问:
Present value of perpetuity with one extra payment
Consider the perpetuity that pays $3$ by the end of the 2nd year and then every $4$ years. I need to calculate the present value of it when $i=0.05$. So the second payment will be at 6th year. I thought about doing it by just discounting the first one and then express the rest of the discounts by a geometric series:
$$PV=3\cdot v^2+\sum_{k=1}^\infty 3\cdot v^{4k+2}=3\cdot \left( v^2+\frac{v^6}{1-v^4}\right)\approx15.348.$$
Is that correct?

答: This is just a deferred perpetuity-due with deferral period of $2$ years, and periodic payments of $4$ years:  $$\require{enclose} PV = 3v^2 \ddot a_{\enclose{actuarial}{\infty} j} = 3\frac{1+j}{(1+i)^2 j}$$ where $v = 1/(1+i)$ is the annual present value discount factor, and $j = (1+i)^4 - 1$ is the $4$-year effective interest rate.  Therefore, the present value is $$PV = \frac{3(1+i)^4}{(1+i)^2 ((1+i)^4 - 1)} = \frac{3(1+i)^2}{(1+i)^4 - 1} = \frac{3(1.05)^2}{(1.05)^4-1} \approx 15.3476.$$


问:
Condition Numbers
I'm working my way through a basic textbook on Numerical Linear Algebra, and one of the problems asks if the fact that an arbitrary invertible matrix $A \in \mathbb{R}^{m \times m}$ is perfectly conditioned in the 1-norm implies that it is perfectly conditioned in the $\infty$-norm. Formally:

cond$_{1}(A) = 1 \Longrightarrow \text{cond}_{\infty}(A) = 1$ ?

I can prove that this holds for symmetric matrices (i.e when $A = A^{T}$), but not for the general case. My intuition was to try to disprove this by contradiction, but I cannot think of/compute any non-symmetric matrix $B$ such that $det(B) \neq 0$ and $\text{cond}_{1}(B) = 1$. I've also been trying to come up with something utilizing $\text{cond}_{1}(A^{T}A)$, $\text{cond}_{1}(AA^{T})$, or $\text{cond}_{1}(A^{T})$, but so far have only succeeded in showing that the lower bound $1 \leq \text{cond}_{\infty}(A)$.

If anyone could offer any help with finding a suitable counterexample or give a hint on how to approach this, I would greatly appreciate it.

答:
$$  B = \begin{pmatrix}0 & \sqrt{10} \\ -\sqrt{10} & 0 \end{pmatrix}  $$

$B$ is antisymmetric.

$\det(B) = 10$.

$\kappa_1(B) = \sqrt{10} \cdot \frac{1}{\sqrt{10}} = 1$.

This is the example you requested.  It is not an example with $\kappa_\infty \neq 1$.

There is no example with $\kappa_\infty = 1$.

$||B||_1$ is the maximum absolute column sum of $B = \left( b_{ij} \right)_{i,j=1}^n$ and $||B||_\infty$ is the maximum absolute row sum of $B$.  Then \begin{align*}
||B||_1 &= \max_{1 \leq j \leq n} \sum_{i=1}^n|b_{ij}|  \\
||B||_\infty &= \max_{1 \leq i \leq n} \sum_{j=1}^n|b_{ij}|  \\
\end{align*}
Observe that $||B^{-1}||_1$ is the quotient of the maximum absolute row sum of $B$ and the determinant of $B$.  Observe that $||B^{-1}||_\infty$ is the quotient of the maximum absolute column sum of $B$ and the determinant of $B$.
\begin{align*}
||B^{-1}||_1 &= \frac{1}{\det B} \max_{1 \leq i \leq n} \sum_{j=1}^n|b_{ij}|  \\
||B^{-1}||_\infty &= \frac{1}{\det B} \max_{1 \leq j \leq n} \sum_{i=1}^n|b_{ij}|  \\
\end{align*}

Then
\begin{align*}
\kappa_1(B) &= ||B||_1 \, ||B^{-1}||_1  \\
    &= \max_{1 \leq j \leq n} \sum_{i=1}^n|b_{ij}| \cdot \frac{1}{\det B} \max_{1 \leq i \leq n} \sum_{j=1}^n|b_{ij}|  \\
    &= \max_{1 \leq i \leq n} \sum_{j=1}^n|b_{ij}| \cdot \frac{1}{\det B} \max_{1 \leq j \leq n} \sum_{i=1}^n|b_{ij}|  \\
    &= ||B||_\infty \, ||B^{-1}||_\infty  \\
    &= \kappa_\infty(B)  \text{.}
\end{align*}

(Since you only ask for a hint, I won't explain why the formulas for the norms of the inverse matrices are correct.)

ContentLicense: CC BY-SA 4.0

Title: Condition Numbers


问:
What does algebraically closed field play the role in Schur's unitary triangulation theorem?
Schur's unitary triangulation theorem said that

Theorem (Schur’s Triangularization Theorem) Every square complex matrix A is unitarily similar to an upper-triangular matrix, i.e., there exists a unitary matrix $U$ such that $T = U^∗AU$ is triangular.

Here, the theorem is about square complex matrix, but in the lecture, professor mentioned that any algebraically closed field matrices can fit the theorem [with similarity instead of unitary similarity] (if there are mistake on the expression, it must be me misunderstanding).

My questions is, why algebraically closed field can work, how to connect the proof with the original square complex matrix proof?

答:
suppose you are working over some algebraically closed field $\mathbb K$.  Just mimic the Schur Triangularization process, but at each stage when you complete to a basis (invertible matrix) you ignore the requirement for orthonormal columns.  The end result is

$A=STS^{-1}$

(Exactly this is walked through step by step in chapter 4 of Artin's Algebra.)

how to connect the proof with the original square complex matrix
proof?

once you have the above, if you are working over $\mathbb C$, you can run QR factorization,

$S=QR\implies A=QRTR^{-1}Q^{-1}=Q\big(RTR^{-1}\big)Q^*$

which recovers standard Schur Triangularization-- recalling that upper triangular matrices form a (semi)group.


问:
Symmetric and self-adjoint unbounded operator
I'm using the definitions here: Distinguishing between symmetric, Hermitian and self-adjoint operators.

Let $\mathcal{D}(D) = \{ (x_n)_n \in \ell^2 : \sum_{n} n^2|x_n|^2 < \infty \}$.

Let $D: \mathcal{D}(D) \rightarrow \ell^2$ be the diagonal operator $De_n=ne_n$.

Show is symmetric. Is $D$ self-adjoint?

My attempt:

To show it's symmetric, we want to show that $\langle x, Dy \rangle = \langle Dx, y \rangle$.

We have $\langle D^* e_n, e_m \rangle = \langle e_n, De_m \rangle = \langle e_n, n e_m \rangle = \bar{n} \delta_{mn} = n \delta{mn}$.

Now the definitions are messing me up.. I'm not sure if I'm proving that it's symmetric or self-adjoint...

Thank you for your help!

答:
The difference is where you are testing your operator.

To see that an operator is symmetric, you proceed as you did and show that given two generic elements of your domain $x,\ y\in\mathcal D\left(D\right)$ you have that $ \langle x, Dy\rangle= \langle Dx,y\rangle $ (with the same $ D $!).

On the other hand, the concept of self-adjoint goes hand in hand with the concept of adjoint; if you have that $ D^* $ is the adjoint of $ D $, then self-adjoint means only that as operators $ D $ and $ D^* $ coincide, i.e., their domains are equal and the action on the domains are equal.

So we are reduced to understand what the adjoint of an operator $ D $ is. To do so, you first assume that your operator $ D $ is densily defined (which is your case) and define the domain
$$
\mathcal D\left(D^*\right):=\left\{x\in\ell^2|\ \exists C\mbox{ s.t. }\left|\langle x,Dy\rangle\right|\le C\left\|y\right\| \forall y\in \mathcal D\left(D\right)\right\}.
$$

Then, for any element in $ \mathcal D\left(D^*\right) $ the mapping $ y\mapsto\langle x,Dy\rangle $ is define on a dense of your Hilbert space $ \ell^2 $ and it is bounded, so it can be extended to an element of the dual, and therefore, via Riesz representation theorem there is a unique element of your space $ z $ such that $ \langle x,Dy\rangle=\langle z,y\rangle $ for any $ y\in\mathcal D\left(D\right) $. By definition, $ D^*x:=z $.

Getting to the exercise. As you noticed, $ \langle e_n, De_m\rangle=n\delta_{nm}=\langle De_n, e_m\rangle $, therefore, given $ x,\ y\in\ell^2 $ we have
\begin{align}
\langle x,D y\rangle
&=\langle\sum_{n\in\mathbb N}x_ne_n,D\sum_{m\in\mathbb N}y_me_m\rangle
=\sum_{n\in\mathbb N}\sum_{m\in\mathbb N}x_n y_m\langle e_n,De_m\rangle
\\
&
=\sum_{n\in\mathbb N}\sum_{m\in\mathbb N}x_n y_mm\delta_{nm}
=\sum_{n\in\mathbb N}\sum_{m\in\mathbb N}x_n y_mn\delta_{nm}
=\langle Dx,y\rangle,
\end{align}
and $ D $ is symmetric.

Now I will show you that $ \mathcal D\left(D^*\right)=\mathcal D\left(D\right) $. The inclusion $ \supseteq $ is trivial, so let's show the other inclusion. Let $ x $ be in $ \mathcal D\left(D^*\right) $. Consider the sequence of elements $ y(k) $ of $ \mathcal D\left(D\right) $ (a sequence of sequences) defined as
$$
y(k)=\sum_{n=0}^k\frac{nx_n}{\left\|\sum_{n=0}^knx_ne_n\right\|}e_n.
$$

With this definition $ \|y(k)\|=1 $ and $ y(k)\in\mathcal D\left(D\right) $ trivially, given that it is a finite sequence. Therefore we get that, given that $ x\in\mathcal D\left(D\right) $, there exists a constant $ C $ such that
\begin{align}
C=C\left\|y(k)\right\|&\ge\left|\langle x,Dy\left(k\right)\rangle\right|
=\left|\sum_{n\in\mathbb N}\overline{x_n}ny(k)_n\right|
=\frac1{\left\|\sum_{n=0}^knx_ne_n\right\|}\left|\sum_{n=0}^kn^2\left|x_n\right|^2\right|
\\
&=\left(\sum_{n=0}^kn^2\left|x_n\right|^2\right)^\frac12.
\end{align}

It follows immediately that we can send $ k $ to infinity and get that the series in the last term is convergent and $ x\in\mathcal D\left(D\right) $ and the two domains coincide.

Finally is an exercise of dialectic to prove that now the actions of $ D $ and $ D^* $ coincide on their (common) domain. Indeed let $ x,y\in\mathcal D\left(D\right)=\mathcal D\left(D^*\right) $, by definition of $ D^* $ and using the symmetry we get
\begin{align}
\langle x,D^* y\rangle
=\langle Dx,y\rangle
=\langle x,D y\rangle.
\end{align}

Given that the domain of $ D $ is dense, this implies $ D^*y=Dy $ for every y in the domain, and therefore $ D=D^* $. Hence, $ D $ is also self-adjoint.


问:
Prove the following limit equals 0
Question: Prove that $$\lim\limits_{x \to 0} \frac{e^{-1/x^2}}{x^2}=0$$ by using the E-D definition.
It suffices to find a $ \delta>0$ such that if $x<\delta \implies \frac{e^{1/x}}{x^2}< \epsilon $. However you need to use Lambert W function to solve this inequality. Is there any other way to approach this question? Any help would be appreciated.

答:
You may  know that
$$\lim_{x\to 0^+}\frac{x^2}{e^{x^2}-1}=1$$

So, you just need proving

$$\lim_{u\to-\infty}u^2e^u=0$$

where $ u=\frac{-1}{x}$.

Title: Prove the following limit equals 0


问:
Elegant way to solve $n\log_2(n) \le 10^6$
I'm studying Tomas Cormen Algorithms book and solve tasks listed after each chapter.

I'm curious about task 1-1. that is right after Chapter #1.

The question is: what is the best way to solve: $n\lg(n) \le 10^6$, $n \in \mathbb Z$, $\lg(n) = \log_2(n)$; ?

The simplest but longest one is substitution. Are there some elegant ways to solve this? Thank you!

Some explanations: $n$ - is what I should calculate - total quantity of input elements, $10^6$ - time in microseconds - total algorithm running time. I should figure out nmax.

答: The first approximation for the inverse of $x\;\text{lg}\; x$ is $x/\text{lg}\; x$.  So begin with the value $10^6/\text{lg}(10^6)$, then adjust by substitution/bisecion.</p>

ContentLicense: CC BY-SA 3.0


问:
Drawing colored balls from a bag
Adam has a bag with $5$ balls that are either black or white.

(i) Suppose that the bag consists of $2$ black balls and $3$ white balls. Adam draws balls from the bag with replacement until he has drawn at least one ball of each color. Let $X$ be the number of draws. Find the probability mass function (pmf) of $X$.

(ii) Suppose that the number of black balls is a random variable $Y$ which can take values $2, 3$ or $4$ with equal probability. If Adam draws $5$ balls with replacement and finds that $3$ of the draws are black and $2$ are white, what is the conditional probability that $Y = 4$?

I am practicing for an exam and found this question. I have tried solving it but I have been unsuccessful and no solutions are available. Any assistance is much appreciated.

答:
The basic idea of probability is counting.  You need to list out the ways something can happen, at least for the first few steps, and then look for the pattern.

For the first question, imagine that you draw the first ball.  There are two possibilities, black or white.  Suppose it is black.  Working from there, what is the distribution?  Then, suppose it is white, and do it again. You already stated that you guessed it was geometric; that is correct.  Work out the first few cases.  Given that the first draw is black, what is the probability that $X = 2$?  That $X = 3$? and so on.

The first branch, black is drawn first, has probability $\frac{2}{5}$.  Given that, $P(X=2) = P($white$) = \frac{3}{5}.$

Given black first, $P(X=3) = P($second was black and third was white$)$. That is $\frac{2}{5} \cdot \frac{3}{5}$. How exactly this spells out as geometric depends on how your class or textbook defines the geometric. For our purposes, $P(X=n$ and first draw was black$) = \frac{2}{5} \cdot \big({\frac{2}{5}}\big)^{n-2} \cdot \frac{3}{5}$

Likewise if the first draw is white, $P(X=n $and first draw was white$) = \frac{3}{5} \cdot {\frac{3}{5}}^{n-2} \cdot \frac{2}{5}$

And $P(X=n) = P(X=n$ and first was black$) + P(X=n$ and first was white).

For the second question, I think we need to assume that the number of black balls is chosen randomly once and then stays fixed.
The goal is $P(Y=4| 3$black, $2$ white).  A conditional probability follows Bayes' rule:
$$P(A|B) = P(B|A)\cdot \frac{P(A)}{P(B)}$$
Since there are only five draws you can map out the probabilities and total up each of them for the formula.  Let $A$ be "$Y=4$" and let $B$ be "$3$ black, $2$ white".

Title: Drawing colored balls from a bag


问:
Intuition for polynomial long division
So I am trying to learn some basic algebra and go over some precalc material and I am terribly confused on why or how polynomial long division works.

I have looked at many sources online and they seem to all suggest that division of numbers is identical to division of polynomials since a number in its decimal expansion can be viewed as a sort of polynomial e.g $a_n10^n+\cdots+a_2100+10a_1+a_0$

What i don't understand is why we only use the term with the highest degree in the denominator as the divisor for the whole polynomial as shown here  Polynomial Long Division whereas with normal division we divide the whole divisor into the dividend for example when calculating $88÷32$ we look at how many times $32$ goes into $88$ and not how many times $30$ does whilst with polynomial division only the highest degree term is considered which corresponds to 30 in this example ignoring 2.

So how are they the same? Some intuition behind why polynomial division works would be greatly appreciated.

Thanks in advance.

答:
Lets just forget how regular long division works for a minute and do it the hard way. Say we're dividing $888/17$ then we could repeatedly subtract $17$ from $888$ and count how many times this occurs until the remainder is less than $17$. This count will be the answer we're looking for.

This would however take a long time. Instead we could use groups of $17$ count how many times we subtract them off then multiply by the size of the groups. So for example I could do groups of $170$, count the number of times then multiply it by $10$. This makes the work a lot easier.

Now if we choose the optimal group size to subtract off at each stage you're now doing long division as usual. You find the highest multiple that fits into the existing number, subtract it off, then repeat until the remainder is less than $17$. You then add the group counts up and you're left with the correct answer.

Polynomial long division is the same, but the optimal size is determined by the highest power and not the coefficients. This is because we lose some information when working with polynomials that we have in base $10$, namely that $x=10$ and the coefficients are less than $10$. It limits how much we can subtract off at each stage reliably but the idea is the same, repeatly removing as much as you can until the remainder is smaller than the divisor.

Title: Intuition for polynomial long division


答:
In long division, we have to pay attention to the entire divisor because multiples of the divisor can result in carries between the various digits (columns in the usual tableau for long division).  There is no equivalent to carries when multiplying polynomials.

There are no carries between columns in polynomial division, so the non-leading degree terms cannot alter the multiple of the divisor that gives matching leading terms.

Consider $40 \div 13$.  The carry from the $3$ in the units place spoils the candidate quotient, $4$: $4 \times 13 = 52$, so the leading digit is too large.  This can't happen in the analogous $(4x+0) \div(x+3)$ -- when we multiply by $4$, $4(x+3) =4x+12$, there are no carries between terms, so the non-leading terms do not interact with the multiple of the leading term that gives the correct term in the quotient.

In long division of numbers, carries allow different columns to talk to each other.  In long division of polynomials, there is no communication between columns, so it is always sufficient to compare just the leading terms.

Title: Intuition for polynomial long division

Body:
So I am trying to learn some basic algebra and go over some precalc material and I am terribly confused on why or how polynomial long division works.

I have looked at many sources online and they seem to all suggest that division of numbers is identical to division of polynomials since a number in its decimal expansion can be viewed as a sort of polynomial e.g $a_n10^n+\cdots+a_2100+10a_1+a_0$

What i don't understand is why we only use the term with the highest degree in the denominator as the divisor for the whole polynomial as shown here  Polynomial Long Division whereas with normal division we divide the whole divisor into the dividend for example when calculating $88÷32$ we look at how many times $32$ goes into $88$ and not how many times $30$ does whilst with polynomial division only the highest degree term is considered which corresponds to 30 in this example ignoring 2.

So how are they the same? Some intuition behind why polynomial division works would be greatly appreciated.

Thanks in advance.


问:
Is ZFC+V=L consistently $\omega$-complete?
Say that an extension of ZFC is $\omega$-complete if any two of its $\omega$-models are elementarily equivalent. While "ZFC+V=L is $\omega$-complete" is easily disprovable in theories only slightly stronger than ZFC, I don't immediately see how to do it in ZFC alone. In particular, my question is:

Is the theory ZFC + V=L + $\omega$-Con(ZFC+V=L) + "ZFC+V=L is $\omega$-complete" consistent?

("$\omega$-Con(ZFC+V=L)" - the statement "ZFC+V=L has an $\omega$-model" - is included to avoid the trivial answer; of course, I could have written "$\omega$-Con(ZFC)" instead, but I thought this was a bit clearer.)

Note that one difficult here is the current paucity of sentences independent of ZFC+V=L which do not involve a jump in consistency strength and are not arithmetic.

答: This question was negatively answered at MO by user Farmer S, using $\omega_1^{CK}$-recursion theory. (I'm posting this here to move this off the unanswered queue, and have made this CW to avoid reputation gain for someone else's work.)


问:
Optimal Strategy for a betting game.
I shall present my question and then add some of the working I have tried, unfortunately I haven't had much luck so far.

The game

The game is played between $3$ players. You and $2$ opponents who we shall call $I$ and $R$ (for intelligent and random) You each select an integer from $1-100$ (inclusive) and then money is exchanged as follows:

Whoever selected the Largest number is the looser, they must pay the two winners what they each selected.

Example: You pick $42$, $I$ picks $12$ and $R$ picks $51$ then $R$ must pay you $\$42$ and he must pay $I$ $\$12$

In the event of a two or more players selecting the largest number then a looser is selected uniformly between them.

$R$ plays uniformly randomly, that is he selects his pick uniformly between the first $100$ integers.

$I$ Plays perfectly to maximise their own E.V (all standard game theory assumptions)

What should your strategy be to maximise your expected value?

My workings so far:

For what it is worth in the 2 player scenario where it is just you v.s $R$ then you should pick $33$ as

$\mathbb{E}[\delta_x ] = \frac{100-x}{100}(x) - \frac{x-1}{100}(-\frac{x}{2})$

(where $\delta_x$ is the strategy of picking $x$)

This function is maximised at $33$ from simple calculus.

No the answer is not $1$ as picking $2$ performs much better when you know someone is picking $1$

I asked my teacher and they said a mixed strategy but I am struggling with formulating this.

My idea was to determine $f(x,y) = \mathbb{E}[\delta_x | \text{other player is picking y } ]$ and then find the maxima of this function but I am struggling.

Any help would be great :)

答: I answered (or at least tried) a question similar to this.The game was the same. Optimal strategy in three-player number-picking game</p>

Title: Optimal Strategy for a betting game.


答:
The fundamental idea is to replace a "digit" in a Euclidean long division's quotient with a "monomial" in a polynomial long division's quotient. The reason we only care about the term of highest degree is because, at each step, we're trying to find the "biggest" (in terms of degree) multiple of the denominator $B(x)$ that allows us to cancel out a monomial from the numerator $A(x)$.

Here is a worked example, and a table of analogies, to give you a better idea.

```<code>    4x^4 + 7x^3 - 4x^2 + 3x + 12 = A(x) | x + 2 = B(x)
  - 4x^4 - 8x^3                         +-------
  _____________                         | 4x^3 - x^2 - 2x + 7 = Q(x)
      0  -  x^3 - 4x^2 + 3x + 12        |
         +  x^3 + 2x^2                  |
           ___________                  |
              0 - 2x^2 + 3x + 12        |
                + 2x^2 + 4x             |
                 __________             |
                     0 + 7x + 12        |
                       - 7x - 14        |
                        ________        |
                           0 - 2        |
</code>```

\begin{array}{ |c|c|c|c| }
 \hline
A(x) & 4x^4 + 7x^3 - 4x^2 + 3x + 12 & a & 51 \\
 \hline
B(x) & x + 2 & b & 10 \\
 \hline
Q(x) & 4x^3 - x^2 - 2x + 7 & q & 5 \\
 \hline
R(x) & -2 & r & 1 \\
 \hline
A(x) = B(x) \times Q(x) + R(x) & \text{Verify by yourself} & a = b \times q + r & 51 = 10 \times 5 + 1 \\
 \hline
\end{array}

Title: Intuition for polynomial long division

Body:
The fundamental idea is to replace a "digit" in a Euclidean long division's quotient with a "monomial" in a polynomial long division's quotient. The reason we only care about the term of highest degree is because, at each step, we're trying to find the "biggest" (in terms of degree) multiple of the denominator $B(x)$ that allows us to cancel out a monomial from the numerator $A(x)$.

Here is a worked example, and a table of analogies, to give you a better idea.

    4x^4 + 7x^3 - 4x^2 + 3x + 12 = A(x) | x + 2 = B(x)
  - 4x^4 - 8x^3                         +-------
  _____________                         | 4x^3 - x^2 - 2x + 7 = Q(x)
      0  -  x^3 - 4x^2 + 3x + 12        |
         +  x^3 + 2x^2                  |
           ___________                  |
              0 - 2x^2 + 3x + 12        |
                + 2x^2 + 4x             |
                 __________             |
                     0 + 7x + 12        |
                       - 7x - 14        |
                        ________        |
                           0 - 2        |

\begin{array}{ |c|c|c|c| }
 \hline
A(x) & 4x^4 + 7x^3 - 4x^2 + 3x + 12 & a & 51 \\
 \hline
B(x) & x + 2 & b & 10 \\
 \hline
Q(x) & 4x^3 - x^2 - 2x + 7 & q & 5 \\
 \hline
R(x) & -2 & r & 1 \\
 \hline
A(x) = B(x) \times Q(x) + R(x) & \text{Verify by yourself} & a = b \times q + r & 51 = 10 \times 5 + 1 \\
 \hline
\end{array}


问:
Convergence test for 3x-x+13
I'm in Calc II and am trying to find what convergence test to use for the following infinite series:

$$\sum_{x=0}^\infty \frac{3x-x^2+13}{\sqrt{x}}$$

Does this series converge?

答:
Hint 1: $\dfrac{(2x-1)}{\sqrt{x^3+1}}\sim \dfrac{2}{\sqrt{x}}, x \to \infty$, where under $f\sim g$ we understand $\lim\limits_{x \to \infty}\frac{f(x)}{g(x)}=1$.

Hint 2 :$\dfrac{(2x-1)}{\sqrt{x^3+1}} \gt \dfrac{(2x-1)}{\sqrt{2x^3}}$

Title: Convergence test for 3x-x+13


答:
But that is how we do long division!

Suppose we are diving $56264377$ by $712$.  I have no idea how many times $712$ goes into $52664377$ so I simply divide it into $52660000$ at first.  But I have no idea how many times $712$ goes into $5266$ either.  So I divide $700$ into $5266$ first and get $700*7$ is $49$ so that's close and $712*7 = 7*(700 + 12) = 4900 + 7*12 = 49+84$ as subtraction I get $300 + (66-84) = 300+(66-86+2)=300 -20 + 2=282$.

Then I need to figure out how many times $712$ divides into $2824$.  Again, no idea but $700$ goes in $4$ times so $4(700 + 12) = 2800 + 48=2824$ is too be big.  SO $3$.  And $3(700+12) = 21 + 36$ so we get remainder $2824 - 2136= 700 + (24-26) = 700 - 12 = 688$

So so fore we have $56264377\div 712 = 74xxxxx$ and we keep going.

Title: Intuition for polynomial long division


问:
Representing a real valued function as a sum of odd and even functions
With $f(x)$ being a real valued function we can write it as a sum of an odd function $m(x)$ and an even function $n(x)$:
      $f(x)=m(x)+n(x)$

Write an equation for $f(-x)$ in terms of $m(x)$ and $n(x)$:

My attempt using the properties - even function if: $f(x)=f(-x)$ and odd function if: $-f(x)=f(-x)$

$f(x)=m(x)+n(x)  \implies f(-x)=m(-x)+n(-x) \implies f(-x)= -m(x)+n(x) \implies   f(-x)=n(x)-m(x)$

I think that is correct but then I need to find equations for both $m(x)$ and $n(x)$ in terms of $f(x)$ and $f(-x)$ so a suggestion on how to tackle that would be great.

答: Add and subtract $f(x) = m(x) + n(x)$ and $f(-x)=n(x)-m(x)$.</p>


问:
Need help solving this ODE with a 3x3 matrix
I have a 3x3 matrix A with the following elements:

$$\begin{pmatrix}
 5 & -4 & 2 \\
4 & -5 & 4 \\
6 & -12 & 9 \end{pmatrix}$$

The ODE is $y' = Ay$ with $y(0) = [1 \quad 0 \quad 1]^T$.

I found the eigenvalue to be 3 with a multiplicity of 3. I found the two eigenvectors to be:$[2 \quad 1 \quad 0]^T$ and $[-1 \quad 0 \quad 1]^T$ (both of these transpose). I feel like I need a 3rd eigenvalue to solve this ODE but not sure what to do. Also not sure how to type matrices so sorry for confusion

答: The third vector is called a generalized eigenvector.  Instead of $(A-\lambda I)v = 0$, we use $(A - \lambda I)^2 v = 0$.  In particular, given an eigenvector $v_1$, you usually want $(A - \lambda I) v_2 = v_1$. This can be repeated if you need more.


答:
Using Taylor's Theorem, $e^t=1+t+\frac12t^2+\frac16t^3+o(t^3)\ge \frac16 t^3$, for $t>0$, we see that

$$\begin{align}
0<\frac{e^{-1/x}}{e^{x^2}-1}&\le 6x
\end{align}$$

Hence, for any given $\varepsilon>0$, we see that

$$\left|\frac{e^{-1/x}}{e^{x^2}-1}\right|<\varepsilon$$

whenever $0<x<\delta=\varepsilon/6$.  And we are done!

Title: Prove the following limit equals 0

Body:
Question: Prove that $$\lim\limits_{x \to 0} \frac{e^{-1/x^2}}{x^2}=0$$ by using the E-D definition.
It suffices to find a $ \delta>0$ such that if $x<\delta \implies \frac{e^{1/x}}{x^2}< \epsilon $. However you need to use Lambert W function to solve this inequality. Is there any other way to approach this question? Any help would be appreciated.


问:
Total orthonormal sequences in a Hilbert space
I'm taking a functional analysis course. I've already been working on orthonormal sets and I'm currently learning about total sets. The definition is from Kreyszig [1]: https://i.stack.imgur.com/AOgxh.png. I want to solve this problem:

Let $(e_n)_n$ a total orthonormal sequence in a Hilbert space H.
i. Prove that ${(e_n-e_{n+1})}_n$ is a total sequence.
ii. Is the previous result true as well if $(e_n)_n$ is just a total
sequence?

I want to use the following Lema (3.3-7) to solve it: "For  any subset $M\neq{\emptyset}$ of a Hilbert space H, the span of M is dense in H if and only if $M^{\bot}=\left\lbrace 0\right\rbrace $". I don't know how to do that.

答:
If $X$ is orthogonal to $e_n-e_{n+1}$ for all $n$ then $ \langle x, e_n \rangle=\langle x, e_{n+1} \rangle$ for all $n$. Since $\sum |\langle x, e_n \rangle|^{2}\leq \|x\|^{2}<\infty$ and all the terms of this sequnece are equal it follows that $\langle x, e_n \rangle=0$ for al $n$. Hence, $x=0$. This proves the first part.

Consider $(x^{n})$ in $L^{2}[0,1]$. This is total by Weierstrass Theorem and the fact that any fucntion in $L^{2}[0,1]$ can be approximated by continuous fucntions. But $(x^{n}-x^{n+1})$ is not total since all these functions vanish at $x=1$.


问:
Does $G/\mathbb{Z} \cong \mathbb{Z}$ imply $G \cong \mathbb{Z}^2$?
Let $G$ be a group. If $G/\mathbb{Z} \cong \mathbb{Z}$, does it follow that $G \cong \mathbb{Z}^2$?

More generally: for a normal subgroup $H$ of $G$, if $G/H \cong H$, does it follow that $G \cong H\oplus H$?

答:
No. We can form a semidirect product $\mathbb Z\rtimes\mathbb Z$ with the action $\varphi_a:x\mapsto (-1)^ax$. The result is a non-abelian group $G$ with a subgroup $H\cong\mathbb Z$ for which $G/H\cong\mathbb Z$.

For a more detailed description of this group, see this related question.


答:
“More generally”: no. If $C_k$ is the cyclic group of order $k$, then $C_4/C_2\cong C_2$, but $C_2\oplus C_2$ is not cyclic, hence not isomorphic to $C_4$.

For $\mathbb{Z}$, again the answer is “no”. For starters, $G$ need not even be abelian: for instance, you can take the group whose underlying set is $\mathbb{Z}\times\mathbb{Z}$, but with group operation given by
$$(a,b)*(c,d) = (a+(-1)^bc, b+d).$$
This is a semidirect product of $\mathbb{Z}$ with itself. The subgroup of all pairs of the form $(a,0)$ is a normal subgroup isomorphic to $\mathbb{Z}$, and the quotient is isomorphic to $\mathbb{Z}$. But the group is not abelian, since $(0,1)*(1,0) = (-1,1)$, whereas $(1,0)*(0,1) = (1,1)\neq (-1,1)$.

What you can say for the special case where the quotient is $\mathbb{Z}$ (or more generally, a free group), is that if $G/N\cong\mathbb{Z}$, then there exists a subgroup $H$ of $G$ that is isomorphic to $\mathbb{Z}$, with $N\cap H=\{e\}$ and $G=NH$; that is, $G\cong N\rtimes\mathbb{Z}$. For your $G$, there are two possibilities: $\mathbb{Z}\oplus\mathbb{Z}$, and the group I described above. But showing both are the only possibilities is a bit more involved (though not terribly hard).

Title: Does $G/\mathbb{Z} \cong \mathbb{Z}$ imply $G \cong \mathbb{Z}^2$?

Body:
Let $G$ be a group. If $G/\mathbb{Z} \cong \mathbb{Z}$, does it follow that $G \cong \mathbb{Z}^2$?

More generally: for a normal subgroup $H$ of $G$, if $G/H \cong H$, does it follow that $G \cong H\oplus H$?


问:
How to establish inequality $U_{n+1} \leq U_{n}+\frac{1}{2^{^n}}$
$\begin{cases}
U_{1} = 1
      \\
U_{n+1} = \sqrt{U_{n}²+\frac{1}{2^{^n}}}
   \end{cases}$

How do I establish the following inequality ?

For $n \geq 1$

$U_{n+1} \leq U_{n}+\frac{1}{2^{^n}}$

I thought about proving that the sequence is increasing so that

$U_{n+1} \geq U_{n}$

But it's pretty useless here, I don't know how to advance any further.

答:
$ U_{n+1} \leq U_n + \frac{1}{2^n} \iff \sqrt{U_n^2 + \frac{1}{2^n}} \leq U_n + \frac{1}{2^n}. $
Squaring both sides, since $U_n$ is clearly positive we want to show that

$$U_n^2 + \frac{1}{2^n} \leq U_n^2 + \frac{U_n}{2^{n-1}} + \frac{1}{4^n}$$

Now it's enough to show that $U_n \geq 1$ for all $n$,  because then we have $\frac{U_n}{2^{n-1}} \geq \frac{1}{2^{n-1}} \geq \frac{1}{2^n}$ so we're done.

$U_n \geq 1$ should be obvious. (Turns out that $U_n$ increasing isn't useless!)


Title: How to establish inequality $U_{n+1} \leq U_{n}+\frac{1}{2^{^n}}$

Body:
$\begin{cases}
U_{1} = 1
      \\
U_{n+1} = \sqrt{U_{n}²+\frac{1}{2^{^n}}}
   \end{cases}$

How do I establish the following inequality ?

For $n \geq 1$

$U_{n+1} \leq U_{n}+\frac{1}{2^{^n}}$

I thought about proving that the sequence is increasing so that

$U_{n+1} \geq U_{n}$

But it's pretty useless here, I don't know how to advance any further.

Body:
A comment:
One can prove that

$$U_{n+1} \leq U_{n}+\frac{1}{2^{\frac{n}{2}}} \qquad (1)$$
Simply
$$U^2_{n+1} \leq U^2_{n}+\frac{1}{2^{^n}}\leq U^2_{n}+\frac{1}{2^{\frac{n}{2}-1}}U_{n}+\frac{1}{2^{^n}}=(U_{n}+\frac{1}{2^{\frac{n}{2}}})^{2}$$

Is (1) enough for your purposes ?


Title: Representing a real valued function as a sum of odd and even functions

Body:
With $f(x)$ being a real valued function we can write it as a sum of an odd function $m(x)$ and an even function $n(x)$:
      $f(x)=m(x)+n(x)$

Write an equation for $f(-x)$ in terms of $m(x)$ and $n(x)$:

My attempt using the properties - even function if: $f(x)=f(-x)$ and odd function if: $-f(x)=f(-x)$

$f(x)=m(x)+n(x)  \implies f(-x)=m(-x)+n(-x) \implies f(-x)= -m(x)+n(x) \implies   f(-x)=n(x)-m(x)$

I think that is correct but then I need to find equations for both $m(x)$ and $n(x)$ in terms of $f(x)$ and $f(-x)$ so a suggestion on how to tackle that would be great.

Body: $$f(x)=\frac{1}{2}\left(f(x)+f(-x)\right)+\frac{1}{2}\left(f(x)-f(-x)\right)$$


问:
Confusion in the solution for this question
Given 4 flags of different colours, how many different signals can be generated. If a signal requires the use of 2 flags one below the other?

Le the colours be R , B , G , O.

My book answers it as  = 4 *3 = 12 which is right answer but not conceptually right .

4 means 4 choices I.e R,B,G,O. Then for 3 choices , we have B,G,O.

Now , if you see . Total combinations it will give  is RB RG RO , BB BG BO , GB GG GO , OB OG OO. Here , OO GG BB are wrong combinations .

So , if I remove them there are total 9 now. But we need 3 more . Those are BR , OR and GR.

Please tell if i am right till here ?

How to consider them then ? Now , all this cutting and then adding made the solution way also long .

Is there a shorter way of doing and how to revive those BR , OR , GR without counting it knowing and also to cut BB , GG ,  OO without counting.

答:
The remaining three colors with be a different set of three colors depending on the color you pick first.

If you pick $R$ first the remaining three colors are $B,G,O$ for three signals: $RB,RG,RO$.

But if you pick $B$ first your remain three colors are $R,G,O$ for tree signals: $BR, BG, BO$.

And if you pick $G$ first your remaining three colors are $R,B,O$ for $GR, GB,GO$.

And if pick $O$ first your remaining three colors are $R,B, G$ for $OR,OB, OG$.  Those are the $12$.

We didn't have to count them.  We knew there where $4$ options for the first color, and we knew once we picked the first color there would be $3$ remaining colors for the second.  That they would be different set of three colors wouldn't matter as long as we knew there'd be $3$ remaining colors.

So we calculate $4\times 3 = 12$.


答: Hint: in the algebra of polynomials there is no concept of "carrying" or "borrowing" when we add two polynomials, because there is no constraint on the size of the coefficients. In the algebra of integers expressed as decimals, when we add or subtract two decimal representations, we have to make adjustments to make the digits (coefficents) lie between $0$ and $9$. The intuition behind polynomial division is the same as the intuition behind integer division (subtract the biggest allowable multiple on each iteration), but the implementation is simpler - because there is no trial and error required to find the the biggest allowable multiple.

Title: Intuition for polynomial long division


问:
How to show that $\mathbb{E}[X\mathbb{1}_{\lbrace X\le a \rbrace}]=\mathbb{P}(X\le a)$?
Let $X$ be a positive countinuous random variable. How do I show that $\mathbb{E}[X\mathbb{1}_{\lbrace X\le a \rbrace}]=\mathbb{P}(X\le a)$?
It seems that I have missed something important.

$\mathbb{E}[X\mathbb{1}_{\lbrace X\le a \rbrace}]=\int\limits_{-\infty}^{\infty}x\cdot f_{X\mathbb{1}_{\lbrace X\le a \rbrace}}(x)dx$ and i want to justify it equals to $\int\limits_{-\infty}^{a}x\cdot f_X(x)dx$.

答:
Thie is very very false. Take $X$ to be  an exponetial r.v. with parameter $1$. Then LHS is $\int_0^{a} xe^{-x}dx$ and RHS is $\int_0^{a} e^{-x}dx$. You can compute these integrals and see that these two are not equal for any $a \in (0,\infty)$.

$EX1_{X\leq a}=\int_0^{a}xf_X(x)dx$ is the correct formula.


问:
Finding minimal polynomial of an element in an extension over $\mathbb{Q}$
What is the minimal polynomial of $\sqrt2 +\sqrt{2}i$ over the field $\mathbb{Q}$.

I proceeded in the following way:

Let $x=\sqrt2+\sqrt{2}i \Rightarrow x^2=2+2i^2+4i \Rightarrow x^2=4i \Rightarrow x^4=-16\Rightarrow x^4+4^2=0$

So for the minimal polynomial (say) $p$ of $\sqrt2+\sqrt2i$ over $\mathbb{Q}$ we have $p|x^4+2^4$.

How do I conclude that this is the minimal polynomial?

答:
In this case the best option is to prove that $\mathbb{Q}(\sqrt{2} + \sqrt{2}\,i) = \mathbb{Q}(\sqrt{2}, i)$. Clearly $\mathbb{Q}(\sqrt{2} + \sqrt{2}\,i) \subset \mathbb{Q}(\sqrt{2}, i)$.

$$x=\sqrt{2}+i\sqrt{2}$$
$$x-\sqrt{2}=i\sqrt{2}$$
$$x^2-2\sqrt{2}+2=-2$$
$$\sqrt{2}=\frac{x^2+4}{2}\in\Bbb{Q}(x)=\Bbb{Q}(\sqrt{2} + \sqrt{2}\,i)$$

So $i\in\Bbb{Q}(\sqrt{2} + \sqrt{2}\,i)$ too. Thus, $\mathbb{Q}(\sqrt{2} + \sqrt{2}\,i) \supset \mathbb{Q}(\sqrt{2}, i)$ and $\mathbb{Q}(\sqrt{2} + \sqrt{2}\,i)= \mathbb{Q}(\sqrt{2}, i)$.
This field has degree 4 over $\Bbb{Q}$, because $[\Bbb{Q}(\sqrt{2})(i):\Bbb{Q}(\sqrt{2})]=2$ (to prove this notice that $x^2+1\in\Bbb{Q}[X]\subset\Bbb{Q}(\sqrt{2})[X]$ the degree can be 1 or 2, but it cannot be 1).


问:
Representing orthogonal projection of a linear operator $A$ on $\mathbb{R}^n$ onto a subspace as $PAP$
I sat on a lecture given recently where the presenter denoted the orthogonal projection of a matrix $A$ (on $\mathbb{R}^n$) onto a subspace as $P_nAP_n$ where $P_n$ is the orthogonal projection onto the subspace of $\mathbb{R}^n$. I understand how to project a vector onto a subspace, but am still unsure how the composition of those matrices gives the orthogonal projection of $A$ onto the subspace.

答:
I think I have figured this out. Given vectors {$v_1 ... v_p$}, we can compute the projection of $A$ onto the subspace spanned by the vectors ($\mathbb{R}^p$) via the matrix

$
A = \begin{bmatrix} <Av_1,v_1> & ... & <Av_1,v_p> \\ \vdots \\ <Av_p,v_1> & ... & <Av_p,v_p> \end{bmatrix}.
$

Then such a matrix of inner products can be represented as $P^TAP$.


问:
How to integrate a function of $y$ over a polygon
I am given the coordinates of the vertices of a polygon and I need to integrate a function of $f$ (shown in the picture) to solve my problem. How can I do it? Btw. I will use it in a software which I am developing as a term project. So, I am looking for a programmable solution

答:
You could make use of Green's theorem.  What is interesting is that the function you are integrating is a function of $y$ only.  So when you examine the form of Green's theorem

$$\iint_D \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right ) dx \, dy = \oint_{\partial D} (P dx + Q dy)$$

where $D$ is the polygon, then clearly $\partial Q/\partial x = 0$.  Clearly we are left with

$$- \frac{\partial P}{\partial y} = f(y)$$

so that, within a constant

$$P(y) = -a \frac{y^2}{b} + a \frac{y^3}{3 b^2}$$

We still do not know $Q$ except that it too must be a function of $y$.  It turns out that this is not important because we are integrating about a closed loop, so ultimately its contribution is zero; thus, we may as well set $Q=0$.  We are then left with the following integral:

$$ \oint_{\partial D} P(y) dx$$

as the integral sought.  To evaluate, simply parametrize each segment of the polygon as some line $y(x) = a x+b$ and integrate accordingly.  The integral will be zero along vertical segments.


答:
The equation you provided is false. The correct one is
$$ \mathbb{E}[\mathbb{1}_{X\leq a}] = \mathbb{P}(X ≤ a) .$$

In fact,
$$ \mathbb{E}[\mathbb{1}_{X ≤ a}] = \int_{-\infty}^{\infty} \mathbb{1}_{X ≤ a} f_X(x) dx = \int_{-\infty}^a f_X(x) dx = \mathbb{P}(A) . $$

Title: How to show that $\mathbb{E}[X\mathbb{1}_{\lbrace X\le a \rbrace}]=\mathbb{P}(X\le a)$?

Body:
Let $X$ be a positive countinuous random variable. How do I show that $\mathbb{E}[X\mathbb{1}_{\lbrace X\le a \rbrace}]=\mathbb{P}(X\le a)$?
It seems that I have missed something important.

$\mathbb{E}[X\mathbb{1}_{\lbrace X\le a \rbrace}]=\int\limits_{-\infty}^{\infty}x\cdot f_{X\mathbb{1}_{\lbrace X\le a \rbrace}}(x)dx$ and i want to justify it equals to $\int\limits_{-\infty}^{a}x\cdot f_X(x)dx$.


问:
Is $e$ (Euler's number) involved in some geometric figure in any way?
Let's take some popular numbers in math: $\pi$, $e$, $\sqrt{2}$ and $\phi$. The number $\pi$ is the ratio between the circumference and the diameter of a circle; $\sqrt{2}$ is the length of a diagonal of a unit square, $\phi$ is the length of a diagonal of a regular unit pentagon. It seems like $e$ is not a part of any reasonably familiar geometric shape. </p>

<p>Is this really so? By 'familiar geometric figure' I mean a geometric figure $F$ that hasn't been artificially constructed so that $e$ is somehow a part of it. By 'part' here I mean the ratio of something in $F$ and something else in $F$, the length of something in $F$, the area of $F$, maybe its perimeter, etc. </p>

答:
It may be a stretch, but a claim can be made that the catenary is a mesmerizing geometrical object that is all about the Euler constant.

Given by its expression:

$$f(x) =  \cosh(x ) = \frac 1 2 \left(e^{x} + e^{-x}\right),$$

it is the average of exponential growth and decay:

And although this implies that calculus is never left behind, it also results in a perfect architectural and engineering structure (with a minus in front): the catenary arc.

Casa Milà, Antoni Gaudí

The equation of a catenary can be derived with a contrained (by the length of the chain) Euler-Lagrange equation assuming uniform physical properties.

It is not surprising that the catenary arc is described as "an arch of uniform density and thickness, supporting only its own weight, the catenary is the ideal curve," when it is (aside from constants) the only function to preserve the ratio between the arc length of any interval and the area subtended:

And although this may sound like it is steering off topic, it is a beautiful counterpoint to $e^x$ being its own integral (up to a constant). Further, this ratio between area under the curve and arc length corresponds to the scaling parameter $a$ in the general equation of the catenary $a \cosh(x/a),$ and can be interpreted as the interest rate in the continuous compounding formula $e^{x/a}.$ In a way a perfect arch is the physical picture of continuously compounded interest.


问:
Extension of the Paris-Harrington principle
Let $[m,n]$ denote the set $\{m,m+1, ... ,n-1,n\}$. $X \to (k)^n_c$ means that whenever $f: [X]^n \to c$ there is a subset $H \subset X$ with cardinality $k$ such that $f$ is constant on $[H]^n$ (The set $H$ here is often described as being homogeneous for $f$.) The Paris-Harrington principle (PH) stipulates further that the homogeneous set should be 'relatively large'. Let $X \to_* (k)^n_c$ mean that whenever $f: [X]^n \to c$ there is $H \subset X$ that is homogeneous for $f$ and $\text{card}(H) \geq \text{max}(k, \text{min}(H))$. PH states that $\forall n,c,k \in \mathbb{N} \; \exists m \in \mathbb{N} \; (m \to_* (k)^n_c)$.

Is the following true? And how do you prove it from PH?

$$ \forall n,c,k,a \in \mathbb{N} \; \exists m \in \mathbb{N} \; ([a, m] \to_* (k)^n_c) $$

答:
The principle

$$\forall n,c,k,a\in\omega\,\exists m\in\omega\,\big([a,m]\to_*(k)_c^n\big)\tag{1}$$

is indeed true; it can be proved by essentially the same proof that Paris and Harrington gave for their principle.

Fix $n,c,k$, and $a$, and suppose that no such $m$ exists. Call $\xi$ a counterexample for $m$ if $\xi$ is a $c$-coloring of $[a,m]^n$ with no relatively large homogeneous set of cardinality at least $k$. Let $T_m$ be the set of all counterexamples for $m$, and let $T=\bigcup_{m\ge a}T_m$; then $\langle T,\subseteq\rangle$ is a finitely branching tree of height $\omega$, so by König’s lemma there is a branch $B$ through it. Let $\xi=\bigcup B$; $\xi$ is a $c$-coloring of $[\omega\setminus a]^n$ such that for each $m\ge a$, $\xi_m=\xi\upharpoonright[a,m]^n$ is a counterexample for $m$.

The infinite Ramsey theorem ensures that there is an infinite $H\subseteq\omega\setminus a$ such that $[H]^n$ is homogeneous for $\xi$. Clearly we can choose $m$ large enough that

$$\big|H\cap[a,m]\big|\ge\max\{k,\min H\}\,,$$

contradicting the assumption that $\xi_m$ is a counterexample for $m$ and establishing $(1)$.

I do not yet see how to prove it directly from PH, however.


答:
The equation is false because $E(X\mathbb{I}_{\{X \le a \}})  \ne P(X \le a )$. In fact, we have rather $P(X \le a ) = E(\mathbb{I}_{\{X \le a \}})$.

For your second question: why we have $E[X \mathbb{1}_{\{X>a\}}]\le \sqrt{E[X^2]}\sqrt{P\{X>a\}}$ ?

In fact, we apply the Cauchy-Schwarz inequality here, for $A,B$ two random variable
$$E(AB) \le \sqrt{E(A^2)}\sqrt{E(B^2)} \tag{1}$$
Just replacing $A = X$ and $B = \mathbb{I}_{\{X \le a \}} $ in $(1)$ and noticing 2 things below:
$$B^2 = B = \mathbb{I}_{\{X \le a \}} $$
and
$$E(\mathbb{I}_{\{X \le a \}})= P(X \le a )$$


问:
What are regular points? (optimization)
I am taking Optimization and my homework question just asked for the definition of a regular point. I have tried googling it. I read multiple definitions but still am confused.
Is a regular point just a point that is in the feasible region?

thanks

答:
Since I can't comment on baxx response to NSP I'll right my own addition.

"For inequality constraints a regular point is one where the gradients of the active constraints are linearly independent." This also applies to equality constraints. See Bertsekas Nonlinear Programming Third Edition pg.346

and

If there is only one constraint $\nabla h_1 \neq 0$

Directly quoting from Bertsekas Nonlinear Programming Third Edition pg. 346:
"... a feasible vector $x$ for which the constraint gradients $\nabla h_1(x),...,h_m(x)$ are linearly independent will be called regular."

Title: What are regular points? (optimization)


问:
Eigen Decomposition of Affine Transforms, for 3D graphics
I am making a 3D graphics processor using VB.Net.

I want to do an airplane game.

I want the plane(s) movements to be scalable by time, so that the appearance is the same regardless of how fast the game-loop is running. And so the amount of matrix & trig operations needed per frame is minimized.

Below is a 2D example of what I am trying to achieve.

I'm trying to do eigen-decomposition so a can scale transform A by time:

I figure out a complex movement (like turning left) using many small
iteration steps at the start of the program.

Solving for the largest eigen-vector & its value using the power iteration

Other stuff to find the other eigen-vector(s) & values (still haven't
nailed that down yet)

Now, with this transform-decomposition for the turn-left operation for T seconds of
game time, I can compute turn-left transforms for any time duration quickly and easily.

Ran into problems fast with a simple translation matrix that does not decompose. Then I saw this: Decomposing an Affine transformation
This guy apparently did it but did not show how. I know its going to involve complex eigenvalues (if you got things rotating). Im not sure how to handle those. And it is going to involve seperating the translation part, which I also don't know how to do.

Here is 2 examples that I would like to see how to solve:

The whole point is to avoid doing hundreds of movement iterations at every game frame.
The example shown has an easy solution for scaling by time, rotating around the arc center by x*dt.
But, when I start doing combinations of 3D rotations like TURNING-UP while also TURNING-RIGHT while also ROLLING (cork-screwing in circle) it is going to be extremely time consuming figuring how to scale these movements by time.

I want the computer to do all the math.

答:
Combining my answer on interpolation with another answer on matrix logarithm / exponential:

```<code>import numpy as np
from scipy.linalg import logm, expm

A = np.array([[1, 0, 0, 30],
              [0, 1, 0,  0],
              [0, 0, 1,  0],
              [0, 0, 0,  1]])
logA = logm(A)
print(logA)
for t in range(0, 11):
  print(expm(t * 0.1 * logA))

B = np.array([[-4.52e-7, 0, -1      , 19.1],
              [ 0      , 1,  0      ,  0  ],
              [ 1      , 0, -4.52e-7, 19.1],
              [ 0      , 0,  0      ,  1  ]])
logB = logm(B)
print(logB)
for t in range(0, 11):
  print(expm(t * 0.1 * logB))
</code>```

https://ideone.com/bnNBqb has the numbers if you want them.

If you want to avoid the matrix logarithm, then you suffer from the problem that the first of the two matrices you gave is not diagonalizable.

Whether computing these matrix exponentials is indeed faster than the transformation you want to avoid is a question you need to answer for your application.

Body:
Combining my answer on interpolation with another answer on matrix logarithm / exponential:

import numpy as np
from scipy.linalg import logm, expm

A = np.array([[1, 0, 0, 30],
              [0, 1, 0,  0],
              [0, 0, 1,  0],
              [0, 0, 0,  1]])
logA = logm(A)
print(logA)
for t in range(0, 11):
  print(expm(t * 0.1 * logA))

B = np.array([[-4.52e-7, 0, -1      , 19.1],
              [ 0      , 1,  0      ,  0  ],
              [ 1      , 0, -4.52e-7, 19.1],
              [ 0      , 0,  0      ,  1  ]])
logB = logm(B)
print(logB)
for t in range(0, 11):
  print(expm(t * 0.1 * logB))

https://ideone.com/bnNBqb has the numbers if you want them.

If you want to avoid the matrix logarithm, then you suffer from the problem that the first of the two matrices you gave is not diagonalizable.

Whether computing these matrix exponentials is indeed faster than the transformation you want to avoid is a question you need to answer for your application.


问:
How to prove two subspaces are complementary
To give some context, I'm continuing my question here.

Let $U$ be a vector space over a field $F$ and $p, q: U \rightarrow U$ linear maps. Assume $p+q = \text{id}_U$ and $pq=0$. Let $K=\ker(p)$ and $L=\ker(q)$. From the previous question, it is proven that $p^2=p$, $qp=0$ and $K=\text{im}(q)$.

Now I want to prove that $U=K \oplus L$.

By definition, I know that if they are complementary, then $K \cap L = \left\{0 \right\}$ and $K + L =U$. Also, from my lecture notes it mentions a certain proposition that when applied to this question, I can say that subspaces $K$ and $L$ of a vector space $U$ are complementary if and only if each vector $u \in U$ can be written uniquely as $u = k + l$, where $k \in K$ and $l \in L$.

How do I go about proving this? It's hard for me to know what's going on. I would appreciate it if someone could help me draw a diagram representing the linear transformations of $p$ and $q$ as it is easier for me to see. Thank you for your time.

答:
From the assumptions you get
$$\def\id{\mathrm{id_U}}\def\im{\mathrm{im}}
p=p\,\id=p(p+q)=p^2+pq=p^2,\qquad
p=\id\, p=(p+q)p=p^2+qp
$$
whence $p^2=p$ and $qp=0$. You also get
$$
\id=(p+q)(p+q)=p^2+qp+pq+q^2=p+q^2
$$
but, since $p+q=\id$ you can also conclude that $q^2=q$.

Let $K=\ker(p)$ and $L=\ker(q)$. Then, for every $x\in U$ you have $x=p(x)+q(x)$. If $x\in K$, then $x=p(x)+q(x)=q(x)$, so $x\in\im(q)$. Therefore $K\subseteq\im(q)$. Conversely, if $x=q(y)\in\im(q)$, we have $p(x)=pq(y)=0$, so $\im(q)\subseteq K$.

Similarly, $\im(p)\subseteq L$, because $qp=0$.

The fact that $x=p(x)+q(x)$ tells now that any element $x\in U$ belongs to $K+L$, because $p(x)\in L$ and $q(x)\in K$. Thus you only need to prove that $K\cap L=\{0\}$.

Let $x\in K\cap L$; then $x\in K$ implies $x=p(x)+q(x)=q(x)$. Now, from $x\in L=\ker(q)$ we get $x=0$.


问:
Issue evaluating the Fourier Transform of a box function at $ w =0 $
In my class we derived that the Fourier Transform of $X[n] = u[n] - u[n-7]$ is

$$\sum_{n=-\infty}^{\infty} X[n] e^{-jwn} = \sum_{n=0}^{6}e^{-jwn}$$
using the finite sum formula we get:

$$ \frac{1-e^{-jw7}}{1-e^{-jw}} =\frac{e^{-jw7/2}(e^{-jw7/2}-e^{-jw7/2})}{e^{-jw/2}(e^{-jw/2}-e^{-jw/2})} = e^{-j3w} \frac{\sin(w7/2)}{\sin(w/2)}$$

if we evaluate $e^{-j3w} \frac{\sin(w7/2)}{\sin(w/2)}$ at $w =0$we get $\frac{0}{0}$. However if we plug in $w = 0 $ into $\sum_{n=-\infty}^{\infty}X[n]e^{-j0n} = 0$
. Since the equality sign was used thrroughout all this, I am confused how one expression evaluates to a number while the other does not. Is there an underlying assumption of using the finite sum formula which causes this discrepancy ?

答: The finite geometric sum formula requires that your ratio not be equal to one, so you will have to consider the cases $w \ne 0$ and $w = 0$ separately. (Luckily, in the case of the ratio being equal to $1$, it is very trivial to get the end result.)


问:
Prove a modular equation
If $a\equiv 1 \pmod n$ and $d\mid n$, then  $a\equiv1 \pmod d$ is true.

That is the question I am trying to understand so I have some understanding of the question but cant place my finger on exactly why it is true.

$n = d \times k$ for $k\in\mathbb{Z}$

$a = k \times n +1$ and $a= k \times d +1$

$n \mid(a-1)$ and $d \mid(a-1)$

答:
If $a$ is congruent to 1 modulo $n$, then $a$ is one more than a multiple of $n$.

If $d$ divides $n$, then $n$ is a multiple of $d$. Hence, $a$ is also one more than a multiple of $d$. So we say $a$ is congruent to 1 module $d$.

Title: Prove a modular equation


问:
Two circles inscribed in a semicircle
Two circles are inscribed in a semi circle. Given the areas of the shaded triangles, what's the radius of the semicircle?
(Note there's a similar but different question here)

答:
Here is a second solution, based on simpler considerations than the first one, involving in particular the concept of power of a point with respect to a circle.

I still refer, for some properties of this problem to this question of mine and its answers containing in particular the following results (see Fig. 1):

Tangency point $C_0(x_0,y_0)$ belongs to the (black) circle $(C)$ with center $S(0,-r)$ and radius $r \sqrt{2}$, with equation:

$$x_0^2+y_0^2+2ry_0-r^2=0\tag{0}$$

Therefore:

$$x_0^2+y_0^2+2ry_0-r^2=0\tag{1}$$

Centers of circles (either $C_1$ or $C_2$) belong to parabola $(P)$ (light blue) with equation:

$$y=\dfrac{1}{2r}{(r^2-x^2)}\tag{2}$$

Points $S,T_1,U_1$, and $S,T_2,U_2$ are aligned.

Fig. 1.

Let $C_1(x_1,y_1)$, $C_2(x_2,y_2)$ be the resp. centers of the two small circles with $x_1<x_2$. As a consequence, the tangency points on diameter $AB$ are $T_1(x_1,0)$, $T_2(x_2,0)$.

The equation of the tangent line to circle $(C)$ in $(x_0,y_0)$ is obtained from (0):

$$xx_0+yy_0+r(y+y_0)-r^2=0\tag{3}$$

As $C_1$ and $C_2$ are the intersection points of this straight line with parabola (P), their abscissas $x_1$ and $x_2$ are the roots of the quadratic equation obtained by plugging $y=\dfrac{1}{2r}(r^2-x^2)$ in ($3$). Rather easy calculations give:

$$\begin{cases}x_1&=&r\dfrac{x_0-\sqrt{2}y_0}{r + y_0},\\ x_2&=&r\dfrac{x_0+\sqrt{2}y_0}{r + y_0}\end{cases}\tag{3}$$

Now, let us use the so-called power of point $T_1$ with respect to the (red) circle with center $(0,0)$ and radius $r$.

$${\frak P}(T_1)=-(T_1U_1).(T_1 S)=-(T_1A).(T_1B) \ \implies \ T_1U_1=\dfrac{r^2-x_1^2}{\sqrt{r^2+x_1^2}}\tag{4}$$

Besides, if we denote by $F_1$ the foot of the altitude issued from $C_0$ in triangle $C_0U_1T_1$, we get (using a classical formula for the distance of a point to a straight line):

$$C_0F_1=\dfrac{|rx_0-x_1(y_0+r)|}{\sqrt{r^2+x_1^2}}$$

As a consequence, the area of triangle $C_0U_1T_1$ can be expressed in two ways  ("half base times height" on the LHS, given value on the RHS), yielding the following equation:

$$\frac12\frac{(r^2-x_1^2)}{\sqrt{r^2+x_1^2}}\frac{(r x_0-r x_1 - x_1 y_0)}{\sqrt{r^2+x_1^2}}=36$$

which is equivalent to:

$$(r^2-x_1^2)(r x_0-r x_1 - x_1 y_0)=72(r^2+x_1^2)\tag{5}$$

In a symmetric way, one gets for the other triangle $C_0U_1T_2$:

$$-(r^2-x_2^2)(r x_0-r x_2 - x_2 y_0)=88(r^2+x_2^2)\tag{6}$$

(the minus sign in front of the LHS accounts for the fact that this triangle is negatively oriented).

Finally, the area of the blue triangle gives rise to equation:

$$\frac12(x_2-x_1)y_0=33\tag{7}$$

We have now the possibility to deduce from equations (3),(1),(5),(6) and (7), by "cascade substitutions", the value of $r$. These computations haven't been explicited because I chose to use the CAS (Computer Algebra System) associated with Matlab for these last computations. Here is the Matlab program I have written:

```<code> syms r x0 y0 x1 x2
 x1= r*(-2^(1/2)*y0 + x0)/(r + y0);
 x2= r*( 2^(1/2)*y0 + x0)/(r + y0);
 [R,X0,Y0]=solve(x0^2+y0^2+2*r*y0-r^2==0,...
          (r^2-x1^2)*(r*x0-r*x1-x1*y0)==72*(r^2+x1^2),...
          (r^2-x2^2)*(r*x0-r*x2-x2*y0)==-88*(r^2+x2^2),...
           (x2-x1)*y0==66,...
          'Real',true,...
           r,x0,y0)
</code>```

giving in particular:

$$r=\dfrac67\sqrt{187\sqrt{2}}\approx 13.9390 \ \ \ \text{and} \ \ \ y_0=6\sqrt{\dfrac{11 \sqrt{2}}{17}}\approx 5.7396$$

($r$ has the same value as that found by g.kov).

Remark: we haven't used the area value $40$ of the red triangle.

Title: Two circles inscribed in a semicircle

Body:
Here is a second solution, based on simpler considerations than the first one, involving in particular the concept of power of a point with respect to a circle.

I still refer, for some properties of this problem to this question of mine and its answers containing in particular the following results (see Fig. 1):

Tangency point $C_0(x_0,y_0)$ belongs to the (black) circle $(C)$ with center $S(0,-r)$ and radius $r \sqrt{2}$, with equation:

$$x_0^2+y_0^2+2ry_0-r^2=0\tag{0}$$

Therefore:

$$x_0^2+y_0^2+2ry_0-r^2=0\tag{1}$$

Centers of circles (either $C_1$ or $C_2$) belong to parabola $(P)$ (light blue) with equation:

$$y=\dfrac{1}{2r}{(r^2-x^2)}\tag{2}$$

Points $S,T_1,U_1$, and $S,T_2,U_2$ are aligned.

Fig. 1.

Let $C_1(x_1,y_1)$, $C_2(x_2,y_2)$ be the resp. centers of the two small circles with $x_1<x_2$. As a consequence, the tangency points on diameter $AB$ are $T_1(x_1,0)$, $T_2(x_2,0)$.

The equation of the tangent line to circle $(C)$ in $(x_0,y_0)$ is obtained from (0):

$$xx_0+yy_0+r(y+y_0)-r^2=0\tag{3}$$

As $C_1$ and $C_2$ are the intersection points of this straight line with parabola (P), their abscissas $x_1$ and $x_2$ are the roots of the quadratic equation obtained by plugging $y=\dfrac{1}{2r}(r^2-x^2)$ in ($3$). Rather easy calculations give:

$$\begin{cases}x_1&=&r\dfrac{x_0-\sqrt{2}y_0}{r + y_0},\\ x_2&=&r\dfrac{x_0+\sqrt{2}y_0}{r + y_0}\end{cases}\tag{3}$$

Now, let us use the so-called power of point $T_1$ with respect to the (red) circle with center $(0,0)$ and radius $r$.

$${\frak P}(T_1)=-(T_1U_1).(T_1 S)=-(T_1A).(T_1B) \ \implies \ T_1U_1=\dfrac{r^2-x_1^2}{\sqrt{r^2+x_1^2}}\tag{4}$$

Besides, if we denote by $F_1$ the foot of the altitude issued from $C_0$ in triangle $C_0U_1T_1$, we get (using a classical formula for the distance of a point to a straight line):

$$C_0F_1=\dfrac{|rx_0-x_1(y_0+r)|}{\sqrt{r^2+x_1^2}}$$

As a consequence, the area of triangle $C_0U_1T_1$ can be expressed in two ways  ("half base times height" on the LHS, given value on the RHS), yielding the following equation:

$$\frac12\frac{(r^2-x_1^2)}{\sqrt{r^2+x_1^2}}\frac{(r x_0-r x_1 - x_1 y_0)}{\sqrt{r^2+x_1^2}}=36$$

which is equivalent to:

$$(r^2-x_1^2)(r x_0-r x_1 - x_1 y_0)=72(r^2+x_1^2)\tag{5}$$

In a symmetric way, one gets for the other triangle $C_0U_1T_2$:

$$-(r^2-x_2^2)(r x_0-r x_2 - x_2 y_0)=88(r^2+x_2^2)\tag{6}$$

(the minus sign in front of the LHS accounts for the fact that this triangle is negatively oriented).

Finally, the area of the blue triangle gives rise to equation:

$$\frac12(x_2-x_1)y_0=33\tag{7}$$

We have now the possibility to deduce from equations (3),(1),(5),(6) and (7), by "cascade substitutions", the value of $r$. These computations haven't been explicited because I chose to use the CAS (Computer Algebra System) associated with Matlab for these last computations. Here is the Matlab program I have written:

 syms r x0 y0 x1 x2
 x1= r*(-2^(1/2)*y0 + x0)/(r + y0);
 x2= r*( 2^(1/2)*y0 + x0)/(r + y0);
 [R,X0,Y0]=solve(x0^2+y0^2+2*r*y0-r^2==0,...
          (r^2-x1^2)*(r*x0-r*x1-x1*y0)==72*(r^2+x1^2),...
          (r^2-x2^2)*(r*x0-r*x2-x2*y0)==-88*(r^2+x2^2),...
           (x2-x1)*y0==66,...
          'Real',true,...
           r,x0,y0)

giving in particular:

$$r=\dfrac67\sqrt{187\sqrt{2}}\approx 13.9390 \ \ \ \text{and} \ \ \ y_0=6\sqrt{\dfrac{11 \sqrt{2}}{17}}\approx 5.7396$$

($r$ has the same value as that found by g.kov).

Remark: we haven't used the area value $40$ of the red triangle.


问:
Is $C^\infty(M, V) $ isomorphic to $C^\infty(M) \otimes V$?
Let $M$ be  a smooth manifold and $V$ be a real vector space . Let's Consider the space $C^\infty(M, V)$ of smooth functions from $M$ to $V$. Can we identifiy the space $C^\infty(M, V)$  with the space $C^\infty(M) \otimes V$ ? And if yes,  why this is true ?

答:
You need a topology on $V$ to make sense to continuity of a function $M\to V$. So I assume $V$ is a real topological vector space.

Then we have the natural identification $$C^\infty(M)\otimes V= \{f\in C^\infty(M,V), \dim span(\{ f(p),p\in M\})<\infty\}$$
$$\sum_{j=1}^J \phi_j \otimes v_j \mapsto \qquad (p\mapsto \sum_{j=1}^J \phi_j(p) v_j)$$

When $V$ is finite dimensional (there is a unique real topological vector space structure on it) then $C^\infty(M)\otimes V$ is clearly the whole of  $C^\infty(M, V)$.

When $V$ is infinite dimensional and $M$ is not compact, no, because we can take infinitely many non-zero compactly supported smooth functions $\phi_j\in C^\infty(M)$ with disjoint supports, with $\bigcup supp(\phi_j)$ closed, and $\dim span(\{v_j\})=\infty$ then  $p\mapsto \sum_{j=1}^\infty\phi_j(p)v_j$ is not in $C^\infty(M)\otimes V$.

When $V$ is infinite dimensional and $M$ is compact, it shall depend on $V$'s topology.


问:
Picking 3 balls from 10 balls, each one with a number from 0 to 9
Given that we pick 3 balls from 10 balls, each one with a number from 0 to 9, I need to define the following event: "Number of balls that we picked, whose number on them is odd, is even".
Meaning for example that (1,3,4) is in the subset that we want to find, but (1,3,5) is not.
I tried to calculate the cardinality of the subset and got $3 \choose 2$$*5*4*5 = 300$. I don't know if it's right. I should add that it's an ordered sample without repetitions. Did I do a mistake ? Thank you for your help

答:
Comment continued with simulation:

```<code>set.seed(2021)
odd = replicate(10^6, sum(sample(0:9, 3) %in% c(1,3,5,7,9)))
mean(odd==0 | odd==2)
[1] 0.49988  @ aprx 1/2
</code>```

Approximate probabilities for 0, 1, 2, or 3 odd numbered balls.

```<code>table(odd)/10^6
odd
       0        1        2        3
0.083837 0.417324 0.416043 0.082796
</code>```

Multiplying by 720 possible outcomes suggests
60, 300, 300, 60 favorable outcomes for 0, 1, 2, 3
odd marbles, respectively. [That's 100 each for
EOO, OEO, OOE.]

```<code>720*table(odd)/10^6
odd
        0         1         2         3
 60.36264 300.47328 299.55096  59.61312
</code>```

Body:
Comment continued with simulation:

set.seed(2021)
odd = replicate(10^6, sum(sample(0:9, 3) %in% c(1,3,5,7,9)))
mean(odd==0 | odd==2)
[1] 0.49988  @ aprx 1/2

Approximate probabilities for 0, 1, 2, or 3 odd numbered balls.

table(odd)/10^6
odd
       0        1        2        3
0.083837 0.417324 0.416043 0.082796

Multiplying by 720 possible outcomes suggests
60, 300, 300, 60 favorable outcomes for 0, 1, 2, 3
odd marbles, respectively. [That's 100 each for
EOO, OEO, OOE.]

720*table(odd)/10^6
odd
        0         1         2         3
 60.36264 300.47328 299.55096  59.61312


问:
Help solving a problem about multivariable chain rule

Hi! I had some questions regarding this exercise. I was wondering if it is possible for me to do it in two different ways:

First way: I calculate the Jacobians for each of the functions, so then I multiply   $D(V(c(t)))\cdot D(c(t))$. But, how do I calculate this:  D(V(c(t))), because I know how to get to $D(V(x,y,z))$ but not to $D(V(c(t)))$.

Second way:
And  the other way I was thinking, but I'm not sure about it, consists on defining:
$$
\begin{align}
x(t) &= e^{t^2} \\
y(t) &= e^{-t^2} \\
z(t) &= \log(1+t^2) \\
\end{align}
$$
And then applying the Chain Rule to $V(x(t),y(t),z(t))$. However, I'm not sure if this is a valid way of doing it.

Thanks a lot !

答:
Both ways work, but (2) gives you immediately that the answer is zero: $x(t) y(t) $ is independent of $t$ so differentiating it gives zero.

For (1) you have
\begin{equation}
DV(x(t),...) = (2x y^2,2x^2y,0)|_{x=x(t),...}=
(2e^{-t^2},2 e^{t^2},0)
\end{equation}
\begin{equation}
DC= (2t e^{t^2}, ,-2t e^{-t^2},...)
\end{equation}
did not compute the last component as we don't need it: DV is  a row vector, DC is a column vector, so multiplying them gives
\begin{equation}
4t -4t+0=0.
\end{equation}


问:
Prove if $a_n$ is non-negative and $\sum a_n$ converges then $a_n\leq\frac{1}{n}$ for all $n\geq N$ for some some N
I believe this is intuitively true but I cannot figure out how to prove it. I was thinking I could use a proof by contradiction. If there is no such $N$ then there are an infinite number of terms of $a_n$ that are greater than or equal to $\frac{1}{n}$. I can't prove that $a_n$ diverges, however. How do I go about proving this statement.

答: $a_n=\frac 2 n$ when $n$ is of the form $m^{2}$ for some $m$ and $0$ otherwise gives a counter-example.


问:
Is it meaningful to define the Dirac delta function as infinity at zero?
I am in discussion with someone online on the subject of the Dirac delta function. This other person wants to say:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

and wants to justify it by saying:

We have that:

$$\delta (x) = \lim_{\epsilon \mathop \to 0} F_\epsilon (x)$$

where:

$$F_\epsilon (x) = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$$

Therefore:

$$\delta (0) = \lim_{\epsilon \mathop \to 0} F_\epsilon (0) = \dfrac 1 {2 \times 0} = \infty$$

and:

$$ \delta {x \ne 0} = \lim_{\epsilon \mathop \to 0} F_\epsilon (x \ne 0) = 0$$

Therefore:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

This comes across as iffy to me. I don't trust $\delta (0) = \infty$ as it is not well-defined exactly what $\infty$ actually means in this context, whereas in fact $\delta$ is precisely defined by means of the limit definition as given above.

I am also seriously unsure about that $\dfrac 1 {2 \times 0}$ in the middle of the exposition, which is at best meaningless and at worst a blasphemous lie.

However, when I consult a number of mathematical works on my shelves and online, there are many of them which give that above definition quite casually, along the lines "As an obvious consequence of the definition: $\delta (0) = \infty$" or some such. Other works heavily italicise the warning that $\delta$ is not a function, and $\delta (0)$ is undefined.

What is the current mode of thought here? I am trying to craft a set of webpages which are mathematically rigorous, but coworkers on the same site are of the mind "It doesn't really matter, we all sort of know what we mean, and hey, $\delta (0) = \infty$ looks really cool!"

When questioning the matter, he is prepared to compromise and say: $\delta (0) \to \infty$ as $x \to 0$, or even:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \to \infty & : x = 0 \end{cases}$$

I am insufficiently mathematically sophisticated as to be able to explain why I hate this approach, but I seriously dislike bandying the $\infty$ sign around, when (once we get past the obvious intuitive meaning we learn in elementary school) we really don't understand what it means.

EDIT:

As requested, I have actually dug up one of my texts which defines the delta function as the limit of the rectangle function as its width goes to zero, as follows. It comes from I.N. Sneddon's "Special Functions of Mathematical Physics and Chemistry", appendix.

If we consider the function
$$\delta_a (x) = \begin {cases} 0 & : |x| > a \\ \dfrac 1 {2 a} & : |x| < a \end {cases}$$
then it is readily shown that
$$\int_{-\infty}^\infty \delta_a(x) d x = 1.$$

...

We now define
$$\delta(x) = \lim_{a \to 0} \delta_a (x).$$
Letting $a$ tend to zero in equations [above] we see that $\delta(x)$ satisfies the relations
$\delta(x) = 0$, if $x = 0$
$$\int_{-\infty}^\infty \delta (x) d x = 1.$$

答:
The answer to the OP question is no, but the reason depends on what definitions you are using for the delta function and for integration.  I don't think any of the existing answers properly address this, so let me try to sketch the spectrum of possibilities:

In the Lebesgue theory of integration (the one most mathematicians use), redefining a function at a point does not affect its integral.  Hence, the value of $\delta$ at $0$ doesn't matter, and so the definition suggested in the OP cannot generate an object that has the right integration properties.

In the Riemann theory of integration, the situation is worse: The upper and lower Darboux sums will not converge at all for the suggested $\delta$ of the OP, so that the proposed delta function definition is not integrable at all.

In non-standard analysis (NSA), there is a way to define a delta function which is zero away from the origin and infinite at the origin, and which has the right integration properties (total mass 1).  There are in fact infinitely many such delta functions, with various infinite values $\delta(0)$ and corresponding infinitesimal widths.  NSA allows this because it incorporates infinite and infinitesimal numbers in a coherent way.  However, because there are many infinite numbers, the symbol $\infty$ is not used.  You have to pick a specific infinite number, not just any arbitrary one.  So in this case again the suggested definition of the OP is meaningless.

There are (to my knowledge) three ways to define a delta function properly:

The standard method is via distributions or measures.

A definition frequently used in physics and engineering is to interpret the delta function as a shorthand for a limiting process involving approximations to identity.  This is sometimes called a "nascent delta function".

In NSA, you can define a delta function in the above-mentioned way as a function with infinitesimal width and infinite height, chosen in such a way that the total mass is 1.


答:
No, this statement is false.

We have the following counter-example

\begin{cases}
\frac{1}{n^2} \qquad \text{if } \color{red}{\not \exists} k \in \Bbb N^* \text{ such that } n= k^4 \\
\frac{1}{\sqrt{n}} = \frac{1}{k^2} \qquad \text{if }  \color{red}{\exists} k \in \Bbb N^* \text{ such that } n= k^4
\end{cases}

Hence, the sum of series $\sum a_n$ converges because

\begin{align}
\sum_{n=1}^{+\infty} a_n & = \sum_{n\ne k^4}\frac{1}{n^2}+\sum_{n= k^4}\frac{1}{\sqrt{n}} \\
& =  \sum_{n\ne k^4}\frac{1}{n^2}+\sum_{k \in \Bbb N^*}\frac{1}{k^2}  <  \sum_{n\in \Bbb N^*}\frac{1}{n^2}+\sum_{k \in \Bbb N^*}\frac{1}{k^2} < +\infty\\
\end{align}

and for all $N$, we have always $n = k^4$ with $k>N$ such that

$$a_n =\frac{1}{\sqrt{n}} > \frac{1}{n}$$

Title: Prove if $a_n$ is non-negative and $\sum a_n$ converges then $a_n\leq\frac{1}{n}$ for all $n\geq N$ for some some N

Body: I believe this is intuitively true but I cannot figure out how to prove it. I was thinking I could use a proof by contradiction. If there is no such $N$ then there are an infinite number of terms of $a_n$ that are greater than or equal to $\frac{1}{n}$. I can't prove that $a_n$ diverges, however. How do I go about proving this statement.


问:
Existence of $\mathbb{Q}$ isomorphism
Let $\alpha\in\Bbb{C}$ algebraic, root of $X^5+2X+6$. The question is exist a $\Bbb{Q}$-isomorphism $\sigma:\Bbb{Q}(\alpha) \to \Bbb{Q}(\alpha)$ such that $\sigma(\alpha)=\alpha ^{-1}$?

I know this is false (I cheked its roots in Maple), and I know that it is true only if $\alpha ^{-1}$ is root of the polynomial too, but how I solve this in an exam? I couldn't find the roots. How do you solve this type of exercises?

Thanks for your time.

答:
Let $f(X)=X^5+2X+6$ and note that $f$ is the minimal polynomial for $\alpha$ since $f$ is irreducible (it's 2-Eisenstein). If  $\sigma(\alpha)=1/\alpha$ then $f(1/\alpha)=0$ as well. Hence
$$
\alpha^5f(1/\alpha)=1+2\alpha^4+6\alpha^5=0,
$$
so you now have $\alpha^5=-(1+2\alpha^4)/6$. But you also know, since $f(\alpha)=0$, that $\alpha^5=-(2\alpha+6)$. Equating these, you can find a degree 4 polynomial whose root is $\alpha$, contradicting the fact that $\alpha$ has degree $5$ over $\mathbb{Q}$.

Title: Existence of $\mathbb{Q}$ isomorphism


问:
Need help with this DS+ proof
I am using SD+ I am new to it

I thought of using Negation but if l can construct  3 I should get K

Derive K

$(W \supset S)\&\lnot M \qquad$  Assume

$(-W \supset H )\lor M \quad$ Assume

$(~S\supset H)\supset K\qquad$ Assume

$W \supset S \qquad\qquad\quad$  1 & E

$\lnot S\supset \lnot W \qquad\qquad$ 4 Trans

$|\quad \lnot S\qquad\qquad\qquad$ Assume

$|\quad \lnot M\qquad\qquad$ 1&E

$|\quad \lnot W\supset H\qquad$   7 DS

$|\quad W \lor H\qquad\quad$ 8 Impl

$|\quad|\quad \lnot W\qquad\quad$ Assume?

$|\quad|\quad H\qquad\qquad$ 9,10 DS

$|\quad|\quad \lnot S\lor H\quad$ 4,11 Impl

$|\quad|\quad \lnot S\supset H\qquad$  9. impl

$|\quad|\quad (\lnot S\supset H)\supset K.\quad$   13 Impl

$|\quad|\quad K.\qquad\qquad$ 4-15 Impl.

Any help would be appreciated.

答:
$\def\fitch#1#2{~~~~\begin{array}{|l}#1\\\hline#2\end{array}}$

You have most of the work you needed done to line 8.   On line 9, you do not need to assume $\lnot W$, you derive it using implication elimination.   Then the path to deriving $K$ is clear.

$$\fitch{~~1.~~(W \supset S)\&\lnot M\\~~2.~~(\lnot W \supset H )\lor M\\~~3.~~(\lnot S\supset H)\supset K}{~~4.~~W\supset S\hspace{9ex}\&\mathsf E~1\\~~5.~~\lnot S\supset\lnot W\hspace{6ex}\textsf{Trans}~4\text{ (Contraposition)}\\\fitch{~~6.~~\lnot S}{~~7.~~\lnot M\hspace{9ex}\&\mathsf E~1\\~~8.~~\lnot W\supset H\hspace{4ex}\textsf{DS}~2,7\hspace{3ex}{\color{green}\checkmark}\\[1ex]\color{red}{~~9.~~\lnot W\hspace{9ex}{\supset}\mathsf{E}~5,6}\\\color{red}{10.~~H\hspace{11ex}{\supset}\mathsf E~8.9}}\\\color{red}{11.~~\lnot S\supset H\hspace{8ex}{\supset}\mathsf I~7{-}10}\\\color{red}{12.~~K\hspace{14ex}{\supset}\mathsf E~3,11}}$$

Note: Much effort could be saved if you may use Hypothetical Syllogism.  $$\lnot S\supset \lnot W, \lnot W\supset H\vdash \lnot S\supset H$$

Title: Need help with this DS+ proof


问:
Bennett's Inequality to Bernstein's Inequality
Bennett's Inequality is stated with a rather unintuitive function,

$$
h(u) = (1+u) \log(1+u) - u
$$

See here.  I have seen in multiple places that Bernstein's Inequality, while slightly weaker, can be obtained by bounding $h(u)$ from below,

$$
h(u) \ge \frac{ u^2 }{ 2 + \frac{2}{3} u}
$$

and plugging it back into Bennett's Inequality.  However, I can't see where this expression comes from.  Could someone point me in the right direction?

答:
Actually, the OP is asking where $\frac{u^2}{2 + \frac23 u} $ comes from.

The answer is: $f(u) = \frac{u^2}{2 + \frac23 u}$ is the Pade $(2,1)$ approximation of $h(u)$ at $u = 0$.

As regards the proof for the Pade $(2,1)$ approximation to be a lower bound, let
$$g(u) = \ln (1 + u) - \frac{1}{1 + u}\left(u + \frac{u^2}{2 + \frac23 u}\right).$$
We have
$$g'(u) = \frac{u^3}{(1 + u)^2(3 + u)^2}.$$
Note that $g'(u) < 0$ on $(-1, 0)$, $g'(u) > 0$ on $(0, \infty)$, and $g(0) = 0$.
Thus, $g(u) \ge 0$ on $(-1, \infty)$. We are done.


问:
How many times to differentiate to obtain differential equation?
Consider the following question:

Find the differential equation of the family of curves, $x^2=4b(y+b)$ ($b$ is a real arbitrary constant).

Now, if we differentiate the given equation twice, we get $xy''=y'$. However, if we only differentiate once, we get $x(y')^2=x+2yy'$

Why is this so?

答:
$\require{cancel}$
Assuming $y=y(x)$ and $y'=\dfrac{d y}{dx}$

$$f(y,x):=\;\;\;x^2=4b(y+b)$$

\begin{align*}
&\dfrac{df(y,x)}{dx}\hspace{1cm}\Rightarrow\hspace{1cm}\dfrac{d}{dx}x^2=\dfrac{d}{dx}4b(y+b)=\dfrac{d}{dx}(4by+4b^2)\\
&\dfrac{df(y,x)}{dx}\hspace{1cm}\Rightarrow\hspace{1cm}2x=4by'+0\\
&\dfrac{d^2f(y,x)}{dx^2}\hspace{1cm}\Rightarrow\hspace{1cm}2=4by''
\end{align*}

The given family of curves seems to be wrong as the ODE we get is drastically different from the given ones.

Let's see if at least the two given ODE are related to each other:

$$g(y,x):=\;\;\;x(y')^2=x+2yy'$$

\begin{align*}
\dfrac{d\;g(y,x)}{dx}\hspace{1cm}\Rightarrow\hspace{1cm}1\cdot(y')^2+x\cdot2y'\cdot y''&=1+2(y'\cdot y'+ y\cdot y'')\\
\cancel{y'y'}+2xy'y''&=1+\cancel{2}y'y'+2yy''\\
xy''&=\dfrac{1+y'y'+2yy''}{2y'}=\dfrac{1}{2y'}+\dfrac{y'}{2}+\dfrac{yy''}{y'}\color{red}{\neq}xy''=y'
\end{align*}

This one is wrong too. What could have the author mean?


问:
Do De Morgan's laws hold in propositional intuitionistic logic?
In Wikipedia page on intuitionistic logic, it is stated that excluded middle and double negation elimination are not axioms. Does this mean that De Morgan's laws, stated $$ \lnot (p \land q) \iff \lnot p \lor \lnot q \\ \lnot (p \lor q) \iff \lnot p \land \lnot q,$$
cannot be proven in propositional intuitionistic logic?</p>

答:
Here's a finite example that basically reproduces what's happening in Zhen Lin's answer/counterexample:

Take a 4-element boolean lattice (labelled 1-2-3-4 above) and a "supertop" to it (labelled 5, above). The elements of the antichain of this lattice (2 and 3) are still pseudocomplements of each other, although no longer complements, which goes hand-in-hand with the "enriched" lattice not being Boolean anymore. (You can check that e.g. $2\land x \le1$ [still] has the greatest solution 3.) It's also easy to see that this lattice is still distributive (because it has exactly 5 elements that's an easy check), so a Heyting lattice because it's finite and distributive.

Now, the meet of the two elements of the antichain is still the bottom element, but because we added the new "supertop" 5, the psedocomplement of the bottom (labelled 1) is going to be 5 in this enriched lattice (because $1 \land x \le 1$ now admits a greater solution (5) than 4 which was the top of the Boolean lattice we started with, so $\neg (2 \land 3) = 5$ now, whereas the "supertop" does nothing for the join of the preudocomplents of the elements of the antichain, i.e. $\neg 2 \lor \neg 3= 3 \lor 2 = 4$.

This is para is not a proof, but note that this supertop doesn't affect the other
De Mogran law: $\neg (2 \lor 3) = \neg 4 = 1$  and $\neg 2 \land \neg 3 = 3 \land 2 = 1$.

(Apologies for using 1 as the bottom element, I now this can be slightly confusing.)

The following part (also) isn't proof of anything, but it does give an additional intuition why only one of De Mogran's laws breaks in a Heyting algebra... and why I added a supertop above, instead of a superbottom, labelled 0 below:

Then, "nothing happens" as far De Morgan's laws are concerned because the definition of a pseudocomplement is relative to the (new) (super)bottom. So $\neg (2 \land 3) = \neg 1 = 0$ but $\neg 2 \lor \neg 3 = 0 \lor 0 = 0$ now as well because adding this superbottom (0) actually changed the pseudocomplements of both 2 and 3 (as well as that of 1).

Likewise this super-bottomed lattice fails to break the other De Morgan law: $\neg (2 \lor 3) = \neg 4 = 0$  and $\neg 2 \land \neg 3 = 0 \land 0 = 0$.


问:
Representing the tensor product of two algebras as bounded operators on a Hilbert space.
Hi Math StackExchange,

Let $A$ be a commutative, infinite dimensional, unital, *-algebra represented by bounded operators on a Hilbert space $H_A$. Next let $B$ be a finite non-commutative *-algebra represented on a Hilbert space $H_B$.

Is it true in general that the product $A\otimes B$ is represented by $\textbf{bounded}$ operators on $H_A\otimes H_B$? How could I show this?

答:
Given two C$^*$-algebras $A$ and $B$ represented on Hilbert spaces $H_A$ and $H_B$, we can always represent their (algebraic) tensor product $A \otimes B$ by bounded operators on $H_A \otimes H_B$. We can do this whether they are commutative or not, infinite-dimensional or not.

Let's start by showing that $a \otimes id_{H_B}$ is a bounded operator on $H_A \otimes H_B$ for $a \in A$. The same proof will show that $id_{H_A} \otimes b $ is a bounded operator on $H_A \otimes H_B$ for $b \in B$. We will then have that
$$
\Vert a \otimes b \Vert \leq \Vert a \otimes id_{H_B} \Vert \cdot \Vert id_{H_B} \otimes b \Vert = \Vert a \Vert \cdot \Vert b \Vert,
$$
for all $a \in A$ and $b \in B$. By the triangle inequality, it will then follow that every operator in the algebraic tensor product can be represented by bounded operators on $H_A \otimes H_B$.

Suppose that $\xi \in H_A \otimes H_B$; we can write it as $\sum_i x_i \otimes y_i$, where the vectors $y_i \in H_B$ are mutually orthogonal and of norm 1, i.e., $(y_i, y_j)=\delta_{ij}$ and $\Vert y_i \Vert_2 =1$.

Let's apply $a \otimes id_{H_A}$ to $\xi$. Then
$$
\Vert (a \otimes id_{H_A})(\sum_i x_i \otimes y_i) \Vert_2= \Vert \sum_i a x_i \otimes y_i \Vert_2,
$$
and, by the definition of the norm on $H_A \otimes H_B$, this is equal to
$$
\Vert \sum_i a x_i \Vert_2.
$$
By the definition of the operator norm, we have
$$
\Vert \sum_i a x_i \Vert_2 \leq \Vert a \Vert \cdot \Vert \sum_i x_i \Vert_2.
$$
We have shown that
$$
\Vert a \otimes id_{H_B} \Vert \leq \Vert a \Vert.
$$
(In fact, it is easy to show that they are equal.) So by the remarks above, we are done.

We have defined a norm on $A \otimes B$, the norm that they have as operators on $H_A \otimes H_B$. This norm is called the spatial tensor product norm or the minimal (C$^*$-algebraic) tensor product. If you search for these terms, you will be able to find a great deal of information about this norm.

By the way, at first glance, it may look like I am assuming that $id_{H_A} \in A$ or $id_{H_B} \in B$; I am not assuming that. I am only using that $a \otimes id_{H_B}$ and $id_{H_A} \otimes B$ make sense as bounded linear operators on $H_A \otimes H_B$.

The above discusses the norm obtained by letting $A \subseteq B(H_A)$ and $B \subseteq B(H_B)$ act in the "natural" way on $H_A \otimes H_B$. (Here $B(H)$ means the bounded linear operators on the Hilbert space $H$.) Given two C$^*$-algebras of bounded operators on a Hilbert space, there are, in general, many different C$^*$-norms that can be placed on $A \otimes B$ and many different ways of representing them as bounded operators on some Hilbert space $K$. However, if $A$ is a commutative C$^*$-algebra or if $B$ is a finite dimensional C$^*$-algebra, then there is only C$^*$-norm that can be placed on $A\otimes B$, the spatial tensor product norm described above. For more on this, search for information about nuclear C$^*$-algebras.


问:
Give a recursive definition of the set: $C:= \{ \frac{1}{3^2}, \frac{1}{6^2}, \frac {1}{9^2}, \frac{1}{12^2}, \frac{1}{15^2} \ldots \}$

Question: Give a recursive definition of the set:
$$C:= \left\{ \frac{1}{3^2}, \frac{1}{6^2}, \frac {1}{9^2}, \frac{1}{12^2}, \frac{1}{15^2}, \ldots \right\}$$

So I see this as the set $\left\{ \frac{1}{(3k)^2}\right\}$ for integers $k$ with $k \geq 1 $.

Attempt:
First the base case will be that $\frac{1}{3^2}$ is in the set.

Then if $x = \frac{1}{(3k)^2}$ is in the set, then

$$k = \sqrt{\frac{1}{9x}}$$

And so I am imagining that for $k+1$ we say that if $x$ is in the set then $$\frac{1}{(3 (\sqrt{1/(9x)} + 1))^2}$$ is also in the set. I see that this works for $x=1/9$ since plugging this value in gives us $\frac {1}{6^2}$.

I'm not at all confident this is the correct solution or if there is an easier way of seeing this problem. Any hints insights appreciated.

答:
Let $a_k$ be the $k$th element of the sequence. Then
$$\frac{a_{k+1}}{a_k} = \frac{1/(3(k+1))^2}{1/(3k)^2}
= \frac{1}{((3k+1)/3k)^2}
= \frac{1}{\left(1+\frac{1}{3k}\right)^2},$$
and you can derive a recurrence formula for $a_{k+1}$ from that.


问:
Weak (coherent) topology determined by a finite family of subsets
Let X be a set and $\{A_j \, : \, j\in J\}$ a family of subsets of $X$ such that $X=\bigcup_{j\in J} A_j$. Assume that

Each $A_j$ is a topological space.

For every $j,k\in J$, the topology on $A_j\cap A_k$ induced from $A_j$ is the same as the one induced from $A_k$.

For every $j,k\in J$, the intersection $A_j\cap A_k$ is closed in both $A_j$ and $A_k$.

Then, the weak topology on $X$ determined by $\{A_j \, : \, j\in J\}$ is the topology on which the closed sets are the sets $F\subseteq X$ such that $F\cap A_j$ is closed in $A_j$ for every $j\in J$. This topology is also known as the coherent topology determined by $\{A_j \, : \, j\in J\}$.

In Rotman's An Introduction to Algebraic Topology the autor says (page 196): "If the index set $J$ is finite, then there is only one topology on $X$ compatible with conditions 1, 2, and 3, and so it must be the weak topology."

I was trying to figure it out: Does it means that if $\tau$ and $\tau'$ are two topologies on $X$ which induce on each $A_j$ its original topology and the collection $\{A_j \, : \, j\in J\}$ satisfies conditions 1, 2 and 3, then $\tau=\tau'$, provided $J$ is finite?

If the answer is no, what does it mean by "compatible  with conditions , 2 and 3"? If the answer is yes, I am unsuccesfully trying to prove that $\tau = \tau'$ (or, which amounts to the same, that if $\tau$ is a topology compatible with conditions 1, 2 and 3, then $\tau$ is the weak topology determined by the finite collection $\{A_j \, : \, j\in J\}$).

Finally, what would be a counterexample of a space $X$ with an infinite collection $\{A_j \, : \, j\in J\}$ and two (or more) distinct topologies compatible with conditions 1, 2 and 3?

答:
The basic question is what "a topology on $X$ compatible with conditions 1, 2, and 3" should be.

I think that Rotman's approach is suboptimal since 1 is not a condition on the family $\{A_j \, : \, j\in J\}$, but means that we are given additional structural components. I would do it as follows:

Let $X$ be a set and $\{A_j \, : \, j\in J\}$ a family of subsets  $A_j \subset X$ each of which is endowed with a topology $\tau_j$ such that $X=\bigcup_{j\in J} A_j$. Assume that

For every $j,k\in J$, the topology on $A_j\cap A_k$ induced from $A_j$ is the same as the one induced from $A_k$.

For every $j,k\in J$, the intersection $A_j\cap A_k$ is closed in both $A_j$ and $A_k$.

We call $\{(A_j,\tau_j) \, : \, j\in J\}$ an admissible family on $X$ (this is just an adhoc notation)

Then, the weak topology on $X$ determined by an admissible family $\{(A_j,\tau_j) \, : \, j\in J\}$ is the topology whose closed sets are the sets $F\subseteq X$ such that $F\cap A_j$ is closed in $A_j$ for every $j\in J$.

The basic features of the weak topology are that each $A_j$ becomes closed subset of $X$ and that each $A_j$ , as a subspace of $X$, retains its original topology.

Now observe that if $(X,\tau)$ is a topological space and $\{A_j \, : \, j\in J\}$ a family of subsets  $A_j \subset X$ as above, then each $A_j$ can be endowed with the subspace topology $\tau_j$ induced by $\tau$. The family $\{(A_j,\tau_j) \, : \, j\in J\}$ trivially satisfies 2, but we cannot expect that 3 is true.

The obscure phrase "a topology on $X$ compatible with conditions 1, 2, and 3" should be replaced by the following definition:

A topology $\tau$ on $X$ is compatible with the admissible family $\{(A_j,\tau_j) \, : \, j\in J\}$ if

(a) Each $\tau_j$ is the subspace topology induced by $\tau$ on $A_j$.

(b) Each $A_j$ is $\tau$-closed in $X$.

Note that condition 3 is an immediate consequence of (b).

Clearly the weak topology determined by $\{(A_j,\tau_j) \, : \, j\in J\}$ is compatible with $\{(A_j,\tau_j) \, : \, j\in J\}$. In fact, it is the finest  topology compatible with $\{(A_j,\tau_j) \, : \, j\in J\}$. To see this, let $\tau$ be any compatible topology on $X$. Then for a $\tau$-closed $F \subset X$ all intersections $F \cap A_j$ are $\tau_j$-closed in $A_j$. That is, $F$ is closed in weak topology. In other words, the weak topology is finer than $\tau$.

Now let us prove that for a finite $J$ the weak topology is the only compatible topology. Let $\tau$ be any compatible topology and let $F \subset X$ be closed in the weak topology. Then all intersections $F \cap A_j$ are $\tau_j$-closed in $A_j$. Since the $A_j$ are closed in $X$, the $F \cap A_j$ are $\tau$-closed in $X$. Hence $F = \bigcup_{j \in J} F \cap A_j$ is $\tau$-closed in $X$. We conclude that $\tau$ agrees with the weak topology.

Let us conclude with an example showing that this is in general false for infinite $J$.

Let $X$ be any set and let $J = X$, $A_x = \{x\}$. Note that singletons always have a unique topology. Then the weak topology on $X$ is the discrete topology. However, any $T_1$-topology on $X$ is compatible. Therefore, if $X$ is infinite (like $X = \mathbb R$), there are many compatible topologies.


问:
Infinite Gamma Derivative Identity
We have
$$
\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}\;dx \tag{1}
$$
We also have
$$
\frac{d^n}{dz^n}\, x^{z-1}e^{-x}=\log(x)^n e^{-x}x^{z-1}, \;\; z>1 \tag{2}
$$
If we create an operator
$$
\hat{O}=\sum_{n=0}^\infty \frac{1}{n!}\frac{d^n}{dz^n} = e^{\frac{d}{dz}}
$$
and apply it to both sides of $(1)$, we get
$$
\hat{O}\Gamma(z)=\hat{O}\int_0^\infty x^{z-1}e^{-x}\;dx
$$
if we allow this under the integral sign, then
$$
\sum_{n=0}^\infty \frac{1}{n!}\frac{d^n}{dz^n}\Gamma(z)=\int_0^\infty \left(\sum_{n=0}^\infty \frac{1}{n!}\frac{d^n}{dz^n}x^{z-1}e^{-x}\right)\;dx
$$
$$
\sum_{n=0}^\infty \frac{1}{n!}\frac{d^n}{dz^n}\Gamma(z)=\int_0^\infty \left(\sum_{n=0}^\infty \frac{\log(x)^n}{n!}\right)x^{z-1}e^{-x}\;dx
$$
$$
\sum_{n=0}^\infty \frac{1}{n!}\frac{d^n}{dz^n}\Gamma(z)=\int_0^\infty x^{z}e^{-x}\;dx
$$
$$
\sum_{n=0}^\infty \frac{1}{n!}\frac{d^n}{dz^n}\Gamma(z)=\Gamma(z+1)
$$
this seems to hold numerically for $z>1$, for $z=1$, the value jumps between $1/2$ and $3/2$, which is curious as the average is right, but then $(2)$ doesn't hold anymore. The $n^{th}$ derivative of $\Gamma(z)$ seems to get increasingly complicated with large $n$.There is always a factor of $\Gamma(z)$ as well so it seems more to be a statement that
$$
\frac{1}{\Gamma(z)}\sum_{n=0}^\infty \frac{1}{n!}\frac{d^n}{dz^n}\Gamma(z)=z \tag{3}
$$

Can this be proven in another way?

Is there any information about the derivative of $\Gamma(z)$ that makes this identity more obvious?

I notice that if we replace $\frac{d^n}{dz^n}\Gamma(z)$ with $\Gamma(z) \log(z)^n$, then the equation would still hold. Is that related at all?

I've noticed that if we replace $\psi^{(t)}(z)$ as $z^t$ then
$$
\frac{1}{\Gamma(z)}\frac{d^n}{dz^n} \Gamma(z) \big|_{\psi^{(t)}(z) \to z^t} = \sum_{k=0}^n \left\{\begin{matrix} n \\ k \end{matrix}\right\}z^{n-k}
$$
with $\left\{\begin{matrix} n \\ k \end{matrix}\right\}$ as Stirling numbers of the second kind.

We end up with a kind of ladder operator
$$
e^{\frac{d}{dz}}\Gamma(z) = \Gamma(z+1)
$$
$$
e^{-\frac{d}{dz}}\Gamma(z) = \Gamma(z-1)
$$

答:
Indeed, you have discovered the shift operator. Note that
$$
e^{\partial_x}x^a=\sum_{k=0}^\infty\frac{1}{k!}\partial_x x^a=x^a\sum_{k=0}^\infty\frac{(-a)_k}{k!}(-x)^{-k}=x^a{_1F}_0(-a;-;-x^{-1})=(x+1)^a,
$$
where $(s)_k$ is the Pochhammer symbol and ${_pF}_q(\mathbf a;\mathbf b;z)$ is the generalized hypergeometric function. Given this identity we can write $t^x$ as a power series to find
$$
e^{\partial_x}t^x=\sum_{k=0}^\infty\frac{\log^kt}{k!}e^{\partial_x}x^k=\sum_{k=0}^\infty\frac{\log^kt}{k!}(x+1)^k=t^{x+1}
$$
and so,
$$
e^{\partial_x}\Gamma(x)=\int_0^\infty e^{\partial_x}t^{x-1}e^{-t}\,\mathrm dt=\int_0^\infty t^xe^{-t}\,\mathrm dt=\Gamma(x+1).
$$
With a little work you can easily extend this result to the more general operator shift operator $e^{t\partial_x}$.

Title: Infinite Gamma Derivative Identity


答:
This looks like the linear family of a Clairaut equation. Identifying the parts you get $\frac{d(x^2)}{dy}=4b\iff x\frac{dx}{dy}=2b$. Now insert back to get
$$
x^2=2xy\frac{dx}{dy}+x^2\left(\frac{dx}{dy}\right)^2
$$
Now revert back the functional dependency
$$
xy'^2=2yy'+x.
$$

Title: How many times to differentiate to obtain differential equation?

Body:
Consider the following question:

Find the differential equation of the family of curves, $x^2=4b(y+b)$ ($b$ is a real arbitrary constant).

Now, if we differentiate the given equation twice, we get $xy''=y'$. However, if we only differentiate once, we get $x(y')^2=x+2yy'$

Why is this so?


问:
Given this limit involving T can I imply that T is injective?
Recently, trying to solve a problem, I got stuck in the following fact: Given a linear transformation $T:\mathbb{R}^m \to \mathbb{R}^n$ such that $$\lim\limits_{v \to 0} T \cdot \frac{v}{|v|}=0.$$ Is it true that $T \equiv 0$? I think I was able to prove it by supposing there is some $w \in \mathbb{R}^m \setminus \{0\}$ such that $T \cdot w \neq 0$. Then the sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ given by $v_k=\frac{1}{k} \cdot w$ is such that $\lim\limits_{k \to \infty} v_k=0$ and$$0=\lim\limits_{v \to 0} T \cdot \frac{v}{|v|}=\lim\limits_{k \to \infty}T \cdot \frac{v_k}{|v_k|}=\lim\limits_{k \to \infty} T \cdot \frac{w}{|w|}=T \cdot \frac{w}{|w|}\neq 0.$$ Hence $T \cdot w =0 $ for all $w \in \mathbb{R}^m$ and $T$ must be the zero transformation.

In weakening the hypothesis, I tried to find an answer to the following problem: Given a linear transformation $T:\mathbb{R}^m \to \mathbb{R}^n$, a sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ such that $\lim\limits_{k \to \infty}v_k=0$ and $$\lim\limits_{k \to \infty} T \cdot \frac{v_k}{|v_k|}=0.$$ Is it true that $Ker(T) \neq 0$?
First I tried to find a counterexample, but I couldn't. And since It is easy to prove the converse (i.e. if $Ker(T) \neq 0$ then there is sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ satisfying the above conditions) I thought this might be true but have no idea how to prove it. Any help with a proof or counterexample is appreciated.

答:
If $\operatorname{Ker}(T) = 0$, then the square matrix $T^\intercal T$ is invertible, as $T^\intercal Tx = 0$ implies $|Tx|^2 = x^\intercal T^\intercal Tx = 0$ and hence $Tx = 0$.

We also have $x^\intercal T^\intercal T x = |Tx|^2 \geq 0$ for any $x\in \Bbb R^m$. Thus $T^\intercal T$ is a positive definite symmetric matrix.

We may write $T^\intercal T = U^\intercal D^2 U$, where $U \in O(m)$ is orthogonal and $D = \operatorname{diag}(\lambda_1, \dots, \lambda_m)$ is diagonal, with all $\lambda_i$ strictly positive. This can be rewritten as $T^\intercal T = (DU)^\intercal DU$.

Letting $\lambda = \min_{1 \leq i \leq m} \lambda_i$, we have $$|Tx|^2 = x^\intercal T^\intercal T x = x^\intercal (DU)^\intercal DU x = |DUx|^2 \geq \lambda^2 |Ux|^2 = \lambda^2 |x|^2$$ for all $x\in \Bbb R^m$, i.e. $|Tx| \geq \lambda |x|$.

This means that for any $x \in \Bbb R^m \backslash \{0\}$, we have $\left|T\cdot \frac x{|x|}\right|\geq \lambda$. Therefore there cannot exist such a sequence $(v_k)_k$ as in your question.


答:
Remark: this has nothing to do with particular $\ast$-subalgebras. Given $S$ bounded on $H_A$ and $T$ bounded on $H_B$, there is a canonical way to define $S\otimes T$ bounded on $H_A\otimes H_B$ with, moreover, $\|S\otimes T\|= \|S\|\,\|T\|$. So if your representations are denoted by $\pi_A:A\longrightarrow B(H_A)$ and $\pi_B:B\longrightarrow B(H_B)$, we obtain in a canonical way a representation $\pi_A\otimes \pi_B:A\odot B\longrightarrow B(H_A\otimes H_B)$ of the algebraic tensor product $A\odot B$ such that $\|(\pi_A\otimes \pi_B)(a\otimes b)=\|\pi_A(a)\otimes \pi_B(b)\|=\|\pi_A(a)\|\| \pi_B(b)\|$.

Let us denote $H_A\odot H_B$ the algebraic tensor product, whose completion with respect to the inner product defined by $(x\otimes y,x'\otimes y'):=(x,x')(y,y')$ is $H_A\otimes H_B$. In particular, note that $\|x\otimes y\|=\|x\|\,\|y\|$.

If $S\in B(H_A)$ and $T\in B(H_B)$, then the bilinear map $(x,y)\longrightarrow Sx\otimes Ty$ factors to a linear map from $H_A\odot H_B$ to $H_A\otimes H_B$ by the universal property of the algebraic tensor product. It is denoted by $S\otimes T$ and it satisfies
$$
(S\otimes T)(x\otimes y)=Sx \otimes Ty\qquad\forall x\in H_A, \forall y\in H_B.
$$

Claim: the operator $S\otimes T: H_A\odot H_B\longrightarrow H_A\otimes H_B$ has norm $\|S\otimes T\|=\|S \|\,\|T\|$. Hence it extends uniquely to a bounded operator $S\otimes T\in B(H_A\otimes H_B)$ with the same norm.

The second part is immediate and is the linear case of the unique extension of uniformly continuous functions from dense sets. So we need only check the first part. Since the unit ball $\{z\in H_A\odot H_B\,;\,\|z\|\leq 1\}$ contains $\{x\otimes y\,;\, \|x\|=\|y\|=1\}$, we have
$$
\|S\otimes T\|=\sup_{\|z\|\leq 1}\|(S\otimes T)z\|\geq \sup_{\|x\|=\|y\|=1}\|(S\otimes T)(x\otimes y)\|=\sup_{\|x\|=1}\|Sx\|\sup_{\|y\|=1}\|Ty\|=\|S\|\,\|T\|.
$$
The other direction is more delicate. The trick is to observe that
$$
S\otimes T=(S\otimes I)(I\otimes T).
$$
So it only remains to show that $\|S\otimes I\|\leq \|S\|$. By symmetry, this will also show that $\|I\otimes T\|\leq \|T \|$ whence $ \|S\otimes T\|\leq \|S\otimes I\|\,\|I\otimes T\|\leq \|S\|\,\|T\|$ as desired.

Take an arbitrary vector $z=\sum_{j=1}^n x_i\otimes y_j$ in $H_A\odot H_B$. By bilinearity, we can replace the $y_j$'s by an orthonormal basis of their span. So we can assume $(y_j)$ is orthonormal, whence $(x_j\otimes y_j)$ and $(Sx_j\otimes y_j)$ are orthonormal as well. Then
$$
\|(S\otimes I)z\|^2=\sum_{j=1}^n \|Sx_j\otimes y_j\|^2=\sum_{j=1}^n \|Sx_j\|^2\|y_j\|^2\leq \|S\|^2\sum_{j=1}^n \|x_j\|^2\|y_j\|^2
$$
$$
=\|S\|^2\sum_{j=1}^n \|x_j\otimes y_j\|^2=\|S\|^2\|z\|^2.
$$
This concludes the proof of the claim.

Title: Representing the tensor product of two algebras as bounded operators on a Hilbert space.

Body:
Hi Math StackExchange,

Let $A$ be a commutative, infinite dimensional, unital, *-algebra represented by bounded operators on a Hilbert space $H_A$. Next let $B$ be a finite non-commutative *-algebra represented on a Hilbert space $H_B$.

Is it true in general that the product $A\otimes B$ is represented by $\textbf{bounded}$ operators on $H_A\otimes H_B$? How could I show this?