答:
$$e^0=1\text{ vs. }0^y=0$$ precludes any solution of the form $$e^x=x^y.$$

(And as shown by @nala, no polynomial can do. We can also add that no fractional polynomial can do. Furthermore, it is know that $e^x=1+x+\frac{x^2}2+\frac{x^2}{3!}+\cdots$ with an infinity of terms.)

The equation

$$e^x=x^y$$ is simply solved by

$$y=\frac{\ln x}x$$ but this is of no use.

Title: Is it possible to rewrite $exp(x)$ with $x$ as the base

Body:
Simple question, rewriting $e^x$ with $x$ as the base- or more generally, trying to remove $x$ from the exponent.
I don't know if it's possible, as I played around with it for a little while and got nothing. If it is, I suspect it has something to do with the Lambert W function. If it's not possible, is there an intuitive reason as to why not or is it just something that can't be done algebraically?

I'm using it to simplify a string of terms into a shorter string of terms with $x$ as the base for each one. It's just convenient if the answer could be given in this polynomial-like form.

Thanks, and sorry if I was a bit vague or missed details.

Edit: It really is what's on the label; $e^x = x^?$- it's meant to be general.

If you're solving for $?$ in the situation above,

$e^x=x^y$

$y=\log_x(e^x)$

$y=x*log_x(e)$

$y=\frac{1}{\log_{e^x}(x)}$

And from there, I don't know how to go about reducing the equation to a single $x$, let alone removing the log and putting it in the form of $x^y$. Any ideas?


问:
Show $\Gamma(z)y^{-z}$ vanishes as $z\to\pm i\infty$
I need to show that $\Gamma(z)y^{-z}$ vanishes as $z\rightarrow\pm i\infty$ if $|\arg(y)|< \frac{\pi}{2}$. Note that $z$ is purely imaginary.

Currently my approach is to use Stirling's Formula to get a representation for this product for large $|z|$. This gives $\sqrt{2\pi}z^{z}e^{-z}y^{-z}$.

I am struggling to make an observation which would cause this this product to vanish for large imaginary $z$. I tried rewriting this as $$\sqrt{2\pi}e^{z \log(z)}e^{-z}e^{-z\log(y)} = \sqrt{2\pi}e^{z\log(z)- z - z\log(y)} = \sqrt{2\pi}e^{z(\log(z) - \log(y) -1)}$$

Am I on the right track here?

答: We have $\Gamma(z)y^{-z}=\Gamma(ix)e^{-ix\operatorname{Log}y}$ and $\operatorname{Log}y=\log|y|+i\arg y$ so $$|\Gamma(z)y^{-z}|=|\Gamma(ix)||e^{x\arg y}|<\sqrt{\frac{\pi e^{\pi x}}{x\sinh\pi x}}\to0$$ as $|x|\to\infty$ using Euler's reflection formula.

ContentLicense: CC BY-SA 4.0


问:
$p = x^2 + xy + y^2$ if and only if $p \equiv 1 \text{ mod }3$?
For a prime number $p \neq 3$, do we have that$$p = x^2 + xy + y^2$$for some $x$, $y \in \mathbb{Z}$ if and only if$$p \equiv 1 \text{ mod }3?$$I suspect this is true from looking at the example$$7 = 2^2  + 2 \times 1 + 1^2.$$Here is a thought I have so far that might be helpful towards solving this: I know that $$\mathbb{Z}[(1 + \sqrt{-3})/2]$$is a PID. But I am not sure what do from here. Could anybody help?</p>

答:
Here is a proof which is long but, I hope, elementary. Let $f(x, y)=x^2+xy+y^2$ and $g(x, y)=x^2+3y^2$. The question is phrased in terms of $f$ but it's easier to work with $g$, so I first prove that the sets of numbers $f$ and $g$ represent are identical.

$f(a, b)=f(b, a)$. Moreover, if $a+b+c=0$, then at least one of $a, b$ and $c$ is even, and $f(a, b)=f(a, c)$ by simple algebra. Thus by using these substitutions as and when needed, if $n$ has the form $f$, then $n$ has the form $f$ with second parameter even. That is, for some $x$ and $y$, $$n=f(x, 2y)=x^2+2xy+4y^2=g(x+y, y).$$ This last chain of equalities also proves that the implication goes the other way.

A number of the form $f$ or $g$ is termed Loeschian. Further, define a primitive first-form Loeschian number as a number of the form $f(a, b)=a^2+ab+b^2$ where $a$ and $b$ are coprime integers,
and a primitive second-form Loeschian number as a number of the form $g(x, y)=x^2+3y^2$ where $x$ and $y$ are coprime integers.

Lemma 1 Every primitive first-form Loeschian number is a primitive second-form Loeschian number.

Proof
Suppose not, and let $n$ be a primitive first-form Loeschian number which is not a primitive second-form Loeschian number.
$n$ is a second-form Loeschian number $n=g(x, y)$, so there is some $u>1$ where $u\mid x$ and $u\mid y$.
$n=g(x, y)=f(a, b)$ where $a=x-y$ and $b=2y$, so $u\mid a$ and $u\mid b$, contradicting the supposition that $n$ is a primitive first-form Loeschian number.

Therefore the result is true.

I now prove a result about the factorisations of primitive Loeschian numbers. My proof adapts a proof of a similar result regarding numbers of the form $x^2+y^2$. I found that, in adapting this proof, I needed the form $g(x, y)=x^2+3y^2$; if the proof could be adapted so as to use the form $f(a, b)=a^2+ab+b^2$, it could be simplified.

Lemma 2
The product of two Loeschian numbers is Loeschian with at least two representations in each form.

Proof
Using representations of the first form:
\begin{align*}
f(a, b)f(c, d)
&= (a^2+ab+b^2)(c^2+cd+d^2)\\
&= a^2c^2+a^2cd+a^2d^2+abc^2+abcd+abd^2+b^2c^2+b^2cd+b^2d^2\\
&=[a^2c^2+2abc^2+b^2c^2+2abcd+2b^2cd+b^2d^2]+{}\\
&\qquad +[a^2cd-abc^2+abcd-b^2c^2+abd^2-b^2cd]+{}\\
&\qquad +[a^2d^2-2abcd+b^2c^2]\\
&=(ac+bc+bd)^2+(ac+bc+bd)(ad-bc)+(ad-bc)^2\\
&=f(ac+bc+bd, ad-bc)
\end{align*}
In addition,
\begin{align*}
f(a, b)f(c, d)
&= (a^2+ab+b^2)(c^2+cd+d^2)\\
&= a^2c^2+a^2cd+a^2d^2+abc^2+abcd+abd^2+b^2c^2+b^2cd+b^2d^2\\
&=[a^2c^2-2abcd+b^2d^2]+{}\\
&\qquad +[a^2cd+abc^2+abcd-abd^2-b^2cd-b^2d^2]+{}\\
&\qquad +[a^2d^2+2abcd+b^2c^2+2abd^2+2b^2cd+b^2d^2]\\
&=(ac-bd)^2+(ac-bd)(ad+bc+bd)+(ad+bc+bd)^2\\
&=f(ac-bd, ad+bc+bd)
\end{align*}

Using representations of the second form:
\begin{align*}
g(t, u)g(x, y)&=(t^2+3u^2)(x^2+3y^2)\\
&=t^2x^2+3(t^2y^2+u^2x^2)+9u^2y^2\\
&=t^2x^2\pm 6tuxy+9u^2y^2+3(t^2y^2\mp 2tuxy+u^2x^2)\\
&=(tx\pm 3uy)^2+3(ty\mp ux)^2\\
&=g(tx\pm 3uy, ty\mp ux)
\end{align*}

Therefore the result is true.

Lemma 3
If $p$ is prime, and $p$ and $pq$ are Loeschian, $q$ is Loeschian.

Proof
Let
\begin{align}
pq &= t^2+3u^2\notag\\
p &= x^2+3y^2\notag\\
\implies p(pq) &= (t^2+3u^2)(x^2+3y^2)\tag{1}
\end{align}
Then
\begin{align*}
p &\mid t^2p - x^2(pq)\\
\implies p &\mid t^2(x^2+3y^2) - x^2(t^2+3u^2)\\
\implies p &\mid 3t^2y^2-3u^2x^2\\
\implies p &\mid 3(ty-ux)(ty+ux)
\end{align*}
$p$ must divide at least one of those three factors.

Suppose $p=3$. Then $3q=t^2+3u^2$, so $3\mid t^2$, so $3\mid t$. Let $t=3s$. Then
\begin{align*}
3q&=(3s)^2+3u^2\\
\implies q&=u^2+3s^2
\end{align*}
so $q$ is Loeschian. The rest of this proof assumes that $p\ne 3$.

Suppose $p \mid ty-ux$. Then $p^2 \mid (ty-ux)^2$. Then, by Lemma 2
\begin{align*}
p(pq)&= g(tx+3uy, ty-ux)\tag{by eqn 1 and lemma 2}\\
\implies p^2q&=(tx+3uy)^2+3(ty-ux)^2\\
\implies p^2 &\mid (tx+3uy)^2\\
\implies p &\mid tx+3uy
\end{align*}
so $q = [(tx+3uy)/p]^2 + 3[(ty-ux)/p]^2$
is Loeschian. Similarly, but with some sign-changes, if $p \mid ty+ux$.

Therefore the result is true.

Lemma 4
If a Loeschian number is divisible by an integer which is not Loeschian, the quotient has a prime factor which is not Loeschian.

This could also be stated as ``if $mn$ and $n$ are Loeschian, $m$ is Loeschian"; a generalisation of lemma 3.

Proof
Suppose $mn$ is Loeschian and the factorisation of $m$ is $p_1\cdot\dots\cdot p_k$. If each of the $p_i$ is Loeschian, we may divide $mn$ successively by $p_1, \dots, p_k$, and, by lemma 3, infer that the quotient is Loeschian. This would prove that $n$ is Loeschian. So therefore if $mn$ is Loeschian but $n$ is not Loeschian, then at least one of the $p_i$ is not Loeschian, that is, $m$ has a prime factor which is not Loeschian.

Therefore the result is true.

Lemma 5
Every factor of a primitive second-form Loeschian number is either 2 or Loeschian.

Proof
Proof by infinite descent.

Suppose not. Let $n$ be the smallest positive integer $\ne 2$ which is a factor of a primitive Loeschian number and is not Loeschian.

Let $n$ divide the primitive Loeschian number $x^2+3y^2$, where $x$ and $y$ are coprime, and let $x = rn+t$, $y=sn+u$ where $r, s$
are chosen so that $|t|<n/2$ and $|u|<n/2$.
(Note that $t, u$ may be of either sign. Note also that $n$ is odd so equality is not possible.) Then
\begin{align*}
x^2+3y^2 &= (rn+ t)^2 + 3(sn+ u)^2\\
&=r^2n^2+ 2rnt+t^2+3s^2n^2+ 6snu+3u^2\\
&= An + t^2+3u^2
\end{align*}
, where
$A = (r^2+3s^2)n + 2rt + 6su$. Thus
\begin{align}
n &\mid t^2+3u^2\notag\\
\implies t^2+3u^2 &= mn\label{cdxy}
\end{align}
for some integer $m$.

Let $h$ be the HCF (GCD) of $t$ and $u$. Then $h^2 \mid t^2+3u^2$. If $p$ is a prime factor of $h$, then $p\mid t$ and $p\mid u$. If $p\mid n$ too, then $p\mid x$ and $p\mid y$, which is false, as $x$ and $y$ are coprime. Thus $p\nmid n$. Thus $h$ is coprime to $n$, so $h^2 \mid m$,
so there are coprime integers $t'=t/h, u'=u/h$ where
$t'^2 + 3u'^2 = ln$ for some integer $l=m/h^2$. Now
$$ln = t'^2 + 3u'^2 \le t^2+3u^2 < (n/2)^2 + 3(n/2)^2 = n^2,$$
so $l<n$. The inequality is strict because $n$ is odd.

If $t$ and $u$ are coprime, we can use $t$ and $u$ directly rather than $t'$ and $u'$.

By definition of $n$, $n$ is not Loeschian. So, by lemma 4, $l$ must have a factor which is not Loeschian. This factor is a factor of the Loeschian $t'^2 + 3u'^2$, but $<n$, contradicting minimality of $n$.

Therefore the result is true.

Lemma 6 Every factor of a primitive first-form Loeschian number is Loeschian.

Proof
Let $n$ be a primitive first-form Loeschian number $n=a^2+ab+b^2$ where $a$ and $b$ are coprime, so at least one of them is odd.
If $a$ and $b$ are both odd, $n$ is the sum of three odd terms and so is odd; if $a$ and $b$ have different parities, $n$ is the sum of two even terms and one odd, and so is odd. In each case, $n$ is odd.

By lemma 1, $n$ is a primitive second-form Loeschian number, so, by lemma 5, $n$'s every factor is 2 or Loeschian. But $n$ is odd, and so 2 is not a factor of $n$.

Therefore the result is true.

And finally, to answer the OP:

Theorem Let $p=6k+1$ be prime. Then $p$ is Loeschian.

Proof
This proof is based on Euler's proof of Fermat's theorem that every prime of the form $4k+1$ is the sum of two squares.

Suppose not. Let $p=6k+1$ be a prime which is not Loeschian. By Fermat's little theorem, $a^{p-1}=a^{6k}=1\mod p$, for all $0<a<p$. Thus the differences $a^{6k}-b^{6k}$ for $a=2, 3,\dots,6k$ and $b=a-1$, are all divisible by $p$. Each of these differences is
$$a^{6k}-b^{6k} = (a^{4k}+a^{2k}b^{2k}+b^{4k})(a^{2k}-b^{2k})$$
$p$ is prime, so, in each case, $p$ divides one of the factors on the right-hand side. Note that the left-hand factor is the Loeschian number $f(a^{2k},  b^{2k})$. If for any $a$, $p$ divides that left-hand factor, then (as $a$ and $b$ are coprime), by lemma 6, $p$ is Loeschian. This contradicts the definition of $p$.

Therefore, for all $a$, $p$ divides the right-hand factor. Then in the sequence 1, $2^{2k},\dots, (6k)^{2k}$, $p$ divides all the first differences, so $p$ divides all the second differences, and so on, and thus $p$ divides the $2k$'th differences. But each of these $2k$'th differences is $(2k)!$, which is not divisible by any prime greater than $2k$, and thus not by $p$, because $p=6k+1>2k$ and is prime. This is also a contradiction.

Therefore there cannot be any such $p$. Therefore the theorem is true.

ContentLicense: CC BY-SA 3.0


问:
fourth moments of truncated unit-variance variables are summable
In an article I found the following:

If $X$ is a r.v. with zero mean and finite variance, then
$$ \sum_N \frac 1 {N^2} \mathbb E\left[ |X|^4 \mathbf 1_{|X|<\sqrt N} \right]<+\infty  $$

and I am struggling to understand how to prove it.
I tried to do the classical estimation, that is
$$
\mathbb E\left[ |X|^4 \mathbf 1_{|X|<\sqrt N} \right] \le N \mathbb E\left[ |X|^2 \mathbf 1_{|X|<\sqrt N} \right] \le N
$$
but it is not enough. I guess I can get $o(N)$, but that is still not enough.

I also tried to come up with some counterexample, but for example a continuous distribution with a density with tail $O(x^{-k})$ needs $k>3$ to have finite variance, that coincides with the condition to get summability.

And  if $X$ has a distribution with compact support, then all the moments are bounded by a same constant, so the summability follows.

答:
Ok, I probably got it.

$$
\sum_{N=1}^{\infty} \frac 1 {N^2} \mathbb E\left[ |X|^4 \mathbf 1_{|X|<\sqrt N} \right] =
\sum_{N=1}^{\infty} \frac 1 {N^2} \sum_{n=1}^{N}\mathbb E\left[ |X|^4 \mathbf 1_{\sqrt{n-1}\le|X|<\sqrt n} \right]\\
=
\sum_{n=1}^{\infty}\mathbb E\left[ |X|^4 \mathbf 1_{\sqrt{n-1}\le|X|<\sqrt n} \right]\sum_{N=n}^{\infty} \frac 1 {N^2} \\
\le \sum_{n=1}^{\infty}\frac 2n\mathbb E\left[ |X|^4 \mathbf 1_{\sqrt{n-1}\le|X|<\sqrt n} \right]\\
\le \sum_{n=1}^{\infty}2\mathbb E\left[ |X|^2 \mathbf 1_{\sqrt{n-1}\le|X|<\sqrt n} \right] = 2 \text{Var}(X) <+\infty
$$


问:
proving $\min(OR,PR,QR)+OR+PR+QR<OQ+QP+PO$
In the figure prove that $$\min(OR,PR,QR)+OR+PR+QR<OQ+QP+PO$$

This is a stronger form of the inequality  $OR+PR+QR<OQ+QP+PO$ that can be easily proven.I have made some constructions in the above figure.From that we can write $$ \frac{RU}{OU}+\frac{RS}{QS}+\frac{RT}{PT} =1$$ I also have tried to use the fact $$RU<\max(RP,RQ),SR<\max(OR,RP) ...$$

Any ideas

答:
Say we have an arbitrary point $R$ in the interior of the triangle $\triangle OPQ$.

WLOG, $\angle PSQ \gt \angle OSQ$, then $\triangle PSQ$ is an obtuse angled triangle and $\angle OSQ \lt 90^0$ (or vice versa). We draw a perp from point $O$ to segment $QS$ which meets it at point $V$.

We also note that if $R$ is between point $S$ and $V$, $\angle ORQ \lt 90^0$ but $3$ angles at point $R \, (\angle ORQ, \angle PRQ, \angle PRO)$ make $360^0$. So at least two of those angles will be $ \gt 90^0$ and we pick those two favorable one's for our construct. In this case we are assuming those are $\angle ORQ$ and $\angle PRQ$.

.

So we have $PR + QR \lt (PS + SR) + QR = PS + SQ \lt PS + QP$ ...(i)

Now, $QR + OR \lt QR + (RV + VO) = QV + VO \lt OQ + OS$ ...(ii)

If $QS$ is perpendicular to $OP$, both inequalities still hold.

So from (i) + (ii), we have

$OR + PR + QR + QR \lt QP + (PS + OS) + OQ = OQ + QP + PO$

$OR + PR + QR + min(OR,PR, QR) \lt OQ + QP + PO$


问:
Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point
If the equation $3{x^2}y + 2x{y^2} - 18xy = 0$  represent the sides of a triangle whose centroid is G, then the equation of the circle  which touches the line $x-y+1=0$ and passing through G and $(1,1)$ is

A:  ${x^2} + {y^2} - x - y = 0$

B:  ${x^2} + {y^2} - 9x - y + 8 = 0$

C:  ${x^2} + {y^2} - 4x + 2y = 0$

D:  ${x^2} + {y^2} + 9x + y + 8 = 0$

I got the answer but had the choice not been provided how would I find the solution.

My steps are elaborated below

$3{x^2}y + 2x{y^2} - 18xy = 0 \Rightarrow xy\left( {3x + 2y - 18} \right) = 0$

The triangle is formed by line x=0, y=0 and 3x+2y-18=0
Centroid  $G = \left( {\frac{{6 + 0 + 0}}{3},\frac{{9 + 0 + 0}}{3}} \right) = \left( {2,3} \right)$

Let the centre of the circle is (h,k)

Circle passes through (2,3) and (1,1), hence the radius is

${r^2} = {\left( {h - 1} \right)^2} + {\left( {k - 1} \right)^2} = {\left( {h - 2} \right)^2} + {\left( {k - 3} \right)^2} \Rightarrow 2h + 4k = 11$

${x^2} + {y^2} + 2gx + 2fy + c + \lambda \left( {x - y + 1} \right) = 0$, represent the equation of the circle touching the line x-y+1=0 and passing through (2,3) and (1,1)

${x^2} + {y^2} + 2x\left( {g + \frac{\lambda }{2}} \right) + 2y\left( {f - \frac{\lambda }{2}} \right) + c + \lambda  = 0$

$Centre:\left( { - \left( {g - \frac{\lambda }{2}} \right), - \left( {f - \frac{\lambda }{2}} \right)} \right)$

Putting it in 2h+4k=1

$ - 2\left( {g - \frac{\lambda }{2}} \right) - 4\left( {f - \frac{\lambda }{2}} \right) = 11 \Rightarrow  - 2g - 4f + \lambda  + 2\lambda  = 11 \Rightarrow 2g + 4f = 3\lambda  - 11$

By Hit and Trial I am getting choice (B) as it satisfies $2h+4k=11$ where $h=\frac{9}{2}$ and $k=\frac{1}{2}$.

Had the option not been given how do I proceed

答:
The centroid $G$ belongs to the line
$\mathcal{L}:$ $y=x+1$.

Since the circle $\mathcal{C}$ must pass through
$G(2,3)$ and $D(1,1)$, it must also pass through
the point $K(4,4)$,
which is a reflection of $D$ wrt the perpendicular through $G$.

Hence, the sought circle is the circumcircle of $\triangle GDK$
with the side lengths

\begin{align}
|GD|=|GK|&=\sqrt5
,\\
|DK|&=3\sqrt2
,
\end{align}

the area
\begin{align}
S_{GDK}&=
\tfrac12\,|DK|\,\sqrt{|GD|^2-\tfrac14\,|DK|^2}
=\tfrac32
\end{align}

and the circumradius

\begin{align}
R&=\frac{|GD|^2|DK|}{4S_{GDK}}
=\tfrac52\,\sqrt2
.
\end{align}

Since in this special simple case
$GO$ is the diagonal
of the $\tfrac52\times\tfrac52$ square,
the coordinates of the center of the circle $\mathcal C$
can be found as

\begin{align}
O&=G+(\tfrac52,\, -\tfrac52)
=(\tfrac92,\, \tfrac12)
.
\end{align}

In case of general $\triangle ABC$
with the side lengths $a,b,c$,
we could find
the center $O$
using a known expression

\begin{align}
O&=
\frac{a^2\,(b^2+c^2-a^2)\,A+b^2\,(a^2+c^2-b^2)\,B+c^2\,(b^2+a^2-c^2)\,C}
{a^2\,(b^2+c^2-a^2)+b^2\,(a^2+c^2-b^2)+c^2\,(b^2+a^2-c^2)}
\\
&=
\frac{a^2\,(b^2+c^2-a^2)\,A+b^2\,(a^2+c^2-b^2)\,B+c^2\,(b^2+a^2-c^2)\,C}
{16S_{ABC}^2}
,
\end{align}

which, of course, works in this special case as well
and provides the same result.


问:
If $A\succ0$ and $J$ is the matrix of ones, then $A \succeq J$ if and only if $\mathrm{trace}(A^{-1}J)\leq 1$
I'm trying to prove that if $A \in \mathbb{R}^{n \times n}$ is positive definite, and $J$ is the $n \times n$ matrix of ones, then $A \succeq J$ if and only if $\mathrm{trace}(A^{-1}J) = \sum_{i,j} A^{-1}_{i,j} \leq 1$.

For the $\Rightarrow$ direction, I'm able to show that $A \succeq J \implies I - A^{-1}J \succeq 0 \implies \mathrm{trace}(I - A^{-1}J) \geq 0 \implies n \geq \mathrm{trace}(A^{-1}J)$, but this isn't the upper bound of 1 I was hoping for. I've also tried the using the eigendecomposition of $A$ but that didn't seem to get anywhere. For the $\Leftarrow$ direction I'm unsure where to start. Any pointers?

答:
For those familiar with Schur complment, one quick way to see the equivalence is to observe that the block matrix
$$
B=\pmatrix{A&\mathbf1\\ \mathbf1^T&1}
$$
is congruent to both $A\oplus(1-\mathbf1^TA^{-1}\mathbf1)$ and $(A-\mathbf1\mathbf1^T)\oplus1$.  Since $A$ and $1$ are positive definite, it follows that $1-\mathbf1^TA^{-1}\mathbf1=1-\operatorname{tr}(A^{-1}J)\ge0$ if and only if $A-\mathbf1\mathbf1^T=A-J\succeq0$, and both inequalities are equivalent to $B\succeq0$.


问:
What is the probability that balls left are white?

From a bag containing $b$ black balls and $a$ white balls, balls are successively drawn without replacement until only those of the same colour are left. What is the probability that balls left are white?

the favourable event occurs when the balls drawn consist of

$0$ white and $b$ black balls

$1$ white and $b$ black balls

$2$ white and $b$ black balls

$\space\space\space$:

$\space\space\space$$a$. $a-1$ white and $b$ black balls

the last ball drawn cannot be white. it has to be black.

the probability of drawing $i$ white and $b$ black balls, successively without replacement from a bag containing $b$ black balls and $a$ white balls, such that the last ball drawn is  black is $$p_i = \frac{\binom{a}{i}\binom{b}{1}\binom{b-1}{b-1}(b-1+i)!}{\binom{a+b}{b+i}(b+i)!} = \frac{\frac{a!}{(a-i)!i!} \cdot b \cdot 1 \cdot (b-1+i)!}{\frac{(a+b)!}{(b+i)!(a-i)!}(b+i)!} = \frac{a! \cdot b \cdot (b-1+i)!}{(a+b)! \cdot i!}$$

$$\text{the probability that the balls left are white } = \sum_{i=0}^{a-1}p_i = \frac{a! \cdot b!}{(a+b)!}\sum_{i=0}^{a-1}\frac{(b-1+i)!}{i!(b-1!)}$$

$$ =  \frac{a! \cdot b!}{(a+b)!}\sum_{i=0}^{a-1} {b-1+i \choose i} = \frac{a! \cdot b!}{(a+b)!}\sum_{i=0}^{a-1} {b-1+i \choose i}{1 \choose 0}{a-i \choose a-i} $$

the given answer is  $\frac{a}{a+b}$

my question is:

is what I have done correct? if yes, how do I proceed further? if no, then what is wrong in my approach?

while I know that there is a very elegant proof for this question, I am curious as to why a very simple approach to solving this question is not working.

Note to myself:
for the summation $\sum_{i=0}^{a-1} {b-1+i \choose i}{1 \choose 0}{a-i \choose a-i}$

The vandermonde identity unfortunately does not apply

since, the summation is from $0$ to $a-1$, the last slot is always chosen. i.e. the last ball chosen is always white.

答:
It seems that your calculation is correct. Using your final formula Mathematica obtained for all $(a,b)\in[100]^2$ the official value ${a\over a+b}$.

In reality it is expected that you solve this problem without any calculation. Just don't stop drawing until you have drawn all $a+b$ balls. The color of the last ball is the "remaining" color you are interested in. If you agree that the probability for the first ball being white is ${a\over a+b}$ then you should also  agree that the probability for the last ball being white is ${a\over a+b}$.


问:
Money in the box - How much would you pay to play this game?
I came across this interview question of Glassdoor, but it seems to be very interesting, yet tricky to me:

There is some amount of money in a box, determined in the following way: 200 fair coins are flipped. Let the number of heads that come up be H. The amount of money put in the box is H*(100-H)/100. How much would you pay for the box?

Considering that X is the number of heads obtained after 200 flips, E[X] = 100, which would yield a profit of 0. However, I feel there is more to this problem. Maybe it's still worth playing and hope for values with a big enough probability (close to 100, but different), which would yield a non-zero payout.

Can anyone shed some light on this one?

答:
Note that $H=100+x$ has the same probability as $H=100-x$, for all applicable $x$. The first is worth $\frac{-x(100+x)}{100}$, the second $\frac{x(100-x)}{100}$, so on average $-\frac{x^2}{100}$. Combining all $x$ (by their respective positive probabilities), the result is certainly negative!

For the exact (negative) value, you need to find out $E[X^2]$ from the variance of $X$. We know
$E[X]=200\cdot \frac12=100$ and $V[X]=200\cdot\frac12\cdot\frac12=50$, and as
$$V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2,$$
we have $E[X^2]=10050$ and

$$ E[X(100-X)/100]=E[X]-\frac1{100}E[X^2]=100-\frac{10050}{100}=-0.5$$


问:
$\lim_{x\to+\infty}\frac{\ln(x)}{x^{p}}$
Let $p>0$. If we know that $\displaystyle\lim_{x\to+\infty}\frac{x^{p}}{e^{x}}=0$, how can we show that $\displaystyle\lim_{x\to+\infty}\frac{\ln(x)}{x^{p}}=0$? My best guess is to use $x=e^{y}$ and $y=\ln(x)$ so I end up with $\displaystyle\lim_{y\to+\infty}\frac{y}{(e^{y})^{p}}$ which is not what I expected. I hope someone can help me out :)

答:
Let $p\gt0$. Using L'Hopital's Rule:

$$\lim_{x\to+\infty}\frac{\ln(x)}{x^{p}} = \lim_{x\to+\infty} \frac{1/x}{px^{p-1}} = \lim_{x\to+\infty} \frac{1}{px^p} = 0$$

For other methods check this.


问:
Function Spaces
What is exactly the difference between $L^2$ space and ${\ell}^1$ space? I believe that one of them is the space of square of square integrable functions.

Does it have to do with one is for series and other for integration?

Thank You.

答:
$L^2(\Bbb R)$ is the space of square-integrable real functions:
$$L^2(\Bbb R) = \left\{ f: \Bbb R \longrightarrow \Bbb R  \mid \int_{-\infty}^\infty |f(x)|^2 \, dx < \infty \right\}.$$
Note that the above is not quite right; we say two functions $f, g \in L^2(\Bbb R)$ are equivalent if they take the same values outside of a set of measure zero.

$\ell^1(\Bbb R)$, on the other hand, is the space of absolutely convergent series:
$$\ell^1(\Bbb R) = \left\{ \{a_n\}_{n = 1}^\infty \mid \sum_{n = 1}^\infty |a_n| < \infty \right\}.$$

In general,
$$L^p(\Bbb R) = \left\{ f: \Bbb R \longrightarrow \Bbb R  \mid \int_{-\infty}^\infty |f(x)|^p \, dx < \infty \right\}$$
for $p \geq 1$ where we identify $f, g \in L^p(\Bbb R)$ if they agree outside of a set of measure zero, and
$$\ell^p(\Bbb R) = \left\{ \{a_n\}_{n = 1}^\infty \mid \sum_{n = 1}^\infty |a_n|^p < \infty \right\}.$$

If you learn measure theory at some point, you will see that there is a unifying definition of $L^p$ spaces that include the above examples as special cases: $L^p(\Bbb R)$ will be the $L^p$ space associated to $\Bbb R$ with the Lebesgue measure, and $\ell^p(\Bbb R)$ will be the $L^p$ space associated to $\Bbb N$ with the counting measure.

Title: Function Spaces


问:
Reversing a modulo with an exponent
So I am doing a challenge to break an encryption and have been stuck on one of the equations.
$$c = K^e\mod n$$
I know $c,e$ and $n$ all of which are very very large numbers. I know I have to get $K$ as it is the only way to reverse the next set of equations but I cannot figure out how to get $K$ from this.

答:
Use Euclidean Algorithm to find the gcd of $e$ and $\varphi(n)$ as their linear combination (which is hopefully $1$ otherwise we don't have a solution guaranteed for all $c$). Suppose $ae\equiv 1 \operatorname{mod} n$. Then take $c$ to the power of $a$ gives you $K$.

Fermat's Little Theorem is the reason behind the mathematics. The difficulty here should be finding $\varphi(n)$, but that's why the encryption works. It is meant to be difficult; but knowing the factorization of $n$ would make it easy. That's why those who know the factorization would be able to decode, and those who don't won't.

Also you might wish to look up RSA algorithm for more information.

Title: Reversing a modulo with an exponent


答:
Can you use continuity? if yes then:
$$\frac{y}{(e^{y})^{p}}=\left(\frac{y^{\frac{1}{p}}}{e^{y}}\right)^p$$

Title: $\lim_{x\to+\infty}\frac{\ln(x)}{x^{p}}$

Body: Let $p>0$. If we know that $\displaystyle\lim_{x\to+\infty}\frac{x^{p}}{e^{x}}=0$, how can we show that $\displaystyle\lim_{x\to+\infty}\frac{\ln(x)}{x^{p}}=0$? My best guess is to use $x=e^{y}$ and $y=\ln(x)$ so I end up with $\displaystyle\lim_{y\to+\infty}\frac{y}{(e^{y})^{p}}$ which is not what I expected. I hope someone can help me out :)


问:
How to find limit of a probability of a continuous random variable with geometric distribution?
Problem 5. A box contains $n$ balls, $k$ of them are white ($k<n$), and $n-k$ are black. Suppose we take balls from the box until we get a white one, each time returning a ball back to the box. Let $X$ be the number of attempts until we get a white ball.

(a) What is the distribution of $X$? What is $E(X_n)$?

(b) Find the probability that we will make more than $n$ attemts, and find the limit of this probability as $n\to\infty$.

(c) Let $Y=X/n$ be the ratio of the number of attempts to the number of balls in the box. For any number $t\ge0$ find the probability $P(Y\le t)$ and find the limit of this probability as $n\to\infty$.

(d) Denote by $F(t)$ the limit that you have found in the previous question. Is $F$ a cumulative distribution function? If so, of what distribution?

I need to solve this problem. I managed to do (a), which was rather obvious ($X\sim\operatorname{Geometric}(k/n), E(X_n)=n/k$), but I struggle to do (b), (c), (d).

It seemed to me that in (b) $P(X>n)=1-P(X<n)=1-F(n)$, but this just leads nowhere, since I am not sure how to derive the cdf.

答:
You are on the right track. Let $p = \frac{k}{n}$ and $q = 1-p$. Then $P(X=x) = q^{(x-1)}p$ and $P(X \leq x) = \sum_{t=1}^xq^{(t-1)}p$.

You just need to work through some algebra. All the best!

P.S.
Here is an excellent reference on typing mathematical formula in your future questions and answers: MathJax Basics


问:
An infinite set of continuous functions doesn't have to be a set of equally continuous functions
I have been studying about sets of equally bounded functions and equally continuous functions. I've found a question which states: Show an example where an infinite set of continuous functions doesn't necessarily need to be set of equally continuous functions.

If someone can give me a simple example I would appreciate it. I've tried googling it but couldn't find the right term for those sets.

答: Assuming by "equally continuous" you mean equicontinuous, just consider the set of functions $f_a(x)=ax\ $ $\forall a \in \mathbb{R}$


问:
Milnor's proof of Homogeneity lemma
I was reading Milnor's topology from the differential  point of view.page 24

In the Homogeneity lemma,which state that for any two interiror point  $y,z$ of smooth connected manifold $N$,exist a diffeomorphism $h:N\to N$ that is smooth isotopic to identity and carries $y$ to $z$.

He prove it as follows:

for $N = S^{n-1}$ the result clear

construct a smooth diffeomorphism $F_t$ on $\mathbb{R}^n$ to itself,such that carries origin to any point inside open disk,nad $F_t$ isotopic to identity map.

Finally we need to complete the proof.I don't know how to use the 2 facts above to complete the proof?

答:
Let $x\in N$, conisder $N_x=\{y:f(x)=y$ and $f$ is isotopy to the identity $\}$.

We are going to show that $N_x$ is open and closed. Let $y\in N_x$ there exists a diffeomorphism $f_y$ connected to the identity such that $f_y(x)=y$. There exists a chart $U_y$ of $N$ containing $y$ which is homeomorphic to $\mathbb{R}^n$. This implies that for every $z\in U_y$, there exists $g_z$ isotopy to the identity, $g_z(y)=z$. $g_z\circ f_y(x)=z$ and is isotopic to the identity. This implies that $N_x$ is open. $N_x$ is closed, if $y=limy_n, y\in N_x$, there exists a neighborhood $U_y$ diffeomorphic to $\mathbb{R}^n$ which contains $y_n$. We deduce that there exists a diffeomorphism $h_n$ isotopic to the identity such that $h_n(y_n)=y$, if $f_n$ is a diffeomorphism isotopic to the identity such that $f_n(x)=y_n$, $h_n\circ f_n$ is isomorphic to the identity. We deduce that $N_x$ is closed.

Title: Milnor's proof of Homogeneity lemma


问:
Calculus: What to do when algebraically stuck, when trying to resolve an equation to find critical point?
My understanding:

g'(x) & g(x) are defined for all real numbers.

However I think I'm confused when I equate $e^{3x}(1+3x)=0$, I'm trying to solve for $e^{3x}=0$ after dividing both sides by $(1+3x)=0$ algebraically. I'm stuck as there is no value of x in $e^{3x}$ that is zero. What do I do next?

答: No. In particular, we clearly have $g'(0) = e^{3\times 0}(1 + 3 \times 0) = e^0(1 + 0) = 1$.


答:
$p>0;$

$(1/p)\dfrac{\log (x^p)}{x^p};$

Set $y:=x^p$ and consider $y \rightarrow \infty.$

($1/p)\lim_{y \rightarrow \infty}\dfrac{\log y}{y}=?$


问:
Show that of x is in the closure of a set, then there exists a sequence that converges to x
Given a metric space $X$ and $E \subseteq X$. How can I show that $x \in \overline{E}$ if and only if there exists a sequence $x_n \in E$ such that $\lim_{n\to\infty} x_n=x$.

答: Hint: The open ball of radius $\frac 1n$ around $x$ intersects $E$


问:
Infinity primes of the form $4k+1$
This is a problem from the book Introduction to algebra by А. И. КОСТРИКИН.

Assume $n,m\in \mathbb Z,\gcd (n,m)=1$. If $p$ is a prime and $p | (n^2+m^2)$, then $p=4k+1$. Use this to prove that there are infinity primes of the form $4k+1$.

My question is, why $p=4k+1$? It seems not so obvious to me. Also, I don't know how to use this to prove the proposition.

Can anyone help?

答:
Hint:

Remember Fermat's Little Theorem:

$$a\not\equiv_p 0\Rightarrow a^{p-1}\equiv_p 1$$

And then consider:

 $$p\mid n^2+m^2\iff (mn^{-1})^2\equiv_p -1$$ This follows since $p\mid n^2+m^2\iff n^2+m^2\equiv_p 0$ by definition. Performing some algebra, we get: $$n^2+m^2\equiv_p 0\\m^2\equiv_p -n^2\\m^2n^{-2}\equiv_p -1\\(mn^{-1})^2\equiv_p -1$$

For the rest of the exercise, suppose that there are only finitely many primes $p$, such that $p=4k+1$, and let us name them $p_1,p_2,\ldots,p_n$. Then, consider:

 $$N=4(p_1p_2\cdot\ldots\cdot p_n)^2+1$$

Title: Infinity primes of the form $4k+1$


问:
Probability stranger in a group
Given a group of n = 100 people made up of Italians and foreigners, two people are chosen at random. Knowing that the probability that only one of them is foreign is p ≈ 0.18, a possible composition of the group is identified (number of Italians nI and number of foreigners nS).

I wrote that $ \binom{100}{1} p_s ^{1} ( 1- p_s)^{99} \binom{99}{1} p_s^{0} {1-p_s }^{99} = 0.18 $ so i found that $ p_s = 0.053 $ where $ p_s $ is the probability that a person is stranger. I dint know if a made mistakes because I dint have the result

答:
If I understand your problem correctly, you should just find ways to choose $2$ people out of $100$ such that one of them is foreigner (and the other is Italian).

Say there are $n$ foreigners. Then we have,

$\displaystyle \frac{{n \choose 1} {100-n \choose 1}}{100 \choose 2} = 0.18$

Leads to quadratic $n^2 - 100n + 891 = 0$. You can solve for possible values taking nearest positive integers.

Title: Probability stranger in a group


答:
A set of functions with their derivatives uniformly bounded is  equicontinuous.

A concrete example:
$$\bigl\{f_n\in\mathcal C\bigl([0,+\infty)\bigr)\mid f_n(x)=\sin(x+4n^2\pi^2),\enspace x\in[0,+\infty)\bigr\}.$$

Title: An infinite set of continuous functions doesn't have to be a set of equally continuous functions

Body:
I have been studying about sets of equally bounded functions and equally continuous functions. I've found a question which states: Show an example where an infinite set of continuous functions doesn't necessarily need to be set of equally continuous functions.

If someone can give me a simple example I would appreciate it. I've tried googling it but couldn't find the right term for those sets.


答:
Let $n_I$ be the number of Italians, and $n_S$ be the number of foreigners.  We know that $$n_I + n_S = n = 100.$$

We are told that if we choose two (presumably different) people at random from the group, the probability that exactly one of the two is foreign is $p = 0.18$.  We count the number of ways to choose two distinct people from the group such that one is foreign and one is Italian.  There are $n_I$ ways to choose one person from the group of Italians; and $n_S$ ways to choose one person from the group of foreigners.  Since these choices may be made independently without restriction, there are a total of $n_I n_S$ ways to select two people from the whole group so that exactly one is foreign.  But there are $$\binom{n}{2} = \frac{n(n-1)}{2}$$ ways to choose any two people from the group of $100$.  So the given condition is equivalent to $$\frac{n_I n_S}{n(n-1)/2} = p = 0.18.$$  We now substitute $n = 100$ and $n_S = 100 - n_I$ to obtain $$n_I(100 - n_I) = 4950(0.18) = 891,$$ and solve the resulting quadratic for $n_I$:  $$n_I = 50 \pm \sqrt{1609} \approx \{9.88766, 90.1123\}.$$  So rounding to the nearest integer gives us the solution set $$n_I \in \{10, 90\}.$$  Notice that both solutions are valid--it is impossible to determine with the information given whether the group comprises $90$ Italians and $10$ foreigners, or $10$ Italians and $90$ foreigners.  This should be obvious because the roles of "Italians" and "foreigners" may be interchanged and the question would remain identical, since the outcome of one foreigner among two people selected is the same as saying there is one Italian among two people selected.

Since we rounded the result, this means that the precise value of $p$ is no longer $0.18$.  If we use $n_I \in \{ 10, 90\}$, what is the resulting value of $p$?

Title: Probability stranger in a group

Body:
Given a group of n = 100 people made up of Italians and foreigners, two people are chosen at random. Knowing that the probability that only one of them is foreign is p ≈ 0.18, a possible composition of the group is identified (number of Italians nI and number of foreigners nS).

I wrote that $ \binom{100}{1} p_s ^{1} ( 1- p_s)^{99} \binom{99}{1} p_s^{0} {1-p_s }^{99} = 0.18 $ so i found that $ p_s = 0.053 $ where $ p_s $ is the probability that a person is stranger. I dint know if a made mistakes because I dint have the result


答:
Elena - in your “binomial speak”, we have two “experiments” and we are looking at probability of seeing one success i.e. one foreigner.

$\binom{2}{1} p_s (1-p_s) = 0.18$

Title: Probability stranger in a group


问:
Finding all normal subgroups of a group
On my homework today, we had to find all the normal subgroups of $D_{n}$, the dihedral group of order 2n. I solved the problem by looking at how the conjugacy classes change based on whether n is even or odd and then constructed the normal subgroups as unions of the conjugacy classes.

I have 2 questions:

(i) Is there a better way to approach the problem than looking at the conjugacy classes?

(ii) Could someone explain why we want to find all the normal subgroups of a particular group?   How does this provide us additional insight into the structure of the group we are studying? (Right now, this exercise feels more like a 'computation' to me than a way of understanding $D_{n}$)

Thanks :)

答: Very vague answer to (ii) (I'm taking this course myself): if you find all normal subgroups of a given group, then you know all its quotient groups, since quotient groups are defined through normal subgroups only - if you consider all left (or right) cosets of a given group, that isn't normal, they may not conform a group. And many theorems that follow in the theory are based on quotient groups.</p>

ContentLicense: CC BY-SA 2.5

Title: Finding all normal subgroups of a group


问:
Equation related to the curve $x^4+3kx^3+6x^2+5$
The sum of all the integral value(s) of 'k' so that the curve $x^4+3kx^3+6x^2+5$ is not situated below any of its tangent line is

(A) 5

(B) 2

(C) 0

(D) -2

I have no idea of how to proceed this question

答: This property is called concave up, so we just solve for $f''(x)>0$ for all $x$ where $f(x)$ is the given curve.


问:
Definition of an algebra
The definition of an algebra says that given a set X, an algebra of subsets of X is a collection $\mathcal{A}$ of subsets of X such that:

if $A \in \mathcal{A}$ then $A^c \in \mathcal{A}$

My question is if it is correct to say that if $A \notin \mathcal{A}$ then $A^c \notin \mathcal{A}$

答:
Yes, the statement
$$A\in \mathcal{A} \implies A^c\in \mathcal{A}$$
immediately implies that the converse holds since $(A^c)^c=A$. Namely, one has
$$A^c\in \mathcal{A}\implies (A^c)^c=A\in \mathcal{A}. \tag{$\ast$}$$
The contrapositive statement of $(\ast)$ is what you want to prove to be true.

Title: Definition of an algebra


问:
Minimal polynomial of $x$ over $K(f(x))$
Let $K$ be a field and $f(x)$ be a one-variable monic polynomial on $K$.

Is the minimal polynomial on $K(f(x))$ of $x$ (with $t$ as a variable)
$f(t) - f(x)$?

I know that the minimal polynomial of $x$ I want to know is a divisor of $f(t)-f (x)$.

I expect the minimal polynomial to be $f(t)-f(x)$.
So, in general, is $f(t)-f(x)$ irreducible on $K(f(x))$?

答:
I think the answer is yes, and hopefully this is a valid proof.

If there's a polynomial $P(t)\in K(f(x))[t]$ that has $x$ as a root with degree less than $\deg f$, then we can write

$$P(t)=\sum_{i=0}^n\frac{p_i(f(x))}{q_i(f(x))}t^i$$

where $p_i$, $q_i$ are polynomials with coefficient in $K$ and $n<\deg f$. Then plugging in $x$ we can write

$$\sum_{i=0}^ns_i(f(x))x^i=0$$

where $s_i$'s are polynomials with coefficient in $K$. Considering this equation $\operatorname{mod} f(x)$ implies that all the $s_i$'s have $0$ constant term. Now divide the equation by $f(x)$ and repeat until you reach a contradiction.


问:
Existential import of particular proposition
I saw that in categorical logic the proposition, some $A$ is $B$ is considered to have existential import meaning we assume there exists a particular in class $A$ but the proposition, "All $A$ is $B$ mean that if $A$ is empty" the proposition is true, why is that saying "some unicorns are pink" is a false statement while saying "All unicorns are pink" is a true statement?

答:
The issue of so-called existential import of categorical proposition has a long history.

In modern logic "Some A is B" is symbolized with $\exists x (Ax \land Bx)$ and thus it is true only when there are $A$s and $B$s.

The universal, instead, is symbolized with $\forall x (Ax \to Bx)$ which is true also when there are no $A$s.

[Why] saying "All unicorns are pink" is a true statement?

Because the conditional $Ax \to Bx$ is true when the antecedent $Ax$ is False.

According to Arsitotle's Logic and the so-called Square of Opposition, we have that the universal imples the corresponding particular, but in modern logic the imference:

$\forall x (Ax \to Bx)$, therefore $\exists x (Ax \land Bx)$

is not valid, exactly due to the counter-examples like yours.

For some references, see:

Joseph Wu, The problem of existential import (1969), and:

John Corcoran & Hassan Masoud, Existential Import Today: New Metatheorems; Historical, Philosophical, and Pedagogical Misconceptions (2015).


问:
Calculating the integral is not giving the same simplified function
I calculated this integral below :
$$
\int_{-\pi/2}^{\pi/2} \cos(x)\cos(nx) dx
$$

And I got this result :
$\dfrac{\left(\left(n-1\right)\sin\left(\frac{{\pi}n+{\pi}}{2}\right)+\left(n+1\right)\sin\left(\frac{{\pi}n-{\pi}}{2}\right)\right)}{n^2-1}$

I had a check at the end at the solution :
$-\dfrac{2\cos\left(\frac{{\pi}n}{2}\right)}{n^2-1}$

So apparently there is some kind of simplification to do but I don't actually know how .

答:
$\sin (\frac {n\pi} 2+\frac {\pi} 2)=\cos (\frac {n \pi} 2)$  and $\sin (\frac {n\pi} 2-\frac {\pi} 2)=-\cos (\frac {n \pi} 2)$.

[$\sin (\frac {\pi} 2+\theta)=\cos (\theta)$ and $\sin (\theta -\frac {\pi} 2)=-\sin (\frac {\pi} 2-\theta)=-\cos (\theta)$]


问:
A lattice of circles and a line which does not cut any of them.

Each lattice point is a center of a circle, all with radius $d$. Let line $y={2\over 5}x+n$ doesn't cut or touch any circle, for some $n$. Find the supremum for $d$.

I was trying to maximize $$d={|2x-5y+5n|\over \sqrt{29}}$$ where $x,y$ runs over all integers and we can assume that $n$ is in $(0,1)$, but I don't know how to do it. Any help?

Body: This diagram might help you..........................


答:
Observe that
$$\sin\left(\frac{\pi n+ \pi}{2}\right)=\sin\left(\frac{\pi n}{2}\right)\cos\left(\frac{\pi}{2}\right)+ \sin\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi n}{2}\right) = \cos\left(\frac{\pi n}{2}\right)$$
$$\sin\left(\frac{\pi n- \pi}{2}\right)=\sin\left(\frac{\pi n}{2}\right)\cos\left(\frac{\pi}{2}\right)- \sin\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi n}{2}\right) = -\cos\left(\frac{\pi n}{2}\right)$$
when $n\in\mathbb{N}$.

Then,
\begin{equation*}
\begin{split}
(n-1)\sin\left(\frac{\pi n+ \pi}{2}\right) & + (n+1)\sin\left(\frac{\pi n- \pi}{2}\right) = (n-1)\cos\left(\frac{\pi n}{2}\right) - (n+1)\cos\left(\frac{\pi n}{2}\right) \\
& = -2\cos\left(\frac{\pi n}{2}\right)
\end{split}
\end{equation*}

Title: Calculating the integral is not giving the same simplified function

Body:
I calculated this integral below :
$$
\int_{-\pi/2}^{\pi/2} \cos(x)\cos(nx) dx
$$

And I got this result :
$\dfrac{\left(\left(n-1\right)\sin\left(\frac{{\pi}n+{\pi}}{2}\right)+\left(n+1\right)\sin\left(\frac{{\pi}n-{\pi}}{2}\right)\right)}{n^2-1}$

I had a check at the end at the solution :
$-\dfrac{2\cos\left(\frac{{\pi}n}{2}\right)}{n^2-1}$

So apparently there is some kind of simplification to do but I don't actually know how .


问:
How to solve $2^{53x} \equiv 3 \pmod{23}$?
I tried to solve this congruence, but I don't understand how to find “?”

$2^{11} \equiv 1 \pmod{23}$ so

$53x \equiv ~?  \pmod{11}$

I know that “?” is equal to $8$, but I don't know how to find it, thank you!

答:
Finding the ???, or equivalently finding $y$ such that $2^y\equiv3\pmod{23}$, is an instance of a discrete logarithm problem. There is no universal efficient method for solving it. Thankfully $11$ is a small number, so you can brute force it.

If you want to see a trick then an ad hoc answer can be derived from the observation that
$$
3^3=27\equiv 2^2\pmod{23}.
$$
Leaving that as a hint :-)

 Plug in $3=2^y$, and solve $y$ from the resulting congruence modulo $11$.

Observations like the one above are the basis of an algorithm called index calculus. For a (barely) pen-and-pencil-doable example of index calculus you may take a look at this answer of mine, partly prepared for referrals like this. That example works in the field $\Bbb{F}_{2^{10}}$. Meaning that the arithmetic is done modulo a polynomial (with coefficients modulo two) instead of modulo some prime $p$ like here. The principles are the same.


问:
Calculate the volume enclosed by the paraboloid $z=x^2+y^2$ and the plane $z=2y$.

Calculate the volume enclosed by the paraboloid $z=x^2+y^2$ and the plane $z=2y$.

When $x = 0$ it seems that the region of interest lies between $y=2$ and $z=4$. If I'm not mistaken I would need to calculate $$\int_D(x^2+y^2-2y) \ dA.$$

Also $D$ seems to be given by $D=\{(x,y) \in \mathbf{R}^2 \mid x^2+y^2 \le 2y \}$

My question is how can I found the bounds for the integrals? Any hints would be appreciated.

答:
I would begin by converting to polar coordinates as they are much easier to work with when we have rotational symmetry. We can do this by making the substitution $x = r\cos\theta$ and $y= r\sin\theta$. That would mean $dA = dxdy = rdrd\theta$. (the extra $r$ comes from the jacobian for polar coordinates).

After making this transformation, the region becomes much easier to describe. The paraboloid with equation $z = x^2 + y^2$ can be written as $z = r^2$ and the plane can be written as $z = r\sin\theta$. The new integral becomes $\int (2r\sin\theta - r^2) rdrd\theta$.

Now for the limits. r begins at $0$, but to find its endpoint, we must find the intersection between the plane and the paraboloid. We can easily do this by solving the equation $r^2=2r\sin\theta$ which turns out to be where $r = 2sin\theta$. For the limits on $\theta$, we can see that it is $\pi$ by observation as we are only taking values where $y \ge 0$.

This leaves the final integral to be:

$$ \int_{0}^{\pi}\int_{0}^{2\sin\theta}(2r^2sin\theta - r^3) drd\theta $$


问:
Finding homomorphism between congruences $\bmod 18$ and $\bmod 3$
The (cyclic) multiplicative group $(\mathbb{Z}/18\mathbb{Z})^\times=\mathbb{Z}^\ast_{18}=\{1,5,7,11,13,17\}=<5>$ has an order $ord(\mathbb{Z}^\ast_{18})=6$ and based on Euler's theorem we can derive the following congruences from $5^j5^{6n+6-j}\equiv1\pmod{18}$:

\begin{array}{cc}
j=0 & [1]_{18}\cdot5^{6n+6}\equiv1\pmod{18}\\
j=1 & [5]_{18}\cdot5^{6n+5}\equiv1\pmod{18}\\
j=2 & [7]_{18}\cdot5^{6n+4}\equiv1\pmod{18}\\
j=3 & [17]_{18}\cdot5^{6n+3}\equiv1\pmod{18}\\
j=4 & [13]_{18}\cdot5^{6n+2}\equiv1\pmod{18}\\
j=5 & [11]_{18}\cdot5^{6n+1}\equiv1\pmod{18}\\
\end{array}

Now let us consider the following congruences:

\begin{array}{c}
[1]_{18}\cdot4\equiv1\pmod{3}\\
[5]_{18}\cdot8\equiv1\pmod{3}\\
[7]_{18}\cdot16\equiv1\pmod{3}\\
[17]_{18}\cdot2\equiv1\pmod{3}\\
[13]_{18}\cdot4\equiv1\pmod{3}\\
[11]_{18}\cdot2\equiv1\pmod{3}
\end{array}

Having in mind that, if $m\mid n$ (in our case $3\mid18$) the map $r\bmod n \rightarrow r\bmod m$ is a homomorphism $(\mathbb{Z}/n\mathbb{Z})^\times\rightarrow(\mathbb{Z}/m\mathbb{Z})^\times$, what is the morphism between both congruences shown above - the one $\bmod{18}$ and the other one $\bmod3$? Maybe considering only the element $5$ as a generator of $\mathbb{Z}^\ast_{18}$ is not sufficient?

答:
If a natural number divides another, $m\mid n$, as in our case $3\mid18$, then for two integers $a,b$ the following implication holds:

\begin{equation}
 \label{eq:reduce_modulus}
 a\equiv b\pmod n\rightarrow a\equiv b\pmod m
\end{equation}

This means in our case $w\cdot5^{6n+6}\equiv 1\pmod{18}\rightarrow w\cdot5^{6n+6}\equiv 1\pmod 3$. Fundamentally we can state that $a\equiv b\pmod m$ implies $(a+m)\equiv b\pmod m$. Let us set $a=w\cdot2^{2n+2}$ and $b=1$ and $m=3$, then we obtain $(w\cdot2^{2n+2}+3)\equiv1\pmod3$ and using a factor $i\in\mathbb{N}$ we obtain the more general congruence $(w\cdot2^{2n+2}+3i)\equiv1\pmod3$. As a consequence the congruences are true, while $w\cdot5^{6n+6}=w\cdot2^{2n+2}+3i$ or rather while $3\mid(5^{6n+6}-2^{2n+2})$. These conditions continue $3\mid(5^{6n+5}-2^{2n+3})$ and $3\mid(5^{6n+4}-2^{2n+4})$ and so forth.

\begin{array}{llllll}
 j=0, & w\in [1]_{18} & \hspace{1em} w\cdot5^{6n+6}&\equiv1\pmod{18}&\hspace{4em}w\cdot2^{2n+2} &\equiv1\pmod{3}\\
 j=1, & w\in [5]_{18} & \hspace{1em} w\cdot5^{6n+5}&\equiv1\pmod{18}&\hspace{4em}w\cdot2^{2n+3} &\equiv1\pmod{3}\\
 j=2, & w\in [7]_{18} & \hspace{1em} w\cdot5^{6n+4}&\equiv1\pmod{18}&\hspace{4em}w\cdot2^{2n+4} &\equiv1\pmod{3}\\
 j=3, & w\in [17]_{18} & \hspace{1em} w\cdot5^{6n+3}&\equiv1\pmod{18}&\hspace{4em}w\cdot2^{2n+1} &\equiv1\pmod{3}\\
 j=4, & w\in [13]_{18} & \hspace{1em} w\cdot5^{6n+2}&\equiv1\pmod{18}&\hspace{4em}w\cdot2^{2n+2} &\equiv1\pmod{3}\\
 j=5, & w\in [11]_{18} & \hspace{1em} w\cdot5^{6n+1}&\equiv1\pmod{18}&\hspace{4em}w\cdot2^{2n+1} &\equiv1\pmod{3}
\end{array}


问:
Test for polynomial reducibility with binary coefficients
I'm learning about Galois Fields, in particular $GF(2^8)$, as they are applied to things like the AES algorithm and Reed-Solomon codes.  Each of these rely on an irreducible 8th degree polynomial with binary coefficients to serve as a modulus for generating the particular field instance.  For example AES uses $x^8+x^4+x^3+x^1+x^0$.

Is there a test I can apply to an 8th degree polynomial with binary coefficients I am presented with to determine whether or not it is irreducible?

答:
A non-irreducible polynomial $p(x)$ of degree eight has an irreducible factor of degree $\le 4$. This suggest the following test. Check divisibility by all of $x$, $x+1$, $x^2+x+1$, $x^3+x+1$, $x^3+x^2+1$, $x^4+x+1$, $x^4+x^3+1$ and $x^4+x^3+x^2+x+1$. That list contains all the irreducible polynomials of degree $\le 4$ with coefficients in $GF(2)$. If your polynomial is not divisible by any of these, it is irreducible.

That is a little bit of work. Of course testing divisibility by either $x$ or $x+1$ is trivial. A polynomial is divisible by $x$, iff its constant term is zero. And a binary polynomial is divisible by $x+1$, iff it has an even number of terms. The remaining six are a bit trickier. Often checking for divisibility by $x^2+x+1$ is aided by the observation that $x^3+1=(x+1)(x^2+x+1)$ is divisible, hence so are all binomials of the form $x^\ell+x^{\ell+3}$. This allows you to replace a high degree term with a lower degree term. Similarly the cubic irreducible ones are both factors of $x^7+1$, but that won't be nearly as useful in the calculations.


Title: How to solve $2^{53x} \equiv 3 \pmod{23}$?

Body:
I tried to solve this congruence, but I don't understand how to find “?”

$2^{11} \equiv 1 \pmod{23}$ so

$53x \equiv ~?  \pmod{11}$

I know that “?” is equal to $8$, but I don't know how to find it, thank you!

Body:
$$2^8\equiv 3 \mod23$$

$$2^{11}\equiv 1 \mod23$$

$$2^{11t}\times2^{8}\equiv 3 \mod23$$

$$2^{53x}\equiv 2^{11t+8}\equiv 3 \mod23$$

$$53x\equiv 8 \mod11$$

$$9x\equiv 8 \mod11$$

$$x\equiv 7 \mod11$$

$$x=7+11k$$

where $k \in Z$


问:
Prove that if $x^TAx = 0 \Rightarrow Ax = 0$
A is $n \times n$ symmetric and positive semidefinite. So we can use $A^T = A$ and $x^TAx \geq 0$ for all x $\in \mathbb{R}^n$.

Proof by contradiction comes to mind, or $x^TAx = 0 \; \land \; Ax \neq 0$. This might seem like a given at first but I'm thinking what if there is some linear dependence between $x^T$ and the $Ax$ vector?

Edit: Ideally, I'm supposed to prove this using the rank of a matrix.

If I were to prove it by contradiction, $x^TAx = 0 \; \land \; Ax \neq 0$, I'd start off by saying that:

$x$ cannot be a zero vector in this case. The original statement $x^TAx = 0 \Rightarrow Ax = 0$ is trivial for a zero $x$ vector and needs no proof.

from $Ax \neq 0$ we have $rank(Ax) = 1$ since $Ax$ is also a non-zero vector.

problem: how to show that the $x^T$ vector and the $Ax$ vector cannot be orthogonal? (this would be the point of the contradiction proof)

problem 2: the only idea I have for a proof barely utilizes the rank of a matrix, which is my assignment.

答:
Consider the EVD of the matrix $A=UDU^T$.

Then we can rewrite $x^TAx=x^TUD^\frac12D^\frac12U^Tx=\|D^\frac12U^Tx\|^2$

Hence if $x^TAx=0$, we must have $D^\frac12Ux=0$ and hence $$UD^\frac12(D^\frac12U^Tx)=(UD^\frac12D^\frac12U^T)x=Ax=0$$


问:
Could we be biased toward positive integers throughout mathematics?
What I am getting at is that throughout math we have to refer to things using positive integers. In a a polynomial, we talk about the first (that is, the "positive oneth") term and so forth. There is no such thing as the "negative oneth" term of a polynomial. Even in 2*3 = 6, I don't know what you call the two, but maybe it is also referred to as the first term or something.
Anyway, perhaps this underlying "bias" towards positive integers is why we consider the product of two negatives to be a positive -- if not that, some other intuition that we might not have if we had objects to represent equations etc. that were not based on positive integers.

Forgive me if this is a dumb question but I have thought about this for a while and I have some experience with mathematics.

答:
An essential building block of mathematics are the natural numbers, which arose by the universal need to count. Negative counts make non sense. [Please don't bombard me with comments explaining how negative counts can be given sense.] So the bias towards positive is obvious.

The case of the product of two negatives is certainly not related to this bias. It comes from the necessity to have a coherent multiplication operation when acting on negatives.

$$(-1)(1-1)=(-1)+(-1)(-1)=0$$ demands $$(-1)(-1)=1.$$

Side remark: the Python language smartly defines a list indexing system such that $0$ and positives denote the first elements, while negatives allow access to the last ones.


问:
What logic can be used express Hilbert's axiomatization of Euclidian geometry?
Can the first order logic be used to express Hilbert axioms for Euclidean geometry as well as all theorems that follow from them? If it is not the case, what logic can be used? Does it have a name? In what way is it more expressive than the first order logic?</p>

答:
The first four groups of Hilbert Axioms can be expressed in first order logic, the occurrences of sets are inessential in the first four groups. An example of a formalization is given in GeoCoq.

The fifth group consists of Continuity axioms:

Axiom of Archimedes:

If AB and CD are any segments then there exists a number n such that n segments CD constructed contiguously from A, along the ray from A through B, will pass beyond the point B.

Axiom of line completeness:

An extension (An extended line from a line that already exists, usually used in geometry) of a set of points on a line with its order and congruence relations that would preserve the relations existing among the original elements as well as the fundamental properties of line order and congruence that follows from Axioms I-III and from V-1 is impossible.

Both axioms of group V require second order logic, where one can quantify over predicates.

The line completeness axiom contains a quantification over the models of I-III and from V-1, so to express it one needs to says "forall predicates incidence, betweeness, congruence, which verify ..."

These two axioms (line completeness and axiom of Archimedes) can also be replaced  equivalently by Dedekind cut axiom.

Tarski studied a variant of the Dedekind cut axiom which can be expressed in first order logic using an axiom schema:

Let φ(''x'') and ψ(''y'') be first-order formulae containing no free  instances of either ''a'' or ''b''. Let there also be no free instances of ''x'' in ψ(''y'') or of ''y'' in φ(''x''). Then all instances of the following schema are axioms:
:
$\exists a \,\forall x\, \forall y\,[(\phi(x) \land \psi(y)) \rightarrow Baxy] \rightarrow \exists b\, \forall x\, \forall y\,[(\phi(x) \land \psi(y)) \rightarrow Bxby].$

In fact,  most of Hilbert's theorems about geometric figures can be developed in FOF. In the proof of Pascal's theorems, he may use second order logic when it makes use of congruence classes of segment or angles. But this use is inessential because the statement is first-order.

Hilbert needs these second order axioms only to characterize the models of its axioms as cartesian planes over the reals.

Using Tarski's first order variant of Dedekind's cut, the models are the cartesian planes over a real closed field.

If one only wants to prove theorems within ruler and compass geometry, the first order axioms of line/circle or circle/circle continuity are enough.


Title: Why $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$?

Body: How we can easily show that $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$. These conditions continue $3\mid(5^{6n+5}-2^{2n+3})$ and $3\mid(5^{6n+4}-2^{2n+4})$ and $3\mid(5^{6n+3}-2^{2n+1})$ and $3\mid(5^{6n+2}-2^{2n+2})$ and $3\mid(5^{6n+1}-2^{2n+1})$.

Body:
By binomial theorem$$5^{6n+6}-2^{2n+2}$$
$$={(3+2)}^{6n+6}-2^{2n+2}$$
$$=3k+2^{6n+6}-2^{2n+2}$$
$$=3k+2^{2n+2}({(3-1)}^{4n+4}-1)=3k+2^{2n+2}(3m)$$


问:
How can I minimize $\|A*w\|$ for all w by minimizing $\|A\|$? (dimension of $A: n\times n$, dimension of $w: n\times1$)
Question: Could I use problem 2 instead of problem 1? In other words, do problem 1 and problem 2 result in a similar optimal value for matrix K?

Problem 1: minimizing norm-2 of $(A+B*K)*(w)$ for all possible $w$; decision variable: $K$
Problem 2: minimizing matrix norm of $(A+B*K)$; decision variable: $K$

Symbols:
There are two matrices $A$ and $B$ with known real values; dimension of $A$: $n$ by $n$, the dimension of B: n by 1
$K$ is a matrix which is not known and in fact, it is a variable; dimension of $K$: $1$ by $n$
w is a vector with real values which belongs to the compact and closed set W; dimension of $w$: n by $1$

My sense: Could I benefit from the following property to prove the equality of problem 1 and problem 2?
$$\|(A+B*K)*(w)\|\le\|A+B*K\|*\|w\|$$

答:
I think some constraints should be added, such as $\|w\|=1$, otherwise, the optimal solution must be $w=0$, and $\|(A+BK)w\|=0$.

$\min\limits_{\|w\|=1}\|(A+BK)w\|^2=\min\limits_{\|w\|=1} w^T(A+BK)^T(A+BK)w$

So, the minimal is $\lambda_{\min}(A+BK)^T(A+BK)$, where $\lambda_{\min}$ represents the minimal eigenvalue.

Indeed, problem 1 is equivalent to minimizing the smallest eigenvalue, which is different from minimizing the norm.


问:
When multiplication is defined on objects other than the real numbers, is there an attempt to define this operation in terms of addition?
I do not understand "abstract" multiplication. If for integers it is repeated addition (in n*k, n times where n is an integer), then what could it even mean for something as simple as multiplying two irrational numbers? Or for more exotic things like vectors? Why for example, is vector addition called addition rather than vector multiplication and vice versa? Perhaps this is not the best expressed question but I hope people know what I am asking.</p>

答:
This is, in a way, more about the language of math than about math itself. But language also carries a lot of mathematical meaning. In general, we have many different algebraic structures with two binary operations defined on them. One of those is usually commutative and serves as the "base" operation. We call that one a sum, or addition, and write it as $+$ or some variation thereof like $\oplus$. Then there is another operation which plays nicely with the base operation $+$. Playing nicely means that it distributes over $+$. We call that one a product, or multiplication, and write it in a variety of different ways ($\cdot,\times,(\cdot,\cdot),\langle\cdot,\cdot\rangle$, etc.). The fact that it distributes over addition is a direct generalization of the "repeated addition" thing in the integers. The fact that multiplication distributes over addition means that $n\cdot m=(1+\dots+1)\cdot m=m+\dots+m$. Distributivity is a way to enforce this "repeated addition" concept by saying that if $n$ can be obtained by repeatedly adding $1$, then $nm$ can be obtained by repeatedly adding $m$. But the distributive property holds for numbers which can't be obtained by repeatedly adding $1$, so it's a generalization. Though we could still say that if $a$ can be obtained by repeatedly adding $b$ (so $a=b+\dots+b$), then $ax$ can be obtained by repeatedly adding $bx$. Of course, we have to define what $bx$ is, but in a way, the distributive property still enforces a certain concept of repeated addition.

Anyway, now that distributivity over addition has been identified as a core feature of products, we tend to call everything that distributes over a commutative, associative binary operation ($+$) a product or multiplication. The product of a field distributes over the addition of the field, the scalar multiplication of a vector space distributes over its addition, the inner product and the vector products do, too. Matrix products distribute over matrix sums. And so on.


Body:
\begin{align}
5^{6n+6}-2^{2n+2} &\equiv 2^{6n+6}-2^{2n+2} \pmod{3} \\
&\equiv  2^{2n+2}(2^{(2n+2)2}-1) \pmod{3}\\
&\equiv2^{2n+2}(1-1) \pmod{3}\\
&\equiv 0 \pmod{3}
\end{align}

Notice that I used  $\gcd(3, 2^{2n+2})=1$ and Fermat's theorem to conclude that $2^{(2n+2)2}\equiv1 \pmod{3}$.


问:
Faithful functor preserves monics
I've been trying to solve a problem and it all reduces to prove that not all faithfull functors preserves monomorphisms. But I haven't been able to come up with an example.
I was thinking about using the fact that $$f\in Hom(C) \text{ is monic}\iff f^{op}\in Hom(C^{op}) \text{ is epic}$$
Then if I have for example
$$f \in Hom(\textbf{Set}) \text{ s.t. } f \text{ is monic but not epic} \Rightarrow f^{op}\in Hom(\textbf{Set}^{op}) \text{ is epic but not monic}$$
So if I consider the functor that sends a category into its opposite then that would do the work.

Is my reasoning correct?
If not can you help me with an especific example?

答: A simple counterexample to the claim that all faithful functors preserve mononmorphisms is to consider functors from the category $$\bullet\to\bullet$$ with only one non-identity arrow. The only non-trivial arrow is a monomorphism, but you can faithfully map it onto any non-monomorphism in any category.

Title: Faithful functor preserves monics


问:
Invertible Matrix Proof by Given that Other is invertible
Given that $A$ is antisymmetric matrix, and $(I + A)$ it is invertible.

I had to prove that $(I - A)$ is also invertible, and I managed to do so,
But now given that $$C = (I - A)(I + A)^{-1}$$
I have to prove that $$C^\intercal C = I$$

Without determinants.

I tried many different ways, First tried to prove that C is invertible and to find it's inverse matrix, also tried to find the inverse of the transposed matrix, I have no clue how to continue...

Thanks.

答:
Lemma: If $AB=I$, then $BA=I$. see here

The antisymmetric matrix $A$ means $A^T=-A$.

We first check the value of $CC^T$:
\begin{equation}
\begin{aligned}
CC^T&=(I-A)(I+A)^{-1}(I+A^T)^{-1}(I-A^T)\\
&=(I-A)(I+A)^{-1}(I-A)^{-1}(I+A)\\
&=(I-A)(I-A^2)^{-1}(I+A)
\end{aligned}
\end{equation}

It follows from $(I+A)(I-A)(I-A^2)^{-1}=(I-A^2)(I-A^2)^{-1}=I$ and the above lemma that$$(I-A)(I-A^2)^{-1}(I+A)=I.$$

Therefore, $CC^T=I$. Furthermore, $C^TC=I$.


问:
A relation between gcds and determinants
While working on a proof, I stumbled on the following fact.

Let $M:=(x_{i,j})_{i,j}\in \mathrm {GL}_{n,d}(\mathbb Z)$. Let's denote by $\alpha_i:=\gcd(x_{1,i},\ldots,x_{n,i})$ for all $i\in \{1,\ldots,d\}$.

Let's also define $N:=\binom nd$ and $\eta_1,\ldots,\eta_N$ be all the minor of $M$ of size $d\times d$.

Then

$$\alpha_1\cdots\alpha_d=\gcd(\eta_1,\ldots,\eta_N).$$

What I tried.

Let $k$ be an integer.

Assume $k$ divides $\alpha_1\cdots\alpha_d$. Let $j\in \{1,\ldots,N\}$. Then $\eta_j$ is a minor of the matrix $M$ of size $d\times d$, extracted by taking only the lines $i_1,\ldots,i_d$. So we have

$$\eta_j=\sum_{\sigma\in \mathfrak S_d}\epsilon(\sigma)\prod_{\ell=1}^d x_{i_\ell,\sigma(\ell)}.$$

But $k\mid \alpha_1\cdots\alpha_d$, so for all $d$-uple $(i_1<\cdots<i_d)$, we have

$$k\mid x_{i_1,1}\cdots x_{i_d,d}= x_{i_1,\sigma(1)}\cdots x_{i_d,\sigma(d)}$$

for all permutation $\sigma\in\mathfrak S_d$. So $k$ divides $\eta_j$ for all $j$, so

$$ k\mid \gcd(\eta_1,\ldots,\eta_N).$$

Reciprocally, I am unable to do it.

Any help would be much appreciated.

答:
This statement seems to be very connected to the theorem in this recent publication:

http://nntdm.net/volume-26-2020/number-3/5-7/
"Equalities between greatest common divisors involving three coprime pairs"

that goes as:

For $a_i$ and $b_i$ positive integers such that $\gcd(a_i, b_i) = 1$ for $i ∈ \{1, 2, 3\}$ and $d_{ij} = |a_ib_j − a_jb_i|$, then $\gcd(d_{32}, d_{31}) = \gcd(d_{32}, d_{21}) = \gcd(d_{31}, d_{21})$.

Basically $a_i$ and $b_i$ can form a 3x2 matrix (your $x$)  and $d_{ij}$ are its minors (your $\eta$). I think this can be helpful.


问:
Is the image of a continuous functions always a topological manifold
Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ be continuous.  Then, for some sufficiently large $M\geq m$, can we always guarantee that $f(\mathbb{R}^n)$ is locally-homeomorphic to some $\mathbb{R}^N$, where $N\leq n$?

My intuition is no, since I can imagine a curve crossing itself; but I can't write down an explicit example.

答:
Another counterexample with a different issue: Let
$$f:\mathbb R^2\to\mathbb R^2,(x,y)\mapsto\begin{cases}(x,y)&x<0\\ (x,(1-x)y)&x\in[0,1]\\(x,0)&x>1\end{cases}$$
For $x<1$, the image is locally homeomorphic to $\mathbb R^2$. For $x>1$, it is locally homeomorphic to $\mathbb R$. At the boundary of both regions, $x=1$, the image can't be locally homeomorphic to either.


问:
Proof that matrix multiplication is continuous
I would like to show that the Circle group $SO(2)$ is a topological group under multiplication. The requirements are

$\alpha$) The group operation is continuous $m:G \times G \rightarrow G $

$\beta$) The inverse is continuous: $x \rightarrow x^{-1}$ is continuous

I'm aware that the group is invertible and if I would think of the $SO(2)$ group as a group of angles rather than a group of matrices, it seems obvious that the group is continuous. However, I would like to show that the group is continuous under multiplication for a set of matrices of the form $ \left( \begin{array}{rr} a & b \\ -b & a\end{array}\right)$.

How can one show continuity for a matrix?

答: You can write matrix multiplication as polynomials in the entries, and those are continuous. For the inverse, you normally also need the fact that $\rm det$ is continuous (and, well, nonzero on the thing you're looking at) but in this case perhaps you don't since it's $1$.


问:
Trying to show that $E[1/X] = 1/E(X)$ is not true in general.
I know that one can use Jensens inequality and such, however I'm wondering if there is a more elementary proof not involving measure theory and Lebesgue-integration. For example something like this:

If we use LOTUS for the discrete case,
\begin{align}
    \mathbb{E}[g(X)] = \sum_{x}g(x)f_X(x),
\end{align}
for $g(X)=1/X$ we obtain
\begin{align}
    \mathbb{E}\left[\frac{1}{X}\right]&=\sum_{x}\frac{f_X(x)}{x}\\
    \frac{1}{\mathbb{E}[X]} &= \frac{1}{\sum_{x}xf_X(x)}.
\end{align}
Setting the two expressions above equal we obtain the equation
\begin{align}
    \left(\sum_x\frac{f_X(x)}{x}\right)\left(\sum_xxf_X(x)\right) = \sum_xf_X(x)^2 = 1.
\end{align}

So the sum of the pmf squared equals to 1. Is there any conclusion we can draw from here?

答:
You could continue by writing

$$ 1 = \sum_x f_X(x)^2 = \left(\sum_x f_X(x)\right)^2 - 2\sum_{x\neq y}f_X(x)f_X(y)$$

And since $\sum_x f_X(x)=1$, you find that it must be

$$\sum_{x\neq y}f_X(x)f_X(y) = 0$$

Since $f_X(x) \ge 0$ for every $x$, the only way for this to be true is that all the terms in the sum are 0. Now let's assume that there are 2 points $x_1$ and $x_2$ where $f_X$ is strictly greater than $0$. This would mean that
$$\sum_{x\neq y}f_X(x)f_X(y) \ge f_X(x_1)f_X(x_2) > 0$$ which is a contradiction. So we know that the pdf can be non-zero in at most one point $x_0$. But it also have to sum to 1, so there is precisely one point where the pdf is non zero, and in that point ($x_0$) it is equal to 1.

This implies that the random variable is a constant and equal to $x_0$.


问:
Proving a parabola and polynomial intersect exactly 3 times
the equations are : $P(x)=5-8(x-1)^2$ for the parabola and the other equation is $E(x)=(x-1)^5$ we need to prove that they intersect exactly 3 times.

What I did was compare them at first  $5-8(x-1)^2 = (x-1)^5 $ and then make a new function $f(x)$ according to that we get $f(x)=(x-1)^5 -5+8(x-1)^2$and the function is continuous because its a polynomial and we can also have a derivative.
I tried applying some numbers and i got $f(-1)=-5$ , $f(0)=2$ , $f(-5)=-7493$ , $f(2)=4$  so according to the Intermediate value theorem we have at least 3 roots but what we need to prove is that it has exactly 3 roots. and this looked like rolle's theorem

What I tried then was we want to show that it has exactly 3 roots according to Rolle's theorem so if the original equation has 4 roots the first derivative needs to have exactly 3

$f'(x)=5(x-1)^4 +16(x-1)$

And I got stuck here I could not figure a way to continue from here ..

Thanks for any tips and help all is appreciated and I would also like to know if anyone can confirm that what I did was right until the last step.

Sorry for my english mistakes hope it is understandable.

答:
shifting the origin :$x-1=t$ we have to find the number of roots of $$f(t)=t^5-8t^2+5=0=(t-a)(t-b)(t-c)(t-d)(t-e)$$ here $a,b,c,d,e$ are its roots.Also one can evaluate $$a^2+b^2+c^2+d^2+e^2=0$$
which is not possible unless  atleast one is complex or all are zero(which in this case is impossible)  ; WLOG $e$ be complex then by $a+b+c+d+e=0$  we must have some other root must alos be complex hence $f(t)$ has minimum of of two complex roots, you have already found 3 real roots so...

by comparing coefficients:
$s=a+b+c+d+e=0$
$$w=ab+ac+ad+ae+bc+bd+be+cd+ce++de=0$$
also $$s^2-2w=a^2+b^2+c^2+d^2+e^2$$
hence $a^2+b^2+c^2+d^2+e^2=0$


问:
How do I solve a nonlinear function with Newton-Raphson assuming that my variable is discrete?
Let's say I want to approximate $\sqrt 2$.

I, therefore, write my function
$$
f(x) = x^2-2 = 0
$$

I define $x$ to be a discrete variable which can take values $\color{brown}{0, 0.1, 0.2, 0.3, 0.4, 0.5}$, etc.

Is there a Newton-Raphson method for solving this?

Kindly also explain the solution procedure by working out the problem.

Regards.

答:
Solve the equation with infinitely small step interval using Newton-Raphson etc. and choose the root making smaller residue with finite interval.

If step is $h=0.1$ then

$$|(1.4^2 - 2)|< ( 1.5^2 - 2)|, $$

and so the nearest root is $x=1.4$;


问:
Equation related to the curve $x^4+3kx^3+6x^2+5$
The sum of all the integral value(s) of 'k' so that the curve $x^4+3kx^3+6x^2+5$ is not situated below any of its tangent line is

(A) 5

(B) 2

(C) 0

(D) -2

I have no idea of how to proceed this question

答:
The curve $y=x^4 + 3 k x^3 + 6 x^2 + 5$ is concave up if $y''>0$ that is

$$6 \left(2x^2+3 k x+2\right)>0 ;\;\forall x\in\mathbb{R}$$
this happens when the discriminant $(9 k^2-16)$ is negative, that is when $-\frac{4}{3}<k<\frac{4}{3}$ and the sum of the integer values in this interval is $0$.


Title: How do I solve a nonlinear function with Newton-Raphson assuming that my variable is discrete?

Body:
Let's say I want to approximate $\sqrt 2$.

I, therefore, write my function
$$
f(x) = x^2-2 = 0
$$

I define $x$ to be a discrete variable which can take values $\color{brown}{0, 0.1, 0.2, 0.3, 0.4, 0.5}$, etc.

Is there a Newton-Raphson method for solving this?

Kindly also explain the solution procedure by working out the problem.

Regards.

Body:
The derivative of $f(x)$ is $f'(x)=2x$.

Let $x_0=0.1$.

step 1: $x_1=x_0-\frac{f(x_0)}{f'(x_0)}=0.1-\frac{-1.99}{0.2}=10.05$

step 2: $x_2=x_1-\frac{f(x_1)}{f'(x_1)}=10.05-\frac{99.0025}{20.1}\approx5.1245$

step 3: $x_3=x_2-\frac{f(x_2)}{f'(x_2)}=5.1245-\frac{24.2605}{10.249}\approx2.75739$

step 4: $x_4=x_3-\frac{f(x_3)}{f'(x_3)}=2.75739-\frac{5.6032}{5.51478}\approx1.74136$

step 5: $x_5=x_4-\frac{f(x_4)}{f'(x_4)}=1.74136-\frac{1.03233}{3.48272}\approx1.44494$

step 6: $x_6=x_5-\frac{f(x_5)}{f'(x_5)}=1.44494-\frac{0.0878516}{2.88988}\approx1.41454$

step 7: $x_7=x_6-\frac{f(x_6)}{f'(x_6)}=1.41454-\frac{0.000923412}{2.82908}\approx1.41421$

step 8: $x_8=x_7-\frac{f(x_7)}{f'(x_7)}=1.41421-\frac{-0.0000100759}{2.82842}\approx1.41421$

Therefore $x\approx 1.41421$.


问:
Counter-example to this property in euclidean geometry?
I have to prove that this property is false, finding a counter-example.

If $\widehat{A}=\widehat{C}=90^{\circ}$ then the convex quadrilateral $ABCD$ is a rectangle.

I can't find any counter-example with this measure of angle because this property seems to be true. I took a right trapezoid $ABCD$ but it's $\widehat{A}=\widehat{D}=90^{\circ}$...

I'm probably blind...

Thanks in advance !

答: Take two right angles as follows, rotate one to some small angle then make them intersect.


问:
Not defined Or zero
let's say we play a game 9 times and I was asked to find the probability of $F_K$ which means the probability to succeed only at the $k^{th}$ try.

What can I say about $F_K$ for $k<0$ or $k>9$ is it 0 or not defined?

My professor said only one is true but I forgot which one of them and why is that.

答:
The probability that rolling a single standard die produces a $7$ is $0$. More generally, the probability of an impossible event is $0$. It may however depend on local definitions on what an event is in the given context.

In short, we consider a set of outcomes and an event is a subset of the set of outcomes. For a die, we may let $\{1,2,3,4,5,6\}$ be the set of outcomes and the events "the roll is a six" and  "the roll is an odd number" then are the subsets $\{6\}$ and $\{1,3,5\}$, respectively. Is "the roll is a seven" an event? As stated, it certainly is not $\{7\}$ as this is not a subset of $\{1,2,3,4,5,6\}$. But just as "the roll is odd" was interpreted as $\{1,3,5\}$ and not as $\{1,3,5,7,8,11,\ldots\}$, it is then correct to state that "the roll is a seven" is the empty event $\{\}$. As such it has probability $0$.

Alternatively, we could allow $7$ (and other values) as outcomes but with individual probability $0$. The result would not change - "roll is seven" now becomes the event $\{7\}$ but still has probability $0$.

With this philosophy in mind, the event "succeed only at $k$th try" for some impossible $k$ (negative, too big, not even an integer, ...) is an impossible event and hence has the (defined) probability $0$.

Title: Not defined Or zero


问:
Irreduciblity of polynomial in $\Bbb Q[x]$
I am trying to prove that $x^5 + 16$ is irreducible in $\Bbb Q[x]$. (Hint: Consider $(x-1)^5 + 16$)

Not sure how to use the hint, thanks for any help in advance.

答: Second hint: $x^5+16$ is irreducible over $\Bbb Q$ if, (and only if), $(x-1)^5+16$ is irreducible over $\Bbb Q$.


答:
Let's assume four points on a circle such that a pair of them are endpoints of a diameter and other vertices are on different part of semi circles.

These four points can determine infinitely many quadrilaterals which are not rectangles

Title: Counter-example to this property in euclidean geometry?

Body:
I have to prove that this property is false, finding a counter-example.

If $\widehat{A}=\widehat{C}=90^{\circ}$ then the convex quadrilateral $ABCD$ is a rectangle.

I can't find any counter-example with this measure of angle because this property seems to be true. I took a right trapezoid $ABCD$ but it's $\widehat{A}=\widehat{D}=90^{\circ}$...

I'm probably blind...

Thanks in advance !


问:
Calculate the volume enclosed by the paraboloid $z=x^2+y^2$ and the plane $z=2y$.

Calculate the volume enclosed by the paraboloid $z=x^2+y^2$ and the plane $z=2y$.

When $x = 0$ it seems that the region of interest lies between $y=2$ and $z=4$. If I'm not mistaken I would need to calculate $$\int_D(x^2+y^2-2y) \ dA.$$

Also $D$ seems to be given by $D=\{(x,y) \in \mathbf{R}^2 \mid x^2+y^2 \le 2y \}$

My question is how can I found the bounds for the integrals? Any hints would be appreciated.

答:
You found the intersection correctly. Here is a 2D cross section of how it would look when seen from the side on $YZ$ plane. It is bound below by the paraboloid $z = x^2+y^2$ and above by the
plane $z = 2y$,

From $z = x^2 + y^2, z = 2y,$ you get $x^2 = 2y-y^2 \implies x = \pm \sqrt{2y-y^2}$

So your integral becomes,

$\displaystyle \int_{0}^{2}\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}(2y - x^2 - y^2) dx \, dy$

In cylindrical coordinates,

$x = r \cos \theta, y = r \sin \theta$

So $z = x^2 + y^2 =r ^2 = 2y = 2r \sin \theta \implies r = 2 \sin \theta$

So your integral becomes,

$\displaystyle \int_{0}^{\pi}\int_{0}^{2\sin\theta}(2r\sin \theta - r^2) \, r \, dr \, d\theta$


问:
Can the order of inverse limit and direct sum be changed for the symmetric polynomial rings?
I am reading the book Symmetric Functions and Orthogonal Polynomials by I.G. Macdonald. On page 1 of chapter 1, the book defines the ring of symmetric functions as
$$\Lambda=\oplus_{r\ge0}\lim_{\overleftarrow{n}}\Lambda_n^r,$$
where $\Lambda_n^r$ denotes the additive group of homogeneous symmetric polynomials of degree $r$ in $x_1,\cdots,x_n$, and the $\displaystyle\lim_{\leftarrow}$ denotes the inverse limit.

I don't quite sure if the order of direct sum and inverse limit really matters here. And I guess it does no effect with the order changing. However, I have no idea how to prove or disprove.

All in all, my question is whether the equality
$$\oplus_{r\ge0}\lim_{\overleftarrow{n}}\Lambda_n^r=\lim_{\overleftarrow{n}}\oplus_{r\ge0}\Lambda_n^r$$
holds with the notation declared above. Could anyone help?

答:
It does actually matter. In the RHS ring, the degree can increase with $n$, for instance there is an element that deserves to be called $\sum_n x_n^n$, whereas the same cannot be said on the LHS.

An element of the LHS has a maximum degree, and all the monomials appearing in the series representing the element have to have degree bounded by that single degree, which is not the case on the RHS, where the only requirement is that if you fix $n$, then the monomials involving variables $x_i, i\leq n$ have a maximal degree.

Making that argument precise shows that the canonical map LHS $\to$ RHS is not surjective


问:
Determine necessary and sufficient conditions on $x, N$ so that the following holds: for any $a, b$, if $ax ≡ bx \pmod N$, then $a ≡ b\pmod N$
Been working on it for a couple of hours with no success. My intuition tells me that it has to do something about $\gcd(n,x) = 1$
I have managed to prove that if $\gcd(x,n) = 1$ the claim holds, now I try to prove the other way around.
Any hint or a solution will help. thanks!

答:
You're right, this holds if and only if $\gcd(n, x) =1$.

So, as you've already proved the first part, let's show the second one.

Assume for all $a, b \in \mathbb{Z}, ax\equiv bx \mod n \implies a \equiv b \mod n$.

Let $d := \gcd(x, n)$ and $(x', n') := (\frac{x}{d}, \frac{n}{d}) \in \mathbb{Z}^2$.

Take $(a, b) = (n', 0)$. One has $xn' = x'n'd = x'n \equiv 0 = x 0 \mod n$.

Then one should have $n' \equiv 0 \mod n$, which means $n | n'$. As one also has $n' | n$, this implies $|n'| = |n|$ thus $d = \gcd(n, x) = 1$.


问:
Interpret "the sum of all natural numbers equal $-1/12$" without complex analysis skills
is it possible to show if there is a link between the sum of all natural numbers and the value $ - \frac {1} {12} $, through simple real analysis intuitions?

In some contexts of physics this result seems to be applyed in agreement with the experimental results. is it possible to give some examples in detail and explain if there are contexts in which $ 1 + 2 + 3 + \dots = - \frac {1} {12} $?

答:
The $\underline{\text{Riemann zeta}}$ $ \zeta (s) $ function is defined in the region $ \Re (s)> 1 $ by the convergent series
$$ \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^{s}}=1+\frac{1}{2^s}+\frac{1}{3^s}+\dots \tag{1} $$

For $ \Re (s) \le 1 $, the series is no longer convergent, however $ \zeta $ can be extended into this region via $\underline{\text{analytic continuation}}$. This allows us to assign a value to $ (1) $ even when it would formally diverge. For example, after analytic continuation, we have $ \zeta (0) = - 1/2 $ and $ \zeta (-1) = - 1/12 $.

We will therefore show the link between the value obtained from the analytic continuation of $ \zeta $ and the limit $ N \to \infty $ of $ \sum_ {n = 1} ^ N 1 / n^s $. We will then clarify why the statement "the sum of all natural numbers equals $ -1 / 12 $" is wrong and why this misunderstanding exists.
$$ 1 + 2 + 3 + \dots \neq- \frac {1} {12} = \zeta (-1). $$

Classically a value is assigned to "infinite sums" (converging or diverging) as the result of an arithmetic operation applied infinite times. A value obtained after analytical continuation is instead linked to a complex analysis operation that cannot be obtained from the repetition of arithmetic operations on natural numbers. The values obtained through the infinite application of arithmetic operations and the values obtained through analytical continuation are however linked and in some cases may coincide. For this we will make the example of the $ \underline {\text {Casimir effect}} $.

$ \ $

$$\underline{\text{Basic knowledge,  informally expressed}}$$

Indicator Function The indicator function of a subset $ A $  of $ X $ is a function
$$ \mathbb{1}_A:X\to \{0,1\} $$
defined as
$$ \mathbb{1}_A(x)  =
\begin{cases}
1  & \text{if }x \in A \\
0 & \text{if } x \notin A
\end{cases} $$
Eg. $\mathbb{1}_{[0,1]}(x)$ is a function that returns $ 1 $ for all $ 0 \le x \le 1 $ otherwise $ 0 $.

Cutoff function A cutoff function is defined as a $ \eta: \mathbb {R} \to \mathbb {R} $ function such that
$$\eta \in C^{\infty}(\mathbb{R}) \ \ \ \ \eta=\begin{cases}
1  & \text{if }|x|\le 1 \\
0\le \eta(x) \le 1 & \text{elsewhere} \\
0 & \text{if } x \ge 1+\epsilon
\end{cases}$$

For our purposes the $ \eta (x) $ function is "smoother" indicator function. $ \eta (x) $ is a $\underline {\text{Schwartz function}}$.

Example to compare an indicator function (blue) with a cutoff function (red)

$ \ $

Schwartz function A  function $ f\in C^{\infty} ( \mathbb{R}^n) $ is a Schwartz function if it goes to zero when $ | x | \to \infty $ faster than any inverse power of $ x $ ($ x ^ {1 / n} $), as well as its derivatives.

That is: $ f \in \mathcal {S} (\mathbb {R} ^ n) $ if there exists a constant $ C _ {\alpha, \beta} $ such that
$$\underset{x \in \mathbb{R}^n }{\sup}|x^{\alpha}\partial^{\beta}f(x)|\le C_{\alpha, \beta} $$
where $ \alpha, \beta $ are $\underline {\text{multi-index}}$.

Fourier transform The Fourier transform of a Schwartz function is still a Schwartz function denoted by $ \hat {f} (t) $ defined as
$$ \hat{f}(t) = \frac{1}{2 \pi}\int_{\mathbb{R}} f(x) e^{ i t x} dx$$
by the $\underline {\text{Fourier inversion theorem}}$
$$ f(x) = \int_{\mathbb{R}} \hat{f}(t) e^{- i t x} dt.$$

Holomorphic and meromorphic functions A holomorphic function is a complex function that can be differentiated (in a complex sense) at any point of the set on which it is defined.
A meromorphic function is a holomorphic function on the whole set on which it is defined with the exception of a set of isolated points which are the $\underline {\text{poles}}$ of the function itself.

Pole The pole of a meromorphic function $ f (z) $ is an isolated singularity $ z_0 $ such that
$$ \lim_{z \to z_0} |f(z)|= +\infty.$$

The $\underline {\text{order of the pole}}$ is a natural number $ m> 0 $ such that if $ z_0 $ is a pole, the limit
$$ \lim_{z \to z_0} f(z)(z-z_0)^m $$

exists, is finite and is not equal to 0. If $ m = 1 $, $ z_0 $ is called $\underline{\text{simple pole}}$.

Residue The residue of a meromorphic function $ f (z) $ at the pole $ z_0 $, denoted by $ \text {Res} _ {z_0} (f) $, is that only value $ a_ {z_0 } $ such that $ f (z) -a_ {z_0} / (z-z_0) $ has analytic derivative in $ 0 <| z-z_0 | <\ delta $.

The residual $ a_ {z_0} $ with $ z_0 $ of order $ m $ can be calculated as

$$ \text{Res}_{z_0}(f)=\frac{1}{(m-1)!} \ \lim_{z \to z_0} \frac{d^{m-1}}{d x^{m-1}} [f(z)(z-z_0)^m]. $$
If $ z_0 $ is a simple pole we have
$$ \text{Res}_{z_0}(f)= \lim_{z \to z_0}f(z)(z-z_0). $$

Residue theorem Let $ U \in \mathbb {C} $ be a simply connected open set (without "holes"), let $ f: U \to \mathbb {C} $ be a meromorphic function. Let $ \gamma: [a, b] \to U $ be a closed and regular curve and $ A = \{z_0, z_1, z_2, \dots \} $ the set of poles of $ f $ contained within the contour of $ \gamma $
$$ \int_{\gamma} f(s)ds =2 \pi i \sum_{z_j \in A}\text{Res}_{z_j}(f) . $$

This theorem tells us that the value of a contour integral, for any contour in the complex plane, corresponds to the sum of the residue of $ f $ at the poles inside the boundary of $ \gamma $.

$ \ $

$ \ $

$$\underline{\text{Sugarcoat it}}$$
Let's start by considering the function
$$ g(N):= \sum_{n=1}^N \frac{1}{n^2}=1+\frac{1}{4}+\frac{1}{9}+\dots+\frac{1}{N^2}.  $$
Estimating the limit of $ (2) $  is a famous problem known as the $\underline {\text{Basel problem}}$ which was solved by Euler, who shows that $ g (N) $ converges to $ \pi ^2/6 $ when $ N \to \infty $, or in other words
$$ g(N)=\frac{\pi^2}{6}+o(1) $$
where $ o(1) $ denotes a quantity that goes to zero when $ N \to \infty $. Using the $\underline {\text{integral test}}$ we obtain a more precise result
$$ \sum_{n=N+1}^{\infty}\frac{1}{n^2} \le \int_N^{\infty} \frac{dx}{x^2} =\frac{1}{N} \Rightarrow  \sum_{n=1}^{N}\frac{1}{n^2}= \frac{\pi^2}{6}+O(\frac{1}{N}) $$
where $ O (1 / N) $ denotes a quantity that goes to infinity such as $ 1 / N $. Thus we can interpreted $ \pi^ 2/6 $ as the degree zero coefficient of the asymptotic series expansion of $ g (N) $.

$ \ $

All problems arise here, due to the discrete nature of the sum

$$ \sum_{n=1}^N \frac{1}{n^s} \tag{2} $$

which has discontinuities of the first kind (jump) for each positive integer value of $ N $.
Such discontinuities produce various artifacts when trying to expand $ (2) $, generating a polynomial in $ N $, to evaluate its trend. This also occurs in the Basel problem, but in that case the artifacts are included in the $ O (1 / N) $ error, while for $ \Re (s) \le 1$ they become large enough to cause problems. Therefore, for $ \Re (s) <1 $, to set $ (2) $ equal to a constant coefficient plus an error that tends to zero, evidently it is no longer sufficient

It is possible to solve this problem, that is to find the missing terms to the expansion of $ (2) $ when $ \Re (s) <1 $, by intervening on the discontinuous nature of $ (2) $.

We can rewrite $ (2) $ as
$$ \sum_{n=1}^{\infty}\frac{1}{n^s} \mathbb{1}_{[0,1]}\big(\frac{n}{N}\big) $$
in fact when $ 0 \le n \le N $ we have $ \mathbb {1} _ {[0,1]} \big(n / N \big) = 1 $ while for $ n> N $ we will have $ \mathbb {1} _ {[0,1]} \big(n / N \big) = 0 $.

Using a cut off function $ \eta $ it is possible to obtain a "sweeter" version of $ (2) $ while preserving its asymptotic behavior.
$$ \sum_{n=1}^{\infty}\frac{1}{n^s}\eta\big(\frac{n}{N}\big) \tag{3} $$
We will now study $ (3) $ to obtain the expansion of $ (2) $ and show how the coefficient of lower degree is related to $ \zeta (s) $. We could therefore say that for $ N \to \infty $ and $ \Re (s <1) $ it is not possible to equal $ \sum 1 / n^ 2 $ to $ \zeta (s) + O (1 / N) $ which instead is the definition $ (1) $ in the region $ \Re (s> 1) $.

$ \ $

$ \ $

$$\underline{\text{Do the math}}$$

We show how - using Fourier analysis and complex analysis - it is possible to determine the asymptotic behavior of $ \sum \frac {1} {n ^ 2} $ thanks to the meromorphic properties of the $ \zeta $ function.

$ \ $

Let $ s $ be a complex number such that $ \Re (s) <1 $, and $ \eta: \mathbb {R}^{+} \to \mathbb {R} $ a cutoff function that returns one near origin, and consider the expression
$$ \sum_ {n = 1} ^ {\infty} \frac {1} {n ^ s} \eta \big( \frac {n} {N} \big) \tag {3} $$
where $ N $ is a large positive integer. Then let $ A> 0 $ be an integer such that $ \Re (s + A)> 1 $ e $ f_A (x): = e^ {A x} \eta (e ^ x) $, we can rewrite $ (3 ) $ like
$$ N ^ A \sum_ {n = 1} ^ {\infty} \frac {1} {n ^ {s + A}} f_A (\ln (n / N)). $$

Note that $ \eta (x) $ is a Schwartz function whose behavior "dominates" any polynomial, than $ x ^ A \eta (x) $ is still a Schwartz function, so $ f_A (x) \in \mathcal {S} (\mathbb {R}) $. From the Fourier inversion formula, it has representation

$$f_A(x)=\int_{-\infty}^{+\infty}\hat{f}_A(t)e^{-i x t}dt $$
where
$$ \hat{f}_A(t):= \frac{1}{2 \pi}\int_{-\infty}^{+\infty}f_A(x)e^{ixt}dx$$
by definition $ \hat {f}_A (x) $ is also a Schwartz function. We rewrite $ (3) $ as
$$ N^A \sum_{n=1}^{\infty}\frac{1}{n^{s+A}}\int_{-\infty}^{+\infty}\hat{f}_A(t)e^{-i \ln(n/N) t}dt=\int_{-\infty}^{+\infty}\Big(\hat{f}_A(t)N^{A+it} \frac{1}{n^{it}}\sum_{n=1}^{\infty}\frac{1}{n^{s+A}}\Big)dt $$
Since $ \Re (s + A + it)> 1 $ we can use the definition $ \zeta (z) = \sum \frac {1} {n ^ z} $ and write
$$ \sum_{n=1}^{\infty}\frac{1}{n^s}\eta\big(\frac{n}{N}\big)= \int_{-\infty}^{+\infty}\hat{f}_A(t)N^{A+it} \zeta(s+A+it)dt.$$

now we have
$$ \hat{f}_A(x)=\frac{1}{2 \pi}\int_{-\infty}^{+\infty}\eta(e^x)e^{(A+it)x}dx $$
integrating by parts:
$$ \hat{f}_A(x)=\frac{1}{2 \pi}\Big[ \frac{\eta(e^x)e^{(A+it)x}}{A+it} \Big]^{+\infty}_{-\infty}-\frac{1}{2 \pi}\int_{-\infty}^{+\infty}\frac{e^x\eta'(e^x)e^{(A+it)x}}{A+it} $$
having that $ \eta(e^x)(e^x)^c \to 0$ for $ x \to +\infty$ and that $(e^{x})^c\to 0$ for $x \to -\infty$  and $\eta(0)=1$

$$ \hat{f}_A(t)=-\frac{1}{2 \pi(A+it)}\int_{-\infty}^{+\infty}\eta'(e^x)e^{(A+it+1)x}dx=-\frac{1}{2 \pi(A+it)}\int_0^{+\infty}y^{A+it}\eta'(y)dy $$
we define
$$ F(A+it)=\frac{1}{(A+it)}\int_0^{+\infty}y^{A+it}\eta'(y)dy.  \tag{4}$$
We have that $ F (A + it) $ is a function entirely of $ A + it $ and furthermore,   $ \hat {f} _A (t) \in \mathcal {S} (\mathbb {R}) $, $ \hat {f} _A (t) = - 1 / (2 \pi) F (A + it) $ therefore fixed $ A $, $ F (A + it) $ decreases faster than any polynomial when $ t \to \infty. $

We can rewrite $ (3) $ as
$$ -\frac{1}{2 \pi}\int_{-\infty}^{+\infty}F(A+it)N^{A+it} \zeta(s+A+it)dt $$

set $z=s+A+it$
$$ -\frac{1}{2 \pi i}\int_{s+A-i\infty}^{s+A+i\infty}\zeta(z)F(z-s)N^{z-s} dz.  $$
It is known that fixed $ \sigma $, $ \zeta (\ sigma + it) $ has polynomial growth in $ t $ and also that $ \zeta (z) $ has a simple pole at $ z = 1 $ with expansion in that point
$$ \zeta(z)^2=\frac{1}{(z-1)^2}+2 \gamma\frac{1}{z-1}+O(1)\Rightarrow \zeta(z)\sim \frac{1}{z-1} \text{ per } z \to 1 $$
We have from $ (4) $ that $ F (z-s) $ has a simple pole for $ z = s $ so the integrating function $ J (z): = \zeta (z) F (z-s) N ^ {z-s} $ has two simple poles. Let's proceed by calculating the residuals:
$$ \text{Res}_{z=s}[J(z)]=\lim_{z \to s} [J(z) (z-s)]=\zeta(s) \int_0^{+\infty}\eta'(y)dy =\zeta(s)\Big[ \eta(y)\Big]^{+\infty}_0=-\zeta(s) $$

To calculate $ \text{Res}_{z = 1} [J (z)] $ we integrate by parts
$$F(1-s)= \frac{1}{(1-s)}\int_0^{+\infty}y^{1-s}\eta'(y)dy $$

$$F(1-s)= \frac{1}{(1-s)}\Big[\eta(y)y^{1-s} \Big]^{+\infty}_0 -\int_0^{+\infty}y^{-s}\eta(y)dy \tag{5}$$

$ \ Re (s) <1 $ and therefore $ \frac{1}{0^s} = 0 $:
$$F(1-s)= -\int_0^{+\infty}y^{-s}\eta(y)dy$$

the integral of $ y^ {- s} \eta (y) $ is known and is the $\underline{\text{Mellin transform}}$ of $ \eta $, which we will denote by $ \mathcal {M}_{\eta} ( -s) $ and has a non-negative value
$$\text{Res}_{z=1}[J(z)]=\lim_{z \to 1} [J(z)(z-1)]=N^{1-s}F(1-s) =-N^{1-s}\mathcal{M}_{\eta}(-s).$$

From the residue theorem we have that if $ \gamma $ is a contour of the complex plane that contains the poles $ z_1 = 1 $ and $ z_2 = s $ of $ J (z) $, we can evaluate
$$ \int_{\gamma} \zeta(z)F(z-s)N^{z-s}dz=2\pi i \Big(-N^{1-s}\mathcal{M}_{\eta}(-s)-\zeta(s)\Big).  $$

Since $ \zeta (\sigma + it) $ has polynomial growth and $ F (\sigma + it) $ decreases faster than any polynomial, $ J (z) $ will go to zero for $ t \to \infty $.

It is therefore possible to choose a $ \gamma $ contour as in the figure, bringing the horizontal paths to infinity, making them negligible. Then applying the residual theorem on $ \gamma $ composed of two lines that include the poles, one from minus infinity to plus infinity and the other from plus infinity to minus infinity. $ \forall B \ | \ \Re (s-B) <\Re (s): $

$$ -\frac{1}{2 \pi i}\Big(\int_{s+A-i\infty}^{s+A+i\infty}J(z) dz+\int_{s-B+i\infty}^{s-B-i\infty}J(z) dz \Big)=\zeta(s)+N^{1-s}\mathcal{M}_{\eta}(-s)$$

$$ -\frac{1}{2 \pi i}\int_{s+A-i\infty}^{s+A+i\infty}J(z) dz=-\frac{1}{2 \pi i}\int_{s-B-i\infty}^{s-B+i\infty}J(z) dz+\zeta(s)+N^{1-s}\mathcal{M}_{\eta}(-s)$$
performing a change of variables

$$ \sum_{n=1}^{\infty}\frac{1}{n^s}\eta\big(\frac{n}{N}\big)=\zeta(s)+N^{1-s}\mathcal{M}_{\eta}(-s)-\frac{N^{-B}}{2 \pi}\int_{-\infty}^{+\infty}\zeta(s-B+i t)F(-B+it)N^{i t}dt \tag{6}$$
by the considerations on the behavior of $ \zeta $ and $ F $, the integral on the right of $ (6) $ converges and has a constant value.

We can finally write for $ \Re (s) <1 $
$$ \sum_{n=1}^{N}\frac{1}{n^s}\eta\big(\frac{n}{N}\big)=\zeta(s)+N^{1-s}\mathcal{M}_{\eta}(-s)+O(N^{-B}) $$
$$ \sum_{n=1}^{N}n=1+2+3+\dots+N=-\frac{1}{12}+N^{2}\mathcal{M}_{\eta}(-1)+O(N^{-B}) $$
We note that for $ \Re (s) <1 $ there is a link between the values obtained by applying algebraic operations $ 1 + 2 + \dots + N $ and the values obtained by analytic continuation $ \zeta (s) $.

We have shown how in this domain it is no longer sufficient to expand the summation - as a polynomial in $ N $ - only to the zero degree term (which can also be negative) but in this case it is necessary to expand up to the second degree term,  obtaining the term $ N ^ {2} \mathcal {M}_{\eta} (- 1) $ (which goes to infinity) and therefore the correct asymptotic trend.
$ \ $

By the $\underline{\text{Euler-Maclaurin formula}}$ it is possible to prove that $ \{\forall s \in \mathbb{C}: \Re (s) <1 \} $, the expression
$$ \sum_ {n = 1} ^ {\infty} \frac {1} {n ^ s} \eta \big ( \frac {n} {N} \big) -N ^ {1-s} \mathcal{M}_ {\eta} (- s) $$
converges smoothly for $ N \to \infty $. Moreover from the $\underline {\text{Morera's theorem}}$ it is possible to prove that this limit is complex-analytic therefore we can define $ \zeta $ in $\Re (s) <1$ region as:

$$\zeta(s)=\lim_{N \to \infty} \sum_{n=1}^{\infty}\frac{1}{n^s}\eta\big(\frac{n}{N}\big)-N^{1-s}\mathcal{M}_{\eta}(-s). \tag{7}$$

$ \ $

$ \ $

$$\underline{\text{The Casimir Effect}}$$

In physics, the Casimir effect consists in the attractive force that is exerted between two extended bodies located in the vacuum, for example two parallel plates, due to the presence of the $\underline {\text{zero-point quantum field}}$. This phenomenon is interesting both because the values obtained through the infinite application of arithmetic operations coincide with the values obtained through analytical continuation and because it offers experimental evidence that the vacuum is not really empty.

Classically to calculate the electromagnetic energy contained in an empty box (of dimension $ L \times L \times a $) with conductive walls at $\underline{\text{absolute zero}}$, we have to sum the energy
$$ E=\frac{1}{2}  \hbar \omega=\frac{1}{2}\hbar c k$$
(obtained from the quantization of the harmonic oscillator at zero point) for all possible harmonic oscillators, i.e. for all wave numbers $ k $ allowed inside the box:
$$E=2\frac{1}{2}\sum_{x=1}^{\infty}\sum_{y=1}^{\infty}\sum_{n=1}^{\infty}\sqrt{\frac{x^2 \pi^2}{L^2}+\frac{y^2 \pi^2}{L^2}+\frac{n^2 \pi^2}{a^2}} $$
for simplicity we have setted $ \hbar = c = 1 $. Assuming that the $ box $ is one-dimensional with length $ a $ we have:
$$\frac{a}{\hbar c \pi}E=\sum_{n=1}^{\infty}n. \tag{8}$$
From the considerations made in the previous paragraph we can conceptually write $ (8) $ as:
$$\frac{a}{\hbar c \pi}E=-\frac{1}{12}+\infty $$
in this context the term $ \infty $ is exactly canceled from the energy density of the $\underline{\text{ground state}}$ which is outside the box, in the region of space called universe.

From this it could conclude that in this context

$$ \sum_{n=1}^{\infty}n=-\frac{1}{12} $$
but so we will neglect the effect that the universe operates on the box which is what leads to the cancellation of the term $ \infty $.

To make this effect explicit, it is possible to define the $\underline {\text{Casimir energy}}$ in a certain region of space as:

"the difference between the allowed wavelengths, in a region bounded by conductive walls, and the allowed wavelengths if the walls weren't there".

In other words:
$$\frac{a}{\hbar c \pi}E= \sum_{n=1}^{\infty}n-\int_0^{\infty}n\ dn \tag{9}$$
we see that to the right side of $ (9) $ you get $ \infty- \infty $ so we must $\underline{\text{regularize}}$:

$$\frac{a}{\hbar c \pi}E=\lim_{\alpha\to0}\bigg[ \sum_{n=1}^{\infty}n e^{-\alpha n}-\int_0^{\infty}n  e^{-\alpha n} dn \bigg]. \tag{10} $$

Regularization is not a mysterious mathematical trick, but it is a physically realistic model that represents the fact that metal is not a good conductor for sufficiently small wavelengths ($ 1 / k $). Also, we performed an exponential regularization, but we would have achieved the same result using any cutoff function. Thus we have that in $ (10) $ the definition of $ \zeta $ expressed in $ (7) $, allowing us to write

$$\frac{a}{\hbar c \pi}E= \zeta(-1)=-\frac{1}{12}.$$ In the three-dimensional case of two parallel plates of unitary area, distant $ a $:

$$\frac{6 a^3}{\hbar c \pi^2}E= \zeta(-3)=\frac{1}{120}.$$
The attraction force will be:
$$ F_c=\frac{d}{d a} E=-\frac{\hbar c \pi^2}{2 a^4}\zeta(-3).$$

Hendrik Casimir predicted this result in 1948. It was only in 1997 that S. Lamoreaux quantitatively measured the force within 5 % of the theoretically predicted value.


问:
How to integrate $\int_0^\infty \frac{1}{1+y^4} dy$
I tried the trigonometric substitution $y^2 = \tan \theta, sec^2\theta = 1 + y^4$

But now I'm stuck with $\frac12 \int \frac{\sqrt{\sin \theta}}{(\cos\theta)^{\frac92} } d \theta$

I ran out of imagination as what to try now

答:
$$\int_0^\infty \frac{dx}{x^4+1} = \frac{1}{2}\int_{-\infty}^\infty \frac{dx}{x^4+1}$$

The latter integral is trivial by means of contour integration.

Let $C$ be the canonical semicircle contour, along the real axis from $-R$ to $-R$ and around the semicircle $Re^{i\theta}$ for $\theta \in [0,\pi]$.  Letting $R \to \infty$, we consider the function on the arc ($|z| = R$):

$$\left|\frac{1}{x^4+1}\right|\le  \frac{1}{|x^4+1|} \le \frac{1}{|x^4|-1} \le \frac{2}{R^4} \to 0$$

so by the estimation lemma, the integral around the arc disappears.

$$\oint_C \frac{dz}{z^4+1} = \int_{-\infty}^\infty \frac{dx}{x^4+1} =2 \pi i \sum \operatorname*{Res}\frac{1}{z^4+1}$$

where the residues are of poles in the upper half plane.  These poles are $z_1=e^{i\pi/4}$ and $z_2=e^{3i\pi/4}$.  It follows that

$$b_1=\operatorname*{Res}_{z=z_1}\frac{1}{z^4+1} = -\frac{1}{4} e^{i\pi/4}$$
$$b_2=\operatorname*{Res}_{z=z_1}\frac{1}{z^4+1} = -\frac{1}{4} e^{3 i\pi/4}$$

Then

$$\int_{-\infty}^\infty \frac{dx}{x^4+1} =2 \pi i (b_1+b_2) = \frac{\pi}{\sqrt{2}}$$

and finally

$$\int_0^\infty \frac{dx}{x^4+1} =  \frac{\pi}{2\sqrt{2}}$$