问:
Formula to calculate logarithmic curve in the context of IT security

Background

I'm actually a dev, but I think this question fits here as its about maths. I'm implementing a site wide throttle on too many failed requests as protection against distributed brute force attacks.

The question I am stuck with is, after how many failed login requests should I start to throttle?

Now one reasonable way is, as mentioned here "using a running average of your site's bad-login frequency as the basis for an upper limit". If the site has an average of $100$ failed logins, $300$ (puffer added) might be a good threshold.

Now I don't have a running average and I don't want someone having to actively increase the upper limit as the user base grows. I want a dynamic formula that calculates this limit based on the active users amount.

The difficulty is that if there are only a few users, they should have a much higher user to threshold ratio than let's say $100,000$ users. Meaning that for example for $50$ users the limit could be set at $50\%$ of the total user count which means allowing $25$ failed login requests site-wide in a given timespan. But this ratio should decrease for 100k users, the threshold should be more like around $1\%$. $1000$ failed login requests in the same let's say hour, is a lot (probably not accurate at all I am not a security expert, the numbers are only examples to illustrate).

The Question

I was wondering, is there any mathematical formula that could archive this in a neat way?

This is a chart of what I think the formula should be calculating approximately:

Here is what I have now (I know it's terrible, any suggestion will be better I'm sure):

```<code>$threshold = 1;
if ($activeUsers <= 50) {
    // Global limit is the same as the total of each users individual limit
    $threshold *= $activeUsers; // If user limit is 4, global threshold will be 4 * user amount
} elseif ($activeUsers <= 200) {
    // Global requests allows each user to make half of the individual limit simultaneously
    // over the last defined timespan
    $threshold = $threshold * $activeUsers / 2;
} elseif ($activeUsers <= 600) {
    $threshold = $threshold * $activeUsers / 2.5;
} elseif ($activeUsers <= 1000) {
    $threshold = $threshold * $activeUsers / 3.5;
} else { // More than 1000
    $threshold = $threshold * $activeUsers / 5;
}
return $threshold;
</code>```

答:
TL;DR
$$y = 659.113 \log\left(\frac{x}{240.399} + 1\right) + \frac{-1213.555 x}{x + 739.845}$$

Let $f: [0, +\infty) \to [0, +\infty)$ be the function that maps the number of users to threshold, i.e. the green curve

$f$ should satisfy these properties:

$f = \Theta{(\log(x))}$ as $x \to +\infty$ by assumption

$f$ is infinitely differentiable (we take the derivatives at $x = 0$ to be right-handed derivatives at $0$, and say it is differentiable at $0$ if right-handed derivatives exist at $0$)

$f(0) = 0$ from the graph

$f'(0) \approx 1$ since $f$ more or less coincides with the blue line of slope $1$ as $x \to 0^+$

$f'(x) > 0$ on $[0, +\infty)$ since $f$ is strictly increasing

$f''(x) < 0$ on $[0, +\infty)$ since $f$ is strictly decelerating

To fit the green curve, we first make some measurements (blue dots). Then, we use R to perform regression analysis (using the nls_multstart function from the nls.multstart package).

Because of the $\Theta{(\log(n))}$ requirement, the obvious choice is to assume $y = a \log (\frac{x}{b} + 1)$ for some unknown $a, b \in \mathbb{R}$. Unfortunately, this gives us a poor fit.

We need to improve our model. To proceed, let's assume there is some correction term such that $y = a \log(\frac{x}{b} + 1) + \mathcal{O}(1)$. A commonly used $\mathcal{O}(1)$ nonlinear model is $\frac{rx}{x + s}$ for some unknown $r, s \in \mathbb{R}$. So let's add it to our model
$$y = a \log\left(\frac{x}{b} + 1\right) + \frac{rx}{x + s}$$

nls_multstart finds us the following best fit (rounded to 3 decimal places):
$$y = 659.113 \log\left(\frac{x}{240.399} + 1\right) + \frac{-1213.555 x}{x + 739.845}$$

From the above plot, we can see the best fit curve closely approximates the data points.

The above residual plot is also roughly symmetric with respect to the $x$-axis.

The above normal quantile-quantile plot is also roughly a straight line.

Next, we need to show properties $1$ - $6$ holds. Property $1$ obviously holds because $y = \Theta(\log x) + \mathcal{O}(1) = \Theta(\log x)$ as $x \to \infty$. To see property $2$ holds, observe

$$f'(x) = \frac{a}{x + b} + \frac{rs}{(x + s)^2} = \frac{a(x + s)^2 + r s (x + b)}{(x + b)(x + s)^2}$$

Now we can obtain $f^{(k)}$ by keep applying the quotient rule. Because of how quotient rule works and $f'$ being a rational function, $f^{(k)}$ can only be non-differentiable at $-b = -240.399$ or  $-s = -739.845$, which means $f$ is infinitely differentiable on $[0, +\infty)$.

Property $3$ holds by simple computation. Property $4$ holds because $$f'(0) = \frac{659.113}{0 + 240.399} + \frac{-1213.555}{(0 + 739.845)^2} = 1.101... \approx 1$$

Now $f'(x) = 0$ exactly when $$a(x + s)^2 + r s (x + b) = 659.113 x^2 + 77440.315994 x + 1449386311.619988 = 0$$ which has negative discriminant, implying $f'$ has no real roots and hence never crosses the $x$-axis. From above, $f'(0) = 1.101... > 0$ and $f'$ is differentiable hence continous on $[0, \infty)$. By the Intermediate Value Theorem, $f'(x) > 0$ on $[0, +\infty)$ holds (property $5$). Similar arguments show $f''$ has no non-negative real roots, continous, take a negative value at $x = 0$, and hence property $6$ holds again by the IVT.

Here is the R code I use for regression and plots:

```<code>## To the extent possible under law, the author(s) have dedicated all
## copyright and related and neighboring rights to this software to the
## public domain worldwide. This software is distributed without any
## warranty.

## You should have received a copy of the CC0 Public Domain Dedication
## along with this software.
## If not, see <https://creativecommons.org/publicdomain/zero/1.0/>.

library('nls.multstart')
options(digits=8) # show 3 decimal places in summary

## draw the residual plot using the x data points and residuals
residual_plot <- function(x, res)
{
    plot(x, res, ylab='Residuals')
    abline(0, 0)
}

## draw the normal quantile-quantile plot using the residuals
norm_q_q_plot <- function(res)
{
    qqnorm(res)
    qqline(res)
}

## draw the best fit curve using the x, y data points
## and a function f taking x data points to best fit value
best_fit_curve <- function(x, y, f)
{
    plot(x, y)
    lines(x, f(x), col='red')
}

## data points
x <- 0:41 * 25
y <- c(  0,  24,  48,  68,  84, 102, 114,
       128, 142, 152, 164, 174, 182, 194,
       202, 212, 218, 230, 236, 244, 252,
       260, 268, 274, 282, 288, 296, 302,
       308, 316, 322, 328, 336, 342, 348,
       354, 360, 366, 372, 378, 384, 390)

## best fit model of the equation a * log(x / b + 1) + r * x / (x + s)
model <- nls_multstart(y ~ a * log(x / b + 1) + r * x / (x + s),
                       iter=500,
                       start_lower=-rep(max(x), 4),
                       start_upper=rep(max(x), 4),
                       supp_errors='Y')

## compute best fit value from x data points using the best fit model
f <- function(x)
{
    coefs <- round(coef(model), 3)
    coefs['a'] * log(x / coefs['b'] + 1) +
        coefs['r'] * x / (x + coefs['s'])
}

## residuals
res <- y - f(x)
</code>```

Body:
TL;DR
$$y = 659.113 \log\left(\frac{x}{240.399} + 1\right) + \frac{-1213.555 x}{x + 739.845}$$

Let $f: [0, +\infty) \to [0, +\infty)$ be the function that maps the number of users to threshold, i.e. the green curve

$f$ should satisfy these properties:

$f = \Theta{(\log(x))}$ as $x \to +\infty$ by assumption

$f$ is infinitely differentiable (we take the derivatives at $x = 0$ to be right-handed derivatives at $0$, and say it is differentiable at $0$ if right-handed derivatives exist at $0$)

$f(0) = 0$ from the graph

$f'(0) \approx 1$ since $f$ more or less coincides with the blue line of slope $1$ as $x \to 0^+$

$f'(x) > 0$ on $[0, +\infty)$ since $f$ is strictly increasing

$f''(x) < 0$ on $[0, +\infty)$ since $f$ is strictly decelerating

To fit the green curve, we first make some measurements (blue dots). Then, we use R to perform regression analysis (using the nls_multstart function from the nls.multstart package).

Because of the $\Theta{(\log(n))}$ requirement, the obvious choice is to assume $y = a \log (\frac{x}{b} + 1)$ for some unknown $a, b \in \mathbb{R}$. Unfortunately, this gives us a poor fit.

We need to improve our model. To proceed, let's assume there is some correction term such that $y = a \log(\frac{x}{b} + 1) + \mathcal{O}(1)$. A commonly used $\mathcal{O}(1)$ nonlinear model is $\frac{rx}{x + s}$ for some unknown $r, s \in \mathbb{R}$. So let's add it to our model
$$y = a \log\left(\frac{x}{b} + 1\right) + \frac{rx}{x + s}$$

nls_multstart finds us the following best fit (rounded to 3 decimal places):
$$y = 659.113 \log\left(\frac{x}{240.399} + 1\right) + \frac{-1213.555 x}{x + 739.845}$$

From the above plot, we can see the best fit curve closely approximates the data points.

The above residual plot is also roughly symmetric with respect to the $x$-axis.

The above normal quantile-quantile plot is also roughly a straight line.

Next, we need to show properties $1$ - $6$ holds. Property $1$ obviously holds because $y = \Theta(\log x) + \mathcal{O}(1) = \Theta(\log x)$ as $x \to \infty$. To see property $2$ holds, observe

$$f'(x) = \frac{a}{x + b} + \frac{rs}{(x + s)^2} = \frac{a(x + s)^2 + r s (x + b)}{(x + b)(x + s)^2}$$

Now we can obtain $f^{(k)}$ by keep applying the quotient rule. Because of how quotient rule works and $f'$ being a rational function, $f^{(k)}$ can only be non-differentiable at $-b = -240.399$ or  $-s = -739.845$, which means $f$ is infinitely differentiable on $[0, +\infty)$.

Property $3$ holds by simple computation. Property $4$ holds because $$f'(0) = \frac{659.113}{0 + 240.399} + \frac{-1213.555}{(0 + 739.845)^2} = 1.101... \approx 1$$

Now $f'(x) = 0$ exactly when $$a(x + s)^2 + r s (x + b) = 659.113 x^2 + 77440.315994 x + 1449386311.619988 = 0$$ which has negative discriminant, implying $f'$ has no real roots and hence never crosses the $x$-axis. From above, $f'(0) = 1.101... > 0$ and $f'$ is differentiable hence continous on $[0, \infty)$. By the Intermediate Value Theorem, $f'(x) > 0$ on $[0, +\infty)$ holds (property $5$). Similar arguments show $f''$ has no non-negative real roots, continous, take a negative value at $x = 0$, and hence property $6$ holds again by the IVT.

Here is the R code I use for regression and plots:

## To the extent possible under law, the author(s) have dedicated all
## copyright and related and neighboring rights to this software to the
## public domain worldwide. This software is distributed without any
## warranty.

## You should have received a copy of the CC0 Public Domain Dedication
## along with this software.
## If not, see <https://creativecommons.org/publicdomain/zero/1.0/>.

library('nls.multstart')
options(digits=8) # show 3 decimal places in summary

## draw the residual plot using the x data points and residuals
residual_plot <- function(x, res)
{
    plot(x, res, ylab='Residuals')
    abline(0, 0)
}

## draw the normal quantile-quantile plot using the residuals
norm_q_q_plot <- function(res)
{
    qqnorm(res)
    qqline(res)
}

## draw the best fit curve using the x, y data points
## and a function f taking x data points to best fit value
best_fit_curve <- function(x, y, f)
{
    plot(x, y)
    lines(x, f(x), col='red')
}

## data points
x <- 0:41 * 25
y <- c(  0,  24,  48,  68,  84, 102, 114,
       128, 142, 152, 164, 174, 182, 194,
       202, 212, 218, 230, 236, 244, 252,
       260, 268, 274, 282, 288, 296, 302,
       308, 316, 322, 328, 336, 342, 348,
       354, 360, 366, 372, 378, 384, 390)

## best fit model of the equation a * log(x / b + 1) + r * x / (x + s)
model <- nls_multstart(y ~ a * log(x / b + 1) + r * x / (x + s),
                       iter=500,
                       start_lower=-rep(max(x), 4),
                       start_upper=rep(max(x), 4),
                       supp_errors='Y')

## compute best fit value from x data points using the best fit model
f <- function(x)
{
    coefs <- round(coef(model), 3)
    coefs['a'] * log(x / coefs['b'] + 1) +
        coefs['r'] * x / (x + coefs['s'])
}

## residuals
res <- y - f(x)


问:
Pictorial puzzle
Can someone suggest a pictorial representation of a mathematical puzzle or problem that could be represented in coloured tiles on an orangery floor $4489\times7525$mm with an insert of $907 \times1996$ for steps  This has to be assembled so simple but complex is preferred</p>

答:
A Penrose tiling is a must. But there is a copyright on them if I remember well.

Nice to ask !

ContentLicense: CC BY-SA 3.0

Title: Pictorial puzzle


问:
Approximate solutions to a transcendental equation of two variables
Say I have an equation of the form:
$$ y = A + B\sin (x) +C\sin (x+Dy) $$ on the domain $0<x<2\pi$.

I want to get $y$ as a function solely of $x$, i.e. remove the $y$-dependence from the R.H.S. This cannot be done analytically, but are there any useful tricks to analytically obtain approximate solutions? The trouble is that $x$ and $y$ can take on arbitrary values, i.e. I can't take them to be small and make a first-order approximation of $\sin$, for example.

答:
Start expanding the last sine and use the tangent half-angle substitution
$x=2 \tan ^{-1}(t)$ to mka a quadratic equation in $t$
$$(a+c \sin (d y)-y)+2  (b+c \cos (d y))t+ (a-c \sin (d y)-y)t^2=0$$ which will show real solutions if
$$(b^2+c^2-a^2)+2 a y+2 b c \cos (d y)-y^2 >0$$ If this is the case, you have two solutions $t_\pm$ and $x_\pm=2 \tan ^{-1}(t_\pm)$.

There is no way to get $y(x)$.

ContentLicense: CC BY-SA 4.0


问:
Getting generating functions for a weirdly-defined weight function
Let $n\in\mathbb{N}$. For a permutation $\sigma:[n]\rightarrow[n]$, we ue the notation $(\sigma(1)\sigma(2)\cdots\sigma(n))$ to describe the mapping. A pair of integers $(i,j)$ is called an inversion of $\sigma$ if $i<j$ and $\sigma(i)>\sigma(j)$. For example, the permutation $(32415)$ on $[5]$ has 4 inversions: $(1,2),(1,4),(2,4),(3,4)$. Define the weight function $w$ on a permutation $\sigma$ to be the number of inversions in $\sigma$. Let $S_n$ be the set of all permutations of $[n]$.

(a) Determine the generating functions for $S_1$, $S_2$, $S_3$ with respect to the weight function $w$.

(b) Prove that for $n\geq 2$, $\Phi_{S_{n}}(x)=(1+x+\cdots+x^{n-1})\Phi_{S_{n-1}}(x)
 $. You may use the following non-standard notation: if $\sigma$ is a permutation of $[n]$, then $\sigma '$ is the permutation of $[n-1]$ obtained from $\sigma$ by removing the element $n$.

(c) Prove that the number of permutations of $[n]$ with $k$ inversions is

$$[x^{k}]\frac{\prod_{i=1}^{n}(1-x^{i})}{(1-x)^{n}}$$

(a) is doable by brute-forcing. (b) seems to be doable by the product lemma if only the weight function was something more tractable, since I need to prove that the weight is equal to adding the weights of some product. (c) is completely over my head since I can't get any general formula for the number of inversions in a permutation.

答:
If you have (a) and (b), you have (c):

$$\begin{align*}
\frac1{(1-x)^n}\prod_{i=1}^n(1-x^i)&=\prod_{i=1}^n\frac{1-x^i}{1-x}\\\\
&=\prod_{i=1}^n\left(1+x+x^2+\ldots+x^{i-1}\right)\;,
\end{align*}$$

so you need only verify that $\Phi_{S_1}(x)=1$.

One way to prove that

$$\Phi_{S_n}(x)=\prod_{i=1}^n\left(1+x+x^2+\ldots+x^{i-1}\right)$$

is by induction on $n$: let $$c_n(k)=[x^k]\Phi_{S_n}(x)\;,$$ the number of $n$-permutations with $k$ inversions, and $$a_n(k)=[x^k]\prod_{i=1}^n\left(1+x+x^2+\ldots+x^{i-1}\right)\;,$$ and show that

$$c_n(k)=\sum_{i=\max\{0,k-n+1\}}^kc_{n-1}(i)\tag{1}$$

and

$$a_n(k)=\sum_{i=\max\{0,k-n+1\}}^ka_{n-1}(i)\;.\tag{2}$$

$(2)$ is pretty straightforward. For $(1)$, think about starting with a permutation of $[n-1]$ with $i$ inversions and inserting $n$ so that it precedes the last $k-i$ elements of the permutation; how many inversions does that add to the permutation? (The non-standard notation mentioned in (b) appears to be designed to let you talk easily about the inverse of this operation.)


问:
Integral that represents the volume of the solid obtained by rotating an ellipse about a line.
I was given this question to answer:

x2+ y2 = 4 rotates about y = 2

I know how to find volume when I have 2 functions and their the outer and inner radius (distances from the rotation axis). What is the approach to this question? Thank you!

答:
The approach is very similar. For example, if we use shell method,

Shell width at $y$ (distance from x-axis) is $2 \sqrt{4-y^2}$.

Radial distance of the shell from $y = 2$ is $(2 - y)$ where $ - 2 \leq y \leq 2$.

Now, can you you complete the working?


问:
Prove that if a finite group $G$ is soluble and has more than one $17$-Sylow subgroup, then it has more than $100$ $17$-Sylow subgroups.
Prove that if a finite group $G$ is soluble and has more than one $17$-Sylow subgroup, then it has more than $100$ $17$-Sylow subgroups.

My main reason for asking: Do people find this easy or hard?

There is a somewhat obscure theorem that yields a one-sentence proof, but I'm not sure whether there is an easy way without using that theorem.

I'd be interested in an elementary proof, if there is one.

NEW: Actually, I just thought of a not-too-hard proof that doesn't use that theorem, but it does use chief series.  So I guess my question amounts to whether there is any reasonable way of proving this without using chief series.

答:
Let $G$ be a counterexample of smallest order, let $P \in {\rm Syl}_{17}(G)$, and let $N$ be a minimal normal subgroup of $G$.

Then, since $G$ is solvable, $N$ is abelian and, since its Sylow subgroups are normal in $G$, $N$ must be a $p$-group for some prime $p$, and hence $|N| = p^k$ for some $k>0$.

If $G/N$ has more than $100$ Sylow $17$-subgroups then so does $G$ so, since $G$ is a minimal counterexample, $G/N$ has a unique Sylow $17$-subgroup. So $PN/N \unlhd G/N$ and hence $PN \unlhd G$.

Now $P$ cannot be the unique Sylow $17$-subgroup of $PN$ or else it would be normal in $G$, and the number of Sylow $p$-subgroups of $PN$ must divide $p^k$ and be equal to $1 \bmod 17$

The smallest number satisfying those conditions is $103$.


问:
Spherical Harmonics Expansion of Analytic Function derivation
Someone has led me to understand that the following spherical harmonic expansion of an analytic function is completely general, however I am having trouble seeing how one would derive it. As far as I understand the following formula is the general solution to Laplace's equation inside a ball centered at the origin – a linear combination of the spherical harmonics multiplied by the scale factor $r^{l}$. I have no trouble seeing how this is derived from Laplace's equation, but am struggling to see how one obtains this as a general expansion for an analytic function.
$$
f(r,\theta ,\varphi ) = \sum\limits_{\ell  = 0}^\infty  {\sum\limits_{m =  - \ell }^\ell  {f_\ell ^m r^\ell  Y_\ell ^m (\theta ,\varphi )} } .
$$
Might anyone illuminate how one obtains this expansion if one is not to obtain it from Laplace's equation?

答:
I didn't fully understand your question, but I guess the confusion starts with your (someone's ;) first statement: The formula you are referencing is not completely general, this can be seen as follows: The spherical harmonics $Y_l^m$ form a  basis on the sphere, whereas the polynomials $r^k$ form a basis in the 'radial part' of the space of functions $f(\mathbf r)$. Since this function space is the (tensor) product of radial functions 'times' angular functions, the basis in this space is the tensor product of the respective 'partial' basis functions (This holds for any tensor product of two vector spaces). Hence the complete basis functions read $r^k Y_l^m$ and a generic function is expanded as $f(\mathbf r)= \sum_{klm} f_{klm}r^k Y_l^m(\hat r)$.

If now $f(\mathbf r)$ is supposed to solve the Laplace equation, there is $\Delta f = (\Delta_r + \Delta_S/r^2)f = \sum_{klm} f_{klm}\Delta_r r^k Y_l^m(\hat r)+f_{klm}r^{k-2}\Delta_S Y_l^m(\hat r) = 0$
where the Laplace operator (in spherical coordinates) was split into a radial and angular part. Since $\Delta_r r^k = (k(k-1) + 2k)r^{k-2}$ and $\Delta_S Y_l^m = -l(l+1)Y_l^m$ were are left with $\Delta f  = \sum_{klm} f_{klm}((k(k-1) + 2k - l(l+1))r^{k-2}Y_l^m(\hat r)= 0 $. One can easily see that the parenthesis is zero for $k=l$, but not otherwise, hence the (generic) expansion coefficient $f_{klm}$ must vanish for $k\ne l$, such that we only need to take coefficients $f_{lm}$ into account and finally arrive at the formula you have given above. (Only valid for harmonic functions).

Also note that a analytic function $f$ is 'automatically' harmonic only if it is holomorphic, i.e. $f: \mathbf{C} \rightarrow \mathbf{C}$ interpreted as $f: \mathbf{R}^2 \rightarrow \mathbf{C}$, this equivalence does not hold for functions with more than two real variables (like here: $f: \mathbf{R}^3 \rightarrow \mathbf{C}$)

Side note: Outside the ball where the above series converges one uses $r^{-(k+1)}, k=0,1,2,..$ as radial functions, they do not form a basis but have the (physical) property that $f(\mathbf r) \rightarrow 0$ for $|\mathbf r|\rightarrow \infty$. A similar calculation as above shows that $k=l$ for harmonic functions.

The above is written from a physicist's perspective, mathematicians are welcome to comment/correct!


问:
Formal Proof of "No Largest Number"
I need to create a formal proof of there not being the largest number, with only the definition of "less than" being given with Peano arithmetic. $s(x)$ means successor of $x$ or $x + 1$.

$$
\text{Premise}\\
\forall x\forall y (x \lt y \iff \exists z(x + s(z) = y)\\
\text{Prove}\\
\forall x \exists y (x < y)
$$

I've tried proof by contradiction, but the resources online don't seem to translate well into a formal proof. In particular, the strategy of "assume there is a largest number, add one, done" doesn't really translate with formal proof steps.

答:
Solved!

The key to solving this was with "creating the right half" of the original premise: $\exists z (x + s(z) = y)$

Title: Formal Proof of "No Largest Number"


问:
If $f_n\to f$ uniformly and $f$ bounded, then $f_n$ is uniformly bounded proof method question

Let $f_n\to f$ uniformly as $n\to\infty$ and $f$ bounded. Show that $f_n$ is uniformly bounded.

This statement is clearly false. But, to provide a counter example to this statement, I passed by a quite strange way using contraposition (because didn't find a counter example for the original statement :( ).

By contraposition, if $f_n\to f$ uniformly as $n\to\infty$ and $f$ bounded $\implies$ $f_n$ is uniformly bounded, then the following must be true as well: not ($f_n$ is uniformly bounded)$\implies$ not$(f_n\to f$ uniformly as $n\to\infty$ and $f$ bounded) or equivalently :

If $f_n$ is not uniformly bounded, then $f_n\to f$ pointwise as $n\to\infty$ or $f$ unbounded.

My question is the following: If I give a counter example to the contraposition statement, does it show as well that the original statement can't hold as well?

If yes, I have the following counter example to the contraposition statement:

Let $f_n(x)=\frac{x}{n}$. Clearly $f_n(x)$ is not uniformly bounded and $f_n\to 0$ pointwise. But, $f$ is bounded. So the contraposition statement is false and the original statement as well.

答:
The contraposition would be, in this case: if $(f_n)_{n\in\Bbb N}$ is not uniformly bounded, then either $(f_n)_{n\in\Bbb N}$ does not converges uniformly to $f$ or $f$ is unbounded. Your counter-exampl is not a counter-example to this statement.

That example at the end of your post is not an example, since the convergence is not uniform. (I am assuming that this is a sequence of functions with domain $\Bbb R$. If you meant something else, please say so.)

A counter example (again, with domain $\Bbb R$) would be$$f_n(x)=\begin{cases}x&\text{ if }n=1\\0&\text{ otherwise.}\end{cases}$$The sequence $(f_n)_{n\in\Bbb N}$ converges uniformly to the null function (which is bounded), but it is not uniformly bounded.

Note that if a sequence $(f_n)_{n\in\Bbb N}$ of functions converges uniformly to a bounded function, then, for some $N\in\Bbb N$, $(f_n)_{n\geqslant N}$ is uniformly bounded.


答:
One key flaw in your question is that the negation of $\{f_n\}$ doesn’t uniformly converge towards $f$ is not that $\{f_n\}$ converges towards $f$ pointwise. Which seems to be what you state in your question.

Also, it should be good that you state the definition set of all your maps.

For example on $I=[0,1]$, the sequence $f_n(x)=n$ doesn’t converge uniformly to the always vanishing map, nor does it pointwise.

Now if a sequence of maps $\{f_n\}$ converges uniformly to $f$ and that $f$ is supposed to be bounded, then $\{f_n\}$ is eventually uniformly bounded.

Note: also the initial statement of the question is true if each of the $f_n$ is supposed to be bounded.

Title: If $f_n\to f$ uniformly and $f$ bounded, then $f_n$ is uniformly bounded proof method question

Body:
Let $f_n\to f$ uniformly as $n\to\infty$ and $f$ bounded. Show that $f_n$ is uniformly bounded.

This statement is clearly false. But, to provide a counter example to this statement, I passed by a quite strange way using contraposition (because didn't find a counter example for the original statement :( ).

By contraposition, if $f_n\to f$ uniformly as $n\to\infty$ and $f$ bounded $\implies$ $f_n$ is uniformly bounded, then the following must be true as well: not ($f_n$ is uniformly bounded)$\implies$ not$(f_n\to f$ uniformly as $n\to\infty$ and $f$ bounded) or equivalently :

If $f_n$ is not uniformly bounded, then $f_n\to f$ pointwise as $n\to\infty$ or $f$ unbounded.

My question is the following: If I give a counter example to the contraposition statement, does it show as well that the original statement can't hold as well?

If yes, I have the following counter example to the contraposition statement:

Let $f_n(x)=\frac{x}{n}$. Clearly $f_n(x)$ is not uniformly bounded and $f_n\to 0$ pointwise. But, $f$ is bounded. So the contraposition statement is false and the original statement as well.


问:
Prove that the diophantine equation $(xz+1)(yz+1)=az^{3} +1$ has no solutions in positive integers $x, y, z$ with $z>a^{2} +2a$.
Let $a$ be a positive integer that is not a perfect cube. From experimental data, it appears all solutions to $(xz+1)(yz+1)=az^{3} +1$ in positive integers $x, y, z$ occur when $z \le a^{2} +2a$ i.e it appears there are no solutions in $x, y,z$ with $z> a^{2} +2a$. Can this observation be proved?

To motivate the question, we shall prove that on the contrary if $a$ is a perfect cube, there are infinitely many positive integer solutions in $x, y, z$.

Proof.
Let $a=m^{3} $ for some integer $m$. Using the identity $n^{3} +1 =(n+1)(n^{2}-n+1)$, we see that $az^{3} +1=(mz)^{3} +1= (mz+1)((mz)^{2}-mz+1) $.

A family of solutions is then given by $x=m$, $y=m^{2}z - m$  where $z$ takes on any positive integer.

How do I go about proving the striking observation: There are no positive integer solutions  $x, y, z$ with $z>a^{2} +2a$ when the integer $a$ is not a perfect cube? Is there any counterexample?

Body:
Below is $a=47,$  I told it to write small $t$ first. Concentrating on $t=1$ we can see how $x+y=z$

This does not hold at all for $t\geq 2,$  where we see immediately that $x>z$

==================================

 a: 47  t: 1 w 8837   x 95   y 93 z 188
 a: 47  t: 1 w 9166   x 116  y 79 z 195
 a: 47  t: 1 w 9871   x 139  y 71 z 210
 a: 47  t: 1 w 10012  x 143  y 70 z 213
 a: 47  t: 1 w 11657  x 185  y 63 z 248
 a: 47  t: 1 w 13631  x 231  y 59 z 290
 a: 47  t: 1 w 17767  x 323  y 55 z 378
 a: 47  t: 1 w 21997  x 415  y 53 z 468
 a: 47  t: 1 w 30551  x 599  y 51 z 650
 a: 47  t: 1 w 39152  x 783  y 50 z 833
 a: 47  t: 1 w 56401  x 1151 y 49 z 1200
 a: 47  t: 1 w 108242 x 2255 y 48 z 2303

 a: 47  t: 2 w 2305   x 59   y 39 z 49
 a: 47  t: 2 w 26980  x 1124 y 24 z 574

 a: 47  t: 3 w 11894  x 743  y 16 z 253

 a: 47  t: 4 w 568    x 28   y 20 z 12
 a: 47  t: 4 w 803    x 53   y 15 z 17
 a: 47  t: 4 w 1555   x 119  y 13 z 33
 a: 47  t: 4 w 6584   x 548  y 12 z 140

 a: 47  t: 6 w 2732   x 340  y 8 z 58

 a: 47  t: 7 w 994    x 140  y 7 z 21

 a: 47  t: 8 w 1324   x 218  y 6 z 28

 a: 47  t: 10 w 104   x 14   y 6 z 2
 a: 47  t: 10 w 245   x 45   y 5 z 5

 a: 47  t: 12 w 59    x 7    y 5 z 1
 a: 47  t: 12 w 200   x 44   y 4 z 4

 a: 47  t: 14 w 61    x 11   y 3 z 1

 a: 47  t: 15 w 156   x 42   y 3 z 3

 a: 47  t: 17 w 64    x 15   y 2 z 1

 a: 47  t: 22 w 304   x 130  y 2 z 6

 a: 47  t: 23 w 916   x 435  y 2 z 19

 a: 47  t: 24 w 71    x 23   y 1 z 1

 a: 47  t: 45 w 1079  x 989  y 1 z 22

 a: 47  t: 46 w 2161  x 2069 y 1 z 45

==================================

Sat Mar 20 21:24:04 PDT 2021


问:
Maxima of a given set of integers

If $a,b,c$ and $d$ are four different positive integers selected from $1$ to $25$, then what is the highest possible value of the following?$$\frac{a+b+c+d}{a+b+c-d}$$

I assumed $a+b+c =x$ and then :-

$\frac{(a+b+c+d)}{(a+b+c-d)} \\ \Rightarrow \frac{x+d}{x-d} \\ \Rightarrow  \frac{1+d/x}{1-d/x}  \\ \Rightarrow Assuming \ d/x \ to \ be \ 'r' \ hence \frac{1+r}{1-r}$

From here I am kind of lost and couldn't figure out a way to proceed also using derivatives didn't help much.
Can someone please help me on this ?

Update : If I take $d = 25$ in order to maximize the numerator and minimize the denominator  then we'll get  $\frac{x+25}{x-25}$.

Thanks in advance !

答:
$\frac {x+d}{x-d} = \frac {(x-d) + 2d}{x-d} = 1 + \frac {2d}{x-d}$.

Now if $x-d = 0$ the whole thing is undefined.  Otherwise $x-d \ge 1$ or $x-d < 0$ and $0< d\le 25$.

If $x-d < 0$ then as $d > 0$ we have  $1 + \frac {2d}{x-d} < 1$.

If $x-d \ge 1$ then

So $1 +\frac {2d}{x-d} \le 1 + \frac{2d}1 \le 1 + \frac {2\cdot 25}1 = 51$ is the absolute maximum.

Which can be achieved if $d=25$ and ${x-d} = a+b+c-d = a+b+c - 25 = 1$ or if $a+b+c = 26$

(Obviously $a+b+c = 26$ is possible.  We can let $a=1; b= 2$ and $c = 23$ or just about any other combination.  The requirement that $a,b,c,d$ be distinct my put limits on generalizing solutions [It'd be irritating to prove abstractly that $1 \le a \le 24; 1\le b \le 25- a;b\ne a; c= 26-a-b; c\ne b;c\ne a$ has solutions] but this is so general and easy that it isn't a serious burden... for this problem.)

Title: Maxima of a given set of integers


答:
What you've done so far is reasonable.  Derivatives won't help, because the variables are required to be integers.  Since $\lim_{r\to1}\frac{1+r}{1-r}=\infty$, to get a large value, we want to make $r$ close to $1$, or $a+b+c$ close to $d$.  At the same time, we want to make $a+b+c+d$ as large as we can.

Let's try $a+b+c-d=1$, which is as small as we can make the denominator, and still have a positive value.  Then the numerator is $2d+1$ which is largest if we make $d=25$.  We can take $a,b,c$ to be any distinct positive integers such that $a+b+c=26$, say $a=7,\ b=9,\ c=10.$  This gives the largest fraction, $51$.

If we allow the denominator to be a little bigger in the hopes of getting a larger, we quickly see that it doesn't work.  If $a+b+c-d=2$, then $a+b+c+d =2d+2$, and even with $d=25$, we don't get more than $26$.

Title: Maxima of a given set of integers

Body:
If $a,b,c$ and $d$ are four different positive integers selected from $1$ to $25$, then what is the highest possible value of the following?$$\frac{a+b+c+d}{a+b+c-d}$$

I assumed $a+b+c =x$ and then :-

$\frac{(a+b+c+d)}{(a+b+c-d)} \\ \Rightarrow \frac{x+d}{x-d} \\ \Rightarrow  \frac{1+d/x}{1-d/x}  \\ \Rightarrow Assuming \ d/x \ to \ be \ 'r' \ hence \frac{1+r}{1-r}$

From here I am kind of lost and couldn't figure out a way to proceed also using derivatives didn't help much.
Can someone please help me on this ?

Update : If I take $d = 25$ in order to maximize the numerator and minimize the denominator  then we'll get  $\frac{x+25}{x-25}$.

Thanks in advance !


问:
Compute the limit $\lim_{x \to 0}\frac{\sin 2x}{x}=2$
I'm trying to compute the following limit, though I'm trying to cement the understanding of it using the definitions of limit, so would really appreciate your support in understanding this.

$\lim_{x \to 0}\frac{\sin 2x}{x}=2$

I know that $\frac{\sin x}{x}=1$

and $\sin 2x = 2 \cos x \sin x$

Hence:

$|\frac{\sin 2 x}{x}-2|=\frac{|2\cos x \sin x-2x|}{|x|}=2\cdot \frac{\cos x \sin x}{x}-1$

This should then be $2-1 = 1$, which I can compute fine using logic, as $\frac{\sin x }{x} = 1; 2\cos(x) = 2;$ when $x = 0$

Although, I'm nervous in where I should take my understanding from above towards $\epsilon$ and $\delta$ definitions to interpret the limit with clearness. Perhaps, an example and steps would help alleviate this anxiety. Although, whatever the help I'm really appreciative.

答: You can just write$$\lim_{x\to0}\frac{\sin 2x}{x}=2\lim_{x\to0}\frac{\sin 2x}{2x}=2\lim_{y\to0}\frac{\sin y}{y}=2\times1=2.$$Or if you insist on using the double angle formula,$$\lim_{x\to0}\frac{\sin 2x}{x}=\lim_{x\to0}\frac{2\sin x\cos x}{x}=2\lim_{x\to0}\frac{\sin x}{x}\lim_{x\to0}\cos x=2\times1\times1=2.$$


问:
Determine convergence of sequence
Does the following sequnce converge or diverge? $$\sum_{n=1}^\infty\frac{n}{3^n}$$

How are these kinds of problems written down and proven? From my understanding, this will converge to $0$ because the denominator grows at a faster rate than the numerator. Is that all I can do about this problem or is there some more detailed way to solve it?

答:
1)Root test:

https://en.m.wikipedia.org/wiki/Root_test

2)Recall:

$e^n=$

$1+n+n^2/2!+n^3/3!+... > n^3/3!$

$\dfrac{n}{3^n} <\dfrac{n}{e^n} <\dfrac{3!}{n^2};$

$\sum \dfrac {1}{n^2}$ converges (comparison test).

Title: Determine convergence of sequence


问:
Congruence relation of an Algebra
Let $2=\{0,1\}$, $+$ denote addition modulo $2$, $g\left(x\right)=x+1$. Consider the following Algebra $A=\langle 2,+,g\rangle$. I want to calculate congruence relation on $A\times A$. Can someone explain how to do it!

Congruence relation $\theta$ is an equivalence relation on algebra together with property that if $x_i\theta y_i$ for $i=1,2,3..,n$ then $f\left(x_1,x_2,...,x_n\right)\theta f\left(y_1,y_2,...,y_n\right)$ for all $f$ $n$-ary operations on algebra.

答:
Your operation $g$ is irrelevant to compute the congruences of $A \times A$ because it's one of the projections of $+$.

So the congruence lattice of $A \times A$ is the same as the one of the group $\mathbb Z_2 \times \mathbb Z_2$, which is the lattice $M_3$.

Here, having the linked diagram as reference,
\begin{align}
0 &= \{((a,b),(a,b)):a,b\in A\},\\
1 &= \{((a,b),(c,d)): a, b, c,d \in A\},\\
x &= \{((a,b),(a,b')): a,b,b' \in A\},\\
y &= \{((a,b),(a',b)): a,a',b \in A\},\\
z &= \{((a,a),(b,b)): a,b \in A\}.
\end{align}

The $0$ and $1$ congruences hold for all algebras: in the $0$, each element is related with itself and no other; in the $1$, every element is related with every other one.

In all algebras of the type $A \times A$, where $A$ is an algebra, we always get the congruences which are kernels of the projections, that is, we relate elements which have a common coordinate.

In this case, these are $x$ (where the first coordinate is fixed) and $y$ (where the second coordinate is fixed).

The last case, $z$, is still a congruence relation here, but it's not for all algebras $A$.

To see that these are precisely the congruence relations of $A \times A$, given that it is equivalent to a group, it is enough to see that that lattice is isomorphic to the subgroup lattice of $\mathbb Z_2 \times \mathbb Z_2$ (notice that, since $\mathbb Z_2$ is Abelian, so is $\mathbb Z_2 \times \mathbb Z_2$, whence all of its subgroups are normal).

Indeed, $x$ corresponds to the subgroup generated by $(0,1)$; $y$ to the subgroup generated by $(1,0)$ and $z$ to the subgroup generated by $(1,1)$.

Title: Congruence relation of an Algebra


问:
Finding extremum of quadratic function using factorisation as symmetry operation.
In this book in Section 3.2.3, the author shows how to find extremum of a quadratic function using an invariant quality and symmetry operation, which preserves it.

E.g. to find the extremum of $f(x)=6x-x^2$ we can notice that $f(x)=6x-x^2 = x(6-x) = (6-x)x$. This operation maps every value $x$ to $6-x$ through the axis of symmetry and vice-versa (e.g. 0 is mapped to 6 and 6 mapped to 0). This operation preserves symmetry for $x=3$ - the axis of symmetry and hence the extremum of the function.

What about the function $f(x)=6x+x^2 = x(6+x) = (6+x)x$?
We know that the minimum of this function is at $x=-3$ but $6+(-3)=3$.
Also this operation maps e.g $4$ to $10$, but $10$ to $16$! Hence, this approach fails with this example (symmetry is not preserved).

Why is that so? Why is this approach not working for all the quadratic functions?

答:
It will work, but you need to find the correct symmetry transformation.  Let's talk more about the first example, $f(x) = x(6-x)$.  The symmetry transformation here is $x\mapsto 6-x$ because if we replace every $x$ in $x(6-x)$ with $6-x$ we get $$(6-x)(6-(6-x)) = (6-x)(6-6+x) = (6-x)x = x(6-x) = f(x).$$  In other words, $f(6-x) = f(x)$.  We do the transformation, in this case replace $x$ with $6-x$, and arrive at the original function.

Now for your second example, $f(x) = x(6+x)$.  To use the same argument we need to find the right transformation.  In this case, the  transformation is $x\mapsto -x-6$.  We can verify this:
$$f(-x-6) = (-x-6)(6-x-6) = (-x-6)(-x) = (x+6)x = x(6+x) = f(x).$$  Furthermore, the transformation $x\mapsto -x-6$ maps $4$ to $-10$ and $-10$ to $4$, and we can verify symmetry in general as well.  In this case the only value that doesn't change is $x = -3$, because $-(-3) - 6 = -3$.  So, the axis of symmetry, as well as the minimum of the function, is at $x = -3$.


问:
convergence of series conditionally

to solve these problems is it good enough to write out a couple of the terms and see how it behaves or do i have to take the limit of the series to see what happens?

and if i do have to take the limit i can just compare them to other series or integrals that i know, right?

I think i have a general idea of how to solve these problems, just want to see if im right.

答:
Hint:

First series: Yes and no

Use the alternate test to show the convergence of the series and since
$$\frac{1}{\sqrt{5n^2-3n}}\sim_\infty\frac{1}{\sqrt{5}n}$$
hence....

Second series: Yes and yes

$$\frac{|\cos (n)| n^2}{4^n+4^{-n}}\leq\frac{n^2}{4^n}=_\infty o\left(\frac{1}{n^2}\right)$$

Title: convergence of series conditionally


问:
What is the mechanics of the union-intersection along a sequence of sets?
This question came about after I got stuck trying to understand this answer.

When $n\to \infty$ I don't understand why in a countable family of sets $(A_n)_{n\geq 0}$ the operation

$$\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $$

depends on whether the sequence is monotonous or not.

It is likely that I don't understand the symbols, or that I am just too dumb. Let's see... Should I read this as a composition of operations? If so, when I start from the inside, i.e.

$$\bigcup_{k \geq n} A_k$$

I necessarily end up with all the elements in all the sets combined into a single set. At that point this entire set containing all possible elements in any one of the sets $A_k$ is intersect-ioned with all of the sets $A_k,$ looking for elements that are in all sets. Is that it? If this is correct, then why does the order of the monotony of the sequence matter? Can I get an example?

Post-mortem:

'
For a finite collection of sets:

$$
\begin{align}
A_0&=\{a,b\}\\
A_1&=\{a,x\}\\
A_3&=\{a,x,y\}\\
A_4&=\{a,c,d,y\}\\[3ex]
\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k &=\{a,b,x,y,c,d\}\cap\{a,x,y,c,d\}\cap\{a,x,y,c,d\}\\
&=\{a,x,y,c,d\}\\[3ex]
\bigcup_{n\geq 0} \bigcap_{k \geq n} A_k
&=\{a\}\cup\{a\}\cup\{a,y\}=\{a,y\}
\end{align}
$$

In pseudo-code $\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $:

for  $n$ in $n=0$ to $\infty:$

        for  $k$ in $n$ to $\infty:$

             $B_k \leftarrow A_k \cup A_{k+1} \cup A_{k+2} \cdots$

      $B_0 \cap B_1 \cap B_2 \cap \cdots$

The pictorial intuition is that for an element $\color{blue}\spadesuit$ to be in $\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $ it would have to keep on appearing so that no matter how far we go into infinity it appears in some subset to as to be part of all $B_k$'s, and hence, in the intersection:

$$\{\cdots\},\cdots,\{\cdots\},\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\{\cdots\},\{\color{blue}\spadesuit,\cdots\},\cdots,\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\cdots$$

On the other hand, $\bigcup_{n\geq 0} \bigcap_{k \geq n} A_k $ is more demanding, requiring that $\color{blue}\spadesuit$ appears consistently after a certain point:

$$\{\cdots\},\cdots,\{\cdots\},\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\{\cdots,\color{blue}\spadesuit\},\{\color{blue}\spadesuit,\cdots\},\cdots,\{\color{blue}\spadesuit,\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\cdots$$

答:
Read it from left to right, like a double summation.  First, we choose $n$.  Then $\bigcup_{k\geq n}A_k$ is the set $B_n$, say, of all elements that are in one of the $A_k$ with index $\geq n$.
$$B_n = A_n\cup A_{n+1}\cup A_{n+2}\cup\cdots$$
and $\bigcap_{n\geq0}\bigcup_{k\geq n}A_k$ is the set of elements in all of the $B_n$
$$\bigcap_{n\geq0}\bigcup_{k\geq n}A_k = B_0\cap B_1\cap B_2\cap\cdots$$
The intersection is not redundant.  Consider an element $x$ that is in $A_0$ but none of the other $A_n$.  Then $x\in B_0, x\notin B_1,\ x\notin B_2, \dots$ so $x\notin C:=\bigcap_{n\geq0}\bigcup_{k\geq n}A_k$

In fact, C is precisely the set of elements in infinitely many of the $A_n$.  (If $x$ is in only finitely many of the $A_n$ then there is some $n$ such that $n\geq N\implies x\notin A_n$.  Consequently, $x\notin B_N$.)

You might want to convince yourself that if we interchange the union and intersection, then $$\bigcup_{n\geq0}\bigcap_{k\geq n}A_k$$
is the set of all elements is all but finitely many of the $A_n$


问:
Can the infinite jungle gym surface be expressed by an exhaustion of compact surfaces with one boundary component?
Is it possible to write the infinite jungle gym surface as the increasing union of compact surfaces whose boundaries consist of only one simple closed curve? I believe that this is true. This surface has only one end. Therefore the result follows from the existence of canonical exhaustions (made of compact surfaces whose components of the complement have one boundary component) (as explained in Ahlfors-Sario book). If anyone could help me, I would like to "see" such a curve. Thanks.</p>

答:
It'll be quite hard to see in fine detail, but here's a somewhat course visualization.

Let $J$ be the infinite jungle gym surface.

Start with the sphere $S_r$ of arbitrarily large radius $r>0$ centered on the origin $\mathcal O$, which is the boundary of the ball $B_r$. You can perturb $r$ by an arbitrarily small value so that $S_r$ is transverse to $J$, and hence $J \cap S_r$ consists of a union of circles. The complement $J - S_r$ has a unique unbounded component, coinciding with $J - B_r$. To prove this one has to show that for each point $x_0 \in J$ there is a continuous path $x_t$, $t \in [0,\infty) \to J$ which starts at $x_0$, such that the distance $d(\mathcal O,x_t)$ is nondecreasing and diverges to $+\infty$.

The intersection $J \cap S_r$ consists of a zillion circles, which I'll enumerate as $C_i$, $i=1,...,$ (a zillion). What "a zillion" really means here is that the number of these circles grows quadratically as a function of $r$, being asymptotically comparable to the area of $S_r$ which is equal to $4 \pi r^2$.

Now start connecting them up by a network of pairwise disjoint paths in $J - B_r$ (this where is an explicit "visualization" really begins to break down). To be precise, take a properly embedded arc $\gamma_{1} \subset J$ with one endpoint on each of $C_1$ and $C_2$. This arc $\gamma_1$ does not separate $J$, because $C_1 \cup \gamma_1 \cup C_2$ has a regular neighborhood in $J$ whose boundary is a single circle. Thus we can take a properly embedded arc $\gamma_2 \subset J$ with one endpoint on each of $C_2$ and $C_3$ and disjoint from $\gamma_1$. Again the union $\gamma_1 \cup \gamma_2$ does not separate $J$, because $C_1 \cup \gamma_1 \cup C_2 \cup \gamma_2 \cup C_3$ also has a regular neighborhood in $J$ whose boundary is a single circle.

Continuing inductively, we obtain a system of circles and arcs
$$C_1 \cup \gamma_1 \cup C_2 \cup \gamma_2 \cup C_3  \cup \cdots \cup C_{\text{a zillion}}
$$
which has a regular neighborhood in $J$ whose boundary is the single circle that you have asked for.

What you can visualize here is a long string of disjoint little circles in the plane, of equal radius, strung out with their centers on the $x$ axis, and connected up by arcs of the $x$-axis. Also you should visualize a big circle which encloses this whole string of centers and arcs. The union of the little circles is the boundary of $J$, the portion of the plane outside these little circles can be thought of as embedded in $J$ (that's the final regular neighborhood), and the final circle you want is any circle in the plane whose inside contains all of the little circles.


答:
They are answering the specific question "wouldn't the $\bigcup_{k \geq n} A_k$ part just include everything, leaving the intersection part redundant?" The answer is no, only if the $A_k$ are monotone increasing to $\mathbb{R}$.

Your interpretation of $\bigcap_{n \geq 0} \bigcup_{k \geq n} A_k$ is exactly correct. The comments in the other post are just pointing out that $\bigcup_{k \geq n} A_k$ is not necessarily $\mathbb{R}$ (although it is in their example, which is perhaps why the comments are confusing).

Title: What is the mechanics of the union-intersection along a sequence of sets?

Body:
This question came about after I got stuck trying to understand this answer.

When $n\to \infty$ I don't understand why in a countable family of sets $(A_n)_{n\geq 0}$ the operation

$$\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $$

depends on whether the sequence is monotonous or not.

It is likely that I don't understand the symbols, or that I am just too dumb. Let's see... Should I read this as a composition of operations? If so, when I start from the inside, i.e.

$$\bigcup_{k \geq n} A_k$$

I necessarily end up with all the elements in all the sets combined into a single set. At that point this entire set containing all possible elements in any one of the sets $A_k$ is intersect-ioned with all of the sets $A_k,$ looking for elements that are in all sets. Is that it? If this is correct, then why does the order of the monotony of the sequence matter? Can I get an example?

Post-mortem:

'
For a finite collection of sets:

$$
\begin{align}
A_0&=\{a,b\}\\
A_1&=\{a,x\}\\
A_3&=\{a,x,y\}\\
A_4&=\{a,c,d,y\}\\[3ex]
\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k &=\{a,b,x,y,c,d\}\cap\{a,x,y,c,d\}\cap\{a,x,y,c,d\}\\
&=\{a,x,y,c,d\}\\[3ex]
\bigcup_{n\geq 0} \bigcap_{k \geq n} A_k
&=\{a\}\cup\{a\}\cup\{a,y\}=\{a,y\}
\end{align}
$$

In pseudo-code $\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $:

for  $n$ in $n=0$ to $\infty:$

        for  $k$ in $n$ to $\infty:$

             $B_k \leftarrow A_k \cup A_{k+1} \cup A_{k+2} \cdots$

      $B_0 \cap B_1 \cap B_2 \cap \cdots$

The pictorial intuition is that for an element $\color{blue}\spadesuit$ to be in $\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $ it would have to keep on appearing so that no matter how far we go into infinity it appears in some subset to as to be part of all $B_k$'s, and hence, in the intersection:

$$\{\cdots\},\cdots,\{\cdots\},\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\{\cdots\},\{\color{blue}\spadesuit,\cdots\},\cdots,\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\cdots$$

On the other hand, $\bigcup_{n\geq 0} \bigcap_{k \geq n} A_k $ is more demanding, requiring that $\color{blue}\spadesuit$ appears consistently after a certain point:

$$\{\cdots\},\cdots,\{\cdots\},\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\{\cdots,\color{blue}\spadesuit\},\{\color{blue}\spadesuit,\cdots\},\cdots,\{\color{blue}\spadesuit,\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\cdots$$


问:
Proof of the product formula for sine function
I am looking for a simple way to prove $$\frac{\sin \pi z}{\pi z}=\prod_{n=1}^\infty \left(1-\frac{z^2}{n^2}\right)$$ using mainly on the fact that the entire function has simple zeros at $n=\pm 1, \pm 2,\cdots$.</p>

答:
We will use Hardamard Factoriztation theorem to prove it (see https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem#Hadamard_factorization_theorem).

Observe $|\sin(\pi z)|\le e^{\pi |z|}$,  hence it has order of growth less than equal to one. Further it has simple zero at every integer $n \in Z$. Hence by Hardamard Factorization theorem we get
$$ \sin (\pi z)= z e^{az+b}\displaystyle \prod_{0 \ne n \in Z}\left(1-\frac zn\right)e^{\frac z n}, \text { for some } a, b \in  C.$$ Pairing $n$ and $-n$ together in the product  we get
\begin{equation}
 \sin (\pi z)= z e^{az+b}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).
\label{on}
\end{equation}
From above equation we will get
$$\frac {\sin (\pi z)}{\pi z}= \frac{e^{az+b}}{\pi}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$
Taking $z \to 0$, in both side we get $\frac{e^{b}}{\pi}=1$.
From above equation we will get
$$\frac {\sin (\pi z)}{\pi z}= {e^{az}}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$ Again using the fact that
$\frac {\sin (\pi z)}{\pi z}$ is an even function, it can be easily seen that $a=0$.
Hence we get $$\frac {\sin (\pi z)}{\pi z}= \displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$
Thus  $${\sin (\pi z)}= \pi z\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$


问:
My son's Sum of Some is beautiful! But what is the proof or explanation?
My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation:
$$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$

Why is this so? What is the proof or explanation? Is it true for any $n$?

His own presentation of his finding:

Every one of these numbers is two times the number before it.
  $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.

  I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that.

  If I then skip three ones and read the fourth, that one equals my sum!

答:
It works because the number

$$
128
$$
has two special properties: it is a power of $2$, and every digit is a power of $2$.  This means that we can write it in two separate ways:

\begin{align}
128&=2^7\\
128&=100\times2^0+10\times2^1+1\times2^3
\end{align}

Multiplying both sides by $2^n$ then gives:

$$
2^{n+7}=(100\times2^n)+(10\times2^{n+1})+2^{n+3}
$$

Edit: others have done a better job of generalizing this, but I feel I should point out another obvious related sequence.

$128$ is not the only power of $2$ whose digits are all powers of $2$ (or $0$).  For instance:

$$
1024 = 2^{10}
$$

So another, similar kind of relation is given by:

$$
2^{n+10} = (1000\times2^n)+(10\times2^{n+1})+2^{n+2}
$$

In your son's notation, that becomes:

Every one of these numbers is two times the number before it.

$1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.

I pick any one of them, times $1000$.  Then I add the next one, times $10$.  Then I add the next one.

If I then skip seven (!) ones and read the eighth, that one equals my sum!

See if he likes that one.


问:
Does the function $\sqrt{z}=|z|^{1/2} e^{i\frac{Arg(z)}{2}}$ extend continuously on $(-\infty, 0)$?
Generally, $z\mapsto Arg (z)\in (-\pi,\pi]$ is not a continuous function on $(-\infty ,0]$. However, we can extend it continuously on $(-\infty,0)$ by making a correction on the definition of the function on the left open complex plane as
$$f(z)=Arg(z),\ z=x+iy, x<0, y\geq 0 $$ and $$f(z)=Arg(z)+2\pi,\  z=x+iy, x<0, y< 0 .$$
So, I was wodering if the function $$z\mapsto \sqrt{z}=|z|^{1/2} e^{i\frac{Arg(z)}{2}}$$ can extend continuously on $(-\infty,0)$ exactly as  $z\mapsto Arg (z)$. I thought to make a correction only on the left open complex plane as
$$g(z)=\sqrt{z}, \ z=x+iy, x<0, y\geq 0 $$ anfd $$g(z)=-\sqrt{z},\ z=x+iy, x<0, y<0 .$$
Eventually, g is a continuous function on $(-\infty,0)$. What can I say now about the continuity of $g$ on $\mathbb{C}\setminus (-\infty,0)$? Is it possible the previous trick destroyed the continuity of $\sqrt{z}$ over $\mathbb{C}\setminus (-\infty,0)$? Thanks

答: If you define $\sqrt z$ that way, then, since $\operatorname{Arg}(-1)=\pi$, you have $\sqrt{-1}=\exp\left(\frac{\pi i}2\right)=i$. But$$\lim_{\theta\to-\pi}\sqrt{e^{i\theta}}=\lim_{\theta\to-\pi}\exp\left(\frac{i\theta}2\right)=\exp\left(-\frac{\pi i}2\right)=-i.$$Therefore, the answer is negative.


问:
Legendre polynomial expansion
The known magnetic field of a simple loop $(B_z,B_r)$ is expressed as a combination of constant current I and the 2 Elliptic integrals K(k), E(k), k being related to the cylindrical position $(z,r)$ and loop radius. I am not aware of any expression of the magnetic field for a large loop with a current variable upon the loop radius, except for $r=0$.

Therefore I have been integrating the known loop magnetic field across the radius, and got as a result the following sum with no obvious generating function :

$$\sum_{n=0}^\infty {(-1)^n\frac{(2n){!}{!}}{(2n+3){!}{!}}t^nP_n(x)}$$

I tried to combine the function $(1+1/t)^\frac{1}{2}Arctanh((1+1/t)^\frac{-1}{2})-1$ and the generating function of Legendre polynomial 1st kind $P_n(x)$, without success.

When I define the Polynomial $M_n(x)=(-1)^n\frac{(2n){!}{!}}{(2n+3){!}{!}}P_n(x)$

the recursion is $M_{n+1}(x)=-2x\frac{(2n+1)}{(2n+5)}M_n(x)-\frac{4n^2}{(2n+3)(2n+5)}M_{n-1}(x)$

then how to find out its' generating function ?

答: An (elliptic) integral representation easily follows from $$\frac{(2n)!!}{(2n+3)!!}=\frac{2^n\Gamma(n+1)\sqrt\pi}{2^{n+2}\Gamma(n+5/2)}=\frac12\mathrm{B}\left(n+1,\frac32\right)=\frac12\int_0^1 y^n\sqrt{1-y}\,{\rm d}y$$ and the generating function of the Legendre polynomials: $$\sum_{n=0}^\infty(-1)^n\frac{(2n)!!}{(2n+3)!!} t^n P_n(x)=\frac12\int_0^1\sqrt\frac{1-y}{1+2xty+t^2 y^2}\,{\rm d}y.$$

Title: Legendre polynomial expansion


问:
$\sum_{m = 0}^n\binom{n}{2m}(-1)^m = 2^{\frac{n}{2}}\cos\left(\frac{n\pi}{4}\right)$
How can I prove that $S_1(n)=\sum_{m = 0}^n\binom{n}{2m}(-1)^m$ is equal to $2^{\frac{n}{2}}\cos\left(\frac{n\pi}{4}\right)$ using the binomial expansion $(1+z)^n=\sum_{r=0}^n\binom{n}{r}z^r$?

答:
Hint

As $-1=i^2,$ let $z=i$

Use $$(1+z)^n+(1-z)^n=?$$

and $1+i=\sqrt2e^{i\pi/4}$ using Intuitive explanation of Euler's formula $e^{it}=\cos(t)+i\sin(t)$


问:
Tripos Type Integral
This is an old calculus question:

Integrate
$$\int x^{-2n-2}(1-x)^n(1-cx)^ndx$$ it comes from Edwards Treatise on Integral Calculus Vol I, pg. 65 and is designated as [Oxford I.P. 1917].

I guess my question is what do they have in mind as a solution to this ? Are they expecting a closed form ?

My solution is to expand the $1-cx$ factor to get $(-1)^{k}\binom{n}{k} \int\frac{x^k(1-x)^n}{x^{2n+2}}dx$ as the coefficient of $c^k$. This in turn can be integrated by the binomial theorem as
$\sum_{l=0}^n(-1)^{k+l}\binom{n}{k}\binom{n}{l}\frac{1}{2n-l-k+1}x^{-2n+k+l-1}$.
This almost seems too simple minded. But maybe it was what they intended. Any opinions ?

答:
Mathematica says:

$$\frac{(c-1) (1-x)^{n+1} x^{-2 n} (1-c x)^{n-1} \left(\frac{(c-1) x}{c x-1}\right)^{2 n}
   \, _2F_1\left(n+1,2 (n+1);n+2;\frac{x-1}{c x-1}\right)}{n+1}$$

I am not sure if this is the sort of thing they would want.

Title: Tripos Type Integral


问:
On Kunen's Set Theory, Exercise I.16.7
Exercise I.16.7 of Kunen's Set Theory, is as follows:

($\mathsf{ZFC}^-$) Let $\gamma > \omega_1$ be a limit ordinal. Prove that there is a countable transitive $M$ and ordinals $\alpha,\beta \in M$ such that $M \equiv R(\gamma)$ and $(\alpha \approx \beta)^M$ is false but $(\alpha \approx \beta)^{R(\gamma)}$ is true.

Hint. By the Downward Lowenheim-Skolem-Tarski Theorem, get a countable $A$ with $\omega,\omega_1 \in A \preccurlyeq R(\gamma)$. Then, let $M$ be the Mostowski collapse of $A$; let $\alpha = \mathrm{mos}(\omega) = \omega$, and $\beta = \mathrm{mos}(\omega_1)$. Then $\beta$ will be a countable ordinal that $M$ "thinks" is uncountable.

My attempt so far is as follows:

Let $A$ be the model as described in the hint with the additional requirement that $\omega \subseteq A$, and let $M$ be its Mostoski collapse. Define $\alpha = \mathrm{mos}(\omega) = \omega$ (last equality holds because $\omega \subseteq A$), and $\beta = \mathrm{mos}(\omega_1)$. Both are clearly transitive and well-ordered by $\in$, and hence are ordinals. Since $\beta \subseteq M$ is countable, we have that $(\alpha \approx \beta)^{R(\gamma)}$ is true, as they are both countably infinite ordinals. It remains to show that $(\alpha \approx \beta)^M$ is false.

This is where I get stuck. My first thought is to prove by contradiction, by assuming there exists a bijection between $\beta$ and $\omega$ in $M$, and show that $(\alpha \approx \beta)^{R(\gamma)}$ is true, but I'm not sure how to proceed.

Any help is appreciated (I would also appreciate any comments on my attempt thus far).

答: Since there's no bijection between $\omega$ and $\omega_1$ (and since being a bijection is an absolute property), $R(\gamma)$ satisfies $\omega\not\approx\omega_1$. So the elementary submodel $A$ satisfies the same thing. Finally, since the Mostowski collapse mos is an isomorphism from $A$ onto $M$, it follows that $M$ satisfies mos$(\omega)\not\approx$ mos$(\omega_1)$.

Title: On Kunen's Set Theory, Exercise I.16.7


问:
Proving $\frac{\sin x}{x} =\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right) \left(1-\frac{x^2}{3^2\pi^2}\right)\cdots$
How to prove the following product?
$$\frac{\sin(x)}{x}=
\left(1+\frac{x}{\pi}\right)
\left(1-\frac{x}{\pi}\right)
\left(1+\frac{x}{2\pi}\right)
\left(1-\frac{x}{2\pi}\right)
\left(1+\frac{x}{3\pi}\right)
\left(1-\frac{x}{3\pi}\right)\cdots$$</p>

Body:
We will use Hardamard Factoriztation theorem to prove it (see https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem#Hadamard_factorization_theorem).

Observe $|\sin(\pi z)|\le e^{\pi |z|}$,  hence it has order of growth less than equal to one. Further it has simple zero at every integer $n \in Z$. Hence by Hardamard Factorization theorem we get
$$ \sin (\pi z)= z e^{az+b}\displaystyle \prod_{0 \ne n \in Z}\left(1-\frac zn\right)e^{\frac z n}, \text { for some } a, b \in  C.$$ Pairing $n$ and $-n$ together in the product  we get
\begin{equation}
 \sin (\pi z)= z e^{az+b}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).
\label{on}
\end{equation}
From above equation we will get
$$\frac {\sin (\pi z)}{\pi z}= \frac{e^{az+b}}{\pi}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$
Taking $z \to 0$, in both side we get $\frac{e^{b}}{\pi}=1$.
From above equation we will get
$$\frac {\sin (\pi z)}{\pi z}= {e^{az}}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$ Again using the fact that
$\frac {\sin (\pi z)}{\pi z}$ is an even function, it can be easily seen that $a=0$.
Hence we get $$\frac {\sin (\pi z)}{\pi z}= \displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$
Thus  $${\sin (\pi z)}= \pi z\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$


问:
$L^2$-norm of sum smaller than $L^2$-norm of either summand
Suppose $f$ and $g$ are real-valued functions in $L^2(\mathbb{R}^n)$, and $\Vert f \Vert_2 = \Vert g \Vert_2$. If $\Vert f + g \Vert_2 \leqslant \Vert f \Vert_2$, does it necessarily follow that for almost all $x \in \mathbb{R}^n$, either $g(x)=-f(x)$ or $g(x)=f(x)$?

This looks like a question where the answer should be known, but I haven't been able to find it anywhere. If $f$ and $g$ are such functions, the assumptions imply that they satisfy a strong version of the triangle inequality, namely

\begin{align}
\Vert f + g \Vert_2 \leqslant \tfrac{1}{2} (\Vert f \Vert_2 + \Vert g \Vert_2).
\end{align}

Body:
Is false. Consider

$$f(x,y)=\begin{cases} \sqrt{x} & \text{for }(x,y) \in [0,1] \times [0,1], \\ 0 & \text{for } (x,y) \in \mathbb{R}^{2} \setminus [0,1] \times [0,1],\end{cases}$$

and

$$g(x,y)=\begin{cases} -\sqrt{y} & \text{for }(x,y) \in [0,1] \times [0,1], \\ 0 & \text{for } (x,y) \in \mathbb{R}^{2} \setminus [0,1] \times [0,1].\end{cases}$$

Notice that

$$||f||^{2}=\int_{0}^{1}\int_{0}^{1}xdxdy=\frac{1}{2} \quad \text{and} \quad ||g||^{2}=\int_{0}^{1}\int_{0}^{1}ydxdy=\frac{1}{2},$$

that is, $||f||=||g||=\frac{1}{\sqrt{2}}.$ Also,

$$||f+g||^{2}=\int_{0}^{1} \int_{0}^{1}(\sqrt{x}-\sqrt{y})^{2}dxdy=\int_{0}^{1} \int_{0}^{1}(x+y-2\sqrt{xy})dxdy=\frac{1}{9},$$

so that $$||f+g||=\frac{1}{3}<\frac{1}{\sqrt{2}}=||f||.$$

Therefore, these functions satisfies all the conditions but $f \neq g$.


问:
Formula for number of surjective functions in case where codomain is greater than domain
We were deriving the formula for number of surjective functions from f: $X \to Y$, where $|X| = n$ and $|Y| = m$. The formula that we have is:

$$\sum_{k=0}^m(-1)^k (m-k)^n {m \choose k}$$

Now, if $m>n$ we have no surjective functions, as codomain is greater than domain and the answer should be zero. The formula gives the right answer, but I cannot figure out why does it work algebraically.

Any hints on why does this sum equals to zero when $m > n$?

Body:
We have

$$\begin{align*}
\sum_{k=0}^m (-1)^k (m-k)^n {m\choose k}
&= n! [z^n] \sum_{k=0}^m (-1)^k \exp((m-k)z) {m\choose k}\\
&= n! [z^n] \exp(mz) \sum_{k=0}^m (-1)^k \exp(-kz){m\choose k}\\
&= n! [z^n] \exp(mz) (1-\exp(-z))^m\\
&= n! [z^n] (\exp(z)-1)^m.
\end{align*}$$

Now $\exp(z)-1 = z + \cdots$ so $(\exp(z)-1)^m = z^m + \cdots$ and hence $n! [z^n] (\exp(z)-1)^m = 0$ when $n\lt m.$


问:
Function that is closed but not open and not continuous
So, I am trying to find a function with requirements of being closed but not open and not continuous which is defined on the real numbers and the usual topology.

I was able to find many piecewise functions that meets the requirements but it feels very cheesy and not mathematical. Does the function x -> $\sqrt x$ meet the requirements.

Why I think it does:
It is not continuous since it is not defined in the negative numbers.
It is not open because (-b,b)-> [0, $\sqrt b$)
It is closed because function of any form including the real numbers is mapped either to [0, $\sqrt b$] or [$\sqrt a$ ,$\sqrt b$]

am I correct? If not can someone give an example of a function that isn't piecewise?

答:
No, you are not correct. That is a continuous function from $[0,\infty)$ into $\Bbb R$. The fact that's undefined on the negative numbers does not make it discontinuous.

You can take, say,$$f(x)=\lim_{n\to\infty}\left(\frac1{1+x^2}\right)^n=\begin{cases}1&\text{ if }n=0\\0&\text{ otherwise.}\end{cases}$$


问:
Given a group of order $pq$ with $p < q$ primes, the index of the commutator subgroup is a multiple of $p$.

Let $G$ be a finite group with order $pq$, where $p < q$ are primes. Let $$[G, G] = \langle xyx^{-1}y^{-1} \mid x, y \in G \rangle$$ denote the commutator subgroup.

Then, the index of $[G, G]$ in $G$ is a multiple of $p$.

I have a solution to the above using representation theory. (Written below.)

Is the above a "standard"/"basic" fact of group theory that one shows using class equations/Sylow theorems? I have not seen it before and a search on this site didn't yield any results. Is there an "elementary" proof of the above?

Usually, with class equations and Sylow theorems, one focuses on the centre of a group or an element. The commutator subgroup does not usually show up, which is why I am curious.

Proof. Let $d_1, \ldots, d_s$ denote the degrees of the irreducible representations of $G$. Then, $$pq = d_1^2 + \cdots + d_s^2$$ and $d_i \mid pq$ for all $i$. The two facts show that for each $i$, $d_i = 1$ or $p$.

Letting $m$ ($n$, resp.) denote the number of degree $1$ ($p$, resp.) representations gives $$pq = m + np^2$$ and thus, $p \mid m$.

However, $m$ is precisely the order of $G/[G, G]$ or the index of $[G, G]$ in $G$. $\Box$

答:
This follows very easily from knowing that the $q$-subgroup is normal (the existence of a $q$-subgroup follows from Cauchy; normality is a trivial consequence of the Sylow theorems, but can be proven by counting arguments on the action of the $p$-subgroup).

If $Q\leq G$ has $|Q|=q$, then $G/Q$ is of order $p$, hence abelian. This means that $G’=[G,G]\leq Q$. Hence
$$[G:G’] = [G:Q][Q:G’] = p[Q:G’],$$
and thus a multiple of $p$. (Of course, it is either $p$ if $G$ is nonabelian, or $pq$ if $G$ is abelian).


答:
$4$ doesn't get mapped to $10$.  $4$ gets mapped to $-10$.  And $-10$ gets mapped to $4$.  It works.

The axis of symmetry is $x=-3$.  So if you have a point $s = -3 + M$ it will be mapped to $-3 -M$.

So $4 = -3+M$ so $M = 7$.  So $4 \mapsto -3 - 7 = -10$.   And $-10 = -3 + M'$ so $M' = -7$ so $-10 \mapsto -3 -(-7) = 4$.  Everything is good.

More generally if $x = -3 + M$ then $M = x+3$ so so $x \mapsto -3 -(x+3) = -6-x$.

And so $4 \mapsto -6 - 4 = -10$.  And $-10\mapsto -6 -(-10) = 4$.

It's all good, man.

Title: Finding extremum of quadratic function using factorisation as symmetry operation.

Body:
In this book in Section 3.2.3, the author shows how to find extremum of a quadratic function using an invariant quality and symmetry operation, which preserves it.

E.g. to find the extremum of $f(x)=6x-x^2$ we can notice that $f(x)=6x-x^2 = x(6-x) = (6-x)x$. This operation maps every value $x$ to $6-x$ through the axis of symmetry and vice-versa (e.g. 0 is mapped to 6 and 6 mapped to 0). This operation preserves symmetry for $x=3$ - the axis of symmetry and hence the extremum of the function.

What about the function $f(x)=6x+x^2 = x(6+x) = (6+x)x$?
We know that the minimum of this function is at $x=-3$ but $6+(-3)=3$.
Also this operation maps e.g $4$ to $10$, but $10$ to $16$! Hence, this approach fails with this example (symmetry is not preserved).

Why is that so? Why is this approach not working for all the quadratic functions?


问:
Inequality of sum
I am trying to solve a inequality consisting of the sums.

QUESTION:

$\text{Suppose that } a_1, \dots, a_n\ \in \mathbb{R}. \text{Show that:}$

$$\left(\frac{1}{n}\sum _{k=1}^na_k^2\right)^{\frac{1}{2}}\le \left(\frac{1}{n}\sum _{k=1}^na_k^4\right)^{\frac{1}{4}}$$

WHAT I HAVE DONE SO FAR:

I substitute $a_k^2=b_k$ and raise the entire inequality to the power of two giving me:

$$\left(\sum _{k=1}^nb_k\right)^2\le n\sum _{k=1}^nb_k^2$$

And since the left side of the inequality consists of the terms $b_k^2$ and the terms $2b_ib_j$, we can subtract n of the square terms from both sides of the inequality:

$$\sum _{i<j}^{ }2b_ib_j\le nb_1^2+nb_2^2+nb_3^2+...+nb_n^2-b_1^2-b_2^2-b_3^2-...-b_n^2$$
$$\sum _{i<j}^{ }2b_ib_j\le nb_1^2-b_1^2+nb_2^2-b_2^2+nb_3^2-b_3^2+...+nb_n^2-b_n^2$$
$$\sum _{i<j}^{ }2b_ib_j\le \left(n-1\right)b_1^2+\left(n-1\right)b_2^2+\left(n-1\right)b_3^2+...+\left(n-1\right)b_n^2$$
$$\sum _{i<j}^{ }2b_ib_j\le \left(n-1\right)\sum _{k=1}^nb_k^2$$

But after this I seem to get stuck. According the answer-sheet, I am supposed to get here from the previous expression:

$$0\le \sum_{i<j}(b_i-b_j)^2$$

which would mean the original inequality is true, because these are equivalent. Can someone please explain to me why that is. Any help would be very much appreciated!

答:
After some more thinking, I managed to figure it out:

SOLUTION:

I substitute $a_k^2=b_k$ and raise the entire inequality to the power of two giving me:

$$\left(\sum _{k=1}^nb_k\right)^2\le n\sum _{k=1}^nb_k^2$$

And since the left side of the inequality consists of the terms $b_k^2$ and the terms $2b_ib_j$, we can subtract n of the square terms from both sides of the inequality:

$$\sum _{i<j}^{ }2b_ib_j\le nb_1^2+nb_2^2+nb_3^2+...+nb_n^2-b_1^2-b_2^2-b_3^2-...-b_n^2$$
$$\sum _{i<j}^{ }2b_ib_j\le nb_1^2-b_1^2+nb_2^2-b_2^2+nb_3^2-b_3^2+...+nb_n^2-b_n^2$$
$$\sum _{i<j}^{ }2b_ib_j\le \left(n-1\right)b_1^2+\left(n-1\right)b_2^2+\left(n-1\right)b_3^2+...+\left(n-1\right)b_n^2$$
$$\sum _{i<j}^{ }2b_ib_j\le \left(n-1\right)\sum _{k=1}^nb_k^2$$

We rearrange the inequality:

$$0\le \left(n-1\right)\sum _{k=1}^nb_k^2-\sum _{i<j}^{ }2b_ib_j$$

Which when expanded becomes:

$$0\le \left(n-1\right)\left(b_1^2+b_2^2+...b_n^2\right)-\left(2b_1b_2+2b_1b_3+....+2b_{n-1}b_n\right)$$
$$0\le \left(n-1\right)b_1^2+\left(n-1\right)b_2^2+...\left(n-1\right)b_n-\left(2b_1b_2+2b_1b_3+....+2b_{n-1}b_n\right)$$
$$0\le \left(n-1\right)b_1^2+\left(n-1\right)b_2^2+...+\left(n-1\right)b_n-2b_1b_2-2b_1b_3-....-2b_{n-1}b_n$$

To make a simple conclusion from this, I might just decide to see what happens when $n=3$.  Then the inequality would become:

$$0\le 2b_1^2+2b_2^2+2b_3^2-2b_1b_2-2b_1b_3-2b_2b_3$$

Which is the same as:

$$0\le \left(b_1^2-2b_1b_2+b_2^2\right)+\left(b_1^2-2b_1b_3+b_3^2\right)+\left(b_2^2-2b_2b_3+b_3^2\right)$$
$$0\le \left(b_1-b_2\right)^2+\left(b_1-b_3\right)^2+\left(b_2-b_3\right)^2$$

And from here I would make the conclusion that:

$$0\le\sum_{i<j}^{ }\left(b_i-b_j\right)^2$$

And since this inequality is equivalent with the original inequality, we would then know that the original inequality is true.

Title: Inequality of sum


问:
Solve $(\cos^2{x}+y \sin{2x})\frac{{\rm d}y}{{\rm d}x}+y^2=0$
Solve
$$(\cos^2{x}+y \sin{2x}) \frac{{\rm d}y}{{\rm d}x}+y^2=0$$
Please help me to solve this. It is not possible for me to separate.

Attempt : If $M=\cos^2{x}+y \sin{2x}, N=y^2$ then $M_y=\sin{2x}, N_x=0$ are not equal.

答:
$$(\cos^2{x}+y \sin{2x}) \frac{{\rm d}y}{{\rm d}x}+y^2=0$$
Divide by $\cos^2 x$ (The DE becomes exact):
$$(1+2y \tan x) y'+\dfrac {y^2}{\cos^2 x}=0$$
$$y'+\color {red}{2yy' \tan x +\dfrac {y^2}{\cos^2 x}}=0$$
$$y'+\color{red}{(y^2\tan x)'}=0$$
Integrate.
$$y+y^2\tan x=C$$


答: We can assume that $G$ is not abelian. By Cauchy $G$ has an element of order $q$, say $x$, hence the subgroup $N=\langle x \rangle$ has order $q$, and $|G:N|=p$ is the smallest prime dividing $|G|$, so $N$ is normal. It follows that $G/N$ is cyclic whence abelian, so $G' \subseteq N$, and $G$ is not abelian. So $G’=N$ and you are done.

Title: Given a group of order $pq$ with $p < q$ primes, the index of the commutator subgroup is a multiple of $p$.

Body:
Let $G$ be a finite group with order $pq$, where $p < q$ are primes. Let $$[G, G] = \langle xyx^{-1}y^{-1} \mid x, y \in G \rangle$$ denote the commutator subgroup.

Then, the index of $[G, G]$ in $G$ is a multiple of $p$.

I have a solution to the above using representation theory. (Written below.)

Is the above a "standard"/"basic" fact of group theory that one shows using class equations/Sylow theorems? I have not seen it before and a search on this site didn't yield any results. Is there an "elementary" proof of the above?

Usually, with class equations and Sylow theorems, one focuses on the centre of a group or an element. The commutator subgroup does not usually show up, which is why I am curious.

Proof. Let $d_1, \ldots, d_s$ denote the degrees of the irreducible representations of $G$. Then, $$pq = d_1^2 + \cdots + d_s^2$$ and $d_i \mid pq$ for all $i$. The two facts show that for each $i$, $d_i = 1$ or $p$.

Letting $m$ ($n$, resp.) denote the number of degree $1$ ($p$, resp.) representations gives $$pq = m + np^2$$ and thus, $p \mid m$.

However, $m$ is precisely the order of $G/[G, G]$ or the index of $[G, G]$ in $G$. $\Box$


问:
A (not necessarily continuous) function on a compact metric space attaining its maximum.
I am studying for an exam and my study partners and I are having a dispute about my reasoning for $f$ being continuous by way of open and closed pullbacks (see below).  Please help me correct my thinking.  Here is the problem and my proposed solution:

Let $(K, d)$ be a compact metric space, and let $f: K \rightarrow \mathbb{R}$ be a function satisfying that for each $\alpha \in \mathbb{R}$ the set {$x \in K: f(x) \ge \alpha$} is a closed subset of $K$.  Show that $f$ attains a maximum value on $K$.

Proof: Notice that $A :=$ {$x \in K: f(x) \ge \alpha$} is precisely $f^{-1}[\alpha, \infty)$.  Since $[\alpha, \infty)$ is closed in $\mathbb{R}$ and $A$ is assumed to be closed in $K$, then it follows that $f$ is continuous on $A$.  On the other hand, $K-A = f^{-1}(-\infty, \alpha)$ is open in $K$ since $A$ is closed in $K$.  And since $(\alpha, \infty)$ is open in $\mathbb{R}$ and $K - A$ is open in $K$, then if follow that $f$ is continuous on $K - A$, hence $f$ is continuous on $K$.  Since $K$ is compact and $f$ is continuous, then $f(K)$ is compact in $\mathbb{R}$.  Compact sets in $\mathbb{R}$ are closed and bounded intervals.  Thus $\sup{f(K)} = \max{f(K)} = f(x_0)$ for some $x_0 \in K$.  Thus $f$ indeed attains its maximum value on $K$.  $\blacksquare$

答:
Consider the function $f: [0,1] \to \mathbb R$
\[f(x) = \begin{cases}
 1 &\text{if }x \le 1/2 \\
 0 &\text{otherwise}
\end{cases}\]

$f$ is defined from a compact metric space, and is certainly not continuous (e.g. because the preimage of $(0,2)$ is not open, or because the preimage of $\{0\}$ is not closed). However, the set $\{ x \in [0,1] : f(x) \ge \alpha \}$ is:

empty if $\alpha > 1$,

$[0,1/2]$ if $0 < \alpha \le 1$,

$[0,1]$ if $\alpha \le 0$.

In all three cases it is closed. So it is possible for $f$ not to be continuous, so your argument can't possibly work.

In order to prove that $f$ is continuous on $A$, you want to prove that for any open (or closed) subset of $A$, the corresponding preimage $f^{-1}(A)$  is open (or closed). You've only proven that for $A$ itself, but in general $A$ may have many open/closed subsets that you will need to also check.

As for actually answering your question: let's say $C_\alpha = \{ x : f(x) \ge \alpha \}$. Every $C_\alpha$ is closed. Now in a compact space, if a collection of closed sets is such that every finite subcollection has non-empty intersection, then the whole collection has non-empty intersection (this is equivalent to "every open cover has a finite subcover", by taking complements and the contrapositive). Apply this to the collection of non-empty $C_\alpha$.


答:
Your function is not defined on the real numbers: it is defined only on the set of non-negative real numbers. Thus, it does not meet the requirements of the problem. Moreover, it is continuous on its domain.

There is absolutely nothing wrong with a functions that is defined piecewise. Perhaps the simplest example that meets the requirements of the problem is the function $\chi_{\{0\}}$, the indicator (or characteristic) function of the set $\{0\}$: it is defined on all of $\Bbb R$; it is not continuous at $0$; it takes every subset of $\Bbb R$ to one of the four subsets $\varnothing,\{0\},\{1\}$, and $\{0,1\}$, all of which are closed; and it takes the open set $\Bbb R$ to the set $\{0,1\}$, which is not open in $\Bbb R$. For that matter, $\chi_A$ works for any non-empty proper subset $A$ of $\Bbb R$. So do the floor and ceiling functions, $f(x)=\lfloor x\rfloor$ and $f(x)=\lceil x\rceil$. All of these are well-known, useful functions

Title: Function that is closed but not open and not continuous

Body:
So, I am trying to find a function with requirements of being closed but not open and not continuous which is defined on the real numbers and the usual topology.

I was able to find many piecewise functions that meets the requirements but it feels very cheesy and not mathematical. Does the function x -> $\sqrt x$ meet the requirements.

Why I think it does:
It is not continuous since it is not defined in the negative numbers.
It is not open because (-b,b)-> [0, $\sqrt b$)
It is closed because function of any form including the real numbers is mapped either to [0, $\sqrt b$] or [$\sqrt a$ ,$\sqrt b$]

am I correct? If not can someone give an example of a function that isn't piecewise?


答:
There are two proofs I can think of. The first uses exponential generating functions. Note that
$$
\frac{(m-k)^n}{n!}=[x^n]e^{(m-k)x}
$$
Here, $[x^n]$ is the "coefficient-of" operator, so $[x^n]f(x)$ is the coefficent of $x^n$ in the power series $f(x)$. Therefore,

$$
\begin{align}
\sum_{k=0}^m(-1)^k(m-k)^n\binom{m}k
&=n!\sum_{k=0}^m \binom{m}k(-1)^k[x^{n}]e^{(m-k)x}
\\&=n![x^{n}]\sum_{k=0}^m \binom{m}k(-1)^k(e^{x})^{m-k}
\\&=n![x^n](e^x-1)^m
\end{align}
$$
Finally, note $e^x-1=x+x^2/2+\dots$, so that $(e^x-1)^m=x^m+$ higher order terms, meaning the coefficient of $[x^n]$ in $(e^x-1)^m$ is zero when $n<m$.

The second is to note that $\sum_{k=0}^m(-1)^k(m-k)^n\binom{m}k$ is the zeroth term of the $m^{th}$ forward difference of the sequence $a_i=i^n$. Furthermore, for a nonzero polynomial of degree $n$, the forward difference has degree at most $n-1$. For example,
\begin{aligned}
\Delta(i^n)&=(i+1)^n-i^n &&\xrightarrow{i=0}1^n-0^n
\\
\Delta(\Delta(i^n))&=(i+2)^n-2(i+1)^n+i^n &&\xrightarrow{i=0}2^n-2\cdot 1^n+0^n
\\
\Delta(\Delta(\Delta(i^n)))&=(i+3)^n-3(i+2)^n+3(i+1)^n-i^n &&\xrightarrow{i=0}3^n-3\cdot 2^n+3\cdot 1^n-0^n
\end{aligned}
Note that on the right, we exactly have the summations for $m=1$, $m=2$, and $m=3$ (respectively), and in the middle column, there are polynomials with degree at most $n-1,n-2$ and $n-3$ (respectively).

Title: Formula for number of surjective functions in case where codomain is greater than domain

Body:
We were deriving the formula for number of surjective functions from f: $X \to Y$, where $|X| = n$ and $|Y| = m$. The formula that we have is:

$$\sum_{k=0}^m(-1)^k (m-k)^n {m \choose k}$$

Now, if $m>n$ we have no surjective functions, as codomain is greater than domain and the answer should be zero. The formula gives the right answer, but I cannot figure out why does it work algebraically.

Any hints on why does this sum equals to zero when $m > n$?


答:
The characteristic function of the rational numbers:

Is closed as the image of any subset of the reals is in $\{\emptyset, \{0\}, \{1\}, \{0,1\}\}$ which are all closed subsets.

The image of $\mathbb R$ is $\{0,1\}$ is not open.

It is clearly not continuous.

Hence the characteristic function of the rationals fulfills all of your requirements.


答:
More generally, the sum is $0$ if you replace $(m-k)^n$ with any polynomial in $k$ with degree $<m$:

Summation of series involving binomial coefficients and polynomial of degree at most n-1


问:
Put functions in ascending order by their growth rate
I need to put the given functions in an ascending order by their growth rate. So that every function is $O(other- function)$ or to show they are equal in growth rate.

The functions are as follows:

$f_1(n) = n^2$

$f_2(n) = 4^\sqrt{n}$

$f_3(n) = \frac{n^3}{\log_{2}n}$

$f_4(n) = 100n^2$

$f_5(n) = n\log_{2}^2n$

As I understand, I need to find the limits of ratios of these functions which are equal to 0.
However, I am not sure which functions I should be comparing and I don't really know how to find these limits.

Any help would be very appreciated!

Thank you in advance!

答:
Hint:

Using that $1<\log n<\sqrt n<n$, where $<$ is understood as "grows slower than", we have

$$n<n\log_2^2n<n^2<\frac{n^3}{\log_2n}<n^3$$


问:
$L^q([\mathbb{R}])\nsubseteq L^p([\mathbb{R}])$ and $L^p([\mathbb{R}])\nsubseteq L^q([\mathbb{R}])$
Let $1\leq p<q\leq \infty$ I have to how that

$L^q([\mathbb{R}])\nsubseteq L^p([\mathbb{R}])$ and $L^p([\mathbb{R}])\nsubseteq L^q([\mathbb{R}]) $

I was trying to come up with counterexamples for the inclusion but unfortunately, I am stuck! I would appreciate any kind of help!

答:
If you consider the function $f(x)= x^{-2p/(p+q)}, x \in (0,1]$ where $1<p<q<\infty $ then you can show that $f \in L^p $ but $f \not\in L^q.$
There is a nuiance here though that you are looking at Lebesgue integrable functions on $\mathbb{R}$ but we only did it for the interval $(0,1]$. You can restrict your functions to $[\epsilon, 1]$ for small $\epsilon >0 $ and set the function to zero everywhere else. You can then apply the monotone convergence theorem and show the limit function is $L^p (\mathbb{R}) $ but not $L^q(\mathbb{R})$.
To show the opposite inclusion doesn't hold, do the same but using the functions restricted to $[1,n]$ instead and zero everywhere else.


问:
A question about vectorial spaces of infinite dimension.
Let $V$ be a $k$-vectorial space of finite dimension $n$, every subspace $U \subseteq V$, can be seen as a direct summand of $V$, it is sufficient choose a basis of $U$ and by completing to a basis of $V$, the elements that don't belong to the basis of $U$, generate a subspace $U'$ such that $U \oplus U'=V$. My question is: is this also true when $V$ is an infinite dimensional vectorial space?

答: The existence of a basis for any infinite-dimensional vector spaces is equivalent to Zorn's lemma and the Axiom of Choice. So "yes", if you accept AC, and "it depends", if you don't.</p>


答:
Using integration by parts, it can be shown that the integrals $I_n$ have the following property: $$I_n= (1-x)^n(1-cx)^n(\frac{x^{-2n-1}}{-2n-1}) + \frac{nc+n}{-2n-1}\int x^{-2n-1}(1-x)^{n-1}(1-cx)^{n-1}dx - \frac{2nc}{2n-1}I_{n-1} $$
That leftover integral does not exactly match the form of an $I_k$, but it is close. To me, it's concievable that $J_n = \int x^{-2n-1}(1-x)^{n-1}(1-cx)^{n-1}dx$ has a decomposition which will look similar to the above but with one of the $I_n$ appearing in the middle and $J_{n-1}$ at the end.

Maybe it is possible to solve the two recurrances, or perhaps the two relationships themselves would suffice for an answer.

Title: Tripos Type Integral

Body:
This is an old calculus question:

Integrate
$$\int x^{-2n-2}(1-x)^n(1-cx)^ndx$$ it comes from Edwards Treatise on Integral Calculus Vol I, pg. 65 and is designated as [Oxford I.P. 1917].

I guess my question is what do they have in mind as a solution to this ? Are they expecting a closed form ?

My solution is to expand the $1-cx$ factor to get $(-1)^{k}\binom{n}{k} \int\frac{x^k(1-x)^n}{x^{2n+2}}dx$ as the coefficient of $c^k$. This in turn can be integrated by the binomial theorem as
$\sum_{l=0}^n(-1)^{k+l}\binom{n}{k}\binom{n}{l}\frac{1}{2n-l-k+1}x^{-2n+k+l-1}$.
This almost seems too simple minded. But maybe it was what they intended. Any opinions ?


问:
Relations on the power set of $X=\{a,b,c\}$.
$$R=\{(A,B) | \ \ |A|=|B|\} \ \ \ \text{and} \ \ \ S=\{(A,B)| \ \ |A|<|B|\}.$$

Are $S$ and $R$ equivalence relations ?

Can someone help me in writing a proof for this? Like I have the idea that $R$ is an equivalence relation and S is not through the general relation and set knowledge but how to give a formal proof of it?

答:
To prove that something is an equivalence relation, you will need to verify the corresponding properties from the definition. A,B, and C are any elements of the given set.

Reflexivity: A ~ A

Transitivity: A ~ B and B ~ C implies A ~ C

Symmetry: If A ~ B then B ~ A

I will show you how to prove the second property:
If A~B and B~C, then |A| = |B| and |B| = |C| by the definition of your first relation. Therefore we have |A| = |C|. By the definition of the relation the last equality means that A ~ C, which is what we needed to prove.

To prove that something is an order relation you will have to verify slightly different properties. All of them can however be verified in a similar way as shown above.


问:
Rationals of the form $\frac{p}{q}$ where $p,q$ are primes in $[a,b]$
Consider the closed interval $[0,1]$, there is $\frac{2}{3} \in [0,1]$ where $p=2$ and $q=3$. Similarly consider $[2,3]$, one can have $\frac{5}{2} \in [2,3]$ where $p=5$ and $q=2$. Does every interval of the form $[a,b]$, where $a,b \in \mathbb{R}$ contain a rational of this kind. If yes, how can we prove it?</p>

答:
Roughly speaking this asks whether the quotients of two primes
are dense in the positive reals. The answer is yes.

Let $0 < a 1.$$
For all large enough $q$, $\pi(bq)/\pi(aq) > 1$ as required.

OwnerDisplayName: anonymous

LastEditorDisplayName: anonymous

ContentLicense: CC BY-SA 2.5

Body:
Roughly speaking this asks whether the quotients of two primes
are dense in the positive reals. The answer is yes.

Let $0 < a < b$ and let $q$ be a prime.
Then there will a a prime $p$ with $a < p/q\le b$ if and only if
$\pi(bq) > \pi(aq)$ where $\pi$ is the prime-counting function.
But by the prime number theorem, as $q\to\infty$,
$$\frac{\pi(bq)}{\pi(aq)}\sim\frac{b\log(aq)}{a\log(bq)}
=\frac{b(\log q+\log a)}{a(\log q+\log b)}\sim\frac ba>1.$$
For all large enough $q$, $\pi(bq)/\pi(aq) > 1$ as required.


问:
Closed form for a generating function of the form $f(t) = t^k$ or more general: $f$ some polynomial
I'm looking for a closed form for
$G(x) = \sum\limits_{t=0}^{\infty} f(t) x^t$ for f(t) of the form $f(t) = t^a$
or more general $f(t) = \sum\limits_{i=0}^{d} a_i t^i$.

I know there are forms for $f(t) = t$ and $f(t) = t^2$ and $f(t) = t^3$.

bg,
Johannes

答:
See e.g. Wilf's "generatingfunctionology". For the generating function for the sequence $< p(k) a_k >$, given the generating function $A(z) = \sum_{n \ge 0} a_n z^n$, you get:
$$
\sum_{n \ge 0} p(n) a_n z^n
  = p(z D) A(z)
$$
where $D$ is the derivative. I.e., to get the generating function for $< n^2 >$,
start with:
$$
G(z) = \frac{1}{1 - z} = \sum_{n \ge 0} z^n
$$
and then:
$$
z \frac{d}{d z} \left( z \frac{d}{d z} \frac{1}{1 - z} \right)
 = \frac{z + z^2}{(1 - z)^3}
$$</p>


问:
What will be the value of this summation, given the recurrence relation?
Let $a_1,a_2,a_3,…,a_n$ be a sequence of real numbers which satisfies the relation $a_{n+1}=\sqrt{a^2_n+1}$∀n∈N. Suppose that there exists a positive integer $n_0$ such that $a_{2{n_0}}=3a_{n_0}$. Then, find the value of $\lfloor \sum_{n=1}^{49} \sqrt{\frac{8}{8a^2_n +7}} \rfloor $ where the $\lfloor . \rfloor$ denotes the greatest integer function.

My attempt:
For $n_0 =1$, we get the value of $a^2_1 = 1/8$.

Also using the relation, $a_n$ simplifies to $a^2_n = a^2_1 + n -1
=\frac {8n-7}{8}$.

Now substituting this value into the required sum, it reduces to $\lfloor \sum_{n=1}^{49} {\frac{1}{\sqrt n}} \rfloor $.
Using Python I have verified that this is 12, which is the correct answer, however the question requires to do it manually, and I am unable to do so.

Any hints on how to approach solving further, or any other methods of going about the problem are appreciated.

答:
Hints:

Consider $b_n=a_n^2$ and rewrite the question in terms of $b_n$

You need to find which values of $n_0$ work and give a credible value of $a_1$.  It look to me as if $n_0=1$ does.

Added:

You have since found $a_n=\frac{n}{8}-\frac{7}{8}$.  This makes
$\lfloor \sum_{n=1}^{49} \sqrt{\frac{8}{8a^2_n +7}} \rfloor = \sqrt{1}+\sqrt{\frac12}+\sqrt{\frac13}+ \cdots+\sqrt{\frac{1}{48}}$ rounded down  which is not easy by hand, but  $\int\limits_{1}^{49} \frac 1 {\sqrt{x}}\,dx = 12$ and $\int\limits_{0.25}^{49} \frac 1 {\sqrt{x}}\,dx = 13$ so the answer is going to be $12$ when rounded down, using $\int\limits_{n}^{n+1} \frac 1 {\sqrt{x}}\,dx < \frac1{\sqrt{n}} < \int\limits_{n-0.75}^{n+0.25} \frac 1 {\sqrt{x}}\,dx$


问:
Stirling Numbers of second kind recurrence
While learning the stirling numbers of second kind, we see a recursive relation :
$$S(n, r) = S(n-1, r-1) + k\cdot S(n-1, r)$$

And we get its solution as :
$$S(n, r) = (1/{r!})[r^n - C(r,1)\cdot (r-1)^n + \cdots + (-1)^{r-1}\cdot C(r,r-1)\cdot 1^n]$$

Can you please tell me how the above solution of the recurrence written is obtained?
Thanks in advance.

答:
A set partition into $k$ parts is unordered partition of a set $S$ in unordered collection pairwise disjoint sets $S_1,S_2,...,S_k$. Number of set partitions of ${1,2,...,n}$ into $k$ parts is $S(n,k)$. Also a set partition of ${1,2,...,n}$ is equivalent to $ \cdot \times Seq\{1\} \times \cdot \times seq\{1,2\} \times \cdot \times \ldots \times seq\{1,2,\ldots,k\}$ where $\cdot$ and all numbers $j$ have size $1$. To see this if a number $l$ is at position $j$ then you put $j$ in $S_l$ and if $i^{th}$ dot is at position $j$ then you put $j$ in $S_i$. So
$$ \sum_{n\geq 0 } S{(n,r)}x^n = \dfrac{x}{1-x}\times\dfrac{x}{1-2x}\times\ldots\times\dfrac{x}{1-rx}$$
Extracting coefficient of $x^n$ you get your formula.


问:
Finding equivalent polynomials (mod n)
During some casual investigation of polynomials over an integer ring $\mathbb{Z}_n$ (or $\mathbb{Z}/n\mathbb{Z}$ if you prefer), I noticed that some polynomials induce the same map. I'm curious about how one could tell if two polynomials are equivalent in this way without checking directly.

For example, in $\mathbb{Z}_8$, $f(x) = 2x^3 + 5x + 3$ is the same as $g(x)=x^4 + 3x^2 + 3x + 3$. They are bijective and act on $\mathbb{Z}_8$ as the permutation $(3, 2, 5, 0, 7, 6, 1, 4)$.

My main question is: What are the criteria for two polynomials to induce the same map on $\mathbb{Z}_n$?

I'm also interested in other information about this, such as: For a given polynomial, are there infinitely many equivalent polynomials? Will the lowest-degree polynomial in an equivalence class always have a degree less than $n$? Does it matter what kind of number $n$ is (e.g. prime or composite)? Are there unique polynomials with $deg\geq1$ having no equivalent?

Without knowing much about this situation, my guess is that the Chinese Remainder Theorem, Euler's Theorem, and/or Fermat's Little Theorem will come into play. I'm exploring a bit outside of my mathematical comfort zone and I have very little experience with number theory, so this is where I get kind of lost.

答:
As proved in another answer, two polynomials $f$ and $g$ induce the same function on $\mathbb{Z}/n\mathbb{Z}$ precisely when, written in the form
$$f(x) = \sum_{k} b_k x^{\underline k}$$
and
$$g(x) = \sum_{k} c_k x^{\underline k},$$
they satisfy the property that
$$b_k \equiv c_k \mod {\frac{n}{\gcd(n, k!)}}$$ for each $k$.

Here, $x^{\underline k}$ is the "falling factorial": $x^{\underline k} = x(x-1)\cdots(x-k+1)$. Any polynomial can be written in this form (i.e. using the "falling factorial basis" instead of the monomial basis) using the identity $x^n = \sum_k {n \brace k} x^{\underline k}$ where ${n \brace k}$ is a Stirling subset number ("Stirling number of the second kind"). See Wikipedia on Stirling numbers, for example.

Title: Finding equivalent polynomials (mod n)


问:
What is the mechanics of the union-intersection along a sequence of sets?
This question came about after I got stuck trying to understand this answer.

When $n\to \infty$ I don't understand why in a countable family of sets $(A_n)_{n\geq 0}$ the operation

$$\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $$

depends on whether the sequence is monotonous or not.

It is likely that I don't understand the symbols, or that I am just too dumb. Let's see... Should I read this as a composition of operations? If so, when I start from the inside, i.e.

$$\bigcup_{k \geq n} A_k$$

I necessarily end up with all the elements in all the sets combined into a single set. At that point this entire set containing all possible elements in any one of the sets $A_k$ is intersect-ioned with all of the sets $A_k,$ looking for elements that are in all sets. Is that it? If this is correct, then why does the order of the monotony of the sequence matter? Can I get an example?

Post-mortem:

'
For a finite collection of sets:

$$
\begin{align}
A_0&=\{a,b\}\\
A_1&=\{a,x\}\\
A_3&=\{a,x,y\}\\
A_4&=\{a,c,d,y\}\\[3ex]
\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k &=\{a,b,x,y,c,d\}\cap\{a,x,y,c,d\}\cap\{a,x,y,c,d\}\\
&=\{a,x,y,c,d\}\\[3ex]
\bigcup_{n\geq 0} \bigcap_{k \geq n} A_k
&=\{a\}\cup\{a\}\cup\{a,y\}=\{a,y\}
\end{align}
$$

In pseudo-code $\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $:

for  $n$ in $n=0$ to $\infty:$

        for  $k$ in $n$ to $\infty:$

             $B_k \leftarrow A_k \cup A_{k+1} \cup A_{k+2} \cdots$

      $B_0 \cap B_1 \cap B_2 \cap \cdots$

The pictorial intuition is that for an element $\color{blue}\spadesuit$ to be in $\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $ it would have to keep on appearing so that no matter how far we go into infinity it appears in some subset to as to be part of all $B_k$'s, and hence, in the intersection:

$$\{\cdots\},\cdots,\{\cdots\},\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\{\cdots\},\{\color{blue}\spadesuit,\cdots\},\cdots,\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\cdots$$

On the other hand, $\bigcup_{n\geq 0} \bigcap_{k \geq n} A_k $ is more demanding, requiring that $\color{blue}\spadesuit$ appears consistently after a certain point:

$$\{\cdots\},\cdots,\{\cdots\},\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\{\cdots,\color{blue}\spadesuit\},\{\color{blue}\spadesuit,\cdots\},\cdots,\{\color{blue}\spadesuit,\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\cdots$$

答:
Let me suggest a way of thinking about

$$\bigcap_{n\ge 0}\bigcup_{k\ge n} A_k$$

and

$$\bigcup_{n\ge 0}\bigcap_{k\ge n} A_k$$

that may be helpful.

How does something get into either of these sets?

$x\in\bigcap\limits_{n\ge 0}\bigcup\limits_{k\ge n} A_k$ if and only if $x\in\bigcup\limits_{k\ge n}A_k$ for each $n\ge 0$, which is the case if and only if for each $n\ge 0$ there is a $k\ge n$ such that $x\in A_k$. And this is the same as saying that $x$ is in infinitely many of the sets $A_k$: no matter how far out in the sequence of sets you go, there is one further out that contains $x$.

$x\in\bigcup\limits_{n\ge 0}\bigcap\limits_{k\ge n}A_k$ if and only if there is an $n\ge 0$ such that $x\in\bigcap\limits_{k\ge n}A_k$, which is the case if and only if there is an $n\ge 0$ such that $x\in A_k$ for every $k\ge n$. This just says that $x$ is in every $A_k$ from some point on: it is in all but finitely many of the sets $A_k$.

To sum up:

$x\in\bigcap\limits_{n\ge 0}\bigcup\limits_{k\ge n} A_k$ if and only if $x$ belongs to infinitely many of the sets $A_k$, and

$x\in\bigcup\limits_{n\ge 0}\bigcap\limits_{k\ge n} A_k$ if and only if $x$ is in all but finitely many of the sets $A_k$ or, equivalently, $x$ is in every $A_k$ from some point on.


问:
Equality of the transpose of a vector-matrix product
This is a rather trivial question, but in my lecture notes for two vectors $x,y\in \mathbb{R}$ and a symmetric matrix $C$ I have given
$$
x^TCy=y^TC^Tx
$$
and I do not understand why this is true. I get that
$$
(x^TCy)^T=y^TC^Tx
$$
but why is $(x^TCy)^T=x^TCy=y^TC^Tx$

答:
It is because $C$ is a symmetric matrix (i.e., $C^T = C$)

\begin{align*}
(x^T C y)^T = (y^T C^T (x^T)^T)^T = (y^T C x)^T = x^T C^T (y^T)^T = x^T C y.
\end{align*}


问:
Search for a counterexample for a Pell’s equation conjecture
Let the constant $d$ in Pell’s equation $x^2-dy^2=1$ be a prime such that $d=n(n+1)(n+2)+1$.
Now compare the - lowest - solution for $d$ and $d-1$. My conjecture is that the solution for $d$ is much higher than for $d-1$, i.e. for $n(n+1)(n+2 )$. This came up in my attempts to understand the special case of $d=61$ being $3\times 4 \times 5+1$ with its high solution.

Someone told me that in some cases the reverse would be true, but he didn’t offer an example. He did however make some calculations showing that my statement is true for $d < 1 500 000$. While this is a large number it still corresponds to a relatively modest number of cases, only 38 (up to $n=115$).

Body:
Interesting conjecture! If my calculation is correct, it goes unchallenged for quite some time before failing at $d=34965457$ (corresponding to $n=326$). The smallest $x$ for $(d-1)$ is

25095664791553218669437437017106409143490137922195
56605243569408784102967806403079182710704502809602
76272541590853007432744322926326067853070023625443
07998378733697133994169449962329077423177517193117
73182806455

while the smallest one for $d$ is "just"

27069794535634534410598514149614146407632935632204
746423979023754891934010659911327241899199134849

The conjecture also seems to fail for $d=45118012441$ ($n=3559$) and $d=63760295881$ ($n=3994$) and I believe these are the only counter-example for $n\leq 5000$ (which produce $829$ cases of prime $d$).


问:
What is the real analysis version of this complex analysis Weierstrass theorem?
I'm reading Gong Sheng's Concise Complex Analysis, where it introduced a Weierstrass Theorem

Theorem 3.1 (Weierstrass Theorem) Suppose $\{f_n(z)\}$ is a sequence
of functions where each $f_n(z)$ is defined and holomorphic in a
region  $U\subseteq \mathbb C$. Assume that $\sum_{n=1}^\infty f_n(z)$
converges uniformly to $f(z)$ on every  compact subset of $U$.
Then $f(z)$ is holomorphic on $U$ and for every $k\in \mathbb N$,
$\sum_{n=1}^\infty f_n^{(k)}(z)$ converges uniformly to $f^{(k)}(z)$
on every compact subset of $U$.

Then it mentions:

This is a profound result. The reader can compare it with the theorem
of  the derivative of function series in calculus.

So what is the corresponding real analysis version of this complex analysis Weierstrass Theorem, and what is the difference? -- I suppose the difference would show some distinct properties in complex analysis.

答:
We find in section 10.2.3 Uniformly Convergent Series of Concise Calculus by Sheng Gong the following theorem:

Given a function series $\sum_{n=1}^\infty u_n(x)$ on $[a,b]$. If every term of a function series is differentiable, every derivative function is continuous, and the series of derivative functions is uniformly convergent, then the series is termwise differentiable.

That is, if a functional series of $\sum_{n=1}^\infty u_n(x)$ is convergent, the derivative function $u_n^{\prime}(x)$ of each term is continuous and the function series $\sum_{n=1}^\infty u_n^{\prime}(x)$ is uniformly convergent, then
\begin{align*}
\frac{d}{dx}\left(\sum_{n=1}^\infty u_n(x)\right)=\sum_{n=1}^\infty \frac{d}{dx}u_n(x)=\sum_{n=1}^\infty u_n^{\prime}(x).
\end{align*}
In other words, the order of the derivative $\frac{d}{dx}$ and the sum $\sum$ can be exchanged.

Conclusion: Here we see how strong the concept of a holomorphic function is. In the Weierstrass theorem we consider functions $f_n$ being holomorphic on a compactum. This implies also the existence of derivatives of the functions $f_n$ of arbitrary order.

Armed with that it is sufficient to require uniform convergence of the holomorphic functions $f_n$ on a compactum and it follows not only that $f(z)=\sum_{n=1}^\infty f_n(z)$ is holomorphic on this compactum but also the uniform convergence of all $k$-th derivatives $f^{(k)}$ of $f$ on it.

On the other hand, in the real case we have to require that all the functions $u_n(x)$ are $C^1$, i.e. differentiable and the derivatives are continuous. Contrary to the complex case we have to require that the derivatives are uniformly convergent on $[a,b]$  and we cannot conclude anything for higher derivatives of the functions $u_n(x)$, since we even don't know if they exist.