问:
Cooley-Tuckey Algorithm, Inverse and Forward.
I'm trying to make some code in MatLab that takes a vector of length $2^n$ (a sequence) and spits out it's Discrete Fourier Transform, or it's inverse Discrete Fourier Transform depending on another input. Now I have the regular Discrete Fourier Transform working, however I am stuck on the inverse. After messing around a bit I noticed that it seems like the Cooley-Tuckey Algorithm seems to work in reverse with some minor differences. So suppose I have a sequence $\{F_k\}_{k \in \mathbb{N}}$, and I want to find it's inverse $f_k$, then I'd use the following,

$$f_k=\frac{1}{N}\sum_{j=1}^{N}F_j\exp\left(\frac{2\pi i}{N}(j-1)(k-1)\right).$$

The $k-1$ because I want to start from $f_1$ as opposed to $f_0$. Which is essentially the same as the regular Discrete Fourier Transform without the minus sign and an extra $1/N$ factor. As such I attempt to do the following ( with $\omega_N=\exp(2\pi i/N)$ ).

$$I_k :=Nf_k=\sum_{j=1,\text{ odd}}^{N}F_j\omega_N^{(j-1)(k-1)}+\sum_{j=1,\text{ even}}^{N}F_j\omega_N^{(j-1)(k-1)}.$$

$$I_k=\sum_{j=1}^{N/2}F_{2j-1}\omega_N^{(2j-2)(k-1)}+\sum_{j=1}^{N/2}F_{2j}\omega_N^{(2j-1)(k-1)}.$$

$$I_k=\sum_{j=1}^{N/2}F_{2j-1}(\omega_N^2)^{(j-1)(k-1)}+\omega_N^{k-1}\sum_{j=1}^{N/2}F_{2j}(\omega_N^2)^{(j-1)(k-1)}.$$

$$I_k=\sum_{j=1}^{N/2}F_{2j-1}(\omega_{N/2})^{(j-1)(k-1)}+\omega_N^{k-1}\sum_{j=1}^{N/2}F_{2j}(\omega_{N/2})^{(j-1)(k-1)}.$$

Then we are basically done because we have $I_k$ in terms of Fourier Transforms of the odd and even sequences. Therefore, the inverse is just the same as the Cooley-Tuckey Algorithm except with $\omega_N$ and dividing every entry by $1/N$ should get you the result you want.

Then if this is correct then the only problem is in my code, though I'm not entirely sure if I should be asking such a question here, this is the code I'm using.

```<code>function F = FourierTransInv(vec,arg)
     N=length(vec);
     F=ones(1,N);
     switch arg
         case 1
             w=exp(-2*pi*1i/N);
         case 0
             w=exp(2*pi*1i/N);
     end
     switch N
         case 1
             F(1)=vec(1);
         otherwise
             Feven=FourierTransInv(vec(2:2:end),1);
             Fodd=FourierTransInv(vec(1:2:end),1);
             for k=1:N/2
                F(k)=Fodd(k)+w^(k-1)*Feven(k);
                F(k+N/2)=Fodd(k)-w^(k-1)*Feven(k);
             end
     end
     if arg==0
        F=F/N;
     end
end
</code>```

As such I've implemented that if we want the Inverse then simply $\omega_N$ is as above and at the end $F$ is divided by $N$. The strange thing is when I run this code, every odd entry agrees with the true result we want, e.g when I run the Fourier Transform of $[1,2,3,4,5,6,7,8]$, then run it back through the inverse, it agrees with $1,3,5,7$ only, and the rest are just gibberish.

Just for reference the output for this input specifically is,
$[1+0i,6+2i,3+0i,4-2i,5+0i,6-2i,7-0i,4+2i]$.

If the regular Fourier Transform is working as intended (I tested it for many sequences using online calculators and they all seem to agree), I don't see a reason why the inverse shouldn't work assuming what I did before showing the code is correct, since I basically reuse what was already working.

答:
I don't know whether there are other issues, but here is one bug that I find.

In the recurrence calls Feven = ... and Fodd = ..., you should pass arg instead of 1.

That is,

```<code>Feven=FourierTransInv(vec(2:2:end),arg);
Fodd=FourierTransInv(vec(1:2:end),arg);
</code>```

Body:
I don't know whether there are other issues, but here is one bug that I find.

In the recurrence calls Feven = ... and Fodd = ..., you should pass arg instead of 1.

That is,

Feven=FourierTransInv(vec(2:2:end),arg);
Fodd=FourierTransInv(vec(1:2:end),arg);


问:
prove $\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$

Prove if $a,b,c$ are positive $$\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$$

My proof:After rearranging we have to prove $$\sum_\text{cyc} \frac{b}{b^2+2b} \le \sum_\text{cyc} \frac{a}{b^2+2b}$$

As inequality is cyclic:

let $a\ge b\ge c$ then $$\frac{1}{a^2+2a}\le \frac{1}{b^2+2b}\le \frac{1}{c^2+2c}$$.The rest follows by rearrangement inequality.

The case $a\ge c\ge b$ is analogous.

Thus Proved!

Is it correct?...And any other alternative ways possible?

答:
Your application of rearrangement is correct, in either case, $(a, b, c)$ and $(a^2+a, b^2+b, c^2+c)$ are similarly ordered, so
$$\sum_{cyc} \frac{a}{a^2+2a} \leqslant \sum_{cyc} \frac{a}{b^2+2b}$$

For another way, which generalises, consider
$$f(x) = \sum_{cyc} \frac{a+x}{b+x}, \quad f'(x) = \sum_{cyc} \frac{b-a}{(b+x)^2} = \sum_{cyc} \frac{b}{(b+x)^2} - \sum_{cyc}\frac{a}{(b+x)^2} \leqslant  0$$
again by Rearrangement.  Hence $f$ is decreasing, and $f(0) \geqslant f(2)$

ContentLicense: CC BY-SA 4.0


问:
Hartshorne Proposition II.6.5
The statement of part (a) of this proposition is as follows:

Let $X$ be a noetherian integral separated scheme which is regular in codimension 1. Let $Z$ be a proper closed subset of $X$, and let $U=X\setminus Z$. Then there is a surjective homomorphism $\mathrm{Cl}(X)\to\mathrm{Cl}(U)$ defined by $\sum n_i Y_i\mapsto\sum n_i(Y_i\cap U)$, ignoring empty intersections.

Now, this seems like it should be blatantly obvious, and the proof in the book more or less treats it as such. I'm having trouble with the first line:

If $Y$ is a prime divisor on $X$, then $Y\cap U$ is either empty or a prime divisor on $U$.

Clearly the former possibility is trivial. For some reason, I'm having trouble with the latter. A prime divisor is defined to be a closed integral subscheme of codimension one. Now, obviously $Y\cap U$ is closed in $U$. But to refer to it as a closed subscheme, don't I need to know what the subscheme structure is? Earlier they refer to the induced reduced closed subscheme structure; I guess I'm supposed to assume this is what they mean, but I'm not 100% sure (though I don't know what else makes sense).

Moving past this for now, I need to know that $Y\cap U$ is integral in $U$ and that $\mathrm{codim}(Y\cap U,U)=1$. Maybe I'm just confusing myself here; since $Y$ is integral (it's a prime divisor on $X$) and $Y\cap U$ is open in $Y$, we know $Y\cap U$ is integral as a subscheme of $Y$; that is, $(Y\cap U, \mathcal{O}_Y|_{Y\cap U})$ is an integral scheme. But is this the same as the closed subscheme structure of $Y\cap U\subseteq U$? Does it even matter?

Finally, the codimension; $\mathrm{codim}(Y,X)=1$ means that there are no irreducible closed subsets of $X$ strictly between $Y$ and $X$ (both are integral, hence irreducible themselves). I think I can see at least why this is true. Assuming we know $Y\cap U$ is an integral (hence irreducible) closed subscheme of $U$, either its codimension is one or it equals $U$. But the latter would force $U\subseteq Y$, which would make $X=Y\cup Z$ not irreducible, which would be a contradiction.

I'd greatly appreciate it if someone could point me in the right direction here; am I way overthinking this? Or are there really some subtleties that I'm not seeing?

答:
Let X be a an integral scheme, Y a closed integral subscheme of codimension 1, and U a nonempty open subscheme.  With the topology induced from U,  $Y \cap U \subset U$ is closed, and with the reduced subscheme structure induced from U, it becomes a reduced subscheme of U.  With the topology induced from Y, $Y \cap U \subset Y$ is irreducible, as a nonempty open subset of an irreducible space.  Since Y and U induce the same topologies on $Y \cap U$, one sees that as a scheme $Y \cap U$ is integral (reduced and irreducible).

Now $\mathrm{codim}(Y \cap U, U) = \mathrm{codim}(Y, X) = 1$ follows from the fact that there is a bijection $Z \mapsto \overline{Z}$ from the set of closed irreducible subspaces of U to the set of closed irreducible subspaces of X intersecting U, which in fact maps irreducible components to irreducible components.  (See (EGA, 0_IV, 14.2.3).)

ContentLicense: CC BY-SA 3.0

Title: Hartshorne Proposition II.6.5


问:
Does ZF prove the existence of a cardinal that is not reachable (through set union) from a given cardinal, for every cardinal?
Its consistent with ZF that every cardinal strictly bigger than $\aleph_0$ is singular (Moti Gitik).

Is it provable in ZF that for every cardinal (initial von Neumann ordinal with respect to equinumerousity) number $\kappa$ there is a cardinal $\lambda$ such that every $\in$-confinal subset of $\lambda$ is strictly bigger in cardinality than $\kappa$?

答:
The cofinality of an ordinal $\lambda$ is the least ordinal $\kappa$ such that there is an unbounded subset of order type $\kappa$ in $\lambda$.

An immediate consequence of the definition is that if $\kappa$ is the cofinality of any ordinal $\lambda$, then the cofinality of $\kappa$ is $\kappa$.

And an immediate consequence of that is that cofinality is always a cardinal. So we can replace "order type $\kappa$" by "cardinality $\kappa$".

And an immediate corollary to all of that is that if every uncountable cardinal is singular, then the only possible cofinality is the only cardinal that $\sf ZF$ proves to be regular: $\aleph_0$. Therefore, in this case, every limit ordinal must have a countable cofinal sequence.


问:
Show that $(E|X|^{p_1})\leq (E|X|^{p_2})^\frac{p_1}{p_2}$, where $0<p_1<p_2<\infty$
Let $X$ be a random variable on $(\Omega,\mathcal{F},\mu)$. for $h\in L^1(\Omega,\mathcal{F},\mu)$, define $Eh(X)=\int h(X)d\mu$.

Show that $(E|X|^{p_1})\leq (E|X|^{p_2})^\frac{p_1}{p_2}$, where $0<p_1<p_2<\infty$

Further show that equality holds iff $|X|$ is constant a.e.

I honestly don't have a good handle on how to start this one.

I sort of want to use jensen's inequality, but $x^{p_1},x^{p_2}$ aren't convex if $p_1$ or $p_2$ are less than $1$. Even if this could be remedied by multiplying  by a negative, it's not clear how to proceed. The fact that $\frac{p_1}{p_2}<1$ seems relevant, but I don't see where exactly that could be relevant to this problem.

I considered trying to show that $(E|X|^{p_1})^{p_2}\leq (E|X|^{p_2})^{p_1}$, but that didn't really get me very far.

Any thoughts suggestions or hints would be appreciated.

Thanks in advance.

答: This is also a special case of the generalized Hölder inequality $\|XY\|_r\le\|X\|_s\|Y\|_t$ ($r^{-1} = s^{-1}+t^{-1}$) with $Y=1$, $r=p_1$, and $s=p_2$ (so that $t=sr/(s-r)>0$).


问:
Losing Solutions to a Rational Equation
To solve a rational equation, say $\dfrac{3}{x-2}=\dfrac{1}{x-1}+\dfrac{7}{(x-1)(x-2)}$, the usual strategy is to multiply both sides of the equation by the least common denominator (LCD). In this case, the LCD is $(x-1)(x-2)$ and multiplying both sides of the equation yields $3(x-1)=(x-2)+7$. The question is now reduced to solving a linear equation and the solutions to this linear equation will be the solutions to the rational equation (assuming they don't create a zero in any of the denominators).

However, this method of multiplying both sides by the LCD seems funny to me. How are we 100% certain that in doing so we don't "lose" any potential solutions to the rational equation? Is there any rigorous way to prove that this method gives us $\textbf{all}$ of the solutions to a rational equation? I apologize if this question is trivial.

Intuitively, it seems as though when we multiply both sides by the LCD, we are just getting rid of the "rejected solutions" that will cause a zero in any of the denominators.

答:
At worse, multiplying an equation by a common expression can introduce alien solutions (when you multiply by zero), but not lose any: if $a=b$, $ac=bc$ remains true.

Your equation has meaning only when $x\ne1$ and $x\ne2$, so if you multiply by $(x-1)(x-2)$, you will not introduce any if you keep this condition.

Title: Losing Solutions to a Rational Equation


答:
Losing or gaining solutions to an equation happens when you perform an operation which is either not uniquely invertible, or an inverse of such an operation. For example, with squaring we see:
$$y=x\to y^2=x^2\to y=\pm x$$

The function $x^2$ is not injective and does not have a unique inverse, and so that step produces extra solutions.

However, multiplying by a given polynomial is fine at any point where the polynomial is non-zero, as the inverse of that operation is just dividing by said polynomial.

So, in your question, multiplying by $(x-1)(x-2)$ is fine, but you must discount $x=1$ and $x=2$ if they arise as solutions. In this case, they aren't, so you haven't gained any solutions.

Title: Losing Solutions to a Rational Equation

Body:
To solve a rational equation, say $\dfrac{3}{x-2}=\dfrac{1}{x-1}+\dfrac{7}{(x-1)(x-2)}$, the usual strategy is to multiply both sides of the equation by the least common denominator (LCD). In this case, the LCD is $(x-1)(x-2)$ and multiplying both sides of the equation yields $3(x-1)=(x-2)+7$. The question is now reduced to solving a linear equation and the solutions to this linear equation will be the solutions to the rational equation (assuming they don't create a zero in any of the denominators).

However, this method of multiplying both sides by the LCD seems funny to me. How are we 100% certain that in doing so we don't "lose" any potential solutions to the rational equation? Is there any rigorous way to prove that this method gives us $\textbf{all}$ of the solutions to a rational equation? I apologize if this question is trivial.

Intuitively, it seems as though when we multiply both sides by the LCD, we are just getting rid of the "rejected solutions" that will cause a zero in any of the denominators.


问:
Prove that $l^2(I)$ is a Hilbert Space.
This problem is taken from Conway's A Course in Functional Analysis section 1.1 problem 2

Verify that $l^2(I)$ (Example 1.7) is a Hilbert Space.

Below is my attempt at a solution. Any feedback is greatly appreciated!

To show completeness, suppose we had a Cauchy sequence $\{x_n\}\in l^2(I)$. We see that:
\begin{align}
    ||x_n - x_m||^2 =\sum_{i} ||(x_n(i) - x_m(i)||^2
    \end{align}

So that $||x_n - x_m|| \rightarrow 0$ as $m,n \rightarrow \infty$ must also imply $||x_n(i) - x_m(i)|| \rightarrow 0$

Thus, $x_n(i)$ is a Cauchy sequence over the real/complex numbers, we it must converge to a limit,  say $x(i)$. Let us construct $x(i)$ this way for all $i \in I$ and show that $x(i) \in l^2(I)$.

We first show that $x(i) = 0$ for all but a countable number of $i$. Suppose the contrary. Then for any $x_n$, $||x - x_n||^2$ being a sum of uncountably many nonzero elements, which cannot be bounded. The contradiciton proves the desired result.

Now We try to show that $\sum_{i\in I}|x(i)|^2 < \infty$. But we know that for any $\epsilon>0$, there is sufficiently large $n$ such that we have $||x - x_n|| < \epsilon$. Thus

\begin{align}
        \sum_{i \in I}|x(i)|^2 &= ||x||^2\\
        &= ||x - x_n + x_n||^2
        \leq \left(||x-x_n|| + ||x_n||\right)^2 \quad \text{ by triangle inequality}\\
        &< \left(\epsilon + ||x_n||\right)^2\\
        &< \infty
    \end{align}

where we noted in the last line that $||x_n||< \infty$ since $x_n \in l^2(I)$. Thus we have shown $x \in l^2(I)$ as desired.

答:
You don't know that $\lVert x - x_n\rVert < +\infty$ until you have shown that $x \in \ell^2(I)$.

The trick is to take a finite $F \subset I$, then $$\sum_{i \in F} \lvert x(i)\rvert^2 = \lim_{n \to \infty} \sum_{i \in F} \lvert x_n(i)\rvert^2 \leqslant M := \sup_n\: \lVert x_n\rVert < +\infty\,.$$
Since the bound $M$ is independent of $F$, it follows that
$$\sum_{i \in I} \lvert x(i)\rvert^2 \leqslant M\,,$$
hence $x \in \ell^2(I)$. Then it remains to show that $\lVert x - x_n\rVert \to 0$, which we do with the same trick,
$$\sum_{i \in F} \lvert x(i) - x_n(i)\rvert^2 = \lim_{m \to \infty} \sum_{i \in F} \lvert x_m(i) - x_n(i)\rvert^2 \leqslant \limsup_{m \to \infty} \lVert x_m - x_n\rVert^2\,.$$
This again holds for all finite $F \subset I$ with the bound not depending on $F$, hence
$$\lVert x - x_n\rVert^2 \leqslant \limsup_{m \to \infty} \lVert x_m - x_n\rVert^2\,.$$
Since $(x_n)$ is a Cauchy sequence, the right hand side tends to $0$ as $n \to \infty$.

Title: Prove that $l^2(I)$ is a Hilbert Space.


答: Declare $x\ne 1$ and $x\ne 2$ to avoid the division by zero. As you do ordinary algebraic manipulation, you get $3(x-1)-(x-2)-7 =0 \implies x=4$ which does not contradict the declarations made, this solution is correct. One can also put it in the original equation to confirm that you get $3/3=3/2$. So all is well, here in your question, you have not missed anyting.

Title: Losing Solutions to a Rational Equation


问:
Prove that $(x^n - 1)$ can divide $(x^{kn} - 1)$ without any remainder.
I would like your help with proving that $(x^n - 1)$ can divide $(x^{kn} - 1)$ without any remainder.
I understand that both of these functions can be translated to a similar form such as $(x^2-1)=(x-1)(x+1)$.
But I'm not really sure how to do so.
Thank you.

答: Notice that $x^{kn} - 1 = (x^n)^k - 1.$ This implies that $x^{kn} - 1 = (x^n - 1)P(x).$


答:
$$(x^n-1)|(x^{kn}-1)$$ can be written

$$(y-1)|(y^k-1).$$

Now by long division,

$$y^k-1=p(y)(y-1)+r$$ where $p$ is a polynomial and $r$ is a scalar constant (zeroth degree polynomial, if you prefer). And plugging $y=1$,

$$0=p(1)\,0+r.$$

You can conclude.

Title: Prove that $(x^n - 1)$ can divide $(x^{kn} - 1)$ without any remainder.

Body:
I would like your help with proving that $(x^n - 1)$ can divide $(x^{kn} - 1)$ without any remainder.
I understand that both of these functions can be translated to a similar form such as $(x^2-1)=(x-1)(x+1)$.
But I'm not really sure how to do so.
Thank you.


Body: $x^{kn}-1=(x^n)^k-1=(x^n-1)(x^{n(k-1)}+x^{n(k-2)}+\dots+x^n+1) $


问:
A clarification of first-order axiomatizability
I have been told that certain classes of structures are not first-order axiomatizable, like the class of finite groups. However, in ZFC set theory, we can define both finiteness and groups, and therefore finite groups, by a very complicated formula. But isn't ZFC itself a first-order theory? I am just a bit confused on this point. I would like a clarification of this little paradox.</p>

答:
Finiteness is not first-order axiomatizable. That can be shown by using the compactness theorem of first-order logic. So there are models of ZFC in which sets that satisfy a definition of finiteness within ZFC are not finite. Just as, for example, there are countable models of ZFC in which the theorem asserting the uncountability of the set of all sets of finite cardinal numbers is true.

Imagine a set $[n]=\{0,1,2,\ldots,n\},$ where $n$ is a finite cardinal number within a model of ZFC. A theorem within ZFC states that for every one-to-one correspondence between that set and some subset of $[n+3] = \{0,1,2,\ldots,n,n+1,n+2,n+3\}$ there are exactly three elements of the latter set that are not within that subset. But if $[n]=\{0,1,2,\ldots,n\}$ is an infinite set, then we know there are one-to-one correspondences between $[n]$ and some of its subsets with infinite complements. But none of those latter correspondences is a member of the model, thus the theorem saying there are always three left over can remain true within the model. That is how an infinite set can satisfying the definition-within-ZFC of finiteness. Similarly, a countable set may satisfy a definition-within-ZFC of uncountability because none of its enumerations is a member of the model.


问:
Finite sum of binomial coefficients
Is there a way to calculate:$\sum_{i=0}^{k} {2k+1\choose k-i}$ using only:

symmetry;

pascal's triangle;

one of these sums: $$\sum_{i=0}^{k} {n+i\choose i}={n+k+1\choose k}$$ and $${p\choose p}+{p+1\choose p}+\dots+{n\choose p}={n+1\choose p+1}.$$ I am not sure, because I do not see a way to get same lower number in all collectors so I can use the second sum. Also I tried using symmetry but not helpful really.

答:
From symmetry you have got

$$\sum_{i=0}^k \binom{2k+1}{k-i} = \sum_{i=1}^{k+1}\binom{2k+1}{k+i}$$

Summing them up we have

$$\begin{align}&\quad\sum_{i=0}^k \binom{2k+1}{k-i} + \sum_{i=1}^{k+1}\binom{2k+1}{k+i}\\&=\binom{2k+1}{k}+\binom{2k+1}{k-1}+\cdots + \binom{2k+1}{0}+\binom{2k+1}{k+1}+\binom{2k+1}{k+2}+\cdots+\binom{2k+1}{2k+1}\\&=\sum_{i=0}^{2k+1}\binom{2k+1}{i}\end{align}$$

The last sum is a famous sum involving the binomial theorem.

Title: Finite sum of binomial coefficients


问:
Var(X_2months) = 2 Var(X_1month) with independent random variables?
My instructor insists that
$$\mathrm{Var}(X_{2\text{months}}) = 2 \mathrm{Var}(X_{1\text{month}}) $$
with independent random variables $X_{1\text{month}}$ and $X_{2\text{months}}$, but generally with a random variable $Y$
$$\mathrm{Var}(2Y) = 2^2\mathrm{Var}(Y).$$
So
$$\mathrm{Var}(X_{2\text{months}}) = \mathrm{Var}(2X_{1\text{month}}) = 2^2 \mathrm{Var}(X_{1\text{month}}).$$
Which one is right and why?</p>

答:
This is a typical confusion. For any independent random variables $A$ and $B$ the following two equations hold:

$Var(k A) = k^2 Var(A) \Rightarrow Var(2 A) = 4 Var(A)$

$Var(A + B) = Var(A) + Var(B)$

The problem arises when one tries to use the second to find $Var(2 A)$, by replacing $B$ for $A$:

$Var(2 A) = Var(A + A) = Var(A) + Var(A) = 2 Var(A)$

which contradicts the first equation. But this is wrong, because the second equation (variance of sum equals sum of variances) does not necessary hold if the variables are not independent -and certainly A is not independent of A...

Added: To get the complete picture, the general formula is easy to get:

$Var(A + B) = Var(A) + Var(B) + 2 Cov(A, B)$

This holds for any $A,B$, and it reduces to $Var(A + B) = Var(A) + Var(B)$ if $A$ and $B$ are independent (uncorrelated is enough). In the other extreme case (extreme correlation), when $B=A$, $Cov(A,B)=Var(A)$ and $Var(A+A)=4 Var(A)$. (You could also check the other extreme case: $B = -A$, correlation coeffient = -1)


问:
Given a bijection $f: A \to B$,then show that $\forall g \subseteq f \subseteq A \times B:g$ is a bijection.
Given a bijection $f: A \to B$,then show that  $\forall g \subseteq f \subseteq A \times B:g$ is a bijection.

If $P$ is a property for which the elements  of $f$ satisfy then for every element $(a,b) \in g$ it's clear from the definition of $\subseteq$ that  $(a,b) \in f$ and so such elements  of $g$ satisfy the property $P$.

This is what I know, but does there exist a better way to show that $g$ is a bijection?

答:
The statement you want to prove is, in itself, extremely questionable (some would say "false"). First, it is clear that if $f\subseteq A\times B$ is a function $A\to B$ and $g\subsetneq f$, then $g$ isn't a function $A\to B$ because $\operatorname{dom}g\subsetneq A$.

Secondly, even if we want to inspect what $g:\operatorname{dom}g\to B$ is, it is a fact that if $f$ is an injective function $A\to B$ and $g\subsetneq f$, then $g[\operatorname{dom} g]\subsetneq f[A]$. Therefore it is impossible for $g$ to be surjective onto $B$.

After peeling off all impossibilities, the ony thing that is possible to prove is that if $f:A\to B$ is a bijection and $g\subseteq f$, then $g:\operatorname{dom} g\to \operatorname{range}g$ is bijective, which is quite a weird way to phrase the fact that $g:\operatorname{dom}g\to B$ is injective.


问:
Proof verification for Cesàro convergence
I've recently encountered in this other question the following statement:

Let $(x_n)$ be a bounded non-negative real sequence, then
$$
\lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^n x_i=0 \Longleftrightarrow \lim_{n\to \infty}\frac{1}{2^n}\sum_{i=2^n}^{2^{n+1}-1} x_i=0
$$

for which I've provided the following partial proof.

(I've already posted that in the previous question but then deleted because the asker was not seeking for a proof but only for a previous reference).

Firstly note that by $2^n=N$ the following holds

$$\lim_{n\to \infty}\frac{1}{2^n}\sum_{i=2^n}^{2^{n+1}-1} x_i=0 \iff  \lim_{N\to \infty}\frac{1}{N}\sum_{i=N}^{2N-1} x_i=0$$

then we proceed with the proof separetely for the two implications.

1) Proof for $\implies$

$$\frac{1}{n}\sum_{i=n}^{2n-1} x_i=\frac{1}{n}\sum_{i=1}^{2n-1} x_i-\frac{1}{n}\sum_{i=1}^{n-1} x_i=\frac{2n-1}{n}\frac{1}{2n-1}\sum_{i=1}^{2n-1} x_i-\frac{n-1}{n}\frac{1}{n-1}\sum_{i=1}^{n-1} x_i\to 0$$

2) Proof for $\impliedby$

$$\frac{1}{n}\sum_{i=1}^{n} x_i=\frac{1}{n}\sum_{i=1}^{2n-1} x_i-\frac{1}{n}\sum_{i=n+1}^{2n-1} x_i=\frac{2n-1}{n}\frac{1}{2n-1}\sum_{i=1}^{2n-1} x_i-\frac{1}{n}\sum_{i=n+1}^{2n-1} x_i$$

$$\iff \frac{2n-1}{n}\frac{1}{2n-1}\sum_{i=1}^{2n-1} x_i-\frac{1}{n}\sum_{i=1}^{n} x_i=\frac{1}{n}\sum_{i=n+1}^{2n-1} x_i=\frac{1}{n}\left(-x_n+\sum_{i=n}^{2n-1} x_i\right)\to 0$$

I'm stuck here: how can we conclude from the last identity that $\frac{1}{n}\sum_{i=1}^{n} x_i \to 0$. That is: how can we assume that this latter limit exists? It is trivially true when the limit for $x_i$ exists but what about the general case? Can we conclude the proof by this way or we need to use a completely different approach?

EDIT

I've added the condition that $x_i\ge 0$ after an update for the linked question.

答:
You write a chain of (correct) identities where the last one is:
$$
\frac{2n-1}{n}\frac{1}{2n-1}\sum_{i=1}^{2n-1} x_i-\frac{1}{n}\sum_{i=1}^{n} x_i=\frac{1}{n}\left(-x_n+\sum_{i=n}^{2n-1} x_i\right).
$$
Now, you wrote that the last term above $\frac{1}{n}\left(-x_n+\sum_{i=n}^{2n-1} x_i\right)$ tends to $0$ by using the fact (by a your previous comment in the deleted answer) that
$$
\lim_{n\to \infty}\frac{1}{2^n}\sum_{i=2^n}^{2^{n+1}-1}x_i=\ell \Longleftrightarrow \lim_{N\to \infty}\frac{1}{N}\sum_{i=N}^{2N-1}x_i=\ell.\,\,\,\,\,\,\,\,(*)
$$
However, this is false.

A counterexample to (*) is the $\{0,1\}$-valued sequence, where $x_n=1$ if and only if $3\cdot 2^k\le n<3\cdot 2^{k+1}$ for some $k$ even. Now, the "if" implication is trivial. However, the "only if" does not hold: indeed, fix $k$ even. Then, for $N=2^k$ we have $\frac{1}{N}\sum_{i=N}^{2N-1}x_i=\frac{1}{2}$, while for $N=3\cdot 2^k$ we have $\frac{1}{N}\sum_{i=N}^{2N-1}x_i=1$.


问:
The set of matrices is a ring
First problem:
I have set of matrices
\begin{pmatrix}
x & y \\
ry & x
\end{pmatrix}
where, $x,y \in R$, $R$ is a ring, and $r$ is a fixed element from $R$. I need to proof that this set is a ring with respect to matrix multiplication and addition.

My attempt:

I need to show that it is an abelian group with respect to addition.

1.1) Associativity is quite simple

1.2)I need to determine the Identity element. The identity element here is
$$\begin{pmatrix}
x & y \\
ry & x
\end{pmatrix} + \begin{pmatrix}
e & e \\
re & e
\end{pmatrix}= \begin{pmatrix}
x & y \\
ry & x
\end{pmatrix} $$

where $e$ is identity element from $R$

1.3) Need to determine the inverse element

$$\begin{pmatrix}
x & y \\
ry & x
\end{pmatrix} + \begin{pmatrix}
-x & -y \\
r(e-y) & -x
\end{pmatrix} = \begin{pmatrix}
e & e \\
re & e
\end{pmatrix}$$
1.4) the group with respect to + is abelian

To proof that multiplication is distributive with respect to addition we need just doing matrix multiplication, nothing special here.

Am I missing something in the first case?

Second problem:
I have another set of matrices:
$$\frac{1}{2} \begin{pmatrix}
x & y \\
ay & x
\end{pmatrix}$$

where $a \in Z$ and not divisible by square of prime, and $x,y \in Z$ have the same parity. How these conditions effect the proof? I don't see any differences with proof if $x,y,a \in Z$.

答:
Your first proof seems to be correct (You could add that is clossed under addition and multiplication).

For the second problem, you have that:
$$\frac{1}{2}
\begin{pmatrix}
x & y\\
ay & x
\end{pmatrix}\frac{1}{2}
\begin{pmatrix}
x' & y'\\
ay' & x'
\end{pmatrix}=\frac{1}{2}
\begin{pmatrix}
\frac{xx'+ayy'}{2} & \frac{xy'+yx'}{2}\\
a\frac{xy'+yx'}{2} & \frac{xx'+ayy'}{2}
\end{pmatrix}
$$

You must check that $\frac{xx'+ayy'}{2}\in\mathbb{Z}$ and $\frac{xy'+yx'}{2}\in \mathbb{Z}$, so for that, you need to use that $a$ is not divisible by square of prime, and $x$, $y$ have the same parity.

Title: The set of matrices is a ring


问:
Simplify $\sqrt[3]{\sqrt[3]{A}-B}$ to $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$
I'm wondering if there is a way to simplify the nested cubic radicals

$$\sqrt[3]{\sqrt[3]{A}-B}$$ into its denested form
$$\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$$

Examples include
$$\sqrt[3]{7\sqrt[3]{20}-1}=\sqrt[3]{\frac {16}{9}}+\sqrt[3]{\frac {100}{9}}-\sqrt[3]{\frac {5}{9}}$$
$$\sqrt[6]{7\sqrt[3]{20}-19}=\sqrt[3]{\frac {5}{3}}-\sqrt[3]{\frac {2}{3}}$$
which seem to have the denested form. Is there a systematic procedure?

答:
The answer is yes, curtesy of Ramanujan.

For nested cubic radicals of the form $\sqrt[3]{\sqrt[3]{A}-B}$, denote $C =\sqrt[3]{B^3-A}$ and construct the corresponding cubic polynomial below
$$R(x)=x^3 + \frac{B+2C}3x^2 - \frac{(B-C)(2B+C)}{27}x+ \frac{(B-C)^3}{729}$$
which, per Ramanujan, has the property that its roots $x_1$, $x_2$ and $x_3$ satisfy
$$\sqrt[3]{x_1}+ \sqrt[3]{x_2 }+  \sqrt[3]{x_3 } =\sqrt[3]{\sqrt[3]{A}-B} $$
Thus, denesting $\sqrt[3]{\sqrt[3]{A}-B}$ reduces to solving the cubic equation $R(x)=0$, a  straightforward exercise. Take the example of
$\sqrt[3]{21\sqrt[3]{6}-17}$, with $B=17$ and $C= -37$. Its denesting polynomial is
$$x^3 -19x^2+6x + 216 =(x-18)(x-4)(x+3)$$
which results in
$$\sqrt[3]{21\sqrt[3]{6}-17}=\sqrt[3]{18}+\sqrt[3]{4}-\sqrt[3]{3}$$

Listed below are other examples admitting rational detestation, along with their respective polynomials:
$$
\begin{align}
\sqrt[3]{\sqrt[3]{2}-1}&=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49},
&\>\>\> &x^3 -\frac13 x^2-\frac2{27}x + \frac8{729}\\
\sqrt[3]{39\sqrt[3]{12}-19}&=\sqrt[3]{48}+\sqrt[3]{9}-\sqrt[3]{4},
&&x^3 -53x^2+204x + 1728\\
 \sqrt[3]{7\sqrt[3]{20}-1} &=\sqrt[3]{\frac {100}9}+\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59},
&& x^3 -\frac{37}3x^2+\frac{340}{27}x + \frac{8000}{729}\\
\sqrt[3]{93\sqrt[3]{30}+19}&=\sqrt[3]{180}+\sqrt[3]{25}-\sqrt[3]{6},
&&x^3 -199 x^2+3270 x + 27000 \\
\sqrt[3]{13\sqrt[3]{70}-17}&=\sqrt[3]{\frac{245}9}+\sqrt[3]{\frac{50}9} -\sqrt[3]{\frac{28}9},
 &&x^3 -\frac{89}3 x^2+\frac{1330}{27} x + \frac{343000}{729}\\
\end{align}
$$


问:
Do the premises logically imply the conclusion?
$$b\rightarrow a,\lnot c\rightarrow\lnot a\models\lnot(b\land \lnot c)$$

I have generated an 8 row truth table, separating it into $b\rightarrow a$, $\lnot c\rightarrow\lnot a$ and $\lnot (b\land\lnot c)$. I know that if it was

$$\lnot c\rightarrow\lnot a\models\lnot(a \land \lnot c)$$

I would only need to check the right side with  every value that makes the left side true to make sure the overall statement is true. How do I deal with more than one premise?

答:
The second premise $\neg c\to\neg a$ implies that $a\to c$.

The first premise $b\to a$ leads to $b\to a\to c$, which implies $\neg c\to\neg b$.

The two last statements clearly prevent $b\land\neg c$ from being true.


问:
Is induction needed for this argument?
This question is about Exercise 13 in Section 1-1 of Number Theory by George E. Andrews. The book says:

The Lucas numbers $L_n$ are defined by the equations $L_1 = 1$, and $L_n = F_{n + 1} + F_{n - 1}$ for each $n \geq 2$. We also have that

$\textbf{Claim: }\ L_n = L_{n - 1} + L_{n - 2}\ \text{ for }\ n \geq 3$.

$\textit{My attempt at a proof: }$
We approach by mathematical induction. For the base case, consider when $n = 3$. We have that $L_3 = F_4 + F_2 = 3 + 1 = (2 + 1) + 1 = (F_3 + F_1) + 1 = L_2 + L_1$ so, the claim holds when $n = 1$. Now suppose the claim holds for an integer $k > 1$. Then,
\begin{align*}
    L_{k + 1} &= F_{k + 2} + F_k && \text{definition of Lucas number}\\
    &= (\underline{F_k} + \underline{\underline{F_{k + 1}}}) + (\underline{F_{k - 2}} + \underline{\underline{F_{k - 1}}}) && \text{Fibonacci recurrence}\\
    &= L_{k - 1} + L_k && \text{definition of Lucas number}.
\end{align*}
Hence, the claim also holds for the integer $k + 1$. By the principle of mathematical induction, the claim holds for all $n \geq 3$.

The problem I'm having is that I haven't used the inductive hypothesis. So, I am confused whether my proof is wrong or if the argument doesn't require induction, etc.

I have done a quick search and it seems that the Lucas numbers are usually defined by $L_0 = 2$, $L_1 = 1$, and $L_n = L_{n - 1} + L_{n - 2}$ for  $n \geq 3$. And the problem is usually to show $L_n = F_{n + 1} + F_{n - 1}$ for each $n \geq 2$. Would this be a typo in the book or a misunderstanding on my end?

答:
You are correct when you say that your proof does not use induction. It's a straightforward argument using the definition given for the Lucas numbers and the property of the Fibonacci numbers.

The conclusion is that the Lucas and Fibonacci numbers satisfy the same recursive condition.

Then to recover the definition you found for the Lucas numbers you need just check the original conditions.

Since the definitions are in fact equivalent, there's no reason to assume that there's a typo in the book.

Title: Is induction needed for this argument?


问:
Find the least integral value of $n$ for $f(n)={\left(\frac{9}{10}\right)}^n<\frac{1}{2}$
I was doing a probability question where it was required to find out the least integral value of n with $$f(n)=\left(\frac 9 {10}\right)^n<\frac{1}{2}$$

A little calculation(by calculator of course) yields $\min(n)=7$.

as:

$f(5)=0.59$

$f(6)=0.53$

$f(7)=0.43$

My question is there a good way to approximately tell which integer $n$ can be without computational software and log tables (even $\log 2,\log 3$ etc.)vCalculus seems to be an unlikely option

答:
The informal heuristic you're looking for is called the "Rule of 70". In its most typical form, it says that if a quantity $A(t)$ increases by a percentage rate $r$ every unit of time, or $$A(t+1) = (1+r) A(t),$$ then the time it takes to double is approximately $70/r$. It can also be used for constant percentage decreases: if $$A(t+1) = (1 - r)A(t),$$ where $r$ is the percent decrease per unit of time, then the "half-life" of $A(t)$ (i.e. the time $\lambda$ it takes for $A(\lambda) = A(0)/2$) is again approximately $70/r$.

In this case, you have a constant percent decrease of $r = 0.10$, or 10%, so the Rule of 70 predicts the halving time should be $$70/10 = 7,$$ which is exactly what we found by checking each case.


问:
prove $\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$

Prove if $a,b,c$ are positive $$\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$$

My proof:After rearranging we have to prove $$\sum_\text{cyc} \frac{b}{b^2+2b} \le \sum_\text{cyc} \frac{a}{b^2+2b}$$

As inequality is cyclic:

let $a\ge b\ge c$ then $$\frac{1}{a^2+2a}\le \frac{1}{b^2+2b}\le \frac{1}{c^2+2c}$$.The rest follows by rearrangement inequality.

The case $a\ge c\ge b$ is analogous.

Thus Proved!

Is it correct?...And any other alternative ways possible?

答:
Your proof is nice and right.

Another way.

Let $c=\min\{a,b,c\}$.

Thus, we need to prove that:
$$\frac{a}{b}+\frac{b}{a}-2+\frac{c}{a}-\frac{b}{a}+\frac{b}{c}-1\geq\frac{a+2}{b+2}+\frac{b+2}{a+2}-2+\frac{c+2}{a+2}-\frac{b+2}{a+2}+\frac{b+2}{c+2}-1$$ or
$$\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}\geq\frac{(a-b)^2}{(a+2)(b+2)}+\frac{(c-a)(c-b)}{(a+2)(c+2)},$$ which is obvious.


问:
Isomorphism between quotient spaces adjoining elements
I've been studying the way to adjoin an element that satisfies some polynomial equation $p$ (meaning it's a root) to a commutative ring with unity $R$, which is done by considering the quotient space $R[x]/(p(x))$. To avoid any possible problems, let's consider the polynomials we'll talk about monic and of degree greater or equal to 1, so each polynomial from $R[x]$ has a representative of degree strictly less than $p$ and each constant element from $R$ is preserved.

Now I'm considering how to adjoin multiple elements, each one satisfying one polynomial equation. There are two ways of doing this: directly, and inductively. My real question is about the equivalence of the two (see the end), but I would like a fast-check of the two ways I got. For the sake of simplicity, let's consider the case of two elements satisfying polynomials $p$ and $q$ respectively:

Directly: Consider $p(x)$ and $q(y)$ as elements of $R[x,y]$. Then to adjoin the two elements we can consider $R[x,y]/(p(x),q(y))$, since $p(\bar x)=\overline{p(x)},q(\bar y)=\overline{q(y)}\in(p(x),q(y))$.

Inductively: First we add the elemet satisfying $p$ the same way as always; by considering $R[x]/(p(x))$. Now we add the other element to the ring $R[x]/(p(x))$ by doing the same; $\Big(\big(R[x]/(p(x))\big)[y]\Big)/(q(y))$.

I really think this is correct, but I wrote it anyway in case I'm missing something. Anything to add?

Now, my real question is whether $R[x,y]/(p(x),q(y))\simeq\Big(\big(R[x]/(p(x))\big)[y]\Big)/(q(y))$ or not. Of course, intuitively, it has to, but I'm struggling a little bit justifying it. Can I use somehow that $R[x,y]\simeq\big(R[x]\big)[y]$? How would you do it?

Thanks in advance.

答:
We have a natural immersion $R[x]\hookrightarrow R[x,y]$. This gives the composite map $R[x]\hookrightarrow R[x,y]\to R[x,y]/(p(x))$, whose kernel is precisely $(p(x))$. This gives you an induced map $R[x]/(p(x))\to R[x,y](p(x))$.

From here, the universal property of the polynomial ring gives you a morphism $(R[x]/(p(x)))[y]\to R[x,y]/(p(x))$, given by sending $y$ to $y$. Now compose with the quotient map to $(R[x,y]/(p(x)))/(\overline{q(y)})$, which by the isomorphism theorems is isomorphic to $R[x,y]/(p(x),q(y))$. This gives you a morphism
$$\left(\frac{R[x]}{(p(x))}\right)[y] \longrightarrow \frac{R[x,y]}{(p(x),q(y))}.$$
The kernel of this map is precisely $(q(y))$, so you get a one-to-one morphism from $\Bigl(\bigl(R[x]/(p(x))\bigr)[y]\Bigr)/(q(y))$ to $R[x,y]/(p(x),q(y))$.

Now it’s just a matter of making sure the map is surjective. This will follow because the map includes at least the (equivalence classes of) $R$, $x$, and $y$ in its image, which generate.


问:
Confused on proof that every group of order $p^2$, $p$ prime is isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_{p}\oplus \mathbb{Z}_{p}$
Every group of order $p^2$, $p$ prime is isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_{p}\oplus \mathbb{Z}_{p}$

I am confused about two parts of this proof.

Proof:Assume every nonidentity element of this group $G$ has order $p$. Then $\langle a\rangle$ is normal otherwise there is an element $b$ in $G$ such that $bab^{-1} \notin \langle a\rangle$

Here is my first source of confusion. if $\langle a\rangle$ is not normal I would suspect there is an element $b$ in $G$ with $ba^{i}b^{-1} \notin \langle a\rangle,\text{for some}\space i \in \mathbb{Z}$. Why does $bab^{-1} \notin \langle a\rangle$ necessarily hold with $a$?

Next part of confusion
Since $\langle a\rangle \cap \langle bab^{-1}\rangle =\{1\}$ the distinct left cosets of $\langle bab^{-1}\rangle$ are $\langle bab^{-1}\rangle,a\langle bab^{-1}\rangle,...,a^{p-1}\langle bab^{-1}\rangle$

is this because there must be $p$ distinct cosets and there union must be $G$, so this must be all of the cosets ?

答:
For your first question, note that $\langle a\rangle$ is disjoint from $\langle bab^{-1}\rangle$ (except for the identity) if it is different by Lagrange's theorem. Thus, if $ba^ib^{-1}\notin \langle a\rangle$ for any $i$, the same is true of $a$, because $\langle a\rangle\cap\langle bab^{-1}\rangle=\{1\}$.

The reason all of those cosets are distinct is because coset $i$ contains $a^i$, which none of the others contain. By cardinality, since there are $p$ disjoint cosets it must be all of them.


Title: Finite sum of binomial coefficients

Body:
Is there a way to calculate:$\sum_{i=0}^{k} {2k+1\choose k-i}$ using only:

symmetry;

pascal's triangle;

one of these sums: $$\sum_{i=0}^{k} {n+i\choose i}={n+k+1\choose k}$$ and $${p\choose p}+{p+1\choose p}+\dots+{n\choose p}={n+1\choose p+1}.$$ I am not sure, because I do not see a way to get same lower number in all collectors so I can use the second sum. Also I tried using symmetry but not helpful really.

Body:
$$S=\sum_{i=0}^{k} {2k+1 \choose k-i}=\sum_{j=k}^{0} {2k+1 \choose j}=\sum_{l=0}^{k} {2k+1 \choose l} \tag1$$
We know that $$2^{2k+1}=\sum_{p=0}^{2k+1} {2k+1 \choose p}
=\sum_{p=0}^{k} {2k+1 \choose p}+ \sum_{p=k+1}^{2k+1} {2k+1 \choose p}=S+S'\tag2$$
In the second sum let $q=p-k-1$, then $$S'=\sum_{q=0}^{k} {2k+1 \choose q+k+1}=\sum_{q=0}^{k} {2k+1 \choose k-q} \tag3.$$ Let $j=k-q$, then
$$S'=\sum_{j=0}^{j=k} {2k+1 \choose j}=S \tag4$$
From (2), we get $$S+S'=2S=2^{2k+1} \implies S=2^{2k}=4^k$$


问:
There is no $n$ such that the digits of $16^n$ form a decreasing sequence.
Show that there is no $n$ such that the digits of $16^n$ in base $10$ form a (non necessarily strictly) decreasing sequence. For example $64$ or $9331$ are decreasing sequences.

You can bash this mod $10^k$ for some big (but fixed) $k$ but as this was an exam problem I would like to see a more computation free approach.

答:
Observation 1: The last digit of $16^n$ is always 6.

Observation 2: The last 2 digits of $16^n$ is divisible by 4, so have to be 76 or 96 in order for the string to be decreasing.

Observation 3: $n \geq 7$ by checking the first few cases. (In fact, we just need to check $n = 3, 5 $ due to observation 2.)

Observation 4: $16^n > 10^n$, so we can talk about the last $n$ digits without worrying about an initial string of 0's. (This is helpful with the subtraction later.) This last $n$ digits must be a multiple of $2^n$.

Observation 5: If $16^n \equiv 96 \pmod{100}$, then for a decreasing string we must have $ 16^n \equiv 10^n - 4 \pmod{10^n}$, which is not a multiple of $8$. So $ n\leq 2$, which we have ruled out.

We now focus on $16^n \equiv 76 \pmod{100}$. For a decreasing string, it looks like     $\underbrace{9\ldots9}_{x} \underbrace{8\ldots8}_{y} \underbrace{7\ldots7}_{z} 6$.

Then $ 16^n \equiv 10^n - \underbrace{1\ldots1}_{a} \underbrace{2\ldots2}_{b}4 \pmod{10^n}$.

Set $R = \underbrace{1\ldots1}_{a} \underbrace{2\ldots2}_{b}4$, which is a multiple of $2^n$. (Note that $ a + b \leq n-1$, though we will not use it.)

Observation 6: For $\underbrace{1\ldots1}_{a} \underbrace{2\ldots2}_{b}4$ to be a multiple of 4, it must be $\underbrace{1\ldots1}_{a} \underbrace{2\ldots2}_{b-1}24$.

Observation 7: For $\underbrace{1\ldots1}_{a} \underbrace{2\ldots2}_{b-1}24$ to be a multiple of 8, since 24 is divisible by 8, the prior digit must be 2, so it must be $\underbrace{1\ldots1}_{a} \underbrace{2\ldots2}_{b-2}224$.

Observation 8: For $\underbrace{1\ldots1}_{a} \underbrace{2\ldots2}_{b-1}24$ to be a multiple of 16, since 224 is divisible by 16, the prior digit must be 2, so it must be $\underbrace{1\ldots1}_{a-1}\underbrace{2\ldots2}_{b-3}2224$.

Observation 9: For $\underbrace{1\ldots1}_{a-1}\underbrace{2\ldots2}_{b-3}2224$ to be a multiple of 32, since 2224 is NOT divisible by 32, the prior digit must be 1, so it must be $\underbrace{1\ldots1}_{a-1}12224$.

Observation 10: $112224$ is not a multiple of 64. So $n \leq 6$.

Notes

As you'd have realized, the leading string of 1, 2 is initially not restrictive. If we had a choice of even and odd, then we can always continue the string.

However, once we're forced to choose 1 (odd), then our only remaining choices are 1 (odd), and it cannot always be continued. That's the motivation for Observation 6-10.


问:
Describing $\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(x, n) \; \mathrm{d}n$ as a contour integral.
My question essentially has to do with the derivative of a Contour Integral's parameterized curve.
$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n$$ to be exact. Where $\partial \Omega(x)$ is a Jordan curve which is differentiable for any $x \in \mathbb{C}$, and $f(n, x): \mathbb{C}^2 \to
\mathbb{C}$ integrable around the curve $\partial \Omega(x)$ in respect to $n$. Define $\gamma$ as the parameterized curve of $\partial \Omega$, and the terminology $f_x(n, x) = \frac{\partial f(n, x)}{\partial x}$ is used.

My work has essentially gotten down to these steps.

STEP 1: Turning the contour integral into the usual integral.

$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n = \frac{\partial}{\partial x} \int_{0}^{2\pi} \gamma_\theta(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 2: Using the Liebniz rule.

$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n = \int_{0}^{2\pi} \frac{\partial}{\partial x} \gamma_\theta(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 3: Taking the derivative.

$$= \int_{0}^{2\pi} \gamma_\theta(\theta, x) f_x(\gamma(\theta, x), x) + \gamma_x(\theta, x) \gamma_\theta(\theta, x) f_n(\gamma(\theta, x), x) + \gamma_{\theta x}(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 4: Separating the integrals.

$$= \int_{0}^{2\pi} \gamma_\theta(\theta, x) f_x(\gamma(\theta, x), x) \; \mathrm{d}\theta$$
$$+ \int_{0}^{2\pi} \gamma_x(\theta, x) \gamma_\theta(\theta, x) f_n(\gamma(\theta, x), x) \mathrm{d}\theta$$
$$+ \int_{0}^{2\pi} \gamma_{\theta x}(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 5: Simplifying the first integral into a contour integral.

$$\int_{0}^{2\pi} \gamma_\theta(\theta, x) f_x(\gamma(\theta, x), x) \; \mathrm{d}\theta = \oint_{\partial \Omega(x)} f_x(n, x) \; \mathrm{d}n.$$

STEP 6: Plugging in the first integral to get the final answer.

$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n$$

$$= \int_{\partial \Omega(x)} f_x(n, x) \; \mathrm{d}n+\int_{0}^{2\pi} \gamma_x(\theta, x) \gamma_\theta(\theta, x) f_n(\gamma(\theta, x), x) \; \mathrm{d}\theta$$
$$+\int_{0}^{2\pi} \gamma_{\theta x}(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

I am unsure of how to simplify this further or if this is even a decent approach. Does anybody have a good resource for this?

My goal is to write this derivative as multiple contour integrals, bar any $\gamma$-parameterized functions.

(Disclaimer: This same question has been posted by myself to MathOverflow)

答:
Your computation is correct (although at the very beginning I would write $d/dx$, since your contour integral is a function of $x$ only). You need to think of $\gamma_x$ as a variational vector field along the curve $\Gamma_x = \partial\Omega(x)$ and then the second integral is a contour integral over $\Gamma_x$ as well.

EDIT: In particular, we have the contour integral of the function $(f_n\gamma_x)(n,x)$ along the curve. As I suggested, this appears to depend on the parametrization of $\Gamma_x$, but you can think of watching a point on the curve move as a function of $x$ and take the velocity vector of this trajectory (thinking of $x$ as time). This is in fact not independent of the parametrization because you need to watch the point $\gamma(\theta,x)$ move to nearby points with the same $\theta$ value.

The third term seems more interesting. You want to think of $\gamma_{\theta x}$ instead as $(\gamma_x)_\theta$, and then integrate by parts. I believe this gives you another copy of the second term.

EDIT: Here is a more conceptual (and more sophisticated) approach. We want to integrate the $1$-form $\omega = f(n,x)\,dn$ over a curve $\Gamma$ in $\Bbb C$. Choose a variational vector field $X$ along $\Gamma$ (in the calculus of variations one often chooses it to be normal to the curve, but that isn't necessary). You can think of this vector field as giving $\partial\Gamma/\partial x$. We ask how the integral varies with $x$.

Let's reinterpret this by mapping a rectangle $R_\epsilon = [0,2\pi]\times [x,x+\epsilon]$ to $\Bbb C$. This is your map $\gamma$, and for fixed $x$, the image is the curve $\Gamma_x$. My variation vector field is $X=\gamma_x=\dfrac d{d\epsilon}\Big|_{\epsilon=0}\gamma(n,x+\epsilon)$. We are trying to compute $$\dfrac d{d\epsilon}\Big|_{\epsilon=0} \int_{\Gamma_{x+\epsilon}} \omega.$$
Now we recognize this derivative as the integral of $\mathscr L_X\omega$ and apply the famous Cartan formula
$$\mathscr L_X\omega = \iota_X(d\omega) + d(\iota_X\omega).$$
Integrating these over $\Gamma_x$ should give you intrinsic formulations of what we were doing. (Without the Cartan formula, you can use Stokes's Theorem to rewrite that integral over $\partial R_\epsilon$ as a double integral and then do the derivative limit with that.)


问:
Matrix addition - can a 3x2 matrix + a 2x3 matrix be added together?
I have two matrices;

$$ 1)
\left(
\begin{matrix}
    3 & -2 \\
    1 & 2 \\
    0 & -3\\
\end{matrix}
\right)
$$

And
$$ 2)
\left(
\begin{matrix}
    -2 & 1 & 3 \\
    4 & 3 & 2 \\
\end{matrix}
\right)
$$

But can these two be added together, is this possible?
I found from this PDF the following quote:

The sum A + B of two matrices A, B (which
must have the same number of rows, and the
same number of columns) is the matrix (also
with the same shape) given by adding
corresponding elements of A and B.

This suggests to me that I cannot add together Matrix 1 and 2, but I can however add together Matrix 2 and a third matrix with the same dimentions, for example:

$$ 3)
\left(
\begin{matrix}
    2 & 1 & -1 \\
    -1 & 0 & 3 \\
\end{matrix}
\right)
$$

I've come to believe that I can add up Matrix 2+3 but not Matrix 1+2 or 1+3 for instance.
Am I correct in my assumptions here?

And would this differ in any way if i try to multiply the matrices?

答:
First of all, you can't add two matrices if they don't have the same dimension. In your case,
$$\left(
\begin{matrix}
    -2 & 1 & 3 \\
    4 & 3 & 2 \\
\end{matrix}
\right) + \left(
\begin{matrix}
    2 & 1 & -1 \\
    -1 & 0 & 3 \\
\end{matrix}
\right)=\left(
\begin{matrix}
    -2+2 & 1+1 & 3+(-1) \\
    4+(-1) & 3+0 & 2+3 \\
\end{matrix}
\right)=\left(
\begin{matrix}
    0 & 2 & 2 \\
    3 & 3 & 5 \\
\end{matrix}
\right)$$

As you can see, adding two matrices is the same as adding its elements. If we talk about multiplication, things change considerably. I recommend you this link, which is very instructive.


问:
Find $\sum_{n=1}^{\infty}\left({\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}}\right)$
This question comes from "adjusting the order of an infinite series".

We are talking about whether a series' value would change or not by changing the position of some terms.
For example, we have

$$\ln{2}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\sum\limits_{n=1}^{\infty}\left({\frac{1}{2n-1}-\frac{1}{2n}}\right)$$

By changing the order, we get

$$H=1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots$$

and
$$H=\sum\limits_{n=1}^{\infty}\left({\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}}\right)$$

And to generalize, we construct a series which has firstly, k positive terms and then 1 negative term, and the same story repeats. We write it as $H_{k}$.

But how to calculate the values of all these $H_{k}$ becomes a problem, in fact, I'm stuck in $H_{2}$.

Thank you for your attention.

答:
For finite upper bound of the summation there exist for each summand a well known presentation:

S_N=

1/4 (-PolyGamma[0, 1/4] + PolyGamma[0, 1/4 + N])

-1/4 (-PolyGamma[0, 3/4] + PolyGamma[0, 3/4 + N])

+HarmonicNumber[N]/2

The PolyGamme function with 0 at the first argument are HarmonicNumber functions.

Evidently, these do not compensate to zero. So there remains some HarmonicNumber behavoir in the solution for very large N.

.

This part must vanish to give a finite solution, but

For N-> infinity use an asymptotic series expansion:

Log[8]/2=3Log2/2


问:
finite elements-exercice
We consider in $\mathbb{R}^2$ the set of points $$\{M_1(-1,1),M_2(0,1), M_3(2,1),M_4(-1,0),M_5(1,0),M_6(2,0)\}$$

Let $\Omega$ a rectangular structure consisting of the heads $\{M_4(-1,0),M_6(2,0), M_3(2,1),M_1(-1,1)\}$

Let the two polygons $\sum_1 = \{M_4,M_5,M_2,M_1\}$ and $\sum_2 = \{M_5,M_6,M_3,M_2\}$

let the mesh $\{(e_1,\sum_1,Q_1),(e_2,\sum_2,Q_2)\}$ such $$Q_1=\{f : \mathbb{R}^2 \rightarrow \mathbb{R} , f(x,y) = a + bx + cy + d xy , a , b, c , d \in \mathbb{R}\}$$

1- Prove that $(e_1,\sum_1,Q_1)$ and $(e_2,\sum_2,Q_1)$ are two finite elements of Lagrange.

2- Let $V$ the finite-dimensional space that corresponds to the mesh. Prove that $V \nsubseteqq H^1(\Omega)$

My problem is to prove 2.

答:
For any piecewise polynomial $v\in V$, $v$ restricted in each polygon has the form of $Q_1$:
$$
v = \begin{cases}
a_1+b_1 x+ c_1y + d_1 x y \quad\text{ in } \Sigma_1
\\
a_2+b_2 x+ c_2y + d_2 x y \quad\text{ in } \Sigma_2
\end{cases}
$$

Now we wanna argue that there exits such $v\notin H^1(\Omega)$. For $v$ is definitely continuous in both $\Sigma_1$ and $\Sigma_2$ respectively, and it has $H^1$-regularity in both polygons. You can check $v$ restricted on $\Sigma_1$ is $L^2$ and so is its gradient, the same applies to $v$ restricted on $\Sigma_2$.

The only possible problematic part is where these two polygons intersect with each other: $\partial \Sigma_1 \cap\partial \Sigma_2 = M_2 M_5$.

The construction is pretty straight forward, we set $v = 1$ at node $M_2$, and zero on all other nodes, see if we can get a function that is continuous across the interface $M_2 M_5$ between these two polygons.

In $\Sigma_1$, we have:
$$
\begin{cases}
&a_1+&b_1 (-1)+ &c_1 1 + &d_1 (-1) &= v(M_1) = 0
\\
&a_1+ & &c_1 1  & &= v(M_2) = 1
\\
&a_1+ &b_1 1 &  & &= v(M_5) = 0
\\
&a_1+ &b_1 (-1) &  & &= v(M_4) = 0
\end{cases}
$$
solving this leads to:
$$
v\big|_{\Sigma_1} = y+xy.
$$

In $\Sigma_2$, we have:
$$
\begin{cases}
&a_2+&b_2 2+ &c_2 1 + &d_1 2 &= v(M_3) = 0
\\
&a_2+ & &c_2 1  & &= v(M_2) = 1
\\
&a_2+ &b_2 1 &  & &= v(M_5) = 0
\\
&a_2+ &b_2 2 &  & &= v(M_6) = 0
\end{cases}
$$
solving this leads to:
$$
v\big|_{\Sigma_2} = y-\frac{xy}{2}.
$$

Therefore, for a $v\in V$ satisfying that $v(M_2) = 1$ and zero at other nodes:

$$
v = \begin{cases}
y+xy \quad\text{ in } \Sigma_1
\\
y-xy/2 \quad\text{ in } \Sigma_2
\end{cases}
$$

Now it is up to you to check this function is not in $H^1$.

Update:

please help me! i need an reference (book ) who we can found the definition of $V$ ($v\in V$ iff $v\in H^1$ on the two polygones and $v$ is continuous on $M_2$ and $M_5$ please.

I already gave you the reference in the comments. You are not satisfied with its presentation, yet I think it is crystal clear: Ciarlet's The Finite Element Method for Elliptic Problems, please refer to the proof of Theorem 2.1.1 on pg40.

The result is commonly used for finite elements w/o citing any references, and it is very straightforward to prove. It can be formulated as follows:

Lemma: Suppose $\Sigma_1$ and $\Sigma_2$ are Lipschitz polygons, and $\partial\Sigma_1 \cap \partial \Sigma_2 = F$. Let $v_1\in H^1(\Sigma_1)$, $v_2\in H^1(\Sigma_2)$, and $v \in L^2(\Sigma_1\cup\Sigma_2\cup F)$ so that $v = v_1$ in $\Sigma_1$ and $v=v_2$ in $\Sigma_2$. Then if $v_1 = v_2$ on $F$, we have $v\in H^1(\Sigma_1\cup\Sigma_2\cup F)$.

Proof: Let $\Omega = \Sigma_1\cup\Sigma_2\cup F$. It suffices to show that the weak gradient of $v$ on the joint domain coincides with $\nabla v_1$ on $\Sigma_1$ and $\nabla v_2$ on $\Sigma_2$. Let $\phi \in C^{\infty}_c(\Omega)$,
$$
\int_{\Omega} \frac{\partial v}{\partial x_i} \,\phi :=
- \int_{\Omega} v \,\frac{\partial \phi}{\partial x_i} .
$$
This is the definition. The right side now can be divided into two parts:
$$
- \int_{\Omega} v \,\frac{\partial \phi}{\partial x_i} =
- \int_{\Sigma_1} v_1 \,\frac{\partial \phi}{\partial x_i}
- \int_{\Sigma_2} v_2 \,\frac{\partial \phi}{\partial x_i}
\\
= \int_{\Sigma_1} \frac{\partial v_1}{\partial x_i} \,\phi - \color{blue}{\int_{\partial \Sigma_1}  v_1 \phi \,(n_{\partial \Sigma_1})_i\,dS}+ \int_{\Sigma_2} \frac{\partial v_2}{\partial x_i} \,\phi
- \color{blue}{\int_{\partial \Sigma_2}  v_2 \phi \,(n_{\partial \Sigma_2})_i\, dS},
$$
$n_{\partial \Sigma_k}$ is the unit outer normal vector to $\Sigma_k$ ($k=1,2$). Blue terms will cancel because $n_{\partial \Sigma_1} = -n_{\partial \Sigma_2}$ pointwisely, and $v_1 = v_2$ pointwisely as limit from their defined domain respectively (or use a trace argument if you like). Hence
$$
\int_{\Omega} \frac{\partial v}{\partial x_i} \,\phi =  \int_{\Sigma_1}\frac{\partial v_1}{\partial x_i} \,\phi + \int_{\Sigma_2} \frac{\partial v_2}{\partial x_i} \,\phi.
$$
This tells us $\nabla v = \nabla v_1$ on $\Sigma_1$ and $\nabla v = \nabla v_2$ on $\Sigma_2$. Hence
$$
|\nabla v|_{H^1(\Omega)}^2 = \|\nabla v\|_{L^2(\Omega)}^2
= \|\nabla v_1\|_{L^2(\Sigma_1)}^2 + \|\nabla v_2\|_{L^2(\Sigma_2)}^2 < \infty,
$$
and the lemma follows.

Title: finite elements-exercice


问:
Turning an observation about triangles into a formal proof
I realize that some of this is going to come off as pretentious and I'm sorry if so. I'm just genuinely curious. Recently I've made an observation regarding isosceles triangles.

Take some number $\omega$ and $\triangle ABC$, where $C$ is the top vertex. If $\overleftrightarrow{AC}$ and $\overleftrightarrow{BC}$ both have a length of $\omega \sqrt5$, and $\overleftrightarrow{AB}$ has a length of $2\omega$, then there exists an inscribed square $\square LMNO$ with sides of length $\omega$ where two of the vertices are the midpoints of $\overleftrightarrow{AC}$ and $\overleftrightarrow{BC}$ respectively. Here is an example $\triangle ABC$ to illustrate what I mean.

I don't think this is life-changing or anything, I just think it's really cool and would like some help writing a proof for it.

答:
Let us denote the midpoint of $AB$ (which is the origin in your diagram) by $D$.

Define $L, M, N, O$ to be the midpoints of $DA, AC, CB, BD$ respectively. We need to show that $LMNO$ is indeed a square with side length $\omega$.

By midline/midpoint theorem, we have $ML // CD$ and $ML = \frac12CD$.

Similarly $NO // CD$ and $NO = \frac12CD$.

By the properties of an isosceles triangle (or SSS congruence), $AB \perp CD$.

By Pythagoras Theorem, $CD = \sqrt{AC^2 - AD^2} = \sqrt{5\omega^2 - \omega^2} = 2\omega$.

Hence $ML = NO = \omega$ and $ML // NO$. This shows that $LMNO$ is a parallelogram.

Since $LO = 2(\frac12\omega) = \omega = ML = NO$, $LMNO$ is a rhombus.

Since $LM // CD$, $LM \perp AB$. Hence $LMNO$ is a square.


问:
Spectral decomposition of $-\Delta$ the Laplacian
I am currently trying to self learn about an interesting idea that caught my eye in spectral geometry, which is the whole idea of hearing the shape of the drum by solving the wave equation $\Delta \psi = k^2 \psi_{tt}$ on a compact Riemannian manifold $(M,g)$ with dirichlet boundary constraint $\psi|_{\partial M}=0$. The thing is, I don't think I have that of a solid background in functional analysis and PDEs, and I am looking for some help or a rather self contained source which proves that such decomposition exists
$$ -\Delta(\cdot)= \sum_{l=0}^\infty \lambda_l \left \langle \cdot,\phi_l \right \rangle_{L^2(M)}\phi_l$$
What I do know: I wasn't able to find many sources which dive deep into this problem, but the first thing I was able to realize is this decomposition probably isn't on $L^2(M)$ but rather some subspace of the form $V=\left \{ f\ \text{nice enough}|\ f|_{\partial M}=0\right \}$, the reason for this is that we probably need the laplacian to be self adjoint, and looking at one of Green's identities
$$\int_M u\Delta v - v\Delta u   =\int_{\partial M}u\frac{\partial v}{\partial n}-v \frac{\partial u}{\partial n}$$
We need to zero the RHS with the constraint. What is $V$? I saw alot of references to a "Sobolev space $H_0^1(M)$" but couldn't find a coherent definition of it, in addition to the fact that it is equipped with a different inner product than of $L^2(M)$ which makes things more confusing for me.

Anyhow, this tells us automatically why eigenfunctions must be orthogonal, since if we have
$$ -\Delta \phi_1 = \lambda_1 \phi_1,\ -\Delta \phi_2 = \lambda_2 \phi_2$$
for $\lambda_1,\lambda_2 \neq 0$ different then
$$\left \langle \lambda_1\phi_1,\phi_2 \right \rangle_{L^2(M)}=-\left \langle \Delta\phi_1,\phi_2 \right \rangle_{L^2(M)}=-\left \langle \phi_1,\Delta\phi_2 \right \rangle_{L^2(M)}=\left \langle \phi_1,\lambda_2\phi_2 \right \rangle_{L^2(M)}$$
and we get that $\left \langle \phi_1,\phi_2 \right \rangle_{L^2(M)}=0$

I also know that $-\Delta$ is positive definite since $$\int_M u (-\Delta) u=\int_M \left \| \nabla u \right \|_g^2\geq 0$$

What I'm looking for: This topic is pretty weird for me right now and I would like to learn about it thoroughly.

Are my proofs correct or relevant at all?

As I mentioned what is $V$? it's hard for me to see why $C^2(M) \cap C(\bar{M})$ needs to be restricted further.

How do we know that $-\Delta$ is compact if at all? This means that if $f_k \rightarrow f$ in $V$, then $-\Delta f_k$ has a convergent subsequence of functions. I know according to functional analysis this should imply the spectral decomposition but I don't know whether this is true. Edit: appearantly not, $-\Delta^{-1}$ is. Then I am looking at the moment for a proof of that

Assuming 3 is true, we also get for free that the eigenvalues are discrete $0\leq \lambda_0 \leq \lambda_1 \leq \cdots$. However, I saw claims that $\lambda_k \rightarrow \infty$ (when according to functional analysis it should approach to $0$). This doesn't make sense to me, how sums like the spectral decomposition $\Delta f=\sum_{l=0}^\infty  \lambda_l\left \langle f,\phi_l \right \rangle_{L^2(M)}\phi_l  $ converge?

How does one show $\left \{ \phi_l\right \}_{l=0}^\infty$ is complete? meaning we can express any $f\in V$ as a fourier series $f=\sum_{l=0}^\infty \left \langle f,\phi_l \right \rangle_{L^2(M)}\phi_l  $

Sorry for the mess of a question, In the probable case I said something wrong I would love to hear. If anyone knows of a free-access source which adresses this problem please refer me to it!

EDIT: I have read several notes and articles which address this issue. Right now, it all boils down for me to find proofs of the Rellich-Kondrachov theorem and Poincaré inequality for Riemannian manifolds, and I can take it from there using the theory of weak solutions.

答:
A confession: I can only really answer this when the manifold in question is an open set $U$ of $\mathbb{R}^n$, where $\partial U$ is smooth.  (I would imagine this case is used, or at least very instructive for, the case on a general Riemannian manifold.) But I'll try to answer in this setting as best I can. I apologize in advance for the length of this answer.

2) The first thing that must be done is to understand what $V$ is, because functional analytic machinery needs to be applied on a particular (Hilbert) space. Indeed, in this case $V = H^1_0(U)$. I don't know if there is any way to get around having to deal with this object, but you can think of $H^1_0(U)$ as the space of all function $f$ which are zero on the boundary (this is what the "$0$" in the subscript refers to), are square-integrable, and whose derivatives exist and are also square-integrable. This is a Hilbert space. The inner product is not just the $L^2$ inner product anymore, because the inner product on $L^2$ does not care about the derivatives, but our space does. So we define the inner product by
$$
\langle f, g \rangle = \int_U f(x)g(x)\, dx + \int_U \nabla f(x) \cdot \nabla g(x)\, dx.
$$
Note that $\langle f, g \rangle = \langle f, g\rangle_{L^2} + \langle \nabla f, \nabla g\rangle_{L^2}$. The last thing that must be done is to note that this space is not complete if we restrict to the usual defintion of derivatives, and therefore the formal definition is that $f, g$ have square-integrable "weak" derivatives. This is morally perhaps unimportant, but important in the rigorous proofs. This is essentially why you cannot take $V = C^2(U) \cap C^2(\bar{U})$, since this space is not complete under the norm induced by the inner product above, and so if we take a limit of functions in this space, the result won't necessarily be $C^2$.

1) Your calculations are essentially correct, but can be (and in the general theory are) extended to functions with these "weak" derivatives. It turns out not to be important for your calculations involving $\phi_i$ which are eigenvalues of $-\Delta$, since elliptic regularity actually implies all eigenfunctions of $-\Delta$ are smooth.

3) This is where the full machinery comes in. As rubikscube09 mentioned in the comments, I don't believe there is a way to do this without the Rellich-Kondravich theorem (all of this is in Chapter 5, 6 of Evans) and abstract functional analysis. Essentially, the R-K theorem says that the embedding $H^1_0(U) \to L^2(U)$ is compact, in that bounded sequences in $H^1_0(U)$ have an $L^2$-convergence subsequence. Therefore we may think of $(-\Delta)^{-1}$ as mapping from $L^2$ to $H^1_0$ defined by $(-\Delta)^{-1}f = u$ is the unique function such that $-\Delta u = f$. Then $u$ lives in $H^1_0(U)$ (and again, there is something being swept under the rug, since $u \in H^1_0(U)$ only implies $u$ has first order derivatives, not second-order), but the embedding $H^1_0(U) \to L^2(U)$ identifies $u$ as an $L^2$ function in a compact way and thus we may think of $(-\Delta^{-1}) : L^2 \to L^2$. By R-K this is a compact operator.

4) You addressed this in your comment.

5) This is a computation that is a little involved, but not too long. I don't know of free-source materials, but Evans ch. 6 deals with this (in more generality) and has detailed proofs there. There may be copies online?


问:
Necessary but not sufficient
This question has already been asked here twice, namely here and here, but none of the answers address my specific question, except probably for this answer, which comes close.

So, using the notation of the close answer, I don't understand why I have to rule out the tuples $(T,T)$ and $(F,F)$.

Let's call the sentence "Q is necessary but not sufficient for P" R.

As for $(F,F)$, if P is false when Q is also false, this should result in R = true; since Q is necessary for P, so the absence of Q should imply the absence of P. Why would I want R to be false in this case?

And for $(T,T)$, I will imagine a more complete picture. Let's say that P depends on Q and some other factors, collectively named W. Now, we should split the row $(T,T)$ into 2, one with W false, and another with W true. In the case with W true, R should evaluate to T, and in the case with W false, R should evaluate to false. On what basis, then, should we decide to rule out $(T,T)$ in the original statement! In my opinion, the row with $(T,T)$ should be undecidable.

I would be grateful if someone could explain to my why the correct answer is $¬(¬r∧¬p)→¬q∧¬((¬r∧¬p)→q)$ in a way other that

"is necessary" translates to so and so and "is sufficient" translates
to so and so, so the conjunction of the first with the negation of the
second gives the correct answer.

Thanks

答:
The problem here is that you’re treating material implication, symbolized by the logical connective $\to$, as if it were the everyday notion of implication; it isn’t.

‘$Q$ is sufficient for $P$’ means precisely that if $Q$ is true, then $P$ must be true; we formalize this as $Q\to P$. If $P$ and $Q$ are both false, this is vacuously true, since it says nothing about the truth of $P$ when $Q$ is false. That is simply how material implication works. Thus, $R$ must be false when $P$ and $Q$ are both false, since in that case $Q$ is sufficient for $P$.

The implication is also true when both $P$ and $Q$ are true, simply because $P$ is true; again, this is just how material implication works. In intuitive terms, the implication $Q\to P$ has the truth value $F$ only when its falsity can actually be demonstrated from the truth values of $P$ and $Q$, and that is the case only when $Q$ is true and $P$ is false. In all other cases its truth value is $T$. Thus, $R$ must also be false when $P$ and $Q$ are both true.

To say the same thing in slightly different words, neither $(F,F)$ nor $(T,T)$ rules out the possibility that $Q$ is sufficient for $P$, in the one case because we don’t have $Q$, and in the other case because we do have $P$. The only case that actually rules out the possibility that $Q$ is sufficient for $P$ is $(F,T)$. If we assert, as $R$ does, that $Q$ is not sufficient for $P$, we are ruling out the possibility that $Q$ is sufficient for $P$, so we are asserting that $P$ is false and $Q$ is true.

Title: Necessary but not sufficient


问:
A question with infinity
I am a sophomore in high school and my math teacher did a very short lesson on infinity, here's how it went:  (Try to solve each part yourself the first two are easy)

Part 1

You have an inf. number of boxes each labeled like so

```<code>[1][2][3][4].....
</code>```

and that there is a person in each one of these boxes.  But you have another set of five boxes like so

```<code>[1][2][3][4][5]
</code>```

You want to only have one set of boxes what do you do?  You must be able to tell each person what box they should goto next.

Spoiler

Each person is told to move down five boxes.  Or n+5 is you new box, and the five people from the other boxes are able to move into the first five boxes or just n+0.

Part 2

Well you actually have two of these inf. box sets, but you really only want one.  What now? (same rules apply)

Each person from set one is told to go to box 2n and each person from set two is told to box 2n-1

Part 3

Now you have an inf. number of these inf. box sets arranged like so

```<code>1: [1][2][3][4][5]...
2: [1][2][3][4][5]...
3: [1][2][3][4][5]...
4: [1][2][3][4][5]...
5: [1][2][3][4][5]...
</code>```

and you want to fit them all into one inf. box set, how do you do it.  She refused to tell us the answer and no one in my class figured it out, but I think I just did (correct me if I'm wrong, I'm also looking for alternate solutions).

How I went about solving this problem

Step 1

I assumed that there had to be some sort of pattern so I tried a few

Step 2

I picked the following patter.

1: [1][3][6][10][15]...

2: [2][5][9][14][]...

3: [4][8][13][][]...

4: [7][12][][][]...

5: [11][][][][]...

The numbers in the box represent the new box of each person

Step 3

I started to create a formula for the people to find there new room based of there current room number and box set number.  I found the following sequence in for people in the first row, 1 3 6 10 15.  Or +2 +3 +4 +5.  Which told me that it was quadratic so the equation had to be ax^2 + bx + c = new room number

Step 4

I found three equation by plugging in room numbers and their correct new room numbers to find these three equations:

(1^2)a + 1b + c = 1 or a + b + c = 1

(2^2)a + 2b + c = 3 or 4a + 2b + c = 3

(3^2)a + 3b + c = 6 or 9a + 3b + c = 6

Step 5

After solving this system I got the following equation 1/2x^2 + 1/2x + 0 = new room number where x = current room number.  This of course only tell the people in the boxes on the bottom where to go. Now I needed to find how to place the rest of the people.  The first thing I noticed was that people on the same diagonal as seen bellow were very much related to each other when you look at the rooms they will end up in.

[x-4][][][][]

[][x-3][][][]

[][][x-2][][]

[][][][x-1][]

[][][][][x]

The solution

Working this knowledge into my equation was fairly easy (I was able to do it while going for a run so I won't explain how I got my equation).  The equation is:

 (1/2(r+c-1)^2 + 1/2(r+c-1) + 0)-r+1 = new room number | r = row | c = column

Now my question to all of you is

Is my equation correct?

Are there any alternate solutions to this problem?

How could you fit a 3D array of boxes (inf boxes of people in all direction) into a 1D line of boxes?

Can you find a way to fit a n dimensional array of boxes into a 1D line of boxes?

For any of you into JAVA here's a a short snippet of code I made to prove my answer

```<code>    package main;

public class Main {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int f = 2;
        int r = 1;
        int c = 1;
        while(true){
            double x = (.5*(r+c-1)*(r+c-1) + .5*(r+c-1) + 0)-r+1;
            System.out.print(x +", ");
            c++;
            r--;
            if(r < 1){
                r = f;
                f++;
                c = 1;
            }
        }
    }
}
</code>```

答:
First, kudos to you for solving this!

As to your question about possible generalizations to $n$ dimension. Well, I suppose you could repeat the work you did for $n=3,4,\ldots$. There is, however, a simpler way. Take a step back and look at what you have achieved. Your function $$
  f(r,c) = \frac{1}{2}(r+c-1)^2 + \frac{1}{2}(r+c-1)-r+1
$$
takes a pair of numbers (positive integers, to be precise) and maps them to just a single positive integer. And, and that's the point, it does so in a reversive way. After all, when rearranging your boxes according to your function, you don't add or lose any box. Thus, given an positive integer $n$, there's only one pair $(r,c)$ for which $f(r,c)=n$.

Now, say you want to combine pairs of three integers into one, which corresponds to turning an infinite cube of boxes into a straight line. You can simply start with two of the three integers, and map them to a single integer, using your function. Then, you take that value and the last of the original integers, and invoke your function again, ending up with only a single integer. You thus get a function $$
  g(r,c,l) = f(f(r,c),l) \text{.}
$$
And that function is again reversible. Starting from an $n$ you can first figure out $f(r,c)$ and $l$, and then proceed to figure out $r$ and $c$.

So by solving the problem for two dimensions, you have actually solved it for all finite dimensions!

Now image the following situation. You're faced with all the finite-dimensional cases together. In other words, there's a line of boxes, a 2-dimensional array of boxes, a 3-dimensional cube of boxes, a 4-dimensional um, well, whatever of boxes, and so on. In the language of sequences of numbers, you're faced with all single numbers, all pairs, all triples, all quadruples, ... Can you map all of those to a single number in a reversible way? To solve that, remember that you've already solved this separately for each of the cases. Thus, each thing that you want to map already falls into one of the solved cases. You can re-use the solution for that case, you just have to make sure that when undoing the mapping, you have a way to figure out which case that was. Luckily, the case itself is handily represented by a positive integer itself...

If you have solved that, it's again time to step back and reflect. Have you then solved things for the infinite dimensional case? (i.e., have you managed to map arbitrary sequences of numbers to a single number?). Consider for example the sequence of numbers $1,2,3,\ldots$. Have you mapped that to a single number? For that to be true, the sequence $1,2,3,\ldots$ would have to fall into one of the cases single number, pair, tripple, ... Does it?

 The answer is no! You have succeeded in mapping all finite sequences of numbers to a single number, i.e. all sequences with stop at some point. They may be arbitrarily long, but not infinitely long - that's the crucial difference.

Now lets assume for a moment that you have solved things for the infinite-dimensional case, i.e. you've managed to map arbitrary sequences of numbers to a single numbers, reversibly. You can then make a list, each row containing an infinite sequence and the number its mapped to. Your list will then be complete in the sense that every target number and every sequence appears in the list exactly once. Your list might like this $$
\begin{array}{l|l}
 \textrm{target}& \textrm{sequence} \\
\hline \\
1& (1,2,3,4,\ldots) \\
2& (2,3,4,5,\ldots) \\
3& (1) \\
4& (100, 101) \\
\ldots&\ldots
\end{array}
$$

You can imagine such a list to be a function $g(n,i)$ which gives you the $i$-th entry in the infinite sequence mapped to the positive integer $n$. This makes it possible to define the following infinite sequence of positive numbers $$
  g(1,1)+1,\,g(2,2)+2,\,g(3,3)+3,\,\ldots
$$
Can you see a problem with that?

 Try to find this sequence in your original table. If you table is complete, i.e. really reversibly maps infinite sequences to integers and back, it has to be there somewhere. Yet the sequence cannot be the first sequence in your list (i.e., the one mapped to the number 1), because their first element differs - the first sequence in your list has $g(1,1)$ as its first element, and the sequence we just defined has $g(1,1)+1$. But the same argument applied to the second sequence.. and the third... So, it seems that the table cannot have been complete! So, it turns out, the infinite dimensional case has no solution. You cannot map all finite and infinite sequences of positive integers to a single positive integers!

You can also try something different. Try mapping finite and infinite sequences of positive integers to real numbers. Can you find a way which does that reversibly, and hits each real number exactly once?

Title: A question with infinity


问:
On the behavior of transcendental functions
Given that single-variable algebraic functions take on algebraic values when the input is algebraic, and take on transcendental values when the input is transcendental, and knowing that transcendental functions are all those functions that are not algebraic,

how do I determine the nature of the output of a transcendental function in a similar fashion?

For example, in the case of $\sin : \mathbb R \to \mathbb R $, there are transcendental numbers (i.e. integer multiples of $\pi$) that map to algebraic numbers; and what can I say about the image of natural numbers or rational numbers ($\sin (\mathbb N)$, $\sin (\mathbb Q)$)?

答:
An injective continuous transcendental function $f\colon D$ open $\subseteq\mathbb{C}\to\mathbb{C}$ has algebraic and transcendental values.

1.) Examples for algebraic values at algebraic points

Each elementary function can be represented as composition of $\exp$, $\ln$ and/or finitary algebraic functions. For the elementary functions, we can therefore apply Schanuel's conjecture or the theorems it includes. see e.g.:

Bauer, N.; Slobin, H. K.: Some transcendental curves and numbers. 1913

Anai, H.; Weispfenning, V.: Deciding Linear-Trigonometric Problems. ISSAC 2000

Achatz, M.; McCallum, S.; Weispfenning, V.: Deciding Polynomial-Exponential Problems. ISSAC 2008

McCallum, S.; Weispfenning, V.: Deciding polynomial-transcendental problems. 2012

For the theorem of Lin and the theorem of Chow see point 3.).

Let $E$ denote the  exceptional set of the given function.

Examples:

$E=\{\}\colon$ $\cot(z)$, $\csc(z)$, $\text{arccot}(z)$, $e^{1+\pi z}$, $e^{e^z}$, $e^z+e^{1+\pi z}$

$E=\{0\}\colon$ $e^z$, $\sin(z)$, $\cos(z)$, $\tan(z)$, $\sec(z)$, $\arcsin(z)$, $\arctan(z)$, $\sinh(z)$, $\cosh(z)$, $\tanh(z)$, $\text{sech}(z)$, $\text{arcsinh(z)}$, $\text{arctanh(z)}$

$E=\{1\}\colon$ $\ln(z)$, $\arccos(z)$, $\text{arcsec}(z)$, $\text{arccosh}(z)$, $\text{arcsech}(z)$

$E=\mathbb{Z}\colon$ $\sin(\pi z)e^z$

$E=\mathbb{Q}\setminus\{0\}\colon$ $z^z$, $z^{\frac{1}{z}}$

$E=\mathbb{Q}\colon$ $e^{2i\pi z}$, $a^z$ ($a\in\mathbb{Q}\setminus\{0,1\}$)

$e^{P(z)}$ with $P\in\mathbb{C}[z]\colon$ $E$ is the set of zeroes of $P$.

Lindemann-Weiserstrass theorem implies: $\sin(a)$, $\cos(a)$, $\tan(a)$, $\sinh(a)$, $\cosh(a)$, $\tanh(a)$, $\arcsin(a)$ are transcendental for any non-zero algebraic $a$.

If $a$ is an algebraic number and $r$ is a rational number, $\sin(r\arcsin(a))$ is algebraic.

Trott, M.: Algebraic Values of Trigonometric Functions of Inverse Trigonometric Functions. 2011. Wolfram Demonstrations Project

But finding the exceptional set of a function can be a difficult mathematical problem because there are only a few transcendence theorems and conjectures known.

For the number of algebraic values of transcendental functions at algebraic points, we have Schneider-Lang theorem for meromorphic functions and Siegel–Shidlovsky theorem for Siegel E-functions.

Now there is an algorithm that gives the algebraic values of a given E-function and the corresponding algebraic points. Adamczewski/Rivoal 2018:

"The celebrated Siegel-Shidlovskii theorem deals with the algebraic (in)dependence of values at algebraic points of E-functions solutions of a differential system. However, somewhat paradoxically, this deep result may fail to decide whether a given E-function assumes an algebraic or a transcendental value at some given algebraic point. Building upon André’s theory of E-operators, Beukers refined in 2006 the Siegel-Shidlovskii theorem in an optimal way. In this paper, we use Beukers’ work to prove the following result: there exists an algorithm which, given a transcendental E-function f(z) as input, outputs the finite list of all exceptional algebraic points α such that f(α) is also algebraic, together with the corresponding list of values f(α). This result solves the problem of deciding whether values of E-functions at algebraic points are transcendental."

References:

Diamond, J.: The Schneider-Lang theorem for functions with essential singularities. Proc. Amer. Math. Soc. 80 (1980) (2) 223-223

Waldschmidt, M.: Algebraic values of analytic functions. J. Comput. Appl. Math. 160 (2003) (1–2) 323-333

Beukers, F.: A refined version of the Siegel-Shidlovskii theorem. Annals Math. 163 (2006) (1) 369-379

Jingjing Huang; Marques, D.; Mereb, M.: Algebraic values of transcendental functions at algebraic points. Bull. Austral. Math. Soc. 82 (2010) (2) 322-327

Herblot, M.: Algebraic points on meromorphic curves. 2012

Marques, D.; Lima, F. M. S.: Some transcendental functions with an empty exceptional set. 2012

Fischler, St.; Rivoal, T.: Arithmetic theory of E-operators. J. École polytechn. Math. 3 (2016) 31-65

Waldschmidt, M.: Irrationality and transcendence of values of special functions. 2017

Adamczewski, B.; Rivoal, T.: Exceptional values of E-functions at algebraic points. Bull. London Math. Soc. 50 (2018) (4) 697-708

2.) Examples for rational values at rational points

Reddy, R.: Irrationality of basic trigonometric and hyperbolic functions at rational values

3.) Examples for algebraic values at elementary points

If you ask for the transcendental values at algebraic/rational/integer points or for the algebraic/rational/integer values at transcendental points, you need to represent this transcendental numbers in closed form.

Wikipedia: Niven's theorem

Wikipedia: Trigonometric constants expressed in real radicals

You can also ask for representing the transcendental numbers as elementary numbers. This is the question of solving an equation of one unknown by elementary numbers. see e.g.:

[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50

[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448

Khovanskii, A.: Topological Galois Theory. Solvability and Unsolvability of Equations in Finite Terms. Springer, 2014. and other publications of this author


答:
The statement "$P$ is necessary for $Q$" means "in order to have $Q$, we must have $P$" or $Q \to P$, though we can also write the contrapositive $\neg P \to \neg Q$.

The statement "$P$ is sufficient for $Q$" means "if we have $P$, we definitely have $Q$" or $P \to Q$, though we can also write the contrapositive $\neg Q \to \neg P$.

So, the statement "$P$ is necessary but not sufficient for $Q$" can be written as $$(\neg P \to \neg Q) \land \neg (P \to Q).$$ In the example you've given, $Q = q$ while $P = \neg r \land \neg p$ due to specifics of the other question, and if we substitute those for $P$ and $Q$, we get the statement you're quoting.

If you look at the logical statement carefully, it turns out that
$(\neg P \to \neg Q) \land \neg (P \to Q)$ is only true in one case: when $P$ is true, but $Q$ is false. Why is that? Because in order to observe $P$ not being sufficient for $Q$, $P$ has to happen, and $Q$ has to still fail to happen.

This doesn't match our intuitions for what "$P$ is necessary but not sufficient for $Q$" means. We want to say something like:

There are some cases where $P$ happens, and $Q$ does not, because $P$ is not sufficient for $Q$. However, in all cases where $Q$ happens, $P$ also happens: $P$ is necessary for $Q$.

To say things like this, the language of logical statements doesn't suffice! We have to have quantifiers for talking about "some cases" and "all cases".

Let $P(x)$ and $Q(x)$ denote "in case $x$, $P$ holds" and "in case $x$, $Q$ holds". Then:

"$P$ is necessary for $Q$" means $\forall x\, Q(x) \to P(x)$.

"$P$ is sufficient for $Q$" means $\forall x\, P(x) \to Q(x)$. Its negation simplifies to $\exists x\, P(x) \land \neg Q(x)$.

The statement "$P$ is necessary but not sufficient for $Q$" has the more sophisticated interpretation
$$
    (\forall x\, Q(x) \to P(x)) \land (\exists x\, P(x) \land \neg Q(x)).
$$
That is: "In all cases $x$ where $Q(x)$ holds, $P(x)$ also holds. However, there is a case where $P(x)$ holds, but $Q(x)$ does not".

A bare statement like $P(x) \land Q(x)$ is neither true nor false, because $x$ isn't quantified. The universal statement $\forall x\, P(x) \land Q(x)$ is false (because otherwise, $P$ would be sufficient for $Q$), but it's possible that $\exists x\,P(x) \land Q(x)$ is true. This is what you want to say when you say "In my opinion, the row with $(T,T)$ should be undecidable", but that's not a thing we can talk about without quantifiers: without quantifiers, rows aren't allowed to be undecidable.

Title: Necessary but not sufficient

Body:
This question has already been asked here twice, namely here and here, but none of the answers address my specific question, except probably for this answer, which comes close.

So, using the notation of the close answer, I don't understand why I have to rule out the tuples $(T,T)$ and $(F,F)$.

Let's call the sentence "Q is necessary but not sufficient for P" R.

As for $(F,F)$, if P is false when Q is also false, this should result in R = true; since Q is necessary for P, so the absence of Q should imply the absence of P. Why would I want R to be false in this case?

And for $(T,T)$, I will imagine a more complete picture. Let's say that P depends on Q and some other factors, collectively named W. Now, we should split the row $(T,T)$ into 2, one with W false, and another with W true. In the case with W true, R should evaluate to T, and in the case with W false, R should evaluate to false. On what basis, then, should we decide to rule out $(T,T)$ in the original statement! In my opinion, the row with $(T,T)$ should be undecidable.

I would be grateful if someone could explain to my why the correct answer is $¬(¬r∧¬p)→¬q∧¬((¬r∧¬p)→q)$ in a way other that

"is necessary" translates to so and so and "is sufficient" translates
to so and so, so the conjunction of the first with the negation of the
second gives the correct answer.

Thanks


答: Alternate proof: The center of a $p$-group is nontrivial.  Then $G/Z(G)$ is cyclic $\implies G$ is abelian.

Title: Confused on proof that every group of order $p^2$, $p$ prime is isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_{p}\oplus \mathbb{Z}_{p}$

Body:
Every group of order $p^2$, $p$ prime is isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_{p}\oplus \mathbb{Z}_{p}$

I am confused about two parts of this proof.

Proof:Assume every nonidentity element of this group $G$ has order $p$. Then $\langle a\rangle$ is normal otherwise there is an element $b$ in $G$ such that $bab^{-1} \notin \langle a\rangle$

Here is my first source of confusion. if $\langle a\rangle$ is not normal I would suspect there is an element $b$ in $G$ with $ba^{i}b^{-1} \notin \langle a\rangle,\text{for some}\space i \in \mathbb{Z}$. Why does $bab^{-1} \notin \langle a\rangle$ necessarily hold with $a$?

Next part of confusion
Since $\langle a\rangle \cap \langle bab^{-1}\rangle =\{1\}$ the distinct left cosets of $\langle bab^{-1}\rangle$ are $\langle bab^{-1}\rangle,a\langle bab^{-1}\rangle,...,a^{p-1}\langle bab^{-1}\rangle$

is this because there must be $p$ distinct cosets and there union must be $G$, so this must be all of the cosets ?


问:
Proof that $\frac{d}{dt} \log |A(t)| = \text{Tr}\left[A(t)^{-1} \frac{d}{dt} A \right]$
For a symmetric $n\times n$  matrix $A(t)$ that depends on the variable $t$, I want to prove that
$$
\frac{d}{dt} \log |A(t)| = \text{Tr}\left[A(t)^{-1} \frac{d}{dt} A(t) \right].
$$

Here, $|A|$ denotes the determinant of $A$ and $\text{Tr}[X]$ denotes the trace of $X$. How can I prove this equality?

The reason I ask this question: The equality is presented in equation C.22 of Pattern Recognition and Machine Learning of C. Bishop and the proof is left "as an exercise to the reader".

My attempt so far (to not clutter the notation, I leave out the dependency on $t$):

We know that $|A|=\prod_{i=1}^n \lambda_i$, where $\lambda_i$ denotes the $i$-th eigenvalue.
Hence, we can write
$$
\frac{d}{dt} \log|A|
=\frac{d}{dt}\log\prod_{i=1}^n\lambda_i
=\frac{d}{dt}\sum_{i=1}^n\log(\lambda_i)
=\sum_{i=1}^n\frac{d}{dt}\log(\lambda_i)
=\sum_{i=1}^n\frac{1}{\lambda_i} \frac{d}{dt}\lambda_i.
$$

The textbook further suggests to make use of the equalities
$$
A = \sum_{i=1}^n \lambda_i u_i u_i^T
\quad \text{and} \quad
A^{-1} = \sum_{i=1}^n \frac{1}{\lambda_i} u_i u_i^T,
$$
where $u_i$ denotes the $i$-th eigenvector.

I do not know how to proceed from here. I also do not know if this is going into the right direction.

答:
Nicholas Higham published a wonderful paper on the Unwinding Function
$$\eqalign{
{\cal U}(z) &= \frac{z-\log(e^z)}{2\pi i}
  = \left\lceil\frac{Im(z)-\pi}{2\pi}\right\rceil \;\in\;{\mathbb Z} \\
}$$
Using this function one can write the following identity
$$\eqalign{
\log\det A
 &= {\rm tr}\log A - 2\pi i\cdot {\cal U}({\rm tr}\log A) \\
 &= {\rm tr}\log A - 2n\pi i \\
}$$
One of the many interesting properties of the unwinding function is that its derivative is zero, which leads directly to your formula, i.e.
$$\eqalign{
\frac{d(\log\det A)}{dt}
 &= \frac{d({\rm tr}\log A)}{dt} \\
 &= \operatorname{tr}\left(A^{-1}\frac{dA}{dt}\right) \\
}$$
The final equality follows from the identity
$$\eqalign{
\frac{d({\rm tr}(f(A)))}{dt}
 &= \operatorname{tr}\left(f'(A)\cdot\frac{dA}{dt}\right) \\
}$$
where $f$ is any analytic function and $f'$ is its derivative.


问:
Integration of a Bessel function times a sine function times a polynomial
I have recently faced the following expression (while trying to compute the Fourier transform of a RKKY-like potential):

\begin{align}
\int_0^{\infty} \text{d} r\;  J_0 (k r) \frac{\sin \left(\alpha \sqrt{1+r^2} \right) }{(1+r^2)^2} \; r = -\frac{1}{\alpha} \int_0^{\infty} \text{d} r\;  J_0 (k r) \frac{1}{(1+r^2)^{3/2}} \frac{\text{d}}{\text d r} \left[\cos \left( \alpha\sqrt{1+r^2} \right) \right]
\end{align}

Although I can solve it numerically, I would like to find a closed-form expression, if there is one.

I found some related questions here, such as,

\begin{align}
\int_0^{\infty} \text{d}r\; J_0 \left( \alpha\sqrt{x^2 +z^2} \right) \, \cos \left(\beta x \right)  = \frac{\cos \left(z\sqrt{\alpha^2 -\beta^2} \right)}{\sqrt{\alpha^2 -\beta^2}}
\end{align}
but did not really helped me.

答:
According to this reference, Table 8.2 , entry (7):

$$\int_0^\infty \dfrac{1}{\left(1+r^2\right)^\frac{3}{2}}J_0(kr)r \; dr = e^{-k}$$

and according to Table 8.2, entry (41):

$$\int_0^\infty \dfrac{\sin\left[\alpha\left(1+r^2\right)^\frac{1}{2}\right]}{\left(1+r^2\right)^\frac{1}{2}}J_0(kr)r \; dr = \dfrac{\cos\left[\left(\alpha^2-k^2\right)^\frac{1}{2}\right]}{\left(\alpha^2-k^2\right)^\frac{1}{2}}\Pi\left(\dfrac{k}{\alpha}-\dfrac{1}{2}\right)$$

where

$$\Pi(t) = \begin{cases} 1 & -\frac{1}{2} < t< \frac{1}{2}\\ 0 & \mathrm{otherwise} \end{cases}$$

Using the convolution property of the Hankel Transform, if the convolution exists

$$\begin{align*} \int_0^\infty \dfrac{1}{\left(1+r^2\right)^\frac{3}{2}}\dfrac{\sin\left[\alpha\left(1+r^2\right)^\frac{1}{2}\right]}{\left(1+r^2\right)^\frac{1}{2}}J_0(kr)r \; dr &= \dfrac{1}{2\pi}e^{-k}**\dfrac{\cos\left[\left(\alpha^2-k^2\right)^\frac{1}{2}\right]}{\left(\alpha^2-k^2\right)^\frac{1}{2}}\Pi\left(\dfrac{k}{\alpha}-\dfrac{1}{2}\right)\\
\\
&= \dfrac{1}{2\pi}\int_0^{2\pi}\int_0^{k\cos\theta+\sqrt{k^2\cos^2\theta-(k^2-\alpha^2)}} e^{-k'}\dfrac{\cos\left[\left(\alpha^2-\left[k^2+k'^2-2kk'\cos\theta\right]\right)^\frac{1}{2}\right]}{\left(\alpha^2-\left[k^2+k'^2-2kk'\cos\theta\right]\right)^\frac{1}{2}}k' \; dk' \; d\theta\\
\\
&\mathrm{or}\\
\\
&= \dfrac{1}{2\pi}\int_0^{2\pi}\int_0^{\alpha^-} e^{-\sqrt{k^2+k'^2-2kk'\cos\theta}}\dfrac{\cos\left[\left(\alpha^2-k'^2\right)^\frac{1}{2}\right]}{\left(\alpha^2-k'^2\right)^\frac{1}{2}}k' \; dk' \; d\theta\\
\end{align*}$$

Neither of which look possible to easily evaluate symbolically.  Although maybe one of them takes less effort to evaluate numerically than your original integral.


问:
Getting lower/upper bounds for $a$ and $b$, when solving an equation in positive integers
I have the following equation:

$$\frac{a(a + 3)(a(r - 5) + (12 - r))}{9}=\frac{b (9 + b (-14 + r) - r)}{3}\tag1$$

I want to solve that equation in positive integers for $a$, $b$, and $r$.

Is there a way to find a lower/upper bound for one ($a$) or two ($a$ and $b$) variables when I set $r$ to be kown? So that I can use software to look between or in these bounds?

Body:
Some solutions $(r,a,b)$ with fixed $r$ (equivalent elliptic equation):

(15, 117, 2340)
(16, 47, 450)
(19, 5, 14)
(24, 3, 6)
(30, 1545, 43860)
(31, 5, 10)
(45, 12, 30)
(46, 9, 20)
(129, 48, 204)
(139, 32, 112)
(177, 9, 18)
(186, 3792, 138336)
(220, 35, 126)
(244, 5, 8)
(553, 45, 180)
(573, 33, 114)
(576, 243, 2214)

Some solutions $(a,b,r)$ with fixed $a$ (equivalent Pell equation):

(5, 8, 244)
(5, 10, 31)
(5, 14, 19)
(9, 18, 177)
(9, 20, 46)
(12, 30, 45)
(32, 112, 139)
(33, 114, 573)
(35, 126, 220)
(45, 180, 553)
(47, 450, 16)
(48, 204, 129)
(63, 294, 3750)
(77, 396, 3889)
(116, 728, 46750)
(117, 2340, 15)
(159, 1166, 6826)
(240, 2156, 2098129)
(243, 2214, 576)
(357, 3906, 72807)
(372, 4154, 2509849)

I think no simple way to find bounds for $a,b$.


问:
Proving $2\cos\left(2\pi ft + \phi\right)\cos(2\pi ft) = \cos(4\pi ft + \phi) + \cos(\phi)$
I need to prove this identity:

$$2\cos\left(2\pi ft + \phi\right)\cos(2\pi ft) = \cos(4\pi ft + \phi) + \cos(\phi)$$

I know I have to use some identity or property but I can't find any to do it.

答:
We use the following trigonometric product formula:

$$2\cos x\cos y=\cos\left(x-y\right)+\cos\left(x+y\right).$$

We take $x=2\pi ft+\phi$ and $y=2\pi ft$. Then

$$2\cos\left(2\pi ft+\phi\right)\cos\left(2\pi ft\right) = \cos\left(2\pi ft+\phi-2\pi ft\right)+\cos\left(2\pi ft+\phi+2\pi ft\right) = \cos\left(\phi\right)+\cos\left(4\pi ft+\phi\right).$$


问:
Solving second order non-linear non-homogenous multi-variable differential equation
I have got a really weird differential that I have already used half of the whole notebook as scratch but was not able to get the solution. It looks like this
$$\frac{d^2y}{dt^2}=-\frac{a}{x}\frac{dx}{dt}-\frac{b}{y^2}$$
I tried attempting separation of variables, so, I get
$$\frac{d^2y}{dt^2}+\frac{b}{y^2}=C\tag{1}$$
$$-\frac{a}{x}\frac{dx}{dt}=C\tag{2}$$
But I am starting to get doubtful about this, I could not remember applying separation of variables on such cases, but only on F(x,y,t). What is the proper way of solving this differential equation?

Note: Both $y$ and $x$ are functions of t. The differential equation is in its unaltered original form, it does not have a couple. So the relationship of x to y and vice versa is dictated by the differential equation.

答:
$$\frac{d^2y}{dt^2}=-\frac{a}{x}\frac{dx}{dt}-\frac{b}{y^2}$$
Chose arbitrary a function $y(t)$

$$\frac{1}{x}\frac{dx}{dt}=-\frac{1}{a}\frac{d^2y}{dt^2}-\frac{b}{a\:y^2}$$
$$\ln|x(t)|=-\frac{1}{a}\int(\frac{d^2y}{dt^2}+\frac{b}{y^2})dt=-\frac{1}{a}\frac{dy}{dt}-\frac{b}{a}\int \frac{1}{y^2}dt$$
$$\boxed{x(t)=C\:e^{-\frac{1}{a}\frac{dy}{dt}-\frac{b}{a}\int \frac{1}{y^2}dt}}$$
Thus your problem has infinity many solutions : For each arbitrary function $y(t)$ you get the related $x(t)$.

For example :

With $y(t)=\frac{1}{sin(t)}\quad;\quad
\frac{dy}{dt}=-\frac{\cos(t)}{\sin^2 (t)}\quad;\quad
x(t)=C\:e^{\frac{1}{a}\frac{\cos(t)}{\sin^2 (t)}-\frac{b}{a}\left(\frac12 t-\frac14\cos(2t) \right)}$


问:
A geometry problem about colinearity

$\textbf{Problem:}$Let $ABC$ be a triangle with circumcircle $\omega$ . Point $D$ lies on the arc $BC$ not containing $A$ of $\omega$ and is different than $B,C$ and the midpoint of arc $BC$. Tangent of $\omega$ on $D$ intersects lines $BC$,$CA$,$AB$ at $A'$,$B'$,$C'$, respectively. Lines $BB'$ and $CC'$ intersect at $E$. Line $AA'$ intersects again the circle $\omega$ at $F$. Prove that points $D,E,F$ are collinear.

I tried to use menelaus theorem on bunch of triangles and in few ways I restated the problem to apply it.But all those attempts failed.I also tried to chase cross ratios but that didn't work out either.

Any help or solution will be appreciated.

Thanks @oldboy for the diagram.

答:
Projective solution (or in modern contest languague, the method of moving points).

Let us fix circle and points $B,C,D$ on it as tangent at $D$ and
let us move $A$ on the circle. Then also $E,F$ and $B',C'$ moves,
but not $A'$. Then composition of projective maps $B'\mapsto A$
and $A\mapsto C'$ is also projective and this map induces projective map of
pencil from $(B)$ to $(C)$: $BB'\mapsto CC'$.

This means that intersection of $BB'$ and $CC'$, that is point $E$, describes
some conic (which goes through points $B$, $C$ and $D$). Now let line $DE$ meet circle at $F'$. Since conic and circle meet at
$D$ we see that map $E\mapsto F'$ is well defined and it is
projective from conic to circle. This also means that composition
of projective maps $A\mapsto B'$, $B'\mapsto E$ and $E\mapsto F'$
i.e. $A\mapsto F'$ is projective map on circle it self.

We want to prove that this is actually an involution of circle
$A\mapsto F$ with center at $A'$. By fundamental theorem of
projective geometry we have to find 3 particular situation for $A$
when $F=F'$ which means that $F=F'$ is always true. But this is
obviously true when $A\in\{B,C,D\}$ and we are done.

Any way here is an Euclidean geometry solution: https://artofproblemsolving.com/community/c6h2205298p16643760

Title: A geometry problem about colinearity