$\int \min \{\frac1{(1+x^2)^2},\frac1{5-x^2}\}$ solution verification $$\int \min \{\frac1{(1+x^2)^2},\frac1{5-x^2}\}$$
I see this integral as the integral of the function which smaller of these two
$$\int\frac1{(1+x^2)^2}=\frac12\arctan(x)+\frac x{2(1+x^2)}+C_1$$
$$\int\frac1{5-x^2}=\frac{\sqrt5}{10}\log|\sqrt5+x| -\frac{\sqrt5}{10}\log|\sqrt5-x| +C_2$$
$$
  \int \min \{\frac1{(1+x^2)^2},\frac1{5-x^2}\}= \cases{        \frac12\arctan(x)+\frac x{2(1+x^2)}+C_1       & $\frac1{(1+x^2)^2}< \frac1{5-x^2}$ \cr
                 \frac{\sqrt5}{10}\log|\sqrt5+x| -\frac{\sqrt5}{10}\log|\sqrt5-x| +C_2 & $\frac1{5-x^2}< \frac1{(1+x^2)^2}$  }
$$
Is the idea correct and what happens in points where $\frac1{(1+x^2)^2}=\frac1{5-x^2}$ ?

For $x^2>5$, the function reduces to $\dfrac1{5-x^2}$ and we can use an indefinite integral
$$F(x)=\int\frac{dx}{5-x^2}+C_0.$$ Notice that you may not cross the borders $x^2=5$ and there can be two distinct constants on either sides.

For $x^2<5$, we first restrict the study to $x\ge0$ for convenience. We switch from one function to the other at $x=1$. So for $x\le1$, we adopt
$$F(x)=\int_0^x\frac{dt}{5-t^2}+C_1$$ and for
$$1\le x\le \sqrt5,$$
$$F(x)=\int_0^1\frac{dt}{5-t^2}+\int_1^x\frac{dt}{(1+t^2)^2}+C_1=\frac1{\sqrt 5}\text{artanh}\frac1{\sqrt 5}+\int_1^x\frac{dt}{(1+t^2)^2}+C_1.$$
Now for the negative domain, we can use by symmetry
$$F(x)= C_1-F(-x).$$

What did you do to determine which values of x make one smaller than the other?
First notice that $\frac{1}{(1+ x^2)^2}$ is always positive while $\frac{1}{5- x^2}$ is positive only for $-\sqrt{5}< x< \sqrt{5}$.
For all $x< -\sqrt{5}$ and all $x> \sqrt{5}$, the minimum if $\frac{1}{5- x^2}$.  For $-\sqrt{5}< x< \sqrt{5}$,
$\frac{1}{(1+ x^2)^2}< \frac{1}{5- x^2}$
if and only if $5- x^2< (1- x^2)^2= 1- 2x^2+ x^4$.
So we want to find x such that $x^4- x^2- 4> 0$.
Let $y= x^2$ and solve $y^2- y- 4> 0$.  $y^2- y- 4= y^2- y+ \frac{1}{4}- \frac{1}{4}- 4= (y- \frac{1}{2})^2- \frac{17}{4}> 0$.  $(y- \frac{1}{2})^2> \frac{17}{4}$.
That will be true for $y= x^2< \frac{1- \sqrt{17}}{2}$ and for $y= x^2> \frac{1+ \sqrt{17}}{2}$.  Of course, for real x, $x^2$ is not negative so we are left with $x^2> \frac{1+ \sqrt{17}}{2}$ which means $x< -\sqrt{\frac{1+ \sqrt{17}}{2}}$ and $x> \sqrt{\frac{1+ \sqrt{17}}{2}}$.


How many lattice paths are there from $(0,0)$ to $(2n,2n)$ that avoids odd points How many lattice paths are there from $(0,0)$ to $(2n,2n)$ that do not go through one of the points $(2i-1,2i-1)$ for $i=1,\dots,n$?

My idea is to count the number of total lattice paths one can take from $(0,0)$ to $(2n,2n)$. There are ${4n \choose 2n}$ such paths. Then subtract the number of paths that are not valid. In counting these, I reasoned that we must avoid the "odd points" inside the grid with height and width of $2n$. I counted the number of paths that take these of points to be ${4 \choose 2}^{n-1}{2 \choose 1}{2 \choose 1}$ with the reasoning that from $(0,0)$ to $(1,1)$, there are ${2 \choose 1}$ paths, similarly for $(2n-1,2n-1)$ to $(2n,2n)$. Now, there are a total of $n-1$ "odd points" we consider and the number of paths from say $(1,1)$ to $(3,3)$ is ${4 \choose 2}$, we consider $n-1$ such scenarios. But in comparing my result, it is wrong, I seem to be undercounting the number of invalid paths that I need to subtract from the total paths.

Edit: The result is expected to be the Catalan numbers of the form $C_{2n+1}$.

Edit 2: I've reworked the problem to make the first couple of terms match $C_{2n+1}$, by removing from the total number of lattice paths the invalid paths (a sum of all the possible cases by which we choose how many and which odd points our invalid path has gone through). It seems to be some recursive function, any ideas how to express this recursively?

Yeah, there is a nice way to do it. This looks long, but it's because I stated everything rigorously. If you draw pictures while reading this, it'll make a LOT more sense.

Let $f(2n)$ denote the number of paths from $(0, 0)$ to $(2n, 2n)$ not crossing through a point of the form $(2k+1, 2k+1)$. I claim that $f(2n) = C_{2n}$, where $C_{2n}$ is the $2n$-th Catalan number.

A well known property of Catalan number $C_{n}$ is that it satisfies the following recursion formula:
$$ C_{n+1} = \sum_{i=0}^{n} C_i C_{n-i} \tag{1}$$
Another well known property is that it counts the number of paths from $(0,0)$ to $(2n,2n)$ which never go above the line $y=x$.

I'll prove the result by induction. Notice it's true for a base case of $n = 0$. Now, suppose the result is true for $f(0), f(2), \dots, f(2n-2)$.

To count $f(2n)$, we do casework on the first point of the form $(2k, 2k)$ our path goes through (other than $(0, 0)$). This casework covers all paths since all paths end up at $(2n, 2n)$. Suppose the  first such point is $(2k, 2k)$. WLOG on our first step, we went $(0, 0) \to (1, 0)$, we'll multiply by $2$ in our final count. Then we must also end with $(2k, 2k-1) \to (2k, 2k)$. It remains to count the number of paths which go from $(1, 0)$ to $(2k, 2k-1)$ without passing any point of the form $(2k, 2k)$. This is just $C_{2k-1}$! After this, there are $f(2n-2k)$ ways to finish up the path $(2k, 2k) \to (2n, 2n)$. Therefore, we have
$$f(2n) = \sum_{k=1}^{n} 2 \cdot C_{2k-1} f(2n-2k)$$
By the inductive hypothesis, $f(2n-2k) = C_{2n-2k}$, so we really have
$$f(2n) = \sum_{k=1}^{n} 2 \cdot C_{2k-1} C_{2n-2k} = \sum_{k=1}^n C_{2k-1}C_{2n-2k} + \sum_{k=1}^nC_{2k-1}C_{2n-2k}$$
using $j = n-k$ as the iterator for the second sum, we get
$$f(2n) = \sum_{k=1}^n C_{2k-1}C_{2n-2k} + \sum_{j = 0}^{n-1} C_{2j} C_{2n-2j}$$
The finish is in sight! The first sum is just $C_1C_{2n-2}+C_3C_{2n-4} + \dots C_{2n-1}C_{0}$ (i.e. the odd terms from $(1)$) while the second sum is just $C_{0}C_{2n-1} + \dots C_{2n-2}C_1$ (i.e. the even terms from $(1)$). Therefore, we deduce that $f(2n) = C_{2n}$ as desired.

I'm sure the bijective proof exists, but I have yet to try to find it. But given this, maybe you'll be able to do it :)

Let $C_k$ be the $k$-th Catalan number. If by "do not go through one of the points $(2i-1,2i-1)$" you mean "do not go through any of them", my answer is
$$2C_{2n-1}=2\cdot \frac{1}{(2n-1)+1}\binom{2(2n-1)}{2n-1}=\frac{1}{n}\binom{4n-2}{2n-1}$$
Indeed, in every lattice path from $(0,0)$ to $(2n,2n)$ the first step, like any other step, is either "$\uparrow$" (going $1$ unit up) or "$\to$" (going $1$ unit to the right). WLOG, assume that one initially chooses "$\to$" and goes frim $(0,0)$ to $(1,0)$. Then the last step must be an "$\uparrow$" (i.e. going from $(2n,2n-1)$ to $(2n,2n)$) because it is required that the lattice path does not intersect the diagonal at all. Now the number of lattice paths from $(1,0)$ to $(2n,2n-1)$ that does not pass above the line segment $\ell$ joining those two points is $C_{2n-1}$. (Any lattice path from $(1,0)$ to $(2n,2n-1)$ that passes above $\ell$ will intersect the main diagonal). By symmetry, the number of lattice paths you are looking for is twice as many.


The equation $x^y = y^x$. The following equation has been of interest for a long time. I'm bringing this to your attention again:

$$x^y = y^x$$

How do you attack to find all real and imaginary solutions for this equation? or

What is the nicest solution that you have seen?

The standard way I have seen
to solve $x^y = y^x$
is to define
$\frac{y}{x} = r$.

Then $y = xr$,
so
$x^{xr} = (xr)^x$
or
$x^r = xr$
or
$x^{r-1} = r$
or
$x = r^{1/(r-1)}$.

From this,
$y
=xr
= r^{1+1/(r-1)}
= r^{r/(r-1)}
$.

Any value you put in for $r$
(except $r=1$)
gives you $x$ and $y$
that satisfy $x^y = y^x$,
and conversely.

For example,
if you start with
$r = i$,
$\frac{1}{r-1}
=\frac{1}{i-1}
=\frac{-i-1}{(-i-1)(i-1)}
=\frac{-i-1}{2}
$
and
$\frac{r}{r-1}
=\frac{i}{i-1}
=\frac{i(-i-1)}{(-i-1)(i-1)}
=\frac{1-i}{2}
$.
Then
$x =
i^{(-i-1)/2}
$
and
$y
=i^{(1-i)/2}
$.
(I'll let you work the
standard form for $x$ and $y$.)

y=-x*W(-(ln(x)/x)/ln(x)

W(x) is the Lambert W function


Figuring out variable pairs in an inequality Let $x$ and $y$ be positive integers such that

  $45 < 8x + 5y < 60$

How many $(x,y)$ pairs can be found?
(Ans: 16)
Of course, there is a way to write it one by one. On the other hand there is a geometrical solution which is really nice but I don't want that. Can we write create and algebraic equation system that helps us to find the pairs. One of my friends talked my about the Diophantine equations, I don't have enough information about that so I don't know if it helps. Thank you in advance.

Consider the diophantine equation $$8x+5y=k$$ and choose one solution, say $(x,y)=(1,1)$, so we have $$\begin{cases} 8x+5y=k\\8*1+5*1=13\end{cases}\\$$ Take the difference $$8(x-1)+5(y-1)=k-13$$ We need $$45\lt{k-13}\lt60 \iff 58 \lt{k}\lt73$$
We have then $73-58=15$ solutions which finally gives with the solution $(x,y)=(1,1)$ the required 16 solutions.

NOTE.-The solution $(x,y)=(1,1)$ is a solution of the diophantine equation without restrictions but it must be consider in final computing the total of solutions after the condition $45\lt{k-13}\lt60$ (it is actually an auxiliary solution).


Concrete balanced category A category is called balanced if every bimorphism is an isomorphism.

Consider a concrete category such that every bijective morphism is a isomorphism. Does the category is balanced? Does converse is true?

In the category of rings, every bijective morphism is an isomorphism, but the inclusion of $\mathbb{Z}$ in $\mathbb{Q}$ is a bimorphism that is not an isomorphism, so the category of rings is not balanced.

Generally speaking, whenever you see a definition that says: A thing X is called Y if.....

what they really mean is: A thing X is called Y iff...

In other words the definition is a logical equivalence (a logical biconditional).

The definition of balanced category in ncatlab  is an exanple of this usage


Dual Vector Space Proof I am trying to prove the following:

Let $\phi: V \rightarrow (V^*)^*$ be defined by $v \to (f \to f(v))$, where $v \in V$ and $f \in V^*$. Show that $\phi$ is an injective map which is an isomorphism if and only if $V$ is of finite dimension. Note that $V$ is a vector space.

I am not too sure where to start on this at all, any help or hints would be appreciated.

Suppose for $v, v' \in V$ we have $\phi(v)=\phi(v')$, which means that these are equal as elements of $(V^\ast)^\ast$. So $$\forall f \in V^\ast: \phi(v)(f)= \phi(v')(f).$$

By definition of $\phi$, we know

$$\forall v \in V: \forall f \in V^\ast: \phi(v)(f)=f(v).$$

So $$\forall f \in V^\ast: f(v)= f(v').$$

If $v \neq v'$, find $f \in V^\ast$ such that $f(v) \neq f(v')$.

It then follows that $v=v'$ and $\phi$ is injective.

If $\dim(V)=n$, then $V \simeq \mathbb{K}^n$ then $\dim(V)=\dim(V^\ast)$ so $V^\ast \simeq \mathbb{K}^n$. Repeat for $V^\ast$. This shows one direction.

Clarifications:

I will replace your vector $v$ with $x$.
To the function $f\mapsto f(x)$ I will denote it by $\operatorname{ev}_x(f)=f(x)$.
Also, the dual space of $\textsf{V}^*$ I will denote it by $\textsf{V}^{**}$.

I hope everything that follows is helpful.

Lemma : Let $\textsf{V}$ a finite-dimensional vector space and let $x\in \textsf{V}$. If $\operatorname{ev}_x(f)=0$ for all $f\in \textsf{V}^*$, then $x=0$.

Or equivalently, if $x\neq 0$ then $\operatorname{ev}_x(f)\neq 0$ for some $f\in \textsf{V}^*$.

Proof : Let $x\neq 0$. Let $$\beta =\{v_1,v_2,\dots,v_n\}$$ be a basis for $\textsf V$ such that $v_1=x$. Now, let $$\beta^* =\{f_1,f_2,\dots,f_n\}$$ the dual basis of $\beta$. Then $f_1(v_1)=1\neq 0$. That is $$\operatorname{ev}_x(f_1)\neq 0$$
Finally, pick $f:=f_1$. $\blacksquare$

Now, with this lemma, suppose that $\textsf V$ is finite-dimensional and we will prove that $\operatorname{ev}_x$ is an injective map from $\textsf V$ to $\textsf{V}^{**}$.

Theorem : Let $\textsf V$ a finite-dimensional vector space and define the function $\phi : \textsf{V} \to \textsf{V}^{**}$ by $$\phi (x)=\operatorname{ev}_x$$
Then, $\phi$ is an isomorphism.

Proof : First, we see that $\phi$ is linear :

Let $x,y\in \textsf V$ and $a$ an arbitrary scalar. For any $f\in \textsf{V}^*$ we have

$$\phi(x+y)(f)=f(x+y)=f(x)+f(y)=\phi(x)(f)+\phi(y)(f)=\left( \phi(x)+\phi(y)\right)(f)$$

and $$\phi(ax)(f)=f(ax)=af(x)=\left(a\phi(x)\right) (f)$$
So, $\phi$ is linear. Next, $\phi$ is one-to-one :

Suppose $\phi(x)$ is the zero functional on $\textsf{V}^*$ for some $x\in \textsf V$ (it is the same as saying that $\phi(x)$ is the zero vector on $\textsf{V}^{**}$). Then $\phi(x)(f)=\operatorname{ev}_x(f)=0$ for all $f\in \textsf{V}^*$. By the previous lemma we conclude that $x=0$.
That is $\phi$ is injective.

And since $\textsf V$ and $\textsf{V}^{**}$ has the same dimension, $\phi$ is also surjective. $\blacksquare$


Functions sequence and lebesgue measure I'm stuck on a problem which seems quite difficult.
I need a functions sequence $(f_n)$ (each fn is continuous) defined in $[0,1] ×[0,1] $ such that $(f_n)$ doesn't converge puntually in any point but
$$\int fn\ \rightarrow 0$$
If anyone can help me I would be thankful. Any idea can be helpful.

It should work if you first try to find a decent discontinuous example for a function defined in $[0,1]$.
What you can do is create a sequence of subsets of $[0,1]$ that exists of intervals of decreasing length that repeatedly cover up the whole interval.
For example intervals of the form $[\frac{k-1}{m},\frac{k}{m}]$ for $m=1,2,3...$ and $k=1,2,...,m$ should do the job if you numerate them like $I_1=[0,1]$, $I_2=[0,\frac{1}{2}]$, $I_3=[\frac{1}{2},1]$, $I_4=[0,\frac{1}{3}]$, $I_5=[\frac{1}{3},\frac{2}{3}]$ and so on.
If you now take the squence of characteristic functions $\chi_{I_n}$ the sequence is not pointwise convergent for any point on the Interval $[0,1]$ but satisfies $\lim\limits_{n\to\infty}\int_{[0,1]}\chi_{I_n}\mathrm{d}\mathcal{L}^1=0$.
You can then try to make the functions continuous by defining them on a ball slightly greater than the original interval for example $\tilde{I}_n:=[\min(I_n)-\frac{1}{n}, \max(I_n)+\frac{1}{n}]$.
Now you define
\begin{equation}
f_n(x):=\begin{cases} 1 & \text{ if } x\in I_n\\
0 & \text{ if } x\in\tilde{I}_n^c\\
nx-n\min(\tilde{I}_n) & \text{ if } x\in[\min(\tilde{I}_n),\min(\tilde{I}_n)+\frac{1}{n}]\\
-nx+n\max(\tilde{I}_n) & \text{ if } x\in[\max(\tilde{I}_n)-\frac{1}{n}, \max(\tilde{I}_n)]
 \end{cases}
\end{equation}
Now the only thing left to do is take $\tilde{f}_n:=f_n\big|_{[0,1]}$ and you are done.
Now to get a sequence of functions defined on $[0,1]^2$ you take $g_n(x,y):=\tilde{f}_n(x)\chi_{[0,1]}$.
This sequence should then satisfy all your wanted conditions.
I'd recommend drawing the functions if you have problems picturing them.
The idea is that you get trapezoid-like shapes that have a volumes that converge to $0$.
There is also probably a simpler way to do this but this is at least what I could come up with so let me know if it helped.
I also just assumed that you where looking for positive functions.

Choose $f_n(x, y) := \sqrt{n}\cos(nx)$. Then, using Fubini:
$$
\int_{[0, 1] \times [0, 1]} f_n(x, y)~\mathrm{d}(x, y) = \int^1_0 \int^1_0 \cos(nx)~\mathrm{d}y~\mathrm{d}x = \int^1_0  \cos(nx)~\mathrm{d}x = \frac{1}{\sqrt{n}}\sin(n) \rightarrow 0
$$
But note that for all $(x, y) \in [0, 1] \times [0, 1]$, the sequence $( \sqrt{n}\cos(nx))_{n \in \mathbb{N}}$ diverges
.


Show that if $g \circ f$ is injective, then so is $f$. The Problem:

Let $X, Y, Z$ be sets and $f: X \to Y, g:Y \to Z$ be functions.

(a) Show that if $g \circ f$ is injective, then so is $f$.

(b) If $g \circ f$ is surjective, must $g$ be surjective?

Where I Am:

So, I really have trouble with these, for some reason. I can draw pictures and make sense of the problems, but writing down proofs is very difficult for me.

Basically, for (a), I ended up with some complicated statement involving an implication implying another implication and then tried to derive a contradiction. It just got so convoluted that I couldn't make sense of it anymore, and I know there's a quick, elegant way to show it.

For (b), I know that $g$ need not be surjective. Once again, though, proving it directly from definitions has given me a bit of a headache.

Any help here would be appreciated. Thanks in advance.

The Proofs!

Ok, I did it. Thanks for the help, everyone! Let me know if there's anything wrong with these proofs, or if they could bet any better.

(a) Suppose $f$ is not injective. Then
$$ f(x_1)=f(x_2) \implies x_1 \ne x_2 \text{        }(*).$$
Let $f(x_1)=y_0=f(x_2)$ and let $g(y_0)=z_0$. Then $$ (g \circ f)(x_1) = (g \circ f)(x_2) = g(y_0) = z_0. $$
Since $g \circ f$ is injective,
$$ (g \circ f)(x_1) = (g \circ f_2) \implies x_1 = x_2. $$
However, this contradicts $(*)$. Therefore, $f$ must be injective.

(b) Suppose $g$ is not surjective. Then

$$ \forall y \in Y, \exists z \in Z \text{ such that } g(y) \ne z \text{         }(**).$$

Since, $g \circ f$ is surjective,

$$ \forall z \in Z, \exists x \in X \text{ such that } g(f(x)) = z \text{       } (***). $$

Let $f(x) = y$. Then,

$$ g(f(x)) = g(y) = z. $$

Because of $(***)$, this is true for all $z \in Z$, which contradicts $(**)$. Therefore, $g$ must be surjective.

(a) Let $f(x_1)=f(x_2)$. Then, $g(f(x_1))=g(f(x_2))$, but since $g\circ f$ is injective...

(b) $g(Y)\supseteq (g\circ f)(X)=g(f(X))$. Hence, if $(g\circ f)(X)=Z$...

(a) Proof:

Since the composite function $g\circ f$ is injective, we have $(g\circ f)(x_{1})=(g\circ f)(x_{2})\implies x_{1}=x_{2}.$ Thus, it follows that $f$ must be injective; otherwise, it would contradict the fact that $(g\circ f)$ is. $\square$

(b) Proof:

If $g \circ f$ is surjective, $\forall z\in Z,\exists x\in X : (g\circ f)(x)=z.$ Let $y=f(x)$, so $y\in Y$ (since $f:X\to Y$). Then, $g(y)=g(f(x))=(g\circ f)(x)=z$, thus, $g$ is surjective. $\square$


"Normalize" values to sum 1 but keeping their weights I am not really sure what this operation might be called, but I have some numbers, for example:

40

10

I need to format these numbers so that they form the sum 1, but they should keep their "weight".

In this specific case

40 would become 0.80, and 10 would become 0.2

But if I have more values (like 40, 10, 25, 5 for example), I am really lost because I don't know the formula.

If anybody can help, could he please reply in words (for example: "Sum up all values then divide through...", and not in a formula? I am really not good at reading formulas at all.

Thank you so much!

Why not just divide each number in your sample by the sum of all the numbers in your sample?

Just divide all the numbers through by 50. Should normalise it to the scale you want 0 - 1.
For example:
40/50 = 0.8
25/50 = 0.5
5/50 = 0.1

etc.

I think you get the idea!


Homotopy equivalence an retractions I have some questions about homotopy. Before starting here a definition:

A topological space $X$ is called contractible if $X$ is homotopy equivalent with a one-point-space
Suppose $X$ a toplogical space, $Y\subset X$. A retraction $r:X\rightarrow Y$ is a continu function such that $r_{|Y}=id_Y$, thus r(y)=y $\forall y\in Y$ and $f(x)\in Y\ \forall x\in X$

Now the questions:

Suppose $X\equiv X'$ and $Y\equiv Y'$ homotopy equivalences, prove that $X\times Y\equiv X'\times Y'$
Prove that every star-shaped subspace $X\subset\Bbb{R^n}$ is contractible
$id_{S^{n-1}}$ not homotopic to constant map (thus $S^{n-1}$ not contractible) $\Longrightarrow$ non-existence of retractions $D^n\rightarrow S^{n-1}$
Brouwers Fix Point Theorem $\Longrightarrow$ Poincare-Miranda Theorem

I have no idea how to solve the questions and implications ... Can someone maybe help me?
Thank you :)

(1)If X,Y is homotopy equivalent to X',Y', then there exists $f_1(f_2)$ from X to X' (X' to X), so that the compositions of them are equivalent to identity. And there exists $g_1(g_2)$ from Y to Y' (Y' to Y), and the compositions of these two maps are equivalent to identity. You just check that $f_1×g_1$:X×Y→X ′ ×Y ′ , $f_1×g_1$(x,y)=($f_1(x)$,$g_1(y)$) is a homotopy inverse of  $f_2×g_2$.
(2)X is star-shaped, thus there is a point x of X, such for any other point x' in X, tx+(1-t)x' is also in X for any $0<t<1$. And the map I×X→X,F(t,x')=tx+(1-t)x' is a homotopy from the identity map on X and the constant map on X.
(3)If there is a retraction $D^n →Sn−1$ , then because $D^n$ is contractible, you would have a contration of S^n-1(first inclusion into D^n, then composite to the contraction homotopy of D^n, finally retract to S^n-1).


Proving an injective map exists from $V$ to $W \iff$ dim $V$≤ dim$W$ I'm reading Axler's Linear Alg Done Right and working through problems. For this one, I proved the forward direction already using Fundamental Theorem of Linear Maps.

And for the backwards direction I assumed dim $V$≤ dim $W$. Let $v_1,\ldots, v_n$ be a basis for $V$ and  $w_1,\ldots, w_n$ be a basis for $W$.  Then $T(a_1v_1 \ldots a_nv_n)=(a_1w_1\ldots a_mw_m)$.

I know I first I have to prove this is indeed a linear map, which I can do. Then I prove this is injective. I know an injective map null($T$)=$0$.

I'm stuck on  proving this map is injective.  Any help/guidance is appreciated.

The map $T$ is defined by
$$
T(a_1v_1+a_2v_2+\dots+a_nv_n)=a_1w_1+a_2w_2+\dots+a_nw_n
$$
(you forgot the + signs). It is injective by the fundamental theorem, because the image is spanned by $\{w_1,\dots,w_n\}$, which is linearly independent. Thus the image has dimension $n$.

1) The statement is wrong because there are no restriction on map exept injectivity. So we can find injective map from $\mathbb R^2$ to  $\mathbb R$ because they are have the same cardinality.

2) The statement will be true if a linear injective mapping is implied


Show that $\int_0^1 h(t)dt\geq\left(\int_0^1 f(t)dt \right)^a\left(\int_0^1 g(t)dt \right)^{1-a}$ Let $ a \in (0,1)$ and $f,g,h :[0,1]\to (0,\infty)$ be continuous functions satisfying
$h(ax+(1-a)y)\geq f(x)^ag(y)^{1-a}$ for all $x,y \in [0,1]$.
Prove that
$$ \int_0^1 h(t)dt\geq\left(\int_0^1 f(t)dt  \right)^a\left(\int_0^1 g(t)dt  \right)^{1-a} $$
I tried integrating left side with respect to $x$ and $y$ separately, but it didn't work.
Edit: Number '$a$' is chosen only once, the condition does not have to hold for values of '$a$' other then the one we chose.

This is the Prékopa–Leindler inequality (see e.g. here). Also here
Edit:
The proof that can be found in the paper of Gardner, in the setting of the OP, is at the level of a Calculus class.
Let us consider the functions
$$
F(u) := \int_0^u f(x)\, dx,
\qquad
G(u) := \int_0^u g(x)\, dx,
\qquad u \in [0,1].
$$
Since $f,g$ are continuous and strictly positive functions, we have that $F$ and $G$ are continuously differentiable and strictly increasing, with $F' = f$ and $G' = g$.
If we set $F_1 := F(1)$, $G_1 := G(1)$, then $F$ is a bijection from $[0,1]$ to $[0, F_1]$ and $G$ is a bijection from $[0,1]$ to $[0, G_1]$.
Let $u,v\colon [0,1] \to [0,1]$ be the functions defined by
$$
u(t) := F^{-1}(F_1\, t),
\qquad
v(t) := G^{-1}(G_1\, t),
\qquad t\in [0,1].
$$
Since $u'(t) = F_1 / f(u(t))$ and $v'(t) = G_1 / g(v(t))$ for every $t\in [0,1]$, using the AM-GM inequality we deduce that
$$
w'(t) := a\, u'(t) + (1-a)\, v'(t)
\geq [u'(t)]^a [v'(t)]^{1-a} =
\frac{F_1^a}{f(u(t))^a}\cdot \frac{G_1^{1-a}}{g(v(t))^{1-a}}
$$
and finally
$$
\int_0^1 h(x)\, dx = \int_0^1 h(w(t))\, w'(t)\, dt
\geq \int_0^1 f(u(t))^a g(v(t))^{1-a} \frac{F_1^a}{f(u(t))^a}\cdot \frac{G_1^{1-a}}{g(v(t))^{1-a}}\, dt
= F_1^a G_1^{1-a}.
$$

We assume here that the inequality (1) below holds for all $a \in\, ]0,1[$. And we prove that conclusion is valid for all $a \in\, ]0,1[$.
Observe that $f,g>0$ on $[0,1]$ implies that, for each fixed $x,y\in [0,1]$, the
exponential functions $a\mapsto f(x)^{a}$, $a\mapsto g(x)^{1-a}$ are well-defined. Recall that the exponential functions are continuous on $\mathbb{R}$. Now, since $h$ is also continuous on $[0,1]$, then both sides of the inequality
$$h(ax+(1-a)y)\geq f(x)^{a}\,g(y)^{1-a}\qquad \qquad (1)$$
are continuous in $a$. This justifies taking the limit of both sides of (1) as $a\rightarrow 0^{+}$ and as $a\rightarrow 1^{-}$. So, we have that
$$h(x)\geq f(x),\forall x\in[0,1]\Longrightarrow
\int_{0}^{1}h(t)dt\geq \int_{0}^{1}f(t)dt\Longrightarrow
\left(\int_{0}^{1}h(t)dt\right)^{a}\geq \left(\int_{0}^{1}f(t)dt\right)^{a},\qquad \qquad (2)$$
and analogously,
$$h(x)\geq g(x),\forall x\in[0,1]\Longrightarrow
\int_{0}^{1}h(t)dt\geq \int_{0}^{1}g(t)dt\Longrightarrow
\left(\int_{0}^{1}h(t)dt\right)^{1-a}\geq \left(\int_{0}^{1}g(t)dt\right)^{1-a},\qquad \qquad (3)$$
The assertion follows from the inequalities (2) and (3).
Remark, we can not use the implication $\int_{0}^{1}\phi(t)dt\geq \int_{0}^{1}\psi(t)dt\Longrightarrow
\left(\int_{0}^{1}\phi(t)dt\right)^{\alpha}\geq \left(\int_{0}^{1}\psi(t)dt\right)^{\alpha}$
without the assumption
$\int_{0}^{1}\psi(t)dt>0$, and $\alpha >0$.
We here have $\int_{0}^{1}f(t)dt,\int_{0}^{1}g(t)dt>0$
by the assumption that both $f,g$ are positive on $[0,1]$.


$⊢p \land q \to (p\to q)$ - Natural deduction proof confusion I have the following:
$$⊢p \land q \to (p\to q)$$

I'm having a difficult time trying to figure out where to begin. I believe that I am supposed to assume p and q and then somehow use the copy rule to construct the equation, however I am not quite sure.

Can someone help me out?

Attempt taken from a comment: "...this is my attempt... first assume p, assume q, copy p, copy q, introduce →, introduce ∧, then introduce → between both... in that order"

To prove $A \to B$ in a natural deduction system you assume $A$ and derive $B$, and then apply Conditional Proof. OK? That's the standard way of proving a conditional.

To prove $(p \land q) \to (p \to q)$ in a natural deduction system, you need to assume $(p \land q)$ as a temporary assumption and aim for $(p \to q)$.

So assume  $p \land q$. That gives you immediately $p$ and $q$.

Can you now get to $p \to q$?

Well there's a standard trick for getting from $q$ to $A \to q$ for any $A$ at all. Do you know it? The fine details of how you set it out will depend, however, on your exact system. (Basically, in an application of Conditional Proof, we are usually allowed "vacuous discharge" of premisses that were never in fact used. You may need to check out how this is handled in the system you are working with! It will be, so to speak, in the fine print!)

Assuming the conjunction operator has higher precedence than the conditional operator, what needs to be proved is the same as this:

$$⊢(p∧q)→(p→q)$$

Here is a proof:

The OP made the following attempt:

  this is my attempt... first assume p, assume q, copy p, copy q, introduce →, introduce ∧, then introduce → between both... in that order

The above proof uses these steps:

Assume $P∧Q$ as the antecedent of the desired conditional.
Assume $P$ as the antecedent of the conditional in the conclusion.
Derive $Q$ from the first assumption using conjunction elimination.
Discharge the assumption on line 2 by rewriting the subproof in lines 2-3 as $P→Q$ on line 5 with conditional introduction as the justification.
Discharge the assumption on line 1 by rewriting the subproof in lines 1-4 as $(P∧Q)→(P→Q)$ on line 5 with conditional introduction as the justification.

That completes the proof.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

"Operator Precedence" Introduction to Logic http://intrologic.stanford.edu/glossary/operator_precedence.html


Adaptive step size Runge Kutta: getting a specific value A system of differential equations can be solved using a numerical integrator with adaptive stepsize $ h$, e.g. Runge Kutta Fehlberg

Now suppose I want to integrate a set of differential equations $\{\dot{x}, \dot{y}, \dot{z} \}$ numerically (e.g. the 3D trajectory of some particle). Each differential equation uses the same adaptive stepsize $h$.

I want to evaluate $x,y$ at some specific target value of $z = z_t$

Now, due to the adaptive stepsize, it is unlikely that $z$ will ever be exactly $z_t$. That is to say, I might get for integration step $n$ and step $n+1$, $z_n=z_t + \epsilon$, $z_{n+1}=z_t - \delta$ for small values $\epsilon, \delta$

Obviously, by setting a more strict tolerance on the integrator, I can get close to $z_t$, but this would be at the expense of speed.

Given that I know a priori that I want to evaluate $x,y$ at $z=z_t$, is there a method which permits both an adaptive stepsize, and an exact $z=z_t$ result?

Thanks

Once you choose a method, it has an order which approximates how the error depends on stepsize.  The popular fourth order Runge-Kutta has an error that scales as $h^4$.  You can get an approximation to the accuracy by doing your step once as a full step and once as two half steps.  You would expect the error to be reduced by a factor $16$ so the difference between them is about $15$ times the error of the two step approach.  You can use this to update your stepsize.  There is a discussion in chapter 17.2 of Numerical Recipes and probably in many other numerical analysis texts.


Is it possible to represent every irrational number as a (limit of) an infinite sum of rational numbers? For instance, we can certainly represent π in this fashion.

$$
\frac{\pi}{4} \;=\; \sum_{n=0}^\infty \, \frac{(-1)^n}{2n+1} .\!
$$

$\ln(2)$ is also irrational. And even that can be represented as an infinite sum of a sequence of rational numbers:

$$
\ln (1+x) \;=\; \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n.
$$
with $x=1$.

And also, $\sqrt2$:

$$
\sqrt2 \;=\; \sum_{k=0}^\infty\frac{(2k-1)!!}{4^kk!}\tag{2}
$$

I'm curious if this applies to all irrational numbers? Is so, how do you go about proving it?

Since that series
$$
\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}
$$
converges conditionally, the Riemann Rearrangement Theorem says that we can get every real number, rational or irrational, by rearranging the terms of that series

So, yes, every irrational number can be written as the limit of the sum of rational numbers.

The rational numbers are dense in the real numbers.

Therefore every irrational number x greater than 0, there exists a rational number y such that

$$0 < y < x $$

and

$$y + \frac{x-y}{2} + \frac{\frac{x-y}{2}}{2} + \frac{\frac{\frac{x-y}{2}}{2}}{2} ...$$

converges to x.

(This can easily be generalized for x < 0.)

QED


Find all solutions for $e^z = -2$ Find all solutions for $e^z = -2$. And we know there is no solution.

My trivial work:

Since if we take $\ln$ both sides, we simply have $z = \ln(-2)$ which isn't defined.

But, since my work looks very trivial, I want to add more.

Thank you.

By taking the natural log of both sides, one obtains $$ z = \log(2) + i\pi(2n+1)$$ $\forall n \in \mathbb{Z}$
Try to see if you can figure out why this is so.

If it truly is for a complex analysis course the following will be helpful:
$e^z = e^{x+iy} = e^xe^{iy} = e^x(\cos(y) + i\sin(y))$


Defining the probability distribution function Suppose $X$ is a random variable that takes values in $[0,1].$ Is it that the probability distribution of $X$ is defined on $[0,1]$ or whole $R$?

The probability distribution is defined on the whole of $\mathbb{R}$. Also, you need to understand that, this is given by an integral or sum depending on what $X$ is, whose limits are usually taken to be from $-\infty$ to $x \in \mathbb R$.

In general, any function $F$ is a distribution function if it satisfies the following properties:

$0 \leq F(x) \leq 1$ for all $x \in \mathbb{R}$
$F$ is a non-decreasing function of $x$.
$F(-\infty) := \lim_{x \to -\infty}{F(x)}=0$ and $F(\infty) := \lim_{x \to \infty}{F(x)}=1$
$F$ is right continuous, i.e. $F(x+)=F(x)$ for all $x \in \mathbb R$.

And, it can be shown that, given $F$ with above properties, there exists a probability space $(\Omega, \mathscr A, P)$ on which a random variable $X$ exists  such that $X$ has $F$ as its distribution.

There is not a lot of difference.

With a probability space $(\Omega, \mathscr A, P)$, the sample space $\Omega$ must contain all possible events; whether it includes impossible ones too does not matter much so long as you are clear.

If the distribution is defined on the whole of $\mathbb{R}$ but values cannot be less than 0 or more than 1, then you can ignore both of these possibilities.  If you keep them then $\Pr(X\lt 0)=0$ and $\Pr(X\gt 1)=0$.


Can this primality test be optimized to runs in polynomial time? Theorem :
Let $k$  be positive integer such that : $k \in [1,\lfloor \sqrt p \rfloor ]$
$\binom{p}{k} \equiv 0 \pmod p$ for all $k$ iff $p$ is a prime number .

Test written in Maple :

Test is general , unconditional , deterministic but it runs in exponential time .
Can it be optimized to runs in polynomial time ?

As a first optimization you could keep the current value of the binomial each time and update it with a multiplication and a division rather than recomputing at each step.

As a next optimization you might realize that each division is exact, so you can use Jebelean-Krandick bidirectional division.

Then you might realize that each step essentially tests if k divides p, and so replace the binomial calculation with a modulus test.

Then you could run the test over only the primes k up to $\sqrt p$.

So after all these optimizations, you are left with trial division, which runs in exponential time.

You can use dynamic programming I mean instead of compute binomial(p,k) in each loop use this one : temp := temp * (p - k+1) / k;. and at the start of the program set temp to 1

I used this fact : binomial(p,k) = binomial(p,k-1) * (p-k+1) / k.


How to define the $0^0$? Possible Duplicate:
  Zero to zero power

According to Wolfram Alpha:

  $0^0$ is indeterminate.

According to google:
$0^0=1$

According to my calculator: $0^0$ is undefined

Is there consensus regarding $0^0$? And what makes $0^0$ so problematic?

This question will probably be closed as a duplicate, but here is the way I used to explain it to my students:

Since $x^0=1$ for all non-zero $x$, we would like to define $0^0$ to be 1. but ...

since $0^x = 0$ for all positive $x$, we would like to define $0^0$ to be 0.

The end result is that we can't have all the "rules" of indices playing nicely with each other if we decide to chose one of the above options, it might be better if we decided that $0^0$ should just be left as "undefined".

Define $f(x)=x^x$ and compute $\lim_{x\to 0} f(x)$.


Show that the function floor-log is primitive recursive I have been stuck on this problem for a while and I was hoping someone could help me with it. This is for my computer science automata and formal languages class.

Given an integer $b$ greater than or equal to $2$ and a positive integer $x$, let $n$ be the greatest natural number such that be $b^n\le  x$. I basically have to prove that a function floor-log that takes two inputs $b$ and $x$ and outputs $n$ is primitive recursive.

Additionally, Given any input that doesn't meet the preconditions, a value of 0 or 1 for the first input(b), or a value of 0 for $x$ should output 0.

I realize I have to use some combination of bounded sums and raise-to-power function but I can't figure out how to do that.

Here is one method. Given $b> 1$ and $n \geq 2$, start computing $b^0, b^1, b^2, \ldots, b^n$. At some point, you will find a $k \in \{0, \ldots, n\}$ such that $b^k \leq n$ and $b^{k+1} > n$. Then return $k$.

The underlying mathematical fact is that $0 \leq \operatorname{floorlog}(b,n) < n$, so we can use $n$ itself as a convenient bound for a bounded search.

The operation I described is a bounded search (bounded by $n$), and the test at each stage is primitive recursive, so the overall function is primitive recursive. In particular, for $b > 1$ and $n > 2$, we can define the function using the bounded $\mu$ operator and a primitive recursive predicate:
$$
\operatorname{floorlog}(b,n) = (\mu. k < n) [ b^k \leq n \land b^{k+1} > n].
$$

To prove that a certain function is primitive recursive, as you probably know, you need to be able to describe the function using only the basic primitive recursive functions.

This gives us a definition that could look like:
$$\mathrm{floorlog}(b,x) = \left\{\begin{array}{ll} \mathrm{null}(x) & b = 0 \\ g(b-1, \mathrm{floorlog}(b-1, x), x) & b > 0 \end{array}\right.$$
NOTE: $\mathrm{null}(x) = 0$.
$$g(k,l,m) = \left\{\begin{array}{ll} \mathrm{succ}(\mathrm{null}(\mathrm{proj}^1_2(l,m))) & k = 0 \\ h(k-1, g(k-1,l,m), l, m) & k > 0 \end{array}\right.$$
NOTE: $\mathrm{succ}(\mathrm{null}(\mathrm{proj}^1_2(l,m))) = 1$.
$$ h(q,r,s,t) = f(proj^4_4(q,r,s,t), proj^3_4(q,r,s,t), \mathrm{succ}(\mathrm{succ}(proj^1_4(q,r,s,t))))$$
NOTE: this isn't anything else as $h(q,r,s,t) = f(t, s, q + 2)$.
$$f(x,y,z) = \left\{\begin{array}{ll} \mathrm{null}(\mathrm{proj}^1_2(y,z)) & x = 0 \\ etc\ldots & x > 0 \end{array}\right.$$

then you just need to check whether $b^n > x$ and increment the $n$ accordingly. The point I'm trying to make is that proving this using the given definitions is quite a lot of work (and I must admit I stopped trying because I didn't see it anymore for a second).

Another way to prove this is to write a program that uses only a clean for-loop. With clean, I mean that you don't mess with the loop variables (because actually you're writing a while-loop in disguise then). I suppose this shouldn't be too much of a problem now...

Hope this helps


Showing product of two transversal maps is a transversal Let $X_1,X_2,M$ be finite dimensional manifolds. Consider maps $f_i:X_i\to M$ such that $f_1,f_2$ are transversal with respect to each other, i.e, for $x_1\in X_1, \ x_2\in Y_1$ with $f_1(x_1)=f_2(x_2)=m$, we have $T_{x_1}f_1(T_{x_1}X_1)+T_{x_2}(T_{x_2}X_2)=T_{m}(T_{m}M)$. Show that $f_1,f_2$ are transversal if and only if $f_1\times f_2: X_1\times X_2\to M\times M$ is a transversal to $\Delta_M=\{(m_1,m_2)\in M\times M: m_1=m_2\}$.


Pointwise convergence in limit of conditional probabilities implies almost sure convergence? Consider two sequences of real valued random variables $\{X_n\}_{n\in \mathbb{N}}$ and $\{Y_n\}_{n\in \mathbb{N}}$, with, $\forall n \in \mathbb{N}$, $X_n$ and $Y_n$ defined on the probability space $(\Omega, \mathcal{F}, P), $ respectively with support $\mathcal{X}, \mathcal{Y}$. Consider a sequence of real numbers $\{a_n\}_{n\in \mathbb{N}}$.

Assume that for some $b\in \mathbb{R}$
$$
\lim_{n\rightarrow \infty} | \overbrace{P(X_n\leq b| Y_n=y)}^{\text{scalar}}-a_n|=0 \text{ }\text{ $\forall y \in \mathcal{Y}$}
$$

Does this imply
$$
\overbrace{P(X_n\leq b| Y_n)}^{\text{Random variable because of $Y_n$ (unless $X_n\perp Y_n$)}}-a_n \rightarrow_{a.s.}0 \text{ as $n\rightarrow \infty$}
$$
?

Given:

$$\forall y \in \mathcal Y, \lim_n \frac{E[1_{B_n} 1_{Y_n=y}]}{P(Y_n=y)}= \lim_n a_n$$

Conjecture:

$$\lim_n E[1_{B_n}| Y_n]= \lim_n a_n$$

Ugh, the fact that we can say '$Y_n=y$' means $Y_n$ is discrete? In that case:

$$E[1_{B_n}| Y_n] = \sum_{y \in \mathcal Y} E[1_{B_n}| Y_n = y]1_{\{Y_n=y\}}$$

$$\to \lim E[1_{B_n}| Y_n] = \lim \sum_{y \in \mathcal Y} E[1_{B_n}| Y_n = y]1_{\{Y_n=y\}}$$

$$= \sum_{y \in \mathcal Y} \lim E[1_{B_n}| Y_n = y]1_{\{Y_n=y\}}$$

$$ = \sum_{y \in \mathcal Y} [\lim E[1_{B_n}| Y_n = y]][\lim[1_{\{Y_n=y\}}]]$$

$$ = \sum_{y \in \mathcal Y} [\lim a_n] \lim[1_{\{Y_n=y\}}]$$

$$ = [\lim a_n] \sum_{y \in \mathcal Y} \lim[1_{\{Y_n=y\}}]$$

Well if I proved for the discrete case, then it would not have been sufficient to disprove as a whole. However, I disproved for the discrete case which is sufficient to disprove as a whole.

Observation: In the discrete case, this sounds like:

$$\lim_n b_{n,k} = \lim_n a_n \to \lim_n \sum_{k} b_{n,k}c_k = \lim_n a_n$$


Resources for self-study of general topology Could you point me to good resources for self-study of general topology. I want to learn the basics and how to prove theorems about structures like polyhedra by myself.

A great resource for self-study in Topology is James Munkres' Topology. You can "preview" the text and it contents at the link given. It's divided into two sections, the second being algebraic topology.

You'll also get an overview of Topology (Wikipedia), its branches and the topics related to those branches/subfields, to better hone in on your area of interest. There are a number of references cited, some with links, along with helpful resources that may aid in your self-study.

A good follow-up to Munkres, and more in the direction of your interests, I presume, is then (of course) Algebraic Topology by Allen Hatcher (freely and legally available online, courtesy of the author! The link takes you to Hatcher's home page, where you can access his text, as well as supplementary resources, e.g. notes in point-set-topology.) You might want to peruse Hatcher's text to get a better idea what preliminary material you should cover in Munkres; that might cut down on the amount of Munkres you'd want to cover.

Another sound intro to Topology is A First Course in Topology by J. McCleary (again, you can preview the text at the given link).

Introduction to Topology: Third Edition (Dover Books on Mathematics) Bert Mendelson


Evaluate this limit of integration in Complex analysis I am trying assignments of Complex analysis and I couldn't evaluate this particular integral.

Find the limit $\int_{|z|=R} |\sin(z)/z | |dz| $ as $R\to\infty$.

I am unable to see which result I should use and also there is confusion due to dz being in modulus and also $\sin(z) /z$.
Can you please tell significance of modulus here and how to compute the limit.
Thanks!!

First note that
$$
 \int_{|z|=R} \frac{|\sin(z)|}{|z|} |dz| = \int_0^{2 \pi} \frac{|\sin(R e^{it})|}{|R e^{it}|} R \, dt = \int_0^{2 \pi} |\sin(R e^{it})| \, dt \, .
$$
Generally, for $z= x+iy$,
$$
 |\sin(z)|^2 = \sin^2(x) + \sinh^2(y) \implies |\sin(z)| \ge |\sinh(y)| \, .
$$
If we restrict the integration to the arc $z = R e^{it}$ with $\pi/2 \le t \le 3\pi /2$ then
$$
|\sin(R e^{it})| \ge \sinh(R \sin(t)) \ge \sinh(\frac{R}{\sqrt 2}) \, .
$$
(This is what @reuns meant with the comment that “$|\sin(z)|$ is large near $iR$”).
It follows that
$$
\int_0^{2 \pi} |\sin(R e^{it})| \, dt \ge
\int_{\pi/2}^{3 \pi/2} |\sin(R e^{it})| \, dt \\
\ge \int_{\pi/2}^{3 \pi/2} \sinh(\frac{R}{\sqrt 2}) \, dt = \frac \pi 2 \sinh(\frac{R}{\sqrt 2}) \, ,
$$
and that grows to infinity for $R \to \infty$.


For what $n$ can $\pm 1\pm 2\pm 3 ... \pm (n-1) \pm n = n+1$? More explicitly,
for what values of $n$
can the signs be chosen
in the equation
$$\pm 1\pm 2\pm 3 ... \pm (n-1) \pm n = n+1$$
so the equation is true?

For all $n$,
if the equation cannot be satisfied,
prove it;
if the equation can be satisfied,
exhibit a solution.

Note:
I have a solution,
and will present it in two days
if no one else submits a similar one.

Equivalently, we want to divide the set $\{1,2,3,\dots, n+1\}$ into two parts, such that the sums of the two parts are the same. The $n+1$ is somewhat unattractive, so let us call it $m$.

A necessary condition is that the sum $1+2+3+\cdots+m=\frac{m(m+1)}{2}$ is even. We show that condition this is also sufficient.

Our necessary condition is met precisely when $m$ is of the form $m=4k$ or of the form $m=4k-1$.

Note that $(4j+1)+(4j+4)=(4j+2)+(4j+3)$. That takes care of all cases where $m$ is of the form $4k$. For any group $\{4j+1,4j+2,4j+3,4j+4\}$ can be divided into $2$ parts with equal sum.

For $m$ of the form $4k-1$, think pf our sum as starting at $0$, and use as groups $\{4j,4j+1,4j+2,4j+3\}$, dividing each group by using the fact that $4j+(4j+3)=(4j+1)+(4j+2)$.

Remark: I feel guilty about these $j$ and $k$ and $m$ and $n$. It all comes down to two sample cases:
$$(1-2-3+4)+(5-6-7+8)+(9-10-11+12)=0,$$
and
$$(0-1-2+3)+(4-5-6+7)+(8-9-10+11)=0.$$

all n such that $n\equiv 3 \mod4$ can satisfy this.  and $\forall n$ that satisfy it $\exists k\in\mathbb{N}$ s.t. $n+4*k$ also satisfies it.  I say that because I am not 100% certain $n\equiv3 \mod 4$ is the entire set of solutions.

$\pm1\pm2...\pm n =n+1 $ gives us $\pm1\pm2...\pm(n+1)=0$
by induction if $n-3$ satisfies the second equation
$(n-2)-(n-1)-n+(n+1)=0 \implies
[\pm1\pm2...\pm(n-3)]+[\pm(n-2)\pm(n-1)\pm n\pm(n+1)]=0$


$\int {x^{1/2}\over {x^{1/2}} - 3} dx$ The given problem is:
\begin{aligned}
\int   {x^{1/2}\over {x^{1/2}} - 3} dx
\end{aligned}

The textbook (Larson, Edwards) 9th edition suggests to use u-substitution and let u be the denominator.

So, I made $u =x^{1/2} -3$. So, $u+3 = x^{1/2}$ Thus, du is $\frac{1}{2\sqrt{x}}$. So, $2x^{1/2} du = dx$, and that follow $2(u+3) du = dx$.

So, I then put the original integral in terms of u:
\begin{aligned}
2\int   {u+3{}\over {u}} du
\end{aligned}

When I carry out the integration, I do not get the correct answer. I noticed something was a bit fishy when I moved the u up and expanded to get $u^0 + 3u^{-1}$ which seemed a bit odd for this problem. I expect something to be wrong with my numerator when rewriting the function, but I cannot seem to find any errors.

Any help would be appreciated.

You have to substitute for the $\sqrt x-3$ in the denominator, the $\sqrt x$ in the numerator, and the $dx$. I think you only did two of the three.

Your problem appears to be in changing over from x to u.

$$\int \dfrac{x^\frac12}{x^\frac12-3}dx=2\int\dfrac{x}{x^\frac12-3}\left(\dfrac{x^{-\frac12}dx}2 \right)=$$

$$\int\dfrac{(u+3)^2}{u}du$$


Regression with error coming from rounding I am looking at the following model:

$c$ is a  fixed vector in $\mathbb{R}_+^n$ and for any $x \in \mathbb{R}_+^n$
we obtain a value $y =[c^Tx]$, i.e. rounding $c^Tx$ to the nearest integer.

I want to determine c based on observations of $x_1, \ldots ,x_k$ and $y_1, \ldots, y_k$. (Before EDIT $x_1, \ldots ,x_n$ and  $y_1 \ldots, y_n$.)

Of course I could simply do a least squares, but I do not know how to get a good
measure of error then.
I think one could somehow exploit the fact that error is not really random but comes from rounding.
And I think it should be possible to measure of quality of the resulting vector $c$.

So here is the question:

Do you know a method to determine $c$ together with a bound on the error in $c$, i.e. an
interval for each coefficient of $c$ where the true value is guaranteed to be contained.
(Of course I'd like to know methods which small error bounds)?

EDIT: I had a typo fixed, namely I may have many more observations $k$ than the dimension of the vectors $n$. My hope is to be able to get better bounds by increasing $k$.

There are several methods to approach the problem of adjusting estimated regression coefficients for rounding errors in the data. Most of these techniques perform an adjustment to the main diagonal of the sample covariance matrix of the variables. One of the most commonly used method is the so-called Sheppard's correction. This is based on a Taylor expansion of the likelyhood function, which finally leads to the subtraction of a quantity (depending on the intervals used in rounding) in the diagonal of the covariance matrix.

To apply the correction: let us consider a regression model with $M$ independent variables and $N$ observations given by

$$y_k=\alpha +\beta_1x_{1k} + \beta_2x_{2k} .... + \beta_M x_{M k} +\epsilon_k$$

with $k=1,2,3.... N$, and where the errors $\epsilon_k$ are normally distributed.  The uncorrected least square estimate of the regression coefficients is given by $\displaystyle \hat \beta=(SS_{xy})(SS_{xx})^{-1}$, where  $S_{xx}$   represents  the $M\times M$ covariance matrix of the independent variables, $S_{xy}$ is the $M\times 1$ matrix relating independent variables with the dependent variable, and $\hat \beta=(\hat \beta_1, \hat \beta_2, \hat \beta_3... \hat \beta_k)^T$. In $SS_{xx}$, any term in the row $i$ and column $j$ of the matrix is given by $\displaystyle s_{ij}=\frac{1}{N} \sum_{k=1}^N x_{ik}x_{jk}$, whereas in $SS_{xy}$ the terms are given by
$\displaystyle r_{i}=\frac{1}{N} \sum_{k=1}^N x_{ik}y_k$.

If we have a rounding error, each of the observations $x_{ik}$ can be rewritten as  $\displaystyle x_{ik}=\mathring {x}_{ik}+z_{ik}$, where $\displaystyle \mathring x_{ik}$ and $\displaystyle z_{ik}$ are the rounded value and the corresponding rounding error referring to the $\displaystyle k^{th}$ item of the $\displaystyle i^{th}$ independent variable, respectively. Similarly, each of the observations $\displaystyle y_{k}$ can be rewritten as $\displaystyle y_{k}=\mathring {y}_{k}+z_{0k}$, where $\displaystyle \mathring y_{k}$ and $\displaystyle z_{0k}$ are the rounded value and the corresponding rounding error of the $\displaystyle k^{th}$ item of the dependent variable. Let us consider the covariance matrix of the independent variables $\displaystyle \mathring {s}_{ij}=\frac{1}{N} \sum_{k=1}^N \mathring {x}_{ik}{x}_{jk}$, obtained using the rounded data. Calling $\displaystyle w_i$ the bin interval  widths used for rounding variables, if we assume that rounding errors are independent of the rounded data and uniformly distributed over the symmetric interval of rounding precision $\displaystyle (-\frac{w_i}{2}, \frac{w_i}{2})$, we get that the correct terms  $\displaystyle s_{ij}$ of the covariance matrix can be estimated from $\displaystyle \mathring {s}_{ij}$ as

$$\displaystyle s_{ij}=\frac{1}{N} \sum_{k=1}^N \mathring {x}_{ik}    \mathring {x}_{jk} +    x_{ik}    z_{jk} +  x_{jk}   z_{ik} +   z_{ik}   z_{jk}$$

which leads to

$$s_{ij}= \left\{ \begin{array}{rl}
 \mathring s_{ij} -\frac{1}{12}w_i^2 + \mathcal{O}(1) &\mbox{ if $i=j$} \\
 \mathring s_{ij} + \mathcal{O}(1) &\mbox{ otherwise}
\end{array} \right.$$

Applying the same procedure to the matrix of dependent vs independent variables, we get

$$\displaystyle r_i=\mathring r_i + \mathcal{O}(1)$$

where $\displaystyle \mathring r_i=\frac{1}{N} \sum_{k=1}^N \mathring {y}_{k}{x}_{ik}$ is the covariance matrix of dependent vs independent variables obtained using the rounded data.

The formulas above allow to obtain the correction for rounding  directly from rounded data. In practice, to perform your analysis:

build an initial covariance matrix $\displaystyle SS'_{xx}$ for the independent variables using rounded data, where each term $\displaystyle \mathring s_{ij}$  is given by $\displaystyle \frac{1}{N} \sum_{k=1}^N x_{ik}x_{jk}$;
build the covariance matrix $\displaystyle SS_{xy}$ for the dependent vs independent variables using rounded data, where each term $\displaystyle \mathring {r}_{i}$  is given by $\displaystyle \frac{1}{N} \sum_{k=1}^N y_{k}x_{ik}$;
using the Sheppard's corrections, modify the initial matrix $\displaystyle SS'_{xx}$ by subtracting $\displaystyle \frac{1}{12}w^2$ to each term of the main diagonal, so that you  get the corrected matrix $\displaystyle SS_{xx}$;
calculate regression coefficients using the standard method $\displaystyle \hat \beta=\frac{SS_{xy}}{SS_{xx}}$, as in ordinary least square regression.

Since this correction simply implies the use of a corrected matrix, we can  estimate the standard error of each regression coefficient by the usual procedure followed in ordinary LSE. The possibility of calculating a measure of  error for the regression parameters may be important to
measure of quality of the resulting vector, and to obtain an interval for each coefficient  where it is  highly probable that the true value is contained (as correctly pointed out in the question). In this regard,  we have to remind that, in linear regression, parameter estimates are normally distributed with mean equal to the true regression parameter and covariance matrix given by $\displaystyle \Sigma =  {s^2}{A'A}^{-1}$, where $\displaystyle s^2$ represents the residual variance and $\displaystyle A'A$ is the design matrix ($\displaystyle A'$ is the transpose of $A$, and $A$ is defined by the regression model equation). Using this formula, the standard error of each regression coefficient can be calculated, and from this we can easily get the corresponding 95% confidence intervals.

Note that, as already highlighted, other methods can be used. For example, if we do not assume that rounding errors are independent of rounded data, then the same calculations as above lead to an alternative correction (the so-called BRB correction) that is similar to the Sheppard's one, but where the term $\frac{1}{12}w^2$ has to be added, and not subtracted, to the main diagonal of the $SS{xx}$ matrix.  However, current evidences suggest that the Sheppard's correction performs better than other techniques. Moreover, it has to be pointed out that, whatever the method used, adjustments often lead to considerably larger 95% CI  for regression coefficients as compared with those obtainable by unrounded data. Lastly, while the sampling standard deviations decreases  in proportion with $\sqrt{N}$  when the sample size $N$ increases, the rounding error adjustment errors usually do not decrease when $N$ increases.

for a limited number of levels of y , try using multinomial probit. the idea is that for n-0.5 < y < n+0.5, you see n. Treat this as a categorical dependent variable.

hope this helps!


How to prove that a revursive sequence is convergent or not, and if it is, how to find its limit? Suppose $a_1 = 2$, and
  $$ a_{n+1} = \frac{1}{2}\left(a_n + \frac{2}{a_n}\right) $$
  if $n \ge 1$.
  Is the sequence convergent or divergent?
  If it's convergent, find
  $$\lim_{n\rightarrow\infty}{a_n}$$

Usually, what's the approach we use for recursive sequence like this?
A friend of mine told me that any recursive sequence is divergent, but after I checked on calculator, it seems that the series is approaching one particular value.

First of all, Welcome to the site !

If the limit exists, it is given by
$$L = \frac{1}{2}\left(L + \frac{2}{L}\right)\implies L=\pm \sqrt 2$$

Now, your turn !


Is the solution from wolfram alpha for $x^4+bx^3+f=0$ only wrong for $b=0$? edit, tldr: it seems like the question is interpreted as "for which combinations of a and f is there a solution to the expression"? I'm only interested in "for which combinations of a and f is the wolfram formula wrong". When the wolfram formula takes square roots of negative numbers or divides by zero, I consider it still correct if there was no solution anyway. But it has divisions by zero for b=0 and f=-1, while there should be solutions then. Does that also hold for some nonzero values of b?

For my hobby computer science project, I need to solve $x$ for $x^4+bx^3+f=0$. ($x, b, f\in \mathbb{R}$). I put the formula in wolfram alpha:

http://www.wolframalpha.com/input/?i=solve+x%5E4+%2B+b+x%5E3+%2B+f+%3D+0+for+x

and implemented the resulting four formula's from that page in Java. This works pretty well, for random values of $a$ and $f$ it always generates a good approximation of the solution set. However, for the "non-random" input {$b=0, f=-1$}, the wolfram formula (and hence my program) fail. It should have two solutions, $x=1$ and $x=-1$, but it gives none. I took a close look at the formula. Each of the four solutions has the same part $\frac{b^3}{\psi}$ (note the 3). When substituting $b=0$ and $f=-1$ into $\psi$, you'll get the formula here (removing the 4 and the square root):

http://www.wolframalpha.com/input/?i=cbrt(sqrt(3)+sqrt(256))+%2F+(cbrt(2)+3%5E(2%2F3))+%2B+4+(cbrt(-2%2F3))+%2F+cbrt(sqrt(3)+sqrt(256))

Indeed, $\psi=0$ and $\frac{b^3}{\psi}$ is a division by zero.

Is the solution given by wolfram wrong (1st link)?
I can easily solve this in my program by checking if $b=0$. But is that enough or are there other combinations of $b$ and $f$ where this formula goes wrong?

The formulas given by Wolfram Alpha are not wrong, they're "generic".  (Which is a very specific form of wrong which means "only wrong on a set so small that you shouldn't ever care if you're picking random numbers, but always contains the point your boss randomly asks about".  Okay.  Not really, but some days...)

The not generic versions can be generated via Mathematica's Reduce[] function because Reduce[], unlike Solve[] and Integrate[], does not give generic solutions.

Simplify[
  Reduce[x^4 + b x^3 + f == 0, x, Reals],
  TransformationFunctions -> {Automatic, ToRadicals}
]

(*  (b < 0 && (
      (f == 0 && (x == Root[b #1^3 + #1^4 &, 1] || x == Root[b #1^3 + #1^4 &, 4])) ||
      ((0 < f < (27 b^4)/256 || f < 0) && (x == 0 ||
                                           x == -b )))) ||
    (x == Root[f + b #1^3 + #1^4 &, 1] && (27 b^4 == 256 f || (b >= 0 && 256 f < 27 b^4))) ||
    (x == Root[f + b #1^3 + #1^4 &, 2] && b >= 0 && 256 f < 27 b^4)
*)

It turns out the behaviour you notice requires $f = 0$ and $b < 0$.  You already have the Root[] expressions for the other cases from Wolfram Alpha.  The conditional ladder/tree shown above may not be optimal.

There's a real root if and only if $256f \leq 27b^4$. I derived this inequality from the derivative to locate the local minimums.

The case is trivial if $b = 0$, so assume $b \neq 0$. Check that the end behavior of the function $g(x) = x^4 + bx^3 + f$ satisfies $g(x) \rightarrow \infty$ as $x \rightarrow \pm \infty$. Thus, there is no guarantee that there exists a real root. Consider the derivative, that is $g'(x) = 4x^3 + 3bx^2 = x^2(4x + 3b)$. It must be the case that $x = 0$ is an inflection point and $x = \frac{-3b}{4}$ is a local minimum since $g''\big(\frac{-3b}{4}\big) = \frac{9b^2}{4}$. Therefore, the function attains a real root if and only if $g\big(\frac{-3b}{4}\big) = f - \frac{27b^4}{256}$ is non-positive.


How many days are there in 70 years? How to calculate the total no. of days in 70 years (or any other no. of years) considering that this period also includes leap years?

It depends on which 70 years.  There might be anywhere from 16 through 18 leap years.  For example, the 70-year period from the beginning of 1901 through the end of 1970 contains 17 leap years (1904, 08, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68), but the period from the beginning of 1904 through the end of 1973 contains 18 leap years (the same as the other, plus 1972).

To get 16 leap years, one observes that (for example) there was an exception in 1900, which was not a leap year, so that the period from 1897 through 1966 contains only 16 leap years. (1904 through 1964, the one in 1900 having been skipped.)

The solution to the general problem is something like this:  Let $Y$ be the number of years.  Calculate $Y\cdot 365$ and then add the leap days.  You can estimate the leap days by computing $Y\div 4$ and rounding off, but to get an exact count  the simplest thing to do is to check the years one at a time.  How you check the years depends on the calendar: in the Julian calendar, 1900 was a leap year, but in the Gregorian calendar, it was not.

In a different calendar, the answer will be completely different.  For example, in the Islamic calendar, the common year has not 365 but 355 days, and the leap year has 356 days; the Hebrew calendar is different again.

Actually, in the Gregorian calendar, it is a little more complicated than that!  For example, the year 1600, though divisible by 4, was NOT a leap year because it is also divisible by 400.  The year 2000, on the other hand, WAS a leap year because it was also divisible by 4000.  (It was the first time that particular provision was used.)


Random variables and independence of $\sigma$-algebras If a random variable $X$ is independent from the $\sigma$-algebra $F_t$ for every $t$ in a collection of indexes, is it true that $X$ in independent from the $\sigma$-algebra generated by all the $F_t$?

No.  For example, you can have three random variables $X, Y, Z$ such that
$X$ is independent of $Y$ and independent of $Z$ but completely determined by $Y$ and $Z$.  The standard example involves two coin-tosses.

Afaik, if we have the additional assumption that $\mathscr F_{t_1}$ and $\mathscr F_{t_2}$ are independent for all distinct indices $t_1$ and $t_2$, we don't have the independence of $X$ and $\sigma(\bigcup_{t \in I} \mathscr F_t)$.


Isomorphic encryption or homomorphic encryption? Many encryption functions are said to be homomorphic:

http://en.wikipedia.org/wiki/Homomorphic_encryption

As encryption functions are invertible, they can be considered one-to-one and onto on properly defined domains and ranges.

So, my basic question is why we don't use the term "isomorphic encryption" rather than "homomorphic encryption"?

First, consider the definition of group isomorphism:

  Given two groups $(G, \otimes)$ and $(H, \odot)$, a group isomorphism from $(G, \otimes)$ to $(H, \odot)$ is a bijective function $f : G \to H$ such that for all $u$ and $v$ in $G$ it holds that
   $f(u \otimes v) = f(u) \odot f(v)$.

Now, consider a homomorphic encryption such as ElGamal cryptosystem: It takes a message from a cyclic group $G$, and outputs a pair $(c_1, c_2) \in G^2$. That is, $\mathcal{E} \colon G \to G^2$.

Notice that under this definition, $\mathcal{E}$ is not a bijection from the message space $G$ to the ciphertext space $G^2$. However, the decryption of ElGamal is unique.

Homomorphic Encryption does not necessarily mean that the encryption function is a group homomorphism.

Let $E: M \rightarrow C$ be the encryption function where $M$ and $C$ are the plaintext and ciphertext space respectively. Now $E$ is said to be homomorphic if $E(m_1 * m_2)=E(m_1)\cdot E(m_2)$, where $*$ and $\cdot$ are binray operations on $M$ and $C$ respectively. However, $M$ and $C$ are not necessarily possessing an algebraic structure like groups, rings etc. Thus, even though the encryption function $E$ is invertible, the question of isomorphism does not arise.


find $u,v\in \mathbb Z$ such that $231u+45v=1$. I have to find $u,v\in \mathbb Z$ such that $231u+45v=1$. By Euclide algorithm,
\begin{align*}
231&=5\cdot 45+6\\
45&=6\cdot 7+3\\
7&=3\cdot 2+1
\end{align*}
The first equation gives $$6=231-5\cdot 45.$$
We put this in the second equation, and we get
$$45=(231-5\cdot 45)\cdot 7+3\implies 3=36\cdot 45-7\cdot 231.$$
Now I replace $3$ in the last equation, and obtain
$$7=2\cdot (36\cdot 45-7\cdot 231)+1\implies 7-72\cdot 45+14\cdot 231=1,$$

but the 7 disturbs me, did I do my algorithm wrong ?

Let's calculate the greatest common divisor:
$\left( \begin{array}{lcr}
231 & 1 & 0 \\
45  & 0 & 1 \\
\end{array}
\right) \rightarrow \left( \begin{array}{lcr}
6 & 1 & -5 \\
45  & 0 & 1 \\
\end{array}
\right) \rightarrow \left( \begin{array}{lcr}
6 & 1 & -5 \\
3  & -7 & 36 \\
\end{array}
\right) \rightarrow \left( \begin{array}{lcr}
0 & 15 & -77 \\
3  & -7 & 36 \\
\end{array}
\right) \rightarrow$
Then the greatest common divisor is 3 and its beonzout equation  $3=(231)(-7)+(45)(36)$, then $d=g.c.d(231,45)=3\rightarrow d|(231x+45y)$, which it is imposible, 3 does not divide to 1, then does not exist $u\ and\  v$


Prove $f$ and $g$ are not bijections if $f \circ g$ is a bijection. I know that if $A$ is an finite set, and $f : A → A$, and $g : A → A$, then if in addition  $f ◦ g : A → A$ is a bijection, then both $f$ and $g$ are also bijections.

However, is there an example to prove that if $A$ is an infinite set then both $f$ and $g$ are not bijections if $f \circ g : A \rightarrow A$ in addition is a bijection.

Let $f,g \colon \mathbf N \to \mathbf N$ be given by
$$ g(n) = n+1, $$
and
$$ f(n) = \begin{cases} n - 1 & n \ge 1\\
                         27   & n = 0 .\end{cases}
$$
then neither $f$ nor $g$ is bijective, but $f \circ g = \mathrm{id}_{\mathbf N}$ is.

Another example is $f, g: \mathbb{R} \to \mathbb{R}$ where $g(x) = e^x$ and $f(x) = \ln |x|$ (with $f(0) = 0$).

However, what is true is,

  Proposition. If $f \circ g$ is a bijection then $f$ is surjective and $g$ is injective.

and,

  Corollary. If $f \circ g$ and $g \circ f$ are bijections, then $f$ and $g$ are bijections.


PUGS is a rectangle. If the equation of PU is $y=\frac{2}{3x} + 4$. What is the slope of SP? I don't get the answer to this problem, can somebody please tell me what the answer is.

It might help to re-draw the rectangle so that it is slanted. Do you mean the equation of PU is y=2/3x+4? That means the slope is 2/3. Since SP is at a right angle to PU (it's a rectangle so it has to be), then it's slope is the negative reciprocal of that. That would be -3/2 (flip the fraction and multiply by -1), so the answer is C. Hope this helps.

The answer is C, assuming that the equation of PU is $y = \frac{2}{3}x+4$.


Does the internal category forgetful functor have both adjoints? Let $\mathcal{E}$ be a category with finite limits, and let $Cat(\mathcal{E})$ be the category of internal categories in $\mathcal{E}$, with forgetful functor $Ob : Cat(\mathcal{E}) \to \mathcal{E}$, sending each internal category in $\mathcal{E}$ to its $\mathcal{E}$-object of objects. A colleague of mine has conjectured/stated that this functor $Ob$ has left and right adjoints - is there a simple conceptual argument for this, and/or an appropriate reference in the literature somewhere?

From our experience in the category of sets, we can see that the left adjoint should be a "discrete internal category" functor, and the right adjoint should be an "indiscrete internal category" functor. The discrete case is easier: we can define $L(X)$ to be the internal category $X\rightrightarrows X$ in which the source, target, identity-arrow, and composition morphisms are all the identity of $X$. One easily checks that this has the right universal property; this relies critically on the fact that the source and target morphisms have a common splitting, and wouldn't work at all for internal semicategories.
Now, what should the indiscrete internal category $R(X)$ be? If $X$ is a set, then the set of arrows in the indiscrete internal category is simply $X\times X$, so let's try defining $R(X)$ as $X\times X\rightrightarrows X$ in general. The source and target morphisms are simply the first two projections, while the identity-arrow morphism is the diagonal. The pullback $(X\times X)\times_X (X\times X)$ is just $X\times X\times X$, which the composition morphism the projection away from the middle factor.
Now given any morphism $f:A\to X$ in $\mathcal E$ and a category structure $(s,t):M\rightrightarrows A$ on $A$, we get a unique morphism $(f\circ s,f\circ t):M\to X\times X$ commuting with the source and target morphisms on either side, which one checks also respects the identity-arrow morphisms in both categories.
So both adjoints exist in the same manner as in the category of sets. It might be enlightening to observe that these adjoints are simply the adjoints to the evaluation-at-zero functor from the category of reflexive graphs internal to $\mathcal E$ to $\mathcal E$. Such evaluation functors always have adjoints on both sides, given by Kan extensions, if $\mathcal E$ admits enough limits and colimits, so one interesting thing that happens here is that the indexing diagram for reflexive graphs is so nice that no colimits are needed to construct the left adjoint to that evaluation functor, where normally some coproducts would be required. (This feels, morally, related to the fact that reflexive coequalizers are nicer than general coequalizers.)
The other interesting thing that happens is that the free and cofree reflexive graph on an object magically come equipped with canonical category structures. This is really a rather rare phenomenon, which is why you can't name any other algebraic structures for which the forgetful functor has a right adjoint.

Not 100% sure that this works, but I would argue as follows.
Your friend's claim amounts to say that every object defines a discrete and a codiscrete internal category; discreteness is the easiest: the object of morphisms $C_1$ is just the object of objects, and all structure is trivial.
Codiscreteness can be argued similarly, abstracting the request that in the "maximaly connected groupoid" over a set there is exactly one arrow between any two objects.
It might help to recall that an internal category in $\cal E$ is just a truncated simplicial object $\Delta^o \to \cal E$ satisfying a Segal condition, so $Cat(\cal E)$ is a suitable category of functors; $Ob$ now is the "evaluation at $[0]$" functor.


Show that the series does not converge uniformly on the interval $I=(-r,0]$ Let $x\in\mathbb{R}$ and consider the power series: $$\sum_{n=0}^\infty (-1)^n(n^2+3n+2)x^n.$$
a) Find the radius of convergence $r$ for the series.

b) Show that the series is the Taylor series for $f(x)=\frac{2}{(1+x)^3}$ in $x=0$.

c) Show that the series does not converge uniformly on the interval $I=(-r,0]$. (You can, without proof, use the fact that the series is convergent towards $f$ on the interval of convergence.)

...In a) I've found the radius of convergence to be $r=1$. Can anybody help me with question c)?

To show that a serie is uniformly convergent, you generally show that

$$R_N(x) = \sum_{n=N}^\infty (-1)^n(n^2+3n+2)x^n \xrightarrow[{N\to\infty}]{} 0 \qquad \text{uniformly}$$

Namely, because for $x < 0$ the sum is always positive, if we note $R_N(x)$ this sum, that $\forall \epsilon, \exists N_0 > 0, \forall N > N_0, \forall x \in ]-1;0], R_N(x) \leq \epsilon$. But here we are going to show the opposite, namely, $\exists \epsilon, \forall N_0 > 0, \exists N > N_0, \exists x \in ]-1;0], R_N(x) > \epsilon$. which can be written $\exists \epsilon, \forall N > 0, \exists x \in ]-1;0], R_N(x) > \epsilon$. Any $\epsilon$ can work, so:

Fixing $A > 0$, no matter how big you choose $N$, you will always be able to find $x \in ]-1;0]$ such that it is bigger than $A$ (because it still tends to infinity). This shows the opposite of uniformly convergence, hence it is not uniformly convergent.

To show this:
Fix $\epsilon = 1$, $N > 0$. The function $x \mapsto R_N(x)$ is decreasing with $R_N(x) \xrightarrow[{x\to -1}]{} +\infty$. So, by definition we can find a $x$ such that $R_N(x) > \epsilon$.

And to show that $R_N(x) \xrightarrow[{x\to -1}]{} +\infty$, it is sufficient to show, that is not upper bounded (by monotonicity). If it was upper bounded by $K$. Then the sum from $N$ to $M$ would be upperd bounded by $K$ (for all $x$). So you make $x$ tend to $-1$ and then $M$ tend to infinity and you shouw that $R_N(-1) < +\infty$ which is not true.

For the question c : it's simple , you see it's an alternating series, meaning it's something-something+something-something+something ....
I'll call $U_n=(-1)^n(n^2+3n+2)x^n$
In order for the series to converge as an alternating series,
$$\lim_{n\to\infty}\lvert\frac{U_{n+1}}{U_n}\rvert<1$$
As well as having $\lim_{n\to\infty}U_n=0$

Let's first evaluate $\frac{U_{n+1}}{U_n}$ :
$$\frac{U_{n+1}}{U_n}=\frac{(-1)^{n+1}((n+1)^2+3(n+1)+2)x^{n+1}}{(-1)^n(n^2+3n+2)x^n}=-x\times\frac{n^2+5n+6}{n^2+3n+2}$$
And so :
$$\lim_{n\to\infty}\frac{U_{n+1}}{U_n}=-x\times\lim_{n\to\infty}\frac{n^2+5n+6}{n^2+3n+2}$$

we see that in the second limit the highest powers are 2, and the coefficients of both are equal to $1$ so the limit is equal :
$$\lim_{n\to\infty}\frac{U_{n+1}}{U_n}=-x$$
since $x\in(-1,0]$, then $0\leq\lim_{n\to\infty}\frac{U_{n+1}}{U_n}<1$ And so the first test is aqcuired and returned with success. The second tests now consits of proving $\lim_{n\to\infty}U_n=0$ so let's do that :
$$\lim_{n\to\infty}U_n=\lim_{n\to\infty}(-1)^n(n²+3n+2)x^n$$
We have $x^n$ an exponential function thus we treat it first. We know that $x\in(-1,0]$ so $x^n$ is a decreasing function meaning that $\lim_{n\to\infty}x^n=0$ and so $\lim_{n\to\infty}U_n=0$ and the second test is passed.
That means that the series $\sum_{n=0}^{\infty}(-1)^n(n²+3n+2)x^n$ for $I=(-r,0]$ converges.

I hope that helped, it's called the Alternating series test.


Proof by induction that $\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$ I am a bit new to logical induction, so I apologize if this question is a bit basic.

I tried proving this by induction:

$$\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$$

Starting with the base case $n=1$:

$$1^2\ge1^2$$

Then to prove that $P(n)\to P(n+1)$:

$$\left(\sum_{k=1}^{n+1}k\right)^2\ge\sum_{k=1}^{n+1}k^2$$
$$\left((n+1)\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$$
$$(n+1)^2\left(\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$$

Since they're both multiplied by $(n+1)^2$, it's easy to see that if $\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$, then $(n+1)^2\left(\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$. But if the $\ge$ were replaced with a $\le$, the proof would still be valid, even though it's demonstrably not true.

I can see that it would take strong induction would fix this problem, but if I didn't know by observation that $\left(\sum_{k=1}^nk\right)^2\le\sum_{k=1}^nk^2$ isn't true by observation, how would I know to use strong induction?


Expected sum if a fair die is rolled continuously The problem statement is :

  What sum can I expect in rolling a fair die n times?

How to determine the mean in a convenient way? Consider the fair die is 6 sided

If you've used dice a lot, you may have noticed that the pips on opposite sides always add up to $7$. Since this is the same for all three pairs of opposite sides and each pair has the same chance of being rolled and both sides of each pair have the same chance of being rolled, the expected value of a single six-sided die is $7/2$. Then by linearity of expectation the expected value of $n$ six-sided dice is $n\cdot7/2$.

The problem can be easily solved by using the concept of indicator variables. Let $X_i$ be the random variable which counts the number of the times $i$ comes up in $n$ rolls of dice. So, we have events $X_1, X_2, X_3, X_4, X_5$ & $X_6$ representing all possible scenario (for e.g. $X_1$ counts the number of times the face $1$ came up in $n$ rolls of dice).
Note that: $$ E[X_i] = P[X_i] = \frac n{6} $$

Also let's assume that random variable $X$ correspond to the sum of $n$ dice rolls.
Then, $E[X]$ is given by:

$$E[X] = \sum_{i=1}^{i=6} i*E[X_i] = \sum_{i=1}^{i=6} i*P[X_i] = \sum_{i=1}^{i=6} i*\frac n{6} = 3.5n $$


How to prove if $u\in W^{1,p}$, then $|u|\in W^{1,p}$? How to prove if $u\in W^{1,p}$, then $|u|\in W^{1,p}$?

Since $|u|\in L_p$, I only need to show weak derivative of $|u|$ exists and $D|u| \in L_p$.

Can anyone give me some hint?
Thanks!

This is done by the celebrated Stampacchia Theorem. It can be found in books from Brezis, Kinderlehrer-Stampacchia and many more.
However, let me give the quick idea on how to show this.

You need the following chainrule: Let $f\in C^1(\mathbb{R})$ with bounded derivative. For $u\in W^{1,p}(\Omega)$ there holds: $D (f\circ u)= f'(u) D u$. This can be shown by approximating $u$ with smooth functions.
Choose a certain approximation $f_\varepsilon(\cdot) \to (\cdot)_+$. Particularly, the approximation is such that $f_\varepsilon'(x)=0$ for $x\leq 0$ and $f_\varepsilon'(x)\leq C$. More precisely, one choice is $$f_\varepsilon(x) =\begin{cases}\sqrt{x^2+\varepsilon^2}-\varepsilon, &x>0\\ 0,& x\leq 0\end{cases}.$$ Then you can prove the chain rule for $f(\cdot)=(\cdot)_+$ passing with $\varepsilon\to 0$. You will obtain $D(u)_+=\chi_{u} Du$, where $$\chi_{u}(x)=\begin{cases}1,& u(x)>0\\0,&\text{elsewise}\end{cases}$$ (After the next step, you will see that $Du=0$ a.e. on the set where $u(x)=0$.)
Perform Step 2 for $f(\cdot)=(\cdot)_-$ and note that $|\cdot|=(\cdot)_+ + (\cdot)_-$

Recall that $|u|=u_+ + u_-$. Since Sobolev spaces are closed under addition, it suffices to show that $u_+,u_-\in W^{1,p}$. Now to show $u_+\in W^{1,p}$, just observe that $\|Du_+\|_p\leq \|Du\|_p$ to conclude.


Definition of a topological property "A topological property or topological invariant is a property of a topological space which is invariant under homeomorphisms. That is, a property of spaces is a topological property if whenever a space X possesses that property every space homeomorphic to X possesses that property. Informally, a topological property is a property of the space that can be expressed using open sets." (I copied it from Wikipedia)

Now my question is: What is the definition of a topological property ? Of course you can define it as wiki defines it. But I am more concerned about the the part of wiki's "definition" which says that "Informally, a topological property is a property of the space that can be expressed using open sets."

Is there a definition of a topological property that says which well formed formulas are well formed formulas of topological properties and which are not ?

Because of what I read in wikipedia, I was expecting to see a definition of a topological property that talks about the internal structure of the well formed formula of the property. Then, I also expected that there was a theorem that says that if $(X_1,T_1),(X_2,T_2)$ are any two homeomorphic topological spaces and the well formed formula $\phi(X,T)$ is a topological property, then:

$\phi(X_1,T_1)$ iff $\phi(X_2,T_2)$

Is there such a definition and such a theorem ?

Such a definition and such a theorem will enable one to spot many topological properties easily.

Here is a similar question: Can you characterize all properties of topological spaces which are preserved by homeomorphisms

We can formalise topology using only the language of set theory. [For instance, a topological space is a pair $\langle X, \tau \rangle$ where $X$ and $\tau$ are sets satisfying various properties, and we can define a homeomorphism $\langle X, \tau \rangle \to \langle Y, \sigma \rangle$ as a function $X \to Y$ (which is itself a set) satisfying some conditions, etc. All this can be formalised.]

So we can define a unary predicate $\text{TS}$ defined by
$$\forall x[\text{TS}(x) \leftrightarrow x\ \text{is a topological space}]$$
where '$x\ \text{is a topological space}$' is shorthand for...
$$\begin{align}\exists X \exists \tau( \hspace{53pt}\\
x= \langle X, \tau \rangle \wedge &\tau \subseteq \mathcal{P}(X) \wedge \varnothing \in \tau \wedge X \in \tau\\
\wedge & \forall U\ \forall V\ [U \in \tau \wedge V \in \tau \to U \cap V \in \tau]\\
\wedge & \forall A\ [A \subseteq \tau \to \bigcup A \in \tau]\\
) \hspace{78pt} \end{align}$$

Now suppose $\phi$ is a formula with one free variable, $x$ say. Then $\phi$ is a topological property (i.e. is preserved under homeomorphism) if
$$\forall x [\text{TS}(x) \wedge \phi(x) \rightarrow \forall y[\text{TS}(y) \wedge x \cong y \rightarrow \phi(y)]]$$
That is, if $\phi$ holds for any space $x$ then for any space $y$ homeomorphic to $x$, $\phi$ holds for $y$.

Here I've used $x \cong y$ as shorthand for the formula expressing that $x=\langle X, \tau \rangle$ and $y=\langle Y, \sigma \rangle$ are homeomorphic.

Is this what you were after?

Frankly, I don't see how it's any more enlightening to put yourself through all this than it is to just say "a topological property is one that is preserved by homeomorphism", as so succinctly put by Thomas Andrews in the comments.

There is a fundamental misunderstanding about what topology is actually capabable of doing. An invariant is a way to calculate whether two space could be homeomorphic. Take the Euler characteristic. Just because given two spaces that have the same Euler characteristic doesn't mean that they are homeomorphic. However, if two space don't have the same Euler characteristic then the are definitely not homeomorphic. There is very much a zen thing that happens in topology. To answer your question more directly you could make an exhaustive list of invariants but you would not know if it was complete.


Is there any method to get a finite sum for $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$? As we can see on Wikipedia, there are some algebraic methods that give us finite sums for the Grandi's series

$$1-1+1-1+1-1+1-1+\cdots$$

Let $S$ be the sum of the Grandi's series. Then

$S=(1-1)+(1-1)+\cdots=0+0+\cdots=0$
$S=1+(-1+1)+(-1+1)+\cdots=1+0+0+\cdots=1$
$1-S=S$ so that $S=1/2$

The algebraic manipulations above are not allowed because the Grandi's series doesn't converges.

Is there something similar related to the harmonic series? In other words, is there any "incorrect algebraic way" to get a finite sum for

$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots \;?$$

Thanks.

Let $~\zeta^\star(x)=\displaystyle\lim_{h\to0}\dfrac{\zeta(x+h)+\zeta(x-h)}2.~$ Then $\zeta^\star(1)=\gamma\simeq\dfrac1{\sqrt3}.~$ See the Riemann $\zeta$ function

and the Euler-Mascheroni constant for more details. Another possible alternative would be $\ln2$, since it appears in many expressions as a substitute for where one might expect to encounter $\zeta(1)$.

Actually, if you only consider the normal harmonic series, with only positive terms, you can't find a rearrangement that will converge to a certain value since all terms are positive, and changing the order is not possible.

However you can prove that, if you re-arrange correctly the terms of the series of general term: $a_n = \frac{(-1)^{n-1}}{n}$ , you can make it converge towards any real x.

Roughly: $a_1 = 1$ , if x<1, add negative terms until $x> S_{f(n)} $ where this partial sum is made with odd terms of the series. Then once you have this condition, you will add positive terms until : x < $S_{f(n)} $ . You do that undefinitely, and putting it properly enough you can prove that this series converges towards x. Of course this in only sketches, but the idea is here


Comparing $\pi^e$ and $e^\pi$ without calculating them How can I

compare (without calculator or similar device) the values of $\pi^e$ and $e^\pi$ ?

Another proof uses the fact that $\displaystyle \pi \ne e$ and that $e^x > 1 + x$ for $x \ne 0$.
We have $$e^{\pi/e -1} > \pi/e,$$
and so
$$e^{\pi/e} > \pi.$$
Thus,
$$e^{\pi} > \pi^e.$$
Note: This proof is not specific to $\pi$.

$$e^\pi>\pi^e$$ because if we subtract $e$ from both exponents we get $$e^{\pi-e}>1$$ which is true because $e$ is is greater than $1$ so when it 8s raised to that power it is equal to $$\frac{e^\pi}{e^e}$$ and that has to be greater than 1 because the top is greater than the bottom.


Given two adjacency matrices, how can I find if they're isomorphic? Matrix 1:
\begin{matrix}
0&1&1&0\\
1&0&1&0\\
1&1&0&1\\
0&0&1&0
\end{matrix}

Matrix 2:
\begin{matrix}
0&1&1&1\\
1&0&0&0\\
1&0&0&1\\
1&0&1&0
\end{matrix}

I've looked on google to find out how to do this, but I can't find an answer that makes sense to me. As far as I can tell, there is no efficient algorithm to do this, so you need to check all the permutations... but I don't even know how to start. Any help would be great, Thanks.

As isomorphims have to preserve degree, there are only 2 possible ones, the first given by
$$ \phi_1 \colon 1\mapsto 3, 2 \mapsto 4, 3 \mapsto 1, 4 \mapsto 2 $$
and
$$ \phi_2 \colon 1\mapsto 4, 2 \mapsto 3, 3 \mapsto 1, 4 \mapsto 2 $$
$\phi_1$ maps the first matrix to (that means the following matrix has $a_{\phi_1^{-1}(i), \phi_1^{-1}(j)}$ as $(i,j)$-th entry.
$$
\begin{pmatrix}
0&1&1&1\\
1&0&0&0\\
1&0&0&1\\
1&0&1&0
\end{pmatrix}
$$
and so $\phi_1$ is an isomorphism.

I think there is a better way of finding a solution to your question.
As the question is given in MATRIX form. I think it's better to do total solution in Matrices.

Concept : Matrix  M be the matrix we get by multiplying adjacency matrix and its TRANSPOSE (A').

  Matrix M will be = [a ij] : where a ij is the number of edges from Vi to Vj if i is not equal to j and a ii is the number of edges incident to i(V is the vertices set of the given matrix).

So if we find Matrices M1 and M2 for the two given matrices..and checking whether identical terms are there are not. If identical numbers are there in Matrices, then both graphs are ISOMORPHIC.
and from these matrices (if they are ISOMORPHIC)we find which element of first graph points to which element of second graph.


maximum number of collinear points? I know this is a very standard question widely popular in the Internet and the Mathworld.
I myself have solved the above problem is N^2 Log N avoiding floating arithmetic.However, can anyone give me a good resource/detailed explanation of how it can be solved in O(N^2) using point line duality concepts.

Sorry to create such a confusion regarding the statement.
The text would be:
"N different points with integer coordinates are given in a plane. You are to write a program that finds the maximum number of collinear points (they all belong to the same line)."

You may also refer to this site which is the problem I have solved using a N^2Log N approach.

By point-line duality, this is equivalent to the question 'given a configuration of $n$ lines in the plane, what is the maximum number of them which intersect at one point?'; since there are $O(n^2)$ pairwise intersection checks, it becomes a question of whether there's some bucketing/perfect hashing scheme that allows for finding a point in a dynamically-built list in $O(1)$ time.  While I don't know of any specific bucketing schemes applicable to this problem, AFAIK it's very common to have $O(1)$ or at least $O(\alpha(n))$ approaches for this sort of thing ($\alpha(n)$ being the inverse-Ackermann function).

The maximum number of collinear points among N points in a plane is N (i.e., if they happen to be collinear.)


Expectation of Double Dice Throw I was wondering about finding expected value of dice throw. I know that in case of single dice throw,the expected value is 3.5 = ((1+2+3+4+5+6)/6). So for purpose of betting the base price could be 3.5 to ensure no profit and no loss scenario. But how can I extend this logic for double dice throw or N dice throw. I though that for case of double dice out of the 36 cases, I could take the maximum value for each pair which would give 6 eleven times, 5 nine times, 4 seven times, 3 five times , 2 three times and 1 once. But this gave expected value to be 4.47, whereas the correct answer in book is 4.25. How can I determine the fair value in this case?

The number of rolls where $h$ is the highest of $n$ dice is the number of ways for $k$ dice to show less than $h$ and $n-k$ to show $h$
$$
\begin{align}
\sum_{k=0}^{n-1}\binom{n}{k}(h-1)^k
&=\overbrace{\color{#C00}{(1+(h-1))^n}\vphantom{\binom{n}{n}}}^{\text{sum for $k=0\dots n$}}-\overbrace{\color{#090}{\binom{n}{n}(h-1)^n}}^{k=n}\\
&=\color{#C00}{h^n}-\color{#090}{(h-1)^n}\tag1
\end{align}
$$
which makes sense since it is the number of possible rolls showing at most $h$ minus the number of rolls showing at most $h-1$ with $n$ dice.

Assuming $n\gt0$, the total number of possible rolls of $n\times d$-sided dice is the sum of $(1)$ for all $h$
$$
\sum_{h=1}^d\left(h^n-(h-1)^n\right)=d^n\tag2
$$
The expected high die is then
$$
\begin{align}\newcommand{\stirtwo}[2]{\left\{#1\atop#2\right\}}
\frac1{d^n}\sum_{h=1}^dh\left(h^n-(h-1)^n\right)
&=\frac1{d^n}\sum_{h=1}^d\left(h^{n+1}-(h-1)^{n+1}-(h-1)^n\right)\\
&=d-\frac1{d^n}\sum_{h=1}^d(h-1)^n\\
&=d-\frac1{d^n}\sum_{h=1}^{d-1}h^n\tag{3a}\\
&=d-\frac1{d^n}\sum_{j=0}^{n}\binom{d}{j+1}\stirtwo{n}{j}j!\tag{3b}
\end{align}
$$
where $\stirtwo{n}{j}$ is a Stirling Number of the Second Kind.

Let $X_1$ be the value on the ﬁrst die and let $X_2$ be the value on the second
 die. Since $X = X1 + X2$ we have $E(X) = E(X_1) + E(X_2)$. Earlier we computed that
 $E(X_1) = E(X_2) = 3.5$, therefore $E(X) = 7$.


How to plot $f(x,y)=\dfrac{1}{x^4+2x^2y^2+y^4+1}$ in MATLAB? How do I plot or how does the following function look?

$$f(x,y)=\frac{1}{x^4+2x^2y^2+y^4+1}$$

I could rewrite this $$\frac{1}{r^4+1}.$$

Have you tried using mesh/meshgrid functions? In your case it should look like this

[x y]=meshgrid(from:step:to);
z=1./(x.^4+2x.^2y.^2+y.^4+1);
mesh(x,y,z);

There is the function called fsurf() to plot the surface. The usage is:

f1 = @(x,y) 1/(x^4+2*x^2*y^2+y^4+1);
fsurf(f1,[-5 5 -5 5])

It generates the graph show here -- Output of fsurf(). The same can also be done using mesh() as mentioned by @polko14.

[x, y]=meshgrid(-5:0.1:5);
z=1./(x.^4+2*x.^2.*y.^2+y.^4+1);
mesh(x,y,z);

which generates the same plot -- Output of mesh().


Convergence of tetration sequence. This question arose from here. I am interested to find a nice proof about the convergence of $${^n}a=\underbrace{a^{a^{\ .^{\ .^{\ .^a}}}}}_{n\ \text{times}}.$$
I find with google a necessary and sufficient condition to have the convergence is $\frac{1}{e^e} \leq a \leq e^{1/e}$ but the part for $a\le1$ need some ugly work.

  Does anyone have an elegant/slick proof ?

Does anyone have an elegant/slick proof ?

Like in the answer given to the link you quote by Kiril, the fastest proof is given by fixed point iteration. The function you iterate, is:

$$g_c(x)=c^x$$

The fixed points of $g_c(x)$ are given in terms of the Lambert $W$ function as:

$$x_0=\frac{W(-\ln(c))}{-\ln(c)}$$

The fixed point condition (for attraction hence convergence) is therefore:

$$|g'(x_0)|\le 1\Rightarrow$$
$$|-W(-\ln(c))|\le 1 (*)$$

Now consider the function:

$$m(x)=x\cdot\exp(x)$$

$$(*)\Rightarrow W(-\ln(c))\in [-1,1]\Rightarrow$$
$$m(W(-\ln(c)))\in m([-1,1])\Rightarrow$$
$$m(W(-\ln(c)))\in [-e^{-1},e]\Rightarrow$$
$$-\ln(c)\in[-e^{-1},e]\Rightarrow$$
$$c\in[e^{-e},e^{1/e}]$$

Fixed point iteration, therefore, takes care of all cases where $c\in(e^{-e},e^{1/e})$ (open interval). The two end points have to be checked separately, because the fixed point condition is inconclusive there, being possibly $\pm1$.

The upper end point ($e^{1/e}$) is easy using standard Calculus (Consider the sequence $a_n=g_{e^{1/e}}^{(n)}(1)$ and prove that it converges to $e$, etc.). The lower end point ($e^{-e}$), is a bit tricky, but can also be done using Calculus, and you are done.

Firstly if $$a \le -1$$ then we are guaranteed $$1 \le |a^a| \le 0$$  so if the value converges it is definitely bounded

For $$-1 < a \le 0$$ we have that $$ |a^a| \rightarrow 1 $$ so again boundedness isn't an issue

For $$0 < a \le 1$$ the same reasoning holds to ensure boundedness

Now for convergence notice the following. If we assume the tower converges to a finite value, that is:

$$ a^{a^{a^{a^{\vdots}}}} = n$$

Then we can take logarithms on both sides and use the power to product rule to find

$$ a^{a^{a^{a^{\vdots}}}}\log(a) = \log(n)$$

In other words:

$$n \log(a) = \log(n)$$

Exponentiate both sides

$$ a^n = n$$

Yielding:

$$a = n^{\frac{1}{n}}$$

Now we can use calculus to find the maximum and minimal values that $n^{\frac{1}{n}}$ take on and this gives us the range of possible $a$ among the real numbers if we want $n$ to be among the real numbers.


$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma =1$ Let be $\alpha, \beta, \gamma$  the angles between a generic direction in 3D and the axes $x,y,z$, respectively.
Prove that
$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma =1$.

PS: the 2D case is trivial. But I can't prove the 3D case.

Consider the unit vector $$(u,v,w).$$

The cosines of the angles it forms with the axis are given by the dot products

$$\cos\alpha=(1,0,0)\cdot(u,v,w)=u,\\\cos\beta=(0,1,0)\cdot(u,v,w)=v,\\\cos\gamma=(0,0,1)\cdot(u,v,w)=w,$$

and as it is unit

$$u^2+v^2+w^2=1.$$

Hint: If $u$ and $v$ are vectors making an angle $\theta$ then $$u\cdot v=||u||\,||v||\cos(\theta).$$


Simultaneous triangularization of matrices Let $\mathcal{F}=\{A_1,A_2,\ldots,A_r\}$ be a triangulable commuting family of $n\times n$ matrices (that is, each $A_i$ is triangulable and $A_iA_j=A_jA_i$ for every $i,j$). I know that $\mathcal{F}$ can be simultaneous triangularization, but what is the algorithm of finding the invertible matrix $P$ such that $P^{-1}A_iP$ is triangular? As a working example consider the matrices
$$
A=
\begin{pmatrix}
-3 & 2 & -4 \\
-1 & 0 & -1\\
 2 & -2 & 3
\end{pmatrix}\qquad
B=
\begin{pmatrix}
3 & -2 & 2 \\
-1 & 2 & -1\\
-2 & 2 & -1
\end{pmatrix}\qquad
C=
\begin{pmatrix}
1 & 0 & 1 \\
1 & 0 & 1\\
0 & 0 & 0
\end{pmatrix}
$$
Here $AB=BA$, $AC=CA$ and $BC=CB$. In addition, the characteristic polynomials of these matrices are
$$
f_A(x)=x(x-1)(x+1)\\
f_B(x)=(x-2)(x-1)^2\\
f_C(x)=x^2(x-1)
$$
so each one of them is triangulable. Thanks!

Here, it is easy. Since $A$ has $3$ distinct eigenvalues and $B,C$ commute with $A$, we can deduce that $B,C$ are polynomials in $A$ and it suffices to triangularize $A$.

In the general case.

Step 1. Find a common eigenvector of $A,B,C$.

Step 2. Proceed by recurrence.

Moreover, you can choose $P$ as an orthogonal matrix.

More explicitly: find a common eigenvector v1 among your given matrices. Now complete a basis arbitrarily *u2,...,un. Let Q = [v,u1,...,un ] a matrix. Now Q-1AQ, Q-1BQ, and Q-1CQ should all have an (n-1) x (n-1) submatrix in the bottom right with only zeros in the entries of first column directly to the left of it. You then want to find a common eigenvector w among all of these submatrices. Let U =[u1,...,un]. Then v2 =Uw. With this update *Q = [v1,v2,u3,...,un ] *  where u3,...,un may have been rechosen in order to complete the space. You rinse and repeat with n-2, n-3 and so on.

If you need more information check out the algorithm in the paper by Dubi:


Triangle geometry - synthetic proof I'm looking for a nice synthetic proof of the following fact.

Consider a non-isosceles triangle, pick a vertex. Assume that the median and the altitude passing through this vertex are isogonal conjugates (i.e. symmetric w.r.t. the bisector of the angle). Prove that the triangle is right-angled in this vertex.

Let $\triangle ABC$ have the desired property at $A$. Let $M$, $P$, $N$ be the points where the median, angle bisector, and altitude from $A$ meet side $\overline{BC}$.

Since
$$\angle BAP \cong \angle CAP \qquad \text{and} \qquad \angle MAP \cong \angle NAP$$
we have
$$\angle BAM \cong \angle CAN = C^\prime \qquad \text{and} \qquad
\angle CAM \cong \angle BAN = B^\prime$$
where $B^\prime$ and $C^\prime$ are respective complements of $B$ and $C$.
Invoking the Law of Sines in $\triangle BAM$ and $\triangle CAM$,
$$\underbrace{|\overline{AM}|\;\frac{\sin C^\prime}{\sin B} = |\overline{BM}|}_{\triangle BAM} = \frac{1}{2} |\overline{BC}| = \underbrace{|\overline{CM}| = |\overline{AM}|\; \frac{\sin B^\prime}{\sin C}}_{\triangle CAM}$$
Therefore,
$$\begin{align}
\sin B\sin B^\prime = \sin C \sin C^\prime \quad&\implies\quad \sin B\cos B = \sin C \cos C \\[4pt]
&\implies\quad \sin 2 B = \sin 2 C \\[4pt]
&\implies\quad B = C \quad\text{or}\quad 2B = \pi - 2C \\[4pt]
&\implies\quad B = C \quad\text{or}\quad B + C = \pi/2 \\[4pt]
&\implies\quad B = C \quad\text{or}\quad A = \pi/2
\end{align}$$

First prove it conversely: that the condition holds for a right triangle.

   If the other two angles are $\alpha\le \beta=90^\circ-\alpha$, then one leg with the altitude makes angle $\alpha$ and the other leg with the median also makes angle $\alpha$, because of an isosceles triangle.

Now if we have an arbitrary triangle $ABC$ with the given property at $A$, find the intersection $A'$ of the Thales circle raised over segment $BC$ with the half-line of side $BA$. Now if $A\ne A'$, the angle of side $AB$ with the altitude remains the same in triangles $ABC$ and $A'BC$ but the other angle in question will differ.


Why do linear and quadratic equations form straight line and parabola? I've been trying to figure this out. I searched on the internet but couldn't find the answers either. Why exactly does a quadratic equation form a parabolic shape? and why does a linear equation form a straight line?

It might help you to go back to the original Euclidean geometry.  For example, one theorem in 'The Elements' is: A straight line is the locus of all points equidistant from two (distinct) given points" ('locus of points' just means 'the shape all of the points fall upon and/or trace out').  If you transpose to a Cartesian coordinate system, and accept the obvious translation of the Pythagorean Theorem into an equation for distance between points as a function of coordinates, then you can translate this theorem into an equation between two formulae defining the distance of each 'locus' point from each of the given points.  That equation can be simplified and rearranged into the usual linear expression.

Similarly, for Euclid a parabola is -defined- to be 'the locus of all points equidistant from a given line ('directrix') and a given point ('focus') which is not on that line'.  Restrict the case to a horizontal line [greatly simplifying the equation for the distance from a point to that line] and a point above the line, then translate the definition into an equation of two distance formulae, and you get a form of the quadratic equation.

This is an interesting question.  However, it is currently not answerable.  I will need to put additional constraints on it.

First, from the comments above, you want to know why it is that the linear equation, when plotted, produces a straight line, and why the parabola, when plotted, produces...a parabola.  However, it's not enough to just say "when plotted".  We need to specify coordinate axes.

Let's begin by working in Euclidean space, where the x and y axes are perpendicular to each other, the origin is at (0,0), and the scales on the x and y axes are the same and the "ticks" on the axes are evenly spaced.  Under these restrictions, then yes, we get straight lines and other things.  Let's go a bit deeper.

We'll begin by stating a fact without proof:

Any differentiable function in the above space that has a constant derivative is representable by a linear equation of the form y = mx+b.

Now, we will think about something: what is the derivative?  Well, it's the rate of change of y with respect to x (in this case, yes it's more complex in other cases, but we don't care at this point).  So, what is a constant derivative?  It would mean that our function's y values change by a constant amount for some given change in the function's x values.  So far, however, this isn't very satisfying - why would this produce the straight line when graphed?  Is there some other way to describe the derivative that might help out?

In fact, one great way to describe it is as the direction of the line that is tangent to the function at a point.  This is just another way to say "slope", but calling it a "direction" allows us to define the derivative geometrically as follows:

Let's say that the derivative at a point $x_0$ is $f'(x_0) = C$.  What this means is this:  If you begin at the point $x_0$ and, initially, you are facing in the same direction as the positive x axis, and you turn clockwise by an angle of $\tan^-1(C)$, then you will be facing in the same direction as the tangent line at that point.  If you now never change your direction and keep walking forever, then a few things happen:

1) you walk straight, never turning, and therefore you trace out a line
2) You are walking along the tangent line at that point.

So the question for why a linear equation traces out a line is that its tangent line (and hence the direction you have to walk) never changes, no matter where you start.

Now, for the parabola, unfortunately the best I can tell you is that it is curved, and that's because the direction of the tangent line DOES change.  However, I can't tell you why it specifically traces out that particular shape.  I'm sure someone else can, but it's not as easy as saying "because the tangent changes", since many other functions have changing tangents as well