gradient Ascent.

I'm not sure if I have understood everything, but in this derivative I see have the derivative from f function disappears (f function is the sigmoid function).

But in Udacity's nanodegree they continue using the sigmoid's derivative in their gradient descent.

The difference between first formula and second formula is the derivative term.

Are the two formula equivalents?
Where can I find all partial derivative steps for that partial derivative?

Please take a look at this part of Machine learning course on Coursera which can help you with your question:
https://www.coursera.org/learn/machine-learning/lecture/MtEaZ/simplified-cost-function-and-gradient-descent
In this part, the lecturer is showing the result of derivative in gradient descent for logistic regression.


Mean value theorem of a function in [a,b] Is Mean Value Theorem (Rolle's Theorem) applicable for the following function:

$$\log \frac{x^2 + ab}{(a+b)x}$$

in the interval $(a,b)$

My text says that it's applicable. But isn't the function discontinuous when $a<0$ and $b>0$? As this interval include $0$ at which the function is not defined and hence is not continuous.

yes this function fulfills the conditions of Rolle's theorem.
$f(x)$ is continous at the interval $[a,b]$ as a composition of continously functions and is differentiable. Also we can prove that $f(a)=f(b)=0$.
Sonnhard.


The polynomial $x^{2k}+1+(x+1)^{2k}$ is not divisible by $x^2+x+1$. Find the value of $k \in \mathbb N$. The polynomial  $x^{2k}+1+(x+1)^{2k}$ is not divisible by $x^2+x+1$. Find the value of $k\in \mathbb{N}$.

I tried finding out the roots of $x^2+x+1$ which were $\dfrac{-1±\sqrt{3}i}{2}$ but in vain. I got no result other than making the polynomial more complicated.

Here's what I got : $$\left(\frac{-1±\sqrt{3}i}{2}\right)^{2k}+1+\left(\frac{-1±\sqrt{3}i}{2}+1\right)^{2k}.$$

Now, I don't know what to do next.

Any help would be appreciated.

Let $\omega=\frac12(-1+i\sqrt3)$. Then $\omega^2=\overline\omega$ and $\omega^3=1$.
Also $1+\omega=\frac12(+1+i\sqrt3)=\exp(\pi i/3)$ so that
$(1+\omega)^2=\exp(2\pi i/3)=\omega$. So you are interested in the quantity
$$a_k=\omega^{2k}+1+\omega^k$$
(and its complex conjugate $\overline{a_k}$). As $\omega^3=1$, then
$a_{k+3}=a_k$: the sequence $a_k$ repeats with period $3$. So all you need
to compute are $a_0$, $a_1$ and $a_2$ etc.

Using the remainder theoram, putting $x^2 + x + 1 = 0$ in the polynomial $P(x) =x^{2k} + 1 + (x + 1)^{2k}$ we will get the remaimder.
$$
P(x) = 1 + x^{2k} + (x^2 + 2x + 1)^k = 1 + x^{2k} + x^k
$$
We can also prove that if $x^2 + x + 1 = 0$, $x^3 = 1$.
We can now clearly see that for $k$ is not equal to $3q$, $P(x)=0$
But given it should not be divisible by $x^2 + x + 1$, Hence $k$ must be $3q$.


If you score 55th percentile on an exam, which of these are true? I'm not sure where I went wrong here... I've tried multiple combinations but still can't find the right combination.  Technically 3 and 5 are the ones that should be true right?  I've also tried 1,3,5, and 6 as well thinking maybe it was a trick question since 45% of a population of 100 could also be in the lower 55th percentile?  I'm lost on this. Any help/explanation is greatly appreciated. This is for statistics class.

45 % of the class scored worse than I did
I earned a 55% on the exam
55% of the class scored as good or worse than I did
I earned an F
45% of the class scored better than I did
45% of the class scored as good or worse than I did

Thanks

Example from R statistical software.  I generate a random sample of size 100
from $Norm(\mu=150, \sigma=20).$ Then round to integers, and sort.
The brackets in the output show the index of the first observation on each line.

x = sort(round(rnorm(100, 150, 20)))
> x
 [1] 117 118 120 125 126 127 127 128 128 129 130 130 131 132 133 133 133 135 135 136
[21] 137 137 139 140 140 141 142 142 142 142 143 143 144 144 144 145 146 146 146 146
[41] 147 148 149 151 151 151 151 151 152 152 153 153 153 154 154 155 155 155 155 156
[61] 156 156 156 157 157 158 158 158 159 159 159 159 160 160 161 161 163 165 165 166
[81] 167 167 167 167 167 167 168 170 170 171 171 173 174 180 180 181 181 185 187 215

The 55th and 56th ordered observations are 154 and 155, respectively.

> x[55]
[1] 154
> x[56]
[1] 155

According to the default method of finding quantiles in R, the 55th percentile
is 154.45 (and there are two correct answers to your question).

> quantile(x, .55)
   55%
154.45

If you want to see particulars of several more methods of finding quantiles,
type ? quantile at the prompt in the Session window of R, and look at Types.
The default type in R is 9. (There may be more information here than is worth knowing.)

However, if we specify type 1, then we get a
different result. This is the inverse of the ECDF (empirical CDF), a
definition used in some textbooks. (Here, you would have only one correct answer.)

> quantile(x, .55, type=1)
55%
155

Each of the nine methods mentioned on the R documentation page has
optimal properties in some particular situation or for some particular
purpose. There seems no possibility of general agreement.

The good news is that percentiles are usually used for very large datasets
and for descriptive purposes. All of the definitions give answers that lie in a closed
interval between two consecutive sorted observations. So in practice
(except for poorly-constructed multiple choice questions),
it does not much matter which definition is used.

Percentiles can be a mite confusing, so this is how I remember it: there is no 100th percentile. That means if I score in the 99th percentile, then 99% of the scores are below mine. Extrapolating, if you score in the 55th percentile, then 54% are below you, and 45% did as well as you or better $(99 - 54 = 45)$. Using that definition, then, none of the answers are true -- except for #6, as $45 < 54$, but it is misleading and shouldn't have gotten past the review stage. Of course, the use of "good" in place of "well" means it probably wasn't reviewed in the first place.


Why is a graph an ordered pair? From the source of all knowledge

  a graph is an ordered pair G = (V, E) comprising a set V of vertices
  or nodes together with a set E of edges or lines, which are 2-element
  subsets of V

Why must it be an ordered pair? It seems irrelevant if you mention V or E first. Must V come first since E is made up of V?

You don't even need a pair. A graph may be defined simply as a set $G$ of singleton or two-element sets. The set of vertices is then $V=\bigcup G$, while $E$ is the set of two-element members of $G$, and $G\setminus E$ is the set of isolated vertices. However, this will not do if you need to specify loops, which are edges that connect a vertex to itself, or multiple edges, although these types of edge would be regarded as illegitimate or irrelevant in the mainstream of graph theory. And of course ordered pairs are needed to define a directed graph.


Percent Change Question If 150 were increased by 60% and then decreased by y%, the result would be 192. What is the value of y?
My solution is to use percent change. Here are my steps and yet my answer is incorrect. Can someone please advise why my answer is incorrect? Thanks.

$150 * 0.60 = 90$
$150 + 90 = 240$
Percent change = ((New value − Old value) /Old value) $∗ 100$
New value = $240$, Old Value = $192$
$((240-192)/192)*100 = 25$

"Increasing by 60%" is multiplying by $1.60$.
"Decreasing by $y$%" is mutiplying by $1-\tfrac{y}{100}$.
You know that $150(1.60)(1-\tfrac{y}{100}) = 192$.
So, solving for $y$, the desired percentage is $y = 100\left(1 - \dfrac{192}{150(1.60)}\right) = \boxed{20}$.

to increase by 60% you multiply by 1,6 so you get 240 wich you did
to decrease bei y% yo u multiply by 1-y/100
so you have 240*(1-y/100)=192
1-(y/100)=192/240=0.8 so y is 0.2*100=20
your kind of reasoning 240-192=48 but 240 is the old value, 192 the new, and 48 is 20% of 240  or 48/240=0.2=20%


why a^2 + b^2 = c^2 in right-angled triangle a^2 + b^2 = c^2

what is the demonstration of this rule with triangle which has 90 deg?
can be proofed using geometry?

Which notions of geometry can you recall for the demonstration?

Here you can find some different proofs of the Pythagorean Theorem.


Find a sequence $a_n$ such that $\lim_{n \to \infty}|a_{n+1} - a_n | = 0$ while the sequence does not converge Find a sequence $a_n$ such that $\lim_{n \to \infty}|a_{n+1} - a_n | = 0$ while the sequence does not converge.

I am thrown for a loop. For a sequence not to converge it means that for all $\epsilon \geq 0$ and for all $N$, when $n \geq N$, $|a_n - L| \geq \epsilon$ as $n \longrightarrow \infty$

or I could change some terms and state it in terms of a Cauchy sequence, but either way how can the terms have a distance of 0 but the sequence not be consider converging?  hints??

Take $a_n=\log(n)$; then

$$
|a_{n+1}-a_n|=\log(n+1)-\log(n) = \log(1+\frac{1}{n})\to 0,
$$

but the sequence diverges. In the definition of Cauchy sequences it is asserted that $|a_m-a_n|\to 0$, but you are free to choose any such $m,n$ independent of each other. In your case $m=n+1$.

How about $a_n = \sum_1^n \frac{1}{m}$? The sequential differences are of course equal to $\frac{1}{n}$, but the limit of the sequence is the sum of all such fractions.

Of course, that's just one concrete example of what @Jake said.


(Solved) Deriving Pi from Euler's Identity I was tinkering with Euler's Identity and I come to wonder if it was possible to derive $\pi$ from it.

I know $\pi$ can't be expressed as a fraction of two rational numbers but neither $i$ nor $e$ is rational.

$e^{\pi i} = -1$ (square both sides)

$e^{2\pi i} = 1$ (logarithm both sides)

$2\pi * i * \log e = \log 1 = 0$

This is how far I got before meeting a contradiction as this the left side equals roughly $9.1 i$.
Is it even possible to derive $\pi$ from Euler's Identity and where have I messed up?

Let us write
$$\log(-1) = \pi i\tag{1}$$
We can make sense of this by defining
$$
\log(-1) := \int_1^{-1}\frac{dz}{z}
$$
where (since we cannot integrate through the pole at $z=0$) we integrate by convention along a curve in the upper half-plane.  For example along the sides of a rectangle.  Using numerical integration, we can compute $\pi$ numerically this way.

Historical: ($1$) appeared in the literature earlier than Euler's identity.

Note: If we use curves not remaining in the upper half plane, we may get values other than $\pi i$ for the integral.

Well, if we know that

$e^{i\pi} = -1$

Then we can use log base e, or just ln() to take out the e and just get the pi*i, so:

$\ln(-1) = i\pi$

That was the hard part -- now we just have to divide by i:

$\frac{\ln(-1)}{i} = \pi$

And now we have pi!


How can the solution of an equation be a collection of expressions? Wikipedia says that a solution is an expression or a collection of expressions(one for each unknown) such that, when substituted for the unknowns, the equation becomes an identity.
So far, as I can remember, I've only came across the case of an expression to be a solution. What are some examples where a collection of expressions is a solution to an equation?

One example is the classic parameterization for Pythagorean triples, $a=p^2-q^2$, $b=2pq$, $c=p^2+q^2$, which solve the equation $a^2+b^2=c^2$.

If you have fewer equations than unknowns, you can have expressions for the solution. Say you are given $$x-3y=2z\\ 2x+y=z$$ Given any one variable you can solve for the others, so one way to express the solution is $$y=-3z/7\\ x=16z/7$$


Prove that $a \leq 0$ if $a \leq \frac 1 n$ for all $n$ The full question reads:
Suppose that $a$ is a number that has the property that for every $n \in \mathbb{N}$, $a \leq 1/n$. Prove $a \leq 0$.

Is there anyway to show this using Archimedean Property, or is it something related to the Completeness Axiom?  The problem using the Archimedean Property is that I get up to $a< \epsilon$ but from there I am not able to conclude anything about whether $a \leq 0$ because $\epsilon > 0$.

In other words, prove that $a\not>0$. Assume $a>0$, what does that tell you about some $n$?

$n$ is a natural number, therefore it can be $1,2,3, \ldots$. On the other hand, $1/n$ is a continuous function for every real $x$ distinct to zero.

Applying the following limit
$$
\mathop {\lim }\limits_{n \to \infty } a \le \mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0
$$

$$
a \le 0
$$

$$Q.E.D.$$


Numerical method to find the roots of the derivative of $y = (x^3 - 4x) \cdot \frac{\sqrt{13+4x^2}}{2}$ (Newton's Method) I am working on an Optimization problem and I would like to suggest finding the solution (i.e. the extrema) of the function below using two methods: the first one being calculating the derivative and making it equal to zero and the second one using a numerical method.

The equation of the model is shown below:

\begin{align*}
y = (x^3 - 4x) \cdot \frac{\sqrt{13+4x^2}}{2}
\end{align*}

(the derivative being the following)

\begin{align*}
y' = \frac{16x^4 +7x^2 - 52}{2 \sqrt{13+4x^2}}
\end{align*}

I have two questions about using it:

Based on some research, I found it is possible to use Newton's Method on functions that are not polynomials. Is that correct?
Is the Newton's Method the most adequate in this case?

We want to find the extrema of the function

$$\begin{align*}
y(x) = (x^3 - 4x) \cdot \frac{\sqrt{13+4x^2}}{2}
\end{align*}$$

One approach is to use Newton's Method on $y'(x)$, where

$$y'(x) = \frac{16 x^4+7 x^2-52}{2 \sqrt{4 x^2+13}}$$

We start by plotting $y(x)$ and see two extrema to try and find using Newton's

The iteration formula for Newton's Method is given by

$$x_{n+1} = x_n-\dfrac {f(x_n)}{f'(x_n)} = x_n - \dfrac{16x_n^4  + 7x_n^2 - 52}{2 \sqrt{4 x_n^2 + 13}\left( \dfrac{64 x_n^3 + 14 x_n}{2 \sqrt{4 x_n^2 +13}} - \dfrac{2 x_n (16 x_n^4 + 7 x_n^2 - 52)}{(4 x_n^2 + 13)^{3/2}}\right) }$$

This can be simplified to remove the square roots and nasty divisions as

$$x_{n+1} = x_n - \dfrac{\left(4 x^2+13\right) \left(16 x^4+7 x^2-52\right)}{2 x \left(96 x^4+430 x^2+195\right)}$$

If we initialize Newton's method using $x_0 = -1.4$, we arrive at $x_4 = -1.26382$ in $4$ steps.

If we initialize Newton's method using $x_0 = 1.4$, we arrive at $x_4 = 1.26382$ in $4$ steps.

Update

There is an easier approach to find where $y'(x) = 0$, we need only find the roots of the numerator of $y'(x)$, so the Newton iteration can be simplified to

$$x_{n + 1} = x_n - \dfrac{16 x_n^4 + 7 x_n^2 - 52}{64 x_n^3 + 14 x_n}$$

This gives the same results as before. Also, comparing this with the above iteration, the quadratic term $(4 x^2 + 13)$, leads to imaginary roots that we don't care about.

Lastly, it is worth noting that you can use a simple transform, $t = x^2$, on the iterations' numerator to get $16 t^2 + 7 t - 52 = 0$, and then solve a quadratic formula and eliminate Newton's Method altogether!


Do the axioms of set theory actually define the notion of a set? In Henning Makholm's answer to the question, When does the set enter set theory?, he states:

  In axiomatic set theory, the axioms themselves are the definition of the notion of a set: A set is whatever behaves like the axioms say sets behave.

This assertion clashes with my (admittedly limited) understanding of how first-order logic, model theory, and axiomatic set theories work.
From what I understand, the axioms of a set theory are properties we would like the objects we call "sets" to have, and then each possible model of the theory is a different definition of the notion of a set. But the axioms themselves do not constitute a definition of set, unless we can show that any model of the axioms is isomorphic (in some meaningful way) to a given model.

Am I misunderstanding something? Is the definition of a set specified by the axioms, or by a model of the axioms? I would appreciate any clarification/direction on this.

Update: In addition to all the answers below, I have written up my own answer (marked as community wiki) gathering the excerpts from other answers (to this question as well as some others) which I feel are most pertinent to the question I originally posed.
Since it's currently buried at the bottom (and accepting it won't change its position), I'm linking to it here. Cheers!

This is the commonplace clash between the semi-Platonic view of the laymathematician and the foundational approach for mathematics through set theory.

It is often convenient, when working in "concrete" mathematics, to assume that there is a single, fixed universe of mathematics. And everyone who took a course or two in logic and set theory should be able to tell you that we can assume this universe is in fact a universe of $\sf ZFC$.

Then we do everything there, and we take the notion of "set" as somewhat primitive. Sets are not defined, they are just the objects of the universe.

But the word "set" is just a word in English. We use it to name this fickle, abstract, primitive object. But how can you ensure that my intuitive understanding of "set" is the same as yours?

This is where "axioms as definitions" come into play. Axioms define the basic ground rules for what it means to be a set. For example, if you don't have a power set, you're not a set: because every set has a power set. The axioms of set theory define what are the basic properties of sets. And once we agree on a list of axioms, we really agree on a list of definitions for what it means to be a set. And even if we grasp sets differently, we can still agree on some common properties and do some math.

You can see this in set theorists who disagree about philosophical outlooks, and whether or not some conjecture should be "true" or "false", or if the question is meaningless due to independence. Is the HOD Conjecture "true", "false" or is it simply "provable" or "independent"? That's a very different take on what are sets, and different set theorists will take different sides of the issue. But all of these set theorists have agreed, at least, that $\sf ZFC$ is a good way to define the basic properties of sets.

As we've thrown Plato's name into the mix, let's talk a bit about "essence". How do you define a chair? or a desk? You can't really define a chair, because either you've defined it along the lines of "something you can sit on", in which case there will be things you can sit on which are certainly not chairs (the ground, for example, or a tree); or you've run into circularity "a chair is a chair is a chair"; or you're just being meaninglessly annoying "dude... like... everything is a chair... woah..."

But all three options are not good ways to define a chair. And yet, you're not baffled by this on a daily basis. This is because there is a difference between the "essence" of being a chair, and physical chairs.

Mathematical objects shouldn't be so lucky. Mathematical objects are abstract to begin with. We cannot perceive them in a tangible way, like we perceive chairs. So we are only left with some ideal definition. And this definition should capture the basic properties of sets. And that is exactly what the axioms of $\sf ZFC$, and any other set theory, are trying to do.

Every axiomatic system begins with undefined concepts, definitions, and axioms. For example, in geometry, some of the undefined concepts are "point", "line", and "incidence".

Likewise, in set theory, some undefined concepts might be "element", "set", and "is a member of".  The peculiar thing about undefined concepts is that when you do a dictionary search, you always get back to the thing you are looking up if you look into synonyms of synonyms of synonyms.   There is no way to get a hard definition.   Nevertheless, we all know what element, set, and membership mean.


Should one pay buy a warrenty? (What is the conclusion?) This problem is from khan academy:

  John just bought a brand new cell phone and is considering buying a
  warranty. The warranty costs 200 euros and is worth 1000 euros if his
  phone breaks. John estimates that there is a 10%, percent chance of
  his phone breaking.

  Find the expected value of buying the warranty.

Step by step guide to the solution

Now, the reason why I created this question is because I am unsure of what the conclusion is. I assume that since the expected value is -100 euros, it means that we can expect to lose 100 euros by buying this warrenty; thus, we should not buy the warrenty.

Is my assumption (conclusion) correct?

expected value function, where x = the probability of braking phone

Intersection with the x axis

Would it be correct to assume that:

if the probability of the phone braking is over 20%, we should buy a warrenty, and simiarily avoid buying one if the probability is less than 20%.

I'm not sure if you are asking this in a mathematical way about the "conclusion" (I would say the only mathematical conclusion is that the company can expect to make an average of \$100 on each warranty sold, although they probably make more; to really know this you would need to know how much it costs them to replace John's phone), or about the practical lesson to take regarding John.

For John, you need to consider what happens if you don't have the warranty. In other words, you should compare with the expected loss of buying the phone without the warranty.

But even that isn't the whole story. Of course the expected value of buying a warranty will be less for the consumer than that of not buying the warranty; that's how companies make money selling them. The real issue is whether John can afford to lose \$1000. If you were to lose \$1000 today, how would that affect your life? How would spending \$200 today affect your life? Can John afford to take a 10% chance of losing \$1000? If John is wealthy enough to take these kinds of losses in stride, he is better off not buying warranties. Otherwise it makes sense to spend an amount he can afford to prevent a loss he cannot afford.

The expected overall value of the warranty, as calculated, is -100 Euros.

On the other hand, assuming that without the warranty John will still purchase a new phone if his current one breaks, then there is a 90% chance John will pay nothing and a 10% chance he will have to pay 1000 Euros, making the expected value of not buying the warranty also -100 Euros.

In this case, as it turns out, there is no change in the value between buying and not buying the warranty so based on these calculations it doesn't matter if he does or not. However, if John valued certain outcomes differently (e.g. he doesn't want to have to pay large chunks of money, or he hates filling out forms), then you would need to used a different measure of the expected utility of the two options.


What do mathematicians mean when they say some conjecture can’t be proven using the current technology? When reading about some open problems, a lot of them have quotes by renowned mathematicians that “[the conjecture] cannot be solved using the current technology” or something along these lines. What do they mean by that? Are they talking about the axioms? Or are they generally speaking in terms of intelligence and mathematical abilities?

These ones are just at the top of my mind:

https://terrytao.wordpress.com/2011/08/25/the-collatz-conjecture-littlewood-offord-theory-and-powers-of-2-and-3/

https://m.youtube.com/watch?v=MXJ-zpJeY3E (skip to the end where the talk about the Riemann Hypothesis)

But I’ve seen many, many more examples

This doesn't have any formal meaning.  It just means that they believe the problem can't be solved with the techniques that mathematicians have already developed and instead some big new idea will be necessary to solve it.  That is, "the current technology" refers to the collection of proof methods that mathematicians have discovered already.  It's meant as a sort of metaphor between mathematics and engineering: some engineering problems can be solved by just finding a clever way to put together already existing technologies, while others require major new inventions.  In the same way, some mathematics problems can be solved by just finding clever new uses of ideas that are already known while other mathematics problems require something more novel.

It means that the methods they have don't suffice to solve the problem and the technology they researched wont give back a mathematic answer that satisfy their conjecture.

The problem in general (which is the amount of times “[the conjecture] cannot be solved using the current technology” is cited) its not a factual (some would say 'engineering') problem because the question is generally not a finite and descriptive one (refer to 'How do I ask a good question?' in stack overflow https://stackoverflow.com/help/how-to-ask
 ).

Many mathematic problems of the sort don't seek a concrete answer but rather a general solution. Meanwhile, technology works great to get concrete solutions to specific problems and therefore technology would have to be directed/tailored to solve that specific problem or to give broader answers that can answer such question, which is something not done yet for the problem in question.


How to prove that the additive group of a finite field of order $p^n$ is isomorphic to $Z_p^n$? Let $\mathbb{F}$ be a finite field of cardinality $p^n$ where $p$ is prime. How to prove that the additive group of $\mathbb{F}$ is isomorphic to $\mathbb Z_p^n$?

The point is that you don't need to use the explicit characterization of $\mathbb F$ as the splitting field of the polynomial $X^{p^n}$ over $\mathbb F_p$ - $\mathbb F$ is a vector space over $\mathbb F_p$ (since it's manifestly closed under addition, taking additive inverses and multiplying by elements of the field $\mathbb F_p$); now from linear algebra we know that any finite-dimensional vector space over a field $K$ is isomorphic to $K^n$ (and our field is finite, so it's certainly finite dimensional).
In this case, $K=\mathbb F_p$, and $\left|K^n\right|=p^n$.  So just by considering cardinalities, we can say that $\mathbb F\cong\mathbb F_p^n$ as vector spaces.  Since an isomorphism of vector spaces is a fortiori an isomorphism on additive groups, we have that $(\mathbb F,+)\cong (\mathbb F_p^n,+)\cong \mathbb Z_p^n$.
Note: $\mathbb F$ is not given by $\mathbb F_p[x]/(x^{p^n}-1)$ - the polynomial $(x^{p^n}-1)$ isn't irreducible in general, so this quotient needn't even be an integral domain, let alone a field.  Instead, it's the field generated by all the roots of the polynomial $x^{p^n}-1$; alternatively, if $f\in\mathbb F_p[x]$ is irreducible of degree $n$, then it is a factor of $x^{p^n}-1$ and we have $\mathbb F\cong \mathbb F_p[x]/(f)$.

According to The fundamental theorem of finitely generated abelian groups I, there are additives groups $G_d1,G_d2,...,G_dt$ so the additive group $F$ is isomorphic to $G_d1 x G_d2 x ... x G_dt$ , where $d_1/d_2/d_3/.../d_t$ and $d_1d_2..d_t=p^n$. We know that the characteristic of the additive group F is p ($p*x=0$ for every $ x ε F $). So $d_1=d_2=..=d_t=p$
Furthermore, F is isomorphic to $ (Z_p)^n $


Precal Probability Homework The letters in the word AARDVARK are printed on square pieces of cardboard with one letter per card.
The eight letters are placed in a hat and one letter is chosen at random. Find the following probabilities:
a) P(the letter chosen is a vowel given that the letter falls in the first half of the alphabet)

So this is a homework problem I'm having trouble with... the teacher has provided an answer (different from mine!) but not an explanation, and he'll be gone a while, so I'd like to know what I did wrong.
The probability that the letter falls in the first half of the alphabet is $5/8$, as there are 3 As, 1 D and 1 K.
The probability that the letter chosen is a vowel from AARDVARK is $3/8$.
Then using conditional probability, I get $(5/8)*(3/8) / (5/8) = (3/8)$. However, the answer is apparently $3/5$. What did I do wrong?

The conditional probability $P(A|B)$ is given by the formula $\frac{P(A\cap B)}{P(B)}$. That numerator is the probability of choosing a vowel that is also in the first half of the alphabet $\left(\frac38\right)$. The denominator is simply the probability of choosing a letter in the first half of the alphabet $\left(\frac58\right)$.

It looks like you tried to use the formula $P(A\cap B)=P(A)P(B)$. That only works for independent events, and in this case, choosing a vowel and choosing a letter in the first half of the alphabet are not independent.

You are given that the letter drawn is $A,D,$ or $K$.  If you drew $R$ or $V$ you would draw again and ignore the first draw.  Three of the $A,D,K$ letters are vowels.


Cesaro means $\to \infty$ Let $\{ a_{n}\}_{n}$ be a sequence and let $a\in \mathbb{R}$. Define $\{ c_{n}\}_{n}$ as:

$$c_{n}=\frac{a_{1}+...+a_{n}}{n}.$$

I want to prove the following claim: if $\lim\limits_{n\to +\infty}a_{n}=+\infty$ then $\lim\limits_{n\to +\infty}c_{n}=+\infty$

Approach: Suppose $\lim\limits_{n\to +\infty}a_{n}=+\infty$

Let $M>0$, $\exists\space n_{0}\in\mathbb{N}$ such that $\forall\space n\ge n_{0}, a_{n}>M$

\begin{align}
c_{n} &=\frac{a_{1}+...+a_{n_{0}-1}+a_{n_{0}}+...+a_{n}}{n} \\
&>\frac{a_{1}+...+a_{n_{0}-1}+(n-n_{0})M}{n}\\
&=\frac{a_{1}+...+a_{n_{0}-1}}{n}+\frac{(n-n_{0})M}{n}
\end{align}

So $\lim\limits_{n\to +\infty}c_{n}\ge\lim\limits_{n\to +\infty}\left[\frac{a_{1}+...+a_{n_{0}-1}}{n}+(1-\frac{n_{0}}{n})M\right]=M$

Since this is true $\forall\space M>0$, then $\lim\limits_{n\to +\infty}c_{n}=+\infty$

Is this approach correct?

You were off to a good start. You showed

$$c_{n}>\frac{a_{1}+...+a_{n_{0}-1}}{n}+\frac{(n-(n_{0}-1))M}{n} $$

for $n\ge n_0.$ Thus $\liminf c_n$ is at least the $\liminf$ of the expression on the right. But the expression on the right has a limit, namely $0+M=M,$ and thus its $\liminf$ is the same. So we have shown $\liminf c_n \ge M.$ Since $M>0$ was arbitrary, $\liminf c_n =\infty,$ proving $\lim c_n =\infty.$

$$\lim_{\infty}(1-\frac {n_0}{n})M=M $$

$$\implies \exists n_1 \in \Bbb N \;: $$
$$n>n_1\implies (1-\frac {n_0}{n})M>\frac {M}{2} $$
on the other hand

$$\lim_{\infty}\frac {a_1+...a_{n_0}}{n}=0\implies $$

$$\exists \; n_2\in \Bbb N  \;:$$
$$ n>n_2\implies \frac {a_1+...a_{n_0}}{n}>\frac {-M}{4} $$
thus for $n>\max (n_0,n_1,n_2) $ we have
$$c_n>\frac {M}{4} $$
and this proves that $$\lim_{\infty}c_n=+\infty. $$


Unique solutions of given an initial condition proof ODEs This is on our study guide for an upcoming test, and none of us have an idea how to answer it.

  Show that if $A(t)$ is an $n \times n$ matrix with continuous dependence on $t \in [a, b]$, then for any $\tau \in [a, b]$ and any $\xi \in \mathbb{R}^n$ the solution of $$x'=A(t)x$$ with initial condition $\phi(\tau)=\xi$ exists for all $t$ in $[a,b]$ and is unique.

Thanks in advance.

Apply the theorem of Picard-Lindelöf, first on bounded intervals then using continuation to the whole of $\Bbb R$. Note that also the matrix norm $\|A(t)\|$ is a continuous function in $t$ that is bounded on bounded intervals.

Let $\mathbf \Phi$ be a $n\times n$ invertible matrix.
$$\dfrac{d\mathbf \Phi}{dt} = A(t) \mathbf \Phi$$
Multiplying both sides at right by $\mathbf x^{-1}$
$$\dfrac{d\mathbf \Phi}{dt} \mathbf \Phi^{-1}= A(t) \mathbf I$$

Integrating both sides with respect to $t$
$$\ln \mathbf \Phi = \int A(t) dt \\
\mathbf \Phi = \exp{\bigg(\int A(t) dt \bigg)}$$

Multiplying $\mathbf \Phi(t)$ by a constant vector gives a vector $\mathbf X(t)$ which satisfies your equation, and you can evaluate in $t=\tau$

I hope it helps.
See Boyce, DiPrima. 2012 for more information


How do you find all $n$ such that $\phi(n)|n$ Where $\phi(n)$ is the Euler phi function, how do you find all $n$ such that $\phi(n)|n$?

Notice that $\varphi(1) = \varphi(2) = 1$, so $\varphi(1) \mid 1$ and $\varphi(2) \mid 2$.

If $n > 2$, assume that the prime factorization of $n$ is

$$n = p_1^{a_1} \ldots p_k^{a_k}$$

Then the formula for the totient function gives

$$\varphi(n) = (p_1 - 1)p_1^{a_1-1}\ldots (p_k - 1)p_k^{a_k-1}.$$

Since $n>2$, this is always an even number, so $p_1=2$ must appear as a factor. We next observe that $n$ cannot have two odd prime factors. If $a_2>0$ and $a_3>0$, then both $p_2-1$ and $p_3-1$ are even, so $2^{a_1+1}\mid \varphi(n)$, which is a contradiction.

So $n=2^{a_1}p^{a_2}$ for some prime $p>2$. Here $p-1\mid\varphi(n)\mid n$, so $p-1$ must be a power of two, say $p-1=2^\ell$. Then $2^{a_1-1+\ell}\mid\varphi(n)$, so we must have $\ell=1$ and $p=3$.

In the end we can verify that $n=1$ or $n=2^a3^b$, with $a>0$, $b\ge0$.

Quasi-brute-force approach using Maple :

with(numtheory):
for n from 1 to 100 do
if n mod phi(n) = 0 then
print(n);
end if;
end do;


Show that the boundary of $A$ is empty iff $A$ is closed and open. I'm reading Intro to Topology by Mendelson.

The problem at hand is,

Show that $\text{Bdry}(A)=\emptyset$ if and only if $A$ is closed and open.

This was all the problem statement had, but I'm in the chapter covering closure, interior and boundary with respect to topological spaces.

Here is my attempt at the proof,

Suppose that Bdry$(A)=\emptyset$. Then $\bar{A}=A$ $\cup$ Bdry$(A)=A\cup\emptyset=A$. Hence, $\bar{A}=A$, which implies that $A$ is closed. To show that $A$ is also open we will show that $\overline{C(A)}=C(A)$, that is, $C(A)$ is closed and hence $A$ open. We already know that $C(A)\subset\overline{C(A)}$, thus it suffices to only show that $\overline{C(A)}\subset C(A)$. This is the case since we know that Bdry$(A)=\bar{A}\cap\overline{C(A)}=\emptyset$, which implies that $\overline{C(A)}\subset C(\bar{A})=$ Int$(C(A))\subset C(A)$. Thus, $A$ is open. Suppose now that $A$ is both open and closed. Since $A$ is open we know that Int$(A)=A$. Also, since $A$ is closed $\bar{A}=A=$ Int$(A)$. Now we know that Bdry$(A)=\bar{A}\cup\overline{C(A)}=$ Int$(A)$ $\cup$ $C($Int$(A))=\emptyset.$

I used an the identity $\bar{A}=A\cup\text{Bdry}(A)$, which was asked later in the problem set. I'm wondering if I should use this, since I was able to prove it, or attempt the proof assuming I'm unaware of the identity. I have also been trying to write more concise proofs and this one is definitely not one of them. Would removing some of the words help or is there a cleaner approach that can be pointed out?

Thanks for any feedback!

As you said $\bar{A} = A \cup \operatorname{Bd}(A)$, but there's a more stronger statement that you can and you should prove: $\bar{A} = \operatorname{Int}(A) \cup \operatorname{Bd}(A)$.

Now it should be fairly easy to prove that $\operatorname{Bd}(A) = \emptyset \iff \text{A is closed and open}$:

(1)If $\operatorname{Bd}(A)=\emptyset$

Then $\bar{A}=\operatorname{Int}(A)$ and since $\operatorname{Int}(A)\subset A \subset \bar{A}$ we conclude that $\operatorname{Int}(A) = A = \bar{A}$.

$\operatorname{Int}(A) = A$ shows that $A$ is open and $A = \bar{A}$ shows that $A$ is closed.

(2)If $A$ is closed and open.

A is open, then $A=\operatorname{Int}(A)$ and since $\bar{A} = \operatorname{Int}(A) \cup \operatorname{Bd}(A)$ we see that $\bar{A} = A \cup \operatorname{Bd}(A)$

But $A$ is also closed!, this means that $\bar{A} = A$ therefore $A = A \cup \operatorname{Bd}(A)$.
Now remember that $\operatorname{Int}(A) \cap \operatorname{Bd}(A) = \emptyset$, then $A \cap \operatorname{Bd}(A) = \emptyset$.

Because of $A = A \cup \operatorname{Bd}(A)$ and $A \cap \operatorname{Bd}(A) = \emptyset$ we conclude $\operatorname{Bd}(A) = \emptyset$

(1) If $Bdry(A) = \emptyset$.

Then, $$\bar A = A \cup Bdry(A) = A \cup \emptyset = A$$ Hence $\bar A = A$, which implies A is closed.

Since the $Bdry(A) = Bdry(A^c)$ this implies $$\overline {A^c} = A^c \cup Bdry(\overline {A^c}) = A^c \cup Bdry(A) = A ^c \cup \emptyset.$$ Hence $\overline {A^c} = A^c$, which implies $A^c$ is closed, which implies A is open.

(2) If $A$ is closed

Then $A = \bar A$. So $$Bdry(A) = \bar A \cap \overline {A^c} = A \cap \overline {A^c} = A \cap (Int(A))^c$$

Since $A$ is open, $A$ is a neighborhood of all its points so $A = Int(A)$,

which implies $A \cap C(Int(A)) = A \cap C(A) = \emptyset$


How can I better notice derivatives inside integrals? I was calculating the integral $\int \frac{x}{\sqrt{1-x^2}} \space dx$ and noticed by accident that $\frac{\text{d}}{\text{d}x}\sqrt{1-x^2}=-\frac{x}{\sqrt{1-x^2}}$. This allowed me to calculate the integral, but the only reason I did the differentiation was to try to somehow fit the integral into $\int \frac{f'(x)}{f(x)} \space dx$. Are there some hints I could look for to notice these, or just test/know them?

It doesn't fit in the $f'/f$ scheme but in the chain rule scheme:
$$\int (f\circ g)\cdot g'=F\circ g.$$

First you have to realise that the derivative of sqrtx is 1/2 sqrt(1/x)

and by the chain rule that the derivative of x^2 is 2x

Integrals are, by their nature, difficult


Calculation of $\min$ distance between $x^2+y^2=9$ and $2x^2+10y^2+6xy=1$ Calculation of $\min$ distance between $x^2+y^2=9$ and $2x^2+10y^2+6xy=1$

$\bf{My\; Try::}$ Using the fact that the distance between two curve is independent of shifting.

So put $x=u+v$ and $y=u-v$ in thsese two curves, We get $\displaystyle u^2+v^2=\frac{9}{2}$

and $\displaystyle 2(u^2+v^2-2uv)+10(u^2+v^2-2uv)+6(u^2-v^2) = 1\Rightarrow 18u^2+6v^2-20uv=1$

But here again i get terms of $uv,$ So i did not understand how can i solve it,

Help me, Thanks

The original ellipse is congruent with the ellipse whose equation is $x^2+11y^2=1$.
And all the ellipse which is similar to it has the equation form:$x^2+11y^2=\lambda$ Make it and $x^2+y^2=9$ as a equation group and solve it: $ x = \sqrt \frac{99-\lambda}{10} or -\sqrt \frac{99-\lambda}{10}, y = \sqrt \frac{\lambda -9}{10} or - \sqrt \frac{\lambda -9}{10}$.That means when $ 9<=\lambda <= 99$, the ellipse has intersection points with circle.

Therefore, when $\lambda = 9$ the ellipse tangent with the circle.
The words are so abstract possibly.

Therefore the minimal distance becomes distance between the long axis vertexes. Since one of the long vertexes of $x^2+11y^2=1$ is (-1,0) and one of the long vertexes of $x^2+11y^2=9$ is (-3,0).

Hence, the answer should be 2.

Complement:
1) why elipse: $x^2+11y^2=1$ is congruent with elipse:$2x^2+10y^2+6xy=1$.

Designate  $2x^2+10y^2+6xy=1$ as ellipse II.
Designate  $x^2+11y^2=1$ as ellipse I.

Think one of point$(x,y)$ on the ellipse II. Make it rotate about original point$(0,0)$ and it should fall into the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $. Find the rotation matrix:
$M = \left(\begin{matrix}cos(\theta)&-sin(\theta)\\sin(\theta)&cos(\theta)\end{matrix}\right)$
where $tan(\theta) = \frac{1}{3}$.
Using rotation matrix multiply the column matrix $ \left(\begin{matrix}x\\y\end{matrix}\right)$every point(x,y) on the ellipse II can get every point(u,v) on the ellipse I.
Now find the linear combination of $u^2, v^2$. Since $u^2 = f(x,y), v^2 = g(x,y)$ and the maximal degree of f(x,y) about x is 2. Similarly, the maximal degree of g(x,y) about x is 2. Through observing the coefficient, its equation should be $u^2 + 11 v ^2 = 1$.

2) Why choose $theta$ satisfying $tan(\theta) = \frac{1}{3}$?
Because our target is to make the ellipse formalized. Luckily, the ellipse's reflecting point is $(0,0)$ because there is no monomial item(i.e. x y) in the original equation. Therefore, assume a line: $y = kx$ passing through the ellipse II. Make $ y=kx$ and $ 2x^2+10y^2+6xy=1$ as a equation group. Solve it. Get $ x = \sqrt \frac{1}{10k^2+6k+2} y = \sqrt \frac{k^2}{10k^2+6k+2} $
Since the long vertex point of the ellipse has maximal distance on the ellipse about $(0,0)$, just find the $ \max x^2+y^2$.

$x^2+y^2 = \frac{k^2+1}{10k^2+6k+2}$ By the division law in differentiation, $(x^2+y^2)' = \frac{-18k-5}{(10k^2+6k+2)^2} = 0$. So when $ k = -\frac{1}{3} $, $x^2 + y^2$ has maximal.
Since the rotation angle should be positive, and then $ tan(\theta) = \frac{1}{3}$

In a nutshell, the answer is 2.

If my answer has some errors or short points, please leave a comment.

There is this beautiful property for curves that if you differentiate its equation w.r.t. $x$ (i.e., $y$ is constant) and similarly w.r.t. $y$ (i.e., $x$ is constant), then whatever lines or curves you get will intersect at the given curve's center.

In this case, differentiating w.r.t. $x$ gives $4x+6y=0$ and differentiating w.r.t. $y$ gives $20y+6x=0$. Clearly, they meet at the origin. Hence, the center of given curve is $(0,0)$.

So any point is of the form $(r\cos\theta,r\sin\theta)$ where $r$ is the distance from origin and $\theta$ is the angle with the $x$-axis.

Satisfying the curve's equation, we get,
$r^2(6-4\cos2\theta+3\sin2\theta)=1$ Do it yourself you will get the same.
$r^2=1/(6-4\cos2\theta+3\sin2\theta)$. For $r$ to be maximum, the denominator should be minimum.

$a\sin\Delta+b\cos\Delta$ always greater than equal to $-\sqrt{a^2+b^2}$ which in this case is= $-\sqrt{9+16} or $-5$.

$r^2$ max$=1/6-5=1$. $r$ max is $1$.

Hence radius of the circle is $3$ and r max for the curve is $1$. Both share a common center origin. So the minimum distance between them is $3-1=2$.

$2$ is the answer.


How can a "proper" function have a vertical slope? Plotting the function $f(x)=x^{1/3}$ defined for any real number $x$ gives us:

Since $f$ is a function, for any given $x$ value it maps to a single y value (and not more than one $y$ value, because that would mean it's not a function as it fails the vertical line test).
This function also has a vertical tangent at $x=0$.

My question is: how can we have a function that also has a vertical tangent? To get a vertical tangent we need 2 vertical points, which means that we are not working with a "proper" function as it has multiple y values mapping to a single $x$. How is it possible for a "proper" function to have a vertical tangent?

As I understand, in the graph I pasted we cannot take the derivative of x=0 because the slope is vertical, hence we cannot see the instantaneous rate of change of x to y as the y value is not a value (or many values, which ever way you want to look at it). How is it possible to have a perfectly vertical slope on a function? In this case I can imagine a very steep curve at 0.... but vertical?!? I can't wrap my mind around it. How can we get a vertical slope on a non vertical function?

The tangent line is simply an ideal picture of what you would expect to see if you zoom in around the point.

$\hspace{8em}$

That being said, the vertical tangent line to the graph $y = \sqrt[3]{x}$ at $(0,0)$ says nothing more than that the graph would look steeper ans steeper as we zoom in further around $(0, 0)$.

We can also learn several things from this geometric intuition.

The line is never required to pass through two distinct points, as the idea of tangent line itself does not impose such extraneous condition.

For instance, tangent lines pass through a single point even in many classical examples such as conic sections. On the other extreme, a tangent line can pass through infinitely many points of the original curve as well.
Tangent line is purely a geometric notion, hence it should not depend on the coordinates system.

On the contrary, identifying the curve as the graph of some function $f$ and differentiating it does depend on the coordinates system. In particular, it is not essential for $f$ to be differentiable in order to discuss a tangent line to the graph $y = f(x)$, although it is a sufficient condition.

OP's example serves a perfect showcase of this. Differentiating the function $f(x) = \sqrt[3]{x}$ fails to detect the tangent line at $(0,0)$, since it is not differentiable at this point. On the other hand, it perfectly makes sense to discuss vertical tangent line of the curve

$$ \mathcal{C} = \{(x, \sqrt[3]{x}) :x \in \mathbb{R} \} = \{(y^3, y) : y \in \mathbb{R} \}, $$

and indeed the line $x = 0$ is the tangent line to $\mathcal{C}$ at $(0, 0)$.

As you observed, $\frac{dy}{dx}$ is not defined at $(x,y) = (0,0).$
And there is no straight segment anywhere on the graph where the slope could be exactly measured by taking any two distinct points on that segment.

But those are not the only ways to define the slope of a curve.

One approach is to take a secant line between $(0,0)$ and a nearby point, and find the limit of the slope of that line as the distance goes to zero.
You can also take the angle between the $x$ axis and the secant line between two points on either side of $(0,0),$ taking a limit as the distance between those points goes to zero.
If the limit is defined (it is not for every curve) then it gives you the tangent line.

Another approach is to parameterize the curve: $x = u(t)$ and $y = v(t)$ where we let $t$ range over some (possibly infinite) interval of $\mathbb R.$
Consider the vector $\left(\frac{d}{dt}u(t), \frac{d}{dt}v(t)\right).$
As long as that vector is not $(0,0)$ we can consider it to be in the same direction as the slope of the curve.
Now let $x = u(t) = t^3,$ $y = v(t) = t.$
Then the vector along the curve at $(x,y) = (0,0)$ is $(0,1),$ which points straight up, so the tangent line is vertical.


If I have an open cover, can I choose a countable sub-cover such that the union of the open sets strictly increasing? If I have an open cover, can I choose a countable  sub-cover such that  the union of the open sets strictly increasing?

Actually I want to use this to prove every noetherian space is compact. Thanks!

In general you can't: Consider $X = [0, 1]$ (which is uncountable) with the topology in which all sets are open. The singletons (= sets of the form $\{x\}$, where $x \in [0, 1]$) form an open cover, but there does not exist a countable subcover.

In the case of Noetherian spaces, we can, in fact, find a finite subcover by failing to find a subcover as you described (in fact, we prove that even in Noetherian spaces it can not exist). Let $X$ be a Noetherian space. Pick an open cover $O_\upsilon, \upsilon \in \Upsilon$ of $X$ and pick $O_1 \in \{O_\upsilon | \upsilon \in \Upsilon\}$ arbitrarily, then $O_2$ such that $O_2 \setminus O_1 \neq \emptyset$, then $O_3$ such that $O_3 \setminus (O_1 \cup O_2) \neq \emptyset$ and so on. Finding such an infinite sequence is equivalent to finding a subcover which you described.

Now given $O_1, \ldots, O_k$, it is possible to find $O_{k+1} \in \{O_\upsilon | \upsilon \in \Upsilon\}$ such that
$$
O_{k+1} \setminus \bigcup_{j=1}^k O_j \neq \emptyset
$$
if and only if the sets $O_1, \ldots, O_k$ do not cover $X$. So if we assume that we never arrive at a $k$ where $O_1, \ldots, O_k$ cover $X$, we can find a sequence $U_n$ of open sets in $X$ defined by
$$
U_n := \bigcup_{j=1}^n O_j
$$
which is strictly increasing. Hence, the sequence of closed sets $A_n$ defined by
$$
A_n := X \setminus U_n
$$
is strictly decreasing. But in Noetherian spaces, there is no strictly decreasing countable chain of closed sets.

Note that we used the axiom of dependent choice for our construction of the sequence $O_n$.

Let $(X,\large\tau),$ be a Noetherian Topology then it is compact

Let $(X,\large\tau)$ be a Noetherian topology  and let $\{U_\alpha\}$ be a cover of $M\subset X$. Set $V_1=U_i$ for any $U_i\in \{U_\alpha\}$ and define $V_{i+1}=V_i\cup U_k$ such that $U_k\in \{U_\alpha\}$ satisfying $V_i\subsetneq V_{i+1}$. Then it follows that $\{V_i^c\}$ is a sequence of closed sets such that $V_{i+1}^c\subsetneq V_i^c$. If $\{V_i^c\}$ is infinite, since $(X,M)$ is a Noetherian topology, it follows that there exist an $N\in Z^+$ such that $i\geq N$ implies $V_i=V_{i+1}$ so $\{V_i^c\}$ is finite. Thus for any open over $\{U_\alpha\}$ of any set $M$ in $(X,\large\tau)$, there can only exist a finite collection of open sets $\{U_{k_j}: U_{k_j}\in \{U_\alpha\}\}$ such that $\{V_i: V_{i}=\bigcup U_{k_j}\}$ is a strictly increasing collection of open sets. Thus letting $M=X$, and we see that $(X,\large\tau)$ is compact.


What is the variance of the number of heads that come up when a fair coin is flipped 10 times? So the answer to this is $5/2$.

I get that $E(x)=5$ which means that $\big(E(x)\big)^2 = 25$, therefore meaning that $E(x^2) = 55/2$.

How do I compute $E(x^2)$ in this scenario?

Note that $X = X_1 + \ldots + X_{10}$, where

$X_i=0$ if the $i$th flip is tails, and $1$ if the $i$th flip is heads.

Since the $X_i$'s are independent, we have

$Var(X) = Var(X_1 + \ldots + X_{10}) = Var(X_1) + \ldots + Var(X_{10})$.

Finally,

$Var(X_i) = E(X_i^2) - E(X_i)^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$,

so that

$Var(X_1) + \ldots + Var(X_{10}) = 10*(\frac{1}{4}) = \frac{5}{2}$.


If $f$ is a nonconstant entire function such that $|f(z)|\geq M|z|^n$ for $|z|\geq R$, then $f$ is a polynomial of degree atleast $n$. I have a question in my assignment :

  If $f$ is a nonconstant entire function such that $|f(z)|\geq M|z|^n$
  for $|z|\geq R$ for some $n\in\mathbb N$ and some $M$ and $R$ in
  $(0,\infty)$ show that $f$ is a polynomial of degree atleast $n$.

Now , I defined a function    $\ g(z) = \frac {1}{f(z)}\ $ such that $\ |g(z)| \le \frac{1}{M{|z|}^n}.$

Now , by using the cauchy inequality
$$|g^{(n)}(z)| \le  \frac{n!}{R^n |z|^nM}.$$

Considering that $ g(z) $ is an analytic function , it has a radius of convergence $ \infty $

$ \implies\  g^{(n)}(z) = 0.$

But if we go by this approach , then $ g^{(n)}(z) = 0 \ $ for any n . Also how can we be so sure that $ f(z) \neq 0 $ for any z ?

Is my reasoning correct or is there some other way to solve it ?

As @Brian points out, $f$ has only finitely many zeros. Of course, $f(z)\neq 0$ if $|z|\geq R$. Since the set $B_R=\{z\mid |z|\leq R\}$ is compact, $f$ can only have finitely many zeros in $B_R$(use the identity theorem). Let $a_1,\ldots,a_k$ be the zeros of $f$ counted according to multiplicty. Let $$p(z)=(z-a_1)\cdots(z-a_k)=z^k+b_{k-1}z^{k-1}+\cdots+b_0.$$ For $|z|\geq R,$ we have $$|p(z)|\leq |z|^k\Bigl(1+\frac{|b_{k-1}|}{|z|}+\cdots+\frac{|b_{0}|}{|z|^k}\Bigl)\leq C|z|^k,$$ where $C=1+\frac{|b_{k-1}|}{R}+\cdots+\frac{|b_{0}|}{R^k}.$ Thus we have $$\frac{|z|^n|p(z)|}{|f(z)|}\leq \frac{|p(z)|}{M}\leq \frac{C|z|^k}{M},$$ for $|z|\geq R$.

Suppose that $n=k$. Then, by Liouville, we see that $\frac{p(z)}{f(z)}$ is a constant function and hence $f$ is a polynomial of degree $k=n$.

Suppose now that $n\lt k$. Then it is easy to see that $\frac{p(z)}{f(z)}$ is a polynomial of degree $\leq k-n$ (use the Cauchy's integral formula for derivatives. Click here for a proof.)
But $\frac{p(z)}{f(z)}$ is a nowhere vanishing entire function. So $\frac{p(z)}{f(z)}$ is a constant and hence $f$ is a polynomial of degree $k\gt n$.

Finally, assume $n\gt k$. Then, by Liouville's theorem, $\frac{z^{n-k}p(z)}{f(z)}$ is a constant. So $f(z)=cz^{n-k}p(z)$ for some constant $c$ and degree of $f$ is $n$. But $f$ and $p$ share the same zeros with same multiplicities. So degree of $f$ is equal to degree of $p$, i.e., $n=k$, a contradiction. (One can also use the Rouche's theorem to conclude. See @N. S.'s comment below.)

Let $g(z) = f(1/z), z\in \mathbb C \setminus \{0\}.$ Then $|g(z)|\ge M/|z|^n$ for $0<|z|<1/R.$ Now $g$ blows up at $0,$ so has a nonremovable singularity there. Thus $g$ has an essential singularity at $0,$ or a pole at $0.$ If the former, then Casorati-Weierstrass tells us $g(\{0<|z|<1/R\})$ is dense in $\mathbb C.$ But $ g(\{0<|z|<1/R\})\subset \{|z|>MR^n\},$ contradiction.

Therefore $g$ has a pole at $0.$ Thus there is a polynomial $p$ and an entire $h$ such that $g(z) = p(1/z)+h(z)$ in $\mathbb C\setminus \{0\}.$ Flipping back, we see $f(z)=p(z) + h(1/z)$ on $\mathbb C\setminus \{0\}.$ It follows that as $|z|\to \infty,$ $f(z)-p(z) \to h(0).$ Because an entire function with finite limit $L$ at $\infty$ equals $L$ everywhere (Liouville), we arrive at $f(z) - p(z) = h(0)$ everywhere. The growth rate of $f$ then implies $p$ has degree at least $n$ as desired.


Taylor series for $\sqrt[3]{x+1}$ I'm trying to find the Taylor series for $\sqrt[3]{x}$, but since the n-th derivative of the function $$f(x)=\sqrt[3]{x}$$
not definded at $x=0$, I've swithched to $f(x)=\sqrt[3]{x+1}$, This is my Attempt:
$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)x^n}{n!}$$
$$\sqrt[3]{x+1}=1+\frac{x}{3}-\frac{2x^2}{9\cdot2!}+\frac{10x^3}{27\cdot 3!}-\frac{80x^4}{81\cdot4!}+...$$
but when i plug $x=125 \iff \sqrt[3]{x+1}=\sqrt[3]{126}$ i get something way bigger than the acttual value which is about $5.01...$, you can check it here.
But I can't find the mistake in the equation.

For $x\gt 1$, expand in terms of $\frac1x<1$ to ensure convergence
$$\sqrt[3]{x+1}= \sqrt[3]{x} \sqrt[3]{1+\frac1x}
= \sqrt[3]{x}  (1+\frac{1}{3x}-\frac{2}{9\cdot2!}\frac1{x^2}+\frac{10}{27\cdot 3!}\frac1{x^3}-\frac{80}{81\cdot4!}\frac1{x^4}+...)
$$
Then, set $ x=125$ to obtain the approximation
$$\sqrt[3]{126} =5(1+\frac1{3\cdot 125}-\cdots )=5.0133$$

The binomial series for $(a+x)^\alpha$ is only valid if the absolute value of x is smaller than one, unless $\alpha$ is a non-negative integer. Your value is x=125>>1 and $\alpha$ is not integer, thus the divergence. To get a good approximation you would have to expand around $x_0=125$, which of course requires knowledge of the third root of 126.


Finding the intersect point between between a parabola and a straight line $x^2 −2x = \frac{1}{2}x+1$

So I have a graph with having both of these equations and put the equations as one. I realize I have to turn this into $x^2-2.5x-1=0$ and then I have to factorize it but how does something times into $1$ and adds into $-2.5$

There is a quadratic formula that lets you use the coefficients in a 2nd degree polynomial to find the variable. Let ax^2+bx+c=0

x= -b + or - root of b^2 - 4 a c
divided by 2a.

Excuse the form, I'm not very familiar with formatting on this website.

You have to use the quadratic formula:

$$\frac{2.5\pm\sqrt{10.25}}{2}.$$


Number of probability distributions on a finite set Is the cardinality finite (of the set of probability distributions on a finite set $A$)? Countable? Uncountable?

I think finite gets rule out because the probability distribution on a 2 element set can assign any probability in $[0,1]$ to the second element, so there should be an uncountable number of probability distributions?

This answer Say a probability distribution  over reals has cardinality $2^{\mathbb{N}_0}$.  So it is indeed uncountable?

(Edited in response to comment) Answer is: Uncountable if the set has cardinality of at least two, as shown in the example in the question. If the set has cardinality 1 then there is only one possible probability distribution.


Get prime number identifying function? $$\prod_{prime}(1-\frac{1}{p^s})\; = \frac{1}{\zeta(s)}$$
$$\Rightarrow\;\prod_{n=2}^{\infty}(1-\frac{1}{n^s})^{v(n)}\; = \frac{1}{\zeta(s)}, \;\;\;\;
v(n) = \left\{\begin{array}{l}1,\;n\;is\;prime\\0,\;otherwise\end{array}\right.$$
$$\Rightarrow\;\sum_{n=2}^{\infty}v(n)\cdot \ln(1-\frac{1}{n^s})\;=-\ln\zeta(s)$$
So by letting s = 2, 4, 6, 8, 10,...
$$\ln(1-\frac1{2^2})\cdot v_2\;+\;\ln(1-\frac1{3^2})\cdot v_3\;+\;\ln(1-\frac1{4^2})\cdot v_4\;+\;...\;=-\ln\frac{\pi^2}{6}$$
$$\ln(1-\frac1{2^4})\cdot v_2\;+\;\ln(1-\frac1{3^4})\cdot v_3\;+\;\ln(1-\frac1{4^4})\cdot v_4\;+\;...\;=-\ln\frac{\pi^4}{90}$$
$$\ln(1-\frac1{2^6})\cdot v_2\;+\;\ln(1-\frac1{3^6})\cdot v_3\;+\;\ln(1-\frac1{4^6})\cdot v_4\;+\;...\;=-\ln\frac{\pi^6}{945}$$
and so on.

Here is my point: this is a set of 1st order equations with unknowns
$v_2, v_3,v_4,v_5,...$ And since they are just first-ordered equations, it is actually possible to solve for $v$!

But I have no idea. Anybody that has ideas?

Also tell me if you think there are "obstacles" that can't make this happen. Thanks.

Let $Z_m(s) = -\sum_{i=1}^m \log(1-p_i^{-s})$. You'll recover the primes from $\log \zeta(2k)-Z_m(2k) \sim p_{m+1}^{-2k}$ so that

$$p_{m+1} = \lim_{k \to \infty}(\log \zeta(2k)-Z_m(2k))^{-1/2k}, \qquad Z_{m+1}(s) = Z_m(s)-\log(1- p_{m+1}^{-s})$$

Equivalently let $P(s) = \sum_{p} p^{-s} $ then $$\log \zeta(s)= \sum_{n=1}^\infty \frac{P(ns)}{n} \implies P(s) = \sum_{n=1}^\infty \frac{\mu(n)}{n}\log \zeta(ns)$$

Obtaining that
$$P(2k) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(2kn)=\sum_{n=1}^\infty \frac{\mu(n)}{n} \log[(-1)^{kn+1}\frac{B_{2kn}(2\pi)^{2kn}}{2(2kn)!}]$$

And you'll recover the primes from $$P_m(s) = \sum_{i=1}^m p_i^{-s}, \qquad p_{m+1} = \lim_{k \to \infty} (P(2k)-P_m(2k))^{-1/2k}, \qquad P_{m+1}(s) = P_m(s)+ p_{m+1}^{-s}$$
As you see, this is not very different and much simpler to using the Sieve of Eratosthenes.

Solving a set of linear equations is easy enough, but here you have an infinite set of equations in infinitely many unknowns. Such a problem can in general not be solved.


Proof that sets with 1 element exist from these axioms I have:

Axiom of existence: there exists a set with no elements
Axiom of extensionality (equality basically?): if two sets have the same elements they are identical
Lemma showing the empty set is unique
Axiom of schema of comprehension: $\{x\in A|P(x)\}$ exists where P is a property.
Lemma showing the set given in the comprehension axiom is unique (justifying my notation)
Axiom of a pair: for any A, B there is a set C such that $x\in C\iff[x\in A\text{ or }x\in B]$

The book then goes on to say "We introduce the notation {A,B} for the unordered pair of A and B, in particular if A=B we write {A} instead of {A,A}"

Is this suggesting that 1 element sets exist? Can I justify that with comprehension?

Or is it simply just notation "We're lazy and don't want to write {A,A}" (I'm hoping for this)

I ask because I'm trying to be rigorous, so far I am happy that: sets containing nothing exist, I can have pairs (but I must not think sets containing two things, because {A} is just one thing) right?

I do not like that pair means "set containing two things"

This is how I like to think about it. The first remark uses the axiom of extensionality.

[Sorry for the picture; I prefer to typeset in my LateX editor with \newcommands + autocompletion]

Since $A$ in the axiom of  schema of comprehension, and $A$ and $B$ in the axiom of pair, are sets, nothing in the 5 axioms and lemmas you provided says that there exists a set other than the empty set (that is, pair axiom as you have stated it does not give you a set containing two elements).

So it is no surprise that you can't see it logically implying existence of a set with one element.