Checking Understanding of DFA Regular Operations - Intersection and Star I'm currently taking a Logics course, and trying to understand the regular operations, intersection and star. I have a question regarding the work I have done so far.

Given the following information:

  $L_A$ and $L_B$ are regular languages defined by DFAs $A$ and $B$.
  $N_A$ and $N_B$ are the number of states in DFAs $A$ and $B$.

  What are the HIGHEST number of states one would need in DFAs for the languages

  $L_A \cap L_B$
  ${L_A}^*$ (star)

INTERSECTION operation

I have understood it so that the state of the new automaton A∩B will be the pairs of states in A and B. That is:

Na∩b = Na x Nb.

That is, the HIGHEST number of states needed in the DFA for language A∩B is the product of the number of states in A (Na) and B (Nb). Is this correct?

Another question: how does this number differ from the highest number of states in the CONCATENATION operation?

STAR operation:

For the star operation, I'm at a loss. Any advice on how I can approach this?

Your question is formally incorrect because you are using the same notation for a language and an automaton and also because your notation $Na \cap b$ is not clear at all.
However, the meaning of your question is clear and the answer is known, but not easy.
So let me first reformulate your question in a more rigorous way and then give the answer.

The (state) complexity of a regular language $L$ is the number of states $c(L)$ of its minimal DFA.

Question 1. Estimate $\max\{ c(L_1 \cap L_2) \mid c(L_1) = n_1 \text{ and } c(L_2) = n_2 \}$

Question 2. Estimate $\max\{ c(L_1L_2) \mid c(L_1) = n_1 \text{ and } c(L_2) = n_2 \}$

Question 3. Estimate $\max\{ c(L^*) \mid c(L) = n \}$

Answers.

\begin{align*}
   \max\{ c(L_1 \cap L_2) \mid c(L_1) = n_1 \text{ and } c(L_2) = n_2 \}  &= n_1n_2\\
   \max\{ c(L_1L_2) \mid c(L_1) = n_1 \text{ and } c(L_2) = n_2 \} &= n_12^{n_2} - 2^{n_2-1}\\
  \max\{ c(L^*) \mid c(L) = n \} &= 2^{n-1}+2^{n-2}
\end{align*}
These results hold for languages over an alphabet containing at least two-letters (and even 3 letters for the product). For one-letter alphabet, the values are also known, but are different.

References

S. Yu, Regular languages, in Handbook of language theory, G. Rozenberg et A. Salomaa (ed.), vol. 1, ch. 2, pp. 679–746, Springer, 1997.

For related questions, I recommend reading the articles of H. Gruber, M. Holzer, M. Kutrib, H. Petersen, where you will find plenty of further references.

My naive estimates for the number of states needed:

Concatenation $L_A+L_B$ can be detected by "chaining" $A$ and $B$ automata, thus not more than $N_A+N_B$ states.

Intersection needs running $A$ and $B$ "simultaneously", thus using the product automaton with not more than $N_A N_B$ states.

The Kleene hull ${L_A}^* = L^0 \cup L \cup L^2 \cup \ldots$ would need finite repetition of $A$. We could extend it into a $\epsilon$-NFA with extra $\epsilon$-transitions from the final states to the initial state. The number of states would not increase. Turning this into a DFA would result in a DFA with up to $2^{N_A}$ states.

All those automata could be minimized, so lower upper bounds are possible.


If $abc=1$, then $\frac{a^{n+2}}{a^n+(n-1)b^n}+\frac{b^{n+2}}{b^n+(n-1)c^n}+\frac{c^{n+2}}{c^n+(n-1)a^n} \geq \frac{3}{n} $ Let $a$, $b$ and $c$ be positive real numbers with $abc=1$. Prove that
  $$
\frac{a^{n+2}}{a^n+(n-1)b^n}+\frac{b^{n+2}}{b^n+(n-1)c^n}+\frac{c^{n+2}}{c^n+(n-1)a^n} \geq \frac{3}{n}
$$
  for each integer $n$.

I have used Cauchy-Schwarz inequalities and Jensen inequality. But I am stuck. I need some idea and advice on this problem. Induction would be cruel.

Using AM-GM we get (hereon $\sum$ denotes cyclic sums): $$\sum \frac{a^{n+2}}{a^n+(n-1)b^n} =\sum \left( a^2- (n-1)\frac{a^2b^n}{a^n+(n-1)b^n}\right) \\ \ge \sum\left( a^2- (n-1)\frac{a^2b^n}{n \cdot a\cdot b^{n-1}}\right)= \sum a^2-\frac{n-1}n\sum ab$$

So it is enough to show that
$$n \sum a^2 \ge (n-1)\sum ab+3$$
which follows from $\sum a^2 \ge \sum ab$ and $\sum a^2 \ge 3$ by AM-GM.

You can see that every term has value $\frac{1}{n}$ in case $a = b = c = 1$.
Try to prove that changing one of the variables (let $a > 1$,  $b < 1$,  $c < 1$ be the first case and $a > 1$, $b > 1$, $c < 1$ - the second one and $a > 1$, $b < 1$, $c > 1$ - the third one) increases the total sum.


Searching function starting exactly constant and approaching another constant for the default of an R API parameter i seek a function that has the property of yielding a good guess.

I want the function to be defined for $\mathbb Z^+$ (But no reason not to define it for $\mathbb R^+$, i guess)
I want it to be smooth
It should satisfy both

$$\mathrm f(x_{small}) = 1 \forall x_{small} \in \left(0, k\right]$$
$$\lim_{x \rightarrow \infty} \mathrm f(x) = l$$

edit to be clear: $k$ and $l$ should be constants appearing in the function definition.

E.g. with $k = 1000$ and $l = 0$, the $f(x) = 1 \forall x \in (0, 1000]$. Then it should gradually decline and approach 0.

To simplify, the function can be written as:

$$
\mathrm f(x) = \begin{cases}
1     & \text{if } x \le k \\
[\dots] & \text{otherwise}
\end{cases}
$$

One example is:

$$f(x)=\frac{\sqrt{x-1}}{\sqrt{x-1}+1}$$


How was the quadratic formula created? I have tested the quadratic formula and I have found that it works, yet I am curious as to how it was created. Can anybody please tell me one of the ways that it was created?

We begin with the equation $ax^2 + bx + c = 0$, for which we want to find $x$. We divide through by $a$ first, and then bring the constant term to the other side:

$$x^2 + \frac b a x = - \frac c a $$

Next, we complete the square on the left-hand side. Remember, to complete the square, you take half of the coefficient of the linear term, square it, and add it both sides. This means we add $(b/2a)^2 = b^2/4a^2$ to both sides:

$$x^2 + \frac b a x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac c a$$

The left-hand side factors as a result, and we combine the terms on the right-hand side by getting a common denominator:

$$\left(x + \frac b {2a} \right)^2  = \frac{b^2 - 4ac}{4a^2}$$

We now take the square root of both sides:

$$x + \frac b {2a}   = \pm \sqrt{ \frac{b^2 - 4ac}{4a^2}}$$

Solve for $x$:

$$x =- \frac b {2a} \pm \sqrt{ \frac{b^2 - 4ac}{4a^2}}$$

Recall that $\sqrt{a/b} = \sqrt a / \sqrt b ^{ \; \text{(note 1)}}$. Using this property, the denominator of our root becomes $2a$, giving a common denominator with $-b/2a$. Thus,

$$x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$$

yielding the quadratic formula we all know and love. $^{ \; \text{(note 2)}}$

Footnotes:

Note $(1)$ - The usual properties for roots, and exponents in turn (since $\sqrt[n] a = a^{1/n}$) that most people are familiar with, do not always hold. In particular, for example

$$\sqrt{\frac a b} = \frac{\sqrt a}{\sqrt b} \;\;\;\;\; \sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$$

do not hold if $a,b$ are complex numbers. They hold if $a,b$ are nonnegative real numbers (and, in the first property shown, $b \ne 0$). A well-known example of why the second does not hold involves $i$, the complex number such that $i^2 = -1$. If this property held,

$$-1 = i^2 = \sqrt{-1} \sqrt{-1} = \sqrt{(-1)\cdot (-1)} = \sqrt{1} = 1$$

but $-1 \ne 1$.

In light of this note, note that we do not necessarily have any problems splitting up the root as we do in the proof. Even if $a<0$, $a^2 > 0$ as a result (of course, we also assume $a,b,c$ are real numbers in this derivation).

Note $(2)$ - As a note of interest, you are only guaranteed that $x$ is a real solution whenever the discriminant of the quadratic is nonnegative. The discriminant is the expression under the root; thus

$$ax^2+bx+c = 0 \; \text{has real solutions if and only if} \; b^2-4ac \ge 0$$

In particular, if $b^2 - 4ac = 0$, then the solutions $x$ gives are the same (what is called a "double root" or a "root of multiplicity $2$"). For $b^2 - 4ac >0$, you are ensured two distinct real values for your solutions.

If $b^2 - 4ac < 0$, then your solutions will instead be complex numbers. You will still have two distinct solutions however.

The way to get it is actually a beautiful one. The original proof is different since algebra was still in an early stage of development, but here's how I would do it today:

Let's consider the equation $ax^2+bx+c=0$ If a=0, we have a first degree equation. Otherwise, our equation can be written as $x^2+\frac{b}{a}x +\frac{c}{a}=0$

On the other hand, $(x+\frac{b}{2a})^2=x^2+\frac{b}{a}x+\frac{b^2}{4a^2}$. This looks like an idea completely out of the air, but you can get it by trying different ones and checking which one works. As you can see, the first two terms look like those form the equation, so we can say that $x^2+\frac{b}{a}x = (x+\frac{b}{2a})^2 - \frac{b^2}{4a^2}$

Now go back to the equation and replace $x^2+\frac{b}{a}x$ We get that $(x+\frac{b}{2a})^2 - \frac{b^2}{4a^2} + \frac{c}{a} = 0$ From this equation, it's just basic algebra.

$(x+\frac{b}{2a})^2 = \frac{b^2}{4a^2} - \frac{c}{a}$ Take the square root on both sides and enjoy your formula (unless I made a mistake, which I probably have)


Do we necessarily have that $W^{2, p}(I) \subset C^1(\overline{I})$ with compact injection? Let $I = (0, 1)$ and $p > 1$. Do we necessarily have that$$W^{2, p}(I) \subset C^1(\overline{I})$$with compact injection?

Yes. In fact, we have the following more general result.

  Theorem (Adams, page 168). Let $\Omega$ be a domain in $\mathbb R^n$ and $\Omega_0$ be a bounded subdomain of $\Omega$. Let $j \geq  0$ and $m \geq  1$ be integers, and let $
1 \leq p < \infty$. If $\Omega$ satisfies the strong local Lipschitz condition, then the following embedding is compact:
  $$W^{j+m,p}(\Omega)\subset C^j(\overline{\Omega_0}),\qquad \text{if}\quad mp>n.$$

  Notes:

  If $\Omega$ is bounded, we may have $\Omega_0=\Omega$ in the statement of the theorem.
  The bounded subdomain $\Omega_0$ of $\Omega$ may be assumed to satisfy the cone condition if $\Omega$ does.

The proof of your particular case, can be done as follows.

Fact 1. $W^{2,p}(I)\subset C^1(\overline{I})$

Proof: Take $u\in W^{2,p}(0,1)$. Define
$$h(x)=\int_0^xu''(t)\;dt,\qquad g(x)=\int_0^xu'(t)\;dt.$$
Then,
$$\int_0^1(u'-h)\varphi'=\int_0^1u'\varphi'-\int_0^1h\varphi'=-\int_0^1u''\varphi+\int_0^1u''\varphi=0,\quad\forall\ \varphi\in C^1_c(0,1)$$
and thus $u'-h=c$ for some constant $c$. This imply $u'=h+c\in C(\overline{I})$. Analogously, there is a constant $c_2$ such that $u=g+c_2\in C(\overline{I})$. So, $W^{2,p}(I)\subset C^1(\overline{I})$.

Fact 2. The inclusion $W^{2,p}(I)\subset C^1(\overline{I})$ is continuous.

Proof: Take $u\in W^{2,p}(I)$ and $x\in I$. Then,
$$u'(x)=u'(0)+\int_0^xu''(t)\;dt$$
and thus
$$|u'(x)|\leq |u'(0)|+\|u''\|_{L^1},\quad |u'(0)|\leq |u'(x)|+\|u''\|_{L^1}.\tag{1}$$
From the second inequality in $(1)$,
$$|u'(0)|=\int_0^1|u'(0)|\;dt\leq \|u'\|_{L^1}+\|u''\|_{L^1}$$
So, from the first inequality in $(1)$,
$$|u'(x)|\leq \|u'\|_{L^1}+2\|u''\|_{L^1}\leq c_0\|u\|_{W^{2,p}}$$
for some constant $c_0$.
A similar argument shows that
$$|u(x)|\leq \|u\|_{L^1}+2\|u'\|_{L^1}\leq c_0\|u\|_{W^{2,p}}.$$
As $x\in I$ is arbitrary, we get
$$\|u\|_{C^1}=\sup_{x\in\overline{I}} |u(x)|+\sup_{x\in\overline{I}} |u'(x)|\leq 2 c_0\|u\|_{W^{2,p}}.$$

Fact 3. The inclusion $W^{2,p}(I)\subset C^1(\overline{I})$ is compact.

Proof: Let $(u_n)$ be a bounded sequence in $W^{2,p}(0,1)$, say by a constant $M$.

The sequence $(u'_n)$ is uniformly bounded because, from the proof of the Fact 2, $|u'_n(x)|\leq c_0 M$ for all $n\in\mathbb{N}$ and all $x\in \overline{I}$. Also, $(u'_n)$ is equicontinuous because, from the Hölder's inequality,
$$|u'_n(x)-u'_n(y)|=\left|\int_y^xu_n''(t)\;dt\right|\leq \|u''\|_{L^p}\left(\int_y^x1\;dt\right)^{\frac{1}{p}}\leq M|x-y|^{1/p}$$
for all $n\in\mathbb{N}$ and all $x,y\in \overline{I}$. So, the Arzelà–Ascoli Theorem implies that $(u'_n)$ has a subsequence $(u'_{n_k})$ that converges uniformly to some $g\in C(\overline{I})$. A similar argument shows that $(u_{n_k})$ has a subsequence $(u_{n_{k_m}})$ that converges uniformly to some $u\in C(\overline{I})$.

It follows from Theorem 7.17 in Rudin's book that $u'=g$ and thus $(u_n)$ has a subsequence $(u_{n_{k_m}})$ that converges in $C^1(\overline{I})$.


$f: \mathbb{R^2} \to \mathbb{R}$ a $C^\infty$ function such that $f(x,0)=f(0,y)=0$ then exists $g$ such that $f(x,y)=xy\, g(x,y)$ I'm trying to solve the following exercise

Let $f: \mathbb{R^2} \to \mathbb{R}$ a $C^\infty$ function such that $f(x,0)=f(0,y)=0$ $\forall x,y \in \mathbb{R}$.
Then there exists a $C^\infty$ function $g:\mathbb{R^2} \to \mathbb{R}$ satisfying $f(x,y)=xy\, g(x,y)$ $\forall x,y \in \mathbb{R}$

I thought first about expanding $f$ using the taylor theorem, we would have $$f(x,y)=f(0,0)+f'(0,0)(x.y)+\frac{f''(0,0)(x,y)^2}{2!}+r(x,y),$$
but $f(0,0)=0$ and $f_x(0,0)=f_y(0,0)=0$ since $f$ is zero on the $x$ and $y$ axis.
$f''(0,0)(x,y)^2=f_{xx}(0,0)x^2+f_{xy}(0,0)xy+f_{yy}(0,0)y^2$, now we have a term that has a product of $x$ and $y$. But I don't know how (and if it is possible) to proceed from here.

We are expanding the hint by zhw.
\begin{align}
f(x, y) &= f(x, y) - f(0,0) \\
&= \int_0^1 \frac{d}{dt} f(tx, ty)\, \mathrm dt \\
&= \int_0^1 \left( x f_x (tx, ty) + y f_y (tx, ty) \right)\, \mathrm dt \\
&= x\int_0^1  f_x (tx, ty) \, \mathrm dt  + y  \int_0^1 f_y (tx, ty) \, \mathrm dt
\end{align}
Since $f(x, 0) = 0$, $f_x(tx, 0) = 0$. Thus
\begin{align}
 \int_0^1 f_x (tx, ty)\, \mathrm dt &= \int_0^1 (f_x (tx, ty) - f_x(tx,0)) \,\mathrm dt \\
&= \int_0^1 \int_0^t \frac{d}{ds} f_x (tx , sy)\, \mathrm ds\, \mathrm dt \\
&= y \int_0^1 \int_0^t f_{xy} (tx, sy) \, \mathrm ds\, \mathrm dt \\
&:= y g_1(x, y)
\end{align}
Similarly, using $f(0, y) = 0$, $f_y(0, ty) = 0$. Thus
$$ \int_0^1 f_y (tx, ty)\, \mathrm dt = x \int_0^1 \int_0^t f_{yx} (sx, ty) \, \mathrm ds\, \mathrm dt := x g_2(x, y).$$
Since $f$ is smooth, it is clear that $g_1, g_2$ are smooth. Thus
$$ f(x, y) = xy g(x, y)$$
with $g = g_1 + g_2$.

Define $g(x,y)$ like this :

$x\neq 0 \land y \neq 0 : \frac{f(x,y)}{xy}=g(x,y) $
$g(0,y_0)=\lim_{(x,y) \to (0,y_0)} \frac{f(x,y)}{xy}$
$g(x_0,0)=\lim_{(x,y) \to (x_0,0)} \frac{f(x,y)}{xy}$
$g(0,0)=\lim_{(x,y) \to (0,0)}   \frac{f(x,y)}{xy} $

The above limits must exist. To show this we can use the fact that $f(x,y)$ is continuously differentiable. This means  : $\frac{\partial}{\partial x}f(x,y)=\lim_{h \to 0} \frac{f(x+h,y)-f(x,y)}{h}=\lim_{(x',y') \to (x,y)} \frac{f(x',y')-f(x,y')}{x'-x}$ , the last limit being independent of the chosen path. Making $\frac{\partial}{\partial x}f(x,y)$ a continuous function.

So for example let $y_0 \neq 0$ :
$ g(0,y_0) = \lim_{(x,y) \to (0,y_0)} \frac{f(x,y)}{xy}=
 \lim_{(x,y) \to (0,y_0)} \frac{f(0+x,y)-f(0,y)}{xy}=
 \lim_{(x,y) \to (0,y_0)} \frac{1}{y}\left[\frac{\partial}{\partial x}f(x,y)|_{(0,y_0)}\right]=\frac{1}{y_0} \frac{\partial}{\partial x}f(x,y)|_{(0,y_0)}
$

Or:$ g(0,0) = \lim_{(x,y) \to (0,0)} \frac{f(x,y)}{xy}=
 \lim_{(x,y) \to (0,0)} \frac{f(0+x,y)-f(0,y) -xf_x(x,0)}{xy}=  \lim_{(x,y) \to (0,0)} \frac{\frac{f(0+x,y)-f(0,y)}{x} -f_x(x,0)}{y}  =  \lim_{(x,y) \to (0,0)} \frac{f_x(x,y) -f_x(x,0)}{y} =\frac{\partial}{\partial y}\frac{\partial}{\partial x}f(x,y)|_{(0,0)} $

And we know that $f(x,y)$ is $C^\infty $ so the partial derivatives exist in $x,y=0$ .

So the above definition gives a continuous function $g(x,y)$.

To prove $g(x,y)$ is $C^\infty$ in the limit $x,y \to 0$ we can use similar argument as above knowing that $f(x,y)$ is $C^\infty$ .

Example showing $g(x,y)$ is continuously differentiable using the fact that  $f(x,y)$ is $C^\infty$  :

$g_x(0,y)=\frac{\partial}{\partial x'}g(x',y')|_{(0,y)} = \lim_{(x',y') \to (0,y)} \frac{g(x',y')- g(0,y')}{x'} = \\$
$\lim_{(x',y') \to (0,y)} \frac{g(x',y')- \frac{1}{y'}f_x(0,y')}{x'} = \\$
$\lim_{(x',y') \to (0,y)} \frac{\frac{f(x',y')}{x'y'}- \frac{1}{y'}f_x(0,y')}{x'} = \\$
$\lim_{(x',y') \to (0,y)} \frac{\frac{f_x(x',y')\cdot x'}{x'y'}- \frac{1}{y'}f_x(0,y')}{x'}= \\$
$\lim_{(x',y') \to (0,y)} \frac{1}{y'}\frac{ f_x(x',y')  - f_x(0,y')}{x'}=\frac{1}{y}\frac{\partial^2}{\partial x'^2}f(x',y')|_{(0,y)}$


Two weighted coins, determining which has a higher probability of landing heads A friend of mine asked me the following question, and I am not sure how to solve it:

You are given two weighted coins, $C_1$ and $C_2$.  Coin $C_1$ has probability $p_1$ of landing heads and $C_2$ has probability $p_2$ of landing heads.  The following experiment is preformed:
Coin $C_1$ is flipped 3 times, and lands heads 3 times.
Coin $C_2$ is flipped 10 times, and lands heads 7 times.
Based on this experiment, choose the coin which is more likely to have a higher probability of being heads.  In other words, which is more likely: $p_1>p_2$ or $p_2>p_1$.

Intuition tells me coin $C_1$ is the better choice, but this could be wrong, and I am wondering how do you solve this in general.  Consider the experiment, $C_1$ is flipped $n_1$ times and lands heads $m_1$ times, $C_2$ is flipped $n_2$ times and lands heads $m_2$ times.
Thanks for the help,
Edit: I think this might answer some questions:  Suppose that the probabilities of the coins, $p_1$ and $p_2$ are chosen uniformly from $[0,1]$.

As Henry mentions, I think one needs some information about the prior distributions of the weights.

Denote by $r$ the weight of a coin.  Suppose that the weight has some prior distribution $g(r)$.  Let $f(r|H=h, T=t)$ be the posterior probability density function of $r$ having observed $h$ heads and $t$ tails tossed.  Bayes' Theorem tells us that:
$$
f(r|H=h, T=t) = \frac{Pr(H=h|r, N = h+t)g(r)}{\int_0^1 Pr(H=h|r, N = h+t)g(r)\ dr}.
$$

This should allow you to answer your question.  Once you have the posterior pdf for each coin, just find their respective expected weights.

I agree with the beta approach, but given the question, I think it makes more sense to plot out the results and compare visually:

$x$ <- seq($0,1,$length $= 1000$) #set $x$ from $0$ to $1$ since we're looking at probabilities
$y_1$ <- dbeta$(x,4,1)$ #calculate density based on a prior of $[1,1]$ and a posterior of $[3,0]$.
$y_2$ <- dbeta$(x,8,4)$ #calculate density based on a prior of $[1,1]$ and a posterior of $[7,3]$.
plot($x,y_1$,type = "l") #plot density of coin 1 in black
lines($x,y2$,col $= 2$) #plot density of coin 2 in red

This yields the following:
Think of the densities as representing the "probability of probability".

Coin 1 (represented by the black line) has a higher probability of having a higher probability of showing heads.

Coin 2 (represented by the red line) has a lower probability of having a higher probability of showing heads (or a higher probability of having a lower probability of showing heads...).


A very different, difficult geometry problem that need to be helped- The problem:

  In $\triangle PQR$, $ST$ is the perpendicular bisector of $PR$, and $PS$ is the bisector of $\angle QPR$ and the point $S$ lies on $QR$. If $|QS|=9$ and $|SR|=7$ and $|PR|=x/y$, where $x$ and $y$ are co-primes, then what is $x+y$?

I can come up with these steps only:

$\angle STR$ is equal to $90^\circ$, so $\triangle STR$ is a right triangle. The same with $\Delta STP$. In $\triangle PST$ and $\triangle STR$, $ST\cong ST$, $PT\cong TR$, and $\angle PTS= \angle STR$, so that $\triangle PST\cong\triangle STR$. Therefore, $|PS|=|SR|=7$, and $\angle SPT\cong\angle SRT$.

That's the end.

I have no clue how to solve the next steps, or to solve the problem. I don't even know if the above steps lead toward an answer to the problem!

So please help me see how I can solve this step. I am a 8th grader. So please be careful, so that I can understand the steps.

Sorry, I can't add the figure.

$\angle SRP=\angle SPR = \angle SPQ=\theta,$ say.  Then adding up the angles in $\triangle PQR$ gives $\angle Q = 180^{\circ}-3\theta,$ and adding up the angles in $\triangle PSQ$ gives $\angle PSQ = 2\theta=\angle QPR.$  That is, $\triangle  PQR$ and $\triangle PSQ$ are similar.

Corresponding sides of similar triangles are proportional, so we have $$
\frac{|QP|}{9}=\frac{16}{|QP|}$$  (On the left-hand side we have two sides of $\triangle PQS,$ and on the right-hand side two we have two sides of $\triangle PQR.)$  We conclude that $|QP|=12.$  Now using similarity once again, we have $$\frac{|PR|}{|PQ|}=\frac{7}{9}\implies |PR|=\frac{28}{3}$$

That is, $\boxed{x=28,\space y=3,\space x+y=31}$.

I think this question is contradictory. If you draw a line segment $PR$ and perpendicularly bisect it. Now draw $angle QPR$ with Q as an unfixed point. You will see that the value of $PT,PS,$ $angle QPR$ will remain same but size of $QS$ will differ as $Q$ is an unfixed point along $rayPQ$.
So length of $QS$ is unwanted information.
Hence, it is impossible to find ${x\over y}$.


How many pairs of primes are there such that $pq < 10^6$ I'm trying to implement this function in my program, so I think that I should find that the number of pairs $$(p, q)$$ such that both p, q are prime numbers and: $$p\cdot q < 10^6$$

I'm not really good in number theory, I know that there are $78498$ primes under one million, but I know that only some of those primes form combinations that are less that one million.

Thanks in advance.

There are
$$\sum_{\substack{p\text{ prim}\\p\,\le\, 10^6}}\pi\left(\frac{10^6}p\right)=419902$$

pairs, where $\pi(x)$ is the prime counting function. $10^6$ is not the product of two primes. So it suffices to count for every prime $p\le 10^6$ the number of primes $q$ less or equal $10^6/p$, because then $pq<10^6$. Actually, we only have to check $p\leq 10^6/2$ as the smallest other prime to multiply with is $2$.

I chose this form above because this was actually feasible to calculate on my computer using Mathematica. Mathematica certainly implements an efficient version of $\pi(x)$.

Here is an even faster version to execute (motivated by Roddy):

$$2\cdot\sum_{\substack{p\text{ prim}\\p\,\le\, 1000}}\pi\left(\frac{10^6}p\right)-\pi(1000)^2=419902.$$

It uses that at least one factor has to be at most $1000=\sqrt{10^6}$. So we sum only over primes $p\le 1000$. The factor $2$ then account for the flipped version $(p,q)\to(q,p)$. As this counts everything twice with $p,q\le 1000$, we have to subtract this in the end.

Just make a list of prime numbers upto $5\times10^5$ let it be $l_1$.
Create a list to store $p$ and $q$ let it be $l_2$.
Iterate over the list of prime element  and let the element you get by each iteration be $p$.
Enter $l_1$ by removing all elements less than $p$ to a function along with $l_2$, the work of function is given below -
Iterate over list starting from $p$ and let the element you get each time be $q$.
Check at every step if $pq$ is less than million if it is than store $p$ and $q$ in the list and if it is not than break.


Find the derivatives to $f(x)=4/x^2$ and $g(t)=(t-5)/(1+\sqrt{t}\,)$ I have these two assignments:

  Find the derivatives to (a) $f(x)=4/x^2$ and b) $g(t)=(t-5)/(1+\sqrt{t}\,)$ by using the definition $$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=f'(x)$$

a) This is my attempt at (a); am I correct?
$$\lim_{h \to 0} \frac{\left(\displaystyle\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)}{h}=\frac{1}{h}\left(\displaystyle\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)$$
Then I found the common denominator
$$\begin{align*}
\frac{4}{(x+h)^2}-\frac{4}{x^2}
&=\frac{4x^2}{(x+h)^2x^2}-\frac{4(x+h)^2}{(x+h)^2x^2} \\[6pt]
&=\frac{4x^2-4x^2-4h^2-8xh}{(x+h)^2x^2} \\[6pt]
&=\frac{-4h^2-8xh}{(x+h)^2x^2}
\end{align*}$$
Then I expand the denominator to
$$\frac{-4h^2-8xh}{x^4+2x^3h+h^2x^2}
=\frac{\left(-4h^2-8h\right)}{x^3+2x^3h+h^2x^2}\frac{1}{h}
=\frac{\left(-4h^2-8h\right)}{hx^3+2x^3h+h^2x^2}$$
And $h$ goes out with:
$$\frac{-4h^2-8}{x^3+2x^3h+h^2x^2}
= \lim_{h \to 0}\frac{-4\cdot0^2-8}{x^3+2x^3\cdot 0+0^2\cdot x^2}
= -\frac{8}{x^3}$$
I dont know how to do part (b)

Let us cheat a little and first establish the general formula for the division:
$$\frac{\left(\frac fg\right)(x+h)-\left(\frac fg\right)(x)}h=\frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x+h)}}h\\=\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)h}\\=\frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{g(x+h)g(x)h}\\
=\frac{(f(x+h)-f(x))g(x)-f(x)(g(x+h)-g(x))}{g(x+h)g(x)h}$$

By taking the limit $h\to0$, we have

$$\left(\frac fg\right)'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}.$$

Now let's work on the numerators and denominators alone:
$$(4)':\frac{4-4}h=0\to 0,$$
$$(x^2)':\frac{(x+h)^2-x^2}h=\frac{2xh+h^2}h=2x+h\to 2x,$$
$$(t-5)':\frac{(t+h-5)-(t-5)}h=1\to1,$$
$$(1+\sqrt t)':\frac{1+\sqrt{t+h}-1-\sqrt t}h=\frac{\sqrt{t+h}-\sqrt t}h=\frac{t+h-t}{(\sqrt{t+h}+\sqrt t)h}\to\frac1{2\sqrt t}.$$

Putting all together,
$$\left(\frac4{x^2}\right)'=\frac{0.x^2-4.2x}{(x^2)^2}=-\frac8{x^3}.$$

$$\left(\frac{t-5}{1+\sqrt t}\right)'=\frac{1.(1+\sqrt t)-(t-5)\frac1{2\sqrt t}}{(1+\sqrt t)^2}=\frac{2\sqrt t+t+5}{2\sqrt t (1+\sqrt t)^2}.$$


How many possible orders are there? A tapas bar serves 15 dishes, of which 7 are vegetarian, 4 are fish and 4 are meat. A table of customers decides to order 8 dishes, possibly including repetitions.

a) Calculate the number of possible dish combinations.

b) The customers decide to order 3 vegetarian dishes, 3 fish and 2 meat. Calculate the number of possible orders.

Progress. For a) I think that the answer would be $15^8$ as this would be the number of different ordered sequences of 8 elements from the 15 possible dishes.

a) First, we note that repetition is allowed, and the order in which we order the dishes is unimportant. Therefore, we use the formula ${k+n-1 \choose k}$. In this case, $n=15$ and $k=8$. So, the number of possible combinations of dishes is
$${8+15-1 \choose 8} = {22 \choose 8} = 319770.$$

b) We will need to apply the above formula three times. For the vegetarian dishes, we have $n_V = 7$ and $k_V=3$. For the fish dishes, we have $n_F = 4$ and $k_F=3$. For the meat dishes, we have $n_M = 4$ and $k_M=2$. So, the number of possible combinations of dishes under these conditions is
$${k_V+n_V-1 \choose k_V}{k_F+n_F-1 \choose k_F}{k_M+n_M-1 \choose k_M} = {9 \choose 3}{6 \choose 3}{5 \choose 2}=16800.$$

a) The given that all dishes are different, then
$$
N_{orders}=15^8
$$

b) In this case, we can calculate how many different orders for each type and then combine them:
$$
N_{orders} = 7^3 \times 4^3 \times 4^2
$$


Multivariable function: $x^3 + y^3 + e^{zw}$ - what to do after taking partial derivatives? I'm working on this problem. I'm kinda stuck and would be very grateful for any help. A have to compute the gradient of this function

$$f(x,y,z,w) = x^3 + y^3 + e^{zw}$$

So, from what I know I should take the partial derivative with respect to each variable:

$$\begin{align}
\frac{\mathrm df}{\mathrm dx} &= 3x \\
\frac{\mathrm df}{\mathrm dy} &= 3y \\
\frac{\mathrm df}{\mathrm dz} &= we^z \\
\frac{\mathrm df}{\mathrm dw} &= ze^w
\end{align}$$

My problem is, is that the right approach? what do I do from here?

You are right that you should take the partial derivative with respect to each variable, but all your derivatives are wrong. The correct derivatives are:
$$\frac{\partial f}{\partial x} = 3x^2$$
$$\frac{\partial f}{\partial y} = 3y^2$$
$$\frac{\partial f}{\partial z} = we^{wz}$$
$$\frac{\partial f}{\partial w} = ze^{wz}$$
The gradient is then the vector valued function
$$\nabla f(x,y,z,w) = \begin{bmatrix}
           \frac{\partial f}{\partial x} \\
           \frac{\partial f}{\partial y} \\
           \frac{\partial f}{\partial z} \\
           \frac{\partial f}{\partial w}
         \end{bmatrix} = \begin{bmatrix}
           3x^2 \\
           3y^2 \\
           we^{wz} \\
           ze^{wz}
         \end{bmatrix}$$

$$ \nabla f \stackrel{\text{def}}{=} \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}, \frac{\partial f}{\partial w} \right).$$


find the derivative of an integral Find $f'(x)$ where $f(x)$ is the integral from ${\sqrt{x}}$ to $x$ of $e^x-e^{t^2} dt$

Is there an easy way to do this using the fundamental theorem of calculus because if I try to ingretate w.r.t $t$ then $e^{t^2}$ is a bit of a problem.

Apply differentiation under the integral sign, which is a formula derived from FTC:
\begin{align*}
\frac{d}{dx}\int_{\sqrt x}^x(e^x-e^{t^2})dt&=(e^x-e^{x^2})-(e^x-e^x)(\sqrt x)^\prime+\int_{\sqrt x}^x e^x dt \\&= (e^x-e^{x^2})+e^x(x-\sqrt x)
\end{align*}

Yes there is.  Assuming no pathological situations, the derivative of a definite integral is the difference of the integrand evaluated at the two endpoints.
So $$\frac{d}{dx} \int_\sqrt{x}^x \left( e^x - e^{t^2} \right) dt =
\left.
e^x - \right) dt =
\left. \left( e^x - e^{t^2} \right)
\right|_{t=\sqrt{x}}^{t=x} = e^x - e^{x^2}
$$


Pigeon Hole Principle Algorithm The “pigeonhole principle” states that if n+1 objects (e.g., pigeons) are to be distributed into n holes then some hole must contain at least two objects. This observation is obvious but useful.
Employ the pigeonhole principle to prove the following:
Claim: Let G be an undirected graph with at least two vertices. Then there exist distinct vertices v and w in G that have the same degree.
Thank you so much for all your help, a couple of problems on this homework were unlike anything we've done in class thus far. I think they're supposed to be an easy concept questions but I'm struggling.

It is tacitly assumed that the graph $G$ is finite and contains no loops or double edges. Otherwise it is easy to give counterexamples.

Let $v_0$ be the number of isolated vertices,  and denote by $v'$ the number of vertices of  degree $\geq1$. When $v_0=2$ we are done. Otherwise $v'\geq1$, and $d_\max$, the maximal occurring degree, is $\geq1$. Since a vertex of maximal degree is connected to  $d_\max$ other nonisolated vertices we have
$$v'\geq d_\max+1\ .$$
The $v'$ vertices of degree $\geq1$ can have degrees from $1$ to $d_\max$ inclusive. As   $v'>d_\max$ we can conclude by the pigeon hole principle that two of them have to have the same degree.

You have two pigeons (vertices) so that pigeon can only have one pigeon friend (edge to another vertex). But you also have two pigeons, so two pigeons and one pigeon friend imply they are friends of each other.

Inductively, you have n pigeons with n-1 pigeon friends.


Convergence of the integral $\int _0^\infty \ln^2x\sin(x^2)\,dx$ $$\int _0^\infty (\ln^2x)\sin(x^2)dx$$

Does this converge?

In the answers in our textbook, we've been told that since the inner function limit's in in infinity is not 0 then the integral doesn't converge. I couldn't find a theorem that states this though.
Are there any other good ways to prove that this integral diverges?

There is no real difficulty at $0$, since near $0$ the function $\sin(x^2)$ behaves like $x^2$, so $\lim_{x\to 0^+}\ln^2 x\sin(x^2)=0$. So we examine
$$\int_1^B (\ln^2 x)( \sin(x^2))\,dx.\tag{1}$$
Rewrite as
$$\int_1^B \frac{\ln^2 x}{2x} 2x \sin(x^2)\,dx,$$
and use integration by parts, letting $u=\frac{\ln^2 x}{2x}$ and $dv=2x\sin(x^2)\,dx$. Then  $du=\frac{2\ln x-\ln^2 x}{2x^2}\,dx$ and we can take $v=-\cos(x^2)$. Thus our integral (1) is
$$\left.\left(-\frac{\ln^2 x}{2x}\cos(x^2)\right)\right|_1^B +\int_1^B \frac{2\ln x-\ln^2 x}{2x^2}\cos(x^2)\,dx.$$
The first part gives no problem, indeed it vanishes as $B\to\infty$. The remaining integral has a (finite) limit as $B\to\infty$, because $\cos(x^2)$ is bounded and the $2x^2$ in tthe denominator crushes the $\ln$ terms in the numerator.

It follows that our original integral converges.

$\sin(x^2)>.5$ on $(\sqrt{\pi/6+2k\pi},\sqrt{5\pi/6+2k\pi}),k\in\mathbb{N}$.  When $x>e$, $\log^2(x)>1$, so we can see that the positive part of the integral is greater than $.5\sum_{k=2}^K \left(\sqrt{5\pi/6+2k\pi}-\sqrt{\pi/6+2k\pi}\right)$ for any $K\in\mathbb{N}\setminus\{1\}$.  Now see that

$$
\sum_{k=2}^K \left(\sqrt{5\pi/6+2k\pi}-\sqrt{\pi/6+2k\pi}\right)=\sum_{k=2}^K \left(\frac{2\pi/3}{\sqrt{5\pi/6+2k\pi}+\sqrt{\pi/6+2k\pi}}\right)\geq\sum_{k=2}^K \left(\frac{2\pi/3}{\sqrt{5\pi/6+2k\pi}}\right)
$$

which we see is unbounded.  We conclude that the positive part of the integral is unbounded.  Similarly, we can conclude that the negative part of the integral is unbounded.  Therefore it is not integrable.


Elliptic Curve and Differential Form Determine Weierstrass Equation I am reading Fermat's Last Theorem by Diamond, Darmon and Taylor and they state:

"An elliptic curve E over a field F is a proper smooth curve over F of genus
one with a distinguished F-rational point. If $E/F$ is an elliptic curve and if $\omega$ is a non-zero holomorphic differential on E/F then E can be realised in the projective plane by an equation (called a Weierstrass equation) of the form
$$Y^2Z + a_1XYZ + a_3Y Z^2 = X^3 + a_2X^2Z + a_4XZ^2 + a_6Z^3$$
such that the distinguished point is (0 : 1 : 0) (sometimes denoted $\infty$ because
it corresponds to the “point at infinity” in the affine model obtained by setting $Z=1$) and $\omega =\frac{dx}{2y+a_1x+a_3}$."

My question is how does the choice of $\omega$ determine the Weierstrass for $E$? Why state this in terms of differential forms instead of the usual projective embedding?

To ease notation, the Weierstrass equation is generally written using non-homogeneous coordinates $x = X/Z$ and $y = Y/Z$, then
$$E:y^2 + a_1xy+a_3y=x^3+a_2x^2+a_4x+a6$$
If $char(\mathbb{K}) \neq 2$, then we can simplify the equation by completing the square. Thus the substitution
$$y \to\frac{1}{2}(y-a_1x-a_3)$$
gives an equation of the form
$$E:y^2 = 4x^3 + b_2x^2+2b_4 + b6$$
$$b_2 = a_1^2 + 4a_4, \quad\quad b_4 =2a_4+a_1a_3, \quad\quad b_6=a_3^2+4a_6$$
We also define quantities
\begin{align}
b_8 &= a_1^2a_6+4a_2a_6-a_1a_3a_4+a_2a_3^2-a_4^2\\
c_4 &= b_2^2 - 24b_4\\
c_6 &= -b_2^3 +36b_2b_4-216b_6\\
\Delta &= -b_2^2b_8 - 8b_4^3 - 27b_6^2+9b_2b_4b_6\\
j &=c_4^3/\Delta\\
\omega &=\frac{dx}{2y+a_1x+a_3} = \frac{dy}{3x^2 + 2a_2x+a_4-a_1y}
\end{align}
Where the quantity $\Delta$ is the discriminant of the Weierstrass equation, the quantity $j$ is the $j$-invariant of the elliptic curve, and $\omega$ is the invariant differential associated to the Weierstrass equation.

References:
J. Silverman - The Arithmetic of Elliptic Curves 2-ed. s. $42$ (pdf)
Silverman and Tate's  - Rational Points on Elliptic Curves
Ludwig Bauer - Weierstrass Equations (pdf)

PS: Maybe this is not a clear answer to your questions, but surely the references will be very helpful.


Proving that $\left(\frac{\pi}{2}\right)^{2}=1+\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k-1}}$. Wolfram$\alpha$ says that we have the following identity
$$
\left(\frac{\pi}{2}\right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k}}
$$

but, how does one prove such identity?

It's well known that
$$
\pi\cot(\pi x)=\frac 1x-2\sum_{n=1}^\infty\zeta(2n)x^{2n-1};
$$
taking derivatives we get
$$
\frac{\pi^2}{\sin^2\pi x}=x^{-2}+2\sum_{n=1}^\infty(2n-1)\zeta(2n)x^{2n-2};
$$
in particular for $x=\frac12$
$$
\pi^2=4+2\sum_{n=1}^\infty(2n-1)\zeta(2n)\frac1{2^{2n-2}};
$$
this is your formula (multiplied by 4).

Do not worry about the downvote. It is just an attack on my answers.

The series can have the following integral representation

  $$ \sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k-1}} = \int_{0}^{1}\frac{(t^2+1)\ln t}{t^2-1}  dt = \frac{\pi^2}{4}-1.$$


Show that the ideal $ (2, 1 + \sqrt{-7} ) $ in $ \mathbb{Z} [\sqrt{-7} ] $ is not principal Show that the ideal $ I =  (2, 1 + \sqrt{-7} ) $ in $ \mathbb{Z} [\sqrt{-7} ] $ is not principal.

My thoughts so far:

Work by contradiction. Assume that $ I $ is principal, i.e. that it is generated by some element $ z = a + b\sqrt{-7} \in \mathbb{Z}[\sqrt{-7}] $. I'm really not sure what to consider though - I can't really 'see' what $ I $ looks like.

Any help would be greatly appreciated. Thanks

Put $\rm\, w = 1\!+\!\sqrt{-7}.\,$ By norms $2\,$ is irreducible so, if principal $\rm\,\color{#c00}{ (2,w) = (1)}\ $ [not $(2)$ by $\rm\,2\nmid w$]

so $\rm\ 2\mid 2w,\,ww'\Rightarrow\ 2\mid (2w',ww') = \color{#c00}{(2,w)}(w') =  (w'),\ $ so $\rm\ 2\mid w',\ $ contradiction. $\ \, $ QED

This is a special case of the fact that the failure of an irreducible element to be prime (or a failure of Euclid's Lemma) immediately yields a nonexistent gcd and nonprincipal ideal - see this answer.

To prove this note that $1 \not \in I$ so $I \not = (1)$. Then suppose $I = (\alpha)$, that implies that $\alpha = 2$ or $\alpha = 1 + \sqrt{-7}$ since those are both irreducibles, but neither of them can hold since one irreducible is not a multiple of another.


Uniform continuity of a bounded function on an unbounded interval Are all bounded continuous functions on an unbounded interval uniformly continuous ?

A counterexample is given by $f(x)=\sin(x^2)$.

Let $x_n=\sqrt{2\pi n}$ and $y_n=\sqrt{2\pi n+\pi/4}$. Then $$|f(x_n)-f(y_n)|=1.$$
But $|x_n-y_n|\to0$. So $f$ is not uniformly continuous.

(One way to see that $|x_n-y_n|\to0$ is to apply the Mean Value Theorem to the function $\phi(t)=\sqrt t$.)

The answer to your question is not necessarily. The counter example provided by @DavidC.Ullrich demonstrates this.

A common reason a function is not uniformly continuous is that its derivative is actually unbounded (but a uniformly continuous function can actually have an unbounded derivative). For instance, if $g(x)=\sin(x^2)$ then $g'(x) = 2x \cos(x^2)$ which can take arbitrarily large positive and negative values.

If a differentiable function has a derivative that is bounded over the whole real line, then it would be uniformly continuous. This follows from the mean value theorem.

For instance suppose that $f:\mathbb{R} \to \mathbb{R}$ is differentiable and that $|f'(x)| < M$ for all $x \in \mathbb{R}$. When we consider $x < y$ and $$|f(x) - f(y)| = |f'(\xi)||x-y|$$ where $\xi \in (x,y)$, then we can see that $$|f(x) - f(y)| \le M |x-y|.$$

Now suppose we let $\epsilon > 0$. If we fix $x_0$, and consider all $y \in \mathbb{R}$ such that $|x_0 - y| < \epsilon/M$ we see that $$|f(x_0) - f(y)| < \epsilon.$$ Finally, since $\epsilon/M$ is independent of $x_0$, $f$ is uniformly continuous.


Calculating Expectation and Covariance $X$ is a random variable which is uniformly distributed on the interval $[1,4]$. Define the random variable $Y = g(x)$ with $g(x) = x^2$

How can I calculate $E(g(X))$, $g(E(X))$ and the covariance Cov$[X,Y]$?

I would really appreciate it if someone can show me how to solve this !

The probability density function of $X$ is defined as

$$f_X(x) = \begin{cases}\frac13 & 1\le x\le 4\\0&\text{otherwise}\end{cases}$$

The expectation of $Y$, $E[Y] = E[g(X)]$ can be obtained by
$$\begin{align*}E[g(X)] &= \int_{-\infty}^\infty g(x)\ f_X(x)\ dx\\
&= \int_1^4x^2\cdot\frac13\ dx\\
&= \frac13\int_1^4x^2\ dx\\
&= \frac13\left[\frac{x^3}3\right]_1^4\\
&= 7
\end{align*}$$

By symmetry of uniform distribution, $E[X]$ can be quickly calculated as
$$E[X] = \frac{4+1}2 = 2.5,$$
or not as quickly calculated by
$$E[X] = \int_1^4x\cdot\frac13\ dx = \frac13\left[\frac{x^2}2\right]_1^4 = 2.5.$$
Then $g(E[X]) = 2.5^2$.

The covariance $COV(X,Y) = E[(X-E[X])(Y-E[Y])]$,
$$\begin{align*}COV(X,Y) &= E[(X-E[X])(Y-E[Y])]\\
&= E[XY] - E[X]\cdot E[Y]\\
&= \frac13\int_1^4 x^3\ dx - 2.5\cdot 7\\
&= \frac13\left[\frac{x^4}4\right]_1^4 - 2.5\cdot 7\\
&= 3.75
\end{align*}$$

Hints

You need
$$\mathbb{E}[g(X)] = \mathbb{E}[X^2]\\
g(\mathbb{E}[X]) = \mathbb{E}[X] \times \mathbb{E}[g(X)] \\
\textrm{Cov}(X,Y)
$$
can you get the first two?


Relation between Hilbert's hotel and Cantor's proof of the uncountability of the continuum I am reading the wikipedia page about Hilbert's Grand hotel. There it says:

  Hilbert's paradox is a veridical paradox: it leads to a counter-intuitive result that is provably true. The statements "there is a guest to every room" and "no more guests can be accommodated" are not equivalent when there are infinitely many rooms. An analogous situation is presented in Cantor's diagonal proof.

Now I wonder what Hilbert's Grand hotel has to do with Cantor's diagonal proof, since Cantor's diagonal proof is concerned with showing that the continuum has bigger cardinality than the natural numbers, but Hilbert's hotel seems to be about showing that certain countable sets are equinumerous. Could you clarify?

Cantor's diagonal proof can be imagined as a game: Player 1 writes a sequence of Xs and Os, and then Player 2 writes either an X or an O:

Player 1: XOOXOX

Player 2: X

Player 1 wins if one or more of his sequences matches the one Player 2 writes.
Player 2 wins if Player 1 doesn't win.

Cantor's diagonal proof basically says that if Player 2 wants to always win, they can easily do it by writing the opposite of what Player 1 wrote in the same position:

Player 1:

XOOXOX

OXOXXX

OOOXXX

OOXOXO

OOXXOO

OOXXXX

Player 2: OOXXXO

You can scale this 'game' as large as you want, but using Cantor's diagonal proof Player 2 will still always win. For number of turns equals infinity  ($T_{n}=\infty$):

Player 1:

XOOXOX...

OXOXXX...

OOOXXX...

OOXOXO...

OOXXOO...

OOXXXX...

...

Player 2: OOXXXO...

Since $T_{n}=\infty$, and $\left | T_{n} \right |=\left | P_{1} \right |$, $\left | P_{1} \right |=\infty$ (please note that the lines are the notation for cardinality, not absolute value). However, by using the tactic described earlier, Player 2 ensures that $P_{2}\notin P_{1}$.

We can extend this to Hilbert's paradox by assigning a company to the hotel to clean the rooms. This company has an infinite number of workers, but the way it trains them is strange. If rooms 2, 57, and 2,246 needed cleaning, the company would send the employee whose job it is to clean rooms 2, 57, and 2,246. Each cleaner would have a different set of rooms to be cleaned, including the worker who gets called when none of the rooms need cleaning.

Assume now that hotel wishes to throw a party for its cleaners and give them all a free room on the same night. Since you have already read the Wikipedia article, you know that you can add an infinite group to an already full hotel without kicking anybody out. Since the cleaning schedules of the workers are infinite strings, we can try to use Cantor's diagonal proof for $T_{n}=\infty$. Replacing Xs with no and Os with yes (for whether or not a worker cleans a room), we get this:

Rooms:

Room 1: No, Yes, Yes, No, Yes, No, ...

Room 2: Yes, No, Yes, No, No, No, ...

Room 3: Yes, Yes, Yes, No, No, No, ...

Room 4: Yes, Yes, No, Yes, No, Yes, ...

Room 5: Yes, Yes, No, No, Yes, Yes, ...

Room 6: Yes, Yes, No, No, No, No, ...

...

Excluded Worker: Yes, Yes, No, No, No, Yes, ...

So, the relationship between Hilbert's paradox and Cantor's diagonal proof is that Cantor's diagonal proof is an exception to the rule of Hilbert's paradox that $\infty+\infty=\infty$, and it establishes that there are different, unequal versions of infinity; the transfinite numbers.


Equivalence Relations on Set of Ordered Pairs Let $\mathbb{R}$ be the relation on $\mathbb{Z} \times \mathbb{Z}$, that is elements of this relation are pairs of pairs of integers, such that $((a,b),(c,d)) \in \mathbb{R}$ if and only if $a-d = c-b$. Can anyone give me a start on how to solve it to be transitive, reflexive and symmetric?

Reflexive: $\forall (a,b):\Bigl[(a,b)\in \mathbb{Z\times Z} \to \bigl((a,b),(a,b)\bigr)\in R\Bigr]$
Symmetric: $\forall (a,b,c,d): \Bigl[\bigl((a,b),(c,d)\bigr)\in R \leftrightarrow \bigl((c,d),(a,b)\bigr)\in R\Bigr]$
Transitive:
$\forall (a,b,c,d)\exists (e,f): \Bigl[\bigl((a,b),(e,f)\bigr)\in R\land \bigl((e,f),(c,d)\bigr)\in R \leftrightarrow \bigl((a,b),(c,d)\bigr)\in R\Bigr]$

Show that these properties hold (or not) when $\Bigl[\bigl((a,b),(c,d)\bigr)\in R \Bigr]\iff\Bigl[ a-d=c-b\Bigr]$

Hint: $[a-d=c-b] \iff [a+b=c+d]$

Reflexive: $((a,b),(a,b))\in R$ since $a-b=a-b$.

Symmetric:  Suppose that $((a,b),(c,d))\in R$ since $a-d=c-b$, hence  $c-b=a-d$. So  $((c,d),(a,b))\in R$.

Transitive:  Suppose that $((a,b),(c,d))\in R$ and $((c,d),(e,f))\in R$. Then  $a-d=c-b$ and  $c-f=e-d$, and hence $a-d+c-f=c-b+e-d$ which implies that $a-f=e-b$. So we have that $((a,b),(e,f))\in R$


Solving Differential Equations without separation of variables How can one proceed for solving differential equations in physics without separation of variables? For ex- Take Laplace equation in spherical coordinates, we always assume the solutions of form R(r)Θ(θ)Φ(φ) and then we resolve the differential equation in three differential equations of single variable. Doesn't it restrict the type of solutions. What if other solutions cannot be written in R(r)Θ(θ)Φ(φ) form, how do we find such solutions then.

I have same doubt for other differential equations of central importance in physics. "Schrodinger equation for hydrogen atom", "Wave-equation" "Diffusion-equation" and many more.

Take Laplace equation in spherical coordinates, we always assume the solutions of form $R(r)\Theta(\theta)\Phi(\phi)$ and then we resolve the differential equation in three differential equations of single variable. Doesn't it restrict the type of solutions.

We choose to use spherical coordinates (in your example) precisely because we want to examine situations which have spherical geometries.

We could equally choose e.g. cylindrical coordinates to study systems with cylindrical geometries.  If you search for this you'll see cylindrical coordinates crop up quite often  in physical models.

Also, there are very few equations (or forms of equation) which admit simple closed form solutions or even the ability to decouple using separation of variables.  When we get a chance to use the technique, we grab it. :-)

Often a good choice of coordinate system will allow useful exploration of a problem even if an explicit solution remains out of reach.


Proof by contradiction proving both numbers are not odd. I have to do a proof by contradiction:

Suppose $a,b,\in\mathbb{Z}$. If $4| (a^2 + b^2)$ then a and b are not both odd.

So far I know that I need to prove that if $4|(a^2+b^2)$ then a and b are both odd.

I would use the definition of an odd number $(2k+1), k\in\mathbb{Z}$

I'm a little but unsure of where to go from here. Any help would be appreciated.

If both are odd then $a=2n+1$, $b=2m+1$ and $$a^2+b^2=4(n^2+n+m^2+m)+2$$ which is never divisible by $4$.

Let $a = (2k+1)$ and let $b=2k'+1$ for some k in the naturals. If you assume both are odd, and you don't get a factor of 4, then clearly 4 doesn't divide $(a^2 + b^2)$ when both $a,b$ are odd.


find interval of convergence for series Is it right that the range of convergence is here $1 < x < 3$:

$$\sum_{n= 1}^\infty \frac{e^n + e^{-n}}{n^2} (x-2)^n$$

Just like you do with the geometric series? Or what is this radius of convergence? Thanks!

update:

i got until now:

$$\frac{\frac{(e^{n+1}+e^{-(n+1)})*(x-2)^{n+1}}{(n+1)^2}}{\frac{(e^n+e^{-n})*(x-2)^n}{n^2}}$$

(the middle fractal line should be the main one)
and this should be less the 1

right?

Using the Ratio Test, we have

$\displaystyle\lim_{n\to\infty}\frac{e^{n+1}+e^{-(n+1)}}{(n+1)^2}\lvert x-2\rvert^{n+1}\cdot\frac{n^2}{(e^n+e^{-n})\lvert x-2\rvert^n}=\lim_{n\to\infty}\frac{e^{n+1}+e^{-(n+1)}}{e^n+e^{-n}}\cdot\frac{n^2}{(n+1)^2}\cdot\lvert x-2\rvert$

$\displaystyle=\lim_{n\to\infty}\frac{e+e^{(-2n+1)}}{1+e^{-2n}}\cdot\left(\frac{n}{n+1}\right)^2\cdot\lvert x-2\rvert=e\cdot 1\cdot\lvert x-2\rvert=e\lvert x-2\rvert$,

and $e\lvert x-2\rvert<1 \iff \lvert x-2\rvert<\frac{1}{e}$.

To test convergence at the endpoints of the interval,

A) $\;\displaystyle x=2+\frac{1}{e}$ gives the series $\displaystyle\sum_{n=1}^{\infty}\frac{1+e^{-2n}}{n^2}$,

$\;\;\;$which converges by comparing to $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}$ using the Limit Comparison Test.

B) $\;\displaystyle x=2-\frac{1}{e}$ gives the series
$\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{1+e^{-2n}}{n^2}$,

$\;\;\;$which converges since its absolute value converges.

Therefore the series converges for x in $\displaystyle\left[2-\frac{1}{e},2+\frac{1}{e}\right]$.

According to WolframAlpha, the interval of convergence is as follows.

You can see it plotted as so:

Unfortunately I'm not sure how one would go about finding it, but for the bottom  part you can use the p-series test and because it is n^2, p>2 so it is convergent everywhere, now you just have to decide for the top part what test to use.


Arithmetic error in Feller's Introduction to Probability? In my copy of An Introduction to Probability by William Feller (3rd ed, v.1), section I.2(b) begins as follows:

  (b) Random placement of r balls in n cells. The more general case of [counting the number of ways to put] $r$ balls in $n$ cells can be studied in the same manner, except that the number of possible arrangements increases rapidly with $r$ and $n$.  For $r=4$ balls in $n=3$ cells, the sample space contains already 64 points ...

This statement seems incorrect to me.  I think there are $3^4 = 81$ ways to put 4 balls in 3 cells; you have to choose one of the three cells for each of the four balls.  Feller's answer of 64 seems to come from $4^3$.  It's clear that one of us has made a very simple mistake.

Who's right, me or Feller?  I find it hard to believe the third edition of a universally-respected textbook contains such a simple mistake, on page 10 no less.  Other possible explanations include:

(1) My copy, a cheap-o international student edition, is prone to such errors and the domestic printings don't contain this mistake.

(2) I'm misunderstanding the problem Feller was examining.

It is an error. The answer should is 81. I was really troubled that this book could have gone through 3 editions and a revision reprint and still have this elementary error.

It is easy to see why 81 is the correct number:
The first ball can be placed in 3 ways. The second in three ways. By the rs principle, the first two balls can be placed in 3x3 ways, i.e. 9 ways. The third ball can be paced in 3 ways. By the rs principle, the number of ways of placing the first 3 balls is 27. The 4th ball can be place in any one of three ways, hence the total number of ways the balls can be placed is 81.Repeated application of the rs principle gets you to 81, not 64. 64 would be the right answer if we had 4 cells and 3 balls, since 4x4x4 = 64.

What bothers me is that anyone studying on his own will run into these problems with errata everywhere. The first reaction is always: I must be wrong, because these people with their ivy league degrees cannot possibly be wrong!

The guilty secret is that almost all mathematics books have errors of this type.
Before you buy any book, first get hold of the errata list. If the error count is too dense, buy a different book.

I hope this helps.

JLL


Prove that no such $f$ exists I have come upon this question and could not find a solution for it.

Prove that there is no entire function $f$ such that $\forall z\in \mathbb{C}$ $\quad|f(z)|>|z|$.

Any ideas on how to solve this question?

Clearly there is no $z$ for which $f(z) = 0$. Therefore, $g(z) := \frac{z}{f(z)}$ is entire. Since $|z| < |f(z)|$, $|g(z)| < 1$ for all $z \in \mathbb{C}$. Liouville implies $g(z) \equiv c$ is constant, i.e. $z \equiv cf(z)$. Clearly $c \not = 0$, so $f(0) = 0$, contradiction.

Let $g(z)=\frac z {f(z)}$ for $z \neq 0$. (Note that $f(z) \neq 0$ for  $z \neq 0$). Then $g$ is analytic in $\mathbb C \setminus \{0\}$. Since it is bounded it extends to an entire function $G$. This entire function is bounded. By Liouville's Theorem it must be a constant. This means $f(z)=cz$ for some constant, but then the given inequality does not hold at $0$.


Is $y=\frac{x^2}{x^3+3}$ has a horizontal asymptote? Is $y=\frac{x^2}{x^3+3}$ has a horizontal asymptote?
Since the graph passes through $(0,0)$, hence $y=0$ is not a horizontal asymptote. Can a rational function may not have a horizontal asymptote?

Hints:
An asymptote is a straight line to which the curve approaches while it moves away from the origin. The curve can approach the line from one side, or it can intersect it again and again. Not every curve which goes infinitely far from the origin (infinite branch of the curve) has an asymptote.
For functions given in explicit form  $y=f(x)$,  we know: the vertical asymptotes are at points of discontinuity where the function  $f(x)$  has an infinite jump; the horizontal and oblique asymptotes have the equation:
$$y=kx+b, \space \text{ with } \space \space k=\lim_{x \to \infty}\frac{f(x)}{x}, \space \space b=\lim_{x \to \infty}\left[f(x)-kx \right].$$
Figure:

Yes, of course:
$$\lim_{x\rightarrow+\infty}\frac{x^2}{x^3+3}=\lim_{x\rightarrow+\infty}\frac{\frac{1}{x}}{1+\frac{3}{x^3}}=0,$$
$$\lim_{x\rightarrow-\infty}\frac{x^2}{x^3+3}=\lim_{x\rightarrow-\infty}\frac{\frac{1}{x}}{1+\frac{3}{x^3}}=0.$$
Id est, $y=0$ is a horizontal asymptote.
The function $f(x)=\frac{x^3}{x^2+3}$ has no a horizontal asymptote.


Prove the inequality between integral and summation of multiplicative inverse I want to prove the following inequality:

$$
\ln(n) = \int\limits_1^n{ \frac{1}{x} dx } \geq \sum_{x = 1}^{n}{\frac{1}{x + 1}} = \sum_{x = 1}^{n}{\frac{1}{x}} - 1
$$

I ask this question as I'm trying to calculate the upper bound of a harmonic series:

$$
\sum_{x = 1}^{n}{\frac{1}{x}} \leq \ln(n) + 1
$$

Using the mL (minumum times Length) estimate
$$\int_a^b |f(x)|dx\geq (b-a)\min_{x\in[a,b]}|f(x)|$$
One has
$$\int_1^n\frac{1}{x}dx=\sum_{k=1}^{n-1}\int_{k}^{k+1}\frac{1}{x}dx\geq\sum_{k=1}^{n-1}(k+1-k)\frac{1}{k+1}=\sum_{k=1}^{n-1}\frac{1}{k+1}=\sum_{k=1}^n\frac{1}{k}-1,$$
where I have used that $\frac{1}{x}$ is positive on the positive axis (so the absolute value is irrelevant) and that $\frac{1}{x}$ is decreasing (again, for positive $x$), so that
$$\min_{x\in[a,b]}\frac{1}{x}=\frac{1}{b}$$

PS: my suggestion to you, is to draw the situation: first draw the $\frac{1}{x}$ function, then subdivide the real axis in intervals of length $1$, and then build up the rectangles which fit under the curve, for each of the subintervals (bases). The sum of their areas will be given by the above formula.


Does this integral expression makes sense? I'm wondering whether this expression has some significance:

$$\int_{-1}^1 \frac{dx}{\sqrt{|x|}}$$

And, in general, if expressions in the following form make sense:

$$\int_a^b f(x) dx$$

Where the set $(a, b)$ contains points out of the domain of $f(x)$.

According to the definitions I know, these are neither definite nor improper integrals, so the expression shouldn't make any sense.

But in case the integral exists, could you please tell me:

Which kind is it?
Is the following equation true? $\displaystyle \int_{-1}^1 \frac{dx}{\sqrt{|x|}} = \int_{-1}^0 \frac{dx}{\sqrt{|x|}} + \int_0^1 \frac{dx}{\sqrt{|x|}}$
Is it an area?
If not, what is the area of $\frac1{\sqrt{|x|}}$?

UPDATE 1: I'm saying that $\displaystyle \int_{-1}^1 \frac{dx}{\sqrt{|x|}}$ is not an improper integral is because of the following definition (taken from Wikipedia):

  [...] an improper integral is the limit of a definite integral as an endpoint of the interval(s) of integration approaches either a specified real number or ∞ or −∞ [...]

-1 and 1 are the endpoints of my integral, but here the problem is 0. Perhaps, is that definition wrong or incomplete?

UPDATE 2: I've replaced $\frac1x$ with $\frac1{\sqrt{|x|}}$, so that my questions about the area make more sense.

UPDATE 3: I'm not particularly interested in calculating the value of that integral, what I really want to know is what is that stuff. An ideal answer would include a valid and coherent mathematical definition of integrals of that kind.

Consider the Cauchy principal value

$$ \lim_{\epsilon \to 0}\left(\int_{-1}^{-\epsilon}\frac{1}{x}+ \int_{\epsilon}^{1}\frac{1}{x}\right)=0 $$

Edit: what follows is partly incorrect. The only way to give a meaning to the integral is to use the concept of Cauchy principal value.

Your specific example. First, I don't see any reason for not calling it an improper integral; second, surely the decomposition you write in 2. is perfectly valid:
$$
\int_{-1}^1 \frac1x dx= \lim_{\epsilon\rightarrow 0}\Bigg(\int_{-1}^{-\epsilon}\frac1x dx + \int_\epsilon^1 \frac1x dx\Bigg)
$$
Changing variable from $x$ to $-x$ in the first integral you see that the parenthesis is $0$.
Due to the symmetry of the integrand, we can say that the result is exactly $0$. (maybe I'm being not rigorous here?)

Accordingly, the area is $0$ as well, in fact it is a signed area.

On the contrary, $\displaystyle\int_{-1}^1\left|\frac1x\right|\; dx$ is a divergent integral.

In general, it still makes sense to define integrals of functions which are singular on the path of integration. Then it may well be that the integral is divergent and the best you can do is to define its principal value, taking a limit analogue to the one which I wrote above.


Fractional Exponents powers I am having problems understanding how to answer questions containing fractional exponents to a given power ie $(2x^{1/2})^6$, i do not understand how to go about answering the question. I know this is an easy topic, but i would really appreciate the help

how to answer questions containing fractional exponents

In the exact same manner in which you answer questions containing non-fractional exponents, since the base is obviously a positive number, given the fact that both $2$ and $\sqrt x$ are always $\ge0$. $(ab)^c=a^cb^c$ for all exponents, fractional or not, it doesn't matter. Likewise, $(a^b)^c=a^{bc}$, again, for all exponents. By combining the two, we have $(ab^c)^d=a^db^{cd}$. In this case, $a=2,b=x,c=$ $=\dfrac12$ and $d=6$. And $\dfrac12\cdot6=3$.

The general rule for exponents is $x^{a/b} = \sqrt[b]{x^a}  \forall x \in \mathbb{N}$. For example, $4^{2/3} = \sqrt[3]{4^2} = \sqrt[3]{8}$. You can apply this to your problem to see that $(2x^{1/2})^6 = 64x^3.$


Relatively prime numbers formula Here is a problem I cannot manage with:

Find two relatively prime positive numbers $p$, $q$ that satisfy:
Sequence $ \{pn + q\}_{n=0,1,2,\ldots}$ does not contain any Fibonacci number.

Any ideas how to touch it?

Write down the Fibonacci sequence modulo the first few primes, and note the periods (A060305):

modulo 2  -- period 3
modulo 3  -- period 8
modulo 5  -- period 20
modulo 7  -- period 16
modulo 11 -- period 10

So the Fibonacci sequence modulo 55 has period 20. That is not long enough to hit all of the 40 numbers between 0 and 55 that are relatively prime to 55.

(But in fact already the sequence modulo 11 itself is too short to hit everything, because there are necessarily repetitions in it. Indeed, neither of $4$, $6$, $7$, nor $9$ are in the sequence, so there is no Fibonacci number of the forms $11n+4$, $11n+6$, $11n+7$, or $11n+9$).

P=2-2(N+1)+3-(3(2N+1))+(6N±1)-{(6N±1)•(6n±1)}, where P does not equal 1, and where P equals all prime numbers, and where N equals all natural whole numbers from 1, 2, 3... Etc.

To think that something deemed so complex could be reduced to such a simple and complete formula. My, my.


How to find the value of this limit? $\lim\limits_{n\to\infty}n\int_0^1 nx^{n-1}\left(\frac{1}{1+x}-\frac{1}{2}\right)\mathrm dx.$ How to calculate the following limit:
$$\lim_{n\to\infty}n\int_0^1 nx^{n-1}\left(\frac{1}{1+x}-\frac{1}{2}\right)\mathrm dx.$$

Use the substitution $x^n=t$.
$$n\int_0^1 nx^{n-1}\left(\frac{1}{1+x}-\frac{1}{2}\right)\,dx=n\int_0^1 \left(\frac{1}{1+t^{1/n}}-\frac{1}{2}\right)\,dt$$
$$\Rightarrow \lim_{n\rightarrow \infty} n\int_0^1 \left(\frac{1}{1+t^{1/n}}-\frac{1}{2}\right)\,dt=\lim_{h\rightarrow 0}\dfrac{\displaystyle \int_0^1 \left(\dfrac{1}{1+t^{h}}-\frac{1}{2}\right)\,dt}{h}$$
Use L'Hopital's rule and Leibniz rule to get:
$$\lim_{h\rightarrow 0}\int_0^1 \frac{-t^h\ln t}{(1+t^h)^2}\,dt=-\frac{1}{4}\int_0^1\ln t \,dt=\boxed{\dfrac{1}{4}}$$

The integral is equal to (via Mathematica):

$$I(n)=\int_0^1 nx^{n-1}\left(\frac{1}{1+x}-\frac{1}{2}\right)\mathrm dx.=(1/2)\left(-1 - n\psi( n/2) + n\psi((1 + n)/2)\right)$$

Where $\psi(z):=\frac{\Gamma'(z)}{\Gamma(z)}$ is the PolyGamma function.

The limit is equal to (via Mathematica):

$$\lim_{n\to\infty} n I(n)=\frac{1}{4}$$

-mike


how to prove the chain rule? I have just learnt about the chain rule but my book doesn't mention a proof on it. I tried to write a proof myself but can't write it. So can someone please tell me about the proof for the chain rule in elementary terms because I have just started learning calculus.

Assuming everything behaves nicely ($f$ and $g$ can be differentiated, and $g(x)$ is different from $g(a)$ when $x$ and $a$ are close), the derivative of $f(g(x))$ at the point $x = a$ is given by
$$
\lim_{x \to a}\frac{f(g(x)) - f(g(a))}{x-a}\\ = \lim_{x\to a}\frac{f(g(x)) - f(g(a))}{g(x) - g(a)}\cdot \frac{g(x) - g(a)}{x-a}
$$
where the second line becomes $f'(g(a))\cdot g'(a)$, by definition of derivative.

If I understand the notation correctly, this should be very simple to prove:

Assume $f(x) = g(h(x))$

Then:
$$f'(x) = \frac{df(x)}{dx}$$

This can be expanded to:
$$\frac{df(x)}{dx} = \frac{df(x)}{dg(h(x))} \frac{dg(h(x))}{dh(x)} \frac{dh(x)}{dx}$$

When you cancel out the $dg(h(x))$ and $dh(x)$ terms, you can see that the terms are equal.

Since $f(x) = g(h(x))$, the first fraction equals 1.
($$\frac{df(x)}{dg(h(x))} = 1$$)

If we substitute $h(x)$ with $y$, then the second fraction simplifies as follows:
$$\frac{dg(y)}{dy} = g'(y)$$
Substituting $y = h(x)$ back in, we get following equation:
$$\frac{dg(h(x))}{dh(x)} = g'(h(x))$$
The third fraction simplifies to the derrivative of $h(x)$ with respect to $x$. This can be written as
$$\frac{dh(x)}{dx} = h'(x)$$

Substituting these three simplifications back in to the original function, we receive the equation

$$\frac{df(x)}{dx} = 1g'(h(x))h'(x) = g'(h(x))h'(x)$$

This is the chain rule.


The distance moved by the tip of the hand in clock. The minute hand of a clock is $15$ cm
  long. The distance moved by the tip
  of the hand in $35$ minutes is

$a.)\ 35\pi \\
\color{green}{b.)\ \dfrac{35\pi}{2}} \\
c.)\ \dfrac{5\pi}{4} \\
d.)\ \dfrac{5\pi}{2} $

For minute hand,

$12\ hrs =360^{\circ} \\
35\ min=\left(\dfrac{35}{2}\right)^{\circ} $

Distance$=\dfrac{2\pi 15\times 35}{360\times 2}=\dfrac{35\pi}{24}\ cm $

But that is not in options

I look for a short and simple way.

I have studied maths upto $12th$ grade.

It has moved $\frac{35}{60}$ of a full circle (a full circle consists of $60$ minutes, and it has moved $35$ of those). A full circle is $2\pi\cdot15cm=30\pi cm$. It has therefore moved a total of
$$
\frac{35}{60}\cdot30\pi cm=\frac{35\pi}{2}cm
$$

Your mistake ? You are missing one fact, the minute hand has already went through $\left(360\cdot 12\right)^{\circ}$ every 12 hours.


Intuition for sum of triangular numbers and significance for $3\choose{k}$ Specific question: according to my calculation based on Timbuc's answer to this question,

$$\sum_{k=0}^n\frac{k(k+1)}{2}=\frac{n(n+1)(2n+4)}{12} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; =\frac{n(n+1)(n+2)}{6}$$

[Edit: RHS simplified based on suggestion from herbsteinberg.]

If this is right, is there an intuitive or geometric proof of this?

Background and motivation: I'm trying to relate the concepts of integration and summation using reasoning which I find simple and intuitive.

As part of that, I'm trying to understand the process of summing sequences in this rotated section of Pascal's triangle:

$(0)\;\;01\;\;01\;\;01\;\;01\;\;01\;\;01\;\;...\frac{k^0}{0!} \\ (1)\;\;01\;\;02\;\;03\;\;04\;\;05\;\;06\;\;...\frac{k+0}{1!} \\ (2)\;\;01\;\;03\;\;06\;\;10\;\;15\;\;21\;\;...\frac{k(k+1)}{2!} \\ (3)\;\;01\;\;04\;\;10\;\;20\;\;35\;\;56\;\;...\frac{k(k+1)(k+2)}{3!}$

I see that summing each line seems to increase the degree of the expression by $1$ and I can imagine that rearrangement, simplification and/or a limit process could later reduce these terms to $\frac{k^2}{2}$, $\frac{k^3}{3}$ etc., but for now I'm interested in how/why each line has the exact expression it does.

Line $(1)$ is just counting.

Line $(2)$ I can picture and understand as in Fig. 1 below.

Fig. 1

Line $(3)$ [Edited to add the following, which may start to answer my question] I can picture and understand as a stepped version of the right-angled tetraga in Fig. 2 below.

Fig. 2

Intuitively, you are walking the pascal triangle, in a zigzag way.

$$\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$$

For your example, $\large \sum_{k=0}^{n}{k(k+1)\over 2} = \sum_{k=1}^{n}\binom{k+1}{2} = \binom{n+2}{3}$

$$\large\begin{matrix}
\binom{2}{2} = \binom{3}{3} = 1 \cr
\binom{3}{3} + \binom{3}{2} = \binom{4}{3} = 4 \cr
\binom{4}{3} + \binom{4}{2} = \binom{5}{3} = 10 \cr
\binom{5}{3} + \binom{5}{2} = \binom{6}{3} = 20 \cr
\cdots \cr
\Large\binom{n+1}{3} + \binom{n+1}{2} = \binom{n+2}{3}
\end{matrix}$$

Each row can be written as a sum of combinatorial numbers. This is what is known as the Hockey Stick Identity

A nice way of imaginining the intuition behind these identities is understanding a combinatorial proof for the hockey stick identity


Theorem on continuous function "If f(x) is continuous and f(a) and f(b) are of opposite signs then there exist at least one or an odd number of roots between a and b." Is it true for polynomial equations only or any continuous function?

If $f$ is continuous function, then there exists at least a root between $f(a)$ and $f(b)$. This is true by intermediate value theorem.

However, for a general function, it is possible that the zero set can be uncountable.

For example let

$$f(x) = \begin{cases} x-1 &, x \ge 1 \\ 0 & , x \in (0,1) \\ x+1 & , x \le -1\\ \end{cases}$$

At least one: yes. An odd number: This is not so clear. Of course for polynomials this is true, but we have to count roots "with multiplicity". And as far as I can see if $f$ is just continuous and $f(c)=0$ there's no reasonable definition of the multiplicity.

Example. Let $f(x)=|x|(x-1)$, $a=-1$, $b=1$. Then $f$ is continuous, $f(a)$ and $f(b)$ have different signs, but $f$ has exactly $2$ zeroes between $a$ and $b$.

Similarly for $f(x)=|x|^\alpha(x-1)$ for any $\alpha>0$; if we want to say there are an odd number of zeroes between $-1$ and $2$ we have to  somehow define the multiplicity of a zero in such a way that for every $\alpha>0$ the function $|x|^\alpha$ has a zero of even order at the origin.

in fact, come to think of it, not only is there no standard definition of the order of a zero of a continuous function, it's easy to see that it's impossible to give such a definition, assuming two natural conditions:

Define $\Bbb N = \{0,1,2\dots\}$.

    Prop There does not exist a function $O:C(\Bbb R)\to\Bbb N$ such that (i) $O(f)>0$ if $f(0)=0$ and (ii) $O(fg)=O(f)+O(g)$.

Proof. If $O$ is such a function then for every postitive integer $n$ we have $$O(|x|)=nO(|x|^{1/n})\ge n.$$


Axiom of Regularity allows for this set be an element of itself I'm new to set theory, and the axiom of regularity has been giving me some trouble. It states that every non-empty set A has an element B such that A and B are disjoint sets. Apparently, this axiom implies that a set can't be an element of itself. I've heard (and agree with) arguments like this:

Let A be a set, and apply the axiom of regularity to {A}, which is a set by the axiom of pairing. We see that there must be an element of {A} which is disjoint from {A}. Since the only element of {A} is A, it must be that A is disjoint from {A}. So, since A ∈ {A}, we cannot have A ∈ A (by the definition of disjoint).

However, let's say there's a set A = {{1, 2}, A}. It seems like this set obeys the axiom of regularity, because {1, 2} is an element of A, and {1,2} and A are disjoint sets. However, the axiom of regularity shouldn't allow A to be an element of itself. I suppose there's a flaw in my logic somewhere, and I'm hoping someone knowledgeable can show me why this isn't allowed.

Just because a set appears to obey the axiom of regularity doesn't mean it actually is a set!  The axiom of regularity restricts what sets exist: if a non-empty set exists, then it has an element that is disjoint from it.  The axiom of regularity doesn't say that any putative collection which follows this rule has to actually exist as a set.

So, all you have observed is that if a set $A=\{\{1,2\},A\}$ existed, then $A$ would not be a counterexample to the axiom of regularity.  This in no way proves that such a set actually exists!  And in fact, if such a set did exist, then $\{A\}$ would be a counterexample to the axiom of regularity.  This is a contradiction, and therefore no such set $A$ exists.

I'm actually not very familiar with set theory, but I think I see the issue here: when working with set theory, you adopt a certain definition of "set", which entails some mathematical rules (including the axiom of regularity) that any object has to satisfy in order to be a set. An object $A$ which satisfies $A = \{\{1,2\}, A\}$ may "exist" (whatever that means), but it does not satisfy the rules of "set-hood", and therefore is not a set. Just because the notation you use to define it looks set-like, that doesn't make it a set.

I could probably make a rough analogy to the "number" $0.00\ldots1$, which is written using basically the same notation as many things which are numbers, but does not actually satisfy the rules of "number-hood".

You might also find it useful to read Define $A = \{ 1,2,A \}$, $A$ can not be a set (Axiom of regularity). Can $A$ be a "class" or a "collection" of elements., e.g. Asaf Karagila's answer which points out that the definition $A = \{\{1,2\}, A\}$ is circular. Depending on your frame of mind that might make more sense than what I've said above.


Is $7$ the only prime followed by a cube? I discovered this site which claims that "$7$ is the only prime followed by a cube".  I find this statement rather surprising.  Is this true?  Where might I find a proof that shows this?

In my searching, I found this question, which is similar but the answers seem focused on squares next to cubes.

Any ideas?

This is certainly true. Suppose $n^3 - 1$ is prime, for some $n$. We get that $n^3-1 = (n-1)(n^2 + n + 1)$ and so we have that $n-1$ divides $n^3 - 1$. If $n-1>1$ then we're done, as we have a contradiction to $n^3 - 1$ being prime.

Wouldn't $-2$ also be a prime followed by $-1$ which is a cube of $-1$.
As $x^2 + x +1$ will also equal one for $x=-2$.


Number of positive integral solution of product $x_{1} \cdot x_{2} \cdot x_{3}\cdot x_{4}\cdot x_{5}=1050$ is The number of positive integral solution of product $x_{1} \cdot x_{2} \cdot x_{3}\cdot x_{4}\cdot  x_{5}=1050$ is

$\bf{My\; Try}::$ Given $x_{1}\cdot x_{2}\cdot x_{3}\cdot x_{4}\cdot x_{5} = 2 \times 3 \times 5^2 \times 7$.

Now Let we have $5$ Different Boxes $x_{1}\;,x_{2}\;,x_{3}\;,x_{4}\;,x_{5}$ and $5$ balls in

which prints a numbers $2\;,3\;,5\;,5\;,7$.

Now We have to put balls into boxes , So we will form different cases.

$\bullet \;$ If all the balls are in same boxes(boxes contain all balls or no balls)

Then no. of ways $ = 5\times 5 \times 5\times 5 \times 5 = 5^5$

$\bullet \; $ If all balles are in different boxes, Then no. of ways $\displaystyle  = $

Now I did not understand how can i find in $\bf{(II)}$ cases,

Help me Thanks

You can put $2,3,7$ in the boxes in five ways each. The fives can be placed in $5+\binom 52$ ways.

$5$ represents both of the $5$s in the same box. $\binom 52$ represents them in different boxes.

So you should get $5\times 5 \times 5 \times 15$ ways.

It appears that the first case in the question, where the $5$s are in the same box should give the result $5^4$ rather than $5^5$ because the pair of $5$s acts as a single unit rather than two separate units.

If the $5$s are in separate boxes, they are indistinguishable, and can be swapped without changing the factorisation. The first $5$ can be placed in any of $5$ boxes, and the second in any of the remaining $4$ boxes, but this counts every possibility twice. So the number of ways of placing the $5$s is $\cfrac {5\times 4}2=10$. This second case counts for $5^3\times 10$ possibilities.

The number of ways of choosing $r$ places out of a possible $n$ is so commonly encountered that it is given the (modern) symbol $\binom nr=\frac {n!}{r!(n-r)!}$. Apologies if this confused you, but it is useful to know, as you are likely to encounter it frequently, especially if you ask similar questions on this site.

Each box can contain $5$ elements so it will be toatal $5^5$ cases I think.


Intuitive explanation of entropy I have bumped many times into entropy, but it has never been clear for me why we use this formula:

If $X$ is random variable then its entropy is:

$$H(X) = -\displaystyle\sum_{x} p(x)\log p(x).$$

Why are we using this formula? Where did this formula come from? I'm looking for the intuition. Is it because this function just happens to have some good analytical and practical properties? Is it just because it works? Where did Shannon get this from? Did he sit under a tree and entropy fell to his head like the apple did for Newton? How do you interpret this quantity in the real physical world?

Here's one mildly informal answer.

How surprising is an event? Informally, the lower probability you would've assigned to an event, the more surprising it is, so surprise seems to be some kind of decreasing function of probability. It's reasonable to ask that it be continuous in the probability. And if event $A$ has a certain amount of surprise, and event $B$ has a certain amount of surprise, and you observe them together, and they're independent, it's reasonable that the amount of surprise adds.

From here it follows that the surprise you feel at event $A$ happening must be a positive constant multiple of $- \log \mathbb{P}(A)$ (exercise; this is related to the Cauchy functional equation). Taking surprise to just be $- \log \mathbb{P}(A)$, it follows that the entropy of a random variable is its expected surprise, or in other words it measures how surprised you expect to be on average after sampling it.

Closely related is Shannon's source coding theorem, if you think of $- \log \mathbb{P}(A)$ as a measure of how many bits you need to tell someone that $A$ happened.

Let me try to answer to part of your question where you ask how it's related to the "physical world" taking the same direction as MichaelNgelo's answer.

Let's say you have $N$ objects, out of them there are $n_1$ many $A$s and $n_2$ many $B$s.

How many ways can you order them? For example for $N = 4$, with one $A$ and three $B$s we have:
$$ABBB$$
$$BABB$$
$$BBAB$$
$$BBBA$$

This is a combinatorics problem and we can find the formula: $$\frac{N!}{n_1!n_2!}$$

Note that there are more cases with $N=4$ and a arbitrarly number of $A$s and $B$s ($16$ cases) or with $N=4$ and with two $A$s and two $B$s ($6$ cases).

Given the proportions of $A$s and $B$s, it reduces the number of possibilities and we just need a way to differentiate between them.

This is where the logarithm comes in to tell the least amount of bits to differentiate between these $M$ cases. Explicitly we need $\log_2(M)$ bits.

Now divide by $N$ to find the number of bits needed per symbol in the sequence. So if put all together, the formula becomes:

$$\frac{\log_2(\frac{N!}{n_1!n_2!})}{N}$$

and for convenience instead of talking about $n_1$ and $n_2$ in absolute number, we talk about the fraction of $N$, respectively $f_1$ and $f_2$. Finally tending $N$ to infinity gives the shannon entropy.

Example with $f_1 = 0.25$ and $f_2 = 0.75$:

Someone experienced can probably derive the Shannon entropy formula from the formula above with $N$ tending to infinity.

So a possible interpretation of the Shannon formula would be: The number of bits needed per symbol given the proportions of each symbol in advance in an infinite sequence.


Find the integral surface of linear PDE $(x^2-yz)p+(y^2-zx)q=z^2-xy$ which passes through the straight line $x=1, y=0$ Here $\;p=\frac{\partial z}{\partial x}\;$ and  $\;q=\frac{\partial z}{\partial y}$

Lagrange's equations are
$
\begin{align}
\frac{dx}{x^2-yz} = \frac{dy}{y^2-zx} = \frac{dz}{z^2-xy}
\end{align}$

Let the general solution be $\phi (C1,C2) = 0 $

By Choosing multipliers $\; x,y,z\;$ and $\;1,1,1\;$ we get $\;C1\;$ as below.

\begin{align}
\frac{xdx+ydy+zdz}{x^3+y^3+z^3-3xyz} & = \frac{dx+dy+dz}{x^2+y^2+z^2-xy-yz-zx} \\
\end{align}

\begin{align}
\frac{xdx+ydy+zdz}{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)} & = \frac{dx+dy+dz}{x^2+y^2+z^2-xy-yz-zx} \\
\end{align}

\begin{align}
\frac{xdx+ydy+zdz}{x+y+z} & = dx+dy+dz \\
\end{align}

\begin{align}
xdx+ydy+zdz & = (x+y+z)d(x+y+z) \\
\end{align}

\begin{align}
\frac{x^2}{2}+\frac{y^2}{2}+\frac{z^2}{2} & = \frac{(x+y+z)^2}{2} + C \\
\end{align}

\begin{align}
x^2+y^2+z^2 - (x+y+z)^2 & = C1
\end{align}

Now I am not able to find $C2$. The answer given in the textbook is $(x-y)(xy+yz+zx)+y-z = 0$

Thanks for any hint.

By Choosing multipliers $(y-z),(z-x),(x-y)$ :
$$\frac{dx}{x^2-yz} = \frac{dy}{y^2-zx} = \frac{dz}{z^2-xy} =
\frac{(y-z)dx+(z-x)dy+(x-y)dz}{(y-x)(x^2-yz)+(z-x)(y^2-xz)+(x-y)(z^2-xy)}=\frac{(y-z)dx+(z-x)dy+(x-y)dz}{0}$$
This implies
$$(y-z)dx+(z-x)dy+(x-y)dz=0$$
I suppose that you can take it from here.


Computing relative homology group of the standard $1$-simplex with respect to it's boundary. Find the relative homology group $H_* ([0,1], \{0,1\}).$

Since $[0,1]$ is contractible
it follows from homotopy invariance property that $$H_n ([0,1])  = \begin{cases} \Bbb Z & n = 0 \\   0 & n \geq 1 \end{cases}$$ and since $\{0,1\}$ is discrete it follows that $H_*(\{0,1\}) = H_*(\{0\}) \oplus H_*(\{1\}).$ So we have $$H_n (\{0,1\})  = \begin{cases} \Bbb Z \oplus \Bbb Z & n = 0 \\   0 & n \geq 1 \end{cases}$$ Now we have the following short exact sequence of chain complexes $$0 \rightarrow C_*(\{0,1\}) \rightarrow C_*([0,1]) \rightarrow C_*([0,1], \{0,1\}) \rightarrow 0$$ This will induce a long exact sequence of the homology groups $$\cdots \rightarrow H_n(\{0,1\}) \rightarrow H_n([0,1]) \rightarrow H_n([0,1], \{0,1\}) \rightarrow H_{n-1}(\{0,1\}) \rightarrow H_{n-1} [0,1] \rightarrow \cdots$$
Using exactness of the above long exact sequence and from the fact that higher homology groups (except the $0$-th one) of $[0,1]$ are all trivial it is easy to show that for $n \geq 2$ $$H_n ([0,1], \{0,1\}) = 0.$$ But I find it difficult to find $0$-th and the $1$-th relative homology groups. Since $H_1(\{0,1\}) = 0$ we find that $H_1 ([0,1])$ is a subgroup of $H_1([0,1],\{0,1\}).$ In fact it is a proper subgroup as there is no injective homomorphism from from $\Bbb Z \oplus \Bbb Z$ ($= H_0(\{0,1\})$) to $\Bbb Z$ ($= H_0([0,1])$). But I can't conclude anything more than that. A small hint will be warmly appreciated at this stage.
Thanks for reading.

Hint: You can find how your map $\Bbb Z \oplus \Bbb Z \to \Bbb Z$ looks explicitly, if you know why the groups are of this form. What are the generators of $H_0(X)$?
$H^0([0,1])$ is generated by points of $[0,1]$ regarded as $0$-simplices, all of which are, as you pointed out, homological to each other (i.e. differ by an element of $im \partial$). Both summands in $\Bbb Z \oplus \Bbb Z$ are represented by points and so are mapped to a generator of $H^0([0,1]),$ so your map has (depending on orientation you choose) the matrix $( \pm1,  \pm1)$. It's easy to see that its kernel is $\Bbb Z,$ so this is (by exactness) your $H^0$.


What are spatial functions? I was reading Einstein's paper 'Concerning an Heuristic Point of View Toward
the Emission and Transformation of Light' and read came across this segment:

"While we consider the state of a body to be completely determined by the positions and velocities of a very large, yet finite, number of atoms and electrons, we make use of continuous SPATIAL FUNCTIONS to describe the electromagnetic state of a given volume,and a finite number of parameters cannot be regarded as sufficient for the complete determination of such a
state."

What are they? And what does it mean in the context above?

The above context is described in the book "Einstein: His Life and Universe" by Walter Isaacson (Page 97).

By spatial continous functions he refers to electromagnetic fields of the wave theory of light. He also states that theories of particles (like kinetic theory of gases) involve velocities and positions of finite number of particles and a wave theory cannot give this information.

Above text emphasize differences between wave-particle approaches.


Direct formula for a variation of Josephus problem: What is the direct formula for the following variation of Josephus problem?

  There are $n$ persons, numbered $1$ to $n$, around a circle.. Starting from $k$-th person, every second person is eliminated. Given the $n$ and $k$, determine the index of the $x$-th person who is eliminated.

ADDED:In the first step the $k+1$th is eliminated.For n = 16,k = 6, elimination={7,9,...}

Thanks

If you go to the Wikipedia page, there is a reference at the bottom

I made a video very closely related to this question. There's a link to the video at: http://oeis.org/A006165


solving coupled equation $$\begin{aligned}\dot{x}&=x\left ( 3-x-2y \right )\\
\dot{y}&=y\left ( 2-x-y \right )\end{aligned}$$

The above is a coupled equation.

The fixed point condition requires all $\;x,y\;$ for which $x^{\ast} =0\;$ and $\; y^{\ast}=0.$
Solving, I arrive at $\left ( x^*=0,y^*=0 \right )\;$ and $\;\left ( x^*=1,y^*=1 \right ).$
But there are $2$ other solutions: $\left ( x^*=0,y^*=2 \right )\;$ and $\;\left ( x^*=3,y^*=0 \right ).$

I am unable to determine the last $2$ ordered pairs of fixed points. Any help would be good.

Consider first solving the $\dot{x}$ equation. You should conclude that either $x=0$ or $x=3-2y$.

Now solve the $\dot{y}$ equation in each of these cases i.e. first suppose $x=0$, then substitute this into the $\dot{y}$ equation to get $0=y(2-y)$, so this is satisfied when $y=0$ and when $y=2$.

So you will have found the points $(0,0)$ and $(0,2)$. Repeat this process for the other relation found on $x$ (i.e. $x=3-2y$) to get the other points.

Analytical solution for the third and fourth fixed points will be a bit tricky (messy and cumbersome). However, you can use substitution in any one of the equations (replace $y$ in $\dot x$ or $x$ in $\dot y$) and solve the quadratic equation to get your other two fixed points.


Find a number M such that $\mid x^3 - x^2 +8x \mid\leq M$ for all $-2\leq x \leq 10$ Find a number M such that $\mid x^3 - x^2 +8x\mid \leq M$ for all $-2\leq x \leq 10$

I am not sure my answer is right or not..

I used triangle inequality.

$\mid x^3 - x^2 +8x\mid \leq \mid x^3 \mid + \mid x^2  -8x \mid  \leq \mid x^3 \mid + \mid x^2 \mid + \mid 8x \mid $

$M = 10^3 + 10^2 + 8(10)$

Isn't the best result |$ 10^3 - 10^2 +8*10$|?see the image here


Probability-Has the owner made a smart decision? Problem: A high school locker room owner has a locker room with 1000 spaces. Each space costs Rs. 100 a day. He has already sold monthly permits to 1001 high school members (knowing that it is likely that not all the high school members would want to keep their bags there at the same time).

If someone with a permit arrives to keep their bag and there are no spaces, the owner will refund Rs. 200 for that day (penalty of Rs.100).

Parking on any given day of the month is independent of every other day.

Has he made the smart decision by selling an extra space?

What I am stuck at is the issue that there are no probabilities given in this question. As we are talking about high school members that means everyone will have lectures on different time of the day (6 working days-from monday to saturday).  I think this question will not be requiring any numerical values to be solved. As expected profit is the part of this problem so E[random variable (RV] will be useful in solving this problem. Now If I am not wrong I think binomial RV will also play a major role in getting a general solution for this problem (because if we are interested in that 1 extra member then we will be interested in only two outcomes. 1. if that member will come 2. if that member will not come). That's all I know. I have no idea from where to start. I have no idea how to use CDF(Cumulative distribution function) to my advantage or how to even start this problem and get to the desired results. I am trying to find solution in terms of concepts related to RV not simply probability problems. I would appreciate if anyone can share their thoughts on this problem.

As true blue anil says you need a lot more information to make a model, then you can use the model to decide how many lockers to rent.  It sounds to me like he rents the lockers on a monthly basis, or on some term longer than a day.  Each day, some of the renters show up and use lockers.  The simplest assumption is that each person shows up independently with probability $p$.  If he has rented $1001$ so far and recorded how many people come each day he has a measure of $p$.  We have enough people that the normal approximation is reasonable, so if he rents to $n$ people he would expect a mean of $np$ to show up with a standard deviation of $\sqrt{np(1-p)}$.  If $k$ people show up his revenue for the day is
$$r(n,k)=\begin {cases} 100n&k \le 1000\\100n-200(k-1000)& k\gt 1000 \end {cases}$$  We want to maximize the expected value of the revenue $E(r(n,k))$ by choosing $n$ given the probability distribution of $k$.  We can convert the sum over $k$ to an integral in the continuous approximation and we get
$$E(r(n,k))=100n-200\int_{1000}^\infty(k-1000)\frac 1{\sqrt {2\pi np(1-p)}}e^{-\frac{(k-np)^2}{2np(1-p)}}dk$$
Evaluate the integral, take the derivative with respect to $n$, set to $0 \ldots$

As a mathematical approach this is defensible if the model is right.  Two problems are that there could be special events where $p$ gets much higher.  Say there is a big soccer match in town.  Maybe all you customers come in for the match and your refund loss is much higher.  It could be that somebody watches the clients arrive and depart and discovers that most people show up by $9$ AM and nobody leaves before $11$.  He signs up $50$ friends who rent lockers, then all show up between $9$ and $11$ hoping to collect refunds.

To begin with, a month has thirty days. You have called binomial distribution correctly - this is $X=B(30,p)$ that you're studying, wherein $p$ is the probability that all 1001 spaces are used at the same day. Calculate the latter, then the expected value of your variable times 100 is a good measure of whether the owner made a smart decision (I believe he did).


Question on z-score value in probability I am given that a student counts the number of motor vehicles passing in the street and he gets following values for the following 10 intervals (1 interval = 1 min):

0-1= 25

1-2= 27

2-3= 28

3-4= 20

4-5 = 22

5-6 = 20

6-7= 29

7-8= 28

8-9=22

9-10=23

Now, the question asks me to calculate the probability of getting a mean greater than 2 standard deviations of the mean from this count if the student were to take another set of 10 measurements.

My approach:
I calculated the sample mean from above and got 24.4 cars/min.
The sample standard deviation was 3.44 cars/min.
Now, my new mean  is supposed to be 31.28 cars/min
Then, I tried to use z score and used the formula:

z = (New mean-Old mean)/(Sample deviation/sqrt(20)), but doing so, my z score is 8.94, and I do not know what it means cause it says my probability is almost 0. What is wrong in my approach?

You should have divided the sample deviation by $\sqrt{10}$, not $\sqrt{20}$.

Regardless, the $z$-score is VERY high ($6.32)$ and you will compute a probability close to $0$.

This means that it is SUPER unlikely.


Find the maximum and minimum of the function $f$ Find the maximum and minimum of $f(x, y)=xy-y+x-1$ at the set $x^2+y^2\leq 2$.

I have done the following:

Since the region $x^2+y^2\leq 2$ is closed, $f$ has a maximum and a minimum,  which is either at the boundary or at the critical points of the function.

To find the critical points we do the following:

$$\nabla f=\overrightarrow{0} \Leftrightarrow (y+1, x-1)=(0, 0) \Rightarrow x=1, y=-1$$

So, we have that the only critical point is $(1, -1)$, which satisfies the contraint.

To find the extremas of $f$ at the boundary $x^2+y^2=2$, we will use Lagrange multipliers therem.

The constraint is $g(x, y)=x^2+y^2-2=0$.

We are looking for $x$, $y$ and $\lambda$ such that $$\nabla f(x, y)=\lambda \nabla g(x, y) \tag 1$$ and $$g(x, y)=0  \tag 2$$

$$(1) \Rightarrow (y+1, x-1)=\lambda (2x, 2y) \Rightarrow \lambda =\frac{1}{2}$$
$$\Rightarrow x=y+1$$

$$(2) \Rightarrow y=\frac{-1+\sqrt{3}}{2}, y=\frac{-1-\sqrt{3}}{2}$$

SO, the extremas are the points $\left (\frac{1+\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}\right )$ and $\left (\frac{1-\sqrt{3}}{2}), \frac{-1-\sqrt{3}}{2}\right )$.

Calculating the values of $f$, we have: $$f\left (\frac{1+\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}\right )=\frac{1}{2} \\ f\left (\frac{1-\sqrt{3}}{2}), \frac{-1-\sqrt{3}}{2}\right )=\frac{1}{2} \\ f(1, -1)=0$$

SO, $f$ has a local maximum at the points $\left (\frac{1+\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}\right )$ and $\left (\frac{1-\sqrt{3}}{2}), \frac{-1-\sqrt{3}}{2}\right )$ and an absolute minimum at $(1, -1)$.

Is this correct?? Could I improve something??

If you wanted to check the answer you get from the Lagrange multipliers, you could look for "extrema" along the boundary of the region by looking at the map:

$$F: \mathbb R \rightarrow \mathbb R^{2} \rightarrow \mathbb R$$
$$F(t) = f(\gamma(t)) = f(\sqrt{2} \cdot cos(t), \sqrt{2} \cdot sin(t)), t \in [0, 2\pi]$$

Which is a function on one variable relating every point in the boundary to its corresponding "height" in the graph of $z = f(x, y)$.

The resulting function is:

$$F(t) = f(\gamma(t)) = 2 \cdot cos(t) \cdot sin(t) - \sqrt{2} \cdot sin(t) + \sqrt{2} \cdot cos(t) - 1$$

From here you may find its derivative, etc. with all the tools from functions in one variable.

(To get $\gamma(t)$ you just parametrize the boundary of the region using the identity $cos(t) + sin(t) = 1$.)

I hope this helps, I find it useful to be able to check this sort of problem with a "1D version", so to speak.

Looks good, I have just a few comments:

"So the extras are the points..." at this point we don't know which of these is an extremum. These are possible extreme points. It is only after we plug in all the values that we know which are extrema.

"$f$ has a maximum and minimum" I would phrase as "$f$ achieves its maximum and minimum"


$$\frac{1}{6}\int \frac{2x+3}{1+(u)^2}du=\frac{1}{6}\int \frac{2\color{green}x}{1+(u)^2}du+\frac{1}{6}\int \frac{3}{1+(u)^2}du=\frac{1}{\color{red}6}\int \frac{2\color{green}x}{1+(u)^2}du+\frac{1}{2}\int \frac{1}{1+(u)^2}du$$

But in subsequent working you need to get rid of the $\color{green}x$ term in the integral

Mathematica gives:

$\frac{1}{9} \log \left(9 x^2-12 x+8\right)+\frac{13}{18} \tan ^{-1}\left(\frac{1}{2} (3
   x-2)\right)$


Probability brain teaser with infinite loop I found this problem and I've been stuck on how to solve it.

  A miner is trapped in a mine containing 3 doors. The first door leads to a tunnel that will take him to safety after 3 hours of travel. The second door leads to a tunnel that will return him to the mine after 5 hours of travel. The third door leads to a tunnel that will return him to the mine after 7 hours. If we assume that the miner is at all times equally likely to choose any one of doors, what is the expected length of time until he reaches safety?

The fact that the miner could be stuck in an infinite loop has confused me.
Any help is greatly appreciated.

Let $T$ be the time spent in the mine. Conditioning on the first door the miner chooses, we get
$$ \mathbb{E}[T]=\frac{1}{3}\cdot3+\frac{1}{3}(5+\mathbb{E}[T])+\frac{1}{3}(7+\mathbb{E}[T])$$
so
$$ \mathbb{E}[T]=5+\frac{2}{3}\mathbb{E}[T].$$
If $\mathbb{E}[T]$ is finite, then we can conclude that $\mathbb{E}[T]=15$.

To see that $\mathbb{E}[T]$ is finite, let $X$ be the number of times the miner chooses a door. Then $ \mathbb{P}(X\geq n)=(\frac{2}{3})^{n-1}$ for $n=1,2,3,\dots$, hence
$$ \mathbb{E}[X]=\sum_{n=1}^{\infty}\mathbb{P}(X\geq n)=\sum_{n=1}^{\infty}\Big(\frac{2}{3}\Big)^{n-1}<\infty$$
And since $T\leq 7(X-1)+3$, we see that $\mathbb{E}[T]<\infty$ as well.

A probability tree reveals that the miner can expect to spend 9 hours to reach safety. Suppose he chooses Door 7, with probability 1/3; he returns to the start and chooses Door 5 with p = 1/2; he returns and chooses Door 3 with p = 1. Then his path would take take 7 + 5 + 3 = 15 hours with p = (1/3)(1/2)(1) = 1/6. The other possible paths are 7 + 3 = 10, p = 1/6; 5 + 7 + 3 = 15, p = 1/6; 5 + 3 = 8, p = 1/6; and 3, p = 2/6. If T is the rv "time to exit", then its expectation is E(T) = 15*(1/6) + 10*(1/6) + 15*(1/6) + 8*(1/6) + 3*(2/6) = 54/6 = 9 hours.


Inequality with partial integration in one dimension Is it possible to prove $ \| u \|_{L^2(0,1)} \leq \| u' \|_{L^2(0,1)} $ for $u \in C^1([0,1])$ with $u(0)=0$ by using partial integration?

I don't know what partial integration is, but if you integrate
$$|u(x)|^2 = \left|\int_0^x u'(t)\,dt \right|^2 \le \int_0^x |u'(t)|^2\,dt \le \|u'\|_{L^2}^2$$
over $0\le x\le 1$,  you get the desired inequality.


Number of size 1 partitions of the empty set Disclaimer: This is a homework problem, but I'm just asking for clarification, not a solution.

We're asked to prove $S(0,1) = 1$, where $S(n,k)$ is "the number
of different partitions of [a set of size $n$] into $k$ mutually disjoint subsets", where a partition is defined like so: "Let $X$ be a collection of pairwise disjoint sets and let $Y = \bigcup X$. Then $X$ is called a partition of
$Y$ if either (i) $X = Y = \varnothing$; or (ii) $X \neq \varnothing \land \varnothing \notin X$."

As far as I can tell the only set $X$ with $\bigcup X = \varnothing$ and $|X| = 1$ is $X = \{\varnothing\}$, but this does not satisfy either (i) or (ii) of the defintion of a partition. So it seems to me that there are no size 1 partitions of $\varnothing$, and $S(0,1)$ should be $0$, not $1$.

Am I missing something here?

You are right according to that definition. By the way, it can be simplified to "... is called a partition if $\emptyset\notin X$."

I believe you're mistaken when you say that the set $X=\{\varnothing\}$ satisfies $\bigcup X = \varnothing$. It actually satisfies $\bigcup X = \{\varnothing\}$.

The only $X$ satisfying $\bigcup X = \varnothing$ is $X=\varnothing$, the empty set itself.

Does that help?


Factoring the following polynomials Factorize the following polynomial:

$t^3 -9t +8$
$t^6 -91t^2 +90$

It's not so bad to be baffled... :)

Putting $t=1$ and and $t^2=1$ into the first and second expressions respectively will make them zero, so $(t-1)$ and $(t^2-1)$ are factors for the first and second expressions respectively.

You can work out the other factor(s) by equating coefficients of powers of $t$.

$$\begin{align}t^3-9t+8&=(t-1)(t^2+t-8)\\ \\
t^6-91t^2+90&=(t^2-1)(t^4+t^2-90)\\
&=\left[(t-1)(t+1)\right]\left[(t^2-9)(t^2+10)\right]\\
&=(t-1)(t+1)(t-3)(t+3)(t^2+10)\\
&=(t-3)(t-1)(t+3)(t+3)(t^2+10)\\\end{align}$$

To solve $t^3 -9t +8$ try to find one root by checking different vaule for t.

From the rational root theorem follows that a rational solution of a polynomial with integer coefficients and leading coefficient $1$ must be an integer dividing the constant term of the polynomial. So if the polynomial $t^3 -9t +8$ has a rational solution it must be among the numbers $\{\pm 8, \pm4, \pm2, \pm1\}$. By substituting $1$ for $t$ in the polynomial you see that $1$ is a zero of the polynomial.

From the he factor theorem follows: if $1$ is a zero of $t^3 -9t +8$ then $t-1$ is a divisor of $t^3 -9t +8$. So we divide $t^3 -9t +8$ by $t-1$  using th polynomial division  algorithm to get the quotient $t^2+t-8$. The remainder is $0$ because $t-1$ is a divisor.

So we now know that $$t^3 -9t +8=(t-1)(t^2+t-8)$$. The remaining solutions can be found by solving
$$t^2+t-8=0$$
(Why?)

To solve $t^6 -91t^2 +90$ we start with substituting $u$ for $t^2$ in the equation to get

$$u^3 -91u +90$$
now we have  a third degree polynomial as we had in the former case. We solve it using the samce technic.
After we have found solutions $u$ we can find solution $t$ of the sixth degree polynomial by solving the equation $$t^2=u$$


Let $P_0 \subsetneq P_{1} \subsetneq \dotsb \subsetneq P_n$ be a chain of prime ideals in a Noetherian ring. Let $P_0 \subset P_{1} \subset \dotsb \subset P_n$ be a chain of prime ideals in a Noetherian ring.
Show that if $x \in P_n$ then there exists a chain of prime ideals $Q_{1} \subsetneq \dotsb \subsetneq Q_n = P_n$ such that $x \in Q_1$.

I'm lost on how to tackle this problem. Any help would be appreciated!

This is Lemma 2.5.13 in Qing Liu's book Algebraic Geometry and Arithmetic Curves. The proof here uses Krull's principal ideal theorem to employ induction on $n$.

I find the proof quite non-trivial, definitely not something I would come up with. I don't know if there is an elementary way to prove it. As an aside, Liu then uses this lemma to deduce that Noetherian local rings have finite dimension (in fact the dimension is bounded by the number of generators needed for the unique maximal ideal) in Corollary 5.14.


Measure nonzero implies dense on a rectangle This would be a very handy lemma for me but I have been unable to prove it thus far.

  If $S \in \mathbb{R}^n$ is bounded and is not of measure zero, then there exists a rectangle $R$ such that $S$ is dense in $R$.

Does anyone have any ideas on how to prove it? Can someone provide a counterexample?

Thank you!
Brian

Edit: I should have clarified what I meant (moreover this may not be (or even worse, it could be completely different from) the standard definition of dense): $S$ is dense on $R$ if for every $x \in R$, if $\epsilon > 0$ then there exists $y \in S$: $|x-y| < \epsilon$.

Again thank you for the wealth of responses.

Any fat Cantor set in $\mathbb R$ has positive measure yet is not dense in any segment. See http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set

Maybe I'm misunderstanding the question, but it seems obviously wrong.  A circle x^2 + y^2 <= 1 is obviously not dense in any rectangle.