问:
Understanding concept of proof by induction
I'm having a little bit of trouble understanding the concept of proving by induction in certain cases, and I was hoping you could answer my question, which is broken down into 2 parts. I will use the following example to help illustrate my confusion.

Naturally, the base case is shown to be true. We would then assume that the following holds:

$$f(k) = 8^k - 3^k$$ is divisible by 5. Then we show that:

$$f(k+1) = 8\cdot8^k - 3\cdot3^k$$

Here's my first issue. I managed to solve this by showing that this can be re-written as:

$$f(k+1)=8(8^k-(3/8)3^k)=8\cdot8^k-8\cdot3^k+5\cdot3k^k$$
$$=8f(k)+5\cdot3^k$$

In the solutions, it says that since f(k) is divisible by 5, and the other term is also divisible by 5, then f(k+1) is also divisible by 5. Alternatively, it also shown that:

$$f(k+1) -8f(k) = 5\cdot3^k$$

Here are my questions:

Surely the entire proof rests on the assumption that f(k) is divisible by 5? If, say, it fails at f(3), then we haven't proved anything and the entire proof fails. It just seems a little circular to me, at least in this case.

Shouldn't we be rather aiming to prove the following algebraically?

$$ f(k+1) = 5\cdot (\mathrm{some\:function})$$
or
$$f(k+1) - f(k) = 5\cdot (\mathrm{some\:function})$$

instead of having 8f(k)? Even if using the solution works, I would imagine that working out one of these two alternatives should also be possible? I've seen other examples where this is exactly the case i.e. you show that f(k+1) - f(k) has a factor of the divisor in the answer, so therefore f(k+1) = f(k) + divisor*answer is what you need. I'll show another example to prove my point.

Long story short, you get the following:

$$f(k+1) - f(k) = 4(2(3^2k))$$

So why isn't it doable for the first question I showed?

Thanks!

答:
This is about your point #2.

For $f(k) = 8^k - 3^k$, we have
$$
f(k+1) - f(k) = 7 \cdot 8^k - 2 \cdot 3^k = 5 \cdot 8^k + 2(8^k-3^k) = 5 \cdot 8^k + 2f(k)
$$
and both terms are multiples of $5$.

As you say, ideally, $f(k+1) - f(k)$ would be a clear multiple of $5$. But it is not necessary to consider $f(k+1) - f(k)$. For instance, $f(k+1) - af(k)$ will do, for any $a$. In the case at hand, $a=3$ works, as the computation above shows:
$$
f(k+1) - 3f(k) = 5 \cdot 8^k
$$
You're not expected to guess $a=3$. Simplifying $f(k+1) - f(k)$ is a great starting point.


答:
The base case is important for the same reason.

You showed that it is true for $k=1$

You also showed that if it is true for $f(k)$ then it is also true for $f(k+1)$

Therefore,

It is true for $f(1)$ then it is true for $f(2)$

It is true for $f(2)$ then it is true for $f(3)$ and so on...

Hence it is true for all natural numbers

ContentLicense: CC BY-SA 4.0

Title: Understanding concept of proof by induction

Body:
I'm having a little bit of trouble understanding the concept of proving by induction in certain cases, and I was hoping you could answer my question, which is broken down into 2 parts. I will use the following example to help illustrate my confusion.

Naturally, the base case is shown to be true. We would then assume that the following holds:

$$f(k) = 8^k - 3^k$$ is divisible by 5. Then we show that:

$$f(k+1) = 8\cdot8^k - 3\cdot3^k$$

Here's my first issue. I managed to solve this by showing that this can be re-written as:

$$f(k+1)=8(8^k-(3/8)3^k)=8\cdot8^k-8\cdot3^k+5\cdot3k^k$$
$$=8f(k)+5\cdot3^k$$

In the solutions, it says that since f(k) is divisible by 5, and the other term is also divisible by 5, then f(k+1) is also divisible by 5. Alternatively, it also shown that:

$$f(k+1) -8f(k) = 5\cdot3^k$$

Here are my questions:

Surely the entire proof rests on the assumption that f(k) is divisible by 5? If, say, it fails at f(3), then we haven't proved anything and the entire proof fails. It just seems a little circular to me, at least in this case.

Shouldn't we be rather aiming to prove the following algebraically?

$$ f(k+1) = 5\cdot (\mathrm{some\:function})$$
or
$$f(k+1) - f(k) = 5\cdot (\mathrm{some\:function})$$

instead of having 8f(k)? Even if using the solution works, I would imagine that working out one of these two alternatives should also be possible? I've seen other examples where this is exactly the case i.e. you show that f(k+1) - f(k) has a factor of the divisor in the answer, so therefore f(k+1) = f(k) + divisor*answer is what you need. I'll show another example to prove my point.

Long story short, you get the following:

$$f(k+1) - f(k) = 4(2(3^2k))$$

So why isn't it doable for the first question I showed?

Thanks!


问:
Project topics on functional analysis.
For my final year course I have to do a project  on functional analysis. I just know about the basics like metric space, normed space, banach space, inner product space, Hilbert space and their properties. It would be helpful if some could suggest some interesting topics on I could do the project.</p>

答:
With that kind of background, I think you can start reading about the theory of $C^*$-algebras. You can introduce them abstractly as Banach algebras together with an involution that satisfy nice compatibility properties, or you can introduce them as being closed $*$-invariant subalgebras of $B(H)$ for some Hilbert space $H$. Within this view, your Hilbert space knowledge will come in handy! Many common objects that you already encountered are $C^*$-algebras:

Bounded (compact) operators on a Hilbert space.

Continuous functions on a compact space.

$L^\infty(\Omega, \mu)$ for a measure space $(\Omega, \mu)$.

An excellent starting place would be the book '$C^*$-algebras and operator theory' by Murphy.

You will need some basic notions from topology and topological vector spaces though, but not too much unless you go further in the theory.

Title: Project topics on functional analysis.


答:
Consider a bunch of dominoes which, for all you know, are standing up in a line. These facts are brought to your attention:

The first domino has actually fallen over.

When one domino falls over, the next in line one also falls over.

You are free to conclude that all of the dominos have fallen over. Lets follow this setup for your example.

Since $8^1-3^1 = 5^1$, $f(1)$ is divisible by 5. (thus the first domino has fallen)

We need to show that one domino falling causes the next one to fall. That is, we need to show: if some $f(k)$ is divisible by $5$, then the next one $f(k+1)$ will also be divisible by 5.

You showed that $f(k+1)=8f(k)+5\cdot3^k$ (nice work). So, if $f(k)$ is divisible by 5, and since certainly $5\cdot3^k$ is divisible by 5, we get that the $f(k+1)$ must be divisible by 5. (one domino falling causes the next to fall)

You conclude that all of the dominoes fall.


问:
Why is the FFT faster than the naive implementation of DFT?
I read several explanations about why the FFT algorithm has the complexity $\mathcal{O}(n \cdot \log(n))$ and not just $\mathcal{O}(n^2)$ like if we directly multiplied the input by the Vandermonde Matrix, but I think I am missing something. I understand that we recursively split the problem in two parts, with the odd elements on one side and the even elements on the other side as shown here:

\begin{align}
X_k &= \sum_{n=0}^{N-1} x_n \cdot e^{-i~2\pi~k~n~/~N} \\
    &= \sum_{m=0}^{N/2 - 1} x_{2m} \cdot e^{-i~2\pi~k~(2m)~/~N} + \sum_{m=0}^{N/2 - 1} x_{2m + 1} \cdot e^{-i~2\pi~k~(2m + 1)~/~N} \\
    &= \sum_{m=0}^{N/2 - 1} x_{2m} \cdot e^{-i~2\pi~k~m~/~(N/2)} + e^{-i~2\pi~k~/~N} \sum_{m=0}^{N/2 - 1} x_{2m + 1} \cdot e^{-i~2\pi~k~m~/~(N/2)}
\end{align}

But I don't see why there are less operations when we split it in two. As I understand it, for each $X_k$ we still have to do $N/2$ additions and multiplications for the even part and $N/2$ additions and multiplications for the odd part, which is exactly the same as if we directly did N additions and multiplication for each $X_k$. So, why are there less operations needed using the FFT?

答:
Consider, for instance, that you have a degree-7 polynomial $f$, and you want to evaluate it at every 8th root of unity (primitive and non-primitive).

Naively, this takes 28 multiplications and 7 additions per evaluation, for a total of 308 operations.

However, we can do much better. Split $f$ into a sum
$$
f(x)=g_e(x^2)+xg_o(x^2)
$$
for two degree-3 polynomials $g_e$ and $g_o$. Now if we can evaluate $g_e(x^2)$ and $xg_o(x^2)$ for only the first four $8$th roots (1, as well as the ones with positive imaginary part), then because of the even-ness / odd-ness of the two functions, evaluating $f$ at the remaining four roots come basically for free.

In total (including the squaring $x^2$), with naive polynomial evaluation, this gives 15 multiplications and 7 additions for each of the first four roots, and just a single subtraction for each of the other four. A total of 76 operations rather than 308. However, we aren't done with the optimisations.

Now for the really cool part: evaluating $g_e(x^2)$ and $g_o(x^2)$ for the first four eighth roots beginners evaluating $g_e(x)$ and $g_o(x)$ for all the fourth roots of unity. So exactly the same trick can be applied again! So it turns out that the above paragraph is very pessimistic; it takes much less than 15 multiplications and 7 additions for each of the first four eighth roots.

Even if you, from the start, are clever about how you evaluate $f$ by, say, using your calculated $x^4$ when you find $x^5$ (the reduction from 28+7 to 15+7 operations is of this nature), the savings are considerable, as each step of recursion essentially halves the number of points you need to evaluate (more than a subtraction) at.

Consider, for instance, how this method would calculate its last point, $f(e^{7\pi i/4})$. It would be found as $g_e(e^{14\pi i/4})+e^{7\pi i/4}g_o(e^{14\pi i/2})$. However, we have $e^{14\pi i/4}=e^{6\pi i/4}$, which means that because we previously calculated $$f(e^{3\pi i/4}) = g_e(e^{6\pi i/4}) + e^{3\pi i/4}g_o(e^{6 \pi i/4})$$ we got almost everything we need, and get $$f(e^{7\pi i/4}) = g_e(e^{6\pi i/4}) - e^{3\pi i/4}g_o(e^{6 \pi i/4})$$This is what I mean by $f(e^{7\pi i/4})$ costing only a single subtraction.


问:
Is $S = \{ (x,y): x^2+y^2=\frac {1}{n^2}, n \in \mathbb N \wedge (x \in \mathbb Q \vee y \in \mathbb Q) \}$ countable?
I am working on the following problem:

Is the set $S = \left\{ (x,y): x^2+y^2=\frac {1}{n^2}, n \in \mathbb N \wedge (x \in \mathbb Q \vee y \in \mathbb Q) \right\}$ countable?

My answer is yes.

Reason:

For a fixed $n$, let $S_n = \left\{ (x,y): x^2+y^2=\frac {1}{n^2}, n \in \mathbb N \wedge (x \in \mathbb Q \vee y \in \mathbb Q) \right\}$.

Then $S_n = (\mathbb Q \times \mathbb Q) \cup \bigcup\limits_{r\in \mathbb Q}\left(\pm r, \pm\sqrt {\frac {1}{n^2}-r^2}\right)$ where $r <\frac {1}{n^2} $.

Clearly each set forming $S_n$ is countable, its union ($S_n$) is also countable.
As $S=\bigcup\limits_{n\in \mathbb N} S_n$, $S$ is also countable.

Am I correct? Thanks in advance!

答:
Given that $~S = \{ (x,y): x^2+y^2=\frac {1}{n^2}, n \in \mathbb N \wedge (x \in \mathbb Q \vee y \in \mathbb Q) \}~.$

So there doesn't exist any pair of $(x,y)$ other than $(1/n,0)$ and $(0,1/n)$, for all $n \in \mathbb N$, that satisfies the given condition.

Now we know that $~n \in \mathbb N ~$  is countable and countable union of countable sets is countable. Hence $~S=(1/n,0)\cup(0,1/n)~$ is countable.


问:
Is the complement of a C7 graph planar or non-planar?
I am trying to prove whether the complement of a cycle-7 graph is planar or non-planar.
I tried to use Euler's theorem to prove it;

In a connected simple planar graph with v vertices and e edges, if v ≥ 3, then e ≤ 3v−6.

I have that the complement of a C7 has 7 vertices and 14 edges.. Plugging this into Euler's theorem this comes out as 14 ≤ 15, which obviously holds, so this method does not prove non-planarity.
I've also tried to find K5 and a K3,3 as subgraphs but also had no luck there either.

Any helpful tips on how to go about it from here?

I've included a photo of the only version of the complement of C7 I could find, apologies for the vertices being labelled as weekdays. I just wanted to show the graph more than anything

答: While the question is perfectly answered by citing the Kuratowski theorem, it is amusing to note that this workday graph actually defines a Moebius strip. You can build it with paper consisting of 7 triangles. The fact that it is not planar can be seen now as a consequence that it is not possible to ``flatten" the Moebius strip: it would provide an orientation, but the Moebius strip is non-orientable. Picture of Workday graph as Moebius strip</p>

ContentLicense: CC BY-SA 3.0


问:
Doubt in chapter $6.8$ of Hoffman and Kunze.
Let $T$ be a linear operator such that minimal polynomial of $T$ is the product of first degree polynomial then $T$ can be written as $D+N$ where $D$ is a diagonalizable linear operator and $N$ is nilpotent linear operator.

Now, this is something that is after the primary decomposition theorem. I am really confused  understand if there is some link between the two.

My understanding is that $T$ w.r.t. some basis can be written as a triangulable linear operator and so it can be written as a sum of nilpotent operator and a diagonalizable linear operator.is this ok?

答:
Well, if the minimal polynomial of $T$ is the product of first degree polynomials, then $T$ is diagonalizable. This is an immediate consequence of the Primary Decomposition Theorem.

Therefore

$$T = T + 0$$ where $T$ is diagonalizable and $0$ obviously nilpotent.


问:
Difficult Cauchy Problem
Let $a, b, c>0$ such that $a^{2}+b^{2}+c^{2}=3 a b c$. Prove the following inequality:
$$
\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \geq \frac{9}{a+b+c}
$$

I tried using the fact that $a^2 + b^2 + c^2 = 3abc$ but I could only think of one case where $a=b=c$. I also tried using knowing what is $a^2 + b^2 + c^2$ in terms of $a$, $b$ and $c$ to try and get $a+b_c$ in terms of them using the sum of squares where $(a+b)^2 - 2ab = a^2 + b^2$ but I reached a dead end where I could not simplify any further.

I think assuming $a=b=c$ did get me somewhere but since it is a proof problem I cannot assume but rather have to prove that $a=b=c$. I cannot figure that bit out.

I know Cauchy but very little bit, so I do not now if it can be applied too well here but I think it is a right approach.

答:
[enter image description here][1]

[1]: https://i.stack.imgur.com/VF8FZ.jpg            hi this is my answer is it true?

Title: Difficult Cauchy Problem


问:
Why is -(-x), the negation of a negation, not the product of two negations?
I was told that the negation of a negation is not the product of two negations. I understand this to be so because if we prove $-(-x)=x$, we do so using the property of negation:
$x+(-x)+(-(-x))=0$, which would equal $x$ and $-(-x)$. I get this part.

However, we know that $-1x=-x$. So if we think of $-(-x)$ as $-1\times(-1x)$, is this not the product of two negations?

答:
As a matter of fact, yes, $-(-x) = (-1)(-x)$, so in that sense, you could say that $-(-x)$ is the product of two negations.

If you write down the expression $-(-x)$ on a sheet of paper, you haven't written it as a product of two negations, because you haven't written a multiplication sign, nor have you used any other kind of notation that denotes multiplication. But, as you know, negating is the same thing as multiplying by $-1$, so there isn't any important difference between $-(-x)$ (which is written without multiplication) and $(-1)(-x)$ (which is written with multiplication).

In response to:

I was under the assumption that the parentheses after the - denote multiplication, but is this a poor way to think about negation?

In this case, no, the parentheses after the negation sign don't denote multiplication. You can denote multiplication by writing two expressions (that is, two sequences of symbols that represent numbers) next to each other. However, the expression $-(-x)$ does not mean $-$ multiplied by $(-x)$, because $-$ is not an expression (that is, $-$ is not a symbol that represents a number).

That said, I think that no, this is not a poor way to think about negation. Negation is the same thing as multiplying by $-1$ (even if it's not technically defined that way), so, in my opinion, there's nothing wrong with seeing a negation sign and thinking "that means 'multiply by $-1$.'"


问:
Confusing math problem
How would I solve this question? I came across it and is really confused.

The payment of Jon was bigger by $960$ than the payment of David.
After the payment of David got increased by $10\%$, Jon and David got the same payment amount.

$\text{A}$. We will mark with $X$ the payment of David in the beginning and using the $X$ find the payment of Jon.

$\text{B}$. Find out the payment of David in the beginning.

How can I solve this? Any hints?

答: Let $J$ = payment of John and $X$ = payment of David. Then we know that $J = X + 960$. Furthermore, $1.1X = J$. Now substituting the first equation into the second we obtain: $1.1X=X+960 \iff 0.1X=960 \iff X=9600 \rightarrow$ $J=9600 + 960 = 10560$ (we obtain J from substituting $9600$ into the first equation).</p>

Title: Confusing math problem


问:
Prove that if $a$ and $b$ are odd positive integers, then there are integers s and t such that $a = bs + t$ where |t| < b, and t is either zero or odd
Prove that if a and b are odd positive integers, then there are integers s and t such that $$a = bs + t,$$ where $|t| \lt b$, and $t$ is either zero or odd.

答:
I am also looking at this problem right now. I am a bit confused. We know that both $a$ and $b$ are odd and by the division algorithm we have that $a=b\cdot s + t$ where $b>0$ and $0\le t< b$. Consider $a=b\cdot s + t$. Since $a$ and $b$ are both odd, then we may consider two cases:

$s$ is even

$s$ is odd

If 1., then clearly $t$ has to be either odd or $=0$. If 2., then either $t=0$ or $t$ is even!? Since if we assume that $s$, $a$ and $b$ are odd then we may write $$a=b\cdot s + t=(2n+1)(2j+1)+t=\bigl[ 2\cdot (2nj+n+j)+1 \bigr]+t,$$ so $t$ must be either $=0$ or even whenever $a$ is odd. Is this correct? Because if it is, then it is not true that $t=0$ or $t$ is odd (?).


答:
Proposed inequality can be rewriten like this:$$(a+b+c)(a^3+b^3+c^3)\geq 9a^2b^2c^2$$

By Cauchy inequality we have $$(a+b+c)(a^3+b^3+c^3) \geq (a^2+b^2+c^2)^2$$ and we are done!

Title: Difficult Cauchy Problem

Body:
Let $a, b, c>0$ such that $a^{2}+b^{2}+c^{2}=3 a b c$. Prove the following inequality:
$$
\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \geq \frac{9}{a+b+c}
$$

I tried using the fact that $a^2 + b^2 + c^2 = 3abc$ but I could only think of one case where $a=b=c$. I also tried using knowing what is $a^2 + b^2 + c^2$ in terms of $a$, $b$ and $c$ to try and get $a+b_c$ in terms of them using the sum of squares where $(a+b)^2 - 2ab = a^2 + b^2$ but I reached a dead end where I could not simplify any further.

I think assuming $a=b=c$ did get me somewhere but since it is a proof problem I cannot assume but rather have to prove that $a=b=c$. I cannot figure that bit out.

I know Cauchy but very little bit, so I do not now if it can be applied too well here but I think it is a right approach.


问:
How does Cantor's diagonal argument work?
I'm having trouble understanding Cantor's diagonal argument.  Specifically, I do not understand how it proves that something is "uncountable".  My understanding of the argument is that it takes the following form (modified slightly from the wikipedia article, assuming base 2, where the numbers must be from the set $ \lbrace 0,1 \rbrace $):

\begin{align}
s_1 &= (\mathbf{0},1,0,\dots)\\
s_2 &= (1,\mathbf{1},0,\dots)\\
s_3 &= (0,0,\mathbf{1},\dots)\\
\vdots &= (s_n \text{ continues})
\end{align}

In this case, the diagonal number is the bold diagonal numbers $(0, 1, 1)$, which when "flipped" is $(1,0,0)$, neither of which is $s_1$, $s_2$, or $s_3$.

My question, or misunderstanding, is: When there exists the possibility that more $s_n$ exist, as is the case in the example above, how does this "prove" anything?  For example:

\begin{align}
s_0 &= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\
s_1 &= (\mathbf{0},1,0,\dots)\\
s_2 &= (1,\mathbf{1},0,\dots)\\
s_3 &= (0,0,\mathbf{1},\dots)\\
s_4 &= (0,1,1,\mathbf{1},\dots)\\
s_4 &= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\
s_5 &= (1,0,0,0,\dots)\\
s_6 &= (1,0,0,1,\dots)\\
\vdots &= (s_n \text{ continues})
\end{align}

In other words, as long as there is a $\dots \text{ continues}$ at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it:

For each $m$ and $n$ let $s_{n,m}$ be the $m^{th}$ element of the $n^{th}$ sequence on the list; so for each $n$,

$$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$

...snip...

Otherwise, it would be possible by the above process to construct a sequence $s_0$ which would both be in $T$ (because it is a sequence of 0s and 1s which is by the definition of $T$ in $T$) and at the same time not in $T$ (because we can deliberately construct it not to be in the list). $T$, containing all such sequences, must contain $s_0$, which is just such a sequence. But since $s_0$ does not appear anywhere on the list, $T$ cannot contain $s_0$.

Therefore $T$ cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable.

I'm not sure this definition is correct, because if we assume that $m = (1, \dots)$, then this definition says that "$s_n$ is equal to itself"&mdadsh;there is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible $T$ list above.  An attempt to correct the "diagonalization" and "flipping" problem:

$$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where $m$ is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 & \mathrm{if\ } s_{m,m} = 1\\1 & \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$

This definition doesn't quite work either, as we immediately run in to problems with just $s_1 = (0),$ which is impossible because by definition $s_1$ must be $ = (1)$ if $s_1 = (0)$, which would also be impossible because... it's turtles all the way down!?  Or more generally, with the revised definition there is a contradiction whenever $n = m$, which would seem to invalidate the revised formulation of the argument / proof.

Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable".

What am I missing?

答:
The diagonal argument is best understood by first examining finite instances. Suppose $\rm\:M_{\:\!i\:j}\:$ is a $\rm\ n \times n\ $ matrix whose entries lie in a set $\rm\:T\:$ with at least two elements. Then one can construct a $\rm\ 1\times n\ $ row $\rm\:R\:$ different from any row $\rm\:M_{\:\!i}\:$ of $\rm\:M\:$ by taking the diagonal $\rm\:M_{\:\!i\: i}\:$ and changing each of its entries, viz. $\rm\:R_{\:\!i} := \neg\: A_{\:\!i\: i},\:$ where $\:\neg\:$ is any "change" function on $\rm\:T,\:$ i.e. $\rm\: \neg\: t \ne t\:$ $\rm\:\forall\:t\in T.\:$ Note that $\rm\:R\:$ is not equal to any row $\rm\:M_{\:\!i}\:$ since its $\rm\:i$'th entry differs, i.e. $\rm\:R_{\:\!i} =\: \neg\: M_{\:\!i\:i}\ne M_{\:\!i\:i}\:.\:$

Viewing each row $\rm\:M_{\:\!i}\:$ as a function $\rm\:j\mapsto M_{\:\!i\: j}\:$ from $\rm\:n = \{0\ 1\:\cdots\: n-1\}\:$ to $\rm\:T,\:$ the proof shows that there are more than $\rm\:n\:$ such functions, i.e. $\rm\:|T^{\:\!n}| > n\:$ if $\rm\:|T|\ge 2\:.\:$ Obviously the proof generalizes from $\rm\:n\:$ to any set $\rm\:S,\:$ yielding that $\rm\:|T^S| > |S|\:.\:$ This illustrates the simple essence of diagonalization - which is due not to Cantor, but to du Bois-Reymond (who used it to diagonalize on growth rates of functions, i.e. "orders of infinity").


问:
Is there a direct proof for $\int_0^{2\pi}\frac{r^2+r(\cos t-\sin t)}{1+2r\cos t+r^2}dt=2\pi$
Is there a direct method to prove that, if $r>1$
$$I=\int_0^{2\pi}\frac{r^2+r(\cos(t)-\sin(t))}{1+2r\cos(t)+r^2}dt=2\pi$$

I ask this question because this integral can be thought as an integral of a 1-form along a circle:
$\omega=\frac{x-y}{x^2+y^2}dx+\frac{x+y}{x^2+y^2}dy$
and $\gamma=(1+r\cos(t),r\sin(t))$ for $t\in[0,2\pi]$ and if I'm not mistaken this curve is homotopic to the unit circle centered in $(0,0)$ if $r>1$, then the integral is equal to the integral along this curve that is $2\pi.$ Is my argument correct?

答:
Utilize the Fourier series

$$\frac{r^2-1}{1+2r\cos t+r^2}=1+ 2\sum_{n=1}^{\infty}\frac{(-1)^n}{r^n}\cos(nt )
$$
and observe that only the first two terms in the series survive the integration, i.e.

$$\int_0^{2\pi}\frac{r^2+r(\cos t-\sin t)}{1+2r\cos t+r^2}dt
= \frac1{r^2-1}\int_0^{2\pi}(r^2- 2\cos^2t)dt
=2\pi$$


问:
hint on a solved old exam question on probabilistic methods calcualation
In my note I have some previous exam solved question as follows in Probabilistic methods section:

Example: We have $k$ classes $C_1, C_2,...,C_k$ where each $C_i$ has uniform distribution over
$-(2^{i-2})<x<2^{i-2}$.

by using maximum likelihood ratio test, then $ \lim_{k\rightarrow \infty} $P$(error)= \frac{1}{2}$

$ \frac{1}{2}$ is Calculated by following idea:

**

My challenge is I couldn't understand the logic of this answer. is
there any intuitive idea to better understand here?

**

答:
If we observe a random variable $\mathbf X$ which could be sampled from any of the $k$ distributions $C_1, C_2, \dots, C_k$ with equal probability, then the maximum-likelihood guess for which it was always goes with the narrowest option that could explain the value of $\mathbf X$.

For example, if we observe $\mathbf X = 5$, then this could have come from $C_5$ (distributed on $[-8,8]$), $C_6$ (distributed on $[-16,16]$) or any of $C_7, \dots, C_k$, but it could not have come from $C_1, C_2, C_3, C_4$. The maximum-likelihood guess is $C_5$.

If in fact $\mathbf X$ came from $C_1$ (which happens with probability $\frac1k$) then $C_1$ will always be the narrowest option explaining $\mathbf X$, so the probability of error is $0$.

Otherwise, suppose $\mathbf X$ came from $C_i$ with $i>1$, which is distributed uniformly on $[-2^{i-2}, 2^{i-2}]$.

For half that range, $[-2^{i-2}, -2^{i-3}) \cup (2^{i-3}, 2^{i-2}]$, the narrowest option explaining $\mathbf X$ is $C_i$, so the maximum-likelihood guess is correct.

For the other half of the range, $[-2^{i-3}, 2^{i-3}]$, $C_{i-1}$ could also explain $\mathbf X$, and will be preferentially guessed over $C_i$. (Closer to $0$, $C_1, \dots, C_{i-2}$ could become even better guess, but that doesn't change anything.) The maximum-likelihood guess is wrong.

Therefore in the $i>1$ case the probability of error is $\frac12$.

The overall probability of error is $\frac1k \cdot 0 + (1 - \frac1k) \cdot \frac12 = \frac{k-1}{2k}$. As $k \to \infty$, this approaches $\frac12$.


问:
Chain rule for inner product.
Let $f:\mathbb{R}^n\to \mathbb{R}^n$ and $M\in\mathbb{R}^{n\times n}$.

Am I using the chain rule correctly to calculate $$  \nabla_x\langle f(x),M f(x) \rangle= \nabla_u\langle u,M u\rangle |_{u=f(x)} D_xf(x) ~~~?$$

where $D_x$ is the matrix of first derivatives of $f$.

答:
$h(x) = \langle f(x),M f(x) \rangle$ is the composition $h = g \circ f$ where

$$g(u) = \langle u,M u \rangle.$$

The chain rule states that

$$h^\prime(x) = g^\prime(f(x)) \circ f^\prime(x).$$

So yes, what you wrote is correct, knowing that
$$g^\prime(u).k = \langle u,M k \rangle + \langle k,M u \rangle$$ which leads to

$$h^\prime(x).k = \langle f(x),M (f^\prime(x).k) \rangle + \langle f^\prime(x).k,M .f(x) \rangle$$

Title: Chain rule for inner product.


问:
Given that $p$ is a prime, and the sum of all positive divisors of $p^4$ is a perfect square, find the possible number of primes $p$ .

Given that $p$ is a prime, and the sum of all positive divisors of $p^4$ is a perfect square, find the possible number of primes $p$ .

What I Tried: I know that the only divisors of $p^4$ will be $(1 , p , p^2 , p^3 , p^4)$ . From here I can conclude :-
$$\rightarrow 1 + p + p^2 + p^3 + p^4 = k^2$$ For some positive integer $k$ . Now this is where I get stuck, I can write it as :-
$$\rightarrow \frac{(1 - p^5)}{1 - p} = k^2$$
But I didn't understand how I can find all values of $p$. Next what I did is :-
$$\rightarrow p(1 + p + p^2 + p^3) = (k + 1)(k - 1)$$
And I am stuck at the same problem.

Can anyone help me here?

答:
Just some thoughts. Still editing...

Note that the sum $1+p+p^2+p^3+p^4$ is odd, so $k^2\equiv1\bmod 8$, and hence, $p(1+p+p^2+p^3)$ is a multiple of $8$.

If $p=2$, this isn't the case, so $p$ must be odd, and hence, $1+p+p^2+p^3$ must be a multiple of $8$.

mod $8$, $1+p+p^2+p^3=1+p+1+p$, so $2+2p$ is a multiple of $8$, so $1+p$ is a multiple of $4$. So, $p\equiv3\bmod 4$.

Through similar logic $\bmod{12}$, we have $p\equiv-1\bmod{12}$, or $p=3$.


问:
Results following from Analyticity on a domain
This is part of an old Oxford exam paper (1997 2602 Q2) I'm working on for revision.

Suppose we have a function $f$ which is holomorphic on the disc radius $R$ about $0$. We want to show that there is a sequence $\{p_n\}$ of polynomials such that $\{p_n\} \rightarrow f$ uniformly on the circle centre $0$ radius $r<R$.

My thoughts are that by Taylor's Theorem we can expand $f$ as a power series on $D(0;R)$ so that $f(z)=\sum^{\infty}_{i=0}c_i z^i$ for $z \in D(0;R)$.

This leads me to propose defining $$p_n(z)= \sum^{n}_{i=0}c_i z^i$$

Clearly then $\{p_n\} \rightarrow f$ so we just require the convergence to be uniform.

Can anyone help me show the convergence is uniform?

The questions then continues to asking us to evaluate $\int_{\alpha}z^k dz$ where $\alpha$ is the positively-oriented circle centre $0$ radius $r$.

I think that this integral is simply equal to zero by Cauchy's Theorem, though I'm more than happy to be corrected.

Now, we have to use this to show that there is no sequence of polynomials that converge uniformly to $\frac{1}{z}$ on $\alpha$. Can someone please assist with this part also.

Thanks

答:
Hint: if $r < t < R$, show that $|c_n| t^n $ is bounded, and use that to get uniform estimates for $\left|\sum_{i= n}^\infty c_i z^i\right|$ for $|z| \le r$.

For the second part: yes, the integral is $0$ for nonnegative integers $k$.  And so $\int_\alpha p(z)\ dz = 0$ if $p$ is a polynomial.  For $1/z$, on the other hand, ...


问:
Can any control problem be put in affine form?
For a control problem given by the dynamics $\dot{\mathbf x} = f(\mathbf x, \mathbf u)$, we may extend the state space by declaring $\mathbf y =(\mathbf x, \mathbf u)^\top$ and writing
$$
\dot{\mathbf y}=\begin{bmatrix}f(\mathbf x, \mathbf u)
\\
\tilde{\mathbf u}\end{bmatrix} =h(\mathbf y)+\sum \mathbf e_i \tilde u_i
$$
Where the $\mathbf e_i$ are the standard basis vectors. Now the control inputs are $\mathbf u$ and the problem is control affine. My question is: is this valid? All the books I have seen on optimal control say things such as "if the problem is control affine" instead of saying "assuming the problem is given in affine form". This makes be believe that something is being thrown away or ignored in the above reformulation. Is it the fact that we are in a way smoothing out $\mathbf u$ by taking it to be the integral of $\tilde{\mathbf u}$? What am I missing?

答:
Optimal control quite often searches for the optimal gains given a fixed structure, for example static state feedback.

If you change the structure, you change the solution and the performance you can achieve.

We don't need a nonlinear system to show this. Take the standard LQR problem with the cost function

$$
J = \int_0^\infty (x^T Q x + u^T R u) dt
$$

Now take the system

$$
\dot{x} = A_1 x + B_1 u
$$

and use $A_1 = -1, B_1= 1, Q_1 = 10, R_1 = 1$. The LQR solution is $u = -K_1 x$ and $K_1 = 2.3166$.

Although the system is linear we can still feed $u$ through an integrator. Now you have:

$$
\dot{y} = A_2y + B_2 v
$$

and $A_2 = \begin{pmatrix}-1 & 0 \\ 0 & 0 \end{pmatrix}, B_2 = \begin{pmatrix}0 \\ 1 \end{pmatrix}, Q_2 = \begin{pmatrix}10 & 0 \\ 0 & 0 \end{pmatrix}, R_2 = 1$.

The LQR solution is $v = -K_2 y$ and $K_2 = \begin{pmatrix}1.4559 & 1.7064\end{pmatrix}$.

You can see that the cost function is for both systems the same:

$$
\begin{align}
J_1 &= 10 x^2 + u^2 \\
J_2 &= 10 y_1^2 + v^2
\end{align}
$$

But does that mean we can reach the same performance? No, we can't.

The cost value depends of course on the initial condition:

$$
J = x_0^T S x_0
$$

with $x_0 = x(0)$ and

$$
\begin{align}
S &= \int_0^\infty \Phi^T (Q + K^T R K) \Phi dt \\
\Phi &= e^{(A - B K)t}
\end{align}
$$

You can use this to calculate the cost value for an initial condition.

For example, say we start at $x(0) = y_1(0) = 1$. Then you get:

$$
\begin{align}
J_1 &= 2.3166 \\
J_2 &= 1.7064 y_2(0)^2 + 2.9118 y_2(0) + 3.9402 \\
\end{align}
$$

So $J_2$ (of course) also depends on the inital condition $y_2(0)$ of your new integrator. But maybe we can tune the initial condition of the integrator so we get $J_1 = J_2$?

Again: No, we can't, in general.

You can easily see this by minimizing $J_2$ over $y_2(0)$ which is easy because it is just a quadratic polynomial:

$$
\min_{y_2(0) \in \mathbb{R}} 1.7064 y_2(0)^2 + 2.9118 y_2(0) + 3.9402
$$

You can solve this by setting the derivative w.r.t. $y_2(0)$ to zero:

$$
3.4128 y_2(0) + 2.9118 = 0
$$

So the "best" initial condition for the integrator is:

$$
y_2(0) = -0.8532
$$

If you insert this into the cost value:

$$
\boxed{
\begin{align}
J_1 &= 2.3166 \\
J_2 &= 2.6980
\end{align}}
$$

So you can see that $J_1 < J_2$.

Changing the structure of the controller changed the best possible cost that we can achieve.

If you simulate:

You can see that $x$ and $y_1$ are different although they minimize the same cost function.

Conclusion: In general you can't just change the structure of your system because you will get a different solution. Here the difference is only small but it could be huge of course.


问:
Prove or disprove this statement: For any $y \in \mathbb{Z}$ and $y \neq \pm 1$, $(y+1)^{2}$ is not divisible by $y$.
I want to prove or disprove this statement:

For any $y \in \mathbb{Z}$ and $y \neq \pm 1$, $(y+1)^{2}$ is not divisible by $y$.

The case where $y$ is even can be easily proved. However, I am stuck at the case where $y$ is odd. I try to start with the fact that "$y+1$ is not divisible by $y$" and expect to use something like the Euclid's Lemma. Any idea on how to proceed?

答:
Since you tagged it "polynomials" we highlight their key role here.

Note that $\ y\mid y^2\!+\!1\iff y\mid (\overbrace{y^2\!+\!1\bmod y}^{\large \color{#c00}1})\!\iff y\mid \color{#c00}1$

Generally  $\ y\mid f(y)\iff y\mid \overbrace{f(y)\bmod y}^{\large \color{#c00}{f(0)}}\iff y\mid \color{#c00}{f(0)}\,$ via Polynomial $\rm\color{#c00}{Remainder}$ Theorem, where $\,f(x)\,$ is any polynomial with integer coefficvients.


问:
Matrix Chain Rule Derivative Question
Matrix Partial Derivative and Chain Rule

I don't know how the partial derivative is taken here, and was wondering if all of the steps can be spelled out. The image is linked.

Specific points of confusion are:

Why $V_j^T x$ is taken inside $g'$ -- I know that we're dealing with the jth row of the $V$ matrix, but once we've identified the jth row, why is it necessary to transpose? I thought $V_j$ is a row and $x \in \R^n$

Why is $V_{ji}$ multiplied to the end of the expression, instead of $V_{ji}^T$? I thought the derivative of $Vx$ is $V^T$

答:
$\def\c#1{\color{red}{#1}}$Define the following vectors
$$\eqalign{
w  &= Vx, \qquad h &= g(w), \qquad h' &= g'(w) \\
}$$
where $g$ is a scalar function, $g'$ is its derivative, and the vectors $(h,h')$ are the result of applying these scalar functions element-wise to the $w$ vector.
Because of the element-wise nature of the functions, the differential of $h$ requires a Hadamard (element-wise) product, i.e.
$$\eqalign{
dh &= h'\odot dw \\
   &= {\rm Diag}(h')\, dw \\
   &= H'\, dw \\
}$$
The last line uses the well known "trick" of replacing a Hadamard product by a diagonal matrix.

With the above notation, calculating the gradient of $h$ with respect to $x$ is almost trivial.
$$\eqalign{
dh &= H'\,\c{dw} \;=\; H'\,\c{V\,dx} \\
\frac{\partial h}{\partial x} &= H'\,V \;\;\doteq\;\; Dh(x) \\
  &= \big(h'{\tt1}^T\big)\odot V &\quad\big({\rm Hadamard\,equivalent}\big) \\
}$$
The answer given in your book is the author's (poor) attempt to write this result using neither the Diag() function nor the Hadamard product.

Title: Matrix Chain Rule Derivative Question


问:
$\{a_n\}$ be a sequence such that $ a_{n+1}^2-2a_na_{n+1}-a_n=0$, then $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...
Let $\{a_n\}$ be a sequence of positive real numbers such that

$a_1 =1,\ \  a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$.

Then the sum of the series $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...

(A) $(1,2]$, (B) $(2,3]$, (C) $(3,4]$, (D)$(4,5]$.

Solution attempt:

Firstly, we figure out what $\frac{a_{n+1}}{a_n}$ is going to look like. We get, from the recursive formula, $\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$ (remembering the fact that $a_n>0$, the other root is rejected).

We know that, if $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}>1$, then $\lim a_n \to \infty$.

Further, $(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}>0$. (Again, the other root is rejected due to the same reason).

Hence, $(a_n)$ increases monotonically.

Therefore, the largest value of $\frac{a_{n+1}}{a_n}$ is approximately $1+\sqrt{1+\frac{1}{1}} \approx 2.15$

Now, the sum can be approximated as $\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$ (In actuality, $\mathbb{sum}< 1.3$).

So, option $(A)$ is the correct answer.

Is the procedure correct?

I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.

Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem?

Body:
$a_{n+1}=a_n+\sqrt{a_n^2+a_n}>2a_n\forall n\ge 1$ $a_1=2^0,a_2>2,a_3>2^2,...,a_n\ge 2^{n-1}$ $1\le a_n\Rightarrow a_n+a_n^2\le 2a_n^2\Rightarrow a_{n+1}\le (\sqrt 2+1)a_n$ $a_1=(\sqrt 2+1)^0,a_2=(\sqrt 2+1),a_3<(\sqrt 2+1)^2,...,a_n\le (\sqrt 2+1)^{n-1}$ $\therefore 2^{n-1}\le a_n \le (\sqrt 2+1)^{n-1}$ $\Rightarrow \frac 13(\frac 23)^{n-1}\le {a_n\over 3^n}\le \frac 13({\sqrt 2+1 \over 3})^{n-1}$ $\therefore \frac 13\sum (\frac 23)^{n-1} < \sum {a_n\over 3^n}\le \frac 13 \sum ({\sqrt 2+1 \over 3})^{n-1}$
$\text{Hence}\,\,1<\sum {a_n\over 3^n}\le {1\over 2-\sqrt 2}\approx 1.707$


问:
Lambda Calculus Terms
According to what I've read so far a lambda calculus term is described as :

$\langle term \rangle ::==$  $\langle var \rangle |\space (\langle term \rangle\space \langle term \rangle) \space |\space (λ\langle var \rangle.\langle term \rangle) $

So many of the lambda terms are defined $recursively$ and the part I struggle with is this one :

Say a lambda term is a variable $x$ and another lambda term is a variable $y$.

According to the second part of the definition some other lambda terms are :

$(xy)$

$(xx)$

$(x(xy))$

How can one interpret the terms above and what is their meaning when they stand alone?

Could you please provide a simple context/example that they make sense?

答:
In the (untyped) $\lambda$-calculus, you can think of everything as a function with one argument. So you can take any term $x$ and apply it to any term $y$, i.e. use $y$ as the input to the function $x$.

So your terms have the intuitive meanings:

$x$ applied to $y$.

$x$ applied to itself.

$x$ applied to the result of applying $x$ to $y$.

Title: Lambda Calculus Terms


问:
Finding the number of standard deviations from the mean, only given $P(X<55) = 0.7$

The continuous random variable, $X$, is normally distributed with
$P(X<55) = 0.7$

a) How many standard deviations from the mean is a score of $55$?

b) If the standard deviation of $X$ is $4$, find the mean of the
distribution.

For part a, the textbook solves $P(Z≤a) = 0.7$, giving the answer $0.5244$. I don't understand how this works. From my understanding, the standard normal distribution has mean of $0$ and S.D of $1$. But in this case, we are only given that the probability of obtaining a score of less than $55$ is $0.7$ - how do you find the number of standard deviations if I don't know what the mean is in this particular instance. Don't general normal distributions, not standard, vary in their measure of mean and S.D?

This is a high school question covered under the "normal distribution" topic.

If anyone could provide an explanation, it would be much appreciated!

答:
Let us say that $X$ has mean $\mu$ and standard deviation $\sigma$.

Then $0.7=P\left(X<55\right)=P\left(\frac{X-\mu}{\sigma}\leq\frac{55-\mu}{\sigma}\right)$

Here $U:=\frac{X-\mu}{\sigma}$ has standard normal distribution and
with an appropriate table on standard normal distribution we can find the value $u$ that satisfies:
$$0.7=P\left(U<u\right)$$

(as it seems $u\approx0.5244$, but I will leave that aside)

Then we can conclude that $\frac{55-\mu}{\sigma}=u$ or equivalently
$55=\mu+\sigma u$.

So $u$ is here the answer on question a).

If $\sigma=4$ then we find $\mu=55-4u$ as answer on question b)


答:
Notice that if $p>3$ we get
$$ \Big(p^2+{p-1\over 2}\Big)^2<\underbrace{p^4+p^3+p^2+p+1}_{k^2}<\Big(p^2+{p+1\over 2}\Big)^2$$ implies $$p^2+{p-1\over 2}<k<p^2+{p+1\over 2}$$

So we are left to check if $p=2$ or $p=3$ works...

Body:
Given that $p$ is a prime, and the sum of all positive divisors of $p^4$ is a perfect square, find the possible number of primes $p$ .

What I Tried: I know that the only divisors of $p^4$ will be $(1 , p , p^2 , p^3 , p^4)$ . From here I can conclude :-
$$\rightarrow 1 + p + p^2 + p^3 + p^4 = k^2$$ For some positive integer $k$ . Now this is where I get stuck, I can write it as :-
$$\rightarrow \frac{(1 - p^5)}{1 - p} = k^2$$
But I didn't understand how I can find all values of $p$. Next what I did is :-
$$\rightarrow p(1 + p + p^2 + p^3) = (k + 1)(k - 1)$$
And I am stuck at the same problem.

Can anyone help me here?


答:
We know $P(X<x)=\phi(x)$

So,

$P\bigg(\dfrac{X-\mu}{\sigma}<\dfrac{55-\mu}{\sigma}\bigg)=0.7 \implies \phi\bigg(\dfrac{55-\mu}{\sigma}\bigg)=0.7$

$P(X<a)=\phi(a)=0.7$

So for what value of $a$ area is $0.7$?

As you have written $a$ is $0.5244$

Also : $\dfrac{55-\mu}{\sigma}=0.5244$

"how do you find the number of standard deviations if I don't know what the mean is in this"

Explanation: The standard normal distribution is centered at mean=zero and the degree to which a given measurement deviates from the mean is given by the standard deviation. For the standard normal distribution, $68$% of the observations lie within $Z=1$ standard deviation of the mean; $95$% lie within two standard deviations of the mean, and $99.9$% lie within $Z=3$ standard deviations of the mean.

So your Z-Score values tell you how much away you are from the mean in our case we are $0.5244$ standard deviation away from the mean.

You can play around with normal curve from here: Pic source : Here

Title: Finding the number of standard deviations from the mean, only given $P(X<55) = 0.7$

Body:
The continuous random variable, $X$, is normally distributed with
$P(X<55) = 0.7$

a) How many standard deviations from the mean is a score of $55$?

b) If the standard deviation of $X$ is $4$, find the mean of the
distribution.

For part a, the textbook solves $P(Z≤a) = 0.7$, giving the answer $0.5244$. I don't understand how this works. From my understanding, the standard normal distribution has mean of $0$ and S.D of $1$. But in this case, we are only given that the probability of obtaining a score of less than $55$ is $0.7$ - how do you find the number of standard deviations if I don't know what the mean is in this particular instance. Don't general normal distributions, not standard, vary in their measure of mean and S.D?

This is a high school question covered under the "normal distribution" topic.

If anyone could provide an explanation, it would be much appreciated!


问:
What's the solution for $\lim\limits_{x \to a} \frac{a^nf(x)-x^nf(a)}{x-a}$
What's the solution for

$$\lim_{x \to a} \frac{a^nf(x)-x^nf(a)}{x-a}, n \in \mathbb{N}?$$

答:
Try to use this:

$\frac{a^nf(x)-x^nf(a)}{x-a}=a^n \frac{f(x)-f(a)}{x-a}-f(a)\frac{x^n-a^n}{x-a}$

Show the identity, have some hypothesis of differentiability, and take the limits ;)


问:
Is the following version of the fundamental lemma of the calculus of variations valid?
Let $U$ be an open subset of $\mathbb{R}^n$ with smooth boundary $\partial U$. Consider a function $f$ in $L^2(u)$. Suppose that for every $h$ in the Sobolev space$ H^2_0(U)$ it holds that
$$\int_U f \Delta h=0.$$
Where $\Delta$ is the Laplacian operator.n

Can we conclude that $f=0$ almost everywhere on $U$?

答: I think so. Assume that $f$ is nonnegative and smooth in some open set $S$. Now test against $\Delta h$ where $\Delta h=1$ in an open set $\Omega\subset S$ with homogeneous Dirichlet boundary conditions. Is this ok? Now, one can use the density of $C_c^\infty$ in $L^2$ and some approximation argument, isn't it?</p>


问:
Simple Logical Simplification - Distributive Law
I can't seem to understand the hopefully simple step of getting from :

$(a \land b \land c) \lor (a \land b \land \lnot c)$

to:

$(a \land b) \land (c \lor \lnot c)$

This step is in the answers and only has '(distribution)' written next to it, however even when breaking this down into several parts with Distributive Laws, I got nothing close to the expected solution (most likely my fault rather than teacher's). So any help explaining this step would very much be appreciated. Thank you!

答:
The equivalence is indeed an instance of Distribution ... it just doesn't feel like Distribution.

To be specific, going from:

$(a \land b) \land (c \lor \lnot c)$

to:

$(a \land b \land c) \lor (a \land b \land \lnot c)$

feels like a 'proper' Distribution. However, going the other way around, i.e. going from:

$(a \land b \land c) \lor (a \land b \land \lnot c)$

to:

$(a \land b) \land (c \lor \lnot c)$

feels like something that should be called 'Reverse Distribution' or 'Factoring' or 'Collecting Common Terms' ....but not 'Distribution'.

And yet, since Distribution is an equivalence, you can go both ways, and so it really is Distribution.

I see this as a clear example of how language can have an enormous power over our thinking. Or, more to the point: how maybe we should try and come up with a better name than 'Distribution'.


问:
Accuracy of numerical methods in finding eigenvalues
I have a problem with assesing the accuracy of my numerical calculation. I have a 2nd order ODE. It is an eigenvalue problem of the form:

$ y'' + ay' + \lambda^2y = 0 $

and the boundary condiations are:

$ y(0) = y(1) = 0 $

This equation describes a vibrating string, clamped at x=0 and x=1, with a certain mass distribution. I want to be able to calculate the eigenvalues of this equation numerically. I do this by using the Runge-Kutta method to find solutions to the equation with initial conditions:

$ y(0) = 0, y'(0) = 1 $

with different values of $\lambda$ and then looking for the ones that are zero at x=1.

I terminate my search for eigenvalues when I find a function that has

$y(1) < \delta$

where $\delta$ is a predefined constant.

Now the problem is that I'd like to know how accurate my calculations are, i.e. how do I choose the values of $\delta$ and the stepsize used in computing the values of the potential eigenfunction, so that the error in the eigenvalue is less than, say, $ 10^{-3}$?

答:
Based on you comment stating that you don't want a formal approach, the following engineering approach occurred to me ...

I assume that you do not have the actual value of $\lambda$, therefore we need to find a way around this fact.
An alternative way to asses the "goodness" of your eigenvalue is the residual of the ODE, that is, how good the discretised ODE is satisfied at each node $x_i$.

To compute it, you need to compute the residual at each node,
$$
r_i = D^2[y(x_i)] + a(x_i) D[y(x_i)] + \lambda^2 y(x_i),
$$
where $D^2$ and $D$ are a spatial discretisation of your derivative operators (which is different from your RK algorithm, since you already have the values of the function $y(x)$ at each node).
Then, the residual of the whole ODE could be built like:
$$
r(\lambda) = \frac{1}{N}\sum_i^N \left|r_i\right|,
$$
or any similar sum that aggregates the error at each node.

Now the algorithm would be:

Define multiple values of $H = [h_0, h_1, \ldots, h_N]$ and $\Delta = [\delta_0, \delta_1, \ldots, \delta_M]$.

Perform a parametric sweep in the previous sets and find for each tuple $(h_i, \delta_j)$ a value of $\lambda$.

(Note that you should have a nested loop, so that for constant values of each parameter you change the other one).

Compute for each $\lambda$ the value of the ODE residual $r(\lambda)$.

From this point onwards, realise that each value of $\lambda$ and the tuple $(h_i, \delta_j)$ are interchangeable, since the latter lead you to the former.
Therefore, we have a two-dimensional function $r(h,\delta)$.

You can slice it by planes in the $h$ or $\delta$ space to obtain one-dimensional figures with iso-lines. From this plot you can draw conclusions of how the values of $h$ and $\delta$ influence the accuracy of $\lambda$, as they will show how the residual behaves with different values of $h$ and $\delta$ (you might see a plateau beyond certain values or linear behaviours in a log-scale).

As a closing remark, the spatial discretisation of the derivate operators $D$ and $D^2$ introduced an error in the residual of the ODE, but since you will be benchmarking all of the $(h, \delta)$ combinations against the same discretisation, I think it is fair to say that it should not have an impact in your conclusions.

Sketch of the desired figure : https://i.stack.imgur.com/bvgnE.png


问:
Conjecture regarding trapping rational numbers in some special intervals

Conjecture:

Let $b\in\mathbb{N}_{\geq3}$ and $\{x_i\}$ be a collection of $b−2$ rational numbers greater than $1$. Does there always exist a natural number $a$ such that for all $i$ there exists some natural number $1\leq c\leq a$ such that:

$$\displaystyle \frac{ab-1}{(c-1)b+1}\geq x_i \geq \frac{ab-1}{cb-1}\tag{1}$$

One may think of this as having a UNION of $a$ intervals $\left[\frac{ab-1}{cb-1},\frac{ab-1}{(c-1)b+1}\right]$  on the real line and proving that, for some $a$, these can trap any finite discrete set of rational numbers having cardinality $b-2$.

For example, Let's assume $b=10$, $x=\{\frac{12}{5},\frac{17}{5},\frac{59}{10},\frac{187}{20},15,21,\frac{67}{2},39 \}$, then $(1)$ is true for $a=4$ and $a=5$ and therefore the intervals for $a=4,5$ contain or "trap" all the members of set $x$. In the diagram, the red/green regions are intervals $\left[\frac{ab-1}{cb-1},\frac{ab-1}{(c-1)b+1}\right]$ for all $c<a$.

What mathematical tools can one employ to tackle this?

WHAT I DID:

I solved the inequality $(1)$ for $a$ and got

$$\large x_i\left(1-\frac{2}{b}\right)+\frac{x_i+1}{b}+cx_i\geq a\geq \frac{x_i+1}{b}+cx_i\tag{2}$$

Note that $\frac{x_i+1}{b}+cx_i$ is common on both sides of the inequality.

Now I'll just have to plug in the values of $x_i$ to get $b-2$ inequalities and check for a common value of $a$ for any $c$.

For example, Let $x_i \in \{  \frac{101}{4},\frac{3001}{7} \}$ and $b=4$

then we have inequalities

$\frac{307}{16}+\left(\frac{101}{4}\right)c_1\geq a\geq \frac{105}{16}+\left(\frac{101}{4}\right)c_1$

$\frac{911}{32}+\left(\frac{301}{8}\right)c_2\geq a\geq \frac{309}{32}+\left(\frac{301}{8}\right)c_2$

Put $c=2$ in the first inequality and $c=1$ in the second inequality to get a common $a=58$.

But how should one go about to prove that there will always be some value of $a$ which satisfies the above set of inequalities? I calculated numerically for some random $b$ and $x_i$ and always managed to get some $a$, i.e., there existed some $a$ which trapped random $x_i$ given some random $b$. A counterexample may prove the negative but I was unable to find any.

答:
Here's what I did. I didn't get too far, but perhaps this will be of help.

First, since all of the relevant quantities are non-zero, let us define $y_i\triangleq 1/x_i$ and invert the inequality:
$$\frac{(c_i-1)b+1}{ab-1} \leq y_i \leq \frac{c_ib-1}{ab-1}$$
Note that each $y_i$ is a rational number between $0$ and $1$, exclusive. So define $y_i=p_i/q$, where $q$ is a common denominator across all $i=1,2,\dots,b-2$, and each $p_i\in\mathbb{N}$  is between $1$ and $q-1$, inclusive. So we have
$$\frac{(c_i-1)b+1}{ab-1} \leq \frac{p_i}{q} \leq \frac{c_ib-1}{ab-1}$$
$$q((c_i-1)b+1) \leq p_i(ab-1) \leq q(c_ib-1)$$
$$q(1-b) \leq p_i(ab-1) - qc_ib \leq -q$$
$$q(1-b/2) \leq p_i(ab-1) - qc_ib + qb/2 \leq q(b/2-1)$$
$$\left| 2(ab-1)p_i - 2qb(c_i-1/2) \right| \leq q(b-2)$$
Here's the geometrical interpretation of this expression. The quantity $2qb(c_i-1/2)$ represents a finite grid $$\{qb,3qb,5qb,..,2qba-qb\}.$$ The quantity $2(ab-1)p_i$, on the other hand, represents the $b-2$ chosen points, which must lie on a different finite grid
$$\{2(ab-1),4(ab-1),\dots,2(q-1)(ab-1)\}.$$
So the challenge is to find a way to ensure that $b-2$ points selected from this second grid lie within $q(b-2)$ of the points from the first grid. Note that this is not trivially satisfied because this distance $q(b-2)$ is strictly less than half the spacing between the $c_i$ grid points. So there are indeed spaces that must be avoided when choosing $a$.


问:
What do sinh and cosh have to do with exp?
My friend told me that $\sinh$ and $\cosh$ result from an exponential function, but I can't figure out why

答:
$$\cosh(x) = \frac{1}{2} (\exp(x)+\exp(-x))$$
$$\sinh(x) = \frac{1}{2} (\exp(x)-\exp(-x))$$

$\exp(x)$ is the natural exponential function $\exp(x)=e^x$


问:
Divergent Series Properties
Given $a_n$ a sequence of positive real numbers and

$$
\sum_{k=1}^{\infty}a_k \rightarrow \infty
$$

Is it necessarily true that

$$
\lim_{k\rightarrow\infty}(a_k)(k^{1+\epsilon})\rightarrow \infty
$$

for any positive real $\epsilon$?

It is true for the prime harmonic series, but I'm curious if it is true in general. It feels like it is not but I can't put together a good counter example and was looking for some help. If $a_n$ is strictly decreasing, does that make a difference?

答:
It is not true in general. Consider the sequence $$a_k=\begin{cases}\frac{1}{k}~~~~~~\text{$k$ even}\\\frac{1}{k^2}~~~~~\text{$k$ odd}\end{cases}$$
Then we have $\sum_{k=1}^\infty a_k=\infty$ but for every $0<\varepsilon<1$ we have $a_kk^{1+\varepsilon}\not\to\infty$.
However, notice that in this example we at least have $$\limsup_{k\to\infty}a_kk^{1+\varepsilon}=\infty$$
And this will be true in general. (proof: If $\limsup_{k\to\infty}a_kk^{1+\varepsilon}<\infty$ for some $\varepsilon>0$ there is a constant $M$ such that $a_k\leq \frac{M}{k^{1+\varepsilon}}$ for all $k$, hence $\sum_{k=1}^\infty a_k<\infty$ by the comparison test)

Atm I am not sure about the last part if $a_k$ is strictly decreasing, maybe I will update later or someone else can post an answer to that.

Title: Divergent Series Properties


问:
show that if $\mu(A\bigtriangleup A_n)\rightarrow 0$ then $μ(A_n)\rightarrow\mu(A)$
Let $(X, A, µ)$ be a positive metric space. If $\mu(X) < \infty$ and $(A_n)_{(n \in N^*)},A \in X$

show that if $\mu(A\bigtriangleup A_n)\rightarrow 0$ then $μ(A_n)\rightarrow\mu(A)$

What I have tried so far is

use that $A\bigtriangleup A_n = (A/A_n)\cup(A_n/A)$

and since $(A/A_n)\cap(A_n/A)= \emptyset$ then $\mu(A\bigtriangleup A_n) = \mu(A/A_n)+\mu(A_n/A) \rightarrow 0$

now I am trying to contain $μ(A_n)$ in an inequality where both side converge to $A$

EDIT: thanks to Thorgott's point both $\mu(A/A_n)$ and $\mu(A_n/A)$ converge to $0$ and $(A/A_n)\cap A_n = \emptyset$
then $\mu(A) = \mu(A/A_n) + \mu(A_n) \Rightarrow \mu(A_n) \rightarrow0+\mu(A)$

答:
We have $\mu(A\Delta A_n)\to 0$ and consequently $\mu\left(A\cap A_{n}^{\complement}\right)\to0$.

Then on base of: $$\mu\left(A\right)=\mu\left(A\cap A_{n}\right)+\mu\left(A\cap A_{n}^{\complement}\right)$$

we conclude that: $$\mu\left(A\cap A_{n}\right)\to\mu\left(A\right)$$

Then on base of: $$\mu\left(A\Delta A_{n}\right)+2\mu\left(A\cap A_{n}\right)=\mu A+\mu A_{n}$$
we find that: $$\mu A_{n}\to\mu A$$


问:
If a linear operator maps to open unit ball to a relatively compact set, does it need to be bounded?
Let $X,Y$ be normed vector spaces and $A:X\to Y$ be linear and such that $AB^X_1(0)$ is relatively compact (where $B^X_1(0)$ denotes the open unit ball in $X$).

Why can we conclude that $A$ is bounded?

Since $AB^X_1(0)$ is relatively compact, we know that $\overline{AB^X_1(0)}$ is totally bounded and hence bounded. So, there is a $c\ge0$ with $$\sup_{x\in B^X_1(0)}\left\|Ax\right\|\le\sup_{y\in\overline{AB^X_1(0)}}\left\|y\right\|\le c.\tag1$$

However, in order to conclude we would need to show that $$\sup_{x\in\overline{B^X_1(0)}}\left\|Ax\right\|<\infty.\tag2 \label{eq2}$$

Why do we know that \eqref{eq2} cannot be infinity?

答:
Because $A$ is linear, we have
$$
\sup_{x \in \overline{B^X_1(0)}} \|Ax\| \leq \sup_{x \in B_2^X(0)} \|Ax\| = 2\cdot \sup_{x \in B_1^X(0)} \|Ax\|.
$$


问:
What is the meaning of "well-defined" and how to prove something is "well-defined"
I'm reading the book Analysis 1 by Terence Tao. There is a concept of well-defined which is mentioned a lot of time in the book but I don't fully understand.

For example, in the picture below, in order to define the $subtraction$ $operation$ of 2 integers, he firstly defined the $negation$ $operation$ of integers.

After the definition, he stated that the definition is well-defined and we can prove that it is well-defined.

Could you please explain me what well-defined means and how can we prove something is well-defined or not ? Is there any collection of criteria for something to be well-defined ? ("something" here can be a definition, an operation)

Thank you very much for your help!

[Image]

答:
In order for $-(a\mathbin —b):=(b\mathbin —a)$ to be well-defined we must have $-(a\mathbin — b)=-(c\mathbin —d)$ whenever $a\mathbin—b=c\mathbin—d$.

In general, this terminology is used when defining operations, functions, maps on equivalence classes. It means that the result is the same when a different representative of the same equivalence class is chosen.

Here is an example of an operation that is not well-defined: For a rational number $\frac{p}{q}$ (where $p,q$ are integers and $q\neq 0$) we could try to define
$$
T\left( \frac p q \right) := \frac {p+1}{q+1}.
$$
However, we have $\frac 1 3 = \frac 2 6$ but
$$
T\left( \frac 1 3 \right) = \frac 2 4 \neq \frac 3 7 = T\left(\frac 2 6\right).
$$
Hence, the expression $T(\frac p q)$ is not well-defined.


答:
Hospital:

$a^nf'(a)-na^{n-1}f(a).$

Title: What's the solution for $\lim\limits_{x \to a} \frac{a^nf(x)-x^nf(a)}{x-a}$

Body:
What's the solution for

$$\lim_{x \to a} \frac{a^nf(x)-x^nf(a)}{x-a}, n \in \mathbb{N}?$$

Body:
Hospital:

$a^nf'(a)-na^{n-1}f(a).$


问:
Getting complex solutions with MATLAB's ode45
I'm trying to solve the ODE below using Matlab's ode45:

$$\frac{dy}{dx}=\sqrt{A\cdot \sin(y) + B\cdot \cos(y)}$$

where $0\le x \le 1$.

The Matlab code is:

```<code>A = -2.45;
B = 2.50;
tspan  = 0:0.01:1;
odefun = @(t,y) sqrt(A*sin(y) + B*cos(y));
[t,y] = ode45(odefun, [0 1], 0);
</code>```

The constants $A$ and $B$ are chosen such that inside the square root it is always non-negative. The solution seems fine for most values of $x$, but for the last few points close to $x=1$ I get complex $y$ with very small imaginary parts. I then took the real or the absolute value of $y$, but then I would get negative value inside the square root.

Should I try other ODE solvers? Or is there anything else I could do to eliminate the complex solutions?

答:
In reviewing my previous answer it became apparent that the solution there is not the most simple one.

The most straight-forward solution is to extend the domain of the differential equation without changing it on the given domain, so that reaching and slightly crossing the boundary of the given domain does not lead to non-real values for the derivatives and then the values of the numerical solution. This in the main part is achievable here by replacing the square root with an extension to the whole real line. For the solution to stabilize at the boundary one can take the odd extension
$$
r(u)={\rm sign}(u)\sqrt{|u|}.
$$
Numerical solvers find large or infinite values for derivatives difficult to deal with, leading to very small step sizes. To mollify this one can replace some small segment around $u=0$ with a line like in
$$
r(u)=\frac{u}{\max(\epsilon,\sqrt{|u|})}
$$
or similar constructs. To get large step sizes in the asymptotic segment of the solution one will need implicit solvers. Using the implicit "Radau" solver gives the following picture where the internal steps are noted with crosses:


问:
Limits of this function in the polar system
Given is the following double integral -
$$\iint\limits_{D}\frac{1}{(x^2+ y^2)^2}\,dy\,dx$$

$D$ is the region for $x\in[1,2]$ and $y\in[0,\sqrt{2x-x^2}]$

How would the limits of integration be derived if one was to convert the coordinate system from Cartesian to polar?

So far, I can only understand that $\theta$ would hold an upper bound of $\frac{\pi}{4}$ with a lower bound of $0$ and could derive by algebra that $r$ would hold an upper bound of $2\cos\theta$. By the same algebra, the lower bound for $r$ turns out as $0$.

Region

Here's my go at finding bounds for $r$
$$y = \sqrt{2x - x^2}$$
$$x^2 + y^2 = 2x$$
$$r^2(\cos^2\theta + \sin^2\theta) = 2r\cos\theta$$
$$r^2 - 2r\cos\theta = 0$$
Thus, $r$ is found to be $0, 2\cos\theta$

From a solution that I found, the lower bound for $r$ was given as $\sec\theta$. Could someone please explain how that came to be? I've been at it for a day now

答:
Everything is correct except for the lower bound which is given by

$$x=1\implies r=\sec\theta$$

The integral in polar coordinates is

$$\int_{0}^{\frac{\pi}{4}}\int_{\sec\theta}^{2\cos\theta} \frac{1}{r^3}\:drd\theta$$


问:
Prove Cauchy's inequality: $(\sum_{i=1}^n a_i b_i)^2 \le \sum_{i=1}^n a_i^2 \sum_{i=1}^n b_i^2.$
Prove that for any real number $a_i, b_i, i = 1,2,...,n$:
$$\left(\sum_{i=1}^n a_i b_i\right)^2 \le \sum_{i=1}^n a_i^2 \sum_{i=1}^n b_i^2.$$
Proof in my book goes:
$$\sum_{i=1}^n a_i^2 \sum_{i=1}^n b_i^2-\left(\sum_{i=1}^n a_i b_i\right)^2 = \sum_{i,j=1}^na_i^2b_j^2-\sum_{i,j=1}^na_ib_ia_jb_j =\frac{1}{2}\sum_{i,j=1}^n(a_ib_j-b_ia_j)^2\ge0$$
But I can't see why there is $\frac{1}{2}$ and I would be glad if someone explained it.
Tia.

答:
Take the last term of the equality and expand it:

$$\sum_{i,j=1}^n(a_ib_j-b_ia_j)^2 = \sum_{i,j = 1\dots n} a_i^2b_j^2 + b_i^2a_j^2 - 2a_ib_jb_ia_j = \sum_{i,j = 1\dots n} 2a_i^2b_j^2 - 2a_ib_jb_ia_j $$

And here you can see that this is exactly the double of the term you had before. So in order to get an equality you have to take an $\frac{1}{2}$ factor.

EDIT

Please note that the last passage I did above is possible since we can interchange the indexes. To be more clear:

$$\sum_{i,j = 1\dots n} a_i^2b_j^2 + b_i^2a_j^2 - 2a_ib_jb_ia_j = \\ \sum_{i = 1\dots n}\sum_{j = 1\dots n} \left(a_i^2b_j^2 - a_ib_jb_ia_j\right) + \sum_{i = 1\dots n}\sum_{j = 1\dots n} \left(a_j^2b_i^2 - a_ib_jb_ia_j\right)$$

Now you can swap $i$ with $j$ in the last double sum and you can conclude.


问:
Examples of nonlinear ordinary differential equations with elementary solutions.
I am looking for nice examples of nonlinear ordinary differential equations that have simple solutions in terms of elementary functions. (But are not trivial to find, like, for example, with separation of variables).

A perfect example of what I am looking for is the Lane-Emden equation of index 5:
$$y''+\frac{2}{x}y'+y^5=0,\qquad y(0)=1,y'(0)=0$$
which admits the solution
$$y(x)=\frac{1}{\sqrt{1+x^2/3}}$$

Another very good example is
$$4y^2y'''-18yy'y''+15y'^3=0$$
from which we find
$$y(x)=\frac{1}{(ax^2+bx+c)^2}$$
for some constants $a,b,c$.

Do you have more examples like these? The higher the order, the better. (But I consider, for example, $H(x,y'',y''')=0$ to be of order 1). And it is a plus if it comes from a physics problem.

答:
I would certainly like to mention six Painlevé equations. These are the only nonlinear 2nd order ODEs of the form $w''=F(t,w,w')$ whose solutions cannot have movable critical points. Painlevé equations have a lot of applications in various areas of mathematics, including integrable models, random matrices, algebraic and differential geometry and combinatorics.

It is known (proven rigorously) that the general solutions of Painlevé equations, in a sense, cannot be expressed in terms of classical functions. However, many particular solutions are known for special values of parameters and integration constants. You can check, for instance, NIST Library for explicit examples.

Such solutions include classical special function solutions (for example, for Painlevé II they are expressed in terms of Airy functions, and for Painlevé VI in terms of Gauss hypergeometric functions). Also, e.g. Painlevé VI has a family of elliptic solutions and a lot of algebraic solutions with polynomial degree going up to $72$ and genus up to $7$.

Concerning physics applications, one could look at the long list on p.19 here to get an idea.

Finally, to give an illustration: Painlevé VI equation
\begin{align}
\nonumber\frac{d^2w}{dt^2}=\;\frac{1}{2}\left(\frac{1}{w}+\frac{1}{w-1}
 +\frac{1}{w-t}\right)\left(\frac{dw}{dt}\right)^2 -
 \left(\frac{1}{t}+\frac{1}{t-1}+\frac{1}{w-t}\right)\frac{dw}{dt}\,+\\
 \nonumber +\,
 \frac{w(w-1)(w-t)}{2t^2(t-1)^2}\left((\theta_{\infty}-1)^2-\frac{\theta_x^2 t}{w^2}+
 \frac{\theta_y^2(t-1)}{(w-1)^2}+\frac{(1-\theta_z^2)t(t-1)}{(w-t)^2}\right)
 \end{align}
with parameters $(\theta_x,\theta_y,\theta_z,\theta_{\infty})=\left(\frac25,\frac15,\frac13,\frac23\right)$ has a $5$-branch genus $0$ algebraic solution given by parametric equations
\begin{align}&w=\frac{2(s^2+s+7)(5s-2)}{s(s+5)(4s^2-5s+10)}, \\ & t=\frac{27(5s-2)^2}{(s+5)(4s^2-5s+10)^2}.
\end{align}


问:
ex falso quodlibet in natural deduction (ND)?
My understanding of inference rules is that they should be intuitively acceptable (like axioms). But my only intuition for ex falso quodlibet in natural deduction is not immediate but comes from its proof, which rests on disjunctive syllogism (which apparently holds in classical propositional logic and intuitionistic logic).
I am probably missing something obvious. Is the principle of explosion really basic/intuitive and why?

The closest justification I see, based on the current answers, is that Ex Falso encodes a (metalogical) desire that the proof system be consistent. But also see the references to intuitionism in the answers.

Edit: as pointed out below, it may have been a distraction for me to raise "intuitionism" in the original question, although Ex Falso arises in classical and intuitionistic logic. This does not exclude that I would like axioms and rules to be intuitively acceptable (to me).

" There  is  surprisingly  little  agreement  about  the  exact  status  of EFQ in  the literature.  Gentzen groups the ex  falso rule alongside the other operational rules(1969,  p.  77).   Prawitz  follows  this  practice.   He  treats ⊥ as  a  ‘0-place  sentential operation’ (Prawitz 1978, p. 38), one ‘for which there is no canonical proof’ (Prawitz1977, p. 26). EFQ is then understood as the elimination rule for ⊥. Others, including Dummett, present ⊥ as an elimination rule for ¬ (e.g. Dummett 1991, p. 291).  I contend that both of these approaches are mistaken...." ref.

Ex Falso has an ‘anomalous position [. . . ]  inside the scheme of introduction/elimination rules’, tarnishing the neat symmetry of the intuitionistic system" (Weir 1986, p. 461).

According to here: "Kolmogorov’s criterion whether to keep an axiom was whether a proposition has an “intuitive foundation” or “possesses intuitive obviousness” (van Heijenoort 1967: 421, 422) ... Kolmogorov said that, just like PEM, Ex Falso “has no intuitive foundation” (van Heijenoort 1967: 419). In particular, he says that Ex Falso is unacceptable for the reason that it asserts something about the consequences of something impossible (van Heijenoort 1967: 421)."

... "I submit that these difficulties can be overcome simply by treating⊥as a punc-tuation marker, as Tennant suggests.  Of course this means thatEFQis left withouta  logical  constant  which  it  serves  to  eliminate.   But  on  the  view  I  am  proposingno such constant is needed becauseEFQis not an operational rule at all.  Its roleis  best  accounted  for  by  according  it  the  status  of  astructural  rule.   It  simply  isthe structural rule that tells us that any sentence whatsoever follows from a con-tradiction.  Such a license isnotspecific to any logical constants, but amounts to ablanket policy.  Our reclassification ofEFQthus fits neatly with our characterizationof structural rules as global rules that assign properties to our deducibility relation." ref.

Edit: I came across this reformulation of my initial uneasiness.

"... adopting disjunctive syllogism as an axiom or as a primitive rule of inference is not an option.  Hence, it is ex  falso that is naturally treated as primitive, not the principle of disjunctive syllogism. Therefore, rather than thinking of ex  falso as the product of the mean-ings of disjunction and negation, we must treat disjunctive syllogism as a product of the ex falso rule.  This suggests that the law of disjunctive syllogism is not a con-sequence only of the meanings of the constants involved, but also hinges on one’s stance concerning the general principle expressed by EFQ." ref.

答:
Apparently a justification for Ex Falso in terms of disjunctive syllogism is due to Glivenko’s paper from 1929. Neither Heyting nor Kolmogorov ever justified Ex Falso by giving this traditional argument here.

"While [Ex Falso] can  of  course  be  justified  byappeal to disjunctive syllogism, it should not be overlooked that disjunctive syllo-gism is itself only provable in the presence of EFQ" ref.

Regarding constructivists,

"Kolmogorov said that “As soon as ¬a is solved, then the solution of a is impossible and the problem a→b is without content” (Kolmogorov 1932: 62 [Mancosu 1998: 331]), and proposed that “The proof that a problem is without content [owing to an impossible assumption] will always be considered as its solution” (Kolmogorov 1932: 59: [Mancosu 1998: 329]). Taken together, this yields a justification of Ex Falso ."  here.

Heyting gave the following justification of Ex Falso:

" Axiom X [¬p→(p→q)] may not seem intuitively clear. As a matter of fact, it adds to the precision of the definition of implication. You remember that p→q can be asserted if and only if we possess a construction which, joined to the construction p, would prove q. Now suppose that ⊢¬p, that is, we have deduced a contradiction from the supposition that p were carried out. Then, in a sense, this can be considered as a construction, which, joined to a proof of p (which cannot exist) leads to a proof of q. (Heyting 1956: 102)"here.

(It may be my ignorance, but I have no idea exactly how to "join" these proofs.)

Regarding the latter, as pointed out here.

"In its attempt to provide, “in a sense”, a construction, the explanation is clearly not of the same kind as Kolmogorov’s stipulation from 1932. But it does not fit into Heyting’s original interpretation of logic in terms of intentions directed at constructions and the fulfillment of such intentions either. For to fulfill an intention directed toward a particular construction we will have to exhibit that construction; we will have to exhibit a construction that transforms any proof of p into one of q. But how can a construction that from the assumption p arrives at a contradiction, and therefore generally speaking not at q, lead to q? It will not do to say that such a construction exists “in a sense”. A construction that is a construction “in a sense”, as Heyting helps himself to here, is no construction.""

Elsewhere, Glivenko stated that Ex Falso "is a consequence of the principle (p∨¬q)→(q→p), the admissibility of which he considers “quite evident” here.

"Undeniably, we take it that contradictionsoccur when we have derived two contradictories, a statement and its negation.  Butthere is no reason in principle whyex  falsocould not also play a role in systemsdevoid of a negation operator.  A ‘contradiction’ in this sense is not necessarily theresult of asserting two contradictories.  For example, we could conceive of systems inwhichA,B`⊥would hold, whereAandBare contraries.  So long as our languagecontains  antonyms  like  hot/cold,  soft/hard,  weak/strong,  etc.   or  contraries  likered/green,  triangle/circle,  there  will  be  cases  where  various  sentences  fail  to  bejointly assertible.  This we register with the sign ‘⊥’.  In all such cases, we have achoice of whether or not to adoptEFQ. The fact that there is such a choice, thatthe decision for or againstEFQis not predetermined by the meanings of the logicalconstants in the system—for a system withEFQneed not contain any constants atall—shows thatEFQis indeed a structural rule. As such it expresses a general policyregarding our consequence relation, namely that we allow any statement whatsoeverto follow from a contradiction." ref.

Reasoning here "shows that the standard natural deduction formulations of intuitionistic logic are  misleading.   They  present ex  falso as  an  operational  rule  that  appears  to  in-troduce  an  awkward  asymmetry  into  the  system.   In  fact,  however,  this  is  just  asuperficial feature of the standard system.  The move from minimal logic to intuitionistic logic, represented by the adoption of theex falsorule, turns on structural assumptions rather than on a shift in the meanings of the logical constants.  Like-wise, relevant logic can tidily be characterized as the substructural logic obtainedby dropping the structural rule of weakening on the left from minimal logic" ...

"... natural deduction systems in sequent format most accurately capture the distinction between the structural rules expressing the structural properties of the system as a whole and operator-specific operational rules. In particular, I showed that, contrary to conventional wisdom, the rule of ex falso ought to be understood as a structural rule closely related to the rule of weakening on the right." (ibid).


问:
Projection to positive Fourier coefficients
$\newcommand{\A}{A({\mathbb D})}\newcommand{\C}{C({\mathbb T})}$Let $\C$ the Banach space consisting of all complex
valued, continuous functions on the unit circle $\mathbb T$ under the sup norm.  The disk algebra $\A$ is an intensely
studied  subspace of $\C$ which can be defined in three equivalent ways:

The closed subspace of $\C$ spanned by the functions $z\mapsto z^n$, for $n\geq 0$.

The subset of $\C$ formed by all functions $f$ with vanishing negative
Fourier coefficients,  that is,
$$
  \hat f(n)=0,\quad\forall n<0.
  $$

The subset of $\C$ formed by all functions $f$ admitting a continuous extension $\tilde f$ to the closed unit disk $\mathbb D$,
such that $\tilde f$ is
holomorphic on the interior of $\mathbb D$.

It is a fact that $\A$ is not complemented in $\C$ in the sense that there is no closed subspace $E\subseteq \C$ such that
$$
  \C=\A\oplus E.
  $$

This can be shown to be equivalent to the following statement: There exists a doubly infinite sequence of complex numbers $\{a_n\}_{n=-\infty }^\infty $, such that
$$
  \sum_{n=-\infty }^\infty a_n e^{int}
  $$
is the Fourier series of a continuous, $2 \pi $-periodic function, while the positive part of this series, namely
$$
  \sum_{n=0}^\infty a_n e^{int}
  $$
is not.

Question.  What is a concete example  of a sequence $\{a_n\}_{n=-\infty }^\infty $, as above?

NOTES

See On projection of fourier series for the corresponding result
regarding $L^\infty(\mathbb T)$.

I found two proofs  of the fact that $\A$ is not complemented in $\C$, which would in theory be enough to provide an
answer to my question, but I was unable to make it work.  These proofs can be found in:

Page 137 of Rudin, Walter, Functional analysis, McGraw-Hill Series in Higher Mathematics. New York etc.: McGraw-Hill Book Comp. XIII, 397 p. (1973). ZBL0253.46001.

Page 55 of Hoffman, K., Banach spaces of analytic functions, Prentice-Hall Series in Modern Analysis. Englewood Cliffs, N.J.: Prentice-Hall, Inc. XIII, 217 p. (1962). ZBL0117.34001.

答:
Pick $b_n$ decreasing to zero, $nb_n \to 0, \sum b_n =\infty$; for example $b_n=\frac{1}{n\log n}, n \ge 2$ will do.

Then it is not hard to show that $g(x)=\sum_{n \ge 2} b_n\sin nx$ is continuous on the reals since the series converges uniformly everywhere.

(the result is true but requires a summation by part and a classical estimate on the partial sums of $\sum \sin nx$ in general when $nb_n \to 0$ but not necessarily monotonically, however for the specific case above where $nb_n$ decreases to zero, it immediately follows from the Abel-Dirichlet convergence criterion for series since it is well known that $\sum_{n \ge 2} \frac{\sin nx}{n}$ has partial sums uniformly bounded in $x$)

However $h(x)=\sum_{n \ge 2} b_n \cos nx$ is not continuous since the series is divergent to $\infty$ at $0$ and $h$ is continuous on $(0, 2\pi)$ but unbounded (though integrable of course) at $0$

But then taking $f(x)=\sum_{n \in \mathbb Z, |n| \ge 2}(-i(\text {sign n}) {b_{|n|}})e^{inx}=2g(x)$ is continuous, but its projection is obviously not since its imaginary part is not by the above


问:
Equivalence Relation question - $(a,b)S(c,d) \iff a-b=c-d$ such that $(a,b),(c,d) \in\ \mathbb{N}\times\mathbb{N}$
The relation $(a,b)S(c,d) \iff a-b=c-d$ such that $(a,b),(c,d) \in \mathbb{N}  \times \mathbb{N}$

Need to find equivalent set for ($6,6)$ and $(2,5)$

I found for $[(6,6)] = \{(c,d) \in\mathbb{N} \times \mathbb N \mid c=d \}$

and for $[(2,5)] = \{(c,d) \in\mathbb{N} \times \mathbb N \mid c=d+3 \}$

I think I'm correct so far.. now I need to find the collection of equivalence classes $\mathbb{N} \times \mathbb{N} /S$

I found that the relation condition works when for example in $(2,5)$ its equivalence set all the difference between $5$ and $2$ that for every $c = d+3$ belongs, $(4,7), (10,13)$ i don't know how to put that perspective in mathematic words

答:
First, a given pair $(a,b)$ is equivalent to $(0,b-a)$ if $a\ge b$ and to $(a-b,0)$ if $a>b$, so the pairs with $a$ or $b$ zero form a full representative set.

The equivalence class of $(0,n)$ is $\{(a,a+n):a\in\Bbb N\}$, and similarly the equivalence class of $(n,0)$ is $\{(a+n,a):a\in\Bbb N\}$.

Note that this is the construction of the set of integer numbers $\Bbb Z$ based on the natural numbers, where a pair $(a,b)$ represents the difference $b-a$, so that $[(0,n)]$ represents the natural $n$ and $[(n,0)]$ represents the negative number $-n$ if $n>0$.


答:
Another useful approach
\begin{align}
\bigl|\mu(A_n)-\mu(A)\bigr|&=\biggl|\int (\mathbb{1}_{A_n}-\mathbb{1}_A)\,d\mu\biggr|\\
&\leq \int|\mathbb{1}_{A_n}-\mathbb{1}_{A}|\,d\mu
\end{align}
Now, by considering all possible options we observe that $|\mathbb{1}_{A_n}-\mathbb{1}_{A}|\leq \mathbb{1}_{A\triangle A_n}$. Hence,
$$\bigl|\mu(A_n)-\mu(A)\bigr|\leq \mu(A\triangle A_n)$$
and the result follows.

Body:
Let $(X, A, µ)$ be a positive metric space. If $\mu(X) < \infty$ and $(A_n)_{(n \in N^*)},A \in X$

show that if $\mu(A\bigtriangleup A_n)\rightarrow 0$ then $μ(A_n)\rightarrow\mu(A)$

What I have tried so far is

use that $A\bigtriangleup A_n = (A/A_n)\cup(A_n/A)$

and since $(A/A_n)\cap(A_n/A)= \emptyset$ then $\mu(A\bigtriangleup A_n) = \mu(A/A_n)+\mu(A_n/A) \rightarrow 0$

now I am trying to contain $μ(A_n)$ in an inequality where both side converge to $A$

EDIT: thanks to Thorgott's point both $\mu(A/A_n)$ and $\mu(A_n/A)$ converge to $0$ and $(A/A_n)\cap A_n = \emptyset$
then $\mu(A) = \mu(A/A_n) + \mu(A_n) \Rightarrow \mu(A_n) \rightarrow0+\mu(A)$


问:
change of variable to get a quasi-cartesian laplace equation?
when writing the (vector) laplace equation in cylindrical coordinates in a $(r,\theta)$ plane, we get:
$$ \left( r^2\frac{\partial^2}{\partial r^2} + r\frac{\partial}{\partial r} + \frac{\partial^2}{\partial\theta^2} \right)A_z(r,\theta)=0 $$
the term $r^2$ is multiplied by the two sides of the equation to form the cauchy-Euler equation.

now through a change of variable of $v=\ln(r)$:
$$ \left( \frac{\partial^2}{\partial v^2} + \frac{\partial^2}{\partial\theta^2} \right)A_z(v,\theta)=0 $$
it is like a cartesian laplace operator, which has its advantages of simplification of computations.

however in $(r,z)$ plane the laplace equation becomes as following:
$$ \left( r^2\frac{\partial^2}{\partial r^2} + r\frac{\partial}{\partial r} -1 + \frac{\partial^2}{\partial z^2} \right)A_\theta(r,z)=0 $$
is it even possible through a change of $u=f(r)$ to get a result like the same?:
$$ \left( \frac{\partial^2}{\partial u^2} + \frac{\partial^2}{\partial z^2} \right)A_\theta(u,z)=0 $$

additional notes:it is worth noting that in the above equations, the 3D vector-laplace equation has been reduced to 2D plane due to symmetry of the problem.

答:
it is not possible. a Cauchy-Euler equation in terms of radial component appears in the partial equation which cannot be transformed into a simple second-order derivative.

Dear readers, for proof, please refer to this question:

Reduce the Cauchy-Euler equation to a smaller expression?


问:
Concrete application of Caratheodory extension
I am working on the following problem:

Question:

Let $\Omega$ be an uncountable set. Denote by $\mathfrak{R}$ the ring
of finite subsets of $\Omega$. Consider the measure $\mu: \mathfrak{R}
 \rightarrow [0, + \infty]: E \mapsto 0$. What is in this case the
outer measure $\mu^{*}$ (induced by $\mu$) and the associated
$\sigma$-algebra $\mathfrak{M}_{\mu^{*}}$? What is the
$\sigma$-algebra $\mathfrak{M}$ generated by $\mathfrak{R}$? Is the
extension of the measure $\mu$ from $\mathfrak{R}$ to $\mathfrak{M}$
obtained via the Caratheodory procedure the only possible extension?

Attempt: First recall some definitions. By definition, $\mathfrak{M}_{\mu^{*}} = \left\{ E \subset \Omega \mid E \ \text{is} \ \mu^{*}\text{-measurable} \right\}$. And a set $E \subset \Omega$ is $\mu^{*} \text{-measurable}$  if for all $A \subset \Omega$, we have $$ \mu^{*} (A) = \mu^{*} (A \cap E) + \mu^{*} (A \cap E^c). $$ Also, recall how $\mu^{*}$ is defined: For each $E \subset \Omega$ we have $$ \mu^{*} (E) := \text{inf} \left\{ \sum_n \mu(E_n) \mid (E_n)_n \in \mathfrak{R} \ \text{is a sequence with} \ E \subset \bigcup_n E_n \right\}. $$

In this specific question, I think we have $$  \mu^{*} (E) =  \begin{cases} 0 & \text{if} \ E \ \text{is finite or countable infinite} \\ + \infty & \text{if} \ E \ \text{is uncountable} \end{cases} $$
However, I am stuck with finding $\mathfrak{M}_{\mu^{*}}$ and $\mathfrak{M}$. I know from a theorem that $\mathfrak{M}_{\mu^{*}}$ will contain $\mathfrak{R}$. I also know that in general, $\mathfrak{M}_{\mu^{*}}$ will be a bigger set than $\mathfrak{M}$. By definition, $\mathfrak{M}$ is the smallest $\sigma$-algebra containing $\mathfrak{R}$.

Any help is appreciated.

答:
Let us go step by step

$1$. The $\sigma$-algebra $\mathfrak{M}$ generated by $\mathfrak{R}$.

Since every countable subset of $\Omega$ is the countable union of finite sets (singletons), we have that every countable subset of $\Omega$ is in $\mathfrak{M}$.

Then it is easy to see that
$$\mathfrak{M} =\{ E \subseteq \Omega : E \textrm{ or } E^c \textrm{ is countable }\}$$

$2$. The outer measure $\mu^{*}$.

It is exactly as you wrote: For each $E \subset \Omega$ we have $$ \mu^{*} (E) := \text{inf} \left\{ \sum_n \mu(E_n) \mid (E_n)_n \in \mathfrak{R} \ \text{is a sequence with} \ E \subset \bigcup_n E_n \right\}. $$

In this specific question, we have $$  \mu^{*} (E) =  \begin{cases} 0 & \text{if} \ E \ \text{is finite or countable infinite} \\ + \infty & \text{if} \ E \ \text{is uncountable} \end{cases} $$
because an uncountable subset can not be covered by a countable collection of finite subsets.

$3$. The $\sigma$-algebra $\mathfrak{M}_{\mu^{*}}$.

We have that, for all $E \subset \Omega$, $E \in \mathfrak{M}_{\mu^{*}}$  if and only if for all $A \subset \Omega$, we have $$ \mu^{*} (A) = \mu^{*} (A \cap E) + \mu^{*} (A \cap E^c). $$

For any $E \subset \Omega$, we have:

If $A$ is uncountable, then $A \cap E$ or $ A \cap E^c$ is also uncountable and we get

$$ \mu^{*} (A) =+\infty = \mu^{*} (A \cap E) + \mu^{*} (A \cap E^c). $$

If $A$ is countable, then $A \cap E$ and $ A \cap E^c$ are both countable and we get

$$ \mu^{*} (A) =0 = 0+0 = \mu^{*} (A \cap E) + \mu^{*} (A \cap E^c). $$

So $\mathfrak{M}_{\mu^{*}}= 2^\Omega =\{ E : E \subseteq \Omega\}$.

$4$. Is the extension of the measure $\mu$ from $\mathfrak{R}$ to $\mathfrak{M}$ obtained via the Caratheodory procedure the only possible extension?

Answer: No.

Let $\mu_1$ be the extension of the measure $\mu$ from $\mathfrak{R}$ to $\mathfrak{M}$ obtained via the Caratheodory procedure. Then

$$  \mu_1(E) =  \begin{cases} 0 & \text{if} \ E \ \text{is countable} \\ + \infty & \text{if} \ E^c \ \text{is countable} \end{cases} $$

Another extension would be $\mu_2$ defined by $\mu_2(E)=0$, for all $E \in \mathfrak{M}$.

Note tha, since $\Omega$ is uncountable, $\mu$ is not $\sigma$-finite. So it does not need to have a unique extension to the $\sigma$-algebra $\mathfrak{M}$ generated by $\mathfrak{R}$.


问:
Better bound for integral inequality
Let $f:[a,b] \rightarrow  \mathbb{R}$ with $f \in C^1$ and $f(a)=0$. We want to show that $$\int_{a}^{b}f^2(x)\mathrm{d}x \leq (b-a)^2\int_{a}^{b}[f'(x)]^2\mathrm{d}x$$

There is a hint namely to consider $$f(x)=\int_{a}^{x}f'(t)\mathrm{d}t$$ because $f(a)=0$ and then use the Cauchy-Schwarz inequality for integrals.

So I tried doing the following: $$f^2(x)=\left (  \int_{a}^{x}f'(t)\mathrm{d}t \right )^2\leq\left ( x-a \right )\int_{a}^{x}\left [ f'(t) \right ]^2\mathrm{d}t\leq\left ( b-a \right )\int_{a}^{b}\left [ f'(t) \right ]^2\mathrm{d}t$$

and finally integrate the expression with respect to $x$ over $[a,b]$. My first question is whether my approach has any mistakes (I am mainly worried since in the textbook the last inequality has $x$ as instead of $b$ as the higher bound of integration and I am wondering whether it is a mistake, because when integrating again the result only follows if the integral in the last inequality is independent of $x$) and whether this this strategy works in general with higher powers of functions and their derivatives (for example using some variation of the Hölder inequality or something).

Then there is a challenge without hint or solution that asks us to prove that there is a smaller upper bound, namely $$\int_{a}^{b}f^2(x)\mathrm{d}x \leq \frac{(b-a)^2}{2}\int_{a}^{b}[f'(x)]^2\mathrm{d}x$$

Here I got stuck. I tried the same approach $$f^2(x)=\left (  \int_{a}^{x}f'(t)\mathrm{d}t \right )^2\leq\left ( x-a \right )\int_{a}^{x}\left [ f'(t) \right ]^2\mathrm{d}t$$ and I tried finding a number  $x$ such that $x-a=\frac{b-a}{2}$ which means $x=\frac{b+a}{2}$ so that I can say $$\left ( x-a \right )\int_{a}^{x}\left [ f'(t) \right ]^2\mathrm{d}t\leq\left ( \frac{b-a}{2} \right )\int_{a}^{\frac{b+a}{2}}\left [ f'(t) \right ]^2\mathrm{d}t$$ and then integrate to get the desired result (as integration over $[a,b]$ would just multiply the right hand side with $b-a$ and we would be done). However this only works when $x\in [a,\frac{b+a}{2}]$ and I cannot seem to find a way to prove it for the rest of the interval. So how do I proceed here?

Finally, as I am new to integral inequalities, any suggestions on how one should study to learn how to solve basic problems or where to study from are welcome. Thanks in advance!

答:
Your proof for the first inequality is correct.

For the second one: As you already showed, the inequality
$$f^2(x)=\left (  \int_{a}^{x}f'(t)\mathrm{d}t \right )^2\leq\left ( x-a \right )\int_{a}^{x}\left [ f'(t) \right ]^2\mathrm{d}t$$
holds. Now use this inquality inside the integral and integrate by parts
\begin{align}
\int_{a}^{b}f^2(x)\mathrm{d}x&\leq  \int_{a}^{b}\left (\left ( x-a \right )\int_{a}^{x}\left [ f'(t) \right ]^2\mathrm{d}t\right )\mathrm{d}x\\&
=\frac{(b-a)^2}{2}\int_{a}^{b}\left [ f'(t) \right ]^2\mathrm{d}t-\int_{a}^{b}\frac{(x-a)^2}{2}\left [ f'(x) \right ]^2\mathrm{d}x\\&
\leq \frac{(b-a)^2}{2}\int_{a}^{b}\left [ f'(x) \right ]^2\mathrm{d}x.
\end{align}
Here we have used that one term disappears while partial integration and the integrands of the term with the minus sign are all greater than zero for the last inequality.

Title: Better bound for integral inequality


问:
Integrals with measures on path spaces
Let $X_t$ for $t\in [0,T]$ be a $\mathbb R^d$-valued stochastic process with law $\rho\in \mathcal P(C([0,T];\mathbb R^d))$. Let $f:\mathbb R^d\rightarrow\mathbb R^d$ be a smooth function. Then
\begin{equation*}
\mathbb E_\rho \Bigl[\int_0^T f(X_t)dt \Bigr] = \int_0^T \mathbb E_{\rho_t}\left[f(X_t) \right] dt,
\end{equation*}
where $\rho_t=\mathrm{law}(X_t)$ and $\mathbb E_\nu$ denotes expectation w.r.t. measure $\nu$.

Question: Is this identity true, and if so why?

Motivation: I came across this identity in this paper (https://arxiv.org/pdf/1605.05878.pdf), specifically equation (2.2) where the authors directly move from law of paths to law of time marginals.

Ideas: One possible approach would be to use a Fubini-type result to exchange the time integral and the integral over $\mathcal X:=C([0,T];\mathbb R^d)$. But its still not clear to me how to go from the path measure to law at each time -- maybe using a push-forward $\pi_t:\mathcal X\rightarrow\mathbb R^d$ such that $\pi_t(X)=X_t$? My exposure to path measures (and even their definitions) is quite limited, and so any hints/ideas are greatly appreciated.

答:
It is Fubini's Theorem.  $\mathbb{P}_{\rho}$ is a probability measure on $C([0,T];\mathbb{R}^{d})$, say, with respect to the Borel $\sigma$-algebra $\mathscr{B}$.  Now the family of random variables $\{X_{t}\}_{t \in [0,T]}$ given by $X_{t}(\omega) = \omega(t)$ for $\omega \in C([0,T]; \mathbb{R}^{d})$ forms a jointly measurable function $X : C([0,T]; \mathbb{R}^{d}) \times [0,T] \to \mathbb{R}^{d}$ given by $X(\omega,t) = X_{t}(\omega)$.  (That is, $X$ is measurable with respect to the product $\sigma$-algebra obtained from $\mathscr{B}$ and the Borel subsets of $[0,T]$.  This is immediate from its joint continuity.)

Applying Fubini's Theorem to $C([0,T];\mathbb{R}^{d}) \times [0,T]$ with product measure $\mathbb{E}_{\rho} \otimes \mathcal{L}^{1}$, we find
\begin{equation*}
\mathbb{E}_{\rho} \left[ \int_{0}^{T} f(X_{t}) \, dt \right] = \int_{0}^{T} \mathbb{E}_{\rho}(f(X_{t})) \, dt.
\end{equation*}
Given $t \in [0,T]$, if $\mathbb{P}_{\rho_{t}}$ is the law of $X_{t}$ under $\mathbb{P}_{\rho}$ (i.e. the pushforward of $\mathbb{P}_{\rho}$ under $X_{t}$), then $\mathbb{E}_{\rho}(f(X_{t})) = \mathbb{E}_{\rho_{t}}(f(X_{t}))$ so we get
\begin{equation*}
\mathbb{E}_{\rho} \left[ \int_{0}^{T} f(X_{t}) \, dt \right] = \int_{0}^{T} \mathbb{E}_{\rho_{t}}(f(X_{t})) \, dt.
\end{equation*}

Title: Integrals with measures on path spaces


问:
How to prove a formula for terminal velocity
I have a problem question in here.

The differential equation for an object falling under the influence of
gravity near the surface of the earth with air resistance comparable
to velocity $v$ is$$\frac{\mathrm{d}v}{\mathrm{d}t} = -g - av$$with
$g = 32 \text{ feet per second per second}$ is the gravitation
accleration and $a > 0$ is shear coefficient. Show that:

(a) $v(t) = (v_0 - v_\infty)e^{-at} + v_\infty$, with $v_0 = v(0)$ and
$v_\infty = -\frac{g}{a} = \underset{t \to \infty}{\lim} v(t)$; and

(b) If $y(t)$ is the height, then$$y(t) = y_0 + tv_\infty + \frac{1}{a}(v_0 - v_\infty)(1 - e^{-at})$$

So, I already proved the part of (a)

$\begin{align} \frac{\mathrm{d}v}{\mathrm{d}t} &= -g - av \\ ve^{at} &= \int -ge^{at} \ \mathrm{d}t \\ ve^{at} &= -\frac{g}{a}e^{at} + C \\ \text{Let } C = v_0 + \frac{g}{a} \\ ve^{at} &= -\frac{g}{a}e^{at} + \left(v_0 + \frac{g}{a}\right) \\ v &= (v_0 - v_\infty)e^{-at} + v_\infty \ \blacksquare \text{ Q.E.D.}\end{align}$

For part of (b)

$\begin{align} v(t) &= \frac{\mathrm{d}y}{\mathrm{d}t} \\ \int v(t) \ \mathrm{d}t &= \int \left(v_0 + \frac{g}{a}\right)e^{-at} - \frac{g}{a} \\ y(t) &= y_0 + tv_\infty + \frac{1}{a}(v_0 - v_\infty)e^{-at}\end{align}$

And here is the problem. It should be $y(t) = y_0 + tv_\infty + \frac{1}{a}\left(v_0 - v_\infty\right)(1-e^{-at})$. I tried to fix my mistake here but I even didn't had any hint.

Oh, and please correct me if my works had mistake

答: You didn't apply limits to the exponential term.$$y(t)-y_0=v_\infty[x]_0^t+\frac1{\color{red}-a}(v_0-v_\infty)[e^{-ax}]_0^t$$which will give you the required result.


问:
What is the measure of the length of the tetrahedron?
Four balls of radius 1 are mutually tangent, three resting on the floor and the fourth resting on the others. A tetrahedron, each of whose edges have length s, is circumscribed around the balls. Then what is the measure of s?

I got s = 2root(3) + 2 but the answer said it was s = 2root(6) + 2 and I tried to figure out where I went wrong but I couldn't see the problem. This was the 1980 AHSME problem 26.

答:
The centers of the spheres are $2$ units apart, so consider the equilateral tetrahedron with edge $2$.

In an equilateral triangle of side $2$, each vertex is $\frac2{\sqrt3}$ from the center of the triangle. Thus, the the fourth vertex of the tetrahedron must be at a height $\sqrt{4-\frac43}=\frac23\sqrt6$ above its base. Thus, the center of the tetrahedron is $\frac16\sqrt6$ above its base.

The center of the tetrahedron around the spheres is therefore $1+\frac16\sqrt6$ above the plane on which it sits.

The linear ratio between the tetrahedron with the centers as vertices to the tetrahedron around the spheres is $\frac{1+\frac16\sqrt6}{\frac16\sqrt6}=1+\sqrt6$ and so the edge of the tetrahedron around the spheres must be $2\!\left(1+\sqrt6\right)=2+2\sqrt6$.


问:
Irreducible ideal implies prime ideal in Dedekind Domains?
An ideal is irreducible if it can not be written as the finite intersection of strictly larger ideals. In a Noetherian ring every irreducible ideal is primary, but the converse doesn't hold. I wonder about the situation in a Dedekind Domain. In a Dedekind Domain every ideal can be factored uniquely into a product of prime ideals. And every primary ideal is a power of a prime ideal. Hence in a Dedekind domain every irreducible ideal is a power of a prime ideal. Furthermore, a prime ideal is always irreducible.

Let $R$ be a Dedekind Domain.

If $A$ and $B$ are ideals in a Dedekind domain then $B$ is said to divide $A$ if there exists an ideal $C$ such that $A=BC$. We have that $A\subseteq B$ iff $B$ divides $A$.

Define $M$ as the set of ideals of $R$ that is such that if $P$ divides $AB$ then this implies that $P$ divides $A$ or $B$.

Let $K$ be those ideals in $R$ that do not have any non-trivial factorization.

In the notation of the book I am using (Mollin) the set $M$ is defined as the set of prime ideals of a number field and the set $K$ is defined as the set of irreducible ideals in a Dedekind domain. Furthermore it is stated that an ideal $A$ belongs to $K$ if and only if $A$ belongs to $M$.

So my natural question are:
Is the set of prime ideals in a Dedekind Domain equal to $M$?
Is the set of irreducible ideals in a Dedekind Domain equal to $K$?
I.e, by proving that $A$ belongs to $K$ if and only if $A$ belongs to $M$ do we prove that an ideal in a Dedekind domain is irreducible if and only if it is prime?

答:
You ask about three different classes of nonzero ideals in a Dedekind domain $R$:

(a) Prime ideals.

(b) Unfactorable ideals: if $\mathfrak{p} = IJ$ then $I = R$ or $J = R$.

(c) Irreducible ideals: If $\mathfrak{p} = I \cap J$ then $I = \mathfrak{p}$ or $J = \mathfrak{p}$.

Classes (a) and (b) are the same: their equivalence follows easily from the fact that every nonzero ideal in a Dedekind domain factors uniquely as a product of primes.

Class(c) consists precisely of the prime powers $\mathfrak{p}^a$ for $a \in \mathbb{N}$.  This follows from the following more general result about intersections of ideals in a Dedekind domain:

Lemma. Let $I = \mathfrak{p}_1^{a_1} \cdots \mathfrak{p}_r^{a_r}$, $J = \mathfrak{p}_1^{b_1} \cdots \mathfrak{p}_r^{b_r}$ be nonzero ideals in a Dedekind domain: here $a_i, b_i$ are non-negative integers.  Then $I \cap J = \mathfrak{p}_1^{\max (a_1, b_1)} \cdots \mathfrak{p}_r^{\max (a_r, b_r)}$.

Let me know if you need help in proving this. If you know about localizations, this allows for a very easy proof: you can reduce to the case in which $R$ has exactly one nonzero prime ideal, and in this case the result is trivial.

Finally, one has the following simple result.

Lemma. If $\mathfrak{p}$ is an ideal in a commutative ring $R$, the following are equivalent:

(i) For any ideals $I$, $J$ in $R$, if $\mathfrak{p} \supset IJ$, then $\mathfrak{p} \supset I$ or $\mathfrak{p} \supset J$.

(ii) $\mathfrak{p}$ is prime.

Proof: $\neg$ (i) $\implies$ $\neg$ (ii): The hypotheses give the existence of $x \in I \setminus \mathfrak{p}$ and $y \in J \setminus \mathfrak{p}$ such that $xy \in \mathfrak{p}$.  Thus $\mathfrak{p}$ is not prime.

$\neg$ (ii) $\implies$ $\neg$ (i): If $\mathfrak{p}$ is not prime, then there are $x,y \in R \setminus \mathfrak{p}$ with $xy \in \mathfrak{p}$.  Take $I = (x)$, $J = (y)$.

I believe this answers all of your questions, but let me know if I missed something.