问:
Finding the range of $x$ such that $x^2+\frac{x^{2}}{(x+1)^2}<\frac54$

Find the range of $x$:
$$x^2+\frac{x^{2}}{(x+1)^2}<\frac54$$

I am confused about a good approach to this question. My teacher says a graphical approach is the most convenient one, but isn't there any other method to find the range?

答:
but isn't there any other method to find the range?

First I derived the minimums of $f(x)$. Very easy task finding $x=-2$ and $x=0$

Then by simple inspection (I spent 30'' to do that) immediately find that the range such that $f(x)<1.25$ is given by

$$x \in (-0.5;1)$$

The suggested graphical method is easy too...

the red line is $y=5/4$

the blue line is the graphics of your $f(x)$

to calculate the two bounds I think it is evident that

$$f(1)=1+\frac{1}{4}=\frac{5}{4}$$

$$f(-0.5)=\frac{1}{4}+1=\frac{5}{4}$$


问:
Surd Manipulations.
If a=$-\sqrt {99}+\sqrt {999}+\sqrt {9999}$

b=$-\sqrt {99}-\sqrt {999}+\sqrt {9999}$

c=$-\sqrt {99}+\sqrt {999}-\sqrt {9999}$

Then $\displaystyle\sum_\limits{cyc} \frac{a^4}{(a-b)(a-c)}$ equals??

I first individually calculated (a-b), (b-c), (c-a) and tried to simplify the sum. However, I couldn't do it. Please help!

答:
Hint:

$$a^4(b-c)+b^4(c-a)+c^4(a-b)$$

$$=ab(a^3-b^3)-c(a^4-b^4)+c^4(a-b)$$

$$=(a-b)\{ab(a^2+ab+b^2)-c(a+b)(a^2+b^2)+c^4\}$$

$$=\cdots$$

$$=-(a-b)(b-c)(c-a)(a^2+b^2+c^2+ab+bc+ca)$$

Alternatively, if $f(a,b,c)=a^4(b-c)+b^4(c-a)+c^4(a-b)$

Clearly, $f(a,a,c)=0\implies a-b$ is a factor

Similarly, $(b-c)(c-a)$ are also factors

As $f(a,b,c)$ is symmetric about $a,b,c$

$$f(a,b,c)=(a-b)(b-c)(c-a)(p(a^2+b^2+c^2)+q(ab+bc+ca))$$

Compare the coefficients of $a^4$ to find $p=-1$

Can you find $q=-1?$

Now $a^2+b^2+c^2+ab+bc+ca=(a+b+c)^2-3(ab+bc+ca)$

ContentLicense: CC BY-SA 4.0

Title: Surd Manipulations.


问:
Python provided Feedback Transfer function is not matching with the manually computed Transfer Function
I am new to Control Theory and started learning it recently. I am following Modern Control Engineering by Ogata. I have come across the Obtaining Cascaded, Parallel, and Feedback Transfer Functions with MATLAB Section. Below is the diagram for serial, parallel, and Feedback Transfer functions.

In this, if I take two systems

```<code>>>> sys1

      10
--------------
s^2 + 2 s + 10
>>> sys2

  5
-----
s + 5
</code>```

For series(a) transfer function, the result would be G1(s) X G2(s).

For the below python program,

```<code>import control
sys1 =  control.tf([10],[1,2,10]) # Equivalent to G1(s)
sys2 = control.tf([5],[1,5])      # Equivalent to G2(s)
series_sys = control.series(sys1,sys2)
print(series_sys)
</code>```

I get this output

```<code>          50
-----------------------
s^3 + 7 s^2 + 20 s + 50
</code>```

This output is matching with my hand calculations.

If I try the same thing with parallel transfer functions, the output is matching with G1(s)+G2(s)

```<code>>>> control.parallel(sys1,sys2)

  5 s^2 + 20 s + 100
-----------------------
s^3 + 7 s^2 + 20 s + 50
</code>```

But for the Feedback Transfer functions, the output is not matching with  G1(s)/{1+[G1(s) X G2(s)]} (This is the resultant Transfer function for feedback systems, or is my understanding is wrong ? )

If I calculate the transfer function for this using python, I get the below output. I have compared this with Octave. The results in both are the same.

```<code>>>> control.feedback(sys1 , sys2)

        10 s + 50
------------------------
s^3 + 7 s^2 + 20 s + 100
</code>```

But if I calculate the transfer function using the above formula, I am getting the following result:

```<code>        10 s^3 + 70 s^2 + 200 s + 500
---------------------------------------------
s^5 + 9 s^4 + 44 s^3 + 210 s^2 + 400 s + 1000
</code>```

I am doing anything wrong, or is my understanding of the feedback transfer function wrong?

Kindly help me understand this.

答:
Notice that

$$
\require{cancel}
\frac{10 s^3 + 70 s^2 + 200 s + 500}{s^5 + 9 s^4 + 44 s^3 + 210 s^2 + 400 s + 1000} = \frac{(10 s + 50)\cancel{(s^2 + 2s + 10)}}{(s^3 + 7 s^2 + 20 s + 100)\cancel{(s^2 + 2s + 10)}}
$$

What you see happens because of numeric inaccuracy so the pole/zero cancellation is not done.


问:
Computing a partial sum and finding its sequence
Let $S=\sum_{n=1}^{\infty}a_n$ be an infinite series such that its partial sum is given by $S_n=8-\frac{2}{n^2}$.

Find $\sum_{n=1}^{10}a_n$ and $\sum_{n=4}^{16}a_n$

Find a general formula for $a_n$

I found the first part of (1). I computed $S_{10}$ to get $8-1/50$. I'm struggling with the second part of (1). I figured it was $S_{16}-S_4$, but that didn't work.

For problem (2) I found a couple terms of the sequence, $$6, \frac32, \frac{5}{18}, \frac{7}{72}, \cdots$$I know that $S=\lim_{n\rightarrow \infty}S_n=8$. Assuming this is correct, I see no pattern here to get the sequence.

答:
To be step-by-step explicit for the second part of (1), we have

$$\begin{align}
\sum_{n=4}^{16}a_n
&=a_4+a_5+\cdots+a_{16}\\
&=(a_1+a_2+a_3+a_4+a_5+\cdots+a_{16})-(a_1+a_2+a_3)\\
&=\sum_{n=1}^{16}a_n-\sum_{n=1}^3a_n\\
&=S_{16}-S_3\\
&=\left(8-{2\over16^2}\right)-\left(8-{2\over3^3}\right)\\
&={2\over9}-{1\over128}\\
&={247\over1152}
\end{align}$$

As for part (2), the OP is quite right that there is no obvious pattern that accounts for the first four values of $a_n$. This is because there isn't one! As others have pointed out, there is a general formula,

$$a_n=S_n-S_{n-1}=\left(8-{2\over n^2}\right)-\left(8-{2\over(n-1)^2}\right)=2\left({1\over(n-1)^2}-{1\over n^2}\right)={2(2n-1)\over n^2(n-1)^2}$$

But this formula does not apply to $n=1$. For $n=1$, we simply have $a_1=S_1=8-{2\over1^2}=6$, so the answer comes in two pieces:

$$a_n=\begin{cases}
6&\text{if }n=1\\
\displaystyle{2(2n-1)\over n^2(n-1)^2}&\text{if }n\gt1
\end{cases}$$


问:
Prove that $A - B = A $ iff $A \cap B = \emptyset$
I have a lemma below:

$${A}-{B} = {A} \Longleftrightarrow A \cap B = \emptyset.$$

I start with the $\Longrightarrow$:

$$\text{Suppose}\ x \in {A}-{B},\text{then}\  x \in A\ \text{but}\  x \notin B.$$

$${A}-{B} = {A}\ \text{so}\ x \in A\ \text{but}\ x \notin B \equiv x \in A.$$

So the fact that $x \notin B$ does not matter at all. It's getting confused because this lemma is just proved to prove this theorem:

$$P(A-B) = P(A) \Longleftrightarrow A \cap B = \emptyset.$$

With the lemma, I can prove the theorem. Anyone has any advice on how to prove the lemma? Thank you!

答:
I think you're almost there. You have "$x\in A$ and $x\notin B$" $\Longleftrightarrow$ "$x\in A$". In particular, $x\in A \Longrightarrow x\notin B$, meaning that there is no $x$ for which $x\in A$ and $x\in B$, so $A\cap B$ is empty.

Conversely, if $A\cap B$ is empty then for any $x\in A$ we have $x\notin B$, so $x\in A\Longrightarrow x\notin B\Longrightarrow(x\in A\text{ and }x\notin B)$, and clearly $(x\in A\text{ and }x\notin B)\Longrightarrow x\in A$.


答:
$A-B$ is defined as $\{a \in A : a \notin B\}$. Then,
$$A-B=A \iff \{a \in A : a \notin B\} =A \iff \{a \in A : a \in B\}= \phi$$
Now, $\{a \in A : a \in B\} = A \cap B$ by definition, so we're done.

Title: Prove that $A - B = A $ iff $A \cap B = \emptyset$

Body:
I have a lemma below:

$${A}-{B} = {A} \Longleftrightarrow A \cap B = \emptyset.$$

I start with the $\Longrightarrow$:

$$\text{Suppose}\ x \in {A}-{B},\text{then}\  x \in A\ \text{but}\  x \notin B.$$

$${A}-{B} = {A}\ \text{so}\ x \in A\ \text{but}\ x \notin B \equiv x \in A.$$

So the fact that $x \notin B$ does not matter at all. It's getting confused because this lemma is just proved to prove this theorem:

$$P(A-B) = P(A) \Longleftrightarrow A \cap B = \emptyset.$$

With the lemma, I can prove the theorem. Anyone has any advice on how to prove the lemma? Thank you!


答:
$\Rightarrow$ We prove the contrapositive statement: $A\cap B\neq\emptyset \Rightarrow A-B\neq A$.

$A\cap B\neq\emptyset\Rightarrow $there exists $x\in A\cap B$ i.e. $x\in A$ and $x\in B$ so $x\in A$ and $x\notin A-B$ thus $A-B\neq A$ (there exists one element, namely $x$, that belongs to one set, $A$, but not to $A-B$).

$\Leftarrow$ We first note that $A-B\subset A$. Now $x\in A\Rightarrow x\notin B$ since $A\cap B =\emptyset$ so it must be $x\in A-B$ thus it is also $A\subset A-B$ which implies that $A-B=A$ and this concludes the proof.


答: $$ x^2 +\frac{x^2}{(x+1)^2} \lt \frac 54 \\ \iff x^2(x+1)^2 +x^2 \lt \frac 54 (x+1)^2 \\ \iff x^4 +2x^3 +2x^2 \lt \frac 54 (x^2 +2x +1) \\ \iff x^4 +2x^3 +\frac 34 x^2 -\frac 52 x -\frac 54 \lt 0  \\ \iff 4x^4 +8x^3 +3x^2 -10x -5 \lt 0$$ $1$ is a root, so factor. $$\iff (x-1)(4x^3 +12x^2+15x+5) \lt 0 $$ $-\frac 12$ is a root of the cubic, so factor. $$\iff (x-1)(x+\frac 12) (4x^2+10x+10)\lt 0$$ The quadratic has a negative discriminant  and hence is always positive. $$\iff (x-1)(x+\frac 12) \lt 0 \\ \iff -\frac 12 \lt x \lt 1$$

Title: Finding the range of $x$ such that $x^2+\frac{x^{2}}{(x+1)^2}<\frac54$

Body:
Find the range of $x$:
$$x^2+\frac{x^{2}}{(x+1)^2}<\frac54$$

I am confused about a good approach to this question. My teacher says a graphical approach is the most convenient one, but isn't there any other method to find the range?


问:
Let $A_{\alpha}$ be the $\alpha$-rotation matrix. Prove $A_{\alpha}^T = (A_{\alpha})^{-1}$

Let $A_{\alpha}$ be the alpha-rotation matrix. Prove $A_{\alpha}^T = (A_{\alpha})^{-1}$

In other words, prove $A_{\alpha}$ transpose = $A_{\alpha}$ inverse.

First of all, what is a rotation-matrix? And what does that imply about the $\alpha$-rotation matrix? What does that make $A_{\alpha}$?

答:
Hint: the rotation matrix is given by:

$$M = \begin{bmatrix}
\cos \alpha & -\sin \alpha\\
\sin \alpha&\cos \alpha
\end{bmatrix}$$

and its transpose by:

$$M^{T} = \begin{bmatrix}
\cos \alpha & \sin \alpha\\
-\sin \alpha&\cos \alpha
\end{bmatrix}$$

Now can you show $MM^{T} = I$? (This will imply $M^{T} = M^{-1}$! Also be sure to think about WHY this is the rotation matrix)

ContentLicense: CC BY-SA 3.0


答:
Hint:

Like Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$,

$$x^2+\left(\dfrac x{x+1}\right)^2=\left(x-\dfrac x{x+1}\right)^2+2\cdot\dfrac{x^2}{x+1}$$

If $\dfrac{x^2}{x+1}=a,$ we have $$a^2+2a<\dfrac54\iff(a+1)^2<\dfrac94\iff-\dfrac32<a+1<\dfrac32$$


问:
Are the ordinary least squares regression coefficients uniformly integrable?
In my previous question I asked whether a set of asymptotically normal random variables $\{X_n\}_{n \ge 1}$ are uniformly integrable. In the accepted answer the poster showed that this does not hold in general.

Now what about a specific case. Consider the standard ordinary least squares (OLS) regression coefficients $\hat{\beta}_n = \beta + (X_n^T X_n)^{-1} X_n^T \varepsilon$ where I have used the subscript $n$ to indicate that these are the OLS coefficients for a regression model with $n$ observations. Is $\{\hat{\beta}_n\}_{n \ge 1}$ uniformly integrable? Of course, $\hat{\beta}_n$ is a random vector, so I am asking whether the elements of this vector, which are all asymptotically normal, are uniformly integrable.

Note that the $X_n$ above are stochastic regressors, not fixed.

Edit: Attempt for the case of single regressor with no intercept
Here is an attempt for the case of a single regressor with no intercept, i.e., the regression model is
$$
Y = X \beta + \varepsilon,
$$
where we have $n$ i.i.d observations $Y = [Y_1,Y_2,\dots,Y_n]^T$ and regressors $X = [X_1,X_2,\dots,X_n]^T$, and $\varepsilon = [\varepsilon_1,\varepsilon_2,\dots,\varepsilon_n]^T$. The errors are mutually independent with $E[\varepsilon_i|X] = 0$ and they all have the same finite variance $\sigma^2$.

Let $\hat \beta_n$ denote the OLS estimate for $\beta$ where the subscript $n$ indicates the number of observations in the regression model.

Note that because we only have a single regressor that $(X^T X)^{-1}$ reduces from a matix to a scalar:
$$
(X^T X)^{-1} = \bigg(\sum_{i=1}^N X_i^2\bigg)^{-1}.
$$

From what I understand, to show that $\{\hat \beta_n\}_{n \ge 1}$ is uniformly integrable, it is enough to show that there exists a $\delta >0$ such that $\sup_n E[|\hat\beta_n|^{1 + \delta}] < \infty$.

First, I think that wlog we can consider $\hat\beta_n - \beta$ instead of $\hat\beta_n$ in the above expectation. Let's take $\delta = 1$ and define $A = (X^TX)^{-1}$. Then

\begin{align}
E[|\hat\beta_n - \beta|^2]
&=
E[|A\varepsilon|^2] \\
&=
E[A\varepsilon \varepsilon A^T] \\
&=
E[A E[\varepsilon \varepsilon|X]A^T] \\
&=
\sigma^2 E[AA^T] \\
&=
\sigma^2 E[(X^TX)^{-1}].
\end{align}

It is commonly assumed that $\text{plim}_{n\to \infty} (X^TX/n) \stackrel{p}{\to} Q$ where $Q$ is a positive constant (usually a positive definite matrix, but since we only have a single regressor its a scalar in our case). And thus $$
\text{plim}_{n\to \infty} (X^TX/n)^{-1} \stackrel{p}{\to} Q^{-1}.
$$

Write $(X^TX)^{-1} = n^{-1}(X^TX/n)^{-1}$ and substitute into the expectation above to get
\begin{align}
E[|\hat\beta_n - \beta|^2] \
& = \sigma^2 E[(X^TX)^{-1}] \\
& = \sigma^2 n^{-1} E[(X^TX/n)^{-1}].
\end{align}
Now we know $(X^TX/n)^{-1} \stackrel{p}{\to} Q^{-1}$ and if can convert this convergence in probability into convergence in expectation we will have shown that $E[|\hat\beta_n - \beta|^2] < \infty$ (in fact it will go to zero due to the $n^{-1}$ coefficient), and thus the OLS regression coefficients are uniformly integrable.

So it seems the problem of uniform integrability of the OLS regression coefficients reduces to the uniform integrability of $(X^TX/n)^{-1}$.

So here are my questions now:

Does my analysis seem correct?

Is $(X^TX/n)^{-1}$ uniformly integrable?

If $(X^TX/n)^{-1}$ is not uniformly integrable by default is it reasonable to assume it is uniformly integrable? How strong an assumption would this be?

P.S. We have $(X^TX/n)^{-1} = \bigg(\frac{1}{n} \sum_{i=1}^n X_i^2\bigg)^{-1}$. So is the inverse of the second sample moment uniformly integrable?

答: Yes, OLS regression coefficients are uniformly integrable. I refer you to this article that proves it along the way: https://arxiv.org/pdf/1511.02962.pdf</p>


问:
$y''+(\sin x)y=0$ series solutions - differential equations
Find the first four nonzero terms in each of two powers series solutions about the origin. Show that they form a fundamental set of solutions.

$y''+(\sin x)y=0$

what will be the solutions of the above differential equation????

答:
The other answers to this question aren't incorrect. Here is another way to solve it which is probably what you are used to seeing in your class or text book.

The first step is to recognize when they ask for a power series solution about the origin, they mean a power series solution about the point $x=0$. Since $y'' + sin(x)y = 0$ is already in standard form, we can see that $x=0$ is an ordinary point. The sine is analytic at $x=0$ so we don't have any singular points to worry about.

The next step is to use a Maclaurin series expansion for $\sin(x)$. Remember that Maclaurin series are Taylor series about the point $x=0$. If they asked for a power series solution about a point other then $x=0$, we would use the Taylor series expansion. The Maclaurin series expansion for $$
\sin(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}
$$
You can now continue like you are most likely used to with a standard power series solution method for linear, simple variable coefficient, homogeneous differential equations. Take the second derivative of the power series : $\sum_{n=0}^\infty C_n X^n$
$$
y''(x) = \sum_{n=2}^\infty (n)(n-1)C_n X^{n-2}
$$
Substitute the power series summation for $y''(x)$, $y(x)$ and the Maclaurin series expansion for $\sin(x)$ into the original differential equation.
$$
\left(\sum_{n=2}^\infty (n)(n-1)C_n X^{n-2}\right)
+ \left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}X^{2n+1}\right)\left(\sum_{n=0}^\infty C_n X^n\right) = 0
$$
The next step in a problem you are used to would be to re-index the summations to have the same power of $X$ and have the same starting point in the index. We won't be doing that for this problem. All you need to do is expand each summation and then factor.
$$
(2C_2 + 6C_3X + 12C_4 X^2 + 20C_5X^3 +...) \\+ (X - \frac{X^3}{6} + \frac{X^5}{120} +...)(C_0 + C_1X + C_2X^2 + C_3X^3 +...) = 0.
$$
One of the areas you can run into problems with doing expansions like the above is you don't expand to enough terms. Since there is a lot of typing involved I won't type it all out. When you factor through, group like terms. The minimum amount of terms required are shown below:
$$
(2C_2) + (C_0 + 6C_3)X + (C_1 + 12C_4)X^2 + (\frac{-C_0}{6} + C_2 + 20C_5)X^3
\\+ (\frac{-C_1}{6} + C_3 + 30C_6)X^4 + (\frac{C_0}{120}
\\- \frac{C_2}{6} + C_4 + 42C_7)X^5 +... = 0
$$
What you want to do now is similar to what you have done for constant coefficient solutions to differential equations. The thing to remember here is to put all the higher coefficients in terms of $C_0$ and $C_1$.
\begin{align}
X^0 &:& 2C_2 &= 0  &-> C_2 &= 0  \\
X^1 &:& C_0 + 6C_3 &= 0  &-> C_3 &= -\frac{1}{6}C_0  \\
X^2 &:& C_1 + 12C_4 &= 0  &-> C_4 &= -\frac{1}{12}C_1  \\
X^3 &:& \frac{-C_0}{6} + C_2 + 20C_5 &= 0  &-> C_5 &= \frac{C_0}{120}  \\
X^4 &:& \frac{-C_1}{6} + C_3 + 30C_6 &= 0  &-> C_6 &= \frac{C_1}{180} - \frac{C_3}{30}
\end{align}
Substitute what we already got for $C_3$ in terms of $C_0$.
\begin{align}
C_6 &= \frac{C_1}{180} + \frac{C_0}{180}
\\
X^5 &: \frac{C_0}{120} - \frac{C_2}{6} + C_4 + 42C_7 = 0  ~~~ -> C_7 = -\frac{C_0}{5040} - \frac{C_4}{42} = -\frac{C_0}{5040} + \frac{C_1}{504}
\end{align}

We can now substitute the above coefficients in the general power series expansion.
$$
\sum_{n=0}^\infty C_n X^n = C_0 + C_1 X + C_2 X^2 + C_3 X^3 +...
$$
This becomes:
$$
C_0 + C_1 X - \frac{C_0}{6}X^3 - \frac{C_1}{12}X^4 + \frac{C_0}{120}X^5 + (\frac{C_1}{180} + \frac{C_0}{180})X^6 + (-\frac{C_0}{5040} + \frac{C_1}{504})X^7 +...
$$
Your general solution is :
$$
y(x) = C_0\left\{1 - \frac{1}{6}X^3 + \frac{1}{120}X^5 + \frac{1}{180}X^6 - \frac{1}{5040}X^7 +...\right\} \\+ C_1\left\{X - \frac{1}{12}X^4 + \frac{1}{180}X^6 + \frac{1}{504}X^7 + ...\right\}
$$
Since your question asked for two power series solutions, the general solution is $y(x) = (C_0)y_1(x) + (C_1)y_2(x)$.  These are two linearly independent solutions. Just take the above general solution and have $y_1(x)$ equal the $C_0$ portion, and have $y_2(x)$ equal the $C_1$ portion.


问:
Solving DDE in Matlab using matrix form
I am a beginner in learning DDE and i am trying to understand how to use matrix notation to simulate basic systems of the specific form below:
Let us assume a directed network with three nodes

$$\dot{y_{1}} = a_{11}y_{1} + a_{12}y_{2}(t-\tau_{12}) + a_{13}y_{3}(t-\tau_{13})$$

$$\dot{y_{2}} = a_{21}y_{1}(t-\tau_{21}) + a_{22}y_{2} + a_{23}y_{3}(t-\tau_{23})$$

$$\dot{y_{3}} = a_{31}y_{1}(t-\tau_{31}) + a_{32}y_{2}(t-\tau_{32}) + a_{33}y_{3}$$

The direct. way to code in Matlab for the above, is:

```<code>tau = [1.0 0.5 1.5];
tf = 10;
sol = dde23(@dde,tau,@history,[0 tf]);

t = linspace(0,tf,200);
y = deval(sol,t);

figure
subplot(1,2,1)
plot(t,y)
grid on;

function dydt = dde(t,y,yd)

a11 = -1;a12 = 2; a13 = 1;
a21 = 2; a22 = -1;a23 = 1;
a31 = 1; a32 = 1; a33 = -1;

dydt = [a11*y(1)    + a12*yd(2,1) + a13*yd(3,2)
        a21*yd(1,1) + a22*y(2)    + a23*yd(3,3)
        a31*yd(1,2) + a32*yd(2,3) + a33*y(3)];

end

function y = history(t)

y = [1;0;-1];

end
</code>```

the result is this:

But for a n by n system, the process should be somehow automatized to produce the above $n(n-1)/2$ matrices plus the first one which is the diagonal:

$$\begin{multline}
\begin{bmatrix}
\dot{y_{1}}\\
\dot{y_{2}}\\
\dot{y_{3}}\\
\end{bmatrix} = \begin{bmatrix}
a_{11} & 0 & 0\\
0 & a_{22} & 0 \\
 0& 0  & a_{33}
\end{bmatrix} \begin{bmatrix}
y_{1}\\
y_{2}\\
y_{3}\\
\end{bmatrix} + \begin{bmatrix}
0 & a_{12} & 0\\
a_{21} & 0 & 0 \\
 0& 0  & 0
\end{bmatrix} \begin{bmatrix}
y_{1}(t-\tau _{1})\\
y_{2}(t-\tau _{1})\\
y_{3}(t-\tau _{1})\\
\end{bmatrix}
\\ + \begin{bmatrix}
0 & 0 & a_{13}\\
0 & 0 & 0 \\
a_{31} & 0  & 0
\end{bmatrix} \begin{bmatrix}
y_{1}(t-\tau _{2})\\
y_{2}(t-\tau _{2})\\
y_{3}(t-\tau _{2})\\
\end{bmatrix} + \begin{bmatrix}
0 & 0 & 0\\
0 & 0 & a_{23} \\
0 & a_{32} & 0
\end{bmatrix} \begin{bmatrix}
y_{1}(t-\tau _{3})\\
y_{2}(t-\tau _{3})\\
y_{3}(t-\tau _{3})\\
\end{bmatrix}
\end{multline}$$

The code for the general case, applied in the same 3 by 3 is the following:

```<code>tau = [1.0 0.5 1.5];
tf = 10;
a11 = -1;a12 = 2; a13 = 1;
a21 = 2; a22 = -1;a23 = 1;
a31 = 1; a32 = 1; a33 = -1;

A = [a11 a12 a13;
     a21 a22 a23;
     a31 a32 a33];

sol = dde23(@(t,y,z)(dde(t,y,z,A)),tau,@(t,y)(history(t,A)),[0 tf],A);

t = linspace(0,tf,200);
y = deval(sol,t);

figure
subplot(1,2,1)
plot(t,y)
grid on;

function dydt = dde(t,y,yd,A)

 n = length(A);

 A0 = diag(diag(A));
 m = n*(n-1)/2;
 P = zeros(n,n*m);
 k = 0;
 for i = 1:n
     for j = 1:n
         if j>i
             P(i,j+n*k) = A(i,j);
             P(j,i+n*k) = A(i,j);
             k = k+1;
         end
     end
 end

for c = 0:m-1

    dydt = A0*y + P(:,n*c+1:(c+1)*n)*yd(:,c+1);

end

end

function y = history(t,A)

y = [1;0;-1];

end
</code>```

答:
It is obvious once you have found it, but at a superficial glance seems to have all the correct parts:

```<code>for c = 0:m-1
    dydt = A0*y + P(:,n*c+1:(c+1)*n)*yd(:,c+1);
end
</code>```

does not what you intended it to do. It should be

```<code>dydt = A0*y;
for c = 0:m-1
    dydt = dydt + P(:,n*c+1:(c+1)*n)*yd(:,c+1);
end
</code>```

Title: Solving DDE in Matlab using matrix form

Body:
It is obvious once you have found it, but at a superficial glance seems to have all the correct parts:

for c = 0:m-1
    dydt = A0*y + P(:,n*c+1:(c+1)*n)*yd(:,c+1);
end

does not what you intended it to do. It should be

dydt = A0*y;
for c = 0:m-1
    dydt = dydt + P(:,n*c+1:(c+1)*n)*yd(:,c+1);
end


问:
Show that the limit in $(0,0)$ of $\frac{xy}{x^3-y}$ doesn't exist.
I'm trying to show that the limit below doesn't exist, I think it's quite simples but I couldn't until now.

$$\lim_{(x,y)\to (0,0)}\dfrac{xy}{x^3-y}$$

My approach is try to find two curves in the domain that have different limits in $(0,0)$

I couldn't find any curve that the limit isn't $0$.

Any tips? Thanks.

答: Your idea is fine. Take the curve $y=x^3+x^4$. Then$$f(x,x^3+x^4)=-x-1$$and the fact that $\lim_{x\to0}-x-1=-1$. On the other hand, $\lim_{x\to0}f(x,0)=\lim_{x\to0}0=0$.


问:
Comparing ordinal sizes
I was trying to understand the proof to the following theorem, specifically (iii). (The same theorem is listed here: Prove that ordinals are comparable, but the proofs seem to be different.)

Suppose $\alpha, \beta, \gamma$ are ordinals. Then, (iii) exactly one of $\alpha<\beta, \alpha=\beta, \beta<\alpha$ holds, and (iv) if $X$ is a non-empty set of ordinals, then $X$ has a least element $\delta$ and $\delta=\bigcap X$.

Proof:
(iii) Certainly we already know that no more than one of these holds. Consider $\alpha \cap \beta .$ This is also an ordinal (see (iv), or do as exercise). If, for example $\alpha \cap \beta=\alpha$ then $\alpha \subseteq \beta$ and so $\alpha \leq \beta .$ So for a contradiction, suppose $\alpha \cap \beta$ is a proper subset of both $\alpha$ and $\beta .$ Thus (by a previous lemma) $\alpha \cap \beta \in \alpha$ and $\alpha \cap \beta \in \beta .$ So $\alpha \cap \beta \in \alpha \cap \beta:$ a contradiction.

I'm having a lot of difficulty understanding this proof.

Why does $\alpha \cap \beta \in \alpha$ and $\alpha \cap \beta \in \beta$ imply that $\alpha \cap \beta \in \alpha \cap \beta$ ? Surely, $\alpha \cap \beta \subset \alpha$, $\alpha \cap \beta \subset \beta$ implies that $\alpha \cap \beta \subseteq \alpha \cap \beta$, and not necessarily $\alpha \cap \beta \subset \alpha \cap \beta$? (In this course, $\subset$ means a strict subset)

I don't comprehend what the argument is. I get that if $\alpha \cap \beta=\alpha$, then $\alpha \subseteq \beta$ and so $\alpha \leq \beta .$ But what does this contribute to proving the theorem?

答:
For any $x$, $A$ and $B$, if $x \in A$ and $x \in B$, then $x \in A \cap B$. The statement you are querying is an instance of that (with $x = \alpha \cap \beta$, $A = \alpha$ and $B = \beta$).

I think words "for example" in the text of the proof are a bit misleading. The form of the argument is that one of the following must hold:
$$
\begin{array}{cl}
\mbox{(a)} & \alpha \cap \beta = \alpha\\
\mbox{(b)} & \alpha \cap \beta = \beta \\
\mbox{(c)} & \alpha \cap \beta \subsetneq \alpha \mbox{ and } \alpha \cap \beta \subsetneq \beta
\end{array}
$$

The "for example" statement is saying that to deal with (a) and (b) , it is sufficient, by symmetry, just to consider (a).

Title: Comparing ordinal sizes


问:
Group Actions of $S_n$ and $O(n)$
I have been reading up a bit on group actions and whether they are faithful, free and transitive. A question I found states:

Consider the following group actions

1) The symmetric group $S_n$ acting on an n-element set.

2) The orthogonal group $O(n)$ acting on $\mathbb{R}^n$.

3) The orthogonal group $O(n)$ acting on the $(n-1)$-sphere $S^{n-1}$.

Which of these actions is faithful, transitive or free and what are the group orbits?

An explicit action is not specified, but I assumed that we could take the 'obvious' action: 1) I took $S_n$ as permuting the elements of the set and for 2) and 3) I took $O(n)$ as rotations. Is this correct, or can groups only act on a set in one way anyway?

Going with these actions, I thought that 1) would be faithful as the only permutation which leaves a set unchanged is the identity permutation; transitive as you can always find a bijection to send one permutation to another; and free as the only bijection which leaves a permutation unchanged is the identity permutation. The orbit would be all possible permutations of $n$ elements. My problem with this though is that my reasonings for faithful and free seem identical, so I don't think my train of thinking is correct.

For 2) and 3), I also don't see how what they are acting on seems to affect the characteristics of the group action. What kind of things do I need to think about?

答:
I think your assumption about the "obvious" action is spot on.  Note that, in general, one should specify the action - there are often different actions.  (The branch of representation theory, in some very simplified sense, exists to answer the question "how many ways can a group act on a vector space").

You're right that 1) if faithful and your reason is fine.  But when you say it's transitive because "you can always find a bijection to send one permutation to another", note that the action of $S_n$ does not  "send one permutation to another", but rather, sends one element of your $n$-element set to another.  That said, yes, the action is transitive.

A transitive action is rarely free - in particular, in this case, it's not (except when $n\leq 2$):  Given any $p$ in your $n$ element set, there are plenty of elements of $S_n$ which fix $p$ and just move around the other points.  In fact, the isotropy subgroup at $p$ is isomorphic to $S_{n-1}$.

For your statement about orbits, again, the orbit wont consist of a subset of $S_n$, but instead of a subset of your $n$ element set.  To answer this, try proving that an action is transitive iff the orbit through any point is the whole space being acted on.

For 2) and 3), here's a hypothetical situation highlighting how the set you're acting on affects things.  (If you'd like to keep a concrete example in mind, apply what I'm about to say to $G = \mathbb{Z}/2\mathbb{Z} = \{1,-1\}$ under multiplication acting on the set  $X = \{-1,0,1\}$ by multiplication.)

Suppose $G$ acts on $X$ and you want to check if this action is free - that is, that the only element of $G$ which fixes any point of $X$ is the identity in $G$.  One way to prove this is to look at each element of $X$ one at a time and ask "What $g\in G$ fixes this element?"  As long as the answer is always "only $g = e$", the action is free.

Suppose you check every point but one and you keep finding the answer is "only $g= e$" so the action is looking free.  Then you check that last point ($0\in X$ in the concrete example) and find "every element of g fixes this point!"  Well, then the action is not free, woops.

But, since $G$ fixes this "bad" point, it must also preserve the other points: if $y\in X$ is not the bad point, then $gy$ isn't the bad point either, for any choice of $g\in G$.  Thus, $G$ also acts on a subset of $X$ and the action on this subset is free.  So, a nonfree action, when restricted appropriately, can become free.

In the case of restricting the $O(n)$ action from $\mathbb{R}^n$ to $S^{n-1}$, nothing so drastic happens, but there still is at least two changes - the action on $\mathbb{R}^n$ has a fixed point and is nontransitive, while the action on $S^{n-1}$ doesn't have a fixed point and is transitive.

Title: Group Actions of $S_n$ and $O(n)$


问:
Baby Rudin Chapter 1 Problem 16
I am attempting to self-study through Baby Rudin and I have done every exercise in chapter 1 except for problem 16. Every solution I find on the internet for this problem uses theorems from linear algebra which I have no background in. I don't want to go on to chapter 2 without completing chapter 1. Can anyone provide an explanation of this problem that does not assume prerequisite knowledge of linear algebra?

Suppose $k \geq 3$, $x, y \in R^k$, $|x-y| = d>0$, and $r>0$. Prove:

(a) If $2r>d$ there are infinitely many $z \in R^k$ such that

$|z-x| = |z-y| = r $

(b) If $2r = d $ there is exactly one such $z$.

(c) If $2r < d$ there is no such $z$.

My work so far:

(c) is a trivial consequence of the triangle inequality. In about 2 weeks I have been unable to make any progress on (a) or (b).

答:
It's late, but I'll try to answer it.

WLOG, suppose $k=3$ and $x=0$,then $|y|=d$. Let

$A=${$z\in R^3$$|$ $|z|=|z-y|=r$}, $A'=${$z'=(a_1,a_2,a_3)\in R^3| |z'|=\sqrt{r^2-\frac{d^2}{4}} and <z',y>=0$}

where $<,>$ is the standard inner product in $R^3$. First, we shall prove that

$A=A'+\frac{y}{2}$  (which consists of $z'+\frac{y}{2}$ , $z'\in A'$)

$z\in A \iff |z|=r$ and $|z-y|=r$

$\iff |z|=r$ and $z_1^2+z_2^2+z_3^2-2(z_1y_1+z_2y_2+z_3y_3)+y_1^2+y_2^2+y_3^2=r^2$

$\iff |z|=r$ and $z_1y_1+z_2y_2+z_3y_3=\frac{d^2}{2}$

$\iff |z-\frac{y}{2}|=\sqrt{r^2-\frac{d^2}{4}}$ and $<(z-y),\frac{y}{2}>=<z,y>-<y,\frac{y}{2}>=0$

$*$  ${|z-\frac{y}{2}|}^2=(z_1-\frac{y_1}{2})^2+(z_2-\frac{y_2}{2})^2+(z_3-\frac{y_3}{2})^2=z_1^2+z_2^2+z_3^2-(z_1y_1+z_2y_2+z_3y_3)+\frac{1}{4}(y_1^2+y_2^2+y_3^2)=r^2-\frac{d^2}{2}+\frac{d^2}{4}=r^2-\frac{d^2}{4}$

$\iff z=(z-\frac{y}{2})+\frac{y}{2}\in A'+\frac{y}{2}$

Now (a), if $2r>d$, since $r^2-\frac{d^2}{4}>0$, we can define above $A'$ and suppose $y_3$ is not $0$.Set

$$\alpha=(y_3,0,-y_1)$$$$\beta=(y_1y_2,-(y_1^2+y_3^2),y_2y_3)$$

then $\alpha, \beta$ are not $0$. And define

$$c(z)=\frac{\sqrt{r^2-\frac{d^2}{4}}}{|z|}$$

for $z≠0$, then we can easliy know that $c(\alpha)\alpha, c(\beta)\beta\in A'$ so that $A$ is not empty.For any $0≦t≦1$, define

$$w(t)=\alpha-t(\alpha-\beta)=(1-t)\alpha-t\beta$$

If $w(t)≠0$, $$|c(w(t))w(t)|=\sqrt{r^2-\frac{d^2}{4}}$$$$<c(w(t))w(t),y>=0$$, i.e., $c(w(t))w(t)\in A'$

If we can suppose that if $b_1\alpha+b_2\beta=0$ for some $b_1,b_2 \in R$, then $b_1=b_2=0$, then clealy $w(t)≠0$ and also for any $0≦t_1<t_2≦1$

$$c(w(t_1))w(t_1)≠c(w(t_2))w(t_2)$$

This implies that $A'$ is infinite and so $A$

(b)If $2r=d$, $A'=${$\frac{y}{2}$}, so, $A$ has exactly one element

(c)If $2r<d$, there is no $z'\in R^3$ such that $|z'|=\sqrt{r^2-\frac{d^2}{4}}$.$A$ is the empty set

For $k=1,2$, only (a) has the different result. When k=1, since if $<z',y>=0$ then $z'=0$, $A'$ has no element.and if k=2, $<z',y>=0$ implies that $z'=b(y_2,-y_1)$ for some $b\in R$. $|z'|=|b|d$.If $z'\in A'$, $$|b|=\frac{\sqrt{r^2-\frac{d^2}{4}}}{d}$$.
So $A'$ has only two elements and so is $A$

Title: Baby Rudin Chapter 1 Problem 16


问:
Finding the smallest $\alpha>0$ for which $\exists\beta(\alpha)>0$ so that $\sqrt{1+x}+\sqrt{1-x}\le 2-\frac{x^\alpha}\beta,\forall x\in[0,1]$.

Find the smallest $\alpha>0$ for which there is $\beta(\alpha)>0$ so that the following inequality holds $\forall x\in[0,1]$: $$\sqrt{1+x}+\sqrt{1-x}\le 2-\frac{x^\alpha}\beta$$ and find $\min\beta(a)$ for that particular $a$.

Source: Elezović, N., Odabrani zadatci elementarne matematike, Zagreb, 1992

EDIT: By $\beta(\alpha)$ I denoted a positive $\beta$ depends on $\alpha$. If $\beta>0$ exists for some $\alpha>0$, it doesn't have to be unique.

My thoughts:

Function $f(x)=\sqrt{1+x}+\sqrt{1-x}$ is concave on $(-1,1)$, and particulary on $(0,1)$, which follows from the fact that $$f''(x)=-\frac14\left(\frac1{\sqrt{(1+x)^3}}+\frac1{\sqrt{(1-x)^3}}\right)<0,\space\forall x\in(0,1).$$

Since $f$ is concave, each of its tangents is above the graph $\Gamma(f)$.

My first idea was to observe the polynomial $g(x)=f'(c)(x-c)+f(c),c\in (0,1),\deg g=1,$ which made me suspect that $\alpha\ge 1$.

I noticed that a function $f_2(x)=-\frac{x^\alpha}\beta$ is concave for $\alpha>1,$ convex for $0<\alpha<1$ and both convex and concave for $\alpha=1,$ when $f_2$ is linear. Then,
$$\alpha\ge 1\implies f_3(x)=2-\frac{x^\alpha}\beta\text{ is also concave}.$$

and this polynomial $f_3$ and the function $f$ should have at most one intersection and at $x=0$.

However, I'm not able to justify my claim that this is the only possibility.

Also, $f'$ is strictly decreasing on an open interval, and hence, my attempt to express $\min\beta(\alpha)$ in terms of $f'$ in the second part of the task failed.

May I ask for advice on solving this task? Thank you very much in advance!

答:
For $0 \le x < 1$ we have from the binomial series
$$
f(x) = (1+x)^{1/2} + (1-x)^{1/2} = \sum_{n=0}^\infty \binom{1/2}{n} (x^n + (-x)^n) \\
= 2 \sum_{k=0}^\infty \binom{1/2}{2k}x^{2k} \le 2 - \frac 14 x^2
$$
since $\binom{1/2}{2k}$ is negative for all $k \ge 1$. This shows that $\alpha = 2$ works, with $\beta = 4$ as the smallest possible $\beta$.

It also shows that $\alpha = 2$ is the smallest possible $\alpha$: For $0 < \alpha < 2$ and any $\beta > 0$ is
$$
 f(x) - \left(  2-\frac{x^\alpha}{\beta} \right)
=  \frac {1}{\beta} x^\alpha-\frac 14 x^2  + O(x^4) \\
= x^\alpha \left( \frac 1 \beta - \frac 14 x^{2-\alpha} + O(x^{4-\alpha})\right)
$$
and that is strictly positive for sufficiently small $x$.


问:
If Banach space is countable union of closed sets, Can I conclude space should be one of the closed set?
Let $X$ be Banach space. $F_n$ be closed subsets of X. If $X = \cup F_n$, Can I conclude $X = F_{n_0}$ for some $n_0$? If right, suggest a hint.

Body: $$\mathbb R =(-\infty,0]\cup[0,\infty).$$


答:
By definition of intersection, any element $x$ satisfies $x\in\alpha\cap\beta\iff x\in\alpha\land x\in\beta$. This is applied to $x=\alpha\cap\beta$.

We have proved that $\alpha\cap\beta$ can't be a proper subset of both $\alpha$ and $\beta$, thus either $\alpha\cap\beta=\alpha$, that is, $\alpha\le\beta$ or $\alpha\cap\beta=\beta$, that is, $\beta\le\alpha$ must hold.

Stating otherwise, if $\alpha\ne\beta$ then either $\alpha<\beta$ or $\beta<\alpha$.

Title: Comparing ordinal sizes

Body:
I was trying to understand the proof to the following theorem, specifically (iii). (The same theorem is listed here: Prove that ordinals are comparable, but the proofs seem to be different.)

Suppose $\alpha, \beta, \gamma$ are ordinals. Then, (iii) exactly one of $\alpha<\beta, \alpha=\beta, \beta<\alpha$ holds, and (iv) if $X$ is a non-empty set of ordinals, then $X$ has a least element $\delta$ and $\delta=\bigcap X$.

Proof:
(iii) Certainly we already know that no more than one of these holds. Consider $\alpha \cap \beta .$ This is also an ordinal (see (iv), or do as exercise). If, for example $\alpha \cap \beta=\alpha$ then $\alpha \subseteq \beta$ and so $\alpha \leq \beta .$ So for a contradiction, suppose $\alpha \cap \beta$ is a proper subset of both $\alpha$ and $\beta .$ Thus (by a previous lemma) $\alpha \cap \beta \in \alpha$ and $\alpha \cap \beta \in \beta .$ So $\alpha \cap \beta \in \alpha \cap \beta:$ a contradiction.

I'm having a lot of difficulty understanding this proof.

Why does $\alpha \cap \beta \in \alpha$ and $\alpha \cap \beta \in \beta$ imply that $\alpha \cap \beta \in \alpha \cap \beta$ ? Surely, $\alpha \cap \beta \subset \alpha$, $\alpha \cap \beta \subset \beta$ implies that $\alpha \cap \beta \subseteq \alpha \cap \beta$, and not necessarily $\alpha \cap \beta \subset \alpha \cap \beta$? (In this course, $\subset$ means a strict subset)

I don't comprehend what the argument is. I get that if $\alpha \cap \beta=\alpha$, then $\alpha \subseteq \beta$ and so $\alpha \leq \beta .$ But what does this contribute to proving the theorem?


问:
Probability question of die throwing
How many 6 show up when throwing 5 dice?

This is expressed by
$\Pr(X=x)=(5/6)^{5-x}\cdot (1/6)^
x\cdot\binom{5}{x}$

Then I got the following

$0\cdot \Pr(X=0)=0$

$1\cdot \Pr(X=1)=0.4$

$2\cdot \Pr(X=2)=0.32$

$3\cdot \Pr(X=3)=0.09$

$4\cdot \Pr(X=4)=0.0012$

$5\cdot \Pr(X=5)=0.0065$

$E(X)=0.825$, but my book says $E(x)=1$.  Where is the mistake?

答:
The answer of your textbook is wrong! And the proof is very simple:

Your random variable is a binomial $X\sim Bin\left(5;\frac{1}{6}\right)$ with expectation

$$\mathbb{E}[X]=5\cdot \frac{1}{6}=\frac{5}{6}$$

Title: Probability question of die throwing


问:
Number of ways to arrange 7 boys and 4 girls around a circular table such that no two girls are together
Find the number of ways to arrange 7 boys and 4 girls such that no two girls are together.

My approach:

First fix 7 boys. This can be done in $(7-1)!=6!$ ways.
This creates 7 slots for girls.

Now these 4 girls can be arranged in $~_7P_4 = {7\choose 3}\cdot 4!=\frac{7!}{3!}$.

So total number of ways to arrange boys and girls according to given condition is $$\frac {6!\cdot 7!}{3!}$$

Is it correct? If not...please tell me how to do it.

P.S. i'm new here. I know the moderation here is pretty strict and people won't like my not using Mathjax to typeset my question.i'm sorry. I will learn it as soon as possible.

Thank you!

答:
Your working and explanation are both correct.

As an aside, to further explain why when using circular permutations for the boys we only use $6!$ rather than $7!$, due to the circular nature of the table we don't try to keep track of which way was north or south, etc...  To avoid overcounting we can take a specific boy, for example the youngest one, and allow him to sit anywhere at the table first before anyone else.  From that point on, we can organize our thoughts about where everyone else sits in reference to him.  After having seated the youngest boy, there remain only $6$ other boys to arrange around the table after him, hence the result.


问:
Evaluate $\lim_{x\to \infty} \frac{ \cot^{-1} (\sqrt {x+1} -\sqrt x}{\sec^{-1} (\frac{2x+1}{x-1})^x}$
I can simplify the numerator by converting into $\arctan$ And then rationalising inside the function which would give me $\tan^{-1} \infty =\frac{\pi}{2}$, but i don’t know how to deal with the denominator

答:
$$\sqrt{x+1} -\sqrt{x} = \sqrt x ( \sqrt{1+1/x} - 1 ) \approx \sqrt x(1+\frac{1}{2x} -1) = \frac{1}{2\sqrt x} \to 0 $$ So, the top is $\cot^{-1} (0) = \frac{\pi}{2}$.

Also, $$\frac{2x+1}{x-1} \to 2$$ and when you raise $2$ to a huge power, you get $\infty$. So, the bottom is $\sec^{-1} (\infty)= \frac{\pi}{2}$

The limit then evaluates to $\frac{\pi/2}{\pi/2} =1 $.


答:
Generally in problems like this, you often want to homogenize the degrees of $x$ and $y$ on the denominator if possible.

In this case you can see that if we take $y=ux^3$ ($u$ not constant, see it as $u(x,y)=\frac y{x^3}$) then you can factor $x^3$ out.

Then $\dfrac{xy}{x^3-y}=\dfrac{x^4}{x^3(1-u)}=\dfrac{u}{1-u}\times x$

Now it is easier to see that to make it divergent, it suffices to take $\frac{u}{1-u}=\frac 1{x^2}$ to get an overall $\dfrac 1x$.

Substituting back into $y$ gives $f(x,\frac{x^3}{1+x^2})=\dfrac 1x$ is divergent.

Another possibility is to take $\frac u{1-u}=O(x)$ to cancel out with the numerator.

Notice that $\frac u{1-u}=\dfrac Cx$ leads to $f(x,\frac {Cx^3}{x+C})\to C$ for an arbitrary $C$, making the limit non-existent, because multiple paths lead to multiple limit values.

On the other hand when you arrive to $f(x,y)=g(u)\times x^\alpha$ and $g(u)$ is bounded then this time $|f(x,y)|<M|x|^\alpha\to 0$ has a limit in $(0,0)$.

For instance the problem $f(x,y)=\dfrac{xy^2}{x^4+y^2}$ with $y=ux^2$ leads to $g(u)=\frac{u^2}{1+u^2}<1$

Title: Show that the limit in $(0,0)$ of $\frac{xy}{x^3-y}$ doesn't exist.

Body:
I'm trying to show that the limit below doesn't exist, I think it's quite simples but I couldn't until now.

$$\lim_{(x,y)\to (0,0)}\dfrac{xy}{x^3-y}$$

My approach is try to find two curves in the domain that have different limits in $(0,0)$

I couldn't find any curve that the limit isn't $0$.

Any tips? Thanks.


答: A rotation matrix is an element of the class of orthogonal matrices. Since all orthogonal matirices' transpose is equal to their inverses, this applies to all elements of the set, including the rotation matrix.</p>

Title: Let $A_{\alpha}$ be the $\alpha$-rotation matrix. Prove $A_{\alpha}^T = (A_{\alpha})^{-1}$

Body:
Let $A_{\alpha}$ be the alpha-rotation matrix. Prove $A_{\alpha}^T = (A_{\alpha})^{-1}$

In other words, prove $A_{\alpha}$ transpose = $A_{\alpha}$ inverse.

First of all, what is a rotation-matrix? And what does that imply about the $\alpha$-rotation matrix? What does that make $A_{\alpha}$?


问:
Show that $ab \mod m = ((a \bmod m)(b \bmod m)) \bmod m.$
With modular definition in mind that two integers $a,b \in \mathbb{Z}$, are congruent module $m$ meaning that $a\equiv b \pmod{m}$, where $m$ is a positive integer.

Details:

$x \equiv y \pmod{m}$ is by definition equivalent to  $m|(x−y)$.

$x \equiv y \pmod{m}$ , $x,y\in \mathbb{Z}, m \in \mathbb{Z^+}$ iff $a \pmod {m} = b \pmod{m}$.

if $a | b$ and $a|c$, then $a|(b+c)$.

if $a | b$ and $b|c$, then $a|c$.

if $a | b$ then $a|bc$, for all integers $c$.

Question: Show that $ab \pmod{m} = ((a \pmod{m} )(b \pmod{m})) \mod m.$

答:
Let $x, y \in \mathbb{Z}_n$ and $l,k \in \mathbb{Z}$.

Further let
\begin{align*}
a &\equiv x \mod n  \implies a = ln +_n x \\
    b &\equiv y \mod n \implies  b = kn +_n y
\end{align*}

If we now multiply $a\cdot b$ we get:

\begin{align*}
     a\cdot b &= (ln +x)(kn +y) \\
     &=lkn^2 +lny +kxn + xy \\
     &= lkn^2 +n(ly +kx) +xy \equiv x\cdot y \mod n
   \end{align*}


问:
How to check the consistency of OLS estimator in macroeconomic models
Problem:

We have a model $$C_t = a + b Y_t + e_t$$ and $$ Y_t = C_t + I_t$$ It's known that $Cov(I, e)$ is zero.
A student estimates the following model: $$C_t = a + b Y_t + e_t$$
Are the estimators $\hat a_{OLS} = \hat a$ and $\hat b_{OLS} = \hat b$ consistent? If not, then recommend the consistent estimators.

My attempt:

I think that both of the estimators are inconsistent because the model is kind of a recursive one with two equations, while the student regress only one function.

Then I tried to prove it using formulas. I start with the slope estimator and there is a formula for the slope that reminds omitted variable bias:

$$ \hat b = b +  \frac{(1/n)\sum(y - \bar y)(u - \bar u)}{(1/n) \sum(y - \bar y)^2}$$

Then I take $\text{plim}$ of the numerator and denominator.

$$\text{plim}[ (1/n) \sum(y - \bar y)^2] = Var(y) \neq 0$$
$$\text{plim}[ (1/n)\sum(y - \bar y)(u - \bar u)] = Cov(y,u)$$

At this step, I don't know how to evaluate $Cov(y,u)$ is either zero or not zero.

I also assumed that this formula is wrong, and instead tried to use the formula for the OLS estimator of the slope. However, when I tried to substitute $C_t$ I received a recursive equation.

Can you give me help?

答:
I think that your problem is that you considered the model as one equation. However, as you wrote yourself, there are two equation, not one. The problem you faced is called the simultaneous equations.
Your equations are given in the structural form. Try to rewrite them in reduced form and then apply the logic to showed in your attempt.


问:
$\lim _{n\to\infty} (\min \{x_n,y_n\})=\min \{\lim_{n\to\infty} x_n, \lim_{n\to\infty} y_n \}.$

$x_n$ and $y_n$ are bounded number sequences. Prove that
$$\lim _{n\to\infty} \bigl(\min \{x_n,y_n\}\bigr)=\min \left\{\lim_{n\to\infty} x_n,  \lim_{n\to\infty} y_n \right\}.$$

Does that also mean that:

$$\lim _{n\to\infty} \bigl(\max \{x_n,y_n\}\bigr)=\max \left\{\lim_{n\to\infty} x_n,  \lim_{n\to\infty} y_n \right\}?$$

答:
Hint: Note that for any $a, b\in \mathbb R$,

$2\min \{a, b\} =a+b-|a-b|$

Similarly, modify the above for $\max \{a, b\} $


问:
Question on invertible matrices $(E-A)$
Let $A$ be a $n^{th}$ order square and skew-symmetric matrix, if $(E-A)$ is an invertible matrix show that
$(E+A)(E-A)^{-1}$ is an invertible matrix (where $E$ is an identity matrix)

This is a part of a question I have proved that $||(E-A)u|| \geq ||u||$ for any column vector $u \in \mathbb{R^n}$

If we can show that $(E+A)$ is invertible then the proof will be trivial. Any ideas would be appreciated.

答: Hint: $\det(E+A)=0$ if and only if $A$ has an eigenvector with eigenvalue $-1$. Assuming that for some $v\in \mathbb{R}^n\backslash\{0\}$ $$Av=-v,$$ try to find a contadiction with the assumption that $E-A$ is invertible. (Alternatively, in the same way you'll get a contradiction to the part of the question you already proved).

Title: Question on invertible matrices $(E-A)$


问:
Quick Integration Clarification
$ \int_ {-\pi}^{\pi}\ f(x) \ dx  $ is equivalent to integrating $f(x)$ over $-\pi < x < \pi$ (strictly less than) and also $-\pi \le x \le \pi$ (less than or equal). So I just need to reassure myself that the bounds of integration don't change when considering either less than or less than or equal. A quick confirmation will do.

Cheers.

UPDATE:Just applying this to a question to do with fourier series: $\
f(x) = \begin{cases}x+π & -π<x\le 0 \\ π & 0 < x < π\\  \end{cases} $ with 2π periodicity.
Can I proceed to compute:

$a_0= {1\over 2L}[\displaystyle \int \limits_{-π}^{0} \ x+π \ dx + \int \limits_{0}^{π} π\ dx]$

答: With respect to your edit, Yes, you can compute this particular integral (and evaluate "at" the bounds $I(x) \big|_b^a$, but keep in mind Jon Claus's answer: that what you are really doing is evaluating the limit of $I(x)$ as $x \to a^-, x\to b^+$, and that's crucial to remember in the case of improper integrals.

Title: Quick Integration Clarification


问:
Elementary way to prove that the radical of a principal ideal in a UFD has a certain form
Let $R$ be a UFD and $x \in R$. Then we can write $x = x_1^{\alpha_1} \cdot \ldots \cdot x_n ^ {\alpha_n}$ with $x_1, \ldots, x_n$ irreducible. To see that $\sqrt{(x)} = (x_1 \cdot \ldots \cdot x_n)$:

$\supseteq$ is obtained by raising $x_1 \cdot \ldots \cdot x_n$ to the maximum of all $\alpha_i$'s.

$\subseteq$: for this, I could only think of using that $(x_1), \ldots, (x_n)$ are all prime ideals that sit over $(x)$ and the fact that in general $\sqrt I = \bigcap\limits_{\substack{\mathfrak{p} \in spec(R) \\ I \subseteq \mathfrak{p}} } \mathfrak{p}$. But the latter fact has a rather involved proof and it seems like it wouldn't be needed for such a simple statement.

My question is, can "$\subseteq$" be proved more elementarily? How? I don't know where to go from just writing $y ^ l = z x_1^{\alpha_1} \cdot \ldots \cdot x_n^{\alpha_n}$.

答: If $\,k\ge1\,$ then $\,x_i\mid x_1^{a_1}\cdots x_n^{a_n}\mid y^k\,\Rightarrow\, x_i\mid y\,\Rightarrow\, x_1\cdots x_n\mid y,\,$ by $\,x_i\,$ nonassociate primes.


问:
Prove that exists a subspace $U$ of $M_{4\times4}^{\mathbb{R}}$ such that the length of the orthogonal projection of matrix $A$ onto $U$ is 5
Prove that exists a subspace $U$ of $M_{4\times4}^{\mathbb{R}}$ such that the length of the orthogonal projection of matrix $A$ onto $U$ is 5.

$A = \begin{bmatrix} 1 & -1 & 1 & -1 \\ -1  & 1 & -1 & 1 \\ 1 & -1 & 1 & -1 \\ -1  & 1 & -1 & 1\end{bmatrix}$

And the inner product is defined as:

$=Trace(AB^t)$

I did some calculations and I've found that the subspace

$U = Sp\{\begin{bmatrix} 1 + \sqrt{10} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$}

matches the requirements, is that correct? if not, can someone explain how does someone approaches this problem generally?

答:
My approach would start with the calculation of $\|A\|$, where $\|\cdot\|$ as defined above is the Frobenius norm (often called Hilbert-Schmidt norm in the infinite-dimensional situation).
Then recall the general fact that every orthogonal projection $P\neq 0$ satisfies $\|P\|=1$.

In the given situation $A$ were thus required to have norm at least equal to $5$, if such a subspace $U$ indeed exists.

As you are quite parsimonious to present your calculations$-$It is generally considered helpful to include them$-$I'm refraining from presenting more details ...


问:
On finding determinantal representations of polynomials
Consider the ring $\mathbb{Q}[x_1,\ldots,x_n]$ of multivariate polynomials with rational coefficients. Given a positive integer $d$, let $\mathcal{C}_d \subset \mathbb{Q}[x_1,\ldots,x_n]$ be the class of polynomials $f$ which can be written as

$$f = \operatorname{det}(A)$$

where $A$ is a $d$-by-$d$ matrix with entries which are affine in the $x_1, \ldots, x_n$ with rational coefficients. It is of course possible to consider entries which are simply in $\mathbb{Q}[x_1,\ldots,x_n]$, but in the problem I am interested in, I think it is enough to consider entries that are affine in the $x_1, \ldots, x_n$.

If $f \in \mathcal{C}_d$, let us say that $f$ has a degree $d$ determinantal representation (with affine entries).

Question 1: concretely, how to devise an algorithm to tell whether or not a polynomial $f \in \mathbb{Q}[x_1,\ldots,x_n]$ of degree $d$ is actually in $\mathcal{C}_d$ or not?

Question 2: assuming there exists at least one such determinantal representation, corresponding to some $d$-by-$d$ matrix $A$, then there are infinitely many other representations, since for instance we could replace $A$ with $PAQ$, where $P,Q \in \mathrm{SL}(d,\mathbb{Q})$. Is there a way to concretely devise an algorithm that would give the "nicest" such representation? For example, if there is a representation $A$ with entries that are affine with integer coefficients, then this is "nicer" than a representation with rational coefficients which are not integers. It is ambiguous, I agree, but if someone could make this more precise, either in the comments or as an answer, then I would appreciate it.

答:
In https://arxiv.org/abs/1804.00145, there is an algorithm which explains how to write any polynomial $f \in \mathbb{Z}[x_1,\ldots,x_n]$ as

$$ f = \operatorname{det}(A),$$

where $A$ is matrix whose entries are affine in the variables $x_i$. Actually, it is even possible to choose $A$ so that in any column of $A$, at most one of the variables $x_i$ "appears".

I also refer to the related post https://mathoverflow.net/questions/385566/expressing-a-polynomial-as-the-determinant-of-a-matrix-of-linear-forms#comment985383_385566 and the discussion and comments there.

I wonder who first proved the result above, that any $f \in \mathbb{Z}[x_1,\ldots,x_n]$ has a determinantal representation.


答:
I had trouble on this same problem when I was working through Baby Rudin. The easiest solution that doesn't use too complicated linear algebra is this:

Visualize that if $2r>d$ then we're dealing with two spheres that intersect at a circle, and therefore there are infinite many solutions. Now how to prove this?

Let $z = (x+y)/2 + h$ where $h$ is a vector orthogonal to $(x-y)/2$. Note this is a vector, going from the mid point between the spheres up a vector $h$ perpendicularly from the line between between $x$ and $y$ to the circle where they intersect. Using $|x|^2 = x \cdot x$ and expanding $(z-x) \cdot (z-x)=(z-y) \cdot (z-y)=r^2$ we see that this $z$ is a solution iff $h\cdot(y-x)/2 = 0$ and $|h|^2 = r^2 - (|y-x|/2)^2$ that is $|h|^2 = r^2 - d^2/4$. These two equations in dimension three or greater have infinitely many solutions iff $2r>d$, one solution iff $2r=d$, and no solutions iff $2r<d$.

Title: Baby Rudin Chapter 1 Problem 16

Body:
I am attempting to self-study through Baby Rudin and I have done every exercise in chapter 1 except for problem 16. Every solution I find on the internet for this problem uses theorems from linear algebra which I have no background in. I don't want to go on to chapter 2 without completing chapter 1. Can anyone provide an explanation of this problem that does not assume prerequisite knowledge of linear algebra?

Suppose $k \geq 3$, $x, y \in R^k$, $|x-y| = d>0$, and $r>0$. Prove:

(a) If $2r>d$ there are infinitely many $z \in R^k$ such that

$|z-x| = |z-y| = r $

(b) If $2r = d $ there is exactly one such $z$.

(c) If $2r < d$ there is no such $z$.

My work so far:

(c) is a trivial consequence of the triangle inequality. In about 2 weeks I have been unable to make any progress on (a) or (b).


问:
Area of triangle as a determinant
I am having a hard time with the following question

Let $P_0,P_1,P_2$ be points in $\mathbb{R}^2$. We denote $r_i = P_0P_i, i =1,2$. Let P be the 2 by 2 matrix with rows $r_1$ and $r_2$. Let A denote the area of the triangle with vertices $P_0,P_1,P_2$. Explain why $A=\frac{1}{2}|\det(P)|$

Any hints or tips would be greatly appreciated.

答:
There's a small error in notation: the rows are made up of the coordinates of the vectors $\vec{P_0P_i}$.

Hint: it is a basic result on determinants that the absolute value of this determinant is the area of the parallelogram built on these vectors.

Title: Area of triangle as a determinant


问:
Newton's root finding method that copes with stationary points
I am using Newton's method to solve an equation of the form $y=f(x)$ for increasing values of $y$. The value of $x$ found for some $y$ is used for the next $y$ as a starting approximation and this works fine.

$f$ is increasing but may have stationary points, and I do have situations where the range of interest of $y$ starts or ends on an extremum or very close. This causes the algorithm to experience a division by zero or yield very inaccurate roots.

Is there a simple way to cope ? Is there a known modified Newton's method able to deal with this situation ?

I know that double roots can be handled by the modified iterations $$x\leftarrow x-2\frac{f(x)}{f'(x)}$$ but this still does not work at the root, and one needs to know when to activate the factor $2$.

For information, my function $f$ is a cubic polynomial, but I want to avoid the explicit computation of the roots by Cardano's formulas.

Update:

If we restate Newton's method at a stationary point, we get to the next term of Taylor's development,

$$f(x+h)-y\approx f(x)-y+\frac{h^2}2f''(x),$$

so that $$h\approx\sqrt{-\frac{f(x)-y}{f''(x)}}.$$

In fact, the approximation

$$f(x+h)-y\approx f(x)-y+h f'(x)+\frac{h^2}2f''(x)$$ always holds, giving the general formula

$$h\approx \frac{f'(x)\pm\sqrt{f'^2(x)-2f''(x)(f(x)-y)}}{f''(x)}.$$

Due to the increased complexity, I don't want to use that formula all the time.

答:
If I understood correctly, the basic step is to solve the equation $f(x)-y=0$, for given $y$. If we are unsure of the multiplicity of a root, we can apply the modified Newton's method, which corresponds to the application of the usual Newton's method to $g(x)=\frac{f(x)-y}{f'(x)}$. This way, the quadratic convergence is recovered, at least in exact arithmetic. The iteration would be
$$
x_{n+1} = x_n-\dfrac{\frac{f(x_n)-y}{f'(x_n)}}{\frac{f'(x_n)^2 - (f(x_n)-y)f''(x_n)}{f'(x_n)^2} } = x_n - \dfrac{f'(x_n)(f(x_n)-y)}{f'(x_n)^2-f''(x_n)(f(x_n)-y)}
$$
If, by any chance, $f'(x_0)=0$, just move it  a bit.


答:
We can choose a coordinate system so that the origin is at one angle of the triangle and the x-axis lies upon one side of the triangle.

In that coordinate system, the vertices of the triangle are (0, 0), (b, 0), and (a, h).  The are of the triangle is 1/2 height times base= bh/2.

Taking P0= (0, 0), P1P0= (b, 0) and P2P0= (a, h).  The determinant is $\left|\begin{array}{cc}b & 0 \\ a & h\end{array}\right|= bh$

Title: Area of triangle as a determinant

Body:
I am having a hard time with the following question

Let $P_0,P_1,P_2$ be points in $\mathbb{R}^2$. We denote $r_i = P_0P_i, i =1,2$. Let P be the 2 by 2 matrix with rows $r_1$ and $r_2$. Let A denote the area of the triangle with vertices $P_0,P_1,P_2$. Explain why $A=\frac{1}{2}|\det(P)|$

Any hints or tips would be greatly appreciated.


问:
Prove that $0<11 - 6 \sqrt{3} <1$
This point was used as part of a solution I was seeing. A way to do it with approximation is using taylor's theorem:

$$ \sqrt{3} \approx \sqrt{4} - \frac{1}{2\sqrt{4} } = 1.75$$ and hence we find that $$ 11-6 \sqrt{3}\approx.5$$

But, are there ways to prove the above without appealing to taylor's theorem?

P.s: I would prefer a proof which is not based on contradiction methods.

Title: Prove that $0<11 - 6 \sqrt{3} <1$

Body:
We can solve this problem by using a suitable rational approximation to $\sqrt 3$, which can be derived from its continued fraction expansion and Pell's equation.

The primitive solution to the Pell's equation for $\sqrt3$ is
$$2^2 - 3\cdot 1^2 = 1$$
Thus
$$(2+\sqrt3)(2-\sqrt3)=1$$
and for any $n$,
$$(2+\sqrt3)^n(2-\sqrt3)^n=1$$
This gives us all of the best rational approximations for $\sqrt3$. If
$$p^2 -3q^2 = 1$$
then
$$\left(\frac pq\right)^2 = 3 + \frac 1{q^2}$$
Let
$$p_n + \sqrt3 q_n = (2+\sqrt3)^n$$
Then
$$p_{n+1}=2p_n + 3q_n$$
$$q_{n+1}=p_n + 2q_n$$
Let $p, q$ be any such solution pair. Then
$$\sqrt3<\frac pq$$
and
$$3<\sqrt3\frac pq$$
so
$$\frac{3q}p<\sqrt3<\frac pq$$

We can solve the OP's problem by looking for a sufficiently large $p,q$ pair.

$$\frac{-6p}q<-6\sqrt3<\frac{-18q}p$$
$$\frac{11q-6p}q<11-6\sqrt3<\frac{11p-18q}p$$

Let $a=11q-p$ and $b=11p-18q$.

We need to find $p,q$ such that $0<a/q$ and $b/p<1$. The first such pair occurs at $n=2$, with $p=7, q=4$.
$$p^2-3q^2=49-3\cdot16=1$$
$$\frac{44-42}4<11-6\sqrt3<\frac{77-72}7$$
$$0<\frac 12<11-6\sqrt3<\frac 57<1$$

We can get tighter bounds on $11-6\sqrt3\approx 0.6076951546$ by using larger $p, q$ pairs. Here's a table of the first eight results:

$p$
$q$
$a/q$
$b/p$
$a/q$
$b/p$

2
1
-1
2
-1.00000000000000
2.00000000000000

7
4
1/2
5/7
0.500000000000000
0.714285714285714

26
15
3/5
8/13
0.600000000000000
0.615384615384615

97
56
17/28
59/97
0.607142857142857
0.608247422680412

362
209
127/209
110/181
0.607655502392344
0.607734806629834

1351
780
79/130
821/1351
0.607692307692308
0.607698001480385

5042
2911
1769/2911
1532/2521
0.607694950188938
0.607695358984530

18817
10864
3301/5432
11435/18817
0.607695139911635
0.607695169261838


问:
Author's assumption that seems to lose generality of his proof
I am reading a book title "A walk through Combinatorics" by Miklos Bona (fourth edition). The exercise I am talking is on page 3. Here is the cited passage.

Example 1.3. A chess tournament has $n$ participants, and any two players play one game against each other. Then it is true that in any given point of time, there are two players who have finished the same number of games.

His solution:

Solution. First we could think that the Pigeon-hole Principle will not be applicable here as the number of players ("balls") is $n$, and the number of possibilities for the number of games finished by any one of them ("boxes") is also  $n$.
Indeed, a player could finish either no games, or one game, or two games, and so on, up to and including $n-1$ games.

The fact, however, that two players play their games against each other,  provides the missing piece of our proof. If there is a player $A$ who has completed all his $n-1$ games, then there cannot be any player who completed zero games because at the very least, everyone has played with $A$. Therefore, the values $0$ and $n-1$ cannot both occur among the numbers of games finished by the players at any one time. So the number of possibilities for these numbers ("boxes") is at most $n-1$ at any given point of time, and the proof follows.

Question

It seems to me, the assumption

If there is a player $A$ who has completed all his $n-1$ games, then there cannot be any player who completed zero games because at the very least, everyone has played with $A$.

makes the proof lose its generality because at any given point of time, there may be none has played with all other players. Could you elaborate his proof? (I am not a native English speaker so I might have misunderstood his sentences.)

答:
To make a comment to answer.

The argument of the author shows that two events: "somebody has played all $n-1$ games" and "somebody played no game" are not consistent because the former player has played with all others. Therefore at any moment only a proper subset of the numbers of played games: $\{0,1,\dots,n-1\}$ (pigeonholes) is available for $n$ players (pigeons).


问:
Order topology related question
Say I have two topological spaces $X,Y$ s.t $Y$ has the order topology, and $f,g$ are two continuous functions $g,f:X\rightarrow Y$. I want to show that the set $O=\{x:f(x)> g(x)\}$ is open in $X$. So my attempt was defining $h=f-g$ and looking at $O=\{x:h(x)>0\}$ and proving it's open in $X$, but I'm not even sure if I'm allowed to use the concept of $0$ since I don't know anything about $Y$.

Say I can use this concept, I want to show that there exists such open $U\subseteq Y$ s.t $h^{-1}(U)=O$, and by continuity I'm done. So I thought about defining $U=Im(h)\cap V$ where $V=\{y\in Y:y>0\}$ which is simply the ray $(0,\infty)$ in $Y$ and obviously open. So $U$ is open in the subspace topology of $Im(h)$ and therefore $O$ is open in $X$.

Is my approach correct?

答:
In this answer I show that $\{x\mid f(x) \le g(x)\}$ is closed when $f,g$ are continuous. Your set is its complement, thus open.

We only need the order and product topology to see this.

Title: Order topology related question


问:
Solving system of polynomial matrix equations over $\Bbb Z_2$
Let $A, B, C, D$ be $4 \times 4$ matrices over $\mathbb{Z}_2$.

Suppose they satisfy
\begin{equation}
    \begin{cases}
      A^2+ BC+ BCA+ ABC+ A= I_4\\
      AB+ BCB+ ABD=0_{4 \times 4} \\
      CA+ DCA+ CBC= 0_{4 \times 4} \\
      DCB+ CBD= I_4
    \end{cases}\,.
\end{equation}

Does anyone have an idea on how to find matrices $A, B, C, D$ either theoretically or numerically? I'm thinking that we can find a solution since there are four equations and four unknowns.

Hints would suffice. Thank you so much.

Edit: I found out that the 148 solutions is not yet the total number of complete solutions. To possibly decrease the number of solutions I added another constraint:

$$\sum_{k=1}^4\sum_{l=1}^4 A_{1,k}A_{k,l}A_{l,1}+ \sum_{k=1}^4\sum_{l=1}^4 B_{1,k}C_{k,l}A_{l,1}+ \sum_{k=1}^4\sum_{l=1}^4 A_{1,k}B_{k,l}C_{l,1}+ \sum_{k=1}^4\sum_{l=1}^4 B_{1,k}D_{k,l}C_{l,1}=2Y_{5,1,1}.$$

I added this constraint to the last part of the code:

```<code>proc optmodel;
   num n = 4;
   set ROWS = 1..n;
   set COLS = ROWS;
   set CELLS = ROWS cross COLS;

   var A {CELLS} binary;
   var B {CELLS} binary;
   var C {CELLS} binary;
   var D {CELLS} binary;

   var Y {1..5, CELLS} >= 0 integer;

    /* M[i,j,k] = A[i,k]*A[k,j] */
   var M {1..n, 1..n, 1..n} binary;
   con Mcon1 {i in 1..n, j in 1..n, k in 1..n}:
      M[i,j,k] <= A[i,k];
   con Mcon2 {i in 1..n, j in 1..n, k in 1..n}:
      M[i,j,k] <= A[k,j];
   con Mcon3 {i in 1..n, j in 1..n, k in 1..n}:
      M[i,j,k] >= A[i,k] + A[k,j] - 1;

    /* X[i,j,k] = B[i,k]*C[k,j] */
   var X {1..n, 1..n, 1..n} binary;
   con Xcon1 {i in 1..n, j in 1..n, k in 1..n}:
      X[i,j,k] <= B[i,k];
   con Xcon2 {i in 1..n, j in 1..n, k in 1..n}:
      X[i,j,k] <= C[k,j];
   con Xcon3 {i in 1..n, j in 1..n, k in 1..n}:
      X[i,j,k] >= B[i,k] + C[k,j] - 1;

     /* O[i,j,k,l] = B[i,k]*C[k,l]*A[l,j] */
   var O {1..n, 1..n, 1..n, 1..n} binary;
   con Ocon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      O[i,j,k,l] <= B[i,k];
   con Ocon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      O[i,j,k,l] <= C[k,l];
   con Ocon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      O[i,j,k,l] <= A[l,j];
   con Ocon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      O[i,j,k,l] >= B[i,k]+C[k,l]+A[l,j] - 2;

      /* P[i,j,k,l] = A[i,k]*B[k,l]*C[l,j] */
   var P {1..n, 1..n, 1..n, 1..n} binary;
   con Pcon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      P[i,j,k,l] <= A[i,k];
   con Pcon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      P[i,j,k,l] <= B[k,l];
   con Pcon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      P[i,j,k,l] <= C[l,j];
   con Pcon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      P[i,j,k,l] >= A[i,k]+B[k,l]+C[l,j] - 2;

   /* A^2 + BC + BCA + ABC + A = I_4 */
   con Con1 {<i,j> in CELLS}:
      sum {k in 1..n} M[i,j,k]
    + sum {k in 1..n} X[i,j,k]
    + sum {k in 1..n, l in 1..n} O[i,j,k,l]
    + sum {k in 1..n, l in 1..n} P[i,j,k,l]
    + A[i,j]
   = 2*Y[1,i,j] + (i=j);

   /* E[i,j,k] = A[i,k]*B[k,j] */
   var E {1..n, 1..n, 1..n} binary;
   con Econ1 {i in 1..n, j in 1..n, k in 1..n}:
      E[i,j,k] <= A[i,k];
   con Econ2 {i in 1..n, j in 1..n, k in 1..n}:
      E[i,j,k] <= B[k,j];
   con Econ3 {i in 1..n, j in 1..n, k in 1..n}:
      E[i,j,k] >= A[i,k] + B[k,j] - 1;

    /* F[i,j,k,l] = B[i,k]*C[k,l]*B[l,j] */
   var F {1..n, 1..n, 1..n, 1..n} binary;
   con Fcon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      F[i,j,k,l] <= B[i,k];
   con Fcon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      F[i,j,k,l] <= C[k,l];
   con Fcon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      F[i,j,k,l] <= B[l,j];
   con Fcon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      F[i,j,k,l] >= B[i,k]+C[k,l]+B[l,j] - 2;

      /* G[i,j,k,l] = A[i,k]*B[k,l]*D[l,j] */
   var G {1..n, 1..n, 1..n, 1..n} binary;
   con Gcon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      G[i,j,k,l] <= A[i,k];
   con Gcon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      G[i,j,k,l] <= B[k,l];
   con Gcon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      G[i,j,k,l] <= D[l,j];
   con Gcon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      G[i,j,k,l] >= A[i,k]+B[k,l]+D[l,j] - 2;

   /* AB + BCB + ABD = 0 */
   con Con2 {<i,j> in CELLS}:
      sum {k in 1..n} E[i,j,k]
    + sum {k in 1..n, l in 1..n} F[i,j,k,l]
    + sum {k in 1..n, l in 1..n} G[i,j,k,l]
   = 2*Y[2,i,j];

   /* H[i,j,k] = C[i,k]*A[k,j] */
   var H {1..n, 1..n, 1..n} binary;
   con Hcon1 {i in 1..n, j in 1..n, k in 1..n}:
      H[i,j,k] <= C[i,k];
   con Hcon2 {i in 1..n, j in 1..n, k in 1..n}:
      H[i,j,k] <= A[k,j];
   con Hcon3 {i in 1..n, j in 1..n, k in 1..n}:
      H[i,j,k] >= C[i,k] + A[k,j] - 1;

    /* Q[i,j,k,l] = D[i,k]*C[k,l]*A[l,j] */
   var Q {1..n, 1..n, 1..n, 1..n} binary;
   con Qcon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      Q[i,j,k,l] <= D[i,k];
   con Qcon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      Q[i,j,k,l] <= C[k,l];
   con Qcon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      Q[i,j,k,l] <= A[l,j];
   con Qcon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      Q[i,j,k,l] >= D[i,k]+C[k,l]+A[l,j] - 2;

    /* R[i,j,k,l] = C[i,k]*B[k,l]*C[l,j] */
   var R {1..n, 1..n, 1..n, 1..n} binary;
   con Rcon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      R[i,j,k,l] <= C[i,k];
   con Rcon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      R[i,j,k,l] <= B[k,l];
   con Rcon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      R[i,j,k,l] <= C[l,j];
   con Rcon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      R[i,j,k,l] >= C[i,k]+B[k,l]+C[l,j] - 2;

   /* CA + DCA + CBC = 0 */
   con Con3 {<i,j> in CELLS}:
      sum {k in 1..n} H[i,j,k]
    + sum {k in 1..n, l in 1..n} Q[i,j,k,l]
    + sum {k in 1..n, l in 1..n} R[i,j,k,l]
   = 2*Y[3,i,j];

   /* W[i,j,k,l] = D[i,k]*C[k,l]*B[l,j] */
   var W {1..n, 1..n, 1..n, 1..n} binary;
   con Wcon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      W[i,j,k,l] <= D[i,k];
   con Wcon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      W[i,j,k,l] <= C[k,l];
   con Wcon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      W[i,j,k,l] <= B[l,j];
   con Wcon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      W[i,j,k,l] >= D[i,k]+C[k,l]+B[l,j] - 2;

      /* T[i,j,k,l] = C[i,k]*B[k,l]*D[l,j] */
   var T {1..n, 1..n, 1..n, 1..n} binary;
   con Tcon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      T[i,j,k,l] <= C[i,k];
   con Tcon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      T[i,j,k,l] <= B[k,l];
   con Tcon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      T[i,j,k,l] <= D[l,j];
   con Tcon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      T[i,j,k,l] >= C[i,k]+B[k,l]+D[l,j] - 2;

   /* DCB + CBD = I_4 */
   con Con4 {<i,j> in CELLS}:
      sum {k in 1..n, l in 1..n} W[i,j,k,l]
    + sum {k in 1..n, l in 1..n} T[i,j,k,l]
   = 2*Y[4,i,j] + (i=j);

   /* U[1,1,k,l] = A[1,k]*A[k,l]*A[l,1] */
   var U {1..n, 1..n} binary;
   con Ucon1 {k in 1..n, l in 1..n}:
      U[k,l] <= A[1,k];
   con Ucon2 {k in 1..n, l in 1..n}:
      U[k,l] <= A[k,l];
   con Ucon3 {k in 1..n, l in 1..n}:
      U[k,l] <= A[l,1];
   con Ucon4 {k in 1..n, l in 1..n}:
      U[k,l] >= A[1,k]+A[k,l]+A[l,1] - 2;

    /* V[1,1,k,l] = B[1,k]*D[k,l]*C[l,1] */
   var V {1..n, 1..n} binary;
   con Vcon1 {k in 1..n, l in 1..n}:
      V[k,l] <= B[1,k];
   con Vcon2 {k in 1..n, l in 1..n}:
      V[k,l] <= D[k,l];
   con Vcon3 {k in 1..n, l in 1..n}:
      V[k,l] <= C[l,1];
   con Vcon4 {k in 1..n, l in 1..n}:
      V[k,l] >= B[1,k]+D[k,l]+C[l,1] - 2;

   con
   sum {k in 1..n, l in 1..n} U[k,l]
    + sum {k in 1..n, l in 1..n} V[k,l]+ sum {k in 1..n, l in 1..n} O[1,1,k,l] + sum {k in 1..n, l in 1..n} P[1,1,k,l]
   = 2*Y[5,1,1] ;

   solve noobj with milp / maxpoolsols=100 soltype=best;
   create data out(drop=s) from [s]=(1.._NSOL_) {i in 1..n, j in 1..n} <col('A_'||i||''||j)=A[i,j].sol[s]> {i in 1..n, j in 1..n} <col('B'||i||''||j)=B[i,j].sol[s]> {i in 1..n, j in 1..n} <col('C'||i||''||j)=C[i,j].sol[s]> {i in 1..n, j in 1..n} <col('D'||i||'_'||j)=D[i,j].sol[s]>;

   print A;
   print B;
   print C;
   print D;
quit;
</code>```

The software found 100 solutions, but when I checked the output data, it is the same solution replicated 100 times. Is there something wrong with the syntax of the constraint that I added to the code?

答:
I have made an algebraic attempt that might help you somewhere. Lets start with the second and third equation:
$$AB(I+D) = -BCB \rightarrow AB = -BCB(I+D)^{-1}$$
$$(I+D)CA = -CBC \rightarrow CA = -(I+D)^{-1}CBC$$
Now plug these results in the first equation:
$$A^2 + BC - B(I+D)^{-1}CBC - BCB(I+D)^{-1}C+A = I$$
From this, two things can be observed:
$$A = -B(I+D)^{-1}C, ~~ \text{and} ~~  BCA = ABC$$
Next, lets postmultiply the second equation with $C$ and the premultiply the third with $B$:
$$ABC + BCBC + ABDC = 0, ~~ BCA + BDCA + BCBC = 0$$
$$BCA + BDCA + BCBC = ABC + BCBC + ABDC \rightarrow BDCA = ABDC$$
Now if we substitute the definition for $A$ in this result the following can also be deduced:
$$BDCA = ABDC \rightarrow -BDCB(I+D)^{-1}C = -B(I+D)^{-1}CBDC$$
$$DCB(I+D)^{-1}=(I+D)^{-1}CBD$$
All these commutations are starting me to believe that $A$ is something like $\alpha I$, and therefore $-(I+D)^{-1} = \alpha (CB)^{-1}$. Hope this does help you get to an answer.


问:
Recovering a group from its quotient group
Suppose I've a group $G$ and a normal subgroup $H$. I know the structure of $G/H$ as well as the structure of $H$. Is it possible to recover the original structure of G from this?</p>

答:
Unfortunately, no; the simplest example is $G_1=\mathbb{Z}/4\mathbb{Z}$ and $H_1=2\mathbb{Z}/4\mathbb{Z}$, as compared with $G_2=(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$ and $H_2=(\mathbb{Z}/2\mathbb{Z})\times\{\overline{0}\}$. We have
$$H_1\cong H_2\cong\mathbb{Z}/2\mathbb{Z},\qquad G_1/H_1\cong G_2/H_2\cong\mathbb{Z}/2\mathbb{Z}$$
but $G_1\not\cong G_2$. The general problem of determining what groups $G$ could arise, given a specified normal subgroup and the corresponding quotient, is called the group extension problem.</p>


问:
Modifying the covariance of Brownian motion, what Gaussian process do we get?
If $(B_t)$ is a Brownian motion, then $Cov(B_t,B_s)=min(t,s)$.

Take a Gaussian process $(X_t)$ with mean $0$ and covariance $Cov(X_t,X_s)=f(min(t,s))$ for a given function $f$ such that the covariance is still positive definite. Is $X$ related to Brownian motion ?

For example, let's take an easy function: $f(x)=x+1$, then clearly $X_t=B_{t+1}$, or more generally, if $f$ is monotone increasing, then $X_t=B_{f(t)}= \int_0^t \sqrt{f'(s)} \, dB_s$ (where the last equality is valid under integrability/differentiability conditions).

But what happens when $f$ is not monotone ? For example, $Cov(X_t,X_s)=min(t,s)(1-min(t,s))$. This is a covariance function on $[0,1]$. Can we describe $X$ using Brownian motion on $[0,1]$?

Idea: decompose $f$ on intervals on which it is increasing and decreasing. Use the above for the increasing parts. But what happens when $f$ is decreasing ? For our example that would be $f(x)=x(1-x)$ on $[0.5,1]$ ?

答:
The function $f(x)=x(1-x)$ applied to $\min(x,y)$ is not a covariance function since it is not positive definite on $[0.5,1]$. To see this, just calculate the determinant of the "covariance" matrix $C$ of the points $v=0.6$ and $w=0.8$. You find $C(v,v)=0.6*0.4=0.24$, $C(w,w)=0.8*0.2=0.16$ and $C(v,w)=f(0.6)=0.24.$ Now $\det C= 0.24 * 0.16 - (0.24)^2<0.$

This example gives already a good idea of what is going on or better going wrong with non-monotonic $f$. In fact I claim:

If $f$ is such that there exist points $v,w$ with $v< w$ and $f(v)> f(w)$ then $C(x,y)=f(\min(x,y))$ is not positive definite.

The first necessary condition is $$C(x,x)=f(x)>0$$ for all $x$ in the domain of $f$. We assume this holds and calculate
$$C(v,v)C(w,w) - C(v,w)^2=f(v)f(w) - f(v)^2 = f(v)( f(w) - f(v))<0$$ by assumption on $f$.


问:
Why they say commutative field?
sometimes I read the term ,,commutative field". But every field is commutative. Why do they say this? For example the english wikipedia page of the Jacobson–Bourbaki theorem.

Edit: Why does my ,,Hey,..." at the beginning of every post disappear?

答: In English, the word field implies commutativity, but in French, corps is used for both commutative and non-commutative division rings (also known as skew fields).

Title: Why they say commutative field?


答:
You can formulate this as an integer linear programming problem, and here is a feasible solution:
\begin{align}
A&= \begin{pmatrix}
0 &0 &0 &1 \\
1 &0 &1 &0 \\
1 &1 &1 &1 \\
1 &0 &0 &1 \\
\end{pmatrix} \\
B&= \begin{pmatrix}
1 &1 &0 &0 \\
0 &1 &1 &0 \\
1 &0 &1 &1 \\
0 &0 &1 &0 \\
\end{pmatrix} \\
C&= \begin{pmatrix}
1 &1 &0 &1 \\
1 &1 &0 &0 \\
1 &0 &0 &0 \\
1 &0 &1 &0 \\
\end{pmatrix} \\
D&= \begin{pmatrix}
0 &0 &1 &1 \\
0 &0 &0 &1 \\
0 &0 &0 &0 \\
0 &0 &0 &0 \\
\end{pmatrix}
\end{align}
There are 64 constraints, 64 binary decision variables $(A_{i,j},B_{i,j},C_{i,j},D_{i,j})$, and 64 nonnegative integer variables ($Y_{t,i,j}$ for $t\in\{1,2,3,4\}$ and all $(i,j)$).  Explicitly, the constraints  are
\begin{align}
    \sum_{k=1}^n \left(A_{i,k} A_{k,j}
    + B_{i,k} C_{k,j}\right)
    + \sum_{k=1}^n \sum_{\ell=1}^n \left(B_{i,k} C_{k,\ell} A_{\ell,j}
    + A_{i,k} B_{k,\ell} C_{\ell,j}\right)
    + A_{i,j}
   &= 2 Y_{1,i,j} + [i=j] \\
      \sum_{k=1}^n A_{i,k} B_{k,j}
    + \sum_{k=1}^n \sum_{\ell=1}^n \left(B_{i,k} C_{k,\ell} B_{\ell,j}
    + A_{i,k} B_{k,\ell} D_{\ell,j}\right)
   &= 2 Y_{2,i,j} \\
      \sum_{k=1}^n C_{i,k} A_{k,j}
    + \sum_{k=1}^n \sum_{\ell=1}^n \left(D_{i,k} C_{k,\ell} A_{\ell,j}
    + C_{i,k} B_{k,\ell} C_{\ell,j}\right)
   &= 2 Y_{3,i,j} \\
      \sum_{k=1}^n \sum_{\ell=1}^n \left(D_{i,k} C_{k,\ell} B_{\ell,j}
    + C_{i,k} B_{k,\ell} D_{\ell,j}\right)
   &= 2 Y_{4,i,j} + [i=j]
\end{align}
These can be linearized by replacing each product with a new binary variable, together with additional linear constraints, as shown in https://or.stackexchange.com/questions/37/how-to-linearize-the-product-of-two-binary-variables

As requested, here is the SAS code.  The NOOBJ option specifies that there is no objective, and the LINEARIZE option automatically performs the linearization.

```<code>proc optmodel;
   num n = 4;
   set ROWS = 1..n;
   set COLS = ROWS;
   set CELLS = ROWS cross COLS;

   var A {CELLS} binary;
   var B {CELLS} binary;
   var C {CELLS} binary;
   var D {CELLS} binary;

   var Y {1..4, CELLS} >= 0 integer;

   /* A^2 + BC + BCA + ABC + A = I_4 */
   con Con1 {<i,j> in CELLS}:
      sum {k in 1..n} A[i,k]*A[k,j]
    + sum {k in 1..n} B[i,k]*C[k,j]
    + sum {k in 1..n, l in 1..n} B[i,k]*C[k,l]*A[l,j]
    + sum {k in 1..n, l in 1..n} A[i,k]*B[k,l]*C[l,j]
    + A[i,j]
   = 2*Y[1,i,j] + (i=j);

   /* AB + BCB + ABD = 0 */
   con Con2 {<i,j> in CELLS}:
      sum {k in 1..n} A[i,k]*B[k,j]
    + sum {k in 1..n, l in 1..n} B[i,k]*C[k,l]*B[l,j]
    + sum {k in 1..n, l in 1..n} A[i,k]*B[k,l]*D[l,j]
   = 2*Y[2,i,j];

   /* CA + DCA + CBC = 0 */
   con Con3 {<i,j> in CELLS}:
      sum {k in 1..n} C[i,k]*A[k,j]
    + sum {k in 1..n, l in 1..n} D[i,k]*C[k,l]*A[l,j]
    + sum {k in 1..n, l in 1..n} C[i,k]*B[k,l]*C[l,j]
   = 2*Y[3,i,j];

   /* DCB + CBD = I_4 */
   con Con4 {<i,j> in CELLS}:
      sum {k in 1..n, l in 1..n} D[i,k]*C[k,l]*B[l,j]
    + sum {k in 1..n, l in 1..n} C[i,k]*B[k,l]*D[l,j]
   = 2*Y[4,i,j] + (i=j);

   solve noobj linearize;

   print A;
   print B;
   print C;
   print D;
quit;
</code>```

Another approach to this feasibility problem is to introduce nonnegative surplus and slack variables and minimize their use.  The original problem is feasible if and only if all surplus and slack variables are 0.

```<code>   var Surplus {1..4, CELLS} >= 0;
   var Slack   {1..4, CELLS} >= 0;

   min Infeasibility = sum {t in 1..4, <i,j> in CELLS} (Surplus[t,i,j] + Slack[t,i,j]);

   /* A^2 + BC + BCA + ABC + A = I_4 */
   con Con1 {<i,j> in CELLS}:
      sum {k in 1..n} A[i,k]*A[k,j]
    + sum {k in 1..n} B[i,k]*C[k,j]
    + sum {k in 1..n, l in 1..n} B[i,k]*C[k,l]*A[l,j]
    + sum {k in 1..n, l in 1..n} A[i,k]*B[k,l]*C[l,j]
    + A[i,j]
    - Surplus[1,i,j] + Slack[1,i,j]
   = 2*Y[1,i,j] + (i=j);

   /* AB + BCB + ABD = 0 */
   con Con2 {<i,j> in CELLS}:
      sum {k in 1..n} A[i,k]*B[k,j]
    + sum {k in 1..n, l in 1..n} B[i,k]*C[k,l]*B[l,j]
    + sum {k in 1..n, l in 1..n} A[i,k]*B[k,l]*D[l,j]
    - Surplus[2,i,j] + Slack[2,i,j]
   = 2*Y[2,i,j];

   /* CA + DCA + CBC = 0 */
   con Con3 {<i,j> in CELLS}:
      sum {k in 1..n} C[i,k]*A[k,j]
    + sum {k in 1..n, l in 1..n} D[i,k]*C[k,l]*A[l,j]
    + sum {k in 1..n, l in 1..n} C[i,k]*B[k,l]*C[l,j]
    - Surplus[3,i,j] + Slack[3,i,j]
   = 2*Y[3,i,j];

   /* DCB + CBD = I_4 */
   con Con4 {<i,j> in CELLS}:
      sum {k in 1..n, l in 1..n} D[i,k]*C[k,l]*B[l,j]
    + sum {k in 1..n, l in 1..n} C[i,k]*B[k,l]*D[l,j]
    - Surplus[4,i,j] + Slack[4,i,j]
   = 2*Y[4,i,j] + (i=j);

   solve linearize;
</code>```

To find more than one solution, you can use the MAXPOOLSOLS= option.

Here is an example of how you can manually linearize the first product:

```<code>   /* M[i,j,k] = A[i,k]*A[k,j] */
   var M {1..n, 1..n, 1..n} binary;
   con Mcon1 {i in 1..n, j in 1..n, k in 1..n}:
      M[i,j,k] <= A[i,k];
   con Mcon2 {i in 1..n, j in 1..n, k in 1..n}:
      M[i,j,k] <= A[k,j];
   con Mcon3 {i in 1..n, j in 1..n, k in 1..n}:
      M[i,j,k] >= A[i,k] + A[k,j] - 1;
</code>```

You would then replace A[i,k]*A[k,j] with M[i,j,k] throughout the original nonlinear constraints.

You can optionally declare the new variables as >= 0 rather than binary.

To exclude a given set of solutions, first read them in:

```<code>set SOLUTIONS;
num Asol {CELLS, SOLUTIONS};
num Bsol {CELLS, SOLUTIONS};
num Csol {CELLS, SOLUTIONS};
num Dsol {CELLS, SOLUTIONS};
read data out into SOLUTIONS=[_N_]
  {i in 1..n, j in 1..n} <Asol[i,j,_N_]=col('A_'||i||''||j)>
  {i in 1..n, j in 1..n} <Bsol[i,j,_N_]=col('B_'||i||''||j)>
  {i in 1..n, j in 1..n} <Csol[i,j,_N_]=col('C_'||i||''||j)>
  {i in 1..n, j in 1..n} <Dsol[i,j,_N_]=col('D_'||i||''||j)>;
</code>```

Then impose a no-good constraint:

```<code>con NoGood {sol in SOLUTIONS}:
  sum {<i,j> in CELLS} (
     (if Asol[i,j,sol] = 0 then A[i,j] else 1 - A[i,j])
   + (if Bsol[i,j,sol] = 0 then B[i,j] else 1 - B[i,j])
   + (if Csol[i,j,sol] = 0 then C[i,j] else 1 - C[i,j])
   + (if Dsol[i,j,sol] = 0 then D[i,j] else 1 - D[i,j])
  ) >= 1;
</code>```

Then call the solver:

```<code>solve noobj;
</code>```

If you do this for the 148 solutions you found and the resulting solve comes back infeasible, then you have found all solutions.

Title: Solving system of polynomial matrix equations over $\Bbb Z_2$

Body:
Let $A, B, C, D$ be $4 \times 4$ matrices over $\mathbb{Z}_2$.

Suppose they satisfy
\begin{equation}
    \begin{cases}
      A^2+ BC+ BCA+ ABC+ A= I_4\\
      AB+ BCB+ ABD=0_{4 \times 4} \\
      CA+ DCA+ CBC= 0_{4 \times 4} \\
      DCB+ CBD= I_4
    \end{cases}\,.
\end{equation}

Does anyone have an idea on how to find matrices $A, B, C, D$ either theoretically or numerically? I'm thinking that we can find a solution since there are four equations and four unknowns.

Hints would suffice. Thank you so much.

Edit: I found out that the 148 solutions is not yet the total number of complete solutions. To possibly decrease the number of solutions I added another constraint:

$$\sum_{k=1}^4\sum_{l=1}^4 A_{1,k}A_{k,l}A_{l,1}+ \sum_{k=1}^4\sum_{l=1}^4 B_{1,k}C_{k,l}A_{l,1}+ \sum_{k=1}^4\sum_{l=1}^4 A_{1,k}B_{k,l}C_{l,1}+ \sum_{k=1}^4\sum_{l=1}^4 B_{1,k}D_{k,l}C_{l,1}=2Y_{5,1,1}.$$

I added this constraint to the last part of the code:

```<code>proc optmodel;
   num n = 4;
   set ROWS = 1..n;
   set COLS = ROWS;
   set CELLS = ROWS cross COLS;

   var A {CELLS} binary;
   var B {CELLS} binary;
   var C {CELLS} binary;
   var D {CELLS} binary;

   var Y {1..5, CELLS} >= 0 integer;

    /* M[i,j,k] = A[i,k]*A[k,j] */
   var M {1..n, 1..n, 1..n} binary;
   con Mcon1 {i in 1..n, j in 1..n, k in 1..n}:
      M[i,j,k] <= A[i,k];
   con Mcon2 {i in 1..n, j in 1..n, k in 1..n}:
      M[i,j,k] <= A[k,j];
   con Mcon3 {i in 1..n, j in 1..n, k in 1..n}:
      M[i,j,k] >= A[i,k] + A[k,j] - 1;

    /* X[i,j,k] = B[i,k]*C[k,j] */
   var X {1..n, 1..n, 1..n} binary;
   con Xcon1 {i in 1..n, j in 1..n, k in 1..n}:
      X[i,j,k] <= B[i,k];
   con Xcon2 {i in 1..n, j in 1..n, k in 1..n}:
      X[i,j,k] <= C[k,j];
   con Xcon3 {i in 1..n, j in 1..n, k in 1..n}:
      X[i,j,k] >= B[i,k] + C[k,j] - 1;

     /* O[i,j,k,l] = B[i,k]*C[k,l]*A[l,j] */
   var O {1..n, 1..n, 1..n, 1..n} binary;
   con Ocon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      O[i,j,k,l] <= B[i,k];
   con Ocon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      O[i,j,k,l] <= C[k,l];
   con Ocon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      O[i,j,k,l] <= A[l,j];
   con Ocon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      O[i,j,k,l] >= B[i,k]+C[k,l]+A[l,j] - 2;

      /* P[i,j,k,l] = A[i,k]*B[k,l]*C[l,j] */
   var P {1..n, 1..n, 1..n, 1..n} binary;
   con Pcon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      P[i,j,k,l] <= A[i,k];
   con Pcon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      P[i,j,k,l] <= B[k,l];
   con Pcon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      P[i,j,k,l] <= C[l,j];
   con Pcon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      P[i,j,k,l] >= A[i,k]+B[k,l]+C[l,j] - 2;

   /* A^2 + BC + BCA + ABC + A = I_4 */
   con Con1 {<i,j> in CELLS}:
      sum {k in 1..n} M[i,j,k]
    + sum {k in 1..n} X[i,j,k]
    + sum {k in 1..n, l in 1..n} O[i,j,k,l]
    + sum {k in 1..n, l in 1..n} P[i,j,k,l]
    + A[i,j]
   = 2*Y[1,i,j] + (i=j);

   /* E[i,j,k] = A[i,k]*B[k,j] */
   var E {1..n, 1..n, 1..n} binary;
   con Econ1 {i in 1..n, j in 1..n, k in 1..n}:
      E[i,j,k] <= A[i,k];
   con Econ2 {i in 1..n, j in 1..n, k in 1..n}:
      E[i,j,k] <= B[k,j];
   con Econ3 {i in 1..n, j in 1..n, k in 1..n}:
      E[i,j,k] >= A[i,k] + B[k,j] - 1;

    /* F[i,j,k,l] = B[i,k]*C[k,l]*B[l,j] */
   var F {1..n, 1..n, 1..n, 1..n} binary;
   con Fcon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      F[i,j,k,l] <= B[i,k];
   con Fcon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      F[i,j,k,l] <= C[k,l];
   con Fcon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      F[i,j,k,l] <= B[l,j];
   con Fcon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      F[i,j,k,l] >= B[i,k]+C[k,l]+B[l,j] - 2;

      /* G[i,j,k,l] = A[i,k]*B[k,l]*D[l,j] */
   var G {1..n, 1..n, 1..n, 1..n} binary;
   con Gcon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      G[i,j,k,l] <= A[i,k];
   con Gcon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      G[i,j,k,l] <= B[k,l];
   con Gcon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      G[i,j,k,l] <= D[l,j];
   con Gcon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      G[i,j,k,l] >= A[i,k]+B[k,l]+D[l,j] - 2;

   /* AB + BCB + ABD = 0 */
   con Con2 {<i,j> in CELLS}:
      sum {k in 1..n} E[i,j,k]
    + sum {k in 1..n, l in 1..n} F[i,j,k,l]
    + sum {k in 1..n, l in 1..n} G[i,j,k,l]
   = 2*Y[2,i,j];

   /* H[i,j,k] = C[i,k]*A[k,j] */
   var H {1..n, 1..n, 1..n} binary;
   con Hcon1 {i in 1..n, j in 1..n, k in 1..n}:
      H[i,j,k] <= C[i,k];
   con Hcon2 {i in 1..n, j in 1..n, k in 1..n}:
      H[i,j,k] <= A[k,j];
   con Hcon3 {i in 1..n, j in 1..n, k in 1..n}:
      H[i,j,k] >= C[i,k] + A[k,j] - 1;

    /* Q[i,j,k,l] = D[i,k]*C[k,l]*A[l,j] */
   var Q {1..n, 1..n, 1..n, 1..n} binary;
   con Qcon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      Q[i,j,k,l] <= D[i,k];
   con Qcon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      Q[i,j,k,l] <= C[k,l];
   con Qcon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      Q[i,j,k,l] <= A[l,j];
   con Qcon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      Q[i,j,k,l] >= D[i,k]+C[k,l]+A[l,j] - 2;

    /* R[i,j,k,l] = C[i,k]*B[k,l]*C[l,j] */
   var R {1..n, 1..n, 1..n, 1..n} binary;
   con Rcon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      R[i,j,k,l] <= C[i,k];
   con Rcon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      R[i,j,k,l] <= B[k,l];
   con Rcon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      R[i,j,k,l] <= C[l,j];
   con Rcon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      R[i,j,k,l] >= C[i,k]+B[k,l]+C[l,j] - 2;

   /* CA + DCA + CBC = 0 */
   con Con3 {<i,j> in CELLS}:
      sum {k in 1..n} H[i,j,k]
    + sum {k in 1..n, l in 1..n} Q[i,j,k,l]
    + sum {k in 1..n, l in 1..n} R[i,j,k,l]
   = 2*Y[3,i,j];

   /* W[i,j,k,l] = D[i,k]*C[k,l]*B[l,j] */
   var W {1..n, 1..n, 1..n, 1..n} binary;
   con Wcon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      W[i,j,k,l] <= D[i,k];
   con Wcon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      W[i,j,k,l] <= C[k,l];
   con Wcon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      W[i,j,k,l] <= B[l,j];
   con Wcon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      W[i,j,k,l] >= D[i,k]+C[k,l]+B[l,j] - 2;

      /* T[i,j,k,l] = C[i,k]*B[k,l]*D[l,j] */
   var T {1..n, 1..n, 1..n, 1..n} binary;
   con Tcon1 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      T[i,j,k,l] <= C[i,k];
   con Tcon2 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      T[i,j,k,l] <= B[k,l];
   con Tcon3 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      T[i,j,k,l] <= D[l,j];
   con Tcon4 {i in 1..n, j in 1..n, k in 1..n, l in 1..n}:
      T[i,j,k,l] >= C[i,k]+B[k,l]+D[l,j] - 2;

   /* DCB + CBD = I_4 */
   con Con4 {<i,j> in CELLS}:
      sum {k in 1..n, l in 1..n} W[i,j,k,l]
    + sum {k in 1..n, l in 1..n} T[i,j,k,l]
   = 2*Y[4,i,j] + (i=j);

   /* U[1,1,k,l] = A[1,k]*A[k,l]*A[l,1] */
   var U {1..n, 1..n} binary;
   con Ucon1 {k in 1..n, l in 1..n}:
      U[k,l] <= A[1,k];
   con Ucon2 {k in 1..n, l in 1..n}:
      U[k,l] <= A[k,l];
   con Ucon3 {k in 1..n, l in 1..n}:
      U[k,l] <= A[l,1];
   con Ucon4 {k in 1..n, l in 1..n}:
      U[k,l] >= A[1,k]+A[k,l]+A[l,1] - 2;

    /* V[1,1,k,l] = B[1,k]*D[k,l]*C[l,1] */
   var V {1..n, 1..n} binary;
   con Vcon1 {k in 1..n, l in 1..n}:
      V[k,l] <= B[1,k];
   con Vcon2 {k in 1..n, l in 1..n}:
      V[k,l] <= D[k,l];
   con Vcon3 {k in 1..n, l in 1..n}:
      V[k,l] <= C[l,1];
   con Vcon4 {k in 1..n, l in 1..n}:
      V[k,l] >= B[1,k]+D[k,l]+C[l,1] - 2;

   con
   sum {k in 1..n, l in 1..n} U[k,l]
    + sum {k in 1..n, l in 1..n} V[k,l]+ sum {k in 1..n, l in 1..n} O[1,1,k,l] + sum {k in 1..n, l in 1..n} P[1,1,k,l]
   = 2*Y[5,1,1] ;

   solve noobj with milp / maxpoolsols=100 soltype=best;
   create data out(drop=s) from [s]=(1.._NSOL_) {i in 1..n, j in 1..n} <col('A_'||i||''||j)=A[i,j].sol[s]> {i in 1..n, j in 1..n} <col('B'||i||''||j)=B[i,j].sol[s]> {i in 1..n, j in 1..n} <col('C'||i||''||j)=C[i,j].sol[s]> {i in 1..n, j in 1..n} <col('D'||i||'_'||j)=D[i,j].sol[s]>;

   print A;
   print B;
   print C;
   print D;
quit;
</code>```

The software found 100 solutions, but when I checked the output data, it is the same solution replicated 100 times. Is there something wrong with the syntax of the constraint that I added to the code?

Body:
You can formulate this as an integer linear programming problem, and here is a feasible solution:
\begin{align}
A&= \begin{pmatrix}
0 &0 &0 &1 \\
1 &0 &1 &0 \\
1 &1 &1 &1 \\
1 &0 &0 &1 \\
\end{pmatrix} \\
B&= \begin{pmatrix}
1 &1 &0 &0 \\
0 &1 &1 &0 \\
1 &0 &1 &1 \\
0 &0 &1 &0 \\
\end{pmatrix} \\
C&= \begin{pmatrix}
1 &1 &0 &1 \\
1 &1 &0 &0 \\
1 &0 &0 &0 \\
1 &0 &1 &0 \\
\end{pmatrix} \\
D&= \begin{pmatrix}
0 &0 &1 &1 \\
0 &0 &0 &1 \\
0 &0 &0 &0 \\
0 &0 &0 &0 \\
\end{pmatrix}
\end{align}
There are 64 constraints, 64 binary decision variables $(A_{i,j},B_{i,j},C_{i,j},D_{i,j})$, and 64 nonnegative integer variables ($Y_{t,i,j}$ for $t\in\{1,2,3,4\}$ and all $(i,j)$).  Explicitly, the constraints  are
\begin{align}
    \sum_{k=1}^n \left(A_{i,k} A_{k,j}
    + B_{i,k} C_{k,j}\right)
    + \sum_{k=1}^n \sum_{\ell=1}^n \left(B_{i,k} C_{k,\ell} A_{\ell,j}
    + A_{i,k} B_{k,\ell} C_{\ell,j}\right)
    + A_{i,j}
   &= 2 Y_{1,i,j} + [i=j] \\
      \sum_{k=1}^n A_{i,k} B_{k,j}
    + \sum_{k=1}^n \sum_{\ell=1}^n \left(B_{i,k} C_{k,\ell} B_{\ell,j}
    + A_{i,k} B_{k,\ell} D_{\ell,j}\right)
   &= 2 Y_{2,i,j} \\
      \sum_{k=1}^n C_{i,k} A_{k,j}
    + \sum_{k=1}^n \sum_{\ell=1}^n \left(D_{i,k} C_{k,\ell} A_{\ell,j}
    + C_{i,k} B_{k,\ell} C_{\ell,j}\right)
   &= 2 Y_{3,i,j} \\
      \sum_{k=1}^n \sum_{\ell=1}^n \left(D_{i,k} C_{k,\ell} B_{\ell,j}
    + C_{i,k} B_{k,\ell} D_{\ell,j}\right)
   &= 2 Y_{4,i,j} + [i=j]
\end{align}
These can be linearized by replacing each product with a new binary variable, together with additional linear constraints, as shown in https://or.stackexchange.com/questions/37/how-to-linearize-the-product-of-two-binary-variables

As requested, here is the SAS code.  The NOOBJ option specifies that there is no objective, and the LINEARIZE option automatically performs the linearization.

proc optmodel;
   num n = 4;
   set ROWS = 1..n;
   set COLS = ROWS;
   set CELLS = ROWS cross COLS;

   var A {CELLS} binary;
   var B {CELLS} binary;
   var C {CELLS} binary;
   var D {CELLS} binary;

   var Y {1..4, CELLS} >= 0 integer;

   /* A^2 + BC + BCA + ABC + A = I_4 */
   con Con1 {<i,j> in CELLS}:
      sum {k in 1..n} A[i,k]*A[k,j]
    + sum {k in 1..n} B[i,k]*C[k,j]
    + sum {k in 1..n, l in 1..n} B[i,k]*C[k,l]*A[l,j]
    + sum {k in 1..n, l in 1..n} A[i,k]*B[k,l]*C[l,j]
    + A[i,j]
   = 2*Y[1,i,j] + (i=j);

   /* AB + BCB + ABD = 0 */
   con Con2 {<i,j> in CELLS}:
      sum {k in 1..n} A[i,k]*B[k,j]
    + sum {k in 1..n, l in 1..n} B[i,k]*C[k,l]*B[l,j]
    + sum {k in 1..n, l in 1..n} A[i,k]*B[k,l]*D[l,j]
   = 2*Y[2,i,j];

   /* CA + DCA + CBC = 0 */
   con Con3 {<i,j> in CELLS}:
      sum {k in 1..n} C[i,k]*A[k,j]
    + sum {k in 1..n, l in 1..n} D[i,k]*C[k,l]*A[l,j]
    + sum {k in 1..n, l in 1..n} C[i,k]*B[k,l]*C[l,j]
   = 2*Y[3,i,j];

   /* DCB + CBD = I_4 */
   con Con4 {<i,j> in CELLS}:
      sum {k in 1..n, l in 1..n} D[i,k]*C[k,l]*B[l,j]
    + sum {k in 1..n, l in 1..n} C[i,k]*B[k,l]*D[l,j]
   = 2*Y[4,i,j] + (i=j);

   solve noobj linearize;

   print A;
   print B;
   print C;
   print D;
quit;

Another approach to this feasibility problem is to introduce nonnegative surplus and slack variables and minimize their use.  The original problem is feasible if and only if all surplus and slack variables are 0.

   var Surplus {1..4, CELLS} >= 0;
   var Slack   {1..4, CELLS} >= 0;

   min Infeasibility = sum {t in 1..4, <i,j> in CELLS} (Surplus[t,i,j] + Slack[t,i,j]);

   /* A^2 + BC + BCA + ABC + A = I_4 */
   con Con1 {<i,j> in CELLS}:
      sum {k in 1..n} A[i,k]*A[k,j]
    + sum {k in 1..n} B[i,k]*C[k,j]
    + sum {k in 1..n, l in 1..n} B[i,k]*C[k,l]*A[l,j]
    + sum {k in 1..n, l in 1..n} A[i,k]*B[k,l]*C[l,j]
    + A[i,j]
    - Surplus[1,i,j] + Slack[1,i,j]
   = 2*Y[1,i,j] + (i=j);

   /* AB + BCB + ABD = 0 */
   con Con2 {<i,j> in CELLS}:
      sum {k in 1..n} A[i,k]*B[k,j]
    + sum {k in 1..n, l in 1..n} B[i,k]*C[k,l]*B[l,j]
    + sum {k in 1..n, l in 1..n} A[i,k]*B[k,l]*D[l,j]
    - Surplus[2,i,j] + Slack[2,i,j]
   = 2*Y[2,i,j];

   /* CA + DCA + CBC = 0 */
   con Con3 {<i,j> in CELLS}:
      sum {k in 1..n} C[i,k]*A[k,j]
    + sum {k in 1..n, l in 1..n} D[i,k]*C[k,l]*A[l,j]
    + sum {k in 1..n, l in 1..n} C[i,k]*B[k,l]*C[l,j]
    - Surplus[3,i,j] + Slack[3,i,j]
   = 2*Y[3,i,j];

   /* DCB + CBD = I_4 */
   con Con4 {<i,j> in CELLS}:
      sum {k in 1..n, l in 1..n} D[i,k]*C[k,l]*B[l,j]
    + sum {k in 1..n, l in 1..n} C[i,k]*B[k,l]*D[l,j]
    - Surplus[4,i,j] + Slack[4,i,j]
   = 2*Y[4,i,j] + (i=j);

   solve linearize;

To find more than one solution, you can use the MAXPOOLSOLS= option.

Here is an example of how you can manually linearize the first product:

   /* M[i,j,k] = A[i,k]*A[k,j] */
   var M {1..n, 1..n, 1..n} binary;
   con Mcon1 {i in 1..n, j in 1..n, k in 1..n}:
      M[i,j,k] <= A[i,k];
   con Mcon2 {i in 1..n, j in 1..n, k in 1..n}:
      M[i,j,k] <= A[k,j];
   con Mcon3 {i in 1..n, j in 1..n, k in 1..n}:
      M[i,j,k] >= A[i,k] + A[k,j] - 1;

You would then replace A[i,k]*A[k,j] with M[i,j,k] throughout the original nonlinear constraints.

You can optionally declare the new variables as >= 0 rather than binary.

To exclude a given set of solutions, first read them in:

set SOLUTIONS;
num Asol {CELLS, SOLUTIONS};
num Bsol {CELLS, SOLUTIONS};
num Csol {CELLS, SOLUTIONS};
num Dsol {CELLS, SOLUTIONS};
read data out into SOLUTIONS=[_N_]
  {i in 1..n, j in 1..n} <Asol[i,j,_N_]=col('A_'||i||''||j)>
  {i in 1..n, j in 1..n} <Bsol[i,j,_N_]=col('B_'||i||''||j)>
  {i in 1..n, j in 1..n} <Csol[i,j,_N_]=col('C_'||i||''||j)>
  {i in 1..n, j in 1..n} <Dsol[i,j,_N_]=col('D_'||i||''||j)>;

Then impose a no-good constraint:

con NoGood {sol in SOLUTIONS}:
  sum {<i,j> in CELLS} (
     (if Asol[i,j,sol] = 0 then A[i,j] else 1 - A[i,j])
   + (if Bsol[i,j,sol] = 0 then B[i,j] else 1 - B[i,j])
   + (if Csol[i,j,sol] = 0 then C[i,j] else 1 - C[i,j])
   + (if Dsol[i,j,sol] = 0 then D[i,j] else 1 - D[i,j])
  ) >= 1;

Then call the solver:

solve noobj;

If you do this for the 148 solutions you found and the resulting solve comes back infeasible, then you have found all solutions.


问:
Quadratic and Linear Function relationship
I just stumbled across this problem and would appreciate some help with it, as I'm not getting far with it.

Problem: You are given a quadratic function $f(x)=ax^2+bx+c$
and a linear function $g(x)$.

The two functions intersect at $x=0$ and at also at an $x$ with $g(x)=f(x)=0$ and where $x<0$.

Which of the two could, for some values of $a,b,c$, be an expression for $g(x)$:

$g(x)=bx+c$

$g(x)=ax+c$

My Progress thus far:
We know that the $y$-intercept must be $c$, because $f(x)$ and $g(x)$ intercept at $x=0$. So that makes perfect sense. However I fail to see how for some values of $a,b,c$  the gradient of $g(x)$ could either be $a$ or $b$.
Any help would greatly be appreciated.

答:
Given f(x)=$ax^2$+bx+c ($a \neq0$),

if g(x)=bx+c for some x=$x_1$<0,

f($x_1$)=${ax_1}^2$+$bx_1$+c (I)

g($x_1$)=b$x_1$+c  (II)

(I)-(II)  ${ax_1}^2$ = 0, a=0 , it is not allowed.

So, this is not the right choice, only option is (2)  g(x)=ax+c.

You can prove that from here.


问:
Prove equality vector spaces
Let $V$ be a finite dimensional $K$-vector space, let $\varphi \in$ End$_K(V)$ be a $K$-endomorphism of $V$, and let $S$ be a $K$-basis of $V$. Prove that for every polynomial $f \in K[X]$ one has the equality:

$$M_{S,S}(f(\varphi))=f(M_{S,S}(\varphi)).$$

My question is do I have to show that $M_{S,S}$ is linearly independent? And how do I do that?

答:
Let $f = \sum a_i x^i$. Then:
$$M_{S,S}(f(\varphi)) = M_{S,S}\left(\sum a_i \varphi^i\right) = \sum a_i M_{S,S} (\varphi^i) = \sum a_i M_{S,S}(\varphi)^i = f(M_{S,S}(\varphi))$$

Title: Prove equality vector spaces


问:
If A and B are Hermitian Matrices, $\text{tr}(ABAB) \geq 0$.
Can anybody help me with this problem.
If $A$ and $B$ are Hermitian Matrices, then $\text{tr}(ABAB) \geq 0$.

It is easy to show that the trace is a real number, but I cannot prove that it must be non-negative.

答:
You need further conditions. Take
$$ A = \pmatrix{1 & 0 \\ 0 & -1},  B=\pmatrix{0 & 1\\ 1 & 0}.$$
Then $AB$ is a rotation by 90 degrees and $(AB)^2=-I$ has trace -2.


问:
Value of $\sum\limits^{\infty}_{n=1}\frac{\ln n}{n^{1/2}\cdot 2^n}$
Here is a series:

$$\displaystyle \sum^{\infty}_{n=1}\dfrac{\ln n}{n^{\frac12}\cdot 2^n}$$

It is convergent by d'Alembert's law. Can we find the sum of this series ?

答:
$f(n)=\dfrac{\ln n}{n^{1/2}\cdot 2^n}>0~\forall~n>1\\e^nf(e^n)=e^{n/2}\dfrac{n}{2^n}$

$[e^nf(e^n)]^{1/n}=\dfrac{\sqrt e n^{1/n}}{2}\to\sqrt e/2<1$

$\displaystyle \sum^{\infty}_{n=1}e^nf(e^n)$ is convergent implies $\displaystyle \sum^{\infty}_{n=1}f(n)$ is convergent.


问:
Computing the rank of $\mathbb Z_p \oplus \mathbb Z_q$
I am following the chapter on Morse Theory of Ambrosetti and Malchiodi's Nonlinear Analysis and Semilinear Elliptic Problems, and they define the rank of an abelian group as follows:

Definition. The rank of an abelian group $G$ is the maximal number $k$ for which $\sum_1^k n_i g_i = 0$ with $(n_i)_i \subset \mathbb Z$ and $(g_i)_i \subset G$ implies $g_i = 0$ for every $i$.

The book then gives the example that if $G = Z^p \oplus Z_q$ for some integers $p$ and $q$, then rank $G = p$.

Regarding the example, I am assuming that they mean $\mathbb Z_p \oplus \mathbb Z_q$ with $p < q$ primes.

My doubt is that I am failing to see why the assertion is true. If we consider $\mathbb Z_3 \oplus \mathbb Z_5$, isn't
$$
(1, 0) + (2, 3) + (0, 2)= (0, 0)
$$
a counterexample? In general, wouldn't $\{(1, 0), (0, 1)\}$ generate this group?

I am sure I am missing some elementary detail, but nonetheless I will appreciate some help with this because I am truly confused.

Thanks in advance.

答:
Let me try to clarify the definition. Let's say a set $g_1, \ldots g_k$ of elements of $G$ is linearly independent if the equation:
$$
n_1 g_1 + n_2 g_2 + \ldots + n_k g_k = 0
$$
where each $n_i \in \Bbb{Z}$ only holds if each $n_i = 0$. Note that if $g_1, \ldots, g_k$ are linearly independent, then the $g_i$ are necessarily distinct, since if $g_i = g_j$ for $i \neq j$, you have $g_i + (-1)g_j = 0$, contradicting linear independence. The rank is the maximum size of a linearly independent set. This can be seen to be well-defined and finite using the structure theorem for finitely-generated abelian groups, which implies that any finely-generated abelian group $G$ is isomorphic to a direct sum $\Bbb{Z}^m \oplus H$, where $H$ is finite and $m$ (the rank) is uniquely determined by $G$.