答: Yes, $\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$ is a UFD because it is norm-Euclidean.

Body:
Consider the integral domain  $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. Is $(4+\sqrt{5})$ a prime ideal of  $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$?

I know the following elementary facts. We have
\begin{equation}
\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right] = \left\{ \frac{m + n \sqrt{5}}{2} : m, n \in \mathbb{Z} \text{ are both even or both odd} \right\}.
\end{equation}

For every $\frac{m + n \sqrt{5}}{2} \in \mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$, define its norm as usual:
\begin{equation}
N\left(\frac{m + n \sqrt{5}}{2}\right)=\frac{m^2-5n^2}{4}.
\end{equation}
Since $m, n$ are both even or both odd, it is easy to see that the norm is an integer. From this fact it is easily seen that $\frac{m + n \sqrt{5}}{2}$ is a unit of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ if and only if $m^2 - 5n^2=4$ or $m^2 - 5n^2=-4$. Now since $N(4+\sqrt{5})=11$ we easily get that $4+\sqrt{5}$ is an irreducible element of  $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. If $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ were a unique factorization domain, we could conclude that $(4+\sqrt{5})$ a prime ideal of  $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. But I do not know if $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ is a unique factorization domain. Does someone know if it is?

Thank you very much in advance for your attention.


问:
(Non) equivalence of regular cardinal definitions
The usual definition of a regular cardinal is "$\kappa$ is regular if $cf(\kappa) = \kappa$", which, assuming the axiom of choice, is equivalent to this definition: "$\kappa$ is regular iff it cannot be expressed as a union of a less than $\kappa$ sets, all of which are of cardinality less than $\kappa$."

Now consider Definition 2: a cardinal $\kappa$ is regular if for every subset $X \subset \kappa$ such that $|X| < \kappa$ we have $\bigcup X < \kappa$.

I think these definitions are not equivalent in ZF, because $X$ and $\bigcup X$ can be well-ordered without invoking the axiom of choice. In particular, I guess one could prove in ZF that with Definition 2 all successor cardinals are regular, which is not true with usual definition.

Questions: Is my intuition correct? How can I show the non-equivalence?

答:
Let me rewrite the answer completely.

You wrote three definitions for a regular cardinal, and you made two false claims. First let me write the definitions, so we'll be clear on those.

Let $\kappa$ be an infinite initial ordinal (i.e. a cardinal in the context of $\sf ZFC$).

$\kappa$ is $1$-regular if every unbounded subset of $\kappa$ has order type $\kappa$.

$\kappa$ is $2$-regular if every union of less than $\kappa$ subsets of $\kappa$, each of cardinality less than $\kappa$, is itself of cardinality less than $\kappa$.

$\kappa$ is $3$-regular if every $X\subseteq\kappa$ such that $|X|<\kappa$ has the property that $\bigcup X<\kappa$.

You also claimed that $1$ and $2$ are equivalent if we assume the axiom of choice, and you wanted to show that $3$ is also inequivalent to those without the axiom of choice. But the truth is that the axiom of choice is not needed in order to show that all three are equivalent.

If $\kappa$ is $1$-regular, then whenever $X\subseteq\kappa$ and $|X|<\kappa$ the order type of $X$ (as a well-ordered set) has to be less than $\kappa$, so $\bigcup X=\sup X<\kappa$ by the assumption that it is $1$-regular.

If $\kappa$ is $3$-regular, let $A$ be an unbounded subset of $\kappa$ then $\bigcup A=\sup A=\kappa$, then by the definition $3$-regular we have to have that $|A|=\kappa$, but since $\kappa$ is an initial ordinal it does not have subsets of size $\kappa$ whose order type is not $\kappa$ itself. Therefore $A$ has order type $\kappa$.

This establishes the equivalence between $1$ and $3$. There was absolutely no use of the axiom of choice. $\bigcup X$ is an ordinal because it is the union of ordinals.

Now to show that these are equivalent to $2$-regularity.

If $\kappa$ is $2$-regular, let $X\subseteq\kappa$ such that $|X|<\kappa$. Since $\kappa$ is an initial ordinal, all the members of $X$ have cardinality strictly less than that of $\kappa$. Therefore we take union of less than $\kappa$ sets of size less than $\kappa$, and by $2$-regularity $|\bigcup X|<\kappa$. But this means that $\sup X<\kappa$, as wanted.

If $\kappa$ is $3$-regular, let $P$ be a set of less than $\kappa$ subsets of $\kappa$, each of cardinality $\kappa$. Then for every $A\in P$ we have that $\bigcup A=\sup A<\kappa$ by the definition $3$-regular. Let $X=\{\sup A\mid A\in P\}$, then $|P|\leq|X|<\kappa$ and therefore $\bigcup X<\kappa$. It follows that $\bigcup P<\kappa$ as well, because there is some $\alpha>\bigcup X$ and that means that $\alpha\notin A$ for every $A\in P$. Therefore $\kappa$ is $2$-regular as wanted. $\square$

In the comments you remarked that your teacher challenged with the following task, if $X\subseteq\aleph_{\alpha+1}$ and $|X|<\aleph_{\alpha+1}$ then you are to find an injection from $\bigcup X$ into $\aleph_\alpha\times\aleph_\alpha$.

But that means that you have to assume that $\aleph_{\alpha+1}$ is regular. Whereas in $\sf ZFC$ every successor cardinal is indeed regular, it is consistent with $\sf ZF$ that successors are not regular. In fact, assuming very large cardinals are consistent, we can construct a model of $\sf ZF$ in which there are no regular cardinals except $\aleph_0$.

So in order to find that injection one has to assume that $\aleph_{\alpha+1}$ is regular, which is an assumption we cannot prove without the help of the axiom of choice, and therefore we cannot write an explicit injection from $\sup X$ into $\aleph_\alpha\times\aleph_\alpha$.

ContentLicense: CC BY-SA 3.0


问:
Solving for point P within the edge AC, given points A, B and C, and length from B to P?

(EDIT: one of the comments noted that AC is too short in the image, this is completely my fault for not checking before rounding, therefor replace the magnitude of AC to, say 10)

So given points A B and C we can calculate the magnitudes of AB, AC, and CB, and
we know there is a point in the line of AC that has the distance of, say 60 like in the image,
to the point B. What are the necessary steps to solve for point P?

答:
Note
\begin{align}
AP &= AB \cos A -BP \cos \angle BPC \\
&= AB \cos A -BP \sqrt{ 1- \sin^2 \angle BPC}\\
&= AB \cos A -BP \sqrt{ 1- \sin^2 \angle BPA} \\
 &= AB \cos A -BP \sqrt{ 1- \frac{AB^2}{BP^2}\sin^2A}\>\>\>\>\>\>\>(sine \>rule)\\
 &= AB \cos A -\sqrt{ BP^2 - {AB^2}(1-\cos^2A)}\\
\end{align}
where $AB$ and $BP$ are given and from the cosine rule
$$\cos A =\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}
$$

ContentLicense: CC BY-SA 4.0


问:
Where $Arg(z)$ is differentiable?
Let $f(z)=Arg(z)$ denotes the principal argument of $z$. If $f(z)=u(x,y)+iv(x,y)$, then for $x\neq0$, we obtain that $u(x,y)=\arctan(y/x)+\{-\pi,0,\pi\}$ and $v(x,y)=0$, so
$$
u_x(x,y)=-\frac{y}{x^2+y^2}\quad
u_y(x,y)=\frac{x}{x^2+y^2}
$$
and $v_x(x,y)=v_y(x,y)=0$. Hence, the Cauchy Riemann equations does not hold unless $(x,y)=(0,0)$. Therefore $f$ is not differentiable at every point with $x\neq0$.

Comment: we say that $f$ is differentiable at $z_0$ if the following limit exists
$$
\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}
$$

(a) What about differentiable in points with $x=0$ ?

(b) How to prove that $f(z)=Arg(z)$ is not differentiable without using the Cauchy Riemann equations?

答:
This is just a partial solution, but maybe it is of interest: Fix $z_0$ and consider the difference quotient

$$\frac{f(z_0+h)-f(z_0)}{h} = \frac{\text{Arg}(z_0+h)-\text{Arg}(z_0)}{h}$$

If we let $h \rightarrow 0$ along the line (or ray) $\text{Arg}(z) = \text{Arg}(z_0)$, this expression clearly tends to $0$. This at leasts shows that $f$ cannot be differentiable at any open set: If it were, the derivative would be zero and $f$ would be contant on that open set. Obviously $\text{Arg}(z)$ is not constant on any open set. So if $\text{Arg}(z)$ is differentiable somewhere, it will be in isolated points.

Title: Where $Arg(z)$ is differentiable?


问:
Isomorphism between $R = \mathbb{R}[X, Y]/(X^2 + Y^2 - 1)$ and $IJ$ where $I = (x - 1, y)$ and $J = (y, x - 1)$
Let $R = \mathbb{R}[X, Y]/(X^2 + Y^2 - 1)$ and $I = (x - 1, y)$, $J = (x, y - 1)$, where $x = X + (X^2 + Y^2 - 1)$ and $y = Y + (X^2 + Y^2 - 1)$. I have managed to prove that $I + J = R$ and $IJ = (x + y - 1)$. I tried to use some isomorphism theorems, but they don't seem to work. Any suggestions how to prove that $R \cong (x + y - 1) = IJ$?

答: This is true in far more generality: Let $R$ be a commutative ring, and let $x \in R$ be any element. Then we have an $R$-module homomorphism $R \to (x)$ defined for each $r \in R$ by $r \mapsto rx$. This map is surjective by the definition of $(x)$, and is injective if and only if $x$ is not a zero-divisor. In other words: any principal ideal generated by a non-zero-divisor is isomorphic as an $R$-module to $R$.


问:
How to prove 'if $A\models B$, then $\models A\Rightarrow B$ ' by contradiction?
I need to find a book or an article with a proof (by contradiction) of the following theorem:

'if $A\models B$, then $\models A\Rightarrow B$'

I need it because I need to check if my proof is ok.

It has to be proof of semantic version, not syntactic.

EDIT: PLEASE CHECK MY PROOF :)

Let's (1) assume that $\alpha \models \beta$. Let's assume (2) that $\not\models \alpha\Rightarrow \beta$. From (2) we have: 'exists interpretation $v$ such that $v(\alpha)=1$ and $v(\beta)=0$'. From (1) we have: 'for any interpretation $v$: if $v(\alpha)=1$, then $v(\beta)=1$'. We remember that from (2) we have 'there exists interpretation $v$ such that $v(\alpha)=1$', so adding this information to (1) we get that for any $v$: $v(\beta)=1$. So we have contradiction: 'there exists $v$ such that $v(\beta)=0$' contradicts 'for any $v$: $v(\beta)=1$'.

答:
I can't offer you a book or an article that offers a proof of that specific result by contradiction. Here is a proof by contradiction:

If $\not\models A \Rightarrow B$, then there is a structure $M$ such that $M \models A$ but $M \not\models B$. So $A \models B$ (which means that $B$ holds in any model in which $A$ holds) must be false. Hence if $A \models B$, then $\models A \Rightarrow B$.

A direct proof is (to my mind) easier and more natural: if $A \models B$, then $B$ holds in any model in which $A$ holds, but then in any model, either $A$ does not hold or $B$ does, i.e., $\models A \Rightarrow B$.


问:
Multiplication of determinants
Show that for any vectors $\bf{a}$,$\bf{b}$,$\bf{c}$,$\bf{u}$,$\bf{v}$,$\bf{w}$ in $\mathbb{R}^3$,

$$[\bf{a},\bf{b},\bf{c}][\bf{u},\bf{v},\bf{w}]=\begin{vmatrix}\bf{a}\cdot\bf{u}&\bf{a}\cdot\bf{v}&\bf{a}\cdot\bf{w}\\
\bf{b}\cdot\bf{u}&\bf{b}\cdot\bf{v}&\bf{b}\cdot\bf{w}\\
\bf{c}\cdot\bf{u}&\bf{c}\cdot\bf{v}&\bf{c}\cdot\bf{w}\\
\end{vmatrix}$$

My question: is there a more elegant way to prove that without actually delving into tedious direct approach by definition?

I was considering that "summing" approach, using the fact that

$$[\textbf{a},\textbf{b},\textbf{c}][\textbf{u},\textbf{v},\textbf{w}]=\bigg(\sum^3_{i,j,k=1}\epsilon_{ijk}a_ib_jc_k\bigg)\bigg(\sum^3_{i,j,k=1}\epsilon_{ijk}u_iv_jw_k\bigg)$$

But I am very weak with manipulations of sums and can't get much further. Any hints?

Note This problem occured when the text book formalized the corkscrew rule and discussed orientation of vectors in $\mathbb{R}^3$, meaning even matrices has not been introduced yet, as the chapter itself was analyzing vectors in the given space.

答:
First note that if $\bf{a},\bf{b},\bf{c}$ are not linearly independent,
then both sides are zero and we are done.

So we may assume that $(\bf{a},\bf{b},\bf{c})$ is a basis of ${\mathbb R}^3$. For the same reason we may further assume that $(\bf{u},\bf{v},\bf{w})$ is a basis of ${\mathbb R}^3$.Say that a sextuple of vectors $(\bf{a},\bf{b},\bf{c},\bf{u},\bf{v},\bf{w})$ is $\bf good$ if the desired equality holds.

Note that if we multiply one of the vectors in the sextuplet by a constant $\lambda$, then both sides of the equality are multiplied by $\lambda$, so this transformation does not affect goodness.

Similarly, if we make interchange $a$ and $b$, or interchange $u$ and $v$ (but do not
mix up any of $a,b,c$ with any of $u,v,w$), then both sides of the equality are unchanged,
so this transformation does not affect goodness.

Also, if we replace $b$ with $b+a$, or replace $v$ with $v+u$ (but do not
mix up any of $a,b,c$ with any of $u,v,w$), then both sides of the equality are unchanged,
so this transformation does not affect goodness.

Now it is well-known that iterating those transformations allows us to go from
the canonical  basis $(e_1,e_2,e_3)$ of ${\mathbb R}^3$ to any basis, and vice versa (those transformations are all revertible); this is the Gauss method to solve linear systems.

So we can assume without loss of generality that both $(\bf{a},\bf{b},\bf{c})$ and
$(\bf{u},\bf{v},\bf{w})$ are in fact the canonical basis, in which case both sides
of the equality are clearly equal to $1$. This concludes the (computation-free) proof.

Title: Multiplication of determinants


问:
Maclaurin series of $\frac{1}{\sqrt{4x^2 + 1}}$
I am trying to find the MacLaurin 6th degree polynomial for the function $ f(x) = \frac{1}{\sqrt{4x^2 + 1}}$.

My first idea was to derive the series expansion of $\frac{1}{2} \operatorname{arsinh}(2x)$, but that seems too complicated. The next idea was the binomial series. As far as I know, the binomial coefficient was not yet introduced in the lecture.

I am slowly running out of ideas because the task explicitly requires the modification of known series.

答:
For $f(x) = \frac{1}{{(1 + 4x^2)}^{1/2}}=\left(1+ 4x^2 \right)^{-\frac{1}{2}}$ is possible to use Newton's_generalized_binomial_theorem for real power $-\frac{1}{2}$:

$$\left(1+ 4x^2 \right)^{-\frac{1}{2}} = \sum\limits_{k=0}^{\infty}\binom{-\frac{1}{2}}{k}4^{k}x^{2k}$$
where $\binom{n}{k}=\frac{n(n-1)\cdots (n-k+1)}{k!}$


答:
As $4x^2\rightarrow 0$ for $x\rightarrow 0$, you can use the Maclaurin polynomial of third degree of $(1+t)^{-\frac{1}{2}}$ (which is very easy to calculate, also using the definition, as it's not difficult to write the third derivative of $(1+t)^{-\frac{1}{2}}$) and substitute $4x^2$ to $t$.
From
$$
(1+t)^{-\frac{1}{2}}=1-\frac{t}{2}+\frac{3}{8}t^2-\frac{5}{16}t^3+o(t^3)
$$
you get
$$
\frac{1}{\sqrt{4x^2+1}}=(1+4x^2)^{-\frac{1}{2}}=1-2x^2+6x^4-20x^6+o(x^6)
$$

Title: Maclaurin series of $\frac{1}{\sqrt{4x^2 + 1}}$

Body:
I am trying to find the MacLaurin 6th degree polynomial for the function $ f(x) = \frac{1}{\sqrt{4x^2 + 1}}$.

My first idea was to derive the series expansion of $\frac{1}{2} \operatorname{arsinh}(2x)$, but that seems too complicated. The next idea was the binomial series. As far as I know, the binomial coefficient was not yet introduced in the lecture.

I am slowly running out of ideas because the task explicitly requires the modification of known series.


问:
Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$?
Consider the integral domain  $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. Is $(4+\sqrt{5})$ a prime ideal of  $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$?

I know the following elementary facts. We have
\begin{equation}
\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right] = \left\{ \frac{m + n \sqrt{5}}{2} : m, n \in \mathbb{Z} \text{ are both even or both odd} \right\}.
\end{equation}

For every $\frac{m + n \sqrt{5}}{2} \in \mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$, define its norm as usual:
\begin{equation}
N\left(\frac{m + n \sqrt{5}}{2}\right)=\frac{m^2-5n^2}{4}.
\end{equation}
Since $m, n$ are both even or both odd, it is easy to see that the norm is an integer. From this fact it is easily seen that $\frac{m + n \sqrt{5}}{2}$ is a unit of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ if and only if $m^2 - 5n^2=4$ or $m^2 - 5n^2=-4$. Now since $N(4+\sqrt{5})=11$ we easily get that $4+\sqrt{5}$ is an irreducible element of  $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. If $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ were a unique factorization domain, we could conclude that $(4+\sqrt{5})$ a prime ideal of  $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. But I do not know if $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ is a unique factorization domain. Does someone know if it is?

Thank you very much in advance for your attention.

答:
Call $A = \mathbb Z \left[ \frac{1 + \sqrt 5}2\right]$. We can show that $A / (4+\sqrt 5) \cong \mathbb Z/11 \mathbb Z$, so that the ideal $(4 + \sqrt 5)$ is maximal.

As $N(4 + \sqrt 5) = 11$, it is clear that the elements $0, 1, \ldots, 10$ are pairwise incongruent modulo $4 + \sqrt 5$.

Every element of $A$ is congruent to an integer modulo $4 + \sqrt 5$: indeed, if it is of the form $a + b \sqrt 5$ with $a, b \in \mathbb Z$ we can subtract a suitable integer multiple of $4 + \sqrt5$ to land in $\mathbb Z$. If it is of the form $(a+b\sqrt5)/2$ with $a, b$ odd, we can subtract
$$\frac{1 + \sqrt5}2 \cdot (4 + \sqrt5) = \frac{9 + 5\sqrt5}2$$
to land in $\mathbb Z + \mathbb Z\sqrt5$.

Consider the ring homomorphism
$$\mathbb Z / (11) \to A / (4+\sqrt5) \,.$$
By the first observation, it is injective. By the second, it is surjective.


问:
Can a function have a branch cut along the real and also imaginary axis?
Is it possible for a complex function to have a branch cut along the real axis and also the imaginary axis, that cross over like a + sign?</p>

答: Sure, why not?  Try $f_1(z) + f_2(z) + f_3(z) + f_4(z)$ where $f_1$, $f_2$, $f_3$, $f_4$ have branch cuts on the positive real, positive imaginary, negative real and negative imaginary axes respectively.


问:
Rotation transformation matrix, question on $\sin\theta$ and $\cos\theta$
Does anyone know how $\sin$ and $\cos$ contribute to the transformed $x', y'$ coordinates?  From elementary geometry it's my understanding if a circle's $r = 1$, then $\cos(\theta)$ represents the $x$ component of the right triangle, while $\sin(\theta)$ represents the $y$.

Perhaps this is something obvious I'm overlooking.

$$\left[\begin{array}{c}
x' \\
y' \\
\end{array}\right] =
\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta \\
\end{array}\right]
\left[\begin{array}{c}
x \\
y \\
\end{array}\right]$$

After the matrix product:

$$\begin{cases}x' = x \cos (\theta) + y \sin (\theta)\\
y' = -x \sin (\theta) + y \cos (\theta)\end{cases}$$

答:
You have signs of the sines wrong (EDIT: if you refer to a counterclockwise rotation). The matrix for a rotation through an angle $\theta$ is

$$\begin{bmatrix}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta) \end{bmatrix}$$

You can see this from the fact that a linear tranformation is uniquely determined by what it does to the standard basis vectors (if you don't feel confident about this you should look closer into it). If you consider the standard basis vectors in $\mathbb{R}^2$ $\mathbf{e}_1 = (1,0)$ and $\mathbf{e}_2 = (0,1)$, basic trigonometry will yield that when rotated they become $(\cos(\theta), \sin(\theta))$, as suggested by yourself in your question, and $(-\sin(\theta), \cos(\theta))$. Adding these as columns in the matrix, we get the matrix above.


问:
Conditions to apply Gradient Descent?
I was reading "An introduction to optimization" by Edwin KP, there is a chapter where they explain the basics of Neural Networks and as example; they use gradient descent with sigmoid function as activation function for a Neural network of 3 layers (input, hidden and output layer).

Also, all the material that i found (book included), to ensure the convergence of the algorithm (theorems), the function to minimize have to satisfy some conditions as: to be convex, have lipschitz gradient, to be strongly convex.

My questions are: In practice, have to we ensure that the function to minimize have those propierties? In particular, if a use least squares with sigmoid funtion (as the example in the book), will gradient descent still working? why?

PD: In advance, im thankful for your help and i apologize for my english.

答:
I think you might be mixing some things here. In general, gradient descent is probably the simplest optimization algorithm there is, and it requires very few assumptions. You do have to assume that the function is continuously differentiable. However, if you seek to ensure convergence to a global minimum or you wish to derive the complexity of the algorithm, things get a bit more complicated.

Given that you make no assumptions on the function, then gradient descent can only be guaranteed to advance towards a stationary point (local min/max or saddle point), if such a point exists. The other assumptions will give you additional guarantees:

If we assume the function has a Lipschitz gradient - then we can derive iteration complexity (how much the algorithm "progresses" from one iteration to the other).

If we assume the function is convex - then in case it will stop, it must be at a local minimum (however that minimum need not exist).

If we assume the function has bounded level sets, then we are guaranteed it will converge to a stationary point. Bounded level sets can be attained in various ways, most notably if the domain is compact, or if the function is coercive or strictly convex.

Strictly convex functions admit a unique global minimum (while convex functions might have a range of minimizers).

Strongly convex functions give an improved iteration complexity.

Note that neural networks optimize cost functions that do not meet any of these assumptions. Moreover, they may not even be continuously differentiable. Therefore neural networks have no theoretical guarantees, but in practice they will converge to a local minimum or saddle point. The rate of convergence will be unknown, and in fact they will use a variant of gradient descent called stochastic gradient descent (SGD). SGD processes the data in batches at every iteration, since calculating and storing the gradients might be too expensive from the perspective of memory and computation resources required.

Title: Conditions to apply Gradient Descent?


问:
Drawing numbers from an interval
We draw $10$ numbers from an interval $[0,5]$. Compute probability, that at least two of these numbers will land in $[1,3]$.

So if $A-$ at least two numbers land in an interval $[1,3]$, then:

$A'-$ none of the numbers or only one will land in $[1,3]$, and it gives me:

$\Bbb P(A')=\frac{\lambda_{10}([0,1)\cup(3,5])+10 \cdot\lambda_{9}([0,1)\cup(3,5])\cdot\lambda_1([1,3]) }{\lambda_{10}([0,5])}$

where $\lambda_a$ is a Lebesgue measure for a  cartesian product of intervals  in $\Bbb R^a.$

Then

$\Bbb P(A)=1-\Bbb P(A')=1-\frac{\lambda_{10}([0,1)\cup(3,5])+10 \cdot\lambda_{9}([0,1)\cup(3,5])\cdot\lambda_1([1,3]) }{\lambda_{10}([0,5])}=1-\frac{3^{10}+10*3^9*2}{5^{10}}=0.9536$

Am I right? Thanks!

Title: Drawing numbers from an interval

Body:
Yes, it's correct.  Here's another way to express it:
$$\sum_{k=2}^n \binom{n}{k}p^k (1-p)^{n-k}
=1-\sum_{k=0}^1 \binom{n}{k}p^k (1-p)^{n-k}
=1-(1-p)^n-n p (1-p)^{n-1}
$$

Now take $n=10$ and $p=(3-1)/(5-0)=2/5$.


问:
Calculating limit of series.
Given a series $\sum_{n=1}^{\infty}\frac{4n+1}{2n(2n-1)(2n+1)(2n+2)}$, how do I find the limit? I understand I need to find the sequence of partial sums, which goes something like this $s_n=\{\frac{5}{24}, \frac{7}{30}, \frac{27}{112}....\}$which will probably converge at 0.25, but I am having trouble finding a description of the sequence on which I could evaluate the limit.

Title: Calculating limit of series.

Body:
As a hint:$$\sum_{n=1}^{\infty}\frac{4n+1}{2n(2n-1)(2n+1)(2n+2)}=\\
\sum_{n=1}^{\infty}\frac{4n+1}{(2n-1)(2n+0)(2n+1)(2n+2)}=\\
\sum_{n=1}^{\infty}\frac{(2n-1)+(2n+2)}{(2n-1)(2n+0)(2n+1)(2n+2)}=\\
\sum_{n=1}^{\infty}\frac{(2n-1)}{(2n-1)(2n+0)(2n+1)(2n+2)}+\sum_{n=1}^{\infty}\frac{(2n+2)}{(2n-1)(2n+0)(2n+1)(2n+2)}=\\
\sum_{n=1}^{\infty}\frac{1}{(2n+0)(2n+1)(2n+2)}+\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+0)(2n+1)}=\\$$ and now you have two telscopic series.can you take over now ?


答:
The partial fraction decomposition of the summand is
$$\frac1{2(2n-1)}-\frac1{2(2n+1)}-\frac1{4n}+\frac1{4(n+1)}$$
The first and second parts, and the third and fourth, telescope. In the infinite sum all parts except the $n=1$ instances of the first and third parts will cancel, so the infinite sum is
$$\frac1{2(2×1-1)}-\frac1{4×1}=\frac14$$

Title: Calculating limit of series.

Body: Given a series $\sum_{n=1}^{\infty}\frac{4n+1}{2n(2n-1)(2n+1)(2n+2)}$, how do I find the limit? I understand I need to find the sequence of partial sums, which goes something like this $s_n=\{\frac{5}{24}, \frac{7}{30}, \frac{27}{112}....\}$which will probably converge at 0.25, but I am having trouble finding a description of the sequence on which I could evaluate the limit.


问:
Is my understanding with integration correct here?
Let’s say $y = F(x)+c$.

If I take $\frac{d}{dx} (F(x)+c )=f(x)$, we get
$$d(F(x)+c))=f(x)dx $$

Now what I got to understand from my textbook is that

$d(F(x)+c)) = dy$, i.e. when we take the slope of small quantity of $y$ with $x$, then that rate of change of $y$ w.r.t.  $x$ is $dy$.

On the other hand , $f(x)$ is the slope that you got $\frac{dy}{dx} $

So can I say $\frac{dy}{dx} dx= dy$ ?

Q is also that what is meant by $\frac{dy}{dx} dx $  if I want to imagine it.

答:
From the definition of the differentials,

$$\frac{dy}{dx}=f(x)$$ where $dx$ is a nonzero variable.

Hence

$$\frac{dy}{dx}dx=dy$$ holds as with any fraction.


问:
$n$-th derivative of $f(x)=x\cos(x)$
Can someone help me prove that the $n$-th derivative of $f(x)=x\cos(x)$ is:

$$f^n(x)=x\cos\left(x+\frac{n\pi}{2}\right)+n\cos\left(x+(n-1)\frac{\pi}{2}\right)?$$

Thanks, and stay safe.

答:
Just using induction using the product rule:

$f(x) = x\cos x$ so

$f'(x) = \cos x + x(-\sin x) = \cos x - x\sin x$

$f^2(x) = -\sin x - \sin x - x\cos x= -2\sin x - x\cos x$

$f^3(x) = -2\cos x-\cos x - x(-\sin x) = -3\sin x +x\sin x$.

By pattern recognition we sort of see that $f^n = \pm n \operatorname{trig} x \pm x\operatorname{other\ trig} x$.  Bearing in mind that $\sin x = \cos (\frac \pi 2 - x)$ and that the pattern of the $\pm$ has a periodicity of $4$ and so on we can see that

$f^n(x) =\pm n \operatorname{trig} x \pm x\operatorname{other\ trig} x=$

$x\cos(x+\frac{n\pi}{2})+n\cos(x+(n-1)\frac{\pi}{2})$

and we can feel grateful that someone else hammered the details and fine polished it for us.

But we have to verify it.

For $n= 0$ this fits

$f(x)  = x\cos x = x\cos(x+\frac{0\pi}{2})+0\cos(x+(0-1)\frac{\pi}{2})$

No question of that.

And for $n=1$ we have

$f'(x) =  \cos x - x\sin x= x\cos(x+\frac{1\pi}{2})+1\cos(x+(1-1)\frac{\pi}{2})$

$=1\cos (x + 0) + x\cos(x+\frac {\pi}2) = \cos x- x \sin x$.

It works.  Now to verify in general we must use induction.

If $f^k(x) =x\cos(x+\frac{k\pi}{2})+k\cos(x+(k-1)\frac{\pi}{2})$

then

$f^{k+1}(x) = \cos(x+\frac{k\pi}{2}) + x(-\sin(x+\frac{k\pi}2) -k\sin(x+(k-1)\frac \pi 2)=$

$=\cos(x+\frac{k\pi}{2}) +k\cos(x+k\frac {\pi}2) + x(\cos(x+\frac {(k+1)\pi}2))=$

$x\cos (x+\frac {(k+1)\pi}2)+ (k+1)\cos(x+k\frac \pi 2)$

So the result follows by induction.

Title: $n$-th derivative of $f(x)=x\cos(x)$


问:
For the given function and sets $A = \{0,1\}, B = (0,1), C = [1, + ∞)$ find counterimages
For the given function and sets $A = \{0,1\}, B = (0,1), C = [1, + ∞)$  find $ f^{-1}(A), f^{-1}(B), f^{-1}(C)$.

Function: $ f:[0,1]\times[0,3]\rightarrow R,\\f(x,y)=\max(x,y)$

How can I do it? Any tips?

答:
What gives you $A$ when plugged into $f$?

$f(0,0) = 0 \in A$, so the point $(0,0)$ is in the preimage of $A$. $f(1,[0,1]) = 1$, so the set of points $\{1\}\times[0,1]$ is in the preimage of $A$.

Keep asking this question until you've exhausted $[0,1]\times[0,3]$ for each of the sets $A,B,C$


问:
Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?
Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?

We know the answer is of the form $ a + b \sqrt{2}$. Since $(a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2 = 1 + \sqrt{2}$, the system we need to solve is
\begin{align*}
2ab &=  1 \\
a^2 + 2b^2 &=  1
\end{align*}
We write $b = \frac{1}{2a}$ and substitute this in the second equation.
\begin{align*}
a^2 + 2\left(\frac{1}{2a}\right)^2 &=  1 \\
a^2 + \frac{1}{2a^2} &=  1 \\
2a^4 - 2a^2 + 1 &= 0
\end{align*}
Let $z = a^2$, so $z^2 = a^4$. The equation is then
\begin{equation}
2z^2 - 2z + 1 = 0
\end{equation}
Using the quadratic formula we find $z = \frac{1 \pm i}{2}$. This worked out when checked. Thus $a = \sqrt{\frac{1 \pm i}{2}}$.
We then find $b$ using $a$ in our original system of equations.
\begin{align*}
\frac{1 \pm i}{2} + 2b^2 &=  1 \\
1 \pm i + 4b^2 &=  2 \\
\pm i + 4b^2 &=  1 \\
4b^2 &=  1 \pm i \\
2b &=  \sqrt{1 \pm i} \\
b &=  \frac{\sqrt{1 \pm i}}{2} \\
\end{align*}

Substituting $a$ and $b$ into the equation $2ab = 1$, leads to inconsistent solutions.
What do I need to reconsider? How can I improve my answer?

答:
The truth is that $\sqrt{1+\sqrt2}$ cannot be expressed as $a+b\sqrt2$ where $a,b$ are integers or rationals. If they were reals then the problem becomes trivial.

(Another way to see the first fact above is that the minimal polynomial of $\sqrt{1+\sqrt2}$ is degree-$4$, whereas if it were expressible as $a+b\sqrt2$ with $a,b\in\mathbb Q$ the minimal polynomial would only be quadratic.)


问:
About concept: how to pull back a manifold function that is not injective?
Given $F:M\rightarrow N$ that is not injective, I have a great confusion on  the pull-back map $F^*:T^*_{F(p)}N\rightarrow T^*_pM$.  For better communication, let's simply take $M=\mathbb{R}^2$, $N=\mathbb{R}$. Take $(x_1,x_2)\in M$, $F(x_1,x_2)=x_1\in N$. Let $f\in C^\infty(N)$.

(1) When we pull back $f^\prime(x)d x\in T_x^*N$, I expect we get $T^*_{(x_1,x_2)}M\ni f^\prime (x_1)dx_1+0dx_2=F^*(f^\prime(x)d x)$. Then excuse me, how do we determine the value of $x_2$? This is basically a question on how to deal with the inverse map when a map is not injective.

(2) Also the resource I am reading says "(a) $F^*$ is NOT a map
from $T^*N$ to $T^*M.$
(b) It is a map from $C^\infty(N;T^*N)$ to $C^\infty(M;T^*M)$ which is quite a
different thing."
How should I understand this remark, especially in the combination with the first question? (I don't think neither (a) or (b) is a map at this moment, and I think (b) somehow would imply (a).)

Little remark: Maybe the answer to my question is simply to take $x_2$ to be any/all $x_2\in \mathbb{R}$?

Been confused on this for a while, hope I can get some enlightenment from you, thanks!

答:
If $M$ and $N$ are two manifolds and $f : M \to N$ is smooth, any vector bundle $E$ on $N$ can be pulled-back on $M$ by the following construction:
$$
f^*E=\left\{(x,v) \in M\times E ~|~ v \in E_{f(x)} \right\}
$$

which basically says that we glue the vector space $E_{f(x)}$, the fiber of $E$ over $f(x)$, to $x$.

This way, if two points $x_1$ or $x_2$ in $M$ have the same image, then the fibers over $x_1$ and $x_2$ of $f^*E$ are exactly the same vector space. But this is not a confusing definition: for example, $M\times \mathbb{R}$ is a vector bundle over $M$ where all the fibers are exactly the same.

Now, if $s$ is a section of $E$ as a vector bundle over $N$, one can pull back this section as a section $f^*s$ over $M$ with the following constuction:
$$
(f^*s) : x \mapsto s_{f(x)}
$$
that is the value of the section $f^*s$ at a point $x\in M$ is the value of a the section $s$ at the point $f(x)\in N$. Again, if $f(x_1)=f(x_2)$, the fibers of $f^*E$ over $x_1$ and $x_2$ are the same, and the value of the pulled-back section are the same at $x_1$ and $x_2$ as seen in the fibers.

If, now, $T$ is a tensor over $N$, then one can ask how to evaluate such a tensor. This construction cannot allow you to talk about the pullback of a vector field on $N$ as a vector field on $M$! Indeed, if $X$ is a vector field on $N$, then $f^*X$ is a section of $f^*(TN)$ which is not $TM$.

But if $T$ is a covariant vector field, there is no problem here. For example, if $\alpha$ is a $1$-form on $N$, then one can decide that $f^{\alpha}$ is a $1$-form on $M$ by saying
$$
(f^*\alpha)_x V_x = \alpha_{f(x)}(\mathrm{d}f(x)V_x)
$$
And similarly, for a $p$-covariant tensor,
$$
(f^*T)(V_1,\ldots,V_p)_x = T_{f(x)}\left(\mathrm{d}f(x){V_1}_x,\ldots,\mathrm{d}f(x){V_p}_{x}\right)
$$
This defined a $p$-covariant tensor on $M$.

Briefly, any smooth map $f : M \to N$ allows you to pullback any vector bundle and any section of this vector bundle. But these pulled-back bundles are not generally usual vector bundles as the tensor bundles.
Any covariant tensor bundle $(T^*N)^{\otimes p}$ can pulled back as a canonical sub-bundle $f^*(T^*N)^{\otimes p}\subset (T^*M)^{\otimes p}$ using $\mathrm{d}f$ (thanks to the covariance). But $TN$ cannot be canonically pulled-back as a subbundle of $TM$, neither are $TM^{\otimes q}$. One can outpass this if $\mathrm{d}f$ is an isomorphism between $T_xM$ and $T_{f(x)}N$ for all $x$.

In you example, where $f(x,y) = x$ from $\mathbb{R}^2 \to \mathbb{R}$, if $g$ is a smooth function on $\mathbb{R}$, then $\mathrm{d}g = g' \mathrm{d}t$ is a $1$-form on $\mathbb{R}$, and $f^* (\mathrm{d}g)$ is the one form on $\mathbb{R}^2$ defined by the folliwing, if $X=(X_1,X_2)$ is a vector field on $\mathbb{R}^2$
$$
f^*(\mathrm{d}g) = \mathrm{d}g(f(x,y)) \mathrm{d}f(X_1,X_1) = \mathrm{d}g(x)X_1
$$


问:
Proof that the minimum of a bounded differentiable real function occurs at a stationary point or at an endpoint
I couldn't find a proof for this well-known result on this site...

The minimum of a bounded differentiable real function occurs at a stationary point or at an endpoint.

I prove the contrapositive to this statement in my answer below.

Alternative proofs are most welcome.

答:
If the domain is unbounded, say $(0,+\infty)$, then the question needs some clarification as the minimum of $x\to1/x$ is not attained.

If the domain is bounded, say $[0,1]$, the minimum exists by continuity of $f$ and a consequence of Bolzano-Weierstrass ("the image of a segment by a continuous function is a segment"). Let $c\in[0,1]$ be the point where the minimum is attained. If $c\notin\{0,1\}$, we look at
$$\frac{f(x)-f(c)}{x-c}$$
which is always non-negative if $x>c$ as $f(x)\ge f(c)$ holds. This proves that the limit $\lim_{x\to c^+}$ of the above quotient is non-negative. Similarly by looking at $x<c$, we see that $\lim_{x\to c^-}$ of the the above quotient is non-positive. Hence the limit is both non-positive and non-negative, i.e., $f'(c)=0$.


答:
Your start of the question Find $\sqrt{1+ \sqrt 2}$ is misleading.

You're supposing that $\alpha = \sqrt{1+ \sqrt 2}$ belongs to the field extension $\mathbb Q(\sqrt 2) / \mathbb Q$... And you proved that it's not the case.

In fact $\alpha$ is an element of degree $2$ over $\mathbb Q(\sqrt 2)$ because $\alpha$ is a root of the polynomial $p \in \mathbb Q(\sqrt 2) [x]$

$$p(x) = x^2 - (1 +\sqrt 2).$$

And $p$ is irreducible over $\mathbb Q(\sqrt 2)$.

Body:
Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?

We know the answer is of the form $ a + b \sqrt{2}$. Since $(a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2 = 1 + \sqrt{2}$, the system we need to solve is
\begin{align*}
2ab &=  1 \\
a^2 + 2b^2 &=  1
\end{align*}
We write $b = \frac{1}{2a}$ and substitute this in the second equation.
\begin{align*}
a^2 + 2\left(\frac{1}{2a}\right)^2 &=  1 \\
a^2 + \frac{1}{2a^2} &=  1 \\
2a^4 - 2a^2 + 1 &= 0
\end{align*}
Let $z = a^2$, so $z^2 = a^4$. The equation is then
\begin{equation}
2z^2 - 2z + 1 = 0
\end{equation}
Using the quadratic formula we find $z = \frac{1 \pm i}{2}$. This worked out when checked. Thus $a = \sqrt{\frac{1 \pm i}{2}}$.
We then find $b$ using $a$ in our original system of equations.
\begin{align*}
\frac{1 \pm i}{2} + 2b^2 &=  1 \\
1 \pm i + 4b^2 &=  2 \\
\pm i + 4b^2 &=  1 \\
4b^2 &=  1 \pm i \\
2b &=  \sqrt{1 \pm i} \\
b &=  \frac{\sqrt{1 \pm i}}{2} \\
\end{align*}

Substituting $a$ and $b$ into the equation $2ab = 1$, leads to inconsistent solutions.
What do I need to reconsider? How can I improve my answer?


问:
Looking for a source of an infinite trigonometric summation and other such examples.
Question:

If $x \neq 0$, then prove that  $\displaystyle \sum_{n=1}^{\infty}\dfrac1{2^n} \tan\left(\dfrac{x}{2^n}\right) = \dfrac1{x} - \cot x.$

My answer:

I proved this result by using the following identity:

$$ \prod_{k=1}^n \cos\left(\frac{x}{2^k}\right) = \frac{\sin x}{2^n\sin \frac{x}{2^n}}$$

I took natural log on both sides of the above equation and then differentiated both sides to get

$$\sum_{k=1}^n \frac1{2^k} \tan\left(\frac{x}{2^k}\right) = \frac{1}{2^n}\cot \left(\frac{x}{2^n}\right) - \cot x.$$

Now taking the limit of $n$ to $\infty$ I get the required identity. $\blacksquare$

I have never seen the above identity before in my life. I was amused and surprised by this identity. That's because I have never seen an infinite trig series summing up to a rational function like $\dfrac1{x}$ before.

So my questions are the following:

A) First of all, is my derivation correct? Secondly, does anyone know a source for this problem?

B) Are there other 'elementary 'derivations that have an infinite trig series on one side and a rational function on the other?  [I say 'elementary' to avoid Fourier series. I am guessing Fourier series must be full of such results.]

答:
The identity
$$
\cot(x)+\frac12\tan(x/2)=\frac12\cot(x/2)
$$
sort of leads you to the solution. Iterating gives your sum.</p>


问:
Find the components of this tensor
It is give to me a tensor type (2,0) and a tensor type (1,0), that is, a vector. Now i need to find a tensor of type (1,1), as follow:
$$X^{\mu v} = \begin{bmatrix}
2 & 0 & 1 & -1\\
-1 & 0 &  3& 2\\
 -1&  1& 0 & 0\\
 -2&  1& 1 & -2
\end{bmatrix}$$
$$ V^{\mu} = (-1,2,0,2)$$
$$X^{\mu}_{v}?$$

I am a little stuck. Normally i do not see tensor as matrix. What i would do normally is find the metric tensor and so lower the indices of the first tensor, but it is not the case here. What do you recommend?

答: Like $\color{red}{X^{\mu\nu}}$, the tensors $\color{limegreen}{X^\mu_{\:\nu}}$, $\color{blue}{X^{\:\mu}_\nu}$ and $\color{orange}{X_{\mu\nu}}$ look like square matrices if you write them in components. Note for asymmetric $\color{red}{X^{\mu\nu}}$ this list includes two $(1,\,1)$ tensors, the green and blue ones. The difference between these four tensors is actually in how those components transform under a general coordinate transformation. They're also related via metric tensors viz.$$\color{red}{X^{\mu\nu}}=g^{\rho\nu}\color{green}{X^\mu_{\:\rho}}=g^{\mu\rho}\color{blue}{X^{\:\nu}_\rho}=g^{\mu\rho}g^{\nu\sigma}\color{orange}{X_{\rho\sigma}}.$$I'll leave it to you to write down some more such equations, including those that use the index-lowering metric tensor.

Title: Find the components of this tensor


问:
Operation of NOT and Z gate on the qubit in the state $|H\rangle = a|0\rangle + b|1\rangle$?
I have this last question for an assignment and I've been stuck on it for hours.

Pauli operators for a two-level system(qubit),
$$
\sigma_x = \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix},
\sigma_z = \begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
$$
$\sigma_x$ is called the quantum NOT gate and $\sigma_z$ is the Z gate. We have a qubit in the state $|H \rangle = \alpha |0\rangle + \beta|1\rangle$. Which state is obtained after operating with the NOT gate and which state with the Z gate? What happens if we apply operation twice?

答:
Well, $\sigma_x |H\rangle$ is given by matrix vector multiplication:

$$\pmatrix{ 0 & 1\\ 1 & 0 } \pmatrix{ \alpha\\ \beta} =\pmatrix{\beta\\\alpha} $$

The state is $\beta |0\rangle + \alpha|1\rangle$. Similar for the 2nd operation.

Why is \ket in Latex not working?


问:
Finding Modular Multiplicative Inverses (Quickly!)
as part of an upcoming number theory exam I will need to find the modular multiplicative inverse of every element of ${Z_n}$ (the ones that exist anyway) very quickly. The only way I know is using the Extended Euclidean Algorithm. Is there no way that is faster? The question isn't worth many marks but will take an inordinate amount of time.</p>

答:
Solve $ax \equiv 1 \pmod{n}$

I would tackle these problems keeping this trick in mind and applying the basic tactic of making $a$ smaller by 'multiplying to the modulus'.

Example 1: Solve $12x \equiv 1 \pmod{19}$.

$\; 12x \equiv 1 \pmod{19} \; \text{ iff}$

$\; -7x \equiv 1 \pmod{19}$

and

$\;-7 \cdot (3) \equiv -21 \equiv -2 \pmod{19}$

$\;-2 \cdot (9) \equiv -18 \equiv 1 \pmod{19}$

and therefore the solution is given by $x \equiv 3 \cdot 9 \equiv 8 \pmod{19}$.

Example 2: Solve $67x \equiv 1 \pmod{97}$.

$\; 67x \equiv 1 \pmod{97} \; \text{ iff}$

$\; -30x \equiv 1 \pmod{97}$

and

$\;-30 \cdot (3) \equiv -90 \equiv 7 \pmod{97}$

$\;7 \cdot (14) \equiv 98 \equiv 1 \pmod{97}$

and therefore the solution is given by $x \equiv 3 \cdot 14 \equiv 42 \pmod{19}$.

Example 3: Solve $7x \equiv 1 \pmod{23}$.

$\;7 \cdot (4) \equiv 5 \pmod{23}$

$\;5 \cdot (5) \equiv 2 \pmod{23}$

$\;2 \cdot (12) \equiv 1 \pmod{23}$

and therefore the solution is given by $x \equiv 4 \cdot 5 \cdot 12 \equiv 10 \pmod{23}$.

Example 4: Solve $6x \equiv 1 \pmod{10}$.

$\; 6x \equiv 1 \pmod{10} \; \text{ iff}$

$\; -4x \equiv 1 \pmod{10}$

and

$\;-4 \cdot (2) \equiv 2 \pmod{10}$

$\;2 \cdot (5) \equiv 0 \pmod{10}$

If $6x \equiv 1 \pmod{10}$ then $5 \cdot (6x) \equiv 5 \pmod{10}$, so there are no solutions.


问:
Proving that Geometric Brownian Motion is positive
An Ito process is defined as:

$$X_t = X_0 + \int_{h=0}^{h=t}\left(a(X_h,h)\right)dh + \int_{h=0}^{h=t}\left(b(X_h,h)\right)dW_h$$

Where $a(X_t, t)$ and $b(X_t, t)$ are square integrable processes.

If one chooses $a(X_t,t)=\mu X_t$ and $b(X_t,t)=\sigma X_t$, with $\mu$ and $\sigma$ being constant, we obtain the well known Geometric Brownian Motion (where $\mu$ can be regarded as the "drift" and $\sigma$ as the "volatility"):

$$X_t = X_0 + \int_{h=0}^{h=t}\left(\mu X_h\right)dh + \int_{h=0}^{h=t}\left(\sigma X_h\right)dW_h$$

The above is of course still an Ito process. Ito's lemma states that for any (well-behaved), twice-differentiable function $F()$ of $X_t$ and $t$, the function $F(X_t, t)$ follows the process:

$$F(X_t,t)= F(X_0,t_0) + \int_{h=0}^{h=t}\left(\frac{\partial F}{\partial t}+\frac{\partial F}{\partial X}a(X_h,h)_{=\mu X_h}+\frac{1}{2}\frac{\partial^2 F}{\partial X^2}b(X_h,h)^2_{=\sigma^2 X_h^2}\right)dh + \int_{h=0}^{h=t}\left(\frac{\partial F}{\partial X}b(X_h,h)_{=\sigma X_h}\right)dW_h$$

Using the Ito's lemma above, we can easily find the solution to the GBM equation for $X_t$ by taking $F(X_t,t)=ln(X_t)$ to obtain:

$$X_t=X_0e^{\left(\mu t -0.5 \sigma^2 t + \sigma W_t\right)}$$

Question: I just got asked how come we can take the log of $X_t$, without knowing whether the equation $X_t = X_0 + \int_{h=0}^{h=t}\left(\mu X_h\right)dh + \int_{h=0}^{h=t}\left(\sigma X_h\right)dW_h$ potentially produces negative values for $X_h$ along some specific paths.

Looking at the solution $X_t=X_0e^{\left(\mu t -0.5 \sigma^2 t + \sigma W_t\right)}$, it is obvious that for any initial positive $X_0$, $X_t$ cannot be negative. However, using this argument to justify taking the log is a bit sketchy because we are using the solution to validate the method we use to derive that solution.

Is there a way to show that the GBM SDE for $X_t$ produces strictly positive values if the initial value $X_0>0$ for all $X_h: 0\leq  h \leq t < \infty$?

I suppose we could start by writing out the integrals by definition:

$$X_t = X_0 + \int_{h=0}^{h=t}\left(\mu X_h\right)dh + \int_{h=0}^{h=t}\left(\sigma X_h\right)dW_h= \\= \mu \lim_{n\to\infty}\sum_{h=t_0}^{h=t_{n-1}}X_h((h+1)-h)+\sigma \lim_{n\to\infty}\sum_{h=t_0}^{h=t_{n-1}}X_{h}(W_{h+1}-W_{h})=\\=\mu \lim_{n\to\infty} \left(x_0(t_{1}-t_{0})+...+x_{t_{n-1}}(t_n-t_{n-1}) \right) +\sigma \lim_{n\to\infty}\left(x_0(W_{t_1})+...+x_{t_{n-1}}(W_{t_n}-W_{t_{n-1}})\right)$$

Not sure where to go from here and whether the above is a good idea?

答:
Let $X$ solve the GBM SDE with $X_0>0$. Let $\tau$ be the first time $t$ at which $X_t=0$. Clearly $\tau>0$. By Ito's formula,
$$
\log(X_t/X_0) = (\mu-\sigma^2/2)t+\sigma B_t,\qquad 0\le t<\tau,
$$
almost surely. On the event $\{\tau<\infty\}$, the left side of this equality goes to $-\infty$ as $t\uparrow \tau$ while the right side converges to the finite quantity $(\mu-\sigma^2/2)\tau+\sigma B_\tau$. This contradiction means that $P[\tau<\infty]=0$. That is, a solution $X$ satisfies $P[X_t>0,\,\forall t]=1$.

(Alternatively, you can exhibit one strictly positive solution, and the GBM SDE has a unique solution.)


问:
Ordinal numbers arithmetic $((1 + \omega) + \omega) \cdot \omega$ Cantor Normal Form
how do you get the Cantor Normalform of
$$((1 + \omega) + \omega) \cdot \omega$$
This would be $$((1 + \omega) + \omega) \cdot \omega = (\omega + \omega) \cdot \omega = (\omega \cdot 2) \cdot \omega$$ right? Is the attempt correct like this? How to proceed?

答: Here's a hint: Remember that ordinal multiplication is associative, so $(\omega\cdot2)\cdot\omega=\omega\cdot(2\cdot\omega)$, so you can simplify a bit more.


问:
How to show that there is an $x$ such that $f'(x)=2$?
So I have that $f: [1,4] \to \mathbb R$ is differentiable and $f(1)=0, f(2)=1$ and $f(4)=8$. How can I show that there is an $x$ such that $f'(x)=2$?

I have shown, by the mean value theorem, that there exists an $x$ such that $f'(x)=1$ since $\frac{f(2)-f(1)}{2-1} =1$ but I dont know how to show that there is an $x$ such that $f'(x)=2$.

答:
Try the mean value theorem again but for other points: $$\frac{f(4)-f(2)}{4-2} = \frac{8-1}{4-2} = \frac 7 2$$

The mean value theorem for first interval $[1,2]$ gives that there must exist one point $\xi_1$ on that interval for which $f'(\xi_1)=1$.

The mean value theorem for second interval $[2,4]$ gives that there must exist one point $\xi_2$ on that interval for which $f'(\xi_2) = \frac 7 2$

Now because of Darboux's theorem (every function that results from the differentiation of another function has the intermediate value property), we can apply the intermediate value theorem between these two points.Since $1<2<\frac 7 2$, the value $2$ must be assumed somewhere on the interval given by $[\xi_1,\xi_2]$


答:
As $x$ increases from $2$ to $4$:

the quantity $f(x)-f(1)$ varies continuously from $f(2)-f(1)=1$ to $f(4)-f(1)=8$;

the quantity $x-1$ increases continuously from $2-1=1$ to $4-1=3$;

and so the quantity $F(x) = \frac{f(x)-f(1)}{x-1}$ varies continuously from $F(2) = \frac{1}{1}=1$ to $F(4) = \frac{8}{3}$.

Applying the Intermediate Value Theorem, since $F(2) = 1 < 2 < \frac{8}{3} = F(4)$, there is a value of $x$ between $2$ and $4$ such that $F(x) = \frac{f(x)-f(1)}{x-1} = 2$.

Applying the Mean Value Theorem to the interval $[1,x]$ over which the equation $\frac{f(x)-f(1)}{x-1} = 2$ is true, one finds $a \in (1,x)$ such that $f'(a)=2$.

Title: How to show that there is an $x$ such that $f'(x)=2$?

Body:
So I have that $f: [1,4] \to \mathbb R$ is differentiable and $f(1)=0, f(2)=1$ and $f(4)=8$. How can I show that there is an $x$ such that $f'(x)=2$?

I have shown, by the mean value theorem, that there exists an $x$ such that $f'(x)=1$ since $\frac{f(2)-f(1)}{2-1} =1$ but I dont know how to show that there is an $x$ such that $f'(x)=2$.


问:
A confusion about UFDs
So I'm new to this concept and when I was reading the definition.
It says that if R is a UFD then every non zero element can be written as the product of irreducible elements.
So here's my question which is stupid I know.
Are these irreducible elements necessarily distinct??</p>

答: No, the integers are a unique factorization domain and $4=2 \cdot 2$.  You have to include the possibility of a unit, which for the integers can only be $-1$, to get the negatives.

Title: A confusion about UFDs


答: You know that exists $a$ with $f'(a) = 1$. Similarly there is $b$ with $f'(b) = \frac{f(4) - f(1)}{4-1} = 8/3 > 2$. Now apply Darboux's theorem.


问:
Euler's theorem multiplicative inverse Z26
Euler's theorem get the inverse of 2 in Z26...~
i tried to solve it but the result is not correct:
a=2,  φ (26) = 12, m=26
gcd(2,26)=21
multiplicative inverse of 2 = 1/2
2 * 1/2 = 1

gcd(2,26)=2

a^-1= a^11mod26=2^11mod26= 2046 mod 26= 20

a=2     a^-1=20
aa^-1= 220=40

when doing the verification, 40 mod 26 = 14 not 1...

or for this group the Euler theorem is not applicable ?

答:
Welcome to MSE!

A nonzero element $a\in\Bbb Z_n$, $n\geq 2$, is invertible if and only if $gcd(a,n)=1$; otherwise $a$ is a zero divisor.

In your case $gcd(2,26)=2$ and so $2$ is not invertible.


问:
How to understand sentence about Sobolev space of functions on compact interval?
In the paper https://projecteuclid.org/download/pdf_1/euclid.aos/1176348252 on page 1351 it is written: "Let $W_2^r$ be the Sobolev space of functions on $[0,1]$ for which f^{(r-1)} is absolutely continuouse and $\int_0^1 |f(x)|^2 dx$ is finite." where the $L^2$-norm is computed on the compact interval $[0,1]$.

How can I understand this sentence?

I thought Sobolev spaces can only be defined on open intervals.

答:
Actually, $W^{r,2}(0,1)$ is a space of equivalence classes of function, i.e, they are only defined a.e. and you can simply change the function on a set of measure zero but it identifies the same function in $W^{r,2}(0,1)$.

Now, note that $W^{1,1}(a,b)$ is embedded in the space $AC([a,b])$, e.g., see here. That means you can identify the function with its continuous representation on the closed interval, which makes the function in $W^{1,1}(a,b)$ really identifiable.

Since $f \in W^{r,2}(a,b)$, it means $f \in C^{r-1}([a,b])$.


问:
if $A\times B = A$, is $B ={} $Identity matrix?
Let $A, B ∈ M_{n×n}$

If   $A\cdot B = A  $, then $B$ is the identity matrix.

I can't find a theorem that proves this statement. Could it be false?

Edit:

A is NOT the zero matrix.

答: It's false. $A$ could be the zero matrix, then $B$ could be absolutely any matrix.


问:
Without loss of generality let's assume that sequence is strictly decreasing...
I have a question considering 'without loss of generality'. In one proof on my lecture we assume that sequence is strictly decreasing. We know that $\lim_{n \to \infty}{a_{n}}=-\infty$ so I think that the other option here is that the sequence is nonincreasing. The thing is I can't show that we can assume this without losing generality, it's kind of natural but I can't form any formal thought.

EDIT:
We also know that
$a_{n} \in \mathbb{R}$

答:
You can have a sequence which increases and decreases infinitely often but still has a limit of $-\infty$.

For example with alternating steps of $+1$ and $-2$:

$$-1, 0,-2,-1,-3,-2,-4,-3, -5, -4,\ldots$$

where the $n$th term is close to $-\frac{n}{2}$, the limit is clearly $-\infty$.

Whether this matters in a particular case, or whether there is no loss of generality in assuming the sequence is strictly decreasing, will depend on the particular circumstances.


问:
Solving for point P within the edge AC, given points A, B and C, and length from B to P?

(EDIT: one of the comments noted that AC is too short in the image, this is completely my fault for not checking before rounding, therefor replace the magnitude of AC to, say 10)

So given points A B and C we can calculate the magnitudes of AB, AC, and CB, and
we know there is a point in the line of AC that has the distance of, say 60 like in the image,
to the point B. What are the necessary steps to solve for point P?

答:
This is the Stewart’s theorem
in its puriest form:

\begin{align}
|BC|^2\cdot|AP|+
|AB|^2\cdot(|AC|-|AP|)
=|AC|\cdot(|BP|^2+|AP|\cdot(|AC|-|AP|))
\tag{1}\label{1}
,
\end{align}

\begin{align}
|AP|&=
\frac{|AB|^2+|AC|^2-
|BC|^2\pm\sqrt{\Big(|AB|^2-|BC|^2\Big)^2-|AC|^2\cdot\Big(2|BC|^2+2|AB|^2-|AC|^2-4|BP|^2\Big)}}
{2|AC|}
\tag{2}\label{2}
,
\end{align}

for given $|AB|=61$, $|BC|=58$, $|BP|=60$, $|AC|=10$

\eqref{2} gives

\begin{align}
|AP|&=\tfrac1{20}(457\pm\sqrt{160449})
\end{align}

with the only valid root

\begin{align}
|AP|&=\tfrac1{20}(457-\sqrt{160449})
\approx 2.82
.
\end{align}

Edit

Note that for some set of side lengths
there are indeed two valid solutions,
for example if
we have $|AB|=61$, $|BC|=58$, $|BP|=55$, $|AC|=60$,
the expressions

\begin{align}
|AP_+|&=\tfrac1{40}\,(1319+\sqrt{626161})
\approx 52.7576
\\
\text{and }\quad
|AP_-|&=\tfrac1{40}\,(1319-\sqrt{626161})
\approx 13.1924
\end{align}
are valid and give two points as the answer:

\begin{align}
P_+&=A+\frac{|AP_+|}{|AC|}\cdot(C-A)
,\\
P_-&=A+\frac{|AP_-|}{|AC|}\cdot(C-A)
,
\end{align}

see the picture:


答:
Thanks! $\alpha$ and $\beta $ are constants. Shouldn't they remain outside the matrix? For example,  in ket notations
$
|0\rangle = \begin{pmatrix}
1 \\
0
\end{pmatrix}, \quad |1\rangle = \begin{pmatrix}
0 \\
1
\end{pmatrix}
$

The operation of NOT gate on these would be something like this

$
\sigma_x |H\rangle = \alpha \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} \begin{pmatrix}
1 \\
0
\end{pmatrix} + \beta \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} \begin{pmatrix}
1 \\
0
\end{pmatrix} = \alpha \begin{pmatrix}
0 \\
1
\end{pmatrix} + \beta \begin{pmatrix}
1 \\
0
\end{pmatrix} = \alpha |1\rangle + \beta |0\rangle
$
Am I correct?

Title: Operation of NOT and Z gate on the qubit in the state $|H\rangle = a|0\rangle + b|1\rangle$?

Body:
I have this last question for an assignment and I've been stuck on it for hours.

Pauli operators for a two-level system(qubit),
$$
\sigma_x = \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix},
\sigma_z = \begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
$$
$\sigma_x$ is called the quantum NOT gate and $\sigma_z$ is the Z gate. We have a qubit in the state $|H \rangle = \alpha |0\rangle + \beta|1\rangle$. Which state is obtained after operating with the NOT gate and which state with the Z gate? What happens if we apply operation twice?


问:
Solving differential system with repeated eigenvalue
I am trying to solve $$ \frac{dx}{dt}=\begin{bmatrix}
1 &-2  & 0\\
2 & 5 & 0\\
2 &1  &3
\end{bmatrix}x$$
and find that it has only one eigenvalue $3$ of multiplicity $3$. Also, $ \begin{bmatrix}
0\\
0\\
1\end{bmatrix}$ is an eigenvector to $3$ and so, $ \begin{bmatrix}
0\\
0\\
1\end{bmatrix}e^{3t}$ is a solution to the system. Now in my book, if an eigenvalue is repeated twice, then it says how to achieve another solution. But, this eigenvalue is repeated thrice and I have no idea how to proceed further.

答:
$$x'=Ax$$
You have a first eigenvector $v_1=\pmatrix {0,0,1}^t$ and a solution $x_1=v_1e^{3t}$
.Then for a second vector you have to calculate $v_2$:
$$(A-3I)v_2=v_1$$
$$\pmatrix {-2 & -2 &0 \\ 2&2&0 \\ 2&1&0} \pmatrix {v_{21} \\v_{22} \\v_{23}}=\pmatrix {0 \\0\\1}$$
You can take $v_2=\pmatrix {1 ,-1,0}^t$. Then a second solution is:
$$x_2=(v_2+tv_1)e^{3t}$$
For the third eigenvector solve for $v_3$:
$$(A-3I)v_3=v_2$$
Then the solution is:
$$x_3=e^{3t} \left(\dfrac {t^2}2v_1+tv_2+v_3\right)$$
And the complete solution is:
$$x(t)=c_1x_1+c_2x_2+c_3x_3$$


问:
Some question on localization of polynomial ring
Let $S=A[x_1,\dots, x_r](r \geq 2)$ be a polynomial ring where $A$ is a commutative ring.
Then is it true that $S=\bigcap_{i=1}^r S_{x_i}$? If $S$ is $A$-algebra and $x_i$ are not zero divisors, then is it true?</p>

答:
First of all you want to take the intersection inside of the fraction field, so I believe you need that $A$ be at least a domain. Now the result you ask for I believe holds at least in the case that $A$ is a UFD. Here is how we prove this result. Suppose you have some element $g$ that is in $S_{x_i}$ and $S_{x_j}$ where $i \neq j$. Then we can write $$g = f/x_i^k= h/x_j^l$$

where $f,h \in A[x_1,\ldots,x_n]$. Now since $A$ is a UFD, $A[x_1,\ldots,x_n]$ is a UFD so we may assume wlog that $f$ and $x_i$ have not common factor, similarly for $x_j$ and $h$. It will now follow that $fx_j^l = hx_i^k$ and so $x_i$ has to divide $x_j^l$. But then necessarily $ l = k = 0$ and so $g$ is just a plain old element of $A[x_1,\ldots,x_n]$.


问:
Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?
Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?

We know the answer is of the form $ a + b \sqrt{2}$. Since $(a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2 = 1 + \sqrt{2}$, the system we need to solve is
\begin{align*}
2ab &=  1 \\
a^2 + 2b^2 &=  1
\end{align*}
We write $b = \frac{1}{2a}$ and substitute this in the second equation.
\begin{align*}
a^2 + 2\left(\frac{1}{2a}\right)^2 &=  1 \\
a^2 + \frac{1}{2a^2} &=  1 \\
2a^4 - 2a^2 + 1 &= 0
\end{align*}
Let $z = a^2$, so $z^2 = a^4$. The equation is then
\begin{equation}
2z^2 - 2z + 1 = 0
\end{equation}
Using the quadratic formula we find $z = \frac{1 \pm i}{2}$. This worked out when checked. Thus $a = \sqrt{\frac{1 \pm i}{2}}$.
We then find $b$ using $a$ in our original system of equations.
\begin{align*}
\frac{1 \pm i}{2} + 2b^2 &=  1 \\
1 \pm i + 4b^2 &=  2 \\
\pm i + 4b^2 &=  1 \\
4b^2 &=  1 \pm i \\
2b &=  \sqrt{1 \pm i} \\
b &=  \frac{\sqrt{1 \pm i}}{2} \\
\end{align*}

Substituting $a$ and $b$ into the equation $2ab = 1$, leads to inconsistent solutions.
What do I need to reconsider? How can I improve my answer?

答:
With a couple of corrections, your answer seems consistent:

$z=\dfrac{1\pm i}2\implies a=\color{red}\pm\sqrt{\dfrac{1\pm i}2}$

$\pm i+4b^2=1\implies 4b^2=1\color{red}\mp i\implies b=\color{red}\pm\dfrac{\sqrt{1\color{red}\mp i}}{2}$


问:
Is $f$ uniformly continuous in $A \cup B$?
Let $A$ and $B$ be only closed in $\mathbb R$, and $f$ uniformly continuous function in $A$ and $B$, is $f$ still uniformly continuous on the union of these two sets?

It is not specified whether $A$ and $B$ are limited, so I could't say that are compact.

答:
I think you put your finger of the issue.  $A$ and $B$ don't have to be compact.

Let $$A= \bigcup_{n=1}^\infty\left[2n+1,2n+2-\frac1n\right]\\
B= \bigcup_{n=1}^\infty\left[2n,2n+1-\frac1n\right]
$$ and left $f$ be identically $0$ on $A$ and identically $1$ on $B$. There are points in $A$ and $B$ arbitrarily close to one another.


问:
Primes of the form $x^2+31y^2$ and $x^2+xy+8y^2$ and quadratic forms
While working on an exercise in algebraic number theory, I encountered the following problem:

A prime number $p$ can be represented as $x^2+31y^2$ if and only if it can be represented  as $x^2+xy+8y^2$.

In this answer by Will Jagy, he claims that those quadratic forms represent the same odd numbers (where odd is in bold).

A while ago, I encountered a similar problem, with $x^2+3y^2$ and $x^2-xy+y^2$, which I managed to solve using some ad-hoc reasoning:

Claim. There exist $a,b\in \Bbb Z$ such that $p=a^2-ab+b^2$ if and only if there exist $x,y\in \Bbb Z$ such that $p=x^2+3y^2$. Proof. $\boxed{\Leftarrow}$ Let $a=x-y$ and $b=x+y$.  $\boxed{\Rightarrow}$ If $a$ and $b$ have the same parity, let $x=\frac{a+b}{2}$ and $y=\frac{a-b}{2}$. In the case where $a$ and $b$ have different parity, wlog $a$ is odd and $b$ is even. Write $a=2n+1$ and $b=2m$ with $n,m\in \Bbb Z$. Let $x=m-2n-1$ and $y=m$. $\blacksquare$

This is not very satisfying, I would like to know if there is some general method for tackling this. Here I found some explanation, pointing me into the direction of equivalent quadratic forms. It suggests to find a matrix $T\in GL_2(\mathbf{Z})$ with determinant $\pm 1$ such that $T^\top \begin{bmatrix}1 & \frac12 \\ \frac12 & 8 \end{bmatrix}T=\begin{bmatrix}1 & 0 \\ 0 & 31 \end{bmatrix}$. Letting $T=\begin{bmatrix} a & b \\ c& d \end{bmatrix}$, we get $1=\begin{bmatrix} a & b\end{bmatrix}\begin{bmatrix}1 & \frac12 \\ \frac12 & 8 \end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix}=a^2+ab+8b^2$, which has unique integer solutions $(a,b)=(\pm 1,0)$. Similarly,  $31=c^2+cd+8d^2$ has unique integer solutions $(c , d)=(\pm 1,\mp 2)$. Hence, such $T$ does not exist. I suspect this has to do  with what Will Jagy is underlining.

Clearly, my knowledge of quadratic forms is too weak. Could someone shed a light on this problem?

答:
It turns out I was thinking too difficult. "Completing the square" gives $$x^2+xy+8y^2=(x+\tfrac12 y)^2+31(\tfrac12 y)^2.$$
Let $n$ be an odd integer such that $n=x^2+xy+8y^2$ for some integers $x,y$. Reducing modulo 2 gives $1\equiv x(x+y)\bmod{2}$, hence $x$ and $x+y$ are odd, so in particular $y=(x+y)-x$ is even.

Conversely, if $n=z^2+31w^2$ for some integers $z,w$, then we can $y=2w$ and $x=z-2w$ to see that $n=x^2+xy+8y^2$.

Hence, we see that an odd number can be represented by the quadratic form $x^2+xy+8y^2$ if and only if it can be represented by the quadratic form $x^2+31y^2$.


问:
$g \colon [0,1] \to [0,1]$ be a continuous map and consider the iteration $x_{n+1}=g(x_n)$.
I came across the following problem:

Let $g \colon [0,1] \to [0,1]$ be a continuous map and  consider the iteration $x_{n+1}=g(x_n)$.Then Which of the following maps will yield a fixed point for $g$?

The options are as follows:

a. $g(x)=\frac{x^2}{4}$,

b. $g(x)=\frac{x^2}{32}$,

c. $g(x)=\frac{x^2}{8}$,

d. $g(x)=\frac{x^2}{16}$.

Can someone point me in the right direction? Thanks in advance for your time.

答:
The answer is certainly all of them, according to a very general and powerful theorem called the Brower fixed point theorem.  You'll probably get some funny looks if you turn that in, though, as I think it's quite a bit easier than that.

Don't all the maps have an obvious common fixed point?  Is that really the whole problem?


问:
Counter example on Riemann-Stieltjes theorem

Suppose $f$ is boounded on $[a,b]$, $f$ has only finitely many points of discontinuity on $[a,b]$ and $ \alpha $ is continuous at every point of discontinuity. Then $f \in \Re(\alpha)$

Is there any example that if $f$ is bounded on $[a,b]$ and discontinuous at $ x=c  \in $[a,b], $ \alpha(x) $ is discontinuous at $ x=c $ as well, but  $ f \in \Re(\alpha)$?

答:
An example where both the integrand and integrator are discontinuous but the Riemann-Stieltjes integral exists is$$f(x) = \begin{cases}0, & a \leqslant x < c \\ 1, & c \leqslant x \leqslant b \end{cases}\quad \alpha(x) = \begin{cases}0, & a \leqslant x \leqslant c \\ 1, & c < x \leqslant b \end{cases}$$

For a partition with subinterval $I_c =[c,c+\delta]$ we have both the upper and lower Darboux-Stieltjes sums equal to $1$ since $\sup_{x\in I_c} f(x) = \inf_{x \in I_c} f(x) = 1$ and $\alpha(c+\delta) - \alpha(c) = 1$. This proves that $f \in \mathcal{R}(\alpha)$ since for any $\epsilon > 0$ there is a partition such that $U(P,f,\alpha) - L(P,f,\alpha) < \epsilon$.


问:
Can we take the exponential function of a vector?
Main question:

If $x=(x_1,x_2)\in \mathbb R^2$, can we have the exponential function of the vector, i.e.
$$
e^{i \pi x}=e^{i \pi (x_1,x_2)} \quad? \tag 1
$$

Follow-up question:

Can we in some way "split" $e^{i \pi (x_1,x_2)}$?
I'm thinking of the rule the $e^{ia}e^{ib}=e^{i(a+b)}$, where $a,b\in \mathbb R$.

A test, if we use the standard basis and write the vector as
$$
x=(x_1,x_2)=\hat e_1 x_1+\hat e_2x_2 \tag 2
$$
Can we now write $(1)$ as
\begin{align}
e^{i \pi (x_1,x_2)} &=
e^{i \pi (\hat e_1 x_1+\hat e_2x_2)}  \tag 3\\
&=
e^{i \pi (\hat e_1 x_1)}
e^{i \pi (\hat e_2 x_2)}
\quad? \tag 4
\end{align}
Is this legitimate?

答:
As pointed out in the comments and the answer by Hermis14, the exponential function is usually generalized via its infinite series. And this is a problem since the square of a vector is not usually defined. However, in Geometric Algebra/a Clifford algebra, we usually have $\mathbf v^2=\Vert\mathbf v\Vert^2$ (the square of the length/magnitude/norm), and may add scalars and vectors independently (kind of like real and imaginary parts of a complex number).

In that sort of nonstandard setting, we have:
\begin{align}\exp(\mathbf v)&=1+\mathbf v+\frac{\Vert\mathbf v\Vert^2}{2!}+\frac{\Vert\mathbf v\Vert^2}{3!}\mathbf v+\cdots
\\&=\begin{cases}1 & \text{if }\mathbf{v}=\boldsymbol{0}\\1+\Vert\mathbf v\Vert\frac{\mathbf v}{\Vert\mathbf v\Vert}+\frac{\Vert\mathbf v\Vert^2}{2!}+\frac{\Vert\mathbf v\Vert^3}{3!}\frac{\mathbf v}{\Vert\mathbf v\Vert}+\cdots & \text{otherwise}\end{cases}
\\&=\begin{cases}1 & \text{if }\mathbf{v}=\boldsymbol{0}\\\cosh\left(\Vert\mathbf v\Vert\right)+\frac{\mathbf v}{\Vert\mathbf v\Vert}\sinh\left(\Vert\mathbf v\Vert\right) & \text{otherwise}\end{cases}
\end{align}
where $\cosh$ and $\sinh$ are the hyperbolic trig functions. You can then insert $i\pi$ if you wish.


答:
A  counterexample:
$$A=\begin{bmatrix}1&1\\0&0\end{bmatrix},\quad B=\begin{bmatrix}0&1\\1&0\end{bmatrix}.$$
$AB=A$ only means that $B-I$ belongs to the annihilator of $A$ in the ring $M_{n\times n}$.

Title: if $A\times B = A$, is $B ={} $Identity matrix?

Body:
Let $A, B ∈ M_{n×n}$

If   $A\cdot B = A  $, then $B$ is the identity matrix.

I can't find a theorem that proves this statement. Could it be false?

Edit:

A is NOT the zero matrix.


答:
for this group the Euler theorem is not applicable ?

Euler's theorem assumes the base and modulus are relatively prime, but that is not the case here:

$\gcd(2,26)=2\ne1$.

Title: Euler's theorem multiplicative inverse Z26

Body:
Euler's theorem get the inverse of 2 in Z26...~
i tried to solve it but the result is not correct:
a=2,  φ (26) = 12, m=26
gcd(2,26)=21
multiplicative inverse of 2 = 1/2
2 * 1/2 = 1

gcd(2,26)=2

a^-1= a^11mod26=2^11mod26= 2046 mod 26= 20

a=2     a^-1=20
aa^-1= 220=40

when doing the verification, 40 mod 26 = 14 not 1...

or for this group the Euler theorem is not applicable ?


问:
Prove that integer $n$ is divisible by $65$
Can anybody help me with this problem?

Prove that for every integer $n$, $$p(n) = n^{18} - n^{14} - n^6 + n^2$$ is divisible by $65$.

Thanks!

答:
Hints:

Factor it as $n^2(n^{4}-1)(n^{12}-1)=(n^5-n)(n^{13}-n)$.

Note that $65=5\times13$.

Now use Fermat's little theorem.


答:
Note
\begin{align}
\sqrt{\sqrt{2}+1}&=\sqrt{(\sqrt2-1)(\sqrt2+1)} \cdot\sqrt{\sqrt{2}+1}
= \sqrt{\sqrt{2}-1} (1+\sqrt2)\\
\end{align}
with $a=b = \sqrt{\sqrt{2}-1}$, thus no rational simplification.

Title: Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?


问:
Turning a linear problem (FDA) into quadratic optimization problem
I would like to transform this problem into a quadratic optimization problem with a quadratic constraint:
$$g = x'Ax / x'Bx$$

with constraint
$$ x'Bx = 1$$

I know that a quadratic optimization problem can have such  a form:

Quadratic optimization problem image

I am stuck at how to apply this to my original formula.
How can I transform my function into this quadratic form?

答: Minimization of $$\frac{x^tAx}{x^tBx}$$ subject to $$x^tBx = 1$$ is the same as minimization of $$x^tAx$$ subject to $$x^tBx = 1$$The latter problem is a QCQP (Quadratically Constrained Quadratic program). This particular QCQP is necessarily non-convex due to the nonlinear equality constraint.


问:
How to solve a third order nonlinear ODE (Falkner-Skan wedge equation)
How can I solve the following ODE numerically
$$F'''+FF''+1-F'^2=0$$
$$F(0)=F'(0)=0\qquad F'(\infty)=1$$
Thank you.

答:
Similar to How do you solve this kind of third order ODE?,

Let $U=(F')^2$ ,

Then $\pm\sqrt U\dfrac{d^2U}{dF^2}+F\dfrac{dU}{dF}-2U+2=0$


问:
How to show that series converges absolutely?
The problem:

Show that $$\sum_{n=2}^{\infty} \frac{\sin(\frac{n\pi}{12})}{\ln(n)}$$ converges Conditionally or Absolutely.

Showing that it converges without the absolute value, I have done by using Dirichlet's Theorem. Showing that $\frac{1}{\ln(n)}$ is decreasing and that the partial sum of $\sin(\frac{n\pi}{12})$ is bounded. However how can one show that it converges or diverges in absolute value?

Thank you very much.

答:
Note that infinite sum of (positive) reciprocals of the sequence $(\ln(an+b))_{n\in\mathbb{N}}$ is divergent for all $a, b>0$ (by comparison test or integral test). Since $\sin(\pi+\theta)=-\sin(\theta)$ for any $\theta,$ lets check $\sin\left(\dfrac{n\pi}{12}\right)$ for $n=0, 1, 2, \cdots, 11.$

\begin{array}{|c|c|}
\hline
  n & \sin\left(\dfrac{n\pi}{12}\right) \\
\hline
   0, 12 & 0 \\
\hline
   1, 11 & \dfrac{\sqrt6-\sqrt2}{4} \\
\hline
   2, 10 & \dfrac12 \\
\hline
   3, 9 & \dfrac{\sqrt2}2 \\
\hline
   4, 8 & \dfrac{\sqrt3}2 \\
\hline
   5, 7 & \dfrac{\sqrt6+\sqrt2}{4} \\
\hline
   6, 30 & 1 \\
\hline
\end{array}

Assuming the absolute convergence we can rearrange the series into $12$ divergent series, according to values in the above table, which is a contradiction.


问:
Finding $m$ given an arithmetic sequence
Suppose we have an arithmetic sequence $\{a_n\}$ so that it has $m$ terms. Given that $m$ is odd, the sum of the odd index terms $(a_1, a_3, \cdots)$ is $44,$ and the sum of the even index terms $(a_2,a_4,\cdots)$ is $33,$ find the value of $m.$

First, I noted that the last addend of the odd index term sum must be $a_m,$ since $m$ is odd. However, from here, I didn't really have any idea where to go because setting $a$ equal to the first term and $d$ being the common difference seemed too complicated. Can someone help please?

答:
Note there are $\frac{m+1}{2}$ odd index terms and $\frac{m-1}{2}$ even index terms.

These sequences separately form AP with common difference $2d$. So

$$ \dfrac{\tfrac{m+1}{2}(a_1+a_m) }{\tfrac{m-1}{2}(a_2+a_{m-1}) }=\dfrac{44}{33} $$

And $a_1+a_m=a_2+a_{m-1}$


答:
Hint:

Use lil' Fermat and the Chinese remainder theorem:

Modulo $5$, we have $n^5\equiv n$ for all $n$, hence
$$n^{18}=n^{15}n^3=(n^5 n)^3\equiv (n^2)^3=n^6\equiv n^2 \\
n^{14}=n^{10}n^4=(n^5n^2)^2\equiv(n^3)^2=n^6\equiv n^2\\
\text{so that }\quad  n^{18} - n^{14} - n^6 + n^2\equiv n^2-n^2-n^2+n^2=0 \mod 5.$$

Similarly, modulo $13$, we have $n^{13}\equiv n$ for all $n$. Can you proceed?

Title: Prove that integer $n$ is divisible by $65$

Body:
Can anybody help me with this problem?

Prove that for every integer $n$, $$p(n) = n^{18} - n^{14} - n^6 + n^2$$ is divisible by $65$.

Thanks!


问:
Basic question about Euler trails
In my lectures, we proved the following theorem:

A graph $G$ has an Euler trail iff all but at most two vertices have odd degree, and there is only one non-trivial component.

Moreover, if there are two vertices of odd degree, these are the end vertices of the trail. Otherwise, the trail is a circuit.

I am struggling with a small point in the $\leftarrow $ direction.

The proof as presented in the lectures (having already proven that if G has only even degree vertices then it has an Euler circuit) was: suppose our two vertices of odd degree are x and y. If they are not adjacent, then we can add an edge between them, and we can construct out Euler circuit. Removing the dge gives the result.

But suppose that x and y are adjacent. In my lectures, we claim that we can construct a new path between x and y using all new edges.

But I don't see how this can be. For a path of length 2, we need x and y to have a common non-neighbour, which we cannot guarantee. For a path of length general length, we need some i'th degree non-neighbour of x to be a jth degree non-neighbour of y. Where I define an i'th degree non-neighbour to be such that you can contruct a trail with i completely new edges between x and this vertex, z say.

If anyone has any hints that would be much appreciated.

答: If $x$ and $y$ are adjacent and you are not allowed to go into multigraphs, you add a completely new vertex $z$ connected only to $x$ and $y$, then construct the Euler circuit. Since $z$ is only connected to $x$ and $y$, the edges $x-z-y$ must be adjacent in the circuit thus constructed; removing these edges gives an Euler trail as required.

Title: Basic question about Euler trails


问:
What is the minimal value of n such that the symmetric group Sn that contains an even permutation of order 312?
What is the minimal value of $n$  such that the symmetric group $S_n$ contains an even permutation of order 312?

what I know:
There is an 8-cycle, a 3-cycle and a 13-cycle.
Any help would be appreciated.

答:
Let $s$ be a permutation in a group $S_n$ having order $312=2^3\cdot 3\cdot 13$. We write it as a product of disjoint (thus commuting) cycles, let $N\ge 1$ be the number of cycles,
$$
s =c_1\cdot c_2\cdot c_3\cdot\; \dots \;\cdot c_N\ .
$$
Let $d_1, d_2,d_3,\dots,d_N$ be respectively the orders of these cycles.

Then the order of $s$ is the lcm of these orders.
Also, each order $d_j$ divides $312=2^3\cdot 3\cdot 13$.

As a digression, let us give some examples of realizations of permutations of order $312$. (Even or odd permutations so far.) Please skip this block with examples, if it feels annoying, and go directly to the restated Problem.

For $n=312$ we have in $S_n$ the one cycle of order $312$, $c_1=(1,2,3,\dots,311,312)$, and take $s=c_1$, an odd permutation, there are $312-1=311$ inversions, $k<312$ "goes to" $k+1=s(k)>s(312)=1$.

We may waste space and use $n=312+312$, and consider $s$ the product of the two cycles $c_1=(1,2,\dots,312)$ and $c_2=(312+1,312+2,\dots, 312+312)$. In such cases, i will denote the "second cycle" using a distinguishing character, e.g. $c_2=(1',2',\dots,312')$.

For the representation $n=8+39$ we can take $c_1=(1,2,3,4,5,6,7,8)$ and $c_2=(1',2',\dots,39')$. This leads to an odd $s=c_1c_2$, since $c_1$ is odd, $c_2$ is even.

To make the above example work, and deliver an even permutation, we may enlarge the space of $8+39$ symbols, and use $n=8+39+2$ symbols, where we can take $c_1=(1,2,3,4,5,6,7,8)$ and $c_2=(1',2',\dots,39')$ and $c_3=(1'',2'')$. Then the product of the three cycles is even, $c_1$ is odd, $c_2$ is even, the transposition $c_3$ is odd.

For the representation $n=24+13$ we can take $c_1=(1,2,\dots,24)$ and $c_2=(1',2',\dots,13')$. This leads to an odd $s=c_1c_2$, since $c_1$ is odd, $c_2$ is even. As above, enlarge $n$ to the new value $n=24+13+2$ to produce an even permutation.

For the representation $n=3+8+13$ we obtain again a solution by using
$$
\begin{aligned}
c_1 &= (1,2,3)\ ,\\
c_2 &= (1',2',3',4',5',6',7',8')\ ,\\
c_3 &= (1^*,2^*,3^*,4^*,5^*,6^*,7^*,8^*,9^*,10^*,11^*,12^*,13^*)\ ,\\
s &=c_1c_2c_3\ ,
\end{aligned}
$$
but in this case the sign $(-1)^s$ (in my notation) of $s$ is
$$
(-1)^s=(-1)^{c_1}(-1)^{c_2}(-1)^{c_3}
=(+1)(-1)(+1)=-1\ ,
$$
so this $s$ is odd.

To have an even permutation of order $312$, with potential of being minimal, we can slightly adjust the above situation, by using $n=3+8+13+2$ and
$$
\begin{aligned}
c_1 &= (1,2,3)\ ,\\
c_2 &= (1',2',3',4',5',6',7',8')\ ,\\
c_3 &= (1^*,2^*,3^*,4^*,5^*,6^*,7^*,8^*,9^*,10^*,11^*,12^*,13^*)\ ,\\
c_4 &= (1^a, 2^a)\ ,\\
s &=c_1c_2c_3c_4\ ,
\end{aligned}
$$
and in this case the sign $(-1)^s$ of $s$ is
$$
(-1)^s=(-1)^{c_1}(-1)^{c_2}(-1)^{c_3}(-1)^{c_4}
=(+1)(-1)(+1)(-1)=+1\ ,
$$
so this $s$ is a solution for the problem, disregarding the minimality.

Is this $s$ above minimal?

The orders $d_1,d_2,d_3,\dots,d_N$ of the cycles $c_1,c_2,c_3,\dots,c_N$ involved in $s$ already determine (up to isomorphism) the permutation $s$ and the minimal $n=\sum d=d_1+d_2+d_3+\dots+d_N$ needed to realize this cycle in $S_n$.

So the problem can be restated:

Problem: Let $N\ge 1$ be a natural number, and let $d=(d_1,d_2,d_3,\dots,d_N)$ be a list of numbers $\ge 2$, with possible repetitions, all of them dividing $312=2^3\cdot 3\cdot 13$ such that:

the lcm of these numbers is $312$.

the number of the even entries in this list (those corresponding to odd cycles) is even.
Which is the minimal value for $\sum d:= d_1+d_2+d_3+\dots+d_N$?

We show that the minimal value is obtained for the list $(8,3,13,2)$.

Proof: Let $d$ be a possible list as above. The lcm of the entries in the list is $2^3\cdot 3\cdot 13$, so there exist indices $i,j,k$ with:

$8$ divides $d_i$, and $3$ divides $d_j$, and $13$ divides $d_k$.

If $d_i>8$, then $d_i=8d_i'$ with an odd $d'$, and we replace the list $d$ by a list $D$, where we copy $d$, all entries but $d_i$, and instead of $d_i$ we use $d_i'$ in $D$, and introduce a new entry $d_{N+1}=8$ in $D$. The number of even entries in the list is the same. The lcm is the same. Observe now that:
$$
\sum D-\sum d = d_i'+8-8d_i'=8-7d_i'\le 8-7\cdot 2<0\ .
$$
So $D$ is "better". When searching for the minimal $d$, we can thus restrict and consider only lists $d$ with an entry equal to $8$, up to reordering let us assume $d_1=8$.

The number of even entries in the list is even, so there is at least one further even entry, let this be $d_2$.

The entries $d_j,d_k$ may "collide" with $d_1,d_2$, well only with $d_2$. (Since $d_1=8$ is not divisible by 3, also not by $13$.) Let's see if such collisions can lead to a minimal $\sum d$.

If $j=2$, then $d_2$ is divisible by $2$, and by $3$. So $d_2\ge 6$. We replace then $d_2$ in $d$ by two entries, $2$ and $3$, obtain a new list $D$. The lcm is the same in $d$ and in $D$, the number of even entries is also the same. But $\sum D-\sum d=2+3-d_2\le 2+3-6<0$, so such a collision cannot deliver a minimal $d$.

If $k=2$, then $d_2$ is divisible by $2$, and by $13$. Same argument. So $d_2\ge 26$. We replace then $d_2$ in $d$ by two entries, $2$ and $13$, obtain a new list $D$. The lcm is the same in $d$ and in $D$, the number of even entries is also the same. But $\sum D-\sum d=2+13-d_2\le 2+13-26<0$, so such a collision cannot deliver a minimal $d$.

The collision $j=k$ is excluded similarly. We replace for instance $d_j$ in $d$ by two entries, $3$ and $d_j':=d_j/3$ (or $13$ and $d_j/13$) obtaining a new list $D$, and
$\sum D-\sum d=3+d_j'-3d_j'=3-2d_j'\le 3\cdot 2\cdot 2<0$, so the minimal $\sum d$ does not occur in case of such a collision.

After all these exclusions, the indices $1,2,j,k$ differ when realizing a minimal $\sum d$, which implies:
$$
\sum d\ge d_1+d_2+d_j+d_k\ge 8+2+3+13=26\ .
$$
$\square$


问:
Euler's theorem multiplicative inverse Z26
Euler's theorem get the inverse of 2 in Z26...~
i tried to solve it but the result is not correct:
a=2,  φ (26) = 12, m=26
gcd(2,26)=21
multiplicative inverse of 2 = 1/2
2 * 1/2 = 1

gcd(2,26)=2

a^-1= a^11mod26=2^11mod26= 2046 mod 26= 20

a=2     a^-1=20
aa^-1= 220=40

when doing the verification, 40 mod 26 = 14 not 1...

or for this group the Euler theorem is not applicable ?

答:
$2$ cannot have an inverse mod $26$ because $13 \cdot 2 \equiv 0$.

Indeed, if $2a \equiv 1$, then $ 0 = 13 \cdot 2 a \equiv 13$, but $13 \not\equiv 0$.


问:
Calculating limit of series.
Given a series $\sum_{n=1}^{\infty}\frac{4n+1}{2n(2n-1)(2n+1)(2n+2)}$, how do I find the limit? I understand I need to find the sequence of partial sums, which goes something like this $s_n=\{\frac{5}{24}, \frac{7}{30}, \frac{27}{112}....\}$which will probably converge at 0.25, but I am having trouble finding a description of the sequence on which I could evaluate the limit.

答:
Let $\dfrac{4n+1}{(2n-1)2n(2n+1)(2n+2)}=f(n-1)-f(n)$  where $f(r)=\dfrac{ar+b}{(2r+1)(2r+2)}$

so that $$\sum_{n=1}^m\dfrac{4n+1}{(2n-1)2n(2n+1)(2n+2)}=\sum_{n=1}^m(f(n-1)-f(n))=f(0)-f(m)$$

We need

$f(n-1)-f(n)=\dfrac{(2n+1)(2n+2)(an-a+b)-(an+b)(2n-1)2n}{(2n-1)2n(2n+1)(2n+2)}$

$\implies4n+1=n^2(4a)+2n(?)+2b-2a$

Comparing the coefficients of $n^2,4a=0\iff a=?$

Comparing the constants, $2b-2a=1\implies b=\dfrac12$

These values of  $a,b$ satisfy the coefficients of $n$

Now set $m\to\infty$ and $\lim_{m\to\infty}f(m)=?$

Title: Calculating limit of series.


问:
Proof P by premises $(\neg P\vee Q)\to P$
I have to proof next question by a formal proof that when the premise is equal to:
$(\neg P\vee Q)\to P$ has the result ,$P$, I first though that I must write:

$\neg P\vee Q\to P$, as:
$\neg(\neg P\vee Q)\vee P$ and work further from there.

The problem is that I only may use Reit ,introduction and Elimination. Can someone help me?

答:
Hint: using Boolean algebra,

$$\overline{\left( \overline{p} + q \right)} + p = p \overline{q} + p = p \,(\overline{q} + 1) = p$$


问:
Partial linear relaxation yields an integer solution
Consider a binary integer program
\begin{align}
\min \quad &\sum _{j \in J}f_j x_j +\sum _{i \in I} c_i y_i \notag \\
\mbox{s.t.} \quad &\sum _{j \in N_i} x_j \ge 1-y_i, \quad \forall i\in I \notag \\
 &x_j, y_i \in \{0,1\} \notag \\
\end{align}
where $f_i,c_i\ge0$. How do I prove that a linear relaxation of only $x$ variables (i.e. $x_j\ge0$) yields a binary solution? Any suggestion is greatly appreciated.

Note: This is a maximal covering location problem, where $N_i$ is a set of nodes. So, if at least one facility is located in $N_i$ (i.e. $\sum _{j \in N_i} x_j \ge 1$), node $i$ is considered covered (i.e. $y_i=0$). Otherwise, a cost $c_i$ is incurred.

答:
This doesn't seem to be true.

Take $|I| = |J| = 3$, and pick three neighbourhoods, namely $\{1,2\}, \{1, 3\},$ and $\{2,3\}$.  Take $c_i = 50$ for all $i$ and $f_i = 1$ for all $i$.

The optimal solution if we let $x$ be fractional is $x = (\frac12, \frac12, \frac12)$, $y = (0, 0, 0)$, which has objective value $1.5$.  You can check that, if $x$ and $y$ must both be integral, the optimal solution has objective value $2$.

EDIT:  However, for a fixed integral $x$, it is quite easy to find an optimal $y$; set $y_i$ to 1 whenever $c_i$ is negative or the corresponding constraint would otherwise be violated; set $y_i$ to zero otherwise.


问:
On the effect of action of product of permutation group on vectors
I am confused with the following permutation. Suppose $\sigma,\tau\in S_n$ the permutation group of order $n$. Define $\sigma.(x_1,x_2,\dots, x_n)=(x_{\sigma(1)},x_{\sigma(2)},\dots, x_{\sigma(n)})$. (it affects on the indices directly, no matter where is it) Using this definition what is the flaw of the following?
$$(\tau\sigma).(x_1,x_2,\dots, x_n)=\tau.(\sigma.(x_1,x_2,\dots, x_n))=\tau.(x_{\sigma(1)},x_{\sigma(2)},\dots, x_{\sigma(n)})=(x_{\tau(\sigma(1))},x_{\tau(\sigma(2))},\dots, x_{\tau(\sigma(n)}))$$

On the other hand, one can compute it as follows (the correct one):

$$(\tau\sigma).(x_1,x_2,\dots, x_n)=\tau.(\sigma.(x_1,x_2,\dots, x_n))=\tau.(x_{\sigma(1)},x_{\sigma(2)},\dots, x_{\sigma(n)})\stackrel{\text{let } y_i=x_{\sigma(i)}}{====}(y_{\tau(1)},y_{\tau(2)},\dots, y_{\tau(n)})=(x_{ \sigma(\tau(1))},x_{ \sigma(\tau(2))},\dots, x_{\sigma(\tau(n)})).$$

I think the confusion point is because of I am acting the $\tau$ and $\sigma$ in wrong order but I don't know how to justify it just using the definition $\sigma.(x_1,x_2,\dots, x_n)=(x_{\sigma(1)},x_{\sigma(2)},\dots, x_{\sigma(n)})$.

答:
The subtlety is that if you want to act on the subscript indices "directly", this needs to be a right action (not a left action). If you write it like

$$ (x_1,\dots,x_n) \cdot \sigma = (x_{\sigma(1)},\dots,x_{\sigma(n)}) $$

Then I think it will work out.

If you want to keep it a left action, you should use the inverse permutation:

$$ \sigma \cdot (x_1,\dots,x_n) = (x_{\sigma^{-1}(1)},\dots,x_{\sigma^{-1}(n)}) $$

Now when you evaluate $y_{\tau^{-1}(1)} = y_{\sigma^{-1}(\tau^{-1}(1))}$, this works, because $\sigma^{-1} \circ \tau^{-1} = (\tau \circ \sigma)^{-1}$.


问:
Examples of functions with a removable discontinuity
I recently came across the following example of a function which is discontinuous at a point:

The cost of a sushi buffet is calculated by weighing your plate, where you are then charged £10 per pound. As part of a promotion, if you manage to fill your plate so that the weight is precisely 1lbs, you get your meal for free.

This is a nice way to motivate a function like
$$f(x) = \begin{cases}10\,x\hfil&\text{$x\neq 1$}\\\hfil0\hfil&\text{$x=1$}\end{cases}$$
which are nice for illustrating the idea of a limit to students.

Are there any other real-world examples of functions like this, for the sake of explaining limits? Ideally one where you would actually want the limiting value, and not the assigned value, to motivate the concept further. (Apart from the obvious difference quotient.)

答:
Suppose there is a carpark that charges \$5 (Singapore dollars) for the first hour and then \$1 for every subsequent half-hour, including parts of half-hours. Thus if you parked for $2.5-\epsilon$ hours you would pay \$8, but parking for $2.51$ hours would incur \$9.
$$f(x)=5+\max(0,\lfloor2x-1\rfloor)$$
This function is discontinuous at infinitely many points.

For a two-sided limit, suppose there is a long, straight fence and a slit in that fence at $x=0$. The function is whether you can see beyond the fence – you cannot ($0$), even if you creep up to it (taking the limit), unless you are actually peering through the slit ($1$).
$$f(x)=[x=0]$$


问:
True or False statements of Splitting Fields
Verify if the following statements is true or false:

A: The splitting field of $X^2-X-1$ is equal to $\mathbb{Q}[\sqrt{5}]$

B: The splitting field of $X^4-5$ is equal to $\mathbb{Q}[\sqrt[4]{5}]$

What have I tried so far?

I have factorised A to give the splitting field $\mathbb{Q}[\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}]$

And for B we get $\sqrt[4]{5}$

Hence I think A is false and B is true.

答: For B:  note that $X=i\sqrt[4]5$ satisfies $X^4-5=0$, but $i\sqrt[4]5\not\in \Bbb Q(\sqrt[4]5)\subset\mathbb R$.


问:
degree of the splitting field of $x^{16}-1$ over $\Bbb F_9$

Compute $[E:\Bbb F_9]$ where $E/\Bbb F_9$ is a splitting field of $x^{16}-1$ over $\Bbb F_9$.

I want to find $\text{Gal}(E/\Bbb F_9)$. Since $E/\Bbb F_9$ is a cyclic extension, it suffices to compute the above degree. What I've done so far is that $x^{16}-1 = (x^8+1)(x^8-1)$ and as $\Bbb F_9$ is a splitting field of $x^9-x\in \Bbb F_3[x]$, $x^8-1$ splits in $\Bbb F_9$. And $x^8+1 = (x^4+2x^2+2)(x^4+x^2+2)$. I don't know how to get further. Could you give any hint?

答:
Hint: Find the splitting field over $\Bbb F_3$. Prove that $x^4+2x^2+2$ and $x^4+x^2+2$ are irreducible over $\Bbb F_3$.

Alternatively, since $z^8+1=0$ implies that $z$ has order $16$ in $E^\times$, consider the smallest $n$ such that $16$ divides $3^n-1$.


答:
First of all ,you should know that implication connector does not have disribution property on right hand side.

$(\neg P \lor Q)\rightarrow P$ is equal to $(\neg P \rightarrow P)\land (Q\rightarrow P)$

$(\neg P \rightarrow P)\land (Q\rightarrow P)$ $\equiv$ $( P \lor P)\land (\neg Q\lor P)$

So, $P \land (\neg Q\lor P)$ , by elimination

$\therefore P$

Title: Proof P by premises $(\neg P\vee Q)\to P$

Body:
I have to proof next question by a formal proof that when the premise is equal to:
$(\neg P\vee Q)\to P$ has the result ,$P$, I first though that I must write:

$\neg P\vee Q\to P$, as:
$\neg(\neg P\vee Q)\vee P$ and work further from there.

The problem is that I only may use Reit ,introduction and Elimination. Can someone help me?


问:
A sum of a continuous random variable and a discrete random variable is a continuous random variable?
Let $ X,Y $ be random variable such that $ X $ is continuous and $ Y $ is discrete.

Define $ Z=X+Y $.

I'm trying to prove that $ Z $ is continuous, but I cant really see why would it even be true. Any hints/ideas would be helpful.

Here's my proof:

In order for us to prove that $Z $ is a continious random variable, we want to show that $ F_z $, the CDF, is continious. We already know that $F_z $ is right continious, so its enough to show that its also left-continious.

Let $R_y $ be the range of the discrete variable $ Y $ and $f_Y $ be the density of $ y $ and $F_x $ the CDF of $ X $.

Now notice the following :
For any $t\in \mathbb{R} :

$ F_{z}\left(t\right)=\mathbb{P}\left(Z\leq t\right)=\mathbb{P}\left(X+Y\leq t\right)=\mathbb{P}\left(X\leq t-Y\right)=\mathbb{P}\left(\left(X\leq t-Y\cap Y\in R_{y}\right)\uplus\left(X\leq t-Y\cap Y\notin R_{y}\right)\right) $

$ =\mathbb{P}\left(X\leq t-Y\cap Y\in R_{y}\right)=\mathbb{P}\left(\biguplus_{K\in R_{y}}\left(X\leq t-Y\cap Y=K\right)\right)=\sum_{k\in R_{y}}\mathbb{P}\left(X\leq t-Y\cap Y=k\right)=\sum_{k\in R_{y}}\mathbb{P}\left(X\leq t-Y|Y=k\right)\mathbb{P}\left(Y=k\right) $

$ =\sum_{k\in R_{y}}\mathbb{P}\left(X\leq t-k\right)\mathbb{P}\left(Y=k\right)=\sum_{k\in R_{y}}Fx\left(t-k\right)f_{Y}\left(k\right) $

All in all we have that for any $ t \in \mathbb{R} $ the following equality holds:

$ F_{z}\left(t\right)=\sum_{k\in R_{y}}Fx\left(t-k\right)f_{Y}\left(k\right) $

Now Im gonna use the following result from calculus:

let $ f: \mathbb{R} \to \mathbb{R} $ be a continious function such that both of the limits :

$ \lim_{x\to\infty}f\left(x\right)=a\in\mathbb{R} $

$\lim_{x\to-\infty}f\left(x\right)=b\in\mathbb{R} $

exists. then $ f $ is absolutly continious function.

The proof is pretty easy, for any $\varepsilon >0 $ there exists $x_1<x_2 \in \mathbb{R} $ such that for any $ x>x_2 $ it follows that $ |f\left(x\right)-a|<\frac{\varepsilon}{3} $

and for any $x<x_1 $ it follows that $ |f\left(x\right)-b|<\frac{\varepsilon}{3} $

Now in the interval $ [x_1-1,x_2+1] $, $ f $ is continious so by Heine-Cantor theorem there exists $0<\delta<\frac{1}{2}|x_{2}+1-\left(x_{1}-1\right)|$ such that for any $ s_1,s_2 \in [x_1-1,x_2+1] $ it follows that $ |f(s_1)-f(s_2)|<\varepsilon/3 $.

One can check that this $ \delta $ proves the absolute continuity of $ f $.

So, since $ F_x $ is such a function, ( $ F(\infty)=1 $ and $ F(-\infty)=0 $ )

We have that $ F_x $ is an absolutly continious function. So for fixed $ \varepsilon > 0 $  there exists $ \delta > 0 $ such that if $ |s-t|<\delta $ then $ |F_x(s)-F_x(t)|<\varepsilon $

Now fix some $ t_0 \in \mathbb{R} $ , we'll prove that $ F_z $ is left-continious at $ t_0 $.

Indeed for any $ t \in (t_0-\delta/2,t_0)  $ we have :

$ |F_{z}\left(t_{0}\right)-F_{z}\left(t_{0}-\frac{\delta}{2}\right)|\underset{F_{z}\thinspace\thinspace is\thinspace\thinspace increasing}{=}F_{z}\left(t_{0}\right)-F_{z}\left(t_{0}-\frac{\delta}{2}\right) $

$ =\sum_{k\in R_{y}}f_{y}\left(k\right)\left(F_{x}\left(t-k\right)-F_{x}\left(t-k-\frac{\delta}{2}\right)\right)\underset{F_{x}\thinspace is\thinspace\thinspace increasing}{=}\sum_{k\in R_{y}}f_{y}\left(k\right)|F_{x}\left(t-k\right)-F_{x}\left(t-k-\frac{\delta}{2}\right)| $

$ <\varepsilon\sum_{k\in R_{y}}f_{Y}\left(k\right)=\varepsilon $

Where I used the fact that CDF function is non-decreasing, and that $ \sum_{k\in R_{y}}f_{Y}\left(k\right)=1 $.

Im totally aware that there are much easier ways to prove it, but it requires additional tools that I do not have yet (I only know the definitions and the basic things that I used in my proof, also I saw other proofs that assumed differntiability and assume much less).

So by the way I noticed that any CDF function of a continious random variable is absolutly continious. Thats nice

答: Note that the measure of the set of possible values of a discrete variable is $0$ because it is countable. Can you proceed from here?


答:
Hint

Define the PDFs of $X,Y,Z$ as $f_X(x),f_Y(y),f_Z(z)$. Start by proving that
$$
f_Z(z)=f_X(z)*f_Y(z)
$$
where $*$ denotes the convolution operator.

Title: A sum of a continuous random variable and a discrete random variable is a continuous random variable?

Body:
Let $ X,Y $ be random variable such that $ X $ is continuous and $ Y $ is discrete.

Define $ Z=X+Y $.

I'm trying to prove that $ Z $ is continuous, but I cant really see why would it even be true. Any hints/ideas would be helpful.

Here's my proof:

In order for us to prove that $Z $ is a continious random variable, we want to show that $ F_z $, the CDF, is continious. We already know that $F_z $ is right continious, so its enough to show that its also left-continious.

Let $R_y $ be the range of the discrete variable $ Y $ and $f_Y $ be the density of $ y $ and $F_x $ the CDF of $ X $.

Now notice the following :
For any $t\in \mathbb{R} :

$ F_{z}\left(t\right)=\mathbb{P}\left(Z\leq t\right)=\mathbb{P}\left(X+Y\leq t\right)=\mathbb{P}\left(X\leq t-Y\right)=\mathbb{P}\left(\left(X\leq t-Y\cap Y\in R_{y}\right)\uplus\left(X\leq t-Y\cap Y\notin R_{y}\right)\right) $

$ =\mathbb{P}\left(X\leq t-Y\cap Y\in R_{y}\right)=\mathbb{P}\left(\biguplus_{K\in R_{y}}\left(X\leq t-Y\cap Y=K\right)\right)=\sum_{k\in R_{y}}\mathbb{P}\left(X\leq t-Y\cap Y=k\right)=\sum_{k\in R_{y}}\mathbb{P}\left(X\leq t-Y|Y=k\right)\mathbb{P}\left(Y=k\right) $

$ =\sum_{k\in R_{y}}\mathbb{P}\left(X\leq t-k\right)\mathbb{P}\left(Y=k\right)=\sum_{k\in R_{y}}Fx\left(t-k\right)f_{Y}\left(k\right) $

All in all we have that for any $ t \in \mathbb{R} $ the following equality holds:

$ F_{z}\left(t\right)=\sum_{k\in R_{y}}Fx\left(t-k\right)f_{Y}\left(k\right) $

Now Im gonna use the following result from calculus:

let $ f: \mathbb{R} \to \mathbb{R} $ be a continious function such that both of the limits :

$ \lim_{x\to\infty}f\left(x\right)=a\in\mathbb{R} $

$\lim_{x\to-\infty}f\left(x\right)=b\in\mathbb{R} $

exists. then $ f $ is absolutly continious function.

The proof is pretty easy, for any $\varepsilon >0 $ there exists $x_1<x_2 \in \mathbb{R} $ such that for any $ x>x_2 $ it follows that $ |f\left(x\right)-a|<\frac{\varepsilon}{3} $

and for any $x<x_1 $ it follows that $ |f\left(x\right)-b|<\frac{\varepsilon}{3} $

Now in the interval $ [x_1-1,x_2+1] $, $ f $ is continious so by Heine-Cantor theorem there exists $0<\delta<\frac{1}{2}|x_{2}+1-\left(x_{1}-1\right)|$ such that for any $ s_1,s_2 \in [x_1-1,x_2+1] $ it follows that $ |f(s_1)-f(s_2)|<\varepsilon/3 $.

One can check that this $ \delta $ proves the absolute continuity of $ f $.

So, since $ F_x $ is such a function, ( $ F(\infty)=1 $ and $ F(-\infty)=0 $ )

We have that $ F_x $ is an absolutly continious function. So for fixed $ \varepsilon > 0 $  there exists $ \delta > 0 $ such that if $ |s-t|<\delta $ then $ |F_x(s)-F_x(t)|<\varepsilon $

Now fix some $ t_0 \in \mathbb{R} $ , we'll prove that $ F_z $ is left-continious at $ t_0 $.

Indeed for any $ t \in (t_0-\delta/2,t_0)  $ we have :

$ |F_{z}\left(t_{0}\right)-F_{z}\left(t_{0}-\frac{\delta}{2}\right)|\underset{F_{z}\thinspace\thinspace is\thinspace\thinspace increasing}{=}F_{z}\left(t_{0}\right)-F_{z}\left(t_{0}-\frac{\delta}{2}\right) $

$ =\sum_{k\in R_{y}}f_{y}\left(k\right)\left(F_{x}\left(t-k\right)-F_{x}\left(t-k-\frac{\delta}{2}\right)\right)\underset{F_{x}\thinspace is\thinspace\thinspace increasing}{=}\sum_{k\in R_{y}}f_{y}\left(k\right)|F_{x}\left(t-k\right)-F_{x}\left(t-k-\frac{\delta}{2}\right)| $

$ <\varepsilon\sum_{k\in R_{y}}f_{Y}\left(k\right)=\varepsilon $

Where I used the fact that CDF function is non-decreasing, and that $ \sum_{k\in R_{y}}f_{Y}\left(k\right)=1 $.

Im totally aware that there are much easier ways to prove it, but it requires additional tools that I do not have yet (I only know the definitions and the basic things that I used in my proof, also I saw other proofs that assumed differntiability and assume much less).

So by the way I noticed that any CDF function of a continious random variable is absolutly continious. Thats nice


问:
$g \colon [0,1] \to [0,1]$ be a continuous map and consider the iteration $x_{n+1}=g(x_n)$.
I came across the following problem:

Let $g \colon [0,1] \to [0,1]$ be a continuous map and  consider the iteration $x_{n+1}=g(x_n)$.Then Which of the following maps will yield a fixed point for $g$?

The options are as follows:

a. $g(x)=\frac{x^2}{4}$,

b. $g(x)=\frac{x^2}{32}$,

c. $g(x)=\frac{x^2}{8}$,

d. $g(x)=\frac{x^2}{16}$.

Can someone point me in the right direction? Thanks in advance for your time.

答:
Graph each of these functions in a square $[0,1]\times[0,1]$. A fixed point is exactly where the graph crosses the diagonal of the square. Where does this happen for each function?

Alternately, in the vein of using too-powerful technology, since $[0,1]$ is a complete metric space, you can use the contraction mapping theorem (bearing in mind that differentiable functions on compact spaces are globally Lipschitz, with global Lipschitz constant the maximum of the derivative.)


问:
Find volume using triple integration with cylindrical coordinates
Find volume of $E$ using triple integration and cylindrical coordinates, when $E$ is bounded by $$x^2+y^2=x,\quad y=0,\quad y=x,\quad z=0,\quad z=\sqrt{x^2+y^2}$$
I know that in cylindrical coordinates $$x=r\cos\varphi,\quad y=r\sin\varphi,\quad z=h,\quad \text{ where } r\geq0 \text{ and } 0\leq\varphi\leq2\pi$$
but I'm very confused how to set up this integral. Cylindrical coordinates are quite new for me and it's hard to understand, how to make this conversion. So I would be grateful if anyone can help me with this.

答:
$x^2+y^2=x$ in polar form is just $r=\cos\varphi$. The lines $y=0$ and $y=x$ intersect this curve at $\varphi=0$ and $\varphi=\frac\pi4$ respectively. So our outer integrals are
$$\int_{\varphi=0}^{\pi/4}\int_{r=0}^{\cos\varphi}r\,dr\,d\varphi$$
The bounds of $z$ are just $z=0$ to $z=r$ in cylindrical coordinates, so our final answer is
$$\int_{\varphi=0}^{\pi/4}\int_{r=0}^{\cos\varphi}\int_{z=0}^rr\,dz\,dr\,d\varphi$$


问:
$\exists g: \Bbb R \to \Bbb R$ such that $f(x) = xg(x)$
In an exercise I'm asked to prove the following:

Let $f: \Bbb R \to \Bbb R$ be differential at $x = 0$ and $f(0)=0$. Show that there exists a continuous function $g: \Bbb R \to \Bbb R$ at $x = 0$ such that $f(x) = xg(x), \forall x \in \Bbb R$.

What I did was the following:

Let $g: \Bbb R \to \Bbb R$ such that $f(x) = xg(x) ,\forall x \in \Bbb R$. Then, because $f$ is differentiable at $x = 0$:

$$\lim_{x \to 0} \frac{f(0) - f(x)}{-x} = \lim_{x \to 0} \frac{-xg(x)}{-x} = \lim_{x \to 0} g(x)$$

Because $\lim_{x \to 0} \frac{f(0) - f(x)}{-x}$ exists then $\lim_{x \to 0} g(x)$ also exists.

So, I proved that $\lim_{x \to 0} g(x)$ exists but this is not enough conlcude that $g$ is continuous at $x = 0$ because I still need to show that $\lim_{x \to 0} g(x) = g(0)$ but I'm not being able to do so. How can I prove this?

Edit:

I think I figured it out:

$f(x) = xg(x)$ so $g$ must be differentiable at $0$, Because if it's not then $f$, would be the product of a $x$ with a non-differentiable function and therefore would not be differentiable.

$f'(0) = g(0) + 0 \cdot g'(0) = g(0)$

But $f'(0) = \lim_{x \to 0} \frac{f(0) - f(x)}{-x} = \lim_{x \to 0} g(x)$.

So we conclude that: $\lim_{x \to 0} g(x) = g(0)$

Is this correct?

答: Define $g(x)$ such that $g(x) = \frac{f(x)}{x}$ if $x \neq 0$. And $g(0) = f'(0)$. Note that $\lim_{x \to 0} g(x) =\lim_{x \to 0} \frac{f(x) - f(0)}{x-0} = f'(0)$ so $g$ is continuous at $x = 0$, and continuous elsewhere as $f(x)$ and $x$ are continuous. And $0g(0) = 0 = f(0)$ and everywhere else clearly $xg(x) = f(x)$.


问:
Difference between x(2t-3) and x(2(t-3)).
Ofcourse x(2(t-3))=x(2t-6), but my question is in terms of operations on signals. In what order does the shifting and scaling happens in the case of x(2(t-3)) and how is it different from the ones that happens with x(2t-3). And is this different in terms of systems ?

In my head I always have the image of x(2(t-3)) as shift three units and then scale . But I am just confused with the signals and systems jargon .

Thank you in advance

答:
All is about the order of operations and nothing more. Consider operation 1 to be shifting 3 units to right and operation 2 to be compressing the signal by a factor of 2.

Signal $x(2t-3)$ is first shifted 3 units to right and then the result is compressed by a factor of 2.

Signal $x(2(t-3))$ is first compressed by a factor of 2 and then the result is shifted 3 units to right.