问:
Resources, references, or examples for logics with finitely many sentences
Are there any interesting logics with only finitely many sentences total? Is there a reason why potential logics with this property would be trivial or badly behaved?

I'm looking for a reference to read more or an example of such a logic that's interesting as a mathematical object.

I was recently looking at some examples of interesting logics that are fragments of better-known logics, such as two-variable logic and a list of intermediate logics that are interesting logics between intuitionistic propositional logic and classical propositional logic.

One thing that I haven't come across yet, though, is a modern logic (possibly described as a fragment of another logic) with structural constraints on well-formed formulas that are so strong that only finitely many sentences are possible or only finitely many sentences up to renaming of variables are possible. Traditional term logic, even with the addition of non for modifying predicates, has finitely many sentences up to renaming of variables.

I can think of a few weird properties of such finite-number-of-sentences logics off the top of my head like the fact that ordinary introduction rules like $A, B \vdash A \land B$ will sometimes fail if the conclusion is too big to be well-formed.

NOTE: removed reference to $\alpha$-equivalence, replaced with up to renaming of variables which is more clear.

答:
If the total number of formulas is finite, then you can simply write down an exhaustive list of all theorems of the logic. All you need to know about the logic is then contained in this list.

That is, even if you come up with some seemingly complex semantics or proof theory for such a logic, in the end you cannot express more than the problem of whether a formula is contained in a given finite list. And this problem is trivial.

EDIT:
I extend this answer to reply Gregory's question below. So why is checking containment in a finite list fundamentally different from checking containment in a countably infinite list? Let us look at an example:

$L$: all prime numbers below $10$.

One way to check whether $n\in L$ is to first verify $n<10$ and then look at the divisors of $n$. But you can also check that $n\in L\iff n\in\{2,3,5,7\}$. The point is now that you can check $n\in L$ without knowing what a prime number is, simply by comparing $n$ to the $4$ possible values in $\{2,3,5,7\}$. In this sense, the problem whether $n\in L$ is not related to factorization at all.

Nothing similar holds for infinite lists, say for the list $P$ of all primes. Of course you can try to say $n\in P\iff n\in\{2,3,5,7,9,11,\ldots\}$, but then in order to explain what the "$\ldots$" means, you'll have to explain what a prime number is.

The same observation as above applies to any finite problem (by this I mean a problem with only finitely many solution instances). Even if you have an initial description of the problem which sounds meaningful (such as being prime and $<10$), there is always a meaningless but equivalent description (such as being either $2,3,5$ or $7$).

For this reason it is also hard to imagine a reasonable logic with only finitely many sentences in total.

(small disclaimer: Artificially, you can create logics where the set of theorems is unknown even if it is finite. For example, your logic could contain as formula only a single atom $p$, and you declare $p$ to be a theorem if the Collatz conjecture is true)


问:
Explanation of the Vandermonde matrix
This is a section from Gilbert Strang's Linear Algebra book:

Although I understand the uniqueness and existence of that matrix, I am having trouble understanding what he means by, "The only such polynomial that vanishes at... No other polynomial of degree $n-1$ can have $n$ roots." What does he mean by vanishing and why does no other polynomial have $n$ roots?

答:
We say that the polynomial $P(t)=x_1+x_2t+\ldots+x_nt^{n-1}$ vanishes at some point $t=t_0$ if $t_0$ is a "root" or a "zero" of that polynomial, i.e. if $P(t_0)=0$.

What the author is trying to say is that a polynomial of degree at most $n-1$ can have at most $n-1$ different zeros (barring the case where the polynomial itself is a zero polynomial - i.e. all its coefficients are zero and therefore it "vanishes" everywhere). This result is a well-known result from algebra, and is a consequence of unique factorization for polynomials.

ContentLicense: CC BY-SA 4.0

Title: Explanation of the Vandermonde matrix


问:
Pointwise and uniform convergence of sum
For $x\in \mathbb{R}$ we have $$f_n(x)=\sum_{k=1}^n\frac{e^{-kx}}{k}$$

Show that $(f_n)_n$ converges pointwise on $(0,+\infty)$.

Let $0<a\in \mathbb{R}$. Show that $(f_n)_n$ converges uniformly on $[a,+\infty)$.

Show that $\displaystyle{f(x)=\lim_{n\rightarrow +\infty}f_n(x)}$ is continuous on $(0, +\infty)$.

Show that $f$ is differentiable on $(0, +\infty)$ with derivative $f'(x)=\frac{-e^{-x}}{1-e^{-x}}$.

$$$$

At 1 do we have to tae the limit $n\rightarrow +\infty$ ? But how can we calculate that series?

$$$$

EDIT :

At question $4.$ do we do the following?

Each of the functions $\{f_n\}$ is continuously differentiable. It holds that $$f_n'(x)=\sum_{k=1}^n -e^{-kx}$$

Let $x\in (0,\infty)$ fixed.

We have that $$g(x)=\lim_{n\rightarrow +\infty}\sum_{k=1}^n -e^{-kx}=-\sum_{k=1}^\infty e^{-kx} = -e^{-x}\sum_{k=0}^\infty e^{-kx}=\frac{-e^{-x}}{1-e^{-x}}$$
So $\{f_n'\}$ converges pointwise to the function $g$ on $(0,\infty)$.

Let $\alpha > 0$ fixed.

Since $-e^{-kx}$ is an increasing function for $x\in [0, \alpha)$, then $-e^{-kx}\leq -e^{-\alpha k}=-(\frac{1}{e^{\alpha}})^k$.

Since $\alpha > 0$, then $\frac{1}{e^{\alpha}}< 1$ and so we have $$\sum_{k=1}^\infty -\Big(\frac{1}{e^{\alpha}}\Big)^k<\infty$$
From Weierstrass M-test we get that the series $\displaystyle{\sum_{k=1}^\infty -e^{-kx}}$ converges uniformly on $[0, \alpha)$ since $$\sum_{k=1}^\infty -e^{-kx} \leq \sum_{k=1}^\infty -e^{-k\alpha}\leq \sum_{k=1}^\infty -\Big(\frac{1}{e^{\alpha}}\Big)^k$$

Since $\alpha>0$ is arbitrary, this holds for each $\alpha$, and so also for $\bigcup_{\alpha> 0}[0, \alpha) = (0,\infty)$.

So we have that $\{f_n'\}$ converges uniformly to $\displaystyle{g(x)=\frac{-e^{-x}}{1-e^{-x}}}$ that is continuous on $(0,+\infty)$.

Then $f$ is also continuously differentiable on $(0,+\infty)$ and it holds that $f'=g$, so $f'(x) = \frac{-e^{-x}}{1-e^{-x}}$.

答: You don't necessarily need to evaluate the series to prove that it converges. To prove that  $( f_n(x) )_n$ converges, you could prove that it is an increasing and bounded sequence.


问:
Define a sketch $s_{\mathbf{Grp}}$ such that $\mathbf{Grp}\backsimeq \mathbf{Mod}(s_{\mathbf{Grp}},\mathbf{Set})$
I have the following

(a) Define a sketch $s_{\mathbf{Grp}}$ and a equivalence functor $$E: \mathbf{Grp}\to \mathbf{Mod}(s_{\mathbf{Grp}},\mathbf{Set})$$ (b) Knowing that finite limits commute with filtered colimits in $\mathbf{Set}$, use the result in (a) to prove that they also commute in $\mathbf{Grp}$.

(c) Prove that $\mathbf{Ab} \backsimeq \mathbf{Mod}(s_{\mathbf{Grp}},\mathbf{Grp})$

This is my first exercise working with sketches. I've been told that even simple examples with sketches gives us a big amount of work. In fact, I couldn't do (a) at a first glance. But I found a useful example at nlab's sketch article. Example 3.2 especifies the directed graph, diagrams, cones and cocones of a sketch which has unital magmas as models (sets with a binary operation which has a two sided unit).

So I thought taking this same sketch and "interpreting" the arrows $e$ as the identity of a group and $m$ as its multiplication, all this via the equivalence functor the exercise request us to build. But I don't even know how to finish the construction of $E$ and in fact I don't see why it will be an equivalence functor at all.

Could you help me? Also is there any result that I'm missing on (b)? Because I think this shouldn't be so difficult.

答: The case of semigroups is spelled out in all details in Sec 2 of Chap 7 of Barr and Wells's Category theory for Computer Science.


问:
Prove complex logarithm problem involving Euler Mascheroni constant
I want to prove/disprove the following equation:
$$\text{Log}{\frac{1}{\sqrt{z^2+1}}} = -\frac12 \ln\left(\frac{1}{z^2}+1\right)-\ln\lvert z\rvert - \gamma$$

Which $\gamma$ is the Euler Mascheroni constant.

This problem arise when my textbook said
$$\mathcal{L} \Bigg\{\frac{\cos(t)}{t}\Bigg\} = \text{Log}\left(\frac{1}{\sqrt{s^2+1}}\right)\tag{1}$$
But, Wolfram gave me
$$\mathcal{L} \Bigg\{\frac{\cos(t)}{t}\Bigg\} = -\frac12 \ln\left(\frac{1}{s^2}+1\right)-\ln\lvert z\rvert - \gamma\tag{2}$$

My attempt:
Here, my working start is deal with my laplace transform:
$$\begin{align}
\mathcal{L} \Bigg\{\frac{\cos(t)}{t}\Bigg\} &= \int_0^{\infty}\frac{e^{-st}\cos(t)}{t}\,\Bbb dt\\
&= \int_0^{\infty} \frac{e^{-st}\left(e^{it}+e^{-it}\right)}{2t}\,\Bbb dt\\
&= \int_0^{\infty} \frac{e^{t(s-i)}}{2t}\,\Bbb dt + \int_0^{\infty} \frac{e^{-t(s+i)}}{2t}\,\Bbb dt
\end{align}$$

I'm stuck there. If i continue, maybe i would get some exponential integral function. But how to produce equation $(1)$ or $(2)$??

I need a help. Thanks!

P.S. haven't tried to use $\text{Log}(z) = \ln|z| + i\text{Arg}(z)$ since i don't know what $\text{Arg}\left(\dfrac{1}{\sqrt{z^2+1}}\right)$ is.
And btw, i'm not sure Wolfram use $\log$ as $\text{Log}$ or $\ln$.
Here is the screenshot:

答:
Here's one way the Laplace transform can be derived:

$$
\begin{align}
f(t) &= cos(t)/t \\
t f(t) &= cos(t) \\
\mathcal{L} \Bigg\{t f(t)\Bigg\} &= \mathcal{L} \Bigg\{ cos(t) \Bigg\} \\
-\frac{d}{ds} \mathcal{L} \Bigg\{ f(t)\Bigg\} &= \frac {s}{s^2 + 1} \\
\mathcal{L} \Bigg\{ f(t)\Bigg\} &= - \int \frac {s}{s^2 + 1} ds \\
u &= s^2 + 1 \\
\mathcal{L} \Bigg\{ f(t)\Bigg\} &= - \frac{1}{2} \int \frac {1}{u} du \\
 &= - \frac{1}{2} ln(u) + C \\
 &= - \frac{1}{2} ln(s^2 + 1) + C \\
 &= ln(\frac{1}{\sqrt{s^2 + 1}}) + C
\end{align}
$$

IVT/FVT to find the value of the integration constant.

You'll notice that both of your forms are very close to each other:

$$
-\frac{1}{2} ln (\frac{1}{s^2}+1) - ln(s) \\
-\frac{1}{2} ln (\frac{s^2 + 1}{s^2}) - ln(s) \\
\frac{1}{2} ln (\frac{s^2}{s^2+1}) - ln (s) \\
ln (\frac{s}{\sqrt{s^2+1}}) - ln(s) \\
ln (\frac{1}{\sqrt{s^2+1}})
$$

So the question comes down to the integration constant $-\gamma$


问:
Stuck on solving 2 simultaneous equations with variables. [Serge Lang - Basic Mathematics]
This problem from Serge Lang's Basic Mathematics in Chapter 2, question 9a.

Let $a,b,c,d$ be numbers such that $ad-bc \neq 0$. Solve the following systems of equations for $x$ and $y$ in terms of $a,b,c,d$.
a) \begin{align*}
ax + by & = 1\\
cx + dy & = 2
\end{align*}

I'm fine with solving these sorts of equations with numbers in place of $a,b,c,d$ but trying to solve it with just variables has been a problem. I assume that the information "$ad-bc \neq 0$" is some sort of hint, possibly that I'm able to divide by $ad-bc$ at some point but I don't know how to apply it.

答:
Hint

Multiply the 1st eqn by $c$ and the 2nd eqn by $a$.  Then the two altered eqn's will each have, as their first term, $(ac)x$.  Therefore, by subtracting the 2nd altered eqn from the 1st altered eqn, you will obtain an eqn that has only the single variable $y$.


问:
How can I know if an equilibrium is stable in a difference equation?
Suppose that I have the following equation:
$$
y_{t+1} = \frac{2+y_t-y^3_t}{2y_t}
$$

Then
$$
\bar{y} = \frac{2+\bar{y} -\bar{y}^3}{2\bar{y}}
$$

And so I obtain these solutions: $$y_1 = -2, \quad y_2 = -1, \quad y_3 = 1 $$

Now how can I know if these equilibria are stable?

I believe that it is useful to compute the derivative with respect to the solutions found.

Therefore I obtain the derivative:
$$
\frac{d}{dy}
=
-\frac{(y^3+1)}{y^2}
$$

For $y_1 = -2$ the derivative becomes: $\frac{7}{4}$

For $y_2 = -1$ the derivative becomes: $0$

For $y_3 = 1$ the derivative becomes: $-2$

So how can I know if the equilibria found are stable?

答:
What you have is not a differential equation, but rather a difference equation. The criterion for stability of the fixed points is then slightly different:

Let $x_{n+1}=F(x_n)$ be a map for which we find fixed points by solving $F(x^{*})=x^{*}$.There are three cases you need to check (as with differential equations):

If $|F'(x^{*})|<1$,the fixed point is stable

If $|F'(x^{*})|>1$,the fixed point is unstable

If $|F'(x^{*})|=1$,we cannot decide whether it is stable or unstable.This is called a "border" case.

For difference equations in 2D/3D, you can analogously check for stability, but the derivative is replaced by a Jacobian matrix, where you linearize and check whether the complex modulus of the eigenvalues is less than 1.

Edit: To answer your question, then the only fixed point which is stable is $y_2$. The other ones are unstable.

Edit: Perhaps I should provide a justification of why this is true. You can prove it just like the case when you have a differential equation.

Let $\Delta x_{n}=x_n-x^{*}$ be a small displacement of the fixed point. Taylor expand the map about $x^{*}$ to get:

$$x^*+\Delta x_{n+1}=F(x^*+\Delta x_n)=F(x^*)+F'(x^*)\Delta x_n+... $$

Because $F(x^*)=x^*$, then you can actually simplify the above to:

$$\Delta x_{n+1}=F'(x^*)\Delta x_n+... $$

The idea is then that for arbitrarily small $\Delta x_n$, you can negate the higher order terms. You can then see that the 3 cases I listed above are what we need to check for stability.

Edit: What I mean by 2D, for instance, is a system of the form:

$$\begin{cases} x_{n+1}=F(x_n,y_n) \\ y_{n+1}=G(x_n,y_n) \end{cases}$$

To study stability in this case, you set $F(x^*,y^*)=x^*$ and $G(x^*,y^*)=y^*$ to find the fixed points.

You then find the Jacobian matrix evaluated at that point:

$$J(x^*,y^*) = \begin{bmatrix}
    \frac{\partial F}{\partial x} &  \frac{\partial F}{\partial y} \\
    \frac{\partial G}{\partial x} &  \frac{\partial G}{\partial y}
\end{bmatrix}\Bigg\rvert_{(x^*,y^*)}$$

To study stability, you find the eigenvalues of this matrix and then you check whether the complex modulus is less than 1. That is, both roots must lie inside the unit circle in order for the fixed point to be stable. Note that you have to repeat this process for different fixed points.

You can do the same for higher dimensional systems. However, you typically don't go higher than 3D for physical applications.


问:
How to translate $x$ when moving from uniform distribution to order statistics with beta distribution?
Let $Y_{i}$ be the number of days until package $i\in[1,7]$ comes (independent). We know that $Y_{i}\sim Uni\left(0,10\right)$. Find the probability so it will take more than 8 days for the last package to come.

So I defined $Y_{(i)}$ be order statistics. So we get $Y_{(7)}\sim Beta\left(7,1\right)$. I would like to calculate $P\left(Y_{(7)}>8\right)$ but $x$ in $P(X>x)$ should be $x\in[0,1]$ by def of beta distribution. So how I translate $8$?

答:
In the following, I assume that $Y_i$ is a continuous random variable (in particular, the 'number of days' does not have to be an integer).

I assume the result you have used is that if $X_1,\dots,X_n$ is a random sample drawn from $U(0,1)$ then the $k$-th order statistic $X_{(k)}$ follows the distribution $\mathrm{Beta}(k,n+1-k)$ for all $k\in \{1,2,\dots,n\}$.

However, note that this result only holds for uniform distributions on the standard unit interval, whereas $Y_i\sim U(0,10)$. To solve this problem, define $Z_i:= Y_i/10$, then $Z_i\sim U(0,1)$. Clearly, the ordering of the random variables is preserved by division by $10$, so one also has that $Z_{(i)}=Y_{(i)}/10$. Then one can use the theorem you have used to conclude that $Z_{(7)}\sim \mathrm{Beta}(7,1)$. Therefore the result is
$$\mathbb{P}(Y_{(7)}>8)=\mathbb{P}(Z_{(7)}>0.8)=\int_{4/5}^1 \frac{\Gamma(7+1)}{\Gamma(7)\Gamma(1)}x^{7-1} (1-x)^{1-1}~dx=x^7\Big|_{4/5}^1=1-(4/5)^7.$$


问:
How many 11-digit numbers are there which contain all of the digits 0-9 at least once?

How many $11$-digit numbers are there which contain all of the digits $0$-$9$ at least once?

Note that in this question a number cannot begin with a digit of $0$.

My instructor explained this as a case problem, but rushed through it without pausing for questions. The answer ended up being $99 \times (10!/2!)$, but I have no idea how to get there!

答:
I disagree with your formula and the other answers.
Select a nonzero digit for the leftmost place and than position all $10$ digits on the remaining $10$ places. These are $$9\cdot 10!$$ possibilities. But these are only the $11$-digit numbers where the digit on the leftmost place appears a second time in the decimal representation. To get the remaining numbers, replace this digit by one of the $9$ different digits. So you get $$9\cdot9\cdot 10!$$ numbers. But each of them appears twice, because the digit that appears twice can have found its place by the initial generation of the number or by replacing the digit equal to the leading digit. So we get only
$$9\cdot9\cdot 10!/2!$$
additional numbers. The sum of these two quantities is
$$9\cdot 10!+9\cdot9\cdot 10!/2!=495\cdot9!$$

To illustrate this process:
First select a leading digit from $1,2,3,4,5,6,7,8,9$, e.g. $7$. Then select a permutation of the $10$ digits $0,1,2,3,4,5,6,7,8,9$, e.g.
$2504769183$. This gives us a number were the leading digit is the digit that is contained twice:
$$7\ 2504\ 7\ 69183$$
Now we want to construct a number where a digit different, to the leading digit is contained twice. So select a digit different to the leading digit, this means one of $0,1,2,3,4,5,6,8,9$ There are $9$  possibilities, we select $8$. Now we replace the digit $7$, which is not on the leftmost place , by the digit $8$ and get
$$7\ 2504\ 8\ 69183$$
but the same number
$$7\ 2504\ 8\ 691\ 8\ 3$$
we get if we replace $7$ by $8$ in the number
$$7\ 2504\ 8\ 691\ 7\ 3$$.

ContentLicense: CC BY-SA 3.0


答:
Calling it the "number of days" seems to suggest it's an integer, i.e. you have a discrete uniform distribution. Moreover, the number of days is not with in the interval from $0$ to $1,$ so you don't have a beta distribution, although you could have a re-scaling of a beta distribution.

The probability that all packages arrive in no more than $8$ days is the $7\text{th}$ power of the probability that a particular package arrives in no more than $8$ days, so it is $(9/11)^7$ (since there are nine numbers in the set $\{0,1,2,3,4,5,6,7,8\}$ and $11$ in the set $\{0,1,2,3,4,5,6,7,8,9,10\}.$)

Title: How to translate $x$ when moving from uniform distribution to order statistics with beta distribution?

Body:
Let $Y_{i}$ be the number of days until package $i\in[1,7]$ comes (independent). We know that $Y_{i}\sim Uni\left(0,10\right)$. Find the probability so it will take more than 8 days for the last package to come.

So I defined $Y_{(i)}$ be order statistics. So we get $Y_{(7)}\sim Beta\left(7,1\right)$. I would like to calculate $P\left(Y_{(7)}>8\right)$ but $x$ in $P(X>x)$ should be $x\in[0,1]$ by def of beta distribution. So how I translate $8$?


问:
Distribution from probability generating function
Let $X$ be a discrete random variable with probability generating function $G(t)=\frac1 {2-t}$, what distribution has $X$?

I know that to get the $G(t)$ I need to apply the definition of probability generating function which is:

\begin{align}
G(t)=E(t^X)=\sum_{x=0}^{\infty} t^x P(X=x)
\end{align}

And in this case $P(X=x)$ is just $X$, my question is: is there an opposite function to apply to $G(t)$ so I can get the probability function and see what kind of distribution appears to be? My guess is that $X$ is continuous uniform distribution but I don't know how to prove it.

答:
Plug in $t=0$ to get $\mathbb P(X=0)=\frac{1}{2}$. So then
$$\frac{1}{2-t}=G(t)=\frac{1}{2}+\sum_{k=1}^\infty\mathbb P(X=k)\cdot t^k.$$
Subtract $\frac{1}{2}$ from each side, divide by $t$ and plug in $t=0$ (strictly speaking we should be taking the limit $t\to0$) again to get $\mathbb P(X=1)=\frac{1}{4}$. So then
$$\frac{1}{4-2t}=\frac{G(t)-1/2}{t}=\frac{1}{4}+\sum_{k=2}^\infty\mathbb P(X=k)\cdot t^{k-1}.$$
You can extract the whole probability mass function by continuing like this.

A trick for this specific PGF: you might know that $\frac{1}{1-t}=1+t+t^2+\dots$ (e.g. from the geometric series formula, or Taylor expanding), so then
$$G(t)=\frac{1}{2-t}=\frac{1}{2}\left(\frac{1}{1-t/2}\right)=\frac{1}{2}+\frac{t}{4}+\frac{t^2}{8}+\dots$$
and we can instantly read off $\mathbb P(X=k)=(1/2)^{k+1}$.

You might recognise this as the geometric distribution -- in this case, the number of tails you flip before you flip your first head, for a fair coin.


问:
Minimum number of partitions required for obtaining all numbers from $1$ to $n$
I took part in a coding contest organized by a club in my college for freshers, In one of the problems it was asked to find out the minimum number of partitions of a number $n$, so that all the integers from $1$ to $n$ can be formed such that the partitions are minimum. For such a problem, I couldn't find anything in searching, and I don't know what to search exactly.

The solution is not satisfactory to me at all, in short, it uses the fact that Strong Induction Proof: Every natural number = sum of distinct powers of 2 and then eerily uses $(7)_{10} = (111)_2$ as an example, saying, from $1, 2, 4$ we can make all integers till $7$, which is true. But this just added more to my confusion.

At the beginning, I got confused with binary representation and since the question on which the answer is based, I couldn't refrain from asking Why not base 3? (I got the answer: because base 3 won't have distinct powers, so it won't be minimum) (1)

Since it is hard to check for even a small test cases, I wrote a program to get more insights, and indeed, minimum number of partitions are the digits in binary representation of $n$, and that the numbers seem arbitrary. Therefore, I think that there might be some link to base $2$, but the way it is being used and explained to me by my peers and seniors might be incorrect.

My approach:

We could use generating function for an arbitrary number of partitions say $k$,  (2)

so the generation function would be:
$$ (1 + x^{p_1})(1 + x^{p_2})\cdots(1 + x^{p_k}) = 1 + a_1x + a_2x^2+\ldots+a_kx^k$$
where each $p_i$ is the $i^{\text{th}}$ number and $p_1 + p_2 +\ldots+p_k = n$

If we somehow proof that each $a_i > 0$, and then find the minimum condition, we'll be done.

on solving the expression, Differentiation can be hectic, if try to find a pattern by expanding it comes out to be
$$ 1 + \sum_{i = 1}^kx^{p_i} + \sum_{i,j = 1}^kx^{p_i + p_j} + \ldots + x^{p_1 + p_2 +\ldots+p_k}$$
...which doesn't give any lead. And of course, numbers can  repeat, which can be even more problematic.

Hence, I'm stuck at the very beginning itself.

My questions are:

There is obviously some link to powers of $2$, but why? and why only $2$?

would my approach (finding a pattern first, then trying to find the minimum condition) actually work?

Some partitions that satisfy the conditions:

$\begin{align} 8 &= 4 + 2 + 1 + 1 \\ &= 3 + 3 + 1 + 1\\ &= 3 + 2 + 2 + 1 \end{align}$

$ \begin{align} 9 &= 5 +2 +1+ 1\\
&=4 +3 +1 +1 \\
&=4+ 2+ 2+ 1\\
&=3 +3+ 2+ 1 \end{align}$

$\begin{align} 14 &= 7+4+2+1 \end{align}$

Note: I know minimum partitions are $\lfloor \log_2(n) \rfloor + 1$, I want to know HOW.

答:
Let's say your number $n$ is between $2^{k-1}$ and $2^k-1$.

On one hand, you can make the partition of size $k$. This can be proven by induction on $k$. Namely, $k=1$ means $n=1$, which is its own partition of size $1$. The inductive step (sketch): "halve" $n$, i.e. take $m=\lceil n/2\rceil$, which gives $n-m=\lfloor n/2\rfloor$; prove that $2^{k-2}\le n-m\le 2^{k-1}-1$. Then, partition $n-m$ into a partition with $k-1$ terms (as per inductive hypothesis) and add $m$ to that partition to get a partition of $n$ into $k$ terms. Prove you can now make all the sums from $1$ to $n$.

(This logic can be easily turned into a simple iterative algorirhm, by the way.)

On the other hand, you cannot make a shorter partition. Imagine you had a partition with $l$ terms, where $l\lt k$. How many different sums would you be able to make out of those terms? Each term can be either added to the sum or not, so it is at most $2^l$ sums (some sums may be equal!), however we don't want the "empty" sum (no terms chosen), so in effect you can have at most $2^l-1\le 2^{k-1}-1\lt 2^{k-1}\le n$ sums, i.e. you cannot possibly cover all numbers from $1$ to $n$.

The last part is, for me, providing the intuition for "why powers of $2$?" . Namely, if the number of terms in your partition is too small, you just cannot recreate all the sums you want. The powers of $2$ naturally arise when you try to count the number of different sums you can make, simply because, for every term in your partition, you have exactly two choices: you can either put the term into the sum - or not!


答:
For $1)$ fix $x\in (0,\infty)$ and notice that $\{f_n(x)\}$ is a strictly increasing sequence of real numbers. Since $$0\leq f_n(x)=\sum_{k=1}^n \frac{e^{-kx}}{k}\leq \sum_{k=1}^\infty e^{-kx} = e^{-x}\sum_{k=0}^\infty e^{-kx}=\frac{e^{-x}}{1-e^{-x}} $$
We know that this geometric series converges because $e^{-x}<1$ since $x\in (0,\infty)$. Thus $\{f_n(x)\}$ is an increasing sequence of real numbers which is bounded above, thus there exists some number $f(x)$ such that $f_n(x)\to f(x)$. Hence $\{f_n\}$ converge pointwise to some function $f$ on $(0,\infty)$.

For $2)$ let $\alpha > 0$ be fixed. Since $e^{-kx}$ is a decreasing function for $x\in [\alpha, 0)$, then $e^{-kx}\leq e^{-\alpha k}=(\frac{1}{e^{\alpha}})^k$. Since $\alpha > 0$, then $\frac{1}{e^{\alpha}}< 1$ so that the geometric series
$$\sum_{k=1}^\infty \Big(\frac{1}{e^{\alpha}}\Big)^k<\infty $$
Hence by the Weierstrass M-test we get that the series $\sum_{k=1}^\infty \frac{e^{-kx}}{k}$ converges uniformly on $[\alpha, 0)$ via
$$ \sum_{k=1}^\infty \frac{e^{-kx}}{k} \leq \sum_{k=1}^\infty \frac{e^{-k\alpha}}{k}\leq \sum_{k=1}^\infty \Big(\frac{1}{e^{\alpha}}\Big)^k$$

For $3)$, use that each of the functions $\{f_n\}$ is continuous. Fix $\alpha>0$. By $2)$ we know that $f_n\to f$ uniformly on $[\alpha,\infty)$. Now any sequence of continuous functions will converge uniformly to a continuous function. Thus $f$ must be a continuous function on $[\alpha, \infty)$. This holds for any $\alpha > 0$. Thus $f$ is a continuous function on $\bigcup_{\alpha> 0}[\alpha, \infty) = (0,\infty)$.

For $4)$ observe that
$$f_n'(x)=\sum_{k=1}^n -e^{-kx} $$
You can use the same argument we used above to show that $\{f_n'\}$ converge uniformly to some function
$$g(x)=-\sum_{k=1}^\infty e^{-kx} = -e^{-x}\sum_{k=0}^\infty e^{-kx}=\frac{-e^{-x}}{1-e^{-x}} $$ which is continuous on $(0,\infty)$. Since we can find a point $x_0$ such that $f_n(x_0)\to f(x_0)$, then the function sequence of functions $\{f_n\}$ converge uniformly to a differentiable function $f$ where $f'=g$. Thus
$$f'(x) = \frac{-e^{-x}}{1-e^{-x}}$$

Title: Pointwise and uniform convergence of sum

Body:
For $x\in \mathbb{R}$ we have $$f_n(x)=\sum_{k=1}^n\frac{e^{-kx}}{k}$$

Show that $(f_n)_n$ converges pointwise on $(0,+\infty)$.

Let $0<a\in \mathbb{R}$. Show that $(f_n)_n$ converges uniformly on $[a,+\infty)$.

Show that $\displaystyle{f(x)=\lim_{n\rightarrow +\infty}f_n(x)}$ is continuous on $(0, +\infty)$.

Show that $f$ is differentiable on $(0, +\infty)$ with derivative $f'(x)=\frac{-e^{-x}}{1-e^{-x}}$.

$$$$

At 1 do we have to tae the limit $n\rightarrow +\infty$ ? But how can we calculate that series?

$$$$

EDIT :

At question $4.$ do we do the following?

Each of the functions $\{f_n\}$ is continuously differentiable. It holds that $$f_n'(x)=\sum_{k=1}^n -e^{-kx}$$

Let $x\in (0,\infty)$ fixed.

We have that $$g(x)=\lim_{n\rightarrow +\infty}\sum_{k=1}^n -e^{-kx}=-\sum_{k=1}^\infty e^{-kx} = -e^{-x}\sum_{k=0}^\infty e^{-kx}=\frac{-e^{-x}}{1-e^{-x}}$$
So $\{f_n'\}$ converges pointwise to the function $g$ on $(0,\infty)$.

Let $\alpha > 0$ fixed.

Since $-e^{-kx}$ is an increasing function for $x\in [0, \alpha)$, then $-e^{-kx}\leq -e^{-\alpha k}=-(\frac{1}{e^{\alpha}})^k$.

Since $\alpha > 0$, then $\frac{1}{e^{\alpha}}< 1$ and so we have $$\sum_{k=1}^\infty -\Big(\frac{1}{e^{\alpha}}\Big)^k<\infty$$
From Weierstrass M-test we get that the series $\displaystyle{\sum_{k=1}^\infty -e^{-kx}}$ converges uniformly on $[0, \alpha)$ since $$\sum_{k=1}^\infty -e^{-kx} \leq \sum_{k=1}^\infty -e^{-k\alpha}\leq \sum_{k=1}^\infty -\Big(\frac{1}{e^{\alpha}}\Big)^k$$

Since $\alpha>0$ is arbitrary, this holds for each $\alpha$, and so also for $\bigcup_{\alpha> 0}[0, \alpha) = (0,\infty)$.

So we have that $\{f_n'\}$ converges uniformly to $\displaystyle{g(x)=\frac{-e^{-x}}{1-e^{-x}}}$ that is continuous on $(0,+\infty)$.

Then $f$ is also continuously differentiable on $(0,+\infty)$ and it holds that $f'=g$, so $f'(x) = \frac{-e^{-x}}{1-e^{-x}}$.


问:
Solutions of $\frac{\lfloor \sqrt{n}\rfloor(1 + \lfloor\sqrt{n}\rfloor)}{2 n} = \frac{\lfloor \sqrt{k}\rfloor(1 + \lfloor\sqrt{k}\rfloor)}{2 k}$
Testing the

$$\frac{\lfloor \sqrt{n}\rfloor(1 + \lfloor\sqrt{n}\rfloor)}{2 n} = \frac{\lfloor \sqrt{k}\rfloor(1 + \lfloor\sqrt{k}\rfloor)}{2 k}$$

And found that for Oblong numbers (numbers of form $(n+1)n$) this equation is true. In other words this equation has infint solutions for those numbers. While for other ones just has discovered that it has finite solutions (experiment).

Trying to prove the last statement.

答:
Let $t$ be a fixed number not equal to $1$.For $$\frac{\lfloor \sqrt{n}\rfloor(1 + \lfloor\sqrt{n}\rfloor)}{ n} = t$$ we require there to be an integer $k$ such that $$k^2\le n<(k+1)^2, k(k+1)=nt.$$
Then $$\frac{k}{k+1}\le\frac{1}{t}<\frac{k+1}{k}.$$

As $k$ tends to infinity, $\frac{k}{k+1}$ and $\frac{k+1}{k}$ both tend to $1$ and so only finitely many $k$ can satisfy the equations. Therefore there are only finitely many $n=\frac{k(k+1)}{t}$.


问:
Equivalence class for the following relation
A relation of $\mathbb{R}$ is defined as $a\sim b : a^4-b^2=b^4-a^2$

Show that $\sim$ is equivalence relation (I have done this part)

Determine the equivalence class $[-1]_\sim$

Prove or disprove: Every equivalence class in $\mathbb{R}/\sim$ contains exactly 2 real numbers.

I am facing difficulty in second and third part of the question.

For the second one I think it is $[-1]_\sim=\{-1,1\}$ as $-1\sim x: (-1)^4-x^2 =x^4-(-1)^2 $ so both $-1$ and $1$ are proving the equality.

I think the third statement is true but how can I prove it?
Please help.

答:
Follow definitions.

$[-1] = \{b\in \mathbb R|(-1)^2 - b^2=b^4 -(-1)^2\}=$

$\{b\in \mathbb R| 1-b^2 = b^4 -1\}=$

$\{b\in \mathbb R| b^4+b^2-2 = 0\}=$

$\{b\in \mathbb R| (b^2+2)(b-1)(b+1)=0\}=$

$\{b\in \mathbb R| b^2 = -2\lor b=1\lor b=-1\}=$

$\{1,-1\}$

As for the second:

$[a] = \{b\in \mathbb R| a^4-b^2 = b^4 -a^2\}=$

$\{b \in \mathbb R|b^4+b^2 -(a^4+a^2) = 0\}$

Now potentially $b^4 + b^2-(a^4 + a^2)=0$ may have up to four solutions.

Indeed in the case of $[-1]$ we found $b \in [-1]$ if $(b^2 + 2)(b+1)(b-1)=0$ had two elements because $b^2 + 2=0$ is impossible.

We can use quadratic formula to solve $b^4 + b^2 -(a^4+a^2)$ for $b$ but it's probably easier to factor:

$b^4 + b^2-(a^4 + a^2)=0$

$(b^4 -a^4) + (b^2 - a^2) = 0$

$(b^2 +a^2)(b^2 - a^2)+(b^2 -a^2) =0$

$(b^2 + a^2 + 1)(b^2-a^2)=0$

$(b^2  +a^2 + 1)(b-a)(b+a)=0$.

Well, $b^2 + a^2 + 1 > 0$ so the only solutions are $b=\pm a$ and that will always be the case that $[a] = \{a,-a\}$ so.....

but what if $a = 0$.  Then $[0] = \{0,-0\} =\{0\}$ has only one element.

And we can show that directly $0^4 -b^2 = b^4 - 0^2$

$b^4 + b^2 = 0$ so $b^2(b^2 + 1)=0$ so $b^2 = 0$ or $b^2 =-1$ so $b = 0$.

So the third is false but only because $[0]$ has one element. For all others it is true.

Not  $\sim$ is exactly the same as the relationship $a = \pm b$.

Note:  $a^4 -b^2 = b^4 - a^2 \iff a =\pm b$

In hindsight had we proven that from the very beginning we wouldn't have need to do so much work.

Live and learn.


问:
Are a square matrix's columns and rows either both(separately) linearly independent or both(separately) linearly dependent?
Prove or disprove:

Given a square matrix $A$,the columns of $A$ are linearly independent iff. the rows of $A$ are linearly independent.

答:
A more intuitive argument
(why for a square (nxn) matrix ,if it's rows are linearly dependent it implies that so are also it's columns)

Given that rows are linearly dependent,performing Gauss elimination must produce  row of all zeroes (possibly more than one)  .

Therefore the  columns of the row reduced echelon form  matrix are linearly dependent.
That's so because all have zero in the same entry,thus have only n-1 "free" variables ,non-zero entries ,thus are n-1 dimensional,and n vectors in only n-1 dimensional space- can't  be linearly independent.

And since columns of the reduced matrix are linearly depependent , so must  be also the columns of the original matrix,which is our claim.

That's so because  Gauss elimination  preserves relation between columns!
More specifically if some column of the  original matrix is some  linear combination of other columns, then the corresponding columns  of the   row reduced ,echelon form,matrix mirror this precisely.  (otherwise it wouldn't be possible ,for example ,to find  null space of the  original matrix ,by examining the much simpler columns of it's row reduced echelon form)


问:
Analytic continuation of a series involving $\zeta(s)$?
Context:

In this this paper, Juan Arias de Reyna introduces on p12 the following equation for a new power series coefficient:

$$A_m = \lambda_m + \frac{\gamma + \log(4\pi)}{2}-\frac{1}{m}-\frac{1}{m}\sum_{j=2}^{m}(-1)^j{m \choose j} \left(1-\frac{1}{2^j}\right)\zeta(j) \qquad m \in \mathbb{N} \tag{1}$$

Let's multiply (1) by $m$, isolate the last three terms and continue these to $s \in \mathbb{C}$. This gives:

$$f1(s) =s\cdot\frac{\gamma + \log(4\pi)}{2}-1- \sum_{j=2}^{\infty}(-1)^j{s \choose j} \left(1-\frac{1}{2^j}\right)\zeta(j) \qquad s \in \mathbb{C}, \quad \Re(s) > 0 \tag{2}$$

I've found ways to analytically continue both $A_m$ and $\lambda_m$ (= Keiper/Li coefficients) into entire functions and this provided data to assess what the analytic continuation of (2) would look like. After some experiments, I managed to numerically derive the following simple function that appears entire:

$$f2(s) = s\cdot\frac{\gamma + \log(\pi)}{2}-\sum_{j=2}^{\infty}{-s \choose j} \frac{\zeta(j)}{2^j} \qquad s \in \mathbb{C}\tag{3}$$

Question:

Is there a way to prove that the entire function (3) is indeed the full analytic continuation of (2) ?

P.S.

A brief inspection of (3) showed it has at least two real zeros and also a couple of complex zeros, but all with $\Re(s)$ negative.

答:
For $\Re(s)>1$

$$\sum_{j=2}^{\infty}(-1)^j{s \choose j} \left(1-\frac{1}{2^j}\right)\zeta(j)=(1+(-1))^s-1+ s -\sum_{j=2}^{\infty}(-1)^j{s \choose j} \frac{1}{2^j}\zeta(j) $$
$$-1+ s -\sum_{j=2}^{\infty}(-1)^j{s \choose j} \frac{1}{2^j}\zeta(j) $$
The latter is entire as $\frac{1}{2^j}\zeta(j) $ has exponential decay and locally $${s \choose j}= \prod_{m=1}^j (\frac{s+1-m}{m})=O(e^{|s| \sum_{m\le j}1/m})= O(e^{|s|\log j})$$
which means local uniform convergence and analyticity.


问:
Schröder–Bernstein Theorem proof help
In Understanding Analysis, there's is a guided Schröder–Bernstein Theorem proof as an exercise.

I'm trying to prove it on my own but I believe I'm confused on several points and would appreciate some aid and corrections where necessary. Here's what I have so far:

a) By partitioning the domains, if we prove that $f$ from $A$ to $B$ and $g$ from $B'$ to $A'$ are bijective, then we can write a function:
$$h(x)=\begin{cases} f(x)\text{ if}\ x\in A \\ g^{-1}(x)\text { if}\ x\in A'\\ \end{cases}$$
that is bijective and concludes the proof.

b) We know that $A_1=X\backslash g(Y)$ and all subsequent $A_n$s are $A_n\subseteq g(Y)$. Thus $A_1\ne A_2$ and inductively, as $f$ is 1-1 and so $f(Z)\ne f(Z')$ when $Z\ne Z'$ and the same goes for $g$, we have $g(f(Z))\ne g(f(Z'))$ when $Z\ne Z'$ and so $g(f(A_n))\ne g(f(A_{n-1}))$.
Thus the $A$s are disjoint pairwise and with the same logic the $B$s are also disjoint.

c) I don't get that, it seems true by definition to me...

d)For every $x\in B'$, $x$ can't be in any $f(A_n)$ for all $n$ and for all $g(x)$, as one generation of $A$ feeds the next and $g(x)$ not in $A_1$, then $g(x)$ not in any $A_n$.
(I'm not very happy about d. to be honest).

答:
I wish I could have made this clearer. But it is difficult. Do let me know if there are confusions. Also, please correct me if I am wrong.

Part A

Since

\begin{eqnarray*}
\forall a'\in\mathbb{A}'\exists b'\in\mathbb{B}'\text{ s.t. }g(b')=a'\land\forall b_1,b_2\in\mathbb{B}'\exists a_1=g(b_1),a_2=g(b_2)\in\mathbb{A}'\text{ s.t. }b_1\neq b_2\rightarrow a_1\neq a_2
\end{eqnarray*}

there exists a function $g^{-1}:\mathbb{A}'\rightarrow\mathbb{B}'$. Since

\begin{eqnarray*}
\forall b'\in\mathbb{B}'\exists a'=g(b')\in\mathbb{A}'\text{ s.t. }g^{-1}(g(b'))=b'
\end{eqnarray*}

it is onto. Since
\begin{eqnarray*}
\forall a_1,a_2\in\mathbb{A}'\exists b_1,b_2\in\mathbb{B}'\text{ s.t. }g(b_1)=a_1\neq a_2=g(b_2)&\rightarrow&b_1\neq b_2\\
g^{-1}(a_1)&=&g^{-1}(g(b_1))\\
&=&b_1\\
&\neq&b_2\\
&=&g^{-1}(g(b_2))\\
&=&g^{-1}(a_2)
\end{eqnarray*}

it is one-to-one. Define
h(x) =
\begin{cases}
f(x) &x\in\mathbb{A}\\
g^{-1}(x) &x\in\mathbb{A}'
\end{cases}

Since both functions are one-to-one and onto, the statement holds.

Part B

Trivially, if $A_1=\emptyset$, then $g$ is bijective and $\mathbb{X}\sim\mathbb{Y}$. Besides,
\begin{eqnarray*}
\forall x\in f(\mathbb{A}_1)&\rightarrow&x\in\mathbb{Y}\\
&\rightarrow&f(\mathbb{A}_1)\subseteq\mathbb{Y}
\end{eqnarray*}
Suppose $\forall i\in\mathbb{N}\mathbb{A}_1\bigcap\mathbb{A}_i\neq\emptyset$, then,
\begin{eqnarray*}
\exists x\in\mathbb{X}\text{ s.t. }x\in\mathbb{A}_1\bigcap\mathbb{A}_2&\rightarrow&x\in\mathbb{A}_1\land x\in\mathbb{A}_i\\
&\rightarrow&x\in\mathbb{X}\land x\notin g(\mathbb{Y})\land x\in g(\cdots f(\mathbb{A}_1)\cdots)\\
&\rightarrow&x\notin g(\mathbb{Y})\land x\in g(\mathbb{Y})
\end{eqnarray*}
Hence, by contradiction, the statement doesn't hold. Suppose inductively, $\forall i,j\leq k\text{ s.t. }A_i\bigcap A_j=\emptyset$ and $P(k+1)$ doesn't hold. Then,
\begin{eqnarray*}
\exists x\in\mathbb{X}\exists1<i\leq k+1\text{ s.t. }x\in\mathbb{A}_{k+1}\bigcap\mathbb{A}_i&\rightarrow&\exists y_{i-1}\in f(A_{i-1}),y_k\in f(A_k)\text{ s.t. }y_{i-1}\neq y_k\\
&\rightarrow&g(y_{i-1})=x=g(y_k)
\end{eqnarray*}
By contradiction, the induction holds. Since
\begin{eqnarray*}
\forall i\neq j\in\mathbb{N}\mathbb{A}_i\bigcap\mathbb{A}_j=\emptyset\land\forall x_1\neq x_2\in\mathbb{X}f(x_1)\neq f(x_2)&\rightarrow&\forall x_i\in\mathbb{A}_i,x_j\in\mathbb{A}_j\text{ s.t. }f(x_1)\neq f(x_2)\\
&\rightarrow&f(\mathbb{A}_i)\bigcap f(\mathbb{A}_j)=\emptyset
\end{eqnarray*}
Hence, the statement holds.

Part C

\begin{eqnarray*}
\forall b\in\mathbb{B}'\exists n\in\mathbb{N}\text{ s.t. }b\in f(A_n)&\rightarrow&\exists a\in\mathbb{A}_n\text{ s.t. }b=f(a)
\end{eqnarray*}
Hence, the statement holds.

Part D

Suppose $\mathbb{A}'$ is empty. Then,
\begin{eqnarray*}
\mathbb{X}=\mathbb{A}&\rightarrow&\mathbb{X}=\bigcup_{n=1}^{\infty}A_n\\
&\rightarrow&\forall x\in\mathbb{X}\exists\mathbb{A}_n\text{ s.t. }x\in\mathbb{A}_n\\
&\rightarrow&\exists b\in\mathbb{B}'\text{ s.t. } b\in f(\mathbb{A}_n)\\
&\rightarrow&\mathbb{X}=g(\mathbb{Y})\\
g:\mathbb{Y}\rightarrow\mathbb{X}&\rightarrow&\mathbb{X}\sim\mathbb{Y}
\end{eqnarray*}
By contradiction, it isn't empty. Hence,
\begin{eqnarray*}
\forall x\in\mathbb{A}'&\rightarrow&x\in\mathbb{X}\backslash\mathbb{A}\\
&\rightarrow&x\in\mathbb{X}\land x\notin\mathbb{A}_1\cdots\\
&\rightarrow&x\in\mathbb{X}\land (x\not\in\mathbb{X}\lor x\in g(\mathbb{Y}))\land x\in g(f(\mathbb{A}_1))\land\cdots\\
&\rightarrow&x\in\mathbb{X}\land x\in g(\mathbb{Y})\land x\notin g(f(\mathbb{A}_1))\land\cdots\\
&\rightarrow&\forall n\in\mathbb{N},\exists b\in\mathbb{Y}\text{ s.t. }b\notin f(\mathbb{A}_n)\\
&\rightarrow&b\in\mathbb{Y}\land b\in\notin\bigcup_{n=1}^{\infty}A_n\\
&\rightarrow&b\in\mathbb{Y}\backslash\mathbb{B}\\
&\rightarrow&b\in\mathbb{B}'
\end{eqnarray*}
Hence, the statement holds

Title: Schröder–Bernstein Theorem proof help


问:
How can I evaluate $0.01 \cdot \sum^{n}_{t=1}t \cdot (0.99)^{t-1}$?
How can I evaluate the summation $$0.01 \cdot \sum^{n}_{t=1}t \cdot (0.99)^{t-1}$$
I am aware that this is almost in the form of a geometric summation, but I am unsure how to proceed and what to do. Any help would be greatly appreciated.

答:
Hint:

Define the function $f(y)=\sum_{t=1}^n ty^{t-1}$, $y\in[0,1]$

Integrate on $[0,x]$ term by term, take the sum, then differentiate it.

Notes.

$$F(x)=\int_0^x f(y)dy=\sum_{t=1}^n\int_0^xty^{t-1}dy=\sum_{t=1}^nx^t=\textrm{you know how ...}$$

Then
$$
0.01\cdot f(0.99)=0.01\cdot F'(0.99)
$$


问:
interpolation of i.i.d. random variables
Let $\{X_k\}_{k\in \mathbb{Z}}$ be a sequence of iid, zero-mean and unit-variance Gaussian random variables. Let $Y(t), t\in\mathbb{R}$ be a continuous-time random process built by linear interpolation of $\{X_k\}_{k\in \mathbb{Z}}$. What is the first order pdf of $Y(t)$, i.e., $f(y;t)$?

Amendment:

What if the $\{X_k\}_{k\in \mathbb{Z}}$ is an iid sequence of random variables uniformly distributed over $(0,1)$?

答:
$$
Y(t) = (\lfloor t \rfloor + 1 - t)X(\lfloor t \rfloor) + (t - \lfloor t \rfloor) X(\lfloor t\rfloor - 1)
$$
So
$$
Y(t) \sim \operatorname N(0, (\lfloor t\rfloor+1-t)^2 +(\lfloor t\rfloor - 1)^2 ).
$$


答:
We recall the power rule, which states:

$$\frac{d}{dx} \left[x^n\right] = nx^{n-1}$$

Hence:

$$\frac{d}{dx}\left[\sum_{t=1}^n x^t\right] = \sum_{t=1}^n tx^{t-1}$$

Recalling that $\sum_{t=1}^{n} x^t = (x-x^{n+1})/(1-x)$ (see if you can derive this yourself! Hint: multiply both sides by $(1-x)$) to get your desired expression we solve for:

$$0.01\left(\frac{d}{dx}\left[\frac{x-x^{n+1}}{1-x}\right]\right)$$

evaluated at $x=0.99$.

The first time you see term-by-term differentiation or integration to manipulate a series it seems like such an illegal, out-there move, but it's actually a regular tool in the toolbox. Once you see it once, you see it everywhere.

Title: How can I evaluate $0.01 \cdot \sum^{n}_{t=1}t \cdot (0.99)^{t-1}$?

Body:
How can I evaluate the summation $$0.01 \cdot \sum^{n}_{t=1}t \cdot (0.99)^{t-1}$$
I am aware that this is almost in the form of a geometric summation, but I am unsure how to proceed and what to do. Any help would be greatly appreciated.


问:
Explicitly determine the prime ideal such that $\mathbb{Q}_5(\sqrt{5})$ is the completion of some number field
Let $K = \mathbb{Q}_5(\sqrt{5})$. As I learned from my last post, it is possible to write $K$ as the completion of $(k,v)$ where $k$ is a number field (i.e. a finite extension of $\mathbb{Q}$) and a valuation $v$ on $k$ defined by some prime ideal $\mathfrak{p}$ in $\mathcal{O}_k$. While I am still digesting the answer of mentioned post, I think it makes sense to choose $k = \mathbb{Q}(\sqrt{5})$. Now the question remains on how to choose the prime ideal $\mathfrak{p}$ (which is I think not shown in the answer).

I know that $\mathbb{Q}_5$ is the completion of $\mathbb{Q}$ wrt. the $5$-adic value which comes from the prime ideal $(5) \subseteq \mathcal{O}_\mathbb{Q} = \mathbb{Z}$ (cf. also this post of mine). Now I think that the $\mathfrak{p}$ we are looking for has something to do with $(5)$ but I was not able to advance further from this point. Question: How to explicitly describe $\mathfrak{p}$ (say, by finitely many explicit generators)?

答:
$$\Bbb{Z}_5[\sqrt5]= \varprojlim_{n\to \infty} \Bbb{Z}[\sqrt5]/( \sqrt5^n)= \varprojlim_{n\to \infty} \Bbb{Z}[\sqrt5]/(5^n)$$
Well $O_{\Bbb{Q}(\sqrt5)}=\Bbb{Z}[\frac{1+\sqrt5}2]$ but we don't really care here.

The valuation extended to $\Bbb{Q}(\sqrt5)$ is $$v(a)= n\text{ if } \ a\in ( \sqrt5^n),\not \in ( \sqrt5^{n+1}), \qquad v(a/b)=v(a)-v(b), v(0)=\infty$$

$\Bbb{Q}_5(\sqrt5)=\Bbb{Z}_5[\sqrt5][5^{-1}]$ is the completion of $\Bbb{Q}(\sqrt5)$ for $v$.


问:
Find an interval $I$ such that the probability that the parameter belongs to $I$ goes to a particular value.
Let $X_i$ be i.i.d. uniform random variables in $[0,\theta]$, for some $\theta>0$ and let $M_n =\max(X_i)$.

We have to find an interval $I$ of the form $I=[M_n,M_n+c]$, that does not depend on $\theta$ and such that $P[I\ni \theta]\to .95$ as $n \to\infty $.

Now, I calculated the CDF of $M_n$ which is $$F_{M_n}(y) = \begin{cases}
0 & \, \text{if} \ y < 0\\
\frac{y^n}{\theta^n} & \, \text{if} \ 0\leq y \leq \theta\\
1 & \, \text{if} \ y > \theta
\end{cases}$$

Now, after some simple calculations, we get, $E[M_n] = \frac{n\theta}{n+1}$ and $\sigma_{M_n} = \frac{\theta}{n+1}\sqrt{\frac{n}{n+2}}$

So, $\frac{M_n-\frac{n\theta}{n+1}}{\frac{\theta}{n+1}\sqrt{\frac{n}{n+2}}} \sim N(0, 1) $ by CLT.

Now, $P(-z_{\alpha/2} \leq \frac{M_n-\frac{n\theta}{n+1}}{\frac{\theta}{n+1}\sqrt{\frac{n}{n+2}}} \leq z_{\alpha/2}) = 1 - \alpha$ ,  where $\alpha = 0.05$

But I don't think this will give an interval of the form $I=[M_n,M_n+c]$. I mean how can I get the above equation in the form of $P(M_n \leq \theta \leq M_n + c) = 1 - \alpha$ for some $c$?

Thanks!

EDIT 1 : The above calculation is totally wrong as mentioned by StubbornAtom. I corrected myself and did the problem in a different way. Please see the next EDIT.

EDIT 2 : We have to use the fact that $Z_n = n(1-\frac{M_n}{\theta})$ converges  in  distribution  to  an  exponential  random  variable  with parameter  $1$.

So, $Z_n$ has approximately Exp$(1)$ distribution for large $n$.

Now, $P(0\leq Z_n \leq k_{\alpha}) = 1 -\alpha$, where $k_{\alpha}$ is such that $P(Z_n\geq k_{\alpha}) = \alpha$ where $\alpha = 0.05$.

And this ultimately gives us the following condition $$P \bigg(M_n \leq \theta \leq \frac{M_n}{1-\frac{k_{\alpha}}{n}}\bigg) = 0.95$$ where $k_{\alpha}$ comes out to be $2.9957$. So the required interval that I am getting is $I = [M_n, \frac{M_n}{1-\frac{k_{\alpha}}{n}}]$ but this is still not the required form. What should I do now?

答:
Outline:

(1) The maximum $W$ of $n$ independent random variables
from $\mathsf{Unif}(0,1)$ has $W \sim \mathsf{Beta}(n-1, 1),$
which can be proved by looking at the CDF $F_W(w)$ of $W.$

```<code>w = replicate(10^6, max(runif(10)))
hdr = "Simulated Maxima of 10 Obs. from UNIF(0,1) with BETA(9,1) Density"
hist(w, prob=T, col="skyblue2",  main=hdr)
curve(dbeta(x, 9, 1), add=T, col="orange",lwd=2)
</code>```

(2) Then one can show that a confidence interval for $\theta$ based on the maximum $W$ of $n$ random observations from $\mathsf{Unif}(0,\theta)$ arises from $P(W < \theta < W/\alpha^n) = 1 - \alpha.$
for $0 < \alpha < 1).$ (See this Wikipedia page.)

Verification: Here is a simulation in R to illustrate the case where $n=20, \theta = 10, \alpha=.05,$
so that we can see what fraction of 100,000 such 95% confidence intervals actually covers $\theta.$

```<code>set.seed(120)
n = 20;  th = 10;  al = .05
w = replicate(10^5, max(runif(n,0,th)))  # 100,000 maxima
mean((w < th) & th < w/al^(1/n))
[1] 0.94989                              # frac of CIs covering 10
</code>```

Body:
Outline:

(1) The maximum $W$ of $n$ independent random variables
from $\mathsf{Unif}(0,1)$ has $W \sim \mathsf{Beta}(n-1, 1),$
which can be proved by looking at the CDF $F_W(w)$ of $W.$

w = replicate(10^6, max(runif(10)))
hdr = "Simulated Maxima of 10 Obs. from UNIF(0,1) with BETA(9,1) Density"
hist(w, prob=T, col="skyblue2",  main=hdr)
curve(dbeta(x, 9, 1), add=T, col="orange",lwd=2)

(2) Then one can show that a confidence interval for $\theta$ based on the maximum $W$ of $n$ random observations from $\mathsf{Unif}(0,\theta)$ arises from $P(W < \theta < W/\alpha^n) = 1 - \alpha.$
for $0 < \alpha < 1).$ (See this Wikipedia page.)

Verification: Here is a simulation in R to illustrate the case where $n=20, \theta = 10, \alpha=.05,$
so that we can see what fraction of 100,000 such 95% confidence intervals actually covers $\theta.$

set.seed(120)
n = 20;  th = 10;  al = .05
w = replicate(10^5, max(runif(n,0,th)))  # 100,000 maxima
mean((w < th) & th < w/al^(1/n))
[1] 0.94989                              # frac of CIs covering 10


问:
Confused about Burmann series coefficients to approximate $erf(x)$
I've been looking into Burmann series to approximate special functions. I looked at "On Bürmann’s Theorem and Its Application to Problems of Linear and Nonlinear Heat Transfer and Diffusion".  Looking at the example for $erf$ also shown here Integration of $e^{-x^2}$ the authors create a new type of constant called $c_i$ and somehow come up with $c0 = \sqrt{\pi}/2 $

where (32) is

and

I don't understand what these constants are.   They appear to be unrelated to Burmann constants in the corresponding burrman series, but some how a retraction of

How can I get the rest of the constants?

答:
I will be frank, I have no idea what they refer to when deriving the coefficients as the text is unclear. The closest interpretation of

For example, using only $c_1$ and requesting the correct slope at $z = 0$, one gets an approximation of the error function with a relative error smaller than $1.2\%$.

that I have is that they tried to have

$$f(z)=\frac{2\operatorname{sign}(z)}{\sqrt\pi}\sqrt{1-e^{-z^2}}\left(\frac{\sqrt\pi}2+c_1e^{-z^2}\right)$$

satisfy

$$f'(0)=\operatorname{erf}'(0)$$

and that they did similar for $c_2$ by trying to match the third derivative as well (the even order derivatives are vanishing at the origin). This however leads me to get

$$\frac{2\operatorname{sign}(z)}{\sqrt\pi}\sqrt{1-e^{-z^2}}\left(\frac{\sqrt\pi}2+\left(\frac{23}{12}-\sqrt\pi\right)e^{-z^2}+\left(\frac{\sqrt\pi}2-\frac{11}{12}\right)e^{-2z^2}\right)$$

which is completely different from their solution.

Indeed one can easily verify their solution does not match the first derivative.

What really boggles me is how good the result is, and without any involvement with $\pi$. I don't see what kinds of conditions might yield such a result.


问:
Help With An Expression Involving Limits
I am left wondering about part of a proof involving limits.  Here is the math included (which I wish had more details!)

$$ \lim_{R\rightarrow \infty }S(Re^{j \theta})=\lim_{R\rightarrow \infty }\frac{1}{1+L(Re^{j \theta})}=\frac{1}{1+\lim_{s\rightarrow \infty }L(s)R^{-n_{r}}e^{-n_{r} j \theta }} $$

where $ n_{r} $ is the relative degree of $L(s)$.

For context $S(s)$ is the sensitivity transfer function and $L(s)$ is the open loop transfer function, from Control Theory.

I can't figure out how they go from the second term to the third term. Any help would be greatly appreciated.  Thanks

For additional context, here is a link to the paper containing the proof.  Bottom, righthand side is the equation in question.

https://www.sciencedirect.com/science/article/pii/S2405896315025045

答:
Okay, here is what I came up with today.  My confidence in its correctness is somewhere around luke warm.

$$ \lim_{R\rightarrow\infty}{S\left(Re^{j\theta}\right)}=\lim_{R\rightarrow\infty}{\frac{1}{1+L\left(Re^{j\theta}\right)}}=\frac{1}{1+\lim_{s\rightarrow\infty}{L\left(s\right)}R^{-n_r}e^{-n_rj\theta}} $$

To understand what is going on when going from the second term to the third term in the equation above, consider the following general form of the open-loop transfer function.
$$ L\left(Re^{j\theta}\right)=\frac{H\left(R^{n_H}e^{j\theta n_H}\right)}{G\left(R^{n_G}e^{j\theta n_G}\right)} $$
The Open-Loop Transfer Function, L(s), is ultimately a complex function, to some order of z.  The above equation shows that the numerator is of order nH and the denominator is of order nG.  Let us take a closer look at this specific term in the denominator of the original equation in question.
$$ \lim_{R\rightarrow\infty}{L\left(Re^{j\theta}\right)}=\lim_{R\rightarrow\infty}{\frac{H\left(R^{n_H}e^{j\theta n_H}\right)}{G\left(R^{n_G}e^{j\theta n_G}\right)}} $$
For the general transfer function, the numerator and denominator will be in terms of powers of z.  Next, consider that in the limit of R going to infinity, it is ultimately the highest order power of R in the numerator and denominator, along with the coefficient associated with that term, which determines the value of the limit.  We can express that term as the following.
$$ \lim_{R\rightarrow\infty}{\frac{H\left(R^{n_H}e^{j\theta n_H}\right)}{G\left(R^{n_G}e^{j\theta n_G}\right)}}=\frac{A\ast R^{n_H}e^{j\theta n_H}}{B\ast R^{n_G}e^{j\theta n_G}}=\frac{A}{B}R^{\left(n_H-n_G\right)e^{j\theta\left(n_H-n_G\right)}} $$
There are two things to note about that last term.  We are taking the limit of L(z) in the complex plane and arriving at the above result.  Well, what if we also took the limit of L(s) in the s-plane as s goes to infinity.  If you work through the same reasoning we used in the complex plane, you will eventually come to an answer of A/B!  The other thing to note is that (nH - nG) is just the relative degree of L(z), or nr.
That let’s us write all this up as:
$$ \lim_{R\rightarrow\infty}{S\left(Re^{j\theta}\right)}=\frac{1}{1+L\left(\infty\right)\ast R^{-n_r}e^{-n_rj\theta}} $$
In the above, $$ L\left(\infty\right)=\lim_{s\rightarrow\infty}{L\left(s\right)}. $$


问:
Logical Conjuction or Implication?
T(x,y) = x has taken class y.

Statement: Every student has taken at least one math class besides Math 101.

My initial thought was the following:

$$\forall x \exists y : (y\neq Math 101) \rightarrow T(x,y)$$

but discussion has lead to the following:

$$\forall x \exists y: (y\neq Math101) \land T(x,y)$$

From my understanding 2) would be incorrect because it is implying that no student took Math 101 where the statement says nothing about whether or not a student took math 101. Does this make sense.

I could also see 1) being incorrect as the class not being Math 101 may not necessarily mean that the student took the class?

答:
That 2) doesn't say that no student took Math 101; for $y=Math101$, $(y\ne Math101)$ is false and the result of $T(x,y)$ is irrelevant.

Then 2) would mean there exists a different class $y$ such that $(y\ne Math101)\wedge T(x,y)$. $(y\ne Math101)$ is true for the different class $y$, and $T(x,y)$ has to be true for that $y$ too.

That 1) is also incorrect in the way that, if $Math101$ is among the classes that $y$ can be, then $y=Math101$ makes the implication $(y\ne Math101)\to T(x,y)$ true, regardless of whether $T(x,Math101)$. Then $y=Math101$ satisfies the $\exists y$, and other classes $y$ becomes irrelevant.

Title: Logical Conjuction or Implication?


问:
Prove or Disprove $A \subseteq C \land B \subseteq D $ is equivalent to $A \times B \subseteq C \times D$
Can this statement be proved? because if i insert values, LHS is not equal to RHS.
For example,
let  $A=\{1\}, B=\{1,2\}, C=\{1\}, D=\{1\}$ then
$A\times B = \{1,2\}, C\times D=\{1\}$

答:
The Cartesian product makes a set of pairs.  So for example:

$$\{1, 2\} \times \{a, b, c\} = \{(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)\}$$

It is all possible pairs where the first element is from the first set and the second element is from the second set.  Try working out some examples with this Cartesian product definition.


问:
Special binary string
Imagine a binary string of increasing length, up to infinity.

What makes it so special? Well, just a simple "rule":

for any given length (odd or even), if one folds the string in half, there is at least one couple of ones that overlaps.

The question is now about the distribution of ones (zeros), as a function of string length. What are its properties? Are there any?

A trivial example that follow the rule is the string $111010101010....$.

答:
For any string that satisfies the rule, you can change any $0$'s to $1$'s and get a new one that satisfies the rule.  So I think you should look for the lowest density string.  One way is the greedy algorithm.  I assume you don't count the bit on the fold, if there is one, but you are satisfied with one position of overlap.  Using Hagen von Eitzen's characterization of the set, The positions of the $1$'s by this rule are $1,2,3,5,8,11,14,\ldots,3k+2$  We can see the $1,2,3,5,8,11,14$ by inspection.  If the pattern continues up to $3k+2$, we can make $3k+3, 3k+4, 3k+5$ but cannot make $3k+6$ because the only number that is $1 \pmod 3$ is $1$.  So we have to add $3k+5=3(k+1)+2$ to the list and the induction continues.  I didn't find the sequence in OEIS.

Added:  with the clarification that a $1$ on the fold counts, the greedy series becomes $1,2,4,6,8,\ldots 2k$ and the proof is the same as the previous one.  This is a higher density that the previous one, so the greedy algorithm will not give the minimum.  We can make the limiting density arbitrarily low by starting with $n\  1$'s, then putting $n-1\  0$'s and a $1$ repeatedly.

Title: Special binary string


问:
Notion of fractal dimension which distinguishes point and square from square
I am new to fractal dimensions and am looking for a notion of fractal dimension which would help me distinguish between the following two sets on the cartesian plane:

\begin{align}A&=[0,1]\times[0,1] \cup (2,2)\\B&=[0,1]\times[0,1]\end{align}

In other words, I am asking:

Is there some form of a fractal dimension, say $D$, which would give $D(B)=2$, as $B$ is a 2-d square, but $D(A)\not=2$, as $A$ is a 2-d square joined with a 0-d point?

答:
Most generic notions of dimension (Hausdorff, box-counting, Assouad, lower, etc.) attempt to assign a single number to a space.  They are, in some sense, defined locally, but they try to capture some global structure.  Examples such as the disjoint union of a square and point suggest that "dimension" is not the right tool for distinguishing or fully describing these spaces.

A better tool might be the "local dimension," discussed in [1] (I don't have a precise reference of the top of my head, and most of my books are still in boxes—I think that the discussion of local dimensions is in chapter 17 of the third edition).
$\DeclareMathOperator{\locu}{\overline{\dim}_{\text{loc}}}$
$\DeclareMathOperator{\locl}{\underline{\dim}_{\text{loc}}}$
$\DeclareMathOperator{\loc}{\dim_{\text{loc}}}$

Definition.  Let $(X,d,\mu)$ be a metric measure space (i.e. a space with underlying set $X$, metric $d$, and measure $\mu$).  The lower local dimension of $\mu$ at a point $x$ is
$$\locl \mu(x) := \liminf_{r\downarrow 0} \frac{\log(\mu(B(x,r)))}{\log(r)},$$
and the upper local dimension of $\mu$ at a point $x$ is
$$\locu \mu(x) := \limsup_{r\downarrow 0} \frac{\log(\mu(B(x,r)))}{\log(r)}.$$
If these two limits coincide, then the local dimension of $\mu$ at $x$ is the common value, denoted $\loc \mu(x)$.

Very roughly speaking, the local dimension of a measure at a point tells us how balls scale around that point—the local dimension "sees" the scaling exponent about a point.  For the spaces we are considering, these limits coincide (or fail to exist in the same way), so I'm going to go ahead and work with the local dimension, and not worry about the upper and lower dimension.

As an example, in $\mathbb{R}^2$ with the usual two dimensional Lebesgue measure $\lambda^2$, the measure of a ball of radius $r$ is $\pi r^2$.  Hence
$$ \loc \lambda^2(x)
= \lim_{r \downarrow 0} \frac{\log(\lambda^2(B(x,r)))}{\log(r)}
= \lim_{r \downarrow 0} \frac{\log( \pi r^2 ) }{ \log(r) }
= \lim_{r \downarrow 0} \frac{2 \log(r) + \log(\pi) }{ \log(r) }
= 2. \tag{*}$$

There are various tools which can be used to analyze the local dimension(s) of a space, but for our purposes, the image of the function $x \mapsto \loc\mu(x)$ ought to be sufficient.

Now... take
\begin{align}
A &= [0,1]^2 \cup \{(2,2)\} \\
B &= [0,1]^2,
\end{align}
each with the subspace metric, and the two dimensional Lebesgue measure restricted to the space (denoted by $\mu_A$ and $\mu_B$).  For any point $x \in [0,1]^2$, the local dimension of the measure is given by
$$ \loc \mu_A(x) = \loc \mu_B(x) = 2. $$
Note that the computation is identical that in (*) for any interior point (with $r$ sufficiently small), and similar for boundary and "corner" points (the $\pi$ is replaced with $\pi/2$ or $\pi/4$.

On the other hand, for the point $(2,2)$, the local dimension is a slightly more wily beast.  The two dimensional Lebesgue measure of an isolated point is $0$, thus for any sufficiently small $r$, we have $$\mu_A\bigl( A \cap B((2,2), r)\bigr) = 0.$$  In this context, it is (perhaps) reasonable to define $\log(0) = -\infty$.  With respect to this convention,
$$\loc \mu_A((2,2)) = \lim_{r \downarrow 0} \frac{\log( 0 )}{ \log(r) } = -\infty. $$
It is, therefore, reasonable to conclude that
$$ \loc \mu_A(A) = \{2,-\infty\}, \qquad\text{and}\qquad
\loc \mu_B(B) = \{2\}. $$

It is, perhaps, worth noting that this computation indicates that an isolated point in this space is not zero dimensional, but is, instead "negative infinite dimensional".  This is an entirely reasonable interpretation, and has to do with the fact that points are null sets for the Lebesgue measure.  You might consider what would happen if you gave the point counting measure (so that $\mu_A((2,2)) = 1$).

Another (related) approach might be to massage the theory of the complex dimensions of a relative fractal drum (RFD), introduced by Lapidus and collaborators, and described in some detail in chapter 5 of [2].  I am not going to get into this in too much detail here, as it takes a little work to develop the right tools, but I will note that the theory of RFDs gives us much better language for talking about spaces which have "negative" dimension (including negative infinite dimension).

[1] Falconer, Kenneth, Fractal geometry. Mathematical foundations and applications, Hoboken, NJ: John Wiley & Sons (ISBN 978-1-119-94239-9/hbk). xxx, 368 p. (2014). ZBL1285.28011.

[2] Lapidus, Michel L.; van Frankenhuijsen, Machiel, Fractal geometry, complex dimensions and zeta functions. Geometry and spectra of fractal strings, Springer Monographs in Mathematics. New York, NY: Springer (ISBN 978-1-4614-2175-7/hbk; 978-1-4614-2176-4/ebook). xxvi, 567 p. (2013). ZBL1261.28011.


问:
Prove this inequality (please using AM-GM or Cauchy-Schwarz if possible)
Let $a ; b ; c > 0$ such that $a+b+c=3$

Prove: $P=\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{6abc}{ab+bc+ca} \geq 5$

I tried:
$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=\dfrac{a^2c+b^2a+c^2b}{abc}=\dfrac{(a^2c+b^2a+c^2b)(c+a+b)}{3abc} \geq \dfrac{(ab+bc+ca)^2}{3abc}$
Then: $P \geq \dfrac{(ab+bc+ca)^2}{3abc}+\dfrac{6abc}{ab+bc+ca}$

But I think it's a wrong way, cuz $\dfrac{(ab+bc+ca)^2}{3abc}$ seems too weak.

答:
Suppose $c = \min \{a,b,c\},$ we have
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} -3 = \frac{(a-b)^2}{ab} + \frac{(a-c)(b-c)}{ac},$$
$$(a+b+c)(ab+bc+ca)-9abc = 2c(a-b)^2+(a+b)(a-c)(b-c).$$
We write the inequality as
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{18abc}{(a+b+c)(ab+bc+ca)} \geqslant 5.$$
equivalent to
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} - 3 \geqslant 2\left[1 - \dfrac{9abc}{(a+b+c)(ab+bc+ca)}\right],$$
or
$$\frac{(a-b)^2}{ab} + \frac{(a-c)(b-c)}{ac} \geqslant \frac{4c(a-b)^2}{(a+b+c)(ab+bc+ca)}+\frac{2(a+b)(a-c)(b-c)}{(a+b+c)(ab+bc+ca)},$$
or
$$\left[\frac{1}{ab} - \frac{4c}{(a+b+c)(ab+bc+ca)}\right](a-b)^2 + \left[\frac{1}{ac} -\frac{2(a+b)}{(a+b+c)(ab+bc+ca)}\right](a-c)(b-c) \geqslant 0.$$
But
$$(a+b+c)(ab+bc+ca) \geqslant 9abc \geqslant 4abc,$$
and
$$\begin{aligned}(a+b+c)(ab+bc+ca) - 2ca(a+b) & \geqslant (a+b)(ab+bc+ca) - 2ca(a+b) \\&= (a+b)[a(b-c)+bc] \geqslant 0.\end{aligned}$$
The proof is completed.


问:
Expressing the infinite sum $1 + 22 + 333 + 4444 + \dotsb$
I would like to express

$$1+22+333+4444+\cdots$$

using $\Sigma$ notation, and have no clue where to start.

After $999999999$, comes 10 $0$s, then 11 $1$s.

答:
I've derived the formula that gives the sum of the series for $n$ terms. For example the series:

$$1 + 22 + 333 + 4444,$$ has four terms.

Let $n$ be the number of terms. Let $S_n$ be the sum of the $n$ terms. Then

$$S_n = \frac{1}{1458}\left((18n-2)10^{n+1} -81n^2 -81n + 20 \right)$$


问:
Help evaluating an interesting series.
I'm trying to evaluate the sum
$$
s = 1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\cdots.
$$

The hint I was given was $\frac{1}{2n+1} = \int_0^1 x^{2n} \, dx$. I'm familiar with the trick of writing the $n$th term of series with a definite integral and then interchanging the order of integral and sum, but the signs on the terms are making this difficult for me. If this were a normal alternating series, we could do something like
\begin{align*}
\sum_{n=0}^{\infty}(-1)^n\frac{1}{2n+1} &= \sum_{n=0}^\infty\int_0^1 (-1)^n x^{2n} \, dx\\
&= \int_0^1 \left(\sum_{n=0}^{\infty} (-x^2)^{n}\right)dx\\
&= \int_0^1 \frac{1}{1+x^2} \, dx = \arctan(1)-\arctan(0) = \frac{\pi}{4}.
\end{align*}

I can't however seem to proceed with the strange alternating series. I know that $(-1)^{(n^2+n+2)/2}$ gives the right pattern of $++--++--\dots$, but that doesn't combine well with the $x^{2n}$ term in the integrand after exchanging sum and integral.

Any thoughts?

Edit: If I use the imaginary unit $i$, then I can write a related sum
$$
\tilde{s} = 1+\frac{1}{3}i-\frac{1}{5}-\frac{1}{7}i+\frac{1}{9}+\frac{1}{11}\cdots.
$$
Using the above trick gives
$$
\sum_{n=0}^{\infty}(i)^{n}\frac{1}{2n+1} = \sum_{n=0}^{\infty}\int_{0}^{1}(i)^nx^{2n}dx = \int_{0}^{1} \sum_{n=0}^{\infty}(ix^2)^n =\int_{0}^{1}\frac{1}{1-ix^2}dx
$$
Which I imagine I can do with some type of partial fraction decomposition.

My hope is that from this new sum I can recover my old sum by adding real and imaginary parts. My only concern is that this would essentially be a rearrangement of the series (adding the even indexed terms together and then the odd indexed terms separately) and this series doesn't converge absolutely so I'm worried that this isn't rigorously justified.

Title: Help evaluating an interesting series.

Body:
$$  \left( 1+x^2 - x^4 - x^6 \right)  \left(1 + x^8 + x^{16} + x^{24}  + x^{32} + x^{40} + x^{48} \cdots  \right) $$
$$   \frac{1+x^2-x^4-x^6}{1-x^8} =   \frac{(1-x^4)(1+x^2)}{(1-x^4)(1 + x^4)} = \frac{1+x^2}{1 + x^4}$$
which can be integrated by partial fractions; note that
$$ x^4 + 1 = \left(x^2 - x\sqrt 2 + 1 \right)  \left(x^2 + x \sqrt 2 + 1 \right) $$

We are looking for
$$  \int \frac{1}{1+(x \sqrt 2 + 1)^2}  + \frac{1}{1+(x \sqrt 2 - 1)^2} dx $$


Body:
If $S=0.01 \cdot  \sum\limits^{n}_{t=1}t \cdot (0.99)^{t-1}=0.01 \cdot  \sum\limits^{n-1}_{t=0}(t+1) \cdot (0.99)^{t}$

then $0.99S=0.01 \cdot \sum\limits^{n}_{t=1}t \cdot (0.99)^{t} $

so by subtraction $0.01S = 0.01 + 0.01 \cdot \sum\limits^{n-1}_{t=1}  (0.99)^{t} - 0.01 \cdot n \cdot (0.99)^{n}$

i.e. $S = 1  -  n \cdot (0.99)^{n} +  \sum\limits^{n-1}_{t=1}  (0.99)^{t}$

and then $0.99S=0.99  - 0.99 \cdot  n \cdot (0.99)^{n} +  \sum\limits^{n}_{t=2}  (0.99)^{t}$

so by subtraction $0.01S = 0.01 - 0.01  \cdot n\cdot(0.99)^{n} +  0.99 -  (0.99)^{n}$

i.e. $S=1-n\cdot(0.99)^{n}+ 100\cdot  0.99 -100\cdot  (0.99)^{n}$

or more tidily:  $$S= 100 -(100+n) \cdot (0.99)^{n}$$


问:
Extracting distinct sequence
Let $(x_n)$ be a non-constant sequence in $\mathbb{R}$ and $x_n\rightarrow p$ for some $p\in \mathbb{R}$. Can I always extract a subsequence $(x_{n_k})$ whose all elements are distinct? (of course $x_{n_k}\rightarrow p$). Thanks in advance.

By non-constant I mean $\forall a\in\mathbb{R}, x_k=a$ for finitely many $k\in \mathbb{N}$

答: Yes, if by non-constant you really mean that no tail of the sequence is constant, i.e., that it’s not eventually constant. If you could not do so, there would be some finite $F\subseteq\Bbb R$ such that $x_n\in F$ for each $n\in\Bbb N$. But then it must be the case that $p\in F$ and the sequence is eventually constant at $p$.

Title: Extracting distinct sequence


问:
Primes of the form $p=X^2+3Y^2$
I'm trying to work in $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ to show that an odd prime $p\in\mathbb{Z}$, $p\neq 3$ is of the form $p=X^2+3Y^2$ if and only if $p\equiv1$ mod $3$. The hint is to first show that if $p=1$ mod $3$, then both $p=u^2+3v^2$ and $4p=u^2+3v^2$ where this time, $u$ and $v$ are odd.

Using quadratic reciprocity, I showed that $-3$ is a square mod $p$, which shows that $p|x^2+3$ and thus $p$ isn't prime in $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$, which in turn gives me that $p=X^2+3Y^2$ using the usual proof. But I dont understand where the $4p=u^2+3v^2$ comes from. Obviously this looks like elements of $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ after dividing by $4$ but maybe I overlooked something simple?

答:
HINT.- If $p\ne3M+1$ then $p=3M-1$. But then
$$3M-1=X^2+3Y^2\Rightarrow X^2\equiv-1\pmod3$$ which is impossible because $\left(\mathbb Z/3\mathbb Z\right)^2= \{0,1\}$.

COMMENT.-Maybe this could help @Saegusa in his inquiring on $\mathbb Q(\sqrt{-3})$ mainly in $4p=u^2+3v^2$. This problem was conjectured by Fermat and proved by Lagrange and Legendre by other method.

The discriminant of the quadratic form $ax^2+bxy+cy^2$ is the number $\Delta=b^2-4ac$

Note that the discriminant of the form $x^2+3y^2$ is $-12$.

In order to solve this problem, using quadratic reciprocity, prove first that this discriminant $-12$ is a square modulo the prime $p=3m+1$.

In fact $$\left(\frac{-12}{p}\right)=\left(\frac{4}{p}\right)\left(\frac{3}{p}\right)\left(\frac{-1}{p}\right)=\left(\frac{3}{p}\right)\left(\frac{-1}{p}\right)$$ As we have $$\left(\frac{p}{3}\right)\left(\frac{3}{p}\right)=(-1)^{\dfrac{(p-1)(3-1)}{4}}=(-1)^{\dfrac{p-1}{2}}$$ multiplingt both sides by $\left(\frac{p}{3}\right)$ we get

$$\left(\frac{-12}{p}\right)=\left(\frac{p}{3}\right)(-1)^{\dfrac{p-1}{2}}(-1)^{\dfrac{p-1}{2}}=1$$  Then the equation $x^2=-12$ has solution modulo the prime $p=3m+1$.

► If $x$ is even put $Y=x$ so $Y^2\equiv{-12}\pmod p$ and $4\space|\space Y^2$

►If $x$ is odd put $Y=p-x$ so $Y^2\equiv{-12}\pmod p$ and again $4\space|\space Y^2$

Since $Y^2+12\equiv 0\pmod p$ i.e. $p\space| Y^2+12$ so $4p\space| Y^2+12$ because $4\space|\space Y^2$.

Definition.-A quadratic form $ax^2+bxy+cy^2$ with discriminant $\Delta\lt0$ is called reduced if the two following conditions hold:

$(1)|b|\le a\le c\\(2)\text{if either }  |b|=a\space \text{or }a=c \space\text{then } b\ge0$.

PROPOSITION:- The only primitive reduced quadratic form having discriminant $-12$ is the form $x^2+3Y^2$.

The end is given by considering the form $p=px^2+bxy+cy^2$ of an evident solution $(x,y)=(1,0)$ and using Lagrange's algorith of reduccion of forms choose $b$ and $c$ carefully to reduce to the form $p=x^2+3y^2$.

Title: Primes of the form $p=X^2+3Y^2$


问:
Time period of a nonuniform oscillator
I am given the equation of a nonuniform oscillator as
$$\dot\theta=\omega-a\sin(\theta)\tag{1}\label{eq1},$$
where $a<\omega$, and I'm told that the period, $T$, of this is given by
$$T=\frac{2\pi}{\sqrt{\omega^2-a^2}}\tag{2}\label{eq2}.$$
To work this out, I first separate the variables for $\eqref{eq1}$ to get
$$\frac{1}{\omega-a\sin(\theta)}d\theta=dt.$$
I know that for the time to increase from $0$ to $T$, then the oscillator has gone from $0$ to $2\pi$. Therefore
$$\int_0^{2\pi}\frac{1}{\omega-a\sin(\theta)}d\theta=\int_0^Tdt$$
and so
$$T=\int_0^{2\pi}\frac{1}{\omega-a\sin(\theta)}d\theta.$$
This is where the fun begins. I use the Weierstrass substitution of $u=\tan\left(\frac{\theta}{2}\right)$ to evaluate this which changes my integral to
$$T=\int_{\theta=0}^{\theta=2\pi}\frac{2}{\omega u^2+\omega-2ua}du.$$
I take the factor of $2$ outside of the integral and complete the square of the quadratic in the denominator to arrive at
$$T=2\int_{\theta=0}^{\theta=2\pi}\frac{1}{\omega\left(\left(u-\frac{a}{\omega}\right)^2+1-\frac{a^2}{\omega^2}\right)}du.$$
I factor out $\frac{1}{\omega}$ from the integral, and use the substitutions $x=u-\frac{a}{\omega}$ and $y=1-\frac{a^2}{\omega^2}$ to make this
$$T=\frac{2}{\omega}\int_{\theta=0}^{\theta=2\pi}\frac{1}{x^2+y}dx.$$
I then use the substitution $x=\sqrt{y}\tan(v)$, meaning $x^2=y\tan^2(v)$ and $dx=\sqrt{y}\sec^2(v)dv$. This makes my integral
$$T=\frac{2}{\omega}\int_{\theta=0}^{\theta=2\pi}\frac{\sqrt{y}\sec^2(v)}{y\tan^2(v)+y}dv.$$
I take out a factor of $\frac{\sqrt{y}}{y}$ and find the following:
$$T=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}\frac{\sec^2(v)}{\tan^2(v)+1}dv=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}\frac{\sec^2(v)}{\sec^2(v)}dv=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}1 dv=\frac{2\sqrt{y}}{\omega y}\left[v\right]^{\theta=2\pi}_{\theta=0}.$$
Then I start the process of re-inserting my many substitutions. First $v=\tan^{-1}\left(\frac{x}{\sqrt{y}}\right)$, then $x=u-\frac{a}{\omega}$, then $u=\tan\left(\frac{\theta}{2}\right)$, and finally $y=1-\frac{a^2}{\omega^"}$. This gets me
$$T=\frac{2\sqrt{1-\frac{a^2}{\omega^2}}}{\omega\left(1-\frac{a^2}{\omega^2}\right)}\left[\tan^{-1}\left(\frac{\tan\left(\frac{\theta}{2}\right)-\frac{a}{\omega}}{\sqrt{1-\frac{a^2}{\omega^2}}}\right)\right]^{2\pi}_{0}.$$
I can simplify this slightly. Firstly,
$$\frac{2\sqrt{1-\frac{a^2}{\omega^2}}}{\omega\left(1-\frac{a^2}{\omega^2}\right)}=\frac{2}{\omega\sqrt{1-\frac{a^2}{\omega^2}}}=\frac{2}{\omega\sqrt{\frac{\omega^2-a^2}{\omega^2}}}=\frac{2}{\sqrt{\omega^2-a^2}},$$
and secondly
$$\tan^{-1}\left(\frac{\tan\left(\frac{\theta}{2}\right)-\frac{a}{\omega}}{\sqrt{1-\frac{a^2}{\omega^2}}}\right)=\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right).$$
So now I have
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left[\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)\right]^{2\pi}_{0}.$$
I feel like I'm so very close to $\eqref{eq2}$ now, but this is where I become stuck. I know that $\tan\left(\frac{2\pi}{2}\right)=\tan\left(\frac{0}{2}\right)=0$, which means when I evaluate the square brackets between $0$ and $2\pi$ they just cancel out and I'm left with
$$T=0.$$

Could someone please tell me where I have gone wrong, or what I am missing here? Been stuck on this one all day to no avail!

答:
Note that
$$
T=
\int_0^{2\pi}\frac{2e^{-i\theta}}{2i\omega-a(e^{i\theta}-e^{-i\theta}) }ie^{i\theta}\,d\theta=
\int_0^{2\pi}\frac{2}{2i\omega e^{i\theta}-a(e^{2i\theta}-1) }ie^{i\theta}\,d\theta=\int_\gamma f,
$$
with $\gamma\colon[0,2\pi]\to\mathbb C$ given by $\gamma(\theta)=e^{i\theta}$ and $f(z)=2/(-az^2+2i\omega z+a)$. By the Residues' theorem:
$$
T=2\pi i \,{\rm Res}\left(f,\frac{i(\omega+\sqrt{\omega^2-a^2})}{a}\right)=2\pi i\frac{2}{-a}\cdot\frac1{-\frac{2i\sqrt{\omega^2-a^2}}{a}}= \frac{2\pi}{{\sqrt{\omega^2-a^2}}}
$$
since the poles satisfy $|i(\omega+\sqrt{\omega^2-a^2})/a|<1$ and $|i(\omega-\sqrt{\omega^2-a^2})/a|>1$.

Title: Time period of a nonuniform oscillator


问:
Is $\lvert \int_0^t e^{isB_s} ds \rvert^2 \leq \left( \int_0^t \lvert e^{isB_s}\rvert ds\right)^2 = \left( \int_0^t 1 ds \right)^2 = t^2$ correct?
Assuming $B$ is a standard Brownian motion and taking the Riemann integral as below, where $i = \sqrt{-1}$, is there anything wrong with this inequality?

$\lvert \int_0^t e^{isB_s} ds \rvert^2 \leq \left( \int_0^t \lvert e^{isB_s}\rvert ds\right)^2 = \left( \int_0^t 1 ds \right)^2 = t^2$.

I'm just confused, since $B_s$ is a random variable, but does that matter? Assuming the first ''triangle inequality'' holds: $e^{isB_s}$ is a random variable, but it's still a complex exponential, whose modulus is equal to $1$, or? I mean, for any $\omega$, $\lvert e^{isB_s(\omega)} \rvert^2 = 1$, right?

答: Yes, this is perfectly fine.  All you're using is that, for each $\omega$ in  the sample space, $B_s(\omega)$ is a real-valued measurable function of $s$.


问:
Factoring Quadratic Formula Equation
I have a Quadratic Equation that I input into the Quadratic Formula to solve for y.

$$2xy^{2}+8xy-z=0$$

$$y=\frac{-8x\pm \sqrt{(8x)^2 - 4(2x)(-z)}}{2(2x)}$$

I've been told factoring will take me to:

$$y=\sqrt{\frac{{z}}{2x}+4} -2$$

I'm not sure how they got from point A to point B.

Any help?

答:
The numerator and denominator have been divided by $4x$, which becomes $16x^2$ inside the square root. Since $4x$ is the denominator, it disappears and we are left with
$$y=-8x/4x\pm\sqrt{64x^2/16x^2+8xz/16x^2}=\sqrt{z/2x+4}-2$$
Clearly this only works if $x\ne0$.

Title: Factoring Quadratic Formula Equation


问:
Question about notation used for planar vector fields and differential equations
I have learned in my calculus courses that a planar vector field is just a continuous (or smooth) map $\mathbb{R}^2 \to \mathbb{R}^2$, say $V(x,y) = (f(x,y),g(x,y))$ where $f$ and $g$ are continuous (smooth) maps from $\mathbb{R}^2$ to $\mathbb{R}$.

But I've seen an example, in this wiki page on Hilbert's $16^{\text{th}}$ problem, where a planar vector field is written just as a system of $2$ ODEs, in this case

$$ \frac{dx}{dt}=P(x,y), \frac{dy}{dt} = Q(x,y) $$

for polynomials $P,Q$ in two variables.

Could one equivalently write the above vector field as $V(x,y)=(P(x,y),Q(x,y))$? Why is it written with ODEs instead of the other way? What differences are there between these two ways of constructing vector fields?

答:
Yes, they are equivalent. It is just two different ways to look at the same thing.

Using your notation $V=(P,Q)$, you have an ODE in $\mathbb{R}^2$:
$$
\frac{d\vec x}{dt}=V(\vec x)
$$
where $\vec x:=(x,y)$.

One can write an (evolutionary) equation in higher dimensional space component-wise and have a system of equations.

There are lots of such examples. For instance, on this page the system of Navier-Stokes equations is written in the vector form while the official problem description in the Clay Mathematics Institute writes it component-wise.

This is similar to how you write a linear equation. For example,
$$
\begin{pmatrix}
1&2\\
3&4
\end{pmatrix}
\begin{pmatrix}
x\\
y
\end{pmatrix}
=
\begin{pmatrix}
3\\
7
\end{pmatrix}
$$
is equivalent to
$$
x+2y=3,\quad 3x+4y=7.
$$


问:
Which of the following statements is/are true regarding Chi square,T-distribution,F-distribution?
Let $X_1,X_2,X_3$ be i.i.d $N(0,1)$ random variables. Then which of the following statements is/are TRUE?

$(A) \ \ \dfrac{\sqrt2X_1}{\sqrt{X_1^2+X_2^2}}\sim t_2$

$(B) \ \ \dfrac{\sqrt2X_1}{|X_2+X_3|}\sim t_1$

$(C) \ \ \dfrac{(X_1-X_2)^2}{(X_1+X_2)^2} \sim F_{1,1}$

$(D) \ \ \sum_{i=1}^{3}X_i^2 \sim \chi_{2}^2 $

$(A)$ is True because $X_1\sim N(0,1)$ and $Y=X_1^2\sim \chi_1^2$ and we know $\dfrac{\sqrt{1}X_1}{\sqrt{Y}}\sim t_{1}$

$X_1^2+X_2^2\sim\chi_2^2$

so it does follows $t_2$

$(C) $ is TRUE because $(X_1-X_2)$ and $(X_1+X_2)\sim N(0,2)$ and squaring and dividing by $\sqrt{2}$ gives standard normal. So ratio of two chi-square variates divided by their degrees of freedom gives F distribution with digrees of freedom $m$ and $n$ both being $1$ here.

$(D)$ is not TRUE.It should be $\chi_3^2$

$(B)$ I am not sure how to proceed in B's denominator. I know $|x|=\sqrt{x^2}$ but how can I utilize it here?

答:
The previous version of my answer was incorrect.

The reason why (B) is $t$-distributed is because $(X_2 + X_3)/\sqrt{2}$ is a standard normal variable; call this variable $Y$.  Thus $$\frac{|X_2 + X_3|}{\sqrt{2}} = \sqrt{Y^2}$$ is the square root of a chi-squared random variable with one degree of freedom; consequently (B) is $t$-distributed with one degree of freedom.


问:
Integration defined by Riemann–Darboux, or Lebesgue approaches, or more?
Integration defined as a sum can be defined in different ways:

(Top) The Riemann–Darboux approach --
Divide the base of the mountain into a grid of 1 meter squares. Measure the altitude of the mountain at the center of each square. The volume on a single grid square is approximately 1 m2 × (that square's altitude), so the total volume is 1 m2 times the sum of the altitudes.

(Bottom) The Lebesgue approach --
Draw a contour map of the mountain, where adjacent contours are 1 meter of altitude apart. The volume of earth a single contour contains is approximately 1 m × (that contour's area), so the total volume is the sum of these areas times 1 m.

The difference between the Riemann and Lebesgue approaches thus: "to compute the Riemann integral of f, one partitions the domain [a, b] into subintervals", while in the Lebesgue integral, "one is in effect partitioning the range of f ."

Questions

A. Can you give examples that

an integration is well-defined via Riemann–Darboux, but not Lebesgue approaches?

an integration is not defined via Riemann–Darboux, but well-defined via Lebesgue approaches?

an integration is not defined via Riemann–Darboux nor Lebesgue approaches, but can still be defined via other approaches?

B. Are there more other novels ways to evaluate or define the integration, differed from  1. Riemann–Darboux and 2. Lebesgue approaches?

Source: https://en.wikipedia.org/wiki/Lebesgue_integration#Intuitive_interpretation

答:
Theorem. Any function $f:[a,b]\to\mathbb R$ that is Riemann integrable is also Lesbesgue integrable.

However, for improper integrals, it is possible that something is Riemann integrable but not Lesbesgue integrable. The classic example is $$\int_1^\infty\frac{\sin x}{x}\mathop{}\!\mathrm{d}x.$$

The classic example for a function that is Lesbesgue integrable but not Riemann integrable is
$$f(x)=\mathbf1(x\in\mathbb Q)=\begin{cases}1&\text{if }x\in\mathbb Q\\0&\text{if }x\not\in\mathbb Q.\end{cases}$$
In any interval, $\sup f=1$ and $\inf f=0$, so the function is not Riemann integrable. However, it is Lesbesgue integrable (say over $[0,1]$) because $f\neq0$ only on a set of measure $0$, so $\int f\mathop{}\!\mathrm{d}\mu=0$.

As for other types of integrals, there's the Riemann–Stieltjes integral, which is like a half-way point between the Riemann integral and the Lesbesgue integral. To integrate $f$ over $[a,b]$ with respect to $g$, you take a partition $P=\{a=x_0<x_1<\dots<x_n=b\}$ of $[a,b]$ as with the Riemann integral, but you define your upper sum as
$$U(P,f,g)=\sum_{i=1}^n\left(\sup_{[x_{i-1},x_i]}f(x)\right)\cdot\left(g(x_i)-g(x_{i-1})\right),$$
and define the lower sum similarly. If these agree, then the Riemann-Stieltjes integral $\int_a^b f\mathop{}\!\mathrm{d}g$ is defined.

So the Riemann integral is the special case of the Riemann-Stieltjes integral with $g(x)=x$. In general, we don't require $g$ to be continuous -- usually we take $g$ to be monotone/of bounded variation though.

This is very useful in probability, as we can define the expectation of a (real) random variable $X$ (with distribution function $F$) to be
$$\mathbb E[X]=\int_{-\infty}^\infty x\mathop{}\!\mathrm{d}F(x).$$
This lets us calculate the expectation of more random variables -- i.e. those without densities.

As a specific example of a function that is Riemann-Stieltjes integrable, but not Riemann-Darboux integrable, try $f(x)=\mathbf1(\mathbb Q\cap[0,1/2])$. Then $\int_0^1f\mathop{}\!\mathrm{d}g$ exists for any $g$ that is constant on $[0,1/2]$.

Finally, there exists something called an Ito integral, which generalises the Riemann-Stieltjes integral. Here, we set the $f$ and $g$ from the Riemann-Stieltjes integral to be stochastic processes. This e.g. lets us integrate over Brownian motions $B_t$ (which are not of bounded variation, so we can't take the Riemann-Stieltjes integral wrt $B_t$).


问:
Finding the volume of a rotating cross section
I have to find the volume of a bounded region as it rotates around the Y-axis. My points are y=6; x=-2; x=2; and y=0. What would be the steps to solving this and the solution? Thank you!</p>

答: A rectangle formed by vertices $(-2,0), (2,0), (2,6) \text{ and } (-2,6)$ is symmetrically positioned around the y-axis, and when rotated, gives a cylinder. Simply plug in $r=2-(-2)=4$ and $h=6-0=6$ in $V=\pi r h^2 $ and this gives the answer $96\pi$ sq. units.


问:
Infinite binary having infinite decimal
Can a finite decimal has an infinite binary representation? I have come to a conclusion that it may not be possible based on what I have read from the following:

What cannot happen is that the decimal is infinite and the binary is finite.

https://www.quora.com/Is-it-true-that-some-binary-numbers-have-infinite-digits-after-the-point-or-will-they-ever-end-for-example-decimal-0-1-to-binary

Comparing infinite binary fractions to infinite decimal fractions

number with finite binary representation and infinite decimal representation

Am I right or wrong. Any reasons or proofs to facilitate the comprehension will be helpful

答: $\frac13$ – and more generally any unit fraction $\frac1n$ where $n$ contains some prime factor other than $2$ or $5$ – is non-terminating in both binary and decimal.

Title: Infinite binary having infinite decimal


问:
Taylor series with remainder term for $\ln(1+x)$ about 1 with error terms less than $2^{-10}$
I was working on this problem and I think I was able to get the correct Taylor series representation but I'm struggling with the number of error terms. I need an inequality that gives the number of terms that must be taken in order to yield ln(4)with error terms less than $2^{-10}$. For the series I got $$f(x) = \ln(2) + \frac{x-1}{2} + \sum_{n=2}^{\infty} \frac{-1^{n+1}(n-1)!}{2^n}(x-1)^n.$$ For the remainder inequality I'm getting $\frac{3^{n+1}}{(n+1)!} < \frac{1}{2^{-10}}$ but when plugging this into a calculator it states that $n < -1$ which clearly doesn't make sense. Can someone help with the error term ?

答:
Well, your formula for the infinite series is certainly wrong, ’cause as written it’s nowhere convergent. Look again. But I’d do the thing entirely differently.

Using $\log(1+x)=-\sum_1^\infty(-x)^n/n$ and $\log(1-x)=-\sum_1^\infty x^n/n$, you get
$$
\log\left(\frac{1+x}{1-x}\right)=2\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}=2\left(x+\frac{x^3}3+\frac{x^5}5+\cdots\right)\,.
$$
Now of course this isn’t an alternating series, which makes it a little harder to estimate the error when cutting off at a finite number of terms; but since $x=\frac35$ will give you the logarithm of $4$, it looks to me as if you’d only need to go to about $x^{15}/15$, thus eight terms. I leave the details to you.


问:
Square roots -- positive and negative
It is perhaps a bit embarrassing that while doing higher-level math, I have forgot some more fundamental concepts. I would like to ask whether the square root of a number includes both the positive and the negative square roots.

I know that for an equation $x^2 = 9$, the solution is $x = \pm 3$. But if simply given $\sqrt{9}$, does one assume that to mean only the positive root? And when simply talking about the square root of a number in general, would one be referring to both roots or just the positive one, when neither is specified?

答: The positive value is taken because it's more valuable in general, since the domains of elementary functions which aren't defined on the entire real line are often the positive or non-negative numbers. For instance, taking the positive value lets you say $\sqrt{\sqrt{x}} = \sqrt[4]{x}$ or $\ln{\sqrt{x}} = \frac{1}{2} \ln{x}$.

ContentLicense: CC BY-SA 2.5

Title: Square roots -- positive and negative


问:
Effect of transformation on a point.
What is the effect of the transformation $$A(x,y)=(0.6x-0.8y, 0.8x+0.6y)\quad ?$$

I understand how transformations work when the variables are not in the same coordinate. I do not understand how I could predict the effect.

答:
The matrix for this transformation is
$$
\begin{pmatrix}
0.6&-0.8\\
0.8&0.6
\end{pmatrix}
$$
This is an example of rotations.

The effect of your transformation is rotating a point around the origin by an angle $\theta$ where $\cos\theta=0.6$.

Title: Effect of transformation on a point.


问:
Given a directional derivative to find a point
Let $ f(x,y,z)$ be differentiable, and assume that

$$ f(x, y, x^2 + y) = 3x - y $$

(for all values of $x,y$).

Also, given the direct-derivative of the point $A=(0, 12, 12)$ and the direction vector is $(1, 0, 1)$ is equal to $3$, I need to find the gradient of $f$ at $A$.

I have no idea how to even start, because $f$ is not given explicitly, only the derivative — but again, not explicitly for $f(x,y,z)$, but for this weird combination $f(x, y, x^2 + y)$.

What should I do?

答: Let $\Big(\nabla{f}\Big)(0,12,12)=\big$. Notice how $$3=\Big(\nabla{f}\Big)(0,12,12)\cdot \Bigg<\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\Bigg>=\frac{a+c}{\sqrt{2}}$$ We also have from chain rule $$f_{x}(x,y,x^2+y)+2xf_z(x,y,x^2+y)=3$$ $$f_{y}(x,y,x^2+y)+f_z(x,y,x^2+y)=-1$$ Plugging in $x=0,y=12$ into the above expressions produces the two equations $$a=3$$ $$b+c=-1$$ Solving this systems yields $a=3,b=2-3\sqrt{2},$ and $c=3\sqrt{2}-3$ i.e. $$\Big(\nabla{f}\Big)(0,12,12)=\Bigg<3,2-3\sqrt{2},3\sqrt{2}-3\Bigg>$$

Body: Let $\Big(\nabla{f}\Big)(0,12,12)=\big<a,b,c\big>$. Notice how $$3=\Big(\nabla{f}\Big)(0,12,12)\cdot \Bigg<\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\Bigg>=\frac{a+c}{\sqrt{2}}$$ We also have from chain rule $$f_{x}(x,y,x^2+y)+2xf_z(x,y,x^2+y)=3$$ $$f_{y}(x,y,x^2+y)+f_z(x,y,x^2+y)=-1$$ Plugging in $x=0,y=12$ into the above expressions produces the two equations $$a=3$$ $$b+c=-1$$ Solving this systems yields $a=3,b=2-3\sqrt{2},$ and $c=3\sqrt{2}-3$ i.e. $$\Big(\nabla{f}\Big)(0,12,12)=\Bigg<3,2-3\sqrt{2},3\sqrt{2}-3\Bigg>$$


答:
Using the chain rule:

$$\frac{ \partial}{ \partial x} f(x,y,x^2 + y) = f_x(x,y,x^2 + y) \cdot \frac{ \partial}{ \partial x} (x) + f_y (x,y, x^2 + y) \cdot \frac{ \partial }{ \partial x}(y) + f_z(x,y,x^2 + y) \cdot \frac{\partial }{\partial x} (x^2 + y) = f_x(x,y,x^2 + y) + 2x \cdot f_z(x,y,x^2 + y) = \frac{\partial}{ \partial x}(3x -y) = 3
f_x(x,y,x^2 +y) = 3- 2x f_z(x,y,x^2+y) $$

Notice that $A$ fits perfectly to the $x,y, x^2 + y$ pattern ($0^2 + 12 = 12 = A_z$) Thus:

$$ f_x(0,12,12) = 3- 2 \cdot 0 \cdot f_z(x,y,x^2+y) = 3$$

So we have one third of the gradient at the point $A$.

Using the chain rule again but now differentiating on $y$:

$$ f_y(x,y,x^2 + y) = f_x(x,y,x^2 + y) \cdot (x)_y + f_y(x,y,x^2 + y) \cdot (y)_y + f_z(x,y,x^2 + y) \cdot (x^2  +y)_y \\=  f_y(x, y,x^2 +y) + f_z(x,y,x^2 + y) = (3x -y)_y = -1 \\ \Rightarrow \color{red}{f_y(0,12,12) + f_z(0,12,12) = -1} $$

Now using the directional derivative:

First we need to normalize the direction vector, which is given by $$ \frac{1}{||(1,0,1)||} \cdot (1,0,1) = ( \frac{1}{\sqrt{2}} , 0, \frac{1}{\sqrt{2}})$$

Now using the fact that the function is differentiable, you don't need to actually calculate the derivative by definition, we can just use this known formula (which is equal to $3$ according to the question):

$$ \nabla f(0,12,12) \cdot ( \frac{1}{\sqrt{2}} , 0, \frac{1}{\sqrt{2}}) = 3$$

By the definition of the gradient at point $A$, it is equal to all the vector of all partial derivatives at that point:

$$ \nabla f(0,12,12) = \left ( f_x(0,12,12), f_y(0,12,12) , f_z(0,12,12) \right ) $$

Regular dot multiplication gives us:

$$ f_x(0,12,12) \cdot \frac{1}{\sqrt{2}} + 0 \cdot f_y(0,12,12) + f_z(0,12,12) \cdot \frac{1}{ \sqrt{2}} = 3$$

Now, we already know what is the value of $f_x(0,12,12)$ and it is equal to $3$. Substituting back gives us:

$$f_z(0,12,12) \cdot \frac{1}{\sqrt{2}}  = 3 - \frac{3}{\sqrt{2}} \\ f_z(0,12,12) = 3 \sqrt{2} - 3$$

Recall the red equation, we can substitute $f_z$ back and get:

$$ f_y(0,12,12) = -1 -f_z(0,12,12) = -1 -3 \sqrt{2} + 3 = 2 - 3 \sqrt{2} $$

Thus we can finally say that the answer is:

$$ \nabla f(0,12,12) = (3, 2-3\sqrt{2}, 3 \sqrt{2} -3 ) $$

And rest.

Title: Given a directional derivative to find a point

Body:
Let $ f(x,y,z)$ be differentiable, and assume that

$$ f(x, y, x^2 + y) = 3x - y $$

(for all values of $x,y$).

Also, given the direct-derivative of the point $A=(0, 12, 12)$ and the direction vector is $(1, 0, 1)$ is equal to $3$, I need to find the gradient of $f$ at $A$.

I have no idea how to even start, because $f$ is not given explicitly, only the derivative — but again, not explicitly for $f(x,y,z)$, but for this weird combination $f(x, y, x^2 + y)$.

What should I do?


答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
T & = \bbox[5px,#ffd]{\left.\int_{0}^{2\pi}{\dd\theta \over
\omega - a\sin\pars{\theta}}
\,\right\vert_{\, \omega\ >\ a\ \geq\ 0}} =
\int_{-\pi}^{\pi}{\dd\theta \over
\omega + a\sin\pars{\theta}}
\\[5mm] & =
\int_{0}^{\pi}\bracks{{1 \over \omega + a\sin\pars{\theta}} +
{1 \over \omega - a\sin\pars{\theta}}}\dd\theta
\\[5mm] & =
2\omega\int_{0}^{\pi}{\dd\theta \over
\omega^{2} - a^{2}\sin^{2}\pars{\theta}} =
2\omega\int_{-\pi/2}^{\pi/2}{\dd\theta \over
\omega^{2} - a^{2}\cos^{2}\pars{\theta}}
\\[5mm] & =
4\omega\int_{0}^{\pi/2}{\dd\theta \over
\omega^{2} - a^{2}\cos^{2}\pars{\theta}} =
4\omega\int_{0}^{\pi/2}{\sec^{2}\pars{\theta} \over
\omega^{2}\sec^{2}\pars{\theta} - a^{2}}\,\dd\theta
\\[5mm] & =
4\omega\int_{0}^{\pi/2}{\sec^{2}\pars{\theta} \over
\omega^{2}\tan^{2}\pars{\theta} + \omega^{2} -a^{2}}\,\dd\theta
\\[5mm] & =
4\omega\,{1 \over \omega^{2} - a^{2}}\,{\root{\omega^{2} - a^{2}} \over \verts{\omega}}
\ \times
\\[2mm] & \int_{0}^{\pi/2}{\verts{\omega}\sec^{2}\pars{\theta}/\root{\omega^{2} - a^{2}} \over
\bracks{\verts{\omega}\tan\pars{\theta}/\root{\omega^{2} - a^{2}}}^{2} + 1}\,\dd\theta
\end{align}

Sets $\ds{t = \verts{\omega}\tan\pars{\theta}/\root{\omega^{2} - a^{2}}}$:
\begin{align}
T & = \bbox[5px,#ffd]{\left.\int_{0}^{2\pi}{\dd\theta \over
\omega - a\sin\pars{\theta}}
\,\right\vert_{\, \omega\ >\ a\ \geq\ 0}} =
{4\on{sgn}\pars{\omega} \over \root{\omega^{2} - a^{2}}}\
\underbrace{\int_{0}^{\infty}{\dd t \over t^{2} + 1}}
_{\ds{\pi/2}}
\\ & =
\bbx{2\pi \over \root{\omega^{2} - a^{2}}} \\ &
\end{align}

Title: Time period of a nonuniform oscillator

Body:
I am given the equation of a nonuniform oscillator as
$$\dot\theta=\omega-a\sin(\theta)\tag{1}\label{eq1},$$
where $a<\omega$, and I'm told that the period, $T$, of this is given by
$$T=\frac{2\pi}{\sqrt{\omega^2-a^2}}\tag{2}\label{eq2}.$$
To work this out, I first separate the variables for $\eqref{eq1}$ to get
$$\frac{1}{\omega-a\sin(\theta)}d\theta=dt.$$
I know that for the time to increase from $0$ to $T$, then the oscillator has gone from $0$ to $2\pi$. Therefore
$$\int_0^{2\pi}\frac{1}{\omega-a\sin(\theta)}d\theta=\int_0^Tdt$$
and so
$$T=\int_0^{2\pi}\frac{1}{\omega-a\sin(\theta)}d\theta.$$
This is where the fun begins. I use the Weierstrass substitution of $u=\tan\left(\frac{\theta}{2}\right)$ to evaluate this which changes my integral to
$$T=\int_{\theta=0}^{\theta=2\pi}\frac{2}{\omega u^2+\omega-2ua}du.$$
I take the factor of $2$ outside of the integral and complete the square of the quadratic in the denominator to arrive at
$$T=2\int_{\theta=0}^{\theta=2\pi}\frac{1}{\omega\left(\left(u-\frac{a}{\omega}\right)^2+1-\frac{a^2}{\omega^2}\right)}du.$$
I factor out $\frac{1}{\omega}$ from the integral, and use the substitutions $x=u-\frac{a}{\omega}$ and $y=1-\frac{a^2}{\omega^2}$ to make this
$$T=\frac{2}{\omega}\int_{\theta=0}^{\theta=2\pi}\frac{1}{x^2+y}dx.$$
I then use the substitution $x=\sqrt{y}\tan(v)$, meaning $x^2=y\tan^2(v)$ and $dx=\sqrt{y}\sec^2(v)dv$. This makes my integral
$$T=\frac{2}{\omega}\int_{\theta=0}^{\theta=2\pi}\frac{\sqrt{y}\sec^2(v)}{y\tan^2(v)+y}dv.$$
I take out a factor of $\frac{\sqrt{y}}{y}$ and find the following:
$$T=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}\frac{\sec^2(v)}{\tan^2(v)+1}dv=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}\frac{\sec^2(v)}{\sec^2(v)}dv=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}1 dv=\frac{2\sqrt{y}}{\omega y}\left[v\right]^{\theta=2\pi}_{\theta=0}.$$
Then I start the process of re-inserting my many substitutions. First $v=\tan^{-1}\left(\frac{x}{\sqrt{y}}\right)$, then $x=u-\frac{a}{\omega}$, then $u=\tan\left(\frac{\theta}{2}\right)$, and finally $y=1-\frac{a^2}{\omega^"}$. This gets me
$$T=\frac{2\sqrt{1-\frac{a^2}{\omega^2}}}{\omega\left(1-\frac{a^2}{\omega^2}\right)}\left[\tan^{-1}\left(\frac{\tan\left(\frac{\theta}{2}\right)-\frac{a}{\omega}}{\sqrt{1-\frac{a^2}{\omega^2}}}\right)\right]^{2\pi}_{0}.$$
I can simplify this slightly. Firstly,
$$\frac{2\sqrt{1-\frac{a^2}{\omega^2}}}{\omega\left(1-\frac{a^2}{\omega^2}\right)}=\frac{2}{\omega\sqrt{1-\frac{a^2}{\omega^2}}}=\frac{2}{\omega\sqrt{\frac{\omega^2-a^2}{\omega^2}}}=\frac{2}{\sqrt{\omega^2-a^2}},$$
and secondly
$$\tan^{-1}\left(\frac{\tan\left(\frac{\theta}{2}\right)-\frac{a}{\omega}}{\sqrt{1-\frac{a^2}{\omega^2}}}\right)=\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right).$$
So now I have
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left[\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)\right]^{2\pi}_{0}.$$
I feel like I'm so very close to $\eqref{eq2}$ now, but this is where I become stuck. I know that $\tan\left(\frac{2\pi}{2}\right)=\tan\left(\frac{0}{2}\right)=0$, which means when I evaluate the square brackets between $0$ and $2\pi$ they just cancel out and I'm left with
$$T=0.$$

Could someone please tell me where I have gone wrong, or what I am missing here? Been stuck on this one all day to no avail!


问:
if $a$ and $b$ are multiplicative unique identities, show that $a = b$ using field properties
A1 For all $a, b \in R,$ we have $a + b = b + a.$ (commutativity of addition)

A2 For all $a, b, c \in R,$ we have $(a + b) + c = a + (b + c).$ (associativity of addition)

A3 There is a unique element 0 ∈ R such that, for all a ∈ R, a + 0 = 0 + a = a. The
element 0 is called the additive identity.

A4 For any a ∈ R, there exists an element −a ∈ R such that a+ (−a) = 0. The elements
−a is called the additive inverse of a.

M1 For all a, b ∈ R, we have a · b = b · a. (commutativity of addition)

M2 For all a, b, c ∈ R, we have (a · b) · c = a · (b · c). (associativity of multiplication)

M3 There is a unique element 1 ∈ R such that a · 1 = 1 · a = a for all a ∈ R. The element
1 is called the multiplicative identity.

M4 For any a ∈ R, a does not equal to 0, there exists an element a^−1 ∈ R such that aa−1 = 1. The
element a^−1
is called the multiplicative inverse of a.

AM1 For any a, b, c ∈ R, we have (a + b) · c = a · c + b · c. (distributivity)

答: If $a$ and $b$ are multiplicative identities, then $a=ab=b.$ The first equality holds since $b$ is an identity, the second because $a$ is.