Homotopy equivalence of mapping telescope of CW complex On P.138 of Hatcher (P.147 of pdf), Hatcher claims that (paraphrase):

Let $X$ be an infinite-dimensional CW complex. Let $X^n$ denote the $n$-skeleton of $X$, i.e. the cells with $n$ dimensions or below. Identify $X^n$ as a subset of $X^{n+1}$. Let $T = \bigcup_{n \in \Bbb N} X^n \times [n,\infty)$. Then, $X$ and $T$ are homotopy equivalent.

I see a map from $T$ to $X$, but I don't see how one builds a map from $X$ to $T$, let alone showing that they are homotopy-inverses of each other.

In Hatcher's proof, the time $t=1$ of the homotopy is left unspecified, and he never constructed a map from $X$ to $T$.

This is the most nonsense I've ever spewed. I apologize to the readers sincerely.

Review the definition of a CW-complex. We will define finite-dimensional CW-complexes first, and then use that to build our infinite-dimensional CW-complexes.

Let $I_n$ be an indexing set for each $n \in \Bbb N$. We build our stages $X^n$ recursively until we reach the dimension of the complex we want.

$X^0$ is defined to be $I_0$ with the discrete topology.

$X^k$ having been assumed to be defined, we require maps $\varphi^k_\alpha : \partial D^{k+1} \to X^k$ with index $\alpha \in I_k$ to advance to the next stage, wherein we define $X^{k+1}$ to the colimit of the following diagram:
$$X^k\overset {\varphi^k_\alpha} \longleftarrow \partial D^{k+1} \overset {i^k_\alpha} \longrightarrow D^{k+1}$$
Note that there are as many copies of $\partial D^{k+1}$ and $D^{k+1}$ in the diagram as the indexing set $I_k$, but only one copy of $X^k$.

Being a colimit, $X^{k+1}$ comes with these maps:

$iL^k : X^k \to X^{k+1}$
$iR^k_\alpha : D^{k+1} \to X^{k+1}$

It has the property that $iL^k \circ \varphi^k_\alpha = iR^k_\alpha \circ i^k_\alpha$.

It also has the property that for every $T$ with maps $tL^k : X^k \to T$ and $tR^k_\alpha : D^{k+1} \to T$ for each $\alpha$ such that $tL^k \circ \varphi^k_\alpha = tR^k_\alpha \circ i^k_\alpha$, there is a unique map $\pi_{k+1}: X^{k+1} \to T$ such that $tL^k = \pi_{k+1} \circ iL^k$ and $tR^k_\alpha = \pi_{k+1} \circ iR^k_\alpha$.

Now, we have defined finite-dimensional CW complexes.

To obtain an infinite-dimensional CW complexes, require that we have $X^n$ for each $n \in \Bbb N$, and take the direct limit of $X^n$ with the maps being $iL^k : X^k \to X^{k+1}$. Call the direct limit $X$, and by universal properties we have maps $iX^k : X^k \to X$, and the property that if we have maps $\pi_k : X^k \to T$ such that $\pi_k = \pi_{k+1} \circ iL^k$, then there is a unique map $\pi : X \to T$ such that $\pi_k = \pi \circ iX^K$.

Now we wish to build $\pi : X \to T$, where $T = \bigcup_{n \in \Bbb N} X^n \times [n,\infty)$ identified as a subset of $X \times [0,\infty)$, with identifications done with $iL^k$.

Before we proceed, I will build some auxiliary functions $\psi_n : X^n \times [0, n] \to T \cap (X^n \times [0, n])$ by recursion.

$\psi_0$ is the identity function.
Given $\psi_k$, we construct $\psi_{k+1} : X^{k+1} \times [0, k+1] \to T \cap (X^{k+1} \times [0, k+1])$ as follows: By the property of $X^{k+1}$, and that colimit commutes with product if the multiplicand ($[0, k+1]$) is locally compact (reference), we need to construct $tL^k : X^k \times [0, k+1] \to T \cap (X^{k+1} \times [0, k+1])$ and $tR^k_\alpha : D^{k+1} \times [0, k+1] \to T \cap (X^{k+1} \times [0, k+1])$ for each $\alpha$ such that $tL^k \circ (\varphi^k_\alpha \times \operatorname{id}_{[0, k+1]}) = tR^k_\alpha \circ (i^k_\alpha \times \operatorname{id}_{[0, k+1]})$.
To construct $tL^k : X^k \times [0, k+1] \to T \cap (X^{k+1} \times [0, k+1])$, we let it to be $\psi_k$ on $X^k \times [0, k]$ and $\operatorname{id}$ on $X^k \times [k, k+1]$. We will leave an asterisk here and show later that this is well-defined.
To construct $tR^k_\alpha : D^{k+1} \times [0, k+1] \to T \cap (X^{k+1} \times [0, k+1])$, we let $(\vec v, (k+1) s) \in D^{k+1} \times [0, k+1]$. If $\|\vec v\| \le \frac {1 + s} 2$, then send it to $(iR^k_\alpha \left( \frac {2v} {1+s} \right), k+1)$. If $\|\vec v\| \ge \frac {1 + s} 2$, then send it to $tL^k(\varphi^k_\alpha(\vec v/\|\vec v\|), (k+1)(s+2(1-\|\vec v\|)))$. Well-defined: when $\| \vec v \| = \frac {1+s} 2$:
$$ \begin{array}{cl}
& tL^k(\varphi^k_\alpha(\vec v/\|\vec v\|), (k+1)(s+2(1-\|\vec v\|))) \\
=& tL^k(\varphi^k_\alpha(\vec v/\|\vec v\|), k+1) \\
=& (\varphi^k_\alpha(\vec v/\|\vec v\|), k+1) \\
=& (iL^k (\varphi^k_\alpha(\vec v/\|\vec v\|)), k+1) \\
=& (iR^k_\alpha (i^k_\alpha(\vec v/\|\vec v\|)), k+1) \\
=& (iR^k_\alpha (\vec v/\|\vec v\|), k+1) \\
=& (iR^k_\alpha \left( \frac {2v} {1+s} \right), k+1)
\end{array}$$
To resolve the asterisk in step 3, i.e. to prove that $\psi_n(x, n) = (x, n)$ for $x \in X^n$, we use induction (and clear that asterisk by using the induction hypothesis). Now, it is true for $n=0$ since $\psi_0$ is the identity function. For $n=k+1$, let $x \in X^{k+1}$. If $x = iL^k(x')$:
$$\begin{array}{cl}
& \psi_{k+1}(x, k+1) \\
=& \psi_{k+1}(iL^k(x'), k+1) \\
=& tL^k(x', k+1) \\
=& (x', k+1) \\
=& (iL(x'), k+1) \\
=& (x, k+1)
\end{array}$$
If $x = iR^k_\alpha(\vec v)$, note that the casing parameter $\frac{1+s}2$ is equal to $1$ when $s=k+1$, so the first case in $tR^k$ always matches:
$$\begin{array}{cl}
& \psi_{k+1}(x, k+1) \\
=& \psi_{k+1}(iR^k_\alpha(\vec v), k+1) \\
=& tR^k_\alpha(\vec v, k+1) \\
=& (iR^k_\alpha(\vec v), k+1) \\
=& (x, k+1)
\end{array}$$

Now we can finally build $\pi : X \to T$.

By property of $X$, we need $\pi_n : X^n \to T$ such that $\pi_n = \pi_{n+1} \circ iL^n$.
Build $\pi_n$ by induction.
Build $\pi_0$ by sending $x \in X^0$ to $(x,0)$.
We build $\pi_{k+1} : X^{k+1} \to T$ by letting $\pi_{k+1}(x) := \psi_{k+1}(x, 0)$.
To check that $\pi_n = \pi_{n+1} \circ iL^n$, let $x \in X^n$:
$$\begin{array}{cl}
& \pi_{n+1}(iL^n(x)) \\
=& \psi_{n+1}(iL^n(x), 0) \\
=& tL^k(x, 0) \\
=& \psi_k(x, 0) \\
=& \pi_k(x) \\
\end{array}$$


Irreducibility of a polynomial in a finite field The question asks me to show that $\mathbb{Z}_{13}[x]/\langle x^{2014}-x^{1000}+1 \rangle$ is a field.

Now what I know is that since $\mathbb{Z}_{13}$ is a field, so if I show that $x^{2014}-x^{1000}+1 $ is irreducible, then $\langle x^{2014}-x^{1000}+1 \rangle$ will be a maximal ideal and so $\mathbb{Z}_{13}[x]/\langle x^{2014}-x^{1000}+1 \rangle$ will be a field.

But I don't see how to show the polynomial is irreducible in $\mathbb{Z}_{13}[x]$. For the degree $\le 3$, I could use whether there is a root or not but I don't know for such a big degree.

Maybe there is a duplicate for this question, but please help anyway. Thank you.

This is not true.  The polynomial $x^{2014}-x^{1000}+1$ has $x^2+6$ as a factor, and in particular is not irreducible.  You can quickly verify this by observing that if $x^2=-6$ then (working in $\mathbb{Z}_{13}$) $$x^{2014}-x^{1000}+1=(-6)^{1007}-(-6)^{500}+1=(-6)^{11}-(-6)^8+1=0.$$

HINT

Use the fact that for $p=13$ prime, $x^{p-1}\equiv 1 \pmod p$.  The resulting t3nth degree polynomial is much easier to work with.


Is it valid to multiply both sides of an equation by a complex number? In case I'd like to get rid of some complex denominators, could I multiply both sides of the equation by the common denominator?
I understand there's an other way, but i'd like to know if the above mentioned would be a valid operation, and if not, why.

If $\left(\mbox{some expression}\right)=\left(\mbox{another expression}\right)$, this means that both expressions evaluate to the same complex number. Let's call this number $z$. This means that:
$$z=\left(\mbox{some expression}\right)=\left(\mbox{another expression}\right)$$
Well, if you want to multiply both sides by another complex, say $w$, the equality will still hold, because both $w\cdot\left(\mbox{some expression}\right)$ and $w\cdot\left(\mbox{another expression}\right)$ are just the same as $w\cdot z$.

Beware that if $w$ is $0$ (that is, if you want to multiply both sides by zero), then the equality will still hold: it will be just $0=0$. Of course, this is not useful---you lost all the information you had in your original equation, because you can't divide both sizes by $0$ to get back where you started.

This is valid. You can still multiply both sides of an equation by a scalar and still have it hold. It's just like the real numbers.


Counterexamples of second mean value theorem of integral Recently I am having trouble understanding Bonnet's mean value theorem. I looked back and forth on the proof, and find that the condition of $\phi$ being decreasing is essential. However, I am unable to find an example to illustrate how the theorem fails when $\phi$ is NOT decreasing.

Can anyone give an example?

DO NOT LEAVE a comment before even reading the question. The question is asking for an example where the theorem FAILS. It is NOT asking for an example where the theorem applies properly. It is irresponsible to vote or comment before even understanding the question properly.

Let $\phi(x)=x$, $G(x)=1$, and $[a,b]=[0,1]$. Then
$$
\int_0^1 G(x)\phi(x)dx=0.5
$$

But for any $c\in[0,1]$,
$$
\phi(0) \int_0^c G(x)dx=0.
$$
So the theorem fails.


Find the volume of the solid obtained by rotating the region about the line $y=7$ Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

$$y=\left(x-1\right)^{1/2}$$$$y=0$$$$x=5$$

rotate about the line y=7

I used this formula$$\int_1^5π\left[7-\left(x-1\right)^{1/2}\right]^{2}dx$$and got 388π/3, which is not correct. I changed the lower limit to 0 but it is also wrong. Can someone help me?

I think your integral should be $\int_1^5 \pi 7^2 dx - \int_1^5 \pi \left(7 -(x-1)^{\frac{1}{2}}\right)
^2dx$.

If $A$ is the region bounded below by the horizontal line $y = 0$, above by the curve $f(x) = (x-1)^{\frac{1}{2}}$ and on the left and right by the vertical lines $x = 1$ and $x = 5$ respectively,...

and furthermore if $B$ is the region bounded above by the horizontal line $y = 7$, below by the curve $f(x)$, and on the left and right by the same vertical lines $x= 1$ and $x = 5$ as before,...

then $A$ and $B$ together form the rectangle $R$ bounded above and below by the horizontal lines $y = 7$ and $y = 0$ and bounded on the left and right by the vertical lines $x = 1$ and $x = 5$.

If you are asked to revolve $A$ about the axis of revolution $y = 7$. So you should revolve $R$ about this axis to form a cylinder (whose volume is the first integral in the difference written above), then remove from this cylinder the revolution of $B$ about the same axis (the volume of this revolution is the second integral above).


Algebra Trick Question Students in a class were asked about the number of brothers and sisters they had. 17 said they had at least one brother. 18 said they had at least one sister. 6 said they had at least one brother AND one sister. 5 said they had no brothers or sisters. What was the least possible number of students in the class?

Well I actually have 2 answers to this question, which both are probably wrong. My first guess is 11, since it is clearly stated in the question that 5 had no siblings and 6 had atleast one brother and one sister.

My second guess is 1, since there has to be atleast a student in the class.

Out of the 17 who had at least one brother, 11 had no sisters.  Combined with the 18 who had at least one sister, this gives us 29 students.  Adding in the 5 who had no brothers or sisters, we get 34.  So there is exactly 34 students.

$$\text{no sister/brother $+$ no sisters some brothers $+$ sisters, with or w/out brothers}=34$$

It says that there are 18 with at least one sister, and 5 without brothers/sisters. So there have to be at least 23. Now we have to look if something implies that this has to be higher. Because these eighteen said that they had at least one sister, those 17 with at least one brother can all be combined with this, so we get a minimum of 23.


Proof that the limit of a particular sequence of Lebesgue integrals equals zero. Prove that if  $f \in L^{1}(A)$ and {${A_{n}}$} is a sequence of measurable subsets of A with $\lim_{n \to \infty}m(A_{n})=0$
then $lim_{n \to \infty} \int_{A_n}f=0$

I know there exists a very similar post for tackling this particular problem but I cannot find the complete proof of it.
I would be grateful if someone could help.

$\lvert\int f1_{A_n}\,dm\rvert\leq\int\lvert f1_{A_n}\rvert\,dm\leq m(A_n)\int\lvert f\rvert\,dm\to0$ by Holder's inequality.

It's easy. For $n \in{} \mathbb{N}$, $\int_{A_n} f=\int_{A} f \mathbb{1}_{A_n}$. By Holder's inequallity, we have $\left |{\int_{A} f \mathbb{1}_{A_n}}\right |^2 \leq \int_{A} f^2 \int_{A} \mathbb{1}_{A_n}^2= \int_{A} f^2 \int_{A} \mathbb{1}_{A_n}=m(A_n)\int_{A} f^2 \xrightarrow{ n \to \infty } 0$ since $\lim_{ n \to \infty } m(A_n)=0$ and $f \in L^1(A)$. We conclude $\lim_{ n \to \infty } \int_{A} f \mathbb{1}_{A_n}=0$.


Prove: A matrix $A \in \mathbb{C}^{m \times n}$ with $m \geq n$ has full rank if and only if it maps no two distinct vectors to the same vector. Thm. A matrix $A \in \mathbb{C}^{m \times n}$ with $m \geq n$ has full rank if and only if it maps no two distinct vectors to the same vector.

I'm going to write out the proof (given by my text) line by line because there are certain statements I'm not sure about. I'm a little rusty on linear algebra. For each statement of the proof, I'll give a statement from the proof and my analysis immediately below. The converse I will attempt to prove in my own words. Please give constructive feedback on correctness of my interpretation(s).

proof   ($\implies$)

If A is of full rank, its columns are linearly independent, ...

Agreed. Since $m \geq n$ and full rank is defined my the maximum possible dimension spanned by the row or column spaces of $A$, then it must be that the rank of $A$ is $n$ which implies that the $n$ columns of A span $\mathbb{C}^n$. Which is only possible if the $n$ columns are linearly independent.

... so they form a basis for range$(A)$.

I think I agree because, since the columns are L.I., they form a basis for $\mathbb{C}^n$, which is the column space of $A$ which is the range of $A$.

This means every vector $b \in \text{range}(A)$ has a unique linear expansion in terms of the columns of $A$, and therefore has a unique $x$ s.t $b = Ax$.

I have a strong intuition that the expansion is unique, since none of the columns of $A$ can be scaled to each other. But how is this statement proved? Do I consider $x,y$ such that $Ax = b$ and $Ay = b$ and then ($a_i$ are columns of $a$)

$$ y_1a_1 + \cdots + y_na_n = x_1a_1 + \cdots + x_na_n $$
$$(y_1 - x_1)a_1 + \cdots + (y_n - x_n)a_n = 0 $$

And since the columns are L.I. the only possible solution set is

$$\{y_1 - x_1 = 0, \dots , y_n - x_n = 0\} \implies x = y$$

Converse (proof by contraposition).

If $A$ is not of full rank then the columns of $A$ must be be linearly dependent. Therefore there is a nontrivial solution $x$ to the equation

$$Ax = x_1a_1 + \cdots + x_na_n = 0$$

This means that for any other vector $c \in \mathbb{C}^n$ we have

$$Ac = Ax + Ac = A(x+c)$$

so that $x+c$ and $c$ are mapped to the same vector $Ac$. That is, we have shown that if $A$ is not of full rank, then the mapping is not unique. So by contraposition, a unique mapping implies A is of full rank.

Sorry, that wasn't pretty.


Tangent of a circle through a point not on circle Given the equation of a circle: $(x+4)^2+(y-2)^2 = 1$ find the tangent that goes through the point $(-3,3)$…

…but this point IS NOT on this circle. Unfortunately, this is not something that I have seen before and the normal formula doesn't work.

Thank you

A little graphing won't hurt:

By inspection, the tangent lines are $x=-3$ and $y=3$.

You should have a system of equation the first is the circle and the second is the general equation for a line that pass from $(3,3)$ substitute one into the others and apply the tangent condition $b^2-4ac=0$ you will find the two points. Another methods is using the derivation.


Solving Limits with L'Hospital's Rules I am having difficulties solving this limit. I was given the question and equation:

Try using L’Hospital’s Rules to evaluate the follwing limit:

$$\lim\limits_{u \to \infty } \frac{u}{\sqrt{u^2 +1}}$$

How do you solve this with L'Hospital's rules? Do you solve this with them?

$\color{red}{\text{Method 1}}$: Using L-Hospital's Rule for $\frac{\infty}{\infty}$ form as follows $$\lim_{u\to \infty}\frac{u}{\sqrt{u^2+1}}$$ $$=\lim_{u\to \infty}\frac{\frac{d(u)}{du}}{\frac{d}{du}(\sqrt{u^2+1})}$$
 $$=\lim_{u\to \infty}\frac{1}{\frac{2u}{2\sqrt{u^2+1}}}$$ $$=\lim_{u\to \infty}\frac{\sqrt{u^2+1}}{u}$$ $$=\lim_{u\to \infty}\sqrt{\frac{u^2}{u^2}+\frac{1}{u^2}}$$ $$=\lim_{u\to \infty}\sqrt{1+\left(\frac{1}{u}\right)^2}$$ Let, $u=\frac{1}{t}\implies t\to 0 \ as \ u\to \infty$ $$=\lim_{t\to 0}\sqrt{1+(t)^2}$$ $$=\sqrt{1+0}=1$$

$\color{red}{\text{Method 2}}$: Without using L-Hospital's rule

Given, $$\lim_{u\to \infty}\frac{u}{\sqrt{u^2+1}}$$
Let, $u=\frac{1}{t}\implies t\to 0 \ as \ u\to \infty$, hence we have  $$\lim_{t\to 0}\frac{\frac{1}{t}}{\sqrt{\left(\frac{1}{t}\right)^2+1}}$$ $$=\lim_{t\to 0}\frac{1}{\sqrt{1+t^2}}$$ $$=\frac{1}{\sqrt{1+0}}=1$$

$$\lim\limits_{u \to \infty } \frac{u}{\sqrt{u^2 +1}} = \lim\limits_{u \to \infty } \frac{\color{red}{\sqrt{u^2}}}{\sqrt{u^2 +1}}$$

because

   while $\sqrt{u^2} = |u|$, we may assume $u \ge 0$ since $u \to \infty$ so $\sqrt{u^2} = |u| = u$

After that we have

   $$ \lim\limits_{u \to \infty } \frac{\color{red}{\sqrt{u^2}}}{\sqrt{u^2 +1}} = \lim\limits_{u \to \infty } \sqrt{\frac{u^2}{u^2 +1}} =  \sqrt{\lim\limits_{u \to \infty }\frac{u^2}{u^2 +1}}$$

Actually

   It may help to note that for large $u$, $\sqrt{u^2 + 1}$ behaves like $\sqrt{u^2}$


continuity in the strong topology implies continuity in the weak one I have to prove that if $T:(E,\|\cdot\|_E)\rightarrow (F,\|\cdot\|_F)$ is a continuous and linear operator, and $x_h\rightharpoonup x$ in $E$, than $Tx_h\rightharpoonup Tx$ in $F$. So we know that $T$ is continuous with respect to the strong topologies, and we want to prove that it is also continuous with respect to the weak ones.

Is there a simple proof of this fact?

I've just noticed that I didn't understand your question at the first time. So I'll change my answer almost completely.

I'll try to answer it in a general setting. Given a set $X$, a family $\{(X_i,\tau_i):i\in I\}$ of topological spaces and a famlily of functions $\mathcal{F}=\{f_i\colon X\to X_i|i\in I\}$, let $\sigma(X,\mathcal{F})$ be the least topology on $X$ such that all the functions of $\mathcal{F}$ are continuous. This topology is generated by finite intersections of sets of the form $f_i^{-1}(V)$, for $i\in I$ and $V\in \tau_i$.

Notice that the weak topology is a particular case of this type of topology.

In this context, we have the following

Fact 1. Let $(x_n)_{n\in\mathbb{N}}$ be a sequence in $X$ and let $x\in X$. Then $x_n\to x$ in $\sigma(X,\mathcal{F})$ if and only if $f_i(x_n)\to f_i(x)$ in $\tau_i$ for all $i\in I$.

The proof is hidden below:

  Suppose that $x_n\to x$ in $\sigma(X,\mathcal{F})$. If $f_i(x)\in V\in\tau_i$, then $x\in f_i^{-1}(V)$, hence $x_n\in f_i^{-1}(V)$ for $n\gg 0$, from which it follows that $f_i(x_n)\in V$ for $n\gg 0$.

${}$

   Conversely, if $U\in\sigma(X,\mathcal{F})$, then $U=\bigcap_{j\leq n}f_{i_j}^{-1}(U_j)$ for $i_0,\dotso,i_n\in I$ and $U_{i_j}\in\tau_{i_j}$ for each $j\leq n$. If $x\in U$, then $f_{i_j}(x)\in U_{i_j}$ for each $j\leq n$, from which it follows that there exists $N_j\in\mathbb{N}$ such that $f_{i_j}(x_n)\in U_{i_j}$ for all $n\geq N_j$. It is easy to see that for $N=\max_{j\leq n}N_j$ one has $x_n\in U$ for all $n\geq N$.

Now, in your context, $x_n\rightharpoonup x$ is equivalent to $\varphi(x_n)\to \varphi(x)$ for all $\varphi \in E^*$, while $Tx_n\rightharpoonup Tx$ is equivalent to $\psi(Tx_n)\to\psi(Tx)$ for all $\psi\in F^*$. Since you have the former by hypothesis, the latter follows: for $\psi\in F^*$, we have that $\psi\circ T\in E^*$, hence $\psi\circ T(x_n)\to \psi\circ T(x)$.

In order to prove that if $T\colon (E,\|\cdot\|)\to (F,\|\cdot\|)$ is continuous then $T\colon (E,\sigma(E,E^*))\to (F,\sigma(F,F^*))$ is continuous, you can use the next fact, stated in the context of the beginning of my answer:

Fact 2. For any topological space $Z$, a function $f\colon Z\to (X,\sigma(X,\mathcal{F}))$ is continuous if and only if $f_i\circ f\colon Z\to (X_i,\tau_i)$ is continuous for all $i\in I$.

The proof is routine, but you can check it below:

   If $f$ is continuous, then $f_i\circ f$ is trivially continuous. Conversely, if $f_i\circ f$ is continuous for all $i\in I$, we take an open set $V\subset X$, say $V=\bigcap_{j\leq n}f_{i_j}^{-1}(V_{i_j})$ for $i_j\in I$ and $V_{i_j}\in \tau_{i_j}$, and notice that $f^{-1}(V)=f^{-1}\left(\bigcap_{j\leq n}f^{-1}_{i_j}(V_{i_j})\right)=\bigcap_{j\leq n}f^{-1}\left(f^{-1}_{i_j}(V_{i_j})\right)=\bigcap_{j\leq n}(f_{i_j}\circ f)^{-1}(V_{i_j})$.

Now, if $T\colon (E,\|\cdot\|)\to (F,\|\cdot\|)$ is continuous, then for any $\psi\in F^*$ one has $\psi\circ T\colon (E,\|\cdot\|)\to k$ continuous, but this implies that $\psi\circ T\colon (E,\sigma(E,E^*))\to k$ is continuous (by the definition of $\sigma(E,E^*)$). So, by replacing $Z$ by $(E,\sigma(E,E^*))$ in the Fact 2, we got the result.

Although it is not part of your question, I am almost sure that the converse is true, but I don't remember how to prove it.

The linearity is not needed. In fact, what you have is a continuous function $T$ between topological spaces $X$ and $Y$ (they're normed but that's not relevant) and a convergent net (or sequence) $x_h \rightarrow x$ in $X$. Then show that $T(x_h)$ converges to $T(x)$.
(weak topology not needed here.)

And this is quite easy: take an open set $O \subset Y$ that contains $T(x)$. Then by continuity of $T$, $T^{-1}[O]$ is open in $X$, and contains $x$. As $x_h \rightarrow x$, there is some $h'$ such that for all $h \ge h'$, we have that $x_h \in T^{-1}[O]$. But this means that for all $h \ge h'$ we have that $T(x_h) \in O$. So $T(x_h)$ converges to $T(x)$ by definition.


Showing public returns to be greater than private returns mathematically Take this headline

"OECD figures show public benefits more than individuals from tertiary education."

How would I present this mathematically, to show that public returns are greater than private returns?

This would be true if there are positive externalities from tertiary education. That is, the rest of the society also derives some positive benefit from tertiary education. For example, if the benefit to an individual getting tertiary education is \$250,000 (in incremental lifetime earnings and other nonpecuniary benefits) and the benefit to society is another \$50,000, then public benefits exceed private benefits. This would be the case for example, if education lead to a more civic society that others value. Opposite will be true with negative externalities (incremental earnings of educated come at the expense of lower earnings for someone less educated).

The decision of whether tertiary education is social welfare enhancing or not depends not just on externalities but also on cost of providing that education. In the above example with \$250,000 private benefit and \$50,000 externalities, tertiary education is welfare enhancing if the cost of education is \$275,000 but nor if the cost is \$325,000.

If modeling theoretically, suppose value of A is VA, and value of B (everyone other than A) is VB when A is not educated. Next, suppose value of A is VA', and value of B (everyone other than A) is VB' when A is educated and everything else is same. Then, the headline is equivalent to the condition VB' > VB.

Empirically, I think the headline is not surprising and is most likely true. To test it, you need an exogenous change in education. For example, what happens to peers/neighbors/cities when the fraction of education changes exogenously. Maybe change in immigration laws. Or if a natural disaster closes schools, compare the effect to the effect in other places with similar disaster but no effect on education. The problem is that most changes are not completely exogenous.


Surface Area of the simple figure There is a question which states

The solid brick figure shown is made of small bricks of side 1. When the large brick is disassembled into its component small bricks , the total surface area of all small bricks is how much greater than surface area of larger brick

Here is how I am attempting it:
Surface Area of Large Brick = $6x^2=6(2)^2=24$
Surface Area of 1 small Brick = $6x^2=6(1)^2=6$
Total Small Bricks = 12 = So Total Surface Area = $72$
Difference = $72-24=48$
But the difference is suppose to be $40$. How ?
(EDIT : Image Edited)

For the brick you've drawn, the large brick has two $3\times3$ faces and four $2\times 3$ faces, so its surface area is $2(9)+4(6)=42$.

There are $3\times3\times2 = 18$ smaller bricks, each of which has surface area $6$ so the total surface area of the $18$ small bricks is $18(6)=108$.

The difference of the areas, then, is $108-42=66$.

On the other hand, if the brick has dimension $2\times2\times3$, then the large brick has two $2\times2$ faces and four $2\times 3$ faces, so its surface area is $2(4)+4(6)=32$.

There are $2\times2\times3 = 12$ smaller bricks, each of which has surface area $6$ so the total surface area of the $12$ small bricks is $12(6)=72$.

The difference of the areas, then, is $72-32=40$. This is likely what the original problem was, as mentioned in the comments.

SA of larger bricks is $$2LW+2LH+2HW  = 2(3)(2)+2(3)(2)+(2)(2)(2)=12+12+8= 32 $$ because it is not a perfect cube, but it's a rectangular prism.

SA of smaller bricks $= 6\cdot 12 = 72$. Then, the difference = larger - smaller $ = 72-32 = 40$.


Show that this inequality involving expectations holds Let there be two events (which are disjoint and a partition of the sample space) $G$ and $B$ where $p = Pr(G)$ and $1-p = Pr(B)$. Let $X$ be a random variable and $K$ be a positive constant. Let $D = \mathbb{E}[\min(K, X)|G] - \mathbb{E}[\min(K, X)|B]$ and $E = \alpha \mathbb{E}(X|G) - \alpha \mathbb{E}(X|B)$ where $\alpha = \frac{\mathbb{E}[\min(K,X)]}{\mathbb{E}(X)}$.

Show that $D < E$ if and only if
$$\frac{\mathbb{E}(\min(K, X)|G)}{\mathbb{E}(\min(K, X)|B)} < \frac{\mathbb{E}(X|G)}{\mathbb{E}(X|B)} $$

What I've tried so far is to write $\alpha$ as

$$\alpha = \frac{p\mathbb{E}(\min(K, X)|G) + (1-p) \mathbb{E}(\min(K, X)|B)}{p\mathbb{E}(X|G) + (1-p)\mathbb{E}(X|B)} $$ and then I am stuck.

It is only true if we assume something more. Let's $X\sim U(-4,2), K=1$, $G$- $X>=0$, $B$- $X<0$. In this case $D=2.5<4.5=E$, but
$$\frac{\mathbb{E}(\min(K, X)|G)}{\mathbb{E}(\min(K, X)|B)}=-\frac{1}{4} >-\frac{1}{2}= \frac{\mathbb{E}(X|G)}{\mathbb{E}(X|B)}. $$

Let's start with $D<E$
\begin{equation}
\mathbb{E}(\min(K,X)|G) - \mathbb{E}(\min(K,X)|B)<  \frac{\mathbb{E}(\min(K,X))}{\mathbb{E}(X)} \big( \mathbb{E}(X|G) -  \mathbb{E}(X|B)\big)
\end{equation}
Let's assume that $\mathbb{E}(X) > 0$ so we can multiply both sides by $\mathbb{E}(X)$.

Also, as you noticed, $\mathbb{E}(\min(K, X)) = p\mathbb{E}(\min(K, X)|G) + (1-p) \mathbb{E}(\min(K, X)|B)$ and $\mathbb{E}(X)= p\mathbb{E}(X|G) + (1-p)\mathbb{E}(X|B)$. Now we have:
\begin{equation}
\begin{split}
p \mathbb{E}(X|G)\mathbb{E}(\min(K,X)|G)&- &p\mathbb{E}(X|G)\mathbb{E}(\min(K,X)|B) + \\
+(1-p) \mathbb{E}(X|B)\mathbb{E}(\min(K,X)|G)&- (1-p&) \mathbb{E}(X|B)\mathbb{E}(\min(K,X)|B)
\\ <\\
 p\mathbb{E}(X|G)\mathbb{E}(\min(K, X)|G)&- &p\mathbb{E}(X|B)\mathbb{E}(\min(K, X)|G) + \\ +(1-p)\mathbb{E}(X|G)\mathbb{E}(\min(K, X)|B)&-(1-p&)\mathbb{E}(X|B)\mathbb{E}(\min(K, X)|B)
\end{split}
\end{equation}

As you can see, we can subtract some components and we are left with:
\begin{equation}
\begin{split}
(1-p) \mathbb{E}(X|B)\mathbb{E}(\min(K,X)&|G)-p\mathbb{E}(X|G)\mathbb{E}(\min(K,X)|B)
\\ <\\
(1-p)\mathbb{E}(X|G)\mathbb{E}(\min(K, X)&|B)-p\mathbb{E}(X|B)\mathbb{E}(\min(K,X)|G)
\end{split}
\end{equation}

Now we add to both sides $p\bigg(\mathbb{E}(X|B)\mathbb{E}(\min(K,X)|G)+\mathbb{E}(X|G)\mathbb{E}(\min(K,X)|B)\bigg)$ to got:

\begin{equation}
\mathbb{E}(X|B)\mathbb{E}(\min(K,X)|G)
 <
\mathbb{E}(X|G)\mathbb{E}(\min(K, X)|B)
\end{equation}

If we divide by $\mathbb{E}(\min(K, X)|B)\mathbb{E}(X|B)$ we got:
$$\frac{\mathbb{E}(\min(K, X)|G)}{\mathbb{E}(\min(K, X)|B)} < \frac{\mathbb{E}(X|G)}{\mathbb{E}(X|B)}. $$

We only have to assure, that both $\mathbb{E}(\min(K, X)|B), \mathbb{E}(X|B)$ have same sign.

If one or three out of $\mathbb{E}(X),\mathbb{E}(\min(K, X)|B), \mathbb{E}(X|B)$ have negative sign, we won't have the dependence. Otherwise it's ok (if $E(X)<0$ you'll have to change the inequality signs respectively after the multiplication, and then after the division).


$\int_{-\infty}^{0}xe^{x}$ diverge or converge How would one find whether the following improper integral converge or diverge.

$\int_{-\infty}^{0}xe^{x}$

I did the following.

$t\rightarrow\infty$

$\int_{t}^{0}xe^x$

I did the integration by parts.

$u=x$

$dv=e^x$

$xe^x-\int 1e^x$

$xe^x-1xe^x$

$(0)(e^0)-e^0(0)-te^t-te^t$

$0-\infty-\infty$

would this mean there is a divergence.

You've made an error in your integration by parts. $$xe^x-\int 1e^x=xe^x-e^x=(x-1)e^x$$

It can be shown that $$\lim_{x\to-\infty}(x-1)e^x=0,$$ so the improper integral converges.

As a side note, we also have that $$\lim_{x\to-\infty}xe^x=0,$$ so even if your integration by parts hadn't been erroneous, your evaluation of the limit would have been.

Step 1: set the limit $\lim_{t \rightarrow -\infty} $

Step 2: find integral (i got $xe^x - e^x$)

Step 3: plug in the limits and  ftc, so u get
$\lim_{ t-> -\infty} [(0-e^0)-(te^t-e^t)]$

*key: $e^\infty = \infty$,
$e^-\infty = 0$

step 4: evaluate

answer: convergent sum $-1 $


A question on integration I want to compute the following integral:
$$\raise 1ex{\Large\int} \frac{\sqrt{\ln(x+\sqrt{1+x^2}})}{1+x^2}\,dx$$

Why not make the substitution $x=\sinh u?$ Then, $d x = \cosh u d u,$ while $(x+\sqrt{1+x^2})= \exp(u),$ and $1+x^2 = \cosh^2 u,$ so we get our integral equal to

$$\int \frac{\sqrt u}{\cosh u} du,$$ Now, letting $u = v^2,$ this is equal to
$$\frac12\int \operatorname{sech}(v^2) d v.$$ At which point I am flummoxed.

I'd add this as a comment but I don't have enough reputation to do so.  Building off Igor's work, I wondered whether $\frac{2v^2}{\cosh(v^2)}$ could be done by parts.  If we let $f=v$ and $g'=2v\text{sech}(v^2)$, then $fg-\int{f'g}$ works out to $$2v\arctan(e^{v^2})-2\int{\arctan(e^{v^2})dv}$$

Of course, I had to go to Wolfram for the second half:

$$ 2 \int{\arctan(e^{v^2}) dv} = \sqrt{\frac{\pi}{\ln(\arctan(e))}}\text{erfi}\left(v\sqrt{\ln(\arctan(e))}\right)$$

where $\text{erfi}$ is apparently the imaginary error function.  (I know a little about the plain ol' error function, but...)  Unless there's something weird with that error function thing, I'd assume we can just undo the substitution, with $v=\sqrt{u}=\sqrt{\text{arcsinh}(x)}$:

$$2\sqrt{\text{arcsinh}(x)}\arctan(e^{\text{arcsinh}(x)})-\sqrt{\frac{\pi}{\ln(\arctan(e))}}\text{erfi}\sqrt{\text{arcsinh}(x)\ln(\arctan(e))}$$

I'm guessing, though, that most of the people looking at this know what they're doing more than I do, so I wouldn't be surprised if there's some reason we can't do the above.

I'm trying to check this on Wolfram, just by subtracting the original integrand from the derivative of the above, but it's not interpreting the parentheses correctly.  This is what I'm entering:

d/dx{

2sqrt{arcsinh(x)}arctan(e^{arcsinh(x)})-sqrt{pi/ln(arctan(e))}*erfi(sqrt{arcsinh(x)ln(arctan(e))})

}

-sqrt(ln(x+sqrt(x^2+1)))/(1+x^2)

It keeps taking the derivative of the whole thing.  I'm pretty sure I parsed that carefully, but I thought I'd include it in case anyone else wants to try.


Proof that $n!$ is divisible by $(n+1)^2$ for $n=xy+x+y$ I noticed the other day that for $n>8, n\in\mathbb{N}$, the factorial $n!$ seems to be divisible by $(n+1)^2$ when $n$ can be written in the form $xy+x+y$ (where $x,y\geq1$ and $\in\mathbb{N}$). Some examples:

$n=14=2 \times 4+2+4$ (i.e. $x=2$, $y=4$), and we have $14!|15^2$
$n=15=3 \times 3+3+3$ (i.e. $x=y=3$), and we have $15!|16^2$,
$n=19=3 \times 4+3+4$ (i.e. $x=3$, $y=4$), and we have $19!|20^2$,
...

This seems to hold for the first $15$ values I checked, so it seemed natural to try to prove it for all $n=xy+x+y$, however I could not find a proof. Here is my attempt:

Plugging $xy+x+y$ into the expressions above and noticing that $xy+x+y+1=(x+1)(y+1)$, we need to prove that

$$
(xy+x+y)!|(x+1)^2(y+1)^2,
$$

or

$$
\Gamma((x+1)(y+1))|(x+1)^2(y+1)^2.
$$

It is easy to prove that

$$
\Gamma((x+1)(y+1))|(x+1)(y+1)
$$

however I am struggling to prove the same for $(x+1)^2(y+1)^2$. Any hints?

Wlog. $x\le y$.
If we are lucky, we find $x+1,2(x+1),y+1,2(y+1)$ as different elements among the factors $1,2,3,\ldots, n$ and we are done.

How can we fail to be lucky? The four candidates may be too large or not be different after all. As we assume $x\le y$, the following list is exhaustive:

It may happen that $2(y+1)>n=xy+x+y$. This implies $x=1$. And indeed, $x=1$, $y=2$ gives $n=5$ and $36$ does not divide $5!$. More generally if $y+1$ is prime, then $(2y+1)!$ is not divisible by $(y+1)^2$.
It may happen that $x+1=y+1$, which of course means that $x=y$. If $x\le 3$, we have $n\le 8$, which is excluded. If $x\ge 4$,  we can use the factors $x+1,2(x+1),3(x+1),4(x+1)$ instead. These are guaranteed to be different and we have $4(x+1)\le xy+y<n$.
It may happen that $2(x+1)=y+1$. Then we can use $x+1,2(x+1)=y+1, 3(x+1), 2(y+1)$ instead.

Thus

  If $n=xy+x+y>8$ with positive integers $x,y$ then $(n+1)^2\mid n!$ except when $n+1$ is twice a prime.

$\color{Green}{\text{Question}}$ :
Find all integers $n$;
such that $\color{Green}{(n+1)! \mid n!} $ .

$\color{Blue}{\star \ \ \ \ \text{condition}}$ :

Now suppose that $n+1$ can be written
in the form $n+1=ab$;
with $2 < a,b$.
Also assume that one of the following occures:

If $a= b$; then $5 \leq a=b$.
If $a=2b$; then $5 \leq b  $.
If $2a=b$; then $5 \leq a  $.

With the condition as in
$\color{Blue}{\star}$ ;
we claim that
the assertion is $\color{Blue}{\text{true}}.$

Let $n, a, b$ to be fixed as above in $\color{Blue}{\star}$ .
Notice that if we let $x:=a-1$ and $y:=b-1$
then one can easilly see that:
$$n=xy+x+y.$$

$\color{Blue}{\text{Remark(I)}}$ :
With the notation as in $\color{Blue}{\star}$;

If $a=b$ , with $5 \leq a$
then we have:
$$b < 2b < 3b < 4b < ab=n+1 \Longrightarrow b < 2b < 3b <4b \leq n=ab-1 \ ; $$
so we can conclede that:
$$(b)(2b)(3b)(4b)=24(ab)^2=24(n+1)^2 \mid n! ;$$
which implies that: $(n+1)^2 \mid n! \ \ $ .

If $a=2b$ ;  with $5 \leq b$
then we have:
$$b < 2b < 3b < 4b < ab=n+1 \Longrightarrow b < 2b < 3b <4b \leq n=ab-1 \ ; $$
so we can conclede that:
$$(b)(2b)(3b)(4b)=6(ab)^2=6(n+1)^2 \mid n! ;$$
which implies that: $(n+1)^2 \mid n! \ \ $ .

If $2a=b$ ;  with $5 \leq a$
then we have:
$$a < 2a < 3a < 4a < ba=n+1 \Longrightarrow a < 2a < 3a <4a \leq n=ba-1 \ ; $$
so we can conclede that:
$$(a)(2a)(3a)(4a)=6(ab)^2=6(n+1)^2 \mid n! ;$$
which implies that: $(n+1)^2 \mid n! \ \ $ .

If $a \neq b$ and $a \neq 2b$ and $2a \neq b$ ;
then we have:
$$b < 2b < ab=n+1 \Longrightarrow b < 2b \leq n=ab-1 \ ; $$
$$a < 2a < ab=n+1 \Longrightarrow a < 2a \leq n=ab-1 \ ; $$
so we can conclede that:
$$(b)(2b)(a)(2a)=4(ab)^2=4(n+1)^2 \mid n! ;$$
which implies that: $(n+1)^2 \mid n! \ \ $ .

Remark(II): Integer $N$
can not be written
in the form $N=ab$;
with $2 < a,b$;
if and only if
$N$ is prime or twice a prime.

Remark(III):
We can split the integres in three cases:

$\color{Red}{   n+1 =   \text{prime numbers and twice of prime numbers;
for which the assertion is wrong!}   }.$
$\color{Purple}{ n+1= a.a
 \ \ \
\text{for}
\ \ \
a \in \{ 1, 2, 3, 4 \} }.$
This case has only four numbers;
which can you check out that
the assertion is $\color{Blue}{\text{true}}$ only for
$\color{Blue}{n+1=16}$ and
$\color{Blue}{n+1= 1}$ .
$\color{Purple}{ n+1= a.(2a)
 \ \ \
\text{for}
\ \ \
a \in \{ 1, 2, 3, 4 \} }.$
This case has only four numbers;
which can you check out that
the assertion is $\color{Blue}{\text{true}}$ only for
$\color{Blue}{n+1=18}$ and
$\color{Blue}{n+1=32}$ .

$\color{Green}{\text{Remark(IV)}}$ :
Except for the following cases;
for which the assertion is
$\color{Red}{\text{wrong}}$

1.$\color{Red}{   n+1 =   \text{prime numbers and twice of prime numbers
}   }.$

2.$\color{Red}{ n+1= 4, 9, 2, 8 } .$

for all the other cases
the assertion is
$\color{Blue}{\text{true}}$ .

$\color{Red}{\text{Remark(V)}}$ :
Let $p$ to be an odd prime number;
then we have:
$p^2 \color{Red}{\nmid} (2p-1)! \ \ . $

For example let $p$ to be an odd prime number;
and let's define:
$$n:=2p-1
\ \ \
\text{and}
\ \ \
x:=1
\ \ \
\text{and}
\ \ \
y:=p-1.$$

Notice that we have: $n=xy+x+y$

We calaim that $(2p)^2 \color{Red}{\nmid} (2p-1)!$.

[
Suppose on contrary
that $(2p)^2 \mid (2p-1)!$ ,
on the otherhand notice that
$p^2 \mid (2p)^2$.
So we must have
$p^2 \mid (2p-1)!$
;
which has an obvious contradiction
with the above $\color{Red}{\text{Remark(V)}}$.
]


Reordering Jordan Matrix Sub-Blocks Let $J$ and $K$ be $n × n$ Jordan matrices which can be made equal after a reordering of sub-blocks.
Show that $J= S^{-1}KS$ by explaining how to construct $S$.

I understand that these two matrices are similar because changing the order of sub-blocks preserves geometric and algebraic multiplicity of the shared eigenvalues, so we simply have to rearrange those blocks to change between Jordan matrices. However, I don't know how to actually construct the invertible matrix that would accomplish this reordering.

Thanks for any help!

It’s not different from the general case. If you know K and J thus you can calculate S by linear system.

Here is a simple example:

Find the jordan basis for $A$


A simpler non-calculator proof for $17^{69}<10^{85}$ I have proved that $17^{69}<10^{85}$ by using the following inequalities:
$x<\exp\left(\dfrac{2(x-1)}{x+1}\right)$ for all $x\in \left]-1,1\right[$
and $x<{\mathrm e}^{x-1}$ for all $x\in \left] 1,+\infty \right[$, but I am looking for a simpler non-calculator proof.

My proof is the following:
\begin{align*}\frac{17^{69}}{10^{85}}&=\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\left(\frac{5^3}{2^7}\right)^2\cdot\frac{5}{4}<\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\frac{5}{4}=\left(\frac{4913}{5000}\right)^{23}\cdot \frac{5}{4}\\&<\left(\exp\left(\frac{2\left(\frac{4913}{5000}-1\right)}{\frac{4913}{5000}+1}\right)\right)^{23}\cdot\exp\left(\frac{5}{4}-1\right)\\&=\exp\left(-\frac{174}{431}\right)\cdot\exp\left(\frac{1}{4}\right)=\exp\left(-\frac{265}{1724}\right)<1.\end{align*}

Could anyone find a simpler non-calculator proof without using big numbers?


What are some conceptualizations that work in mathematics but are not strictly true? I'm having an argument with someone who thinks it's never justified to teach something that's not strictly correct. I disagree: often, the pedagogically most efficient way to make progress is to iteratively learn and unlearn along the way.

I'm looking for examples in mathematics (and possibly physics) where students are commonly taught something that's not strictly true, but works, at least in some restricted manner, and is a good way to understand a concept until one gets to a more advanced stage.

How about this notorious one I remember from high school?

"$f(x)$ is just a fancy name for $y$."

The existence of Gabriel's Horn.

  That is, there exists an object with a finite volume but with an infinite surface area.

Mathematically, we can show this to be true by taking a 360-degree revolution of the graph $y=\frac{1}{x}$ in the domain $[1,\infty)$ about the $x$-axis, and calculating its volume ($\pi$) and its surface area (infinite).

This is all well and good in theory, but in our physical world, no such object exists.

One of the reasons for this is that the thickness of the horn is limited by the size of the atom, so cannot be arbitrarily small.


Limit of $\ln(t)t^xe^{-t}$ I want to calculate the limit: $\lim\limits_{t\to\infty}\ln(t)t^xe^{-t}$ where $x\in(0,1]$.
For $t>1$ we can find the following upper bound:
$$\ln(t)t^xe^{-t}\leq t^2 e^{-t}.$$
Using the series representation of $e^{-t}$ we find another upper bound:
$$\ln(t)t^xe^{-t}\leq t^2 e^{-t}\leq t^2 \frac{3!}{t^3}=\frac{3!}{t}.$$
So it follows:
$$0\leq \lim\limits_{t\to\infty}\ln(t)t^xe^{-t}\leq \lim\limits_{t\to\infty}\frac{3!}{t}=0\Rightarrow \lim\limits_{t\to\infty}\ln(t)t^xe^{-t} =0.$$
Is this correct?

If $t\geq 1$, then $$\frac{\ln(t)}{e^t}\leq \left|\ln(t)t^xe^{-t}\right|\leq \frac{t\ln(t)}{e^{t}}.$$
Using Bernoulli L'Hospital gives the wishes result.

You can do that almost instantly with asymptotic calculus:

$\ln t=_\infty o(t^\alpha)$ for any $\alpha>0$, therefore $t^x\ln t=_\infty t^xo(t)=o(t^{x+1})$,
also, $t^{x+1}=o(\mathrm e^t)$, and consequently  $\;t^x\ln  t=_\infty o(\mathrm e^t)$,
which means exactly that
$$\lim_{t\to+\infty}\frac{t^x\ln t}{\mathrm e^t}=0.$$


Analysis question limit sin function as n goes to infinity can you help me with the following:

$\lim_{n \rightarrow \infty} \sin^{2} \pi \sqrt{n^2 + n}$

Thanks a lot!

For the limit $\lim_{n\to+\infty}\sin^2\sqrt{n^2+n}$, since $\{e^{in}\}_{n\in\mathbb{N}}$ is dense in the unit circle (it is a consequence of the irrationality of $\pi$ and the Dirichlet box principle), the same holds for the sequence $\{e^{i(n+1/2)}\}_{n\in\mathbb{N}}$ (a translation of the unit circle preserves density).

By taking imaginary parts (the projection preserves density too), we have that the sequence $\{\sin(n+1/2)\}_{n\in\mathbb{N}}$ is dense in the $[-1,1]$ interval.

Since the sine is a Lipschitz function and the distance between $\sqrt{n^2+n}$ and $n+1/2$ is bounded by $\frac{1}{4n}$, the sequence $\{\sin\sqrt{n^2+n}\}_{n\in\mathbb{N}}$ is dense in the interval $[-1,1]$, so your limit does not exist.

For the modified limit,
$$\lim_{n\to +\infty}\sin^2 \pi\sqrt{n^2+n},$$
the fact that the distance between $\sqrt{n^2+n}$ and $n+\frac{1}{2}$ is bounded by $\frac{1}{4n}$, together with the fact that the sine is a Lipschitz function, ensures that the limit is just $1$ (the sine function equals $\pm 1$ in the odd multiples of $\frac{\pi}{2}$).

The sine function doesn't have a limit if $x \rightarrow \infty$. The sine function goes up and down between -1 and 1 infinitely many times .


Betting game: $3$ $20$-sided dice versus $2$ $30$-sided dice Alice and Bob play a dice game. Alice rolls three $20$-sided dice and
  Bob rolls two $30$-sided dice. Whoever has the highest sum wins.

  Who has a better chance of winning?

This was a question asked in an interview. So I don't have much time to answer and I want to find the answer quickly. I did it the long way (enumerate all possibilities), and I figured out that Alice has a better chance of winning. But is there any way I can solve this problem very quickly?

I tried to make a "flipping over the dice" argument but got nowhere.

Thanks

Let the outcomes of the dice rolls be $(A_1, A_2, A_3, B_1, B_2)$ where the $A_i$ are uniform on $\{1, 2, \ldots, 20\}$ and the $B_j$ are uniform on $\{1, 2, \ldots, 30\}$.  You want to know whether $\Delta = A_1 + A_2 + A_3 - B_1 - B_2$ is more likely to be positive (Alice wins) or negative (Bob wins).  Of course it can be zero.

Consider the cases where these take on the values $(a_1, a_2, a_3, b_1, b_2)$ and ("flipping over" the dice) $(21-a_1, 21-a_2, 21-a_3, 31-b_1, 31-b_2)$.

In the first case we get $\Delta = a_1 + a_2 + a_3 - b_1 - b_2$.

In the second case we get $\Delta = (21 - a_1) + (21 - a_2) + (21 - a_3) + (31 - b_1) + (31 - b_2) = 1 - (a_1 + a_2 + a_3 - b_1 - b_2)$.

So this gives a bijection between outcomes where Alice wins by $\Delta$ and outcomes where Alice wins by $1-\Delta$.     In particular 0 and 1 are equally likely; -1 and 2 are equally likely; and so on.   So all the positive values of $\Delta$ have probability adding up to 1/2 - that is, Alice wins with probability 1/2.  There is a tie with some nonzero probability, and Bob wins with probability less than 1/2.

Twenty-sided dice roll 10.5 on average. Thirty-sided dice roll 15.5 on average. So Alice is going to roll 31.5 on average, while Bob only rolls 31 on average.

These are only averages, but the symmetry of the problem ensures Alice wins out (on average).


Help finishing this exercise! Given the following functions:

$$ F(t)= \int_0^\infty e^{-tx}\dfrac{\sin{x}}{x}\,dx, \quad t>0$$
$$ F_s(t)= \int_0^s e^{-tx}\dfrac{\sin{x}}{x}\,dx, \quad t \geq 0, s>0$$

Show that $F$ is well-defined in $(0,+\infty)$ and it is differentiable in that same interval. Compute $F'$ and $\lim_{t \to \infty} F(t)$. Finally, compute $F(t)$ and $\lim_{t \to 0^+} F(t)$
Compute $F_s'$ in $(0, +\infty)$. Show that $F_s$ is a equicontinous family of functions
Show that $\lim_{s \to \infty}F_s(t) = F(t)$ uniformly for $t \in (0,+\infty)$
Show that $\int_0^\infty \dfrac{\sin{x}}{x}\,dx = \dfrac{\pi}{2}$

I have already asked you about (3) (Show that $\lim_{s \to \infty}F_s(t) = F(t)$ uniformly for $t \in (0,+\infty)$). Now I have 1,2 and 3 solved, but I don't quite know how to finish up things in (4)

Using the definition I have

$$ \int_0^\infty \dfrac{\sin{x}}{x}\,dx := \lim_{L \to \infty} \int_0^L \dfrac{\sin{x}}{x}\,dx$$

and I am guessing I should connect this somehow with $\lim_{t \to 0^+} F(t)$ (which I have already computed). Also, I have noticed that I have not used that $\{ F_s \}$ is equicontinous, maybe that is necessary?. Could you give me any hints about how to complete this?

Given as a solution from this post:

This can also be solved using double integrals.

Using the double integral:

$$\int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dy \Bigg)\, dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$
Notice that
$$\int_{0}^{\infty} e^{-xy} \sin x\,dy = \frac{\sin x}{x}$$

We can then substitute this in and say:

$$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$
Now the right hand side can be found easily, using integration by parts.

$$\begin{align*}
I &= \int e^{-xy} \sin x \,dx = -e^{-xy}{\cos x} - y \int e^{-xy} \cos x \, dx\\
&= -e^{-xy}{\cos x} - y \Big(e^{-xy}\sin x + y \int e^{-xy} \sin x \,dx \Big)\\
&= \frac{-ye^{-xy}\sin x - e^{-xy}\cos x}{1+y^2}.
\end{align*}$$
Thus $$\int_{0}^{\infty} e^{-xy} \sin x \,dx = \frac{1}{1+y^2}$$
Thus $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty}\frac{1}{1+y^2}\,dy = \frac{\pi}{2}.$$


Bounded $L^2$ expectation implies convergence in $L^2.$ In probability at least four different types of convergence are considered:

$a)$ Almost sure convergence

$b)$ Convergence in probability

$c)$ Convergence in $L^p$

$d)$ Convergence in distribution

It is known (see e.g. the book of Karr: "Probability", Springer) that $$a)\,\Rightarrow \, b)\,\Rightarrow \, d)$$ and that $$c)\,\Rightarrow \, b).$$
All the other implications are false and it is possible to find counterexamples to them here:
Convergence types in probability theory : Counterexamples

Moreover, convergence in $L^p$ implies convergence in $L^q$ if $p>q$ and the space is finite.

Now, let ${x_n}$ be a family of bounded random variables defined over a given probability space that converge a.s. to a bounded random variable $x.$

The following statements are true or false?

1) If $\mathbb E(x_n^2)\leq \mathbb E(x^2)$ then $x_n$ converges to $x$ in $L^2.$

2) If $\mathbb E(x_n^2)=1$ $\forall n$ then $lim_{n\to \infty}\mathbb E(x_n)=\mathbb E(x).$

3) If $\mathbb E(x_n^2)=1$ and $\mathbb E(x_n^3)=c\in \mathbb R\,$ $\,\forall n$ then $x_n$ converges to $x$ in $L^2.$

I proved that 1) and 2) are true in the following way.

Since $x_n^2$ is a sequence of non-negative random variables, we can apply the Fatou Lemma getting:

$$\int_{\Omega}x^2=\int_{\Omega}lim\,inf_{n\to \infty}x_n^2\leq lim\,inf_{n\to \infty}\int_{\Omega}x_n^2 \leq lim\,inf_{n\to \infty} \int_{\Omega}x^2=\int_{\Omega}x^2 $$
 where the first equality is due to the fact that $x_n^2$ converges a.s. to $x^2.$

Hence we have that $lim_{n\to \infty}\mathbb E(x_n^2)= \mathbb E(x^2).$ This implies that $x_n$ converges to $x$ in $L^2$ by the following well known proposition (see e.g. Jang: "Large Sample Techniques for Statistics", Springer, pag. 33)

"Suppose that $\mathbb E(|x_n|^p)<\infty$ $\forall n$ and that $x_n$ converges to $x$ in probability.
Then the following are equivalent:

(i) $x_n,$ $n = 1, 2, . . .,$ is uniformly integrable in $L^p$;

(ii) $x_n$ converges to $x$ in $L^p$ with $\mathbb E(|x|^p)<\infty$;

(iii) $lim_{n\to\infty}\mathbb E(|x_n|^p)=\mathbb E(|x|^p) < \infty$."

This proves 1).

As a Corollary to the just stated proposition (see Jang: "Large Sample Techniques for Statistics", Springer, pag. 35), the author shows that if $x_n$ converges to $x$ in probability and if $\mathbb E(|x_n|^q)$ is bounded for some $q$ and for all $n$ then $x_n$ converges to $x$ in $L^p$ for all $0<p<q.$

We can apply this Corollary to my second question with $q=2,p=1$ getting that
$x_n$ converges to $x$ in $L^1.$

As a consequence, $$0=lim_{n\to\infty}\mathbb E(-|x-x_n|)\leq lim_{n\to\infty}\mathbb E(x-x_n)\leq lim_{n\to\infty}\mathbb E(|x-x_n|)=0$$
This imply that $lim_{n\to\infty}\mathbb E(x_n)=\mathbb E(x)$ so 2) is also true.

For 3) I found the following counterexample.
Consider the space $\Omega=[0,1]$ with the usual Lebesgue misure. Let
$$x_n(t)=\begin{cases}0& \mbox{ if }0\leq t\leq 1-\frac{1}{n}\\
\sqrt{n}& \mbox{ if } 1-\frac{1}{n}<t\leq 1-\frac{0.5}{n}\\
-\sqrt{n}& \mbox{ if }1-\frac{0.5}{n}< t\leq 1\\
\end{cases}.$$
Now, $x_n$ converges a.s. to the random variable $x$ identically zero over $[0,1].$
Moreover $\mathbb E(x_n^3)=0,$ $\mathbb E(x_n^2)=1,$ $\mathbb E(x)=0.$
Hence the hypotesis are satisfied but $x_n$ does not converges to $x$ in $L^2.$
Note that in 3) it is not possible to use the same technique used in 2) with $q=3$ and $p=2$ because here we know $\mathbb E(x_n^3)=c\in\mathbb R$ and not
$\mathbb E(|x_n|^3)=c\in\mathbb R.$ Actually in the previous counterexample we have $\mathbb E(|x_n|^3)=\sqrt{n}\to \infty.$

For 1) you may consider on $((0,1],B((0,1]),λ_{|(0,1]})$ the sequence$(X_n)$ defined by $\forall n, \  X_n = \imath n 1_{(0,\frac{1}{n}]}(\omega)$. It converges almost surely to $0$, but not in $L^2$ and $\mathbb{E}[X_n^2] = -n^2 \frac{1}{n} = -n \lt 0$.


Proof technique for between any two real numbers is an irrational number I'm proving that between any two real numbers is an irrational, and my first thought is to prove that between any two real numbers is a rational number, and then show that between any two rational numbers is an irrational. However, I don't think this would complete the proof because I will not have shown that between any two real numbers there are two rational numbers. Would the first part of my proof imply that there are an infinite number of rationals between any two reals, or do I need to show that separately?

Also, I'm really trying to figure out this problem on my own, so I ask not for the answer, but a pointer in the right direction.

Thanks!

Here's my proposed solution:

1) First you show that between two rationals there is always an irrational number. Suppose that $q_1<q_2$, then take for ex. $r=\frac{q_1}{\sqrt{2}}+q_2\left(1-\frac{1}{\sqrt{2}}\right)$, which obeys $q_1<r<q_2$. The number $r$ can be written as $a+b\sqrt{2}$, where $a, b\in\mathbb{Q}$, which implies that $r\in\mathbb{R}-\mathbb{Q}$.

2) Given any two real numbers $r_1<r_2$, then $s_1=3r_1/4+r_2/4$ and $s_2=r_1/4+3r_2/4$ are in the order $r_1<s_1<s_2<r_2$. If $s_1$ or $s_2$ is irrational then you're done. Otherwise, go to step 1).

For the infinite part, you could use the mapping $r(\lambda)=\lambda q_1+(1-\lambda)q_2$, where $\lambda\in(0, 1)$. Again $q_1<r(\lambda)<q_2$.

Suppose $x < y$.

Define $f(t) = (1-t)x + t y = x + t(y-x)$.

Let $Q$ be any rational number in the interval $(0,1)$.

Let $Z$ be any irrational number in the interval $(0,1)$.

Then $ x < f(Q), f(Z) < y$ for all $Q$ and $Z$ as defined.

If $x$ and $y$ are both rational, then f(Z) is irrational.

If $x$ and $y$ are both irrational, then $f(t)$ is irrational where

$$t =
\begin{cases}
   Z & \text{If $y-x$ is rational.} \\
   Q & \text{If $y-x$ is irrational.}
\end{cases}
$$

If one of $x,y$ is rational and the other is irrational, then $f(Q)$ is irrational.


For which $n$ are there primitive Pythagorean triples with legs of lengths $a$ and $a+n$? For which n can $a^{2}+(a+n)^{2}=c^{2}$ be solved, where $a,b,c,n$ are positive integers?
I have found solutions for $n=1,7,17,23,31,41,47,79,89$ and for multiples of $7,17,23$...
Are there infinitely many prime $n$ for which it is solvable?

The general primitive solution to $x^2+y^2 = z^2$ is given by: $x=u^2-v^2$, $y=2uv$, $z=u^2+v^2$, with $u,v$ relatively prime and not both odd.

For $(a,a+n,z)$ to be a primitive triple, we'd have to have a $(u,v)$ such that: $|u^2 - v^2 - 2uv| = n$.  We can rewrite that as: $(u-v)^2 - 2v^2 = \pm n$

So, setting $w = u-v$, we want to find $(w,v)$ which are relatively prime and $w$ is odd, with:

$$w^2-2v^2 = \pm n$$

This means that $n$ must be odd.

In fact, we can use unique factorization in $\mathbb{Z}[\sqrt{2}]$ to show that $n$ can be any product of primes of the form $8k\pm 1$.  Since there are infinitely many primes of the form $8k\pm 1$, the answer to your question is, "yes."

(Oh, and once you find one solution $(w,v)$ for a particular $n$, you can find infinitely many solutions for that $n$.)

If you solve expression for $n$ you get

$n=\sqrt{c^2-a^2}-a$, let's denote $b=\sqrt{c^2-a^2}$,so we have that $n=b-a$

Now,take look at picture bellow.Note that $AD=a$,and $BD=b-a=n$

If you change value of $b$ and keep $a$ to be constant you will get a infinite number of right triangles,and therefore infinite number of values of $n=b-a$,so answer is yes, there are infinitely many primes $n$ for which equation is solvable.


A Variation of "rational is dense in $\Bbb R$" I know that the theorem that the set $\Bbb Q$ of rationals is dense in $\Bbb R$ says:

  For every $x\in \Bbb R$ and every $\epsilon>0$, there exist $a$, $b\in \Bbb Z$ with $b\ne0$ such that $$|x-{a\over b}|<\epsilon.$$

But what about if I change the condition "$a,b\in \Bbb Z$ with $b\ne0$" by "$a$ and $b$ are both primes". Does the theorem still hold? In other words, I wonder whether the following is true:

  For every $x\in \Bbb R^+$ and every $\epsilon>0$, there exist $a$, $b\in \Bbb P$, where $\Bbb P$ is the prime numbers set, such that $$|x-{a\over b}|<\epsilon.$$


How to calculate limit without L'Hopital How can I calculate limit without using L'Hopital rule?

$\displaystyle\lim_{x\to 0 }\frac{e^{\arctan(x)}-e^{\arcsin(x)}}{1-\cos^3(x)}$.

Let $\arctan x = t$ so that $x = \tan t$ so that $\sin t = x/\sqrt{1 + x^{2}}$ or $\arctan x = t = \arcsin (x/\sqrt{1 + x^{2}})$.

Now we have

$\displaystyle\begin{aligned}\arctan x - \arcsin x &= \arcsin (x/\sqrt{1 + x^{2}}) - \arcsin x\\
&= \arcsin\left(x\sqrt{\frac{1 - x^{2}}{1 + x^{2}}} - \frac{x}{\sqrt{1 + x^{2}}}\right)\\
&= \arcsin\left\{\frac{x}{\sqrt{1 + x^{2}}}\left(\sqrt{1 - x^{2}} - 1\right)\right\}\\
&= \arcsin\left\{\frac{-x^{3}}{\sqrt{1 + x^{2}}\left(\sqrt{1 - x^{2}} + 1\right)}\right\}\\
&= \arcsin y = f(x) \text{ (say)}\end{aligned}$

Next we have

$\displaystyle\begin{aligned}\lim_{x \to 0}\frac{e^{\arctan x} - e^{\arcsin x}}{1 - \cos^{3}x} &= \lim_{x \to 0}e^{\arcsin x}\cdot\frac{e^{f(x)} - 1}{1 - \cos^{3} x}\\
&= \lim_{x \to 0}1\cdot\frac{e^{f(x)} - 1}{f(x)}\cdot\frac{f(x)}{1 - \cos^{3}x}\\
&= \lim_{x \to 0}1\cdot 1\cdot\frac{f(x)}{(1 - \cos x)(1 + \cos x + \cos^{2}x)}\\
&= \frac{1}{3}\lim_{x \to 0}\frac{f(x)}{1 - \cos x}\\
&= \frac{1}{3}\lim_{x \to 0}\frac{\arcsin y}{y}\cdot \frac{y}{1 - \cos x}\\
&= \frac{1}{3}\lim_{x \to 0}\frac{y}{2\sin^{2}(x/2)}\\
&= \frac{1}{6}\lim_{x \to 0}\frac{y}{(x/2)^{2}}\cdot \frac{(x/2)^{2}}{\sin^{2}(x/2)}\\
&= \frac{1}{6}\lim_{x \to 0}\frac{y}{(x/2)^{2}}\cdot 1\\
&= \frac{2}{3}\lim_{x \to 0}\frac{y}{x^{2}}\\
&= \frac{2}{3}\lim_{x \to 0}\frac{-x}{\sqrt{1 + x^{2}}\left(\sqrt{1 - x^{2}} + 1\right)}\\
&= \frac{2}{3}\cdot 0 = 0\end{aligned}$

No Taylor or L'Hospital is required. We just need algebraic and trigonometric manipulation combined with the use of standard limits.


$\int_{U}f(x)dx$ converge iff $\int_{U}|f(x)|dx$ converge Problem: Let be $U\subset\mathbb{R}^m$ a open set. Show that $\int_{U}f(x)dx$ converge iff $\int_{U}|f(x)|dx$ converge.
It seems to be a simple matter, but I'm not having ideas to solve it. The settings are as follows:

Definition of integral improper: Given a continuous function (limited or not) it says that the improper integral $\int_{U}f(x)dx$ is convergent when, for all exhaustion $U=\cup K_i$, there is the limit $\lim_{i \rightarrow \infty}\int_{K_i}f(x)dx$ and its value is independent of the exhaustion taken.
Definition of Exhaustion: An exhaustion of $U$ is a sequence of J-measurable $K_i\subset U$ compact such that $U=\cup K_i$ and $K_i\subset int(K_{i+1}) \forall i\in\mathbb{N}$.

Is this solution?

Proof. Define $f_+,f_-$ parts positve and negative of $f$. Then, we have that $f=f_++f_-$ and $|f|=f_+-f_-$. Now, suppose that $\int_Uf(x)dx$ converge. Hence  $\lim_{i\rightarrow \infty}\int_{K_i}f(x)dx=\lim_{i\rightarrow\infty}[\int_{K_i} f_+ + \int_{K_i}f_-]$ exist. Then, how $\int_{K_i} f_+,\int_{K_i}f_-$ converge, implies that $\int_{K_i}|f|$ converge too.

Is the reciprocal similar?


What is the area of shaded region which is lies between outer and inner circle. There is a outer circle with radius 2r and another inner circle with radius r whose center is the middle of big circle.As depicted in the following figure.

Foo graph Image

There is a sector of 120 degree in inner circle which leads to shaded part between outer and inner circle. What is the area of shaded circle in terms of r ?

i.e. What is the value of K ?

WLOG let $r=1$.

Let us use a coordinate system centered on the large circle. Then the oblique line on the right has equation $x=\sqrt3(y-1)$ and meets the circle $x^2+y^2=4$.

Solving the quadratic equation, the intersection point is $(\dfrac{\sqrt{39}-\sqrt3}4,\dfrac{\sqrt{13}+3}4)$.

Hence, the aperture angle of the large sector

$$\theta=2\arctan\left(\frac xy\right)=1.19873\cdots \text{rad}=68.6821\cdots°.$$

The requested area is the difference between the large and the small sector and two triangles having base $1$ and height $x$.

$$K=\frac\theta2\,2^2-\frac\pi31^2-\frac22x=4\arctan\left(\frac{\sqrt{39}-\sqrt3}{\sqrt{13}+3}\right)-\frac\pi3-\dfrac{\sqrt{39}-\sqrt3}4=0.222026268\cdots.$$

something like this:
$$
2r^2\left(\int_0^{\frac{\sqrt{3}}2} \left(\sqrt{4-x^2} - 1 - \sqrt{1-x^2} \right)dx + \int_{\frac{\sqrt{3}}2}^{\frac{\sqrt{13}-1}4} \left(\sqrt{4-x^2} - 1-\frac{\sqrt{3}x}2 \right)dx \right)
$$


Which of the following statements are correct? T he order of the smallest possible non trivial group containing elements $x$ and y such that
$x^7=y^2=e$ and $yx=x^4y$ is

(A) $1\space\space$ (B) $2\space\space$ (C) $7\space\space$ (D) $14$

You can just construct it with $\mathbb{Z}_2$ under addition (mod 2) and let $x=0$ and $y=1$.  If you add $0$ seven times you get $0$ and if you add $1$ twice, you get $0$ (mod $2$).  The last one is $$1 + 0 = 0+0+0+0+1 \pmod{2}$$

Not going to find any smaller non-trivial groups than one of order $2$.

Now if you knew the order of $x$ were $7$ and the order of $y$ were two, then things would be different.

If we read the question literally, we might assume that $e,x\ne e,y\ne e\in G$, so $|G|\ge2$. This immediately removes A and B, and we can argue as follows:

It cannot be A because this is the trivial identity group.

It cannot be B because we have $x=y$ and therefore the statement $x^7=y^2=e$ is inconsistent.

It cannot be C because we have $y=e$ and therefore $ex=x^4e$, so $x^3=e$ so $x=e$, which means the group has only $1$ element.

It cannot be D because we have $yx=y$, so $x=e$ but this means the group is order 2, but the claim for D is that the group is order $14$.

But this doesn't leave any correct answer, so we can assume either $x=e$ and/or $y=e$. This removes A and C as we have in both cases $x=y=e$, or $D$ as $x=e$.

If we assume $x=e$ both B and D are correct groups in principle but D claims the group is order $14$ whereas it is order $2$. B is a correct claim, and indeed the smallest non-trivial group.


Number of real roots of a polynomial Find the number of real roots of the polynomial $f(x)=x^5+x^3-2x+1$

( I think similar questions have already been asked. However, I am not sure whether the exact question has already been raised by some user. I urge the members of this mathematical community to mark my question as duplicate if it has already been asked. I tried an attempt below:-)

What I attempted:-  I know a little bit of theory of equations but not all. While I was in the examination, I did not ponder much more on any theorems relating to real roots. Rather I tried to do it with some intuition.
We have $f(1)=1$,$f(0)=1$,$f(2)=1$, $f(\frac{1}{2})=\frac{5}{32}$,$f(-\frac{1}{2})=\frac{59}{32}$,$f(-2)=-35$
It just helped me to trace a root between $-1$ and $-2$. Then I tried to study the nature of the graph.

Here $f'(x)=5x^4+3x^4-2$
Let $f'(x)=5x^4+3x^2-2=0$
    $\Rightarrow 5y^2+3y-2=0$
 (where $x^2=y$)
$\Rightarrow (5y-2)(y+1)=0$
Therefore, $y=\frac{2}{5}$,   or
$y=-1$ (It is not of our interest, as we are dealing with real roots)
  Therefore $x=\sqrt{\frac{2}{5}},-\sqrt{\frac{2}{5}}$
These two values divides the real line into three disjoint subsets $(-\infty,-\sqrt{\frac{2}{5}})$, $(-\sqrt{\frac{2}{5}},\sqrt{\frac{2}{5}})$, $(\sqrt{\frac{2}{5}},\infty)$

It s quite easy to notice that $f$ is decreasing in $(-\sqrt{\frac{2}{5}},\sqrt{\frac{2}{5}})$ and increasing in $(-\infty,-\sqrt{\frac{2}{5}})$, $(\sqrt{\frac{2}{5}},\infty)$

Again, $f(-\sqrt{\frac{2}{5}})=1.911$ and $f(\sqrt{\frac{2}{5}})=0.089$

Looking at all these details it is clear that $f(x)$ comes close to zero in $(-\sqrt{\frac{2}{5}},\sqrt{\frac{2}{5}})$ but is not equal to zero. There is no possibility for $f(x)$ to be zero in $(\sqrt{\frac{2}{5}},\infty)$   as it is increasing here. Again, we have already discovered one root between $-1$ and $-2$. Apart from that there is of course no root as $f$ is again increasing in $(-\infty,-\sqrt{\frac{2}{5}})$.
So, te graph should look something like this

Therefore there is only one real root of the polynomial.

Is my approach correct? Is there any way to do the same thing using some more beautiful as well as advanced method ? (I have just made a rough plot of the graph in my answer sheet). I am not sure if the examiner would expect much better method than this one.

Sturm's theorem is a very powerful tool used to count real roots, but may be overkill here since your method also works and is arguably simpler. We take $f(x)$ and $f^\prime(x)$ and then compute successive (negated) remainders $r_1(x),r_2(x),\dots$ of polynomial long division (similar to the Euclidean algorithm) to get the sequence of polynomials $f(x), f^\prime(x), r_1(x), \dots$.

So if $f(x) = x^5 + x^3 - 2x  +1$ and $f^\prime(x) = 5x^4 + 3x^2 - 2$, then
$$f(x) = \frac{x}{5} \cdot f^\prime(x) - (-\frac{2}{5}x^3+\frac{8}{5}x - 1)$$
$$f^\prime(x) = (-\frac{25}{2}x)(-\frac{2}{5}x^3 + \frac{8}{5}x - 1) - (-23x^2 + \frac{25}{2}x + 2)$$
$$-\frac{2}{5}x^3 + \frac{8}{5}x - 1 = (\frac{2}{115}x + \frac{5}{529})(-23x^2 + \frac{25}{2}x + 2) - (-\frac{1531}{1058}x + \frac{539}{529})$$
$$-23x^2 + \frac{25}{2}x + 2 = (\frac{24334}{1531}x+\frac{5984577}{2343961})(-\frac{1531}{1058}x + \frac{539}{529}) - (+\frac{1409785}{2343961})$$

Now, for really large numbers, this sequence of polynomials gives the signs $+,+,-,-,-,+$, which has $2$ sign changes. For really large negative numbers, the sequence has the signs $-,+,+,-,+,+$, which has $3$ sign changes. The single additional sign change means that $f(x)$ has a single real root.

Since its a fifth degree equation it will have five roots if it has one complex root then it must have its congugate too as a root thus the number of real roots can only be one or three. you can find a real rrot using newton-Raphson method and check the remaining equation.


How much faster is the Trachtenberg system? How much faster are various mathematical operations using a Trachtenberg method rather than a conventional method?

From playing with the system a little myself, 20% sounds close for day to day use and not putting in much dedicated practice.  As with most mental math (and most math in general) speed will increase based on practice.  I enjoy that there are only a small number of techniques used for a large number of applications, rather than a large number of application specific techniques to memorize as in some systems.

The more extreme example mentioned sounds more like a stage demonstration than an average improvement.

Studying and practicing the Trachtenberg System has helped me DRASTICALLY improve my day to day mental arithmetic, especially noticeable when out shopping and working out the cost savings in buying bulk and cost comparisons per kilo.

Your question is hard to answer but I will say that where as before I'd simply give up on trying to do large multiplication in my head (as it would be impossible), now I actually am able to do the large multiplications and check them with good accuracy in my head. 3-4 digit multiplications (ie: 2873 x 231) in my head now generally take me around 15-30 seconds to work out (depending on the numbers) with almost 100% accuracy, about a month ago I would be able to do them in about 30-40 seconds ~80% correct. Give me a pen and paper and I can now do the same sums often in under 10 seconds.

The claims of a boy doing very large multiplications doesn't sound like it's impossible but would take a good few months of practice. Good memory recall is obviously an important skill to have and will need improving if you're going to stand a chance doing sums of that length.

I study once a week for a couple of hours and have been for about 2 months but also practice when out shopping like I said.

A good free site I found helpful to use alongside the book is http://www.trachtenbergsystem.org

...Same info as in the book but easier to understand (for me) and talks you through every step.


What does it mean by 'monotone' when referring to functions? To clarify, I know what it means when a function is surjective, injective, bijective, and what its inverse is. This is something my module notes covers.

However, I have a question (past exam paper) asking to show that a function is monotone. I have no clue what it means, and there is not a single mention of it in the module notes (provided by the professor, and I don't have access to a recommended textbook either).

Please could someone explain what monotone means?

A real valued function $f$ of a real variable is monotonically increasing if
$$
f(a) \ge f(b) \text{ when } a > b.
$$
"Monotone" might be monotonically increasing or monotonically decreasing. Sometimes you want the inequality to exclude equality. Then you use the adjective "strictly".

A monotonic function is one that preserves some sort of well-defined "order". Some examples are:

a monotonically increasing function $f$, such that $f(x)<f(x+\epsilon)$ for all $\epsilon>0$,
a monotonically decreasing function $f$, such that $f(x)>f(x+\epsilon)$ for all $\epsilon>0$.


Decompose sum into reversible pairs Is there any efficient way to find if a sum can be decomposed into reversible pairs?And if it does can we find these numbers?
For example 66 can be decomposed into 24+42 or 66666=12345+54321.
One solution could be brute force but is there any other solution?
Thanks.

You're going to have different techniques for different numbers of digits.

1 digit numbers
$n$ must be even, then there is only one sum: $\frac n2 + \frac n2$

2 digit numbers

For $10 \le n<20$ and $n$ even, there is the simple sum: $\frac n2 + \frac n2$

For other $n$, consider the sum of a number $10a+b$ and its reverse $10b+a$.

The sum is $11a+11b = 11(a+b)$

So the only other 2 digit numbers are those that are multiples of 11. We can call these $n=11k$ for $k=1,2,3,4,5,6,7,8,9$. We can list the pairs of values $a,b$ that give $a+b=k$.

3 digit numbers

Consider first 3 digit numbers that have been created by adding together two 2 digit numbers. So $(10a+b)+(10b+a)=100x+10y+z=n$. Note that $n = 11(a+b)$: $n$ must be a multiple of $11$.

It must be that $x=1$ and that $(a+b) \ge 10$. So $a+b=10+z$

Consider next 3 digit numbers that have been created by adding together two 3 digit numbers. So $(100a+10b+c)+(100c+10b+a)=100x+10y+z=n$.

To prevent a carry into the Thousands column, we need $a+c<10$. This means there is likewise no carry from the Units column to the Tens column, so $2b=y$ or $2b=y+10$. In each case, we require $y$ to be even. We have to check if $x=z$ (which indicates that there was no carry from the Tens so $2b=y$) or if $x=z+1$ (in which case there was a carry from the Tens so $2b+10=y$.

Pseudocode algorithm for 1 to 3 digit numbers.

Let number under consideration be $n$

If number of digits = 1

If $n$ odd then there are no ways to achieve this.

If $n$ even then there is a single sum: $\frac n2 + \frac n2$

If number of digits = 2

If $n$ is a multiple of 11

Let $n=11p$

The sums are: $(10k  + p-k) + (10(p-k) +k)$ for $k = 0, 1, ..., \lfloor \frac p2 \rfloor $

If $10 \le n<20$ and $n$ is even, then there is the sum: $\frac n2 + \frac n2$

Fotr other $n$ it is not possible to write them as a reversed pairs.

If number of digits = 3

Let $n=100x+10y+z$

If $x=1$ and $n$ is a multiple of 11, then sums are: $(10k  + z-k) + (10(z-k) +k)$ for $k = 0, 1, ..., \lfloor \frac z2 \rfloor $

If $x=z$ and $y$ is even, then sums are: $(100k + 10 \frac y2 + z-k) + (100(z-k) +10 \frac y2 +k)$ for $k = 0, 1, ..., \lfloor \frac z2 \rfloor $

If $x=z+1$ and $y$ is even, then sums are: $(100k + 10 \frac {y+10}2 + z-k) + (100(z-k) +10 \frac {y+10}2 +k)$ for $k = 0, 1, ..., \lfloor \frac z2 \rfloor $

Else no sums are possible.

If number of digits = 4

Let $n=1000w+100x+10y+z$

If $w=1$ and $x=z$, then $y$ must be even (or no sums). The sums are: $\left (100(z-k) + (\frac y2) \times 10 + k \right)+\left (100k + (\frac y2) \times 10 + z-k \right)$ for $k = 0, 1, ..., \lfloor \frac z2 \rfloor $

If $w=1$ and $x=z+1$, then $y$ must be odd (or no sums).

The sums are: $\left (100(z-k) + (\frac {y+10}2) \times 10 + k \right)+\left (100k + (\frac {y+10}2) \times 10 + z-k \right)$ for $k = 0, 1, ..., \lfloor \frac z2 \rfloor $

There are other sums that give 4 digit numbers - just need more time to develop that. Is this helping you?


Heat Equation with One Non-Homogeneous Boundary Condition I posed myself the following PDE because it would be interesting to graph:
$$
u_t=u_{xx},\qquad0<x<L,\qquad t>0,\\
\begin{align}
u(0,t)&=\sin^2\frac t2,\\
u_x(L,t)&=0,\\
u(x,0)&=0.
\end{align}$$
Physically, this is a rod with one end insulated and the other being periodically heated and cooled.

Now, since the $x$-dependent problem does not have two homogeneous boundary conditions, I wonder if it is possible to solve this PDE as it stands.

The way I was taught to solve boundary value problems with non-homogeneous boundary value conditions is via the introduction of a second term to satisfy the boundary, i.e. set
$$ u(x,t) = \phi(x,t) + v(x,t)$$
where $\phi(x,t)$ satisfies the boundary conditions.

So, we need
$$\phi_x(0,t)=0 \text{ and }\phi(L,t)=\sin^2\frac{t}{2}.$$
The simplest function that satisfies these conditions is
$$\phi(x,t)=\sin^2\frac{t}{2},$$
which means $u$ is of the form
$$u(x,t)=\sin^2\frac{t}{2}+v(x,t).$$

The pde now required to solve, for $v$ after substituting into your equation is
$$v_t=v_{xx}-\cos\frac{t}{2}.$$
Substituting into the boundary conditions also gives
$$v_x(0,t)=0\text{ and } v(L,t)=0$$
and the initial condition condition
$$v(x,0)=0.$$

This transforms the original problem with non-homogeneous boundary conditions into one with homogeneous boundary conditions, but a non-homogeneous pde, which is easier to solve.

Can you continue from this point?

I correct your differentiation mistake,

$$(\sin^2(\frac{t}{2}))' = 2\sin(\frac{t}{2})(\frac{1}{2})\cos(\frac{t}{2}) = \sin(\frac{t}{2})\cos(\frac{t}{2}).$$


If you draw two cards, what is the probability that the second card is a queen? We had this question arise in class today and I still don't understand the answer given. We were to assume that drawing cards are independent events. We were asked what the probability that the second card drawn is a queen if we take two from the deck. The answer given was 4/52, which seems counter-intuitive to me. How is the probability still 4/52 if there was a card drawn before it? What if the first card drawn was a queen?

There are two cases here:

Case 1: First card chosen is a queen

$$\frac{4}{52}*\frac{3}{51}=\frac{1}{221}$$

Case 2: First card chosen is not a queen.

$$\frac{48}{52}*\frac{4}{51}=\frac{16}{221}$$

Adding both the cases, we get $\frac{17}{221}$ = $\frac{4}{52}$ = $\frac{1}{13}$

I would think it would be slightly below 1/13 since 1/13 is the probability for the first pick being a queen then the second round should account for the possibility of there only being 3 queens in the deck.


Determination of the prime ideals lying over $2$ in a quadratic order Let $K$ be a quadratic number field.
Let $R$ be an order of $K$, $D$ its discriminant.
By this question, $1, \omega = \frac{(D + \sqrt D)}{2}$ is a basis of $R$ as a $\mathbb{Z}$-module.

Let $x_1,\cdots, x_n$ be a sequence of elements of $R$.
We denote by $[x_1,\cdots,x_n]$ the $\mathbb{Z}$-submodule of $R$ generated by $x_1,\cdots, x_n$.

My Question
Is the following proposition correct?
If yes, how do you prove it?

Proposition

Case 1 $D$ is even.

$P = [2, \omega]$ is a prime ideal and $2R = P^2$.

Case 2 $D \equiv 1$ (mod $8$).

$P = [p, \omega]$ and $P' = [p, 1 + \omega]$ are distinct prime ideals and $2R = PP'$.
Moreover $P' = \sigma(P)$, where $\sigma$ is the unique non-identity automorphism of $K/\mathbb{Q}$.

Case 3 $D \equiv 5$ (mod $8$).

$2R$ is a prime ideal.

Since $\omega = (D + \sqrt{D})/2$ has minimal polynomial equal to $x^2 - D x + D(D-1)/4,$ we may write $R = \mathbb Z[x]/(x^2 - Dx - D(D-1)/4),$ and so $R/2R =
\mathbb F_2[x]/(x^2 - D x - D(D-1)/4)$ (identify $\omega$ with the image of $x$).

If $D$ is even (and hence a multiple of $4$), the polynomial $x^2 - D x - D(D-1)/4 $ is congruent to $x^2 + 1 = (x+1)^2$ mod $2$.

If $D \equiv 5 \bmod 8$, then this polynomial is congruent to $x^2 + x + 1$ (an irreducible polynomial) mod $2$.

If $D \equiv 1 \bmod 8,$ then this polynomial is congruent to $x^2 + x = x(x+1)$ mod $2$.

In the first case we see that $(2) = (x,2)^2$, in the second case that $(2)$ is prime, and in third case that $(2) = (x,2)(x+1,2)$.

This proves the proposition. (Actually, the way the proposition is phrased, you have to check that the $R$-modules with the indicated generators, which is what I am writing, are the same at the $\mathbb Z$-modules with these generators.  This is straightforward.)

I realized after I posted this question that the proposition is not correct.
I will prove a corrected version of the proposition.

We need some notation.
Let $\sigma$ be the unique non-identity automorphism of $K/\mathbb{Q}$.
We denote $\sigma(\alpha)$ by $\alpha'$ for $\alpha \in R$.
We denote $\sigma(I)$ by $I'$ for an ideal $I$ of $R$.

Proposition
Let $K$ be a quadratic number field, $d$ its discriminant.
Let $R$ be an order of $K$, $D$ its discriminant.
By this question, $D \equiv 0$ or $1$ (mod $4$).
By this question, $1, \omega = \frac{(D + \sqrt D)}{2}$ is a basis of $R$ as a $\mathbb{Z}$-module.
Let $f$ be the order of $\mathcal{O}_K/R$ as a $\mathbb{Z}$-module.
Then $D = f^2d$ by this question.
We suppose gcd$(f, 2) = 1$.

Case 1 $D$ is even.

Since $D \equiv 0$ (mod $4$), $D \equiv 0, 4$ (mod $8$).
If $D \equiv 0$ (mod $8$), let $P = [2, \omega]$.
If $D \equiv 4$ (mod $8$), let $P = [2, 1 + \omega]$.
Then $P$ is a prime ideal and $2R = P^2$.
Moreover $P = P'$.

Case 2 $D \equiv 1$ (mod $8$).

$P = [p, \omega]$ and $P' = [p, 1 + \omega]$ are distinct prime ideals and $2R = PP'$.

Case 3 $D \equiv 5$ (mod $8$).

$2R$ is a prime ideal.

We need the following lemmas to prove the proposition.

Lemma 1
Let $K, R, D, \omega$ be as in the proposition.
Let $P = [2, b + \omega]$, where $b$ is a rational integer.
Then $P$ is an ideal if and only if $(2b + D)^2 - D \equiv 0$ (mod $8$).
Moreover, if $P$ satisfies this condition, $P$ is a prime ideal.

Proof:
By this question, $P = [2, b + \omega]$ is an ideal if and only if $N_{K/\mathbb{Q}}(b + \omega) \equiv 0$ (mod $2$).
$N_{K/\mathbb{Q}}(b + \omega) = (b + \omega)(b + \omega') = \frac{2b + D + \sqrt D}{2}\frac{2b + D - \sqrt D}{2}
= \frac{(2b + D)^2 - D}{4}$.
Hence $P$ is an ideal if and only if $(2b + D)^2 - D \equiv 0$ (mod $8$)
Since $N(P) = 2$, $P$ is a prime ideal.

Lemma 2
Let $K, R, D, \omega$ be as in the proposition.
Suppose gcd$(f, 2) = 1$ and there exist no prime ideals of the form $P = [2, b + \omega]$,
where $b$ is an integer.
Then $2R$ is a prime ideal of $R$.

Proof:
Let $P$ be a prime ideal of $R$ lying over $2$.
Then $P \cap \mathbb{Z} = 2\mathbb{Z}$.
By this question, there exist integers $b, c$ such that $P = [2, b + c\omega],
c \gt 0, 2 \equiv 0$ (mod $c$), $b \equiv 0$ (mod $c$).
Then $c = 1$ or $2$.
By the assumption, $c = 2$.
Hence $P = [2, 2\omega] = 2R$.

Proof of the proposition

Case 1 $D$ is even.

Let $P = [2, b + \omega]$, where $b$ is an integer.
We may assume that $b = 0$ or $1$.
By Lemma 1, $P$ is a prime ideal if and only if $(2b + D)^2 - D \equiv 0$ (mod $8$).

Suppose $D \equiv 0$ (mod $8$).
Then $(2b + D)^2 - D \equiv 0$ (mod $8$) if and only if $b = 0$.
Hence $P = [2, \omega]$ is an ideal of $R$.
$P' = [2, \omega'] = [2, D - \omega] = [2, -\omega] = [2, \omega] = P$.

Suppose $D \equiv 4$ (mod $8$).
Then $(2b + D)^2 - D \equiv 0$ (mod $8$) if and only if $b = 1$.
Hence $P = [2, 1 + \omega]$ is an ideal of $R$.
$P' = [2, 1 + \omega'] = [2, 1 + D - \omega] = [2, -1 - D + \omega] = [2, 1 + \omega] = P$.

Since gcd$(f, 2) = 1$, $P$ is regular by this quuestion.
Hence $PP' = 2R$ by this question.

Case 2 $D \equiv 1$ (mod $8$).

If $b = 0, 1$, then
$(2b + D)^2 - D \equiv (2b + 1)^2 - 1 \equiv 0$ (mod $8$).
Hence, by Lemma 1, $P = [2, \omega]$ and $Q = [2, 1 + \omega]$ are prime ideals of $R$.
$P' = [2, \omega'] = [2, D - \omega] = [2, - D + \omega] = [2, 1 + \omega] = Q$.
Since gcd$(f, 2) = 1$, $P$ is regular by this question.
Hence $PP' = 2R$ by this question.

Case 3 $D \equiv 5$ (mod $8$).

Consider the following congruence equation.
$(2b + D)^2 - D \equiv (2b + 5)^2 - 5 \equiv 0$ (mod $8$).
Since $b$ does not satisfy this congruence equation when $b = 0$ or $1$,
there exist no ideals of the form $[2, b + \omega]$.
Hence $2R$ is a prime ideal by Lemma 2.


Group which has a nontrivial normal Sylow subgroup but a quotient doesn't. Let $G$ be a group which has a normal Sylow subgroup. Must every nontrivial quotient group
$G/N$ have a nontrivial normal Sylow subgroup?

For example, every group of order up to 23 is a semidirect product of its Sylow subgroups so they all have a nontrivial normal Sylow subgroup, and also their quotients. $S_4$ doesn't have a normal Sylow subgroup. Is there a group which has a nontrivial normal Sylow subgroup but not every quotient has it?

Just finished posting the question and I noted it. Take $G=S_4\times \mathbb{Z}/7\mathbb{Z}$.

You could take $\langle (12)(34), (13)(24)\rangle=K \lhd A_4.$ Since $|A_4|=12,$ and $|K|=4,$ then $K$ is a sylow $2-$subgroup. The quotient has no nontrivial normal Sylow subgroup since it's $\Bbb{Z}/3\Bbb{Z},$ and cyclic groups of prime order are generated by all of their non-identity elements.

Take any group $H$ which doesn't have a normal Sylow subgroup, take $p\nmid|H|$ and let $G=H\times C$ where $C$ is cyclic of order $p$. Then $G$ has a normal Sylow $p-$subgroup, but $G/(1\times C))\cong H$ doesn't. The example can be generalized even more, taking $C$ as any $p-$group.


How do I Prove De Morgan's Law with Inference? I'm trying to get from ~(q -> p) to (q ^ ~p), which is a version of De Morgan's Law with the conditional, in Fitch using only basic logic rules (e.g. ^ intro, ^ elim, etc. No laws or shortcuts).

(Actually specifically I'm trying to get from ~(~q -> p) to (~q ^ ~p), but that's the same idea, right?)

Any help?

$\def\fitch#1#2{\begin{array}{|l}#1\\\hline#2\end{array}}$

You seek to derive $\neg q\wedge\neg p$ from the premise $\neg(\neg q\to p)$.   Well, derive each conjunct separately then introduce the conjunction.

First show that you can derive $\neg q\to p$ from an assumption of $p$.   Therefore you can infer that $\neg p$ may be derived from $\neg(\neg q\to p)$.

Next show that you can derive $\neg q\to p$ from an assumption of $q$.   Therefore you can infer that $\neg q$ may be derived from $\neg(\neg q\to p)$.

Here's a quick Fitch style skeleton of the proof.   Fill in the missing details and you are done.

$$\fitch{\neg(\neg q\to p)}{\fitch{p}{~\vdots\\\neg q\to p\\\bot}\\\neg p\\\fitch{q}{~\vdots\\\neg q\to p\\\bot}\\\neg q\\\neg q\wedge\neg p}$$

DeMorgan's Laws are pretty much your only means of distributing a negation by inference; you can't prove them by the same.

Fortunately, they're both intuitive and can be proven by other means, such as truth tables.

As far as your expression, $!(!q -> p) = !q!p$, that's easily proven if DeMorgan's laws are allowed. $x -> y = !x + y$, so your expression is actually $!(q + p) = !q!p$, which is DeMorgan's law verbatim.


What is $X\cap\mathcal P(X)$? Does the powerset of $X$ contain $X$ as a subset, and thus $X\cap \mathcal{P}(X)=X$, or is $X\cap \mathcal{P}(X)=\emptyset$ since $X$ is a member of the
$\mathcal{P}(X)$, and not a subset?

Define a set X={a,b,c} having 3 elements(let).

So, power set of X={{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c},{}}, [i.e it has 2^(3)=8 elements.]

So, X AND P(X) means the common part in 'X' & p(X)

So, X intersection/and P(X)={}


Solving inequalities involving square roots What is the correct way to solve inequalities like $\sqrt {x+2}>x $   ?

I deduced that $x$ must be greater than$-2$ and also on squaring I get $-1 \leq x \leq 2$ but even on combining these two I'm not getting the correct answer.What am I missing?

First note that $x$ must be greater than $-2$, because of the square root. Then you have two cases :

If $x \in [-2,0[$ the inequality is true since a square root is positive.

If $x \in [0,+\infty)$, then both sides are positive and you can square it to obtain the equivalent inequation $$x+2 > x^2.$$ That is $$x^2-x-2 < 0.$$ The discriminant here is $9$ and you can hence deduce the factorisation $$(x-2)(x+1) <0.$$ Then do a sign table with the two factors to conclude.

Use: $$\sqrt{f(x)}>g(x) \Leftrightarrow$$

$\Leftrightarrow \begin{cases}
g(x)<0\\
f(x)\ge0\end{cases}$
or
$\begin{cases}
g(x)\ge0\\
f(x)>\left(g(x)\right)^2
\end{cases}$

  !$$\sqrt{x+2}>x$$

  !Case 1) $$\begin{cases}
x<0\\
x+2\ge0\end{cases}$$

  !Case 2) $$\begin{cases}
x\ge0\\
x+2>\left(x \right)^2
\end{cases}$$


Max/min problem - show that diagonal is least when rectangle is a square. A rectangle has constant area, show that the length of a diagonal is least when the rectangle is a square.

$$Area (a) = xy$$

$$y=\dfrac{a}{x}$$

$$D^2 = x^2 + y^2$$

$$D^2 = x^2 + (\dfrac{a}{x})^2$$

$$\dfrac{dD^2}{dx} = 2x - \dfrac{2a^2}{x^3}$$

$$2x - (\dfrac{2a^2}{x^3})=0$$

$$x=a^\dfrac{2}{4}$$

$$x=\sqrt{a}$$

$$\dfrac{d^2D^2}{dx^2}=2+\dfrac{6a^2}{x^4}$$

$$2+\dfrac{6a^2}{x^4}>0, minima$$

$$y=\dfrac{a}{\sqrt{a}}$$

$$y=\sqrt{a}$$

Is this correct? I'm assuming it is since the length of $x$ and $y$ are the same..sorry, I don't usually get these questions right and just wanted to make sure this is the way to do it.

Thank you.

the proof above is ok, since $x^2+y^2\geq 2xy$ and this is equivalent to $(x-y)^2\geq 0$ and the equal sign holds if $x=y$ this means we have a square.

$D =\sqrt{x^2+y^2} \geq \sqrt{2xy}=K$, constant. Thus equality occurs when $x = y$, or you have a square.


Prove by induction: $\forall n\in\mathbb{Z}_{\geq1}:3\ |\ (6n^2-12n+3)$ I'm not sure how to start this induction problem.

I  was told that we start doing induction by using a base case $n=1$.  Then we set $n=k$ to prove $n=k+1$.  But how do I prove that $6(k+1)^2-12(k+1)+3$ is divisible by $3$ if $6k^2-12k+3$ is?

I'm confused, can someone give me some help?  Thanks.

Hint:  If you have to use induction, you have the correct idea.  You assume $3\mid6k^2-12k+3$ and want to prove $3\mid6(k+1)^2-12(k+1)k+3$.  You make use of your assumption by finding it in the new expression.  So expand out the $(k+1)$'s and group the terms together to use the assumption.  Now you look at what is left and show that $3$ divides it, too.

You can also proceed as follows:

Prove this for $n=1$.

Assume $3|(6k^2-12k+3)$.

Then $6(k+1)^2-12(k+1)k+3 = 3(2(k+1)^2-4(k+1)+1)$ which is divisible by $3$.

And you're done.

The idea of induction is that you assume $P(n)$ is true for some $k$ and then prove it for $k+1$. If you prove it for $k+1$ without using the induction hypothesis, it is still a valid induction argument since you first assumed $P(n)$ is true for some $k$ and then proved it for $k+1$.


Function defined in coursework has unmapped domain? Hi I was doing my Discrete Math coursework when I realise the function defined in the question has unmapped domain.

Note the definition of Zn*:

and in part c):

So basically you have f([3]) = [6] = [0], which maps to none of the codomain. How should I approach this?

Also, is the function injective? ( if both f( [ 7 ] ) = f( [ 1 ] ) = [ 2 ] ==> [ 7 ] =/= [ 1 ])??

Edit: I should add the domain and codomain do not seem to include [0].

You're correct that the definition fails to define a function $\mathbb Z^*_6\to\mathbb Z^*_6$.

Arguably that is a bug in the exercise. In that situation the best thing you can do is probably to point out the problem, and also explain that if we take the definition to define a function $\mathbb Z_6\to\mathbb Z_6$ instead, then it (is/is not) injective, and (is/is not) surjective.

Note that it is more common to take $\mathbb Z_n^\times$ to mean the group of all residue classes $[a]$ modulo $n$ where $a$ and $n$ are coprime. In that case, however, $\mathbb Z_6^\times$ would be $\{[1],[5]\}$ and the definition in the exercise still wouldn't define a function $\mathbb Z_6^\times \to \mathbb Z_6^\times$.

Okay, f([0])= [2][0]= (2)(0)= 0= [0], f([1])= [2][1]= 2= [2], f([3])= (2)(3)= 6= [0], f[4]= [2][4]=  8= [2], and f[5]= [2][5]= 10= [4].  The "codomain" is {[0], [2], [4]}.  I don't know what you mean by "f([3])= f([6])= [0] maps to none of the codomain".  You just said that it maps to [0]!  What is true is that there is no "x" in the domain that maps to [1] (or to [3] or [5]).  This function is not surjective or "onto" the range space {[0], [1], [2], [3], [4], [5]}.


10% of inmates are innocent. We randomly select 20 inmates. Find the probability that we will find at least 3 innocent inmates. Suppose that 10% of inmates in a large prison are known to be innocent. A non-profit group randomly selects 20 inmates from this prison. Find the probability the group will find at least 3 innocent inmates.
I assume that I will be using Binomial Distribution. Can someone confirm and give the value of the answer to this.

Suppose $p$ is the probability of an inmate being innocent. And suppose we can assume that the probability of each inmate being innocent is independent of the innocence or guilt of the other $19$ inmates (which is approximately true if the $20$ inmates are chosen from a large population). Then the probability of finding exactly $n$ innocent inmates in $20$ is

$P(n)=p^n(1-p)^{(20-n)}\binom{n}{20}$

Now you could work out $P(3), P(4), P(5) \dots$ etc. and add then up to find $P(n \ge 3)$. But is is easier to use the fact the $P(n \ge 3)+P(n \lt 3)=1$ (how do we know this ?) and so

$P(n \ge 3) = 1-P(n\lt 3) = 1-(P(0)+P(1)+P(2))$

How many inmates in total? Let's say it's $n$. So $0.1n$ of them are innocent. The total number to select 20 inmates is $\binom{n}{20}$. There's $\binom{0.9n}{20}$ to select no innocents at all and $\binom{0.9n}{19} \binom{0.1n}{1}$ to select exactly one. Can you take it from here?


How to represent Hadamard product in terms of matrix multiplication? In the case of two vectors $u, v$ with dimensions $n\times 1$, their Hadamard product can be represented by the following matrix multiplication:
$$\mathrm{diag}(u)v = \left[\begin{array}{ccc}
u_{1} &  & 0\\
 & \ddots\\
0 &  & u_{n}
\end{array}\right]\left[\begin{array}{c}
v_{1}\\
\vdots\\
v_{n}
\end{array}\right]=\left[\begin{array}{c}
u_{1}v_{1}\\
\vdots\\
u_{n}v_{n}
\end{array}\right]\equiv u\circ v$$

Is there a way to generalize this for Hadamard products of matrices?

Let $\mathcal M_n$ be the space of $n\times n$ matrices (over some field). Then if $A \in \mathcal M_n$ you can define a linear map:
$ T_A : \mathcal M_n \to \mathcal M_n$ given by $T_A(B) = A\circ B$, where $\circ$ is the Hadamard/Schur product of $A$ and $B$ (note that this definition is basis dependent!). The matrix units are eigenvectors of this linear trasformation:
$$ T_A(E_{ij}) = a_{ij}E_{ij},$$
where $A=(a_{ij})_{i,j=1\dots n}$.

Now $T_A$ is a linear operator, so if you pick a basis for $\mathcal M_n$ you get a matrix representation for it. If you choose the same basis in which you defined the Hadamard product, you get a diagonal matrix of dimension $n^2\times n^2$ whose diagonal entries are the matrix entrires of $A$, i.e. the $a_{ij}$ (ordered according of the order of the matrix units you choose).

Hadamard product (Schur product) of matrices is element-wise product (two matrices dimension have to be same).

Reference: Topics in matrix analysis.


Convergence of given sequence Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a function with:

$$f(x) = x - \arctan{x}$$

We consider the sequence $(x_{n})$ with $x_{0} > 0$ and $x_{n + 1} = f(x_{n})$, for any $n \in \mathbb{N}$.

Prove that $(x_{n})$ is convergent and find its limit.

So far, I've proved that $f(x) \geq 0$ for any $x \geq 0$. From this I've concluded that $(x_{n})$ is a positive sequence. Now I need to find $(x_{n})$'s monotony. I've calculated both $x_{n + 1} - x_{n}$ and $\frac{x_{n + 1}}{x_{n}}$. For the first I've got $- \arctan{x_{n}}$ and the second one gave me $1 - \frac{\arctan{x_{n}}}{x_{n}}$. From this point I don't know what to do next.

Thank you in advance!

Look at the graph: (from Wolfram Alpha)

The way to answer this question is to prove that, say for $x > 0$, $f(x) < x \le 0$. It should be clear that iterating the function only gets you closer to zero.

Does this help?

The derivative is $\frac{x^2}{1+x^2}$and  the function is a contraction mapping on any interval $[0,n]$. The given sequence is used to find the  fixed point x=0.


How can I prove false is not a theorem? Definitions

Let our propositional calculus be defined as follows:

Alphabet $D := \{p_i:i\in \mathbb{N}\}$.

Operators $\Omega := \Omega_0 \cup \Omega_2$ where $\Omega_0 := \{f\}$ and $\Omega_2 := \{\rightarrow\}$.

Axioms:

AX1: $A\rightarrow(B\rightarrow A)$
AX2: $(A\rightarrow(B\rightarrow C))\rightarrow((A\rightarrow B) \rightarrow (A\rightarrow C))$
AX3: $((A\rightarrow f)\rightarrow f) \rightarrow A$

where $A$, $B$ and $C$ can be arbitrary formulas.

Modus ponens is the only inference rule. To be precise, a formula $A$ is a theorem (denoted $\vdash A$) if there exists a sequence of formulas $A_1, ..., A_n$ (called an inference sequence) such that $A_n=A$ and for all $i\in \{1,...,n\}$ the following holds:

$A_i$ is an axiom, or
there exists $j,k\in \{1,...,i-1\}$ such that $A_k=A_j\rightarrow A_i$.

Word and formula

A word is a nonempty and finite sequence of symbols from the set $D\cup \Omega \cup \{(,)\}$.

A word $A$ is a formula if there exists a finite sequence of words $A_1,...,A_m$ such that $A_m=A$ and for all $i \in \{1,...,m\}$ the following holds:

$A_i=p_n$ for some $n\in \mathbb{N}$, or
$A_i=f$, or
$A_i=(A_j\rightarrow A_k)$ for some $j,k \in \{1,...,i-1\}$.

For brevity, the outermost parentheses can be omitted, so $(A \rightarrow B)$ is the same as $A \rightarrow B$.

Problem

Show that $\not\vdash f$.

Comments

This problem seems difficult because I somehow need to prove there exists no inference sequence that would prove $\vdash f$. Intuitively, it seems that for any inference sequence $A_1,...,A_n$, the formula $A_i$ always contains at least one arrow symbol for all $i\in \{1,...,n\}$, but I cannot figure out how to prove it.

(I believe I'm not supposed to use the soundness or completeness theorems in the proof, as they are introduced much later in the lecture notes.)

Let $V(E)$ for a propositional expression $E$ be defined as:

$1$ if every bivalent valuation of $E$ is true
$0$ if any bivalent valuation of $E$ is false

given that $'f'$ valuates to false.  So for example, $V(\ulcorner A \lor (A \to f) \urcorner)$ has 2 valuations:  $\top \lor (\top \to \bot) = \top$ as well as $\bot \lor (\bot \to \bot) = \top$, so its valuation is $1$.

Prove that the valuation of every axiom is $1$ (keep in mind that for example, an instance of AX1 might be $(X \to Y) \to ((M \to N) \to (X \to Y))$, that is the $A$ and $B$ in the axiom are expressions not propositional variables).

Prove that modus ponens preserves positive valuations:  if $A \to B$ and $A$ has positive valuations, then $B$ must have a positive valuation.  Same cavaet, $A$ and $B$ are expressions not propositional variables.

Put that together to make an inductive argument that no provable expression has a valuation of $0$.  Combine that with the valuation of $\ulcorner f \urcorner$.

None of the axioms are f.  So, f could only get deduced by uses of detachment (i. e. modus ponens).
Every detachment of a formula has one of the following forms 1. (B→A) originating from axiom 1.  2. A originating from axiom 1.  3. ((A→B)→(A→C)) originating from axiom 2.  4. (A→C) originating from axiom 2.  5. C originating from axiom 2.  6. A originating from axiom 3.

I'll note that all of the axioms are sound, in that none of them evaluate to f for any valuation of their variables.

(B→A), ((A→B)→(A→C)), and (A→C) do not have the same form of f.  So, it is not possible to derive f such that it has the form indicated by 1., 3., or 4.

Consider the possibility that f has form A originating from axiom 1, or from axiom 3.  In the first case, axiom schema 1 gets instantiated as the axiom (f→(B→f)).  But, then nothing could get detached, since the axioms are sound.  Thus, (B→f) would not be derivable, and neither would f.  So, f cannot get derived that way.  In the second case, axiom schema 3 gets instantiated as the axiom (((f→f)→f)→f).  But, the antecedent of that axiom is ((f→f)→f), which is false.  Since the axioms are sound, that does not permit a deduction of f.  Thus, both possibilities 2., and 6. consider here cannot derive f.

Finally, we come to the possibility that f got derived having form C from axiom 2.  Then axiom schema 2 gets instantiated as an axiom having the form ((A→(B→f))→((A→B)→(A→f))).  In order to get to f, since all of the axioms are true, and the deduction rule of modus ponens is sound, it follows that (A→(B→f)), (A→B), and A all hold true.  But, if A holds true, and (A→B) holds true, so does B.  Then since B is true, (B→f) is false.  Since (B→f) is false, and A is true, that would imply (A→(B→f)) as false.  But, that is not possible since the axioms are sound.

Having eliminated all possible cases of how f could get deduced according to the definition of a proof, we conclude that f is not provable.


Construction of a circle through two interior points of a disk and meeting the boundary at two antipodal points Let $D$ be a disk, $P,Q$ be two distinct points in its iterior.
A geogebra experiment shows that there exists one and only one (possible degenerated, with infinite radious) circle through $P$ and $Q$ and meeting the boundary of $D$ at two antipodal points.

  How to prove it?

Motivation.
I'm looking for a model of the real projective plane in the unit disk $D$ (with antipodal points identified).
By taking as lines of this models the arcs of circles which meets the boundary of $D$ at two antipodal points, we are proving that given two points $P,D$ one and only line of this model passes through them.

A rule-compass construction.
Let $S,T$ be the intersection of $PQ$ with the boundary of $D$, $S',T'$ be the respective antipodal points.

Given two points $X,Y$, let $b_{XY}$ denote the line bisector of the segment $XY$.

Consider $G=b_{T'P}\cap b_{QT}$, $H=b_{SP}\cap b_{QS'}$.
Then $I=GH\cap b_{PQ}$ is the center of the required circle.

Let $D$ be the unit disc in ${\mathbb R}^2$. Embed ${\mathbb R}^2$ via $(x,y)\mapsto(x,y,0)$ into ${\mathbb R}^3$, and consider the stereographic projection $\sigma:\>\dot{S^2}\to{\mathbb R}^2$ from the north pole $(0,0,1)$. This map keeps the points of $\partial D$ fixed, and maps the set of circles in $S^2$ to the circles and lines in ${\mathbb R}^2$. A circle $\ne\partial D$ in $S^2$ is a great circle iff it intersects $\partial D$ in two antipodal points.

Now let $\hat P=\sigma^{-1}(P)$ and $\hat Q=\sigma^{-1}(Q)$. There is exactly one great circle $\gamma$ through $\hat P$ and $\hat Q\in S^2$. It follows that $\sigma(\gamma)$ is the single circle through $P$ and $Q$ that intersects $\partial D$ in two antipodal points.

Based on the work done in the 1st answer to this question (by Christian Blatter) , we can go further and construct the arc of circle through P and Q  by defining
the model of the real projective plane as the stereographic projection of the sphere with
antipodal points identified.
Please find attached below some more details and diagrams.

Let us define the circle model of $\mathbb{R}P^2$   as the the circle D2 obtained via stereographic projection
of  $S^{2}$  modulo the equivalence relation that identifies antipodal points.
Stereographic projection sends lines into lines and circles into circles so that the geodesics on
D2 are arcs of circle which go through antipodal points with respect to the center of D2 (Figure 1)
   1
Given two points D and E not belonging to  $\partial D2$ we want to find the geodesic { C,D,E,B }
going through them (Figure 3 below)  :
 2

So let D and E be points belonging to circle Y (Figure 4)  and pick another circle (W) which $\epsilon$  { Sheaf of circles through D and E } and which intersects Y in the points G and F.

Draw  the line DE and the line GF , let I = GF $\cap$ DE, then I is the center of a Sheaf of lines induced by the { Sheaf of circles through D and E }.
Draw also  the line AI which intersects Y in 2 antipodal points $A_{1}$  and $A_{2}$ (not showed), then the circumference Z ={ $A_{1}$DE $A_{2}$ } is the searched unique geodesic.