问:
How to prove that an irrational number to an irrational number could be irrational. - SOLVED
There exists $a,b \in \mathbb{R}-\mathbb{Q}$ such that $a^b \in \mathbb{R}-\mathbb{Q}$.

I am lost as to where to begin. In my class, the only irrational numbers we've really covered is $\sqrt{2}$ and cubed root of two. Any help would be appreciated!

Edit : How my professor did it.

Proof: Consider $\sqrt{2}^{\sqrt{2}}\cdot\sqrt{2}^{1-\sqrt{2}} = \sqrt{2}$. Since a rational multiplied by an irrational is irrational, either $\sqrt{2}^{\sqrt{2}}$ is irrational or $\sqrt{2}^{1-\sqrt{2}}$ is irrational. We will proceed by cases.

Case 1: $\sqrt{2}^{\sqrt{2}}$ is irrational. Then $a=\sqrt{2},b=\sqrt2$.

Case 2: $\sqrt{2}^{1-\sqrt{2}}$ is irrational. Then $a=\sqrt2,b=1-\sqrt2$.

Thus there exists irrationals a,b where $a^b$ is irrational.

答: Here's the simplest way. $\sqrt2$ is irrational, as you know. $\log_23$ is irrational, since $\log_23=a/b$ is $2^a=3^b$ which contradicts the Unique Factorization Theorem. And $\sqrt2^{\log_23}=\sqrt3$, which can be shown to be irrational by the techniques that work for $\sqrt2$.


问:
Percent-wise decay in exponential function
I've been given a set of points, and I used Geogebra's FitExp tool to perform an exponential regression to find a suitable model for the decay in the values.

I ended up with the function $f(x) = 40308e^{-0.05x}$, which visually seems to fit pretty well.

$x$ is counted in years, which will become important in a moment.

Question

Given this function, what is the percentage decay per year?

My thoughts

I'm thinking, given $-0.05x$ in the exponent, that it decays by 5% each year, but I can't make a solid argument for it.

I've found $\frac{f(1)}{f(0)} = 0.95 = 1-0.05$ if that helps, but still, how do I firmly state AND reason why -5% is the rate of decay for this function?

Thanks in advance for any help!

答:
Another way to  think about it as by trying to conceptualize this in terms of a standard financial growth formula vs the general exponential form you have (I am signifying the difference between the r terms because they are not the same number):

$$y = C(r_f-1)^t$$

$$y = Ce^{rt}$$

These are identical. We know from exponent identities that:

$$(a^b)^c = a^{bc}$$

Therefore we can rewrite your general form as this:

$$y = C(e^r)^t$$

You can see now that

$$ e^r = 1 - r_f $$
$$ r_f = 1 - e^r$$

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Probability of match in three elements choosing from a group
One of my teachers asked all 26 ($t$) of the students in our class to randomly choose 5 ($k$) exercises from a website from a set of 20 ($n$). He then said that no two students in the class should have 3 ($s$) or more of the same exercises solved. I want to know the probability of two students in the class having 3 ($s$) of the same exercises solved. I know if $s$ was equal to $k$, it would be the same as the birthday problem with ($n$ choose $k$) days and a room of $t$, but what about the other case of $s<k$?

答:
Too long for a comment:

Looking at this again, I am now convinced that the probability of satisfying the teacher's conditions is extremely small, and might even be $0$.

A simulation of $10^5$ cases found no examples (the closest had $7$ pairs of students with $3$ or more matches) and the variance of the number of students with too many matches was smaller than the mean, so the distribution looks tighter than a Poisson distribution: the chart below gives shows the results from the simulation, plus a Poisson distribution with the same mean in shown as a grey line

```<code>pairshared <- function(t,k,n,s){
  pair <- cbind(rep(1:t, each=t), rep(1:t, times=t))
  pair <- pair[pair[,1] < pair[,2],]
  selections <- replicate(t, sample(n,k))
  shared <- numeric(nrow(pair))
  for (i in 1:nrow(pair)){
    shared[i] <- sum(selections[,pair[i,1]] %in% selections[,pair[i,2]])
    }
  table(factor(shared, levels = 0:k))
  }

set.seed(2021)
matdat <- replicate(10^5, pairshared(t=26, k=5, n=20, s=3))
pairstoomany <- colSums(matdat[(s+1):(k+1),])
mean(pairstoomany == 0)
# 0
min(pairstoomany)
# 7
mean(pairstoomany)
# 23.61912
var(pairstoomany)
# 21.82561

plot(table(pairstoomany))
minp2m <-  min(pairstoomany)
maxp2m <-  max(pairstoomany)
points(minp2m:maxp2m, 10^5*dpois(minp2m:maxp2m, mean(pairstoomany)),
       col="grey", type="l")
</code>```

Body:
Too long for a comment:

Looking at this again, I am now convinced that the probability of satisfying the teacher's conditions is extremely small, and might even be $0$.

A simulation of $10^5$ cases found no examples (the closest had $7$ pairs of students with $3$ or more matches) and the variance of the number of students with too many matches was smaller than the mean, so the distribution looks tighter than a Poisson distribution: the chart below gives shows the results from the simulation, plus a Poisson distribution with the same mean in shown as a grey line

pairshared <- function(t,k,n,s){
  pair <- cbind(rep(1:t, each=t), rep(1:t, times=t))
  pair <- pair[pair[,1] < pair[,2],]
  selections <- replicate(t, sample(n,k))
  shared <- numeric(nrow(pair))
  for (i in 1:nrow(pair)){
    shared[i] <- sum(selections[,pair[i,1]] %in% selections[,pair[i,2]])
    }
  table(factor(shared, levels = 0:k))
  }

set.seed(2021)
matdat <- replicate(10^5, pairshared(t=26, k=5, n=20, s=3))
pairstoomany <- colSums(matdat[(s+1):(k+1),])
mean(pairstoomany == 0)
# 0
min(pairstoomany)
# 7
mean(pairstoomany)
# 23.61912
var(pairstoomany)
# 21.82561

plot(table(pairstoomany))
minp2m <-  min(pairstoomany)
maxp2m <-  max(pairstoomany)
points(minp2m:maxp2m, 10^5*dpois(minp2m:maxp2m, mean(pairstoomany)),
       col="grey", type="l")


问:
Dot product of a tangent vector and the gradient
I need to evaluate the dot product of a tangent vector and the gradient of $f$ at $(0,3)$. Anyways I found the tangent vector at $(0,3)$ which is $\langle-3, 0\rangle$ and then I found the gradient of $f$ at $(0,3)$ which gave me an answer of $\langle-2, 2\rangle$. So am I just suppose to do $\langle-3, 0\rangle \cdot \langle-2, 2\rangle$ to get my final answer? I ask this because an answer of $6$ does not really make much sense.

答: Well, if everything is defined as you said, your answer is correct. Maybe I could help you more if you told why 6 doesn't make much sense</p>


答:
If you mean the tangent vector to the level curve/set at (0,3), then the tangent vector is perpendicular to the gradient. Thus, their dot product should be 0.

If you mean a tangent vector to the surface at (0,3), there are infinitely many tangent vectors that lie on the tangent plane, but the issue is they are of dimension 3 so you can't dot them with the gradient.

Title: Dot product of a tangent vector and the gradient

Body: I need to evaluate the dot product of a tangent vector and the gradient of $f$ at $(0,3)$. Anyways I found the tangent vector at $(0,3)$ which is $\langle-3, 0\rangle$ and then I found the gradient of $f$ at $(0,3)$ which gave me an answer of $\langle-2, 2\rangle$. So am I just suppose to do $\langle-3, 0\rangle \cdot \langle-2, 2\rangle$ to get my final answer? I ask this because an answer of $6$ does not really make much sense.


问:
Fundamental group of shrinking wedge of spheres.
Let $X$ be the subspace of $\mathbb R^3$ which is union of the spheres of radius $1/n$ and centered at $(1/n,0,0)$. Then $X$ is simply connected.

I had thought for it in this way to attach $2$-cells to a single point, namely the origin, but then I realized the space I will get is a wedge sum of spheres and not the space given in the question.

Please help to figure out the fundamental group of the space in question.

Not a homework problem

答:
Unfortunately there is a flaw in this solution. It is in the statement "To see that $F$ is also continuous when $t=1$, it suffices to prove that for any neighborhood $U$ of the origin $O$ in $X$, $f^{−1}(U)$ contains all but finitely many intervals of $(a_i,b_i)$."

The problem is that if the homotopies are chosen as described above, but otherwise arbitrary, continuity can fail. Example: suppose that the $(a_i,b_i)$ cluster at some point $s_0$. Suppose the initial loop is trivial. Suppose that the homotopy "shrinking" the small loop on $[a_i,b_i]$ sweeps over the sphere $S_1$ before collapsing back down to $O$. (Perverse, but not excluded by the description. N.B.: always the same sphere.) Then for $t$ arbitrarily close to 1 and points $s$ arbitrarily close to $s_0$, we have $F(s,t)$ always at least some fixed distance (say 1/2) from $O$.

To fix this, we need to exert some control on the homotopies. If the small loop on $(a_i,b_i)$ does not have the ''opposite pole'' (the point furthest from $O$) in its range, then shrink the loop by sliding along the meridians from the opposite pole to $O$. This creates a distance decreasing property for these homotopies. If the opposite pole is in the range, then any homotopy will work.

Now we use the continuity of the initial loop. It follows, as in the previous poster's argument, that the $F(\cdot,t_k)$ is continuous for $t_k=1-(1/k)$. Also for $t_k$ sufficiently close to 1, $F(\cdot,t_k)$ is as close as you like to $O$. We now have a dichotomy for loops yet to be shrunk. Either the support of the loop is in a ''small'' sphere, or else the opposite pole is not in the range. Either way, the homotopies will keep us close to $O$.


问:
Hessian matrix of $x^T Q x$ where $Q=Q^T$ and $Q$ is positive-definite
I'm interested in the Hessian matrix $\nabla^2f(x)$ of $f(x) = x^T Q x$ where $Q$ is symmetric and positive definite. I understand that $\nabla f(x) = x^T(Q+Q^T) = 2Qx$, but am not sure how to calculate the Hessian matrix without considering each partial directly.

Thank you!

答:
In the denominator layout (more commonly used), with $\nabla f(x)=2Qx$
$$\begin{split}\nabla^2 f(x) &=2Q^T\\ &=2Q \text { because Q is positive definite}\end{split}$$

That is, $\nabla_x (Ax)=A^T$ in general. The gradient of a matrix vector product wrt the vector is the transpose of the matrix.


问:
How to prove if $a_n$ and $b_n$ converge and $a_n,b_n \ge 0$ then $a_n b_n$ also conveges?.
If $a_n$ and $b_n$ are sequences of non-negative numbers and both converge, how do I argue that $\sum a_n b_n$ also converges?

My attempt:

I tried to argue that $a_n b_n\le(a_n+b_n)^2=a_n^2+2a_nb_n+b_n^2$, but I am stuck on how to proceed from here.. I have a feeling that I should use the comparison test, but I am not too sure.

答: The condition stated in your post isn't clear. You could have $a_n = b_n = \dfrac{1}{\sqrt{n}}\to 0$, but $\sum_{n=1}^\infty a_nb_n = \sum_{n=1}^\infty \dfrac{1}{n} = \infty$. Thus the said conditions about $a_n$ and $b_n$ should be:  $a_n$ converges, $b_n$ converges, and either $\sum_{n=1}^\infty a_n$ or $\sum_{n=1}^\infty b_n$ converges. We can proceed from this point. Let's say $\sum_{n=1}^\infty a_n$ converges. Since $b_n$ converges,it is a bounded sequence, i.e. $|b_n| \le K, \forall n \ge 1$. We have: $0 \le a_nb_n \le Ka_n\implies \sum_{n=1}^\infty a_nb_n \le K\sum_{n=1}^\infty a_n$. This shows the series in question converges by comparison test.


问:
Calculating radii (or difference of radii) of concentric spheres with spherical shell volume
I'd like to calculate the radii or the difference of the radii (thickness of spherical shell) of 2 concentric spheres where the only given value is the volume between the concentric spheres (spherical shell). The inner sphere has a constant volume (non-zero).

I know the volume of the spherical shell's equation would be like this:

$V = 4/3\pi(R^3-r^3)$

I thought this formula might be involved for the cubic differences of the radii

$(x^3-y^3) = (x-y)(x^2+xy+y^2)$

Which would lead to this:

$V = 4/3\pi(R-r)(R^2+Rr+r^2)$

I'm not sure where to go from here. I would also be happy to know if there are any alternative methods for the solution.

答:
Your question is underdetermined. For a given volume, we could simply have the outer sphere have that volume and the inner sphere have volume $0$. Or we could move the outer sphere's surface outward and move the inner sphere's surface outward a somewhat greater amount to keep the volume between them constant.

EDIT: For example, suppose the difference in volumes of the two spheres is $\frac{4\pi}{3}\cdot 16^3$.  This can be accomplished with radii $0$ and $4$, or with any inner radius you want, say $5$, you can calculate what the outer radius must be:$$\frac{4\pi}{3}R^3-\frac{4\pi}{3}5^3 = \frac{4\pi}{3}\cdot 16^3$$
Which you can solve for $R$. Replace $5$ and $16$ by any values you like and solve for the outer radius needed.


问:
Fenchel conjugate of $f(x)=\frac{1}{2}\|x-y\|^2$
Let $f(x)=\frac{1}{2}\|x-y\|^2$ and $\|\cdot\|$ be 2-norm.

$x,y \in R^n$

Let the fenchel conjugate be defined as $f(z)=\sup_x(z^Tx -\frac{1}{2}\|x-y\|^2)$.

Taking the derivative w.r.t. $x$ and setting it $0$:

$z - (x-y)=0$ and $x=z+y$

Then the fenchel conjugate is $f^*(z)=\frac{1}{2}\|z\|^2+z^Ty$.

Could somebody please check my calculations?

答:
Yes this is correct. There is a calculus of convex conjugates that allows you to compute the convex conjugate of a translated function so you could also derive the result as follows: let $h(x) = \frac{1}{2}\|x\|^2$. Then $f(x) = h(x-y)$ and we obtain using the translation rule: $$f^*(z) = h^*(z) + z^{\intercal}y$$
and it is an easy exercise to show that $h^*(z)= h(z)$.


问:
Showing a simple property in statistics
I know this is very elementary but I cannot remember how to show $\hat\alpha$ as below.

答:
You minimize the squared error
$$\epsilon^T\epsilon=(y-X\alpha)^T(y-X\alpha)=y^Ty-2\alpha^TX^Ty+\alpha^TX^TX\alpha$$
This expression can be minimized by setting its derivative w.r.t. $\alpha$ equal to zero:

$$-2X^Ty+2X^TX\alpha=0\Rightarrow \alpha=(X^TX)^{-1}X^Ty$$

Title: Showing a simple property in statistics


问:
How do I figure out maxima and minima after getting the condition relating partial derivatives using Lagrange multipliers?
The problem states: Use Lagrange multipliers to find the absolute minimum and maximum values of the function $f(x,y)=x^2y$ subject to the constraint $3x^2 + 2y^2 = 4$.

My attempt at a solution:

I am using Lagrange multipliers and obtain these equations. These are the equations.
$$f_x=\lambda g_x$$ $$f_y=\lambda g_y $$ $$g(x,y)=4$$ $$f_x=2xy$$ $$g_x=6x$$
$$f_y = x^2$$ $$g_y = 4y$$ $$g(x,y)=3x^2+2y^2$$

Plugging in the equations and coming up with a system of three equations and three unknowns yields.
$$2xy=\lambda 6x$$ $$x^2=\lambda 4y$$ $$3x^2+2y^2=4$$

I cannot figure out how to take it further to find the absolute maximum and minimums.

答:
The function $ \ f(x,y) \ = \ x^2y \ $ is symmetric about the $ \ y-$ axis, but has an anti-symmetry about the $ \ x-$ axis [ $ \ f(x,-y) \ = \ -f(x,y) \ $ ] and is positive for $ \ y \ > \ 0 \ $ . Since the constraint equation represents an ellipse centered on the origin, we should expect there to be four critical points placed symmetrically then, one in each quadrant. Because of the anti-symmetry of the function, the two maxima will occur in the first and second quadrant, while the third and fourth quadrants contain the two minima.  So this can be determined before any calculation is made.

The work by Buraian confirms the presence of the four critical points, in that $ \ x^2 \ = \ 12 \lambda^2 \ $ and $ \ y^2 \ = \ 9 \lambda^2 \ $ .  We have

$$ 54 \ \lambda^2 \ \ = \ \ 4  \ \ \Rightarrow \ \ \lambda^2 \ \ = \ \ \frac{2}{27} \ \ \Rightarrow \ \ x^2 \ \ = \ \ \frac{24}{27} \ \ = \ \ \frac{8}{9} \ \ , \ \ y^2 \ \ = \ \ \frac{2}{3} \ \ . $$

The absolute maxima  are thus located at $ \  \left( \ \pm \frac{2 \sqrt{2}}{3} \ , \ \ \sqrt{\frac{2}{3}} \ \right) $ with   $ \ f(x,y) \ = \ + \frac{8}{9} \sqrt{\frac{2}{3}} \ $  and the absolute minima are at $ \  \left( \ \pm \frac{2 \sqrt{2}}{3} \ , \ \ -\sqrt{\frac{2}{3}} \ \right) $ with   $ \ f(x,y) \ = \ - \frac{8}{9} \sqrt{\frac{2}{3}} \ \ . $


问:
Derivative of $f(x, g(x))$ with respect to $x$
Both $f$ and $g$ are $C^1$ functions and no other assumption. My attempt
\begin{align*}\frac{df}{dx} &= \frac{\partial f}{\partial x}\frac{dx}{dx} + \frac{\partial f}{\partial g}\frac{dg}{dx} \\
&= \frac{\partial f}{\partial x} +  \frac{\partial f}{\partial g}\frac{dg}{dx}\end{align*}
But this does not look correct, how can $\frac{df}{dx}$ is equal to itself plus some other term?

*note: I want to find $\frac{df}{dx}$ exactly which means treat $f$ as a function of $x$ only, not $\frac{\partial f}{\partial x}$.

答: That's not $\frac{df}{dx}$, that's $\frac{df(x,g(x))}{dx}$. Actually $\frac{df}{dx}$ makes no sense, since $f$ is a function of two variables.


答:
Your expression is correct.  The key is that "$d\neq \partial$".  That is to say, $\frac{d}{dx} f(x,g(x))$ means the derivative of the one-variable function $f(x,g(x))$, whereas $\frac{\partial}{\partial x} f$ means the derivative of the two-variable function $f(x,y)$ with respect to its first argument.

It can be helpful to introduce extra symbols for variables when doing these sorts of manipulations.  So e.g. let's write $f=f(u,v)$ when taking partial derivatives.  Then substituting $u=x$ and $v=g(x)$ we have $$\frac{d}{dx} f(x,g(x)) = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} = \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}\frac{\partial g}{\partial x}$$

Alternatively, sometimes you see the notation $D_1 f$ to indicate "partial derivative of f with respect to its first argument", and similarly for $D_2$. In this notation, you could write $$\frac{d}{dx} f(x,g(x)) = D_1 f(x,g(x)) + D_2 f(x,g(x)) g'(x)$$

Title: Derivative of $f(x, g(x))$ with respect to $x$

Body:
Both $f$ and $g$ are $C^1$ functions and no other assumption. My attempt
\begin{align*}\frac{df}{dx} &= \frac{\partial f}{\partial x}\frac{dx}{dx} + \frac{\partial f}{\partial g}\frac{dg}{dx} \\
&= \frac{\partial f}{\partial x} +  \frac{\partial f}{\partial g}\frac{dg}{dx}\end{align*}
But this does not look correct, how can $\frac{df}{dx}$ is equal to itself plus some other term?

*note: I want to find $\frac{df}{dx}$ exactly which means treat $f$ as a function of $x$ only, not $\frac{\partial f}{\partial x}$.


问:
Analysis and formal proofs.
Ever since I started learning formal logic I've had these kind of doubts: Is analysis ever studied in a completely axiomatic/formal proofy way?

What I mean is, given a set of axioms and inference rules, to prove things via formal proofs.

Example:

Let $f: \Bbb R\to \Bbb R: x\mapsto x$.

Theorem. $f(a)=0\iff a = 0$.

Proof:

Assume $f(a)=0$.
\begin{eqnarray*}
\forall x\, f(x)&=&x \quad \text{Definition of $f$.}\tag 1\\
f(a)&=&0\quad \text{Hypothesis.}\tag 2\\
f(a)&=&a\quad \text{Instantiation of 1.}\tag 3\\
a&=&0\quad \text{Substitution Axiom (Lines 2, 3).}\tag 4
\end{eqnarray*}
Thus we have that $f(a)=0\implies a=0$, the other arrow is pretty trivial, as this one was, but this was just an example, as is the "substitution axiom".

Is analysis studied or treated in this way in any book you know?

答:
On the larger question, rather than being all of one or the other, there is actually a gradation between theorem-provers and proof-checkers. In reality, when a pure or applied mathematician lays out a proof, they're actually only listing a series of conjectures or lemmas, each one comprising a "step" in the proof that is small enough to be "obvious without further elaboration". A proof-checker that meets the users of mathematics on this level is then functioning as a proof-checker only to connect the steps, but also as a theorem-prover to prove each "conjecture" or "lemma" - that is to "fill in the missing steps".

As to the issue of Automath, itself: rumors of its demise are exaggerated, and all "was"-s and other past tense should be changed to "is"-s. It's still here, and it's back.

Version 4.2 was written in C in the late 1990's; a reference to the page and software can be found under the "Automath" entry in the nCategory cafe: https://ncatlab.org/nlab/show/Automath

We're locally maintaining a version 4.3 that may be distributed soon, in the future, once we get everything cleared. It will include many of the enhancements and repairs whose current absence are called out below.

So, yes, the infrastructure for Automath actually serves quite well as a foundation for a tractable higher-order formalism powerful enough to embody real analysis. The deficiencies don't lie with the underlying type theory, but mostly with the language itself.

The formalization of real number theory in terms of Automath - the "Grundlagen" file - was heavily labor-intensive, largely because of the cumbersome features of the original versions of the language: particularly its inability to do more intelligent polymorphism, the absence of type inheritance, the lack of prefix/postfix/infix notation (like Prolog), the lack of support for symbolic language (that is, no UTF-8 or other UNICODE support) and the absence of any significant support for equational logic.

All operators in Automath are polymorphic, in a style similar to Hindley-Milner, but the polymorphism parameters are matched only by exact match to their declaration and all parameters are treated as polymorphism parameters! No distinction between Hindley-Milner parameter versus function argument is made.

If there are any parameters that mismatch then all the parameters from that point onward have to be explicitly specified. For instance, assuming we had infix notation available, if we were to define the composition as an operator of the type:
$$(f: B → C, g: A → B) ↦ f∘g: A → C$$
it would have to be declared as an operator taking 5 arguments, not 2!
$$f: B → C, g: A → B ↦ Comp(A,B,C, f,g): A → C,$$
when it is actually meant to be a polymorphic operator $Comp(f,g)$. You couldn't really exploit the advantages of infix notation for the operator.

When called, say on functions
$$a: B → D, b: A → B$$
you could omit the parameters $A$ and $B$, but would have to explicitly cite the parameter $D$ and write it as $Comp(D,a,b)$, instead of as just $a∘b$. Similarly, if it were called on functions
$$a: D → C, b: A → D$$
it would have to be called with both of the last two parameters explicitly cited, as $Comp(D,C,a,b)$, instead of as just $a∘b$.

Cases like this are replete in the Automath rendering of the Grundlagen book, and it cascades, as each layer is built on top of the one underneath it.

The digitized book also force-fit a boolean logic framework, instead of using the native facilities for logic already contained in a dependent types formalism. It was not a constructive or intuitionistic translation of the book, but was grounded in classical logic, with the logic built on the Automath formalism as an extra layer. It would be a much more difficult exercise to try and directly implement Grundlagen in terms of the facilities already natively-present in the Automath type theory or in any higher-order constructive or intuitionistic type theory formalism built on top of it.

Martin-Löf can be more effectively used, but it does not contain the double-negative rule, nor any of the watershed rules that distinguish classical logic from intuitionistic logic. So, that would be a difficult exercise.

As for the relation between the two: there is a widespread misunderstanding of how they are or should be connected. The usual account you will hear is that Automath's type system is a "flattened" version of those seen in contemporary dependent-types formalisms like Martin-Löf; because it uses the same abstraction/application for both objects and types, while Martin-Löf treats object abstractions as lambdas, and type abstractions as pi's - or dependent products.

This is actually the wrong way of looking at it. Automath was designed to be an "assembly language" layer to higher level languages (or "superposed languages" as deBruijn called them) like Martin-Löf. It actually represents a layer beneath Martin-Löf which - when made explicit - actually simplifies the presentation of Martin-Löf's formalism(!)

In Martin-Löf, the product type is defined in a somewhat cumbersome way by conditional rules. If $A$ is a type, and $B(x)$ is a type, for each $x ∈ A$, i.e. if one has a type function $B: x ∈ A → type$, then $(Πx∈A)B(x)$ is the product type, whose members are the lambda abstractions $λb = (λx∈A)b(x)$ of functions $x ∈ A ↦ b(x) ∈ B(x)$, which can be converted back to functions by the "application" operator $Ap(λb,a) = b(a)$.

Automath would actually represent the layer beneath all of this - the functions themselves. So, one would treat $A$ as a type (I'll use $∙$ to denote "type"), and $B$ as of $[x,A]∙$ (which is permitted in some versions of the Automath language, like versions 4.2 and 4.3, and is considered as a subtype of $∙$.) The $[x,A]$ is Automath's native abstraction operator - its version of $λx∈A$.

It's an abstraction operator underlying that of Martin-Löf and which is frequently referred to tacitly in Martin-Löf's presentation. The function application $b(a)$ native to Automath would then be represented as $b$ (or just $b$ in version 4.3), and the product type $(Πx∈A)B(x)$ would just be $ΠB$ - with the prefix operator $Π$ mapping an object $B$ of type $[x,A]∙$ to type $∙$. The $λ$ operator would be a map from a object $b$ of type $B$ to an object $λb$ of type $ΠB$; while the application operator $Ap(c,a)$ would map an object $c$ of type $ΠB$ and $a$ of type $A$ to type $Ap(c,a)$ of type $B$; i.e. it maps the object $c$ back to a function $[x,A]Ap(c,x)$ has type $B$.

So, it's a layer beneath the layer seen in higher-order type theories which can also be used to directly embody all the other operators seen in Martin-Löf or similar theories.

The actual weakness of Automath lies not in its infrastructure, but in its lack of direct support for equational logic. It has no easy way to write out the β-rule $Ap(λb,a)$ = $b$ or the η-rule, $λ([x,A]Ap(c,x))=c$ at the higher layer; or the other reductions rules seen in Martin-Löf. The versions of the β-rule and the η-rule in Automath correspond to the ones tacitly used at the deeper, informal, layer throughout Martin-Löf's presentation.

Title: Analysis and formal proofs.

Body:
On the larger question, rather than being all of one or the other, there is actually a gradation between theorem-provers and proof-checkers. In reality, when a pure or applied mathematician lays out a proof, they're actually only listing a series of conjectures or lemmas, each one comprising a "step" in the proof that is small enough to be "obvious without further elaboration". A proof-checker that meets the users of mathematics on this level is then functioning as a proof-checker only to connect the steps, but also as a theorem-prover to prove each "conjecture" or "lemma" - that is to "fill in the missing steps".

As to the issue of Automath, itself: rumors of its demise are exaggerated, and all "was"-s and other past tense should be changed to "is"-s. It's still here, and it's back.

Version 4.2 was written in C in the late 1990's; a reference to the page and software can be found under the "Automath" entry in the nCategory cafe: https://ncatlab.org/nlab/show/Automath

We're locally maintaining a version 4.3 that may be distributed soon, in the future, once we get everything cleared. It will include many of the enhancements and repairs whose current absence are called out below.

So, yes, the infrastructure for Automath actually serves quite well as a foundation for a tractable higher-order formalism powerful enough to embody real analysis. The deficiencies don't lie with the underlying type theory, but mostly with the language itself.

The formalization of real number theory in terms of Automath - the "Grundlagen" file - was heavily labor-intensive, largely because of the cumbersome features of the original versions of the language: particularly its inability to do more intelligent polymorphism, the absence of type inheritance, the lack of prefix/postfix/infix notation (like Prolog), the lack of support for symbolic language (that is, no UTF-8 or other UNICODE support) and the absence of any significant support for equational logic.

All operators in Automath are polymorphic, in a style similar to Hindley-Milner, but the polymorphism parameters are matched only by exact match to their declaration and all parameters are treated as polymorphism parameters! No distinction between Hindley-Milner parameter versus function argument is made.

If there are any parameters that mismatch then all the parameters from that point onward have to be explicitly specified. For instance, assuming we had infix notation available, if we were to define the composition as an operator of the type:
$$(f: B → C, g: A → B) ↦ f∘g: A → C$$
it would have to be declared as an operator taking 5 arguments, not 2!
$$f: B → C, g: A → B ↦ Comp(A,B,C, f,g): A → C,$$
when it is actually meant to be a polymorphic operator $Comp<A,B,C>(f,g)$. You couldn't really exploit the advantages of infix notation for the operator.

When called, say on functions
$$a: B → D, b: A → B$$
you could omit the parameters $A$ and $B$, but would have to explicitly cite the parameter $D$ and write it as $Comp(D,a,b)$, instead of as just $a∘b$. Similarly, if it were called on functions
$$a: D → C, b: A → D$$
it would have to be called with both of the last two parameters explicitly cited, as $Comp(D,C,a,b)$, instead of as just $a∘b$.

Cases like this are replete in the Automath rendering of the Grundlagen book, and it cascades, as each layer is built on top of the one underneath it.

The digitized book also force-fit a boolean logic framework, instead of using the native facilities for logic already contained in a dependent types formalism. It was not a constructive or intuitionistic translation of the book, but was grounded in classical logic, with the logic built on the Automath formalism as an extra layer. It would be a much more difficult exercise to try and directly implement Grundlagen in terms of the facilities already natively-present in the Automath type theory or in any higher-order constructive or intuitionistic type theory formalism built on top of it.

Martin-Löf can be more effectively used, but it does not contain the double-negative rule, nor any of the watershed rules that distinguish classical logic from intuitionistic logic. So, that would be a difficult exercise.

As for the relation between the two: there is a widespread misunderstanding of how they are or should be connected. The usual account you will hear is that Automath's type system is a "flattened" version of those seen in contemporary dependent-types formalisms like Martin-Löf; because it uses the same abstraction/application for both objects and types, while Martin-Löf treats object abstractions as lambdas, and type abstractions as pi's - or dependent products.

This is actually the wrong way of looking at it. Automath was designed to be an "assembly language" layer to higher level languages (or "superposed languages" as deBruijn called them) like Martin-Löf. It actually represents a layer beneath Martin-Löf which - when made explicit - actually simplifies the presentation of Martin-Löf's formalism(!)

In Martin-Löf, the product type is defined in a somewhat cumbersome way by conditional rules. If $A$ is a type, and $B(x)$ is a type, for each $x ∈ A$, i.e. if one has a type function $B: x ∈ A → type$, then $(Πx∈A)B(x)$ is the product type, whose members are the lambda abstractions $λb = (λx∈A)b(x)$ of functions $x ∈ A ↦ b(x) ∈ B(x)$, which can be converted back to functions by the "application" operator $Ap(λb,a) = b(a)$.

Automath would actually represent the layer beneath all of this - the functions themselves. So, one would treat $A$ as a type (I'll use $∙$ to denote "type"), and $B$ as of $[x,A]∙$ (which is permitted in some versions of the Automath language, like versions 4.2 and 4.3, and is considered as a subtype of $∙$.) The $[x,A]$ is Automath's native abstraction operator - its version of $λx∈A$.

It's an abstraction operator underlying that of Martin-Löf and which is frequently referred to tacitly in Martin-Löf's presentation. The function application $b(a)$ native to Automath would then be represented as $<a>b$ (or just $b<a>$ in version 4.3), and the product type $(Πx∈A)B(x)$ would just be $ΠB$ - with the prefix operator $Π$ mapping an object $B$ of type $[x,A]∙$ to type $∙$. The $λ$ operator would be a map from a object $b$ of type $B$ to an object $λb$ of type $ΠB$; while the application operator $Ap(c,a)$ would map an object $c$ of type $ΠB$ and $a$ of type $A$ to type $Ap(c,a)$ of type $B<a>$; i.e. it maps the object $c$ back to a function $[x,A]Ap(c,x)$ has type $B$.

So, it's a layer beneath the layer seen in higher-order type theories which can also be used to directly embody all the other operators seen in Martin-Löf or similar theories.

The actual weakness of Automath lies not in its infrastructure, but in its lack of direct support for equational logic. It has no easy way to write out the β-rule $Ap(λb,a)$ = $b<a>$ or the η-rule, $λ([x,A]Ap(c,x))=c$ at the higher layer; or the other reductions rules seen in Martin-Löf. The versions of the β-rule and the η-rule in Automath correspond to the ones tacitly used at the deeper, informal, layer throughout Martin-Löf's presentation.


问:
What is the intuition behind matrix splitting methods (Jacobi, Gauss-Seidel)?
Descent Methods, like Gradient and Conjugate Gradient ones, have a nice geometric interpretation and I really love them. What about Jacobi, Gauss-Seidel or other matrix splitting methods? I can't see any intuition behind the formulas.

A posteriori I know that these methods works because of Banach Fixed Point theorem but I still need a more stimulant picture.

Any ideas?

答:
In order to see the idea of how the Jacobi Method works, the straightforward approach is to use the very useful Contraction Mapping principle, which is roughly,
if there is a constant $0\le c<1$ such that $\|F(x_{k+1}) − F(x_{k}) \|\le c\|x_{k+1} – x_{k}\|$, then you can see with a little thought, that the iteration $x_{k+1} = F(x_{k})$ will converge to a fixed point.

Now would be a good time to review the Contraction Mapping principle, since iterative methods are an easy downhill slide from there.

Thus we need to convert $Ax=b$ into a form $x_{k+1} = F(x_{k})$. The most apparent conversion is to extract the diagonal elements of $A$, because a diagonal matrix has a simple inverse,  \begin{align*}
Ax &= b,  \\
(A - D + D)x &= b,   \\
(A - D)x + Dx &= b,   \\
Dx &= b - (A - D)x,   \\
x &= D^{-1}b - D^{-1} (A - D)x,
\end{align*}
and iterate as,

$$x_{k+1}  = D^{-1}b - D^{-1}(A - D) x_k$$

This is the Jacobi iterative method.

Importantly, notice above, that if you have a value for x that satisfies the relation Dx = b - (A - D)x, then x also satisfies Ax = b.

Now there is no guarantee that this will be a contraction mapping, but often it is.  And it will contract when the diagonal elements are dominant.

What would motivate you to do this? An easy-peasy motivation starts with the rudimentary relaxation method. First assume that the diagonal elements of A are much larger than the off-diagonal elements of A.  In this case the inverse of A is approximately the inverse of D, which is simple to calculate.

Compute a residual:

$$r = b - Ax_k$$

Compute an approximate correction:

$$Correction  = D^{-1}r$$

Then,

\begin{align*}
x_{k+1} &= correction + x_k,  \\
x_{k+1} &= D^{-1}r + x_k,  \\
x_{k+1} &= D^{-1}(b - Ax_k) + x_k  \\
x_{k+1} &= D^{-1}b - D^{-1}(A - D)x_k,  \\
\end{align*}

And this is the Jacobi method from the relaxation method.
There you have it.


问:
How can the dual $i^*$ of the inclusion map $i:B_n\rightarrow Z_n$ ever be not surjective?
On page 192 of Hatcher's Algebraic Topology, he shows that cohomology groups for any chain complex $C_n$ fit into a short exact sequence

$$0\leftarrow \text{Ker} \,i^*_n \leftarrow H^n(C;G)\leftarrow \text{Coker} \, i^*_{n-1}\leftarrow 0$$

where $i^*_n$ is the dual of the inclusion $i_n:B_n\rightarrow Z_n$ from boundaries to cycles.

It seems to me like this dual map should always be surjective (and hence, the cokernel should always be zero). Since $B_n$ is a subgroup of $Z_n$, shouldn't any map $B_n\rightarrow G$ have an extension to $Z_n$ just by setting the map to zero for all cycles that aren't boundaries? On the same page, he makes a similar argument to show that maps out of $Z_n$ can be extended to all of $C_n$: why is the same not true for $B_n$?

答:
Oh, figured it out. Such an extension is not in general possible: I was imagining that $B_n$ and $Z_n$ were subgroups generated by individual $n$-simplices, but that's definitely not the case in general. A counterexample to the general idea is actually on the next page of Hatcher - the sequence $$0\rightarrow\mathbb{Z}\xrightarrow{n}\mathbb{Z}\rightarrow\mathbb{Z}_n\rightarrow 0$$

dualizes to

$$0\leftarrow \mathbb{Z}\xleftarrow{n}\mathbb{Z}\leftarrow 0\leftarrow 0$$

and viewing the first map in the original sequence as the inclusion of $n\mathbb{Z}$ into $\mathbb{Z}$, it's clear that the only cases where you can extend maps $n\mathbb{Z}\rightarrow\mathbb{Z}$ to all of $\mathbb{Z}$ are when those maps have degree divisible by $n$ (which is reflected in the fact that $n\mathbb{Z}$ is the image of the dualized map).


问:
Given $2n$ points in the plane, $n$ blue points and $n$ red points, no $3$ are collinear. Prove that we have at least two "balanced lines".
We define as balanced line the line that connect one red point and one blue point that divides the plane with the same amount of $m$ red points and  $m$ blue points in one side and $n$ red points and $n$ blue points in the other side. The image below shows that $r_1$ and $r_2$ are balanced lines, but $r_3$ is not. Note that $m$ is not necessarily equals to $n$.

Exercise: Given $2n$ points in the plane, $n$ blue points and $n$ red points, no $3$ are collinear. Show that we have at least $2$ balanced lines.

I know that the objective of this exercise is to use combinatorics associated with geometry.

I tried two different approachs:

The first one: Make a $s$ line that is not perpendicular to any line that connect one red point to one blue point. We do de projection of these points to the line and try to make some combinatoric, but I couldn't proceed.

The second one: I tried by absurd that doesn't exist any balanced line, but I couldn't reach a contradiction.

答:
At least $3$ of the points must be vertices of their convex hull. Let $p$ be one of these, and let $\ell$ be a line through $p$ that has all of the other $2n-1$ points on one side of it. Without loss of generality we may establish a coordinate system with $p$ at the origin, $\ell$ as the $x$-axis, and all of the other points above the $x$-axis, and we may suppose that $p$ is red. The open half space above $\ell$ has $n$ blue points and $n-1$ red points, so that half space has a surplus of $1$ blue point. Now rotate the coordinate system counterclockwise about the origin $p$. The $x$-axis passes through the other $2n-1$ points one at a time. For $0\le\theta\le\pi$ let $d(\theta)$ be the excess of blue points over red points in the open upper half plane, so that $d(0)=1$, and $d(\pi)=0$. Clearly $d(\theta)$ is piecewise constant, changing only when the $x$-axis hits one of the other $2n-1$ points. Specifically, $d(\theta)$ decreases by $1$ when the $x$-axis hits a blue point and increases by $1$ when it hits a red point.

Let $\theta_0=\min\{\theta\in[0,\pi]:d(\theta)=0\}$. Since $d(0)=1>0$, and $d(\theta)$ can change only in steps of $\pm1$, the $x$-axis at angle $\theta_0$ must pass through $p$ and a blue point $q$ and must therefore be a balanced line.

As noted at the beginning, there must be at least one vertex of the convex hull of the $2n$ points that is neither $p$ nor $q$, and the same argument shows that there is a balanced line through it. Thus, there are at least two balanced lines.


答:
一）If an and bn are sequences of non-negative numbers and both converge, how do I argue that sum ∑anbn  also converges?

you can disprove it by plug in an=1/√n ,  bn=1/√n
∑aₙbₙ=∑1/n⇒+∞, is not a convergent sequence.

二)However, if the problem is:
If ∑an and ∑bn are sequences of non-negative numbers and both converge, how do I argue that sum ∑anbn  also converges?

proof:
for any N
LET ∑aₙbₙ=a៷b៷+...aₙbₙ
<（a៷+a៷₊₁+...aₙ)(b៷+...+bₙ)

because ∑an ,∑bn are convergent sequence,
thus for any ε>0, there exist  N,
∑an=a៷+a៷₊₁+...aₙ+...<ε
∑bn=b៷+...+bₙ+...<ε

hence ∑aₙbₙ=a៷b៷+...aₙbₙ+...<a៷b៷+...aₙbₙ<
（a៷+a៷₊₁+...aₙ)(b៷+...+bₙ)<(a៷+a៷₊₁+...aₙ+...)(b៷+...+bₙ+...)
<ε²=β

So.that.we have conclude that for any β>0, there is N
partial sum ∑aₙbₙ=a៷b៷+...aₙbₙ+...<β
so  ∑aₙbₙ is a Cauchy sequence, and it converge

Title: How to prove if $a_n$ and $b_n$ converge and $a_n,b_n \ge 0$ then $a_n b_n$ also conveges?.

Body:
If $a_n$ and $b_n$ are sequences of non-negative numbers and both converge, how do I argue that $\sum a_n b_n$ also converges?

My attempt:

I tried to argue that $a_n b_n\le(a_n+b_n)^2=a_n^2+2a_nb_n+b_n^2$, but I am stuck on how to proceed from here.. I have a feeling that I should use the comparison test, but I am not too sure.


问:
What is the average latitude of all points on a hemisphere?
I'm looking to calculate the average angular altitude of the sun at noon for everywhere on Earth in order to create a location-agnostic example sun. To start, I would like to calculate what the average latitude is.

Latitudes on a sphere go from 0 to 90 degrees, either North or South, usually with  negative numbers representing South. If we included both hemispheres, the average would be zero, so let's only look at one hemisphere. A naïve guess would be that the average is 45 degrees, but by reasoning in my head, there is far less surface area between 45° and 90° than there is between 0° and 45°. However, I have no idea what the formula is to calculate the area of each range of latitudes. The solution would be to solve some formula to find a value X such that 0° to X° has the same area as X° to 90°.

答:
https://en.wikipedia.org/wiki/Spherical_cap

The above article tells us that formula for the surface area of a spherical cap is $A = 2{\pi}r^2(1-cos\theta)$ where $\theta$ is the angle between the pole and the cap's end. In our case we are dealing with latitude, so what we want to eventually get is $90^{\circ} - \theta$.

The surface area of a sphere is $4{\pi}r^2$. We are working with a hemisphere, so we take half this to get $2{\pi}r^2$, and we want to find the point (line) in the middle where the areas are equal, which means we want half of this area, or $A = {\pi}r^2$. We can now form the equation ${\pi}r^2 = 2{\pi}r^2(1-cos\theta)$. We can divide both sides by ${\pi}r^2$ to get $1 = 2(1-cos\theta)$ then we can get $1/2 = 1-cos\theta$ then $1/2 = cos\theta$ then $acos(1/2) = \theta$ then $\theta = 60^{\circ}$.

This means that the average (mean) latitude on a sphere is $90^{\circ} -  60^{\circ}$ or $30^{\circ}$, which answers the stated question in the title. Additionally, to answer my use case, the average sun angular altitude at noon would be the same as the equinox angular altitude at noon, which is $90^{\circ} -  latitude$ or $90^{\circ} - 30^{\circ}$ or $60^{\circ}$.


问:
show that $f$ is Riemann integrable on $[0,1]$
Let $f(x)=\sin\left(\frac{1}{x}\right)$ if $0<x\le1$ and $f(x)=0$ if $x=0$. Show that $f$ is Riemann integrable on $[0,1]$ and calculate it's integral on $[0,1]$.

I would like to know if my proof holds, please. And, I would like have a hint on how we can calculate the intergal of this such of functions, please.

My attempt is to pass by Darboux upper and lower sums.

To show that $f$ is Riemann integrable, we have to show the following:$\forall \epsilon>0$ $\exists$ a partition $\sigma$: $\overline{S}_{\sigma}(f)<\underline{S}_{\sigma}(f)+\epsilon$.

Let $0<\epsilon<2$ and consider the following intervals $[0,\epsilon/2]$ and $[\epsilon/2,1]$. First of all, as $f(x)$ is continuous on $[\epsilon/2,1]$($\sin(\frac{1}{x})$ is continuous on $[\epsilon/2,1]$), it is integrable on this interval. We would like to show now that $f$ is intergable on $[0,\epsilon/2]$. Consider the partition $\sigma=\{0,\frac{\epsilon}{2}\}$ on $[0,\epsilon/2]$ We have that

$\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}(f)=\sum_{i=0}^{0}M_i(x_{i+1}-x_i)-\sum_{i=0}^{0}m_i(x_{i+1}-x_i)$ with $M_i=\sup\{f(x):x\in[x_i,x_{i+1}]\}$ and $m_i=\inf\{f(x):x\in[x_i,x_{i+1}]\}$.

We have that $M_i\le1$ and $m_i\le 0$. Thus,

$\sum_{i=0}^{0}M_i(x_{i+1}-x_i)-\sum_{i=0}^{0}m_i(x_{i+1}-x_i)\le1\cdot \frac{\epsilon}{2}< \epsilon$. Therefore, as $f(x)$ is integrable on $[0,\epsilon/2]$ and on $[\epsilon/2,1]$, we conclude that $f$ is integrable on $[0,1]$.

Now, I would like to calculate it's integral value... If someone could help with it, I would appreciate it. Honestly, I have no idea how to integrate this such of discontinuous functions at finitely many points.

Edit:

As $f(x)$ is continuous on $[\epsilon/3,1]$, it is integrable on this interval. Therefore, $\forall \epsilon>0 \ \exists$ partition $\tau$ on the interval $[\epsilon/3,1]$: $\overline{S}_{\tau}(f)-\underline{S}_{\tau}(f)<\epsilon/3$

To show that $f$ is integrable on $[0,1]$, we have to show that: $\forall \epsilon>0 \ \exists$ a partition $\sigma$: $\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}<\epsilon$

Suppose $0<\epsilon<3$ and let $\sigma=\tau \ \cup \{0\}$ be a partition on the interval $[0,1]$. We have that (for the following inequality I use the same definition for $M_i$ and $m_i$ as above in the post):

$\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}=\Big(\sum_{i=0}^{0}\underbrace{M_i}_{\le1}(x_{i+1}-x_i)-\sum_{i=0}^{0}\underbrace{m_i}_{\ge -1}(x_{i+1}-x_i)\Big)+\underbrace{\Big(\sum_{i=1}^{n}M_i(x_{i+1}-x_i)-\sum_{i=1}^{n}m_i(x_{i+1}-x_i)\Big)}_{<\epsilon/3}<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$.

Therefore, $f$ is integrable on $[0,1]$.

答:
This is about feedback on your proof.

Your goal is to find a partition $\sigma$ such that $\overline{S}_{\sigma} (f) <\underline{S} _{\sigma} (f) +\epsilon $ but instead you are trying to use that $f$ is integrable on $[\epsilon/2,1]$ and prove  that $f$ is integrable on $[0,\epsilon/2]$.

This is not what you want. You want a partition $\sigma$ which works as expected.

You can just say that since the function is integrable on $[\epsilon/2,1]$ there is a partition $\sigma_1$ of $[\epsilon/2,1]$ which works for this interval and then take $\sigma=\sigma_1\cup \{0\}$ and using your argument in question show that this particular partition works for $[0,1]$.

You have the correct idea but you need to present it in proper manner.


问:
Is $\tan(\alpha/2)=(1-\cos\alpha)/\sin \alpha, \,?$
Is it true that

$$\bbox[5px,border:2px solid #138D75]{\tan\left(\frac {\alpha}2\right)=\frac{1-\cos\alpha}{\sin \alpha}, \quad ?} \tag1$$

My solution:
First, we note that the expression $\tan \left ( \frac{\alpha}{2} \right )$ makes sense when $\alpha \neq \pi + 2k\pi, k \in \mathbb{Z}$: henceforth, we will assume that this condition is met.

By definition of tangent function, we have: $$\tan \left ( \frac{\alpha}{2} \right ) = \frac{ \sin \left ( \frac{\alpha}{2} \right )}{\cos \left ( \frac{\alpha}{2} \right )}$$

When $\alpha \neq 2k \pi, k \in \mathbb{Z}$, we can multiply numerator and denominator by $\sin \left ( \frac{\alpha}{2} \right )$, because this quantity is non-zero. We therefore get:

$$\tan \left ( \frac{\alpha}{2} \right ) = \frac{ \sin^2 \left ( \frac{\alpha}{2} \right )}{\cos \left ( \frac{\alpha}{2} \right ) \cdot \sin \left ( \frac{\alpha}{2} \right )}. $$

For what we saw earlier: $$\sin^2 \left ( \frac{\alpha}{2} \right ) = \frac{1 - \cos \alpha}{2}$$and instead, applying the sine duplication formula: $$\cos \left ( \frac{\alpha}{2} \right ) \cdot \sin \left ( \frac{\alpha}{2} \right ) = \frac{1}{2} \sin \alpha$$

In conclusion we get the following formula: $$\tan \left ( \frac{\alpha}{2} \right ) = \frac{ 1 - \cos \alpha}{\sin \alpha}.$$

If in the expression of the tangent we had multiplied numerator and denominator by $\cos \left ( \frac{\alpha}{2} \right )$ then with steps similar to what we have done we get another formula:$$\bbox[5px,border:2px solid #118D45]{\tan \left ( \frac{\alpha}{2} \right ) = \frac{\sin \alpha}{1 + \cos \alpha}}. \tag 2 $$

We note that it always holds $\cos \left ( \frac{\alpha}{2} \right ) \neq 0$ since we have imposed $\alpha \neq \pi + 2k\pi, k \in \mathbb{Z}$ (and this entitles us to multiply numerator and denominator by this term).

Is there a faster and more immediate proof of the $(1)$ or $(2)$?

答:
Replace $\alpha$ by $2\beta$ so that the right side is
$${{1-\cos 2\beta}\over {\sin 2\beta}}=
{{1-(\cos^2\beta-\sin^2\beta)}\over {2\sin\beta\,\cos\beta}}$$

$$={{2\sin^2\beta}\over{2\sin\beta\cos\beta}}=
{{\sin\beta}\over {\cos\beta}}=\tan\beta$$


问:
How prove this $\int_{a}^{b}[f''(x)]^2dx\ge\dfrac{4}{b-a}$
let $f$ on $[a,b]$ two  continuously differentiable functions,such

$$f(a)=f(b)=0, f'(a)=1,f'(b)=0,b>a>0$$
  show that
  $$\int_{a}^{b}[f''(x)]^2dx\ge\dfrac{4}{b-a}$$

My idea: use Cauchy-Schwarz inequality
$$\int_{a}^{b}[f''(x)]^2dx\cdot\int_{a}^{b}g^2(x)dx\ge\left(\int_{a}^{b}f''(x)g(x)dx\right)^2$$
How find the $g(x)$?

then I think

\begin{align}
\int_{a}^{b}f''(x)g(x)dx &=\int_{a}^{b}g(x)df'(x)=f'(x)g(x)|_{a}^{b}-\int_{a}^{b}f'(x)g(x)dx\\
&=f'(b)g(b)-f'(a)g(a)-\int_{a}^{b}g(x)df(x)\\
&=-g(a)+\int_{a}^{b}g'(x)dF(x)\\
&=-g(a)+F(x)g'(x)|_{a}^{b}-\int_{a}^{b}F(x)g''(x)\\
&=-g(a)+F(b)g'(b)-F(a)g'(a)-\int_{a}^{b}F(x)g''(x)dx
\end{align}
  where $F(x)=\int_{a}^{x}f(t)dt$

Now following I can't find the $g(x)$? can you help me or use other methods solve it? Thank you

and some hours ago: I ask this integral inequality:How prove this $\int_{a}^{b}f^2(x)dx\le (b-a)^2\int_{a}^{b}[f'(x)]^2dx$

答:
Let $g(x)$ be a polynomial, by integration by parts,
\begin{align*}
\int_a^bg(x)f''(x)\mathrm{d}x
&=g(x)f'(x)\bigg|_a^b-\int_a^bg'(x)f'(x)\mathrm{d}x\\
&=-g(a)+\underbrace{\int_a^bg''(x)f(x)\mathrm{d}x}_{=0}
\end{align*}
Only let $g''(x)=0\Rightarrow g(x)=c_1x+c_0$. by the Cauchy-Schwarz inequality,
$$\int_a^b\big(f''(x)\big)^2\mathrm{d}x\int_a^bg^2(x)\mathrm{d}x\geqslant\left(\int_a^bg(x)f''(x)\mathrm{d}x\right)^2\longrightarrow\text{Constants}$$
We have,
$$\int_a^b\big(f''(x)\big)^2\mathrm{d}x\geqslant \frac{g^2(a)}{\int_a^bg^2(x)\mathrm{d}x}=\frac{4}{b-a}$$
Substituting,
$$3c_1(b-a)(c_1a+c_0)^2=4\big[(c_1b+c_0)^3-(c_1a+c_0)^3\big]$$
Simplification, we get
$$\big((a+2b)c_1+3c_0\big)^2=0\Rightarrow (a+2b)c_1=-3c_0$$
So, we get,
$$g(x)=c_1x-\frac{1}{3}(a+2b)c_1$$
Inside,$c_1$ is a constant. $\Box$


问:
How many ways can I permute the letters in the word 'ACCOUCHEUSES' with special property?
How many ways can I permute the letters in the word 'ACCOUCHEUSES' knowing that the C letters must always be consecutive. I thought the answer was $\frac{10!}{2!2!2!} = 453600$, but it seems my answer is not good. Where am I wrong?

答:
Hint

All you have to do is pretend that the 3 C's are glued together into one unit.


问:
Monte Carlo integration &expected value
I'm confused as to how evaluating the Monte Carlo integration is the same as estimating
the expected value. For example, if $x \sim$ unif[0,1], why does $$\int_{0}^{1}f(x)dx = E(f(x)) ?$$

答:
Let me call the function $g$ to avoid confusion with the probability density.

You know that for any variable $X$ with density $f_X(x)$ we have

$$E[X] = \int x f_X(x)dx \tag 1$$
and also

$$E[g(X)] = \int g(x) f_X(x) dx \tag 2$$

In particular, if $X$ is uniformly distributed on $[0,1]$ then

$$E[g(X)] = \int_0^1 g(x) dx \tag 3$$

You also know that, under some conditions, for large $n$ the expectation of a variable can be approximated by the sample average:

$$E[Y] \approx \frac{Y_1 + Y_2 + \cdots Y_n}{n} \tag 4$$

where $Y_1, Y_2 \cdots Y_n$ are samples of $Y$
(for the precise meaning of that $\approx$, and the conditions that $Y$ must fullfill, see
the law of large numbers; in particular, it's sufficient that $Y$ is bounded)

Then, letting $Y=g(X)$ we conclude

$$\int_0^1 g(x) dx \approx \frac{g(X_1)+g(X_2) \cdots g(X_n)}{n}$$

and we can approximate an integral by averaging over random samples of the function (Montecarlo integration).

Title: Monte Carlo integration &expected value


问:
Are indeterminate forms a sign that our mathematics system is flawed?
This is much more philosophical and opinion-based than many of the other questions asked here, but yeah, pretty much what the title says. Sometimes I think that we have this system of mathematics simply because we could count with whole numbers and it sort of snowballed, but then indeterminate forms like $1^\infty$ or $0^0$ being ambiguous make me think that maybe our system has fundamental flaws. I don't know. I'm no mathematician, just an undergrad, so if anyone has some interesting insights on this, I'd be happy to hear them.

答:
Obligatory "not a full answer but too long for a comment". That being said, this is a soft-question so no answer may be 100% satisfying.

Is mathematics flawed? No.

Is mathematics incomplete? Absolutely.

Gödel's incompleteness theorems and the Halting problem are infamous proofs that systems equipped to handle finite logic/arithmetic (where mathematics has "snowballed" from), will never be able to handle the paradoxical nature of infinity & recursion consistently & completely.

It seems to me that indeterminate forms are indeterminate not because mathematics is flawed, but because an infinity arises. Mathematics can often only at best treat the supertask that is infinity, as a limit; which may inherently underestimate the complexity of the true nature of infinity.

A good example are infamous divergent sums, such as $1+2+3+4+\dots$, where our finite math clearly indicates a divergent sum. Yet, looming in shadows is the strange pseudo-convergence to $-\frac{1}{12}$, which could only ever be correct if our understanding of infinity were incomplete.

On top of such, we often don't even know which infinity we're referring to when using it in a sentence or an expression. Do we want to treat $\infty$ in $1^\infty$ being raised to a limit or having a number-like value? Ordinal infinity or cardinal infinity? Countable infinity or uncountable infinity? Do these questions even make sense in this context?


问:
Polar decomposition of a deformation gradient for the simple shear
Consider the following deformation $\boldsymbol{x} = \boldsymbol{f}(\boldsymbol{p})$ definied by

$$
\begin{cases}
x_1 = p_1 + \gamma p_2 \\
x_2 = p_2 \\
x_3 = p_3
\end{cases}
$$

which corresponds to the simple shear.

it's immediate to find the deformation gradient $F$ to be

$$\begin{bmatrix}
1 & \gamma & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
$$

This tensor (it's a matrix) have positive determinant, so we can compute the Polar decomposition, i.e. $F=VR=RU$ where $R$ is a rotation and $U,V$ are symmetric and positive definite.

My goal is to compute $U$, which is known to be $\sqrt{F^T F}$, which is not easy at all. Is there any way to compute it, without computing the square root of the matrix $ F^T F$?

答:
Instead of calculating $\sqrt{F^TF}$, you can construct a U and R with some variables and solve for them individually.

Let's say,

$$
    \boldsymbol{U}=\quad
    \begin{pmatrix}
    a & b & 0 \\
    b & c & 0 \\
    0 & 0 & 1
    \end{pmatrix}
    \quad
    \boldsymbol{R}=\quad
    \begin{pmatrix}
    x & y & 0 \\
    -y & x & 0 \\
    0 & 0 & 1
    \end{pmatrix}
    \quad
$$

Here, U(1,2) = U(2,1), because it is symmetric. Similarly for R, because it is anti-symmetric.

Using, $F=R.U$, we get,
$$
ax-by=1\\
    bx-cy=\gamma\\
    bx+ay=0\\
    cx+by=1\\
    x^2+y^2=1
$$

Solving the above equations, we get (I used Mathematica),
$$
\boldsymbol{U}=\quad
        \begin{pmatrix}
        \frac{2}{\sqrt{4+\gamma^2}} & \frac{\gamma}{\sqrt{4+\gamma^2}} & 0 \\
        \frac{\gamma}{\sqrt{4+\gamma^2}} & \frac{2}{\sqrt{4+\gamma^2}}+\frac{\gamma^2}{\sqrt{4+\gamma^2}} & 0 \\
        0 & 0 & 1
        \end{pmatrix}
        \quad\\
    \boldsymbol{R}=\quad
        \begin{pmatrix}
        \frac{2}{\sqrt{4+\gamma^2}} & -\frac{\gamma}{\sqrt{4+\gamma^2}} & 0 \\
        \frac{\gamma}{\sqrt{4+\gamma^2}} & \frac{2}{\sqrt{4+\gamma^2}} & 0 \\
        0 & 0 & 1
        \end{pmatrix}
        \quad
$$


问:
How to get a approx nature of $f'''(x)$ graph from the given data ( Graph techinque using LMVT etc., or using just calculus both suffices )
The above mentioned title is for giving a rough graph for $f'''(x)$ , such that we can show that there exists some values of $f'''(c) \ge 6$. Problem where i encountered this:  " If $f(x)$ is a function with a continuous third derivative on $[0, 2]$ such that
$f(0) = f'(0) = f''(0) = f'(2) = f''(2) = 0$ and $f(2) = 2$, then minimum number of value(s) of $c$ belongs $[0, 2]$
such that $f'''(c)\ge 6$ ", What I tried is that I draw graphs of $f(x) ,f'(x) , f''(x)$ and tried to use LMVT to conclude something useful for $f'''(x)$ ,but seems like graph way is too much messy and I know there are many cases which the graph can be for every type , so I thought some calculus way might be there leaving the graph technique.

答:
A third derivative is very hard to see from a graph.  It can make a sharp corner on a graph.  Your function starts out very flat at $0$ and ends up very flat at $2$.  If the third derivative is small the function can't rise very quickly from $0$ and won't have time to get to $2$.  You are told the Taylor series at $0$ is $kx^3$ plus higher order terms and the Taylor series at $2$ is $2+k'(x-2)^3$ plus higher order terms.

If there are no points where the third derivative is $6$ or greater, you can bound the second derivative by integrating $6$.  You can integrate twice more and bound the value of the function from above.  You can do the same starting from $2$ and going downwards.  If you can make the values and derivatives match you can exhibit a function that never touches $6$ and claim victory.

If not, you probably need $f'''(x) \ge 6$ over an interval, so the number of points is uncountable.  A more interesting question would be to show you can succeed with just one interval.  Intuitively you could jump with a(n almost) square wave from $0$ to almost $2$ with one bit of $f'''(x) \ge 6$.  Can you slow down enough to arrive at the other end without the third derivative getting large?


问:
What would be the smallest/lowest Big-O set possible of these $2$ functions
Here i have $2$ functions and i was just wanting to know what the smallest big-oh set possible is for these functions, my guess would be the lowest possible of 1 would be the big-ph set $O(n^3)$ but was unsure if it may be $O(1)$, and for the 2nd function i was unsure if it was $O(n \log n)$ or $O(n)$

$\frac {n^3} 4 + 6n^2 – 10000$

$5n\log 2n + 6n$

Any help and explanation is appreciated.

答:
It seems that you have struggles with understanding what Big-O means in the first place, so I've decided to add some further details, although the comments and the already given answer are sufficient.

The Big-O notation was first introduced by Paul Bachmann in "Analytische Zahlentheorie (1894)" for a way of expressing approximation. On page $401$ is the first occurrence of Big-O in history. He states (in german)

[...] (es) findet sich $\tau(n)=n\ log\ n\ +\ O(n)$ wenn wir durch das Zeichen $O(n)$ eine Grösse ausdrücken, deren Ordnung in Bezug auf $n$ die Ordnung von $n$ nicht überschreitet; ob sie wirklich Glieder von der Ordnung $n$ in sich enthält, bleibt [...] dahingestellt.

This translates to

[...] (it) yields $\tau(n)=n\ log\ n\ +\ O(n)$ if we express by the sign $O(n)$ a quantity whose order with respect to $n$ does not exceed the order of $n$; whether it really contains members of order $n$ in itself remains undecided.

By this definition, we could write $1 + n + n^2=O(n^4)$, since the order of the "quantity" on the left hand side does not exceed the order of $n^4$. But furthermore, the following is also valid $1+n+n^2=O(n^3)$ as the "quantity" on the left hand side does not exceed the order of $n^3$. Obviously, the second statement is much stronger than the first statement. Being a little bit informal, we could say that we have introduced a notation which allows us to replace "$\approx$" with "$=$".

Being more precise now in todays terms, I am going the state the definition of Big-O as given by Donald E. Knuth in his famous book "The Art of Computer Programming".

Def.: The symbol $O(f(n))$ stands for the set of all functions $g$ of integers such that there exist constants $M$ and $n_0$ with $|g(n)| \leq M|f(n)|$ for all integers $n \geq n_0$. We usually write $g(n)=O(f(n))$ or $g(n) \in O(f(n))$.

Taking your first function as an example. You need to find constants $M$ and $n_0$ such that for $n \geq n_0$ it holds that $\frac{1}{4}n^3 + 6n^2-10000 \leq Mn^3$, and if we can show that, $\frac{1}{4}n^3 + 6n^2-10000 \in O(n^3)$, and this is by definition the best we can do.

(once we are more familiar with the notion we could prove a much stronger result, which states that given any polynomial $P(n)=a_0+a_1n+\cdots+a_mn^m$, we have $P(n)=O(n^m)$. Then the solution would follow directly.)

For the second function, the following rules might help:

$c \cdot O(f(n)) = O(f(n))$, $c$ is constant

$O(f(n))+O(f(n))=O(f(n))$

$O(f(n)) O(g(n))=O(f(n)g(n))$


问:
For what $z$ does the sequence $z_n=nz^n$ converge?
For what $z$ does the sequence $z_n=nz^n$ converge?

Attempt

Consider $\sum_{n\geq 1}nz^n $ and notice
$$\lim_{n \to \infty}\frac{z_{n+1}}{z_n}=(1+\frac{1}{n})z=z$$

but the serie converges iff $|z|<1$ therefore

$z_n$ converges iff $|z|<1$ and in fact $z_n \to 0$

Is my answer right?  or I should consider

$z^n=\cos n \theta+ i \sin n \theta$ and consider the limit

$$\lim_{n \to \infty} n \left( \cos n \theta+i \sin n \theta \right)$$ and
compare
$$ \lim_{n \to \infty} n \left( \cos n \theta \right)\wedge \lim_{n \to \infty} n \left( i\sin n \theta \right)$$

There exists a general form of attack this kind of problem, when they request see for what values of $z$ a given sequence converges?

答: Suggestion: Use $a_n = nz^n$. To prove $a_n \to 0$, you can prove: $|a_n| \to 0$. Thus $|a_n| = n|z|^n, |z| < 1$ as indicated in your post. You have: $|a_n| = \dfrac{n}{b^n}$, $b = \dfrac{1}{|z|} > 1$. Thus put $b = x+1, x > 0\implies |a_n| = \dfrac{n}{(1+x)^n} < \dfrac{n}{1+nx+\frac{n(n-1)x^2}{2}} < \dfrac{2}{(n-1)x^2}\to 0$ as $n \to \infty$ while $x$ is fixed, proving the assertion !


问:
Testing if a Point is in the Circumcircle of a Triangle
I am currently reading about methods of Delaunay triangulation. In particular, this paper by Leach: Improving Worst-Case Optimal Delaunay Triangulation Algorithms (1992)

In it, Leach describe a method of testing whether a point $D$ is in the circumcircle of a triangle $\triangle{}ABC$. This is contrasted with the usual determinant method (see the Delunay Triangulation Wikipedia page if you want to read more on that one).

The test is described as follows

We instead use a test based on $y$-coordinates of circumcircle centres. As shown in Figure 4, for two fixed sites $A$ and $B$ the centre of the circumcircle of $A$, $B$ and a third point lies on the bisector of $A$ and $B$.

As the merge step of the algorithm always seeks the best
candidate site above the topmost cross edge, a site $D$ is a better site than a site $C$ if its circumcircle centre lies lower on the bisector. The best candidate site is the one whose circumcircle centre is the lowest. Substituting $(x_A, y_A)$, $(x_B, y_B)$ and $(x_C, y_C)$ into
$$(x - p)^2 + (y - q)^2 = r^2$$
and solving for $q$ gives
$$q = \frac{1}{2} \cdot{} \frac{
  (x_A^2 + y_A^2)(x_C - x_B) - (x_B^2 + y_B^2)(x_C - x_A) + (x_C^2 + y_C^2)(x_B - x_A)
}{
  (x_B - x_A)(y_C - y_A) - (x_C - x_A)(y_B - y_A)
}.$$
This calculation requires 23 operations (ignoring the division by 2 and assuming common subexpression elimination). Two such calculations are needed to compare the first two candidate sites; thereafter only one is needed per candidate site, for which the cost of
the InCircle test is reduced by approximately half.

I do not follow what's happening here. I understand why the circumcenters fall on the line between $A$ and $B$, but it is not immediately obvious to me:

Why a lower $q$ is better for the Delaunay triangulation.

What the method should do when the $q$ falls below the center of $AB$. Is lower still better, or is it distance from the center we really care about?

If someone could explain (1) and answer (2), this would be greatly helpful to me, thanks.

答:
After some additional thought, I have worked it out.

Recalling the inscribed angle theorem, we have that

and from this, we can see that the angle of the triangle at $D$ can be ascertained from the rise above the midpoint of $AB$, $q$, and half the distance of $AB$, i.e.
$$\theta = \mathrm{atan}\left(\frac{|AB|}{2q}\right)$$
and moreover, as $\frac{|AB|}{2}$ is fixed, $\theta$ will decrease with larger $q$.

This also reveals the answer to the second question: the quantity we need to consider for this method is the distance from the midpoint of $AB$.


问:
Derivative of $f(x, g(x))$ with respect to $x$
Both $f$ and $g$ are $C^1$ functions and no other assumption. My attempt
\begin{align*}\frac{df}{dx} &= \frac{\partial f}{\partial x}\frac{dx}{dx} + \frac{\partial f}{\partial g}\frac{dg}{dx} \\
&= \frac{\partial f}{\partial x} +  \frac{\partial f}{\partial g}\frac{dg}{dx}\end{align*}
But this does not look correct, how can $\frac{df}{dx}$ is equal to itself plus some other term?

*note: I want to find $\frac{df}{dx}$ exactly which means treat $f$ as a function of $x$ only, not $\frac{\partial f}{\partial x}$.

答:
I also had the same confusion before : )

Note $f$ has two inputs, let us define the partial derivatives with respect to the first and second entry as $\partial_1 f$ and $\partial_2 f$.

The reason I like these notations above is that as you have noticed, $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ would be confusion here.

Thus, we have
$$\frac{d}{dx}f(x, g(x)) = \partial_1 f(x, g(x))\frac{dx}{dx}(x) + \partial_2 f(x, g(x)) \frac{dg}{dx}(x).$$


问:
Why isn't the derivative of $e^x$ equal to $xe^{(x-1)}$?
When we take a derivative of a function where the power rule applies, e.g. $x^3$, we multiply the function by the exponent and subtract the current exponent by one, receiving $3x^2$. Using this method, why is it that the derivative for $e^x$ equal to itself as opposed to $xe^{x-1}$? I understand how the actual derivative is derived (through natural logs), but why can't we use the original method to differentiate? Furthermore, why does the power rule even work? Thank you all in advance for all of your help.</p>

答:
In applying the power rule, chain rule and other rules for derivatives, it critically matters how the independent variable $x$, with respect to which the derivative is being taken, appears in the formula.

The power rule does not simply assert that we look at the syntax of the formula, find powers, and then bring them down as coefficients and decrement them by one!

The power rule says that we do this for terms where the derivation variable $x$ is being raised to something. If we are differentiating with respect to $x$, and there is a term of the form $x^a$, then the power rule applies. It does not apply to some $a^b$ where $x$ does not even occur (basically a constant term with respect to $x$) nor to some $a^x$ where $x$ occurs, but not in the correct position for the rule to apply.

The power rule is that $x^a$ differentiates into $ax^{a-1}$.

More generally, in combination with the chain rule, $f(x)^a$ goes to $af(x)^{a-1}f'(x)$.  In the special case of $f(x) = x$, $f'(x)$ is just 1.

One way to approach an understanding of why the power rule works is to go back to first principles of differentiation and work with the following limit:

$$\lim_{h \to 0}\frac{(x+h)^a - x^a}{h}$$

You can see where the above is headed right away, because the $(x + h)^a$ term, when expanded, will produce an $x^a$ term, which will cancel with the $-x^a$ term, and so the overall polynomial reduces in degree by one.

After the $x^a$ term $(x + h)^a$ produces a ax^{a-1}h term. The other terms don't matter because they produce higher powers of $h$, thus:

$$\lim_{h \to 0}\frac{(x^a + ax^{a-1}h + Khx^{a-2}h^2 + \ldots + h^a) - x^a}{h}$$

We don't care what K is; it's some constant that depends on $a$ (which row of Pascal's triangle we are in to pick the coefficients for the expansion of the binomial). That term and all the others denoted by $\ldots$ will disappear. First, the $x^a - x^a$ cancels:

$$\lim_{h \to 0}\frac{ax^{a-1}h + Khx^{a-2}h^2 + \ldots + h^a}{h}$$

Then we do the division to distribute the denominator $h$ into the numerator:

$$\lim_{h \to 0}ax^{a-1} + Khx^{a-2}h + \ldots + h^{a-1}$$

Now, since this is a limit as $h$ approaches zero, all the terms that have $h$, or a power of $h$ disappear, leaving us with the limit just being $ax^{a-1}$, which is the power rule:

$$\lim_{h \to 0}ax^{a-1} + Khx^{a-2}h + \ldots + h^{a-1} = ax^{a-1}$$

So we did not even have to compute the full expansion of $(x + h)^a$, because we know that the third and subsquent terms all contain $h$ and therefore disappear in the limit.

Another piece of intuition about derivatives is to look at difference series. For instance take the succession of squares: 1, 4, 9, 16, 25, 36, 49, 64 and compute the deltas between them. They are: 3, 5, 7, 9, 11, 13, 15.  Hey look, $n^2$ became $2n + 1$.  The power is reduced from quadratic to linear, and the coefficient doubled.


问:
Example of a quantifier elimination procedure for a simple-but-nontrivial theory
Is there a simple-but-nontrivial example of a concrete quantifier elimination procedure with a concrete theory, especially one that's a standard example of what a constructive argument for quantifier elimination looks like? Also, what does a complete proof of the existence of quantifier elimination look like for the simple theory of the successor operation described in this question?

From what I've read on quantifier elimination, I gather that not every proof that a given theory has it is constructive.

I'm also assuming a) the theory of the successor operation allows quantifier elimination and b) a complete argument that it allows quantifier elimination is reasonably succinct.

What follows is my attempt to understand the material.

The Wikipedia article on quantifier elimination sketches out a way to construct an argument that a particular theory has quantifier elimination, by first considering statements of the form

$$ \exists x \mathop. L_1 \land L_2 \land \cdots \land L_n $$

Where the elements of $L$ are literals, i.e. either atomic formulas headed by a non-logical predicate, equality, or the negation of a positive literal.

From there, disjunctions can be eliminated using the following rule

$$ (\exists x \mathop. (L_1 \land L_2 \land \cdots \land L_n) \lor (K_1 \land K_2 \land \cdots \land K_n)) \\ \text{is equivalent to} \\ (\exists x \mathop. L_1 \land L_2 \land \cdots \land L_n) \,\,\lor\,\, (\exists x \mathop. K_1 \land K_2 \land \cdots \land K_n) $$

The argument for $\forall x \mathop. F$ where $F$ is a quantifier-free formula doesn't make much sense to me.

$$ \forall x \mathop. F \;\; \text{gets sent to} \;\; \lnot \left( \exists x \mathop. \text{negation of $F$ in disjunctive normal form} \right) $$

I think the idea is to eliminate quantifiers in the expression inside the parentheses on the right hand side and then apply negation again to the result, but I'm not sure.

It's also not clear to me why we can't simply strip the $\forall x$ from $F$ directly since free variables are treated identically to universally quantified variables when we're considering which statements are tautologies.

I also read this question, which presents a simple theory with just a successor operation, described below.

It has three axioms $\text{S1}$ through $\text{S3}$ and an axiom schema $\text{S4}$.

$\text{S1}$: Zero has no predescessor.

$$ \forall x \mathop. S(x) \ne 0 \tag{S1} $$

$\text{S2}$: Successor is injective.

$$ \forall x \forall y \mathop. S(x) = S(y) \,\to\, x = y \tag{S2} $$

$\text{S3}$: Every nonzero number has a predecessor.

$$ \forall x \mathop. \left((x \ne 0) \to \left(\exists y \mathop. S(y) = x \right)\right) \tag{S3} $$

$\text{S4}$: The $k$th iterate of successor has no fixed points.

$$ \forall x \mathop. S^k(x) \ne x \tag{Schema S4} $$

The question asks about eliminating quantifiers from sentences of the form $\exists x \mathop. \left( \Delta_1 \land \Delta_2 \land \cdots \land \Delta_n \right)$ where $\Delta$ is a set of positive literals, i.e literals of the form $S^n(x) = S^m(y)$ where $x$ and $y$ are fixed variables.

I suspect that this theory has quantifier elimination as a whole, not just for sentences with existential quantifiers and conjunctions of positive literals.

The answer is somewhat terse, but seems to suggest the following procedure.

Walk each expression in $\Delta$. For every literal $S^n(x) = S^m(y)$, replace $x$ with $S^{(m-n)}(y)$ if $m-n \ge 0$. It doesn't mention explicitly what to do if $n > m$. The answer is also confusing since $x$ can potentially occur in many literals within $\Delta$. I think we just pick one occurrence of $x$ and perform the substitution everywhere in $\Delta_1 \land \Delta_2 \cdots \Delta_n$, but I'm not sure.

It also doesn't mention how to extend the argument to be a full proof of quantifier elimination for the theory of the successor operation, assuming one exists.

答:
This theory does have quantifier elimination, and here's a proof of that fact.

First, here's the definition of a theory having quantifier elimination from Marker's Model Theory: An Introduction.

Definition 3.1.1

We say that a theory $T$ has quantifier elimination if for every
formula $\phi$ there is a quantifier-free formula $\psi$ such that

$$ T \models \phi \leftrightarrow \psi $$

The important thing to note about this definition is that $\phi \leftrightarrow \psi$ is just another well-formed formula, which means that if a free variable $v$ occurs in $\phi$ and $\psi$ then it appears free in $\phi \leftrightarrow \psi$ overall.

First, let me correct an error in my original question. This theory also has a constant symbol $0$ in addition to the unary function symbol $S$.

In order to prove that a theory has quantifier elimination, it suffices to eliminate quantifiers in expressions of the following form, i.e. a conjunction of literals with a single existentially quantified variable shared between them and arbitrary free variables

$$ \exists x \mathop. L_1 \land \cdots \land L_n $$

For the purposes of this question, let $z$ be a variable that does not occur free in the formula being considered.

Let's proceed by cases

Case 1: $n = 0$

$\exists x \mathop. \top$ is equivalent to $\top$.

Case 2: $n > 0$

First let's define $L^2$ in terms of $L$ as follows.

Every conjunct of $L$ is of the following form, where $p$ and $q$ are variables or $0$.

$$S^j(p) = S^k(q) \;\; \text{or} \;\; S^j(p) \ne S^k(q) $$

Let's consider only the conjuncts where $p$ and $q$ are both $0$.

Note that the truth of $S^j(x) = S^k(x)$ does not depend on the value of $x$ at all, so we can replace it with $S^j(z) = S^k(z)$, and likewise for $\cdots \ne \cdots$.

$$ L^2 \\ \textit{is equal to} \\ \{ l \mathop| l \in L \;\;\text{and $l$ does not have $x$ in both terms} \} \cup \\
\{ S^j(z) = S^k(z) \mathop| \text{$l$ is of the form $S^j(x) = S^k(x)$ and $l \in L$} \} $$

With that out of the way, let's consider three cases. Case 2.1 and 2.2 are split for clarity.

Case 2.1: $L^2$ has no terms containing $x$ at all

If $L^2$ has no terms containing $x$, then the following holds

$$ \exists x \mathop. L_1 \land \cdots \land L_n \;\; \text{is equivalent to} \;\; L^2_1 \land \cdots \land L^2_n $$

Case 2.2: $L^2$ has at least one term with $x$ and $\ne$ but none with $x$ and $=$

If $L^2$ has no terms containing $x$ and $=$, but does have a term containing $x$ and $\ne$, then I claim that the following holds.

$$ \exists x \mathop. L_1 \land \cdots \land L_n \;\; \text{is equivalent to} \;\;
\bigwedge \{ l \mathop| l \in L^2 \;\; \text{and $l$ does not contain $x$} \} $$

In order to prove it, note that $\exists x \mathop. L_1 \land \cdots \land L_n$ is true if and only if it holds for all valuations $V$ of the free variables of $L$. Any valuation of $V$ can be extended to include $x$ in the following way. Pick the least element $t$ of $0, S(0), S(S(0)), S^3(0), \cdots$ so that no $L_i$ is false when $x = t$. Every $L_i$ in $L$ either does not contain $x$ or is false for exactly one value of $x$. There are finitely many $L_i$ in $L$ and countably infinitely many elements of the sequence $0, S(0), S(S(0)), S^3(0), \cdots$ so $t$ exists. Therefore, if $\exists x \mathop. L_1 \land \cdots \land L_n$ is true for all valuations then $\bigwedge \{ l \mathop| l \in L^2 \;\; \text{and $l$ does not contain $x$} \}$ is true for all valuations.

If $\bigwedge \{ l \mathop| l \in L^2 \;\; \text{and $l$ does not contain $x$} \}$ is true for all valuations, then adding back the expressions containing $x$ and binding $x$ with an existential quantifier is true for all valuations because only finitely many values of $x$ are ruled out by the new conjuncts.

Case 2.3: $L^2$ has at least one term containing both $x$ and $=$

Here we use the fact that $S$ is an injection in our theory.

Suppose without loss of generality that $L^2_1$ contains $x$ and $=$. Further suppose without loss of generality that $l$ is of the form $S^k(x) = t_2$, i.e. $S^k(x)$ appears on the left rather than the right.

Take every expression from $L^2_2$ through $L^2_n$ and replace apply $S$ $k$ times to both sides, producing $L^3$, with $L^2_1$ unmodified.

$$ L^3_1 = L^2_1 \;\; \text{and} \;\; L^3_{1+n} = \big( L^2_{1+n} \; \text{with $S$ applied $k$ times to both sides} \big) $$

Now note that every occurrence of $x$ occurs inside an expression of the form $S^k(x)$.

Define $L^4$ in the following way, take every occurrence of $S^k(x)$ and replace it with $t_2$.

$$ L^4_i = \big( \text{ $L^3_i$ with $S^k(x)$ replaced with $t_2$ } \big) $$

So the first conjunct in $L^4$ will be $t_2 = t_2$.

$x$ appears nowhere in $L^4$, so

$$ \exists x \mathop. L_1 \land \cdots \land L_n \;\; \text{is equivalent to}\;\; L^4_1 \land \cdots \land L^4_n $$

This completes the argument that the successor theory has quantifier elimination.


问:
Solve the differential equation ${y}'\left ( x+ y^{2} \right )= y$

Solve the differential equation
$${y}'\left ( x+ y^{2} \right )= y$$

by jp.Wolfram|Alpha _the result is $\left ( 2y+ c_{1} \right )^{2}= c_{1}^{2}+ 4x\Leftrightarrow x= y\left ( y+ c_{2} \right )\Leftrightarrow 1= {y}'\left ( 2y+ c_{1} \right ),$ how can I break this cycle ?

答: Hint: Rewrite the DE as  $$\frac{dx}{dy}-\frac{x}{y}=y$$ which is easy by integrating factor method


问:
Do Cauchy-Riemann Equations Hold?
Consider $f(z) = \frac{(\overline{z})^2}{z}$ if $z \neq 0$ and $f(z) = 0$ if $z = 0$

Do Cauchy-Riemann Equations only hold at (0,0)?

I'm confused as to how to start this problem. I have tried converting the function $\frac{(\overline{z})^2}{z}$ into the form U(x,y) + iV(x,y) by letting $z = x+yi$ and $\overline{z} = x-yi$ but that just makes things over complicated and I end up with division by zero at the point (0,0).

Can anyone point me in the right direction?

答:
The Cauchy Riemann Equations are not well-defined. The function is continuous, but its derivative are not. Consider $\operatorname{f}(x+iy) = (x-iy)^2/(x+iy)$. Let $\operatorname{U}(x,y)$ be the real part and $\operatorname{V}(x,y)$ be the imaginary part of $\operatorname{f}(x+iy)$. We have:

\begin{array}{ccc}
\operatorname{f}(x+iy) &=& \frac{(x-iy)^2}{x+iy} \\ \\
&\equiv& \frac{(x-iy)^2(x-iy)}{(x+iy)(x-iy)} \\ \\
&\equiv& \left(\frac{x(x^2-3y^2)}{x^2+y^2}\right) + i\left(\frac{y(3x^2-y^2)}{x^2+y^2}\right) \end{array}

If we approach $z=0$ along different paths then we get different answers. For example, let $\gamma_{\theta}(r) = r\!\operatorname{e}^{i\theta}$, so that $x = r\cos\theta$ and $y = r\sin\theta$. What happens as $r \to 0$? Well,

\begin{array}{ccc}
(\operatorname{U}_x\circ \, \gamma_{\theta})(r) &\equiv& 12\cos^2\theta -8\cos^4\theta-3 \\
(\operatorname{U}_y\circ \, \gamma_{\theta})(r) &\equiv& -8\sin\theta\cos^3\theta \\
(\operatorname{V}_x\circ \, \gamma_{\theta})(r) &\equiv& 8\sin^3\theta\cos\theta \\
(\operatorname{V}_y\circ \, \gamma_{\theta})(r) &\equiv& 8\cos^4\theta - 4\cos^2\theta-1\\
\end{array}

These expressions are independent of $r$, and so as $r \to 0$ we get partial derivatives which depend upon the angle of approach. Moreover, computing the actual Cauchy Riemann equations does not rid us of the problem:

\begin{array}{ccc}
((\operatorname{U}_x-\operatorname{V}_y) \circ \, \gamma_{\theta})(r) &\equiv& 16\cos^2\theta - 16\cos^2\theta -2 \\
((\operatorname{U}_y+\operatorname{V}_x) \circ \, \gamma_{\theta})(r) &\equiv& 4\sin(2\theta)(1-2\cos^2\theta)
\end{array}

Title: Do Cauchy-Riemann Equations Hold?


问:
How to show this limit approaches $0$?
For the sequence $ a_n=\dfrac{2^n}{n!} $, is there a way to show that the $\lim_{n \to \infty }\dfrac{2^n}{n!}=0$?

I know that it is bounded above by $2$ because it is a decreasing sequence. I also know that the closest number it will every get to is $0$ because it won't be negative.

However, is there a way of showing that the limit goes to $0$?

答:
The trick is to see that for suiffiently large $n$, $n!>3^n.$

$$\lim_{n\to\infty}\frac{2^n}{n!}\leq\lim_{n\to\infty}\frac{2^n}{3^n} = \lim_{n\to\infty}\left(\frac{2}{3}\right)^n\to0.$$
See my answer here.

Title: How to show this limit approaches $0$?


问:
Show a function is bounded below by $\sqrt{a}$
I have to show a function $$f(x)= \frac{1}{2}\left(\frac{a}{x}+x\right),$$ where $x>0$, is bounded from below by $\sqrt a$.

The first thing I think of is: if a converges, then it is bounded. So my question: is looking if the limit of $f(x)$ from left and right is $\sqrt a$ the right way to go?

答: Clearly, $f(x)$ goes to infinity when $\,x\to 0^+$ or $x\to\infty.$ Since $f'(x)=0$ iff $x=\sqrt a, \ \ f(\sqrt a)$ is the lower bound.


答:
You have two isometric transformations, $A$ and $B$,
each of which is a combined translation and rotation.

Let's define some useful notation.
Define $T_{X,Y,Z}$ as the matrix for a translation by $X$ units in the $x$ direction,
$Y$ units in the $y$ direction, and $Z$ units in the $z$ direction.
Define $R_{x;\theta}$ as the matrix for a rotation by the angle $\theta$ around the $x$ axis; likewise $R_{y;\theta}$ and $R_{z;\theta}$ for rotations around the $y$ and $z$ axes, respectively.

Also, let's write our angles in radians.

With a little help from WolframAlpha, I think I have deduced that $A$ is actually the transformation whose matrix is
$$
M_A = T_{-0.0036,0.1235,-0.9213} R_{x;2.09819} R_{y;-0.008519} R_{z;-1.52068}.
$$

That is, first you rotate by $-1.52068$ radians around the $z$ axis;
then $-0.008519$ radians around the $y$ axis; then $2.09819$ around the $x$ axis;
and finally you translate by the vector $(-0.0036,0.1235,-0.9213)^T.$
Note that the matrix on the right is the one whose transformation is applied first, because you apply the transformation by putting any coordinates you want transformed in a column vector on the right side of the matrix $M_A,$
and when you expand $M_A$ into a product of component transformations like this the rightmost matrix is closest to the column vector and is the first to be multiplied with it.

We could also say that $A$ is a rotation by $2.40815$ radians ($137.977$ degrees) around the axis through the origin in the direction of the vector
$(0.674519, 0.638394, -0.370778)^T,$
followed by a translation by the vector $(-0.0036,0.1235,-0.9213)^T.$

Similarly, the matrix for $B$ is
$$
M_B = T_{-0.1241,-0.0112,-1.1028} R_{x;-1.01927} R_{y;0.0279253} R_{z;1.60745}.
$$

Now suppose we have a figure $F$ that is transformed by the rotation-translation $A$ into the figure $F_A$, whereas the rotation-translation $B$ transforms the figure $F$ into $F_B.$ Then the transformation that maps $F_B$ to $F_A$ is $B^{-1}$ followed by $A$, and the matrix of that transformation is $M_A M_B^{-1}$ (because, again, the matrix on the right is the matrix for the transformation that occurs first).

The matrix $M_B^{-1} M_A$ is something else altogether. I don't think it does what you think it does.

But since you are asking about a particular matrix you call $M_{BA} = M_B^{-1} M_A$, let's work with that matrix. This matrix works out to the following matrix product:
$$
M_{BA} = T_{-0.0882,-0.1114,0.2131} R_{x;3.12309} R_{y;-0.0248221} R_{z;0.086982}.
$$
Now you should expect that $M_{AB} = M_A^{-1} M_B$ is the inverse of $M_{BA},$
and indeed it is.
You can confirm this by multiplying the two matrices together.
But
$$
M_{AB} = T_{0.0830,-0.1230,0.2088} R_{x;-3.12099} R_{y;-0.0231151} R_{z;0.0874497}.
$$
As you observed, the angles do not match up in any obvious way.

The reason for mismatched angles is that
$$
\left(R_{x;3.12309} R_{y;-0.0248221} R_{z;0.086982}\right)^{-1}
\neq R_{x;3.12309}^{-1} R_{y;-0.0248221}^{-1} R_{z;0.086982}^{-1}.
$$
That is, you don't reverse a sequence of rotations around the $x$ axis, then the $y$ axis, then the $z$ axis by making a sequence of equal (or opposite) rotations around the $x$ axis, then the $y$ axis, then the $z$ axis.
If you want to re-use the angles, write
$$
\left(R_{x;3.12309} R_{y;-0.0248221} R_{z;0.086982}\right)^{-1}
 = R_{z;0.086982}^{-1} R_{y;-0.0248221}^{-1} R_{x;3.12309}^{-1}.
$$
That is, you can undo the sequence of rotations by exactly reversing the last rotation first, and then the second rotation, and then finally the first rotation.
But if you insist on rotating around the $x$ axis first both times, both for $M_{BA}$ and also for the inverse transformation $M_{AB},$ you need different sets of angles.
This will generally be true for all but the simplest cases.
Neither matrix multiplication nor transformations in three-D space are commutative in general; in most cases, it matters in what order you do them.

Notice that the components of the translation also don't match;
this is again because the transformations do not commute.
In most cases, if you want to reverse a rotation-and-then-translation using the inverse of the original translation, you have to apply the inverse translation first.

Title: Transforming two 3D vectors into each other gives different Euler angles

Body:
Lets take two vectors A and B represented in x,y,z(metres) and Euler angles XYZr(deg) as

```<code>A =

   -0.0036    0.1235   -0.9213  120.2176   -0.4881  -87.1283

B =

   -0.1241   -0.0112   -1.1028  -58.4000    1.6000   92.1000
</code>```

Converting them to 4x4 Homogeneous matrix gives

```<code>M_A =

    0.0501    0.9987   -0.0085   -0.0036
    0.5023   -0.0326   -0.8641    0.1235
   -0.8632    0.0390   -0.5033   -0.9213
         0         0         0    1.0000

M_B =

   -0.0366   -0.9989    0.0279   -0.1241
    0.5245    0.0046    0.8514   -0.0112
   -0.8506    0.0458    0.5238   -1.1028
         0         0         0    1.0000
</code>```

Transform B into A and A into B gives

```<code>M_BA = inv(M_B) * M_A
M_AB = inv(M_A) * M_B

M_BA =

    0.9959   -0.0868   -0.0248   -0.0882
   -0.0873   -0.9960   -0.0185   -0.1114
   -0.0231    0.0206   -0.9995    0.2131
         0         0         0    1.0000

M_AB =

    0.9959   -0.0873   -0.0231    0.0830
   -0.0868   -0.9960    0.0206   -0.1230
   -0.0248   -0.0185   -0.9995    0.2088
         0         0         0    1.0000
</code>```

Converting M_BA and M_AB into Euler angles gives

```<code>angles_BA =

  178.9396   -1.4222    4.9837

angles_AB =

 -178.8198   -1.3244    5.0105
</code>```

I was thinking that angles_BA and angles_AB should be same. But, they are not.
Why there is a difference between angles_BA and angles_AB ??


答: OK. If you want to have a proof that may be different from those you've seen so far. Here is one. Observe that: $a_{n+1} = \dfrac{2^{n+1}}{(n+1)!}= \dfrac{2^n}{n!}\cdot \dfrac{2}{n+1}= \dfrac{2a_n}{n+1}$. With this recursive equation for the $a_n$'s coupled with $a_4  = \dfrac{16}{24} = \dfrac{2}{3} < 1$, you can by induction on $n \ge 4$ show that $a_n < 1$. Thus it follows that $a_n < \dfrac{2}{n}, n \ge 4$. So $a_n \to 0$ as desired by the squeeze lemma in analysis if you will.

Title: How to show this limit approaches $0$?

Body:
For the sequence $ a_n=\dfrac{2^n}{n!} $, is there a way to show that the $\lim_{n \to \infty }\dfrac{2^n}{n!}=0$?

I know that it is bounded above by $2$ because it is a decreasing sequence. I also know that the closest number it will every get to is $0$ because it won't be negative.

However, is there a way of showing that the limit goes to $0$?


答:
The issue with looking at the limits is that it will only tell you things (by definition) at the two extreme points. It won't by itself tell you what the infimum is; just that there is one. You'll have to do extra work afterwards to prove that $\sqrt{a}$ is the infimum (here, minimum).

Here is a simple way to show the statement. By the AM-GM inequality, for every $x>0$
$$
f(x) = \frac{\frac{a}{x}+x}{2} \geq \sqrt{\frac{a}{x}\cdot x} = \sqrt{a}\,.
$$

Title: Show a function is bounded below by $\sqrt{a}$

Body:
I have to show a function $$f(x)= \frac{1}{2}\left(\frac{a}{x}+x\right),$$ where $x>0$, is bounded from below by $\sqrt a$.

The first thing I think of is: if a converges, then it is bounded. So my question: is looking if the limit of $f(x)$ from left and right is $\sqrt a$ the right way to go?


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答:
Wedderburn’s Theorem: Every finite division ring is a field.

Why should finiteness imply commutativity???

(Background: The only way a division ring can fail to be a field is if its multiplication is not commutative.)

LastEditorDisplayName: anon


问:
Find an angle in an isosceles triangle.
Question

Given $\triangle ABB'$, where $\angle BAB'=108^{\circ}$.
$\overline{BC}$ is the bisector of $\angle ABB'$, and $\overline{AB}=\overline{AB'}=\overline{B'C}$. Find $\angle{AB'C}$.

I prefer solutions without trigonometric functions, but answers with them are also fine.

I have tried constructing a point $D$ on $\overline{AB}$ such that $\overline{AB}=\overline{BD}$. Therefore, both $\triangle{AB'D}$ and $\triangle{ACD}$ are isosceles. However, the problem remains unsolved after such construction. I can't see the correct auxiliary lines to make.

Any suggestions, hints, or even full solutions (you don't have to) are appreciated.
Thanks for reading my post.

答:
I wanted you to notice it yourself, but I have already received a downvote so I will spoil the fun.

If $BC$ is an angle bisector and $\bigtriangleup ABB'$ an isosceles triangle can you find $\measuredangle ABC=\alpha$? Name the angles, say $\alpha$. The interior angles of $\bigtriangleup ABB'$ should sum to $180°$.
$$4 \alpha + 108°=180° \implies \alpha=18°$$

I was trying to show that this is a special triangle, in particular a golden gnomon. Therefore, the side $EC$ equals: $$EC=DC \cdot \phi$$

Now, apply the law of sines in $\bigtriangleup EFC$ to get $\measuredangle EFC=150°$.

It may help to recall that (which of often comes up in geometry problems, proof ):
$$\sin(18°)=\frac{1}{2\phi}$$

The interior angles of $\bigtriangleup EFC'$ should sum to $180°$. So, $\measuredangle ECF=12°$.

We know that $\measuredangle DCE=36°$, so you can easily find $\measuredangle AB'C=24°$.

Just for fun I completed the diagram to a regular pentagon :)

Title: Find an angle in an isosceles triangle.


答:
$${y}'\left ( x+ y^{2} \right )= y$$
$$xy'-y=-y'y^2$$
$$\left(\dfrac yx\right)'=-y'\dfrac {y^2}{x^2}$$
This DE is separable.
$$\dfrac {x^2}{y^2}d\left(\dfrac yx\right)=-dy$$
Integrate.

Title: Solve the differential equation ${y}'\left ( x+ y^{2} \right )= y$

Body:
Solve the differential equation
$${y}'\left ( x+ y^{2} \right )= y$$

by jp.Wolfram|Alpha _the result is $\left ( 2y+ c_{1} \right )^{2}= c_{1}^{2}+ 4x\Leftrightarrow x= y\left ( y+ c_{2} \right )\Leftrightarrow 1= {y}'\left ( 2y+ c_{1} \right ),$ how can I break this cycle ?


问:
In ZF, if injection from A to B doesn't exist, then does surjection from A to B exist?
Claim : For non-empty sets $A, B$, (if there's no injection from $A$ to $B$, then there exists a surjection from $A$ to $B$.)

I wonder how to prove or disprove that the 1st(2nd)-order logic sentence is generally true in ZF.

That claim is true in ZFC, since we can well-order both A and B, so A and B have cardinalities, the class of cardinalities are linearly ordered by inclusion. So, the claim isn't generally false in ZF.

The converse is false in ZF, since both surjection and injection can exist between the same set A.
And the similar sentence : for non-empty sets A and B, (if there exists a surjection from A to B, then there exists injection from B to A.),  is independent about ZF.

I guess the claim isn't generally true in ZF, and I tried to deduce the sentence which is independent about ZF, but not yet succeeded.
Also, I couldn't construct a model of ZF where the counter example exists.

How can I prove or disprove the claim is independent about ZF?

答: In fact, this statement is equivalent to the axiom of choice.  Given a nonempty set $B$, let $A$ be its Hartogs number, the least ordinal that does not inject into $B$.  Then, by your assumption, there must be a surjection $f:A\to B$.  But since $A$ is well-ordered, we can use this to construct an injection $g:B\to A$, by letting $g(b)$ be the least $a$ such that $f(a)=b$.  This gives a well-ordering of $B$.


问:
Let $a$ be a non zero real number. Evaluate the integral $\int \frac{-7x}{x^{4}-a^{4}}dx$
I hit a wall on this question. Below are my steps

$$\int \frac{-7x}{x^{4}-a^{4}}dx=-7\int \frac{x}{x^{4}-a^{4}}dx$$

Let $u=\frac{x^2}{2}, dx = \frac{du}{x}, x^{4}=4u^{2}.$

$$-7\int \frac{1}{4u^{2}-a^{4}}du=-7\int \frac{1}{(2u+a^2)(2u-a^2)}du$$

Use partial fraction decomposition,
$$\frac{1}{(2u+a^2)(2u-a^2)}=\frac{A}{2u+a^{2}}+\frac{B}{2u-a^{2}}.$$

Solve for $A$ and $B$:

$$\begin{cases}
A=\frac{1}{-2a^{2}}
\\
B=\frac{1}{2a^{2}}
\end{cases}$$

Now $$\int \frac{1}{(2u+a^2)(2u-a^2)}du=\int \frac{1}{-2a^2(2u+a^{2})}+\int \frac{1}{2a^{2}(2u-a^{2})}$$
Factoring out $a$ yields
$$\frac{7}{2a^{2}}(\int \frac{1}{2u+a^{2}}-\int \frac{1}{2u-a^{2}})$$

Evaluate the integral and substitute $u=\frac{x^{2}}{2}$ back.

My final answer is $$\frac{7}{2a^{2}}(\log(x^2+a^2)-\log(x^2-a^2)).$$

Feedback says my answer is wrong. Where did I mess up?

答:
Mistake appeared in evaluating:
$$\int\frac{1}{2u\pm a^2}\,\mathrm{d}u.$$
To solve this you have to substitute $v=2u\pm a^2$, $\mathrm{d}u=1/2 \mathrm{d}v$. We get
$$\frac{1}{2}\int\frac{1}{v}\,\mathrm{d}v.$$
From now on everything should be fine.

There is also a much easier way to evaluate this integral.

Substitute $u=-ix^2/{a^2}$. The differential is $\mathrm{d}x=ia^2/x^2\,\mathrm{d}u$.
$$\int-\frac{7ia^2}{2(-a^4u^2-a^4)}\,\mathrm{d}u = \frac{7i}{2a^2}\int\frac{1}{u^2+1}\,\mathrm{d}u = \frac{7i}{2a^2}\arctan u+C.$$
Substituting back gives us
$$\frac{7i\arctan\left(-\frac{ix^2}{a^2}\right)}{2a^2}+C=\frac{7i\operatorname{arctanh}\left(\frac{x^2}{a^2}\right)}{2a^2}+C.$$

If you want to get a solution with logarithm function involved, you can do that with this identity
$$\arctan(x)= \frac{1}{2i}\log \left( \frac{x-i}{x+i}\right).$$


答:
Let $A$ be an amorphous set; then there is no injection from $\Bbb N$ to $A$, and there is no surjection from $\Bbb N$ to $A$.

Let $f:\Bbb N\to A$. If $f$ were injective, then $A_0=\{f(2n):n\in\Bbb N\}$ and $A\setminus A_0$ would be disjoint infinite subsets of $A$ whose union is $A$, which is impossible, since $A$ is amorphous. And if $f$ were surjective, the function

$$h:A\to\Bbb N:a\mapsto\min\{n\in\Bbb N:f(n)=a\}$$

would be a bijection from $A$ to the infinite set $M=h[A]\subseteq\Bbb N$. If $\varphi:\Bbb N\to M$ is the canonical order-isomorphism, $h^{-1}\circ\varphi$ would be a bijection from $\Bbb N$ to $A$, which again is impossible.

Title: In ZF, if injection from A to B doesn't exist, then does surjection from A to B exist?

Body:
Claim : For non-empty sets $A, B$, (if there's no injection from $A$ to $B$, then there exists a surjection from $A$ to $B$.)

I wonder how to prove or disprove that the 1st(2nd)-order logic sentence is generally true in ZF.

That claim is true in ZFC, since we can well-order both A and B, so A and B have cardinalities, the class of cardinalities are linearly ordered by inclusion. So, the claim isn't generally false in ZF.

The converse is false in ZF, since both surjection and injection can exist between the same set A.
And the similar sentence : for non-empty sets A and B, (if there exists a surjection from A to B, then there exists injection from B to A.),  is independent about ZF.

I guess the claim isn't generally true in ZF, and I tried to deduce the sentence which is independent about ZF, but not yet succeeded.
Also, I couldn't construct a model of ZF where the counter example exists.

How can I prove or disprove the claim is independent about ZF?


问:
Minimize $z=e^{-x_{1}}+e^{-2 x_{2}}$ Subject to $\quad x_{1}+x_{2} \leq 1 \ , x_{1}, x_{2} \geq 0$ without KKT
I want to minimize $z=e^{-x_{1}}+e^{-2 x_{2}}$
subject to $\quad x_{1}+x_{2} \leq 1 \ ,  x_{1}, x_{2} \geq 0$ without using the KKT conditions. The restrain is on regular inequalities like Cauchy or Jensen ( some of the common known and used)
I don't know how to proceed. I tried to lower bound and get that bound to equality.
Any suggestions?

Body: By A.M-G.M inequality, we have $$e^{-x_1}+e^{-2x_2}=\dfrac{e^{-x_1}}{2}+\dfrac{e^{-x_1}}{2}+e^{-2x_2} \geq 3 \left(\dfrac{e^{-2x_1-2x_2}}{4}\right)^{\frac{1}{3}} \geq 3\left(\dfrac{e^{-2}}{4}\right)^{\frac{1}{3}} \;.$$