How do I show that a given lambda value is an eigen value of a matrix? Also how do I show that a matrix is not diagonalizable based on my calculations from the equation $\det(A-\lambda I)=0$

For the first part say the matrix is:

$$
        \begin{pmatrix}
        1 & -2 & 3 \\
        0 & -1 & 3 \\
        -1 & 2 & -2 \\
        \end{pmatrix}
$$
is $\lambda=1$ an eigenvalue for the matrix?

Also I wouldn't mind finding out the code for lambda on here and the code for putting a set vectors in brackets as related to span.

Thanks!

To find the eigenvalues of $A=\begin{bmatrix} 1&-2&3\\0&-1&3\\-1&2&-2\end{bmatrix}$ we compute the roots of the characteristic polynomial, $p_A(t)=det(tI-A)$. Using Laplace Expansion we have:
$$p_A(t)=det\begin{bmatrix} t-1&2&-3\\0&t+1&-3\\1&-2&t+2\end{bmatrix}=(t-1)det\begin{bmatrix} t+1&-3\\-2&t+2 \end{bmatrix}+det\begin{bmatrix} 2&-3\\t+1&-3 \end{bmatrix}=(t-1)[(t+1)(t+2)-6]+[-6+3t+3]=(t-1)(t^2+3t-1)$$
From factored form, we see that the eigenvalues of $A$ are
$\lambda_1=1, \lambda_2=\frac{-3+\sqrt {13}}{2}, \lambda_3=\frac{-3-\sqrt {13}}{2}$. So to answer the first question: yes, the matrix has $1$ as an eigenvalue. Now, because the matrix is $3\times 3$ and it has $3$ distinct eigenvalues, it is diagonalizable.

Note that showing an $n\times n$ matrix is diagonalizable is equivalent to showing that the algebraic multiplicities of its eigenvalues equals their geometric multiplicities. In the special case where it has $n$ distinct eigenvalues, this condition is immediately seen to hold true, and the matrix is diagonalizable.

After subtracting the eigenvalues $\lambda$ from diagonal elements are some rows or columns or after some linear combinations the same? Does it vanish?

$$
        \begin{pmatrix}
        0 & -2 & 3 \\
        0 & -2 & 3 \\
        -1 & 2 & -3 \\
        \end{pmatrix}
$$


Why are these two upper bounds the same? I'm trying to upper bound the quantity $\int_t^{\infty} e^{-x^2} dx$ as a function of $t$. I came up with two ways of doing this:

We have that $(x-t)^2 = x^2 -2tx + t^2 \ge 0 \implies -x^2 \le -2tx + t^2$, so that $\int_t^{\infty} e^{-x^2} dx \le \int_t^{\infty} e^{-2tx}e^{t^2} dx= e^{t^2} \int_t^{\infty} e^{-2tx} dx = e^{t^2} \left(-\frac{1}{2t} e^{-2tx}\bigg\rvert_{x=t}^{x=\infty}\right) = \frac{1}{2t}e^{-t^2}$.

We have, on the domain $x \in [t, \infty)$, that $x \ge t \implies 1 \le \frac{x}{t}$, so that
$\int_t^{\infty}e^{-x^2} dx \le \int_t^{\infty} \frac{x}{t} e^{-x^2} dx = \frac{1}{t} \left(-\frac{1}{2}e^{-x^2} \bigg \rvert_{x=t}^{x=\infty}\right) = \frac{1}{2t} e^{-t^2}$.

Is it a coincidence that these two totally different methods give the same upper bounds? Or is there a connection between these two estimates that I'm not realizing?

The coincedence is that from the point of view of an integral your estimations are the same. Why?
Let's recall the substitution rule for integrals:
For $\varphi \in C^1([a,b])$ it holds
$$\int_{a}^{b} f(\varphi(x)) \cdot \varphi'(x)\,\mathrm{d}x = \int_{\varphi(a)}^{\varphi(b)} f(x)\,\mathrm{d}x$$
If we now have a closer look we can see that both of your estimations are of the form of the LHS:
For $$f(x) = \frac{1}{2}\exp(-tx)$$ and $$\varphi_1(x) = 2x-t$$ we get $$\int_{t}^{\infty} f(\varphi_1(x)) \cdot \varphi_1'(x)\,\mathrm{d}x = \int_t^\infty \exp(-2tx+t^2) \,\mathrm{d}x$$ for your first estimator while taking
$$\varphi_2(x) = \frac{x^2}{t}$$ it gives us $$\int_{t}^{\infty} f(\varphi_2(x)) \cdot \varphi_2'(x)\,\mathrm{d}x = \int_t^\infty \frac{x}{t}\exp(-x^2) \,\mathrm{d}x$$ so your second estimator.
And because it holds $$\varphi_1(t) = \varphi_2(t) = t$$ as well as $$\varphi_1(\infty) = \varphi_2(\infty) = \infty$$ both functions give the same result by using the substitution rule namely $$\int_{\varphi(a)}^{\varphi(b)} f(x)\,\mathrm{d}x = \int_t^\infty \frac{1}{2}\exp(-tx) \,\mathrm{d}x$$
Moreover: Any suitable $\varphi \in C^1([t,\infty))$ with $\varphi(t) = t$ and $\varphi(\infty) = \infty$ will lead to the same result.
So what you did in both cases was actually estimating $\exp(-x^2)$ by $f(\varphi(x)) \cdot \varphi'(x)$ for a suitable $\varphi$

You could also try and use the gaussian integral and taylor series to get that:
$$\int_t^\infty e^{-x^2}dx=\frac{\sqrt{\pi}}{2}+\sum_{n=0}^\infty\frac{(-1)^{n+1}t^{2n+1}}{(2n+1)n!}$$
or if you take the first few terms you get:
$$I\approx\frac{\sqrt{\pi}}{2}+\frac{t^3}{3}-t+O(-t^5)$$


What is the two norm of $\text{diag}\{a_1,a_2,\dots,a_n\}\left[\begin{smallmatrix}\exp(x_1)\\\exp(x_2)\\\vdots\\\exp(x_n)\end{smallmatrix}\right]$ I know that the two norm of a matrix is the largest singular value.

I am presented with an operator $U$ where $U: \mathbb{R}^n \to \mathbb{R}^n$

$U = A\exp(x) = \text{diag}\{a_1, a_2, \ldots, a_n\}\begin{bmatrix} \exp(x_1) \\ \exp(x_2) \\ \vdots \\ \exp(x_n) \end{bmatrix}$ where $a_i > 0$

How do I evaluate  $\|U\|_2$ in this case?

I'm afraid your operator does not have a finite norm. Consider, for example, the case $x_1=x_2=\ldots=x_n=c_1$. Then $\left\|x\right\|_2=c_1\sqrt{n}$. However, by your definition of $U$, we have $$\left\|Ux\right\|_2 = \sqrt{\sum_k a_k^2 {\rm e}^{2c_1}} = C_2 \cdot {\rm e}^{c_1},$$
where $C_2 = \sqrt{\sum_k a_k^2}>0$. Therefore, $$\frac{\left\|Ux\right\|_2}{\left\|x\right\|_2} = \frac{C_2}{c_1} {\rm e}^{c_1},$$
which grows unbounded for $c_1\rightarrow  \infty$. Therefore, the supremum over $x$ does not exist.

You have $U= diag (a_1 \exp(x_1),...,a_n \exp(x_n))$, thus $U^T=U$ and therefore

$U^TU=U^2=diag (a_1^2 \exp(2x_1),...,a_n^2 \exp(2x_n))$

The largest singular value 0f $U$ is given by

$$\max\{a_1^2 \exp(2x_1),...,a_n^2 \exp(2x_n)\}$$


Why does a number like $2^n\bmod 7$ have a repeating pattern? I am struggling with the formal understanding of why a number like $2^n\bmod7$ have a repeating pattern? $1, 2, 4, 1, 2 ,4\ldots$

The repeating pattern show up in many places of modular arithmetic, what is the primary reason for this?

Thank you

Consider a sequence $a^1, a^2, a^3,\ldots$, all taken $\mod{n}$. We only assume a  finite number of values (at most $n$), and the sequence is infinite, so after at most $n+1$ many terms we have a repetition, say $a^i = a^j$ ($\mod{n}$,for some $i <j$). But this means that the sequence will repeat after $j$ again: $a^{j+1} = a^{i+1}$ and so on, till we have again $j-i$ terms, after which we are back again etc. Because you have a prime modulus in your example, $a^{n-1} = 1$, so we actually cycle back to the starting value. In general you might get a start sequence and then the repeating part (between $a^i$ and $a^j$).


Proving that $\widehat{g*f}\in L^1(\mathbb{R})$ with $f(t) = \frac{1}{\sqrt{1 + t^4}}$ and $g(t) = e^{-|t|}$ The question is as follows:

  Verify whether the Fourier of the  convolution: $f * g$ belongs to $L^1(\mathbb{R})$? Where  $$f(t) = \frac{1}{\sqrt{1 + t^4}}~~and ~~~~g(t) = e^{-|t|}$$

$\textbf{Efforts:}$
As we know $L^1(\mathbb{R}) $ is closed under convolution. So is it enough to show that $f(t)$ and $g(t)$ are in $L^1(\mathbb{R})$ for to see that their convolution is also in $L^1(\mathbb{R})$?
Can you please let me know what conditions has to satisfy for the convolution be in $L^1(\mathbb{R})$?

Indeed, see here
\begin{split}
\widehat{g}(x)= \int_{-\infty}^{\infty}e^{-|t|}e^{-ix t}\,dt
&=&\int_{-\infty}^{0}e^{t}e^{-ix t}\,dt+\int_{0}^{\infty}e^{-t}e^{-ix t}\,dt\\
 &=&\left[ \frac{e^{(1-ix)t}}{1-ix} \right]_{-\infty}^0-\left[\frac{e^{-(1+ix)t}}{1+ix} \right]_{0}^{\infty}\\
&=&\frac{1}{1-ix}+\frac{1}{1+ix}\\
&=&\color{blue}{\frac{2}{x^2+1}\in  L^1(\Bbb R)}.
\end{split}

Therefore,  we have  that $\widehat{g}\in L^1$ and we know that  $\widehat{f} \in L^\infty$ (Since $f\in L^1$ ) hence $$\color{red}{\widehat{f*g}=\widehat{f}\displaystyle\widehat{g}  \in L^1}$$

Proof of the claim above
$$|\widehat{f}(a)|=\left|\int f(x)e^{-ixa}\, dx\right| \le \int \left|f(x)\, \right|dx =\|f\|_1 $$

that is $$\|\widehat{f}\|_{\infty}\le \|f\|_1 $$

$f, g , f*g $ are non negative and since $f,g\in L^1\cap L^{\infty}$ we may apply Fubini's theorem to

$$ \int_{-\infty}^{+\infty}(f*g)(t)\,dt = \int_{-\infty}^{+\infty}\int_{-\infty}^{\infty}f(x)\,g(t-x)\,dx\,dt $$
leading to $\|f*g \|_1 = \|f\|_1\cdot \|g\|_1 = \frac{1}{\sqrt{\pi}}\,\Gamma\left(\frac{1}{4}\right)^2.$


How can one know every Cauchy sequence in a complete metric space converges? I am new to Cauchy sequences. I stumbled onto them in the process of learning what a Hilbert space is.  As I understand it, a Cauchy sequence is a sequence whose elements become arbitrarily close to each other as the sequence progresses.

But consider the following statement:

A metric space $X$ in which every Cauchy sequence converges to an element of $X$ is called complete.

How can we know every Cauchy  sequence converges?

$\Bbb R$ is a complicated example.  Let's instead consider a simple example.  Let $S$ be the space $\{7, 8\}$ with the usual distance.  (That is, $d(7,7) = d(8,8) = 0, $ and $d(7,8) = d(8,7) = 1$.)  I claim that $S$ is a complete space.

We are not going to show this by considering every cauchy sequence one at a time.  As you observed, that is impossible.

Let $C$ be a cauchy sequence and the elements of $C$ are $c_0, c_1, \ldots$.  Since $C$ is a cauchy sequence, we have, for any positive $\epsilon$, that elements of the sequence that are far enough out will be closer than $\epsilon$.  Formally, for any given positive $\epsilon$ there is some integer $N$ so that for any $i,j>N$, we have $d(c_i,  c_j) < \epsilon$.

Take, for example, $\epsilon = \frac12$.  Then because $C$ is cauchy, we know there must exist some $N$ such that for any $i,j>N$, we have $d(c_i , c_j) < \frac12$, because that is what it means for $C$ to be cauchy; that is the definition.

But if we have  $d(c_i , c_j) < \frac12$ then $d(c_i , c_j) = 0$, because the only  possible values of $d(\ldots)$ are $0$ and $1$, and it can't be $1$ because $1$ is not less than $\frac12$.  So it must be $0$.

So we conclude that if $C$ is cauchy, then for all sufficiently large
$i$, and $j$ we have $d(c_i, c_j) = 0$ and therefore $c_i = c_j$ for all sufficiently large $i$ and $j$. Thus, every cauchy sequence in $S$ is eventually constant.  It might have a mixture of $7$s and $8$s at the beginning, but there must come some point (that's $N$) after which all the elements are $7$s or all the elements are $8$s.  If this doesn't happen, then $C$ was not a cauchy sequence.

Since $C$ is eventually constant, there is some value, either $7$ or $8$, call it $L$, such that all sufficiently late elements of $c$ are equal to $L$.  That is, there is some $N$ such that for any $i>N$, $c_i = L$, where $L$ is either 7, or 8, depending on which $C$ we actually chose.

Is is then trivial to show that $C$ converges to the value $L$. To show this we need to show that for any positive $\epsilon$, the terms of the sequence are eventually within $\epsilon$ of $L$.  In this case we can do even better; we know that the terms of $C$ are eventually equal to $L$.  $C$  is therefore a convergent sequence.

Therefore, all cauchy sequences in $S$ converge, and $S$ is therefore a complete metric space.

The same argument goes through just as well if we take the more complicated space $\Bbb Z=\{\ldots, -1, 0, 1, 2, 3\ldots\}$ with the usual distance function $d(i,j) = |i-j|$; the proof is exactly the same.

Mathematics is not (usually) about considering a lot of examples.  It is about what properties logically imply which other properties.  Just as we can prove that every square number is positive without considering each square individually, we can also prove that every cauchy sequence of elements of $\Bbb Z$ must converge, even without considering each cauchy sequence individually.

You have your statement wrong:

A metric space $X$ in which every Cauchy sequence that has a limit converges to an element of $X$ is called complete.

And you sense of what "Cauchy" means is a bit vague. For example, in the space $\Bbb{R}$ (the real line), the sequence $H_n \equiv \sum_{i=1}^n \frac{1}{i}$ has arbitrarily small distances between successive members, but is not is Cauchy because there for any given $m$ there is an $H_m$ with $m>n$ that is arbitrarily far from $H_m$.  So this sequence it is not considered in deciding whether $\Bbb{R}$ is a complete space.

One way to prove that all Cauchy sequences that have a limit converge to an element of $X$ is to show that $X$ is compact, for that is a property of a compact space.  This is easy to see:  If $X$ is compact, then it contains all its limit points, and the limit of a Cauchy sequence is a limit point.

It turns out that compactness is a bit of a stronger condition than is needed for completeness. For example, $\Bbb{R}$ is not compact. But since it does contain open sets including each of the points it contains, it is easy to see that $\Bbb{R}$ contains all its limit points, and therefore the limit of a Cauchy sequence of elements of $\Bbb{R}$ is in $\Bbb{R}$ so $\Bbb{R}$ must be complete.

I will admit that since you generally can't just test every possible Cauchy sequence, there are some spaces for which it is tough to prove whether they are complete or not. Sometimes very clever arguments are needed.


Examples of continuous functions $f:(0,1)\to[0,1]$ Hi so the question is asking if if there is a continuous function over $(0,1)$ which will result in the image $[0,1]$. Thank you so much! Explanations are greatly appreciated!

you may consider the function $f(x)=\frac{\cos{(100x)+1}}{2}$. Surely it is continuous and reaches its maximum and minimum several times in $(0,1)$.

Hope I helped.

Consider the piecewise function:
\begin{cases}
2x, \qquad &\text{ for } 0 < x < \frac{1}{2}, \\
1, &\text{ otherwise. }
\end{cases}

EDIT: Oops, thought we wanted the image to be $(0, 1]$. To get the appropriate answer, consider:
\begin{cases}
4x, \qquad &\text{ for } 0 < x < \frac{1}{4}, \\
1 - 4(x - \frac{1}{4}) = 2 - 4x, &\text{ for } \frac{1}{4} < x \leq \frac{1}{2}, \\
0, &\text{ otherwise.}
\end{cases}


Establish the inequality $e^x \geq 1-\ln (1-x)$ for $x \leq 0$. I'm very confused as to where to start on this one.  I assume I would start with some smaller inequalities then work towards this one with algebraic manipulation but I can't figure out where to start.

Define
$$
f(x)=e^x-1+\ln(1-x)
$$
where we hope to show $f(x)\geq 0$ for $x\leq 0$. Note that at $x=0$ we have
$$
f(0)=1-1=0
$$
And we have that
$$
f'(x)=e^x-\frac{1}{1-x}
$$
where we note that
$$
f'(x)=e^x-\frac{1}{1-x}\leq 0\iff e^x\leq \frac{1}{1-x}\stackrel{x\leq 0}{\iff} e^{-x}\geq1-x
$$
a well known equality derived from the mean value theorem or visual obvious-ness. Thus your function is decreasing to $0$ on the left of the $y$ axis.

PLEASE NOTE THERE IS AN ISSUE WITH THIS ANSWER AT THE SERIES EXPANSION STAGE. I CAN'T DELETE IT AS IT WAS ACCEPTED. - Karn

Hint:

Let $f(x)=e^x$ and $g(x)=1-\ln(1-x)$.

We know that $f(0)=g(0)=1$.

We need $f(x)\ge g(x)$ so if $f(x)-g(x)$ is positive and decreasing then our proof is done. You may want to consider the Taylor expansion to help with the proof.

Following this is the full solution but I suggest you do not look at it until you have attempted it yourself.

   We have $f'(x)=e^x$ and $g'(x)=\frac{1}{1-x}$.\begin{align}f'(x)-g'(x)&=e^x-\frac{1}{1-x}\\&=\sum_{r=0}^\infty{\frac{x^r}{r!}}-\sum_{r=0}^\infty{x^r}\\&<0\end{align}


Minimalist proof that the operator norm on $H^*$ satisfies the parallelogram law? Let $H$ be a complex Hilbert space. I am trying to prove that $H^*$ (the set of bounded linear maps from $H$ to $\mathbb C$) is again a Hilbert space, where the inner product induces the operator norm. I have already seen proofs that $H^*$ with the operator norm is a Banach space, and also that a norm is induced by an inner product if and only if the norm satisfies the parallelogram law

$$\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2.$$

So if I can show that the operator norm on $H^*$ obeys this identity, then I am done. But how can I do that?

I would prefer not to invoke the Riesz representation theorem, as I have seen some sources do. (Hence "minimalist" in the question title.)

Letting $\phi, \psi\in H^*$, we have

\begin{align}
\|\phi+\psi\|^2+\|\phi-\psi\|^2
&=\left(\sup_{\|x\|=1}|\phi(x)+\psi(x)|\right)^2 + \left(\sup_{\|x\|=1}|\phi(x)-\psi(x)|\right)^2 \\\\
&=\sup_{\|x\|=1}|\phi(x)+\psi(x)|^2 + \sup_{\|x\|=1}|\phi(x)-\psi(x)|^2 \\\\
&=\sup_{\|x\|=1}\Big(|\phi(x)|^2+|\psi(x)|^2+\overline{\phi(x)}\psi(x)+\phi(x)\overline{\psi(x)}\Big)\\\\
&\qquad + \sup_{\|x\|=1}\Big(|\phi(x)|^2+|\psi(x)|^2-\overline{\phi(x)}\psi(x)-\phi(x)\overline{\psi(x)}\Big)
\end{align}

If we could only combine the two "sup" expressions and cancel cross terms, and then also separate out the "sup" terms again to give $2\|\phi\|^2+2\|\psi\|^2$, we would be done. But I can't see any justification for this.

Here is an idea, how one could prove it. Unfortunately, there is an additional factor of 2, which I cannot get rid of. Maybe someone has an idea, how to improve it.

Let $x,y\in H$. Then
$$\begin{split}
(\phi+\psi)(x)^2 + (\phi-\psi)(y)^2 &= (\phi(x)+\psi(x))^2 + (\phi(y)-\psi(y))^2\\
&=\frac12\Big( \phi(x+y)^2  + \phi(x-y)^2 + \psi(x+y)^2 + \psi(x-y)^2 \\
& \qquad+ 2\phi(x+y)\psi(x-y) + 2\phi(x-y)\psi(x+y) \Big)\\
&\le \frac12\Big( (\|\phi\|^2+\|\psi\|^2)(\|x+y\|^2+\|x-y\|^2)
+ 4\|\phi\|\|\psi\|\|x+y\|\|x-y\|\Big)\\
& \le (\|\phi\|^2+\|\psi\|^2)(\|x+y\|^2+\|x-y\|^2)\\
& = 2(\|\phi\|^2+\|\psi\|^2)(\|x\|^2+\|y\|^2).
\end{split}$$
First inequality is Cauchy-Schwarz, second inequality uses $2ab\le a^2+b^2$ twice.
In the last step, I used the parallelogram identity in $H$.
Taking the supremum over $x$ and $y$ with $\|x\|\le1$, $\|y\|\le1$ of this inequality, yields
$$
\|\phi+\psi\|^2 + \|\phi-\psi\|^2 \le 4(\|\phi\|^2+\|\psi\|^2).
$$
If we could prove this inequality with factor $2$ instead of $4$, we could finish, by substituting $(\phi,\psi)$ by $(\phi+\psi,\phi-\psi)$.

By Riesz's theorem, there is an antilinear isometric bijection $I:H^\ast\to H$ such that $\varphi(x)=(x,I\varphi)$. Antilinear means that $I(\alpha\varphi_1+\beta\varphi_2)=\overline{\alpha}I(\varphi_1)+\overline{\beta}I(\varphi_2)$. As you know, $H^\ast$ is Banach space. We define a function $<\varphi_1,\varphi_2>=(I\varphi_2,I\varphi_1)$ in it. All you need to do is show that $<\cdot,\cdot>$ -- is inner product in $H^\ast$ and $<\varphi,\varphi>=\|\varphi\|^2$. It will follow from here that the space $H^\ast$ is Hilbert and the parallelogram law will turn out automatically.


Ten books are placed in a random order. Find the probability of three given books being side by side As i understand it, this question should be answered by N(A)/N.

I take N to be 10!, im not sure if this is correct.

How do you calculate N(A)?

We can use your $10!$. There certainly are $10!$ equally likely ways to arrange the books.

Now we count the number of orderings in which our $3$ books are together. The leftmost of our $3$ books can be in any of $8$ places, $1$ to $8$. (Write the numbers $!$ to $10$ in a row, and look.)  Once this place is chosen, implicitly the other locations are chosen. Then the $3$ books can be put in the places in $3!$ orders. And for each such arrangement, the other books can be chosen in $7!$ ways, for a total of $(8)(3!)(7!)$. Now divide by $10!$. We get probability $\dfrac{(8)(3!)(7!)}{10!}$. There is a lot of cancellation.

The following is perhaps easier. The set of positions our $3$ books will occupy can be chosen in $\dbinom{10}{3}$ equally likely ways.
Note that we are not placing the books in these places yet, we are just choosing a set of locations.

Of these ways, $8$ have our books in consecutive order. So the probability is $\dfrac{8}{\binom{10}{3}}$.

Well, for that you need to find all possible orders where $3$ books stay together and their own possible orders. so your $N(A)$ is a multiplication of those above which is $8!3!$ ($8$ is for the other $7$ books and, lets say, those $3$ tied together). So your final probability will look like $\frac{8!3!}{10!}= 0.0\overline6$.


Math Riddle in Combinatorics. A blind man is on a strange island and he has 2 red pills and 2 white pills, completely identical and has kept in his pockets, he needs to take 1 red pill and 1 white pill order doesn't matter.  If he takes 2 pills of the same color he dies.  How does he survive.  There is a standard solution for this and I was wondering if there could be a combinatorial way to solve this?

Break each pill in half, and take half of each pill.

Just take a glass of water, fill it to the brim.
Put all the pills inside, they should dissolve well, all pills will dissolve well, drink half the glass of water, even a blind man can tell, problem solved.


Product $\sigma$ algebra Consider $\mathbb{R}$ with the $\sigma$-algebra of Borel sets, and $\mathbb{R}^\mathbb{R}$ with the product $\sigma$-algebra(see p.22 of 'Real Analysis - Gerald B. Folland'). Does $[0,1]^\mathbb{R} \subset \mathbb{R}^\mathbb{R}$ belong to the product  $\sigma$-algebra?

Claim: every set $A$ in $\mathbb{R}^\mathbb{R}$ is of the form $$A =
 \{f : (f(x_1),f(x_2),\dots, f(x_n), \dots) \in
 \mathcal{B}(\mathbb{R}^\mathbb{N})\},$$ where
  $\mathcal{B}(\mathbb{R}^\mathbb{N})$ is the product $\sigma$-algebra
  on $\mathbb{R}^\mathbb{N}$ and $\{x_i\}_{i=1}^\infty$ is countable.

In other words, we can only specify a measurable set at countably many points. This will immediately yield your question by a contradiction argument: if $[0,1]^\mathbb{R}$ is measurable and non-empty, it contains some $f : \mathbb{R} \rightarrow \mathbb{R}$ and is specified only at countably many points. Change $f$ at one of the points not specified so that it does not lie in $[0,1]$ to obtain a contradiction to $f \in [0,1]^{\mathbb{R}}$.

To see the claim, notice that if $\mathcal{E}$ is a family of sets, the  collection $\mathcal{C}(\mathcal{E})$ obtained from $\mathcal{E}$ through finitely many applications of complementation and countable union and intersection is a $\sigma$-algebra. In other words, if we assume w.l.o.g. that $\mathcal{E}$ contains all complements (i.e. $A \in \mathcal{E} \Rightarrow A^c \in \mathcal{E})$, then any $A \in \mathcal{C}(\mathcal{E})$ can be written as
$$\{A : A = \cap_{i_1 \geq 1}\cup_{i_2\geq 1} \dots \cap_{i_n \geq 1} E_{i_1, \dots, i_n}, 1 \leq n < \infty, E_{i_1,\dots, i_n} \in \mathcal{E}\}.$$

In our case, the product $\sigma$-algebra on $\mathbb{R}^\mathbb{R}$ is $\sigma(\mathcal{E})$ with $\mathcal{E} = \{\pi^{-1}_x(A) : x \in \mathbb{R}, A \in \mathcal{B}(\mathbb{R})\}$. Since $\pi_x^{-1}(A) = \{f : f(x) \in A\}$ and $\sigma(\mathcal{E}) \subset \mathcal{C}(\mathcal{E})$, any $E \in \sigma(\mathcal{E})$ is specified by at most countably many points, as in the claim.

This has nothing to do with the closedness of $[0,1]$, but you're right that $[0,1]^\mathbb{R}$ is not in the product $\sigma$-algebra $M$. The generators of $M$ are of the form $\pi_x^{-1}(A)$ for $x\in \mathbb{R}$ and $A\subset \mathbb{R}$, where the projection $\pi_x(f)=f(x)$. What's $\pi_x^{-1}(A)$? Well, it's the set of all functions $f:\mathbb{R}\to\mathbb{R}$ with $f(x)\in A$. We could take the intersection of countably many of these to get, for instance, the functions which map $\mathbb{Q}\to [0,1]$, but $[0,1]^{\mathbb{R}}$ could only be written as an uncountable intersection of such sets, which is not permitted in a $\sigma$-algebra.


Counting Real Numbers Forgive me if this is a novice question. I'm not a mathematics student, but I'm interested in mathematical philosophy.

Georg Cantor made an argument that the set of rational numbers is countable by showing a correspondence to the set of natural numbers. He did this by scanning rational numbers in a zigzag scheme starting at the top left corner of a 2D table of integers representing the numerator vs. denominator of every rational number. He also proved that the set of real numbers is uncountable through his famous diagonalization argument.

My question is, why can't real numbers also be counted in the same fashion by placing them in a 2D table of integers representing the whole vs. decimal parts of a real number i.e. like this:

    0   1   2   3   4  ...
0  0.0 0.1 0.2 0.3 0.4 ...
1  1.0 1.1 1.2 1.3 1.4 ...
2  2.0 2.1 2.2 2.3 2.4 ...
3  3.0 3.1 3.2 3.3 3.4 ...
4  4.0 4.1 4.2 4.3 4.4 ...
.   .   .   .   .   .  .
.   .   .   .   .   .   .
.   .   .   .   .   .    .

and scanning them in a zigzag scheme starting at the top left corner? Negative reals can also be treated the same way as negative rationals (e.g. by pairing even natural numbers with positive real numbers, and odd natural numbers with negative real numbers).

You only counted a subset of the reals, namely, the set including the integers as well as reals with one decimal place. You cannot count the reals, as you would have to count to an infinite number of decimal places, as some reals have no fractional representation.

To 'count' as you propose, you would need the top heading to be reals, as well as the side heading, making it all but impossible to actually count anything, due to the fact you would now have two of the infinite sets, which you are using to count the infinite set itself.

Between any two real numbers lies the entire set of integers (Or so hypothesized. In any case, some infinite quantity), thus you cannot "count" reals, as there are infinite reals between any two reals you can pick to "count".


How to solve $x^4+8x-1=0 $ Any idea how to solve the following equation?

$$x^4+8x-1=0$$

I tried to find some obvious roots of this equation so it could help me to find the other roots (if it has more then $1$) but I had no success, and I would really appreciate some help.

hint see this article from wolramalpha its irreducible over $Q$ http://www.wolframalpha.com/input/?i=is%20x%5E4%2B8%20x-1%20irreducible%3F&lk=2 if you want to find roots they are hard to find . $x\approx -2,0.12,0.95-\pm\sqrt{3}i$


list of convergent series I wanted to know if there is an online reference I can use to find out known results about convergent series. I could not find this one, for example, on wikipedia

$\sum_{k=1}^{+\infty} \left(\frac{1}{2}\right)^k k^2$

There exists an $N$ such that for all $k > N$, $k^2 \le (3/2)^k$. This is just because
$$
\lim_{k \to \infty} \frac{(3/2)^k}{k^2} = \lim_{k \to \infty} \left( \frac{\sqrt{3/2}^k}{k} \right)^2 = \infty, \quad \Longrightarrow \quad \exists N \mbox{ s.t.} \frac{(3/2)^k}{k^2} \ge 1 \mbox{ for all $k<N$}.
$$
Therefore,
$$
\sum_{k=N+1}^{\infty} \frac{k^2}{2^k} \le \sum_{k=N+1}^{\infty} \frac{(3/2)^k}{2^k} = \sum_{k=N+1}^{\infty} \left( \frac 34 \right)^k
$$
which converges.

If you're looking for some library which lists all known patterns of convergence series, you're not trying to understand the mathematics behind those series correctly. The right way to study this (for example, for an exam) is to not only understand the tricks used to show that one series converges/diverges, but also to understand why this was the chosen trick and that it wasn't done purely by luck. Also I suggest trying to apply every elementary tests a few times to know when they work and when they don't.

Hope that helps,

Use the ratio criteria. This is the limit when K tends to infinit, for the term evaluated at K+1 over the term evaluated at K. As the result is 1/2 < 1, then the series converges


Tractrix tangent segment (from Baby Do-Carmo) I need some help with question 4 in section 1.3 in Baby Do-Carmo textbook in DG.

The question asks:
Let $\alpha(t):(0,\pi)\rightarrow R^2 $ be given by:
$$ \alpha(t)= (\cos(t), \cos(t) +\log(\tan(t/2)) $$ its image is called the tractrix.

Question b, asks to prove that the length of the segment of the tangent of the tractrix between the point of tangency and the y axis is constantly 1.

Now the angle between $\alpha$ and the y axis is t.

So basically if I were to use the sine theorem from trig, where $$\frac{S}{\sin(t)} = \frac{|\alpha(t)|}{\sin(\pi-(t+\angle \alpha(t) \alpha '(t)))}$$
Where S is the required line segment I am looking for.

Now I am only left with calculating the angle between $\alpha(t)$ and $\alpha '(t)$, is this about right, or I am way off here?

It's hell of a calculation if I am right (and it's really rare when I am).
Thanks.

Anyway, for completeness: start with the (now correct) parametrization

$$\begin{align*}x&=a\sin\,t\\y&=a\left(\cos\,t+\log\tan\frac{t}{2}\right)\end{align*}$$

(I prefer my tractrices to have the horizontal axis as the asymptote, but oh well...)

It is easy to construct the slope corresponding to any $t$:

$$\frac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}=\frac{-\sin\,t+\csc\,t}{\cos\,t}$$

and thus the equation of the tangent line as well:

$$\frac{y-a\left(\cos\,t+\log\tan\frac{t}{2}\right)}{x-a\sin\,t}=\frac{-\sin\,t+\csc\,t}{\cos\,t}$$

and the expression for the y-intercept of this line is $\log\tan\frac{t}{2}$; it is now easy to see that the distance from the point $(x,y)$ of the tractrix to the point $\left(0,\log\tan\frac{t}{2}\right)$ is $a$.

It is easier to prove this one backwards.  The original definition of the tractrix is the curve created when a constant length "leash" is pulled along a straight line.  Thus the slope of the line is always tangent to the curve and leads directly to the differential equation that creates the tractrix.  Wikipedia has the equations nicely type-set.

That is, the answer to your question is "by definition" of the curve itself.


Probability of choosing bat I got the option A and B by manually checking but after that got stuck .

Having used B1 in the first match he is not allowed to use it in the second match, justifying $P(2)=0$.  As he has not used it in the second match he has three to choose from for the third and B1 is one of them so $P(3)=\frac 13$.  Now for the fourth match he has $\frac 23$ chance of not using bat B1 in the third match.  For that chance he has $\frac 13$ chance of using it for the fourth, so $P(4)=\frac 23 \cdot \frac 13=\frac 29$.  Generally, if he has chance $p$ of using B1 in a match, he has $\frac 13(1-p)$ chance of using it in the next match because if he does use B1 in the match he won't use it the next time and if he does not use it he has $\frac 13$ chance of using it next time.  Thus we get $P(5)=\frac 13(1-\frac 29)=\frac 7{27}, P(6)=\frac 13(1-\frac 7{27})=\frac {20}{81}, P(7)=\frac 13(1-\frac {20}{81})=\frac {61}{243}$

For Your Option C,

Let the matches be P,Q,R,S

Now. The $B_1$ has been used in match P and is to be used in match S. Therefore It shouldn't be used in match R and obviously it can't bes used in match Q . For this, Probability is simply $\frac{2}{3}.\frac{1}{3}$ ($B_1  $ not used in match R and $B_1$ used in match S).

On the similar grounds, I'm getting $\frac{8}{81}$ for option D. Although I'm not pretty sure about it.


Can you give an example of a complex math problem that is easy to solve? I am working on a project presentation and would like to illustrate that it is often difficult or impossible to estimate how long a task would take. I’d like to make the point by presenting three math problems (proofs, probably) that on the surface look equally challenging.  But…

One is simple to solve (or prove)
One is complex to solve (or prove)
And one is impossible

So if a mathematician can’t simply look at a problem and say, “I can solve that in a day, or a week, or a month, how can anyone else that is truly solving a problem? The very nature of problem solving is that we don’t know where the solutions lies and therefore we don’t know how long it will take to get there.

Any input or suggestions would be greatly appreciated.

This isn't exactly what you're asking for, but it should serve the same purpose very nicely.

Hilbert gave a talk in 1920 or so in which he discussed the difficulty of various problems.

He said that great progress had been made in analytic number theory in recent years, and he expected to live to see a proof of the Riemann Hypothesis.

Fermat's Last Theorem, he said, was harder; maybe the youngest members of his audience would live to see a proof.

But the problem of determining whether $2^{\sqrt2}$ is transcendental was so hard that not even the children of the youngest people in the audience would live to see a solution to that one.

With the benefit of hindsight, we can see that Hilbert had it exactly backwards.

$2^{\sqrt2}$ was settled in 1929 - Hilbert lived to see it.

Fermat, as we know, held out until the 1990s.

And the Riemann Hypothesis is still unsettled.

The point of the story is not to make fun of Hilbert. The point of the story is that if even Hilbert, the strongest mathematician of his era, could be so wrong in judging the relative difficulty of various mathematical problems, then it must be a really hard thing to do - which, I think, is the point you are trying to make.

$\zeta(2) = \sum_n \frac{1}{n^2} = \frac{\pi^2}{6}$

$\zeta(1+it) \neq 0$

$\zeta(s)=0$ $\implies$ $\Re(s)= 1/2$, if $s$ is lies in the right half-plane.


How do I draw a Venn diagram for A ∩ B = A ∩ C? How would I draw a (clean) Venn diagram for the statement A ∩ B = A ∩ C?

This is the Venn diagram for sets A and B but how would I draw set C in this? Are Venn diagrams just not suitable for drawing these kind of equivalences?

How about this way. The center is A ∩ B = A ∩ C = B ∩ C

Like this :)

The intersecting areas are all the same size, I think that's all that matters.

!(http://i740.photobucket.com/albums/xx42/saddamhansson/venndiagram.png)


Compute the integral $\int_0^\pi \frac{\sin x}{x}dx$ . How to compute precisely the integral $$\int_0^\pi \frac{\sin x}{x}dx$$ analytically? It is well-known that $$\int_0^{+\infty}\frac{\sin x}{x}dx=\frac{\pi}{2}.$$ One way to compute the above integral using Taylor expansion for $\sin x$, we then obtain a power series value approximately 1.84. Thanks for any useful hints.

$$\color{#00f}{\large{\rm Si}\left(\pi\right)}$$

  $\displaystyle{\rm Si}\left(x\right)$ is the
  Sine Integral Function.

Your number is sometimes called the Gibbs constant and evaluates to

1.85193705198246617036105337015799136334580972898115
  49098047837818769818901663483585327103365029547577

with 100 digit accuracy. It is most likely not expressible in a simple way in terms of other "well-known" constants.


Finding the solution to $V\frac {dV}{dz}=\gamma + \alpha(\beta(R^2-z^2)^{-\frac12}-V)$ The following equation was found by balancing buoyancy with drag in an engineering field.

$V\frac {dV}{dz}=\gamma + \alpha(\beta(R^2-z^2)^{-\frac12}-V)$

$\gamma$,$\alpha$, $\beta$, and $R$ are all constant, and $0\le z \le R$

Obviously, the equation is nonlinear in V. It almost looks like you can put it in the form $\frac {d}{dz}f(z, V)=\gamma + \alpha(\beta(R^2-z^2)^{-\frac12})$  for some $f(z,V)$ but I have not been able to find one, let alone something separable.

I'm thinking of using Runge–Kutta to do a numerical solution, with the constants able to range over a lot of values I'm worried there might be cases of stange behavior.

Is there a way to simply solve this, numerically or in a closed form, for all constant values?

Hint:

Let $V=-\alpha U$ ,

Then $\dfrac{dV}{dz}=-\alpha\dfrac{dV}{dz}$

$\therefore\alpha^2U\dfrac{dU}{dz}=\gamma+\alpha(\beta(R^2-z^2)^{-\frac{1}{2}}+\alpha U)$

$U\dfrac{dU}{dz}=U+\dfrac{\gamma}{\alpha^2}+\dfrac{\beta}{\alpha\sqrt{R^2-z^2}}$

This belongs to an Abel equation of the second kind in the canonical form.

When $\gamma=0$ , this belongs to a special case of Abel equation of the second kind in the canonical form in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=140.

Let $\begin{cases}z=-\dfrac{4RW}{W^2+4}\\U=t-\dfrac{4RW}{W^2+4}\end{cases}$ ,

Then $\pm\dfrac{\beta}{\alpha}\dfrac{dW}{dt}=-\dfrac{tW^2}{4}+RW-t$


What is a good book for math students to learn machine learning in depth? I am a math master student and have done fundamental math courses like probability theory, measure theory, linear algebra and know a little bit about functional analysis. What is good way for me to learn machine learning in depth?

I have read the classical text Pattern-Recognition and Machine Learning last summer; my impression was that it was very ineffective to read the book chapter by chapter like a mathematical text. The book does not go deep enough for many algorithms and skip too many steps considered too technical by engineers.

Is there a machine learning book that maybe does not cover too many topics, but treat each one in depth and takes advantage of math when necessary? It will be great to be able connect fundamental mathematical objects with machine learning (I am thinking about Lp spaces, hilbert space etc).

Hmm, that is a very good question!

I see a couple avenues you could proceed:

In general, once you start doing rigorous machine learning, the distinction with modern statistics really starts to vanish. So you could pick up a book from that literature. I would recommend for example, the book by Peter Bühlmann and Sara van de Geer. It essentially only considers one model: The LASSO (as well as slight variations, such as L1-penalized logistic regression), and is a "math" book (lots of definitions, lemmata, theorems, proofs).
Keeping with the idea of only studying one method, you could read the book on boosting by Schapire and Freund. This would give a flavour of rigorous results, but with more of a CS rather than stats perspective. [Caveat: I have not actually read this one, but have heard good things about it.]
The two previous recommendation focus on understanding one specific method (L1 regularization, boosting respectively); if instead you want a unifying framework in terms of a specific mathematical space, then there is a very nice review paper of RKHS (reproducing kernel Hilbert space) machine learning methods in the Annals of Statistics. It's not a textbook, but it seems to perfectly suit your needs.
Larry Wasserman has been teaching an amazing course on Statistical Machine Learning. There he goes through a whole bunch of unrelated methods (what you complained about), but for each one he explains what the main mathematical tools and results are, and proves many of these. The website provides both videos of the lectures, exercises and lecture notes (which I think he is compiling into a textbook). Highly recommended.

There is a good book that is probably easier to start with: An introduction to statistical learning, by hastie and tibshirani.

I also have an online course that will go through the basic topics in machine learning. You can find out more by checking out:

Math for Machine Learning Online Course


Find last two digits of $33^{100} $ Find last two digits of $33^{100}$.

My try:

So I have to compute $33^{100}\mod 100$

Now by Euler's Function $a^{\phi(n)}\equiv 1\pmod{n}$

So we have $33^{40}\equiv 1 \pmod{100}$

Again by Carmichael Function : $33^{20}\equiv 1 \pmod{100}$

Since $100=2\cdot40+20$ so we have $33^{100}=1\pmod{100}$

So last two digits are $01$

Is it right?


Minimum of empty set in $\infty$ why is that? According to a definition a minimum in a set is the smallest number in it.
My professor claimed: the minimum in empty set in $\infty$ why is that? $\infty$ isn't even a member of the empty set which contradicts the definition, plus it's will make more sense to me if it's $- \infty$

As others have noted, the minimum and maximum of the $\emptyset$ are undefined, since a minimum or maximum must belong to the set, but the infimum and supremum are well-defined.
The infimum of a set is the greatest lower bound.  That is, it is the largest number such that no member of the set is smaller than it.  Now if we take any real number $x$, it is surely true that no member of $\emptyset$ is smaller than $x$.  What is the largest $x$?  Well, there's no largest real number, so $\infty$ is a sensible agreement.
Similarly, the supremum of $\emptyset$ is $-\infty$.

You are right, the definition doesn't apply to the empty set. What your professor should have said is that for the empty set it's defined separately as $\infty$.
Why is this a sensitive definition?
You have some properties that you would like to remain true in the case of the empty set.
For example, if $A \subseteq B$ then $\min A \ge \min B$. That's easy to prove for non-empty sets, so you would like to define it for the empty set so this remains true.
But $\emptyset \subseteq B$ for all subsets $B$, so you'll need $\min \emptyset \ge \min B$ for all subsets $B$. In particular if $B=\{x\}$, $\min \emptyset \ge x$ for all $x \in \mathbb R$. Obviously there's no such real number and this clearly motivates the defintion $\min \emptyset = \infty$.
Other properties involving union, intersection or complement of sets also can be used to motivate this definition and in general all works very nice with this definition.
Note: For subsets of $\mathbb R$ in general one should talk about the infimum, not the minimum, since not all sets have a least element.


How to calculate the number of possible connected simple graphs with $n$ labelled vertices Suppose that we had a set of vertices labelled $1,2,\ldots,n$.

There will several ways to connect vertices using edges. Assume that the graph is simple and connected.

In what efficient (or if there is no efficient way, you can just tell me whatever procedure you can think of) way do we be able to calculate the number of possible ways the graph can be made? (even if some graphs are isomorphic to each other, they are counted as separate cases.)

There are $\binom{n}2=\frac12n(n-1)$ pairs of distinct points. If you do not allow loops or multiple edges, each of these pairs determines one possible edge, and you can have any subset of those possible edges. A set with $\binom{n}2$ members has $2^{\binom{n}2}$ subsets, so there are $2^{\binom{n}2}$ possible graphs without loops or multiple edges.

If you demand that the graphs be connected, the problem becomes very much harder. From your final comment I take it that you are in effect counting labelled graphs. This sequence of numbers is A001187 in the On-Line Encyclopedia of Integer Sequences. If $d_n$ is the number of labelled, connected, simple graphs on $n$ vertices, the numbers $d_n$ satisfy the recurrence

$$\sum_k\binom{n}kkd_k2^{\binom{n-k}2}=n2^\binom{n}2\;,$$

from which it’s possible to calculate $d_n$ for small values of $n$. This recurrence is derived as formula (3.10.2) in Herbert S. Wilf, generatingfunctionology, 2nd edition, which is available for free download here.

According to MathWorld, Brendan McKay’s software package nauty includes a routine that efficiently enumerates such graphs; it’s available here.

If you count unlabelled graphs instead, so that you don’t count isomorphic graphs separately, you get the sequence mentioned by Arturo in the comments.

I can't upload image
because I have below 10 reputition.
I have gotten OEIS A001187.
If you want to see my fomula,
please visit below.
https://blog.naver.com/oroblast/220278283168
It is coded with Korean.
I am apartment house security in Seoul.
I'm very much busy.
So, I can't make time for translating above.
I am sorry.!!^^~~
$m_1$, $m_2$, $m_3$, $\dots$ mean integer partition.
$\alpha$, $\beta$, $\gamma$, $\dots$ mean:
if $1 + 2 + 2 + 2 + 3 + 3 = 13$ then $\alpha = 1$, $\beta = 3$, $gamma = 2$.
Above url contains java source code to calculate my fomula.!!^^~~


Isomorphism between two finitley presented groups which are not finite I am looking for an isomorphism between the two following groups (infinite groups) ( probably by GAP to show that there is an isomorphism between $f$ and $g$)

$$f=\langle  a,b,c ~| ~ bab^{-1}a^{-1},cac^{-1}a^{-1}\rangle $$

$$g=\langle a,b,c ~|~ bcb^{-1}c^{-1},cac^{-1}a^{-1}\rangle$$

The generators $a$, $b$, $c$ are the same in two groups, I am not sure but guess, there should be an isomorphism which takes $a$ to $c$?


Approximate $\coth(x)$ around $x = 0$ I'm trying to approximate $\coth(x)$ around $x = 0$, up to say, third order in $x$. Now obviously a simple taylor expansion doesn't work, as it diverges around $x = 0$. I'm not quite sure how to proceed from here. I had a look at
Series expansion of $\coth x$ using the Fourier transform

but this is a bit technical for my purposes, and I also don't need the entire series. Is there an easy way of getting some terms, or do I really have to solve for the entire series? And is there a simpler way to do so than with the Fourier transform?

There is a simple way of approximating $\coth$ by noticing that it is a logarithmic derivative. Since:
$$ \frac{\sinh z}{z}=\prod_{n=1}^{+\infty}\left(1+\frac{z^2}{\pi^2 n^2}\right)\tag{1} $$
by the Weierstrass product for the (hyperbolic) sine function, we have:
$$ \log\sinh z-\log z = \sum_{n=1}^{+\infty}\log\left(1+\frac{z^2}{\pi^2 n^2}\right),\tag{2}$$
so, by differentiating both sides:
$$ \coth z-\frac{1}{z} = \sum_{n=1}^{+\infty}\frac{2z}{\pi^2 n^2+z^2}.\tag{3} $$
By expanding the terms in the RHS of $(3)$ as geometric series we can also derive the full Laurent expansion of $\coth$ in a puntured neighbourhood of the origin, in terms of $\zeta(2m)$. This is also the standard technique for showing that $\zeta(2m)$ just depends on the Bernoulli number $B_{2m}$ (formula $(41)$ here).

My copy of the classic "Handbook of Mathematical Functions" by Abramowitz and Stegun says

$$ \mathrm{coth}(z) = \frac{1}{z} + \frac{z}{3} - \frac{z^3}{45} + \frac{2}{945} z^5 - \ldots + \frac{2^{2n}B_{2n}}{(2n)!} z^{2n-1} $$

where $B_n$ is the $n^{th}$ Bernoulli number.

Note in particular the difference in the $z^3$ term compared with the answer above.


What is the limit of the $a_{n+1}=\frac{1+a_n+{a_{n-1}}^3}{3}$? $a_{n+1}=\frac{1+a_n+a_{n-1}^3}{3}$  with $n \geq 2$  ( $a_1 =0$,  $a_2 = 1/2$)
By mathematical induction, the $0 \leq a_n \leq 1$ and the $a_n$ is monotonically increasing.  If the limit exists, then $1$ or $\frac{-1 + \sqrt{5}}{2}$ would be value.
What is the exact limit of this recurrence relation between two values?

If the limit exists, then let $L = \lim\limits_{n \to \infty} a_n$. We must have $L = \lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} a_{n + 1} = \lim\limits_{n \to \infty} \frac{1 + a_n + a_{n - 1}^3}{3} = \frac{1 + (\lim\limits_{n \to \infty} a_n) + (\lim\limits_{n \to \infty} a_{n - 1})^3}{3} = \frac{1 + L + L^3}{3}$.
Then we must have $L^3 - 2L + 1 = 0$. Note that $L^3 - 2L + 1 = (L - 1)(L^2 + L - 1)$. Then the three possible values are $L = 1$ and $L = \frac{-1 \pm \sqrt{5}}{2}$.
Note that we have $a_3 = \frac{1 + 1/2}{3} = 1/2$. Then it's easy to show that for all $n$, we have $a_{n + 1} \geq a_n$ using strong induction, since the base cases of $n = 1, 2$ are satisfied and for larger $n$, we have $a_{n + 1} = \frac{1 + a_n + a_{n - 1}^3}{3} \geq \frac{1 + a_{n - 1} + a_{n - 2}^3}{3} = a_n$.
So the sequence is increasing. Now, we show that the sequence is bounded. In fact, we see that for all $n$, we have $a_n \leq \frac{-1 + \sqrt{5}}{2}$. Let $w = \frac{-1 + \sqrt{5}}{2}$. This can be shown directly for $n = 1, 2$. To show it for other values of $n$, we use strong induction and note that $a_{n + 1} = \frac{1 + a_n + a_{n - 1}^3}{3} = \frac{1 + w + w_{n - 1}^3}{3} = w$.
The sequence is bounded and increasing, so it must have a limit. Since the sequence is increasing and $a_2 > \frac{-1 - \sqrt{5}}{2}$, that can't be the limit. And since $w < 1$ and the sequence is bounded by $w$, $1$ cannot be the limit either. Thus, the only possibility for the limit is $w$.

Blockquote

solve x^3 + x + 1 = 3 x
Solution
x = 1,
x = 1/2 (-1 - √5),
x = 1/2 (√5 - 1).
Plot the sequence a_n with Mathematica software, with initial condition

$a_0 = 1, a_1 = 0$

$a_0 = 0, a_1 = 1$

Both solution converges to golden ratio r = x = 1/2 (√5 - 1)


Is Competition Problem on Constrained Inequality Relevant in Research? Many inequality problems on variable $x,y,z$ or $a, b$ come with constraints such as $ab = 1,xyz = 1, x+y+z = 1$. For example, Let $a$ and $b$ be positive real numbers with $a + b = 1$ then prove that - $$ \frac{a^2}{a + 1}+ \frac{b^2}{b + 1} ≥\frac1 3$$  (Hungary 1996).

Now I understand that contest-problem is good for exercise but what I am asking, is that are such inequality are useful in research? Or are such inequality found in research level mathematics?

This question is motivated by this post (click here).

Such problems are very useful in research, particularly when we want to optimize some property of a system under constraints.  This occurs when we're designing computer chips (optimize bandwidth consistent with power envelope), internal-combustion motors (optimize power consistent with fuel economy standards), finance (optimize return on investment consistent with portfolio or legal constraints), etc.

By C-S we obtain: $$\frac{a^2}{a+1}+\frac{b^2}{b+1}\geq\frac{(a+b)^2}{a+1+b+1}=\frac{1}{3}.$$
Now we see that $\frac{1}{3}$ is a minimal value of the function $f(a,b)=\frac{a^2}{a+1}+\frac{b^2}{b+1}$

under conditions $a>0$, $b>0$ and $a+b=1$ (because the equality occurs for $a=b=\frac{1}{2}$).

Id est, we solved an optimization problem without standard Lagrange Multipliers method.

Specific this inequality is very easy, but of course  it can be found in research level mathematics.


Show with $\varepsilon$-$\delta$ that the function $f(x) = \frac{2x+3}{5}$ is continuous Show with $\varepsilon$-$\delta$ that the function $$f(x) =
\frac{2x+3}{5}$$ is continuous.

I found this task on the internet but it was without solution, so I'm asking here if I did it correctly.

Let $\varepsilon>0$ and let $\delta = \frac{\varepsilon}{2}$. If $|x-x_{0}|< \delta$, then:

$$\left |\frac{2x+3}{5} - \left(\frac{2x_{0}+3}{5}\right)\right | = \left|\frac{2x+3-(2x_{0}+3)}{5}\right|=\left|\frac{2x+3-2x_{0}-3}{5}\right|$$

$$=\left|\frac{2x-2x_{0}}{5}\right|<|2x-2x_{0}|=\left|2(x-x_{0})\right|<|2\delta|=\varepsilon$$

Therefor, the function is continuous.

Is it alright?

I'm not sure because I have just removed $5$ from the denominator and claimed it's greater without (it really is..). But am I actually allowed to do that?

Your proof is complete. Are you doubting that whenever $A < B$ and $B < C$, then also $A < C$ ? Don't :)


Can there be a new of adding vectors? Suppose we have a vector v=<a,b>. And another vector w=<c,d>. Now, when we do (v+w), we do <a+c.b+d>. I am asking why isn't there also another kind of vector addition such as (v+w)(We may use another appearance) =<a+d,b+c>?
I'm not saying there should be one. But I want some logics for it to not exist.
Reasons not accepted-
1.They are of no use in the real world-Mathematica has its own vast world. And I know there's many things in Mathematics that have no real counterpart...at least yet.
2.They don't have any geometrical interpretation-I don't think many ones have tried to give them geometrical meaning. It's like i before Gauss.
3.One dimension is independent of others- Give me reason why I can't add them above way I told.
Maybe they can open a new chapter of vector addition...

You can define vector spaces with very different kinds of addition, they only have to satisfy the vector space axioms. Therefore, when you choose a basis $\{v_i\}$ of your vector space, every vector needs to be able to be written as
$$ v = \sum_i \lambda_i v_i.$$
When we use component representation of a vector space we simply keep track of the $\lambda_i$, so we would write
$$ v = (\lambda_1,\dots,\lambda_n).$$
Therefore, if $w=(\mu_1,\dots,\mu_n)$, we have
$$v + w = \left(\sum_i \lambda_i v_i \right)+ \left(\sum_i \mu_i v_i \right)= \sum_i (\lambda_i + \mu_i) v_i = (\lambda_1+\mu_1,\dots, \lambda_n+\mu_n).$$
Thus the way that you add vectors in components is a direct consequence of the vector space axioms, and does not depend on the way we choose our addition in the vector space in the first place.


Number of ways to color a cube How many ways can you color the sides of a cube using three distinct
  colors such that no two opposite sides are painted the same color?

I tried this: Consider any side of the cube. You can color it in $3$ ways and then the other side can be colored in $2$ ways, which gives $3\times 2 = 6$ ways for each "pair" of opposing sides. Since there are $4$ such pairs of opposing sides, we get $6 \times 3 = 18$ ways.

However, I think that this overcounts the number of ways since I don't take the symmetries of a cube into account. I know that there aren't too many possibilities, and I can probably enumerate them all, but I was trying to find some clever way to count this quantity. Maybe someone knows a way.

So from each of the axes, exactly one colour of red, green, blue is missing.

If the same colour (say, green) is missing from all axes, then we can always move the cube in such a way that the top, the left, and the front face are red, and the bottom, the right, and the back face are blue. So there is really just one such pattern, but with three choices of missing colour, we count
$$ 3.$$
if one colour (green) is missing from one axis and the same other colour (red) from the other axes, then we can move the cube so that the top face is red and the bottom is blue. Now we can still rotate around the $z$-axis to achieve that the left and the front are green, and the right and the back are blue. Again there is really just one such pattern, but with three choices of once-missing colour and two choisec of twice-missing colour, we count
$$ 6.$$
If each colour is missing from one axis, we can move the green-free axis to the vertical and achieve that top is red and bottom is blue. Then we can still rotate to make the front red and the back green (from the blue-free axis). Now the cube is fixed, but we might have blue on the left and green on the right face or vice versa. Thus we have another
$$2. $$

In total, I count $$3+6+2=11.$$

We can first "fix" one side to be at the bottom. We then have $2\times 3=6$ possibilities to paint the bottom and the top side. For each of these $6$ possibilities, we have another $6$ to paint left and right and for each of those another $6$ for front and back of the cube. This leaves us with $6\times 6\times 6=216$ possibilites.

Now we can take symmetry into account. There are six faces which can be at the bottom and for each of those another 4 ways to rotate the cube. This leaves us with a total of $\frac{216}{6\times 4}=9$ ways of painting the cube.


Prove that $f$ has derivatives of all orders at $x=0$, and that $f^{(n)}(0)=0$ for $n=1,2,3\ldots$ Define that $f(x) = e^{-1/x^2}$ when $x$ is not $0$, and $f(x) = 0$ when $x=0$.

Prove that $f$ has derivatives of all orders at $x=0$, and that $f^{(n)}(0)=0$ for $n=1,2,3,\ldots$

Using induction, it is easy to see/prove for $x \ne 0$, $f^{(n)}(x)$ has the form
$Q_n(\frac{1}{x}) e^{-\frac{1}{x^2}}$ for some polynomial $Q_n$ of degree $3n$ in $\frac{1}{x}$ $\color{blue}{^{[1]}}$.

By definition $f(0) = 0$. Assume we have shown
$$f(0) = f'(0) = \ldots f^{(n-1)}(x) = 0,$$
then by definition again:
$$f^{(n)}(0) := \lim_{h\to 0} \frac{f^{(n-1)}(h) - f^{(n-1)}(0)}{h} = \lim_{h\to 0} \left(\frac{1}{h}Q_{n-1}(\frac{1}{h})\right) e^{-\frac{1}{h^2}}\tag{*}$$

Being a polynomial of degree $3n$, for sufficiently small $h$ and hence large $\frac{1}{h}$, say $|h| < \epsilon$, we can find a $M > 0$ such that $\color{blue}{^{[2]}}$:

$$|Q_n(\frac{1}{x})| < \frac{M}{|x|^{3n}}\quad\text{ for } x \in (-\epsilon,\epsilon) \setminus \{0\}$$

As a result $\color{blue}{^{[3]}}$, $$\limsup_{h\to 0} \left| \left(\frac{1}{h}Q_{n-1}(\frac{1}{h})\right) e^{-\frac{1}{h^2}} \right| \le \lim_{h\to 0}\frac{M e^{-\frac{1}{h^2}} }{|h|^{3n+1}}
=\lim_{r\to+\infty} M r^{3n+1} e^{-r^2} = 0$$

This implies the limit in R.H.S of $(*)$ exists and vanishes. i.e. $f^{(n)}(0)$ also vanishes.

By induction again, we get $f^{(n)}(0) = 0$ for all $n \in \mathbb{N}$.

Notes

$\color{blue}{[1]}$ $Q_n(t)$ can be computed using the recurrence relation:
$$Q_n(t) = \begin{cases}
1, & n = 0\\
2t^3 Q_{n-1}(t) - t^2 Q'_{n-1}(t), & n > 0
\end{cases}$$

$\color{blue}{[2]}$ Recall for any polynomial $\sum_{k=0}^{m} a_k x^k$ of degree $m$, we have
the inequality:
$$ \left|\sum_{k=0}^{m} a_k x^k\right| < \left( \sum_{k=0}^{m} |a_k| \right) \max( |x|^m, 1 )$$

$\color{blue}{[3]}$ Notice when $r$ is positive,

$$e^{r} > 1 + r > r, \forall r
\quad\implies\quad e^{\frac{r}{3n+1}} > \frac{r}{3n+1}, \forall r
\quad\implies\quad e^{r} > \left(\frac{r}{3n+1}\right)^{3n+1}, \forall r
$$
When $r > 2$, we get:
$$r^{3n+1} e^{-r^2} < (3n+1)^{3n+1} e^{r-r^2} < (3n+1)^{3n+1} e^{-r}$$
and the R.H.S tends to $0$ as $r \to \infty$.

Obviously the left side of the function, being a constant zero, has infinitely many zero derivatives coming into zero from the left.  So we have to show it has infinitely many and they are all zero coming in from the right.  The hint nbubis gives is not so much a hint as an answer provided you know that the exponential function will always swamp out the polynomial.

Since the 1/x tends to confuse things, I think there is a clearer way to see this. That is to consider the function

g(x) = f(1/x) = e^(-x^2).

Equivalent to f(x) → zero, we observe g(x) as x → +∞.  These derivatives clearly exist and just as clearly are zero for the same reason as above -- the exponential goes to zero faster than the polynomials can rise.

This is the standard example used to demonstrate that although a function may have infinitely many derivatives at a point, it is not necessarily analytic there.  Viewed simply as a function of a real variable, this means that its Taylor series will not converge; however proving that on the real line could be quite a pain.

Yet it can be shown very easily by going to the complex plane, where f'(z) simply does not exist at zero.  To see that, evaluate f'(z) (or the corresponding limit) along the i-axis. f'(z) = 2z^(-3)*f(z). Since f(i) ≡ 1 (check it out), this gives f'(i)= 2*i^(-3) which certainly is not going to zero as i goes to zero.  So there is no complex derivative at zero.  Various theorems from the study of complex functions state that f(z) is analytic at a point iff it has a derivative there as a function of a complex variable.


$$f(x)\in \left[2m\pi+\dfrac{\pi}{3},2m\pi+\dfrac{2\pi}{3}\right] \text { where m is integer }$$

This isn't true because the range of the inverse sine function is from $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Let $$\theta=\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$$
then
$$\sin{\theta}=\frac{\sqrt{3}}{2}$$
with the restriction that $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. The possible triangles with this ratio of $y$ to hypotenuse (i.e. $\sin$) are shown below

where only $\pi/ 3$ falls within the restricted range for the inverse sine function. Similarly, letting

$$\theta=\sin^{-1}\left(1\right)$$

means

$$\sin{\theta}=1$$

where only $ \pi/ 2$ falls within the restricted range for the inverse sine function.

The domain of $$f(x)=\sin^{-1}(\sqrt{x^2+x+1})$$ is where you get $$0\le x^2+x+1\le 1$$ which is the interval $[-1,0]$

Note that $f(-1/2)=\pi/3$ is the minimum value and $f(-1)=f(0)= \pi/2$ is the maximum.

Thus the range is the interval  $[\pi/3, \pi/2]$


Galois closure of a $p$-extension is also a $p$-extension I'm working on a problem in Dummit & Foote and I'm quite stumped. The problem reads:

  Let $p$ be a prime and let $F$ be a field. Let $K$ be a Galois extension of $F$ whose
  Galois group is a $p$-group (i.e., the degree $[K:F]$ is a power of $p$). Such an
  extension is called a $p$-extension (note that $p$-extensions are Galois by
  definition).

  Let $L$ be a $p$-extension of $K$. Prove that the Galois closure of $L$ over $F$ is a $p$-extension of $F$.

This is what I've done so far:

  Using the tower law we can readily show that $L$ is a $p$ extension of $F$ so we have $[L:F]=p^{\ell}$ for some integer $\ell$.  Then if $M$ is the Galois closure of $L$ over $F$ then $$[M:F]=[M:L][L:F]$$ and therefore $[M:F]=p^{\ell}n$ for some integer $n$ that is not divisible by $p$.  So $[M:L]=n$.

From here it seems like I want to show that either $n=1$ or that $n$ is in fact a power of $p$. I just don't see how to proceed. I've considered using the Sylow Theorems, but I'm not sure how that would really work.  I also realize that this statement depends on $K$ being Galois over $F$ but I can't figure out how to take advantage of that.

First, $L/F$ is a finite and separable extension, thus by the primitive element theorem $L = F(\alpha)$ for some $\alpha \in L$. Let $m_\alpha \in F[X]$ be the minimal polynomial of $\alpha$.  Since $K/F$ is Galois, over $K$, $m_\alpha$ splits into irreducible factors of the same degree, $d$ say. In Dummit-Foote this last statement is Exercise 14.4.4, the exercise right before the one in the question.  Since $d$ divides $p^\ell = [L:F]$, it is a power of $p$ itself.  This proves

Claim 1.  For any root $\beta \in M$ of $m_\alpha$, $K(\beta)/K$ has degree a power of $p$.

Now the Galois closure $M$ is the composite field of the $K(\beta_i)$ where $\beta_i$ runs through the roots of $m_\alpha$.  Reason: The composite field is the splitting field of $m_\alpha$, so it is Galois and contains $M$.  On the other hand $M$ contains the splitting field of $m_\alpha$ because it contains the root $\alpha$.

Claim 2. $K(\beta)/K$ is Galois for every root $\beta$ of $m_\alpha$.

Proof. $L/K = K(\alpha)/K$ is Galois.  Let $\phi : K(\alpha) \to K(\beta)$ be the isomorphism induced by $\alpha \mapsto \beta$ and the identity on $K$. Then $\text{Aut}(K(\alpha)/K) \cong \text{Aut}(K(\beta)/K)$ via $\sigma \mapsto \phi\sigma\phi^{-1}$.

Finally, the Galois group of $M$ (the composite of the $K(\beta)$) over $K$, is a subgroup of the direct product of the Galois groups of $K(\beta)/K$, which are all $p$-groups by Claim 1.  Therefore it is a p-group by Dummit-Foote Proposition 14.12.  Therefore $[M:K]$ is a power of $p$ and since $[K:F]$ is a power of $p$ and the same follows for $[M:F]$ by the tower formula.

We denote by $G(K/F)$ the Galois group of a Galois extension $K/F$.
We need the following lemma.

Lemma
Let $F$ be a field, $\Omega$ its algebraic closure.
Let $K_1$ and $K_2$ be subfields of $\Omega$ each of which is a Galois extension of $F$.
Then $K_1K_2$ is a Galois extension of $F$ and $G(K_1K_2/F)$ is isomorphic to a subgroup of
$G(K_1/F)\times G(K_2/F)$.

Proof.
Let $\sigma$ be an automorphism of $\Omega/F$.
Then $\sigma(K_1K_2) = \sigma(K_1)\sigma(K_2) = K_1K_2$.
Hence $K_1K_2/F$ is a normal extension.
Clearly it is a separable extension.
Hence $K_1K_2/F$ is a Galois extension.
We define a homomorphism $\psi\colon G(K_1K_2/F) \rightarrow G(K_1/F)\times G(K_2/F)$
by $\psi(\sigma) = (\sigma\mid K_1, \sigma\mid K_2)$.
Since it is clearly injective, we are done.

Now let $F$ be a field, $\Omega$ its algebraic closure.
Let $F \subset K \subset L \subset \Omega$ be a tower of fields such that $K/F$ and $L/K$ are $p$-extensions. Let $M/F$ be the Galois closure of $L/F$ in $\Omega$.
Then $M = \sigma_1(L)\cdots \sigma_n(L)$, where $\sigma_i(L), i = 1,\cdots, n$ are all the distinct conjugates of $L$ over $F$ in $\Omega$.
Since each $\sigma_i(L)$ is a Galois extension over $K$, $G(M/K)$ is isomorphic to a subgroup of
$G(\sigma_1(L)/K)\times \cdots \times G(\sigma_n(L)/K)$ by the lemma. Since each $G(\sigma_i(L)/K)$ is a $p$-group, $G(M/K)$ is a $p$-group. Since $K/F$ is a $p$-extension, $(M : F)$ is a power of $p$.
Hence $M/F$ is a $p$-extension and we are done.


Non-textbook Math book recommendation to read to my kids I'm looking for a book to read to my kids.
NOT a kids book, but not too mature for a kid. My youngest kid that reads with me is 6 and the eldest is 10.

I'm looking for a book that is good literature, and is hard to put down.
Also maybe a book that makes the subject in school more interesting. Not just tedious memorization, but presenting it in a way that makes the child excited about what he/she is learning. And not a book that teaches the subject, (math etc....) but a book that teaches how it got started or why it's important.

Are there any books like these you that you have read that made you excited about math? If so please tell me.

When I was a child, I really liked the book The Number Devil by Hans Magnus Enzensberger.

I don't know if it's suitable for under 10 since I've only browsed through it a little, but my girlfriend really liked this one: The Simpsons and their Mathematical Secrets, by Simon Singh

https://www.simonsingh.net/Simpsons_Mathematics/


Explain how to get the right solution of y $dy/dx=y$ When solving the following equation to find y as a function of x:

\begin{equation}
dy/dx=y
\end{equation}

First I divide both sides by $y$ and multiply both sides by $dx$:

$dy/y=dx$

Then I integrate both sides:

$\ln(y)+C1=x+C2$

Then:

$y=e^{x+C2-C1}=e^{x+C}$

I now know my result is problematic becomes $y=-e^x$ also satisfies $dy/dx=y$ but it is included in my solution: $y=e^{x+C}$. So can you help me find what is wrong with my analysis process and provide the right analysis process.

Thanks.

Update:
I said $y=-e^x$ is not included in $y=e^{x+C}$ because $e^C>0$.

It should be noted that $$\int \frac{dy}{y} =\ln|y| + C$$ That absolute value is important.

This means that $|y| = De^x$ for some positive constant $D$.

The really odd trick is that $D=0$ works, as well...

$$\frac{dy}{dx} = y \iff$$
$$\frac{1}{y} dy = dx \iff$$
$$ \int \frac{1}{y} dy = \int dx \iff $$
$$ \ln(y) + C_1 = x+C_2 \iff$$
$$ \ln(y)  = x+C \iff$$
$$e^{\ln(y)}  = e^{x+C} \iff$$
$$e^{\ln(y)}  = e^{x}e^{C} \iff$$

$$y = D e^x$$

where $D$ may be  chosen to satisfy the initial conditions:

For example, if $y(0) = \pi$, then $y(0) = De^0 = \pi \implies D=\pi$


Proof of the angle sum identity for cos using series Let $sin(x)=\sum_{n=0}^{\infty}{(-1)^n\frac{x^{2n+1}}{(2n+1)!}}$ and $cos(x)=\sum_{n=0}^{\infty}{(-1)^n\frac{x^{2n}}{(2n)!}}$.How using above definition prove that $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$?

When you multiply the series for $sin(x)$ by $i$, and add it to the series for $cos(x)$, you get the series for $exp(ix)$. In other words, $cos(x)$, is the real part of $exp(ix)$.

Therefore, $cos(x+y)$ is the real part of $exp(i(x+y))$. By laws of exponents, this is $$real( exp(ix)exp(iy))$$

By the first observation, this is $$real( (cox(x)+i sin(x)) (cox(y)+i sin(y)))$$

Multiply it out and select the real part: $$cos(x)cos(y) - sin(x)sin(y)$$

Very bad answer.  One cannot use complex analysis to prove the sum of angles formula, one must in fact use the sum of angles formula to prove all of the useful properties of the "complex exponential" (which is based on the definitions of sine and cosine!).  Otherwise you are just playing with definitions and giving circular reasoning.  The proof should be self-contained, involving multiplying the partial sums of the series and re-arranging the terms in a clever way.


The solutions to $x^2+5=y^2$. Consider the equation $$x^2+5=y^2.\tag{1}$$

If $x,y\in\mathbb{Z}$, what are solutions to (1)?
If $x,y\in\mathbb{Q}$, what are solutions to (1)?

Note: $\mathbb{Z}$ is the set of all integers and $\mathbb{Q}$ is the set of all rational numbers.

My attempt: For problem 1, I run the following Mathematica program:
Solve[x^2 + 5 == y^2, {x, y}, Integers]
The output is
{{x -> -2, y -> -3}, {x -> -2, y -> 3}, {x -> 2, y -> -3}, {x -> 2,
  y -> 3}}
I do not know if there is any other solutions. If the above result is correct, I would like to know why there is no other solutions.

For problem 2, I run the following Mathematica program:
Solve[(x/a)^2 + 5 == (y/b)^2 && a != 0 && b != 0 && GCD[x, a] == 1 &&
  GCD[y, b] == 1, {a, b, x, y}, Integers]
The output is the same as the input, it means that Mathematica cannot solve this equation.

Any comments and answers are welcome. Thank you very much.

For solutions over $\Bbb Z$ see comments by others.

Finding solutions over $\Bbb Q$ is equivalent to solving
$$x^2+5z^2=y^2$$
over $\Bbb Z$, with $x,y,z$ coprime.  Without loss of (much) generality assume that $x,y,z$ are positive.  Two points first:

$x,z$ cannot both be even as then $y$ is even, contradicting coprimality;
$x,z$ cannot both be odd as then $y^2\equiv2\pmod4$ which is impossible.

Thus one of $x,z$ is even , the other odd, so $y$ is odd.  We have
$$5z^2=(y-x)(y+x)\ .$$
First consider the case $x$ odd, $z$ even.  Then $y-x$ and $y+x$ are both even and
$$5\Bigl(\frac{z}{2}\Bigr)^2=\frac{y-x}{2}\frac{y+x}{2}\ .$$
The two factors on the RHS are coprime, so one of them is a square and the other is $5$ times a square.  So there are two possibilities:

$(y+x)/2=5a^2$, $(y-x)/2=b^2$, so $x=5a^2-b^2$, $y=5a^2+b^2$, $z=2ab$.
$(y-x)/2=5a^2$, $(y+x)/2=b^2$, so $x=b^2-5a^2$, $y=5a^2+b^2$, $z=2ab$.

In both of these $a,b$ must be coprime; and one is even, one odd (since $x$ is odd).  This gives positive rational solutions of the original equation as
$$\left|\frac{5a^2-b^2}{2ab}\right|\ ,\quad \frac{5a^2+b^2}{2ab}\ .$$

Similarly in the case $x$ even, $z$ odd, $y-x$ and $y+x$ are coprime; similar working leads in fact to the same formulae but with with $a,b$ both odd.  So the complete solution is as above, with $a,b$ having no common factor.  (And of course if you do allow them to have a common factor you get repetitions of the same solutions.)

Why do you need computer/software  help for this small problem?  You are looking for two squares differing by 5. Expand $(x+1)^2$. See how much it differs from $x^2$. This would give you an idea about the difference about consecutive squares.


Derivative of $\; y={(1+e^x)}^{0.5}\; $ using the definition of the derivative $$y={(1+e^x)}^{0.5} =f(x)$$

$$\frac{dy}{dx}= \lim_{h\to0}\frac {f(x+h)-f(x)}{h}$$

My attempt I got down to
$$\lim_{h\to0}\frac{(1+e^xe^h)^{0.5}-(1+e^x)^{0.5}}{h}$$

I can't see where to go from here

Provided you know

$$\lim_{h\to0}{e^h-1\over h}=1$$

you can take what you got to and write

$$\begin{align}
{(1+e^xe^h)^{0.5}-(1+e^x)^{0.5}\over h}&={(1+e^xe^h)^{0.5}-(1+e^x)^{0.5}\over h}\cdot{(1+e^xe^h)^{0.5}+(1+e^x)^{0.5}\over(1+e^xe^h)^{0.5}+(1+e^x)^{0.5}}\\
&={(1+e^xe^h)-(1+e^x)\over h\left((1+e^xe^h)^{0.5}+(1+e^x)^{0.5}\right)}\\
&={e^x\over(1+e^xe^h)^{0.5}+(1+e^x)^{0.5}}\cdot{e^h-1\over h}
\end{align}$$

When you take the limit now, you get $e^x/2(1+e^x)^{0.5}$.

The multiplication step is a standard trick for working with a square root.  Computing the limit of $(e^h-1)/h$ requires more than the definition of the derivative; it requires some definition of the exponential function.


Definition for Covariant Derivative What is simple definition of the covariant derivative that looks like the definition of the derivative of a function in calculus?

definition of the derivative of a function in calculus is:

$$\frac {df(x)}{dx}=\lim_{\Delta x\to o}\frac {f(x+\Delta x)-f(x)}{\Delta x}$$

what about Covariant derivative, what is definition of the Covariant derivative of a function in calculus?
$$\frac {{\mathcal D}f(x)}{dx}$$

Remark: ok lets say that:

$$\frac {{\mathcal D}f(x)}{dx}=\frac {df(x)} {dx} +\delta f(x)$$ where the covariant derivative is broken into two parts, the extrinsic normal component and the intrinsic covariant derivative component . Now, how do we define a simple definition for the intrinsic covariant derivative component $\delta f(x)$ (this small addition is the result of parallel translation)

There is a very intuitive way to understand the covariant derivative (for the Levi-Civita connection) for the special case of isometrically embedded submanifolds in $\mathbb R^n$. Roughly speaking, first take the usual derivative in $\mathbb R^n$, and then project the answer onto the tangent plane of the submanifold. The image of the projection is then the covariant derivative.

In fact, this is how the concept of covariant derivative if often introduced to undergraduates in their first differential geometry course. For instance, if you consider the sphere $S^2 \subset \mathbb R^3$, then it is easy (both analytically and visually) to prove that the tangent vector field to a great circle has zero covariant derivative and therefore that great circles must be geodesics.

Sounds like it is Geodesic equation:
$$\frac {d^2x^a}{ds^2}+\Gamma^a_{bc}\frac {dx^b}{ds}\frac {dx^c}{ds}$$
i really love this equation. this $\Gamma^a_{bc}\frac {dx^b}{ds}\frac {dx^c}{ds}$ is $\delta f$


Sylvester rank inequality: $\operatorname{rank} A + \operatorname{rank}B \leq \operatorname{rank} AB + n$ If $A$ and $B$ are two matrices of the same order $n$, then
  $$ \operatorname{rank} A + \operatorname{rank}B \leq \operatorname{rank} AB + n. $$

I don't know how to start proving this inequality. I would be very pleased if someone helps me. Thanks!

Edit I.  Rank of $A$ is the same of the equivalent matrix $A' =\begin{pmatrix}I_r & 0 \\ 0 & 0\end{pmatrix}$. Analogously for $B$, ranks of $A$ and $B$ are $r,s\leq n$. Hence, since $\operatorname{rank}AB = \min\{r,s\}$, then $r+s\leq \min\{r,s\} + n$. (This is not correct since $\operatorname{rank} AB \leq \min\{r,s\}$.

Edit II. A discussion on the rank of a product of $H_f(A)$ and $H_c(B)$ would correct this, but I don't know how to formalize that $\operatorname{rank}H_f(A) +\operatorname{rank}H_c(B) - n \leq \operatorname{rank}[H_f(A)H_c(B)]$.

Subtract $2n$ from both sides and then multiply by $-1$; this gives the equivalent inequality $\def\rk{\operatorname{rank}}(n-\rk A)+(n-\rk B)\geq n-\rk(AB)$. In other words (by the rank-nullity theorem) we must show that the sums of the dimensions of the kernels of $A$ and of $B$ gives at least the dimension of the kernel of $AB$. Intuititively vectors that get killed by $AB$ either get killed by $B$ or (later) by $A$, and the dimension of $\ker(AB)$ therefore cannot exceed $\dim\ker A+\dim\ker B$. Here is how to make that precise.

One has $\ker B\subseteq \ker(AB)$, so if one restricts (the linear map with matrix) $B$ to $\ker(AB)$, giving a map $\tilde B:\ker(AB)\to K^n$, one sees that $\ker(\tilde B)=\ker(B)$ (both inclusions are obvious). Since the image of $\tilde B$ is contained in $\ker A$ by the definition of $\ker(AB)$, one has $\rk\tilde B\leq\dim\ker A$. Now rank-nullity applied to $\tilde B$ gives $$\dim\ker(AB)=\dim\ker\tilde B+\rk\tilde B\leq\dim\ker B+\dim\ker A,$$ as desired.

In response to Mr. Albanese's correct critique. I submit the following. The rank of A,B is in the set of integers [0,n]. Assume we know the relatively straigtfoward proofs that rank(AB)<= min(rank(A),rank(B)), -> rank(AB) = n if rank(A) and rank(B) have rank n. There are three rank possibilities. Let the rank of A = p, the rank of B = q with p,q integers. Then (p or q < n) or (p and q < n) or (p=q=n). The first two cases imply (rank(A) + rank(B) < rank(AB) + n) or (p+q < min(p,q) + n) as p+q < 2n and (rank(A) + rank(B)  > rank(AB) + n) is a contradiction. This is a contrapositive proof of the strict inequality in the Sylvester theorem (rank(A) + rank(B) < rank(AB) + n).  For the equality case p=q=n. Then (rank(A) + rank(B)  = rank(AB)) + n) or n + n = n + n which is true because rank(A) and rank(B) have full rank, n, implies rank(AB)=n.


Bounds of sum by using integral Is it correct to say?

$$\int _1^{n+1}\:\frac{1}{x^3}dx\le \sum _{k=1}^n\:\frac{1}{k^3}\le 1+\:\int _1^n\:\frac{1}{x^3}dx$$

Therefore:

$$\sum _{k=1}^n\:\frac{1}{k^3}\le 1+\:\int _1^n\:\frac{1}{x^3}dx=\frac{3}{2}-\frac{1}{2n^2}\le \frac{3}{2}$$

And the same thing for the other side.

Yes, that's right. Since the system won't allow such a short answer, I'll add that indeed

$$\sum_{k=1}^\infty\frac1{k^3}=\zeta(3)\approx1.202\lt\frac32\;.$$


Prove that every $k$-chromatic graph has size $m\geq \binom k2 $ Prove that every $k$-chromatic graph has size $m\geq \binom k2$.

Here is what I know:

Let $G$ be a $k$-chromatic graph, that mean $\chi(G)=k$. Thus $G$ must have a subgraph of a complete $k$-partite graph, say $H$. Hence $m_G \geq m_H$.

Here are my questions

1) I'm kinda confused between $k$-coloring, $k$-colorable and $k$-chromatic.

I know that $G$ is $k$-colorable means $\chi (G) \leq k$ and $\chi(G)$ is the minimum number of colors needed for a legal coloring of $G$. Then the book said if $G$ is $k$-chromatic then there exists a $k$-coloring but no $(k-1)$-coloring.
I feel like the book is just kicking me around.

2) I know $\binom k2$ is the size of a complete graph of order $k$, but I can't see how a $k$-partite graph has that size.

Let G be a $k$ chromatic graph of order $n$ and size $m$.  Let there be given a coloring of $G$ using $k$ colors.  Denote the color classes of G by $C_1,C_2,...C_k$.
We're gonna run the ol' proof by contradiction here.  I'm not sure if it's necessary, but whatever, I like it.  So, we assume that $m<\binom{k}{2}$ Now, doesn't this already seem weird?  $G$ has less edges than $K_k$?  We get our contradiction from this last curiosity.  Consider the graph $H$, formed by associating each color class, $C_i$, with a vertex, $u_i$ where$\; 1\le i\le k$ and $u_iu_j\in E(H), i\ne j, $ if and only if $\exists$ an edge between any vertex in $C_i$ and any vertex in $C_j$.  Since $|E(G)|=m<\binom{k}{2}$ and since $H$ necessarily has less edges than $G$, it follows that $H$ is not a complete graph.  Therefore, there exists 2 color classes, $C_m$ and $C_n$  $(m\ne n),$ with no edge between them.  Change the color of all vertices in $C_m$ to the color of the vertices in $C_n$.  This produces a coloring of $G$ with $k-1$ colors, a contradiction.  Therefore, it must be the case that $m\ge \binom{k}{2}$

Basically for a $k$ chromatic graph with $n$ vertices, there  are $k$ edges such that all are adjacent to each other which mean every 2 vertices of those $k$ vertices are connected, i.e, there are at least $_kC_2$ (or $\binom k2$) edges in the graph.


Can we find a formula defining a recursively enumerable set? By Post's Theorem we know that a set $A\subseteq\mathbf{N}$ is recursively enumerable iff it is definable by a $\Sigma_1$-formula, i.e. there exists a $\Sigma_1$-formula $\varphi(x)$ with $x$ free such that for every number $n$:
\[
n\in A\longleftrightarrow \mathfrak{N}\vDash\varphi(\overline{n})
\]
where $\mathfrak{N}$ is the standard model of the first-order language of Peano Arithmetic.

I have the following question: given a r.e. set $A$ can we always find a $\Sigma_1$-formula defining it?

The definition of a recursive enumerable set is that it is the domain of some partial recursive function.

There is a recursive primitive function $\psi$, such that $\psi(n,t,x)=0$ if and only if $\phi_n(x)$ (the recursive function $n$ on entry $x$) halts in less than $t$ steps, else $\psi(n,t,x)=1$. Any recursive primitive function can be defined by a $\Delta_0$-formula. Hence

$$\phi_n(x)\mbox{ halts } \Leftrightarrow \exists t\; \psi(n,t,x)=0$$

The existence of $\psi$ is a consequence of Kleene T-predicate

By the axiom of choice there is a function that maps each r.e. set to one of its definitions. We can 'find' the definition in that sense. This function cannot be too nice, however, because that would decide the extensional equivalence of two definitions.

Let $f$ be a function that maps each set to one of its definitions. For each arithmetical statement $p$ there is a set $\{n|p\}$. The equality of definitions $f(\{n|p\}) = f(\mathbb N)$ is way to decide if $p$ is true. G\"odel's incompleteness theorems say we can't do that.


Proving something is a real inner product space I have this question here that I'm trying to solve. I need to show that $S$ is equipped with an inner product and is thus an inner product space.
I've started out by trying to prove this with the axiom of symmetry, which it satisfied, but where it gets confusing for me is trying to prove it using the remaining axioms of additivity, homogeneity and positivity.
How would I go about proving $S$ is an inner product space using the remaining axioms, if that is in fact the correct way of going about it?

In order to show that this is an inner product, you do indeed need to check every property from the definition. Let me show an easy way to do it: first, you can write this operation as:
$$\left<u,v\right> = \begin{pmatrix} u_1 & u_2 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$$
It now suffices to show that this matrix $A = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$, which is the form's associated matrix in the standard basis, is symmetric and positive definite. You can clearly see it's symmetric. For positive-definiteness, you can use Sylvester's Criterion. You can check that the principal minors of the matrix are $2$ and $2 - (-1)(-1) = 1$, both positive, and hence the bilinear form is positive definite, and, with everything we've seen so far, an inner product.


For which $y$ is $3 x^4-4 x^2 y+y^2<0$ How does one show that the function $f(x,y)=3 x^4-4 x^2 y+y^2$ is negative for $x^2\lt y<3x^2$?

I do not understand what exactly to do to determine this. I have studies linear inequalities before.

My poor attempt:

$3 x^4-4 x^2 y+y^2<0 \Leftrightarrow y^2-4x^2y<-3x^4 \Leftrightarrow y(y-4x^2)<-3x^4$

But i'm not sure how to proceed from here. Any help is appreciated


Why is $\cos (90)=-0.4$ in WebGL? I'm a graphical artist who is completely out of my depth on this site.

However, I'm dabbling in WebGL (3D software for internet browsers) and trying to animate a bouncing ball.

Apparently we can use trigonometry to create nice smooth curves.

Unfortunately, I just cannot see why.

I can accept this diagram:

However, running some calculations just do not make sense to me:

Let's set $\alpha$ to 45 (around where it appears to be in the diagram) and find the cosine value, thus giving us the green line.  $$\cos(45) = 0.5$$

Fair enough. $\cos(\alpha)$ / the green line is $0.5$ units.

But now this is where it all falls apart. I would have thought if we set $\alpha$ to $90$, $\cos$ would become $0$. Do you see why I think this? Look at the digram, isn't that reasonable to think? Similarly, $\cos(0)$ I would have said should equal $1$ (twice that of $\cos(45)$ )

While $\cos(0)$ does equal $1$, this does not check out: $$ cos(90) = -0.4$$

Just do not get that $0.4$? Could someone explain? That just makes no sense to me. None.

I'm using the google calculator and I would stress I have not touched maths for about $6$ years (ever since I left school!) so please lots of examples and words to explain!

The cosine function takes radians as its argument in most computational languages.

Indeed, $\cos 45^{\circ} = \sqrt{0.5}$, and $\cos 90^{\circ} = 0$. However, in math, for various reasons, we don't like working with degrees. We work with radians, where $2\pi\ \textrm{radians} = 360^{\circ}$.

In fact, $\cos 90\ \textrm{radians} \approx -0.44807$ and $\cos 45\ \textrm{radians} \approx 0.525$.

90 radians would be about 5156 degrees, or about 14.3 turns around the circle!

Wrong unit.
You talk about deg, while the function obviously expects rad, which gives -0.4.


Is C is basis of vector space V If $\{e_1, e_2, e_3\}$ are basis of vector space $V$ over $\Bbb{R}$ then is
$C=\{ 2e_1, 3e_{1}+e_{3}, 6e_{1}+3e_{2}+e_{3} \} $ a basis of $V$?
Can anyone help ?
$C$ can be reduce to trivial basis given by some column operation so I think it must be basis .
But my textbook says its not!!

The equation
\begin{align}
x_1(2e_1)+x_2(3e_1+e_3)+x_3(6e_1+3e_2+e_3)=0
\end{align}
is equivalent to
\begin{align}
(2x_1+3x_2+6x_3)e_1+3x_3e_2+(x_2+x_3)e_3=0
\end{align}
Now $e_1,e_2,e_3$forms a basis, thus we have
\begin{align}
2x_1+3x_3+6x_3&=0\\
3x_3&=0\\
x_2+x_3&=0
\end{align}
This implies the family $C$ is linear independent, hence a basis.

Hint :

You can write the matrix of passage from $\{e_1, e_2, e_3\}$ to C. Your problem is then: is that matrix invertible ?
I think you're right. Can you show us exactly how the result is given in your textbook ?


9-Bits Game, a Brain Teaser on Information Theory or Cryptography This question was asked in a recently interview, I didn't solve it.

Suppose there are two very smart people Alice and Bob, that participate in a game. The game is set as follows.

  Some computer generates 9 random 0/1 bits in a sequence $C_i, i=1,2,3,...,9$.
  Before the rounds start, Alice takes a look at the sequence and remembers the entire sequence.
  The game has nine rounds.
  At the start of the i-th round, Bob enters a bit (0 or 1) $B_i$, then Alice enters another bit $A_i$. If $A_i=B_i=C_i$ then they win the round, else they lose.
  Alice and Bob know $A_i,B_i,C_i$ just after the result of the round.
  Round i ends, round i+1 starts, go to step 4.

Alice and Bob can develop a strategy before the game starts, but during the game they are not allowed to communicate with each other.

Q1. Is there a strategy for them to win at least 6 rounds?

Q2. What is the optimal solution measured by the expectation of the winning rounds?

EDIT:

For Q1, I have some idea. Bob can receive information during the mismatched rounds.

$P_k$ denote the guaranteed winning rounds of totally $k$ round.

Obviously, $P_k \ge \dfrac k2$ when $k$ is even.

Strategy is simple, $A_i := C_{i+1},B_{i+1} := A_i$, when i is odd. $A_i := C_i$, when i is even.

And we should have the following relationships.

$P_{k+1} \ge P_k,P_{k+1} \le P_k+1$

Easy to prove, $P_1=0, P_2=1$.

When $k=3, 1=P_2 \le P_3 \le P_2+1=2$

We want to testify whether $P_3$ could be 2.

Considering the extremely bad case, the first round doesn't match. Bob only have 1-bit information, it is impossible to cover every 2-bit case.

So $P_3=1$

$2=\dfrac42 \le P_4 \le P_3+1=2$, so $P_4=2$.

$2=P_4 \le P_5 \le P_4+1=3$ , we testify if $P_5$ could be 3.

I come up with a rather complex strategy.

Let $B_1=1$, if $C_1=1$, then everything done.

If $C_1=0$, $A_1 := \text{most frequent bit in} \{C_2, C_3, C_4\}, B_2=B_3=B_4 := A_1$.

If $C_2=C_3=C_4$ then done.

If they are not same, suppose $C_2=1,C_3=C_4=0$. Let $A_2:=C_5, B_5:=A_2$, problem solved.

Another post of just thoughts (for now):  The default idea, as has been noted, is basically for Alice to use the first bit to tell Bob whether there are more 1s or more 0s.  Now, Bob may be wrong multiple times if he just keeps guessing $A_1$.  But whenever Bob gets it wrong, Alice's bit can be taken as a bit of information given to Bob.

The interesting observation then is that since Alice knows when Bob will get it wrong, Alice could do as above, but she could also purposely get it wrong early to transfer some extra information.  Then, for example (based on the strategy) Bob might know that he was about to get the next bit wrong (so we haven't lost anything) but also some additional information based on the fact that Alice chose to get it wrong early rather than waiting for Bob to get it wrong.  For example, this could be used to tell Bob that he was about to get the next TWO bits wrong (even given the information that Alice would have given in the first bit Bob would get wrong).  Then, for example, Alice could get it wrong again on purpose during that next bit or two to transmit that it is some kind of worst case scenario like 101010101.  This is hard to turn into an actual strategy but it makes me think there is actually some strategy to guarantee 6.  A naive upper bound on the amount of information given by Alice is something like $2^3 {8 \choose 2}$ since she gives 3 bits of information and can get those bits wrong in two of the last 8 bits.  And this is a decent amount larger than $2^6$.

Probably not the correct answer, but you wrote:

  Before the game start, Alice take a look at the sequence and remember them all

and

  Alice and Bob could develop a strategy before the game start

Since they are both before the game starts, Alice could tell Bob the sequence.

The expected win would be 9 out of 9.


Upper bound of coefficient of variation computed with mean average deviation From $n$ data $x_i$ in $[0,1]$, I am computing a coefficient of variation $C=d/\mu$ where $d$ is the mean average deviation $d=1/n \sum_{i=1}^n |x_i-\mu|$ and $\mu$ is the standard mean $\mu=1/n \sum_{i=1}^n x_i$.

I am looking for an analytical expression of the upper bound of $c$.

I found that the case of largest deviation happens when $n-1$ data equals 0 and 1 equals 1. In this case, $\mu=1/n$, $d=2(n-1)/n^2$, so $C=2(n-1)/n$

However, I am not sure how to show that the maximum deviation happens when all but 1 data equal 0. Moreover, I think this result (if it is right) must have been shown properly before, but I cannot find by who and where.

Can you help me regarding the demo and the 2 latter points ?

Thanks !

EDIT

I think the following could be a demonstration, but I still have a doubt, as indicated below.

First, let us consider the data are sorted in ascending order and denote $l$ the number of points lesser or  equal than $\mu$, i.e. $l=|\{x_i \text{ s.t. } x_i\leq\mu\}|$. Then we define $\theta$ the sum of their values i.e. $\theta=\sum_{i=1}^l x_i$ and $\theta'=\sum_{i=l+1}^n x_i$, so:
\begin{align*}
C & =\frac{d}{\mu}\\
 & =\frac{\sum_{i=1}^{n}\left|x_{i}-\mu\right|}{\sum_{i=1}^{n}x_{i}}\\
 & =\frac{\sum_{i=1}^{l}\left(\mu-x_{i}\right)+\sum_{i=l+1}^{n}\left(x_{i}-\mu\right)}{\theta+\theta'}\\
 & =\frac{l\mu-\theta+\theta'-\left(n-l\right)\mu}{\theta+\theta'}\\
 & =\frac{2l\mu-\theta+\theta'-\theta-\theta'}{\theta+\theta'}\text{ since }n\mu=\theta+\theta'\\
 & =2\frac{l\mu-\theta}{\theta+\theta'}\\
 & =2\left(\frac{l}{n}-\frac{\theta}{\theta+\theta'}\right)
\end{align*}

Moreover, increasing $\theta'$ and decreasing $\theta$ of the same value keeps $\mu$ constant and increases $d$, hence for $l$ and $\mu$ given, the  largest $C$ is reached with the smallest $\theta$ and the largest $\theta'$. In the best case, $\theta=0$ if the  $l$ values equal 0 and $\theta'=n-l$ if the $n-l$ values equal 1 and:
$$
C=\frac{2l}{n}
$$

The largest $C$ is not reached for $l=n$ since in this case $d=0$ and so does $C$. Therefore, the largest $C$ is reached for  $l=n-1$, so the upper bound of $C$ is $2(n-1)/n$, or approximately 2 when $n$ is large.

Note
I think that even though the result may be correct, I am not sure about the step from "$\theta$ should be min and $\theta'$ should be max" to "let us consider now series made of 0 and 1 only", since in this case, $\mu$ is not constant anymore, even though is is assumed in the previous step.

Anyone has a "clean" version for this ?

You're right in identifying the problem: we shouldn't assume that $\mu$ is constant. The solution seems to be a tiny tweak to your excellent work:

[1] Note that $0 \leq \theta / [\theta + \theta'] <=1$

[2] Note that $2(l/n - \theta / [\theta + \theta'])$ is maximised when $\theta / [\theta + \theta']$ is minimised.

[3] In particular note that, despite the resulting change in $\mu$, the $l / n$ is constant because $l$ doesn't change as we reduce the first $l$ values towards $0$ or increase the $n-l$ values towards $1$.

[3'] I just spotted for this to be true we need to define l to be the number of items  $\lt \mu$ not $\leq \mu$. Everything still seems to work.

[4] Whatever the value of $\theta'$ we can obtain the $0$ minimum by setting all the values below $\mu$ to $0$.

[5] So the maximum is $2l/n$ as you say.

[6] And this is maximised by choosing $l = n-1$ ($l=n$ is all the same which minimises)

[7] The final point to note is that with the first $n-1$ items as zero the expression is maximised when the $n$th value is $1$.

OK, let me try a slightly different way. There are two related challenges here:

1) What is the maximum value of d / $\mu$ where we have full flexibility over x values subject to $0 \leq x<=1$?  NB In this case $\mu$ varies as the x values vary.

2) What is the maximum value of d / $\mu$ where flexibility over the x values is additionally limited because their mean may not vary but must be equal to the fixed value $\mu$?

I think you're trying to solve 1) but you seem to suggest you're muddled because you're moving between fixing $\mu$ in your mind, but then noticing that the calculated value of the mean changes, because you're not in fact constraining the x values to have a fixed mean.

The "new" material was supposed to be [2] above. Let me try to be clearer. I pointed out that $0 \leq \theta / [\theta + \theta'] <=1$. Now because you're not constraining the x values to result in a fixed mean the denominator i.e. $\theta + \theta'$ will vary as the x values vary. Nonethless the ratio will still be minimised when $\theta=0$ - as easy as that! From this it followed that you should set the $l$ x values to zero. Then that l = n-1 and that the top c value is 1. That's all in your proof already, so it's just a slightly different thought pattern which (I thought) avoids the muddle you thought you were in.

My other bit was my own concern, namely that in changing the x values you changed $\mu$ and hence potentially $l$. My solution was a tweak:
* Define $l$ so that it referred to values less than the mean
* Then simulataneously move all these $l$ xs to zero. The mean changes but exactly the same xs are less than the mean. what I was trying to prevent was a constructive approach where we reduced one x value to zero only to find one x value which was just below the old mean is now just above it. Perhaps this doesn't matter, but it felt clearer to me.

The second problem I mentioned above - where we constrain the changes in the x values so that the "new" mean after changing the x values is equal to the old mean $\mu$, a constant, is also interesting. In this case the maximum we can achieve is:

dmax = [ H * (1-$\mu$) + (n-H-1) * $\mu$ + 1 * ($\mu$-f) ] / n

where $0\leq f \lt 1$ and H is the largest integer such that:

$\mu$ = (H+f)/n

In this case we put:

H of the x values at 1
n-H-1 of the x values at 0
1 of the x values at f (note we could have f=0)

As a special case of this take $\mu$ = 1/n. Then:

f=0
H=1
dmax/$\mu$ = 2*(n-1)/n as before


Combinatorics problem on $k$ sets and partionning a set of numbers consider $2k+1$ real numbers satisfying the property such that any $2k$ of them can be partitioned into two $k$ sets which exist that the sum of the numbers in each of the sets are equal.
Prove that all the numbers are equal.

First you can see easily all numbers are even or all numbers are odd. If you take $a+k_1+k_2$ is even, its mean a is even because $k_1=k_2$ thus $k_1 + k_2 = 2\cdot k_1$ which is even. The sum $a+ k_1 +k_2$ is the sum of all numbers and must be all time even or all time odd. Let say $a$ was odd, than the hole sum is odd, so when you take $b + k'_1+k'_2$ you see $b$ is also odd.

Now regardless to that, lets say they are not all equal. So there is at least one pair on numbers such that $a < b$. Lets take the ones that minimize the subtraction, I mean look for the minimum $a-b$ while $a-b>0$. Now you get $a, k_1=k_2$ and $b,k'_1=k'_2$.

Now I may not explain it well, and English is not my native so excuse me..
In the first configuration, b is in one of the groups $k_1$ or $k_2$, lets say $b\in k_1$. Swap $b$ and $a$ so you get the configuration $b, k_1\cup a,k_2$. Now try to move numbers from $k_1 \cup a$ to $k_2$ and in the other way to make them into 2 groups equal to each other (suppose to get $k'_1$ and $k'_2$ or other configuration that can hold it).
But because you took $a-b$ to be the minimum, you won't be able to do it. The proof from here is on you :)


CDF of sum of 3 dependent random variables Given three dependent random variables, $S_1, S_2$ and $S_3$, such that $0 < S_1, S_2, S_3 < \infty$ and assuming known their joint PDF $f_{S_1,S_2,S_3}(s_1,s_2,s_3)$ I would like to find the CDF of their sum, $E := S_1 + S_2 + S_3$. Basically, I want to find $F_E (e) = P(E \leq e)$.

I know that

$$F_E(e)=P(E \leq e)=P(S_1+S_2+S_3\leq e)=∫^?_0∫^?_0∫^{e−s_1-s_2}_{0}f_{S_1,S_2,S_3}(s_1,s_2,s_3)dS_3dS_2 dS_1$$
but I'm unsure of what is the upper limit in the other two
integrals (those on $S_1$ and $S_2$): is it ∞ or $e$?

I think we can solve it in this way:

$F_{E}(e)=P(E<=e)=\int_{-\infty}^{+\infty}ds_{1} \int_{-\infty}^{E-s_{1}}ds_{2} \int_{-\infty}^{E-s_{1}-s_{2}}f_{S_{1},S_{2},S_{3}}(s_{1},s_{2},s_{3})ds_{3}$

but I'm not sure.
I like to know the solution too.


Limits $\sqrt{2n^2-1}/(n+1)$ and $1 = 0.9999...$ $1)$  Find the limit (if it exists) of the following sequence:
$$\frac{\sqrt{2n^2-1}}{n+1} = x_n$$

Attempt: Rewrite as $$\frac{\sqrt{n^2(2 - \frac{1}{n^2})}}{n+1} = \frac{n\sqrt{(2-\frac{1}{n^2})}}{n+1} = \frac{\sqrt{2 - \frac{1}{n^2}}}{\frac{1}{n} + 1}$$ So as $n \rightarrow \infty,\,\,x_n \rightarrow \sqrt{2},$ using the limit law $\operatorname{lim}_{n \rightarrow \infty} \sqrt[n]{f(x)} = \sqrt[n]{\operatorname{lim}_{n \rightarrow \infty} f(x)} $and clearly $f(x) \geq 0 \,$for $n \neq 0$

Is the above calculation rigorous enough to determine the limit?

2)Interpret a decimal expansion $0.a_1a_2a_3....$ as $$0.a_1a_2a_3.... = \operatorname{lim}_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{a_k}{10^k}.$$
Show that $1 = 0.9999...$

Attempt: $$0.9999... =\operatorname{lim}_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{a_k}{10^k} = 9 \cdot \sum_{k=1}^{\infty} \frac{1}{10^k} = 9 \cdot \frac{1/10}{9/10} = 9 \cdot \frac{1}{9} = 1$$ Done?

In my book, they also include the following:Let $ y_n = \sum_{k=1}^{\infty} \frac{9}{10^k}$ and we want $|1-y_n| = |1 -  \sum_{k=1}^{\infty} \frac{9}{10^k}|$ As shown above the RHS tends to 0 so $|1-y_n| \rightarrow 0 $as $n \rightarrow \infty$
What does this add and is it required?

Many thanks.

Another way for second question.

Let $x=0.999999999999\dots $

$10x=9.999999999999\dots$

subtract  both equations

$9x=9$

$x=1;$


Infinite rings with lots of zero divisors Today I was trying to find an infinite ring $R$ whose all nonzero and nonidentity elements were zero divisors and actually found one: $\mathcal R =\text{Fun}(\mathbb N, \mathbb Z/2\mathbb Z)$. Given a function $f \in \mathcal R$ we can define another function $g \in \mathcal R$ by $g(x) = 1 - f(x)$. As everyone can easily check, $$(fg)(x) = f(x) - f(x)^2 = 0.$$

One can of course replace $\mathbb N$ with any infinite set. Now I'm curious if there are more examples of such strange behaving ring, that aren't similar to $\mathcal R$. Is there a notion of such rings in abstract algebra?

A very obvious example is any Boolean ring. Since $x^2=x$ for every element in the ring, $x(1-x)=0$. Thus if $x$ is neither $0$ nor $1$, $x$ must be a zero divisor.

These rings are very easy to come by. You can take any subring of a direct product of any number of copies of the field of two elements.

In [Malik, Mordeson, Sen, "Fundamentals of Abstract Algebra"] and some other books, they give following definition:

  Let $(R,+)$ be a commutative group. Define multiplication on $R$ by $ab=0$ for all $a,b \in R$, where $0$ denotes the identity element of the group $(R,+)$. Then $(R,+,.)$ is a ring called the "zero ring". If $R$ contains more than one element, then $R$ is a commutative ring without $1$ and every nonzero element of $R$ is a zero divisor.

In this sense, $(\mathbb Z(p^{\infty}),+,.) $ seems to satisfy this property by definition.