问:
Given $\cotθ = 1$ and angle θ lies in the third quadrant, determine the value of $\secθ$ and $\cscθ$
Recently learned about the reciprocal trig ratios and I'm still don't fully understand them. Wondering if I can get a little help not only with this specific question but the calculating reciprocal trig ratios in general please and thank you.</p>

答:
For these types of problems, the hardest thing to do is finding what $\theta$ actually is. As for calculating these "other" trigonometric functions just use the definition, that is $$ \csc\theta = \frac{1}{\sin\theta}, \\ \sec\theta = \frac{1}{\cos\theta}, \\ \cot\theta = \frac{1}{\tan\theta}.$$

Back to your problem. By definition $$\cot\theta = \frac{1}{\tan\theta} = \frac{1}{\frac{\sin\theta}{\cos\theta}} = \frac{\cos\theta}{\sin\theta}.$$ You are given that $\cot\theta = 1$, which means that $\frac{\cos\theta}{\sin\theta} = 1$ and so $\cos\theta = \sin\theta$. Since $\theta$ lies in the third quadrand this implies that $\theta = \frac{5\pi}{4}$. I will calculate $\sec\theta$ for you: $$\sec\theta = \sec\frac{5\pi}{4} = \frac{1}{\cos\frac{5\pi}{4}} = \frac{1}{-\frac{\sqrt{2}}{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}.$$ I will leave it to you to calculate $\csc\theta.$


问:
Three digit number even and greater than $500$
I am reading the following problem:

The digits $1$,$2$,$3$,$4$,$5$ are randomly arranged to form a three
digit number (digits are not repeated). What is the probability that
the number is even and greater than $500$?

My solution:

There are $5\cdot 4 \cdot 3$ ways to form the number in total.

In order to be greater than $500$ the first digit must be $5$ and in order to be even the last digit must be either $2$ or $4$. So there is $1$ way to select the first digit, $2$ ways to select the last and $2$ ways to select the middle one (either $3$ or $4$) i.e. $2\cdot 2\cdot 1$ = $4$ ways

Therefore the probability is: $\frac{4}{60}=\frac{1}{15}$

But the solution mentions $\frac{1}{10}$.

What am I doing wrong here?

答:
J Moravitz' comment addresses the flaw.

The easiest alternative approach is that there are $D = 5!$ numbers possible, of which $4!$ of the numbers start with the digit $5.$  By symmetry, there are $N = \frac{4!}{2}$ such numbers that are even.  So the desired probability is $\frac{N}{D}.$

Addendum

Per OP's request, I will combine his analysis with the analysis inherent in the comments of J Moravitz to provide an alternative direct approach to the problem.

As in the start of my answer, the computation will be

$$\frac{N}{D}, ~~\text{where}~~ D = 5!.$$

The leftmost digit must be $5$.  The rightmost digit must be either $2$ or $4$. I will let $N_1$ denote the number of possibilities where the rightmost digit is $2$.  Then, by symmetry, $N = 2 \times N_1$.

With the leftmost and rightmost digits being $5$ and $2$ respectively, the other 3 digits, $1,3,4$ can be permuted in $3!$ ways to fill in the 3 middle digits.

Therefore, $N_1 = 3!$ and $N = 2 \times 3!$.

ContentLicense: CC BY-SA 4.0


问:
Suppose $f(x) = x^5 + 2x^4 − x^3 + 3x^2 + 5x + 1$. Find a polynomial $g$ of degree $2$ or less such that $f(x) \equiv g(x) \pmod 3$.
Suppose $f(x) = x^5 + 2x^4 − x^3 + 3x^2 + 5x + 1$. Find a polynomial $g$ of degree $2$ or less such that $f(x) \equiv g(x) \pmod 3$.

答: By employing Fermat's little theorem: $x^3 = x\pmod 3\implies x^5=x^3=x\pmod 3, 2x^4 = 2x^2\pmod 3, -x^3=-x\pmod 3\implies x^5+2x^4-x^3=x+2x^2-x=2x^2\pmod 3$. Thus we can take $g(x) = 2x^2+3x^2+5x+1= 5x^2+5x+1$.


问:
System of equations in x and y
Solve for $x, y \in \mathbb{R} $

$$ 5x \left(1+\frac{1}{x^2+y^2}\right) =12$$
$$ 5y \left(1-\frac{1}{x^2+y^2}\right) =4$$

I need a Different Approach apart from what i posted..Thank You

ContentLicense: CC BY-SA 3.0

Title: System of equations in x and y

Body:
I Called... $ x^2+y^2=t$, So

$$ 5x= \frac{12t}{t+1} $$
$$ 5y= \frac{4t}{t-1} $$

Squaring and Adding:

$$ 25t= \left(\frac{12t}{t+1}\right)^2+\left(\frac{4t}{t-1}\right)^2 $$ $\implies$

$$ 25= \left(\frac{144t}{t^2+2t+1}\right)+\left(\frac{16t}{t^2-2t+1}\right)$$ $\implies$

$$ 25= \left(\frac{144}{t+\frac{1}{t}+2}\right)+\left(\frac{16}{t+\frac{1}{t}-2}\right)$$
Let $$t+\frac{1}{t} =z$$, Then

$$ 25=\left(\frac{144}{z+2}\right)+\left(\frac{16}{z-2}\right)$$ $\implies$

$$ 25z^2-160z+156=0$$ $\implies$

$$ z=\frac{6}{5}, \frac{26}{5} \implies t=5, \frac{1}{5} \implies x=2,y=1  $$


问:
Proving $\mathbb{Q\times Z}$ is countable
I want to prove that  $\mathbb{Q\times Z}$ is countable. I found bijection from $\Bbb N\times \Bbb N$ to $\Bbb N$. Can I use it?

答: We have $\Bbb Q\times \Bbb Z=\cup_{n\in\mathbb{Z}}(\Bbb Q\times \{n\}).$ But $\Bbb Q\times\{n\}$ and $\Bbb Z$ are both countable. The union of countably many countable sets is countable.


问:
Can you critique this proof that a vector space is infinite-dimensional?
I'm self-studying Linear Algebra using Axler's Linear Algebra Done Right, and I'm not entirely sure the following proof is correct. I've already seen another solution on StackExchange ( Suppose $V$ is finite-dimensional, Will $L(V,W)$ be infinite-dimensional?) that is clearly cleaner, but I wanted to know how (if at all) the approach I took fails.

Suppose $V$ is finite-dimensional with $\dim V > 0$, and suppose $W$ is infinite-dimensional. Prove that $\mathcal{L}(V, W)$ is infinite-dimensional.

Proof: Let $\dim V = n > 0$, and let $\mathbf{B} = \{v_1, v_2, \ldots, v_n\}$ be a basis of $V$. Now assume for the sake of contradiction that $\mathcal{L}(V, W)$ is finite dimensional. Then we can produce a basis for $\mathcal{L}(V, W)$:$\{T_1, T_2, \ldots, T_m\}$, where $T_i \in \mathcal{L}(V, W) \, \forall i \in \{ 1, \ldots, m \}$. Now consider $\\S = span(T_1 v_1, \ldots, T_1 v_n, \ldots, T_m v_1, \ldots, T_m v_n)$. $S$ is the set of all linear combinations of a finite set of vectors, so it must be finite dimensional. Then we can produce a basis $\{w_1, \ldots, w_k\}$ of $S$, where each $w_i \in W$. But $W$ is infinite-dimensional, so we know there must exist a $w_{k+1} \in W, w_{k+1} \notin span(w_1, \ldots, w_k)$. Now consider $T_{m+1} \in \mathcal{L}(V, W)$ defined by $$T_{m+1}v = \begin{cases}
w_{k+1} &\text{if } v = v_1\\
0 &\text{otherwise}
\end{cases}$$

Note that any linear combination of $\{T_1, T_2, \ldots, T_m\}$ must map to a vector in $S$. However, the image of $T_{m+1}$ includes $w_{k+1} \notin span(w_1, \ldots, w_k) = S$, so $T_{m+1} \notin span(T_1, T_2, \ldots, T_m)$. But this contradicts our assumption that $\{T_1, T_2, \ldots, T_m\}$ is a basis for $\mathcal{L}(V, W)$. It follows that $\mathcal{L}(V, W)$ is infinite dimensional.

答:
I think your proof is generally fine.

You should fix a typo when you define the $T_i$ where you say $i\in\{1,\dots,n\}$.
It should be $i\in\{1,\dots,m\}$.

You should make it clear that your definition of the linear transformation $T_{m+1}$ is with regards to the base $B$.

Moreover, you don't need to introduce a basis of $S$, since $S\subset W$ is the span of a finite set and hence must be finite-dimensional.
The rest of your argument works perfectly well just from the fact that $\text{Im}(T_{m+1})\not\subset S$.


问:
Extrema of a functional given weak and weak lower semicontinuity
Assume $X$ is Hilbert space and let $\bar B_r=\{u \in E: \|u\|\le r\}$. Also, Let $J \in C(X, \mathbb{R})$
I want to show that if $J$ is weak lower semi continuous then there exist a $z$ in $\bar B_r$ which minimizes the functional.

$X$ must be reflexive since it is a Hilbert space. We also have boundedness of the functional from weak lower semi continuity, so coercivity is not needed. The conditions of Weierstrass theorem are therefore satisfied and $J$ has a minimum.

Next, I want to show that if $J$ is weakly continuous then there exist a $z \in \bar B_r$ that maximizes $J(z)$.
If we can show that $-J(z)$ is a minimum then this will be enough to conclude.
Now weak continuity implies boundedness and again, $X$ is reflexive. But I don't have weak lower semicontinuity so how do I show the existence of a maximum?

Am I  on the right track for the first part?

答:
First part is correct. Let $x_n$ be such that $f(x_n) <m+\frac  1 n$ where $m$ is the infimum of $f$ on $\overline {B_r}$. If $x$ is a limit point of $(x_n)$ in the weakly compact set $\overline {B_r}$ then weak lower-semi continuity gives $f(x) \leq m$. But $m \leq f(x)$ so $f(x)=m$.

If $J$  is weakly continuous then so is $-J$ which implies that $-J$ is weakly (upper and) lower semi continuous. Hence, $-J$ attains its minimum which means $J$ attains its maximum on $\overline {B_r}$.


问:
Maximum value of function without root of derivative
Given the function

$$f(x) = \frac{x^{p-k} a_k}{\sum_{i=0}^p (x^{p-i} a_i)}$$

, where $x>0, a_i>0$ for all $i$. $p>0, k=1,...,p-1$ are integers,

is it possible to characterize the maximum value of $f$ for a given $k$? Can we say anything about the maximum of $f$ at all?

In particular, is there a closed-form expression for the maximum value of $f$, or if not, can we find the maximum value of $f$ as the root of a polynomial or algebraic equation (in terms of $p,k,a_i$ for all $i$, i.e. eliminating $x$)?

I note the usual method of differentiating wrt $x$, solving the resulting equation yields only one root, the solution of

$$\sum_{i=0}^p (i-k)x^{p-i}a_i = 0$$

However, that root cannot be obtained in closed-form, so doesn't help us say anything about the maximum value of $f$. I am interested in whether that max $f$ can be calculated, e.g. directly as the root of a polynomial, without first calculating the $x$ which maximizes it.

答:
Upgraded from "sometimes yes" to "frequently yes".  I suspect there may be failure for certain parameter combinations, but haven't found such a combination after some searching over generic systems with small $p$.  (OP observes that a promising potential obstruction, an inseparable $r(x)$ was shown impossible in a previous Question, the answer to which used Descartes's rule of signs to argue that there was exactly one positive root of $r$.)

I don't have a full solution, but I do have a "sometimes yes".

You want to maximize the ratio of a monomial of degree less than $p$ to a polynomial of degree $p$ with the constraint that a different polynomial of degree $p$ is zero.

Let's give a concrete example.  Suppose $a_0 = a_1 = a_5 = 1$ and $a_i = 0$ otherwise.  (Stricly, this does not match the specification of the problem, since we have $a_2 = a_3 = a_4 \not > 0$, but making these coefficients positive utterly obscures the illustrated operations.  Alternatively, reducing the degree to "squeeze out" these zero coefficients yields an example of too low degree to be illustrative.  Consequently, we proceed with this minor violation of the specification.)  Then, with $k = 1$, you want to maximize
$$  \frac{x^4}{1+x^4+x^5}  $$
and we know this maximum is attained (using your result from the first derivative) when $x^5 - 4 = 0$.

We can treat this as an attempt at reduction of
$$  \frac{x^4}{1+x^4+x^5}  $$
with the relation $x^5 = 4$.  This gives the sequence of reductions \begin{align*}
\frac{x^4}{1+x^4+x^5} &\rightarrow \frac{x^4}{5+x^4} \ \left[ {}\cdot \frac{x}{x} \right]  \\
    &= \frac{x^5}{5x+x^5}  \\
    &\rightarrow \frac{4}{4 + 5x}
\end{align*}

If you plot the first, second and fourth of these (the second and the third give the same curve), you see that they intersect at the maximal point of $\frac{x^4}{1+x^4+x^5}$.

(If we continue, reducing and multiplying by "fancy forms of $1$" (e.g. $x/x$), we get the second and fourth expressions forever.  If we had more nonzero $a_i$, we would get more expressions in the cycle.)

As a happy coincidence, we can take the function given by the fourth expression
$$  y = \frac{4}{4+5x}  $$
and solve for $x$.  (We should expect to be able to do this if any of our rational functions has small degrees in the numerator and denominator and, in very rare coincidences, otherwise.)
$$  x = \frac{4(1-y)}{5y}  $$
Now plug that into either other expression.  (It doesn't matter which one, we get the same result with either choice, but the amount of work to get there varies.)
$$ y = \frac{x^4}{5+x^4} = \frac{256(y-1)^4}{625 \left( 5+\frac{256(y-1)^4}{625y^4} \right) y^4}  \text{.}  $$
Solving the resulting polynomial for $y$,
$$  3381 y^5 - 1280 y^4 + 2560 y^3 - 2560 y^2 + 1280 y- 256 = 0  \text{,}  $$
yields the maximum value for the expression in question as its one real (positive) root.  (It's
$$  \frac{2^{8/5}}{5+2^{8/5}} = 0.37744{\dots}  \text{.)}  $$

Admittedly, this method relies on a couple of happy coincidences.  But, when it works, it works.

There is some theory in play here.  Suppose we wish to maximize the rational function
$$  \frac{n(x)}{d(x)}  $$
in $\Bbb{R}(x)$, the ring of rational expressions with positive coefficients.  Here, $n(x)$ is a polynomial of degree in the interval $[1,p-1]$ and $d(x)$ is a polynomial of degree $p$.  We can work in the ring $\Bbb{R}[x,y]$ by rewriting as
$$  y \cdot d(x) - n(x) = 0  \text{.}  $$
This generates an ideal in $\Bbb{R}[x,y]$ (of all polynomials in $x$ and $y$ which are $0$ when $y \cdot d(x) - n(x)$ is and possibly at additional points in the plane).

We have the relation (from solving for the stationary point using the derivative) $r(x) = 0$ for some polynomial of degree $p$.  When we quotient by $\langle r(x) \rangle$, we get the quotient ideal
$$  I = \frac{\langle y \cdot d(x) - n(x) \rangle}{\langle r(x) \rangle} \subset \frac{\Bbb{R}[x,y]}{\langle r(x) \rangle}    \text{.}  $$
(The reduction steps above move from element to element of this quotient ideal.)

We would like to know if there is a polynomial in $y$ in $I$.  That is, we want to know if $I \cap \Bbb{R}[y]$ is nonempty.  If it is, since $\Bbb{R}$ is a field, $\Bbb{R}[y]$ is a PID, and $I \cap \Bbb{R}[y]$ is generated by a single polynomial of least degree in the intersection.  This polynomial is the one requested in the Question.

Whether the intersection is nonempty can be determined by constructing a reduced Groebner basis of $\{ y \cdot d(x) - n(x), r(x) \}$ with the variable ordering $x, y$.  Such a Groebner basis will have an initial segment of polynomials in $x$ and $y$ and a terminal segment of polynomials in only $y$.  If the terminal segment is empty, the intersection is empty.  If the terminal segment is nonempty, the polynomial (as observed above, there is only one if the basis is reduced) is the one sought in the Question.

I don't know of any simple to perform test on $d$, $n$, and $r$ that will tell if this process will produce a polynomial in $y$.  The best that can be said is that this is a finite process guaranteed to complete and either find such a polynomial or fail to do so.  And if it fails to do so, that constitutes a proof that the intersection is empty.

One down side: there is little reason to believe that the degree of the polynomial in $y$ is low.  So there may be a solution that is valid in principle, but not computationally useful in practice.

As an explicit example of the Groebner basis method, take
$p = 6$, \begin{align*}
a_0 &= 3  \\
a_1 &= 5  \\
a_2 &= 1  \\
a_3 &= 17  \\
a_4 &= 1/2  \\
a_5 &= 31415/10000  \\
a_6 &= 2  \\
\end{align*}

Then the generators for the ideal $\langle d(x)y-n(x), r(x) \rangle$ are
\begin{align*}
k&=1 :& \left\langle \left(3 x^6+5 x^5+x^4+17 x^3+\frac{x^2}{2}+\frac{6283 x}{2000}+2\right) y-5 x^5,-3 x^6+x^4+34 x^3+\frac{3 x^2}{2}+\frac{6283 x}{500}+10 \right\rangle  \\
k&=2 :& \left\langle \left(3 x^6+5 x^5+x^4+17 x^3+\frac{x^2}{2}+\frac{6283 x}{2000}+2\right)y-x^4,-6 x^6-5 x^5+17 x^3+x^2+\frac{18849 x}{2000}+8 \right\rangle  \\
k&=3 :& \left\langle \left(3 x^6+5 x^5+x^4+17 x^3+\frac{x^2}{2}+\frac{6283 x}{2000}+2\right)y-17 x^3,-9 x^6-10 x^5-x^4+\frac{x^2}{2}+\frac{6283 x}{1000}+6 \right\rangle  \\
k&=4 :& \left\langle \left(3 x^6+5 x^5+x^4+17 x^3+\frac{x^2}{2}+\frac{6283 x}{2000}+2\right) y-\frac{x^2}{2},-12 x^6-15 x^5-2 x^4-17 x^3+\frac{6283 x}{2000}+4 \right\rangle  \\
k&=5 :& \left\langle \left(3 x^6+5 x^5+x^4+17 x^3+\frac{x^2}{2}+\frac{6283 x}{2000}+2\right) y-\frac{6283 x}{2000},-15 x^6-20 x^5-3 x^4-34 x^3-\frac{x^2}{2}+2 \right\rangle  \\
\end{align*}

Then using software to compute the Groebner bases using elimination order ("lexdeg") we obtain the intersection ideals generated by the following:
\begin{align*}
k&=1 :& \left\langle 6138919015691079101529909209 y^6 + 2240405737753079415080897600 y^5 + 2081288581773650633878448000 y^4 + 238526373367786274502272000 y^3 - 64178056050703070776320000 y^2 - 163849551050401173321216000 y + 16000000000000000000000000 \right\rangle  \\
k&=2 :& \left\langle 6138919015691079101529909209 y^6+24739953076838110263114400 y^5+51094587902899422481781600 y^4+6732609568373285915340800 y^3+311050507863667790521600 y^2-8437109555351601157120 y-503316480000000000000 \right\rangle  \\
k&=3 :& \left\langle 6138919015691079101529909209 y^6-48819875712398124502379959200 y^5+151515077917012145269151616000 y^4-233975590117705242864677760000 y^3+191217615919976688245964800000 y^2-79039391285214164366714880000 y+12973391069921280000000000000 \right\rangle  \\
k&=4 :& \left\langle 6138919015691079101529909209 y^6+165028651650769632816068000 y^5+30361674369652778778480000 y^4+1372726062269954208000000 y^3+26035777224447040000000 y^2-726516916346880000000 y-17694720000000000000 \right\rangle  \\
k&=5 :& \left\langle 6138919015691079101529909209 y^6+3417936660792812819590016586 y^5+1348264686861445252149012055 y^4+269194306035102012141441740 y^3-114653545686145512007201545 y^2-67879726900560328135508534 y+4982959727226654732330489 \right\rangle  \\
\end{align*}

Inspecting the polynomial for $k =1$, it has two real, positive roots.  The zero at $y = 0.318{\dots}$ is the maximum you want corresponding to $x = 2.37{\dots}$.  The zero at $y = 0.0965{\dots}$ corresponds to a local minimum at $x = -0.496{\dots}$.  Of course, nothing in the above restricts to $x > 0$ or to $y > 0$, so even if there is a polynomial in the intersection, you may have some filtering of its positive real roots to find the one that corresponds to the maximum.

For small $p$, we can generate the generic polynomial in $y$.  For $p = 1$, there are no valid choices of $k$.

\begin{align*}
p = 2 &, k = 1 &:& a_2 \left(a_1^2 y^2-4 a_0 a_2 y^2-2 a_1^2 y+a_1^2\right)  \\
\\
p = 3 &, k = 1 &:& a_3 \left( \left(4 a_1^3a_3- a_1^2a_2^2-18 a_0 a_1 a_2 a_3 +4 a_0 a_2^3+27 a_0^2 a_3^2\right) y^3 + \left(-12 a_1^3a_3 +2 a_1^2a_2^2 +18 a_0 a_1a_2 a_3 \right) y^2 + \left(12 a_1^3 a_3-a_1^2 a_2^2\right) y-4 a_1^3 a_3 \right)  \\
p = 3 &, k = 2 &:& a_3 \left( \left(4 a_1^3a_3 -a_1^2a_2^2 -18 a_0 a_1 a_2 a_3 +4 a_0 a_2^3+27 a_0^2 a_3^2\right) y^3 + \left(-12 a_0 a_2^3+2 a_1^2 a_2^2+18 a_0 a_1 a_2a_3 \right) y^2+\left(12 a_0 a_2^3-a_1^2 a_2^2\right) y - 4 a_0 a_2^3 \right)
\end{align*}

Of course, we now get a criterion for the polynomial in $y$ to have degree less than $p$ : make the leading coefficient zero.  For instance, in the $p = 3$, either $k$ cases, taking (a random example) $a_0 = a_1 = 1$, $a_2 = 5/32$, and $a_3 = (-76+14\sqrt{34})/3456$ makes both polynomials in $y$ quadratic.  We can work to find simple criteria for this reduction in degree.  For instance, this can only happen if $a_3 \in \left( 0, a_2^2 / (3 a_1) \right]$.


问:
Bayesian network calculations
I'm new to probability theory and Bayesian networks so I really don't understand how to calculate some probabilities based on this network:

$$\require{enclose}\boxed{\begin{array}{c|cc}X\backslash Y&\mathrm c&\mathrm d\\\hline\mathrm a&0.85&0.15\\\mathrm b&0.65&0.35\end{array}}\quad\begin{array}{c}\boxed{\begin{array}{c|cc}X&\mathrm a&\mathrm b\\\hline0&0.2&0.8\end{array}}\\\enclose{circle}{\rm X}\\\swarrow~~\searrow\\\enclose{circle}{\rm Y}\qquad\quad\enclose{circle}{\rm Z}\end{array}\quad\boxed{\begin{array}{c|cc}X\backslash Z&\mathrm e&\mathrm f\\\hline\mathrm  a&0.70&0.30\\\mathrm b&0.25&0.75\end{array}}$$

Network

I want to caclute:

$\qquad\text{Calculate:}\\~(i)~~\mathsf P(X=\mathrm a\mid Y=\mathrm c, Z=\mathrm e)\\(ii)~~\mathsf P(Z=\mathrm e\mid Y=\mathrm c)$

Calculate

答:
The Diagram expresses the factorisation (for any $x,y,z$) of: $$\mathsf P(X{=}x, Y{=}y, Z{=}z)=\mathsf P(X{=}x)\,\mathsf P(Y{=}y\mid X{=}x)\,\mathsf P(Z{=}z\mid X{=}x)$$

The table gives you: $${\qquad\quad\mathsf P(X{=}\mathrm a)=0.2\\ \mathsf P(Y{=}\mathrm c\mid X{=}\mathrm a)=0.85\\ \mathsf P(Z{=}\mathrm e\mid X{=}\mathrm a)=0.70}$$

The definition of conditional probability, and Law of Total Probability say: $$\begin{align}\mathsf P(X{=}\mathrm a\mid Y{=}\mathrm c, Z{=}\mathrm e)&=\dfrac{\mathsf P(X{=}\mathrm a, Y{=}\mathrm c, Z{=}\mathrm e)}{\mathsf P(Y{=}\mathrm c, Z{=}\mathrm e)}\\[2ex]&=\dfrac{\mathsf P(X{=}\mathrm a, Y{=}\mathrm c, Z{=}\mathrm e)}{\mathsf P(X{=}\mathrm a, Y{=}\mathrm c, Z{=}\mathrm e)+\mathsf P(X{=}\mathrm b, Y{=}\mathrm c, Z{=}\mathrm e)}\end{align}$$

Put it together.

Title: Bayesian network calculations


问:
What are the directions generated by the Newton method for $f(Ay)$?
Suppose $A \in R^{n \times n}$ and that $A$ is invertible, derive the directions generated by the Newton Method for the following problem

$\min_{y \in R^n} f(Ay)$

I know what it means now but am stuck. I think I have found the gradient but am stuck with the Hessian. Gradient should be $\nabla f(Ay)^T A$ but what about the Hessian?

答:
Let $g(y) = f(Ay)$. The Newton direction is $\Delta x = -H_g(y)^{-1} D_g(y)$. By application of the chain rule, $D_g(y) = A^TD_f(Ay)$ and $H_g(y) = A^TH_f(Ay)A$, so

$$\Delta x = -(A^TH_f(Ay)A)^{-1} (A^TD_f(Ay))=-A^{-1}H_f(Ay)^{-1}D_f(Ay).$$


问:
Identity concerning complete elliptic integrals
It can be easily checked that both the complete elliptic integrals $K(k), K'(k)$ satisfy the same second order differential equation $$kk'^{2}\frac{d^{2}y}{dk^{2}} + (1 - 3k^{2})\frac{dy}{dk} - ky = 0$$ and hence from the theory of second order differential equations there is a relation of the form $$K'(k) = cK(k)\cdot\log k + f(k)$$ where $c$ is some constant and $f(k)$ is some analytic function of $k$. The exact relation between $K(k)$ and $K'(k)$ is given by $$K'(k) = \frac{2K(k)}{\pi}\log\left(\frac{4}{k}\right) - 2\left[\left(\frac{1}{2}\right)^{2}\left(\frac{1}{1\cdot 2}\right)k^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{2}\left(\frac{1}{1\cdot 2} + \frac{1}{3\cdot 4}\right)k^{4} + \cdots\right]$$ It can be verified with some patience that the RHS does satisfy the differential equation given above and thereby the relation between $K'(k)$ and $K(k)$ can be established.

However is there an alternative proof based on the definition of $K'(k)$ and $K(k)$ as complete elliptic integrals or using the hypergeometric relation $$\frac{2K(k)}{\pi} =\,_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; k^{2}\right)$$ which can be presented to someone unaware of the theory of second order differential equations?

答:
We will use the hypergeometric representation of the elliptic integral
\begin{equation}
 K'(k)=\frac\pi2\,_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; 1-k^{2}\right)
\end{equation}
and the linear transformation of the variable
\begin{align}
 \mathbf{F}\left({a,b\atop a+b+m};z\right)=&\frac{1}{\Gamma\left(a+m\right)
\Gamma\left(b+m\right)}\sum_{s=0}^{m-1}\frac{(a)_{s}(b)_{s}(m-s-1)!}{s!}(z-1)^{s}-\\
&\hspace{1cm}-\frac{(z-1)^{m}}{\Gamma\left(a\right)\Gamma\left(b\right)}\sum_{s=0}^{
\infty}\frac{(a+m)_{s}(b+m)_{s}}{s!(s+m)!}(1-z)^{s}\\
&\hspace{2cm}\left[\ln\left(1-z\right)
-\psi\left(s+1\right)-\psi\left(s+m+1\right)+\right.\\
&\hspace{2cm}\left.+\psi\left(a+s+m\right)+\psi\left(
b+s+m\right)\right]
\end{align}
which is valid if $m$ is a non negative integer and if $\left|z-1\right|<1,\left|\arg(z-1)\right|<\pi$. Here $\psi$ is the digamma function and $\mathbf{F}$ is regularized Olver's hypergeometric function. When $a+b+m=1$, it is identical to the standard hypergeometric function and if $m=0$, the first summation cancels. By taking $a=1/2,b=1/2,c=1,m=0$, we obtain
\begin{align}
 K'(k)=-\frac\pi2\frac{1}{\Gamma(1/2)^2}&\sum_{s=0}^{
\infty}\left( \frac{(1/2)_s}{s!} \right)^2 k^{2s}\left[\ln\left(k^2\right)
-2\psi\left(s+1\right)+2\psi\left(s+1/2\right)\right]
\end{align}
or
\begin{equation}
 K'(k)=-\sum_{s=0}^{
\infty}\left( \frac{(1/2)_s}{s!} \right)^2 k^{2s}\left[\ln\left(k\right)
-\psi\left(s+1\right)+\psi\left(s+1/2\right)\right]
\end{equation}
From the properties of the digamma function, for $s\ge1$,
\begin{align}
 \psi\left(s+1\right)&=\sum_{p=1}^{s}\frac{1}{p}-\gamma\\
 \psi\left(s+\tfrac{1}{2}\right)&=-\gamma-2\ln 2+2\sum_{p=1}^{s}\frac{1}{2p-1}
\end{align}
we deduce
\begin{equation}
 \psi\left(s+\tfrac{1}{2}\right)-\psi\left(s+1\right)=-2\ln2+2\sum_{p=1}^{s}\frac{1}{2p(2p-1)}
\end{equation}
for $s\ge1$ and $\psi\left(\tfrac{1}{2}\right)-\psi\left(1\right)=-2\ln2$.
Recognizing the series expansion for the elliptic integral, we deduce
\begin{equation}
 K'(k)=\frac{2}{\pi}\ln\left( \frac{4}{k} \right)K(k)-2\sum_{s=1}^{
\infty}\left( \frac{(1/2)_s}{s!} \right)^2 k^{2s}\sum_{p=1}^{s}\frac{1}{2p(2p-1)}
\end{equation}
which is the proposed result.


问:
Proving that $f$ is continuous at $3$
We have the function $f:\mathbb{R}\to\mathbb{R}$, defined by $f(x)=x^2+3x-5$ and want to show that it is continuous at $3$. This is how I've done it

Proof:

Let $\varepsilon>0$. Take $\delta=\mbox{min}\{1,\frac{\varepsilon}{10}\}$ and suppose that $|x-3|<\delta$. We note that for $|x-3|<1$, we have $|x+6|=|x-3+9|\leq|x-3|+9<1+9=10$. Therefore, we have the following:

\begin{align*}
|f(x)-f(3)|&=|x^2+3x-18|\\
&=|x-3||x+6|\\
&<\delta\cdot 10\\
&=\frac{\varepsilon}{10}\cdot10\\
&=\varepsilon\\
\end{align*}
$\blacksquare$

I'm wondering if this proof is sound. If somebody could verify this for me I would greatly appreciate it. I'm still quite new to epsilon-delta proofs. Any additional advice or pointers are always appreciated as well. Thanks

答:
From your second to last line... $\delta \cdot 10 \neq \frac{\varepsilon}{10} \cdot 10$.
$$  \delta \cdot 10 = \begin{cases}
\frac{\varepsilon}{10} \cdot 10 ,& \varepsilon/10 \leq 1  \\
1 \cdot 10 ,& \varepsilon/10 > 1
\end{cases}  \text{.}  $$
The first case is, of course, $\varepsilon$, but the second is not.

In the second case, from $\varepsilon/10 > 1$, you have $\varepsilon > 10$, so $\delta \cdot 10 = 1 \cdot 10 < \varepsilon$ in this case.

If your audience would automatically understand this detail, you could correctly write
$$  \delta \cdot 10 \leq \min\{1, \varepsilon/10\} \cdot 10 = \min\{10,\varepsilon\} \leq \varepsilon  \text{.}  $$

Title: Proving that $f$ is continuous at $3$


答:
I regard Blaze's answer as superior to the answer that I am giving, because he focuses on identifying the specific angle.  This is the natural way of attacking the problem.  The alternative approach is to compute $\cos(\theta)$ and $\sin(\theta)$ without identifying $(\theta)$.  Once this is done, you can use the formulas $\sec(\theta) = \frac{1}{\cos(\theta)}, \csc(\theta) = \frac{1}{\sin(\theta)}.$

Alternative approach:

Since $\theta$ is in the 3rd quadrant, both $\cos(\theta)$ and $\sin(\theta)$ are negative.

Since $\frac{\cos(\theta)}{\sin(\theta)} = \cot(\theta) = 1$, $\cos(\theta) = \sin(\theta)
 \implies \cos^2(\theta) = \sin^2(\theta).$

You also have the general formula that $\sin^2(\theta) + \cos^2(\theta) = 1.$

Therefore, since $\cos^2(\theta) = \sin^2(\theta), 2\sin^2(\theta) = 1 \implies $

$\sin^2(\theta) = \frac{1}{2} \implies \sin(\theta) = \pm \frac{1}{\sqrt{2}}.$

Therefore, since it is known that $\sin(\theta)$ is negative,

$\sin(\theta)$ must be $- \frac{1}{\sqrt{2}}.$

Then, since it is known that $\cos(\theta) = \sin(\theta)$

$\cos(\theta)$ must also be $- \frac{1}{\sqrt{2}}.$

Title: Given $\cotθ = 1$ and angle θ lies in the third quadrant, determine the value of $\secθ$ and $\cscθ$

Body: Recently learned about the reciprocal trig ratios and I'm still don't fully understand them. Wondering if I can get a little help not only with this specific question but the calculating reciprocal trig ratios in general please and thank you.</p>


问:
Proof of the Riesz Representation Theorem
Theorem: Let $F$ be a continuous linear functional on the Hilbert space $H$, then $\exists !$ (exists one and only one) $y \in H$ such that $F(x) = (x,y)$ for $x\in H$.

Proof:

Uniqueness: $$F(x)=(x,y)=(x,y') \Rightarrow (x,y-y')=0 \space \forall x \in H \Rightarrow ||y-y'||^2=0$$

Existence: if for any $x\in H, \space F(x) =(x,y)= 0 $ we'll set $y=0$. Let that $F \neq 0. $ and define a closed subspace $H_0 = \{x \in H: F(x)=0\}$. We know that $H \neq H_0$ because $H_0 = F^{-1}(\{0\})$.

Q1: "$H_0$ is a subspace by the linearity of $F$, closed by continuity of $F$." What does this mean? I understand what's a subspace, but don't understand what does it have to do with linearity and continuity.

$\exists z\in H$ such that $||z||=1$. For $u \in H_0, \space F(u)=(u,z) =0$.

Q2: Does this imply that $u \perp z \space \forall u \in H_0, \forall z \in H$?

Let's define $u=F(x)z - F(z)x, \space x \in H$

$$F(u)=(F(x)z - F(z)x,z)=F(x)F(z)-F(z)F(x)=0=F(x)(z,z)-F(z)(x,z)$$
Since we know that $||z|| = \sqrt{(z,z)}=1$, then
$$0=F(x)(z,z)-F(z)(x,z) \Rightarrow F(x) =F(z)(x,z) = (x,\bar{F(z)z)} $$
By defining $y= \bar{F(z)}z$ we've completed the proof.

By the way, the bar on the top denotes the complex conjugate, as is common in Hilbert spaces. i.e. $\alpha (x,y)= (\alpha x,y)=(x,\bar{\alpha} y) $

Corollary: If $H=L^2(\mu)$ and $F$ is a continuous linear functional on $L^2(\mu)$, then $\exists ! \space g \in L^2(\mu)$ such that $$F(f)=\int_X f \bar{g} d\mu = (f,g)$$

In the context of the Radon-Nikodym theorem, $g$ is the R-N derivative.

答:
Question 1: You defined $H_0 = F^{-1}(\{0\})$, where $F \in H^*$ (the continuous linear functionals on $H$). In order for $H_0$ to be a subspace, you have to check that $\lambda x + \mu y \in H_0$ for all $x,y \in H_0$, $\lambda, \mu \in \mathbb K$, right? But
$$ \lambda x + \mu y \in H_0 \iff F(\lambda x + \mu y) = 0 $$
As $F$ is linear (and $x,y \in H_0$), we have
$$ F(\lambda x + \mu y) = \lambda F(x) + \mu F(y) = 0 $$
So $H_0$ being a subspace follows from the linearity of $F$. In order for $H_0$ to be closed, we must show that for all sequences $(x_n)$ in $H_0$ converging to some $x \in H$ we must have $x\in H_0$. But, as $F$ is continuous, we have that $x_n \to x$, implies $F(x_n) \to F(x)$. As $F(x_n) = 0$ for every $n$, we get $F(x) = 0$, so $x \in H_0$. So the closedness of $H_0$ follows from the continuity of $F$.

Question 2: No, it doesn't. As $H_0$ is a proper closed subspace of $H$ ($H_0 \subsetneq H$ as $F \ne 0$), there is some $z \in H$, with $z \mathrel\bot H_0$ and $\|z\| = 1$. This means that for this $z$, not for all we have
$$ (z, u) = 0, \quad \text{all } u \in H_0. $$


问:
Find law of linear combination of random vectors
Let $X, Y$ be independent centered Gaussian random variables with variance $1$ and $a,b$ be real constants. Is it possible and how would one in general compute the law of the gaussian vector $(aX-bY,bX+aY)$?

答: $(aX-bY,bX+aY)$ has  a two dimensional normal distribution. All you have to do is to find the variance covariance matrix (since the means are both $0$). $Var (aX-bY)=a^{2}+b^{2}$, $Var (bX+aY)=a^{2}+b^{2}$ and $cov((aX-bY,bX+aY))=ab-ab=0$.


问:
A quartic whose Julia set includes two quadratic Julia sets
The abstract for "Quartic Julia sets including any two copies of quadratic Julia sets" in
Discrete & Continuous Dynamical Systems - A,36,4,2103,2112,2015-9-1,
by Koh Katagata, states

[...] for any two quadratic Julia sets, there exists a quartic polynomial whose Julia set includes copies of the two quadratic Julia sets.

I am struggling with the construction, particularly as I have not found a copy of the paper to read.  For example, I choose two quadratic polynomials $f_1(z) = z^2-2$ and $f_2(z) = z^2-1$, and let the quartic be $F(z) = z^4 + az^3 + bz^2 + cz + d$ with $a,b,c,d$ to be determined with $F$ behaving like $f_1$ near $z_1$ and like $f_2$ near $z_2$.

The quadratic Julia sets look like this:

In particular I've tried "behaving like" to mean something like affine conjugacy, and end up with things like $F(z_k + \lambda_k z) = z_k + \lambda_k f_k(z) + O(z^3)$ where $k \in\{1,2\}$.  I get (using Maxima):
$$\lambda_{1}^2\,b+3\,z_{1}\,\lambda_{1}^2\,a+6\,z_{1}^2
 \,\lambda_{1}^2 = \lambda_1\\
\lambda_{1}\,c+2\,z_{1}\,\lambda_{1}\,b+3\,z_{1}^2\,\lambda_{1}\,a+4\,z_{1}^3\,\lambda_{1} = 0 \\
d+z_{1}\,c+z_{1}^2\,b+z_{1}^3\,a+z_{1}^4=-2\,\lambda_{1}+z_{1}$$
and
$$\lambda_{2}^2\,b+3\,z_{2}\,\lambda_{2}^2\,a+6\,z_{2}^2\,\lambda_{2}^2 = \lambda_{2}\\
\lambda_{2}\,c+2\,z_{2}\,\lambda_{2}\,b+3\,z_{2}^2\,\lambda_{2}\,a+4\,z_{2}^3\,\lambda_{2} = 0\\
d+z_{2}\,c+z
 _{2}^2\,b+z_{2}^3\,a+z_{2}^4=-\lambda_{2}+z_{2}$$

Which is 6 nonlinear equations in 8 unknowns.  Arbitrarily setting $z_1 = -1, z_2 = 1, \lambda_k \ne 0 $ gives
$$
      (b - 3 a + 6) \lambda_1 = 1 \\
(c - 2 b + 3 a - 4) \lambda_1 = 0 \\
d - c + b - a + 2 + 2\lambda_1 = 0\\
      (b + 3 a + 6) \lambda_2 = 1 \\
(c + 2 b + 3 a + 4) \lambda_2 = 0 \\
d + c + b + a + \lambda_2 = 0 \\
$$
which has three solutions according to WolframAlpha.  Their Julia sets look like this:

of which the middle one looks most like what I expected, albeit somewhat distorted on the right.  It has the $\lambda_k$ about a factor of 2 apart, the other two have $\lambda_k$ about a factor -35 apart.

Is this the right way to construct the quartic?

答:
I can't offer any insights on your construction, but the paper you wanted is available here.

Just for fun here's an implementation of the quartic given in Figure 3 of the paper (it is interactive, so you can pan and zoom and modify the parameter in the upper left).  The construction really delivers the goods, because the resultant Julia set is composed of copies of the quadratic Julias.


答:
From Pythagoras' identity: $\:\cos^2\theta+\sin^2\theta=1$, are instantly deduced the relations

$1+\tan^2\theta=\sec^2\theta$, whence $\;\sec^2\theta=\dfrac{\cot^2\theta+1}{\cot^2\theta}$

$\cot^2\theta+1=\csc^2\theta$.

If $\cot\theta=1$, this means $\sec^2\theta=2=\csc^2\theta$.

Furthermore, in the third quadrant, $\sin\theta$ and $\cos\theta$, hence $\csc\theta$, and $\sec\theta$ are negative, so that
$$\csc\theta=\sec\theta=-\sqrt{2^{\vphantom n}}.$$


问:
Identify the set of points which satisfy $ |z|=\text{Re}z+1 $
Let $ z=x+iy $

I want to Identify the set of points that satisfy $ |z|=\text{Re}z+1 $.

Here's what I have found:

$ \sqrt{x^{2}+y^{2}}=x+1\thinspace\thinspace\thinspace\iff $

$ x^{2}+y^{2}=\left(x+1\right)^{2}\thinspace\thinspace\thinspace\iff $

$ y^{2}-2x-1=0\thinspace $

Now how can I identify the set of points based on the equation? I tried to use desmos, it looks like hyperbola, but its not. So im not sure...

Thanks in advance

答: Geometrically, the equation $|z|=\text{Re}(z)+1 $ indicates that the horizontal distance from $z$ to the vertical line $x=-1$ is equal to its distance to the origin, which defines the parabola with the focus at the origin and the directrix $x=-1$.


问:
Proof of Folland Lemma 3.2
On Folland page 86 Lemma 3.2 states that the union of any countable family of positive sets is positive, and the proof is as follows:

If $P_1, P_2, ... $ are positive sets let $Q_n = P_n \setminus \bigg( \bigcup_{i=1}^{n-1}P_i\bigg)$.  Then $Q_n \subset P_n$ so $Q_n$ is positive.  Hence if $E \subset \bigcup_{i=1}^{\infty}P_i$ then

$$
\nu(E) = \nu \bigg(\bigcup_{i=1}^{\infty}E \cap Q_i \bigg) = \sum_{i=1}^{\infty}\nu(E \cap Q_i) \geq 0
$$

My question is on the last equality, I assume that we know the infinite sum is greater than $0$ since each of the terms $\nu(E \cap Q_i)$ will be greater than $0$, but we only know that each $Q_i$ is a positive set and not necessarily $E$.

How do we know that for each $i$ that $\nu(E \cap Q_i) \geq 0$?

In other words how do we know that $E \cap Q_i$ is a positive set?

答:
Let $E$ be measurable and $E \subset \cup_{n=1}^{\infty}P_n$.

Then $E \cap Q_i \subset Q_i \subset P_i$ for all $i = 1,2,\ldots$. Now $Q_i$ is measurable for all $i=1,2,\ldots$ since it is constructed from complements and unions of measurable sets, and since $E$ is measurable, $E \cap Q_i$ is measurable for all $i=1,2,\ldots$. Since $E \cap Q_i \subset P_i$ then $E \cap Q_i$ is positive for all $i=1,2,\ldots$.

Title: Proof of Folland Lemma 3.2


问:
On Absolutely Continuous Functions
I would like to know if we can extend the concept of absolute continuity to functions $f:[a,b]\to X$, where $X$ is a topological vector space. I browsed some books on Topological Vector Spaces but can't find the definition of absolutely continuous functions defined on $[a,b]$ and take values on $X$. I would be greatful if someone can provide me a definition (if possible) of absolute continuous function $f:[a,b]\to X$. Thanks in advance...</p>

答:
For general Topological Vector Spaces: no.

There are two definitions of absolute continuity: one is for functions, the other for measures.

Definition (AC for functions): Let $I$ be an interval in $\mathbb{R}$. A function $f:\mathbb{R}\to\mathbb{R}$ is said to be absolutely continuous if for every $\epsilon > 0$ there exists $\delta > 0$ such that if $(a_k,b_k) = I_k \subset I$, $k \in \{1\ldots n\}$ are pairwise disjoint subintervals of $I$ satisfying
  $$ \sum_1^n |I_k|  = \sum_1^n |b_k - a_k| < \delta $$
  then
  $$ \sum_1^n |f(b_k) - f(a_k)| < \epsilon.$$

This definition only requires that we can evaluate
$$ \sum_1^n |f(b_k) - f(a_k)| $$
and naturally we can extend this to the case of the codomain being any metric space $(X,d)$ by replacing
$$ \sum_1^n |f(b_k)- f(a_k)| \implies \sum_1^n d(f(b_k),f(a_k)) $$

A general TVS may not even be metrizable, and I don't see how you can have a reasonable generalisation of the notion of absolute continuity if you don't have a reasonable replacement for $\sum_1^n|f(b_k) - f(a_k)|$.

See also the EOM entry.

Definition (AC for measures): Let $(\Omega,\mathcal{S},\mu)$ be a measure space (in the interval case we usually take $\Omega = [a,b]$, $\mathcal{S},\mu$ corresponding to the Lebesgue measure). A measure $\lambda$ is said to be absolutely continuous with respect to $\mu$ if every $\mu$-null set is $\lambda$-null.

The usual generalisation of this concept is to the case where $\lambda$ is a vector measure (a measure "taking values" in a Banach space). See, e.g., Chapter 29 of Schecter's HAF.

Title: On Absolutely Continuous Functions


答:
Let $g(x)=ax^2 +bx +c$

Then $f(x) =x^5 + 2x^4 -x^3 +3x^2 + 5x + 1\equiv$

$x^5 -x^4 -x^3-x+1\equiv ax^2 + bx + c\pmod 3$.

Which means

If $x \equiv 0\pmod 3$

Then $1 \equiv c\pmod 3$.

If $x \equiv 1$

then $2\equiv a+b +c \equiv a+b + 1$ so $a+b\equiv 1$.

And if $x\equiv -1$ then

$-1-1+1+1+1 \equiv a-b + c \equiv a-b +1$ so $a-b\equiv 0\pmod 3$

So $a\equiv b$ and $a+b\equiv 2a \equiv -a \equiv 1\pmod 3$ so $a \equiv b \equiv -1\pmod 3$ and $c\equiv 1\pmod 3$.

So we can have $-x^2-x + 1$ or any other coefficients where $a,b,c \equiv -1,-1,c$.

....

One thing to not.

If $g(x) = -x^2 -x^2 +1$ then $f(x) -g(x) = x^5 + 2x^4 -x^3 +4x^2 + 6x =$

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

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

$x^2(x^2(x-1)+3x^2 -x +4) +6=$

$x^2(x^2(x-1) + 3x(x-1) + 2x + 4) + 6=$

$x^2(x^2(x-1)+3x(x-1) + 2(x-1) + 6) + 6=$

$x^2(x-1)(x^2 + 3x + 2) + 6x^2 + 6 =$

$x^2(x-1)(x+1)(x+2) + (6x^2 + 6)$.

For $f(x)-g(x)$ to be equiv $0$ for all $x\in \mathbb Z$ then $x,x-1,x+1$ must be roots, plus or minus polynomial multiple of $3$.  And that is exactly the case here.

....

we could have solved it from the get go by synthetically dividing but $x(x+1)(x-1) = x^3 - x$. and taking the negative of the remainder.

$x^5 + 2x^4 -x^3 + 3x^2 + 5x + 1 =$

$x^2(x^3-x) +2x^4 +3x^2 + 5x +1=$

$x^2(x^3 -x) + 2x(x^3 - x) + 5x^2 + 5x + 1=$

So $5x^2 + 5x + 1 \equiv -x^2 -x +1\pmod 3$ would work.

Title: Suppose $f(x) = x^5 + 2x^4 − x^3 + 3x^2 + 5x + 1$. Find a polynomial $g$ of degree $2$ or less such that $f(x) \equiv g(x) \pmod 3$.


问:
Is there a name for a monounary algebra with an injective unary operation? Is there a research of these algebras.
To put is short, a unary algebra is a universal algebra with several unary operations and a monounary algebra is a unary algebra with only one operation. I am interested in a special kind of monounary algebras - those whose sole operation $f$ is injective, i.e. it has the property for any $x$ and $y$, if
$f(x)= f(y)$, then $x = y$. Possibly, a term different from "injective" is customarily used.

答:
Such algebras are rather simple objects, and as such there isn't much to say about them. Specifically, an injective monounary algebra is determined up to isomorphism by a few cardinal invariants:

How many cycles of size $n$ it has, for each finite $n$.

How many "$\mathbb{Z}$-chains" it has.

How many "maximal $\mathbb{N}$-chains" it has.

(A bit more precisely, if $\mathcal{A}$ is an injective monounary algebra with underlying set $A$ and function symbol $f$, let $\approx$ be the transitive symmetric closure of the relation $y=f(x)$. A cycle of size $n$ is an $\approx$-class of size $n$; a $\mathbb{Z}$-chain is an infinite $\approx$-class on which $f$ is a bijection; and a maximal $\mathbb{N}$-chain is an infinite $\approx$-class on which $f$ is not a bijection.)

Given this, I don't think they have been studied on their own.


答:
1) As with any other (linear) vector space, the kernel of a linear transformation is always a vector subspace of the definition domain vector space, and in this case we have that $\;H_0:=\ker F\;$ , with $\,F:H\to\Bbb F\;,\;\;\Bbb F=\Bbb R\;\,\;\Bbb C\,$ . This is also denoted by $\,F\in H^*:=$ the dual space of $\,H\,$

The conditions about closedness and being a subspace of Hilbert space then follow from conditions other than the linear ones (i.e., continuity of a linear functional), which hopefully are already clear.

(2) About your second question: the answer's no, since the only elements that is orthogonal to all the elements in the space is the zero vector.

What is written there follows from the fact that $\,H_0\neq H\,$ (in fact, since $\,F\,$ is a non-trivial functional we have that $\;H_0\;$ is a maximal proper subspace of $\,H\,$ . This much is true also in any general vector space ,whether dinite or infinite dimensional over its definition field) . and then $\,H_0^\perp\neq0\;$ , and from here that there exists such $\,z\in H\,$ as implied just before your second question

Title: Proof of the Riesz Representation Theorem

Body:
Theorem: Let $F$ be a continuous linear functional on the Hilbert space $H$, then $\exists !$ (exists one and only one) $y \in H$ such that $F(x) = (x,y)$ for $x\in H$.

Proof:

Uniqueness: $$F(x)=(x,y)=(x,y') \Rightarrow (x,y-y')=0 \space \forall x \in H \Rightarrow ||y-y'||^2=0$$

Existence: if for any $x\in H, \space F(x) =(x,y)= 0 $ we'll set $y=0$. Let that $F \neq 0. $ and define a closed subspace $H_0 = \{x \in H: F(x)=0\}$. We know that $H \neq H_0$ because $H_0 = F^{-1}(\{0\})$.

Q1: "$H_0$ is a subspace by the linearity of $F$, closed by continuity of $F$." What does this mean? I understand what's a subspace, but don't understand what does it have to do with linearity and continuity.

$\exists z\in H$ such that $||z||=1$. For $u \in H_0, \space F(u)=(u,z) =0$.

Q2: Does this imply that $u \perp z \space \forall u \in H_0, \forall z \in H$?

Let's define $u=F(x)z - F(z)x, \space x \in H$

$$F(u)=(F(x)z - F(z)x,z)=F(x)F(z)-F(z)F(x)=0=F(x)(z,z)-F(z)(x,z)$$
Since we know that $||z|| = \sqrt{(z,z)}=1$, then
$$0=F(x)(z,z)-F(z)(x,z) \Rightarrow F(x) =F(z)(x,z) = (x,\bar{F(z)z)} $$
By defining $y= \bar{F(z)}z$ we've completed the proof.

By the way, the bar on the top denotes the complex conjugate, as is common in Hilbert spaces. i.e. $\alpha (x,y)= (\alpha x,y)=(x,\bar{\alpha} y) $

Corollary: If $H=L^2(\mu)$ and $F$ is a continuous linear functional on $L^2(\mu)$, then $\exists ! \space g \in L^2(\mu)$ such that $$F(f)=\int_X f \bar{g} d\mu = (f,g)$$

In the context of the Radon-Nikodym theorem, $g$ is the R-N derivative.


问:
Open maps of spaces vs open maps of locales
A continuous function between topological spaces $f:X\to Y$ is called open, if $f[U]\in\mathcal{O}(Y)$ for all $U\in\mathcal{O}(X)$.

A morphism of locales $f:X\to Y$ is called open, if the associated morphism of frames $f^*:\mathcal{O}(Y)\to\mathcal{O}(X)$ has a monotone left adjoint $f_!$ which satisfies the Frobenius condition $f_!(U)\wedge V\leq f_!(U\wedge f^*(V))$ for all $U\in\mathcal{O}(X)$ and $V\in\mathcal{O}(Y)$.

It is easy to see that every continuous $f$ that is open in the topological sense is also open in the localic sense, but the converse is false in general. For example,  the maps $1\to\nabla(2)$ and $\mathsf{cofin}(\mathbb{N})\to\mathsf{sobrify}(\mathsf{cofin}(\mathbb{N}))$ are open in the localic, but not in the topological sense, since they're not surjective, but the induced locale-morphisms are iso.

These two examples involve non-sober spaces, so the natural thing to ask is: Do the two notions coincide for sober spaces? Or more generally, does the canonical functor from locales to spaces preserve open maps?

Or are there other criteria for when the two notions coincide? For example, I think I have convinced myself that continuous maps that are open in the localic sense are open in the topological sense whenever the codomain is $T_1$.

答:
Here is a counterexample with sober spaces.  Let $X=\mathbb{N}\times\mathbb{N}$, equipped with a topology defined as follows.  Fix an injection $i:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$.  Say that a set $U\subseteq X$ is open if for each $(a,b)\in U$, $U$ contains $(i(a,b),c)$ for all but finitely many $c\in\mathbb{N}$.

I claim that $X$ is Hausdorff, and thus sober.  Let $p,q\in X$ be distinct points.  Start with $U_0=\{p\}$ and $V_0=\{q\}$.  Recursively define $U_{n+1}=U_n\cup ((i[U_n]\times\mathbb{N})\setminus\{q\})$ and $V_{n+1}=V_n\cup((i[V_n]\times\mathbb{N})\setminus\{p\})$.  It is clear that $U=\bigcup U_n$ and $V=\bigcup V_n$ are open, since if $(a,b)\in U_n$ then $U_{n+1}$ contains $(i(a,b),c)$ for all but finitely many $c$, and similarly for $V$.  I claim that $U_n$ and $V_n$ are disjoint for all $n$, and thus $U$ and $V$ are disjoint.  Indeed, note that $U_n$ has the property that if $(a,b)\in U_n$ then either $(a,b)=p$ or $a=i(x)$ for some $x\in U_{n-1}$, and similarly for $V_n$.  Thus $U_n$ and $V_n$ are disjoint by induction on $n$: if we already know $U_{n-1}$ and $V_{n-1}$ are disjoint, then all the points of $U_n$ (except $p$) have distinct first coordinate from all the points of $V_n$ (except $q$).

Now let $Y=\mathbb{N}\cup\{g\}$ where $\mathbb{N}$ has the cofinite topology and $g$ is the generic point.  Define $f:X\to Y$ by $f(a,b)=b$.  For any nonempty open $U\subseteq X$, $f[U]$ is cofinite in $\mathbb{N}$, and so there is a smallest open set of $Y$ that contains it, namely $f_!(U)=f[U]\cup\{g\}$.  To verify the Frobenius condition, suppose $x\in f_!(U)\cap V$ for $U$ open in $X$ and $V$ open in $Y$.  If $x\in\mathbb{N}$, then $(a,x)\in U$ for some $a\in U$ and then this $(a,x)$ is in $U\cap f^{-1}(V)$, witnessing that $x\in f_!(U\cap f^{-1}(V))$.  If $x=g$, then note that $U$ and $V$ are both nonempty; let $(a,b)\in U$.  Then $(i(a,b),c)\in U$ for all but finitely many $c$, and also $(i(a,b),c)\in f^{-1}(V)$ for all but finitely many $c$ since $V$ is cofinite.  Thus $U\cap f^{-1}(V)$ is nonempty, and thus $g\in f_!(U\cap f^{-1}(V))$.

So, the map $f:X\to Y$ is open in the localic sense.  However, it is not open in the topological sense, because $f[X]=\mathbb{N}$ is not open in $Y$.

On the other hand, you are correct that the two notions coincide if $Y$ is $T_1$.  First, the existence of the left adjoint $f_!$ just means that for each $U\in\mathcal{O}(X)$, there is a smallest open set $f_!(U)\in\mathcal{O}(Y)$ that contains $f[U]$.  If $Y$ is $T_1$, this immediately implies $f_!(U)=f[U]$ so $f$ is open in the topological sense: if $x\in f_!(U)\setminus f[U]$ then $f_!(U)\setminus\{x\}$ still contains $f[U]$ and is open, contradicting the definition of $f_!(U)$.

A bit more generally, they coincide if $Y$ is $T_D$, meaning that for each $y\in Y$, $\{y\}$ is open as a subset of $\overline{\{y\}}$.  To prove this, note that a similar argument as in the paragraph above shows that every point of $f_!(U)$ must be a generization of a point of $f[U]$.  So, to show that $f$ is open in the topological sense, it suffices to show that $f[U]$ is specialization-open for any $U\in\mathcal{O}(X)$.  Suppose $y\in Y$ is a generization of a point of $f[U]$.  Since $Y$ is $T_D$, there exists $V\in \mathcal{O}(Y)$ that contains $y$ but does not contain any other specialization of $y$.  Now the Frobenius condition says $y\in f_!(U)\cap V\subseteq f_!(U\cap f^{-1}(V))$, so there is a point $x\in U\cap f^{-1}(V)$ such that $y$ is a generization of $f(x)$.  Since $f(x)\in V$, this means $f(x)=y$, so $y\in f[U]$, as desired.


问:
Questions about the proof of Riesz-Fischer Theorem in Measure Theory
I'm learning from Stein's Real analysis, Chapter 2. When reading the Riesz-Fischer (pp70), I found part of the proof confusing:

Suppose$\{f_n\}$ is a sequence of integrable functions，and is Cauchy
in $L_1$ space$\to \forall \epsilon > 0, \exists N \in \mathbb{N}^{+}$ s.t.
$\forall n, m > N(\epsilon), ||f_n - f_m|| < \epsilon$ .

Then choose a subseq ${f_{n_k}}$ s.t. $||f_{n_{k + 1}} - f_{n_k}||
 \leq 2^{-k}, \forall k \geq 1$.

Define $f := f_{n_1}(x) + \sum_{k = 1}^\infty (f_{n_{k+1}}(x)
 -f_{n_{k}}(x)) $,

$g := |f_{n_1}(x)| + \sum_{k = 1}^\infty (|f_{n_{k+1}}(x) -f_{n_{k}}(x)|)$

Easy to see that $\int_{\mathbb{R}}g < \infty$, so $g$ is integrable。Since
$|f| \leq g$, $f$ is integrable as well.

Also, the series defining $f$ converges almost everywhere...

What does it mean by the series defining $f$ converges almost everywhere? How can we see that?

答:
Since $\int_{\mathbb{R}}g < \infty$, $g$ is finite almost everywhere.

It means that $|f_{n_1}(x)| + \sum_{k = 1}^\infty (|f_{n_{k+1}}(x) -f_{n_{k}}(x)|)$ converges  almost everywhere.

So, $f_{n_1}(x) + \sum_{k = 1}^\infty (f_{n_{k+1}}(x)
 -f_{n_{k}}(x)) $ converges (absolutely) almost everywhere.


问:
Vector identity to rewrite triple product with gradient into divergence
Is there a vector identity that can rewrite a product of type

$$ \mathbf{a}\cdot(\mathbf{a}\cdot\nabla\mathbf{a}), $$

where $\mathbf{a}$ is a vector, into a divergence, for example of the form

$$ \mathbf{a}\cdot(\mathbf{a}\cdot\nabla\mathbf{a})= \nabla\cdot(|\mathbf{a}|^2 \mathbf{a}) + ... ? $$

I have had no luck so far in trying to employ any of the well known vector identities, but it feels like something like this should be possible

答:
I am interpretting your product as $$\mathbf{a}\cdot\left[\left(\mathbf{a}\cdot\nabla\right)\mathbf{a}\right]=\mathbf{a}\cdot\left(\sum_i a_i\partial_i \mathbf{a}\right)=\sum_{ij}a_i a_j\partial_i a_j.$$
The divergence you use as an example can be written as
$$\nabla\cdot\left(|\mathbf{a}|^2\mathbf{a}\right)=\sum_{ij}\partial_i\left(a_j^2 a_i\right)=2\sum_{ij}a_ia_j\partial_ia_j+\sum_{ij}a_j^2\partial_ia_i.$$
Identifying
$$\sum_{ij}a_j^2\partial_ia_i=|\mathbf{a}|^2\left(\nabla\cdot\mathbf{a}\right),$$ your desired identity could be
$$\mathbf{a}\cdot\left[\left(\mathbf{a}\cdot\nabla\right)\mathbf{a}\right]=\frac{1}{2}\left[\nabla\cdot\left(|\mathbf{a}|^2\mathbf{a}\right) - |\mathbf{a}|^2\left(\nabla\cdot\mathbf{a}\right)\right].$$

Here I have used the shorthand $\partial_i=\partial/\partial i$.


问:
Backing into mixture from weighted average of 2 items
Item1 Price: $1

Item2 Price: $2

WAP (weighted average price): $1.5

Correct Answer: 50% of item 1 and 50% of item 2 would have a WAP of $1.50

But how would I calculate the percentage of each needed to back into a weighted average price of $1.79 for example (where the percentages of each aren't obvious)?

答:
Let the price of item 1 be $p_1$ and the price of item 2 be $p_2$.  Then if the weighted average price is $w$, we want to solve for the weight $\lambda$ such that $$w = \lambda p_1 + (1 - \lambda) p_2.$$  This gives us $$(p_1 - p_2) \lambda = w - p_2,$$ or $$\lambda = \frac{w - p_2}{p_1 - p_2}.$$

Note that $0 \le \lambda \le 1$ if and only if $p_1 \le w \le p_2$.

So for your example, $p_1 = 1$, $p_2 = 2$, and $w = 1.79$.  Hence $$\lambda = \frac{1.79 - 2}{1 - 2} = 0.21,$$ which means that the weight is $21\%$ on item 1, and $79\%$ on item 2.


问:
Maple - LinearSolve evaluates seemingly equivalent augmented matrix differently
I am trying to get myself acquainted with Maple and was fooling around with some linear algebra.

I noticed that the following commands

```<code>restart;
with(LinearAlgebra);

P:=<0.6,0.3;0.4,0.7>;
Id := IdentityMatrix(2);

A:=P-Id;
b:=ZeroVector(2);

Aug1:=<A|b>;
LinearSolve(Aug1,free='x');
</code>```

produce the unexpected result
$$
\begin{bmatrix}
-0. \\
-0. \\
\end{bmatrix}
$$

Entering the augmented matrix manually, however

```<code>Aug2:=<-0.4,0.3,0;0.4,-0.3,0>;
LinearSolve(Aug2,free='x');
</code>```

does produce the result I was expecting
$$
\begin{bmatrix}
0.750000000000000x_1 \\
x_1 \\
\end{bmatrix}
$$

The only difference I see, is that Aug1 looks like
$$
\begin{bmatrix}
-0.400000000000000 & 0.300000000000000 & 0 \\
0.400000000000000 & -0.300000000000000 & 0 \\
\end{bmatrix}
$$
while Aug2 looks like
$$
\begin{bmatrix}
-0.4 & 0.3 & 0 \\
0.4 & -0.3 & 0 \\
\end{bmatrix}
$$

Using whattype() I already determined that both Aug1 and Aug2 have the Matrix datatype and their elements have the float datatype.

Any idea how I could produce the expected result using Aug1 ($=P-Id$)?

答:
When you add your Matrix A (whose entries are all of type numeric, and at least one is a float) to your Id the 2x2 identity Matrix, then the result has datatype=float[8] and is stored in so-called hardware double precision.

That is a base-2 format, in which the floats 0.3 and 0.7 cannot be represented exactly. Notice how the following values p1 and temp differ by a very small (but nonzero!) amount. I use Maple's evalhf mode here, to mimic the double precision operations in question.

```<code>evalhf(proc() local p1,p2,res,temp;
   p1:=0.3; p2:=0.7;
   temp:=p2-1.0; res:=p1+temp;
   eval(lprint([p1,temp,res]));
   return NULL; end proc()):

   [.299999999999999989, -.300000000000000044,
     -.555111512312578270e-16]

HFloat(0.3), HFloat(0.7)-HFloat(1.0):
lprint(%);

  HFloat(.299999999999999989),
  HFloat(-.300000000000000044)

`+`(%);

                        -17
    -5.55111512312578 10
</code>```

Notice the res is not exactly 0.0.

At the default values for the environment variables Digits (10) and UsehardwareFloats (deduced) the following LinearAlgebra computations are done in hardware double precision.

The result from LinearSolve is indeed a single, valid solution, but the parametric answer is not obtained. That is because the linear-solving algorithm does not detect that the two rows are "close enough" to being merely elementwise negations of each other.

```<code>restart;
with(LinearAlgebra):
P:=<0.6,0.3;0.4,0.7>:
Id := IdentityMatrix(2):
b:=ZeroVector(2):

UseHardwareFloats; # default is `deduced`

             deduced

A:=P-Id;

       [-0.400000000000000   0.300000000000000]
  A := [                                      ]
       [ 0.400000000000000  -0.300000000000000]

lprint(A);

  Matrix(2,2,{(1, 1) =
  HFloat(-.400000000000000022),
  (1, 2) = HFloat(.299999999999999989),
  (2, 1) = HFloat(.400000000000000022),
  (2, 2) = HFloat(-.300000000000000044)},
  datatype = float[8],storage = rectangular,
  order = Fortran_order,shape = [])

Aug1:=<A|b>:

LinearSolve(Aug1);

              [-0.]
              [   ]
              [-0.]
</code>```

The following provides some insight as to how that was computed.

```<code>LUDecomposition(Aug1);

  [1  0]  [1.0   0.]  [-0.400000000000000        0.300000000000000  0.]
  [    ]  [        ]  [                                               ]
  [0  1], [-1.  1.0], [                                        -17    ]
                    [                0.  -5.55111512312578 10     0.]

RowOperation(Aug1,[2,1],1);

     [-0.400000000000000        0.300000000000000  0]
     [                                              ]
     [                                        -17   ]
     [                0.  -5.55111512312578 10     0]

ReducedRowEchelonForm(Aug1);

          [ 1.  -0.  -0.]
          [             ]
          [-0.   1.  -0.]
</code>```

Notice the similarity with the earlier scalar computation. Notice that the bottom row of the results from ReducedRowEchelonForm and RowOperation are not all zero, and neither is the L Matrix returned by LUDecomposition.

One way to work around this issue is to perform the computations in so-called "software floats", which in Maple means a base-10 floating-point format in which the values 0.3 and 0.7 can be stored "exactly".

```<code>SFloat(0.3), SFloat(0.7)-SFloat(1.0):
lprint(%);

  .3, -.3

`+`(%);

            0.
</code>```

With the default value of deduced for UseHardwareFloats then software floats will get used if Digits is greater than 15. Or UseHardwareFloats could be set to false and then software floats would get used, for any valid setting of Digits.

Let's re-do all those computations, with UseHardwareFloats set to false.

```<code>UseHardwareFloats:=false:

A:=P-Id;

           [-0.4   0.3]
      A := [          ]
           [ 0.4  -0.3]

lprint(A);

  Matrix(2,2,{(1, 1) = -.4, (1, 2) = .3,
  (2, 1) = .4, (2, 2) = -.3},
  datatype = sfloat,storage = rectangular,
  order = Fortran_order,shape = [])

Aug1:=<A|b>:

LinearSolve(Aug1);

         [0.7500000000 _t0[1]]
         [                   ]
         [      _t0[1]       ]
</code>```

The following provides some insight as to how that was computed.

```<code>LUDecomposition(Aug1);

     [1  0]  [          1.  0.]  [-0.4  0.3  0.]
     [    ], [                ], [             ]
     [0  1]  [-1.000000000  1.]  [  0.   0.  0.]

RowOperation(Aug1,[2,1],1);

         [-0.4  0.3  0]
         [            ]
         [  0.   0.  0]

ReducedRowEchelonForm(Aug1);

     [1.  -0.7500000000  -0.]
     [                      ]
     [0.             0.   0.]
</code>```

Another way to get at a parametric solution for your example is to compute the NullSpace (Kernel) of your A Matrix, and form a parametrized linear combination of the (here, one) Vector in the basis.

This question is about Maple programming and not about mathematics per se, and as such is off-topic in this forum.  More suitable would be stackoverflow.com or www.mapleprimes.com the Maplesoft user forum.

Body:
When you add your Matrix A (whose entries are all of type numeric, and at least one is a float) to your Id the 2x2 identity Matrix, then the result has datatype=float[8] and is stored in so-called hardware double precision.

That is a base-2 format, in which the floats 0.3 and 0.7 cannot be represented exactly. Notice how the following values p1 and temp differ by a very small (but nonzero!) amount. I use Maple's evalhf mode here, to mimic the double precision operations in question.

evalhf(proc() local p1,p2,res,temp;
   p1:=0.3; p2:=0.7;
   temp:=p2-1.0; res:=p1+temp;
   eval(lprint([p1,temp,res]));
   return NULL; end proc()):

   [.299999999999999989, -.300000000000000044,
     -.555111512312578270e-16]

HFloat(0.3), HFloat(0.7)-HFloat(1.0):
lprint(%);

  HFloat(.299999999999999989),
  HFloat(-.300000000000000044)

`+`(%);

                        -17
    -5.55111512312578 10

Notice the res is not exactly 0.0.

At the default values for the environment variables Digits (10) and UsehardwareFloats (deduced) the following LinearAlgebra computations are done in hardware double precision.

The result from LinearSolve is indeed a single, valid solution, but the parametric answer is not obtained. That is because the linear-solving algorithm does not detect that the two rows are "close enough" to being merely elementwise negations of each other.

restart;
with(LinearAlgebra):
P:=<0.6,0.3;0.4,0.7>:
Id := IdentityMatrix(2):
b:=ZeroVector(2):

UseHardwareFloats; # default is `deduced`

             deduced

A:=P-Id;

       [-0.400000000000000   0.300000000000000]
  A := [                                      ]
       [ 0.400000000000000  -0.300000000000000]

lprint(A);

  Matrix(2,2,{(1, 1) =
  HFloat(-.400000000000000022),
  (1, 2) = HFloat(.299999999999999989),
  (2, 1) = HFloat(.400000000000000022),
  (2, 2) = HFloat(-.300000000000000044)},
  datatype = float[8],storage = rectangular,
  order = Fortran_order,shape = [])

Aug1:=<A|b>:

LinearSolve(Aug1);

              [-0.]
              [   ]
              [-0.]

The following provides some insight as to how that was computed.

LUDecomposition(Aug1);

  [1  0]  [1.0   0.]  [-0.400000000000000        0.300000000000000  0.]
  [    ]  [        ]  [                                               ]
  [0  1], [-1.  1.0], [                                        -17    ]
                    [                0.  -5.55111512312578 10     0.]

RowOperation(Aug1,[2,1],1);

     [-0.400000000000000        0.300000000000000  0]
     [                                              ]
     [                                        -17   ]
     [                0.  -5.55111512312578 10     0]

ReducedRowEchelonForm(Aug1);

          [ 1.  -0.  -0.]
          [             ]
          [-0.   1.  -0.]

Notice the similarity with the earlier scalar computation. Notice that the bottom row of the results from ReducedRowEchelonForm and RowOperation are not all zero, and neither is the L Matrix returned by LUDecomposition.

One way to work around this issue is to perform the computations in so-called "software floats", which in Maple means a base-10 floating-point format in which the values 0.3 and 0.7 can be stored "exactly".

SFloat(0.3), SFloat(0.7)-SFloat(1.0):
lprint(%);

  .3, -.3

`+`(%);

            0.

With the default value of deduced for UseHardwareFloats then software floats will get used if Digits is greater than 15. Or UseHardwareFloats could be set to false and then software floats would get used, for any valid setting of Digits.

Let's re-do all those computations, with UseHardwareFloats set to false.

UseHardwareFloats:=false:

A:=P-Id;

           [-0.4   0.3]
      A := [          ]
           [ 0.4  -0.3]

lprint(A);

  Matrix(2,2,{(1, 1) = -.4, (1, 2) = .3,
  (2, 1) = .4, (2, 2) = -.3},
  datatype = sfloat,storage = rectangular,
  order = Fortran_order,shape = [])

Aug1:=<A|b>:

LinearSolve(Aug1);

         [0.7500000000 _t0[1]]
         [                   ]
         [      _t0[1]       ]

The following provides some insight as to how that was computed.

LUDecomposition(Aug1);

     [1  0]  [          1.  0.]  [-0.4  0.3  0.]
     [    ], [                ], [             ]
     [0  1]  [-1.000000000  1.]  [  0.   0.  0.]

RowOperation(Aug1,[2,1],1);

         [-0.4  0.3  0]
         [            ]
         [  0.   0.  0]

ReducedRowEchelonForm(Aug1);

     [1.  -0.7500000000  -0.]
     [                      ]
     [0.             0.   0.]

Another way to get at a parametric solution for your example is to compute the NullSpace (Kernel) of your A Matrix, and form a parametrized linear combination of the (here, one) Vector in the basis.

This question is about Maple programming and not about mathematics per se, and as such is off-topic in this forum.  More suitable would be stackoverflow.com or www.mapleprimes.com the Maplesoft user forum.


问:
Prove inequality $\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right) \tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le\frac13 $
How to determine the range of the function

$$f(x)=\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)
$$

It it is straightforward to verify that $f(x)$ is even and

$$f(0)= \frac13, \>\>\>\>\> \lim_{x\to\pm \infty} f(x) \to -\infty$$

which implies $f(x) \in (-\infty,\frac13]$, i.e.

$$\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le \frac13
$$

and is visually confirmed below

However, it is not obvious algebraically that $f(x)$ monotonically decreases away from $x=0$. The standard derivative tests are not viable due to their rather complicated functional forms.

So, the question is how to prove the inequality  $f(x) \le \frac13$ with rigor.

Note that it is equivalent to proving

$$\cot \left(\frac{\pi(1+x)}{3+x^2}\right)
\cot \left(\frac{\pi(1-x)}{3+x^2}\right)\le \frac13
$$

答:
Note that
$$f(x)=\tan\left(\frac{\pi}{2}\frac{(1+x)^2}{3+x^2}\right)\tan\left(\frac{\pi}{2}\frac{(1-x)^2}{3+x^2}\right)=\frac{2\cos(2\pi/(3+x^2))}{\cos(2\pi x/(3+x^2))-\cos(2\pi/(3+x^2)} +1,$$
which is slightly easier to take the derivative:

$$f'(x)=\frac{8 \pi x \cos(\frac{2 \pi x}{3 + x^2}) \sin(\frac{2 \pi}{3 + x^2}) - 4 \pi (x^2-3) \cos(\frac{2 \pi}{3 + x^2}) \sin(\frac{2 \pi x}{3 + x^2})}{(3 + x^2)^2 \left(\cos(\frac{2 \pi}{3 + x^2}) - \cos(\frac{2 \pi x}{3 + x^2})\right)^2}$$

Obviously the denominator is positive, so we need to show that for $x \geq  0$:
$$8 \pi x \cos\left(\frac{2 \pi x}{3 + x^2}\right) \sin\left(\frac{2 \pi}{3 + x^2}\right) - 4 \pi (x^2-3) \cos\left(\frac{2 \pi}{3 + x^2}\right) \sin\left(\frac{2 \pi x}{3 + x^2}\right) \leq 0.$$

This can be further simplified to
$$-2 \pi \left((x-1)(x+3) \sin\left(\frac{2 \pi(x-1)}{3 + x^2}\right) + (x+1)(x-3) \sin\left(\frac{2 \pi(x+1)}{3 + x^2}\right)\right)\leq 0\tag{1}$$
ant thus we need to show that for $x\geq 0$
$$(x-1)(x+3) \sin\left(\frac{2 \pi(x-1)}{3 + x^2}\right) \geq -(x+1)(x-3) \sin\left(\frac{2 \pi(x+1)}{3 + x^2}\right)$$
It seems we have to split it into intervals (probably two $x\in(0,1)$ and $x\in(1,\infty)$), and investigate the inequalities on the intervals, but I am stuck here at the moment.

UPDATE:

We see that (1) is an odd function, so it sufficient to investigate only one part, e.g. we need to show that for $x\geq 0$,
$$g(x)=(x-1)(x+3)\sin\left(\frac{2\pi(x-1)}{3+x^2}\right)\geq 0$$
We see that for $x\geq 0$, the equation has exacly 2 roots, $x=0$ and $x=1$ and therefore we can look at the sign in the subintervals. For example,plugging $x=0.5$, $g(0.5)=1.44022$ and $g(2)=3.909$. Thus the function $g(x)\geq 0$, and therefore $f(x)$ is non-increasing for $x\geq0$ (and by the oddness of (1), non-decreasing for $x\leq 0$).

If we look at the extrema $x=0$, and $x=1$, we see that $f(0)=\frac{1}{3}$ and $\lim_{x\to 1}f(x)=0$, therefore $f(0)=\frac{1}{3}$ is global maximum.


问:
What's the dimension of this vectorspace?
Consider the concatenation operator, e.g. $(1,2)+(3,4)=(1,2,3,4)$. With the conceit that adjacent numbers can "cancel out", e.g. $(1,2)+(-2,3)=(1,3)$, this is a group.

We can make it a vectorspace by the standard componentwise multiplication: $cx=(cx_1,cx_2,\dots)$.

Note that any element in this vector space can be written as $\sum_i c_i(1)$ for some $c_i$.

I'm wondering if this means that this vector space has dimension one, since a basis seems to be $\{(1)\}$? I have read that a vector space with a field $F$ and dimension $n$ is isomorphic to $F^n$ which would mean that my vectorspace is isomorphic to $\mathbb R$, which I don't think is true.

答:
This is not a vector space. Vadim123 and Tobias Kildetoft already pointed out that the underlying additive group is not abelian. Also in a vector space the axiom
$$
(a+b)X=aX+bX
$$
is required to hold for all scalars $a,b$ and vectors $X$. But here if $X=(1,2)$, $a=3$, $b=4$, then
$$
(a+b)X=7X=(7,14)\qquad\text{but}\qquad aX+bX=(3,6)+(4,8)=(3,6,4,8).
$$
Allowing permutations won't help.</p>


问:
When the fundamental group of a space is Infinite
I am not sure how to proceed with this problem:

Let $X$ be a path-connected and locally path-connected space. Suppose $p:X\to X$ is a non-injective covering map.
Prove that $\pi _1(X) $ is infinite for any choice of base point $a\in X$.

Pick $a\in X$ arbitrarily. I know that since $X$ is path connected, $| p^{-1}(a)| \leq |\pi _1(X,a) | $. Then since $p$ is a non-injective covering map, $|p^{-1}(a)|>1 $. (In fact, all fibers have the same cardinality.)

Is there some way to show $|\pi _1 (X,a)| $ is infinite by using deck transformations?

Or, is it possible to show that if a space covers itself, then its fundamental group is either trivial or infinite?

Any suggestions would be helpful.

答: Let $F$ be the fibre of the fibration. The exact sequence defined by the fibration gives $1=\pi_1(F)\rightarrow\pi_1(X)\rightarrow\pi_1(X)\rightarrow\pi_0(F)\rightarrow\pi_0(X)=1$. It result that $\pi_1(X)$ has a subgroup isomorphic to itself which is not itself since $\pi_0(F)$ is not trivial. This implies that the cardinality of $\pi_1(X)$ is infinite, since a subset of a finite set distinct of itself has a strictly inferior cardinality.


问:
Inner Product is Less than Zero
Let $V$ be an inner product space over $\mathbb R.$ Let $u_1, u_2, u_3, u_4 \in V$ such that $\langle u_i,u_j\rangle < 0,$ for $i,j = 1, 2, 3, 4 i\neq j.$

Show that $u_1, u_2, u_3$ are linearly independent.

Please advise the ways to prove the result.

答:
Example: $V=\mathbb{R}^4$, $u_1=(1,0,0,0)$, $u_2=(-1,2,0,0)$, $u_3=(-1,-1,3,0)$, and $u_4=(-1,-1,-1,1)$.
This are all linearly independent because they generate $\mathbb{R}^4$. However, the inner products between each two are negative.

PS: the inner product is zero iff the vectors are orthogonal (that is the definition, but think of it geometrically). Also, a finite dimensional vector space has always a basis, and it has also an inner product, but that basis does not have to be an orthogonal basis. The example I gave was cooked so that the inner products are all negative, but imagine that they are all positive: $\{(1,0,0,0), (1,1,0,0), (1,1,1,0), (1,1,1,1)\}$ is a linearly independent set of vectors, and non is orthogonal to another.

Title: Inner Product is Less than Zero


问:
Prove that functional set is a basis
Question:
$p_1(f)=f(0), p_2(f)=f'(0),p_3(f)=f(1)$ are linear functionals
$f \in V=\Bbb R[X]_{\le 2}$
Prove that $p_1,p_2,p_3$ is a basis for $V^*$

What I did:
So I understand I need to prove that $p_1,p_2,p_3$ are linearly independent so I need to show:
$a_1p_1+a_2p_2+a_3p_3=0 \Rightarrow a_1p_1(f)+a_2p_2(f)+a_3p_3(f)=0$$
 \Rightarrow a_1f(0)+a_2f'(0)+a_3f(1)=0$
so I need to show that f(0), f'(0) and f(1) are linearly independent.

If f is of degree 2 then if $f(1)=f(0)$ then $f=x^2-x$ and then $f'(0)=-1$ and then the solution must be trivial.

If f is of degree 0 then the solution may not be trivial (for example $f(0)=f(1)=c),f'(0)=0$.

If f is of degree 1 then $f=cx: f'(0)=c ,f(1)\ne f(0)$ Can I now say that the only a's for that are zeros? I can find for example $f(x)=3x$ where $f(0)=0, f(1)=3,f'(0)=3$ and this will not be linearly independent. This is where I'm stuck...

答:
I'll discuss two (equivalent) methods. The first one is a bit more direct.

To show that $p_1, p_2$ and $p_3$ are linearly independent, suppose
$$
a_1p_1 + a_2p_2 + a_3p_3 = 0.
$$
This means
$$
a_1p_1(f) + a_2p_2(f) + a_3p_3(f) = 0
$$
for all $f \in V$.
If $f = 1$, then $p_1(f) = 1$, $p_2(f) = 0$ and $p_3(f) = 1$, so
$$
a_1 + a_3 = 0.
$$
If $f = x$, then $p_1(f) = 0$, $p_2(f) = 1$ and $p_3(f) = 1$, so
$$
a_2 + a_3 = 0.
$$
If $f = x^2$, then $p_1(f) = 0$, $p_2(f) = 0$ and $p_3(f) = 1$, so
$$
a_3 = 0.
$$
We now have a system of $3$ equations with $3$ unknowns.
Since this system is non-singular (the matrix $\begin{pmatrix}1 & 0 & 1\\0 & 1 & 1\\0 & 0 & 1\end{pmatrix}$ is non-singular), $a_1 = a_2 = a_3 = 0$. By dimension counting, $p_1, p_2$ and $p_3$ span $V^*$.

Another method is via the isomorphism to the known vector space $\mathbb R^3$ and its corresponding space of functionals. Define $\varphi:\mathbb R^3 \to V$ by
$$
\varphi(a_1e_1 + a_2e_2 + a_3e_3) = a_1 + a_2x + a_3x^2
$$
where $e_1, e_2, e_3$ are standard basis vectors of $\mathbb R^3$.
$\varphi$ induces a map $\varphi^*:V^* \to (\mathbb R^3)^*$ defined by $\varphi^*(f) = f \circ \varphi$.
We will let $\mathbb R^3$ be the space of column vectors and $(\mathbb R^3)^*$ be the space of row vectors. They act by the conventional matrix multiplication.
Then, $p_1, p_2$ and $p_3$ are mapped by $\varphi^*$ to
\begin{align*}
\varphi^*(p_1) & =
\begin{pmatrix}p_1(1) & p_1(x) & p_1(x^2)\end{pmatrix} =
\begin{pmatrix}1 & 0 & 0\end{pmatrix}\\
\varphi^*(p_2) & =
\begin{pmatrix}p_2(1) & p_2(x) & p_2(x^2)\end{pmatrix} =
\begin{pmatrix}0 & 1 & 0\end{pmatrix}\\
\varphi^*(p_3) & =
\begin{pmatrix}p_3(1) & p_3(x) & p_3(x^2)\end{pmatrix} =
\begin{pmatrix}1 & 1 & 1\end{pmatrix}
\end{align*}
These three elements are linearly independent (the matrix $\begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\1 & 1 & 1\end{pmatrix}$ is non-singular), so they form a basis of $(\mathbb R^3)^*$. Since $\varphi$ is an isomorphism, $\varphi^*$ is also an isomorphism.

Title: Prove that functional set is a basis


问:
The national lottery
When playing the lottery you have to pick 6 numbers out of 45 possibilities. Since the order of the numbers don't matter, the number of possible combinations for the jackpot (and hence the 6 correct numbers) is given by:

$
\dbinom{45}{6} = C^6_{45} = \frac{45!}{(45-6)!6!} = 8145060
$

Assuming the numbers are picked at random out of a uniform distribution the chance of winning the jackpot is given by:

$P(win) = \frac{1}{8145060}\approx 0.0000123 \%$

Now I can figure two possible scenarios:

You play every week with the same numbers hoping that one day you'll get the jackpot.

You play every week with a different set of numbers hoping that one day you'll get the jackpot.

Now I was wondering if there is a difference in the chance of winning between the two methods. I was thinking that in the first case the chances of winning are larger because if you stick to your number the alternatives might run out. While if you switch you don't have the other options that become eliminated. But then I started wondering, because in the Monty Hall paradox the chances become larger if you switch.

For the first case I believe the chance of winning in the $n^{th}$ drawing is simply calculated as:

$1-(1-P(win))^n$

But in the second case I wouldn't know how to succeed.

答:
I was thinking along the same lines except not choosing numbers that were in last weeks winning pool. I know the actual ticket you buy and the specific numbers I choose have exactly the same odds as last weeks ticket and that is undisputable.

But its the notion that my choice, my strategy to formulate my number choices based on a pool of numbers feels healthier than random pick or staying the same. After all that pool is a healthier pool from which to choose my numbers statistically even if the odds of me matching the exact number are the same. The odds a number from from each pool respectively.  Pool 1-5=5/50   Pool 6-50=45/50   5/45  As such the decision to choose from that pool seems 9 times healthier than even if statistically the specific numbers themselves provide no advantage.

Title: The national lottery


问:
$V(x,xy)=V(x)$ but also $V(x,xy)=V(xy)$?
In $\operatorname{Spec}k[x,y]$, we have
$$V(x,xy)=V(x)$$
but if we expand the LHS we also get
\begin{align}
V(x,xy)&=V(x) \cap V(xy)\\
&=V(x) \cap (V(x) \cup V(y))\\
&=(V(x) \cap V(x))\cup (V(x) \cap V(y))\\
&=V(x) \cup V(xy)\\
&=V(xy).
\end{align}

$V(x)$ corresponds to the $y$-axis, but $V(xy)$ corresponds to the union of the $x$ and $y$ axes.

What is going on here?

答: Your third and fourth lines aren't equal:  $V(xy) = V(x) \cup V(y)$, not $V(x) \cap V(y)$.  The latter consists of the origin.


问:
Calculate this integral
I am having trouble calculating this integral.

Let $1<p<q<\infty$ and $a,b\geq 0$. Calculate:

$$ \int_{0}^{x} \frac{a}{(1+bt^{\frac{q}{p-1}})^\frac{q}{q-p}}dt $$

It would take 'a' out of the integral. But she wouldn't know how to go on. Could you help me?

Title: Calculate this integral

Body:
Given $0\le a,0\le b,1<p<q,0<x$, we have $0<\frac{q}{p-1}\land0<\frac{q}{q-p}$. Then,

$$\begin{align}
\int_{0}^{x}\mathrm{d}t\,\frac{a}{\left(1+bt^{\frac{q}{p-1}}\right)^{\frac{q}{q-p}}}
&=\int_{0}^{x^{\frac{q}{p-1}}}\mathrm{d}u\,\frac{a\left(\frac{p-1}{q}\right)u^{\frac{p-1}{q}-1}}{\left(1+bu\right)^{\frac{q}{q-p}}};~~~\small{\left[t^{\frac{q}{p-1}}=u\right]}\\
&=a\mu\int_{0}^{x^{1/\mu}}\mathrm{d}u\,\frac{u^{\mu-1}}{\left(1+bu\right)^{\nu}};~~~\small{\left[\mu:=\frac{p-1}{q}\land\nu:=\frac{q}{q-p}\right]}\\
&=a\mu\int_{0}^{y}\mathrm{d}u\,\frac{u^{\mu-1}}{\left(1+bu\right)^{\nu}};~~~\small{\left[y:=x^{1/\mu}>0\right]}\\
&=a\mu\,y^{\mu}\int_{0}^{1}\mathrm{d}t\,\frac{t^{\mu-1}}{\left(1+byt\right)^{\nu}};~~~\small{\left[u=yt\right]}\\
&=a\mu\,y^{\mu}\operatorname{B}{\left(\mu,1\right)}\,{_2F_1}{\left(\nu,\mu;\mu+1;-by\right)}\\
&=ay^{\mu}\,{_2F_1}{\left(\nu,\mu;\mu+1;-by\right)}\\
&=ax\,{_2F_1}{\left(\frac{q}{q-p},\frac{p-1}{q};\frac{p+q-1}{q};-bx^{\frac{q}{p-1}}\right)},\\
\end{align}$$

where we've employed Euler's integral representation for the Gauss hypergeometric function which states

$$\operatorname{B}{\left(\beta,\gamma-\beta\right)}\,{_2F_1}{\left(\alpha,\beta;\gamma;z\right)}=\int_{0}^{1}\mathrm{d}t\,\frac{t^{\beta-1}\left(1-t\right)^{\gamma-\beta-1}}{\left(1-zt\right)^{\alpha}};~~~\small{\left(\alpha,\beta,\gamma,z\right)\in\mathbb{R}^{4}\land0<\beta<\gamma\land z<1}.$$


问:
Binomial identity involving central binomial coefficient
I came across this nice binomial identity

$${\sum}_{k=0}^{2n} \frac{(-1)^k {2n \choose k} {2k \choose k}}{n+k \choose k} = 1$$

This is equivalent to the hypergeometric function ${}_2 F_1(-2n, 1/2; n+1; 4)$. I am having a hard time trying to prove this. Can someone help me with a hint to prove this?

答:
In trying to verify the identity

$$\sum_{k=0}^{2n} (-1)^k
{n+k\choose k}^{-1} {2n\choose k} {2k\choose k} = 1$$

we see that

$${n+k\choose k}^{-1} {2n\choose k}
= \frac{(2n)! / (2n-k)!}{(n+k)! / n!}
= {3n\choose n}^{-1} {3n\choose 2n-k}$$

so that we seek to prove

$$\sum_{k=0}^{2n} (-1)^k {3n\choose 2n-k} {2k\choose k}
= {3n\choose n}.$$

The LHS is

$$\sum_{k=0}^{2n} (-1)^k {3n\choose k}
{4n-2k\choose 2n-k}
= [z^{2n}] (1+z)^{4n}
\sum_{k=0}^{2n} (-1)^k {3n\choose k}
\frac{z^k}{(1+z)^{2k}}$$

Here the coefficient extractor enforces the range of the sum and we
find

$$[z^{2n}] (1+z)^{4n}
\sum_{k\ge 0} (-1)^k {3n\choose k}
\frac{z^k}{(1+z)^{2k}}
= [z^{2n}] (1+z)^{4n}
\left(1-\frac{z}{(1+z)^2}\right)^{3n}
\\ = [z^{2n}] \frac{1}{(1+z)^{2n}}
(1+z+z^2)^{3n}.$$

Expanding the second powered term

$$[z^{2n}] \frac{1}{(1+z)^{2n}}
\sum_{q=0}^{3n} {3n\choose q} (1+z)^{3n-q} z^{2q}$$

The coefficient extractor sets the upper limit of the sum to $n$ and we
get (note that the powers of $1+z$ do not have a pole at zero hence the
expansion about zero starts with $z^{2q}$ and there is no  contribution
to $[z^{2n}]$ when $q\gt n$):

$$[z^{2n}] \sum_{q=0}^{n} {3n\choose q} (1+z)^{n-q} z^{2q}
= \sum_{q=0}^{n} {3n\choose q}
{n-q\choose 2n-2q} = {3n\choose n}.$$

Observe that the the power $n-q$ to which $1+z$ is raised is a
non-negative integer and hence we are justified  in writing $[z^{2n}]
z^{2q} (1+z)^{n-q} = [z^{2n-2q}] (1+z)^{n-q} = {n-q\choose 2n-2q}.$  The
only $q$ in the range $0\le q\le n$ where this binomial coefficient  is
not zero is $q=n$, producing a contribution of ${3n\choose n}$ and we
have the claim.


答: You know that $| p^{-1}(a)| \leq |\pi _1(X,a) | $. Thus, if the fiber $p^{-1}(a)$ is infinite, we are done. So let us consider the case of a finite fiber with $n$ elements. We can compose $p$ with itself arbitarily many times to get surjections $p^m : X \to X$. It is well-known that they are coverings; see Question on covering spaces . The fiber of $p^m$ has $n^m$ elements. It is well-known that $p^m_*(\pi_1(X,a))$ is a subgroup of $\pi_1(X,a)$ of index $n^m$. Since $m$ is arbitrarily large, we conclude that $\pi_1(X,a)$ must be infinite.

Title: When the fundamental group of a space is Infinite

Body:
I am not sure how to proceed with this problem:

Let $X$ be a path-connected and locally path-connected space. Suppose $p:X\to X$ is a non-injective covering map.
Prove that $\pi _1(X) $ is infinite for any choice of base point $a\in X$.

Pick $a\in X$ arbitrarily. I know that since $X$ is path connected, $| p^{-1}(a)| \leq |\pi _1(X,a) | $. Then since $p$ is a non-injective covering map, $|p^{-1}(a)|>1 $. (In fact, all fibers have the same cardinality.)

Is there some way to show $|\pi _1 (X,a)| $ is infinite by using deck transformations?

Or, is it possible to show that if a space covers itself, then its fundamental group is either trivial or infinite?

Any suggestions would be helpful.


问:
Help proving that there exists $a^i\equiv a^j (mod \space n)$ with $i\ne j$
I have to prove the following theorem for my course on Number Theory,

Let $a$ and $n$ be natural numbers. Then there exist natural numbers $i$ and $j$, with $i\ne j$, such that $a^i\equiv a^j (mod \space n)$ with $i\ne j$

I have absolutely no idea how to prove this. Please note that I have not learned Fermat's Little Theorem yet and cannot use it in the proof. Could I receive some generous help and hints please?

答:
Every integer number $m$ can be written uniquely in the form $m = qn + r$, where $0 \leq r < n$. So, one can group together all the integer numbers with the same residue $r$ after division by $n$. Then, we get the set $[0]=\{kn:k\in \mathbb{Z}\}, \space [1]=\{kn+1:k\in \mathbb{Z}\}$, ... , $[n-1]=\{kn + (n-1):k\in \mathbb{Z}\}$.

Consider the set $A = \{a^i:i=0,\dots,n\}$. This set has cardinality $n+1$.

But there are only $n$ residue classes: $[0],[1],...,[n-1]$.

Then, by the pigeonhole principle, we conclude that there are two elements in this set $A$, say $a^i,a^j$, that belong to the same class $[b]$. In other words, there exist $i,j=0,\dots,n$, $i\neq j$, such that $a^i\equiv_n b \equiv_n a^j$.


问:
Considering there exists a homomorphism between two rings show that a condition must exist.
$R$ is defined as \begin{bmatrix}a&b\\b&a\end{bmatrix} such that $a,b \in \mathbb{Z}$ where R is a subring of all 2-by-2 matrices with integer inputs.

There exists a homomorphism $θ :R→Z$ such that $θ(R) = ra + sb$.

You are asked to prove that must $r = 1$ and $s = ± 1$.

I know the 2 points that must be satisfied for a homomorphism to happen between two rings and don't think these help to proving the question. I think you need to start with using the idea that $\ker(θ)$ is an ideal of $R$ and the $im(θ)$ is a subring of $\mathbb{Z}$. Am I right or is this wrong?

Thank you.

答:
Assume $\theta:R\to\Bbb Z$ is a homomorphism of unital rings, mapping
$\theta\pmatrix{a&b\\b&a}=ra+sb$.

Then specifically it maps the identity element ($I=\pmatrix{1&0\\0&1}$, i.e. when $a=1,\ b=0$) to $1$, which means $r=1$.

On the other hand, observe that for $B:=\pmatrix{0&1\\1&0}$ we have $B^2=I$, so we must also have $\theta(B)^2=\theta(I)=1$, where $\theta(B)=s$.


问:
Help proving there exists a natural number $k$ such that $a^k\equiv 1 (mod \space n)$
I need to prove the following theorem for my Number Theory class,

Let $a$ and $n$ be natural numbers with $gcd(a,n)=1$. Then there exists a natural number $k$ such that $a^k\equiv 1 (mod \space n)$

I'm immensely awful at proofing so I have absolutely no idea how to prove this theorem. I haven't learned Fermat's Little Theorem yet so I can't use it in the proof either (if it's even applicable). But I do know how we define a residue systems modulo $n$, if that's useful. Could I please receive some guidance?

答:
Consider the powers $a^1, a^2, a^3,\dots$.  They can't all be different, because there are only finitely many residue classes, so there exist $0<j<k$ with $a^j=a^k$.

Can you take it from here?

Hmmm.  I guess I'm assuming you've learned about the extended Euclidean algorithm, or Bezout's theorem.


问:
2nd Order Differential equation with Unit Step and initial conditions
I have some work but I can't get it to work quite yet, I need to solve this so I can plot it versus time.

The function is $90y''+30y'+1000(y-z(t))=0$,

$$z(t)= \begin{cases}0 & 0\leq t\leq5/9 \\ -\frac18 & 5/9<t\leq40/9\\0 & 40/9\leq t\leq 20/3\end{cases}$$
with $y(0)=0$ and $y'(0)=0$. I know $z(t)$ is a heaviside function but I am having difficulty moving forward. I've tried to simplify the main function to $90/1000y''+30/1000y'+y=z(t)$, and the heaviside notation I have is $z(t)=-\frac18u_{5/9}(t)$.

Help would be greatly appreciated. I apologize for the poor notation, I am new here and I do not know how to properly put in my functions.

答:
If you're interested in an analytical solution . . .

Your ODE is
\begin{align}
\ddot{y}+\alpha \dot y+\beta y=\beta z,
\end{align}
where I've defined
\begin{align}
\alpha&=\frac{3}{9}\\
\beta&=\frac{100}{9}
\end{align}
for brevity.
Substituting
\begin{align}
x(t)=\gamma y+\dot y
\end{align}
where $\gamma$ satisfies
\begin{align}
\gamma^2-\alpha \gamma+\beta=0
\end{align}
reduces your ODE to
\begin{align}
\dot x+\frac{\beta}{\gamma}x=\beta z.
\end{align}
This has the solution
\begin{align}
x=e^{-\beta t/\gamma}\left(k_1+\int\beta e^{\beta t/\gamma}z\mathrm dt\right).
\end{align}
For the right integral
\begin{align}
-\frac{1}{8}\int \beta e^{\beta t/\gamma}(\mathrm H_{5/9}-\mathrm H_{40/9})\mathrm dt,
\end{align}
using integrating by parts
\begin{align}
-\frac{\gamma}{8}\left((e^{\beta t/\gamma}-e^{5\beta/9\gamma})\mathrm H_{5/9}-(e^{\beta t/\gamma}-e^{40 \beta/9\gamma})\mathrm H_{40/9}\right).
\end{align}
So
\begin{align}
\dot y+\gamma y=k_1e^{-\beta t/\gamma}+\frac{\gamma}{8}\left((\mathrm{exp}\left[\frac{\beta}{\gamma}\left(\frac{5}{9}-t\right)\right]-1)\mathrm H_{5/9}-(\mathrm{exp}\left[\frac{\beta}{\gamma}\left(\frac{40}{9}-t\right)\right]-1)\mathrm H_{40/9}\right)\equiv f(t)
\end{align}
and
\begin{align}
y=e^{-\gamma t}\left[k_2+\int e^{\gamma t}f(t)\mathrm dt\right],
\end{align}
where you could solve the right integral using integration by parts again.


答:
I refer to you're other question Help proving that there exists $a^i\equiv a^j (mod \space n)$ with $i\ne j$

There are $0\leq i<j$ such that $a^i\equiv_n a^j \equiv_n a^{j-i} a^i$.

On the other hand, $a^i$ and $n$ are coprime, so there are $x,y$ such that $a^i x + n y = 1$. Looking to the residues mod $n$, we get $a^i x \equiv_n 1$.

Therefore, $1 \equiv_ n a^i x \equiv_n a^j x \equiv_n a^{j-1} a^i x \equiv_n a^{j-i}$.

Title: Help proving there exists a natural number $k$ such that $a^k\equiv 1 (mod \space n)$

Body:
I need to prove the following theorem for my Number Theory class,

Let $a$ and $n$ be natural numbers with $gcd(a,n)=1$. Then there exists a natural number $k$ such that $a^k\equiv 1 (mod \space n)$

I'm immensely awful at proofing so I have absolutely no idea how to prove this theorem. I haven't learned Fermat's Little Theorem yet so I can't use it in the proof either (if it's even applicable). But I do know how we define a residue systems modulo $n$, if that's useful. Could I please receive some guidance?


答:
Take $n$ numbers $a^1,...,a^n$.

Because $\gcd(a,n)=1$ then $a^i \not \equiv 0 \pmod n $ for all $i=1,...,n$. Hence there are $n-1$ possible values from $1$ to $n-1$ for $n$ numbers $a^1,...,a^n$. According to the Pigeonhole principle, there exists $i,j \in \{1,...,n\}$ such that $i \ne j$ and
$$a^j \equiv a^i \pmod n\implies a^i(a^{j-i}-1) \equiv0 \pmod n \implies a^{j-i} \equiv 1 \pmod n$$

Let's take $k = j-i$, then $a^{k} \equiv 1 \pmod n$


问:
Find a vector in the column space of both matrices?
Question taken from Strang's Introduction to Linear Algebra:

The floor $\boldsymbol{V}$ and the wall $\boldsymbol{W}$ are not orthogonal subspaces, because they share a nonzero vector (along the line where they meet). No planes V and W in $\mathbf{R}^{3}$ can be orthogonal. Find a vector in the column spaces of both matrices:

$$
A=\left[\begin{array}{ll}1 & 2 \\ 1 & 3 \\ 1 & 2\end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{ll}5 & 4 \\ 6 & 3 \\ 5 & 1\end{array}\right]
$$

This will be a vector $A \boldsymbol{x}$ and also $B \widehat{\boldsymbol{x}}$. Think 3 by 4 with matrix $\left[\begin{array}{ll}A & B\end{array}\right]$.

Text Solution: $A \boldsymbol{x}=B \widehat{\boldsymbol{x}}$ means that $\left[\begin{array}{ll}A & B\end{array}\right]\left[\begin{array}{r}\boldsymbol{x} \\ -\widehat{\boldsymbol{x}}\end{array}\right]=\mathbf{0}$. Three homogenous equations in four unknowns always have a nonzero solution. Here $\boldsymbol{x}=(3,1)$ and $\widehat{\boldsymbol{x}}=(1,0)$ and $A \boldsymbol{x}=B \widehat{\boldsymbol{x}}=(5,6,5)$ is in both column spaces. Two planes in $\mathbf{R}^{3}$ must share a line.

My Question: I'm not following the problem at all.  I understand the point with regard to the wall and floor.  I understand $ n + (n - r) $ has to span all of n, but I do not follow how these two matrices deal with that.  Where does $A x=\widehat{B x}$ even come from and how does it lead to the expression $\left[\begin{array}{ll}A & B\end{array}\right]\left[\begin{array}{r}x \\ -x\end{array}\right]=0$?

Edit: If a vector is in the column space of both matrices, then I understand that the vector would have to exist as the right hand side of Ax and Bx? (It would be a different vector x I presume.)

答:
When finding common vectors in the column spaces of the matrices $A$ and $B$ (since in this case you only have two vectors(columns) per matrix you know they are not linearly dependent because they are not multiples of one another)

you are looking for common vectors in the spans of the form $a_1 \begin{bmatrix} 1\\1\\1\end{bmatrix}+a_2\begin{bmatrix} 2\\3\\2\end{bmatrix} = A\begin{bmatrix} a_1\\a_2\end{bmatrix} = A\vec{x} = b_1\begin{bmatrix} 5\\6\\5\end{bmatrix} + b_2\begin{bmatrix} 4\\3\\1\end{bmatrix} = B\begin{bmatrix} b_1\\b_2\end{bmatrix}=B\vec{y}$

(can a linear combination of the column vectors of $A$ equal a linear combination of the column vectors of $B$, i.e. the spaces share a common vector)

and $[\: A \:\: B \:] \begin{bmatrix} \vec{x}\\ -\vec{y}\end{bmatrix} = \vec{0}$ means essentially the same thing :=

$\begin{bmatrix} 1&2&5&4\\1&3&6&3\\1&2&5&1\end{bmatrix}\begin{bmatrix} a_1\\a_2\\-b_1\\-b_2\end{bmatrix} = a_1 \begin{bmatrix} 1\\1\\1\end{bmatrix}+a_2\begin{bmatrix} 2\\3\\2\end{bmatrix} - b_1\begin{bmatrix} 5\\6\\5\end{bmatrix} - b_2\begin{bmatrix} 4\\3\\1\end{bmatrix} = A\vec{x} -B\vec{y}=\vec{0}$

Since you are in $3$-space and you have two linearly independent vectors for each column space of $A$ and $B$, the subspaces generated by the span of those column vectors will be planes (if you had just one vector, the subspace would be a line, three linearly independent vectors, all of $3$-space), and as mentioned by the textbook answer, there are exists at least one nonzero solution (every homogeneous set of equations is consistent, has the zero solution) to the homogeneous set of equations

$$[\: A \:\: B \:] \begin{bmatrix} \vec{x}\\ -\vec{y}\end{bmatrix} = \vec{0} $$

because there are more unknowns than equations. So the planes either intersect in a line, or completely overlap.

This means the column spaces are not orthogonal by the definition of orthogonal because since they intersect, you can find a nonzero vector $\vec{v}$ s.t. $\vec{v}\in \text{col}(A),\text{col}(B)$ and $\vec{v} \cdot \vec{v} = ||\vec{v}||^2 \not= 0$.


问:
$\lbrace S_n\rbrace_{n\in \mathbb{N}}$ is bounded , $\lim Sup \, S_n=\lim Inf\, S_n=s$ then $\lbrace S_n\rbrace_{n\in \mathbb{N}}$ converges to $s$.
If $\lbrace S_n\rbrace_{n\in \mathbb{N}}$  is bounded and $\lim Sup \, S_n=\lim Inf\, S_n=s$ then $\lbrace S_n\rbrace_{n\in \mathbb{N}}$ converges to $s$.

Proof

Let us denote by $S$ be the set of subsequential limits, since $\lim Sup \, S_n=\lim Inf\, S_n=s$ then  by definition of infimum.
$\forall l\in S$ we have that $l\geq s$ and if $s^{\prime}$ is other number such that $l\geq s^{\prime}$ then $s^{\prime} <s$

And by definition of supremum
$\forall l\in S$ we have that $l \leq s$ and if $s^{\prime}$ is other number such that $l\leq s^{\prime}$ then $s^{\prime} >s$.

From here $\forall l\in S$ $l=s$ and hence every subsequence of $\lbrace S_n \rbrace $ have limit $s$ hence $\lbrace S_n \rbrace $ is convergent sequence and in fact their limit is $s$

What is wrong with this proof?
Any comment was very useful, thanks in advice

答:
HINT

Here it is another way to approach it for the sake of curiosity.

Notice that $m_{j}\leq S_{j} \leq M_{j}$, where
\begin{align*}
\begin{cases}
m_{j} = \inf\{S_{n}\in\mathbb{R}\mid n\geq j\}\\\\
M_{j} = \sup\{S_{n}\in\mathbb{R}\mid n\geq j\}
\end{cases}
\end{align*}

Then apply the squeeze theorem.


问:
Show that $e^x\ge x^e$ for all $x\ge 0$
Show that $e^x\ge x^e$ for all $x\ge0.$

I was playing around with splitting a number into parts (that add to the original number) and then multiplying the parts to get the result, and I found that the product is highest when the parts tended towards $e$. This inequality is a consequence of that. Is there a proof or intuitive explanation for this?

答: It suffices to prove for all $x\neq 0$ $$f(x)=x-e\log x\ge 0$$ $$f'(x)=0\to 1-\frac{e}{x}=0\to x=e$$ also  here`$f''(e)>0$ so $$f(x)\ge f(e)=0$$ note that $$\lim_{x\to 0^+}f(x),\lim_{x\to \infty} f(x)>0$$

Title: Show that $e^x\ge x^e$ for all $x\ge 0$


问:
set notation when showing that $X_n$ is a Markov chain
In Durrett's probability theory exercise 5.1.1,:

Let $\xi_1, \xi_2, \dots$ be i.i.d. $\in \{ 1, 2, \dots, N\}$ and taking each value with probability $1/N$. Show that $X_n = |\{\xi_1, \dots, \xi_n\}|$ is a Markov chain and compute its transition probability.

I don't understand what $X_n = |\{\xi_1, \dots, \xi_n\}|$ is. Is it the sum of $\xi_i$'s or is it the cardinality of the set?

答: It's the cardinality of the set. In other words, $X_n$ counts the number of unique values among $\xi_1, \dots, \xi_n$.


问:
Confused about mathematically (formally) finding a conditional expectation in a double expectation problem versus in words
Textbook example: Losses on an insurance policy are uniformly distributed on [0, 10000]. A policy covering these losses has a 1000 deductible. Calculate the average payment made by this policy for each loss.

Textbook solution: Let $X$ be the loss, $Y = \text{max}\{0, X - 1000\}$ be the payment. Condition on the loss being above or below 1000. Then

$$E[Y] = P[X \leq 1000]E[Y| X \leq 1000] + P[X \geq 1000]E[Y|X \geq 1000].$$

If the loss is below 1000, then the payment is 0, so the mean payment is 0. If the loss is greater than 1000, then the payment is uniform on [0, 9000], so the mean payment is 4500.

$$E[Y] = P[X \leq 1000](0) + P[X \geq 1000](4500)$$

The probability that $X$ is greater than 1000 is $1 - F(1000) = 1 - 1000/10000 = 0.9$. Therefore

$$E[Y] = (0.9)(4500) = 4050.$$

My solution: Now if I was doing this on my own I would have proceeded as follows:

$X$ follows a uniform distribution on [0, 10000] so that the pmf of $X$ is 1/10000.  The payment amount $Y$ is a piecewise function defined in terms of $X$.  It is $0$ if $X \leq 1000$ and it is $X - 1000$ if $X > 1000$.

Using double expectation we have that

$$E[Y] = E_X[E[Y|X]] = P[X \leq 1000]E[Y| X \leq 1000] + P[X > 1000]E[Y|X > 1000].$$

The probabilities $P[X \leq 1000]$ and $P[X > 1000]$ are gotten by integrating 1/10000 over the intervals $[0,1000]$ and $[1000, 10000]$, respectively, to get $1/10$ and $9/10$.

Question: But now I'm confused about how to formally find the remaining conditional expectations.  By my definition of conditional expectation in the continuous case, which is new to me and have never used for inequalities, I would have that

$$E[Y| X \leq 1000] = \int_0^{9000} y \cdot f_{Y|X}(Y|X \leq 1000)\, dy.$$

And for $f_{Y|X}(Y|X \leq 1000)$ I have by my definition of conditional distribution that

$$f_{Y|X}(Y|X \leq 1000) = \frac{f(y,x)}{f_X(x \leq 1000)}.$$

If this is correct then how would I find the numerator?  The denominator is just $\int_0^{1000} 1/10000 \, dx = 1/10$.

答:
If you are going to integrate, don't use Total Expectation unless needed.  You don't need it.

$Y$ is a function of $X$, so by the Law of the Unconscious Statastitian:

$$\begin{align}\mathsf E(Y) &=\mathsf E(\max\{0,X-1000\})\\[1ex]&= \int_0^{10000} \max\{0, x-1000\} \cdot\dfrac 1{10000}\,\mathrm d x\\[1ex]&=\int_0^{1000} 0\mathrm d x+\int_{1000}^{10000}\dfrac{x-1000}{10000}\,\mathrm d x\\[1ex]&=4050 \end{align}$$

$\blacksquare$


问:
Is this integral increasing?
I have a function $f(t)$ decreasing and a function $g(t)$ nonnegative.

I know for grant that $\int_0^t f(t-\theta)g(\theta)d\theta$ is increasing (I mean, this is part of hypothesys).

Now, I have $h(t)=f(t)k(t)$ where both $f(t) $ and $k(t)$ are decreasing, but $k(t)>1$.

I could say that is $\int_0^t h(t-\theta)g(\theta)d\theta$ increasing?

I know that $\int_0^t h(t-\theta)g(\theta)d\theta>\int_0^t f(t-\theta)g(\theta)d\theta$, but how can I prove that there is not oscilation?

Thank you so much.

答:
It's not true. Here is a counterexample. Let $f(t)=g(t)=e^{-t}$, and $k(t)=1+e^{-t}$. Then $f$ is decreasing, $g$ is non-negative, $k$ is decreasing, and $k>1$. Then,

$$\int_0^t f(t-\theta)g(\theta)d\theta =\int_0^t e^{-t+\theta}e^{-\theta}d\theta = te^{-t} $$

and

$$\int_0^t h(t-\theta)g(\theta)d\theta =\int_0^t (1+e^{-t+\theta})e^{-t+\theta}e^{-\theta}d\theta $$
$$ = \int_0^t (e^{-t}+e^{-2t+\theta}) d\theta $$
$$ = te^{-t} + e^{-2t} \int_0^t e^{\theta} d\theta $$
$$ = te^{-t} + e^{-2t} (e^t-1) $$
$$ = (t+1)e^{-t}-e^{-2t} $$
The derivative wrt $t$ is
$$ \frac{d}{dt} \int_0^t h(t-\theta)g(\theta)d\theta = 2e^{-2t}-te^{-t} $$
Then
$$ 2e^{-2t}-te^{-t}=0  \iff te^{t}=2 $$
We have that $\int_0^t h(t-\theta)g(\theta)d\theta$ is increasing for $0<t<W(2)$ and decreasing for $t>W(2)$, where $W$ is the Lambert W-function and $W(2)\approx 0.853...$

Title: Is this integral increasing?


问:
Graph : Node addressing for easier path finding
I'm looking for a way of naming the node of my graph such that knowing only the name of the node where I stand and the name of its direct neighbor, the name of the node I want to go and an algorithm I can find which node to hop to get closer to my destination node such that if I did this enough time I would effectively end up on the arrival node while crossing the least amount of node.

Here is an example that work well for a specific topology : I generate my graph by adding n node to the previously generated nodes and repeat the process as many times as we need. We end up with a fractal like graph. I would then name the central node 0. The first layer of node 0.0 - 0.n the second 0.0.0 - 0.n.n etc. I would then have to find the node that have the most matching digit with mine (with respect to their place) and if none is closer get one hop closer to the center. I am 0.3.4 and I want to go to 0.2.6. I would first choose 0.3, then 0 then 0.2 the 0.2.6

how could i generalize this to different topology like a small world network ? Or alternatively have a proof that it is not possible.

please excuse my clumsy way of describing the question.

答:
In a general graph $G = (V,E)$, there's an obvious way of doing this : for each vertex $v$, precompute its distance to every other vertex, and store it in the "name" of the vertex (in graph theory we rather speak of labels but the idea is there). Then, when you want to find your path to a destination, compare your neighbourg's distance to it, and choose the best.

This is good, but it actually makes long names, and you have to precompute your distances, which is long (about $\Theta(|V||E|)$).

But unless you have better properties in your graph, there's probably nothing better to do. It's hard to prove though.