问:
Algebra problem (problem from Swedish 12th grade ‘Student Exam’ from 1932)
The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.

The sum of two numbers are $a$, the sum of the 3rd powers is $10a^3$. Calculate the sum of the 4th powers, expressed in $a$.

Is there a shorter/simpler solution than the one presented below? It feels there is some ‘trick’ to it. The solution presented below is more a ‘straight forward’ one.

Solution

We have
\begin{gather*}
\left\{
\begin{aligned}
x+y&=a\\
x^3+y^3&=10a^3
\end{aligned}
\right.
\quad\Leftrightarrow\quad
x^3+(a-x)^3=10a^3
\quad\Leftrightarrow\quad
x^2-ax-3a^2=0
\end{gather*}
which has the solutions
$$
x_{1,2}=\tfrac{1}{2}(1\pm\sqrt{13}\,)a
\qquad \Rightarrow \qquad
y_{1,2}=\tfrac{1}{2}(1\mp\sqrt{13}\,)a.
$$

Since
$$
(1+z)^4+(1-z)^4=2(1+6z^2+z^4)
$$
we have
\begin{align*}
x_1^4+y_1^4
&
=\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}
=\tfrac{a^4}{16}\cdot2\bigl(1+6z^2+z^4\bigr)\big|_{z=\sqrt{13}}
\\&=\tfrac{a^4}{8}(1+6\cdot13+13^2)
=\tfrac{a^4}{8}\cdot248
=31a^4
\end{align*}
and, as above,
$$
x_2^4+y_2^4
=\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}
=31a^4.
$$

Hence, the answer is $31a^4$.

The original exam

答:
Yes, there is a shorter solution that relies on the identity $$(x+y)(x^n + y^n) = (x^{n+1} + y^{n+1}) + xy(x^{n-1} + y^{n-1}),$$ which is easily verified by multiplication.  Then if we let $f_n = x^n + y^n$, this may be written $$f_{n+1} = (x+y)f_n - xy f_{n-1} = f_1 f_n - xy f_{n-1}.$$

Next, observe that $f_0 = 2$ for any nonzero choice of $x, y$.  In addition, we are given $f_1 = a$.  Then $$f_2 = f_1^2 - xy f_0 = a^2 - 2xy,$$ and $$f_3 = f_1 f_2 - xy f_1 = a(a^2 - 2xy - xy) = a(a^2 - 3xy).$$  Since we are also given $f_3 = 10a^3$, it follows that $10a^3 = a(a^2 - 3xy)$ and if $a \ne 0$, we obtain $$xy = -3a^2.$$  This gives us the information we need to compute $$f_4 = f_1 f_3 - xy f_2 = a (10a^3) - (-3a^2) (a^2 - 2(-3a^2)) = 31a^4$$ and we are done.  If $a = 0$, then $f_n$ is trivially $0$ for all $n \ge 1$.

Note this method furnishes the more general recursion $$f_{n+1} = a f_n + 3a^2 f_{n-1}, \quad f_0 = 2, \quad f_1 = a,$$ for which the non-recursive solution is $$f_n = \left(\frac{a}{2}\right)^n \left( (1 - \sqrt{13})^n + (1 + \sqrt{13})^n \right).$$


问:
Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$
Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$

The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank you!

答:
Let $x=5/u$ with $0\lt u\lt5$ (there are no solutions with $x\lt0$ since $35/12\gt0$). The equation becomes

$${1\over u}+{1\over\sqrt{25-u^2}}={7\over12}={1\over3}+{1\over4}$$

The solutions $u=3$ and $u=4$ (corresponding to $x=5/3$ and $x=5/4$) are easy to see by inspection.  To show there are no other solutions, it suffices to note that

$$\begin{align}
f(u)=u^{-1}+(25-u^2)^{-1/2}
&\implies f'(u)=-u^{-2}+u(25-u^2)^{-3/2}\\
&\implies f''(u)=2u^{-3}+(25-u^2)^{-3/2}+3u^2(25-u^2)^{-5/2}\gt0
\end{align}$$

so the curve is convex on $(0,5)$ and thus $f(u)$ cannot take any value more than twice.

ContentLicense: CC BY-SA 4.0


答:
A very nice question.

You can think in this way: if we denote
$$x+y = s\\
x^3 + y^3= t\\
x^4 + y^4 = u$$
$s$, $t$, $u$ depend on only two parameters $x$, $y$, so they have only two degrees of freedom. Therefore, there must be some relation between $s$, $t$, $u$. It can be obtained by eliminating $x$, $y$ from the above equation. The relation obtained will be a polynomial one. I used WA and got this relation
$$s^6 - 8 s^3 t - 2 t^2 + 9 s^2 u=0$$
You can  check it directly by substituting $s=x+y$, $t=x^3 + y^3$, $u=x^4 + y^4$.

Title: Algebra problem (problem from Swedish 12th grade ‘Student Exam’ from 1932)

Body:
The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.

The sum of two numbers are $a$, the sum of the 3rd powers is $10a^3$. Calculate the sum of the 4th powers, expressed in $a$.

Is there a shorter/simpler solution than the one presented below? It feels there is some ‘trick’ to it. The solution presented below is more a ‘straight forward’ one.

Solution

We have
\begin{gather*}
\left\{
\begin{aligned}
x+y&=a\\
x^3+y^3&=10a^3
\end{aligned}
\right.
\quad\Leftrightarrow\quad
x^3+(a-x)^3=10a^3
\quad\Leftrightarrow\quad
x^2-ax-3a^2=0
\end{gather*}
which has the solutions
$$
x_{1,2}=\tfrac{1}{2}(1\pm\sqrt{13}\,)a
\qquad \Rightarrow \qquad
y_{1,2}=\tfrac{1}{2}(1\mp\sqrt{13}\,)a.
$$

Since
$$
(1+z)^4+(1-z)^4=2(1+6z^2+z^4)
$$
we have
\begin{align*}
x_1^4+y_1^4
&
=\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}
=\tfrac{a^4}{16}\cdot2\bigl(1+6z^2+z^4\bigr)\big|_{z=\sqrt{13}}
\\&=\tfrac{a^4}{8}(1+6\cdot13+13^2)
=\tfrac{a^4}{8}\cdot248
=31a^4
\end{align*}
and, as above,
$$
x_2^4+y_2^4
=\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}
=31a^4.
$$

Hence, the answer is $31a^4$.

The original exam


问:
Solution without truth tables
Find all the estimations that show that the following is false.
$(x \land y)\lor (x\land z)\lor(y\land z)\lor(u\land v)\lor (u\land w)\lor(v\land w)\lor(\neg x\land\neg u)$.
I know how to solve this by using truth tables however, I need 64 lines.

How can I simplify it, or use another way to solve it?

答:
Since this is a disjunction, for it to be false you need each of the propositions in the disjunction to be false.

Each of the propositions is a conjunction, so in order to be false, you just need one of the two to be false.

The first three and fourth through sixth are symmetric, so the analysis for those for $x,y,z$ will be the same as for $u,v,w$ before we get to the last one.

For $x\wedge y$, $y\wedge z$, and $x\wedge z$ to all be false, you need at least two of $x,y,z$ to be false; at most one can be true. Likewise for $u$, $v$, and $w$.

Now, in the last one you have $\neg x\wedge\neg u$. For this to be false, you need either $x$ to be true (which forces $y$ and $z$ to be false), or $u$ to be true (which forces $v$ and $w$ to be false), or both.

So, when is this full proposition false?

When $x$ is true and $u$ is false, in which case $y$ and $z$ are false, and at most one of $v$ and $w$ is true (three possibilities total).

When $x$ is false and $u$ is true, in which case $v$ and $w$ are both false and at most one of $y$ and $z$ is true (three possibilities total).

When both $x$ and $u$ are both true, in which case $y$, $z$, $v$, and $w$ are all false.

That gives 3 assignments for case 1, three assignments for case 2, and one assignment for case 3, giving you all seven situations in which the proposition is false.

Title: Solution without truth tables


Body:
Since $x+y=a$,

$$10a^3=x^3+y^3=(x+y)^3-3xy(x+y)=a^3-3axy\,,$$

and $xy=-3a^2$.

$$\begin{align*}
x^4+y^4&=(x+y)^4-2xy\left(2x^2+3xy+2y^2\right)\\
&=(x+y)^4-2xy\left(2(x+y)^2-xy\right)\\
&=a^4+6a^2\left(2a^2+3a^2\right)\\
&=a^4+30a^4\\
&=31a^4\,.
\end{align*}$$


问:
Calculate path of vehicle with two wheels parallel to each other
There's a vehicle that starts at $P(0|0)$ oriented in positive y direction. This vehicle has two wheels that are a certain distance $d$ apart, have a diameter of $\frac{10}\pi cm$, are parallel to each other and can move at different speeds ($v_r$ and $v_l$) backwards and forwards. $d$, $v_l$ and $v_r$ are known constants.

I'm wondering if it is possible to calculate the vehicles coordinates at any point in time.

A concrete example would be for $d=8cm; \; v_l=-1\frac{rev}{s}; v_r=3\frac{rev}{s}$

Here's a visualization of the above description:

Note that this is only a model representation and the scale might not be perfectly correct.

Edit: The black dots in the first and in the second figure represent the center of the vehicle and not the point $P(0|0)$. I just marked the point in the first figure to clarify, that the vehicle's center is starting at $(0|0)$.

I've noticed that there always seems to be a point where both wheels are rotating around. If this is true finding this point could help also finding the location of the vehicle but I'm not 100% sure

答:
With the help of this sketch:

We know that the length of an arc is $l=\beta R$

Let the center of rotation be O, located in the x-axis, at distance s from the y-axis

In a time $dt$ the right wheel goes $l_1=v_1dt$, while the left wheel moves $l_2=v_2dt$

So, $l_1=v_1dt=d\beta (\frac{L}{2}+s)$  and $l_2=-v_2dt=-d\beta (\frac{L}{2}-s)$

Notice the "-" sign for $v_2$ as it goes as "-y".

The angle $d\beta$ is the same for both wheels, so:

$\frac{v_1dt}{\frac{L}{2}+s } \; =\;\frac{-v_2dt}{\frac{L}{2}-s }$

$v_1\frac{L}{2} - v_1s \;=\; -v_2\frac{L}{2} - v_2s$

$(v_1+v_2)\frac{L}{2} \;=\; (v_1-v_2)s$

$s= \frac {v_1+v_2}{v_1-v_2}\frac{L}{2}$

That formula works for all values of $v_1$ and $v_2$, except for the case $v_1=v_2$ when the vehicle draws a straight (radius=$\infty$) and not a curve.

Also  notice that if both wheels spin in the same direction then $s>\frac{L}{2}$, out of AB zone.

You see, the distance "s" is the same over time, which means that the path of every point in the wheel-axis is a circumference, centered at point O.

Then, the path for the middle point of the wheel-axis (your black box) is also a circumference centered at $(-s,0)$ and radius $R=s$

$(x+s)²+y²=s²$

Expressed with parametrics and time:

$x=s·cos (\frac{v_1·t}{\frac{L}{2}+s}) - s\;\;\;$  and $\;\;\; y=s·sin (\frac{v_1·t}{\frac{L}{2} +s})\;\;\;$ if $v_1 \neq 0$

or

$x=s·cos (\frac{v_2·t}{\frac{L}{2}-s}) - s\;\;\;$  and $\;\;\; y=s·sin (\frac{v_2·t}{\frac{L}{2} -s})\;\;\;$ if $v_2 \neq 0$

In the above formulas the velocities are expressed in linear units per time. If you have rev/s you need the radius a of the wheel to find $v_i= v_{rev}·a$


问:
Notation for set of equivalence classes
Suppose we define an equivalence relation $\sim$ on a set $S$. The exact details of the relation and checking that it verifies reflexivity, symmetry, and transitivity is less important, so let's assume it satisfies those properties. Then $S/\sim$, read "S mod $\sim$ denotes the set of equivalence classes.

I am trying to make sense of this notation, specifically the use of "mod." In a quotient $R/S$, we say that $r_1, r_2 \in R$ are equivalent if $r_1 - r_2 \in S$. Similarly, in $\mathbb{Z}/2\mathbb{Z}$, $a$ is equivalent to $b$ if $a - b \in 2\mathbb{Z}$. I do not see any parallel in general to $S/\sim$. The only similarity is that they are equivalent "with respect to $\sim$."

Am I missing something? Is there a better way to think of this notation or even what it means to "quotient by $\sim$"?

答:
When you defined $a \equiv b \pmod n$ we use the notation $\pmod n$ BECAUSE $a- b\in n\mathbb Z$ is an equivalence relation that means that the relation $a-b\in n\mathbb Z$ "partitions" $\mathbb Z$ into equivalence classes and $\mathbb Z/n\mathbb Z$ is the set of equivalence classes it is partitioned into.

$\mathbb Z/n\mathbb Z = \{[0],[1],......., [n-1]\}$ where $[k]=\{m\in \mathbb Z| m\equiv k \pmod n\} = \{m \in \mathbb Z| m-k \in n\mathbb Z\} = \{....,k-n, k, k+n, k+2n,.....\}$.

But there is nothing magical about the relation $a\equiv b$.  EVERY equivalence relation will partition a set, $S$ into equivalence classes $U_\alpha$ where each and every $s\in S$ is in exactly one, and only one, $U_\alpha$ and all the elements in $U_\alpha$ are related to each other.  And $S/\sim$ is the set of all these equivalence classes.

So for instance; take any equivalence relation $\sim$ an a set $S$.  For example let say $S= \mathbb N$ and $a \sim b\iff $ the highest power of $3$ that divides $a$ is the same highest power of $3$ that divides $b$.

for example $3\sim 6 \sim 12 \sim 51 etc. $ because $3|6, 12, 51$ but $3^2 \not\mid 6,12,51$.

If we let $U_1 = \{3, 6, 12, 15, 21, 24, .....\}=\{k\in \mathbb Z| 3\mid k, 9\not \mid k\}$ then that is one equivalence class.  If we let $U_2=\{9,18,36,....\} = \{k\in \mathbb Z| 9|k, 27\not \mid k\}$ that's another equivalence class.  And we can let $U_m =\{k\in \mathbb Z| 3^m|m, 3^{m+1}\not \mid k\}$ can be any other.

Also note $\{0\} = U_\infty = \{k\in \mathbb N| 3^j\mid k$ for all $k\in \mathbb N\}$ is the final equivalence class..

Then $\mathbb Z/\sim = \{U_\infty, U_1, U_2, U_3......\}$ is the set of all equivalence classes.

That is entirely equivalent to $\mathbb Z/(\pmod n) = \{[0],[1],.....[n-1]\}$ where $[k] = \{.... k-2n, k-n, k, k+n,...\} = \{k+mn|m\in \mathbb Z\} = \{z\in \mathbb Z| a\equiv z\pmod n\} = \{z\in \mathbb Z| a-z\in n\mathbb Z\}$ are the set of all equivalence classes.

=====

I do not see any parallel in general to S/∼. The only similarity is that they are equivalent "with respect to ∼."

Thats all the similarity that there is. but that's that's enough similarity for anyone.  That is a HUGE similarity.

Title: Notation for set of equivalence classes


答:
Note that $x>1$. To avoid squaring equations, factorize directly as follows

\begin{align}
0 &=x+\frac{x}{\sqrt{x^2-1}}-\frac{35}{12}\\
&= \left(x-\frac{35}{12} + \frac{25}{12x}\right)
 + \left( \frac{x}{\sqrt{x^2-1}} -\frac{25}{12x}\right)\\
&=\frac1x \left( x^2-\frac{35}{12} x+ \frac{25}{12}\right)
 + \frac{x^2-\frac{25}{12}\sqrt{x^2-1}}{x\sqrt{x^2-1}}\\
& =\frac1x \left( x^2-\frac{35}{12} x+ \frac{25}{12}\right)
 + \frac{x^4-\frac{25^2}{12^2}({x^2-1})}{x\sqrt{x^2-1}\left( x^2+\frac{25}{12}\sqrt{x^2-1}\right)}\\
& =\frac1x \left( x-\frac{5}{4} \right)\left(x- \frac{5}{3}\right)
 + \frac{(x^2-\frac{25}{16})(x^2-\frac{25}{9} )}{x\sqrt{x^2-1}\left( x^2+\frac{25}{12}\sqrt{x^2-1}\right)}\\
& =\frac1x \left( x-\frac{5}{4} \right)\left(x- \frac{5}{3}\right)
\left( 1+ \frac{(x+\frac{5}{4})(x+\frac{5}{3} )}{\sqrt{x^2-1}\left( x^2+\frac{25}{12}\sqrt{x^2-1}\right)}\right)\\
\end{align}
which leads to the solutions $x= \frac54,\>\frac53$. (The last factor is always positive.)

Title: Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$

Body:
Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$

The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank you!


问:
Finding the value of a coefficient given the roots are equal?
I've had some trouble with this question:

"$P(x)$ denotes the quadratic polynomial $kx^2+(k-1)x-(2k-1)$, where $k$ is a rational, real number. Find the value of $k$ for which the roots of $P(x)=0$ are equal."

How do I approach this?
Any help would be appreciated.

ContentLicense: CC BY-SA 3.0

Body:
for a quadratic poly. ($ax^2+bx+c\,$)  roots are equal when $b^2-4ac=0$

so in ques.
$$(k-1)^2+4\cdot k\cdot (2k-1)=0$$
$$k^2+1-2k+8k^2-4k=0$$
$$9k^2-6k+1=0$$
$$9k^2-3k-3k+1=0$$
$$3k(3k-1)-1(3k-1)=0$$
$$(3k-1)^2=0$$
$$3k-1=0$$
$$k=\dfrac13$$


问:
How to solve this four variables system of equations?
$$(x-1)^2+(y-1)^2+(z-1)^2=(t-1)^2,$$

$$(y+1)^2+(z+1)^2=(x+1)^2+(t+1)^2$$

Ηow to solve this for $(x,y,z,t)$ if $x<y<z<t$,  $2\leq x, y, z, t\leq 16$ and if $x, y, z$ and $t$ are all natural numbers?

I don't how to start to solve this. Please help!

答:
The problem could be solved in different ways:

The first one as mentioned by Gribouillis, could be doing an exhaustive search. In order to do so, different types of programming languages or software could be used. But you can also use an Access database and some Excel. The procedure will be as follows:

Create a table in Access, for example, name it Table1, and fill it with numbers from 2 to 16. So, one of the constraints will be achieved. Then use query designer to create the following query

```<code>SELECT    Table1.Field1
        , Table1_1.Field1
        , Table1_2.Field1
        , Table1_3.Field1
FROM  Table1
    , Table1 AS Table1_1
    , Table1 AS Table1_2
    , Table1 AS Table1_3
WHERE
         Table1.Field1      <= Table1_1.Field1
    AND  Table1_1.Field1    <= Table1_2.Field1
    AND  Table1_2.Field1    <= Table1_3.Field1
;
</code>```

The last part of the query, the WHERE section or conditions, makes sure that constraint $x<y<z<t$ holds.  If done properly the resulting set would have 3060 records.

Finally use Excel to check every record and find, if any, hold both constraints

$$
(x-1)^2+(y-1)^2+(z-1)^2=(t-1)^2 \\
(y+1)^2+(z+1)^2=(x+1)^2+(t+1)^2
$$

And we arrived at the solution

```<code>x   y   z   t
3   7   10  12
3   6   15  16
</code>```

A second way could be if we classified the problem as a Mixed Integer Nonlinear Programming. The Integer part comes from the constrain: "must be only natural numbers", the Nonlinear part of the problem is given by the constraints

$$
(x-1)^2+(y-1)^2+(z-1)^2=(t-1)^2 \\
(y+1)^2+(z+1)^2=(x+1)^2+(t+1)^2
$$

In order to solve it this way an objective function must be created, and a simple one could be

$$
MIN: x+y+z+t
$$

So the optimization model would be complete and it would look like the following

$$
\begin{align}
Min  &: x+y+z+t \\
s.t. & \\
(x-1)^2+(y-1)^2+(z-1)^2&=(t-1)^2 \\
(y+1)^2+(z+1)^2&=(x+1)^2+(t+1)^2 \\
x&\leq y \\
y&\leq z \\
z&\leq t \\
x&\geq 2 \\
y&\geq 2 \\
z&\geq 2 \\
t&\geq 2 \\
x&\leq 16 \\
y&\leq 16 \\
z&\leq 16 \\
t&\leq 16 \\
\end{align}
$$

$$
x,y,z,t \in \mathbb{N}
$$

Then use some solver software like Premium Solver, What'sBest!, Lingo, etc.

Finally, as we've already classified the problem as MINLP several techniques could be used to solve it, it could be also possible to program a Genetic Algorithm to solve the problem.

I hope to give you some ideas

Body:
The problem could be solved in different ways:

The first one as mentioned by Gribouillis, could be doing an exhaustive search. In order to do so, different types of programming languages or software could be used. But you can also use an Access database and some Excel. The procedure will be as follows:

Create a table in Access, for example, name it Table1, and fill it with numbers from 2 to 16. So, one of the constraints will be achieved. Then use query designer to create the following query

SELECT    Table1.Field1
        , Table1_1.Field1
        , Table1_2.Field1
        , Table1_3.Field1
FROM  Table1
    , Table1 AS Table1_1
    , Table1 AS Table1_2
    , Table1 AS Table1_3
WHERE
         Table1.Field1      <= Table1_1.Field1
    AND  Table1_1.Field1    <= Table1_2.Field1
    AND  Table1_2.Field1    <= Table1_3.Field1
;

The last part of the query, the WHERE section or conditions, makes sure that constraint $x<y<z<t$ holds.  If done properly the resulting set would have 3060 records.

Finally use Excel to check every record and find, if any, hold both constraints

$$
(x-1)^2+(y-1)^2+(z-1)^2=(t-1)^2 \\
(y+1)^2+(z+1)^2=(x+1)^2+(t+1)^2
$$

And we arrived at the solution

x   y   z   t
3   7   10  12
3   6   15  16

A second way could be if we classified the problem as a Mixed Integer Nonlinear Programming. The Integer part comes from the constrain: "must be only natural numbers", the Nonlinear part of the problem is given by the constraints

$$
(x-1)^2+(y-1)^2+(z-1)^2=(t-1)^2 \\
(y+1)^2+(z+1)^2=(x+1)^2+(t+1)^2
$$

In order to solve it this way an objective function must be created, and a simple one could be

$$
MIN: x+y+z+t
$$

So the optimization model would be complete and it would look like the following

$$
\begin{align}
Min  &: x+y+z+t \\
s.t. & \\
(x-1)^2+(y-1)^2+(z-1)^2&=(t-1)^2 \\
(y+1)^2+(z+1)^2&=(x+1)^2+(t+1)^2 \\
x&\leq y \\
y&\leq z \\
z&\leq t \\
x&\geq 2 \\
y&\geq 2 \\
z&\geq 2 \\
t&\geq 2 \\
x&\leq 16 \\
y&\leq 16 \\
z&\leq 16 \\
t&\leq 16 \\
\end{align}
$$

$$
x,y,z,t \in \mathbb{N}
$$

Then use some solver software like Premium Solver, What'sBest!, Lingo, etc.

Finally, as we've already classified the problem as MINLP several techniques could be used to solve it, it could be also possible to program a Genetic Algorithm to solve the problem.

I hope to give you some ideas


问:
Maxima and minima of $\frac{x^2-3x+4}{x^2+3x+4}$ without calculus
How to find the minimum and max values of $y=\frac{x^2-3x+4}{x^2+3x+4}$ for all real values of $x$ without using calculus?

Perhaps it could be done graphically by noting the fact that the numerator and denominator are a pair of parabolas symmetric about the $x$ axis, but I do not know how to continue.

Thanks!

答:
They did not specify that calculus mustn't be used, but I was curious
if it could be solved in a simpler way – Monocerotis Nov 20 at 8:19

thanks man, you saved me from a lot of differentiation and
substitution – Monocerotis Nov 20 at 8:46

The most simple way to solve your problem is by using calculus:

By applying the quotient rule, you get: $y'(x)=(\frac{x^2-3x+4}{x^2+3x+4})'=\frac{(2x-3)(x^2+3x+4)-(2x+3)(x^2-3x+4)}{(x^2+3x+4)^2}$.

After putting the condition $y'(x)=0$ and expanding the numerator of $y'(x)$, you obtain:

$x^2-4=0$, whose solutions are:

$x_1=2$ and $x_2=-2$.

In conclusion:

$y_{max}=7$ (for $x=-2$) and $y_{min}=\frac{1}{7}$ (for $x=2$).


问:
Quotient Space Metric with Nice Equivalence Classes
Quotient Space Metric: The quotient metric for arbitrary quotient spaces is defined as

If $M$ is a metric space with metric $d$, and $\sim$ is an equivalence
  relation on $M$, then we can endow the quotient set $M/{\sim}$ with the
  following (pseudo)metric. Given two equivalence classes $[x]$ and
  $[y]$, we define $$
     d'([x],[y]) = \inf\{d(p_1,q_1)+d(p_2,q_2)+\dotsb+d(p_{n},q_{n})\} $$ where the infimum is taken over all finite sequences $(p_1, p_2,
 \dots, p_n)$ and $(q_1, q_2, \dots, q_n)$ with $[p_1]=[x], [q_n]=[y],
 [q_i]=[p_{i+1}], i=1,2,\dots, n-1$.

Often we encounter "nice" spaces where the quotient metric can actually be defined as $d([x],[y]) = \inf\{d(p,q)|p\in[x], q\in [y]\}$ (i.e. the shortest path between equivalence classes $[x],[y]$ on the quotient space). This is not true in general, as dicussed here:
Why are quotient metric spaces defined this way?

However, it seems that there are many useful quotient spaces where this holds true.
Examples:

$R^2$ under the identification of vertical lines ($x,y \in \mathbb{R}^2$, $x \sim y$ if $x-y \in \{0\} \times \mathbb{R}$).

$R^2$ under the identification of circles ($x,y \in \mathbb{R}^2$, $x \sim y$ if $\|x\|_2^2=\|y\|_2^2$).

Special Euclidean Group $SE(2)$ under identification of $SO(2)$ ($x,y \in SE(2)$, $x \sim y$ if $x-y \in \{0,0\} \times SO(2)$).

Other spaces like $SE(3) / SO(3)$

That begs the question: What are the conditions under which $d([x],[y]) = \inf\{d(p,q) | p\in[x], q\in [y]\}$ actually holds true?

答:
Here is a sufficient condition that covers all your examples.  If the quotient $M/{\sim}$ is the quotient of $M$ by a group $G$ of isometries $M\to M$, then the quotient metric is given by $d([x],[y]) = \inf\{d(p,q) | p\in[x], q\in [y]\}$.

To prove this, it suffices to show that the definition $$d([x],[y]) = \inf\{d(p,q) | p\in[x], q\in [y]\}$$ satisfies the triangle inequality (since that implies the distance along any finite chain is bounded below by this simple distance).  So, let $x,y,z\in M$, and suppose $p\in[x],q,q'\in[y],$ and $r\in [z]$.  We wish to show that $d([x],[y])$ as defined above is less than or equal to $d(p,q)+d(q',r)$.  To prove this, note that there is some $g\in G$ such that $q'=gq$.  We then have $d(q',r)=d(gq,r)=d(q,g^{-1}r)$ since $g$ is an isometry.  Thus $$d(p,q)+d(q',r)=d(p,q)+d(q,g^{-1}r)\geq d(p,g^{-1}r).$$  Since $g^{-1}r\in [z]$, $d(p,g^{-1}r)\geq d([x],[y])$, as desired.

For citations: This result can be quotient from Theorem 2.1 of this paper.


问:
Example of calculation of Frechet derivative considering Hilbert space
Let $\mathbb R$ a Hilbert space in their canonic form. We define the function $f(x)=a\|x-x^{o}\|^{2}+\langle b,x \rangle $, $\forall x \in \mathbb R^{n}$, where $a$ is a constant, while $x^{o}$ and $b$ are vectors of $\mathbb R^{n}$.

Find Df(x).

答:
My answer is the following (I'm not sure if my answer is correct):

From the definition of Frechet derivative:

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

$Df(x)=\lim_{h \to 0}\frac{a\|(x+h)-x^{o}\|^{2}+\langle b,x+h \rangle - a\|x-x^{o}\|^{2}-\langle b,x \rangle}{h}$

$Df(x)=\lim_{h \to 0}\frac{a\|(x+h)-x^{o}\|^{2}+\langle b,h \rangle - a\|x-x^{o}\|^{2}}{h}$

If we consider the euclidean norm, we have:

$Df(x)=\lim_{h \to 0}\frac{a[(x+h)-x^{o}]+\langle b,h \rangle - a[x-x^{o}]}{h}$

$Df(x)=\lim_{h \to 0}\frac{a[h]+\langle b,h \rangle}{h}$

$Df(x)= a+b$


问:
Remainder of Polynomial Division of $(x^2 + x +1)^n$ by $x^2 - x +1$
I am trying to solve the following problem:

Given $n \in \mathbb{N}$, find the remainder upon division of $(x^2 + x +1)^n$ by $x^2 - x +1$

the given hint to the problem is:

"Compute $(x^2 + x +1)^n$ by writing $x^2 + x +1 = (x^2 - x +1) + 2x$. Then, use the uniqueness part of the division algorithm."

If I take $a = x^2 - x +1$ I have
$$(x^2 + x +1)^n = (a + 2x)^n = a^n + \binom{n}{1}a^{n-1} 2x+ \binom{n}{2}a^{n-2} (2x)^2 + \dots + (2x)^n$$
but how do I proceed further?

答: You're trying to divide by $a$ with remainder. You almost have it, since in your expression for $(a+2x)^n$ all terms but the final $(2x)^n$  are multiples of $a$ so can be dropped in getting the remainder. You now can focus just on $(2x)^n.$


答: Great job so far! Notice that all terms of the form $\binom{n}{i}a^{n-i}$ are divisible by $a$ if and only if $i<n$. This means that the remainder is simply $(2x)^n$.

Title: Remainder of Polynomial Division of $(x^2 + x +1)^n$ by $x^2 - x +1$

Body:
I am trying to solve the following problem:

Given $n \in \mathbb{N}$, find the remainder upon division of $(x^2 + x +1)^n$ by $x^2 - x +1$

the given hint to the problem is:

"Compute $(x^2 + x +1)^n$ by writing $x^2 + x +1 = (x^2 - x +1) + 2x$. Then, use the uniqueness part of the division algorithm."

If I take $a = x^2 - x +1$ I have
$$(x^2 + x +1)^n = (a + 2x)^n = a^n + \binom{n}{1}a^{n-1} 2x+ \binom{n}{2}a^{n-2} (2x)^2 + \dots + (2x)^n$$
but how do I proceed further?


答: Hint: every term but the last is divisible by $a$, so your remainder is congruent to $(2x)^n$. But that's not quite enough either, because that's still too 'large' if $n\gt 1$. You need to figure out how to divide $(2x)^n$ by $x^2-x+1$. The easiest way (IMHO) to do that is this: modulo $x^2-x+1$, we can just say that $x^2\equiv x-1$. This means that every polynomial is equivalent to one of the form $ax+b$; if you have $(2x)^n\equiv a_nx+b_n$, then you can multiply the RHS by $2x$ and use the equivalence to find $a_{n+1}$ and $b_{n+1}$ in terms of $a_n$ and $b_n$.


答:
Since every term but the last is divisible by $a$, so we only need to deal with $(2x)^n \bmod \left(x^2-x+1\right)$. Let's start from small ones, and try to find patterns:
$$
\begin{aligned}
((2 x)^1 &\bmod \left(x^2-x+1\right))=\color{gray}{2x} \\
((2 x)^2&\bmod \left(x^2-x+1\right)) =\color{red}{-4+4 x} \\
((2 x)^3 &\bmod \left(x^2-x+1\right)) = \color{blue}{-8}\\
((2 x)^4 &\bmod \left(x^2-x+1\right)) = \color{green}{-16x}\\
((2 x)^5 &\bmod \left(x^2-x+1\right)) = 32-32x\\
((2 x)^6 &\bmod \left(x^2-x+1\right)) = 64\\
((2 x)^7 &\bmod \left(x^2-x+1\right)) =\color{gray}{128x}\\
((2 x)^8 &\bmod \left(x^2-x+1\right)) = \color{red}{-256+256x}\\
((2 x)^9 &\bmod \left(x^2-x+1\right)) = \color{blue}{-512} \\
((2 x)^{10} &\bmod \left(x^2-x+1\right)) =\color{green}{-1024 x}
\end{aligned}
$$
What do you find? Hint: Divide into groups by $n \bmod 6$. Prove by induction separately.


问:
Proving that if $a$ is not a quadratic residue, then $a^{p'}\equiv -1\pmod p$: Is my proof correct?
On Courant/Robbins' "What is mathematics?", it says that:

Due to Fermat's Little Theorem, we have $a^{p-1}\equiv 1 \pmod p$. This suggests that we define $p'=\frac{p-1}{2}$ and factor the previous congruence as:

$$(a^{p'}-1)(a^{p'}+1)\equiv 0 \pmod p \tag{$\color{red}{☭}$}$$

And then we must have $a^{p'}\equiv 1 \pmod p$ or $a^{p'}\equiv -1 \pmod p$. Now, If $a$ is a quadratic residue, that is, for some $x$, we have:

$$a\equiv x^2\pmod p$$

From this, we have:

$$a^{p'}\equiv x^{p-1} \equiv 1\pmod p$$

That is: If $a$ is a quadratic residue, then $a^{p'}\equiv 1\pmod p$ and he says that it can easily be proven that if $a$ is not a quadratic residue, then $a^{p'}\equiv -1\pmod p$. I suppose I know how to prove it:

If $a$ is a quadratic residue, then $a^{p'}\equiv 1\pmod p$. With contrapositive, we obtain that $a^{p'}\not\equiv 1\pmod p$ then $a$ is not a quadratic residue.

Also, if $a^{p'}\not\equiv 1\pmod p$ due to the Fermat's little theorem, we have $(\color{red}{☭})$ and as $a^{p'}-1\not\equiv 0 \pmod p$ we must have $a^{p'}+1\equiv 0 \pmod p$.

Is my proof correct?

答:
Your proof is not complete.

It has been shown that, if $a$ is a quadratic residue, then $a^{p'}\equiv1\pmod p$.

Furthermore, it has been shown that, if $a$ is not a quadratic residue, then $a^{p'}\equiv\pm1\pmod p$.

To show that, if $a$ is not a quadratic residue, then $a^{p'}\equiv-1\pmod p$,

it remains to be shown that, if $a$ is not a quadratic residue, then $a^{p'
}\not\equiv1\pmod p$.

To show that, consider how many quadratic residues there are,

and how many solutions $a^{p'}-1\equiv0\pmod p$ could have.


问:
Given $a,b,k,l$, solve $k\vec{xa}+l\vec{xb}=m\vec{xc}$ for $m$ and $c$
Given two points $a$ and $b$ in a Euclidean space, and two nonzero real numbers $k,l\in\Bbb R-\{0\}$, how to find a point $c$ in this space and a nonzero real number $m\in\Bbb R-\{0\}$ such that $k\vec{xa}+l\vec{xb}=m\vec{xc}$ for all $x$ in the space? Proof of existence of the pair $c$ and $m$ is also ok if a closed-form particular solution is impossible.

I just drew a picture and got stuck by this question, because I could not see a fixed point $c$ that the vector combination $k\vec{xa}+l\vec{xb}$ always points to. Thank you for any clue for me to proceed.

答: The equations are linear, so if it is true for $x=(1,0)$ and $x=(0,1)$ it is true everywhere.  If you write it componentwise you have two equations in two unknowns, which you should be able to solve most of the time.  Do your conditions prohibit the singular case?


问:
$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$

Q. If $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$, prove that $a = b = c = 0$ ($a$, $b$, $c$ and $p$ are rational and $p$ is not a perfect cube.)

My approach:

Solving the quadratic, I get:

$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$

Case 1: If the $b^2 - 4ac$ is a perfect square, I get the LHS as irrational and the RHS as rational, which is a contradiction.

Case 2: If $b^2 - 4ac$ is not a perfect square, $b =  \pm \sqrt{b^2 - 4ac} - 2cp^\frac{1}{3}$

Here, the LHS is rational and the RHS is irrational, contradiction again. (Edit: The answer of @GNUSupporter has the proper proof.)

So the equation is not quadratic and $c = 0$.

$a + bp^\frac{1}{3} = 0$

$-\dfrac{a}{b} = p^\frac{1}{3}$

This is a contradiction and hence $b = 0$ and $a = 0$

Is there any other way to solve this?

答:
Since $q=p^{1/3}$ is irrational it follows that minimal polynomial of $q$ over $\mathbb{Q} $ is of degree greater than $1$. If we have $a+bq+cq^2=0$ for some rational $a, b, c$ with $c\neq 0$ then $f(x) =(a/c)+(b/c)x+
x^2\in\mathbb{Q}[x]$ is the minimal polynomial of $q$.

And therefore it must divide the polynomial $g(x) =x^3-p\in\mathbb{Q} [x] $ (which has $q$ as a root) leading to a linear factor of this polynomial $g(x)$. But $g(x) $ has no rational root so we get a contradiction.

We must have thus $c=0$ and then it is easy to conclude $a=b=0$ just from irrationality of $q$.

The fact that we are dealing with cube roots and polynomials of degree $3$ makes the algebra a lot simpler. If a polynomial of degree $3$ is reducible it must have a linear factor and hence a rational root. And thus if a third degree polynomial with rational coefficients has no rational roots then it is irreducible.

The same can't be said of a general polynomial of degree greater than $3$. But for polynomials of the form $x^n-p$ the same result holds.

Let $n$ be a positive prime and $p$ be a positive rational number which is not an $n$-th power then $x^n-p$ is irreducible over $\mathbb{Q} $.


答:
As another approach, factor the two quadratics as

$$x^2 + x + 1 = (x-\phi)(x+\psi)$$ $$x^2 - x + 1 = (x + \phi)(x - \psi)$$

Where $\phi$ and $\psi$ are solutions to $\phi\psi=-1$, $\psi - \phi=1$

So problem reduces to solving for $A$ and $B$

$$(x-\phi)^n(x+\psi)^n = P(x)(x+\phi)(x-\psi) + Ax + B$$

Evaluate this for $x=-\phi$ and $x=\psi$ to get
$$(-2\phi)^n(\psi - \phi)^n = -A\phi + B$$
$$(\psi - \phi)^n(2\psi)^n = A\psi + B$$

Remember that $\psi - \phi = 1$ and the rest should be straightforward


答:
Let $a=\left( a_1,a_2,\cdots ,a_n \right) ,b=\left( b_1,b_2,\cdots ,b_n \right) ,c=\left( c_1,c_2,\cdots ,c_n \right), x=\left( x_1,x_2,\cdots ,x_n \right)
$. Then $k\vec{xa}+l\vec{xb}=m\vec{xc}$ is equivalent to say $$k\left( a_1-x_1,a_2-x_2,\cdots ,a_n-x_n \right) +l\left( b_1-x_1,b_2-x_2,\cdots ,b_n-x_n \right) =m\left( c_1-x_1,c_2-x_2,\cdots ,c_n-x_n \right) $$

EDIT: I was wrong. The correct way to proceed is as the comment shows. $m$ should be $k+l$, in order that $$[m-(k+l)]x_i=mc_i-(ka_i+lb_i)$$ holds for all $x$.

Title: Given $a,b,k,l$, solve $k\vec{xa}+l\vec{xb}=m\vec{xc}$ for $m$ and $c$

Body:
Given two points $a$ and $b$ in a Euclidean space, and two nonzero real numbers $k,l\in\Bbb R-\{0\}$, how to find a point $c$ in this space and a nonzero real number $m\in\Bbb R-\{0\}$ such that $k\vec{xa}+l\vec{xb}=m\vec{xc}$ for all $x$ in the space? Proof of existence of the pair $c$ and $m$ is also ok if a closed-form particular solution is impossible.

I just drew a picture and got stuck by this question, because I could not see a fixed point $c$ that the vector combination $k\vec{xa}+l\vec{xb}$ always points to. Thank you for any clue for me to proceed.


问:
Any open subset of $\Bbb R$ is a countable union of disjoint open intervals

Let $U$ be an open set in $\mathbb R$. Then $U$ is a countable union of disjoint intervals.

This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.

答:
These answers all seem to be variations on one another, but I've found each one so far to be at least a little cryptic. Here's my version/adaptation.

Let $U \subseteq \mathbb{R}$ be open and let $x \in U$. Either $x$ is rational or irrational. If $x$ is rational, define
\begin{align}I_x = \bigcup\limits_{\substack{I\text{ an open interval} \\ x~\in~I~\subseteq~U}} I,\end{align}
which, as a union of non-disjoint open intervals (each $I$ contains $x$),  is an open interval subset to $U$. If $x$ is irrational, by openness of $U$ there is $\varepsilon > 0$ such that $(x - \varepsilon, x + \varepsilon) \subseteq U$, and there exists rational $y \in (x - \varepsilon, x + \varepsilon) \subseteq I_y$ (by the definition of $I_y$). Hence $x \in I_y$. So any $x \in U$ is in $I_q$ for some $q \in U \cap \mathbb{Q}$, and so
\begin{align}U \subseteq \bigcup\limits_{q~\in~U \cap~\mathbb{Q}} I_q.\end{align}
But $I_q \subseteq U$ for each $q \in U \cap \mathbb{Q}$; thus
\begin{align}U = \bigcup\limits_{q~\in~U \cap~\mathbb{Q}} I_q, \end{align}
which is a countable union of open intervals.


答:
You expand the quadratic to get
\begin{eqnarray}
f(x+h) &=& a (\langle x-x_0+h, x-x_0+h \rangle + \langle b, x+h \rangle \\
&=& f(x) +2a \langle x-x_0, h \rangle + \langle b, h \rangle+ \langle h, h \rangle
\end{eqnarray}
Dropping the $\|h\|^2$ term we get $Df(x)h = \langle 2a(x-x_0)+b, h \rangle = (2a(x-x_0)+b)^T h$.

Title: Example of calculation of Frechet derivative considering Hilbert space

Body:
Let $\mathbb R$ a Hilbert space in their canonic form. We define the function $f(x)=a\|x-x^{o}\|^{2}+\langle b,x \rangle $, $\forall x \in \mathbb R^{n}$, where $a$ is a constant, while $x^{o}$ and $b$ are vectors of $\mathbb R^{n}$.

Find Df(x).


问:
Why does $k \mid (p-1) /q$?
In M. Ram Murty's paper Artin's Conjecture on Primitive Roots I am not able to understand a statement.

Let $k$ be the order of $a \bmod p$ where $p$ is a prime and also we have $a^{p-1} \equiv 1 \mod p$ then $k \mid (p-1)$ if $k \neq p-1$ then $k\mid(p-1)/q$ for some prime divisor $q$ of $p-1$.

I am not able to understand how the author was able to make the second claim i.e if $k \neq p-1$ then $k\mid(p-1)/q$ for some prime divisor $q$ of $p-1$.

答:
This is a special case of a more general statement: if $k\mid n$ and $k\ne n$ then $k\mid \frac nq$ for some prime divisor $q$ of $n$.

Indeed, this statement holds for any prime $q$ that divides $\frac nk$, which I invite you to verify.

Title: Why does $k \mid (p-1) /q$?


问:
Implementing the sign function with basic math
The $\operatorname{sign}(x)$ function returns:
$$
\begin{cases}
-1 & \text{if } x < 0 \\
\phantom{-}0 & \text{if } x = 0 \\
\phantom{-}1 & \text{if } x > 0
\end{cases}
$$

How do I implement this with only basic math principles (addition, subtract, multiplication, division, powers, square roots)?

答: Since $\text{sgn}(x)$ is not a continuous function on $\mathbb R$, but all the elementary operations are continuous on $\mathbb R$(except the division, it may be undefined at some certain points) and the composite of continuous functions are also continuous, you cannot find a way to represent $\text{sgn}(x)$ in elementary functions.


问:
$\lim_{x→0^+} \frac{\sin(6x)}{\sqrt{\sin(2x)}}$
I'm trying to prove and compute the limit of this function.

  $\lim_{x→0^+} \frac{\sin(6x)}{\sqrt{\sin(2x)}}$

I've tried converting it into different functions like $\cos(\pi/2-2x)$ or multiplying by the inverse function and so on, but it keep getting back to $0/0$.

I'm sure that the limit does in fact exist because using L'Hôpital's rule it is fairly easy to prove it, but I can't use it because the course I'm attending still hasn't defined the rule's properties, but I couldn't find a different way aside from that.

答:
You can do this with regular old trig identities. Note that:

$$\sin\left(6x\right) = 6 \sin x \cos^5 x - 20 \sin^3 x \cos^3 x + 6 \sin^5 x \cos x$$

$$\sqrt{\sin\left(2x\right)} = \sqrt{2 \sin\left(x\right) \cos\left(x\right)}$$

For shorthand, let me write: $$\sin\left(x\right) = S \hspace{2.54cm} \cos\left(x\right) = C$$

So your ratio is: \begin{align*}\lim_{x \rightarrow 0^+} \frac{\sin\left(6x\right)}{\sqrt{\sin\left(2x\right)}} &= \frac{6SC^5 - 20S^3C^3 + 6S^5C}{\sqrt{2SC}}  \\ & = \frac{6\sqrt{S}C^5 - 20S^2\sqrt{S}C^3 + 6S^4\sqrt{S}C}{\sqrt{2C}}\end{align*}

Finally, we have a form where the denominator won't blow up. Note that $$\lim_{x \rightarrow 0^+} \sin\left(x\right) = 0 \hspace{2.54cm} \lim_{x \rightarrow 0^+} \cos\left(x\right) = 1$$ to seal the deal: $$\lim_{x \rightarrow 0^+} \frac{\sin\left(6x\right)}{\sqrt{\sin\left(2x\right)}} = \frac{6\sqrt{0}C^5 - 20\cdot 0^2\sqrt{0}C^3 + 6\cdot 0^4\sqrt{0}C}{\sqrt{2C}} = 0$$


问:
Algebra problem (problem from Swedish 12th grade ‘Student Exam’ from 1932)
The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.

The sum of two numbers are $a$, the sum of the 3rd powers is $10a^3$. Calculate the sum of the 4th powers, expressed in $a$.

Is there a shorter/simpler solution than the one presented below? It feels there is some ‘trick’ to it. The solution presented below is more a ‘straight forward’ one.

Solution

We have
\begin{gather*}
\left\{
\begin{aligned}
x+y&=a\\
x^3+y^3&=10a^3
\end{aligned}
\right.
\quad\Leftrightarrow\quad
x^3+(a-x)^3=10a^3
\quad\Leftrightarrow\quad
x^2-ax-3a^2=0
\end{gather*}
which has the solutions
$$
x_{1,2}=\tfrac{1}{2}(1\pm\sqrt{13}\,)a
\qquad \Rightarrow \qquad
y_{1,2}=\tfrac{1}{2}(1\mp\sqrt{13}\,)a.
$$

Since
$$
(1+z)^4+(1-z)^4=2(1+6z^2+z^4)
$$
we have
\begin{align*}
x_1^4+y_1^4
&
=\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}
=\tfrac{a^4}{16}\cdot2\bigl(1+6z^2+z^4\bigr)\big|_{z=\sqrt{13}}
\\&=\tfrac{a^4}{8}(1+6\cdot13+13^2)
=\tfrac{a^4}{8}\cdot248
=31a^4
\end{align*}
and, as above,
$$
x_2^4+y_2^4
=\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}
=31a^4.
$$

Hence, the answer is $31a^4$.

The original exam

答:
This answer provides some motivation for a high school student learning the binomial theorem and problem solving techniques; we also get a 'feeling' for homogeneous polynomials.

Employing the binomial theorem we naturally write out (including the quadratic),

$\tag 1 x  + y \color\red{ = a} $

$\tag 2 x^2 + y^2 = (x+y)^2- (2xy) $

$\tag 3 x^3 + y^3 = (x+y)^3 - (3 x^2 y + 3 x y^2) \color\red {= 10a^3}$

$\tag 4 x^4 + y^4 = (x+y)^4 - (4 x^3 y + 6 x^2 y^2 + 4 x y^3) $

Using $\text{(1)}$,

$\tag 1 x  + y \color\red{ = a} $

$\tag 2 x^2 + y^2 = \color\red {(a)^2} - (2xy) $

$\tag 3 x^3 + y^3 = \color\red {(a)^3} - (3 x^2 y + 3 x y^2) \color\red {= 10a^3}$

$\tag 4 x^4 + y^4 = \color\red {(a)^4} - (4 x^3 y + 6 x^2 y^2 + 4 x y^3) $

It looks likes we have a good shot at writing $xy$ in terms of $a^2$ and from $\text{(3)}$,

$\quad \text{(3)} \implies -3xy(x+y) = 9a^3 \land \text{(1)} \implies xy = -3a^2$

Checkpoint,

$\tag 1 x  + y \color\red{ = a} $

$\tag 2 x^2 + y^2 = \color\red {7a^2}$

$\tag 2 xy = \color\red{ -3\, a^2}$

$\tag 3 x^3 + y^3  \color\red {= 10a^3}$

$\tag 4 x^4 + y^4 = \color\red {a^4} - (4 x^3 y + 6 x^2 y^2 + 4 x y^3) $

Always hopeful, we tackle the remaining part,

$\quad - (4 x^3 y + 6 x^2 y^2 + 4 x y^3) =$

$\quad -2xy(2x^2+3xy+2y^2) =$

$\quad -2xy\bigr(2(x^2+y^2) + 3xy\bigr) =$

$\quad 6a^2\bigr(14a^2 -9a^2\bigr) = 30a^4$

and so

$\tag{ANS} x^4 + y^4 = 31a^4$

Bonus Section

Keep going?

$\quad x^5 + y^5 = (x+y)^5 -(5 x^4y + 10 x^3 y^2 + 10 x^2 y^3 + 5 x y^4 ) = $

$\quad \quad \quad \quad \quad \quad a^5 -xy(5 x^3 + 10 x^2 y^1 + 10 x^1 y^2 + 5  y^3 ) =$

$\quad \quad \quad \quad \quad \quad  a^5 +3a^2(50 a^3 + 10 x^2 y^1 + 10 x^1 y^2  ) =$

$\quad \quad \quad \quad \quad \quad  a^5 +3a^2\bigr(50 a^3 + 10 xy(x + y )\bigr) =$

$\quad \quad \quad \quad \quad \quad  a^5 +3a^2(50 a^3 -30a^3) =$

$\quad \quad \quad \quad \quad \quad  61a^5$

More?

$\quad x^6 + y^6 = \,?$


答:
What you mean to call the "basic math principles" (where I'd include trigonometric functions and logarithms too) are ultimately the result of compositions of functions which are continuous on their domains. Therefore any resulting function must be continuous on its domain as well, which isn't the case for $\operatorname{sgn}$.

If you are one of the people who believe that the map $(x,y)\mapsto x^y$ should be defined on $((0,\infty)\times \Bbb R)\cup (\{0\}\times [0,\infty))$ by imposing $0^0=1$ (I don't exactly count myself in that number), then the considerations above no longer apply, and with the help of $\sqrt{x^2}=\lvert x\rvert$ you can write $$\operatorname{sgn}(x)=\frac{x}{0^{\sqrt{x^2}}+\sqrt{x^2}}$$

Title: Implementing the sign function with basic math

Body:
The $\operatorname{sign}(x)$ function returns:
$$
\begin{cases}
-1 & \text{if } x < 0 \\
\phantom{-}0 & \text{if } x = 0 \\
\phantom{-}1 & \text{if } x > 0
\end{cases}
$$

How do I implement this with only basic math principles (addition, subtract, multiplication, division, powers, square roots)?


问:
subring not graded
Definition 1. A pair $(A, \{A_{i}\}_{i\in\mathbb{Z}})$ is a graded ring if $A$ is a commutative ring with unit, and $\{A_{i}\}_{i\in\mathbb{Z}}$ is a family of $\mathbb{Z}$-submodules of $A$, such that

(1) $A = \bigoplus_{i}A_{i}$ as $\mathbb{Z}$-module.

(2) $\forall i$, $j$, $A_{i}A_{j}\subseteq A_{i + j}$.

Definition 2. Let $A = \bigoplus_{i}A_{i}$ be a graded ring. A graded ring $B = \bigoplus_{i}B_{i}$ is a graded subring of $A$, if $B$ is a subset of $A$, and the inclusion satisfies $\forall i$, $B_{i}\subseteq A_{i}$.

Remark. The definition of graded subring above is equivalent to that, a subring $B$ of $A$ is graded if and only if $B = \bigoplus_{i}(B\cap A_{i})$ as $\mathbb{Z}$-module. It seems that the only difference between subrings and graded subrings is whether $B\subseteq \bigoplus_{i}(B\cap A_{i})$ or not, i.e. closed or not in homogeneous decomposition.

Question. Is there an example of a subring of a graded ring that is not graded?

答: Take $A$ to be a polynomial ring $k[x]$ with the usual grading $x^i \in A_i$ and take $B$ to be the subring generated by, say, $x^2 + x$. $B$ does not contain any homogeneous polynomials of degree $1$ or $2$ and so is not equal to $\bigoplus_i (B \cap A_i)$.

Title: subring not graded


问:
Bernoulli shift is a measure-preserving transformation
Let (for simplicity) $(\{0,1\}^\mathbb{N},\mathcal{B},\mu,T)$ be a Bernoulli scheme, when $\mathcal{B}$ is the $\sigma$-algebra generated from the cylinder sets, $\mu$ is $(0.5,0.5)^\mathbb{N}$, and $T$ is "right shift", which means $\forall(x_n)_n\in\{0,1\}^\mathbb{N}: T((x_n)_n)=(x_{n+1})_n$

$T$ is a measure-preserving transformation, but why?

To prove that $T$ is measurable, let $A\in\mathcal{B}$, $T^{-1}(A)=\{x_n\mid (x_{n+1})_n\in A, x_1=0$ or $1\}$.

Is this set measurable? If it is, why $T$ is a measure-preserving transformation? I am struggling to prove it.

Thanks in advance.

答:
If $A \subseteq \{0,1\}^{\mathbb{N}}$ is a cylinder set, then $T^{-1}(A)$ is also a cylinder set.  Define $\mathcal{G}$ by
\begin{equation*}
\mathcal{G} = \{A \subseteq \{0,1\}^{\mathbb{N}} \, \mid \, T^{-1}(A) \in \mathcal{B} \}.
\end{equation*}
An exercise shows that $\mathcal{G}$ is a $\sigma$-algebra.  Since it contains the cylinder sets, $\mathcal{G} \supseteq \mathcal{B}$.  That is, if $A \in \mathcal{B}$, then $T^{-1}(A) \in \mathcal{B}$, so $T$ is measurable.

If $A \subseteq \{0,1\}^{\mathbb{N}}$ is a cylinder set, then $\mu(T^{-1}(A)) = \mu(A)$ follows by a direct computation.  The family of all cylinder sets if a $\pi$-system (since the intersection of two cylinders is a cylinder) and the family $\Lambda = \{A \in \mathcal{B} \, \mid \, \mu(A) = \mu(T^{-1}(A))\}$ is a $\lambda$-system.  Therefore, Dynkin's $\pi-\lambda$ theorem implies that $\Lambda \supseteq \mathcal{B}$.  In particular, $\mu(A) = \mu(T^{-1}(A))$ for all $A \in \mathcal{B}$, so $T$ is measure-preserving.


问:
How to reduce lower one number, while another number increases goes up increments
This is such a basic question I'm sure, but I've been trying to find a robust solution to it for a while.

Lets say I have the numbers 0-100 in series.

As the number increases, I want another number say 1000, to decrease. so that when you got to 100, it was at 0

0 = 1000

50 = 500

100 = 0

What is this kind of series called?

答: There is no way to uniquely answer your question without knowing more about what class of functions you permit to interpolate these points. Notice that all of your points are divisible by 50, so we may simplify a bit by scaling down by 50, yielding the points $\rm (x,y) = (0,20), (1,10), (2,0)$. Since the intermediate slopes agree $(10-20)/(1-0) = -10 = (0-10)/(2-1)\;$ it follows that a linear function interpolates these points, viz. $\rm\;y = 20-10x,\;$ or $\rm\;Y = 1000 - 10X\;$ when scaled up by 50 to $\rm (X,Y) = (50x,50y)$. But one could easily interpolate many other curves between these points. Perhaps if you tell us more about your intended application we can elaborate further.</p>

ContentLicense: CC BY-SA 2.5


问:
Has anything useful come from Ackermann's Function?
Here is the function:

```<code>if (m == 0)
    return n + 1;
else if (n == 0)
    return A(m-1, 1);
else
    return A(m-1, A(m, n-1));
</code>```

This seems like an interesting function, especially since its values grow quite rapidly (my computer crashes if I try to run A(4,2) or A(3,10)).

From the wikipedia page it seems like it was only invented to show that a total computable function does not have to be primitive recursive. Has anything practical come from this function, other than homework problems in a computer science class?

答:
The inverse of Ackermann's function is useful in computational complexity, for instance in line arrangements and other problems in computational geometry. It's appearance in the analysis of some concrete problems and algorithms is somewhat surprising. See

Davenport–Schinzel sequences and their role in lower envelopes

Davenport-Schinzel Sequences and Their Geometric Applications


问:
Inequality for a two variables function
I have tried to solve this exercise but I am having difficulties.

I have to find a function greater than $$\frac{|y|^\alpha \sin(xy)}{(x^2+y^2)^{3/2}}$$ as the parameter $\alpha$ changes.
That's what I tried:

Since $$sin(xy)\leq1$$

$$\frac{|y|^\alpha \sin(xy)}{(x^2+y^2)^{3/2}} \leq \frac{|y|^\alpha}{(x^2+y^2)^{3/2}}$$

Then:

$$(x^2+y^2)^{3/2} \geq y^3 $$ so that:

$$\frac{|y|^\alpha}{(x^2+y^2)^{3/2}} \leq \frac{|y|^\alpha}{y^{3}}$$
Now I have problems and I don't know what I did wrong. Following my reasoning I should have:
$$\frac{|y|^\alpha}{y^{3}}=|y|^{\alpha - 3}$$
But I Know it's wrong.

Title: Inequality for a two variables function

Body: $(x^2+y^2)^{3/2} >= y^2 $ is not correct.  Take  $x=0, y=0.1.$


问:
A cyclic inequality $\sum\limits_{cyc}{\sqrt{3x+\frac{1}{y}}}\geqslant 6$
Given positive real numbers $x,y,z$ satisfying $x+y+z=3$, prove that $$ \sqrt{3x+\frac{1}{y}}+\sqrt{3y+\frac{1}{z}}+\sqrt{3z+\frac{1}{x}}\geqslant 6. $$

答:
Remark: @chenbai gave a nice solution using the Buffalo Way (BW).

Here, I give an alternative solution using BW:

By Holder, we have
\begin{align}
\left(\sum_{\mathrm{cyc}} \sqrt{3x + 1/y}\right)^2 \sum_{\mathrm{cyc}} \frac{(6x + y + 3z)^3}{3x + 1/y}
\ge \left(\sum_{\mathrm{cyc}} (6x + y + 3z)\right)^3.
\end{align}
It suffices to prove that
$$\left(\sum_{\mathrm{cyc}} (6x + y + 3z)\right)^3 \ge 36 \sum_{\mathrm{cyc}} \frac{(6x + y + 3z)^3}{3x + 1/y}.$$
After homogenization, it suffices to prove that, for all $x, y, z > 0$,
$$\left(\sum_{\mathrm{cyc}} (6x + y + 3z)\right)^3 \ge 12(x+y+z) \sum_{\mathrm{cyc}} \frac{(6x + y + 3z)^3}{3x + \frac{(x+y+z)^2}{9y}}.$$

The Buffalo Way kills it. We are done.


问:
The direct limit of $A_{f}$ for $f \notin P$ is $A_{P}$
Let $A$ be a commutative ring with identity, and $P$ a prime ideal of $A$. I want to show that $\lim_{f \notin P} A_{f} = A_{P}$. $A_f$ is the localization of $A$ at $\{1\} \cup \{f^{n}: n \in \mathbb N\}$, and $A_P$ is the localization of $A$ at $A \setminus P$. There is a natural map from $A_{f} \to A_P$ if $f$ is not in $P$, so by the universal property of the direct limit, we get a map from $\Phi : \lim_{f \notin P} A_{f} \to A_{p}$. Then it remains to show that this canonical map is an isomorphism. Surjectivity can be proven just using the universal property without knowing what $\Phi$ looks like. However I cannot prove the injectivity using the universal property. It seems that one has to use the explicit form of $\Phi$ in order to achieve that.

答: Take an element $\xi$ of the kernel of $\Phi$, represented by $\frac a{f^n}$ for some $f\not\in P$. We know that $$\frac a{f^n}=0\quad\text{in the ring $A_P$}$$whence there exists $g\not\in P$ such that $ga=0$, so that $$\frac a{f^n}=\frac {ag^n}{(fg)^n} = 0\quad\text{in $A_{fg}$}$$This means that $\xi=0$ (in $\varinjlim A_f$).


问:
A couple has two children. At least one is a boy. What is the probability that the couple has one child of each sex?
I assume that the question assumed that the probability of having a boy is 50% and the probability of having a girl is 50%

So I'm confused because this seems like a really easy probability question. If they already have a boy. The probability that they have a girl is just $.5$ isn't it?

But the book says that the probability is $2/3$

This seems completely illogical to me because then that would assume that the boy to girl birth ratio is $1$ to $2$ which doesn't seem right at all.

What am I misinterpreting here? Is the book wrong?

答: We know the situation is one of these three equally likely cases: , , . So what can you conclude from that?</p>


问:
Area of triangle with incircle tangent to semicircle whose diameter is on one triangle side
Triangle $\triangle ABC$ with incenter $I$ and inradius $r$ has the following property: the incircle is tangent to the semicircle with diameter $AB$ and is within the semicircle. Find the area of $\triangle ABC$ as a function of the side $AB$ and the inradius $r$.

So far I've noticed that $\angle AEF=\angle BEF=\frac{\pi}{4}$ but I don't think that that's useful in any way.

答:
The area of the triangle ABC is

\begin{align}
Area & = \frac12 r(AB + BC +CA)
= \frac12 rAB \frac{\sin A+\sin B +\sin C}{\sin C} \\
&= \frac12 rAB \frac{4\cos \frac A2 \cos \frac B2\cos\frac C2 }{2\sin \frac C2\cos \frac C2}
= rAB \frac{\cos \frac A2 \cos \frac B2 }{\cos \frac{A+B}2}
=  \frac{rAB }{1- \tan\frac A2\tan\frac B2 }
\end{align}

Apply the Pythagorean theorem to the right triangle IFO, with $AB = 2R$

$$d^2 = (R-r)^2 - r^2 = R^2 - 2rR$$
$$\tan\frac A2\tan\frac B2= \frac r{R+d} \frac r{R-d}=\frac {r^2}{R^2-d^2} = \frac {r^2}{2rR} = \frac r{AB}
$$
Thus
$$Area =  \frac{r AB }{1- \frac r{AB} } = \frac {rAB^2}{AB -r}$$


问:
Computing the centroid of some $n$ points in 3-dimensions
Solving for the centroid of some set of points in 2-dimensions is trivial. I'd like, however, to determine the centroid in 3-dimensions, given an arbitrary set of $n$ points. How can this most easily be done?

Offhand, it seems that this should require some sort of integral. I'm not sure though.

Note that I speak of the centroid which minimizes the sum of the
squared distances.

答: Just as the squared distance between two points in 2D space is of the form $x^2+y^2$, so is the squared distance between two points in $n$-D space of the form $x_1^2+x_2^2+\cdots+x_n^2$. Thus the centroid in any dimension of Euclidean space can be computed in exactly the same way as the two-dimensional centroid – average per-coordinate.


问:
$\frac{\partial F}{\partial y}\neq0\implies$ continuous contour line? (Implicit Function Theorem)
I have a function $F(x,y)=z$ and two points $(x_1,y_1),(x_2,y_2)$ s.t. $F(x_1,y_1)=F(x_2,y_2)=c$, $x_1<x_2$. I know that $\frac{\partial F}{\partial y}<0$ in $ [x_1,x_2]\times\mathbb{R}$.

I'd like to prove that there is a continuous contour line between the two points.

I know that there's a rectangle $V\times W $ that contains $(x_1,y_1)$ s.t. $F^{-1}(c)\cap V\times W $ is the graphic of a function, i.e., I have in there a continuous contour line.

I'd like to know if the fact that I have $\frac{\partial F}{\partial y}<0$ in the entire interval $[x_1,x_2]$ allows that I consider $V=[x_1,x_2]$ and $W=\mathbb{R}$, so I can prove the statement.

Many thanks!

==========

Edit for comment on 12/26

Edit for new comments

答:
There may not be a contour line between the two points. Indeed, consider the function
$$F(x,y) = (x^2+1) \arctan y$$
with $\frac{\partial F}{\partial y}(x,y) = \frac{x^2+1}{y^2+1}> 0$ on $\mathbb{R}^2$.
At  the points $(-1,1)$ and $(1,1)$ the functions takes the value $c=\frac{\pi}{2}$, but the level curve $F(x,y) = c$, that is
$$(x^2+1)\arctan y = \frac{\pi}{2}$$
does not intersect the vertical axis $x=0$.

The idea is simple: for every fixed $x$ , the function $F(x,y)$ is strictly increasing in $y$, with image an open interval $I_x$.  As $x$ varies, this interval varies too. It may be  that the interval $I_x$ contains $c$ for $x=x_1$, $x_2$, but, for some intermediate point $x_3$,  $I_{x_3}$ does not contain $c$. The example is chosen so that the $c$ will be one end of the interval $I_{x_3}$.

Note: The level curve $F(x,y) = \frac{\pi}{2}$ is the graph of the function $y = \tan \frac{\pi}{2(x^2+1)}$ defined on $\mathbb{R} \backslash \{0\}$.

$\bf{Added:}$ I changed the sign so that $\frac{\partial F}{\partial y} > 0$, with $F\mapsto -F$.


答:
Given $S = \{y_1, y_2, y_3, ..., y_n\}$, you want to find a point $P$ to minimize $$\sum_{\boldsymbol x \in S}||P-\boldsymbol x||^2$$

Let $Y_k$ be the $k$th value of the point $Y$ (i.e. $Y_k$ would be the $x$ coordinate, $Y_2$ would be the $y$ coordinate, etc).

Then, the sum of the squared distances is
$$\sum_{\boldsymbol x \in S} \sum_{k=1}^d\left(P_k-\boldsymbol x_k\right)^2$$

Reversing the order of summation yields $$\sum_{k=1}^d\sum_{\boldsymbol x \in S} \left(P_k-\boldsymbol x_k\right)^2$$

From this, each coordinate can be minimized separately, so for $k = 1...d$, you need to find $P_k$ to minimize $\sum_{\boldsymbol x \in S} \left(P_k-\boldsymbol x_k\right)^2$. Taking the derivative with respect to $P[k]$ yields $$\sum_{\boldsymbol x \in S}\left(2P_k-2\boldsymbol x_k\right) = 0 \to P_k = \frac{\sum_{\boldsymbol x \in S}\boldsymbol x_k}{|S|}$$

This means that $P$ is given by the average of the points in $S$.

Title: Computing the centroid of some $n$ points in 3-dimensions

Body:
Solving for the centroid of some set of points in 2-dimensions is trivial. I'd like, however, to determine the centroid in 3-dimensions, given an arbitrary set of $n$ points. How can this most easily be done?

Offhand, it seems that this should require some sort of integral. I'm not sure though.

Note that I speak of the centroid which minimizes the sum of the
squared distances.


Body:
I'm trying to prove and compute the limit of this function.

  $\lim_{x→0^+} \frac{\sin(6x)}{\sqrt{\sin(2x)}}$

I've tried converting it into different functions like $\cos(\pi/2-2x)$ or multiplying by the inverse function and so on, but it keep getting back to $0/0$.

I'm sure that the limit does in fact exist because using L'Hôpital's rule it is fairly easy to prove it, but I can't use it because the course I'm attending still hasn't defined the rule's properties, but I couldn't find a different way aside from that.

Body:
Hint

$$\frac{\sin (6 x)}{\sqrt{\sin (2 x)}}=\frac{6x\,\frac{\sin (6 x)}{6x} } {\sqrt{2x}\,\sqrt{\frac{\sin (2 x)}{2 x}}}$$


问:
A strange occurrence in the decimal digits of $\pi$?
I was messing around with various ways to calculate $\pi$ with my computer, and I noticed something a bit strange. I was using $\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+... = \frac{\pi}{4}$, and computing this sum to varying degrees of accuracy. Using pi = 4*sum([((-1)**i)/(2*i+1) for i in range(n)]), I got the following results for different values of n (I have highlighted in bold the digits which are incorrect):

n
result

1,000
3.140592653839794

10,000
3.141492653590034

25,000
3.141552653589803

80,000
3.1415801535897496

100,000
3.1415826535897198

125,000
3.141584653589728

160,000
3.1415864035897374

200,000
3.1415876535897618

250,000
3.141588653589781

300,000
3.141589320256464

500,000
3.141590653589692

600,000
3.1415909869230147

1,000,000
3.1415916535897743

It seems that when n has $2$ and $5$ as its only prime factors, there is a strange collection of correct digits after the first incorrect digit. Furthermore, it seems that the number of incorrect digits in the first string of incorrect digits increases in some way with the difference between the power of $2$ and the power of $5$. (I haven't tested enough values of n to deduce much more about this though)

I can't tell if this an artefact of the way the decimal computation is done by the computer, or if it the result of some theorem about the convergence of this series. Does anyone have any insights as to why this might be happening?

答:
This is a known property of the Leibniz–Gregory series, and has been used to actually compute $\pi$ to many digits using this series. It arises from the Euler–Maclaurin formula:
$$\frac{\pi}{2} - 2 \sum_{k=1}^\frac{N}{2} \frac{(-1)^{k-1}}{2k-1} \sim \sum_{m=0}^\infty \frac{E_{2m}}{N^{2m+1}}$$
where $E_n$ are the Euler numbers. When $N$ (the number of terms) factors into twos and fives only (such as powers of $10$), the error (last sum) becomes a sum of finite decimals.


问:
Concise description of why rotation quaternions use half the angle
I'm currently writing the report on my master thesis project, where I use Android sensors to perform inertial navigation in a heavy industrial environment. In my application, I make use of quaternions to represent the rotation of the device, and I feel it necessary to include a short description of what quaternions are and why they are suitable to represent rotations.

When defining a rotation quaternion (with vector notation) representing a rotation of $\theta$ around a vector $u=(u_x, u_y, u_z)$ as:

$$
q=\left(\cos\left(\frac{\theta}{2}\right), u_x\sin\left(\frac{\theta}{2}\right), u_y\sin\left(\frac{\theta}{2}\right), u_z\sin\left(\frac{\theta}{2}\right)\right),
$$

I think it is necessary to explain why there is a $\frac{1}{2}$ tied to the angle everywhere. I found some suggestions in the answer to this question, but I feel the answers given are either too simple:

The actual rotation is defined by the map $x↦qxq^*$. You get a $θ/2$ from $q$ on the left, and another $θ/2$ from $q^∗$ on the right, which adds up to a $θ$.

and

If it were $\cosθ+a\sinθ$ instead of $\cos(θ/2)+a\sin(θ/2)$, then rotation of $\pi$ about any axis would give you the same result.

... or way too in depth (especially the accepted answer). I feel like both of the quotes above are relevant to what I want to convey, but not quite there.

Related to the first quote, one answer to the question linked above suggests that the reason for using $\theta/2$ is to, given a vector $p$ to rotate expressed as a pure imaginary quaternion, keep the resulting rotated vector after a conjugation operation $qpq^*$ in the pure imaginary 3D space (essentially, the multiplication from the left rotates it $\theta/2$, but also moves it out of the imaginary 3D space - then the multiplication from the right moves it back into the imaginary 3D space and rotates it $\theta/2$ again for a full rotation of $\theta$).

There is also this proof on Wikipedia that shows that using $\theta/2$ makes the conjugation operation equivalent to Rodrigues's rotation formula. However, I feel like this proof is too much to include in my report.

Is there a concise (at most a few lines of text) way of conveying why one has to use $\theta/2$ in rotation quaternions?

It could be something along the lines of:

This is necessary to prevent overlap of positive and negative rotations around the same axis.

This is because the rotation is, in effect, applied twice during the conjugation operation described in [...].

...

答:
I will put into algebraic terms using interesting commutation properties, the really nice explanation of Edin and Mad Physicist above.

The short answer is that for perpendicular inputs either way works. If we take the full angle approach of a single quaternion with no conjugate, it will rotate by the full angle. But it won't leave parallel inputs unaltered. Only the half-angle approach with the conjugate works for both, so that is the correct formulation. Read on for the long answer.

Consider the two cases of the input vector v being parallel and perpendicular to the rotation axis u, in the below formula.

$e^{\frac{\theta}{2} \mathbf{u}}\mathbf{v}e^{-\frac{\theta}{2} \mathbf{u}}$

This covers all possible cases as any input v can be resolved as the sum of a parallel and perpendicular component to u. First we deal with the parallel case v= ku, where we can set the scalar factor k to 1, without loss of generality.

Now since v = u, the expression commutes! (as there is only one vector in play, the order does not matter). So it cancels out as the exponents add up as follows:

$e^{\frac{\theta}{2} \mathbf{u}}\mathbf{u}e^{-\frac{\theta}{2} \mathbf{u}} = e^{\frac{\theta}{2} \mathbf{u}}e^{-\frac{\theta}{2} \mathbf{u}}\mathbf{u}=\mathbf{u}$

This implies that for vectors parallel to the rotation axis, the final result is unaltered as desired.

Now for the perpendicular case, we use the commutation property of perpendicular vectors only, ab = -ba, which can be seen in the basis rules ij=-ji, jk=-kj etc. Let x be perpendicular to u, then:

$e^{\frac{\theta}{2} \mathbf{u}}\mathbf{x}e^{-\frac{\theta}{2} \mathbf{u}}=e^{\frac{\theta}{2} \mathbf{u}}\mathbf{x} \left({cos\left(\frac{\theta}{2}\right)-\mathbf{u}sin\left(\frac{\theta}{2}\right)}\right)=e^{\frac{\theta}{2} \mathbf{u}} \left({cos\left(\frac{\theta}{2}\right)\mathbf{x}+\mathbf{u}\mathbf{x}sin\left(\frac{\theta}{2}\right)}\right)=e^{\frac{\theta}{2} \mathbf{u}} e^{\frac{\theta}{2} \mathbf{u}} \mathbf{x}= e^{\theta\mathbf{u}}\mathbf{x}$

This means that it amounts to the same approach as rotating by a single full angle quaternion and rotating by the angle theta around the axis u, just like in the case of complex numbers.


问:
how to find the distribution density of the variable X + Y?
X and Y are independent and have the same exponential distribution wih density: $$\lambda e^{-\lambda x} 1_{(0,\infty)} (x) $$ The density of the X + Y distribution can be calculated by convolution of the functions: $$f*f(t)=\int_{-\infty}^{\infty} f(x) f(t-x) \, dx $$ but I still have no idea how to solve it, during the lecture we only gave an example for a uniform distribution on (0,1)

答:
\begin{align}
& \int_{-\infty}^\infty \lambda e^{-\lambda x} 1_{(0,\infty)} (x) \cdot \lambda e^{-\lambda (t-x)} 1_{(0,\infty)} (t-x) \, dx \\[12pt]
= {} & \int_0^\infty \lambda e^{-\lambda x}(x) \cdot \lambda e^{-\lambda(t-x)} 1_{(0,\infty)}(t-x)\,dx \\
& \text{because } 1_{(0,\infty)}(x)=0 \text{ when } x<0 \\[12pt]
= {} & \int_0^t \lambda e^{-\lambda x}(x) \cdot \lambda e^{-\lambda(t-x)} \,dx \\
& \text{because } 1_{(0,\infty)}(t-x)=0 \text{ when } x>t \\[12pt]
= {} & \int_0^t \lambda^2 e^{-\lambda t} \,dx
\end{align}
and then observe that $\lambda^2 e^{-\lambda t}$ does not change as $x$ goes from $0$ to $t,$ i.e. you're integrating a constant function.


问:
Finding the height of a triangle without using trigonometry functions
```<code>             A
            /|\
           / | \
          /  |  \
         /   |   \
        c    h    b
       /     |     \
      /      |      \
  B  /___a1__|___a2__\ C
</code>```

Given a triangle ABC as shown above.
By using the Pythagoras theorem, a1= { c^2-b^2+(a1+a2)^2 } / 2(a1+a2) .
How can I express h using a1+a2, b, and c?

答:
By Heron's formula(It has algebraic proof using the Pythagorean theorem), we have
$$
S = \sqrt{p(p-(a_1+a_2))(p-b)(p-c)}, \text{ where } p = \dfrac{a_1+a_2+b+c}{2}.
$$
Also,
$$
S = \dfrac{1}{2} (a_1+a_2)\times h
$$
Can you proceed from here?

$$
\dfrac{1}{2} a\times h = \sqrt{p(p-a)(p-b)(p-c)}\\
\dfrac{1}{4}a^2 h^2= p(p-a)(p-b)(p-c)\\
h^2=4p(p-a)(p-b)(p-c)/a^2
$$


问:
Invariant hermitian forms and irreducible representations
Let $V$ be a vector space over $\mathbb{C}$ of finite dimension $n$, $G$ is a finite group and $T:G\rightarrow GL(V)$ its irreducible representation that sends each $g$ into $T_g$.
Let $E:V^{\bigoplus 2}\rightarrow \mathbb{C}$ be an arbitrary hermitian form on $V$. It is easy to verify $H$ that sends a pair of vectors $(x,y)$ into $\sum_{g\in G} E(T_g x,T_g y)$ is a $G$-invariant hermitian form, i.e. each $T_g$ respects $H$. It means that an invariant hermitian form always exists.

I am trying to prove that each $G$-invariant hermitian form $H_i$ is $\alpha_i H$ for some $\alpha_i \in \mathbb{C}$. Could you give me any hints or help to develop the following idea?

There is a widely known result that is called Schur's lemma: each endomorphism of irreducible complex representation is scalar (has a form $x\rightarrow \lambda x$). It is easy to deduce the following lemma:

If $T_1$ and $T_2$ - isomorphic irreducible representations then each isomorphism $\phi$ has a form $x \phi_0$ for some $x\in \mathbb{C}$ and fixed isomorphism $\phi_0$.

Is it possible to solve my problem using this lemma? Or somehow otherwise?

答:
I think it is most useful here to interpret any Hermitian form$~h$, which I'm supposing conjugate-linear in its first argument, as associated to a conjugate linear mapping $f_h:V\to V^*$ defined by $f_h(v)=h(v,\cdotp)\in V^*$. Now a first thing to check is that $H$, which is nonzero because $H(v,v)$ is an average over positive values, makes $f_H$ invertible: the kernel of $f_H$ is easily seen to be a $G$-invariant subspace of $V$ (becuase $H$ is $G$-invariant) and not being the whole space it can only be $\{0\}$, and then $f_H$ is invertible because $\dim V=\dim V^*$.

Now if $H_1$ is another $G$-invariant Hermitian form, the composition $(f_H)^{-1}\circ f_{H_1}$ is a $\Bbb C$-linear map $V\to V$ that can be checked to intertwine with the $G$-action. This allows you to apply Schur's lemma.


答:
Applying the convolution formula to your example you immediately get

$$f_Z(z)=\int_0^{z}\lambda e^{-\lambda x}\lambda e^{-\lambda (z-x)}dx=\lambda^2 z e^{-\lambda z}$$

$z\geq 0$

Proof:

By Definition you get

$$F_Z(z)=\int_{-\infty}^{\infty}f_X(x)\Bigg[\underbrace{\int_{-\infty}^{z-x}f_Y(y)dy}_{F_Y(z-x)}\Bigg]dx$$

Derivating you get the density

$$f_Z(z)=\frac{d}{dz}F_Z(z)=\int_{-\infty}^{\infty}f_X(x)f_Y(z-x)dx$$

To realize that the integral bounds are $[0;z]$ just do a drawing of the transformation function

Without any calculations, $Z$ density could be found using the properties of Gamma distribution. As $Exp(\lambda)=Gamma(1;\lambda)$, thus $Z=X+Y\sim Gamma(2;\lambda)$

Title: how to find the distribution density of the variable X + Y?

Body: X and Y are independent and have the same exponential distribution wih density: $$\lambda e^{-\lambda x} 1_{(0,\infty)} (x) $$ The density of the X + Y distribution can be calculated by convolution of the functions: $$f*f(t)=\int_{-\infty}^{\infty} f(x) f(t-x) \, dx $$ but I still have no idea how to solve it, during the lecture we only gave an example for a uniform distribution on (0,1)


问:
If I have an MST, and I add any edge to create a cycle, will removing the heaviest edge from that cycle result in an MST?

If I have an MST, and I add any edge to create a cycle, will removing the heaviest edge from that cycle result in an MST?

MST means minimum spanning tree of a graph. I came across these two posts:

https://stackoverflow.com/questions/13437702/how-to-get-the-new-mst-from-the-old-one-if-one-edge-in-the-graph-changes-its-wei

https://sudeepraja.github.io/MST/

and I follow both until the case where $w_{old}>w$ and $e\notin T$. They both say  that deleting the heaviest edge will guarantee an MST, but I don't see how to prove that. The cycle property just says that IF you have an MST, it can't have an edge which is the heaviest edge in a cycle of the original graph $G$; it is NOT saying that IF you have a tree that doesn't contain an edge that happens to be the heaviest edge of some cycle in the original graph $G$, you are an MST.

To make the question more explicit in terms of the problem it was trying to solve, I will copy a part of the first link:

If its weight was reduced, add it to the original MST. This will create a cycle. Scan the cycle, looking for the heaviest edge (this could select the original edge again). Delete this edge.

I don't understand why this guarantees that we find an MST. Sure, we get a spanning tree but why does deleting this heaviest edge yield a MINIMUM spanning tree?

答: If you already have an MST, then any edge e you add to it will itself be guaranteed to be a maximal weight edge on the resulting cycle. If the resulting cycle had an edge that had higher weight than e, then you would be able to remove it and keep e, which contradicts your claim that we began with an MST.</p>


问:
Why doesn't the Weierstrass approximation theorem imply that every continuous function can be written as a power series?
I hope that my question in the title is well formulated.

I am a little bit confused with the next exercise from a book:

Argue that there exists functions $f \in C[0, \frac{1}{2}]$ that cannot be written on the form
$$
f(x) = \sum_{k=0}^{\infty}c_kx^k, x\in(0, \frac{1}{2}).
$$

But Weierstrass' theorem states that $||f(x)-P(x)||_{\infty} < \epsilon$, where $P(x)$ is a polynomial for $x\in[a,b]$.

The statement of the exercises refers to an open and bounded interval, and in Weierstrass to a closed and bounded interval for the polynomial. Is this the key? to be honest, I do not see why.

答:
Weierstrass theorem tells that on a compact interval, given a continuous function, there is a sequence of polynomials converging uniformly to this function. But the polynomials are not necessarily of the form $\sum_{j=0}^{N_n}c_jx^j$, because the coefficients depend on $n$. They are indeed of the form $\sum_{k=0}^{N_n}c_{n,k}x^k$.

To get a concrete example, consider $\left|x-\frac 12\right|$. This cannot be written as a power series, because this function would be differentiable at $1/2$.


问:
What does "subgroup of $F_n$ consisting of all words of exponent sum $k$" mean?
I am currently reading the following paper by Birman and Hilden,  https://www.jstor.org/stable/1970830?casa_token=7y0vMR6x_lsAAAAA:AObJdZWkgbTOaDbo5woR_Zm01B2xYRNo1aW1j605DqwmFyZfT83cJIngRHDZtAmGR2YymSASh4xx19f8lEaZxlx1Ox2yIR6aS1grOkgTOB5Kc9euIEVc
On page 436 after Lemma 7.1 the author says "Let $H$ be a subgroup of index $k$ in $F_n$ consisting of all words of exponent sum $k$ in $x_1,x_2\dots x_n$." What subgroup is being talked about? Is the "exponent sum" taken over the entire word or each generator individually? I think they mean sybgroup generated by words of exponent length $k$ but I have not been able to prove that it is a index $k$ subgroup. Any help would be appreciated.

答: It seems that it should say "exponent sum divisible by $k$".  This is a subgroup of index $k$ because it is exactly the kernel of the surjective homomorphism $F_n\to\mathbb{Z}/(k)$ which sends each generator to $1$ (so it sends a word to the sum of its exponents mod $k$).


问:
Differentiating multiple products/quotients
I have a function

$$y(x) = \frac{u’(x)}{u(x)r(x)}$$

I want to differentiate this function to find $y’(x)$  in terms of $u’(x)$, $u(x)$, $r(x)$.

I am trying to decide whether to use the product or quotient rule but I have never seen a situation where there were three different functions of x all involved at the same time.

Any suggestions please on what approach I should take?

Thanks

答:
Don't use any of them but logarithmic differentiation instead
$$y(x) = \frac{u’(x)}{u(x)r(x)}\implies \log(y(x))=\log(u'(x))-\log(u(x))-\log(r(x))$$
$$\frac{y'(x)}{y(x)}=\frac{u''(x)}{u'(x)}-\frac{u'(x)}{u(x)}-\frac{r'(x)}{r(x)}$$
$$y'(x)=y(x) \times\frac{y'(x)}{y(x)}$$ and simplify.


问:
Confused about free variables
What is the difference between a free variable and a basic variable (the opposite of a free variable) when the free variable only appears once in only one equation in the system. It seems to me that the purpose of getting row reduced echelon form is to only get one equation for each basic variable (correct me if I'm wrong), but what happens when a free variable also only appears once? Why can't the free variable be considered a basic variable?</p>

答:
Note that what you refer to as a 'basic variable' is more commonly referred to as a bound variable, which you might find useful in learning more about linear algebra. In a sense which variables are free and which are bound is just a question of choice. Consider the somewhat degenerate example of the equation
$$
x_1+2x_2 = 0
$$
Which has the corresponding matrix
$$
\begin{bmatrix}
1&2&0
\end{bmatrix}
$$
We could think of this as $x_1$ being a bound variable and $x_2$ being free and that would be correct, but we would be equally correct (if a bit more confusing) to write our original equation $2x_2+x_1 = 0$, which has RREF
$$
\begin{bmatrix}
1&\frac{1}{2}&0
\end{bmatrix}
$$
The general idea that you can take from this is that as long as one is careful you can permute the columns of your matrix (or equivalently change the order of your variables) and still faithfully encode a linear system.

Title: Confused about free variables


问:
Inequality for a two variables function
I have tried to solve this exercise but I am having difficulties.

I have to find a function greater than $$\frac{|y|^\alpha \sin(xy)}{(x^2+y^2)^{3/2}}$$ as the parameter $\alpha$ changes.
That's what I tried:

Since $$sin(xy)\leq1$$

$$\frac{|y|^\alpha \sin(xy)}{(x^2+y^2)^{3/2}} \leq \frac{|y|^\alpha}{(x^2+y^2)^{3/2}}$$

Then:

$$(x^2+y^2)^{3/2} \geq y^3 $$ so that:

$$\frac{|y|^\alpha}{(x^2+y^2)^{3/2}} \leq \frac{|y|^\alpha}{y^{3}}$$
Now I have problems and I don't know what I did wrong. Following my reasoning I should have:
$$\frac{|y|^\alpha}{y^{3}}=|y|^{\alpha - 3}$$
But I Know it's wrong.

答:
$x^2, y^2 \geq 0$ thus $x^2 + y^2 \geq 0$ and $x^2+y^2\geq y^2$ because $x^2 \geq 0$.

Then put both sides to the power $\frac{3}{2}$

$(x^2+y^2)^\frac{3}{2}\geq (y^2)^\frac{3}{2}=|y|^3$ instead of $y^2$

and we can do this because $y^2 \geq 0$ and $x^2 + y^2 \geq 0$ and $\frac{d}{dz}[z^\frac{3}{2}]=\frac{3}{2}z^\frac{1}{2}\geq 0$ for $z\geq 0$, i.e. $z^\frac{3}{2}$ is an increasing function, so if $z_2\geq z_1$ then $z_2^\frac{3}{2} \geq z_1^\frac{3}{2}$.

Other than that, it doesn't look like the typo really affected your answer.

Title: Inequality for a two variables function


答:
think about $f(x) = \frac 1x$.

On any closed interval $[a,b]$ with $a>0$ the function is bounded and I can approximate it very well with a polynomial.

But on the open interval $(0,1)$ it's unbounded.  The reason I can get away with this is because it's not defined at $0$.  Any continuous function on a closed interval is bounded.  That's not true on open intervals.

That's why Weierstrass can't help with open intervals. For the answer to the homework problem I'll refer you to Davide Giraudo's answer which has just popped up. (Which is much easier than proving that $\frac 1x$ can't be written as a polynomial.)

Title: Why doesn't the Weierstrass approximation theorem imply that every continuous function can be written as a power series?

Body:
I hope that my question in the title is well formulated.

I am a little bit confused with the next exercise from a book:

Argue that there exists functions $f \in C[0, \frac{1}{2}]$ that cannot be written on the form
$$
f(x) = \sum_{k=0}^{\infty}c_kx^k, x\in(0, \frac{1}{2}).
$$

But Weierstrass' theorem states that $||f(x)-P(x)||_{\infty} < \epsilon$, where $P(x)$ is a polynomial for $x\in[a,b]$.

The statement of the exercises refers to an open and bounded interval, and in Weierstrass to a closed and bounded interval for the polynomial. Is this the key? to be honest, I do not see why.


问:
Complex Integral $\int_{-\infty}^{+\infty}\frac{e^{(-A\sqrt{x^2+B}+Cix)}}{\sqrt{x^2+B}}$
I am confused calculating the below integral:

$$\int_{-\infty}^{+\infty}\frac{e^{(-A\sqrt{x^2+B}+Cix)}}{\sqrt{x^2+B}}dx$$

Where A, B, C are real and A and B are positive. I don't think I can use roots of the denominator and calculate residue at those points. Please let me know if you need further information. Thank you !

Edit: Bases on what suggested (Thanks to Zachary), I changed my focus on calculating below integral:
$$\int_{-\infty}^{+\infty}e^{(-A\sqrt{x^2+B}+Cix)}dx$$
where I considered $Cix=cos(Cx)+isin(Cx)$ considering even/odd function properties, I ended up with:
$$2\int_{0}^{+\infty}e^{(-A\sqrt{x^2+B})}cos(Cx)dx$$
At this point, I applied a change of variables as $x'=\sqrt{x^2+B}$, doing so, I come up with below:
$$2\int_{0}^{+\infty}e^{(-A\sqrt{x^2+B})}cos(Cx)dx=2\int_{0}^{+\infty}e^{(-Ax')}cos(C\sqrt{(x'^2-B)})\frac{x'}{\sqrt{x'^2-B}}dx'$$
Then, I was thinking of changing another variable $x''=sin(C\sqrt{(x'^2-B)})$ to get:
$$2\int_{0}^{+\infty}e^{(\frac{-A}{\sqrt{C}}(\sqrt{arcsin(x'')^2+BC}))}dx''$$
Not sure, if it makes sense, please advise, Many thanks.

答:
This is not a complete answer, but a long comment on an upper  bound that I found for your integral. Let $B=u^2$. Indeed, substituting $ x\mapsto u\sinh z$ (as done similarly here), $-I'(A)$ reduces to
$$2u\int_0^\infty\cos\left(uC\sinh z\right)e^{-Au\cosh(z)}\cosh(z)\,dz$$
This is roughly the Fourier Transform of the second modified Bessel function's  "density". Indeed, this function has the integral representation
$$K_1(x)=\int_{0}^\infty e^{-x\cosh(z)}\cosh(z)\,dz.$$
The triangle inequality tells us that
$$-I'(A)\le 2u\,K_1(Au).$$
We know that $-I'(A)>0$ from the integral, as well as $\lim_{A\to\infty} I(A)=0$. Hence, integrating, you find that
$$I(A)\le 2K_0(Au),$$
where $K_0(x)$ is the second modified Bessel function of null order. This shows you (although we could have seen this above) that $I$ is best viewed as a function of the dimensionless parameter $s=Au$ (as well as of $uC$ but we will ignore this dependence here).
In particular, you now know that
$$I(s)=\mathcal{O}\left(\frac{1}{\sqrt{s}}e^{-s}\right)\qquad \text{as } s\to\infty$$
thanks to the asymptotics of the Bessel functions.

I am not sure about the existence of a closed-form, but if I find one I will edit my answer.


问:
Am I correct about my intuitive understanding of Neyman-Pearson theorem?
I've been studying the Neyman-Pearson theorem and I want to know if my understanding about the relationship between test statistics and the significance level $\alpha$ is correct.
My problem to understand it's how we can come up with different cutoffs with the same $\alpha=0.05$. What I didn't realize is the cutoffs are related to the tests and not to the p.d.f. of the distribution (right?). Since the tests $T1$ and $T_2$ are different I can have different cutoffs $C_1$ and $C_2$ with the same $\alpha$.
In the context of the Neyman-Pearson theorem, I choose the most powerful test. Am I right?

答:
Yes, of course.

$T_1$ and $T_2$ tests have their own distribution that you have to calculate. These distribution can obvioulsy be different from the  pdf of the original rv

Limiting the reasoning to a system of simple hypothesis (both simple), Neymann Pearson Lemma grants you to find the Most powerful of all possible tests, say a test with a higher power with respect to any other test with the same (or lower) size $\alpha$

You will see that in many case the distribution of the test can be very hard to find thus you will have to reason on some function of the distribution in order to find the cut off bound(s)

Example

Let's have the following population density

$$f(x,\theta)=\theta x^{\theta-1}$$

where $x \in [0;1]$,$\theta>0$,

Using a simple random sample $(X_1,X_2,X_3)$ let's prove the following system

$$\begin{cases}
H_0: & \theta=1 \\
H_1: & \theta=2
\end{cases}$$

First Test

I do not apply NP lemma, I choose a test by myself. I choose the following test

$T_1$: reject $H_0$ if $X_1>0.95$

(I do not mind I have 3 observation, I base my test only on the first observation $X_1$)

Let's calculate $\alpha$:

$$P(X_1>0.95|\theta=1)=\int_{0.95}^{1}dx=0.05$$

So my test has a Type I error of $5\%$...nice. Now let's calculate the power:

$$\gamma=P(X_1>0.95|\theta=2)=\int_{0.95}^{1}2x dx=0.0975\approx 10\%$$

Second Test

I apply NP lemma:

$$\frac{L(\mathbf{x}|\theta_0)}{L(\mathbf{x}|\theta_1)}\leq k$$

$$\frac{1}{8X_1X_2X_3}\leq k$$

taking the logarithm,

$$\Sigma_i(-log X_i)\leq c$$

Observing that $-log X_i\sim Exp(\theta)$

We have that the critical region is

$$Y\leq c$$

where $Y\sim Gamma(3;\theta)$ that is $2\theta Y\sim \chi_{(6)}^2$

Now to compare the two test, let's fix $\alpha=0.05$ also fot $T_2$

$$0.05=P(\chi_{(6)}^2<c) \rightarrow c=1.64$$

that is $Y=0.82$

Finally let's calculate $T_2$ power

$$\gamma=P(Y<0.82|\theta=2)=P(\chi_{(6)}^2<3.27)\approx 22.6\%$$

whichi is greater than $T_1$ power...


问:
Using the Maclaurin series of $\arctan(x)$ to evaluate $\int_{0}^{\frac{\pi}{2}} \arctan( a \sin x) \, \mathrm dx$
I want to use the Maclaurin series of $\arctan (x)$ to show that
\begin{align}\int_{0}^{\pi/ 2} \arctan (a \sin x) \,  \mathrm dx &= 2 \sum_{k=0}^{\infty} \frac{\left(\frac{\sqrt{1 + a^{2}}- 1}{a}\right)^{2k+1}}{(2k+1)^{2}} \\ &= \operatorname{Li}_{2} \left(\frac{\sqrt{1+a^{2}}-1}{a} \right) -  \operatorname{Li}_{2} \left(\frac{1-\sqrt{1+a^{2}}}{a} \right).
\end{align}

I guess we should first impose the restriction $  |a| \le 1$.

Then we have

$$ \begin{align} \int_{0}^{\pi/ 2} \arctan (a \sin x) \ dx  &= \int_{0}^{\pi /2} \sum_{k=0}^{\infty} (-1)^{k} \, \frac{(a \sin x)^{2k+1}}{2k+1} \, \mathrm  dx \\ &= \sum_{k=0}^{\infty} (-1)^{k} \, \frac{a^{2k+1}}{2k+1} \int_{0}^{\pi /2} \sin^{2k+1} (x) \, \mathrm  dx \\ &= \sum_{k=0}^{\infty} (-1)^{k} \frac{a^{2k+1}}{(2k+1)^{2}} \frac{2^{2k}}{\binom{2k}{k}}. \end{align}$$

But how do we proceed from here?

答:
Sangchul Lee showed that the problem reduces to showing that $$\sum_{n=0}^{m} (-1)^{n} \, \frac{(m+n)!}{(m-n)!} \left( \frac{n!2^{2n}}{(2n+1)!} \right)^{2} = \frac{1}{(2m+1)^{2}}. \tag{1} $$

The following is a way to prove $(1)$ using properties of hypergeometric functions.

Using the duplication formula for the gamma function, along with  the fact that $$\frac{1}{(m-n)!} = \frac{(-1)^{n}(-m)_{n}}{m!}, $$ we can express the left side of $(1)$ as the hypergeometric series $$_{3}F_{2} \left(1,-m,m+1;\frac{3}{2}, \frac{3}{2}; 1 \right). $$

Then using Euler's integral representation of the generalized hypergeometric function,  we have $$_{3}F_{2} \left(1,-m,m+1;\frac{3}{2}, \frac{3}{2}; 1 \right) = \frac{1}{2} \int_{0}^{1} (1-t)^{-1/2}  \, _2F_{1} \left(-m, m+1; \frac{3}{2} ; t \right) \mathrm dt, $$

where $$_2F_{1} \left(-m, m+1; \frac{3}{2} ; t \right) = \frac{\sin \left[(2m+1)\arcsin (\sqrt{t}) \right]}{(2m+1)\sqrt{t}}.  $$

(See my answer here to a previous question.)

Therefore, $$ \begin{align} \sum_{n=0}^{m} (-1)^{n} \, \frac{(m+n)!}{(m-n)!} \left( \frac{n!2^{2n}}{(2n+1)!} \right)^{2} &= \frac{1}{2(2m+1)} \int_{0}^{1} (1-t)^{-1/2} \, \frac{\sin \left[(2m+1)\arcsin (\sqrt{t}) \right]}{\sqrt{t}} \,  \mathrm dt \\ &= \frac{1}{(2m+1)^{2}} \int_{0}^{(2m+1)\pi /2} \sin (u) \,  \mathrm du \\ &= \frac{1}{(2m+1)^{2}}. \end{align}$$


问:
The direct limit of $A_{f}$ for $f \notin P$ is $A_{P}$
Let $A$ be a commutative ring with identity, and $P$ a prime ideal of $A$. I want to show that $\lim_{f \notin P} A_{f} = A_{P}$. $A_f$ is the localization of $A$ at $\{1\} \cup \{f^{n}: n \in \mathbb N\}$, and $A_P$ is the localization of $A$ at $A \setminus P$. There is a natural map from $A_{f} \to A_P$ if $f$ is not in $P$, so by the universal property of the direct limit, we get a map from $\Phi : \lim_{f \notin P} A_{f} \to A_{p}$. Then it remains to show that this canonical map is an isomorphism. Surjectivity can be proven just using the universal property without knowing what $\Phi$ looks like. However I cannot prove the injectivity using the universal property. It seems that one has to use the explicit form of $\Phi$ in order to achieve that.

答:
It seems to me you're assuming the direct limit exists, but justifying that requires some sort of explicit construction anyway, which you're trying to avoid.

I think what you really what you want to do is to show that $A_P$ (with the system of natural maps $\phi_f \colon A_f \to A_P$) satisfies the universal property in question, using only the universal property of localization.

For that, suppose we have $Y$ with natural maps $\psi_f \colon A_f \to Y$. We need a unique map $u \colon A_p \to Y$ where $\psi_f = u \circ \phi_f$. Fix $f \not\in P$. For all $g \not\in P$, the map $\psi_f \colon A_f \to Y$ factors through the map $A_f \to A_{fg}$, and $g$ is a unit in $A_{fg}$, so $\psi_f(g)$ is a unit in $Y$. Hence there exists a unique map $u_f \colon A_P \to Y$ such that $\psi_f = u_f \circ \phi_f$.

All that's left is to show $u_f$ is independent of $f$, which is essentially formal. Write $\alpha_{f,g} \colon A_f \to A_g$ (when this makes sense). For all $f, g \not\in P$,
$$\psi_f = \psi_{fg} \circ \alpha_{f, fg} = u_{fg} \circ \phi_{fg} \circ \alpha_{f, fg} = u_{fg} \circ \phi_f.$$
By the uniqueness of $u_f$, we have $u_{fg} = u_f$. Hence $u_f = u_{fg} = u_{gf} = u_g$.


问:
How can I show this field extension equality?
How can I show this field extension equality   $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$?</p>

答:
One can use Galois theory to show that $\sqrt{2}+\sqrt{3}$ is a primitive
element of $$\mathbb{Q}(\sqrt{2},\sqrt{3})$$ by finding explicitly
what $$G:=Gal(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q})$$ is and showing
that this element is fixed under all elements of $G$</p>


问:
Is checking really needed?
I recently bought Pinter's A book of Abstract Algebra and I came to Examples (Chapter 2). Now, I do not understand why are "Checks" in (iii), (iv) needed because we deductively concluded that neutral element $e$ must be $- 1$ and simillary we showed that each inverse $x'$ of $x$ is $x'= - x - 2$ (of course, we concluded all this after assuming that symbol $+$ means regular adittion with all its properties).

Therefore, Checks are just surplus because we’ve already set up to get what we’re “checking” now. (Probably I am just wrong so please correct me.)

Here is the 1st example in part B:

答:
This is not an answer to the Question.  It is apparently a response to OP's underlying question, which has been asked repeatedly in comments to the other answers.

Let $x \ast y = x^y$ for $x,y > 0$.  Explore: If $x \ast \varepsilon = x^\varepsilon = x$ for all $x$, then $\varepsilon = 1$.  Check: $x \ast 1 = x^1 = x$ is true, but $1 \ast x = 1^x = x$ is false (for all $x \neq 1$).  (This also demonstrates a failure of commutativity.)

What we have shown is that $1$ is a genuine identity on the right, but fails to be an identity on the left.

Let $x \ast y = xy^2$ for real numbers $x$ and $y$.  Explore: If $x \ast \varepsilon = x\varepsilon^2 = x$ for all $x$, then $\varepsilon^2 = 1$, for instance $\varepsilon = -1$.  Check : $-1 \ast y = (-1)y^2 \neq y$ (for all $y \neq -1$).  So this fails.  Alternatively, $\varepsilon = 1$ ... (works as a right-inverse for all $x$ and fails as a left-inverse for $y \neq 1$).

Finally, an example that is commutative and associative, but does not have an identity (on either side) or inverses.

Let $x \ast y = x + y$ for $x,y$ positive integers.  Explore: If $x \ast \varepsilon = x + \varepsilon = x$, then $\varepsilon = 0$, which is not a positive integer.

Title: Is checking really needed?


问:
Determine Lyapunov stability of an ODE
$$
\begin{cases}
 \frac{\mathrm{d}x}{\mathrm{d}t}=-x-3y+2z+yz,\\
 \frac{\mathrm{d}y}{\mathrm{d}t}=3x-y-z+xz,\\
 \frac{\mathrm{d}z}{\mathrm{d}t}=-2x+y-z+xy.\\
\end{cases}
$$
Let Lyapunov function $\mathcal{V}(x,y,z)=x^2+y^2+z^2$, we get
$$
\dot{\mathcal{V}}\left( x,y,z \right) =2x\left( -x-3y+2z+yz \right) +2y\left( 3x-y-z+xz \right) +2z\left( -2x+y-z+xy \right) =-2\left( x^2+y^2+z^2-3xyz \right).
$$
But here $\dot{\mathcal{V}}$ is neither positively definite nor negatively definite, then how to use Lyapunov's theory to determine its stability? Thank you in advance for your help and contribution!

答: I've realized that the question does not necessarily require me to determine the stability upon the whole space, but a neighbourhood of zero. Then it's easy to see that $\dot{\mathcal{V}}(x,y,z)$ is negetively definite in the neighbourhood$\{x^2+y^2+z^2<1\}$, so it enjoys asymptotic stability.

Title: Determine Lyapunov stability of an ODE


问:
Suggestion applications Cauchy's integral formula
if $\int_{\gamma}f(z)dz=0$, then f is analytic at all points within $\gamma$

I consider it to be false since if I consider $f(z)=\frac{1}{z^2}$, and $\gamma:=|z|=1$, by Cauchy's integral formula, is satisfied
$$\int_{\gamma}\frac{1}{z^2}dz=0$$
However $f(z)=\frac{1}{z^2}$ is analytic in $\mathbb{C}-\lbrace 0\rbrace$, that is, the function f is not analytic at all points of $\gamma$.

therefore the result is false.

Find a result of Morera's theorem, which adds the continuity hypothesis, on the contour, which guarantees that the previous result is true.

Find $\displaystyle\int_{\gamma}\frac{\sin{z}}{(z-\zeta)^2}dz$ where $\gamma:=|z|=1$ and $|\zeta|<1$

By Cauchy's integral formula if $\zeta\in int \gamma$, it must be fulfilled that $\zeta=0$, now
$$\displaystyle\int_{\gamma}\frac{\sin{z}}{(z-\zeta)^2}dz=\frac{2\pi i  f^{'}(\zeta)}{1!}=2\pi i\cos{\zeta}$$

I have a doubt regarding, if this is still fulfilled this equation for $|\zeta|>1$, I think that it is not fulfilled since the singularities of $f (z)$ would no longer be in $int\gamma$

If $f$ is analytic within $\gamma$, except in $z_{0}$ and $(z-z_0)f(z)$ is analytic inside $\gamma$, then $\displaystyle\int_{\gamma}f(z)dz\neq 0$

I do not understand this last section, I must verify if it is true false, any suggestion is gratefully received.

In the same way, I would like to know if the other sections are well done, I appreciate your understanding

答: You can prove this using the Residue Theorem. The value of the integral is $2\pi i \lim (z-z_0)f(z)$. If the integral is $0$ then this limit must be $0$. Now define $g(z)=(z-z_0) f(z)$ if $z \neq z_0$ and $0$ if $z=z_0$. Then $g$ is analytic and it has a zero at $z_0$. So we can write $g(z)=(z-z_0)h(z)$ where $h$ is analytic . But this makes $f$ anayltic at $z_0$ Hence, as long as $f$ is not analytic at $z_0$ the integral cannot vanish.


问:
No continuous injective functions from $\mathbb{R}^2$ to $\mathbb{R}$
Which of the following statements is true?

$(a)$ There are at most countably many continuous maps from $\mathbb{R}^2$ to $\mathbb{R}$

$(b)$ There are at most finitely many continuous surjective maps from $\mathbb{R}^2$ to $\mathbb{R}$

$(c)$ There are infinitely many continuous injective maps
from $\mathbb{R}^2$ to $\mathbb{R}$

$(d)$ There are no continuous bijective maps from $\mathbb{R}^2$ to $\mathbb{R}$

My thinking :- $(a)$ and $(b)$ are false since $f(x,y)=kx$ where $k\in \mathbb{R}$ are continous and choosing $k\neq 0$ falsifies $(b)$

For $(c)$ and $(d)$, My best guess is that there is no continuous injective function from $\mathbb{R}^2$ to $\mathbb{R}$ but I can't prove that !.

I was thinking that $\mathbb{R}^2$ is field isomorphic to $\mathbb{C}$ and so the required function is a real-valued function in complex variable but then I am clueless...

Please give hint. Thank you.

答:
Suppose there exists $f:\mathbb R^2\to\mathbb R$ injective and continuous. Since $f$ is continuous, its image, $\operatorname{Im} f$, must be a connected non-empty subset of $\mathbb R$, ie. an interval. Moreover, since $f$ is injective, said interval must contain more than one point and therefore, has an interior.

Let $y$ be an interior point of $\operatorname{Im} f$ and $x=f^{-1}(y)$. The set $A=\mathbb R^2\smallsetminus\{x\}$ is clearly connected in $\mathbb R^2$ but $f(A)$ is not (as it is an interval minus one point, $y$). This contradicts the continuity of $f$.


问:
Prove that $V$ is a direct sum of $V_{-i}$ and $V_{+i}$
I came across this question in an exercise in my linear algebra course and am at a loss as to how to solve it.

Let $V$ be a complex vector space. Let  $\mathcal{J}:V \to V$ be a linear map such that $$\mathcal{J}^2=-I_V$$ Show that  $V=V_{+i} \oplus V_{-i}$, where $$V_{\pm i}=\{v \in V \mid \mathcal{J} v=\pm i v\}$$ Here  $i=\sqrt{-1}$.

I initially began by proving that $V_{+i} \cap V_{-i} = \{0\}$ and subsequently following it up with a proof that for all $v \in V$, $v = v_1 +v_2$ where $v_1 \in V_{+i}$ and $v_2 \in V_{-i}$. However, I was unable to fully prove the latter statement that $V = V_{+i} + V_{-i}$.

A natural step would be to define a map $\mathcal{K} = \frac{1}{2}(\mathcal{J} + iI_V)$, proving that $\ker(\mathcal{K}) = V_{-i}$ and that $\mathrm{im}(\mathcal{K}) = V_{+i}$ and then considering $v - i\mathcal{K}v + i\mathcal{K}v$. However, that line of reasoning hit a dead end since I wasn't able to reach the conclusion.

Can someone guide me to the correct proof?

答:
Suppose $v\in V_i\cap V_{-i}$. Then $Jv=iv=-iv$ and dividing $2iv=0$ gives $v=0$.

Note that $J$ is invertible. Thus, given any $v\in V$ there is a $w\in V$ for which $v=Jw$, and thus

$$ v=Jw=\frac{1}{2}\big(J+i\big)w+\frac{1}{2}\big(J-i\big)w. $$

Show that the image of $\frac{1}{2}\big( J\pm i \big)$ is contained in $V_{\pm i}$.


答:
Your counterexamples for (a) and (b) are fine.

For (c) and (d) one can make a cardinality argument. suppose that $f:\Bbb R^2\to\Bbb R$ is continuous and injective, and consider the subsets $K_a=\{a\}\times[0,1]$ of $\Bbb R^2$ for $a\in\Bbb R$. Each of these sets is compact and connected, and continuous functions preserve compactness and connectedness, so $f[K_a]$ is a compact, connected, subset of $\Bbb R$ for each $a\in\Bbb R$. The compact, connected subsets of $\Bbb R$ are precisely the closed intervals, so for each $a\in\Bbb R$ there are $u_a,v_a\in\Bbb R$ such that $u_a\le v_a$ and $f[K_a]=[u_a,v_a]$.

Finally, $f$ is injective, so the intervals $[u_a,v_a]$ are non-degenerate and pairwise disjoint. Thus, for each $a\in\Bbb R$ there is a $q_a\in\Bbb Q\cap(u_a,v_a)$, and since these intervals are pairwise disjoint, the map $\Bbb R\to\Bbb Q:a\mapsto q_a$ must be injective. But this is impossible, since $\Bbb R$ is uncountable while $\Bbb Q$ is countable, so there is no such function $f$.

Title: No continuous injective functions from $\mathbb{R}^2$ to $\mathbb{R}$

Body:
Which of the following statements is true?

$(a)$ There are at most countably many continuous maps from $\mathbb{R}^2$ to $\mathbb{R}$

$(b)$ There are at most finitely many continuous surjective maps from $\mathbb{R}^2$ to $\mathbb{R}$

$(c)$ There are infinitely many continuous injective maps
from $\mathbb{R}^2$ to $\mathbb{R}$

$(d)$ There are no continuous bijective maps from $\mathbb{R}^2$ to $\mathbb{R}$

My thinking :- $(a)$ and $(b)$ are false since $f(x,y)=kx$ where $k\in \mathbb{R}$ are continous and choosing $k\neq 0$ falsifies $(b)$

For $(c)$ and $(d)$, My best guess is that there is no continuous injective function from $\mathbb{R}^2$ to $\mathbb{R}$ but I can't prove that !.

I was thinking that $\mathbb{R}^2$ is field isomorphic to $\mathbb{C}$ and so the required function is a real-valued function in complex variable but then I am clueless...

Please give hint. Thank you.