问:
Varying payment- with both increasing and decreasing annuity
I am trying to find the present value of an annuity-immediate such that the first payment
is $1000$, and each subsequent payment increases by $100$ until the
payments reach $2000$, but then decreases annually by $200$ until the final
payment of $200$. The effective annual rate of interest is $8$%.

Similar to another question I asked but with now a payment decreasing payment
(annuity- find the present value of varying payments)

so the payments would increases by $100$ until it reaches $2000$ but drops every $200$ until it reaches $200$ as the final:

$$1000, 1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000$$
$$1800, 1600, 1400, 1200, 1000, 800, 600, 400, 200$$

I tried to understand it by using a table, similar to the response shown from my previous problem

$$\begin{array}{c|c|c|c|c}
\text{Year} & \text{Level Payment} & \text{Increasing 1} & \text{Decreasing 2} & \text{Total} \\
\hline
1 & 1000 & 0 & 0 & 1000 \\
2 & 1000 & 100 & 0 & 1100 \\
3 & 1000 & 200 & 0 & 1200 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
11 & 1000 & 1000 & 0 & 2000 \\
12 & 2000 & 0 & 200 & 1800 \\
13 & 2000 & 0 & 400 & 1600 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
20 & 2000 & 0 &1600 & 200 \\
\end{array}$$

then I got

$$\require{enclose}
\begin{align}
PV
&= 1000(v + v^2 + \cdots + v^{11}) + 2000(v^{12} + v^{13} + \cdots + v^{20}) + (100v^2 + 200v^3 + \cdots + 1000v^{11}) \\ &\qquad + (200v^{12} + 400v^{13} + \cdots + 1600v^{20}) \\
&= 1000 a_{\enclose{actuarial}{11} i} + 2000 v^{11} a_{\enclose{actuarial}{9} i} + 100 v (Ia)_{\enclose{actuarial}{11}i} + 200 v^{11}(Da)_{\enclose{actuarial}{9}i}.
\end{align}$$

Did I do the level payment and equation correctly? When I went back to the previous problem I forgot to ask how the level payment was found for the second table (ie. where did 950 and 1450 come from?). How can I restructure this table to make the calculations less messy?

The first payment starts at 1000 then increases by 100, 10 times. Then it decreases by 200, 9 times so
$100(10) - 200(9) = -800 = 200 - 1000$

答:
Since I was the one who answered the previous question you are using as a model for this question, I would like to start by answering this part of your question:

How the level payment was found for the second table?

The main insight is to plan for the increasing annuities to have the same term (number of payments).  Since the first payment is $1000$ and each subsequent payment increases by $50$, we can't just write $\require{enclose} 1000(Ia)_{\enclose{actuarial}{n}i}$ because this would correspond to a cash flow of $1000, 2000, 3000$, etc.  And we can't write $1000 a_{\enclose{actuarial}{n}i} + 50(Ia)_{\enclose{actuarial}{n}i}$ because then the first payment is $1050$, not $1000$.  So we have to write the first payment as the sum of a level annuity of $950$ plus an increasing annuity where the increment is $50$ per payment.  That's where $950$ comes from.  Similarly, because the second series of payments begins at $1550$ and increases by $100$ each period, we have to write this as a level annuity of $1450$ plus an increasing annuity with increment $100$, in order to get the total to match.

That said, your list of payments is correct, but the way you structured the table is not.  For each row, the last column must be the sum of the other columns.  When you chose $2000$ for the level portion, that already exceeds the total amount scheduled to be paid, which would mean the decreasing column would need to be negative, not positive; for instance:
$$\begin{array}{c|c|c|c|c}
\text{Year} & \text{Level Payment} & \text{Increasing 1} & \text{Increasing 2} & \text{Total} \\
\hline
1 & 1000 & 0 & 0 & 1000 \\
2 & 1000 & 100 & 0 & 1100 \\
3 & 1000 & 200 & 0 & 1200 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
11 & 1000 & 1000 & 0 & 2000 \\
12 & 2000 & 0 & \color{red}{-}200 & 1800 \\
13 & 2000 & 0 & \color{red}{-}400 & 1600 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
20 & 2000 & 0 & \color{red}{-}1\color{red}{8}00 & 200 \\
\end{array}$$

This would then correspond to the equation of value (corrections in red)
$$\require{enclose}
\begin{align}
PV
&= 1000(v + v^2 + \cdots + v^{11}) + 2000(v^{12} + v^{13} + \cdots + v^{20}) + (100v^2 + 200v^3 + \cdots + 1000v^{11}) \\ &\qquad \color{red}{-} (200v^{12} + 400v^{13} + \cdots + 1\color{red}{8}00v^{20}) \\
&= 1000 a_{\enclose{actuarial}{11} i} + 2000 v^{11} a_{\enclose{actuarial}{9} i} + 100 v (Ia)_{\enclose{actuarial}{\color{red}{10}}i} \color{red}{-} 200 v^{11}(\color{red}{I}a)_{\enclose{actuarial}{9}i}.
\end{align}$$
Why is the last annuity increasing and not decreasing?  Because that's how you chose to write it in the table!  A decreasing annuity is the same as a level annuity minus an increasing annuity.  Notice how in the table, the payments in the column "Increasing 2" are negative but are increasing in magnitude.

Instead of doing it this way, I would suggest writing the table like this:
$$\begin{array}{c|c|c|c|c}
\text{Year} & \text{Level Payment} & \text{Increasing} & \text{Decreasing} & \text{Total} \\
\hline
1 & 900 & 100 & 0 & 1000 \\
2 & 900 & 200 & 0 & 1100 \\
3 & 900 & 300 & 0 & 1200 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
10 & 900 & 1000 & 0 & 1900 \\
11 & 0 & 0 & 2000 & 2000 \\
12 & 0 & 0 & 1800 & 1800 \\
13 & 0 & 0 & 1600 & 1600 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
20 & 0 & 0 & 200 & 200 \\
\end{array}$$

By now, you should see how this works.  Since the first payment is $1000$ but increases only by $100$ each period, we split this into a level annuity of $1000 - 100 = 900$, plus an increasing annuity with increment $100$.  This structure holds for $10$ payments, until the $10^{\rm th}$ payment, which is $1900$.  Then on the $11^{\rm th}$ year, we begin paying a $10$-year decreasing annuity where the decrement is by $200$ each period, which automatically takes care of the requirement that the $11^{\rm th}$ payment is $2000$, since $10 \times 200 = 2000$.  Using this table, you should easily be able to construct the equation of value.  In both cases, you should get an answer of about $13117.4$.

Finally, to help with your understanding, suppose the last ten payments are not decreasing by $200$, but by $50$.  So the list of payments becomes
$$1000, 1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, \\
2000, 1950, 1800, 1750, 1700, 1650, 1600, 1550, 1500, 1450.$$
If you were to modify my version of the table, how would you do it?  Obviously, you don't need to change the first ten rows, only the last ten rows.  What might you write in the "Level payment" column?  What would you write in the "Decreasing" column?  What is the resulting equation of value?


问:
Is it possible for $\delta$ to increase when $\epsilon$ is decreasing
If $$\lim_{x \to x_0} f(x) = L$$
The $\epsilon-\delta$ definition of limit states that- $\forall \epsilon$, $\exists \delta$, such that:

$0<|x-x_0|<\delta\Rightarrow|f(x)-L|<\epsilon$. For any value of $\epsilon$ if I can provide a $\delta$, then $L$ is the limit of $f(x)$ at $x_0$.

If lesser and lesser values of $\epsilon$ are taken, is it possible for the maximum possible $\delta$ to increase?

i.e.: If $\epsilon=\epsilon_1$, $\delta=\delta_1$ is the maximum value that satisfies the condition, and if $\epsilon=\epsilon_2$, $\delta=\delta_2$ is the respective maximum value that satisfies the condition.

Then if $\epsilon_2<\epsilon_1$, is it possible for $\delta_2>\delta_1$?

If it is, can you give an example.

答:
If a given $\delta$ "works for $\epsilon_1$" and $\epsilon_2>\epsilon_1$ then that $\delta$ also "works for $\epsilon_2$" (since everything within $\epsilon_1$ of $L$ is also within $\epsilon_2$ of $L$). Making $\epsilon$ smaller can only every shrink the set of "working $\delta$s."

So there is no sense in which this can happen. Of course, a given $\epsilon$ may not have a "largest working $\delta$" at all - think about a constant function - but if we assume that such maximal $\delta$s always exist, or allow $\delta=\infty$, then your question makes sense and the situation you ask about cannot occur.

ContentLicense: CC BY-SA 4.0


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:
Your answer is correct. However, the perpetuity formula works for the second term.

$$PV = 3v^2 + v^2\cdot \frac{3}{1.05^4-1} = 15.348$$

The second term is the perpetuity formula discounted the first two years. Saves from using the infinite sum.


问:
Finding general term of sequence
I need to find the general terms of the sequence, $$\{1.25, 2, 3.25, 5, 7.25, \cdots\}$$ and $$\{0, 6, 24, 60, 120, 210, \cdots\}$$

I can't think of any such pattern that would produce these.

答:
With a problem like this there are some standard things to try. One is forming the sequence of differences, as in Light Yagami’s answer. If they’re constant, the original sequence is arithmetic. If not, you can look at the differences of the difference sequence, and if they’re constant, the original sequence is described by a quadratic. And so on: if the $n$-th difference sequence is constant, the original sequence is described by a polynomial of degree $n$.

Instead of differences, you can look at the ratios of consecutive terms; if those seem to be nearly constant, the sequence is likely to be some modification of a geometric sequence, e.g., something like $a_n=3\cdot2^n-4n+1$.

Another possibility is seeing whether multiplying the sequence by some constant, or adding some constant to each term, results in a familiar sequence. The more sequences you’re familiar with, the more useful this can be. For instance, for the second sequence one might also notice that each term is a multiple of $6$. Dividing out that common factor leaves a sequence $\langle 0,1,4,10,20,35,\ldots\rangle$, and if one has done enough with binomial coefficients and Pascal’s triangle, one might notice that this is the sequence

$$\left\langle\binom23,\binom33,\binom43,\binom53,\binom63,\binom73,\ldots\right\rangle\,.$$

Thus, the sequence $\langle 0,6,24,60,120,210,\ldots\rangle$ is

$$\left\langle 6\binom{n}3:n\ge 2\right\rangle=\langle n(n-1)(n-2):n\ge 2\rangle$$

which you can write as

$$\left\langle 6\binom{n+1}3:n\ge 1\right\rangle=\langle n(n+1)(n-1):n\ge 1\rangle$$

if you prefer to index the terms starting at $n=1$, or as

$$\left\langle 6\binom{n+2}3:n\ge 0\right\rangle=\langle n(n+1)(n+2):n\ge 0\rangle$$

if you prefer $0$-based indexing. With $1$-based indexing you get the nice closed form

$$a_n=n(n+1)(n-1)=n^3-n\,,$$

as Light Yagami found by a different route.

Similarly, after converting the first sequence to fractions with a common denominator, one might notice that the sequence $\langle 5,8,13,20,29,\ldots\rangle$ of numerators is $4$ more than a sequence of consecutive squares:

$$\left\langle1^2+4,2^2+4,3^2+4,4^2+4,5^2+4,\ldots\right\rangle\,.$$

The $1$-based general form would therefore appear to be

$$a_n=\frac{n^2+4}4=\left(\frac{n}2\right)^2+1\,.$$

Title: Finding general term of sequence


问:
Mathcounts National 2014 Sprint #23

I have been attempting this problem for some time now and can't seem to figure it out. If someone could help I would be grateful.

My attempt:

$4b^2 + 4b + 1  =  n^2$

$3b^2 + 5b + 1 = (n-2)^2$

$b^2 - b  = n^2 - (n^2+4-4n) = 4n-4$

$b(b-1) = 4(n-1)$

This led me to believe that b = n = 4.
However, for this to be true, $4(16) + 4(4) + 1$ would have to equal $4^2$ .

When this failed, I backtracked.

$b^2-b=4n-4$

$(b-2)^2 = 4n$

This let me find out that n is a square number, and it is positive. I tried substitution various squares but could not find anything that fit the original two equations:

$4b^2 + 4b + 1  =  n^2$

$3b^2 + 5b + 1 = (n-2)^2$

答:
From $4b^2 + 4b + 1 = n^2$, you should be able to conclude that $n = 2b + 1$.

Putting it into the equation $3b^2 + 5b + 1 = (n - 2)^2$, you should get $b^2 - 9b = 0$ and conclude that $b = 9$. And $n = 2b + 1 = 19$.

Title: Mathcounts National 2014 Sprint #23


问:
I don't understand when I should be using Ito's formula.
The question give was "consider the Ito integral: $\int_0^T sin(B_t)dB_t $ ."

I thought I had to use Ito's formula which is: $df(B_t)= f'(B_t)dB_t + \frac{1}{2} f''(B_t)dt $ hence I got: $sin(B_t) = \int_0^T cos(B_t)dB_t  - \frac 12 \int_0^T sin(B_t)dt $

But the answer given was $1 -  cos(t)$ which was done by integrating Sin. So I don't understand why wasn't Ito's formal used?

答: Your answer $1-\cos(t)$ can not be correct as the integral to be calculated is a stochastic integral. Using Ito's lemma with $f(x) = \cos(x)$ you got $$\mathrm{d}\cos(B_t) = -\sin(B_t)\,\mathrm{d} B_t - \frac 12\cos(B_t)\,\mathrm{d}t.$$ Now integrating both sides from $0$ to $T$, can you conclude?


答:
If something is true for all $x: 0< |x-a| < \delta_1$ then it will also be true for all $x: 0< |x-a| < \delta_0 < \delta_1$ because if $0< |x-a| < \delta_0$ then it is also true that $0 < |x-a| < \delta_1$.

So there is no way that trying to fix it for a smaller $\epsilon$ would require a larger $\delta$.  To require a greater $\delta$ would be to say the result fails for smaller values, and we must have the result pass for all smaller values.

Now it is possible that the $\delta_1$ required for a smaller $\epsilon$ doesn't have to be any smaller than the $\delta_0$... but it's illogical for it to be required to be larger.

Title: Is it possible for $\delta$ to increase when $\epsilon$ is decreasing

Body:
If $$\lim_{x \to x_0} f(x) = L$$
The $\epsilon-\delta$ definition of limit states that- $\forall \epsilon$, $\exists \delta$, such that:

$0<|x-x_0|<\delta\Rightarrow|f(x)-L|<\epsilon$. For any value of $\epsilon$ if I can provide a $\delta$, then $L$ is the limit of $f(x)$ at $x_0$.

If lesser and lesser values of $\epsilon$ are taken, is it possible for the maximum possible $\delta$ to increase?

i.e.: If $\epsilon=\epsilon_1$, $\delta=\delta_1$ is the maximum value that satisfies the condition, and if $\epsilon=\epsilon_2$, $\delta=\delta_2$ is the respective maximum value that satisfies the condition.

Then if $\epsilon_2<\epsilon_1$, is it possible for $\delta_2>\delta_1$?

If it is, can you give an example.


问:
Calculate the matrices of $R$ and $R\circ R$ with respect to the basis $(e_1,e_2,e_3,e_4)=(1,x,\frac{1}{2}x^2,\frac{1}{6}x^3)$
I am unsure how to calculate the basis matrices of the linear map defined below. I appreciate your help.

Let $V=\mathbb{Q}[x]_{\le3}$ be the set of polynomials over $\mathbb{Q}$ of degree at most $3$. Define $R:V \rightarrow V$ by $R(f)=xf''-5f$ where $f':=df/dx$. Prove that $R$ is linear (here the base field $K=\mathbb{Q}$). Calculate the matrices of $R$ and $R\circ R$ with respect to the basis $(e_1,e_2,e_3,e_4)=(1,x,\frac{1}{2}x^2,\frac{1}{6}x^3)$.

I proved that it's linear.

Now, $R(1) =-5$, $R(x) =-5x$, $R\left(\frac{1}{2}x^2\right) = x-\frac{5}{2}x^2$, $R\left(\frac{1}{6}x^3\right) = x^2-\frac{5}{6}x^2$.

Then the basis matrix of $R$ is as follows

$$\begin{pmatrix}
-5 & 0 & 0 & 0\\
0 & -5 & 1 & 0\\
0 & 0 & -5/2 & 1\\
0 & 0 & 0 & -5/6
\end{pmatrix}.$$

I want to find the matrices of $R \circ R$, so I thought of multiplying the matrix of R together
$$\begin{pmatrix}
-5 & 0 & 0 & 0\\
0 & -5 & 1 & 0\\
0 & 0 & -5/2 & 1\\
0 & 0 & 0 & -5/6
\end{pmatrix}\begin{pmatrix}
-5 & 0 & 0 & 0\\
0 & -5 & 1 & 0\\
0 & 0 & -5/2 & 1\\
0 & 0 & 0 & -5/6
\end{pmatrix} = \begin{pmatrix}
25 & 0 & 0 & 0\\
0 & 25 & -15/2 & 1\\
0 & 0 & 25/4 & -20/6\\
0 & 0 & 0 & 25/36
\end{pmatrix}.$$

I also tried of $R(R(1)) =25$, $R(R(x)) =25x$, $R\left(R\left(\frac{1}{2}x^2\right)\right) = -10x+\frac{25}{2}x^2$, $R\left(R\left(\frac{1}{6}x^3\right)\right) = -8x-5x^2-\frac{25}{6}x^3$ which results in the matrix

$$\begin{pmatrix}
25 & 0 & 0 & 0\\
0 & 25 & -10 & -8\\
0 & 0 & 25/2 & -5\\
0 & 0 & 0 & 25/6
\end{pmatrix}.$$

Which way is the correct one? Thank you for your time.

答:
The matrix of $R$ is wrong.

For example $R(e_3)=R\left(\frac{1}{2}x^2\right)=x-\frac52x^2=x-5\cdot\left(\frac12x^2\right)=e_2-5e_3$...

ContentLicense: CC BY-SA 3.0


问:
Definite integral $\int_0^{\pi} \sin^{n} x \ln(\sin x) dx$?
Is the integral
$$I(n) = \int_0^{\pi} \sin^{n}(x) \ln(\sin(x)) dx$$
analytically tractable for $n \in \mathbb{N}$? If not, are there good upper and lower bounds?

To clarify, although this has a fairly standard-looking form, I haven't been able to find an answer anywhere and I've tried a few of the usual tricks (although I'm a tad rusty).

I have tried integration by parts to reduce to a simpler or recursive form but this seems to make the expression more complex, and likewise with the substitution $u=\sin(x)$.

Equally, it is relatively easy to see that the expression under the integral is non-positive. Taking a derivative
$$ \frac{d}{dx} \sin^{n}(x) \ln(\sin(x)) = \sin^{n-1}(x) \cos(x) (1 + n \ln(\sin(x))) = 0$$
gives a has minimum at $-\frac{1}{en}$.

Together these give the reasonably simple bounds: $$-\frac{\pi}{en} < I(n) < 0.$$

But numerically these don't appear to be particularly tight, so I'm still hopeful of an exact solution or an improvement.

答:
As was hinted at in @Quanto's answer, we have the Wallis integral
$$W(\alpha) := \int_0^{\pi/2} \sin^\alpha(x)\ dx = \frac{\sqrt \pi}{2} \frac{\Gamma\left(\frac{\alpha+1}{2}\right)}{\Gamma\left(\frac \alpha 2 + 1\right)}, $$
for all $\alpha \geqslant 0$. Differentiating $W$ with respect to $\alpha$, we find (by the symmetry of the integrand)
$$\frac{dW}{d\alpha} (\alpha) = \int_0^{\pi/2} \sin^{\alpha}(x) \log(\sin(x))\ dx = \frac 1 2 I(\alpha), $$
so that, setting $g(\alpha) := \Gamma\left(\dfrac{\alpha+1}{2}\right)\Big/\Gamma\left(\dfrac \alpha 2 + 1\right)$,
$$\begin{split}
\frac{1}{\sqrt{\pi}}I(\alpha) &= \frac{dg}{d\alpha} (\alpha) = \frac{d}{d\alpha}e^{\log g(\alpha)} = e^{\log g(\alpha)} \frac{d}{d\alpha}\log(g(\alpha)) \\
&=g(\alpha) \frac{d}{d\alpha}\left\{ \log \Gamma\left(\frac{\alpha+1}{2}\right) -\log\Gamma\left(\frac \alpha 2 + 1\right)\right\} \\
&= \frac 1 2\frac{\Gamma\left(\frac{\alpha+1}{2}\right)}{\Gamma\left(\frac \alpha 2 + 1\right)} \left\{\psi^{(0)}\left(\frac{\alpha+1}{2}\right) -\psi^{(0)}\left(\frac \alpha 2 + 1\right) \right\},
\end{split}$$
where $\psi^{(0)}(z)=(\log \Gamma)'(z)$ is the digamma function.

Bonus. We can go further. Using the identity $\psi^{(0)}(t+1) = \psi^{(0)}(t) + 1/t$ for all $t>0$, and taking advantage of the fact that
$$\psi^{(0)}\left(\frac{s+1}{2s} \right) - \psi^{(0)}\left(\frac{1}{2s} \right) = \sum_{k=0}^\infty \frac{(-1)^k}{sk+1}$$
for $s>0$, choosing $s = (\alpha+2)^{-1}$ we find
$$\frac{\alpha+2}{2}\left\{\psi^{(0)}\left(\frac{\alpha+1}{2}\right) -\psi^{(0)}\left(\frac \alpha 2 + 1\right) \right\} = - \frac{\alpha+2}{\alpha+1} + \sum_{k=0}^\infty \frac{(-1)^k}{k/(\alpha+2)+1},  $$
whence
$$\begin{split}
I(\alpha) &=  \sqrt\pi g(\alpha)\left\{-\frac1{\alpha+1} + \frac{1}{\alpha+2}\sum_{k=0}^\infty \frac{(-1)^k}{k/(\alpha+2)+1}\right\}\\
&=\sqrt\pi g(\alpha) \sum_{k=1}^\infty \frac{(-1)^k}{\alpha+k}.
\end{split}$$
Now we restrict ourselves to nonnegative integers $\alpha = n \geqslant 1$, and we observe that
$$\sum_{k=0}^\infty \frac{(-1)^k}{n+k} = \sum_{j=n}^\infty \frac{(-1)^{j-n}}{j} = (-1)^{n+1} \sum_{j=n}^\infty \frac{(-1)^{j+1}}{j} =(-1)^{n+1}\left[-\tilde H_{n} + \log(2) \right], $$
where $\tilde H_{n} = \sum_{k=1}^{n} \frac{(-1)^{k+1}}{k}$ is the $n$-th alternating harmonic number. we obtain
$$\boxed{I(n) =   (-1)^n\sqrt\pi \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac n 2 + 1\right)} \big( \tilde H_n - \log(2)\big).}$$
One may then distinguish even-$n$ and odd-$n$ cases and employ the duplication formula for $\Gamma$ to express $I(n)$ in terms of factorials and double factorials.


问:
Finitely generated module over $A[[x]]$
I have a question about module over power series ring.

Let $A$ is a local ring with maximal ideal $\mathfrak m$, I'm more interested in the case that $A=\mathbb Z_p/p^n\mathbb Z_p$

If $M\neq0$ is a finitely generated torsion module over $A[[x]]$, then is $M$ finitely generated over $A$?

I think it suffices to assume that $M$ is generated by one element $m$ and $m$ is killed by some series $f(x)=\sum_{i=0}a_ix^i$. But if $a_0\in\mathfrak m$, I can't deduce anything.

Thanks.

答:
You say you know the structure theory for f.g. torsion modules over the Iwasawa algebra $\Lambda := \mathbb Z_p[[x]]$. But then the result follows easily.

Namely, the ring $(\mathbb Z_p/p^n \mathbb Z_p) [[x]]$ you look at is just $\Lambda / (p^n)$, and like over any ring, a $\Lambda / (p^n)$-module is the same as a $\Lambda$-module on which $p^n$ operates as zero. Further, such a module $M$ being torsion and f.g. as $\Lambda / (p^n)$-module implies that it is f.g. and torsion as $\Lambda$-module. So we can apply said structure theory and now there is a homomorphism with finite kernel and cokernel

$$ M \rightarrow \bigoplus_i\mathbb{Z}_p[\![x]\!]/(p^{\mu_i})\oplus\bigoplus_j\mathbf{Z}_p[\![x]\!]/(f_j^{m_j})$$

Since being f.g. over $\mathbb Z_p$ will not be changed by those finite kernel and cokernel, w.l.o.g. we can assume this is an isomorphism; then, to be torsion as $\Lambda/(p^n)$-module means that the first sum must vanish (because it's not torsion). But to have $p^n$ operate as $0$, the second sum must vanish as well. Meaning that up to something finite, $M =0$; meaning that $M$ is finite; meaning that it's finitely generated as module over whatever ring you want.

Title: Finitely generated module over $A[[x]]$


答:
The basic recipe for using Ito's formula to solve $\int_0^T f(B_t) dB_t$ when $f$ is a $C^1$ function is to look at $F(B_t)$ where $F$ is an antiderivative of $f$ and then compute the differential of this process by Ito's formula. What you get is

$$dF(B_t)=f(B_t) dB_t + \frac{1}{2} f'(B_t) dt.$$

Therefore by moving a term over:

$$\int_0^T f(B_t) dB_t = F(B_T)-F(0)-\frac{1}{2} \int_0^T f'(B_t) dt.$$

Title: I don't understand when I should be using Ito's formula.

Body:
The question give was "consider the Ito integral: $\int_0^T sin(B_t)dB_t $ ."

I thought I had to use Ito's formula which is: $df(B_t)= f'(B_t)dB_t + \frac{1}{2} f''(B_t)dt $ hence I got: $sin(B_t) = \int_0^T cos(B_t)dB_t  - \frac 12 \int_0^T sin(B_t)dt $

But the answer given was $1 -  cos(t)$ which was done by integrating Sin. So I don't understand why wasn't Ito's formal used?


问:
Rearranging Cauchy Riemann equations
I've used the Cauchy-Riemann equations to find the analytic function:

$$y^2−x^2−2y+2+i(2x(1−y))$$

But I'm having a slight rearranging problem and need to write it in terms of $z$, where $z=x+iy$.

Any suggestions would be much appreciated

答: Given a function in terms of $x$ and $y$ you can introduce formally new variables $z:=x+iy$ and $\bar z:=x-iy$, i.e., $x=(z+\bar z)/2$, $y=(z-\bar z)/(2i)$. If the resulting expression in $z$ and $\bar z$ does not contain the variable $\bar z$ your function is actually an analytic function of $z$.</p>

Title: Rearranging Cauchy Riemann equations


问:
Matrix operations - filter out diagonal elements of a matrix using multiplication operation
Is there a way to filter out only the diagonal elements of a matrix $A$, by doing a matrix multiplication with some matrix $B$ like this:

$A*B$ = $D$

where $D$ matrix contains only diagonal elements of $A$?

答: No, unless $A$ is $1\times1$. Try $A=\pmatrix{1&0\\ 1&0}$.</p>


问:
$|(f, Tf)| \leq (f, |T|f)$?
Let $H$ be a separable and complex Hilbelt space.
If $T$ is a bounded linear operator on $H$, then is it true that $|(f, Tf)| \leq (f, |T|f)$ for all $f \in H$ ?

Observation.
If $H = L^2(\mathbb{R}^d)$ and $T_g f:=gf$ with $g \in L^\infty$, the above statement is true because
$$
|(f, T_g f)| \leq \int_{\mathbb{R}^d} |f(x)|^2 |g(x)| dx = (f, |T_g|f).
$$
Note that $|T_g| = \sqrt{(T_g)^* T_g} = \sqrt{T_{\overline{g}} T_g} = \sqrt{T_{|g|^2}} = T_{|g|}$.

答:
Let  $H={\mathbb C}^2$, and
$$
  T= \pmatrix{0 & 0 \cr 1 & 0}.
  $$
Then
$$
  |T| =
  (T^*T)^{1/2} =
  \left(\pmatrix{0 & 1 \cr 0 & 0} \pmatrix{0 & 0 \cr 1 & 0}\right)^{1/2} =
  \pmatrix{1 & 0 \cr 0 & 0}.
  $$
With $f = (z,w)$, we then have
$$
  \langle f,Tf\rangle  =   \langle (z, w),(0,z )\rangle  = \bar w z,
  $$
while
$$
  \langle f,|T|f\rangle  =   \langle (z, w),(z,0 )\rangle  = |z|^2.
  $$
The assertion being made thus becomes
$$
  |\bar wz|\leq  |z|^2,
  $$
which obviously does not always hold.

On the other hand,  in case $T$ is normal,  then the answer is yes and the proof is essentially that given by the OP,
observing that the spectral Theorem implies that every normal operator is conjugate to  a multiplication operator on an
$L^2$ space.

Title: $|(f, Tf)| \leq (f, |T|f)$?


问:
$z=x=y$ in $\Bbb R^3$
It is a simply question. But I want to make sure. The surface $$z=x=y$$ represents  a line in $\Bbb R^3$ that passes  through the origin. Is that correct?

答: Consider the your surface $S=\lbrace(x,y,z)\in \mathbb{R}^3 \mid x=y=z \rbrace$ subspace of $\mathbb{R}^3$ a basis for $S$ is $\beta=\lbrace (1,1,1)\rbrace$ therefore your $S$ is a line that pass throught of $(1,1,1)$. As @JMoravitz say.

Title: $z=x=y$ in $\Bbb R^3$


问:
Show that $r′′(x) = −1$ for some value $x$
Consider a smooth function $r : R → R$ for which $r(0) = r(1) − 2 =
r(2) − 2 = r(3) = r(4)$.
So based on my approach, we know that $r'(c)= \frac{(r(a)-r(b))}{(a-b)}$. and $r''(c)= \frac{(r'(a)-r'(b))}{(a-b)}$, however,I don't know how to obtain -1, could someone help me?

答:
Using the Mean Value Theorem(MVT), we can find $a \in [1, 2]$ and $b \in [3, 4]$ such that $r'(a)=0=r'(b)$. Applying the MVT again we can find a $c \in (a, b)$ such that $r''(c)=0$.

Using MVT once again, we can find $\alpha \in [0, 1]$ and $\beta \in [2, 3]$ such that $r'(\alpha)=2$ and $r'(\beta)=-2$. Applying the MVT again we can find a $\gamma \in (\alpha, \beta)$ such that $r''(\gamma)=\frac{-4}{\beta - \alpha}$. Note that $1 < \beta - \alpha < 3$, so $\frac{-4}{\beta - \alpha} < \frac{-4}{3}$.

Finally, since $\frac{-4}{3} < -1 < 0$, by Darboux's Theorem(or by continuity of $r''$), there is some $x \in )\gamma, c($ such that $r''(x)=-1$.


问:
Finding all rational values of $x$ for which $256(x + 15\{x\})\{x\} = 2021$
Find all rational numbers $x$ that satisfy $$256(x + 15\{x\})\{x\} = 2021$$ where $\{x\}$ is the fractional part of $x.$

I first substituted in $x - \lfloor x \rfloor = \{x\},$ which gave me $$256(16x - 15\lfloor x \rfloor)(x - \lfloor x \rfloor) = 2021$$ after simplifying. However, from here, I got stuck.

答:
Assuming that $x=\lfloor x\rfloor+\{x\}$, this is equivalent to
$$
(\lfloor x\rfloor+16\{x\})\{x\}=\frac{2021}{256}
$$
Let $n=\lfloor x\rfloor\in\mathbb{Z}$ and $t=\{x\}\in[0,1)$. Then we need to solve
$$
16t^2+nt-\frac{2021}{256}=0
$$
which by the quadratic formula gives
$$
t=\frac{-2n\pm\sqrt{4n^2+2021}}{64}
$$
To ensure that $t\ge0$, we need to take the positive square root.

If $\sqrt{4n^2+2021}\in\mathbb{Q}$, then $\sqrt{4n^2+2021}\in\mathbb{Z}$. So we must have a positive $m\in\mathbb{Z}$ so that
$$
(m-2n)(m+2n)=43\cdot47=1\cdot2021
$$
This leaves us with
$$\require{cancel}
\begin{array}{l}
&&t=\frac{m-2n}{64}&x=n+t\\\hline
n=1&m=45&t=\frac{43}{64}&x=\frac{107}{64}\\
n=-1&m=45&t=\frac{47}{64}&x=-\frac{17}{64}\\
n=505&m=1011&t=\frac1{64}&x=\frac{32321}{64}\\[-3pt]
n=-505&m=1011&\hspace{-1.5mm}\cancel{t=\frac{2021}{64}}
\end{array}
$$


答:
In between $x = 1$ and $x = 2$, there must be a $c_1$ such that $r'(c_1) = 0$.

In between $x = 2$ and $x = 3$, there must be a $c_2$ such that $r'(c_2) = -2$.

In between $x = 3$ and $x = 4$, there must be a $c_3$ such that $r'(c_3) = 0$.

So in between $x = 1$ and $x = 3$, there must be a $c$ such that $\frac{-2-0}{3-1}=-1 \ge r''(c)$. The upper bound happens if $c_1=1$ and $c_2 = 3$.

In between $x = 2$ and $x = 4$, there must be a $c$ such that $\frac{0-(-2)}{4-2}=1 \le r''(c)$.

Then you can use the mean value theorem to finish.

Title: Show that $r′′(x) = −1$ for some value $x$

Body:
Consider a smooth function $r : R → R$ for which $r(0) = r(1) − 2 =
r(2) − 2 = r(3) = r(4)$.
So based on my approach, we know that $r'(c)= \frac{(r(a)-r(b))}{(a-b)}$. and $r''(c)= \frac{(r'(a)-r'(b))}{(a-b)}$, however,I don't know how to obtain -1, could someone help me?


问:
Do we use material conditional in regular mathematics?
I have been confused by the definition and truth table of material conditional (MC) for quite a while and the only responses I have received are variations of 'it is correct' and 'I should just get used to it'.

Today I discovered an interesting MC example, which regular mathematicians would never agree to as true, yet according to MC rules it is for most of natural numbers:

$x$ is prime $\implies$ $x + 2$ is prime

According to the definition of MC, this statement is true for all even numbers (except 2), all non-prime odd numbers, as well as for a small fraction of prime numbers (every first prime in a twin-prime pair). Only for the prime numbers that are not the first prime in a twin prime pair, does it fail to hold. This statement is true for about 99% of natural numbers.

Yet it is clear and understood according to regular mathematics that this is statement is simply wrong. By regular mathematics I mean 99% of mathematics, when we're not studying foundational issues, logic, and so on.

My question is, is it correct to state that MC has practically no relevance to regular mathematics? If yes, is it a good idea to insist on 'getting used to the MC definition', 'just treat MC as a truth table', or 'try not to worry too much about MC' when studying logic?

Edit 1: I would like to expand on what I said, in response to the comments mentioning quantified (or first-order) implication. Let's denote it by $_{\implies}^{\forall}$. The following is false in logic as well as regular mathematics:

$x$ is prime $_{\implies}^{\forall}$ $x+2$ is prime

However, there's agreement here only because we have counterexamples (e.g., $x=7$ and many more). We could reach the same conclusion even if we discarded the true parts of the MC truth table, i.e., the three clauses F$\implies$F, F$\implies$T, and T$\implies$T. We could replace their true values with something like, "N/A", "Unknown", or "Inconclusive", and our first-order logic, and the rest of regular mathematics would still work just fine. (Of course in that case, the MC table would no longer be a complete truth-table).

Quantified-implication <s>does not need the MC truth-table, or even for MC to exist. All it</s> requires <s>is</s> the result of a counterexample search (i.e., a case of $A \land \lnot B$, where, e.g., $A$ is '$x$ is prime' and $B$ is '$x+2$ is prime'). If the counterexample search is complete, and a counterexample is found, $_{\implies}^{\forall}$ is false. If the search is complete and no counterexample found, $_{\implies}^{\forall}$ is true. (If a counterexample is not found, and the search is not complete, or we're not sure if the search is complete, the true values of MC only mislead, never help).

In other words, we introduce true clauses in MC, because (maybe) we like a completed truth-table, at the expense of confusing students, when those true values are useless, and infact harmful, once we start using quantified implication with counterexample search.

Edit 2: I should clarify (and maybe correct) myself. Quantified-implication does need the MC truth table. So I stand corrected. However, MC truth table does not make much sense without the use of a quantifier. So what I'm trying to say is the following:

Take the MC formula $\lnot [A \land \lnot B]$ and ignore the MC symbol $\implies$.

The formula $\lnot [A \land \lnot B]$ is combined with $\forall$ to form quantified-implication $_{\implies}^{\forall}$ as follows:

$A(x) {}_{\implies}^{\forall} B(x)$ defined as $\forall x (\lnot [A(x) \land \lnot B(x)])$

What that means is that the MC formula finds its use through first-order quantifiers in mathematics. However, in my opinion, if we try to ascribe 'implication' or 'if ... then' or any other meaning to the MC formula itself, instead of treating it as a mere formula, that causes needless confusion.

Edit 3: After the answer by Alex Kruckman, I should add that that's another helpful test-case, i.e., prove implication mathematically before invoking MC. However, it means the MC truth-table was only consulted after-the-fact. The implication was established mathematically, not through propositional logic. So the two use-cases so far: (1) as part of a quantifier, (2) as a pre-proved implication. My prime example doesn't fall in either, and hence I'm still of the opinion that there are problematic consequences of MC (and I'm not even talking about the paradoxes).

答:
I have been confused by the definition and truth table of material
conditional (MC) for quite a while and the only responses I have
received are variations of 'it is correct' and 'I should just get used
to it'.

The definition of the material conditional usually given as $A\to B \equiv \neg [A \land \neg B]$ as well as each line of its truth table can be formally derived from first principles using a simple form of natural deduction.

$A\to B \equiv \neg [A \land \neg B]~~~$ Formal Proof (19 lines)

$A \land B \to [A\to B]~~~~~~~$  Line 1 of the truth table Formal Proof (6 lines)

$A \land \neg B \to \neg [A \to B]~$ Line 2 of the truth table
Formal Proof (8 lines)

$\neg B \to [A\to B]~~~~~~~~~~~$ Lines 3 and 4 of the truth table
Formal Proof (8 lines)

My question is, is it correct to state that MC has practically no
relevance to regular mathematics?

No. Quite the contrary. MC is unavoidable in "regular" mathematics. You seem to be quite misinformed on this matter. If you want to make a name for yourself, find the error in any of the above proofs. The first is probably the most important. (Only 19 lines!)


问:
$\boldsymbol{2}^{\kappa} +_{\mathscr{C}} \kappa = \boldsymbol{2}^{\kappa}$ where $\kappa$ is an infinite cardinal
I am proving the following formula:

$\boldsymbol{2}^{\kappa} +_{\mathscr{C}} \kappa = \boldsymbol{2}^{\kappa}$, where $\kappa$ is an infinite cardinal and $\boldsymbol{2}^{\kappa}$ is cardinal exponential, $+_{\mathscr{C}}$ is cardinal addition.

The left-hand side of the formula can be expanded as below:
\begin{equation*}
\begin{aligned}
\boldsymbol{2}^{\kappa} +_{\mathscr{C}} \kappa &= \operatorname{Card}\left(\boldsymbol{2}\right)^{\operatorname{Card}\left(\kappa\right)} +_{\mathscr{C}} \operatorname{Card}\left(\kappa\right)\\
&= \operatorname{Card}\left(\boldsymbol{2}^{\kappa}\right) +_{\mathscr{C}} \operatorname{Card}\left(\kappa\right)\\
&= \operatorname{Card}\left(\left(\boldsymbol{2}^{\kappa}\times \left\{\boldsymbol{0}\right\}\right) \cup \left(\kappa \times \left\{\boldsymbol{1}\right\}\right)\right)\\
&\approx \left(\boldsymbol{2}^{\kappa} \times \left\{\boldsymbol{0}\right\}\right) \cup \left(\kappa \times \left\{\boldsymbol{1}\right\}\right),
\end{aligned}
\end{equation*}
and similarly, we have
\begin{equation*}
\boldsymbol{2}^{\kappa} \approx \boldsymbol{2}^{\kappa},
\end{equation*}
where the second $\boldsymbol{2}^{\kappa}$ is the set of all functions from $\kappa$ to $\boldsymbol{2}$.

To finish the proof, we have to establish $\left(\boldsymbol{2}^{\kappa} \times \left\{\boldsymbol{0}\right\}\right) \cup \left(\kappa \times \left\{\boldsymbol{1}\right\}\right) \approx \boldsymbol{2}^{\kappa}$. To establish $\boldsymbol{2}^{\kappa} \preceq \left(\boldsymbol{2}^{\kappa} \times \left\{\boldsymbol{0}\right\}\right) \cup \left(\kappa \times \left\{\boldsymbol{1}\right\}\right)$ is pretty straightforward, but how should we come up with the reverse? I guess we have to utilize the property that $\kappa$ is infinite. Can anyone help establish
\begin{equation}
\left(\boldsymbol{2}^{\kappa} \times \left\{\boldsymbol{0}\right\}\right) \cup \left(\kappa \times \left\{\boldsymbol{1}\right\}\right) \preceq \boldsymbol{2}^{\kappa}?
\end{equation}

答: In general, whenever we have two infinite cardinals $A$ and $B$, we have $A +_{\mathscr{C}} B = \max(A, B)$. So in this case, since $2^\kappa > \kappa$, we have $2^\kappa +_{\mathscr{C}} \kappa = 2^\kappa$.


问:
finding the remainder of a variable n where n is not an integer multiple of 4
I started off by saying that $(n+1)$ must be a multiple of $4$ for $n(n+1)$ to be a multiple of 4. Hence $n$ when divided by $4$ would have a remainder of $3$
i.e $n = 4k+3$

$n(n+1)=(4k+3)(4k+3+1)
= 4(k^2+7k+3)$

Hence $n(n+1)$ is a multiple of 4

I'm not sure if this explanation is adequate and how to progress from here if it is not. Please advise, thanks!

答:
Let $n$ be a positive integer such that $4\mid n(n+1)$ and $4\nmid n$.

Suppose $n$ even : then $n+1$ is odd and thus $4$ and $n+1$ are coprime. By Gauss theorem, we see that $4\mid n$, a contradiction.

As a consequence : $n$ is odd and the same argument shows that $4\mid n+1$. Hence the remainder of $n$ divided by $4$ is $3$.


问:
Vakil exercise 17.4.D: different notions of degree of a finite morphism
We have a finite morphism $\pi: C \rightarrow C'$ between two regular integral curves over a field $k$. On the previous page, Vakil proves that such a morphism has a well defined rank defined as the rank of locally free sheaf $\pi_{*} \mathcal{O}_C$. Now we have to show that the degree can also be computed as follows:

Let $p$ be a closed point in $C'$, and let $\pi^{-1}(p) = \{p_1, ..., p_m\}$. If $t$ is a uniformizer of the DVR $\mathcal{O}_{C',p}$, then show that $$\text{deg} \pi = \sum_{i=1}^m (\text{val}_{p_i} \, \pi^*t) \,\,\text{deg}(k(p_i)/k(p))$$

With standard reductions we have that the map corresponds to injective ring map $f: B \rightarrow A$ where $A$ and $B$ are Dedekind domains and $p$ corresponds to a maximal ideal of $B$ such that its uniformizer $t$ is a regular function of $\text{Spec} B$.  With this terminology, $\pi^* t$ is simply $f(t)$.  I am stuck at this point. This sounds like the situation of splitting of primes in basic number theory, but since Vakil hasn't talked about that I am wondering If there is a straightforward way to do it.

答:
Let $\operatorname{Spec} A\subset C'$ be an affine open subset where $f_*\mathcal{O}_{C}$ is free of rank $r$. Let $\operatorname{Spec} B=f^{-1}(\operatorname{Spec} A)$. Then $(f_*\mathcal{O}_C)(\operatorname{Spec} A)\cong B$, but on the other hand, since $f_*\mathcal{O}_C$ is free of rank $r$ over $\operatorname{Spec} A$, we have that $(f_*\mathcal{O}_C)(\operatorname{Spec} A)\cong A^r$. So $A^r\cong B$ as $A$-modules, and therefore for a closed point $p\in C'$ we have that $A^r\otimes_A k(p)\cong B\otimes_A k(p)$ as $k(p)$-modules. The $k(p)$-dimension of the LHS is $r$, while the dimension of the RHS is exactly the quantity you're interested in.

To see this, suppose $t\in A$ is a uniformizer for $\mathcal{O}_{C',p}$. Then $k(p)\cong A/t$, and $B\otimes_A k(p)\cong B/f^*t$. As $f^*t$ vanishes exactly on the finite set of preimages of $p$, we have that $B/f^*t\cong \prod_{q\mapsto p} \mathcal{O}_{C,p}/f^*t$: the LHS is the global sections of the sheaf $\widetilde{B/f^*t}$ on $\operatorname{Spec} B$, while the RHS is the product of the stalks, and these quantities are equal for sheaves with discrete support. Next, $\dim_{k(p)} \mathcal{O}_{C,q}/f^*t = \dim_{k(p)} k(q)^{\operatorname{val}_q f^*t} = (\operatorname{val}_q f^*t)\cdot[k(q):k(p)]$ and summing, we have the result.


问:
Two stage cyclic queue
Given a cyclic queue of two servers of exponential service rates, if there are N customers at one server at time t, how do i start about showing that N can be modeled as a birth and death process? and then find the BD rates??

Totally have no idea how to begin. what are we supposed to show to prove it can be modeled as a BD process?

答:
Suppose server 1 serves at rate $\mu_1$ and server 2 at rate $\mu_2$ and that each server can only serve a single job at a time. All other jobs at the service node are waiting.

As the system is closed and cyclic, if there are $k$ customers at node 1 we know that there are $N-k$ customers at node 2. Write $X_t$ for the number of customers at node 1 at time $t$. This can take values in $\{0,1,2,3,\ldots,N\}$.

In states $\{0,1,2,3,\ldots,N-1\}$ arrivals can happen as there are jobs at node 2 to be served. These arrivals happen at rate $\mu_2$. In state $N$ no arrivals are possible as all jobs are at node 2.

In state $0$ there can be no departures/services because all the jobs are at node 2, there are no jobs at node 1 to serve. In states $\{1,2,3,\ldots,N\}$ departures are possible (as a job completes service) and happen at rate $\mu_1$. Therefore the transition rate matrix $Q$ is (zeroes in the blanks)
$$Q=\begin{pmatrix}
-\mu_2 & \mu_2 \\
\mu_1 & -\mu_1 -\mu_2 & \mu_2 \\
&\mu_1 & -\mu_1 -\mu_2 & \mu_2 \\
&&\mu_1 & -\mu_1 -\mu_2 & \mu_2 \\
&&&& \ddots \\
&&&&\mu_1 & -\mu_1 -\mu_2 & \mu_2 \\
&&&&& \mu_1 & -\mu_1
\end{pmatrix}.$$

The model $X_t$ is a finite capcity M/M/1 queue. The process is a birth-death process because the transition rate matrix is a tridiagonal matrix; it has non-zero elements only on the main diagonal, on the first diagonal above this and the first below this.

Title: Two stage cyclic queue


问:
Bass: Exercise 2.5, preimages of sets in sigma algebra forms a sigma algebra

Exercise 2.5: Let $(Y, \mathcal{A})$ be a measurable space and let $f$ map $X$ into $Y$, but do not assume that $f$ is one-to-one. Define $\mathcal{B} = \{f^{-1}(A): A \in \mathcal{A}\}$. Prove that $\mathcal{B}$ is a $\sigma$-algebra of subsets of $X$.

My work so far:

First, notice that $f(\emptyset) = \emptyset \in \mathcal{A}$ and since $f$ maps $X$ into $(Y, \mathcal{A})$ we have that $f(X) \in \mathcal{A}$. \

Figure out argument for complement (My problem point)

Now, suppose that $\{B_i\}_{i=1}^\infty \subset \mathcal{B}$, that is $f(B_i) = A_i \in \mathcal{A}_i$ for all $i \in \mathbb{N}$. Then, notice that
\begin{equation*}
 f\biggr(\bigcup_{i=1}^\infty B_i\biggr) = \bigcup_{i=1}^\infty f(B_i) = \bigcup_{i=1}^\infty A_i \in \mathcal{A},
\end{equation*}
since $\mathcal{A}$ is a $\sigma$-algebra. Therefore, $\bigcup_{i=1}^\infty B_i \in \mathcal{B}$. Moreover, due to De Morgan's laws we have
\begin{equation*}
 \biggr(\bigcap_{i=1}^\infty B_i \biggr)^c = \bigcup_{i=1}^\infty B_i \in \mathcal{B},
\end{equation*}
since $\mathcal{B}$ is closed under complement and countable union. Exploting the closure under complements again, we have that $\bigcap_{i=1}^\infty B_i \in \mathcal{B}$.

My Issue: I think I have everything figured out except for the complement. My issue stems from the fact that we may not assume that $f$ is injective. If so, it would be simple, I think
\begin{equation*}
f(B^c) = f(X \setminus B) = f(X) \setminus f(B) = f(X) \cap f(B)^c
\end{equation*}
but since it is not injective, there may be some $b \in B^c$ such that $f(b) \in f(B)$. Any hints will be appreciated!

答:
You seem to be having a bit of notational confusion. You keep referring to the images of sets, but the definition of $\mathcal{B}$ has to do with preimages, not images.

For example, you write

"First, note that $f(\emptyset) = \emptyset \in \mathcal{A}$"

This is totally irrelevant. What you should say is:

"First, note that $\emptyset \in \mathcal{A}$ and $f^{-1}(\emptyset) = \emptyset$. Then $\emptyset \in \mathcal{B}$."

The proof follows quickly when you use pre-images and not images.


问:
Why does the determinant of $(A- \lambda I) $ give its eigenvector?

In question 6, why does determinant of $(A- \lambda I)$ gives its eigenvector ?

答:
Start from the eigenvalue problem: we look for $\vec{v}, \lambda$ such that

$$A \vec{v} = \lambda \vec{v} = \lambda I \vec{v}$$

where $I$ is the identity matrix. Gathering things on the left hand side, we are trying to solve

$$(A - \lambda I) \vec{v} = 0$$

but we aren't interested in the "trivial" solution $\vec{v}=0$. If the matrix $(A - \lambda I)$ is invertible, then we can multiply by its inverse on both sides and we just get $\vec{v}=0$. A matrix is invertible if and only if its determinant is not equal to $0$. So we are looking specifically for the case that

$$\det(A - \lambda I)=0$$

to find the eigenvalues of $A$.


问:
Find a function $g : [−1, 1] \to \mathbb R$ with $|g(x)| < |x|^4$ for which $g ′′(0)$ does not exist.
For this question, I couldn't think of one function for g, based on what I know, $g'(c)= \frac{g(a)-g(b)}{a-b}$. and $g''(c)= \frac{(g'(a)-g'(b))}{(a-b)}$, but how do we exactly show $g''(0)$ doesn't exist?

答:
Consider $f:[-1,1]\to\mathbb{R}$ defined by :

$$f(x)=x^4\sin\left(\frac1{x^2}\right)\text{ if }x\neq0\text{ and }0\text{ otherwise}$$

It is clear that $f$ is $C^\infty$ on $[-1,0)\cup (0,1]$.

Since $\vert f(x)\vert\leqslant x^4$ for all $x\in [-1,1]$, we see that $\displaystyle{\lim_{x\to 0}f(x)=0=f(0)}$. This proves that $f$ is continuous at $0$.

Moreover, for all $x\neq0$ :

$$\frac{f(x)-f(0)}{x}=x^3\sin\left(\frac{1}{x^2}\right)\underset{x\to 0}{\rightarrow}0$$which proves that $f$ is differentiable at $0$ and $f'(0)=0$.

For all $x\neq0$ :
$$f'(x)=4x^3\sin\left(\frac{1}{x^2}\right)-2x\cos\left(\frac{1}{x^2}\right)\underset{x\to 0}{\rightarrow}0=f'(0)$$which proves that $f'$ is continuous at $0$.

Now, for all $x\neq0$ :
$$\frac{f'(x)-f'(0)}{x}=4x^2\sin\left(\frac{1}{x^2}\right)-2\cos\left(\frac{1}{x^2}\right)$$doesn't have any limit at $0$.
This proves that $f$ is not twice differentiable.

EDIT

Finally, if we multiply $f$ by some $A\in(0,1)$ (say $A=0.9$ for example), we get a function with the desired properties (strict inequality for any $x\in[-1,0)\cup (0,1]$).


问:
A beautiful limit involving primes and composites
I observed the following limit empirically. Let $p_n$ be the $n$-th prime and $c_n$ be the $n$-th composite number then,

$$
\lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n}\frac{p_n c_n}{p_n c_n + p_i c_i} = \frac{\pi}{4}.
$$

I am looking for a proof.

答:
It turns out that

The limit is correct, but

It's not saying anything that's very special to primes and composites.

Note that (inspired by DonAntonio's answer)
$$
\lim_{n\to\infty} \frac1n \sum_{k=1}^n \frac{n^2}{n^2 + k^2}
= \lim_{n\to\infty} \frac1n \sum_{k=1}^n \frac{1}{1 + \left(\frac{k}{n}\right)^2}
= \int_{0}^{1} \frac{dx}{1 + x^2} = \frac{\pi}{4}
$$

It just so happens that $p_n$ and $c_n$ are (at a very loose level of approximation) on the order of $n$ each, so that $p_n c_n$ is of the order of $n^2$, and therefore your sum
$$
\frac{1}{n}\sum_{k=1}^{n}\frac{p_n c_n}{p_n c_n + p_k c_k}
\approx \frac{1}{n}\sum_{k=1}^{n}\frac{n^2}{n^2 + k^2},
$$
the approximation turning exact in the limit.

To prove this rigorously, we have from the prime number theorem, that $p_n \sim n\ln n$, or to be precise
$$p_n = n\left(\ln n + \ln\ln n - 1 + O\left(\frac{\ln\ln n}{\ln n}\right)\right) = n\left(\ln n + o(\ln n)\right).$$
Similarly for the $n$th composite number $c_n$, we have
$$c_n = n\left(1 + \frac1{\ln n} + O\left(\frac{1}{\ln^2 n}\right)\right) = n\left(1 + o(1)\right).$$

So $$p_nc_n = n^2\left( \ln n + o(\ln n) \right).$$

Consider a particular value of $\frac{k}{n}$ (say $\alpha$) so that $k = \alpha n$. Then
$$
\frac{p_kc_k}{p_nc_n}
= \frac{k^2 (\ln k + o(\ln k))}{n^2(\ln n + o(\ln n))}
= \frac{k^2}{n^2}\frac{\ln n + \ln \alpha + o(\ln n)}{\ln n + o(\ln n)}
= \frac{k^2}{n^2}(1 + o(1))
$$

Therefore
$$
\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n}\frac{p_n c_n}{p_n c_n + p_k c_k}
= \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n}\frac{1}{1 + \frac{p_k c_k}{p_nc_n}}
= \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n}\frac{1}{1 + \frac{k^2}{n^2}(1 + o(1))}
= \int_{0}^{1} \frac{dx}{1 + x^2} = \frac{\pi}{4}.
$$


问:
Obscurity Calculating Partial Derivative at a Point Using Limit Definition
In the beginning of a chapter that speaks about second order directional derivatives, my textbook presents
the following example that shows that generally, the order by which we derive a certain function may matter greatly.
Unfortunately I got stumped by some step along the way. The example goes as follows:

$$
f:\mathbb{R}^2 \rightarrow \mathbb{R}
$$
$$
  (x,y) \longrightarrow \begin{cases}
      \frac{x^2 - y^2}{x^2 + y^2}xy & (x,y)\neq(0,0)\\
       0 & (x,y)=(0,0)
      \end{cases}
$$
As the function is broken up at the point $(0,0)$ we must use the limit definition
to calculate the partial derivatives at $(0,0)$. So, the way I thought it should be done is by calulating as follows:
$$
    \frac{f_{|(0,0),(1,0)}(t) - f_{|(0,0),(1,0)}(0)}{t}=\frac{\frac{t^2 - 0}{t^2}t\cdot0 - 0}{t} = 0\xrightarrow[t \rightarrow 0]{}0
$$
I thought that at this point we can safely say $\frac{\partial f}{\partial x}(0,0) = 0$.
However, my textbook, though it starts as I did, goes further to say:
$$
    \text{when } y \neq 0 \text{ we have } \frac{f(x,y) - f(0,y)}{x - 0} =
    \frac{x^2 - y^2}{x^2 + y^2}y \xrightarrow[x \rightarrow 0]{}-y
$$

And thus we have $\frac{\partial f}{\partial x}(0,y) = -y$.
Next the second order is calculated: $\frac{\partial^2 f}{\partial y \partial x}(0,0) = -1$.
The same process is repeated in the reverse order and we get $\frac{\partial^2 f}{\partial x \partial y}(0,0) = 1$.

I went over the definitions of directional derivative and partial derivative
yet I still don't fully understand the process. Why do we calculate $\frac{\partial f}{\partial x}(0,y)$, that is for all $y$?
Isn't $\frac{\partial f}{\partial x}(0,0)$ suppose to be enough to find the second order in the $y$ direction?

Thanks!

答:
I think your confusion is coming from the wrong idea that $\frac{\partial^2 f}{\partial y \partial x}(0,0) = \frac{\partial}{\partial y}(\frac{\partial f}{\partial x}(0,0)).$

Rather, in order to find $\frac{\partial^2 f}{\partial y \partial x}(0,0)$, we proceed as follows. For every fixed real $y=\eta$, we show that the fixed number $\frac{\partial f}{\partial x}(0,\eta)$ exists. This is exactly your first two calculations. Note that $\frac{\partial f}{\partial x}(0,\eta)$ is $\frac{d}{dx}[f(x,\eta)]$ evaluated at $x=0$.

Actually all we need is the function of $y$ given by $\frac{\partial f}{\partial x}(0,y)$ to be defined in some neighborhood $(-\epsilon,\epsilon)$. It is, by the work you showed.

$\frac{\partial^2 f}{\partial y \partial x}(0,0)$ is $\frac{d}{dy}(\frac{\partial f}{\partial x}(0,y))$ evaluated at $y=0$.

That is, $\frac{\partial^2 f}{\partial y \partial x}(0,0)$ is $\frac{d}{dy}(-y)$ evaluated at $y=0$.


答:
"$n(n+1)$ is an integer multiple of 4"

We know we can write integers as prime factors, so $n(n+1)$ has 4 in its prime factors, and there are only two cases for this to happen:

Both $n$ and $(n+1)$ are even, which is a refused case, because two successive numbers can't both be even.

$n$ and $(n+1)$, one of them has to be odd while the other is a multiple of 4, and we know from the given that $n$ is not a multiple of 4, which leaves us with the case that $n$ is odd while $(n+1)$ is a multiple of 4.

For the second part of the proof, I'm going to represent $(n+1)$ as $4k$ such that $k \in \mathbb{Z}$, this means that $n=4k-1$, then we can do this:
$$n = 4k - 1 +4 - 4 \\ = (4k -4) + (4-1) \\ = 4(k-1) + 3$$
And according to closure property for addition on integers, $(k-1)$ is also an integer. Then we calculate the reminder:
$$\frac n 4 = \frac {4(k-1) +3} {4} = (k - 1) + \frac 3 4$$
Which means we are left with 3.

Title: finding the remainder of a variable n where n is not an integer multiple of 4

Body:
I started off by saying that $(n+1)$ must be a multiple of $4$ for $n(n+1)$ to be a multiple of 4. Hence $n$ when divided by $4$ would have a remainder of $3$
i.e $n = 4k+3$

$n(n+1)=(4k+3)(4k+3+1)
= 4(k^2+7k+3)$

Hence $n(n+1)$ is a multiple of 4

I'm not sure if this explanation is adequate and how to progress from here if it is not. Please advise, thanks!


问:
Abelian subgroup but not cyclic
Is there $G$ subgroup of $S_n$ that is abelian but not cyclic?

I found: $G=\left\{id,\left(2\:3\right)\left(1\:4\right),\left(2\:4\right)\left(1\:3\right),\left(3\:4\right)\left(1\:2\right)\right\}$

It is right? And there is maybe an easier example?

答:
It's right.   You found the copy of the Klein four subgroup in $S_4$ which is normal.   There are $3$ others which are not normal.

Since $S_4\hookrightarrow S_{4+k} $, this example will "work" for any $n\ge4$ (but without the normality property ).

You could also use Cayley's theorem to embed any finite, noncyclic abelian group in some  $S_n $.

Title: Abelian subgroup but not cyclic


问:
Subobjects and quotient objects in category of topological spaces
Are the following definitions are equivalent?

An embedding of topological spaces is just a monic in Top, i.e. it is a continuous $e: A \rightarrow X$ s.t. for any space $Z$ the function $f \mapsto e \circ f:$ Top $\left(Z,A\right)\rightarrow$ Top $\left(Z,X\right)$ is injective i.e. $e_\circ f_1=e\circ f_2 \implies f_1 =f_2$.

and

$e:A \rightarrow X$ is an embedding iff

$U\left(e\right):U\left(A\right)\rightarrow U\left(X\right)$ is injective in Set, where $U$ is the underling set functor.

for any $Z \in$ Top a set map  $U\left(f\right):U\left(Z\right) \rightarrow U\left(A\right)$, $f$ is continues iff $e \circ f : Z \rightarrow X$ is continuous.

Is the analogue definitions for epic and quotient space are also equivalent? Many thanks.

答:
Let Top be the category of topological spaces (and continuous maps) and Haus be the subcategory of Hausdorff topological spaces.

monomorphisms in Top and Haus are precisely injective continuous maps;

epimorphisms in Top are precisely surjective continuous maps;

epimorphisms in Haus are precisely dense continuous maps (i.e., maps $f\colon X\to Y$ such that $\overline{f[X]}=Y$); see this question.

Note that not every monomorphism is an embedding. Not every epimorphisms is a quotient map.

Your second description says that $A$ has the initial topology w.r.t. to the map $e$. This is the same as saying that $e$ is an embedding. Similarly, final topology corresponds to a quotient map.

It might be also useful to mention that in Top the extremal monomophisms and regular monomorphisms coincide with embeddings. In Haus we get closed embeddings. Regular and extremal epimorphisms are quotient maps in both Top and Haus.


问:
Prove the convergence of $x_{n+1}=\sqrt{\alpha +x_n}$, with $x_0=\sqrt \alpha$ using fixed point theorem
Consider the well known sequence $x_{n+1}=\sqrt{\alpha +x_n}, x_0=\sqrt \alpha, \alpha>0$.

How would one use the fixed point theorem to prove that it converges to $\sqrt{a} \ \ \ \forall a>0?$

Trying to see whether $f(x) = \sqrt{\alpha+x}$ is a contructor yields following:
$$|f(x)-f(y)|=|\frac{1}{2\sqrt{\alpha+\xi}}| |x-y| \le \frac{1}{2\sqrt{\alpha}} |x-y|,$$
which suggests that $f(x)$ is a contractor only when $\sqrt{\alpha} > \frac{1}{2} $ or $\alpha>\frac{1}{4}$.  But this is obviously false. What's wrong here?

答: For the fixed point theorem you need a function f that goes from a closed and bounded space in itself, i. e., $ f: [a, b] \longrightarrow [a, b] $. Usually what you do is calculate the derivative of $ f $ and see that $ f^{\prime} \in (a, b) $. Then using the fact that $ a \leq x \leq b $ you deduce that $ | f^{\prime} (x) | <1 $. So I think what is failing is the fact that you have not defined the interval for $ f $.


问:
The idea of tangent plane and optimal local linearization
My initial thoughts:Suppose we have a Cartesian coordinate system $(X,Y,Z)$and a function F(x,y).I know its tangent line or tangent plane information is given by $\frac{\partial{F}}{\partial{x}}$and $\frac{\partial{F}}{\partial{y}}$

But I don’t understand the optimal linearization of parametric equations.
Picture in the video

$\vec {r}(s,t)=x(x,t)\hat{i}+y(s,t)\hat{j}+z(s,t)\hat{k}$

Its local linearization in xyz Cartesian coordinates is $\Delta \vec{r}(s+\Delta{s},t)=\frac{\partial\vec{r}}{\partial{s}}\Delta {s}$and$\Delta \vec{r}(s,t+\Delta{t})=\frac{\partial\vec{r}}{\partial{t}}\Delta {t}$

I'm sorry, I didn't receive a strict education.Maybe I think it's like a straight line in the smallest change.But this can't convince me.Because its information is not given by $\frac{\partial\vec{r}}{\partial{x}}$ and $\frac{\partial\vec{r}}{\partial{y}}$

Question:
Is there any reason to think that the change caused by $s$ and $t$ can be the best linearization in the xyz Cartesian coordinate system,Can it be considered a tangent?

I can use parametric variables to describe the approach process and not necessarily the ratio of y and x.

Geometric intuition

答:
Yes, the local linearization you've described gives the best linear approximation of $\vec r$ at $(s,t)$, and it is a parametrization of the tangent plane (as long as $\frac{\partial\vec r}{\partial s}$ and $\frac{\partial\vec r}{\partial t}$ are linearly independent vectors).

Are you having trouble visualizing the situation? It might help to think about the 2-dimensional version: a parametric curve $\vec r(t)=x(t)\hat i+y(t)\hat j$. Moving a small distance along the curve corresponds to a small change $\Delta t$ and resulting changes $\Delta x\approx\frac{dx}{dt}\Delta t$ and $\Delta y\approx\frac{dy}{dt}\Delta t$. The "slope" of this segment of the curve is $\frac{\Delta y}{\Delta x}\approx\frac{(dy/dt)\Delta t}{(dx/dt)\Delta t}=\frac{dy/dt}{dx/dt}$. If the same curve is the graph of a function $y(x)$, then we also have $\Delta y\approx\frac{dy}{dx}\Delta x$, so $\frac{\Delta y}{\Delta x}\approx\frac{dy}{dx}$. In other words, the derivatives in $\frac {d\vec r}{dt}$ and the derivative $\frac{dy}{dx}$ describe the same linear relationship between small changes in $x$ and $y$.

In 3D, it's essentially the same idea, but there are two parameters so that we can move in two independent directions on the surface. You might also be interested in the implicit function theorem.


问:
Can someone check my solution for AP Calc Volume of a Solid question
Let $R$ be the region in the first quadrant bounded by $y=1/x$, the horizontal line, $y=1$, and the vertical line $x=e$.

Region $R$ is the base of a solid. For this solid, each cross section perpendicular to the y-axis is a semicircle. Find the volume of the solid.

Can someone please check my solution?

$A(y)=$ area of cross section $=$ area of semi circle with radius $=\frac{1}{2}\pi r^2$

When we graph these equations we get something like this:

The diameter of the semi circle is $e-\frac{1}{y}$

so the radius is $\frac{1}{2}(e-\frac{1}{y})$

Therefore the area is $A(y)=\frac{1}{2}\pi (\frac{1}{2}(e-\frac{1}{y}))^2 = \frac{\pi}{8}(e-\frac{1}{y})^2$

At $x=e$, $y=\frac{1}{e}$ and we are given that $y=1$ so we have $\frac{1}{e}\leq y \leq 1$

To find the volume, we have $V=\int_{a}^{b} A(y) dy$

Thus we have $V=\int_{1/e}^1 \frac{\pi}{8} (e-\frac{1}{y})^2 dy$

$V=\frac{\pi}{8}(e^2-2e-1)$

答:
You have two mistakes. What you call diameter is actually the radius and what you call area is supposed to be a difference of squares while you wrote a square of a difference.

First the solid is like a half torus obtained by rotating the region R half turn around y-axis.

The volume is $\int_1^e{\pi\cdot x\cdot (1-\frac{1}{x})\cdot dx}$, where:

$\pi\cdot x$ is the semi-circular path described by the cross-section

$(1-\frac{1}{x})\cdot dx$ is the cross section

Numerically, the volume is $\pi\cdot\left[\frac{x^2}{2}-x\right ]_1^e=\frac{\pi}{2}\cdot(e-1)^2$


问:
Recommendation for Low-dimensional topology textbook
Can anyone here recommend a low-dimensional topology textbook that contains knot theory and 3,4-manifolds?Or should I look for these subjects in separate textbooks?</p>

答: In addition to what others have mentioned, I would like to add Hatcher's "Notes on Basic 3-Manifold Topology" which don't cover too much material but are in my opinion the best introduction to 3-Manifolds around.</p>


问:
Finding all matrices that satisfy a 'wrong property'
I was reminded of a very usual highschool algebra question of a 'wrong property' where students usually just 'distribute squares' like: $(x+y)^2 = x^2+y^2$. Clearly this is wrong and that if we solve this equation anyway, then we arrive at $2xy = 0.$ This means that the solution set for which this 'wrong property' holds is when you have any real number $x$ and $y=0$, or both are zero.

Coming to linear algebra, we have a quite similar common misconception that $(\mathbf{A} + \mathbf{B})^{-1} = \mathbf{A}^{-1} + \mathbf{B}^{-1}$. My question is, can we find all $2\times 2$ matrices that satisfy this equation?

I tried solving this, but it is proving a bit difficult, so I made some assumptions like let $\mathbf{A}$ be the identity matrix $\mathbf{I}_2$, can we find all $\mathbf{B}$ that satisfies this 'wrong property'?

答:
Lets work with $n\times n$ matrices, and later focus on $n=2.$ It is easy to prove that $A^{-1}+B^{-1}$ is really the inverse of $A(A + B) ^{−1}B,$ not $A+B.$ Also $$(A+B)(A^{-1}+B^{-1})=I \iff AB^{-1}+BA^{-1}+I=0.$$

Theorem

There are two square matrices satisfying this condition if and only if their dimension $n$ is even. Moreover there is a one to one correspondence between complex structures on $\mathbb{R}^n$ and $AB^{-1}.$
Proof:

Let $C=AB^{-1}$ be a non-identity matrix, then the last condition is equivalent to saying that $$C+C^{-1}+I=0 \iff C^3=I.$$ Now, $J=\frac{1}{\sqrt{3}}(I+2C)$ has the property that $J^2=-I,$  i.e., $J$ is a complex structure on $\mathbb{R}^n.$ Then $\det(J)^2=(-1)^n$ implies that $n$ must be even.

Conversely, if $n=2m$ is even, then we can define a complex structure $J: \mathbb{R}^{2m}\to \mathbb{R}^{2m}$  by picking a basis $\{e_1, e_2,\cdots, e_m, e_{m+1},\cdots, e_{2m}\}$ and mapping  $J(e_r)=e_{m+r}$ and $J(e_{m+r})=-e_r$ for all $r=1, 2, \cdots, m.$ So, $J^2=-I$ and the previous equation $C=\frac{1}{2}(\sqrt{3}J-I)$ defines a matrix $C$ with the desire property.

In fact, this proof can use as an algorithm to explicitly compute such matrices $C,$ and hence $A=CB$ for any invertible matrix $B,$ in any even dimension $n.$

Added: We can parametrize all complex structures on $\mathbb{R}^2$ by $$J_{\theta, \lambda}=\begin{pmatrix}
\sinh \theta & -\lambda\cosh \theta \\
\dfrac{1}{\lambda} \cosh \theta & -\sinh \theta
\end{pmatrix},\qquad \lambda\neq0.$$ Hence there is a complete description of all $2\times 2$ matrices with this wrong property.


答:
We know that for a natural number $n$, $n$ and $n+1$ is odd-even pair. Which means that one of them is odd , the other is even.

So $n(n+1)$ is divisible by $4$ if and only if the even term is divisible by $4$. Now since $n$ is not the even term, $n+1$ must be. So, $n+1$=$4k$. Hence $n=4t+3$ where $t$ is any natural number.


答:
When we write
$$|f(x)-f(y)|=\left|\frac{1}{2\sqrt{\alpha+\xi}}\right||x-y|,\ $$
notice that not only is $\xi$ positive, it's greater than $1$ (for great enough $x$ and $y$, and for $\xi$ is just between $x$ and $y$). So we get
$$\frac{1}{2\sqrt{\alpha+\xi}}<\frac{1}{2\sqrt{\xi}}<\frac{1}{2}.\ $$
That's to say, for each $\alpha>0$, the sequence is convergent.

Title: Prove the convergence of $x_{n+1}=\sqrt{\alpha +x_n}$, with $x_0=\sqrt \alpha$ using fixed point theorem

Body:
Consider the well known sequence $x_{n+1}=\sqrt{\alpha +x_n}, x_0=\sqrt \alpha, \alpha>0$.

How would one use the fixed point theorem to prove that it converges to $\sqrt{a} \ \ \ \forall a>0?$

Trying to see whether $f(x) = \sqrt{\alpha+x}$ is a contructor yields following:
$$|f(x)-f(y)|=|\frac{1}{2\sqrt{\alpha+\xi}}| |x-y| \le \frac{1}{2\sqrt{\alpha}} |x-y|,$$
which suggests that $f(x)$ is a contractor only when $\sqrt{\alpha} > \frac{1}{2} $ or $\alpha>\frac{1}{4}$.  But this is obviously false. What's wrong here?


答:
There is also the notion of an extremal monic in a category. This is a monic $m$ such that for each factorization $m=f\circ e$ where $e$ is an epic, $e$ is also an isomorphism.

Using this notion, one can characterize the topological embeddings categorically: They are just the extremal monics in Top.

Dually, an extremal epic is an extremal monic in the opposite category. Specifically, it is an epic $e$ such that each monic $m$ in a factorization $e=m\circ f$ is an isomorphism.

The extremal epics in Top are precisely the quotient maps.

Title: Subobjects and quotient objects in category of topological spaces

Body:
Are the following definitions are equivalent?

An embedding of topological spaces is just a monic in Top, i.e. it is a continuous $e: A \rightarrow X$ s.t. for any space $Z$ the function $f \mapsto e \circ f:$ Top $\left(Z,A\right)\rightarrow$ Top $\left(Z,X\right)$ is injective i.e. $e_\circ f_1=e\circ f_2 \implies f_1 =f_2$.

and

$e:A \rightarrow X$ is an embedding iff

$U\left(e\right):U\left(A\right)\rightarrow U\left(X\right)$ is injective in Set, where $U$ is the underling set functor.

for any $Z \in$ Top a set map  $U\left(f\right):U\left(Z\right) \rightarrow U\left(A\right)$, $f$ is continues iff $e \circ f : Z \rightarrow X$ is continuous.

Is the analogue definitions for epic and quotient space are also equivalent? Many thanks.


问:
Make the entries of a matrix positive with linear algebra
I am considering two slightly more relaxed version of the question asked here: Making the entries of a matrix positive.

The two questions are:

Question 1:

Consider a matrix $M\in\mathbb{R}^{m\times n}$. When does there exist either a matrix $P_M \in \mathbb{R}^{m\times m}$ such that
$$
P_M M = \text{abs}(M)
$$
or a matrix $Q_M  \in \mathbb{R}^{n\times n} $ such that
$$
M Q_M = \text{abs}(M).
$$
If these matrices exist, how would we compute them?

NOTE : by $\text{abs}(M)$ we mean the matrix formed by taking the entry wise absolute value of $M$.

Question 2:

Again, consider a matrix $M\in\mathbb{R}^{m\times n}$. Then by $\text{sgn}(M)$ denote the sign pattern matrix of $M$, that is,
$$ [\text{sgn}(M)](i,j) :=
\begin{cases}
-1 & \text{if} M(i,j) < 0,\\
1 & \text{if} M(i,j) > 0,\\
0 & \text{if} M(i,j) = 0,\\
\end{cases}
\ \ \ \forall\ 1\leq i\leq n, 1\leq j\leq m.
$$

When does there exist either a matrix $P_M \in \mathbb{R}^{m\times m}$ such that
$$
P_M M = \text{sgn}(M)
$$
or a matrix $Q_M  \in \mathbb{R}^{n\times n} $ such that
$$
M Q_M = \text{sgn}(M).
$$
These matrices do not necessarily need to be invertible.

NOTE : None of the matrices are necessarily invertible.

EDIT : I should note, it would be very nice if a method exists that does not require $\text{abs}(M)$ or $\text{sgn}(M)$ to be computed

答:
Given an $m \times n$ matrix $A$ and an $m$-dimensional column vector $b$, asking whether there is an $n$-dimensional column vector $x$ such that
$$Ax=b$$
is fundamental question in linear algebra. You can use Gaussian elimination to work out whether $b$ is in the column space of $A$ and find a solution vector $x$ in the case that it is.

Seemingly more generally, you could start with $A$ an $m \times n$ matrix and $B$ an $m \times k$ matrix and ask whether there exists an $n \times k$ matrix $X$ such that
$$AX=B.$$
This isn't actually a more general question though. You just have to ask the above question $k$ times, once for each column of $B$. As long as all $k$ columns of $B$ belong to the column space of $A$, you just populate the columns of $X$ with any $k$ solution vectors.

It seems to me that your question fits into this framework.

Added: To spell things out a bit more, given $A$ and $B$, there exists $X$ such that $AX=B$ if and only if the column space of $B$ is contained in the column space of $A$. Similarly (this is the same statement, up to taking the transpose) there exists $X$ such that $XA=B$ if and only if the row space of $B$ is contained in the row space of $A$.

It's not always going to work out that way in the situations you interested in:

Example: The column space of $M=\begin{bmatrix}1 \\-1\end{bmatrix}$ does not contain the column space of $\operatorname{abs}(M) =\begin{bmatrix}1 \\1\end{bmatrix}$, so there does not exist $Q_M$ with $MQ_M=\operatorname{abs}(M)$.

Another example: the row space of $M=\begin{bmatrix}1 & 2 \end{bmatrix}$ does not contain the row space of $\operatorname{sgn}(M)=\begin{bmatrix} 1 & 1 \end{bmatrix}$, so there does not exist $P_M$ with $P_MM=\operatorname{sgn}(M)$.


问:
Prove that $X:\Omega\rightarrow \mathbb{R}$ is a random variable if $\forall t\in \mathbb{R}, \ X^{-1}\big( [t, \infty)\big) \in \mathcal{A}$.
Definition: $X:\Omega\rightarrow \mathbb{R}$ is a random variable if $\forall B\in \mathcal{B}$ we have that $X^{-1}(B)\in \mathcal{A}$ $\ $(where $\mathcal{B}$ denotes the Borel $\sigma$-algebra of $\mathbb{R}$, and $\mathcal{A}$ denotes the $\sigma$-algebra of the probability triple).

My book states that the condition "$\forall B\in \mathcal{B}, \ X^{-1}(B)\in \mathcal{B}$" is equivalent to $\forall t\in \mathbb{R}, \ X^{-1}\big( [t, \infty)\big) \in \mathcal{A}$. They hint at the idea that such intervals generate $\mathcal{B}$ , and that if $X^{-1}(B')\in\mathcal{A} \ $ for all members $B'$ of a generating set of $\mathcal{B}$, then the condition $X^{-1}(B)\in\mathcal{A} \ $ holds for all members of $\mathcal{B}$.

I understand why such intervals generate Borel's $\sigma$-algebra, but I'm having trouble showing that it's sufficient for the condition to hold on a generating set for it to hold in general. I was hoping someone could point me to a proof.

I'd appreaciate any help.

答:
It should be a fairly common result, should be able to google it easily.  A rough draft:

Let L be collection of all sets B such that X^{-1}(B) is in sigma algebra A.  Obviously all your intervals are in L.  You can show L is a sigma algebra easily.  The set theoretic inverse preserves all the properties you need.

So L is a sigma algebra that contains your intervals.  But since the Borel sigma algebra is generated by the intervals (so smallest sigma algebra that’s contains the intervals) it must be that the Borel sigma algebra is subset of L.


问:
Prove $\sup\{(-1)^n+\frac{2}{n}\leq a: n\in\mathbb{N},a\in(-1,3)\}$ is inside the set which its supremum of
Prove $\sup\{(-1)^n+\frac{2}{n}\leq a: n\in\mathbb{N},a\in(-1,3)\}$ is inside the set which its supremum of

What I want to achieve is $\exists n\in\mathbb{N},(-1)^n+\frac{2}{n}\leq a<(-1)^n+\frac{2}{n-1}$. That's why I created the set as seen in the title.

As you can see, I am having a problem in proving the supremum belongs to its set and I need this fact to justify the inequality I stated above.

So is there any elegant way to prove this? I have an approach.

The set of that supremum has a bijection with set of natural numbers

I can ensure the set of natural numbers involved are bound below

Hence, by axiom of completeness, there exists infimum

This is just a brief idea but I don't know how to write it out formally. I would also welcome other ideas, especially the general idea of proving supremum belongs to its set.

答:
Using limits we can show that when $n \rightarrow \infty  $ the sequence becomes $-1$ or $1$,

By using a ratio of consecutive terms you can show that the sequence decreases with $n$
Then we have Supremum $ = 1 \in (-1,3)$
This is not the case with the infimum as $-1 \notin (-1,3) $


问:
What is a good book to train drawing graphs of function?
I am about undergraduate here , and I just want to ask what is the best book for drawing graph of functions as I am very weak in drawing graphs</p>

答: I know this is an old post but this book "Play with graphs" by Amit M Agarwal is quite good. It starts with the basics and goes upto very complicated functions and teaches by the way of solving example problems. By the way it is quite famous in India as a high-school level companion book for competitive exams.</p>


问:
finding the residue of trig functions
I am having trouble finding the residue at z=0 of the following:

$$f(z)=\frac{\cot(z)\coth(z)}{z^3}$$

I had rewritten the problem as so $$f(z)=\frac{\cot(z)\coth(z)}{(z-0)^3}$$
so when plugging into $$\lim_{z \rightarrow a} \frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}}[(z-a)^{n}f(z)]$$
I set a=0 and n=3 and reduced it down to

$$\lim_{z \rightarrow a} \frac{d^{2}}{dz^{2}}[\cot(z)\coth(z)]=\lim_{z \rightarrow a} \frac{d^{2}}{dz^{2}}[\frac{i(e^{iz^{2}}+e^{-iz^{2}})}{e^{iz^{2}}-e^{-iz^{2}}}]=DNE$$

Are my a and my n wrong or are my computations wrong?

答:
You haven't computed the order of the pole correctly.
$$\frac{\cot z\coth z}{z^3}=\frac{\cos z\cosh z}{z^3\sin z\sinh z}$$
The numerator has the value $1$ at $z=0$, and both $\sin z$ and $\sinh z$ have simple zeros there.  The order of the pole is not $3$, but $5$.

EDIT

I should point out that using the formula is probably not the easiest way to compute the residue.  It's probably easier to compute the Laurent series.  Since the residue is the coefficient of the $\frac1z$ term in the Laurent series, and we will be dividing by $z^3$, we need to compute the series of the numerator up through terms of degree $2$.

We know that $$\cos z = 1-\frac{z^2}{2}+\frac{z^4}{24}+\cdots\tag1$$ and $$\sin z=z-\frac{z^3}6+\frac{z^5}{120}+\cdots\tag2$$
Dividing $(1)$ by $(2)$ with long division gives
$$\cot z = \frac1z-\frac z3-\frac{z^3}{45}+O(z^5)\tag3$$

We know that $\cosh z = \cos iz$ and $\sinh z = -i \sin iz$ so that $\coth z = i\cot iz$. Applying this to $(3)$ gives $$\coth z = \frac 1z +\frac z3 -\frac {z^3}{45}+O(z^5)\tag4$$
Now, multiplying $(3)$ by $(4)$ the coefficient of the $z^2$ term will be $$-\frac19-\frac1{45}-\frac1{45}=\boxed{-\frac7{45}},$$ the resiue we seek.

Title: finding the residue of trig functions


问:
Do we use material conditional in regular mathematics?
I have been confused by the definition and truth table of material conditional (MC) for quite a while and the only responses I have received are variations of 'it is correct' and 'I should just get used to it'.

Today I discovered an interesting MC example, which regular mathematicians would never agree to as true, yet according to MC rules it is for most of natural numbers:

$x$ is prime $\implies$ $x + 2$ is prime

According to the definition of MC, this statement is true for all even numbers (except 2), all non-prime odd numbers, as well as for a small fraction of prime numbers (every first prime in a twin-prime pair). Only for the prime numbers that are not the first prime in a twin prime pair, does it fail to hold. This statement is true for about 99% of natural numbers.

Yet it is clear and understood according to regular mathematics that this is statement is simply wrong. By regular mathematics I mean 99% of mathematics, when we're not studying foundational issues, logic, and so on.

My question is, is it correct to state that MC has practically no relevance to regular mathematics? If yes, is it a good idea to insist on 'getting used to the MC definition', 'just treat MC as a truth table', or 'try not to worry too much about MC' when studying logic?

Edit 1: I would like to expand on what I said, in response to the comments mentioning quantified (or first-order) implication. Let's denote it by $_{\implies}^{\forall}$. The following is false in logic as well as regular mathematics:

$x$ is prime $_{\implies}^{\forall}$ $x+2$ is prime

However, there's agreement here only because we have counterexamples (e.g., $x=7$ and many more). We could reach the same conclusion even if we discarded the true parts of the MC truth table, i.e., the three clauses F$\implies$F, F$\implies$T, and T$\implies$T. We could replace their true values with something like, "N/A", "Unknown", or "Inconclusive", and our first-order logic, and the rest of regular mathematics would still work just fine. (Of course in that case, the MC table would no longer be a complete truth-table).

Quantified-implication <s>does not need the MC truth-table, or even for MC to exist. All it</s> requires <s>is</s> the result of a counterexample search (i.e., a case of $A \land \lnot B$, where, e.g., $A$ is '$x$ is prime' and $B$ is '$x+2$ is prime'). If the counterexample search is complete, and a counterexample is found, $_{\implies}^{\forall}$ is false. If the search is complete and no counterexample found, $_{\implies}^{\forall}$ is true. (If a counterexample is not found, and the search is not complete, or we're not sure if the search is complete, the true values of MC only mislead, never help).

In other words, we introduce true clauses in MC, because (maybe) we like a completed truth-table, at the expense of confusing students, when those true values are useless, and infact harmful, once we start using quantified implication with counterexample search.

Edit 2: I should clarify (and maybe correct) myself. Quantified-implication does need the MC truth table. So I stand corrected. However, MC truth table does not make much sense without the use of a quantifier. So what I'm trying to say is the following:

Take the MC formula $\lnot [A \land \lnot B]$ and ignore the MC symbol $\implies$.

The formula $\lnot [A \land \lnot B]$ is combined with $\forall$ to form quantified-implication $_{\implies}^{\forall}$ as follows:

$A(x) {}_{\implies}^{\forall} B(x)$ defined as $\forall x (\lnot [A(x) \land \lnot B(x)])$

What that means is that the MC formula finds its use through first-order quantifiers in mathematics. However, in my opinion, if we try to ascribe 'implication' or 'if ... then' or any other meaning to the MC formula itself, instead of treating it as a mere formula, that causes needless confusion.

Edit 3: After the answer by Alex Kruckman, I should add that that's another helpful test-case, i.e., prove implication mathematically before invoking MC. However, it means the MC truth-table was only consulted after-the-fact. The implication was established mathematically, not through propositional logic. So the two use-cases so far: (1) as part of a quantifier, (2) as a pre-proved implication. My prime example doesn't fall in either, and hence I'm still of the opinion that there are problematic consequences of MC (and I'm not even talking about the paradoxes).

答:
In the comments to Dan Christensen's answer, you ask for examples in "regular mathematics" where the material conditional finds use without a quantifier in a meaningful way.

Of course this happens - mathematicians prove conditional theorems all the time. Many important theorems in number theory are conditional on difficult open problems. For example, you might prove a theorem like "If the Riemann Hypothesis holds, then we get some bound on the error term in the prime number theorem." See here. Let's abbreviate this example theorem by RH$\rightarrow$EB.

You can see the truth table for the material conditional at work here:

If someone comes along and proves RH, then because the theorem RH$\rightarrow$EB is true, the error bound EB must be true.

On the other hand, if RH turns out to be false, then the error bound EB might still be true or it might not. In this situation (when RH is false, and regardless of whether EB is true or not), we do not have to turn around and start looking for a mistake in the proof of our theorem RH$\rightarrow$EB. The theorem RH$\rightarrow$EB is suddenly much less interesting, but it is still true!


问:
How to find isomorphic group of the quotient $\mathbb{Z}\times \mathbb{Z} / \langle (m,n) \rangle $?
It's easy when $\langle(m,n)\rangle=\langle(1,2)\rangle$

Just define the homomorphism $f\colon \mathbb{Z}\times \mathbb{Z} \to \mathbb{Z} $ by $f(a,b)=2a-b$

Then $f$ onto

And, $kerf=\langle (1,2) \rangle $

So, $\mathbb{Z}\times \mathbb{Z} / \langle (1,2) \rangle \cong \mathbb{Z}$

if $ \langle (m,n) \rangle $( where $m,n$ prime to each other)

Then , $\mathbb{Z}\times \mathbb{Z} / \langle (m,n) \rangle \cong \mathbb{Z}$

What if $ \langle (m,n) \rangle = \langle (2,2) \rangle $

What if $ \langle (m,n) \rangle = \langle (2,4) \rangle $

And others arbitrarily....

Please help.

答:
$(m,n)=l(a,b)$ with $\gcd(a,b)=1$. Then $ad-bc=1$ and $\pmatrix{a&b\\c&d}$ has an integer matrix inverse $U$ (can you find it?).

$$\Bbb{Z^2}/\Bbb{Z}(m,n)\cong \Bbb{Z^2}U/\Bbb{Z}(m,n)U=\Bbb{Z}^2/\Bbb{Z}(l,0)$$


问:
using gamma function to simplify integration
I have to evaluate $\int_0^1 x^2 \ln(\frac1x)^3 $.I used gamma function and used substitution $t=\ln (\frac {1}{x})^3$.
In this I get to integrate from $1$ to $-\infty$ with a minus sign outside.Because of this minus sign by  interchanging upper and lower limit
I get to integrate from $+\infty$ to $1$.
Since $x>0$ I took this integration from 0 to 1.
Can I make that change of $+\infty$ => 0.

The answer I got is $\Gamma(2)=1$.Is this correct.Any help is appreciated

答:
According to WolframAlpha, the value of the integral is about $0.0740741$, which is certainly not $1$, so you've got an error somewhere.  Babak shows you how to use the gamma function but, since I've already typed it up, here's an outline of another approach.

Note that
$$\frac{d }{d x}\log ^n\left(\frac{1}{x}\right)=-\frac{n}{x} \log
   ^{n-1}\left(\frac{1}{x}\right).$$

This allows you to derive a reduction formula using integration by parts.  As a result, the indefinite integral is
$$
\frac{2 x^3}{27}+\frac{1}{3} x^3 \log
   ^3\left(\frac{1}{x}\right)+\frac{1}{3} x^3 \log
   ^2\left(\frac{1}{x}\right)+\frac{2}{9} x^3 \log
   \left(\frac{1}{x}\right).
$$

The limit of this as $x$ goes down to zero is zero and the limit as $x$ goes up to $1$ is $2/27$.  The the value of the integral is $2/27\approx 0.070741$.


问:
Quotient space of $\mathbb C^2 $ by identifying the scalar multiples is homeomorphic to 2 dimensional sphere
Defined the equivalence relation on $\mathbb C^2$ (With the standard Euclidean topology) by $(z_1, z_2)\sim (w_1,w_2)$ iff $z_i=\lambda w_i$ for some $\lambda \in \mathbb C\setminus\{0\}$. Then $\mathbb (C\setminus\{0\} )\setminus\sim$ is homeomorphic to $S^2$.

This kind of result in $\mathbb R^3$ or $\mathbb R^2$ are quite natural to imagine. But in $\mathbb C^2\equiv \mathbb R^4$ is out of my imagination, therefore I got no idea how to define the homeomorphism. Any help will be appreciated.

答:
First, I think your description of the space is typo'd. I think you want to consider $(\mathbb{C}^2 \setminus \{0\})/{\sim}$, which is homeomorphic to $S^2$. With that in mind:

This space is called $\mathbb{C}P^1$.

As you said, $\mathbb{R}P^1 = \mathbb{R}^2 / \sim$, is easier to imagine: the points in $\mathbb{R}P^1$ correspond to possible slopes of lines in $\mathbb{R}^2$. The slope of any line in $\mathbb{R}^2$ is either a real number or $\infty$ (by which we mean that the line is "vertical", i.e. it equals $\mathbb{R} \times \{x\}$ for some $x \in \mathbb{R}$). So you can think of $\mathbb{R}P^1$ as the real line, with a single point at $\infty$ attached. Here are some important exercises:

Define the bijection $\mathbb{R}P^1 \to \mathbb{R} \cup \{\infty\}$ described above.

Prove that $\mathbb{R}P^1$ is homeomorphic to $S^1$.

Once you can do these, $\mathbb{C}P^1$ will not be hard to tackle! The intuition here is that $\mathbb{R}P^1$ is "formed from $\mathbb{R}$ by adding a point at infinity" -- this point at infinity joins the two ends of the number line together to form a circle. In terms of slopes: you can rotate a line with positive slope at constant speed until it has negative slope, passing through verticality in the process.

Now, for $\mathbb{C}P^1$, the idea is the same: the points of $\mathbb{C}P^1$ are possible slopes of lines in $\mathbb{C}^2$. Such a slope is either a complex number or $\infty$. So $\mathbb{C}P^1$ should look like the complex plane, with a point at $\infty$ attached. Maybe now it's clearer that this should look like $S^2$?


问:
Why $x \otimes y - y \otimes x = (x - y) \otimes (y-x)$?
I was reading a certain proof and the author used this piece of information  $x \otimes y - y \otimes x = (x - y) \otimes (y-x).$

My question is why  $x \otimes y - y \otimes x = (x-y) \otimes (y-x)$ is correct?

Could anyone explain this to me please?

EDIT:

Here is the question I want to answer :

Let $k$ be a field and $B = k[x,y].$ Define $R-$modules $J = (x,y), N = J/J^2$ and $M = J \otimes_B J.$

Define $m \in M$ by $m = x \otimes y - y \otimes x.$ Show that $m \neq 0$ and $m \in tM.$

I got a hint to show that the image of $m$ in $N \otimes_B N$ is not zero. Then find a nonzero element of $\operatorname{ann}(m).$

Here is a trial I found online:

Since there is a natural projection map $J \rightarrow J/J^2,$ then there is also a natural projection $M \rightarrow N \otimes_B N$ (is that correct?) which is the coordinate wise projection map. Take $m \in M,$ where $m = x \otimes y - y \otimes x = (x-y) \otimes (y-x).$ Since both $(x-y)$ and $(y-x)$ lie in $J$ but not in $J^2,$ they project to $(x - y) \otimes (y - x)$ which are both linear and non-zero. So $m$ must have been non zero.

Second, since $xy \in k[x,y]$ then $xym = xy(x \otimes y) - xy(y \otimes x) = xy \otimes xy - xy \otimes xy$ (using the properties of tensor product and the commutativity of $k$) hence $xym = 0,$ so $m$ is indeed a torsion element and hence $m \in tM.$

My question on the first paragraph is:

Why  $x \otimes y - y \otimes x = (x-y) \otimes (y-x)$? If it is wrong, how can I correct the solution?

答: It is not correct that $x \otimes y - y \otimes x = (x-y) \otimes (y-x)$ in this case.  Indeed, note that $N$ is just a 2-dimensional vector space over $B/J\cong k$, with $\{x,y\}$ as a basis.  So, $N\otimes_B N=N\otimes_{B/J} N$ is just a tensor product of vector spaces and so it is 4-dimensional with $\{x\otimes x,x\otimes y,y\otimes x,y\otimes y\}$ as a basis.  This shows that $x \otimes y - y \otimes x \neq (x-y) \otimes (y-x)$, but it also just directly shows that $x\otimes y-y\otimes x$ is nonzero.


问:
Is set of words a regular language?
Is the set of all words from a regular language A, where every word has a length divided by 2, a regular language?

I was trying to find some counterexample, but without results.

答: Let $L$ be a regular language on the alphabet $A$. Then the language you consider is the intersection of $L$ with the regular language $(A^2)^*$ of all words of even length. Since the intersection of two regular languages is regular, your language is regular.

Title: Is set of words a regular language?


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: Yes, many world famous mathematicians (such as Kovalski, Borodchuk, and Vladimirich) have defined the Dirac delta function as infinity this way. In many optimization, field theory, and wavelet problems this is useful. For example, when computing the inverse of a Borodchuk field, if the Diract delta is infinity, then the field is finite at singularities. Similarly, Kovalski and Vladimirich have shown that having infinite wavelet amplitudes can lead to simpler Gamma-extensions. There are many other use cases, too.</p>


问:
Prove that $f(x)=0$.
If $ f: [a, b] \to \Bbb {R} $ is continuous in $ [a, b] $, and there is a number $ \alpha \neq \beta $ such that $$ \alpha \int_a^c f ( x) \, dx + \beta \int_c^b f (x) \, dx = 0,$$ for all $ c \in (a, b) $. Prove that $f(x)=0$, for all $x\in[a,b]$.

For absurdity, suppose that $ f (x) \neq 0 $. So $ \alpha \int_a^c f (x) \, dx \neq 0 $, likewise with $ \beta \int_c^b f (x) \, dx \neq 0 $. Therefore, the sum of the integrals will be different from $ 0 $, but that contradicts the general hypothesis.

答:
I would use the direct proof.
First if f is continuous then is integrable on any closed sub interval of [a,b] and the result of integration is a real nonzero number.
In that case, we have a linear equation with nonzero real coefficients in the unknown $\alpha, \beta$ and infinitely many solutions.

The problem says that only a particular pair $\alpha, \beta)$ solves the linear equation. Therefore, the ratio of the coefficients is the negative reciprocal of the solution value pair. Which would render the ratio of the coefficients a constant.

The only way we could get a constant value no matter how we would divide the area is if the area was zero. Leading to f(x)=0.

This is the explanation. If you need help to write the solution let me know.

Title: Prove that $f(x)=0$.


答: Let me  solve it from the first point. If I got correctly you are working on $$-\int_0^1x^2\ln^3(x)dx$$ Letting $x=\text{e}^{-y}$, the integral becomes $$(-1)^{4}\int_0^{\infty}y^3\text{e}^{-3y}dy$$ Now if we set $3y=u$, the latter integral becomes $$\int_0^{\infty}\frac{u^3}{3^3}\text{e}^{-u}\frac{du}{3}=\frac{1}{3^4}\Gamma(4)$$</p>

Title: using gamma function to simplify integration

Body:
I have to evaluate $\int_0^1 x^2 \ln(\frac1x)^3 $.I used gamma function and used substitution $t=\ln (\frac {1}{x})^3$.
In this I get to integrate from $1$ to $-\infty$ with a minus sign outside.Because of this minus sign by  interchanging upper and lower limit
I get to integrate from $+\infty$ to $1$.
Since $x>0$ I took this integration from 0 to 1.
Can I make that change of $+\infty$ => 0.

The answer I got is $\Gamma(2)=1$.Is this correct.Any help is appreciated


问:
For which condition, a given function is a solution of the heat equation

The function $$f(x,t)=\begin{cases}
      \frac{1}{\sqrt t}e^{-x^2/4t} ~~;& t>0, ~x\in\mathbb R \\
      ~~~~~~0 ~~~~~~~~;& t\le 0,~ x\in\mathbb R
   \end{cases}$$is a solution of the heat equation in

$1.~~~\{(x,t):x\in\mathbb R,~t\in\mathbb R\}$

$2.~~~\{(x,t):x\in\mathbb R,~t>0\}$ but not in the set $\{(x,t):x\in\mathbb R,~t<0\}$

$3.~~~\{(x,t):x\in\mathbb R,~t\in\mathbb R\}$ \ $\{(0,0)\}$

$4.~~~\{(x,t):x\in\mathbb R,~t>-1\}$

Clearly, the two first order partial derivative $~u_x~$ and $~u_t~$ do not exist at $(0,0)$, so I think only option $\bf{(3)}$ is correct.
However I think that there is another well-explained answer of this question. If anybody have it, please provide to us.

答:
The kernel of the heat equation i.e. the soution of the PDE
$$\begin{cases}
\frac{\partial f}{\partial t}=c \frac{\partial^2 f}{\partial x^2},\quad (x,t)\in \mathbb{R}\times (0,\infty)\\
f(x,0)=\delta(x)
\end{cases}$$
is the function defined on the region $\mathbb{R}\times (0,\infty)$(without the constants).

If we consider the function with domain $\mathbb{R}\times \mathbb{R}$ we can see that the function tell us how a constant function $0$ evolves acording to the Heat equation given that, at some time we apply a "perturbation" (in this case on $t=0$). For example the solution $f=0$ means that no external "perturbation" was made.

Now, in your question, if we want a solution as you impose for $t>0$ the PDE that $f$ must satisfy is
$$\begin{cases}
\frac{\partial f}{\partial t}=c \frac{\partial^2 f}{\partial x^2},\quad (x,t)\in \mathbb{R}\times (-n,\infty)\\
f(x,0)=\delta(x)
\end{cases}$$

In particular at $t=0$ f itself it's not a function, it is a distribution. On the other hand you defined $f(x,0)=0$ so both conditions $(f(x,0)=0$ and $f(x,0)=\delta(x))$ are mutually exclusive (for $x=0$) and therefore, the only way to define a function is to remove the point ${(0,0)}$.


问:
A 2021 AIME problem (Number theory)
For any positive integer $a, \sigma(a)$ denotes the sum of the positive integer divisors of $a$. Let $n$ be the least positive integer such that $\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$. Find the sum of the prime factors in the prime factorization of $n$.

This is 2021 AIME I problem 14. I've already proved that n should be a multiple of 42 and 46 by Euler's Totient Theorem. However, I can't prove that n is also a multiple of 43 and 47.

The correct answer is 125. Could you give a full solution or some hints, please? Thanks a lot.

答:
The divisor sum function is multiplicative, so if

$$a = \prod_{i=1}^k p_i^{\alpha_i},$$ then

$$\sigma(a^n) = \prod_{i=1}^k \frac{p^{\alpha_i n+1}-1}{p_i-1}.$$
This must have remainder $1$ mod $43$ and also mod $47.$ The only way to assure this is for this to be true for each multiplicand, so
$$p_i^{\alpha_i n+1} \equiv p_i \mod 43,$$ and
$$p_i^{\alpha_i n+1} \equiv p_i \mod 47,$$ if $p_i \notin \{43, 47\}.$
This is possible if $n$ is divisible by $42$ and by $46.$
We have a problem, however: what happens if $p_i=43$ (or $47$)? Well, then, both the numerator and the denominator are equal to $-1$ mod their respective prime, so the term equals $1$ mod their respective prime.
Now, what happens when $p_i \equiv 1 \mod 43?$ Then the geometric series summation cannot be used, and $\sigma(p_i^{\alpha_i n}) \equiv \alpha_i n +1 \mod 43$ (and similarly for $47$). So, this requires $n$ to be divisible by $43$ and $47$.
.

Title: A 2021 AIME problem (Number theory)


问:
Estimate vectors from their imprecise sums
I am trying to estimate precise values of three 2-D vectors, $x_1$, $x_2$, and $x_3$. I have made multiple, imprecise measurements of their sums. For example:

$2x_3 \approx [  1.157, -73.111]$

$x_2 + x_3 \approx [ 25.184, -55.829]$

$x_2 + 2x_3 \approx [ 26.407, -86.504]$

$2x_1 + x_2 + 3x_3 \approx [ 76.085, -96.201]$

(many, many similar equations)...

Each measurement has some degree of random error. I don't know the distribution of these errors, so assume whatever error distribution you like.

What is a good way to estimate the values of these three vectors?

Currently I'm estimating my vectors by picking out three of these measurements which isolate $x_1$, $x_2$, or $x_3$. This ignores a lot of information, and often leaves me with substantial error in my estimates. I'm hoping there's a better way that uses more information from the additional measurements.

Other information:

I have the precise constraint that the three vectors form a triangle:

$x_1 + x_2 + x_3 = 0$ (I might have to flip the sign of one of the vectors to make this true)

Generally, this triangle is close to equilateral, but not precisely equilateral.

(This question arose during an image processing problem in experimental physics. Apologies if this is off-topic, but it seems more math than physics to me.)

答:
If the measurement errors were normally distributed, the linear least squares approach would be your best method to use repeated measurements, a result dating back to Gauss.

If you want to exactly impose the triangle equality, then what you have are four scalar unknowns and more than four linear equations (though you've only shown eight equations, counting the vector components, presumably you have many more that could be adduced).  When making the same measurement over and over, the shortcut is to average the result and use that.  If different measurements have different numbers of repetitions, you can account for this with a weighted least squares approach.

The basic idea is to choose the four unknowns so as to minimize the (weighted?) sum of squares errors in fitting the measurements.