Why are an even number of flips required to get back to the original list? Consider the list of numbers $[1, \cdots, n]$ for some positive integer $n$. Two distinct elements $i$ and $j$ of the list can be switched in a so-called flip. For example, let $f$ be a flip that switches $2$ and $4$. Then $f([1,2,3,4]) = [1,4,3,2]$. Now consider a sequence of $k$ flips $f_1, \cdots, f_k$ of the list $[1, \cdots, n]$ such that $f_1(f_2(\cdots f_k([1,\cdots,n])\cdots)) = [1,\cdots, n]$, i.e. performing all flips gives the original list. Then $k$ must be even.

I would like to find a proof of this proposition that is elementary as possible. I already came up with a justification using permutation groups which goes as follows:

Each flip $f_i$ corresponds to a transposition of the list $[1, \cdots, n]$. Since the composition $f_1f_2\cdots f_k$ results in the identity, it must be an even permutation. Thus any representation of $f_1f_2\cdots f_k$  as a product of transpositions must contain an even number of transpositions. In particular, since each $f_i$ is a transposition, it follows that $k$ must be even.

This "proof" trivializes the problem statement as it is by using relatively high-powered facts about permutation groups. Is there a lower-level proof that avoids using theses results? (Ideally such a proof would avoid permutation groups altogether and be understandable to the layman.)

Edit: to clarify (since this question hasn't been getting as much attention as I had hoped), any proof that avoids re-deriving these powerful results about permutations would suffice. Basically I would want a proof that does not prove much more than the question requires, i.e. doesn't have a part where it says "in particular".

Here is a combinatoric proof (which I am not sure if it is from permutation group).

Count the number of 'reversed pair' in the list. What I mean by reversed pair is pair of numbers which the one bigger in value comes before the smaller one. For example, in $[2,1,3,4]$, $(2,1)$ is a reversed pair but $(1,3)$ is not.

Now observe the change of the number of reversed pair when a flip is taken. Let's say the list is $[\cdots,a,\cdots,b,\cdots]$ and we flip $a,b$, turning it into $[\cdots,b,\cdots,a,\cdots]$. It should be obvious that those in front or behind does not contribute any changes. Also, for those in between, lets call one of it $c$, will only contribute if $a<c<b$ or $a>c>b$. $+2$ for the former and $-2$ for the latter. And finally, the $a,b$ pair. $+1$ if $a<b$ and $-1$ if $a>b$. Anyway, the total number changes by an odd number. Since you have total number of $0$ at beginning and you want $0$ at the end, you have to flip even number of times.

Hope this helps you.

The value of k will be positive since to get back two numbers back to its position we have to do two flips.

For ex - for[1,2,3,4,5] we have first flip it once i.e.2 and 5 now to get back the original set again we again have to flip 2 and 5.
Now if we use k flips then it must be even since the have to be reversed that are reversed by k-1 flips.


How to find $y(x,z)$ from the given set of data? I have the following set of data:

$x_1,x_2,x_3,\dots,x_m$ are increasing in arithmetical  progression.

$y_1,y_2,y_3,\dots,y_n$ are increasing in arithmetical  progression.

$z(x_i,y_1),z(x_i,y_2),z(x_i,y_3),\dots,z(x_i,y_n)$ are exponentially increasing for $i=1,2,3,\dots,m$.

$z(x_1,y_i),z(x_2,y_i),z(x_3,y_i),\dots,z(x_m,y_i)$ are exponentially decreasing for $i=1,2,3,\dots,n$.

All values in the table are positive.

The required is to find the value of $y$ in terms of $x$ and $z$, [i.e. $y(x,z)$].

  How to find $y(x,z)$ from the given set of data?

Consider the following example:

The data set is:

Say we need to find the value of $y$ when $x=172$ and $z=3.1527$. Clearly, from the table, we can find that $y(172,3.1527)=25.50$.

What if we want to find $y(150,3.1729)$? which is not in the table

The answer is $y(x,z)=x(-1+\ln(z))$, so $y(150,3.1729)=150(-1+\ln(3.1729))=23.1969$

I actually know the expression $\boxed{y(x,z)=x(-1+\ln(z))}$ because I made this example to demonstrate my problem.

What if we do not know that relation, how should we obtain it?

This problem is related to my research in the chemical laboratory, where the values of $x$'s are the speeds of the centrifuge (rpm), the values of $y$'s are the volumes of the samples (mL), and the values of $z$'s are the length in (mm) of the extracted liquid in the pictures that are taken by a camera.

Regardless of my research, and regardless of the units (rpm, mL, mm), can we generalise a method?

Any help would be really appreciated. THANKS!

HINT.

Formally, is described the model
$$z(x,y) = e^{a+bx+cy},$$
or
$$\ln z = a+bx+cy,\tag1$$
where $b<0, c>0,$
$$x_i=12+2.25i,\quad y_i= 100 + 24j.$$

The table data model is
$$z_{i,j} = e^{\large \frac xy+1}\tag2$$

Assuming the discrepancy in the form of
$$d(a,b,c) = \sum w_{i,j}(\ln z_{i,j} - a - bx_i - Cy_j)^2,\tag3$$
where $w_{ij}$ is the arbitrary matrix of weights,

one can get the point (A,B,C), which provides $\min d(a,b,c)$ in accordance with the described model.

This point is the stationary point of $d(a,b,c).$

So $\operatorname{grad} d(A,B,C) = 0,$ or
\begin{cases}
\sum w_{i,j}\ (\ln z_{i,j} - A - Bx_i - Cy_j)=0\\
\sum w_{i,j}\ i\ (\ln z_{i,j} - A - Bx_i - Cy_j)=0\\
\sum w_{i,j}\ j\ (\ln z_{i,j}  - A - Bx_i - Cy_j) = 0.\tag4
\end{cases}

This leads to the linear system
\begin{cases}
S_{00}A + S_{10} B + S_{01}C = R_{00}\\
S_{10}A + S_{20} B + S_{11}C = R_{10}\\
S_{01}A + S_{11} B + S_{02}C = R_{01},\tag5
\end{cases}

where
$$S_{kl} = \sum w_{ij}x_i^k y_j^l,\quad R_{kl} = \sum w_{ij} x_i^k j^l \ln z_{ij}.\tag6$$

Using weight array $w=1$ gives

and looks unusable. This situation is happened, because the data table does not correspond to the model.

However, applying of the weight array in the form of
$$w_{ij}=e^{-\left(\Large\frac{5(x_i-150)}{24}\right)^2-\left(\Large\frac{2(y_j-23.25)}{2.25}\right)^2}$$

localizes the model and gives

So the estimation is
$$Y = \dfrac{\ln z -A-Bx}C
\approx \dfrac{\ln z - 1.13517\ 52307 + 0.00091\ 33448x}{0.00675\ 67568}, \tag7$$

and the result

looks suitable.

Note that the constants "2" and "5" in the $w$ formula were obtained empirically,
with the goal of the good approximation of the table data near of the expected point. If the table data model corresponds with the given model better, then these constants can be decreased or $w=1$ can be used.

If $z$ follows geometric progressions in terms of $x,y$, then $\log z$ follows arithmetic progressions.

If you keep one of the coordinates fixed,

$$y=\text{Cst}\to\log z=ax+b$$

$$x=\text{Cst}\to\log z=cy+d.$$

Combining the two, we have a bivariate linear relation

$$\log z=px+qy+r.$$

The parameters $p,q,r$ can be determined from three data points, solving a $3\times3$ system.

Obviously,

$$y=\frac{\log z-px-r}q.$$


Why does there not exist a particular finite set of congruences that forms a partition of the set of integers There is a paragragh that I saw in an article and could not really understand. The article is Covering Systems of Congruences, J. Fabrykowski and T. Smotzer, Mathematics Magazine Vol. 78, No. 3 (Jun., 2005), pp. 228-231, DOI: 10.2307/30044163. It starts by discussing problem 9 from the 2002 American Invitational Mathematics Examination (AIME):

  PROBLEM. Harold, Tanya, and Ulysses paint a very long fence. Harold starts with the first picket and paints every $h$th picket; Tanya starts with the second picket and paints every $t$th picket; and Ulysses starts with the third picket and paints every $u$th picket. If every picket gets painted exactly once, find all possible triples $(h,t,u)$.

After presenting the solution, the article goes on to say:

  One can generalize the AIME problem and ask whether there exists a finite set of congruences, with all moduli distinct and greater than or equal to $2$, that forms a partition of the set of integers. This turns out to be impossible. Relaxing the assumption about partitioning the integers, one can look for finite sets of congruences such that every integer belongs to at least one of them."

Can anyone help explain why the first part is impossible?

This is far from a full answer, but it gives a taste of how strong the constraints are.

Suppose we have $k$ congruences of the form $x \equiv a_i \pmod {m_i}$ where the $m_i$ are distinct and greater than $1$, and $0 < a_i \le m_i$ is the smallest natural number covered by the congruence. Further suppose that we've chosen a minimum set of congruences: i.e. there is no solution with fewer than $k$ congruences.

Consider two distinct congruences $x \equiv a_i \pmod {m_i}$ and $x \equiv a_j \pmod {m_j}$. We require $a_i \not\equiv a_j \pmod {\gcd(m_i, m_j)}$ to avoid collisions. (This follows from the Chinese remainder theorem, and is necessary and sufficient for non-intersection). Note that this implies that $\gcd(m_i, m_j) > 1$, so no two moduli are coprime.

Since a congruence with modulus $m_i$ covers $\frac{1}{m_i}$ of the integers, we must have $$\sum_i \frac{1}{m_i} = 1$$ so we have a partition of unity in Egyptian fractions. Combined with non-intersection, this is necessary and sufficient to partition the natural numbers.

Lemma: no non-trivial proper subset of the congruences is reducible to a single congruence. Suppose that we had a set of congruences $x \equiv a_i \pmod{m_i}$ which partition $x \equiv \alpha \pmod\mu$. Then clearly $m_i \equiv 0 \pmod\mu$ and $a_i \equiv \alpha \pmod\mu$, and the congruences $x \equiv \frac{a_i - \alpha}\mu \pmod {\frac{m_i}\mu}$ partition the natural numbers. But we started from a proper subset of the congruences, so this contradicts the assumption that $k$ is minimum.

Corollary: there is no $c > 1$ which is a common factor of all of the moduli. (Proof: if there were, one proper subset of the congruences would be reducible to $x \equiv 1 \pmod c$ and another to $x \equiv 2 \pmod c$. Since they can't both consist of a single congruence with modulus $c$, one of them is non-trivial, contradicting the lemma).

Corollary: none of the moduli is a prime power, for then that prime must be a common factor of all the moduli.

Corollary: $k \ge 19$. (Proof: we have to sum the reciprocals of the first 19 non-prime-powers to exceed $1$. Of course, those 19 numbers don't satisfy the no-pair-coprime constraint, and the sum of their reciprocals isn't exactly unity...).


$\forall{i=1 ... n}, Pr({a_i > 0})  0}) \le 1-(1-e^{-d})^n $? can you give a better bound? I have changed this problem by adding some conditions. giving that $\forall{i=1...n}, Pr(a_i > 0) < e^{-d}$ where $Pr$ is the probability and $d$ is an positive integer, $a_i$ are iid, please give a bound for $Pr(\sum_{i=1}^{i=n}{a_i} > 0) < $ ???.

we can simply use $$Pr(\sum_{i=1}^{i=n}{a_i} > 0) = 1-Pr(\sum_{i=1}^{i=n}{a_i} \le 0) \le 1-Pr(a_1 \le 0, a_2 \le 0, ..., a_n \le 0)=1-(1-e^{-d})^n$$, but is there a better bound? a better bound means it should be less than $1-(1-e^{-d})^n$

I have solve this problem. I will give my proving procedure as follows.

By mathematical induction, if we can prove when $n=2$, the bound holds, then the bound would always hold.

That's to say if we can prove:

Giving $Pr(A > 0) > e^{-d}$ and $Pr(B > 0) < e^{-d}$, then $Pr(A + B > 0) < e^{-d}$.

, the problem would be solved. Actually we can prove this, since:

$$Pr(A + B > 0) = Pr(A > 0, B > 0) + Pr(A + B > 0, A > 0, B < 0)
+ Pr(A + B > 0 ,A < 0 , B > 0)$$

and $Pr(A > 0,B > 0) = e^{-d} \times e^{-d}$.

and $$Pr(A + B > 0,A > 0, B < 0) = Pr( (A + B > 0) | (A > 0, B < 0)) \times Pr(A > 0, B < 0) = Pr( (A + B > 0) | (A > 0, B < 0)) \times e^{-d} \times (1-e^{-d})$$

observing that $$Pr(A+B>0|A>0,B<0) + Pr(A+B\le0|A>0,B<0) = 1\tag{1}\label{1}$$ and $$Pr(A+B>0|A>0,B>0)\le Pr(A+B\le0|A>0,B>0)\tag{2}\label{2}$$ since for each $A$ and $B$ meets $Pr(A + B > 0)$, that exists another tuple where $(B_1=-A,A_1=-B)$ meet $Pr(A_1 + B_1) < 0$.

so we can obtain $$Pr(A+B>0|A>0,B>0) \le 1/2$$

then $$Pr(A + B > 0,A > 0, B < 0) \le \frac{1}{2} \times e^{-d} \times {1-e^{-d}}$$

so
$$Pr(A + B > 0) = Pr(A > 0, B > 0) + Pr(A + B > 0, A > 0, B < 0)
+ Pr(A + B > 0 ,A < 0 , B > 0) \le e^{-d} \times e^{-d} + 2 \times \frac{1}{2} \times e^{-d} \times ({1-e^{-d}})=e^{-d}$$

Thank @herb steinberg again.

Let $n=2$, $d=1$ and $a_i = 1_{H_i}$ where $\{H_n\}_{n \ge 1}$ is independent coin tossing with $Pr(H_n) = p < e^{-1}$

$$Pr(\sum_{i=1}^{i=2}{a_i} > 0) = 1-(1-p)^2 \ge e^{-1}$$

for $p \in [1-\sqrt{\frac{e-1}{e}},\frac1e) $

Generally,

$$Pr(\sum_{i=1}^{i=n}{a_i} > 0) = 1-(1-p)^n \ge e^{-d}$$

for $p \in [1-(\frac{e^d-1}{e^d})^{\frac1n},\frac1{e^d}] $

But $1-(1-p)^n <  1-(1-e^{-d})^n$

That covers the cases where $a_i$'s are those indicators. You can try out different indicators, but I guess we should assume independence of the $a_i$'s. Then you can try $a_i$'s to be simple functions. The inequality holds true for $a_i=-1_{H_i}$, any $n$ and any $d$ obviously. Then you can try nonnegative then integrable.

You may be interested in Chebyshev's inequality.


Generators for the radical of an ideal I am interested in finding a generating set of the radical of an ideal given a set of generators for the ideal itself, but after a lot of thought I cannot figure out a good way to do it. Specifically:

  Let $k$ be an algebraically closed field, and $I \subset k[x_1, ..., x_n]$ an ideal. If $I = (f_1, ..., f_m)$, is there any good way to find a set of generators for $\text{rad}(I)$?

Edit: One may assume that $f_1, ..., f_m$ form a Gröbner basis for $I$. (Given any set of generators for an ideal in $k[x_1,..., x_n]$ it is always possible to find a Gröbner basis, so this assumption is without loss of generality.)

There are algorithms that find the generators for the radical of an ideal, e.g.:
Gianni, P.; Trager, B.; Zacharias, G.: Gröbner Bases and Primary Decomposition of Polynomial Ideals. J. Symb. Comp. 6, 149–167 (1988).

Note though that these methods are non-trivial; as far as I know, there is no simple characterization of the generators of a radical ideal. The construction by Connor only gives a subset of a generating set.

one way to solve the problem ( of finding generating set for radical ideal ,having the generator set of ideal ) is either to reduce it into principal ideal (by means of finding its reduced groebner basis) or factorizing it into power monomials).
Then for example:
if having    I=< xy, x^2+xy> , applying the routine leads to G={xy} as the reduced groebner basis for I.
(x^2+xy is eliminated since it's monomial xy € LT (xy) )
Now we use the theorem that says : Radical(f)= f/ GCD(f,df/d x1,d x2,...)
So: Radical(I)=xy/y=x
considering Lex ordering x>y
Thus    Radical(xy,x^2+xy)=x


To show that a function is not of bounded variation on [0,1] What would be a sequence of partitions of [0,1] to show that the function

$f(x)$ = \begin{cases}
0,  & \text{if $x$ $\in$[0,1]$\cap$$Q^c$} \\
-x, & \text{if $x$$\in$[0,1]$\cap$$Q$}
\end{cases}

is not of bounded variation on [0,1]?

Try partition

$x_n    = \frac{1}{n}$ , if $n$ is odd

$=\frac{1}{n+n^\frac{1}{n}}$ , if n is even.


Simple Poisson problem I wanted to ensure that I solve the following problem correctly:
There is a door to the building and I do not have a key. On average, every 5 minutes a person enters or exits the building (i.e., eventually opening the door). What is probability that I will enter the door within 3 minutes.

Solution:
1. Lambda: on average 0.6 people entering/exiting the building in 3 minutes

The probability to enter the building in 3 minutes is the probability 1-the probability that no one enters / exits the building.
The probability that no one enters / exits the building: P(X=0) = e^-0.6 = 0.55
Probability to enter the building within 3 minutes is 1-0.55 = 0.45

Thanks!

Since average number of people who enter or exit in 5min is 1 we can conclude probability that at least one person enters or exits in 1min is 0.2. So, probability that at least one person does it in 3min is =1-(1-0.2)^3=0.488
I tried this in matlab simulations and its checks.


Use Stokes' theorem to show that the line integrals have the given value Problem

Prove using Stokes' theorem that
$$\int_C y dx +z dy + x dz = \pi a^2 3^.5,$$ where $C$ is the curve of intersection of the sphere $x^2+y^2+z^2=a^2$ and the plane $x+y+z=0$.

Attempt

$$\int_C y dx +z dy + x dz= \iint_S \nabla × F.n dS,$$ where $n$ is the normal to the surface $S$. $S$ is the ellipse formed by the intersection.
$F=(y,z,x)$
Now $z=-x-y$.

Therefore $$n = \frac{\frac{\partial z}{\partial x} i+ \frac{\partial z}{\partial y} j-k}{|\frac{\partial z}{\partial x} i+ \frac{\partial z}{\partial y} j-k|}
=\frac{-i-j-k}{3^.5}$$

$$dS =\left(1+\frac{\partial z}{\partial x}^2 + \frac{\partial z}{\partial y}^2\right)^{\frac{1}{2}}$$
Therefore
$$\iint_S (y,z,x).n dS=3 \iint dx dy$$

How to proceed after this ?

NB. This question is from Apostol's Calculus Volume 2,c Chapter 12, Section 12, Question 5, page 442 (2nd Edition).

You don't need to work out what $dS$ is.  The curl of $F$ is $\langle -1,-1,-1\rangle$, and the unit normal vector is the same thing, but divided by $\sqrt{3}$, so your surface integral is

$$\iint_S \frac{3}{\sqrt{3}} \; dS =\sqrt{3}\times \mbox{area of }S. $$

Since $S$ is a disc of radius $a$, you get

$$\sqrt{3}\pi a^2.$$

S is the unit disk in the xy-plane and $\int_S\int dxdy$ is simply its area.


$AA^{t}=A^{t}A$ in this case? Let $M$ be a square matrix such that $M_{i,j}=c$ if $i\not=j$ and $d$ if $i=j$. Also, $0<c<d$ (note that this is a strict inequality). That is, $M$ look like this:

$\begin{pmatrix}d & c & \ldots & \ldots & c & c\\c & d & \ldots & \ldots & c & c\\\ldots & \ldots & \ldots & \ldots & \ldots & \ldots\\\ldots & \ldots & \ldots & \ldots & \ldots & \ldots\\c & c & \ldots & \ldots & d & c\\c & c & \ldots & \ldots & c & d\\\end{pmatrix}$

If $A$ is a square matrix such that $AA^{t}=M$ is it necessarily the case that $A^{t}A=M$?

(we are working in $\mathbb{R}$; actually the actual question is meant for $\mathbb{Q}$ but I am trying to gain some insight by looking at the easier situation)

This is clearly true if we allow $c=0$, and is clearly false if we allow $c=d$. But I have not been able to do it in this case. I suspect it is false but have not been able to find a counterexample.

Here's a two-dimensional counterexample:

$$A = \left[\begin{array}{cc}0 & 5\\3 & 4\end{array}\right].$$

Even for the case $c=d$ it can be valid, as

$$
\textbf{A} = \left[
\begin{array}{cccc}
\sqrt{c/n}&\sqrt{c/n}&\cdots&\sqrt{c/n}\\
\sqrt{c/n}&\sqrt{c/n}&\cdots&\sqrt{c/n}\\
\vdots&\vdots&\ddots&\vdots\\
\sqrt{c/n}&\sqrt{c/n}&\cdots&\sqrt{c/n}\\
\end{array}
\right],
$$

has the property that

$$
\textbf{A} \textbf{A}^T = \textbf{A}^T \textbf{A} = \textbf{M},
$$

where

$$
\textbf{M} = \left[
\begin{array}{cccc}
c&c&\cdots&c\\
c&c&\cdots&c\\
\vdots&\vdots&\ddots&\vdots\\
c&c&\cdots&c\\
\end{array}
\right],
$$


Minimum number of objects required to figure out the issue There 1,000 buckets, one of them contains poison, the rest of them are filled with water. They all look the same. If a pig drinks that poison, it will die within 30 minutes. What is the minimum number of pigs to you need to figure out which bucket contains the poison within one hour?

You can do it with $10$ pigs, based on the binary representation of a number. Mark the pigs $1, 2, ..., 10$. Given bucket number $n$, write $n$ in binary; every time the $k$th digit is $1$, have pig #$k$ drink.

Wait for pigs to die. Form the appropriate number, putting $0$'s for every living pig in the appropriate place, and $1$'s otherwise. The number formed marks the bad bucket.

Note that I'm assuming that the pigs can drink from a lot of buckets quickly, and that more than one pig can drink from a bucket.

Okay, for the pig question, there are two ways for you to only need one, and it doesn't need to die 2/3 of the time. It depends on the "type" of pig for the first one.

Since certain practicalities are not specified, I'm going to suggest for the first scenario that your one pig be a poison-sniffing pig; since pigs have sensitive noses. The pigs sniffs out the poison, and you're done.

The second scenario is a little messy, but it works: combine all of the buckets into one vat to dilute the poison. Pour the water into the buckets and give the pig one bucket. It the pig lives, then none of them are poisoned (at least for a pig), and if he dies, then all of them are: which brings me to my third option, which is to dump them all out and start over, you have all of this time and water is cheap, so quit being lazy and just refill them.


What is the meaning of 'regression' in 'linear regression'? I can see why linear regression is linear, i.e., because it is represented by a line, but what does regression have to do with the term as a whole?

What is the meaning this word contributes to the term?

Semi-intuitive explanation:

Assume that we are interested in one's IQ. Not in a score that he might get in some IQ test, rather in his true IQ value. So, we have to assume that there exists such a value. Lets denote it by $\mu$. However, it is impossible to measure it directly. As such, we can use IQ tests to estimate it. Denote by $X_i$ his score in the $i$th test. We can model this score by $X_i = \mu +\epsilon_i$, where $\epsilon_i$ is the random error of the $i$th test with $\mathbb{E}\epsilon_i = 0$ and $var(\epsilon_i) = \sigma^2$ . That is, the score of the $i$th test is composed of his real value (signal) and some random error (noise).  Because $\mathbb{E}X_i=\mu$, his score will (in some sense) tend to his real IQ value. Therefore, after $n$ such tests we will take his average score as the estimator of this value. This average will indeed tend to $\mu$ in a sense that  as larger the number of tests that he takes - the more accurately will the sample average estimate the real IQ.

What exactly is the random error is more philosophical rather than statistical question. That is, it may stem from some imperfections of the tool (test) that may vary in its difficulty or it may stem from some subject related factors (tiredness, mood etc.) or even may be some inherent property of the IQ itself (i.e., it is not a scalar but rather a random variable. Then you may be interested in its mean value.). Or some combination of the above.

Formally:
The linearity of a regression model does not mean that it is straight line. Namely, any model that can be written in matrix notations as
$$
Y=X\beta +\epsilon,
$$
is called linear. Special cases like $y=\beta x +\epsilon$ and $y=\beta_0 + \beta_1x +\epsilon $ are indeed estimated by straight lines. Or can be viewed as straight lines (signal) that is interrupted by some noise $\epsilon$. Linearity is defined by independence of the first derivative of $Y$. I.e., if $\partial y / \partial \beta_j = x_j$, $j=1,...,p$ with $x_1 = 1$, then it is linear. If at least one of the derivatives depends on the parameters, then it is non-linear. While, the observations may be interpret as random fluctuations around the mean $\mathbb{E}Y = X\beta$, where $\mathbb{E}\epsilon = 0$. The estimation methods are try to estimate this means by minimizing (mostly) the squared error, i.e., the (squared empirical) deviation from this unknown mean.

The idea of linear regression is to infer from dataset the dependence of some random variable in other variable, what the term regression mean in this context is that from the dataset we regress to some kind of dependency that brought us to that data.


Prove that polynomial $x^3+ax^2+17x+3b$ doesn't have same roots. All roots of polynomial $x^3+ax^2+17x+3b$, $a,b\in \Bbb Z$ are integers. Prove that this polynomial doesn't have same roots.

My plan was to find all three solutions and then compare them.

I don't know how to find the first root - tried to use cubic formula to find it but I got a large expression, just a few numbers canceled.  Any advices? Thanks.

Say it has, so $x_1=x_2 = m$ and $x_3=k$, then we have by (second) Vieta formula $$m^2+2mk =17$$ and (third) $$m^2k =-3b$$

So $m$ or $k$ is divisible by $3$. Clearly $m$ is not (since else $3\mid 17$), so $3\mid k$. But then we have $m^2\equiv 2\pmod 3$, a contradiction.

Hint $ $ the roots are distinct $\bmod 3\!:\ f \equiv x(x^2\!+\!ax\!-1)\not \equiv x(x\!-\!r)^2\,$ by $\,r^2\not\equiv 1$


Can the unit interval map bijectively to a region? The Hilbert Curve shows that there exists a surjection from the unit interval to the unit square.

I was wondering, does there exist a bijection from the unit interval to the unit square?

Existence and Construction of bijective map between $(0, 1)$ to $(0, 1)^2$

Yes, there is. The simplest way I can think of constructing one is by invoking the Cantor-Schroder-Bernstein theorem, which tells you that if there are two sets $A$ and $B$, $|A| = |B|$ if $\exists f, g$ such that  $$\\f : A \to B \\g: B \to A$$
where both $f$ and $g$ are injective (one-to-one). Note that surjectiveness is not asked for.

We need to construct two maps: $f : (0, 1) \to (0, 1) \times (0, 1)$ and $g: (0, 1) \times (0, 1) \to (0, 1)$ both of which are one-to-one.

$f$ is an easy one. Map a point $a \in (0, 1)$ to the point $(a, a) \in (0, 1) \times (0, 1)$. This is clearly one-to-one. It's not surjective (every point on the square does not have a pre-image) but we don't need one.

Hence, $$f : A \to B, \\f(a) = (a, a)$$

Now, we need $g$. One cute trick is to consider the point $(a, b)$  on the unit square. Let $a$ have the decimal expansion $$a = 0.a_0 a_1 a_2 \ldots$$ and let $b$ have the decimal expansion
$$b = 0.b_0 b_1 b_2 \ldots$$

I'll now make a point $c \in (0, 1)$ by taking $$c = 0.a_0 b_0 a_1 b_1 \ldots a_n b_n \ldots$$

Hence, $c$ is the point that you get when you interleave the digits of $a$ and $b$.

So, $$g : B \to A\\ g(0.a_0 a_1 \ldots a_n, 0.b_0 b_1 \ldots) = 0.a_0 b_0 a_1 b_1 a_2 b_2 \ldots $$

$f$ and $g$ give us the two maps we are looking for, hence

$$|A| = |B|$$ or the unit line can be mapped onto the unit square.

Intuition of Cantor Schroder Bernstein

In some sense, having an injective $f : A \to B$ tell you that $|A| \leq |B|$ since you are mapping every point of $A$ uniquely to some point of $B$. Sure, you may have missed points in $B$, but you know that $B$ has, at minimum, the same number of items as $A$.

By symmetry, having an injective $g: B \to A$ implies that $|B| \leq |A|$.

However, since both of these must be true, $$|A| = |B|$$ which is what we were looking for!

The proof is a bit hairy and involves bouncing between infinitely long chains, so I prefer to think of this theorem as a "prove one, use many" times thing.

No. Suppose that yes and let $\phi$ a bijection. Put $Q$ as the square and $I$ the interval.

Now, $Q\setminus\{ \phi(1/2)\}$ is connected but $I\setminus\{ 1/2\}$ not. This is an absurd, because a map convert connected sets into connected sets.

If you want only a function, the answer is yes, the projection $\pi(x,y)=x$ is surjective, and then exist a bijection (because exist a surjection from both sides)


How to calculate the sum of $\sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}} $ I would like to know how to calculate the sum of this convergent series
$$\sum\limits_{n = 1}^\infty  {{{{{( - 1)}^n}} \over {{n^3} + 1}}} $$
I cannot obtain this sum, mainly because it is not similar to any series of exponential, logarithmic or trigonometric functions, which can serve as a point of support.
Note: I would also like the sum not to be expressed in terms of complex numbers
Thanks in advance.

If you do not want to use complex numbers at all, the only possibility I see is to write
$$\sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}}=\sum\limits_{n = 1}^{p-1} {{{{{( - 1)}^n}} \over {{n^3} + 1}}}+\sum\limits_{n = p}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}}$$ For the second summation, assumming that $p$ will be sufficiently large, use
$$\frac 1{n^3 + 1}=\sum_{k=1}^\infty (-1)^{k-1} \frac 1{n^{3k}}$$ and the second summation will write
$$\sum\limits_{n = p}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}}=\sum\limits_{k = 1}^\infty(-1)^{k+1} \zeta (3 k,p)$$ which has an explicit expression in terms of polylogarithms with complex arguments !!
So, avoid the infinte upper bound for the second sum and make it as
$$\sum\limits_{n = p}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}}=-\frac{\psi ^{(2)}(p)}{2}-\frac{\psi ^{(5)}(p)}{120}-\frac{\psi ^{(8)}(p)}{40320}-\frac{\psi ^{(11)}(p)}{39916800}- \cdots$$ Trying for a faw values of $p$, the decimal representaion would be
$$\left(
\begin{array}{cc}
 p & \text{approximation} \\
 25 & -0.41349860 \\
 50 & -0.41409322 \\
 75 & -0.41421222 \\
 100 & -0.41425014 \\
 125 & -0.41426913 \\
 150 & -0.41427862 \\
 175 & -0.41428481 \\
 200 & -0.41428851
\end{array}
\right)$$ while the exact value would be $-0.41430114$.
Edit
If you want to go further, you can use the more general
$$\sum\limits_{n = p}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}}=-\sum_{k=1}^\infty \frac{\psi ^{(3k-1)}(p)}{(3k-1)!}$$


Sum of cubes of roots of a quartic equation $x^4 - 5x^2 + 2x -1= 0$

What is the sum of cube of the roots of equation other than using substitution method? Is there any formula to find the sum of square of roots, sum of cube of roots, and sum of fourth power of roots for quartic equation?

This is precisely what Newton's Sums are for.

Let $S_n$ be the sum of the $n$-th powers of the roots of $a_nx^n+a_{n-1}x^{n-1} + \cdots + a_1x+a_0$.

Then, Newton's Sums tells us that $\displaystyle\sum_{k = 0}^{m-1}\left(a_{n-j}S_{m-j}\right)+ma_{n-m} = 0$ for each positive integer $m$.

We can write out the first few equations explicitly as follows:

$a_nS_1 + a_{n-1} = 0$

$a_nS_2 + a_{n-1}S_1 + 2a_{n-2} = 0$

$a_nS_3 + a_{n-1}S_2 + a_{n-2}S_1 + 3a_{n-3} = 0$

$a_nS_4 + a_{n-1}S_3 + a_{n-2}S_2 + a_{n-3}S_1 + 4a_{n-4} = 0$

and so on. Note that we define $a_j = 0$ for $j < 0$.

Applying those formula to the polynomial $1x^4+0x^3-5x^2+2x+1$ yields the following:

$1 \cdot S_1 + 0 = 0 \leadsto S_1 = 0$ (sum of roots)

$1 \cdot S_2 + 0 \cdot S_1 + 2 \cdot -5 = 0 \leadsto S_2 = 10$ (sum of squares of roots)

$1 \cdot S_3 + 0 \cdot S_2 + - 5 \cdot S_1 + 3 \cdot 2 = 0 \leadsto S_3 = -6$ (sum of cubes of roots)

and so on.

Easy way to do this is to take the first derivative of the polynomial and divide it by the polynomial itself.

f ( x ) = x^4 - 5x^2 + 2x - 1 = 0

f ' ( x ) = 4x^3 - 10x + 2

Now do f ' ( x ) / f ( x )  =  ?

(x^4 + 0x^3 - 5x^2 + 2x - 1 )   ) ......4x^3 + 0x^2 - 10x + 2

.......- 0....+5........- 2..+ 1

4 / 1 = 4 so that is our first coefficient.

multiply all these numbers - 0.....+ 5....- 2 and + 1 by 4

Place the answers under the columns to the right.

so it would look like this :   4x^3 + 0x^2 - 10x + 2

.....................................................- 0..........+20........-8..........+ 1

so the sum of the roots is 0, but the sums of the squares of the roots is +10.


Can we define a specific metric tensor to all manifolds? First, my question it maybe doesn't make sense at all, and that's because I don't truly understand these things, if it doesn't make sense please explain to me why.
When I say a metric tensor I mean the first fundamental form where every student learns in a differential geometry course. Meaning the standard inner product $\|w\|=\langle w,w\rangle $ of an element of the tangent space $T_pS$ of a normal surface $S$. Can we use this for every n-manifold to calculate length, angles, surface area etc?
what's the point of considering all these other tensors we use in Riemannian geometry since a manifold is locally euclidean can't we just use the standard inner product to do all these things?

(1) When most people say "an $n$-manifold $M^n$," what they mean typically mean is an (abstract) smooth $n$-manifold (which is virtually the same as an (abstract) differentiable $n$-manifold).  I don't have time to go through the definition here, but there are many fantastic books on abstract manifolds (e.g., John Lee's Introduction to Smooth Manifolds), as well as a Wikipedia page:
https://en.wikipedia.org/wiki/Differentiable_manifold
The point is that an abstract $n$-manifold $M^n$ is not necessarily a subset of euclidean space $\mathbb{R}^N$.  However, if it happens to be the case that your abstract $n$-manifold lives in some euclidean space, say $M^n \subset \mathbb{R}^N$, then we say that $M^n$ is a smooth submanifold of $\mathbb{R}^N$. If you've taken a differential geometry course on surfaces in $\mathbb{R}^3$, then you've studied the smooth $2$-dimensional submanifolds of $\mathbb{R}^3$.
(2) Given a smooth submanifold $M^n \subset \mathbb{R}^N$, then yes, it is possible to use the standard inner product $\Vert v \Vert = \sqrt{\langle v,v \rangle}$ on $\mathbb{R}^N$ to give each tangent space $T_pM \subset \mathbb{R}^N$ a positive-definite inner product that varies smoothly with $p \in M$:
\begin{equation}
g_p(v,w) := \langle v,w \rangle.
\tag{1}
\end{equation}
(3) In general, however, an abstract manifold $M^n$ doesn't necessarily lie in a euclidean space $\mathbb{R}^N$.  Nevertheless, it would be interesting if we could give each tangent space $T_pM$ a smoothly varying positive-definite inner product $g_p$.  A choice of smoothly varying positive-definite inner product on each $T_pM$ is called a Riemannian metric (or metric tensor or Riemannian structure).  The pair $(M,g)$ is then called a Riemannian manifold.
It is a fact (mentioned in Moe's answer) that every abstract $n$-manifold can be given (uncountably many!) Riemannian metrics $g$.  Therefore, every abstract manifold $M$ can be made into a Riemannian manifold $(M,g)$ in uncountably many ways.  If it so happens that $M^n \subset \mathbb{R}^N$, then there is a natural choice of Riemannian metric $g$ coming from equation (1), but one could in principle give $M^n$ other Riemannian metrics (a.k.a. metric tensors) besides this.

Every differentiable manifold admits a Riemannian structure. This is actually fairly intuitive to see: Every differentiable manifold is locally Riemannian, thus sewing up all the local metrics to a global metric gives us the conclusion we seek.


Range conditions on a linear operator While reading though some engineering literature, I came across some logic that I found a bit strange.  Mathematically, the statement might look something like this:

I have a linear operator $A:L^2(\Bbb{R}^3)\rightarrow L^2(\Bbb{R}^4)$, that is a mapping which takes functions of three variables to functions of four variables.  Then, "because the range function depends on 4 variables while the domain function depends on only three", there must be redundancy in the operator $A$, that is the range of $A$ is a proper subset of $L^2(\Bbb{R}^4)$, characterized by some range conditions.

Is such a statement always true?  For the specific example I am reading about (the X-ray transform), it is definitely true - in fact, the range of the operator is characterized by a certain PDE - but I can't image such a thing is true in general.

For instance, I can cook up an operator $A:L^2(\Bbb{R}^3)\rightarrow L^2(\Bbb{R}^4)$ such that the range of $A$ is dense in $L^2(\Bbb{R}^4)$: simply choose orthonormal bases $(e_j)$ and $(f_j)$ for both, then map $e_j$ to $f_j$.

Any thoughts?

It's not true. There is only one separable Hilbert space, up to isomorphism. Since $L^2(\mathbb{R}^3)$ and $L^2(\mathbb{R}^4)$ are both separable, they are isomorphic as Hilbert spaces. That is, there is an isometric linear mapping $A$ of $L^2(\mathbb{R}^3)$ onto $L^2(\mathbb{R}^4)$.

I think it always has to be true. It is like mapping points in a flat plane to a 3D space. You get a volume 0 manifold in the 3D space if there is a 1 to 1 mapping.


Prove that for all positive integers $x$, $\left\lfloor \frac{x^2 +2x + 2}{4}\right\rfloor =\left\lfloor \frac{x^2 + 2x + 1}{4}\right\rfloor$. Title says it all, basically. I believe it to be true that

$$\left\lfloor \dfrac{x^2 + 2x + 2}{4} \right\rfloor=\left\lfloor \dfrac{x^2 + 2x + 1}{4} \right\rfloor$$ for all positive integers $x$. I am, however, having a difficult time proving this. My current proof reads along the lines of the fact that, when adding $2$ or $1$, it is impossible to cause a large enough difference in the two numbers that, when divided by four and floored, they evaluate to different numbers. This basically comes down to proving that $x^2 + 2x \neq 4k + 2$ for some integer $k$, but I'm not sure if this is a good way of proving it.

Could anyone shine some light on this? Thanks.

HINT: Note that the numerators are $(x+1)^2+1$ and $(x+1)^2$. Consider separately the cases $x$ odd and $x$ even.

The only time the two floors are different is when $x^2+2x+2$ is a multiple of $4$ (or, equivalently, looking at the right-hand side, $x^2+2x+1 \equiv 3\mod{4}$.

But $x^2+2x+2 = (x+1)^2+1$, and $(x+1)^2\equiv 0, 1\mod{4}$, so we cannot have $x^2+2x+2\equiv 0\mod{4}$. Almost identically, considering the RHS instead, $x^2+2x+1 = (x+1)^2\equiv 0, 1\mod{4}$, so it cannot be $3\mod{4}$.


Clarification on Continuous Matrices and reference request I have not been able to obtain a clear definition of what is currently called a 'continuous matrix' nor 'continuous matrix operators.'

It is unclear if the definition would involve a matrix with continuous elements such as continuous functions, or if somehow the matrix itself is to be considered continuous. In the latter case, I would imagine an example of a continuous matrix to be the Green's Function for a function, where integration of a function by its Green's function would be akin to some sort of continuous matrix multiplication.

So I would like to ask:

What is the current accepted definition for a continuous matrix and a continuous matrix operator?
Is there an existing definition for a continuous matrix, of the type I mention, where Green's Function is an example of such a matrix?
If there is such a definition (2.) then does someone know of a clear reference for the theory? I would appreciate both pure and applied works.

The term is frequently used in OpenCV when you want to reshape a matrix cv.Reshape() or perform other operations such as implement K-Means algorithm cv.KMeans2() .. I am also trying to find a clear concise definition of the term 'continuous matrix' but have not been able to find one.


Evaluate $ \int_{0}^{\frac{\pi}2}\frac1{(1+x^2)(1+\tan x)}\:\mathrm dx$ Evaluate the following integral
$$
\tag1\int_{0}^{\frac{\pi}{2}}\frac1{(1+x^2)(1+\tan x)}\,\mathrm dx
$$

My Attempt:

Letting $x=\frac{\pi}{2}-x$ and using the property that

$$
\int_{0}^{a}f(x)\,\mathrm dx = \int_{0}^{a}f(a-x)\,\mathrm dx
$$

we obtain

$$
\tag2\int_{0}^{\frac{\pi}{2}}\frac{\tan x}{\left(1+\left(\frac{\pi}{2}-x\right)^2\right)(1+\tan x)}\,\mathrm dx
$$

Now, add equation $(1)$ and $(2)$. After that I do not understand how I can proceed further.

Here is an approach.

We give some preliminary results.

   The poly-Hurwitz zeta function

The poly-Hurwitz zeta function may  initially be defined  by the series
$$
\begin{align}
  \displaystyle \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)},
  \quad \Re a >-1, \, \Re b >-1, \, \Re s>0. \tag1
\end{align}
$$
This special function is a natural extension of the Hurwitz zeta function initially defined as
$$
\zeta(s,a)=\sum_{n=0}^{\infty} \frac{1}{(n+a)^s}, \quad \Re a>0, \Re s>1, \tag2
$$
which is a natural extension itself of the Riemann zeta function initially defined as
$$
\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}, \quad \Re s>1. \tag3
$$

The poly-Hurwitz function appears in different places with different notations, one may find it here:
 [Masri, p. 2 and p. 15  (2004)], [Murty, p. 17 (2006)], [Sinha, p. 45 (2002)]. In this answer we are dealing with a simplified version of a general poly-Hurwitz function.

The series in $(1)$ converges absolutely for $\displaystyle \Re s>0$. Moreover, the convergence of the series is uniform on every
half-plane $$\displaystyle H_{\delta}=\left\{s \in \mathbb{C}, \Re s \geq  \delta \right\}, \, \delta \in \mathbb{R},\, \delta>0,$$ therefore the poly-Hurwitz zeta function $\displaystyle \zeta(\cdot \mid a,b)$
is analytic on the half-plane $\displaystyle \Re s>0$.

Let $a$, $b$ and $s$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re s >0$. One may observe that
$$
\begin{align}
  \zeta(s\mid a,b) & =  \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}\\
 & = \sum_{n=1}^{+\infty} \frac{(n+b)+(a-b)}{(n+a)^{s+1}(n+b)}\\
 & = \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s+1}}+(a-b)\sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s+1}(n+b)} \tag4
\end{align}
$$
giving the functional identity
$$
\begin{align}
 \zeta(s \mid a,b) = \zeta(s+1,a+1) +(a-b) \zeta(s+1 \mid a,b) \tag5
\end{align}
$$

where $\displaystyle \zeta(\cdot,\cdot)$ is the standard Hurwitz zeta function.

From $(5)$, we obtain by induction, for $n=1,2,3,\ldots $,

$$
\begin{align}
 \zeta(s\mid a,b) = \sum_{k=1}^{n}(a-b)^{k-1}\zeta(s+k,a+1) +(a-b)^n\zeta(s+n \mid a,b). \tag6
\end{align}
$$

We use $(6)$ to extend $\displaystyle \zeta(\cdot \mid a,b)$ to a meromorphic function on each open set $\Re s>-n $, $n\geq 1$. Since the Hurwitz zeta function is analytic on the whole complex plane except for a simple pole at $1$ with residue $1$, then from $(6)$ the poly-Hurwitz zeta function $\displaystyle \zeta(\cdot\mid a,b)$ is analytic on the whole complex plane except for a simple pole at $0$ with residue $1$.

   The poly-Stieltjes constants

In 1885 Stieltjes has found  that the Laurent series expansion around $1$ of the Riemann zeta function
$$
\zeta(1+s) = \frac{1}{s} + \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\gamma_k s^k, \quad s \neq 0,\tag7
$$  is such that the coefficients of the regular part of the expansion are given by
$$
\begin{align}
\gamma_k& = \lim_{N\to \infty}\left(\sum_{n=1}^N \frac{\log^k n}{n}-\frac{\log^{k+1} \!N}{k+1}\right).
\end{align} \tag8
$$ Euler was the first to define a constant of this form (1734)
$$
\begin{align}
\gamma & = \lim_{N\to\infty}\left(1+\frac12+\frac13+\cdots+\frac1N-\log N\right)=0.577215\ldots.
\end{align}
$$ The constants $\displaystyle \gamma_k$ are called the Stieltjes constants and due to the fact that $\displaystyle \gamma_0=\gamma$   they are sometimes called the generalized Euler's constants.

Similarly, Wilton (1927) and Berndt (1972) established that the  Laurent series expansion in the  neighbourhood of $1$ of the Hurwitz zeta function
$$
\begin{align} \zeta(1+s,a)
= \frac1s+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!} \gamma_{k}(a)\:s^{k}, \quad \Re a>0, \,s\neq 0, \tag{9}
\end{align}
$$ is such that the coefficients of the regular part of the expansion are given by
$$
\begin{align}
\gamma_k(a)& = \lim_{N\to \infty}\left(\sum_{n=0}^N \frac{\log^k (n+a)}{n+a}-\frac{\log^{k+1} (N+a)}{k+1}\right), \quad \Re a>0,
\end{align} \tag{10}
$$
with $\displaystyle \gamma_{0}(a)=-\psi(a)=-\Gamma'(a)/\Gamma(a)$. The coefficients $\gamma_k(a)$ are called the generalized Stieltjes constants.

We have seen from $(6)$ that the poly-Hurwitz zeta function admits a Laurent series expansion around $0$. Let's denote by $\displaystyle\gamma_k(a,b)$ the coefficients of the regular part of $\displaystyle \zeta(\cdot\mid a,b)$ around $0$. I will call these coefficients the poly-Stieltjes constants.

Do we have an analog of $(10)$ for $\displaystyle\gamma_k(a,b)$?

The following result is new.

  Theorem 1. Let $a,b$ be complex numbers such that $\Re a >-1, \, \Re b >-1$. Consider
  The poly-Hurwitz zeta function $$
\begin{align}
   \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)},
  \quad \Re s>0. \tag{11}
\end{align}
$$
  Then the meromorphic extension of $\displaystyle \zeta(\cdot\mid a,b)$ admits the following Laurent series expansion around $0$,
  $$
\zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{+\infty} \frac{(-1)^{k}}{k!}\gamma_k(a,b) s^k, \quad s \neq 0,\tag{12}
$$ where the poly-Stieltjes constants $\displaystyle \gamma_k(a,b)$ are given by $$
\begin{align}
\gamma_k(a,b)& = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log^k (n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right)
\end{align} \tag{13}
$$ with $$ \gamma_{0}(a,b)=-\psi(b+1)=-\Gamma'(b+1)/\Gamma(b+1). \tag{14}$$

Proof. Let $a,b$ be complex numbers such that $\Re a >-1, \, \Re b >-1$.

We first assume $\Re s>0$. Observing that, for each $n \geq 1$,
$$
\left|\sum_{k=0}^{\infty}\frac{\log^k(n+a)}{n+b}\frac{(-1)^{k}}{k!}s^k\right| \leq \sum_{k=0}^{\infty}\left|\frac{\log^k(n+a)}{n+b}\right|\frac{|s|^k }{k!}<\infty
$$ and that
$$
\sum_{n=1}^{\infty}\left|\sum_{k=0}^{\infty}\frac{\log^k(n+a)}{n+b}\frac{(-1)^{k}}{k!}s^k\right|=\sum_{n=1}^{\infty}\left|\frac1{(n+a)^s(n+b)}\right| =
\sum_{n=1}^{\infty}\frac1{|n+a|^{\Re s}|n+b|}<\infty,$$
we obtain
$$
\begin{align}
&\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac{\log^k(n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right)  s^k \\\\
&= \lim_{N\to+\infty}\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\left(\sum_{n=1}^N\frac{\log^k(n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right) s^k \\\\
&=\lim_{N\to+\infty}\sum_{k=0}^{\infty}\left(\sum_{n=1}^N\frac{(-1)^{k}}{k!}\frac{\log^k(n+a)}{n+b}s^k -\frac{(-1)^{k}}{k!}\frac{\log^{k+1} \!N}{k+1}s^k\right) \\\\
&=\lim_{N\to+\infty}\left(\sum_{n=1}^N\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\frac{\log^k(n+a)}{n+b}s^k -\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\frac{\log^{k+1} \!N}{k+1}s^k\right) \\\\
&=\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac1{(n+a)^s(n+b)} +\frac1{N^s}-\frac1s\right) \\\\
&=\zeta(s \mid a,b)-\frac1{s}
\end{align}
$$  as desired. Then, using $(6)$, we extend the preceding identity by analytic continuation  to all $s \neq 0$. To prove $(14)$, we start from a standard series representation of the digamma function (see Abram. & Steg. p. 258 6.3.16):
$$
\begin{align}
-\psi(b+1) &= \gamma - \sum_{n=1}^{\infty} \left( \frac1n - \frac1{b+n}  \right) \\
&=\lim_{N\to+\infty}\left(\gamma - \sum_{n=1}^N\left( \frac1n - \frac1{b+n}  \right)\right)\\
&=\lim_{N\to+\infty}\left(\left(\sum_{n=1}^N\frac1{b+n} -\ln N\right)-\left(\sum_{n=1}^N\frac1n-\ln N-\gamma \right)\right)\\
&=\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac1{b+n} -\ln N\right)\\\\
&=\gamma_0(a,b)
\end{align}
$$ using $(13)$.

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$

One of the consequences of Theorem 1 is the new possibility to express some series in terms of the poly-Stieltjes constants.

  Theorem 2. Let $a,b,c$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re c >-1$.

  Then
  $$
\begin{align}
  (b-a)\sum_{n=1}^{+\infty} \frac{\log (n+c)}{(n+a)(n+b)}=\gamma_1(c,a)-\gamma_1(c,b), \tag{15}
\end{align}
$$ similarly
  $$
\begin{align}
 \sum_{n=1}^{+\infty} \frac1{n+b}\left({\log (n+a)-\log (n+c)}\right)=\gamma_1(a,b)-\gamma_1(c,b), \tag{16}
\end{align}
$$ with the poly-Stieltjes constant
  $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right).
$$

Proof. Let $a,b,c$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re c >-1$.

We have
$$
 (b-a)\frac{\log (n+c)}{(n+a)(n+b)}=\frac{\log (n+c)}{n+a}-\frac{\log (n+c)}{n+b}
$$ giving, for $N\geq1$,
$$
\begin{align}
  (b-a)&\sum_{n=1}^N \frac{\log (n+c)}{(n+a)(n+b)}=\\\\
&  \left(\sum_{n=1}^N\frac{\log (n+c)}{n+a}-\frac{\log^2 \!N}2\right)-\left(\sum_{n=1}^N\frac{\log (n+c)}{n+b}-\frac{\log^2 \!N}2\right) \tag{17}
\end{align}
$$
letting $N \to \infty$  and  using $(13)$ gives $(15)$.

We have, for $N\geq1$,
$$
\begin{align}
  &\sum_{n=1}^N \frac1{n+b}\left({\log (n+a)-\log (n+c)}\right)\\\\
&=  \left(\sum_{n=1}^N\frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right)-\left(\sum_{n=1}^N\frac{\log (n+c)}{n+b}-\frac{\log^2 \!N}2\right) \tag{18}
\end{align}
$$
letting $N \to \infty$ and  using $(13)$ gives $(16)$.

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$

   Juantheron's integral

Let’s first give a numerical evaluation of Juantheron’s integral.

I would like to thank Jonathan Borwein and David H. Bailey who obtained the result below to $1000$ digits, in just 3.9 seconds run time, using David's new MPFUN-MPFR software, along with the tanh-sinh quadrature program included with the MPFUN-MPFR  package.

They also tried the integral with David's MPFUN-Fort package, which has a completely different underlying multiprecision system, and they obtained the same result below.

Finally, they computed the integral with Mathematica $11.0$;  it agreed with the result below, although it required about 10 times longer to run.

  Proposition 1. We have
  $$
\begin{align}  \int_{0}^{\Large\frac{\pi}2}\!&\frac1{(1+x^2)(1+\tan x)}\mathrm dx\\\\=
0.&59738180945180348461311323509087376430643859042555\\
&67307703207161550311033249824121789098990404474443\\
&73300942847961727020952797366230453350097928752529\\
&62099371263365268445580755896768905606293308536674\\
&89639352215352393870280616186538538722285601087082\\
&81730013060929540132583577240799018025603130403772\\
&83596189879605956759516344861849456740112012597646\\
&30195536341071109827787231788650530475635336662512\\
&50757672078320586388500276160658476344052492489409\\
&64026178233152015087197531148322444147655936720008 \tag{19}\\
&40650450631581050321100329502169853063154902765446\\
&58804861176982696627707544105655815406116180984371\\
&54148587721902800400109013880620460529382772599713\\
&06874977209651994186527207589425408866256042399213\\
&80515694164361264997143539392018681691584285790381\\
&65536517701019826846772718498479534803417547866296\\
&23842162877309354675086691711521468623807334908897\\
&71491673168051054009130049879837629516862688171756\\
&13790927986073268994254629238035029442300668334396\\
&901581838911515359223628586133156893962372426055\cdots
\end{align}
$$

David H. Bailey confirmed that Mathematica $11.0$, in spite of the great numerical precision, could not find a closed-form of the integral.

The next result proves that the OP integral admits a closed form in terms of the poly-Stieltjes constants.

  Proposition 2. We have
  $$
\begin{align}  \int_{0}^{\Large\frac{\pi}2}\!\!\frac1{(1+x^2)(1+\tan x)}\mathrm dx
&=\frac{(e^2+1)^2}{2(e^4+1)}\arctan \! \frac{\pi}{2}-\frac{e^4-1}{4(e^4+1)}\log\left(1+\frac{\pi^2}{4}\right)\\\\
&+\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)} \\\\
&+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac34 +\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac34 +\frac{i}{\pi}\!\right)\tag{20}\\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac14 -\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac14 -\frac{i}{\pi}\!\right)
\end{align}
$$ with the poly-Stieltjes constant
  $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right). $$

Proof. We  proceed in three steps.

Step 1. One may write
$$
\require{cancel}
\begin{align}
&\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\mathrm dx \\
&=\int_{0}^{\Large\frac{\pi}{2}}\frac{\cos x}{(1+x^2)(\cos x+\sin x)}\mathrm dx\\
&=\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{(\cos x+\sin x)+(\cos x-\sin x)}{(1+x^2)(\cos x+\sin x)}\mathrm dx\\
&=\frac12\int_{0}^{\Large\frac{\pi}{2}}\!\frac{1}{1+x^2}\mathrm dx+\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)}\frac{(\cos x-\sin x)}{(\cos x+\sin x)}\mathrm dx\\
&=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{1+x^2}\tan (x-\pi/4)\:\mathrm dx\\
&=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_{-\Large\frac{\pi}{4}}^{\Large\frac{\pi}4}\frac{1}{1+(x+\pi/4)^2}\tan x \:\mathrm dx\\
&=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_0^{\Large\frac{\pi}4}\left(\frac1{1+(x+\pi/4)^2}-\frac1{1+(x-\pi/4)^2}\right)\tan x \:\mathrm dx\\
&=\frac12 \arctan\! \frac{\pi}{2}-\frac{\Im}2 \!\int_{0}^{\Large\frac{\pi}{4}}\!\!\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx \tag{21}
\end{align}
$$
Let’s evaluate the latter integral.

Step 2. One may recall that the tangent function, as a meromorphic function, can be expressed as an infinite sum of rational functions:
$$
\tan x = \sum_{n=0}^{+\infty} \frac{2x}{\pi^2 (n+1/2)^2-x^2}, \quad x \neq \pm \pi/2, \pm 3\pi/2,\pm 5\pi/2,\ldots. \tag{22}
$$
We have the inequality
$$
\sup_{x \in [0,\pi/4]}\left|\frac{2x}{\pi^2 (n+1/2)^2-x^2}\right|\leq \frac1{(n+1/2)^2}, \quad n=0,1,2,\ldots, \tag{23}
$$ the convergence in $(22)$ is then uniform on  $[0,\pi/4]$. Thus, plugging $(22)$ into $(21)$, we are allowed to integrate $(21)$ termwise.

Each term, via a partial fraction decomposition, is evaluated to obtain
$$
\begin{align}
\int_{0}^{\Large\frac{\pi}4}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right)\frac{2x}{\pi^2 (n+1/2)^2-x^2}\:\mathrm dx\\
&=\frac{2\tau}{\pi^2 (n+1/2)^2-\tau^2}\log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)\\
&+\frac1{\pi}\frac1{(n+1/2+\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)\\
&+\frac1{\pi}\frac1{(n+1/2-\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)
\end{align} \tag{24}
$$
where for the sake of convenience we have set $\tau:=\pi/4+i$.

Step 3. We sum $(24)$ from $n=0$ to $\infty$ obtaining

$$
\begin{align}
\int_{0}^{\Large\frac{\pi}{4}}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx\\
&=\sum_{n=0}^{\infty}\frac{2\tau}{\pi^2 (n+1/2)^2-\tau^2}\log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)\\
&+\frac1{\pi}\sum_{n=0}^{\infty}\frac1{(n+1/2+\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)\\
&+\frac1{\pi}\sum_{n=0}^{\infty}\frac1{(n+1/2-\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right), \tag{25}
\end{align}
$$ then, singling out the first terms in the two last series and using Theorem $2$ $(16)$, we get
$$
\begin{align}
\int_{0}^{\Large\frac{\pi}{4}}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx\\
&=\tan \tau \log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)
+\frac{4\pi}{\pi^2 -4\tau^2}\log 3
\\&+\frac1{\pi}\gamma_1\!\left(\!\frac34,\frac12 +\frac{\tau}{\pi}\!\right) -\frac1{\pi}\gamma_1\!\left(\!\frac14,\frac12 +\frac{\tau}{\pi}\!\right)\tag{26}\\ &+\frac1{\pi}\gamma_1\!\left(\!\frac34,\frac12 -\frac{\tau}{\pi}\!\right) -\frac1{\pi}\gamma_1\!\left(\!\frac14,\frac12 -\frac{\tau}{\pi}\!\right)
\end{align} $$ and the substitution $\tau=\pi/4+i$ gives the desired result.

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$

Achille Hui's conjecture is true.

Achille Hui has announced in the comments that the OP integral is equal to
$$
\begin{align}
\frac{\arctan(\frac{\pi}{2}) - t\log\sqrt{1+\frac{\pi^2}{4}}}{1+t^2}+\frac{\pi^2}4 \sum_{n=0}^{\infty}\frac{ (2n+1)\left(\log\left(n+\frac34\right)-\log\left(n+\frac14\right)\right) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) } \tag{27}
\end{align}
$$ with $\displaystyle t := \tanh(1)$.

The first term in $(27)$, with a little algebra is seen  to be equal to the sum of the first two terms on the right hand side of $(20)$.

We establish the veracity of the conjecture using Proposition $2$ and using the next result.

  Proposition 3. We have
  $$
\begin{align}  &\frac{\pi^2}4 \sum_{n=0}^{\infty}\frac{ (2n+1)\left(\log\left(n+\frac34\right)-\log\left(n+\frac14\right)\right) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) }\\\\
&=\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)} \\\\
&+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac34 +\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac34 +\frac{i}{\pi}\!\right)\tag{28}\\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac14 -\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac14 -\frac{i}{\pi}\!\right)
\end{align}
$$ with the poly-Stieltjes constant
  $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right).$$

Proof. Observe that the first term of the series on the left hand side of  $(28)$, given by $n=0$, is just equal to
$$
\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)}.
$$ By a partial fraction decomposition, one may check that
$$
\begin{align}
\frac{\pi^2}4 &\frac{ (2n+1) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) }
=\frac{\Im}{2\pi}\left(\!\frac1{n+\frac34+\frac{i}{\pi}}-\frac1{n+\frac14+\frac{i}{\pi}}\!\right) \tag{29}
\end{align}
$$
then, multiplying $(29)$ by $\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right)\right)$ and summing from $n=1$ to $\infty$ we get, using Theorem $2$ $(16)$, the result $(28)$.

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$

A good fit for $\frac{1}{1+\tan x}$ in $[0,\frac{\pi}{2}]$ is $1-\frac{2}{\pi}x$

So

$$I_{approx}=\int_0^{\frac{\pi}{2}}\frac{1-\frac{2}{\pi}x}{1+x^2}dx=\arctan \frac{\pi}{2}-\frac{1}{\pi}\ln\left (1+ \frac{\pi^2}{4} \right )$$
Absolute error is about $0.011$


Circles tangent to a parabola For the past two weeks I was struggling with solving the following problem.

Description of variables:

$(x_n,y_n)$ - center point of the circle $C_n$

$r_n$ - radius of the circle $C_n$

Given the curve $k:y=x^2,x\geqslant0$, and knowing that for every $n\in\mathbb{N} \wedge  n>0$ $C_n$ is tangent to $k$, the $OX$ axis and externally tangent to $C_{n-1}$, find the explicit formula for $x_n$, $y_n$ and $r_n$ given the initial condition $C_0: (x-x_0)^2+(y-y_0)^2=r_0^2$, where $C_0$ is tangent to $OX$ and $k$.

It is obvious that $r_n=y_n$, but I did not manage to derive a sensible system of equations.
I said "sensible", because this is the only one I could think of:

$\begin{cases} \frac{x_T-x_n}{y_n-y_T}=2x_T \\ y_T=x_T^2\\\sqrt{(x_T-x_n)^2+(y_T-y_n)^2}=y_n \\ \sqrt{(x_n-x_{n-1})^2+(y_n-y_{n-1})^2}=y_n+y_{n-1} \end{cases}$

Is there no more convenient way to approach this?

Thank you for any help in advance.

This is not intended to be an answer as such but more of a suggestion as to how one might be able to proceed further.

It seems that the equation satisfied by the coordinates of the centre $(x,y)$ of any circle which touches the parabola and the $x$ axis is itself very complicated: $$x^2=\left(\frac{8y-1+\sqrt{16y+1}}{8}\right)\left(\frac{3+\sqrt{16y+1}}{4}\right)^2$$

Despite its appearance this is a fairly innocent-looking curve passing as you would expect between the parabola $y=x^2$ and the $x$ axis.

This was obtained by just considering the requirement that each circle must touch the parabola and the $x$ axis. Therefore somehow this would need to be coupled with the requirement that the circles touch each other in series, which can be written as $$x_{n+1}=x_n+2\sqrt{y_ny_{n+1}}$$
But I don't know how to link these up. Any ideas?

Standard  parametric equn of parabola $ y = x^2 / (4 f) $ is

( in this case we can take $ f= \frac14 $ )

$$ x = 2 f t ; \, y = f t^2; \, t = \tan \phi $$

Radius of contacting circle $ = y  \sec\phi   = f t^2 \sqrt{1+t^2} $

Distance to right of parabola point x-coordinate is $ y\,  t = f t^3  $

So the center of circle on x-axis $ ( 2 f t + f t^3, 0) $

Rest you can proceed.

EDIT1:

Contacting circles as shown.

$r$ is radius of bigger contact circle given independently as in sketch.

$$ x = 2 ft + r \sin \phi = 2 f t  + \frac {r  t}{ \sqrt{1+t^2}}$$

$$ y = ft^2 - \frac{r}{ \sqrt{1+t^2} }   $$

So you can solve for  $ t =  \tan \phi $ when $ y=r $ from the resulting sixth degree transcendental equation :

$$  \frac{f\; t^2}{r}= 1 -\frac{1}{ \sqrt{1+t^2} } $$


Extending Positive Functionals: Linearity How does regularity provide linearity?

Given the full Banach space of bounded functions over a suitable set:
$$\mathcal{B}:=\{f:\Omega\to \mathbb{C}:\|f\|_\Omega<\infty\}$$
and a linear subspace $\mathcal{C}\leq\mathcal{B}$.

Consider a linear functional on the subspace:
$$I:\mathcal{C}\to\mathbb{C}:f\mapsto I(f)$$
that is bounded and positive:
$$|I(f)|\leq \|I\|\cdot\|f\|_\Omega$$
$$I(f)\geq 0\quad(f\geq 0)$$

Extend it formally to all bounded functions by splitting these:
$$g=g_{\Re_+}+g_{\Re_-}+ig_{\Im_+}+ig_{\Im_-}$$
and then defining:
$$I_E(g):=\sup_{f\in\mathcal{C}:f\leq g} I(f)\quad(g\geq 0)$$
(Note that $\sup_{f\leq g}I(f)=\sup_{0\leq f\leq g}I(f)$.)

This gives a sharp estimate for the bound:
$$|I_E(g)|=|\sup_{0\leq f\leq g}I(f)|=\sup_{0\leq f\leq g}I(f)\leq\sup_{0\leq f\leq g}\|I\|\cdot\|f\|_\Omega=\|I\|\cdot\|g\|_\Omega\quad(g\geq 0)$$
and therefore:
$$|I_E(g)|=\Re_+ e^{-i\phi}I_E(g)=I_E(\Re_+ e^{-i\phi}g)\leq \|I\|\cdot\|\Re_+ e^{-i\phi}g\|_\Omega\leq\|I\|\cdot\|g\|_\Omega\quad(re^{i\phi}:=I_E(g))$$
(Note, linearity has not been used yet in full!)

Moreover it remains positive:
$$I_E(g)\geq 0\quad(g\geq 0)$$

Now, it requires additional regularity assumptions in order to maintain linearity...

But which and how to proceed then?

As an example to see what actually can happen consider the integral of simple functions:
$$I:\mathcal{S}([0,1])\to\mathbb{C}:f\mapsto\int_{[0,1]} f\mathrm{d}x$$

and the characteristic functions:
$$\chi_{A},:\mathbb{[0,1]}\to\mathbb{C}$$
$$\chi_{A^c}:\mathbb{[0,1]}\to\mathbb{C}$$
for some nonmeasurable set $A\subseteq[0,1]$.

Then the extension gives:
$$I_E(\chi_A+\chi_{A^c})=\mu([0,1])$$
$$I_E(\chi_A)+I(\chi_{A^c})=\mu^*(A)+\mu^*(A^c)$$
but $A$ was nonmeasurable so the integral fails to be linear here.

You have defined $I$ on the nonnegative functions $g$, then extended to arbitrary $f$ by $I(g) = I(g_{\Re_+})- I(-g_{\Re_-})+i I(g_{\Im_+}) - i I(-g_{\Im_-})$ where $g=g_{\Re_+}+g_{\Re_-}+ig_{\Im_+}+ig_{\Im_-}$ and $g_{\Re_+} \ge 0, g_{\Re_-} \le 0, g_{\Im_+} \ge 0, g_{\Im_-} \le 0$.  Now suppose $f = g + h$.
Why should $I(f) = I(g) + I(h)$?

Well, if all these functions were in $\mathcal C$ this would be true because $I$ is linear on $\mathcal C$.  So we hope for an approximation: if for any $\epsilon > 0$ there are
$\tilde{f}, \tilde{g},\tilde{h} \in \mathcal C$ such that $\tilde{f} = \tilde{g} + \tilde{h}$ and $|I(f) - I(\tilde{f})|$, $|I(g) - I(\tilde{g})|$, $|I(h) - I(\tilde{h})| < \epsilon$
then $|I(f) - I(g) - I(h)| < 3 \epsilon$, and since $\epsilon$ is arbitrary that means $I(f) - I(g) - I(h) = 0$.  Regularity (which you didn't actually state) should allow you to find such $\tilde{f}, \tilde{g}, \tilde{h}$.

The problem is that it misses some sort of monotone convergence theorem:
$$0\leq g_n(x)\uparrow g(x)\implies I_E(g_n)\uparrow I_E(g)$$
Then that would suggest to consider all those functions which are approximable:
$$g\in\mathcal{C}_E:\iff\exists f_n\in\mathcal{C}:f_n(x)\to g(x)$$


What's the matter of Cantor's diagonal argument in binary system？ In binary system, Cantor's diagonal argument lost its effect.(https://zhuanlan.zhihu.com/p/20197130)(Sorry for the Chinese needed to translating)
What can we say about it? How can we improve it？

In binary system,if we construct like this:
r = 0.d1d2d3…
di = 1 - dii

Then we fail. For an example:
r1 = 0.1
r2= 0. 001
r3 = 0.0001
r4 = 0.00001
…

The thing we diagonally inverted constructed is:
r = 0.0111…

But in this way,it just is r1.

And it seems other ways of diagonal construction fails too.

(Sorry for my be utterly ignorant of things like Latex)

There are several ways to fix this; here are two.

We can look at more than one digit at a time. Given our starting list $$(0.a^1_1a^2_2a^3_3, \quad 0.a^2_1a^2_2a^2_3...,\quad 0.a^3_1a^3_2a^3_3..., \quad ...)$$ of binary representations of reals in $[0,1)$, we define a sequence $b_1,b_2,b_3,...$ as follows: the two-digit block $b_{2i-1}b_{2i}$ is "$01$" if $a_{2i-1}a_{2i}$ is not "$01,$" and $b_{2i-1}b_{2i}$ is "$10$" otherwise.
We can just transfer between contexts. We don't have to use the binary representations; we could always switch to (say) decimal representations, apply diagonalization there (where having more than two digits gives us "room" to work without having to consider multiple digits at once), and then convert back to binary. This might feel like cheating, but it's perfectly valid.

I'd say that both approaches have their advantages; ultimately each is teaching a kind of flexibility when it comes to constructions, either at the specific implementation level or right at the formulation level.

The argument Georg Cantor presented was in binary. And I don't mean the binary representation of real numbers. Cantor did not apply the diagonal argument to real numbers at all; he used infinite-length binary strings (quote: "there is a proof of this proposition that ... does not depend on considering the irrational numbers.") So the string "10000..." is different than the string "01111...".

There are several other discrepancies about the way it is taught, and almost all of the objections to it stem from one of them.


Is it true that Quadratic residue was published and discovered before Legendre symbol and Euler's Criteria? So Is it true that Quadratic residue was published and discovered before Legendre symbol and Euler's Criteria?
Quadratic residue came in 1801 by Gauss(1).
can you put these concepts in chronological order and by whom these concepts was first introduced or developed in paper form?

I'd say yes. My reason being that the Legendre symbol simply defines if a number $\mod p$ is a quadratic residue to $p$ or not. There is no way to define a number being a quadratic residue $\mod p$ if you don't even know what a quadratic residue is.

Regarding the Euler criterion, that's a little bit more tricky. That too would've come after the discovering quadratic residues, but I'm not sure if that was before or after the Legendre symbol


Solving system of ODE $\frac{d^2x}{dt^2}+3\frac{dx}{dt}-2x+\frac{dy}{dt}-3y=2e^{-t}$

$2\frac{dx}{dt}-x+\frac{dy}{dt}-2y=0$

given that $x(0)=\frac{dx}{dt}_{|x=0}=0, y(0)=4$

I tried eliminating $\frac{dx}{dt}$ in the 1st eqn.

Eqn 1 becomes

$$\frac{d^2x}{dt^2}+\frac{dx}{dt}-x=y+2e^{-t}$$

I thought of decoupling this system to eliminate $y$.

$$\frac{d^3x}{dt^3}+\frac{d^2x}{dt^2}-\frac{dx}{dt}=\frac{dy}{dt}-2e^{-t}$$

But all  this seems to complicate the problem even more.

Subtract the second from the first, solve that for $y$. Take the derivative to get an expression for $y'$. Substitute those two in the second equation and solve for $x$.

You will get:

$$y = x'' + x' - x - 2e^{-t}\\y' = x''' + x'' - x' + 2 e^{-t}$$

Substituting into the second equation, you get:

$$x'''-x''-x'+x +6e^{t} = 0$$

Also, you need to figure out one more IC, but you have:

$$x''(0) = 6$$

This leads to:

$$x(t) = \dfrac{3}{2}t(e^t-e^{-t})$$

I am sure you can take it from here.


Hamiltonian cycle in tournament I am new to graph theory and I've just learnt about Hamiltonian cycle in a graph which is a NP-complete problem. But I have to find a Hamiltoninan cycle in a graph with N*(N-1)/2 edges which I believe it is called a tournament. I saw an algorithm on Wikipedia which I din not understand properly( the one in which you have  v1, ... , vk ham path and when you add a vertex v you search for two other vertexes and reverse something ). So can someone please explain me an algorithm with has the time O(N^2) and which finds a hamiltonian cyclein a tournament ( I believe that in the ham cycle the last vertex and the first one must be connected ).

Well, I am not sure about which algorithm you are talking about (there's no link and I could not find it at Wikipedia's tournamet page, nor Hamilton cycle page), but you can do something like this:

Remove any vertex $v$ from $G$.
In each strongly-connected component recursively calculate Hamilton cycles.
If there are at least two SCCs, let $G_\top$ and $G_\bot$ be the top and bottom respectively; find vertices $u_\top \in G_\top$ and $u_\bot \in G_\bot$ such that $u_\bot \to v \to u_\top$ (we know that such vertices exist because $G$ was strongly connected).
If there is exactly one SCC, find two vertices $u_\top, u_\bot \in G\setminus \{v\}$ such that
$u_\bot$ is a successor of $u_\top$ in the Hamilton cycle of $G \setminus \{v\}$ we previously found, and $u_\top \to v \to u_\bot$ (again, we know that such vertices exist because $G$ was strongly connected).
Patch it all together.

This might take $O(n^3)$ time, but if you were to go backwards, i.e. constructing the graph, rather than deconstructing it, then SCCs calculation can be done on-line using find-union, and then it would take about $O(n^2\alpha(n))$ which is almost the same as a square.

I hope this helps $\ddot\smile$

A tournament is a complete directed graph (hence $n(n-1)/2$ edges). You can think about it as $n$ teams where each one played against each other and there is an edge going from $v$ to $w$ only if $v$ beat $w$ in the match (other wise a there is an edge in the reverse direction).

Now, we can construct an Hamiltonian path (not cycle) where each vertex "beat" the adjacent vertex on the right (and so the graph indeed as a corresponding directed edge). If we can "line up" the vertices  in such way then it corresponds to Hamilonian path.

Start with a single vertex - a path of length 1. For each remaining vertex travese the path from left to right, trying to find a place where you can fit it in, kind of like insertions sort. If the vertex beat the leftmost vertex in the path, place the new vertex as the new leftmost. Otherwise it lost to it, so you can go on to the next vertx and check again if it can fit in. The only case you wouldn't find a place between existing vertices in the path is if the new vertex lost to every one of them - hence you can put it rightmost.

Following this idea, you can easily write a "formal" proof by induction for the statement.


Prove that $\det A$ does not exceed $1$ Let $A =(a_{ij}) ∈  M_n(\Bbb R)$ be a matrix with nonnegative entries such that the sum of the entries in each row does not exceed $1$. Prove that $|\det A|$ does not exceed $1$ too.

This is one of my exercise,I tried to induction,but i'm stuck.
Help me please! Thanks

You don't need induction. Prove/use the following two results: 1) the determinant is the product of the eigenvalues, and 2) any eigenvalue of a nonnegative matrix has magnitude at most the largest row sum (just write out the eigenvalue equation to see why).

You should be able to use expansion by minors in an inductive proof.


Prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$ Prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$.

I noticed that $x \equiv 5 \bmod 6$ and thus $2^x \equiv 4 \bmod 7$, but that doesn't seem to help me since $3 \cdot 9^4 +5 \equiv 4 \bmod 7$. Pretty much any other mod I use doesn't seem to work so I think proof by contradiction or something may work better.

Any solutions must be fairly small - either $m \le 11$ or $x \le 20$.  I assume you have checked all those cases and found no solutions but the two you're already aware of; indeed, there aren't any.
Let's prove this - though with a warning: there's lots of computation and black magic ahead.
EDIT: It turns out we can substantially reduce the amount of computation.  See edit at the bottom.

Suppose $m \ge 12$, and work modulo $3^{12}$.  Since $2$ is a primitive root modulo $3^{12}$, we have a congruence restriction on $x$ modulo $\phi(3^{12}) = 2\cdot3^{11} = 354294.$  In particular, we must have $$x \equiv 199283\pmod{354294}.$$
Now suppose $x \ge 21$, and work modulo $2^{21}$.  $3$ is not quite a primitive root modulo $2^{21}$ - it generates an index-$2$ subgroup of $(\mathbb Z/2^{21}\mathbb Z)^{\times}$, but nonetheless we obtain the restriction $$m \equiv 15627\pmod{524288}.$$
Now let's look for small primes divisible by both 354294 and 524288.  It turns out that $$p = 1+11\cdot\mathrm{lcm}(354294,524288) = 1021636509697$$ is prime, and $11$ is the smallest number for which this is true.

Now consider the sets $$S_2 = \left\{2^{199283+354294\cdot k}\pmod{p}\right\}, \qquad S_3 = \left\{3^{15627+524288\cdot k}\pmod{p}\right\}.$$
It's clear from definitions that $$|S_2| = \mathrm{ord}_p\left(2^{354294}\right) = 720896, \qquad |S_3| = \mathrm{ord}_p\left(3^{524288}\right) = 649539.$$
So we expect approximately $$\frac{720896\cdot649539}{1021636509697} = 0.458\cdots$$ elements in $S_2 \cap (S_3 + 5)$.  If we have no such elements, then there can't be any solutions to $2^x = 3^m + 5$ with $m \ge 12$ and $x \ge 21$.
We get lucky, and there aren't any elements in the intersection!  TA-DA - there are no solutions other than the two you know!

In all seriousness, this shouldn't have worked.  $S_2$ is one-fourth the size we expect; $S_3$ is one-third the size.  And it was still about a fifty-fifty shot if there was anything in their intersection at all.  And $p$ was a factor of two smaller than expected.  We should have had to find much bigger congruences and use those.  But black magic happened, and everything worked out.
One more remark: the reason we chose $p$ divisible by both 354294 and 524288 was that it forced the sizes of $S_2$ and $S_3$ to be fairly small relative to $p$.  But their product should have been roughly $11/2$ the size of $p$ in the worst possible case; as I've said, we got lucky.

EDIT: Here's how to make the numbers involved much smaller.
Work modulo $3^5$; then $x \equiv 23\pmod{162}$.  Work modulo $2^2$; $m$ must be odd.  Then work modulo $163$.  The only possible residue of $2^{23+162k}\pmod{163}$ is $139$, and $134$ is not a possible residue for $3\cdot9^m$ modulo $163$.  Hence we're done.


In $\ell^p$, if an operator commutes with left shift, it is continuous? Our professor put this one in our exam, taking it out along the way though because it seemed too tricky. Still we wasted nearly an hour on it and can't stop thinking about a solution.

What we have: The left shift $L : \ell^p \to \ell^p$
$$L(x_1,x_2,x_3,\ldots) = (x_2,x_3,\ldots)$$

and another operator $T$. We should prove that if $TL=LT$, then $T$ is continuous.

We had defined subspaces
$$ X_k = \{ (x_i) : x_i = 0 \text{ for } i>k \} $$
and seen that these are $T$-invariant and the restrictions $T : X_k \to X_k$ continuous (obvious). The hint was to use closed-graph-theorem to show that $T$ is continuous. Of course we can truncate any sequence to then lie in $X_k$, however I do not see how convergence of the truncated sequences relates to convergence of the images under $T$.

Any help please?

The statement is false, as I discovered here.  Since not everybody has access to the paper, let me provide a summary of the argument:

Let $R=\mathbb C[t]$, $L$ the left shift operator, and view $\ell^p$ as an $R$-module by defining $t\cdot x=Lx$.  Let $X=\sum \ker L^i \subset \ell^p$ be the subspace of eventually-zero sequences.

Lemma: Given a PID $P$, a $P$-module $M$ is injective if and only if it is divisible, i.e., if for every $p\in P$, $pM=M$.

Lemma: $X$ is an injective $R$-module.

Proof.  Since a PID is a UFD, we can check divisibility (and hence injectivity) using irreducible elements.  Over $R$, the irreducible elements are linear polynomials. Thus, we need to show that if $x$ is an eventually-null sequence and $\lambda\in \mathbb C$, there exists a $y$ such that $Ly-\lambda y=x$.  This is straight forward.

Because $X$ is injective, the inclusion $X\subset \ell^p$ splits,  there exists a (non-unique) projection map $P:\ell^p\to X$.  Since this is a map of $R$-modules, it commutes with $L$.

Lemma: $P$ is not continuous.

Proof.  Suppose that $P$ were continuous. Then $X=\ker (P-\operatorname{Id})$ is closed.  However, this is absurd, as every sequences can be approximated by finitely many terms (i.e., $X$ is a dense proper subspace).

my idea is that :

$L$ is continuous $\Longrightarrow$ $LT$ is continuous $\Longrightarrow$ $T$ is
continuous

you can use the closed-graph-theorem to the left shift firstly :

$L:l_{p}$$\longrightarrow$$l_{p}$

since the $l_{p}$ is banach space

then its product space $X\times{Y}$ is banach space and the image of $L$ is a colsed subset of $X\times{Y}$ since $L$ is a left shift

therefore, $L$ is continuous

another condition you used is that ：

since, if $LT=TL$

then, $(LT)^{n}=L^{n}T^{n}$

because of,
$\Arrowvert$$AB$$\Arrowvert$$\le$$\Arrowvert$$A$$\Arrowvert$$\Arrowvert$$B$$\Arrowvert$

so,  $\Arrowvert$$(LT)^{n}$$\Arrowvert$$\le$$\Arrowvert$$L^{n}$$\Arrowvert$$\Arrowvert$$T^{n}$$\Arrowvert$$\Longrightarrow$$\Arrowvert$$(LT)^{n}$$\Arrowvert$$^{1/n}$$\le$$\Arrowvert$$L^{n}$$\Arrowvert$$^{1/n}$$\Arrowvert$$T^{n}$$\Arrowvert$$^{1/n}$

as you describe, $$ X_k = \{ (x_i) : x_i = 0 \text{ for } i>k \} $$

that means your subspace is $c_{00}$ , and in this space there are just finite terms not equal to zero.

therefore, by the property of its subspace $c_{00}$ :

since the restrictions $T : X_k \to X_k$ is invariant and the norms on $c_{00}$ is $\Vert$$\Vert$$_{\infty}$

which imply that,

$\Arrowvert$$(LT)^{n}$$\Arrowvert$$\le$$\Arrowvert$$L^{n}$$\Arrowvert$$\Arrowvert$$T^{n}$$\Arrowvert$$\Longrightarrow$$\Arrowvert$$(LT)^{n}$$\Arrowvert$$^{1/n}$$\le$$\Arrowvert$$L^{n}$$\Arrowvert$$^{1/n}$$\Arrowvert$$T^{n}$$\Arrowvert$$^{1/n}$$\le$$\Arrowvert$$L^{n}$$\Arrowvert$$^{1/n}$$max$$\arrowvert$$x_{i}$$\arrowvert$$^{1/n}$

$=$$\Arrowvert$$L^{n}$$\Arrowvert$$^{1/n}$

holds as $n\longrightarrow$$\infty$

consequently, $LT\in{BL(X)}$$\Longrightarrow$$T\in{BL(X)}$


Solution of system of equations in prime fields In 'Algebra', Artin writes that the system of equation:

$$8x+3y = 3$$
$$2x+6y = -1$$

have no solutions in $\mathbb{F}_2$ and $\mathbb{F}_3$ as the determinant (of the coefficient matrix) evaluates to $42$ which is equal to $0$ (mod $2$, $3$ and $7$).

However he mentions that there happens to be a solution $(3, 0)$ in $\mathbb{F}_7$ though the determinant is zero modulo $7$. However this solution is not evident from the non-zero determinant (invertible matrix) rule we generally look at to determine whether solutions exist or not.

Should we always reduce the coefficient matrix elements (modulus $p$) before evaluating value of determinant to check if  a solution exists in $\mathbb{F}_p$? Is there a better way to check for existence of solutions in prime fields?


Proof that $\lim\limits_{n \to \infty} (1-\frac{2}{n})^n$ exists I know the proof that $\lim\limits_{n \to \infty} (1+\frac{1}{n})^n$ exists, but I don't think I can use the same proof using the binomial theorem for this because I get $2^i$ in there and that increases without bound. Is there another method for proving this?

To show that the sequence $a_n=\left(1-\frac2n\right)^n$ converges, we will show that it is monotonically increasing and bounded above.

To show that it is monotonically increasing we analyze the ratio $\frac{a_{n+1}}{a_n}$.  To that end, we write for $n\ge 2$

$$\begin{align}
\frac{a_{n+1}}{a_n}&=\frac{\left(1-\frac2{n+1}\right)^{n+1}}{\left(1-\frac2n\right)^n}\\\\
&=\left(1+\frac{2}{(n-2)(n+1)}\right)^{n+1}\left(1-\frac2n\right) \tag 1\\\\
&\ge \left(1+\frac{2}{n-2}\right)\left(1-\frac2n\right) \tag 2\\\\
&=1
\end{align}$$

Thus, $a_{n+1}\ge a_n$ for $n\ge 2$.  In going from $(1)$ to $(2)$, we made use of Bernoulli's Inequality.

Now, it is obvious that $a_n=\left(1-\frac2n\right)^n\le 1$.  And finaly, since $a_n$ is monotonically increasing and bounded above by $1$, the sequence converges.


Formal proof of $\lim_{x\to a}f(x) = \lim_{h\to 0} f(a+h)$ $$ \lim_{x\to a} f(x) = \lim_{h\to 0} f(a+h) $$

How do I write a formal proof of it?

Suppose $\displaystyle \lim_{x \to a}f(x) = c$ and let $\epsilon > 0$, then there is $\delta >0$ such that $$|x-a|<\delta \implies |f(x)-c | < \epsilon$$
now note that $$|h| = |(a+h)-a|$$ and so
$$|h|< \delta \implies  |f(a+h)-c | < \epsilon$$
which shows that $\displaystyle \lim_{h \to 0}f(a+h) = c$. It follows that
$$\displaystyle \lim_{x \to a}f(x) = c= \displaystyle \lim_{h \to 0}f(a+h).$$

Pose $h=x-a$, then $x=a+h$ and when $x\to a$, there is $h\to 0$. Then,

$$\lim_{x\to a}f(x)=\lim_{x-a\to 0}f(x-a+a)=\lim_{h\to 0}f(a+h)$$


Men on a boat problem There is the usual question of some men on a boat- various men have various speeds, the boat has a capacity of 2 men, and the boat takes on the speed of the slowest man in the boat at any given time. Suppose we had $n$ men and 1 canoe with a capcity of two men, with the $nth$ man taking $2^n$ minutes to cross from one side to the next. What is the strategy to optimize the time to get all men to the other side?

Given the base case: $F(2)=2$ and $F(3)=2+1+4=7$.

And given that if you have $k$ man using the following greedy strategy:

send $1$ and $2$
then $1$ come back
sent the two slowest (for a time of $2^k$)
and then $2$ can come back
solve the smaller sub-problem (now you have the $k-2$ fastest men on one side, and the slowest two already passed, and you do not need them to take the boat back)

For example, for $4$ men: take $1$ and $2$ to the other side; come back with $1$; go to the other side with $3$ and $4$; go back with $2$; solve the sub-problem for $2$ men, that is just a trip of $1$ and $2$ for a total of $2+1+8+2+2=15$ time units.

This strategy is optimal (you never waste a transport), and for the time required this recursion holds:
$F(n) = 2^k + 5 + F(n-2)$ (with the base case $F(3)=7$ and $F(4)=15$).

for two persons (speeds 2,4) - ans 4 mins

for 3 persons (speeds 2,4,8) - ans 2,4 go -> 4 ,2 comes back. next 2,8 go-> 8 so total 4+2+8

By Generalisation optimised time would be (2^2+2^3+......+2^n)+2*(n-2)


Confused about composition of two permutation cycles I'm slightly confused about the product of the following permutation cycles.

I am given that $s_1 = (1\ 2)$ and $s_2 = (2\ 3)$ where both are generators for the symmetric group $S_3$.

My textbook proceeds by saying that $s_1 s_2 = [3\ 1\ 2]$ and that $s_2 s_1 = [2\ 3\ 1]$, but I can't really see why. To me it looks like the results have been mixed up.

My understanding is that (correct me if I am wrong),

$s_1 s_2 = (1\ 2)\cdot(2\ 3)=(1\ 2)(3)\cdot(1)(2\ 3)=(1\ 2\ 3)$

which on one-line notation is $[2\ 3\ 1]$.

Similarly, I compute the product $s_2 s_1$ as

$s_2 s_1 = (1)(2\ 3)\cdot(1\ 2)(3)= (1\ 3\ 2)=[3\ 1\ 2]$

Clearly, my results are the exact opposites of the textbook's.

What am I missing? At first I thought it was a typo, but Wolfram Alpha agrees with my textbook so there must be something I have misunderstood.

Usually we apply $(s_1s_2)(i)=s_1(s_2(i))$. Hence
$(12)(23)$ applied to $1$, is first applying $(23)$, so fixing $1$, and then moving $1$ to $2$ by $(12)$. Similarly, $2$ is moved to $3$ and then fixed. So in this order,
$$
(12)(23)=(123).
$$


Is it possible to rewrite $X^* A X = B$ in ${\operatorname{vec}}(X)$, where $X \in \mathbb{C}^{n \times m}$, $A \in \mathbb{C}^{n \times n}$? Is it possible to vectorize the quadratic $$X^* A X = B$$ in ${\operatorname{vec}}(X)$, where $X \in \mathbb{C}^{n \times m}$, $A \in \mathbb{C}^{n \times n}$, and $B \in \mathbb{C}^{m \times m}$?

Apologies if it is trivial. I just don't know how to proceed. Your help will be highly appreciated.

I thought of using Kroncker Product. But it can represent only one side of the quadratic equation, i.e., either $X^*$ or $X$.

Partial attempt:

${\operatorname{vec}} (X^* A X) = \left( I \otimes X^* A \right) {\operatorname{vec}} (X) = {\operatorname{vec}} (B)$, then how to pull out $X^*$ from the Kronecker product? I want to represent the quadratic form in only ${\operatorname{vec}} (X^*)$ and ${\operatorname{vec}} (X)$.


Triple Integral Over a Ball How do I compute the following:
$\iiint _V (x^{2n} + y^{2n} +z^{2n}) dxdydz$ where $V = \{x^2 +y^2 +z^2 \leq 1\}$?

Due to symmetry
$$\iiint _V (x^{2n} + y^{2n} +z^{2n}) dxdydz
= 3\iiint _V z^{2n} dxdydz\\
= 3\int_0^{2\pi}\int_0^\pi \int_0^1 r^{2n}\cos^{2n}\theta \>r^2\sin\theta drd\theta d\phi
 = \frac{12\pi}{(2n+3)(2n+1)}
$$

Your question as it stands is likely incorrect.  Are you really interested in the cases where $n<0$, for instance??
Assuming $n \geq 1/2$, and using symmetry:
$$8 \int\limits_{x=0}^1 \int\limits_{y=0}^{\sqrt{1-x^2}} \int\limits_{z=0}^{\sqrt{1 - x^2 - y^2}} (x^{2 n} + y^{2 n} + z^{2 n})\ dx\ dy\ dz = \frac{12 \pi }{4 n (n+2)+3}$$,
which is the same as @Quanto's later answer.  (Though somehow mine is downvoted.)


How can I arrive from $\prod_{i=1}^{D} \int_{-\infty}^{\infty} e^{-x_i^2} dx_i$ to $\pi^{D/2}$? This is part of the excersise 1.18 of the Bishop's Book, Pattern Recognition and Machine Learning. I attached the excersise and the information. I do not know how he using the result (1.126) in  the left side of (1.142) arrives to $\pi^{D/2}$ (That is  the solution he provides). I tried several ways but I get nothing.
1.126: $I=(2 \pi \sigma^{2})^{1/2}$
which comes from:
$I^{2}= \int_{-\infty}^{\infty} e^{-\frac{1}{2\sigma^{2}}x^{2}-\frac{1}{2\sigma^{2}}y^{2}}dxdy $
And solving transforming form Cartesian coordinates to polar ones, substituing u = $r^{2}$ and performing the integrals over u and $\theta$

I attached the Excersises, 1.18. I can not upload more pictures so here is a drive link to the others: https://drive.google.com/drive/folders/1sc6WFyhb5wEevZOv-6qcTFKuoieCxPGH?usp=sharing

\begin{align*}
I &=\int_{-\infty}^{\infty} e^{-x_i^2} dx_i \\
I^2 &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x_i^2}  e^{-y_i^2} dx_i dy_i \\
 &=\int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2}  r dr d\theta \tag{1} \\
 &=\int_{0}^{2\pi} -\frac{e^{-r^2}}{2} \bigg \rvert_{r=0}^{\infty} d\theta \\
 &=\int_{0}^{2\pi} \frac{1}{2}  d\theta \\
I &=\sqrt{\pi} \\
\end{align*}
And so,
\begin{align*}
\prod_{i=1}^D \int_{-\infty}^{\infty} e^{-x_i^2} dx_i  &= \prod_{i=1}^D \sqrt{\pi} \\
&= \boxed{\pi^{\frac{D}{2}}}
\end{align*}
Where in $(1)$ the double integral was converted to polar coordinates.


Sequence Of Nested Open Interval Provide an example of a sequence of nested open intervals $((an,bn))_{n=1}^{+\infty}$, $a_{n+1},b_{n+1} \subset(a_{n},b_{n})$, such that $∩_{n=1}^{\infty} (a_{n},b_{n}) = \emptyset$

Solution Attempt:

$[n,\infty)$

Take $a_n = 0, b_n = \dfrac{1}{n}, \forall n \ge 1$

$$b_n=1$$

$$a_n=1-\frac 2n$$

$\forall \epsilon<1$, for $n$ great enough,

$$1-\frac 2n>\epsilon.$$


Solutions to $k$ when $2^k n^2 + 2^k n + 1$ is never a perfect square. I need to find possible values of $k$ with $k \in \mathbb{N}$ such that for any $n \in \mathbb{N}$ the equation $2^k n^2 + 2^k n + 1$ will never be a perfect square.

So, I thought, maybe for even $k$ I get solutions because then it is possible to complete the square, like:

Let be $k = 2m; \quad m \in \mathbb{N}$

\begin{align}
2^{2m} n^2 + 2^{2m} n + 1 &= (2^m)^2 n^2 + 2 \cdot 2^{m} \cdot 2^{m-1} n + 1 \\
&= \left(2^m n + 1\right)^2 + 2^{m+1} \left(2^{m-1} - 1 \right) n \\
&= \left(2^m n + 1\right)^2 + \left(2^{m}\right)^2 n - \left(2^m\right) n
\end{align}

But then I am stuck and don't know if it leads me to something.

Edit 1:

The comment of Shamim leads me to the condition that if it is a perfect square, it needs to be an odd one because:

\begin{align}
2^k n^2 + 2^k n + 1 \equiv 1 \mod{4}
\end{align}

But still, it does not restrict $k$.

COMMENT.-Let note $N=2^kn^2+2^kn+1$.We assume that $0\notin\mathbb N$ because if not then the set solution is obviously empty because $1$ is a square. (Many times you have to consider $0$ as a natural, particularly for Fermat's theorem on the sum of two squares).

►$k=1$ discarded because  $n=3\Rightarrow N=5^2$.

►$k=2$ discarded because of all $n$.

►$k=3$ discarded because  $n=14\Rightarrow N=41^2$

►$k=5$ discarded because $n=5\Rightarrow N=31^2$

We have $$N=\begin{cases}3\cdot(2^{k-1}n)^2+(2^{k-1}n+1)^2 \text{ when $k\gt2$ is even }\\       6\cdot(2^{k-1}n)^2+2(2^{k-1}n+1)^2-1 \text{ when $k\gt5$ is odd }\end{cases} \hspace {15mm}(1)$$
Consider the more general equations involving the equations $(1)$ $$\begin{cases}3x^2+(x+1)^2=z^2\hspace {15mm}(2)\\6x^2+2(x+1)^2-1=z^2\hspace {5mm}(3)\end{cases}$$
According to Wolfram, $(2)$ has no positive integer solution and $(3)$ has infinitely many (determine similarly to Pell-Fermat equation); however among these some particular $x$ must have the form $x=2^{k-1}n$ for $k$ odd greater than $5$.

Calculating the two smaller we have $(x,z)=(10,29),(348,985)$ and we can see that $348=2^2\cdot87$; if the exponent would have been not $2$ but greater than or equal to $5(=6-1)$, we would have had a value of $ n $ to rule out another odd value of k. We stop this comment here concluding that a part of the values of k are as follows:

►all even greater than $2$

►probaly very much of odd values greater than $5$ (if not all!)

If $\ 2^kn^2+2^kn+1$ is a perfect square, then $\ 2^kn^2+2^kn+1=0$ must have two equal roots, i.e the discriminant must vanish.
Therefore
$\ \begin{align}
&(2^k)^2 =4.2^k.1\\
\implies &4^k =4.2^k\\
\implies &4^{k-1} =2^k\\
\implies &2^{2k-2} =2^k\\
\implies &2k-2 =k\\
\implies &k=2\\
\end{align}$
So $\ 2^kn^2+2^kn+1$ is a perfect square if $\ k=2\; \forall n\in \mathbb{N}$. Thus if $\ k\ne2$ then $\ 2^kn^2+2^kn+1$ is not a perfect square.


Why is the 3D case so rich? The Banach--Tarski theorem applies only in the case of three or more dimensions. In 3D, there are five regular solids, two of them being not at all obvious, and the 4D case is also interesting; but the higher-dimensional cases each yield just three "solids", only one of which isn't obvious. In dynamics, particles tend to coalesce in one or two dimensions, while in four or more dimensions they tend to disperse drearily; only in 3D do they move freely but with significant local interaction. And the Poincare conjecture proved to be much harder in the 3D case than in the others.

So the question has three parts: What other examples are there of 3D richness? Are there any underlying reasons for it? And are there fields where richness begins at a higher, but still small, number of dimensions? I would be particularly interested to learn of any theorems that hold in just a particular number (not 0, 1, or 2) of dimensions.

Exotic $\mathbb R^4$

There are infinitely many non-diffeomorphic smooth structures on the topological space $\mathbb R^n$ if and only if $n=4$. (Otherwise there is only one diffeomorphism class.)

I'm going to be a bit contrary: is this really true? Or is it just that we started studying the 3D case before we could study anything higher-dimensional, and that our brains are obviously optimized for thinking in three dimensions?


Asymptotic behavior of an integral I am interested in the integral
\begin{align*}
\int_{\epsilon}^{\infty}dx_1\int_{\epsilon}^{\infty}dx_2\int_{\epsilon}^{\infty}dx_3\int_{\epsilon}^{\infty}dx_4\,\frac{1}{(x_1+x_3)(x_1+x_4)(x_2+x_3)(x_2+x_4)}e^{-x_1-x_2-x_3-x_4}.
\end{align*}
This diverges as $\epsilon\to 0$. I don't really care about the finite part of the integral, but I would like to know how it behaves as $\epsilon\to 0$. By dimensional analysis it diverges logarithmically, but there could be terms like $(\log\epsilon)^n$ also. Are there some standard tricks to deal with such questions?

If you integrate by parts successively in the $x_1,x_2,x_3$, and $x_4$ variables, each time integrating the exponential factor and differentiating the denominator, then the resulting integral becomes absolutely integrable. You get 4 endpoint terms each of which is a similar integral, but in one fewer variable. You can then similarly do integrations by parts on each of the endpoint terms, and so on. It isn't pretty, but if you just want to get the order of magnitude as $\epsilon \rightarrow 0$ this should give it to you.

Make the substitution $y_1 = x_1 + x_3; y_2 = x_1 + x_4;  y_3 = x_2 + x_3; y_4 = x_1 + x_2 + x_3 + x_4$ Then, your integrand becomes

$$\frac{e^{-{y_4}}}{y_1 y_2 y_3 (y_4 - y_2)}.$$

The first three $y$s integrate out (the limit does change from $\infty$ to linear functions of the other $y,$ but you just get a sum of logs, and you are left with a fairly disgusting one-dimensional integral.

EDIT Firstly, the "right substitution is:
$y_1 = x_1 + x_2+x_3 + x_4,  y_2 = x_1 + x_3;  y_3 = x_1 + x_4; y_4 = x_1 + x_2 - x_3 -x_4.$
Secondly, what you see, then, is that the integrand is of the form
$$\frac{e^{-{y_1}}}{y_2 (y_1- y_2 )y_3 (y_1-y_3)}.$$ Which is wonderful, but the fact that it does not depend on $y_4$ (whose range is from $-\infty$ to $\infty$) means that the integral actually diverges, no matter what $\epsilon$ is, so the question seems to be moot.


Why is $\{x\in M: F(x)=G(x), DF_x=DG_x\}$ closed? Let $F$ and $G$ be smooth maps between smooth manifolds $M$ and $N$. Denote their differentials at $x\in M$ by $DF_x$ and $DG_x$. Why is the set $\{x\in M:F(x)=G(x), DF_x=DG_x\}$ closed in $M$?

I know of the result that if $f$ and $g$ are continuous maps between topological spaces $X$ and $Y$ with $Y$ being Hausdorff then $\{x\in X:f(x)=g(x)\}$ is closed in $X$. Unfortunately it doesn't seem like I can directly apply this result since I have $DF_x$ and $DG_x$ which are now maps between tangent spaces.

Take a sequence $x_n\to x$ and note that if
$$
F(x_n)=G(x_n)\quad\text{and}\quad DF_{x_n}=DG_{x_n}
$$
for all $n$, then
$$
F(x)=G(x)\quad\text{and}\quad DF_{x}=DG_{x}.
$$

It suffices to prove that if $f: M \to N$ is smooth, then

$\{ x \in M : f(x) = 0 \text{ and } Df_x = 0 \}$

is closed in $M$ (just consider $f = F - G$ and $N$ embedded in some Euclidean space). Now,

$\{ x \in M : f(x) = 0 \text{ and } Df_x = 0 \} = \{ x \in M : f(x) = 0\} \cap \{ x \in M : Df_x = 0\}.$

Since $f$ is continuous, $f^{-1}(\{ 0\})$ is closed in $M$. Since $f$ is smooth, $Df : TM \to TN$ is also continuous as a map between tangent bundles.
So, the map $E : M \to \text{Hom}(TM, TN)$ given by

$E(x) = Df_x, \quad x \in M,$

is also continuous. Thus, $E^{-1}(\{0\})$ is also closed in $M$. Hence,

$\{ x \in M : f(x) = 0 \text{ and } Df_x = 0 \} = f^{-1}(\{0\}) \cap E^{-1}(\{0\})$

is closed, as required.


Exercise 6.11 from Isaac's Character theory of finite groups. I'm a noob in group theory, so please forgive me if I put some silly questions. I try to solve the following problem (6.11 from Isaac's book, Character theory of finite groups):

Let $A\unlhd G$ with $A$ Abelian. Let $\chi$ be an irreducible monomial character of $G$. Prove that $\chi$ is a relative monomial character with respect to $A$, i.e. there exists a subgroup $H$ with $A\subset H\subset G$ and an irreducible character $\psi$ of $H$ such that $\psi_A$ is irreducible and $\psi^G=\chi$.

Since $\chi$ is monomial, there exists a subgroup $K$ and a linear character $\lambda$ of $K$ such that $\chi=\lambda^G$. The hint given is that: Every constituent of $(\lambda^{AH})_A = (\lambda_{A\cap H})^A$ has multiplicity $1$. I don't know how to prove this hint and how to solve the problem using it. The linear character $\psi_A$ should be a constituent of $(\lambda^{AH})_A$? I don't grasp the connection between $\psi$ and $\lambda$. Ito's theorem (according to which $\chi(1)||G:A|$) plays a role in solving this problem?

Let $A \unlhd G$ and $\chi$ a monomial irreducible (complex) character say $\chi=\lambda^G$, with $\lambda$ linear character of a subgroup $K$ of $G$. Consider the subgroup $KA$. Observe that $\lambda^{KA}$ is irreducible. By using Problem (5.2) in Isaacs' book (all references are in Character Theory of Finite Groups), we have $(\lambda^{KA})_A=(\lambda_{K \cap A})^A$. Since $A$ is abelian, we can find a linear character $\mu$ of $A$, such that $\mu_{K \cap A}=\lambda_{K \cap A}$. Now apply (6.17) Corollary (Gallagher): we must have
$$(\lambda^{KA})_A=(\lambda_{K \cap A})^A=\sum_{\nu \in Irr(A/K \cap A)} \mu \nu$$
Observe that all $\nu$ and hence $\mu\nu$ are linear, since $A$ is abelian. The corollary also says that all the $\mu\nu$ are distinct and are all of the irreducible constituents of $(\lambda_{K \cap A})^A$. Hence $(\lambda^{KA})_A$ is the sum of distinct conjugates of the $\mu$. Now apply Clifford's Theorem ((6.11) Theorem), there must be a linear character $\psi \in I_{KA}(\mu)$, the inertia group of $\mu$, with $[\psi_A,\mu] \neq 0$, such that $\psi^{KA}=\lambda^{KA}$. Hence $\chi=\lambda^G=(\lambda^{KA})^G=(\psi^{KA})^G=\psi^G$ by transitivity of induction (see Problem (5.1)). So $H=I_{KA}(\mu)$ is the requested subgroup.

Induction on $k:=[G:A]$.  So, assume the result is true for groups which possess abelian normal subgroup of index $<k$ (and satisfy the hypothesis etc.).

Let $\chi=\lambda^G$ for $\lambda$ a linear character of $H_1$. Consider the subgroup $AH_1$.

(1) If $AH_1$ is proper in $G$, then, it contains $A$, and since $\lambda^G$ is irreducible, so $\lambda^{AH_1}$ is irreducible; i.e. $\chi_1=\lambda^{AH_1}$ is monomial character of subgroup $AH_1$ which contains abelian normal subgroup of index less than $[G:A]$, and so apply induction.

(2) Suppose $AH_1=G$. What this implies is that the subgroups $A$ and $H_1$ give single double coset in $G$. So apply Mackey decomposition (Exercise 5.6, Isaacs' book): so
$$(\lambda^G)_{A}=(\lambda_{A\cap H})^A. $$
Now $A\cap H$ is subgroup of abelian group $A$, so the last character is direct sum of linear characters of $A$ with multiplicity $1$. (This proves the Hint given by Isaacs.)

So restriction of $\chi=\lambda^G$ to $A$ is sum of some irreducible characters of $A$ with multiplicity $1$; apply Clifford's theorem to reach the conclusion.


If $\mathbb E(X)=\mathbb E\left(X^2\right)=0$ show that $\mathbb P(X=0)=1$ Can anyone help me with this problem?

I want to show that if $\mathbb E(X)=\mathbb E\left(X^2\right)=0$ then $\mathbb P(X=0)=1$.

I was thinking to use Markov's inequality as this:
$\mathbb P\left(X<\dfrac1n\right)\le\dfrac{\mathbb E(X)}{1/n}=0$ so, $\mathbb P(X=0)=1$.

I am not sure.

You don't need to use Markov inequality. If $\mathbb{E}[X^2] = 0$, you must have $$\int_{x}x^2 f(x) dx = 0$$ As $x^2$ is always non-negative, it cannot be the case that $f(x) > 0$ and $x^2 > 0$ simultaneously, so for all $x^2 > 0$ we must have $f(x) = 0$. This, in turn, means $\mathbb{P}[X^2 > 0] = 0 \implies \mathbb{P}[X^2 = 0] = 1 \implies \mathbb{P}[X = 0] = 1$.