问:
How to find $\operatorname{P.V.}\int_0^1 \frac{1}{x (1-x)}\arctan \left(\frac{8 x^2-4 x^3+14 x-8}{2 x^4-3 x^3-11 x^2+16 x+16}\right) \textrm{d}x$?
The following problem is proposed by Cornel Valean:$$\operatorname{P.V.} \int_0^1 \frac{1}{x (1-x)}\arctan \left(\frac{8 x^2-4 x^3+14 x-8}{2 x^4-3 x^3-11 x^2+16 x+16}\right) \textrm{d}x=\log \left(\frac{5}{4}\right) \arctan\left(\frac{1}{2}\right).$$

The question is how can we prove this equality without using the main result in this paper and without the use of complex numbers?

答:
Here is a full detailed answer respecting the rules of the OP
as seen in the revision #1:

https://math.stackexchange.com/posts/4007626/revisions

The peculiar added condition of "not using complex numbers" came after some hours.

It is easy to add those $39$ characters without suggesting why or how this should work.

But well, even so, there is certainly a way to "avoid $\Bbb C$" below,
by using a messy language.
All i need is a close form for a primitive function
for $\int\frac 1x\arctan F(x)\; dx$ with $F$ rational. One can write this primitive function
as if there would be no complex numbers on this world.
The reader really willing to do this can do it, just write formulas for
the real and for the imaginary part of the logarithm and dilogarithm. (Then work with them.)
For instance, for
$z=r(\cos\theta+i\sin \theta)\in\Bbb C$ we have explicit real functions
$z\to|z|$ and $z\operatorname{arg}z$ and the formula
$\log (r(\cos\theta+i\sin \theta))=\log r +i\theta$ can be written without
using complex numbers, when we need only $\log z$ and $\log^2 z$, but where is the point
in doing so?!

So i will write these primitives using complex numbers,
since my concern is rather on concise ideas, not on typing to avoid $i$,
and since my hand has to type all the stuff,
in a magnitude highly contrasting the OP in detail and precision.
The OP does not even mention any try, and - by the way - does not satisfy (m)any of the current common guide lines of this site.

The accepted answer also uses $\Im$, so i will do it too. We see numbers
and prove the result as stated. Well, with complex numbers.

Outline:

$(1)$ Use Sophie's splitting, which deserves its own line:
$$
\tag{$1$}
\arctan h(x) = \arctan 2{x^2-x-4}-\arctan\frac{4x}{2^2-x-4}\ .
$$

$(2)$ Get rid of the P.V. by resolving the singularities. For this write $\frac 1{x(1-x)}$
as $\frac 1x+\frac1{1-x}$, for the expressions in $\frac 1{1-x}$ use the substitution
$y=1-y$, and resolve the uniqe singularity that comes from $\frac 1x$.
This leads to an expression of the given integral $J$ in the form
$$
\tag{$2$}
J = 2J_1-J_2-J_3\ ,
$$
where $J_1$ is an integral of $\displaystyle\frac 1x\arctan F(x)$ on $[0,1]$,
and $J_2,J_3$, are similar, replacing the rational function $F$ by other rational functions $G,H$.

$(3)$ Use
$$
\tag{$3$}
\arctan t = \frac 1{2i}\log\frac{1+it}{1-it}
$$
to pass from $\arctan$ to $\log$, which can be further splitted.

$(4)$ Write
$$
\tag{4}
\frac {1+iF(x)}{1-iF(x)}=\frac{a_3a_4}{a_1a_2}\cdot\frac{(x-a_1)(x-a_2)}{(x-a_3)(x-a_4)}\ ,
$$
and split the logarithm. Similarly for $G,H$, where the
values $b_1,b_2,b_3,b_4$ and $c_1,c_2,c_3,c_4$ appear. Then we integrate the pieces using the formula:

$(5)$
$$
\tag{$5$}
\int\frac 1x\log(x-a)\; dx
% = \log(x-a)\log \frac xa
% +\operatorname{Li}_2\left(1-\frac xa\right) + C''
% =
% \log(x-a)\log \frac xa
% -\operatorname{Li}_2\left(\frac xa\right) + C'
% -\log\left(1-\frac xa\right)\log\frac xa
% =
% \log(-a)\log\frac xa -\operatorname{Li}_2\left(\frac xa\right) + C'
=
\log(-a)\log x -\operatorname{Li}_2\left(\frac xa\right) + C
\ .
$$

$(6)$ We have explicit real numbers for the integrals $J_1,J_2,J_3$ so far.
We work rather in $i\Bbb R$, seen as $\Bbb C\cong \Bbb R\oplus i\Bbb R$ taken modulo
the first $\oplus$ component $\Bbb R$.
These integrals are sum of terms involving dilogarithms and products of two logarithms,
calculated in points which are expressions in $a_1,a_2,a_3,a_4$ for $J_1$, and similarly for $J_2,J_3$.
We use dilogarithmic identities to get rid of all dilogaritm values.
The properties of the dilogarithm, allowing to express $\operatorname{Li}_2(1-z)$ and $\operatorname{Li}_2(1/z)$ in terms of $\operatorname{Li}_2 z$,
are useful:

dlmf.nist.gov/25.12 and

mathworld.wolfram.com/Polylogarithm.html and

mathworld.wolfram.com/Dilogarithm.html

$(7)$ We exhibit a simple hidden relation between the "$a,b,c$-values",
which immediately leads to the simplification of the logarithm products.

Let's go.

$(1)$ and $(2)$

Let $J$ be the given expression
$$
J = \lim_{\epsilon\searrow 0}\int_{0+\epsilon}^{1-\epsilon} \frac 1 {x(1-x)}\arctan  \underbrace{\frac{-4x^3 +8 x^2 +14 x-8}{2 x^4-3 x^3-11 x^2+16 x+16} }_{h(x)}  \;dx
\ .
$$
I hate P.V. - so we split $\displaystyle \frac 1{x(1-x)}=\frac 1x+\frac 1{1-x}$
and accordingly the integral into two pieces.
The special values of $h$ in zero and one are  $h(0) = -8/16=-1/2$, $h(1) = 10/20=1/2$. So $h(0)+h(1)=0$, giving
$$
\begin{aligned}
J &=
\lim_{\epsilon\searrow 0}
\int_{0+\epsilon}^{1-\epsilon} \frac 1x(\arctan h(x)-\arctan h(0))\; dx
\\
&\qquad\qquad+
\lim_{\epsilon\searrow 0}
\int_{0+\epsilon}^{1-\epsilon} \frac 1{1-x}(\arctan h(x)- \arctan h(1))\; dx
\\
&=
\int_0^1 \frac 1x(\arctan h(x)-\arctan h(0))\; dx
+
\int_0^1 \frac 1{1-x}(\arctan h(x)-\arctan h(1))\; dx
\\
&=
\int_0^1 \frac 1x(\arctan h(x)-\arctan h(0))\; dx
+
\int_0^1 \frac 1x(\arctan h(1-x)-\arctan h(1))\; dx
\\
&=
2
\underbrace{\int_0^1\frac 1x\arctan \frac {x^2-x}{2(x^2-x-5)}\; dx}_{J_1}
\\
&\qquad\qquad-
\underbrace{\int_0^1\frac 1x\arctan \frac {4x}{2x^2-x-4}\; dx}_{J_2}
\\
&\qquad\qquad\qquad\qquad-
\underbrace{\int_0^1\frac 1x\arctan \frac {8x^2-24x}{6x^2+7x-25}\; dx}_{J_3}
\ .
\end{aligned}
$$
At the last step we have used first
Sophie's splitting, which deserves again its own line:
$$
\arctan h(x) = \arctan 2{x^2-x-4}-\arctan\frac{4x}{2^2-x-4}\ ,
$$
to undo the chicanery in the expression of the integral.

(My interest in this problem came from the explicit computation of the separated integrals $J_1$, $J_2$, $J_3$.
Yes, this is possible.)

$(3)$

Now we use $\displaystyle\arctan t = \frac 1{2i}\log\frac{1+it}{1-it}$,
replace each $\arctan$ by $\log$, split every $\log$ by computing roots of equations of
second degree for the three functions
$$
\begin{aligned}
F(x) &=\frac {x^2-x}{2(x^2-x-5)}\ ,\\
G(x) &=\frac {4x}{2x^2-x-4}\ , \\
H(x) &= \frac {8x^2-24x}{6x^2+7x-25}\ .
\end{aligned}
$$
("Periods" may be introduced, but will not, final numerical checks are enough.)

$(4)$:

We have explicitly:
$$
\begin{aligned}
\frac {1+iF(x)}{1-iF(x)} &= \frac{a_3a_4}{a_1a_2}\cdot\frac{(x-a_1)(x-a_2)}{(x-a_3)(x-a_4)}\ ,\\
\frac {1+iF(x)}{1-iF(x)} &= \frac{b_3b_4}{b_1b_2}\cdot\frac{(x-b_1)(x-b_2)}{(x-b_3)(x-b_4)}\ ,\\
\frac {1+iF(x)}{1-iF(x)} &= \frac{c_3c_4}{c_1c_2}\cdot\frac{(x-c_1)(x-c_2)}{(x-c_3)(x-c_4)}\ ,
\end{aligned}
$$
where
$$
\begin{aligned}
a_1 &= \frac 12(1 - \sqrt{17-8i})\ , &
a_3 &= \frac 12(1-\sqrt{17+8i})=\bar a_1\\
a_2 &= \frac 12(1 + \sqrt{17-8i})\ , &
a_4 &= \frac 12(1+\sqrt{17+8i})=\bar a_2\ ,
\\[2mm]
b_1 &= \frac 14(1-4i - \sqrt{17-8i})\ , &
b_3 &= \frac 14(1+4i - \sqrt{17+8i})=\bar b_1\ ,\\
b_2 &= \frac 14(1-4i + \sqrt{17-8i})\ , &
b_4 &= \frac 14(1+4i + \sqrt{17+8i})=\bar b_2\ ,
\\[2mm]
c_1 &= \frac 14(3+4i - \sqrt{17-8i})\ , &
c_3 &= \frac 14(3-4i - \sqrt{17+8i})=\bar c_1\ ,\\
c_2 &= \frac 14(3+4i + \sqrt{17-8i})\ , &
c_4 &= \frac 14(3-4i + \sqrt{17+8i})=\bar c_2\ .
\end{aligned}
$$
One can replace the square root $\sqrt{17-8i}$ by its longer version, using:
$$
\begin{aligned}
2\Re \sqrt{17-8i} & = \sqrt{17-8i}+\sqrt{17+8i}
=\sqrt{\ \left(\sqrt{17-8i} + \sqrt{17+8i}\right)^2\ }\\
&=\sqrt{34+2\sqrt{17^2+8^2}}=\sqrt{34+2\sqrt{353}}\ ,
\\[3mm]
2i\Im \sqrt{17-8i} & = \sqrt{17-8i}-\sqrt{17+8i}
=\frac{(17-8i)-(17+8i)}{\sqrt{34+2\sqrt{353}}}\\
&=-i\sqrt{-34+2\sqrt{353}}\ .
\end{aligned}
$$

$(5)+(6)$:

We compute $J_1$, $J_2$, $J_3$ to some extent. In fact, i wanted to computed $J_1$ explicitly,
and this is possible, we have a close formula for it involving the functions
$\log$ and $\arctan$ at special places. For $J_2$, $J_3$ individually, we have to
add the dilogarithm.

$(5)+(6)$ The computation of $J_1$:

We have collected all ingrediends. The following computations are considered
in $i\Bbb R$ modulo the real part $\Bbb R$, and we will write $\equiv$ for this
modulo computation.

Let $a$ be a number in the list $a_1,a_2;a_3,a_4$. Then
$$
\int\frac 1x\log(x-a)\; dx
% = \log(x-a)\log \frac xa
% +\operatorname{Li}_2\left(1-\frac xa\right) + C''
% =
% \log(x-a)\log \frac xa
% -\operatorname{Li}_2\left(\frac xa\right) + C'
% -\log\left(1-\frac xa\right)\log\frac xa
% =
% \log(-a)\log\frac xa -\operatorname{Li}_2\left(\frac xa\right) + C'
=
\log(-a)\log x -\operatorname{Li}_2\left(\frac xa\right) + C
\ .
$$
The above primitive makes sense in $x=1$. For $x\to 0$ we may have problems with $\log x$,
but there are none after we consider all contributions from $\log\left(\frac{a_3a_4}{a_1a_2}\cdot\frac{(x-a_1)(x-a_2)}{(x-a_3)(x-a_4)}\right)$ to the factor $\log x$:

$$
\lim_{\epsilon\to 0}\int_\epsilon^1\frac1x\log\frac{x-a}{-a}\; dx
=-\operatorname{Li}_2\left(\frac 1a\right)\ ,
$$
since the two $\log\epsilon$ terms finally cancel.

Note that $(a_1,a_3)$ and $(a_2,a_4)$ are pairs of conjugated
complex numbers, $a_1=\bar a_3$, $a_2=\bar a_4$. The result of the integration is thus
in $i\Bbb R$:
$$
\begin{aligned}
2i\; J_1
&= 2i\;\int_0^1\frac 1x\arctan F(x)\; dx \\
&= 2i\;\int_0^1\frac 1x\log\left( \frac{a_3a_4}{a_1a_2}\cdot\frac{(x-a_1)(x-a_2)}{(x-a_3)(x-a_4)}\right)\; dx \\
&=
-\operatorname{Li}_2\left(\frac 1{a_1}\right)
-\operatorname{Li}_2\left(\frac 1{a_2}\right)
+\operatorname{Li}_2\left(\frac 1{a_3}\right)
+\operatorname{Li}_2\left(\frac 1{a_4}\right)
\\
&\equiv
-2\operatorname{Li}_2\left(\frac 1{a_1}\right)
-2\operatorname{Li}_2\left(\frac 1{a_2}\right) \text{ since $a_3=\bar a_1$ and $a_4=\bar a_2$}
\\
&=
 2\underbrace{
 (\operatorname{Li}_2(a_1)
+\operatorname{Li}_2(1-a_1))}_{\pi^2/6 -\log a_1\log a_2}
\\&\qquad\qquad
+\log^2 (-a_1)+\log^2(-a_2)
\\
&\qquad\qquad\text{ and one may STOP HERE, but optionally let's get $J_1$ explicitly...}
\\
&\equiv
-2\log a_1\log a_2
\\
&\qquad\qquad
  +
  \log^2 a_1+\log^2a_2 +2i\pi\log\frac{a_2}{a_1}
\\
&\equiv
\log^2\frac {a_1}{a_2}
+2i\pi\log\frac {a_2a_4}{a_1a_3}
\\
&=
\log^2\frac {a_1^2}{a_1a_2}
-2\pi i\log\frac {a_1^2a_3^2}{a_1a_2a_3a_4}
\\
&=
\log^2\frac {a_1^2}{-4+2i}
-2\pi i\log\frac {a_1^2a_3^2}{20}
\\
&\equiv
\left(
\frac 12
\log\frac
{(1 - \sqrt{34+2\sqrt{353}} + \sqrt{353})^2}
{4^2\cdot 20}
+
i\left(2\operatorname{arg}(a_1)+\arctan\frac 12-\pi\right)\right)^2
\\
&\qquad
-i\pi\;
\log\frac{(1 - \sqrt{34+2\sqrt{353}} + \sqrt{353})^2}{4^2\cdot 20}
\\
&\equiv
i
\log\frac
{(1 - \sqrt{34+2\sqrt{353}} + \sqrt{353})^2}
{4^2\cdot 20}
\cdot
\left(2\operatorname{arg}(a_1)+\arctan\frac 12-2\pi\right)
\ .
\end{aligned}
$$
We have used $a_1+a_2=1=a_3+a_4$, $a_1a_2=-4+2i$, $a_3a_4=-4-2i$, $a_1a_2a_3a_4=20$.

Here, $\operatorname{arg}(a_1)$ is explicit, just use the real part and the absolute value
(or the imaginary part instead) to write this as a value of the $\arcsin$ or $\arccos$ or $\arctan$:
$$
\begin{aligned}
\Re a_1 &=a_1+a_3=1-\frac 12\sqrt{34+2\sqrt{353}})\ ,\\
|a_1|^2 &= a_1\bar a_1 =a_1a_3 = \frac 14(1-\sqrt{34+2\sqrt{353}}+\sqrt{353})\ .
\end{aligned}
$$
Note that we need to take care which sign is taken in each of the cases
$\log(-a_j)=\log(a_j)\pm\log(-1)$. The sign is plus for $a_2,a_3$ and minus for $a_1,a_4$.

The result is an element in $i\Bbb R$.
The $\arctan\frac 12$ comes when getting the argument for
$\displaystyle
\frac{-4-2i}{-4+2i}=
\frac{-2-i}{-2+i}=
\frac{(2+i)^2}{2^2+1^2}$, which is twice the argument of
$2+i$.

For a numerical check, look in the previous version of this edited answer.
(Or in the commented block of this version.)
A lot of work.
$$
% Numerical check for $J_1$. We use sage.
%
%     F = 1/2*(x^2 - x)/(x^2 - x - 5)
%     a1, a2 = ((1+i*F)/(1-i*F)).numerator()  .roots(multiplicities=False, ring=QQbar)
%     a3, a4 = ((1+i*F)/(1-i*F)).denominator().roots(multiplicities=False, ring=QQbar)
%
%     A1, A2, A3, A4 = a1.n(200), a2.n(200), a3.n(200), a4.n(200)     # approximations
%
%     J1 = integral( atan(F(x)) / x, x, 0, 1, hold=True ).n(200)
%     print(f'Numerical value of J1 is:\n{J1}')
%
%     def T(a):    return -dilog(1/a)
%     J1_v1 = 1/(2*i) * ( T(A1) + T(A2) - T(A3) - T(A4) )
%     print(f'Numerical value of J1 rewritten as signed sum of dilog values is:\n{J1_v1}')
%
%     J1_v2 = 1/(2*i) * ( dilog(A1) + dilog(A2) - dilog(A3) - dilog(A4)
%                         + 1/2 * ( log(-A1)^2 + log(-A2)^2 - log(-A3)^2 - log(-A4)^2 ) )
%     print(f'Numerical value of J1 rewritten as signed sum of dilog and log values is:\n{J1_v2.n(200)}')
%
%     J1_v3 = 1/(2*i) * ( -log(A1)*log(A2) + log(A3)*log(A4)
%                         + 1/2 * ( log(-A1)^2 + log(-A2)^2 - log(-A3)^2 - log(-A4)^2 ) )
%     print(f'Numerical value of J1 rewritten using only log values is:\n{J1_v3.n(200)}')
%
%     J1_v4 = 1/(2*i) * 1/2 * ( log(A1/A2)^2 - log(A3/A4)^2 + 2*i*pi*log(A2*A4/A1/A3))
%     print(f'Numerical value of J1 rewritten as signed sum of log values is:\n{J1_v4.n(200)}')
%
%     J1_v5 = 1/(2*i) * ( 1/2*log( A1^2/(-4 + 2*i) )^2 - 1/2*log( A3^2/(-4 - 2*i) )^2
%                         - i*pi * log(A1^2*A3^2 / 20 ) )
%     print(f'Numerical value of J1 rewritten is:\n{J1_v5.n(200)}')
%
%     s = sqrt(353)
%     S = sqrt(34 + 2*s)
%     E = (1 - S + s)^2/16/20
%     J1_v6 = 1/2 * ( log(E) * ( 2*A1.argument() + atan(1/2) - 2*pi ) ).n(200)
%     print(f'Numerical value of J1 from explicit formula is:\n{J1_v6.n(200)}')
%
% This gives the confirmations:
%
%     Numerical value of J1 is:
%     0.048392365330461616
%     Numerical value of J1 rewritten as signed sum of dilog values is:
%     0.048392365330461619161981331382180296201063587714971948595743
%     Numerical value of J1 rewritten as signed sum of dilog and log values is:
%     0.048392365330461619161981331382180296201063587714971948595740
%     Numerical value of J1 rewritten using only log values is:
%     0.048392365330461619161981331382180296201063587714971948595742
%     Numerical value of J1 rewritten as signed sum of log values is:
%     0.048392365330461619161981331382180296201063587714971948595755
%     Numerical value of J1 finally rewritten is:
%     0.048392365330461619161981331382180296201063587714971948595755
%     Numerical value of J1 from explicit formula is:
%     0.048392365330461619161981331382180296201063587714971948595743
%
% Since i am interested in the K-theory of number fields, here the number field involved being the field
% $F$ associated to the polynomial $x^4 - 34x^2 + 353$, the above collapsing of the dilog terms
% is such a pitty... (Well the collapse is an instance of K-theory, we use the symbolic relation in
% $K_2(F)$ written as $[a]+[1-a]=0$, in ad-hoc terms meaning that $\operatorname{Li}_2(a)+\operatorname{Li}_2(1-a)$
% is a sum of "simpler polylogarithms", so i am not so highly disappointed.)
$$

$(5)+(6)$ The computation of $J_2$ and $J_3$:

In the same manner, we get:
$$
\begin{aligned}
2i\; J_2
&= 2i\;\int_0^1\frac 1x\arctan G(x)\; dx \\
&= 2i\;\int_0^1\frac 1x\log\left( \frac{b_3b_4}{b_1b_2}\cdot\frac{(x-b_1)(x-b_2)}{(x-b_3)(x-b_4)}\right)\; dx \\
&=
-\operatorname{Li}_2\left(\frac 1{b_1}\right)
-\operatorname{Li}_2\left(\frac 1{b_2}\right)
+\operatorname{Li}_2\left(\frac 1{b_3}\right)
+\operatorname{Li}_2\left(\frac 1{b_4}\right)
\\
&\equiv
-2\operatorname{Li}_2\left(\frac 1{b_1}\right)
-2\operatorname{Li}_2\left(\frac 1{b_2}\right)
\\
&=
 2\operatorname{Li}_2(b_1)
+2\operatorname{Li}_2(b_2)
\\&\qquad\qquad
+\log^2 (-b_1)+\log^2(-b_2)\ ,
\\[3mm]
2i\; J_3
&= 2i\;\int_0^1\frac 1x\arctan H(x)\; dx \\
&= 2i\;\int_0^1\frac 1x\log\left( \frac{c_3c_4}{c_1c_2}\cdot\frac{(x-c_1)(x-c_2)}{(x-c_3)(x-c_4)}\right)\; dx \\
&=
-\operatorname{Li}_2\left(\frac 1{c_1}\right)
-\operatorname{Li}_2\left(\frac 1{c_2}\right)
+\operatorname{Li}_2\left(\frac 1{c_3}\right)
+\operatorname{Li}_2\left(\frac 1{c_4}\right)
\\
&=
-2\operatorname{Li}_2\left(\frac 1{c_1}\right)
-2\operatorname{Li}_2\left(\frac 1{c_2}\right)
\\
&=
 2\operatorname{Li}_2(c_1)
+2\operatorname{Li}_2(c_2)
\\&\qquad\qquad
+\log^2 (-c_1)+\log^2(-c_2)\ ,
\\[3mm]
2i(J_2+J_3)
&\equiv
 2\cdot
 \underbrace{(
 \operatorname{Li}_2(b_1)
+\operatorname{Li}_2(c_2))}_{\pi^2/6 -\log b_1\log c_2}
\\
&+2\cdot
\underbrace{(
 \operatorname{Li}_2(c_1)
+\operatorname{Li}_2(b_2))}_{\pi^2/6 -\log c_1\log b_2}
\\&\qquad\qquad
+\log^2 (-b_1)+\log^2(-b_2)
+\log^2 (-c_1)+\log^2(-c_2)
\\
&\qquad\qquad\qquad\qquad\text{ since $b_1+c_2=1=b2+c_1$}
\\
&\equiv
-2\log b_1\log c_2 - 2\log c_1\log b_2
\\&\qquad\qquad
+\log^2 (-b_1)+\log^2(-b_2)
+\log^2 (-c_1)+\log^2(-c_2)
\\
&\equiv
-2\log b_1\log b_2
-2\log b_1\log c_2 - 2\log c_1\log b_2
-2\log c_1\log c_2
\\&\qquad\qquad
+\log^2 (-b_1)+\log^2(-b_2) + 2\log b_1\log b_2
\\&\qquad\qquad
+\log^2 (-c_1)+\log^2(-c_2) + 2\log c_1\log c_2
\\
&\equiv
-2(\log b_1 + \log c_1)(\log b_2 + \log c_2)
\\&\qquad\qquad
+\log^2 (-b_1)+\log^2(-b_2) + 2(\log(-b_1) - i\pi)(\log(-b_2) - i\pi)
\\&\qquad\qquad
+\log^2 (-c_1)+\log^2(-c_2) + 2(\log(-c_1) + i\pi)(\log(-c_2) + i\pi)
\\
&\equiv
-2(\log b_1 + \log c_1)(\log b_2 + \log c_2)
\\&\qquad\qquad
+(\log (-b_1)+\log(-b_2))^2 - 2\pi i(\log(-b_1) + \log(-b_2))
\\&\qquad\qquad
+(\log (-c_1)+\log(-c_2))^2 + 2\pi i(\log(-c_1) + \log(-c_2))
\\
&=
-2\log (b_1 c_1)\log (b_2 c_2)
\\&\qquad\qquad
+(\log (b_1b_2)+0\pi i)^2 - 2\pi i(\log(b_1b_2)+0\pi i)
\\&\qquad\qquad
+(\log (c_1c_2)-2\pi i)^2 + 2\pi i(\log(c_1c_2)-2\pi i)
\\
&=
-2\log (b_1 c_1)\log (b_2 c_2)
\\&\qquad\qquad
+\log^2 (-2) - 2\pi i\log(-2)
\\&\qquad\qquad
+\left(\log \frac {-3+4i}2 -2\pi i\right)^2 + 2\pi i\left(\log \frac {-3+4i}2 - 2\pi i\right)
\\
&=
-2\log (b_1 c_1)\log (b_2 c_2)
\\&\qquad\qquad
+(\log 2 +i\pi)^2 - 2\pi i(\log 2+i\pi)
\\&\qquad\qquad
+       \left(\log \frac 52 +i\pi - i\arctan \frac 43 -2\pi i\right)^2
+ 2\pi i\left(\log \frac 52 +i\pi - i\arctan \frac 43 -2\pi i\right)
\\
&\equiv
-2\log (b_1 c_1)\log (b_2 c_2)
\\&\qquad\qquad
+2\pi i \log 2 - 2\pi i\log 2
\\&\qquad\qquad
- 2i       \log \frac 52\left(\pi +\arctan \frac 43\right)
+ 2\pi i   \log \frac 52
\\
&\equiv
-2\log (b_1 c_1)\log (b_2 c_2)
\\&\qquad\qquad
- 2i       \log \frac 52\arctan \frac 43
\\
&\equiv
-2\log (b_1 c_1)\log (b_2 c_2)
\\&\qquad\qquad
- 4i       \log \frac 52\arctan \frac 12
\ .
\end{aligned}
$$
No intermediary step was omitted. The above shows how a piece of the final result comes in.

The above gives is a more or less explicit value for the sum $(J_2+J_3)$.

$(7)$:

We are in position to put all together. For this, the following simple relations
connecting the $(a,b,c)$-values are freeing the door in the wall:
$$
\boxed{\qquad b_1c_1=\frac 12 a_1^2\qquad\text{ and }\qquad b_2c_2=\frac 12a_2^2\ .\qquad}
$$
It can be checked by hand or computer.
This was in my initial answer the missing bridge to get to the end.
(Having it, all the computations can be directed to the target,
so the answer was dramatically changed in form, but not in ideas.)

Using it, and putting all together:
$$
\begin{aligned}
2i J
&=
2i(2J_1-J_2-J_3)
\\
&=2i\cdot 2J_2 - 2i(J_2+J_3)
\\
&\equiv
- 4 \log a_1\log a_2 + 2\log^2 (-a_1)+2\log^2(-a_2)
\\
&\qquad\qquad
+2\log (b_1 c_1)\log (b_2 c_2)
\\&\qquad\qquad
+4i\log \frac 52\arctan \frac 12
\\
&=
- 4 \log a_1\log a_2 + 2\log^2 (-a_1)+2\log^2(-a_2)
\\&\qquad\qquad
+ 2\log \frac {a_1^2}2 \log \frac {a_2^2}2
\\&\qquad\qquad
+4i\log \frac 52\arctan \frac 12
\\
&=
- 4 \log a_1\log a_2 + 2(\log a_1 -i\pi)^2 + 2(\log a_2 + i\pi)^2
\\&\qquad\qquad
+ 2(2\log a_1 - 2\pi i - \log 2)(2 \log a_2 -\log 2)
\\&\qquad\qquad
+4i\log \frac 52\arctan \frac 12
\\
&\equiv
+ 4 \log a_1\log a_2 + 2\log^2 a_1 - 4\pi i\log a_1 + 2\log^2 a_2 + 4\pi i\log a_2
\\&\qquad\qquad
-4\log 2(\log a_1 +\log a_2) -4\pi i(2 \log a_2 - \log 2)
\\&\qquad\qquad
+4i\log \frac 52\arctan \frac 12
\\
&= 2(\log a_1 + \log a_2)^2 -4\pi i(\log a_1 + \log a_2)
\\&\qquad\qquad
-4\log 2(\log a_1 +\log a_2) + 4\pi i \log 2
\\&\qquad\qquad
+4i\log \frac 52\arctan \frac 12
\\
&= 2\log^2 (a_1a_2) -4\pi i\log (a_1a_2)
-4\log 2\log (a_1a_2) + 4\pi i \log 2
\\&\qquad\qquad
+4i\log \frac 52\arctan \frac 12
\\
&\qquad\qquad\text{ and with $\log(a_1a_2)=\log(-4+2i)=\frac 12\log 20+i(\pi-\arctan(1/2))$...}
\\
&\equiv
2i\log 20\left(\pi-\arctan\frac 12\right)
-2\pi i\log 20
-4i\log 2\left(\pi-\arctan\frac 12\right)
+ 4\pi i \log 2
\\&\qquad\qquad
+4i\log \frac 52\arctan \frac 12
\\
&\equiv
-2i\log 20\arctan\frac 12
+4i\log 2\arctan\frac 12
\\&\qquad\qquad
+4i\log \frac 52\arctan \frac 12
\\
&=2i\log \frac 54\arctan \frac 12\ .
\\
&\qquad\qquad\text{ So finally:}
\\[5mm]
J&=\log \frac 54\arctan \frac 12\ .
\end{aligned}
$$
$\square$


问:
Real polynomials forming a basis
How do you show that a sequence of polynomials form a basis? I know that the rref criterion of a basis (no $0$ rows and has a leading 1 in every column) only works for $K^n$. Particularly,

Consider the set Consider the set $V$ of real polynomials $f$ of degree at most
3 which satisfy $f(1) = f(2)$, and $V$ is a vector space over $\mathbb{R}$.

Show that the following polynomials $f_1(x) = 1$, $f_2(x) = x^2-3x$
and $f_3(x) = x^3-7x$ form a basis of V .

Note: I can't use the determinant method to determine independence.

PROVING V is a vector space:

Let $f(1)=f(2)$, $g(1)=g(2)$ and $\alpha, \beta \in \mathbb{R}$

$(\alpha f+ \beta g)(1)$

= $(\alpha f)(1) + (\beta g)(1)$

= $\alpha (f(1)) + \beta (g(1))$

= $\alpha (f(2)) + \beta (g(2))$

= $(\alpha f)(2) + (\beta g)(2)$

= $(\alpha f + \beta g)(2)$

答:
Assuming your vectors are in $V$. There are two conditions.

Linear independence. Let $a,b,c\in\mathbb R$ be such that
$$\forall x\in\mathbb R,\quad a + b(x^2-3x) + c(x^3-7x) = 0. $$
Does it follow that $a=b=c=0$? Note that this equality holds for any $x$. Fix three different $x$ and get 3 equations.

Spanning. Let $p(x) := ax^3 + bx^2+cx + d$ such that $p(1) = p(2)$. Find $u,v,w$ such that
$$\forall x\in\mathbb R,\quad ax^3+bx^2+cx + d = u + v(x^2-3x) + w(x^3-7x). $$

To check $V$ is a vector subspace, take $f,g\in V$ i.e $f(1) = f(2)$ and $g(1) = g(2)$. Check the condition for $V$:
$$(f+g)(1) \overset{?}= (f+g)(2),\qquad \forall \alpha\in\mathbb R,\quad (\alpha f)(1) = (\alpha f)(2)\,?  $$
or you could do this in one go by verifying
$$ \forall \alpha,\beta\in\mathbb R,\quad (\alpha f+\beta g)(1) = (\alpha f+\beta g)(2). $$
Either option suffices.

ContentLicense: CC BY-SA 4.0

Title: Real polynomials forming a basis


问:
Ascending or descending property of a sequence according to the value of its first element
Suppose that $(a_n)_{n≥1}$ is a sequence of real numbers satisfying $a_{n+1}=\frac{3a_n}{2+a_n}$.

Suppose $0<a_1<1$, then prove that the sequence $a_n$ is increasing and hence show that $lim_{n \rightarrow \infty}a_n=1$.

Suppose $a_1>1$, then prove that the sequence $a_n$ is decreasing and hence show that $lim_{n \rightarrow \infty}a_n=1$.

I attempted to write $a_{n+1}-a_n=\frac{a_n(1-a_n)}{2+a_n}$ but I am kind of stuck here.

答:
Consider first term = $\alpha\\$

Then second term = $\dfrac{3\alpha}{2+\alpha}\\$

Third term = $\dfrac{9\alpha}{4+5\alpha}\\$

Forth term = $\dfrac{27\alpha}{8+19\alpha}\\$

Fifth term = $\dfrac{81\alpha}{16+65\alpha}\\$

General term = $\dfrac{3^n\alpha}{2^n+(3^n-2^n)\alpha} , n =0,1,2...\\$

Now put limit n tending to infinity in the general term... you'll get 1 regardless of $\alpha$

Simplify to see that:
$f(n) = \dfrac{3^n\alpha}{3^n\alpha+2^n(1-\alpha)}\\$ If $\alpha$ is between 0 and 1 denominator is greater than numerator and if $\alpha > 1$ denominator is smaller than numerator.

Now to prove  that the general term is increasing or decreasing consider $f(n) = \dfrac{3^n\alpha}{2^n+(3^n-2^n)\alpha} \\$
Differentiate to find that:

$f'(x) = -\dfrac{(ln(3)-ln(2))\alpha(\alpha-1)2^x3^x}{(3^x\alpha+2^x(1-\alpha))^2} \\$

Note the sign $f'(x)$ is solely determined by the value of $\alpha$, because all the other terms are strictly positive. See below:

$f'(x) = -\alpha(\alpha-1) \left(\dfrac{(ln(3)-ln(2))2^x3^x}{(3^x\alpha+2^x(1-\alpha))^2} \right)\\$

We see when $\alpha \in (0,1) f'(x)>0$ and when $\alpha>1 f'(x)<0$... implies $f(x)$ is increasing when $0<\alpha<1$ and decreasing when $\alpha>1. \\$

Comment down below if any doubts.


问:
Compact subset of a space
Consider the topological space $X=\{0,1\}^{\mathbb{Z}}\times \mathbb{Z}$ and let $Y$ be a subset of $X$ given by $Y=\{0,1\}^{\mathbb{Z}}\times\{0,1,2,\cdots\}$. Is $Y$ compact subset of $X$?

Set $U_k=\{0,1\}^{\mathbb{Z}}\times\{k\}$. Then $\{U_k
; k=0,1,2,\cdots\}$ is an open cover of $Y$ having no subcover. So $Y$ cannot be compact. Am I correct?

答: Yes, you are correct. More generally, if $X$ is any nonempty topological space and $Y$ is infinite discrete, then $X\times Y$ is never compact. That's because $\big\{X\times\{y\}\big\}_{y\in Y}$ is an open cover without finite subcover.

Title: Compact subset of a space


问:
Showing that $\{(x, y) \in \mathbb{R}^2 \mid x< 0, y<0 \}$ is open

Let $A = \{(x, y) \in \mathbb{R}^2 \mid xy > 0 \}$. Show that $A$ is a open subset of the plane.

I have that $A = \{(x, y) \in \mathbb{R}^2 \mid x > 0,y>0 \} \cup \{(x, y) \in \mathbb{R}^2 \mid x< 0, y<0 \}$

Letting $(x,y) \in \{(x, y) \in \mathbb{R}^2 \mid x > 0,y>0 \}$ and $r=\min\{x,y\},$$B((x,y), r)$ I have that $$|x-x'| \leqslant \sqrt{(x-x')^2 + (y-y')^2} = d((x,y), (x',y')) \leqslant r < \frac{r}{2}$$ so $x' >\frac{r}2 >0.$ Similar reasoning can be done for $y'$. Which would show that $B((x,y),r) \subset \{(x, y) \in \mathbb{R}^2 \mid x > 0,y>0 \}$.

How should I deal with the set $\{(x, y) \in \mathbb{R}^2 \mid x< 0, y<0 \}$? I cannot seem to show this... If I let $(x,y) \in \{(x, y) \in \mathbb{R}^2 \mid x< 0, y<0 \}$ and again $r = \min\{x,y\}$ and construct a ball $B((x,y), r)$ I can pick $(x',y') \in B((x,y), r)$, but I'm not sure how to proceed from here?

答:
The other answer given definitely works, but I will offer a perhaps more "nuts and bolts" type answer.

Let $D = \{(x,y) \in \mathbb{R}^2: x,y < 0\}$

Fix a point $(x,y) \in D$, and let $\delta = \min\{|x|, |y|\}$

We claim the Ball $B:=B_\delta((x,y))$ of radius $\delta$ centered at $(x,y)$ is contained in $D$.

Let $(s,t) \in B$. We have to show that $s,t < 0$. Since $(s,t) \in B$, we have the $$\sqrt{(s-x)^2 + (t-y)^2} < \delta$$

Note that $$|s-x| = \sqrt{(s-x)^2} \leq \sqrt {(s-x)^2 + (t-y)^2} < \delta $$

Since $|s-x| < \delta$, we have $-\delta < s-x < \delta \iff -\delta + x < s < \delta + x$

But $\delta + x \leq |x| + x = -x + x = 0$ where we have used the fact that $|x| = -x$ since $x < 0$.

This immediately tells us $s < 0$. Showing $t < 0$  is similar.

Hence, $(s,t) \in D \implies B \subset D$, as desired


问:
Is a finite convex function subdifferentiable everywhere?
Let $X$ be a (infinite-dimensional) Hilbert space and $f:X\to\mathbb{R}$ be convex. Then is $f$ subdifferentiable everywhere in $X$?

I have looked at Non-subdifferentiable convex function on a normed space and it is clear that there are counterexamples when $f$ can take value $+\infty$. However, what about finite convex functions?

The answer by Tom suggests that convex functions should be subdifferentiable on their relative interior - does someone have a reference and/or proof for this?

答:
Let $\tilde{\mathbf{x}}\in\text{ri}(\text{dom}(f))$. Since the vector $(\tilde{\mathbf{x}}, f(\tilde{\mathbf{x}}))\subseteq\mathbb{R}^n\times\mathbb{R}$ is on the relative boundary of $\text{epi}(f)$, it follows by the supporting hyperplane theorem that there exists a separating hyperplane between $(\tilde{\mathbf{x}}, f(\tilde{\mathbf{x}}))$ and $\text{epi}(f)$, meaning that there exists a nonzero vector $(\mathbf{p},-\alpha)$ for which \begin{equation}\tag{1}
\mathbf{p}^T\tilde{\mathbf{x}}-\alpha f(\tilde{\mathbf{x}})\geq\mathbf{p}^T\mathbf{x}-\alpha t,\quad \forall (\mathbf{x},t)\in\text{epi}(f).
\end{equation}
Since $(\tilde{\mathbf{x}}, f(\tilde{\mathbf{x}}))$ is on the boundary of $\text{epi}(f)$ it follows that $(\tilde{\mathbf{x}}, f(\tilde{\mathbf{x}})+1)\in \text{epi}(f)$ as well. Hence we can plug $\mathbf{x}=\tilde{\mathbf{x}}$ and $t=f(\tilde{\mathbf{x}})+1$ to the equation above and get
$$\mathbf{p}^T\tilde{\mathbf{x}}-\alpha f(\tilde{\mathbf{x}})\geq\mathbf{p}^T\tilde{\mathbf{x}}-\alpha(f(\tilde{\mathbf{x}})+1 )\quad \iff \alpha\geq 0. $$
After showing that $\alpha\geq 0$ we need to prove that $\alpha\neq 0$. Denote $A_f=\text{aff}(\text{dom}f)$. Since $\text{epi}(f)\subseteq A_f\times\mathbb{R}$ and a supporting hyperplane $(\mathbf{p},-\alpha)$ by definition does not contain $\text{epi}(f)$, it does not contain $A_f\times\mathbb{R}$ as well. Now consider the affine combination $\mathbf{y}_{\lambda}=(1-\lambda)\tilde{\mathbf{x}}+\lambda\mathbf{y},\;\lambda\in\mathbb{R}$. The points in this line all lie in the affine set $A_f$ and $\mathbf{y}_{\lambda}\to\tilde{\mathbf{x}}$ as $\lambda\to 0$. Since $\tilde{\mathbf{x}}\in\text{ri}(\text{dom}(f))$, for small values of $|\lambda|$ the points $\mathbf{y}_{\lambda}$ lie in $\text{dom}(f)$. Hence we can plug $\mathbf{y}_{\lambda}$ into $(1)$ and get (after some  algebra)
$$\lambda\mathbf{p}^T(\tilde{\mathbf{x}}-\mathbf{y})-\alpha(f(\tilde{\mathbf{x}})-f(\mathbf{y}_{\lambda}))\geq 0.\tag{2}$$
Inequality $(2)$ must hold for sufficiently small positive and negative values of $\lambda$. As long as $\alpha>0$ there is no problem and the inequality still holds$^\textbf{**}$. Assume in contradiction that $\alpha=0$. Then $(2)$ becomes $\lambda\mathbf{p}^T(\tilde{\mathbf{x}}-\mathbf{y})\geq 0$ which cannot hold for both positive and negative values of $\lambda$.

Take a deep breath, we're almost done :)

After establishing that $\alpha>0$ we can divide $(1)$ by it to get
$$f(\mathbf{x})\geq f(\tilde{\mathbf{x}})+\mathbf{g}^T(\mathbf{x}-\tilde{\mathbf{x}}),\quad \forall \mathbf{x}\in\text{dom}(f)$$
where $\mathbf{g}=\mathbf{p}/\alpha$. Thus $\mathbf{g}\in\partial f(\tilde{\mathbf{x}})$, establishing the non-emptiness of $\partial f(\tilde{\mathbf{x}})$.

$^\textbf{**}$If you analyze $f(\mathbf{y}_{\lambda})$ you notice it's monotone decreasing in $\lambda<0$ (it increases as $\lambda$ moves away from zero in the negative direction), hence offsetting the change of sign from $\lambda\mathbf{p}^T(\tilde{\mathbf{x}}-\mathbf{y})$.


问:
$\ln{2n+1\over n+1}<{1\over n+1}+{1\over n+2}+\cdots +{1\over 2n}<\ln 2$
Using the fact: ${x\over 1+x}<\ln (1+x)<x$, when $x>0$; deduce that:
\begin{equation}
\ln{2n+1\over n+1}<{1\over n+1}+{1\over n+2}+\cdots +{1\over 2n}<\ln 2
\end{equation}

This problem also asked to prove the fact I have given. I could prove it using MVT, but the deduction which should be easier is out of my hand.

答:
The fact is: $\frac{x}{1+x} \lt \ln(1+x) \lt x$, when $x > 0$. Since we particularly know that $\ln(1+x) \lt x$, we can write:
$$\begin{align} & \ln(2n+1)-\ln(2n)=\ln\Biggl(1+\frac{1}{2n}\Biggl)\lt\frac{1}{2n} \\
& \ln(2n)-\ln(2n-1)=\ln\Biggl(1+\frac{1}{2n-1}\Biggl) \lt \frac{1}{2n-1} \\
& \space \space \space \space \space \vdots \\
& \ln(n+2)-\ln(n+1)=\ln\Biggl(1+\frac{1}{n+1}\Biggl)\lt \frac{1}{n+1} \\ \end{align} $$ If we rewrite these expressions in a single line all at once, we eventually get: $$\ln\frac{2n+1}{n+1}\lt \frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{2n}$$
$\ln()$ is an increasing function where it is defined. We can write for any $x>\frac{1}{2}$ that: $$\ln\Biggl(2-\frac{1}{x}\Biggl)\lt \ln2 \lt \ln\Biggl(2+\frac{1}{x} \Biggl)$$
Therefore, we can deduce that: $$\ln\frac{2n+1}{n+1}=\ln\Biggl(2-\frac{1}{n+1}\Biggl) \lt \ln2 \lt \ln\Biggl(2+\frac{1}{n+1} \Biggl)$$ Finally: $$\ln\frac{2n+1}{n+1} \lt \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} \lt \ln2$$


问:
Prove that A’C + A’B + BC equals C (A XOR B)’ + A’B
Prove
$$A’C + A’B + BC = C (A \oplus B)’ + A’B$$

Here is what i have tried:
$$A’C + A’B + BC = C(A’+B) + A’B$$
Now, need to prove $(A \oplus B)’ = A’+B$

$$(A \oplus B)’ = A'B'+AB = \text{now what??}$$

答:
HINT:
I am unsure whether the question is correct or not, but here is the hint: use truth tables.

And $(A \oplus B)’ = A’+B$ is wrong...(by symmetric relation in A and B)


问:
Seemingly impossible exponential equation
$2 \cdot 3^x + 2 = 5 \cdot 2^x$
I am looking for the exact real solutions of this equation(no Newton's method). Even my professor couldn't solve this. I myself have struggled a lot to find the solutions. Can anyone give me any ideas?
(This is my first question on this forum so I'm sorry for the formatting)

答:
Excluding the trivial solution,inspection shows that function
$$y=2 \cdot 3^x + 2 - 5 \cdot 2^x$$
has a root "close" to $x=-1$.

Build a series expansion around this point to have
$$y=\frac 16+\sum_{n=1}^p \frac{a_n}{n!}\,(x+1)^n+O\left((x+1)^{p+1}\right)$$ where
$$a_n=\frac 2 3 \big[\log(3)\big]^n-\frac 52  \big[\log(5)\big]^n$$ Now, make a series reversion to obtain
$$x=-1+\sum_{n=1}^p b_n \left(\frac{y-\frac{1}{6}}{\frac{2 \log (3)}{3}-\frac{5 \log (2)}{2}}\right)^n+O\left(\left(y-\frac{1}{6}\right)^{p+1}\right)$$ Now, make $y=0$ and you have long explicit forms of more and more accurate solutions.

Here are the results of this process
$$\left(
\begin{array}{cc}
p & x_{(p)} \\
 1 & -0.8334099246 \\
 2 & -0.8389092790 \\
 3 & -0.8385065961 \\
 4 & -0.8385304569 \\
 5 & -0.8385285436 \\
 6 & -0.8385286777 \\
 7 & -0.8385286666 \\
 8 & -0.8385286675 \\
 9 & -0.8385286674
\end{array}
\right)$$

Edit

As I replied to a comment, there is no closed form. But the exact solution is obtained using summations up to infinity.


问:
If a,b are real numbers. Find the least value of $a^2 + ab + b^2 - a - 2b$
I tried simple $AM \geq GM$ inequality But no luck

$a^2 + ab + b^2 - a - 2b \geq \sqrt{a^2 * ab * b^2 * (-a) * (-2b)} = \sqrt{2a^4b^4} = a^2b^2\sqrt{2}$

Anwer is $-1$

答:
Another approach:

$p=a^2+ab+b^2-a-2b=a^2+a(b-1)+b^2-2b$

We solve polynomial p for $a$:

$a=\frac{-(b-1)\pm \sqrt {\Delta}}2$

$\Delta=(b-1)^2-4b^2+8b$

a can be minimum if $\Delta=0$, in this case

we get $b\approx 1/6$, which gives $a\approx 5/12$ and $p\approx-0.48$ .

Also $b\approx -2$ which gives $a\approx 0.2$ and $p\approx 2.2$. We can see that for b=0, 1 and 2 , p is zero and $\Delta<0$ for $b<0$ and $b>2$.

Value $p=-0.48$ deduces that $0>p_{min}>-1$


问:
Computing double integral for expected value
Show that the mean kinetic energy for a one-dimensional particle equals $k_BT/2$. That is, show

$$E[p^2/2] =\frac{1}{Z} \int_{\mathbb{R}^2} \frac{p^2}{2} e^{-\beta(V(x) + p^2/2)} \mathop{dxdp},$$

where $\beta = (k_BT)^{-1}$. I also know that

$$Z = \int_\mathbb{R^2} e^{-\beta(V(x) + p^2/2)} \mathop{dx dp}.$$

I just tried to expand the integral

$$E(p^2/2) = \frac{1}{Z} \int_{-\infty}^{\infty} \frac{p^2 e^{-\beta p^2/2}}{2} \int_{-\infty}^{\infty} e^{-\beta V(x)} \mathop{dx} \mathop{dp},$$

but the innermost integral works out to be $\exp(-\beta V(x)/(-\beta V'(x))$ evaluated from $x = -\infty$ to $x = \infty$, and I'm not entirely sure how to do that. I was thinking maybe to somehow cancel out the $Z$ in the denominator with an integral but I couldn't come up with anything.

Any help is appreciated.

答:
We can use Fubini's theorem to write

$$\left\langle \frac{p^2}{2} \right\rangle = \frac{\int_{\Bbb{R^2}}\frac{p^2}{2}\exp\left(-\beta(V(x)+p^2/2)\right)dxdp}{\int_{\Bbb{R^2}}\exp\left(-\beta(V(x)+p^2/2)\right)dxdp} = \frac{\left(\int_\Bbb{R}\exp\left(-\beta V(x)\right)dx\right)\cdot\left(\int_\Bbb{R}\frac{p^2}{2}\exp\left(- \frac{\beta p^2}{2}\right)dp\right)}{\left(\int_\Bbb{R}\exp\left(-\beta V(x)\right)dx\right)\cdot\left(\int_\Bbb{R}\exp\left(-\frac{\beta p^2}{2}\right)dp\right)}$$

$$= \frac{\int_\Bbb{R}\frac{p^2}{2}\exp\left(- \frac{\beta p^2}{2}\right)dp}{\int_\Bbb{R}\exp\left(-\frac{\beta p^2}{2}\right)dp} = \frac{\frac{1}{4}\sqrt{\frac{8\pi}{\beta^3}}}{\sqrt{\frac{2\pi}{\beta}}} = \frac{1}{2\beta} = \frac{1}{2}k_BT$$

from Feynman's trick and the result of the Gaussian integral.


问:
Deducing a result about entire functions
We need to show that for an entire function $f$ on $\mathbb{C}$, there are constants $a_1,...,a_n$ such that $\int^{2 \pi}_{0}|f(re^{i \theta})|^2 d\theta=2\pi\sum_{n=0}^{\infty}|a_n|^2r^{2n}$.

Thoughts so far are that we can find a power series expansion about $0$ by Taylor's Theorem. Given the result it seems natural to use $f(z)^2$ somewhere but I can't see how.

Any help appreciated.

答: Using the power series representation is a good idea. Take $r$ (strictly) smaller than the radius of convergence of the Taylor series at $0$, say $\sum_na_nz^n$. Use uniform convergence to switch the series and the integral. </p>

ContentLicense: CC BY-SA 3.0


问:
Understanding the definitions of vector and scalar
So I am preparing now to start studying Lagrangian and Hamiltonian mechanics with Marion's book on classical dynamics. It is the first time I encounter the formal definition of vector and scalar, and I found it hard to understand.

First of all, the definition of vector: correct me if I am wrong, but what I have understood is that, if a set of quantities, $A_1, A_2, A_3$ for three-dimensions, transforms as a point under a rotation transformation, then we call $\vec{A}=(A_1, A_2, A_3)$ a vector. In summary, a vector's components transform as a point under a coordinate rotation.

The definition I don't quite understand is the definition of a scalar. It is said that a scalar is a quantity that remains invariant under a coordinate rotation. How can one understand this definition for, for example, temperature? How can a scalar be expressed in terms of the coordinate we are in?

I would appreciate help in understanding these concepts, thanks in advance!

答:
What is a Scalar?

Scalar is a type of ‘number’. Most of the physical, abstract entities(or quantities or items or attribute etc..) can be represented by a single number(or type of number) are called as Scalars. Like Weight of a person(60 kg), height of a person(180 cm), cost of a chocolate(50 cents), number of students(60), age of a person(34 years) etc.

Most of the day to day entities that we deal with are represented as Scalars. We don’t call them explicitly as Scalars in day to day language, because there is no need to call them as Scalars until we need to work with something other than and different from a Scalar, called as Vectors.

Scalar is derived from ‘scale’ which means it can stretch or shrink. i.e. the number can increase or decrease. Mathematically it is a number in a number line axis whose range is between -infinity to + infinity with zero in the middle.

What is a Vector?

Vector is a (another) type of ‘number’. There are certain physical, abstract entities(quantities or items or attributes etc.) which are a combination of more than one entity. Hence can not represented by a single number or Scalars but requires ‘more than one number’ or ‘list of numbers’ called as Vectors for representation. Vector representation is required as a necessity and convenience.

Lets expand one example from Scalars to see why Vectors are needed and does the job.

While the ‘weight’(is a Scalar) of a person is important for the person himself, the combination of ‘weight and height’(is a Vector) is important if you are his Doctor. Remember the combination of weight and height is the Vector. If only weight or only height, then these would still be scalars.

When you know both Weight and Height then you can calculate the Body Mass Index to determine if the person is healthy with in the range or obese. As a doctor your patient’s Weights and heights are Vectors. So your observation of your patients for the day might be Person-1(80kg, 150cm), Person-2(95Kg, 180cm)... Person-N(64Kg, 165cm) as unique Patient vectors.

Now there is nothing to stop at Weight and Height. We can add age, sex in to the mix and it is endless. Person-1(80kg, 150cm, 44 years, Male), Person-2(95Kg, 180cm, 28 years, Male)... Person-N(64Kg, 165cm, 30 years, Female) etc.

Let’s move one-level above, if you as doctor is practising in large hospital which has 10s of other doctors attending to 1000s of patients everyday, the patient vector might be Person-N(64Kg, 165cm, 30 years, Female, Doctor1).

Lets move step above to the level of Country’s Health Care system which has the records of 100s of hospitals, 1000s of doctors and Millions of patients, the patient vector might be Person-N(64Kg, 165cm, 30 years, Female, Doctor1, Hospital2,CityX) etc.. Soon you would realize the usual number line arithmetic and Scalar representation is too complex and not convenient.

If you are from computer science/software back ground, Vectors might remind you the organization of data as tables in MS Excel or database. Each row in a table is a Vector while the columns of the table are the dimensions of the vector.

With Vectors, we can create clusters(clustering procedures already developed by mathematicians) among patient groups i.e. Group1(60–70 kg, 150–160cm, 30–35 years, Male only), Group2(50–60 kg, 160–170cm, 25–30 years, Male only) etc.. We can compare between individual patients, compare between cluster groups, predict who or which group is in danger etc.

Thanks to Mathematicians, Vector representation is advantageous as it is convenient and provides lot of theorems, tools, methods, shortcuts, procedures to work with Vectors.

What other examples can be cited for Vectors?

Virtually any entity which is combination of more than one entity are all valid examples of Vectors.

Directions(Walk100 m straight, turn right and walk 50 m to your destination).
Velocity(Distance in a direction Vs. time ) and Acceleration(Velocity in a direction Vs. time)
Force acting on a body(in each direction), Buying Patterns(essential or luxury items), Budget allocations etc.

Can we live with out Vectors?

Of course, we can.
From the above example, we can continue to treat each of the patient entity(weight, height, age, sex) as separate entities and use Scalars and the number line arithmetic(addition, subtraction, multiplication, division) but it would be tedious and time consuming and good luck!!!

Hope this is clear!!!


问:
If $V$ is $k$-dimensional then show that $\dim \wedge^n V = \binom {k} {n}.$

If $V$ is $k$-dimensional then show that $\dim \wedge^n V = \binom {k} {n}.$

I don't understand this result. If $k \lt n$ then how does the quantity $\binom {k} {n}$ make sense? Can anybody please help me in this regard?

Thanks for your time.

EDIT $:$ Let $V$ be a vector space. Then for any $n \in \Bbb N$ we can identify $V^{\times n}$ by $V^{\{1,2, \cdots, n \}}.$ Since $\{1,2, \cdots, n \}$ is an $S_n$-set it follows that $V^{\{1,2, \cdots, n \}}$ is an $S_n$-vector space. For any $\sigma \in S_n$ and any function $f : \{1,2, \cdots, n \} \longrightarrow V$ the action of $\sigma$ on $f$ is given by $(\sigma \cdot f) (i) = f (\sigma^{-1} (i)).$ This shows that $V^{\times n}$ is a $S_n$-vector space. Let $\mu$ be the action of $S_n$ on $V^{\times n}$ then for each $\sigma \in S_n$ the map $\otimes_V \circ \mu (\sigma, \cdot) : V^{\times n} \longrightarrow V^{\otimes n}$ will be a bilinear map. Then by universal property of tensor products there exists a unique bijective linear map $\rho (\sigma) : V^{\otimes n} \longrightarrow V^{\otimes n}.$ This gives a group homomorphism $\rho : S_n \longrightarrow GL(V^{\otimes n}).$ Hence $V^{\otimes n}$ is also a $S_n$-vector space where for any element $\sigma \in S_n$ and for any $v \in V^{\otimes n}$ the action of $\sigma$ on $v$ is given by $\sigma \cdot v : = \rho (\sigma) (v).$ If $\rho (\sigma) (v) = \left (\text {sign} (\sigma) \right ) v,$ for all $\sigma \in S_n$ and for some $v \in V^{\otimes n}$ then $v$ is said to be an alternating/anti-symmetric tensor. The collection of all alternating/anti-symmetric tensors forms a subspace of $V^{\otimes n}$ which is denoted by $\wedge^n V$ or $V^{\wedge n}.$

答:
We can define a projection from $V^{\otimes n}$ to $\wedge^n V$, the so called antisymmetrization map:

$$\pi:v\mapsto\frac{1}{n!}\sum_{\sigma\in S_n}\operatorname{sign}(\sigma)\cdot\rho(\sigma)(v).$$

You can check for yourself that $\pi(v)$ is antisymmetric for all $v\in V^{\otimes n}$, and that $\pi(v)=v$ if $v$ is antisymmetric. So this is a surjective map $V^{\otimes n}\to\wedge^n V$.

Now consider the fact that surjective linear maps map generating sets to generating sets. In particular, they map bases to generating sets. Now let $B:=\{b_1,\dots,b_k\}$ be a basis of $V$. Then the set $\{c_1\otimes\dots\otimes c_n~\vert~c_i\in B\}$ is a basis of $V^{\otimes n}$. And thus $\{\pi(c_1\otimes\dots\otimes c_n)~\vert~c_i\in B\}$ is a generating set of $\wedge^n V$. Let's call $\pi(v_1\otimes\dots\otimes v_n)$ for $v_1,\dots,v_n\in V$ by another name, just for practicality:

$$v_1\wedge\dots\wedge v_n:=\pi(v_1\otimes\dots\otimes v_n).$$

This is the exterior product of $v_1,\dots,v_k$, and it has the nice property to be linear in each component and antisymmetric. All elements of $\wedge^n V$ can be written as linear combinations of such exterior products, and in particular, $\{c_1\wedge\dots\wedge c_n~\vert~c_i\in B\}$ is a generating set of $\wedge^n V$. It is not linearly independent, but you can choose a linearly independent subset: Throw away all exterior products containing basis vectors multiple times, since those are $0$. Also, among all exterior products containing the basis vectors $b_{i_1},\dots,b_{i_n}$, choose only one, since these are all multiples of each other due to antisymmetry. Once you did this, you have a linearly independent generating set (check this yourself). It consists of elements of the form $b_{i_1}\wedge\dots\wedge b_{i_n}$, where $b_{i_j}$ are all distinct and only one possible ordering is chosen. So among $k$ basis vectors, you choose $n$ distinct ones, without caring about the ordering. So you choose $n$ from $k$: $n$ choose $k$, or $\binom{k}{n}$ of them.

Also note, if $k<n$, there are no antisymmetric tensors except the $0$-tensor. This means the dimension is $0$ in that case. In accordance with the combinatorial definition of $\binom kn$, which makes it $0$ for $k<n$.


问:
Uniform convergence of an integral sequence of functions
I have to study the punctual and uniform domain of
$$f_n(x)=\int_0^{x+n}\dfrac{du}{2e^u+\sin^2u}.$$

First I observed that the integrand function is defined in $\mathbb R$ and it is decreasing $\forall x\ge 0$.

For the punctual convergence $x\in\mathbb R$ is fixed and $n\to \infty$. Since $x+n\sim n\to\infty$ we have
$$\lim_{n\to\infty}\int_0^{x+n}\dfrac{du}{2e^u+\sin^2u}=\int_0^{+\infty}\dfrac{du}{2e^u+\sin^2u}\le\int_0^{+\infty}\dfrac{du}{2e^u}=\dfrac{1}{2}\implies$$
$$\forall x\in\mathbb R,\exists\lim_{n\to\infty}f_n(x).$$
So the domain of punctual convergence should be $D_p=\mathbb R$.

We know that a sequence of functions is uniformly convergent in $U$ iff $\underset{n\to\infty}\lim\underset{x\in U}\sup |f_n(x)-f(x)|=0$.

What I know about $f(x)$ is that $f(x)\le\dfrac{1}{2}$, so I'm not sure about how to proceed to determine the domain of uniform convergence.

PS: Consider $n<m<N$ (we still have to determine $N$) and $x\in \mathbb R^+$, with an arbitary positive $\varepsilon$ fixed.
$$|f_m(x)-f_n(x)|=\Big |\int_0^{m+x} \dfrac{du}{2e^u+\sin^2u}-\int_0^{n+x}\dfrac{du}{2e^u+\sin^2u}\Big |=$$
$$=\int_{n+x}^{m+x} \dfrac{du}{2e^u+\sin^2u}\le \int_{n+x}^{m+x} \dfrac{du}{2e^u}=-\dfrac{e^{-u}}{2}\Big|_{n+x}^{m+x}=-\dfrac{e^{-(m+x)}}{2}+\dfrac{e^{-(n+x)}}{2}=\dfrac{1}{2e^x}\Big(\dfrac{1}{e^n}-\dfrac{1}{e^m}\Big)\le \dfrac{1}{2}\Big(\dfrac{1}{e^n}-\dfrac{1}{e^m}\Big).$$
For the fact that $n<m\implies \dfrac{1}{e^n}>\dfrac{1}{e^m}\implies 0<\dfrac{1}{e^n}-\dfrac{1}{e^m}<1$ we have
$$ \dfrac{1}{2}\Big(\dfrac{1}{e^n}-\dfrac{1}{e^m}\Big)<\dfrac{1}{2}\implies \mathbb N\ni N>2\varepsilon$$

答:
For $U=\mathbb{R^+}$
$$\underset{n\to\infty}\lim\underset{x\in U}\sup |f_n(x)-f(x)|=\underset{n\to\infty}\lim\underset{x\in U}\sup \int\limits_{x+n}^{+\infty}\dfrac{du}{2e^u+\sin^2u}\leqslant\\
\leqslant\lim\limits_{n \to \infty} \int\limits_{n}^{+\infty}\dfrac{du}{2e^u+\sin^2u}=0$$
But for $U=\mathbb{R}$ we have $\sup\limits_{x \in \mathbb{R}} \int\limits_{x+n}^{+\infty}\dfrac{du}{2e^u+\sin^2u}=\int\limits_{-\infty}^{+\infty}\dfrac{du}{2e^u+\sin^2u}$ where last diverges.


问:
Proving that L1 and L2 have the same theorems.
[From "Introduction to Mathmatical logic" by Elliott Mendelson,exercise 1.58]

Axioms of $L_1$:

(A1) $B \rightarrow (C \rightarrow B)$

(A2) $(B \rightarrow (C \rightarrow D)) \rightarrow ((B \rightarrow C) \rightarrow (B \rightarrow D))$

(A3) $(\lnot C \rightarrow \lnot B ) \rightarrow ((\lnot C \rightarrow B) \rightarrow C)$

only rule of inference is $MP$

Axiom system of $L_2$:

(B1) $(\lnot B \rightarrow B) \rightarrow B$

(B2) $B \rightarrow (\lnot B \rightarrow C)$

(B3) $(B \rightarrow C) \rightarrow ((C \rightarrow D) \rightarrow (B \rightarrow D))$

only rule of inference is $MP$

I have to prove that $L_1$ and $L_2$ have the same theorems.For that I have to prove that the axioms of $L_1$ can be proven from axioms of $L_2$ and vice-versa (both systems have the same inferences so I don't have to worry about that).I was able to prove the axioms of $L_2$ from the axioms $L_1$ (in $L_1$) , but failed at proving the axioms of $L_1$ from the axioms of $L_2$ (in $L_2$).

The theorems I was able to derive in $L_2$ so far were:-

(1) $\vdash_{L_2} B \rightarrow B$

(2) $B \rightarrow C, C \rightarrow D \vdash_{L_2} B \rightarrow D$

(3) $B \rightarrow C \vdash_{L_2} ((C \rightarrow D) \rightarrow (B \rightarrow D))$

(4) $\vdash_{L_2}(\lnot B \rightarrow B) \rightarrow (\lnot B \rightarrow C)$

(5) $\vdash_{L_2}B \to (\lnot B \to B)$

(6) $\vdash_{L_2}(B \to C) \to ((\lnot B \to B) \to C)$

(7) $\vdash_{L_2}((\lnot B \to B) \to C) \to (B \to C)$

(8) $\vdash_{L_2}B \to ((B \to D) \to (\lnot B \to D))$

(9) $\vdash_{L_2}\lnot (B \to B) \to (B \to B)$

(10) $\vdash_{L_2} (B\to(\lnot B \to \lnot C)) \to (((\lnot B \to \lnot C) \to (C \to B)) \to (B \to (C \to B)))$

(11) $\vdash_{L_2} ((\lnot B \to \lnot C) \to (C \to B)) \to (B \to (C \to B))$

(12)  $\vdash_{L_2} (B\to(\lnot B \to \lnot C)) \to (((\lnot B \to \lnot C) \to D) \to (B \to D)))$

(13)  $\vdash_{L_2} ((\lnot B \to \lnot C) \to D) \to (B \to D) $

(14)  $\vdash_{L_2} ((\lnot B \to \lnot C) \to C) \to (B \to C) $

But I still am a long way from proving the axioms of $L_2$ from $L_1$ .The problem is that I have ran out of ideas about how I can manipulate the axioms of $L_2$.Can someone give me a hint about this?

Edit:I found another axiom system in another post which I will call $L_3$.

Axioms of $L_3$:

(C1) $B \to (C \to B)$

(C2) $(B \to (C \to D)) \to ((B \to C) \to (B \to D))$

(C3) $(\lnot B \rightarrow B) \rightarrow B$

(C4) $\lnot B \rightarrow (B \rightarrow C)$

idk , maybe I should try to prove (C1) and (C2) from (B3) in $L_2$ as a subgoal/lemma [Not sure if it will help me with my main problem of proving the axioms of $L_1$ from the axioms of $L_2$ (in $L_2$)].

I found a wiki article List of Hilbert systems where, $L_1$ is called "Mendelsons axiom system" and $L_2$ is called "Łukasiewicz First Axiom system".I tried to find a paper,book or article about "Łukasiewicz First Axiom system" instead of the wiki article of Hilbert systems, but the only thing I was able to get that uses "Łukasiewicz First Axiom system" was this MIT OCW lecture slide An example of usage of "Łukasiewicz First Axiom system".There, they derived $\vdash_{L_2} B \to B$ and stopped.I also tried to searching "lukaweicz" to see if someone has a similar question like mine,but the questions I found were actually about "lukaweicz third axiom system", not "lukaweicz first axiom system" an example. I will be glad if someone shows me some other reference that I can read.

答:
Example from Jan Łukasiewicz, Elements of Mathematical Logic (Pergamon Press, 1963), page 42.

Starting from your axiom (B3) $[(p \to q) \to ((q \to r) \to (p \to r))]$ with substitution: $[p \leftarrow (p \to q), q \leftarrow ((q \to r)\to(p \to r)), r \leftarrow s]$ we get:

$[((p \to q) \to ((q \to r)\to(p \to r))) \to ((((q \to r)\to(p \to r)) \to s) \to ((p \to q) \to s))]$

The antecedent is (B3) again; thus, by Modus Ponens, we get:

(T4) $(((q \to r)\to(p \to r)) \to s) \to ((p \to q) \to s)$.

In the same way (from (T4) and a suitable substitution) we get:

(T5) $[(p \to (q \to r))] \to ((s \to q) \to [p \to(s \to r)]).$

The basic "proof technique" is to identify a suitable substitution in some axiom or already proven thesis such that the result is a "big" conditional where the antecedent is an axiom or an already proven thesis and the consequent is the thesis to be proved.

We can see it in the derivation of $p \to p$ [page 45: your (1)].

From (B3) again, with substitution $[q \leftarrow (\lnot p \to q)]$ we get: $[p \to (\lnot p \to q)] \to [((\lnot p \to q) \to r) \to (p \to r)]$.

The antecedent is axiom (B2) and thus by MP we get (T9): $((\lnot p \to q) \to r) \to (p \to r)$ [that seems quite strange...].

Now, from (T9) with substitution $[q \leftarrow p, r \leftarrow p]$ we get: $((\lnot p \to p) \to p) \to (p \to p)$.

The antecedent is axiom is (B1); using MP we get: (T16) $p \to p$.

Going on, we have [page 45]: (T18) $q \to (p \to q)$ [which is (A1) of L1], and "the very important law of commutation" [page 47]: (T21) $(p \to (q \to r)) \to (q \to (p \to r))$.

Then [page 49]: (T35) $(p \to (q \to r)) \to [(p \to q) \to (p \to r)]$ [which is (A2) of L1].

Thesis 35 occurs as an axiom in the first axiom system of the sentential calculus, given by Frege in his Begriffsschrift.

The main difficulty in Łukasiewicz's original book is the lack of the $\text { Deduction Theorem}$.

Regarding axiom (A3) of L1, we can use "the very important law of double
negation" [page 50]:

(T39) $\lnot \lnot p \to p$ and (T40) $p \to \lnot \lnot p$,

as well as "the four laws of transposition" [page 51]:

(T46) $(p \to q) \to (\lnot q \to \lnot p)$, etc.

What we can do now is to use (T46): $(\lnot q \to p) \vdash (\lnot p \to \lnot \lnot q)$ with (T39) and your (2) above to get:

$(\lnot q \to p) \vdash (\lnot p \to q)$.

Using it with $(\lnot q \to \lnot p)$ and (2) again we get:

$(\lnot q \to \lnot p), (\lnot q \to p) \vdash (\lnot q \to q)$.

Finally, using axiom (B1) of L2 and MP we get:

$(\lnot q \to \lnot p), (\lnot q \to p) \vdash q$.


问:
Proof for sum and difference of angles in terms of tan inverse
We have the following formulas for sum of angles when angles are in terms of tan inverse:
$\tan^{-1}(x)+\tan^{-1}(y)$ =

$\tan^{-1}((x+y)/(1-xy))$,  if $xy<1$

$\pi+\tan^{-1}((x+y)/(1-xy))$,  if $x>0,y>0$and $xy>1$

$-\pi +\tan^{-1}((x+y)/(1-xy))$,  if $x<0,y<0$ and $xy>1$

I tried it like this :
Let $\tan^{-1} (x)=A$ and $\tan^{-1}(y)=B$ where $A,B \in(-(\pi/2),(\pi/2))$

So , $\tan(A+B)=(\tan A+\tan B)/1-\tan A \tan B=(x+y)/1-xy$

$\tan^{-1}((x+y)/(1-xy))=\tan^{-1}(\tan(A+B))$

let $A+B=\alpha$

=$\alpha+\pi$,  $-\pi<\alpha<-\pi/2$

$\alpha,  -\pi<\alpha <\pi/2$

$\alpha-\pi,  \pi/2<\alpha <\pi$

Further how to proceed to get the condition xy><1...?

答:
Well what's happening is, we are doing this step below:

$$\tan^{-1} A + \tan^{-1} B = \tan^{-1} (\tan (\tan^{-1} A + \tan^{-1} B))$$

$\tan^{-1} \tan x$ will shift your input by $n \pi \ ; \ n \ \epsilon \ Z$ to bring it into principal range of tan inverse.

So that step we did above is only valid when you know that $\tan^{-1} A + \tan^{-1} B$ lies within the principal range. To account for other ways are the cases you've listed.

I'm not aware of a purely algebraic approach because I dont get how one would proceed by taking expressions like $\tan^{-1} A + \tan^{-1} B > \frac \pi 2$. Ill present a graphical approach:

That's $y = \tan x$ in $(-\pi, \pi)$

Five cases arise corresponding x,y of opposite signs + to the 4 intervals in x axis, made equally in the above graph when they are of same sign:

0. With $A$ and $B$ of opposite signs, we have $\tan^{-1} A + \tan^{-1}B \ \epsilon \ (-\frac \pi 2, \frac \pi 2)$, which corresponds to the principal range.

1. $\tan^{-1} A + \tan^{-1}B < 0$ and $\tan (\tan^{-1} A + \tan^{-1}B) < 0$:
Here since of same sign, both are negative but then since tan is positive you have $1 - AB < 0$

Now $\tan ^{-1} \tan (\tan^{-1} A + \tan^{-1}B)$ gives a value $\pi$ more than what it shoudlve been.

$$\tan^{-1} A + \tan^{-1} B = \tan ^{-1} \left( \frac {A+B}{1-AB} \right) - \pi \ ; \ A,B <0, AB>1$$

Similarly you can do this for the other 3 regions too.

You'll see that regions 0. ($AB<0$), 2.($AB<1; A,B<0$) and 3.($AB<1; A,B>0$) correspond to the same condition of $AB < 1$ and to same result from the expression.


答:
Let $f(z)=\sum_{n\geq 0}a_nz^n$ the power series expansion of $f$ with infinite radius of convergence. Then, for $r\geq 0$ fixed, the function
$$
g:\theta\longmapsto f(re^{i\theta})=\sum_{n\geq 0}a_nr^ne^{in\theta}
$$
is a Fourier series which converges normally as $\sum_{n\geq 0}|a_n|r^n$ converges. In particular, $g$ is $2\pi$-periodic and continuous, whence in $L^2(0,2\pi)$. So the equality you want is Parseval.</p>

Title: Deducing a result about entire functions

Body:
We need to show that for an entire function $f$ on $\mathbb{C}$, there are constants $a_1,...,a_n$ such that $\int^{2 \pi}_{0}|f(re^{i \theta})|^2 d\theta=2\pi\sum_{n=0}^{\infty}|a_n|^2r^{2n}$.

Thoughts so far are that we can find a power series expansion about $0$ by Taylor's Theorem. Given the result it seems natural to use $f(z)^2$ somewhere but I can't see how.

Any help appreciated.


问:
A question regarding an algorithm
I'm a software developer, and I wrote an algorithm for converting any positive integer with at least 2 digits into 9, based on what Scott Flansburg discovered.

For example: $N = 22$

Sum up the digits: $2 + 2 = 4$

Then, subtract the result from the original number: $22 - 4 = 18$

The sum of the digits: $1 + 8 = 9$

It is valid for any positive integer with at least 2 digits.

Why does the algorithm work?

Is there a name for this algorithm?

EDIT:

I'm not a mathematician, only a math enthusiast, so I'd appreciate it if the answers wouldn't be complex

答: If the number is $xy$, it can be written as $10x+y$. So the second step results in $10x+y-(x+y)=9x$, and we know that the sum of digits of a number that is divisible by $9$ is (eventually) $9$.

Title: A question regarding an algorithm


问:
Is there a duality principle for classical first-order logic? (and/or free logic?)
In order theory, there's a variety of duality principles, like:

If a sentence involving only the meet and join operations is a
  consequence of the lattice axioms, then: the dual sentence, obtained by

replacing all meets in the sentence with joins, and

replacing all joins in the sentence with meets

is also a consequence of the lattice axioms.

Is there something similar for classical first-order logic? (and/or free logic that allows for empty domains?)

答:
If ∃xLx, where Lx is the predicate, is a consequence of some axiom set A, then if the all the axioms of A are true, so is ∃xLx.  But, this does not imply that A implies ∀xLx.  That said, there does exist the quantifier exchange rules of predicate logic which read:

$\lnot$∀x$\lnot$...==∃x... and

$\lnot$∃x$\lnot$...==∀x...

Where "==" indicates logical equivalence.


问:
Find $a$ if $y=f(x)$ satisfies $y-x\frac{dy}{dx} = a(y^2 +\frac{dy}{dx})$

If the curve $y=f(x)$ satisfies the equation $y-x\frac{dy}{dx} = a(y^2 +\frac{dy}{dx})$ and always passes through $(1,1)$, then find number of possible values of $a$.

Assume constant of integration to be 0

$$y-ay^2 =(a+x) \frac{dy}{dx}$$
$$\ln (a+x) =\int a (\frac {1}{ay} + \frac{1}{1-ay})$$
$$a+x=\frac{y}{1-ay}$$

At $(1,1)$

$$(1+a)(1-a)=1$$
$$a=0$$

I could find one value of $a$, but the answer is 3. What are the other two?

答: You have to be careful about zero appearing in the denominator. Also, the integral of $\frac  1 x$ is $\ln |x|$ not $\ln x$. Take $a+x=-\frac y {1-ay}$ to get another solution.  Two possible values of $a$ are $0$ and $1$ with solutions $y=x$ and $y=1$ respectively. For the third value put $x=y=1$ in $a+x=-\frac y {1-ay}$. This gives $a =\pm \sqrt 2$ but I will let you find out which one of these values is actually valid.


问:
How many 6 digit positive integers have their digits in weakly decreasing order?
How do I solve this combinatorically? I'm not quite sure how to approach this since each digit is dependent on the digit before it.

For example, if the first digit is a 9, then there are 10 possible choices for the second digit (0-9).
9 _ _ _ _ _.
If we choose 5 to be the second digit, then the third digit has 6 possible choices (0-5).
9 5 _ _ _ _.
If we choose 0 to be the third digit, then the fourth digit has 1 possible choices (0).
9 5 0 _ _ _.
And for the rest of the digits, only 0 is possible.
9 5 0 0 0 0

What formula do I use to solve this? Could I use Stirling numbers or Permutations and Combinations in this situation?

答:
Notice that each number with weakly decreasing digits is completely determined by the number of times each digit appears.  For instance, if the number contains one 5, two 3s, one 2, and two 1s, it must be the number 533211.

Therefore, if we let $x_i, 1 \leq i \leq 9$, be the number of times digit $i$ appears in the number, the number of six-digit strings with weakly decreasing digits which can be formed is the number of solutions of the equation
$$x_0 + x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 = 6 \tag{1}$$
in the nonnegative integers.

A particular solution of this equation corresponds to the placement of $10 - 1 = 9$ addition signs in a row of six ones.  For instance,
$$+ 1 1 + 1 + 1 1 + + 1 + + + +$$
corresponds to the solution $x_0 = 0$, $x_1 = 2$, $x_2 = 1$, $x_3 = 2$, $x_4 = 0$, $x_5 = 1$, $x_6 = x_7 = x_8 = x_9 = 0$ and the number 533211.

Thus, the number of six-digit weakly decreasing strings is the number of solutions of equation 1 in the nonnegative integers.

  The number of such solutions is $$\binom{6 + 10 - 1}{10 - 1} = \binom{15}{9}$$ since we must choose which nine of the $15$ positions required for six ones and nine addition signs will be filled with addition signs.

The astute reader will have noticed that I have referred to six-digit strings with weakly decreasing digits.  These strings include 000000, which corresponds to the nonnegative number 0.  Since we only wish to count positive numbers, we must subtract the zero string from the total.


问:
My paradox in determining the probability of distributing similar balls into different boxes
Inspired by the discussion between @Matthew Pilling and me in this post What is the probability that there more rabbits than chickens in each of these three cages?, I am trying to find and understand the answer to a simple problem I am thinking for a while. (Feel free to read the comments under my answer in that post if you are interested and want to know more.)

Let's say I have two identical balls and I want to distribute them into two different boxes (box1 and box2). What is the probability of having one ball in each box?

Approach 1:
We can talk about the total number of solutions for $n_{1} + n_{2} = 2$, where $n_{1},n_{2} \in W $. And, then, the number of solutions to $n_{1} + n_{2} = 2$, where $n_{1},n_{2} \ge 1$. The probability will be calculated as: $\frac{1}{3}$.

Approach 2:
Or, we can think of this problem as follows: we have two boxes. The probability of choosing the first box is $\frac{1}{2}$. So, there is $\frac{1}{4}$ probability of choosing the first box for both balls. Same reasoning is true for choosing the second box for both balls. So, in total, there is $\frac{1}{2}$ probability to have both balls in only one box. And, thus, the probability of having only one ball in each box is $\frac{1}{2}$.

Could someone please help me understand what's happening here? Are we dealing with two different problems (in these two approaches) or one of the approaches is wrong because ... ?

Thanks for any input you may provide.

Nima

答:
Your problem is a very common one. Your approach 1 is a combinatorial approach, you count solutions to a discrete (integer) problem. Then you use the classical approach that probability equals "good cases" divided by "all cases".

However, this is only true if you a priori know that each one of "all cases" is as likely to come up as any other. In classical "throw a dice" problem, that assumption is coded into the statment that the involved dice are fair.

In your approach 1, you consider the 3 solutions to the problem $n_1+n_2=2, n_1, n_2 \in \mathbb Z, n_1,n_2 \ge 0.$

You are incorectly thinking that each of those 3 cases is equally likely to come up. It isn't, because when you use approach 2 you see that the actual process you distribute balls is:

Put ball #1 into a box, each with probability $\frac12$, then

Put ball #2 into a box  each with probability $\frac12$.

Your approach 2 directly shows that $n_1=0, n_2=2$ and $n_1=2, n_2=0$ both have probabability $\frac14$, while $n_1=1, n_2=1$ has probability $\frac12$.

The main problem is that incorrect believe that just because you can model the result of putting balls into boxes by the three pairs $(0,2), (1,1), (2,0)$, those pair are equally likely.

In a similar case, you the sum of any 2 independent, fair standard 6 sided dice is a number from $2$ to $12$. Those 11 possible values are not equally distributed either, they are centered around the middle, with sum=$7$ having probaility $\frac16$, while $2$ and $12$ have the much lower probability $\frac1{36}$ each.

Again, your main error is that assuming that some repreasentation if the result of the choices is going to be one where each choice is as likely to come up as any other.


问:
Estimation of a function of a random vector
The joint probability distribution of the random variables $X_1$, $X_2$, ..., $X_n$ is available. The objective is to find the distribution of $\epsilon = g(X_1, X_2, ...,X_n)$. In other words, $g$ is unknown. A sample of size $m$ is drawn from the population: $([X_{i1}, X_{i2},..., X_{in}], \epsilon_i)$ is available for $i=1,...,m$. How to find $\epsilon^*$ such that $P(\epsilon>\epsilon^*)<0.05$. On a more general note, how to find an optimized estimator of $g$, namely $\hat{g}$?

Probably a common problem for professionals in this field but I'm a beginner, so your help is highly appreciated.

答:
I have a suggestion for what method you can use. As I said in the comment, you have no assumption about the function $g$, so this sollution is more likely heuristic, assuming that the function $g$ is nice with respect to this method.

When you are given a point $[X_1, X_2, X_3, \ldots]$ and you need to calculate $g(X_1, X_2, X_3, \ldots)$, you take the measurements within radius $R$ and estimate the 0.95-quantile of them. You just need to fit the radius $R$ to your data, so it gives satisfactory results. Or more generally, you weighten all measurements by a kernel function of your choice (a non-increasing function of distance between two points). Again, you have to choose a fitting kernel function.

If you choose a kernel function which is infinite for zero distance (for example inverted square distance), it will have a special effect. If you are given a measurement at a point and then you ask for the same point, the measurement at that point will outweight all other points, so you get the measurement which you got.


答: The sum of the digits of a number is congruent with itself modulo 9 (this comes from the fact that $10\equiv 1 \mod(9)$, therefore $10^k \equiv 1 \mod(9)$, so, for example, $123=1\times 10^2+2\times 10+3\times 1\equiv 1+2+3=6$). This is commonly used to test divisibility by 9, but in your case, what you are doing is basically subtracting the number's remainder when dividing by 9, getting to a multiple of 9, so when you sum the digits again you get 9 or a multiple of 9.

Title: A question regarding an algorithm

Body:
I'm a software developer, and I wrote an algorithm for converting any positive integer with at least 2 digits into 9, based on what Scott Flansburg discovered.

For example: $N = 22$

Sum up the digits: $2 + 2 = 4$

Then, subtract the result from the original number: $22 - 4 = 18$

The sum of the digits: $1 + 8 = 9$

It is valid for any positive integer with at least 2 digits.

Why does the algorithm work?

Is there a name for this algorithm?

EDIT:

I'm not a mathematician, only a math enthusiast, so I'd appreciate it if the answers wouldn't be complex


问:
Question concerning isomorphism between two direct product of a pair of sets.
For the following question: prove that any two direct products of a pair of sets A and B are isomorphic.

The text where this question comes from gives the following definitions and a theorem which the proof was presented in text:

Given two sets $A$ and $B$, we defined their cartesian products as: $A \times B := \{(x,y)\mid x\in A,  y\in B\}$.

We define their corresponding projection maps on sets $A$ and $B$ respectively as:

$\pi_{A}: A\times B \rightarrow A; (x,y) \mapsto x)$ and

$\pi_{B}: A\times B \rightarrow B; (x,y) \mapsto y)$

Theorem 1: Given any set $C$ and maps $f:C\rightarrow A$, $g:C\rightarrow B$, there is a unique map $(f,g):C\rightarrow A\times B$ such that $(f,g)\pi_{A}=f$ and $(f,g)\pi_{B}=g$.

Definition:  The $\textit{direct product } A\times B$ of two sets $A$ and $B$ is a set $A\times B$ equipped with maps $\pi_{A}: A\times B \rightarrow A$ and $\pi_{B}: A\times B \rightarrow B$ such that, for any set $C$ and set maps $f:C\rightarrow A$, $g:C\rightarrow B$, there is a unique map $(f,g):C\rightarrow A\times B$ such that $(f,g)\pi_{A}=f$ and $(f,g)\pi_{B}=g$.

I am confused about the wording of the question.  Is the question asking the following:  Given sets $M$, $N$, $E$ and $F$, the two cartesian products respectively $A= M \times N$ and $B=E\times F$ We define the function $h:A\rightarrow B$
 and the corresponding pairs of projection maps associated with the product set $A$ and the product set $B$ such that
$\pi_{M}: M\times N \rightarrow M$

$\pi_{N}: M\times N \rightarrow N$

$\pi_{E}: E\times F \rightarrow E$

$\pi_{F}: E\times F \rightarrow F$

Show that the function $h$ is an isomorphism.  I am suppose to define direct product for sets $A$ and sets $B$.  I am not sure how to do that properly and also, do I have to show that the function $h$ is homomorphic,  The text at this point have not defined what homomorphism suppose to look like for direct product of sets.  I am really having trouble rephrasing the question in more formal math notations.  Thank you in advance

答:
The notion 'direct product' is a generalization of the Cartesian product of sets, motivated by Theorem 1. The notation $A\times B$ for this generalized version is also motivated by this, but it doesn't a priori mean the Cartesian product in this context.

The advantage of this generalization is not clearly seen in this context, but observe that the pictural definition of direct product only involves dots (as sets), arrows (as functions) and composition of arrows.

Such a framework is called 'category', and it's very general: for instance instead of sets we can take some algebraic structures like groups, vector spaces or rings as dots and take homomorphisms as arrows.

Other categories contain continuous functions as arrows.

The definition of direct product makes perfect sense in each case, and by the way, in most such category the underlying set of the direct product can be taken as the Cartesian product of the underlying sets.

For a more abstract example, if the dots denote positive integers and there's one arrow $a\to b$ iff $a$ is a divisor of $b$, then in this category, the 'direct product' of numbers (dots) $a,b$ is just their greatest common divisor.

(So, actually, the 'direct product' is a common generalization of the $\gcd$ of numbers and the Cartesian product of sets/structures.)

About the exercise, what you need to prove is that Theorem 1. characterizes the Cartesian product. In order to not getting confused, just replace the disturbing notation $A\times B$ with arbitrary letters, unless we're really talking about the Cartesian product.

Let sets $A,B$ be given and suppose that a set $P$ with maps $p_1:P\to A,\ p_2:P\to B$ satisfies the definition of direct product.

Then actually we can apply Theorem 1., so we obtain the unique map $u:P\to A\times B$
$$u(x)=(p_1(x),\,p_2(x))$$
that makes $\pi_A\circ u=p_1$ and $\pi_B\circ u=p_2$.

For the inverse of $u$, apply the hypothesis on $P,p_1,p_2$ with $C=A\times B$ and $f=\pi_A,\,g=\pi_B$, so we obtain (a unique) $v:A\times B\to P$ that makes $p_1\circ v=\pi_A$ and $p_2\circ v=\pi_B$.

Now, by these compositions, we get
$$p_1\circ v\circ u=\pi_A\circ u=p_1=p_1\circ{\rm id}_P\\
p_2\circ v\circ u=\pi_B\circ u=p_2=p_2\circ{\rm id}_P$$
thus, by the uniqueness condition we must have $v\circ u={\rm id}_P$, and similarly $u\circ v={\rm id}_{A\times B}$.

It means that $u,v$ are inverses of each other and define a bijection between $P$ and $A\times B$.

Now, if also $Q,q_1,q_2$ satisfies the definition of direct product, then we receive another bijection $w,w^{-1}$ between $Q$ and $A\times B$, and composing these with $u$ or $v$ gives a bijection between $P$ and $Q$.


问:
Showing that a 2-variable function is decreasing with respect to a given variable on an interval
For all integers $x\geq 7$ and all real numbers $y$ such that $0.249 \leq y \leq 0.25$ define
$$F(x,y)=-\frac{\log\left( \frac{x+3}{x}\right)+\frac{1}{2}\log\left(\frac{x^2}{x^2-1} \right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{-\log\left( \frac{x+3}{x}\right)+2\sqrt{\frac{x+3}{x}}+\log\left( \frac{x+1}{x}\right)-2\sqrt{\frac{x+1}{x}}}.$$

I claim that $$F(x,y)\geq F(7,y)$$ for all $x \geq 7$.

I thought of several ways to try and prove this:

Working directly on this inequality is complicated (even Maple couldn't do it for me).

I plotted $F(x,y)$ for fixed values of $y$ and saw that $F(x,y)$ was increasing with respect to $x$ in all of the plots. However, the derivative of $F(x,y)$ with respect to $x$ is very complicated as well.

Since the values of $x$ are integers, I thought about an inductive argument, like showing that $F(x,y)\leq F(x+1,y)$ for all $x\geq 7$, but again, this is a difficult inequality to work with.

***If it is too difficult to show that $F(x,y)\geq F(7,y)$, then I would also be very happy to figure out how to at least show that $$F(x,y)\geq 1.05-y$$
for all $x\geq 7$ and $y$ sufficiently close to $0.25$.

Any suggestions are greatly appreciated. Thank you.

Body:
$F(x,y)=-\frac{\log\left( \frac{x+3}{x}\right)+\frac{1}{2}\log\left(\frac{x^2}{x^2-1} \right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{-\log\left( \frac{x+3}{x}\right)+2\sqrt{\frac{x+3}{x}}+\log\left( \frac{x+1}{x}\right)-2\sqrt{\frac{x+1}{x}}}$

$=-\frac{\log\left( \frac{x+3}{x}\right)+\frac{1}{2}\log\left(\frac{x^2}{x^2-1} \right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{-\log\left( x+3\right)+2\sqrt{\frac{x+3}{x}}+\log\left( x+1\right)-2\sqrt{\frac{x+1}{x}}}$

$=-\frac{\log\left( x+3\right)-\frac{1}{2}\log\left(x^2-1 \right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{-\log\left( x+3\right)+2\sqrt{\frac{x+3}{x}}+\log\left( x+1\right)-2\sqrt{\frac{x+1}{x}}}$

$=-\frac{\log\left( \frac{x+3}{\sqrt{x^2-1}}\right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$

$=-\frac{\log\left( \frac{x+3}{\sqrt{x^2-1}}\right)+\frac{1}{2}\log\left( \frac{x+1}{x-1} \right)-y\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$

$=-\frac{\log\left( \frac{x+3}{\sqrt{x^2-1}}\frac{\sqrt{x+1}}{\sqrt{x-1}}\right)-y\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$

$=-\frac{\log\left( \frac{x+3}{\sqrt{x+1}\sqrt{x-1}}\frac{\sqrt{x+1}}{\sqrt{x-1}}\right)-y\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$

$=-\frac{\log\left( \frac{x+3}{x-1}\right)-y\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$

$=\frac{\log\left( \frac{x-1}{x+3}\right)+y\log\left( \frac{x+1}{x-1} \right)+2\sqrt{\frac{x+3}{x}}-3+4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$

Now it should be easier to take the derivative or do other simplification steps.

EDIT: As mentioned in the comments, it seems to be decreasing instead of increasing. That could you also show by the last term of above simplification.


问:
Determining when $(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$
Question:

$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2 \qquad \text{ LHS} \\ $

Answer key:

$\implies  \frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \qquad \text{ RHS} $

I verified RHS = LHS. However, to get to the RHS part, they have factored it somehow and I can't figure out an intuitive way to do so. If anybody has a more intuitive solution (does not have to be the same as what is given here), please provide it. From this step onwards we can easily solve the question because:

$\implies  (\sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0\\ $

$ \implies  (\sin(\theta)-\cos(\theta)-1) \leq0 \because  (3 + \sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)) >0 \text{ }\forall \theta \in \mathbb{R} \\ $

Doubt:

I have no idea how the went from the LHS part to the RHS part. Is there an intuitive way to solve $(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2$? In this solution they have simply said:
$$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) -2 \leq 0 \\ \implies
\frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \\$$
Please provide a logical way to solve it. Thanks in advance.

I'm looking for a no calculator solution where each step is motivated. And prove one factor is always positive or negative and use the other factor to find the region would be appreciated. Although, if there is another method, (as long as its intuitive) that works too

答:
We will use

$$(\sin\theta- \cos\theta)^2
= \sin^2\theta - 2\sin\theta \cos\theta + \cos^2\theta
= 1 - 2\sin\theta \cos\theta$$
a few times.

$$(\sin \theta -\cos \theta)(2+\sin \theta \cos \theta)- 2 \le 0 $$

$$2[(\sin \theta -\cos \theta)(2+\sin \theta \cos \theta)- 2] \le 0 $$

$$(\sin \theta -\cos \theta)(4+2\sin \theta \cos \theta)- 4 \le 0 $$

$$(\sin \theta -\cos \theta)(5-(1-2\sin \theta \cos \theta))- 4 \le 0 $$

$$(\sin \theta -\cos \theta)(5-(\sin\theta - \cos\theta)^2)- 4 \le 0 $$

$$-(\sin\theta - \cos\theta)^3 + 5(\sin\theta - \cos\theta)-4 \le 0$$

$$(\sin\theta - \cos\theta - 1)
(-(\sin\theta - \cos\theta)^2 - (\sin\theta - \cos\theta) + 4) \le 0$$

$$(\sin\theta - \cos\theta - 1)
(-1 + 2\sin\theta\cos\theta - \sin\theta + \cos\theta) + 4) \le 0$$

$$(\sin\theta - \cos\theta - 1)
(\cos\theta - \sin\theta + 2\sin\theta\cos\theta + 3) \le 0$$


问:
Why are roots of unity always distinct?
As per the fundamental theorem of algebra, $z^n-1=0$ has $n$ complex solutions--but, for a general polynomial, some of those solutions may have multiplicity $> 1$, they may repeat.

Why is it that roots of unity never repeat, i.e., that the equation above always has $n$ distinct solutions for all $n$?

答:
Suppose a complex root $ \ \alpha \ $ had multiplicity 2 . Then $ \ (z \ - \ \alpha)^2 \ $ would generate a term $ \ -2 \alpha z \ $; similarly, higher multiplicities would produce terms which had powers of $ \ z \ $ other than $ \ x^n \ $ or the "constant term".

The coefficient of the $ \ z^{n-1} \ $ term is zero, which is the negative of the sum of the $ \ n \ $ zeroes of the polynomial.  In order to meet this condition and not produce "unwanted" terms, the zeroes must all be distinct.

We can in fact go further:  the zeroes must be paired in complex conjugates  $ \ a \ \pm \ bi \ $ .  If  $ \ n \ $ is even, then $ \ 1 \ $ and $ \ -1 \ $ are zeroes, and the other $ \ (n-2) \ $ zeroes (an even number) must have "conjugate pairs" with the $ \ a \ $'s in one pair balanced by $ \ -a \ $'s in another pair.  If  $ \ n \ $ is odd, then $ \ 1 \ $ is a zero, and  the other $ \ (n-1) \ $ zeroes (an even number) must have "conjugate pairs" with the $ \ a \ $'s in the pairs summing to $ \ -1 \ $.

Title: Why are roots of unity always distinct?


问:
Evaluate $\int^1_0 x^a (1-x)^b \operatorname{Li}_2 (x)\, \mathrm dx$
For what $a,b$ the integral

$$\int^1_0 x^a(1-x)^b\operatorname{Li}_2 (x)\, \mathrm dx$$

has a closed form solution? I tried to solve it by expanding dilogarithm function, or by reducing it to linear combination of similar integrals, but I was unable to sum up series containing product of harmonic numbers. I wonder whether a closed formula is indeed available.

答:
Reduction. As @Raffaele mentioned, by expanding dilog, integrating term by term (Beta integral) and summing up, one reduces the integral to a generalized hypergeometric function, i.e.
$$\int^1_0 x^a (1-x)^b\operatorname{Li}_2 (x)\, \mathrm dx=B(a+2,b+1) \, _4F_3(1,1,1,a+2;2,2,a+b+3;1)$$

Claim. This hypergeometric function can be expressed by Multiple Zeta Values of level $1,2,4$ (up to Gamma factors) whenever:

$2a\in \mathbb Z, 2b\in \mathbb Z$

$4a\in \mathbb Z, 4b\in \mathbb Z, \text{and at least one of}\ a, b, a+b\in \mathbb Z$

$4a\in \mathbb Z, 4b\in \mathbb Z, a=b$

And can be expressed via polygamma function whenever:

$a=b$

$a+b\in \mathbb Z$

$\text{At least one of}\ a+2b, b+2a \in \mathbb Z$

Proof. For proof of the MZV part (mainly by Fourier Legendre expansion), see Theorem $1, 2$ of this paper (there seems to be no general formula though). For case $1$ of polygamma part, recall reflection formula of dilog one have a general formula that

$\int^1_0 x^{a-1} (1-x)^{a-1}\operatorname{Li}_2 (x)\, \mathrm dx=\frac12\int^1_0 x^{a-1} (1-x)^{a-1}(\operatorname{Li}_2 (x)+\operatorname{Li}_2 (1-x))\, \mathrm dx$
$=\frac12\int^1_0 x^{a-1} (1-x)^{a-1}(\zeta(2)-\log(x)\log(1-x))\, \mathrm dx=\frac12(\zeta(2)B(a,a)-\partial^{(1,1)}B(a,a))$
$=\frac{1}{12} \left(-6 (\psi ^{(0)}(a)-\psi ^{(0)}(2 a))^2+6 \psi ^{(1)}(2 a)+\pi ^2\right) B(a,a)$

For case $2$ of polygamma part, see Theorem $3$ of paper linked above. For case $3$, due to second entry of this page the integral is evaluable whenever $b+2a=-2$. As @pisco mentioned, by repeated IBP and modulo polygamma integrals $\int_0^1 x^* (1-x)^* \log^*(1-x) \mathrm dx$ one may evaluate all $I(a+m,b+n)$ once $I=I(a,b)$ is known; this completes the proof of case $b+2a\in\mathbb Z$. The case $a+2b\in \mathbb Z$ is direct by reflection again (see case $1$).

Examples. Case $1$ of MZV part:

$a=\frac12, b=\frac32: I=\frac{\pi ^3}{96}-\frac{17 \pi }{576}-\frac{1}{8} \pi  \log ^2(2)+\frac{1}{48} \pi  \log (2)$

Case $2$ of MZV part:

$a=\frac54, b=\frac34: \\ I=\frac{\pi  \left(-480 C+35 \pi ^2-178 \pi -134-540 \log ^2(2)+180 \pi  \log (2)+1068 \log (2)\right)}{1536 \sqrt{2}}$

$a=-1, b=\frac14: \\ I=-32 C-48 \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-14 \zeta (3)+\frac{7 \pi ^3}{8}-4 \pi ^2+64+\frac{3}{2} \pi  \log ^2(2)+3 \pi ^2 \log (2)$

Case $3$ of MZV part:

$a=b=-\frac54: I= \frac{\left(-5 \pi ^2+24 \pi +12 \log ^2(2)-12 \pi  \log (2)-48 \log (2)\right) \Gamma \left(\frac{3}{4}\right)^2}{3 \sqrt{\pi }}$

Polygamma part:

$a=b=-\frac56: I=-\frac{\sqrt{\pi } \Gamma \left(\frac{7}{6}\right) \left(\pi ^2+24 \log ^2(2)+8 \pi  \sqrt{3} \log (2)-6 \psi ^{(1)}\left(\frac{1}{3}\right)\right)}{\sqrt[3]{2} \Gamma \left(\frac{2}{3}\right)}$

Notes.

In MZV Case $1$ example I omit the trivial case $a, b\in \mathbb Z$ since they are direct via IBP.

MZV Case $2$ examples are more complicated; for reason of appearance of certain constants, see paper linked above.

MZV Case $3$ example can be established by either Fourier Legendre expansion or method of "polygamma part",  with the former generalizable to trilog case (and above) but the latter not.

Method of polygamma part, case $1,3$ are ad-hoc and not generalizable to trilog. Case $2$ of polygamma part is generalizable since only partial fractions and Beta derivatives are involved.

I only give a polygamma Case $1$ example; further examples are left to readers.

This post is the trilog counterpart of this problem; some more difficult series are evaluated there. For more information, see paper linked above. The theoretical basis of evaluating these integrals and series are given here.


问:
Simple version of asymmetric encryption
I'm searching for an example of a simple, asymmetric encryption algorithm, which could be used to present someone, who is not fluid in maths, how asymmetric encryption works. An obvious example of asymmetric encryption algorithm is RSA, but this algorithm requires far more knowledge than an average person might have.

A trivial example might be f(x, pub) = x * pub = c and g(c, priv) = c * priv = x, under assumption, that priv = 1 / pub.

However, this example is trivial, because pub and priv are directly related to one another and having public key you can easily derive private one. And frankly, this is actually symmetric algorithm stretched to look like asymmetric rather than truly asymmetric.

So I'm searching for a demo asymmetric encryption algorithm, which:

Requires minimum mathematical knowledge to understand fully (in contrast to RSA, where you have to understand chinese rest theorem, small Fermat's theorem, modulo calculus etc.)

Is nontrivial (eg. having public key and cipher you cannot easily find the private key or plaintext and by easily I mean eg. a simple formula)

Does not have to be secure (eg. bruteforce attacks may be easy to perform)

Is there an algorithm, which matches these requirements?

I've been trying to invent such algorithm by myself and my thinking is as following: we need to find a function f(c, pub), which causes some information to be lost in the process, which is not recoverable easily. The same information is in the priv key, which allows to restore the original information.

An example of a function, which causes information loss is obviously modulo, because it doesn't have a direct reverse function. I mean direct, because knowing the integral part of division, you may reverse the modulo. I however cannot think of an algorithm, which could take advantage of this property. Or else other function, which has this property of being "reversible under specific conditions".

答:
''However, this example is trivial, because pub and priv are directly related to one another and having public key you can easily derive private one. And frankly, this is actually symmetric algorithm stretched to look like asymmetric rather than truly asymmetric.''

Your example hits the point:

''A trivial example might be f(x, pub) = x * pub = c and g(c, priv) = c * priv = x, under assumption, that priv = 1 / pub.''

In the RSA algorithm the situation is similar:

Each participant has a public key $(n,e)$ and a private key $d$ such that $d$ is the inverse of $e$ modulo $\phi(n)$, where $\phi$ is the Eulerian phi-function.

An adversary knows $n$ but not the factorization $n=pq$ into distinct primes $p,q$ and so he/she does not know $\phi(n)$ which is $(p-1)(q-1)$.

Title: Simple version of asymmetric encryption


问:
Combining convergence in probability and the means of the positive sequence of r.v. implies convergence in L 1
Let $\{X_n\}$ be a collection of positive random variable with $X_n \rightarrow X$ in probability. Prove that if $E(X_n) \rightarrow E(X)$, then $X_n \rightarrow X$ in $L^1$.

My partial answer:
Let $(\Omega,B,P)$ be the probability space. For any $\varepsilon>0$, we have
\begin{align*}
\int_{\Omega} |X_n-X| dP &= \int_{\{|X_n-X|<\varepsilon\}} |X_n-X| dP+\int_{\{|X_n-X|\geq \varepsilon\}} |X_n-X| dP \\ &<\varepsilon P(|X_n=X|<\varepsilon)+\int_{\{|X_n-X|\geq \varepsilon\}} |X_n-X| dP \\ &<\varepsilon +\int_{\{|X_n-X|\geq \varepsilon\}} |X_n-X| dP
\end{align*}
The problem is I don't know how to estimate the second term in RHS with information convergence in probability and expectation convergence.

Thanks for your help.

答: Let $Y_n:=|X-X_n|+X-X_n$. Then $0\leqslant Y_n\leqslant 2X$. As $E(X_n)\to E(X)$, we just need to show that $E(Y_n)\to 0$. As $X_n\to X$ in measure, we have $Y_n\to 0$ in measure. We can conclude by dominated convergence.</p>


问:
What is a linear transformation $T:\mathbb{R}^3\longrightarrow\mathbb{R}^4$, whose null space is generated by $(0,1,-3)$ and $(0,-3,4)$?
What is a linear transformation to $T:\mathbb{R}^3\longrightarrow\mathbb{R}^4$, whose null space is generated by $(0,1,-3)$ and $(0,-3,4)$?

So, I know that the given pair of vectors generate the subspace $\{(0,y,z):\ y,z\in\mathbb{R}\}$, as they are linearly independent. By the Rank-nullity theorem, the image space in $\mathbb{R}^4$ is of dimension $3-2=1$.

But I can't seem to go further than this. Any help will be appreciated.

答:
Note that in general, given a linear mapping $T\colon V\to W$, $T$ is completely determined by its action on a basis $\beta$ of $V$. (Proof included below).

In your example, given any basis $\beta=\left\{v_1,v_2,v_3\right\}$ of $\mathbb{R}^3$, $T$ is completely determined by $T(v_1),T(v_2)$ and $T(v_3)$.

Note that $v_1=(0,1,-3)$ and $v_2=(0,-3,4)$ are linearly independent and you definitely want $T(v_1)=0=T(v_2)$. Now choose $v_3$ such that $\beta$={v_1,v_2,v_3} is a basis. You don't want $T(v_3)=0$ as this would make the null space larger than you want.

You can now simply plug any values that fit this data!

Proof of claim: Let $v\in V$ be any vector and let $\beta$ be a basis of $V$. As $\beta$ is basis, we can write $$v=\sum_{i=1}^n\lambda_iv_i$$ where $\lambda_i\in \mathbb{R}$ (or whatever field you are working over) and each $v_i$ belongs to $V$. Then $$T(v)=T(\sum_{i=1}^n \lambda_iv_i)=\sum_{i=1}^n\lambda_iT(v_i)$$
is determined by the values $T(w)$ for all $w\in \beta$.


问:
Find all equilibria of this zero sum game
We have to find the equilibria of the zero sum game specified by the following matrix:
\begin{matrix}
\\  & A & B & C
\\ T &-2 & 10 & 4
\\ M & 7 & 8 & 7
\\ B & 8 & 1 & 4
\end{matrix}

Clearly, a saddle point is - 1 plays M, 2 plays C, which gives a value of 7. My question is, are there any other equilibria of this game? Also, in general, if a zero sum game has saddle points - i.e. pure strategy equilibria - can it still have additional mixed strategy equilibria?

答:
In general - yes. For example, take the regular $2\times 2$ matching pennies and add an action to each player, where all payoffs in the added row and column are $0.5$. This game has the previous mixed equilibrium and some new ones in which one or both players play their new action.

Here, we can look for more equilibria using the following theorem for zero-sum games: if $(s_1,s_2)$ is an equilibrium and $(b_1,b_2)$ is another equilibrium, then also $(s_1,b_2)$ and $(b_1,s_2)$ are equilibria. The idea is that in a zero-sum game, an action is "optimal" and then can be played against any action of the opponent.

For example here, we deduced that $C$ is optimal for 2. This means that if there is another equilibrium where player 1 mixes, he can play it against $C$ and get the same result. But any mixture against $C$ is sub-optimal, so Player 1 has no mixed optimal strategy.

You can do the same exercise for player 2 and see that he can mix $A$ and $C$, but he should play $A$ with low enough probability so that $M$ will still be the best response of Player 1. (lower than $0.75$).

To conclude here there are infinitely many equilibria of the form: Player 1 plays $M$, player 2 mixes $A$ w.p. $x\leq 0.75$ and $C$ w.p. $1-x$.


问:
If $\binom{n}{r} = 45$, and $(n-3r)! = 24$, Is there a method of knowing $n$ and $r$ without guessing?
I got the equation:

$n = 4 + 3r.$

So, $\displaystyle\binom{4+3r}{r} = 45$, and by using trial and error, I get $r = 2$, and $n = 10$, which is correct.

Body:
If $r\ge 2$ then $$\binom {4+3r}{r}= \prod_{j=0}^{r-1}\frac {(4+3r-j)}{j+1}\ge \frac {(4+3r)}{1}\cdot  \frac {(3+3r)}{2}\ge\frac {(10)(9)}{2}=45$$ (....because each term $\frac {(4+3r-j)}{j+1}$ is $\ge 1$ ....)

with equality only when $r=2$.

If $r\le 1$ then $$\binom {4+3r}{r}\le \binom {4+3r}{1}= 4+3r \le 7.$$


答:
Here's a very elementary approach. Roughly speaking, the entries in Pascal's triangle grow very quickly, except the $1$'s on the edges, so we can just compute enough of the triangle to be sure we've found all the $45$'s:

```<code>                      1
                    1   1
                  1   2   1
                1   3   3   1
              1   4   6   4   1
            1   5  10  10   5   1
          1   6  15  20  15   6   1
        1   7  21  35  35  21   7   1
      1   8  28   *   *   *  28   8   1
    1   9  36   *   *   *   *  36   9   1
  1  10  45   *   *   *   *   *  45  10   1
1  11   *   *   *   *   *   *   *   *  11   1
</code>```

Here the *'s indicate the value is $ > 45$. The only 45's here are $\binom{10}{2} = \binom{10}{8} = 45$, and only the former is of the correct form. The only other 45's in the triangle will be $\binom{45}{1} = \binom{45}{44} = 45$, neither of which is of the correct form.

Title: If $\binom{n}{r} = 45$, and $(n-3r)! = 24$, Is there a method of knowing $n$ and $r$ without guessing?

Body:
I got the equation:

$n = 4 + 3r.$

So, $\displaystyle\binom{4+3r}{r} = 45$, and by using trial and error, I get $r = 2$, and $n = 10$, which is correct.

Body:
Here's a very elementary approach. Roughly speaking, the entries in Pascal's triangle grow very quickly, except the $1$'s on the edges, so we can just compute enough of the triangle to be sure we've found all the $45$'s:

                      1
                    1   1
                  1   2   1
                1   3   3   1
              1   4   6   4   1
            1   5  10  10   5   1
          1   6  15  20  15   6   1
        1   7  21  35  35  21   7   1
      1   8  28   *   *   *  28   8   1
    1   9  36   *   *   *   *  36   9   1
  1  10  45   *   *   *   *   *  45  10   1
1  11   *   *   *   *   *   *   *   *  11   1

Here the *'s indicate the value is $ > 45$. The only 45's here are $\binom{10}{2} = \binom{10}{8} = 45$, and only the former is of the correct form. The only other 45's in the triangle will be $\binom{45}{1} = \binom{45}{44} = 45$, neither of which is of the correct form.


问:
Subrings and homomorphisms of unitary rings
Let $(R,+,\cdot)$ be a ring (in the definition i use multiplication is associative operation and it's not assumed that there is unity in the ring).

I've seen two definitons of subring.

1) non-empty subset $S \subset R$ is called subring of ring $(R,+,\cdot)$ iff $(S,+,\cdot)$ is a ring

2) Let $(R,+,\cdot)$ be unitary ring with unity $e$. Non-empty subset $S \subset R$ is called subring of ring $(R,+,\cdot)$ iff $(S,+,\cdot)$ is a ring and $e \in S$.

I'm looking for an example of such ring R and its subset S that $(S,+,\cdot)$ is a ring but not a unitary ring.

I will be also very grateful for an example of such UNITARY rings $(R_1,+_1,\cdot_1)$, $(R_2,+_2,\cdot_2)$ and function $f: ~~ R_1 \longrightarrow R_2$ that

$(1) \forall a,b \in R_1 ~~~ f(a+_1 b) = f(a) +_2 f(b) $

$(2) \forall a,b \in R_1 ~~~ f(a\cdot_1 b) = f(a) \cdot_2 f(b)$

$(3) f(e_1) \neq e_2$

where $e_1$ is unity in $R_1$; $e_2$ - in $R_2$.

Thanks in advance.

答:
Possibly the simplest example of the second kind is provided by the rings of matrices
$$
R_1 = \left\{ \begin{bmatrix}a&0\\0&0\end{bmatrix} : a \in \Bbb{Z}
\right\},
\qquad
R_2 = \left\{ \begin{bmatrix}a&0\\0&b\end{bmatrix} : a, b \in \Bbb{Z}
\right\},
$$
with
$$
f\left(\begin{bmatrix}a&0\\0&0\end{bmatrix}\right) = \begin{bmatrix}a&0\\0&0\end{bmatrix}.
$$

The identity of $R_1$ is
$$
e_1 = \begin{bmatrix}1&0\\0&0\end{bmatrix},
$$
that of $R_2$ is
$$
e_2 = \begin{bmatrix}1&0\\0&1\end{bmatrix},
$$
and $f(e_1) = e_1 \ne e_2$.

The same example can also be seen via $f : \Bbb{Z} \to \Bbb{Z} \times \Bbb{Z}$ given by $f(a) = (a, 0)$, but using matrices makes it slightly more elementary.

Also, $R_1$ is a subring of $R_2$, both rings are unitary, but the identity elements are different.


答:
Often physical examples work better than digital for demonstration purposes. Asymmetric encryption would then be a padlock.

Suppose someone wants to send you an item in a chest. If the chest has a regular key lock, the sender needs to have a copy of the key to lock it, and you need a copy to unlock it. So the sender needs to send you a copy of the key in addition to sending you the chest.

Now, if it's a padlock chest, you can send out a bunch of padlocks. Only you have the keys to unlock them, but anyone can lock them. So the sender can simply get one of your padlocks and use it to lock the chest.

Now, suppose someone is intercepting your post. If a key is send, they can simply make a copy, and then use it to open the chest. If they intercept a padlock, making a key that can open it is that much more difficult. Of course, they could swap the padlock - but that's a problem that can happen with digital asymmetric cryptography as well.

Title: Simple version of asymmetric encryption

Body:
I'm searching for an example of a simple, asymmetric encryption algorithm, which could be used to present someone, who is not fluid in maths, how asymmetric encryption works. An obvious example of asymmetric encryption algorithm is RSA, but this algorithm requires far more knowledge than an average person might have.

A trivial example might be f(x, pub) = x * pub = c and g(c, priv) = c * priv = x, under assumption, that priv = 1 / pub.

However, this example is trivial, because pub and priv are directly related to one another and having public key you can easily derive private one. And frankly, this is actually symmetric algorithm stretched to look like asymmetric rather than truly asymmetric.

So I'm searching for a demo asymmetric encryption algorithm, which:

Requires minimum mathematical knowledge to understand fully (in contrast to RSA, where you have to understand chinese rest theorem, small Fermat's theorem, modulo calculus etc.)

Is nontrivial (eg. having public key and cipher you cannot easily find the private key or plaintext and by easily I mean eg. a simple formula)

Does not have to be secure (eg. bruteforce attacks may be easy to perform)

Is there an algorithm, which matches these requirements?

I've been trying to invent such algorithm by myself and my thinking is as following: we need to find a function f(c, pub), which causes some information to be lost in the process, which is not recoverable easily. The same information is in the priv key, which allows to restore the original information.

An example of a function, which causes information loss is obviously modulo, because it doesn't have a direct reverse function. I mean direct, because knowing the integral part of division, you may reverse the modulo. I however cannot think of an algorithm, which could take advantage of this property. Or else other function, which has this property of being "reversible under specific conditions".