问:
How do I find $\operatorname{det} T_Q$?

Let $S$ be the space of all $n \times n$ real skew symmetric matrices and let $Q$ be a real orthogonal matrix. Consider the map $T_Q: S \to S$ defined by $$T_Q(X) = QXQ^T.$$ Find $\operatorname{det} T_Q$.

I thought about diagonalizing $Q$, but I don't think we know it is real diagonalizable. I can show it is an isometry using the Hilbert-Schmidt inner product, but I can't really relate it to the determinant of $Q$ (I've seen posts here that say the determinant should be $\operatorname{det}Q^{n-1}$). So all I know is that $\operatorname{det} T_Q = \pm1$. How would I find $\det T_Q$?

答:
revised proof:

I revised this into $2$ distinct proofs of the result, one analytic and one algebraic.  In both cases, the key insight comes from examining a very simple reflection matrix

$D:= \displaystyle \left[\begin{matrix}-1 & \mathbf 0 \\ \mathbf 0 & I_{n-1}\end{matrix}\right]$

and computing

$T_D\mathbf B = \mathbf B \displaystyle \left[\begin{matrix}-I_{n-1} & \mathbf 0 \\ \mathbf 0 & I_{\binom{n}{2}-(n-1)}\end{matrix}\right]=\mathbf BA$

where $\mathbf B$ is a collection of well-chosen (skew symmetric matrix) basis vectors.  This computation is shown at the very end under "Computing $T_D\mathbf B$"

The conclusion ultimately is that $\det\big(T_Q\big) =\operatorname{det}\big(Q\big)^{n-1}$

1.) analytic proof:

the determinant is a (real) continuous function and integer valued (taking on values in $\big\{-1,+1\big\}$ as identified by OP), hence constant on any connected component. (I.e. if maps in this group are path-connected, then they have the same determinant.)

Case 1:

$\det\big(Q\big)=1$

Then $Q$ is path connected to the identity, so $T_Q$  is path-connected to $T_I$ (where $T_IX = IXI=X$). Thus, $\det\big(T_Q\big) =\det\big(T_I\big) =1$.

Case 2:

$\det\big(Q\big)=-1$

$Q$ is path connected to $D$, so $T_Q$ is path connected to $T_D$ and

$\det\big(T_Q\big)=\det\big(T_D\big)=(-1)^{n-1}$

2.) algebraic proof:

observe that for $Q_1, Q_2 \in O_n\big(\mathbb R\big)$

$T_{(Q_1Q_2)}X = Q_1Q_2XQ_2^T Q_1^T=T_{Q_1}T_{Q_2}X$

and in the case of $Q_1=Q_2^T$  then $T_{Q_1}=T_{Q_2}^{-1}$

(ignoring the trivial $n=1$ case)
for arbitrary $Q \in O_n\big(\mathbb R\big)$, first decompose $Q$ into $r$ Householder matrices $H_j$, for some $1\leq r\leq n$, where we know $r$ is even if $\det\big(Q\big)=1$ and odd if $\det\big(Q\big)=-1$.  In the below $U_k$ is some orthogonal matrix and $W_k$ is an orthogonal matrix as well, each of the appropriate dimension

$T_Q\mathbf B$

$= T_{(H_1H_2\cdots H_r)}\mathbf B$

$= T_{H_1}T_{H_2}\cdots T_{H_r}\mathbf B$

$= T_{U_1DU_1^T}T_{U_2DU_2^T}\cdots T_{U_rDU_r^T}\mathbf B$

$= \big(T_{U_1}T_DT_{U_1}^{-1}\big)\big(T_{U_2}T_{D}T_{U_2}^{-1}\big)\cdots \big(T_{U_r}T_DT_{U_r}^{-1}\big)\mathbf B$

$= \mathbf B \big(W_1 A W_1^{-1}\big)\big(W_2 A W_2^{-1}\big) \cdots \big(W_r A W_r^{-1}\big)$

$\implies \det\big(T_Q\big) = \det\big(W_1 A W_1^{-1}\big)\det\big(W_2 A W_2^{-1}\big) \cdots \det\big(W_r A W_r^{-1}\big)=\det\big(A\big)^r = \Big((-1)^{n-1}\Big)^r$

which is to say $\det\big(T_Q\big)=1$ if $n$ is odd and/or $\det\big(Q\big) = 1$, and

$\det\big(T_Q\big)=-1$ in the case of even $n$ and $\det\big(Q\big) =-1$.

The technique here is to examine some group homomorphism (determinant) by decomposing the group into its generators (Householder matrices) and then examine how the generators look under the homomorphism.

Computing $T_D\mathbf B$:

$D:= \displaystyle \left[\begin{matrix}-1 & \mathbf 0 \\ \mathbf 0 & I_{n-1}\end{matrix}\right]$

now construct a simple basis for your space of real skew symmetric matrices (where $\mathbf e_k$ is the kth standard basis vector in $\mathbb R^n$).

$v_1 := \mathbf e_1\mathbf e_2^T-\big(\mathbf e_1\mathbf e_2^T\big)^T$

$v_2 := \mathbf e_1\mathbf e_3^T-\big(\mathbf e_1\mathbf e_3^T\big)^T$

$v_3 := \mathbf e_1\mathbf e_4^T-\big(\mathbf e_1\mathbf e_4^T\big)^T$

$\vdots$

$v_\binom{n}{2} := \mathbf e_{n-1}\mathbf e_{n}^T-\big(\mathbf e_{n-1}\mathbf e_{n}^T\big)^T$

(the pattern is hopefully clear -- each matrix is all zeros, except for a single 1 and a single negative one)

collect these in

$\mathbf B:= \bigg[\begin{array}{c|c|c|c|c} v_1 & \cdots &v_{n-1}& v_n &\cdots & v_\binom{n}{2}\end{array}\bigg]$

(Artin would refer to this as a 'hyper-vector' though not many other texts use this term)

Now by inspection $T_D$ leaves all vectors (skew matrices) unchanged if they have all zeros in the first column/row.  Conversely $T_D$ negates a vector (skew matrix) that only has non-zero components in its first column/row (where we recall that since the vector space is of real skew matrices, the diagonals are of course zero). Put differently

for $k\in\big\{1,2,...,n-1\big\}$

$T_Dv_k = -v_k$

(i.e. all skew-symmetric matrix basis vectors that have a one in the first row)

and for $k\in\big\{n+1,n+2,...,\binom{n}{2}\big\}$

$T_Dv_k = v_k$

to conclude

$T_D\mathbf B = \mathbf B \displaystyle \left[\begin{matrix}-I_{n-1} & \mathbf 0 \\ \mathbf 0 & I_{\binom{n}{2}-(n-1)}\end{matrix}\right]= \mathbf BA$

$\implies\det\big(T_D\big)=  \det\big(A\big)= (-1)^{n-1}\cdot 1= (-1)^{n-1}$

Title: How do I find $\operatorname{det} T_Q$?


问:
Complex integration with fractional powers
How would one compute
$$
\int_{\gamma}\frac{z^{n + \frac{1}{2}}}{(z^{\frac{1}{2}} + 1)^2}\ dz,
$$
for an integer $n$ and where $\gamma$ denotes the closed unit circle contour, that is using complex analysis. Can residue theory be applied to this?

History

I am interested in computing
$$
\int_{-\pi}^{\pi}\frac{e^{inx}}{1+\cos(x/2)}\ dx, \quad n \in \mathbb{Z}
$$
using complex methods. Setting $z = e^{ix}$, we then obtain that $z^{1/2} + z^{-1/2} = 2\cos(x/2)$, which after factoring yields
$$
\int_{-\pi}^{\pi}\frac{e^{inx}}{1+\cos(x/2)}\ dx = \int_{\gamma}\frac{z^{n + \frac{1}{2}}}{(z^{\frac{1}{2}} + 1)^2}\ dz,
$$
where $\gamma : [-\pi,\pi] \to \mathbb{C}$ such that $\gamma(t) = e^{it}.$

答:
Substitute $z=w^2$ so that your integral becomes

$$\int_{\Gamma}\frac{w^{2n + 1}2w}{(w + 1)^2}\ dw,$$

where $\Gamma$ is one-half of the unit circle.

The choice of $\Gamma$ depends where you choose your square roots.

(If $\Gamma$ goes from $0$ to $\pi,$ then you can return along the X-axis.)

ContentLicense: CC BY-SA 4.0


问:
Music and Maths - Modes of Limited Transposition
In music, 'Modes of limited transposition' are modes that have a limited availability of transpositions. Unlike a major scale that has $12$ possible unique transpositions, the seven modes of limited transposition have fewer than $12$ possibilities of transposition.

French Composer Olivier Messiaen coined this idea, and he wrote, "Their series is closed, it is mathematically impossible to find others, at least in our tempered system of $12$ semitones." I want to try and figure out why. Might anyone have any idea how I could use mathematics to prove this statement?

Edit: See Link in comments for ideas surrounding Group Theory

答:
The reference you provided,
arxiv.org/pdf/1407.5757.pdf,
includes some uses of arithmetic modulo $12$ that
suggest a handy way of identifying a mode of limited transposition.

First, identify the twelve pitch classes of conventional equal temperament with the numbers in $\mathbb Z_{12}$.
A scale is an ordered list of numbers modulo $12$ obtained from the list
$(t, t+1, t+2, \ldots, t+11)$ for some $t \in \mathbb Z_{12}$
by deleting zero or more elements (other than $t$) from that list of $12$ numbers.
A transposition of a scale is the image of the scale under a function
$x \mapsto x+k$ for some constant $k \in \mathbb Z_{12}$;
a mode of a scale is a cyclic permutation (which may be the identity permutation) of the elements of the list.

For example, if we identify the pitch class C with $0 \in \mathbb Z_{12}$,
the C major scale is $(0,2,4,5,7,9,11)$; the A major scale is a transposition of the C major scale, $(9,11,1,2,4,6,8)$; and the natural A minor scale (aka A Aeolian) is a mode of the C major scale, $(9,11,0,2,4,5,7)$.

The subset of elements of $\mathbb Z_{12}$ of a scale is the pitch collection of that scale; all modes of that scale have the same pitch collection.

A mode of limited transposition is a set of scales consisting of one scale and all modes of all its transpositions, such that the set of pitch collections belonging to this entire set of scales has fewer than $12$ members.

(The word "mode" is used in two very different ways here. We call A Aeolian a "mode" of C major because you can play A Aeolian simply by starting at the sixth note of C major and playing the same cycle of pitch classes. In contrast, Messaien's Mode $1$ and Mode $5$ are distinct "modes" because even though they have the same number of notes, there is no way to make Mode $5$ sound like Mode $1$ simply by starting on a different note in the scale. In the first sense, "mode" means "can use the same notes," whereas in the second sense, "mode" means "cannot use the same notes." In order to distinguish the two meanings, I will henceforth use the second sense of "mode" only in the complete phrase "mode of limited transposition.")

These definitions are more mathematical than musical.
They allow the following to be scales:

\begin{align}
& (0), \\
& (0, 6), \\
& (0, 4, 8), \\
& (0, 3, 6, 9).
\end{align}

The first "scale" isn't much of a scale at all; you play it on a piano by striking the C key in each successive octave. The second "scale" goes up by augmented fourths -- C, F$\sharp$, C, F$\sharp$, C -- also not what you would usually consider a musical scale.
The next two scales are the notes of an augmented triad and a diminished seventh chord, respectively, and would sound more like arpeggios than like scales.
But it would take some tweaking of the definitions to eliminate these as scales.

According to the definitions, the last three of these scales belong to modes of limited transposition. Messiaen does not count them, perhaps because (as already noted) they do not appear to be scales in a musical sense.

But if we accept all of these scales as scales (and the last three as modes of limited transposition) and are willing to continue to accept other musically strange scales, it is not hard to show the following:

A scale belongs to a mode of limited transposition if and only if it has the same pitch collection as its image under $x \mapsto x+k$ where $k$ is congruent to a proper divisor of $12$ (that is, where
$k \in \{1,2,3,4,6\} \subset \mathbb Z_{12}$).

To give a notion of how this might be proved, consider that in order for the transpositions of a scale to have fewer than $12$ pitch collections, at least two of those transpositions must have the same pitch collection.
We can transpose those two scales to the original scale and the original scale transposed by $x \mapsto x+n$ for some $n\in\mathbb Z\setminus\{0\}$;
then all the transpositions produced by composing $x \mapsto x+n$ one or more times have the same pitch collection as the original scale, and one of these transpositions is $x\mapsto x+k$ for $k$ congruent to a proper divisor of $12$.

So we can identify a mode of limited transposition (abbreviated MOLT in the text below) by a function $x\mapsto x+k$
for $k$ congruent to a proper divisor of $12$ and a subsequence of the sequence
$(0, 1, 2, \ldots, k-1)$;
the remainder of the scale is repeatedly applying $x\mapsto x+k$ to this sequence until the next application would produce the original sequence,
and concatenating the resulting sequences onto the original subsequence.
Moreover, without loss of generality we can require $0$ to be in the subsequence, because a MOLT will always contain a scale starting at $0$.

We merely need to watch for duplicates, because in some cases a MOLT
can be generated in more than one way according to the preceding paragraph.
In order to help avoid duplication, I will use the convention of always selecting the mode of the representative scale that starts with the longest run of semitones.
Note that this convention means I always delete $k-1$ from the subsequence.

Starting with $k = 1$, the following MOLT has only one unique pitch collection:

$(0),\ x\mapsto x+1$ produces the scale $(0,1,2,3,4,5,6,7,8,9,10,11)$ (the chromatic scale).

For $k = 2$, the following MOLT has two pitch collections:

$(0),\ x\mapsto x+2$ produces the scale $(0,2,4,6,8,10)$. This is Messiaen's Mode $1.$

We do not count $(0,1),\ x\mapsto x+2$ since it produces the same MOLT as
$(0),\ x\mapsto x+1$.

For $k = 3,$ each MOLT has three pitch collections:

$(0),\ x\mapsto x+3$. This is the diminished seventh chord already described.

$(0,1),\ x\mapsto x+3$. This is Messiaen's Mode $2.$

For $k = 4,$ each MOLT has four pitch collections:

$(0),\ x\mapsto x+4$. This is the augmented triad already described.

$(0,1),\ x\mapsto x+4$.

$(0,1,2),\ x\mapsto x+4$. This is a mode of Messiaen's Mode $3$ (that is, $(0,2,3),\ x\mapsto x+4$).

Note that $(0,2),\ x\mapsto x+4$ duplicates $(0),\ x\mapsto x+2.$

For $k = 6,$ each MOLT has six pitch collections. For scales with no semitones:

$(0),\ x\mapsto x+6$; augmented fourths.

$(0,2),\ x\mapsto x+6$ (a mode of $(0,4),\ x\mapsto x+6$).

Note that $(0,3),\ x\mapsto x+6$ is the same scale as $(0),\ x\mapsto x+3$.
For runs of one semitone we have:

$(0,1),\ x\mapsto x+6$.

$(0,1,3),\ x\mapsto x+6$.

$(0,1,4),\ x\mapsto x+6$.

Note that $(0,1,3,4),\ x\mapsto x+6$ is the same scale as $(0,1),\ x\mapsto x+3.$
For runs of two semitones we have

$(0,1,2),\ x\mapsto x+6$. This is a mode of Messiaen's Mode $5$ (that is, $(0,1,5),\ x\mapsto x+6$).

$(0,1,2,4),\ x\mapsto x+6$. This is a mode of Messiaen's Mode $6$ (that is, $(0,2,4,5),\ x\mapsto x+6$).

For runs of three or more semitones:

$(0,1,2,3),\ x\mapsto x+6$. This is a mode of Messiaen's Mode $4$ (that is, $(0,1,2,5),\ x\mapsto x+6$).

$(0,1,2,3,4),\ x\mapsto x+6$. This is Messiaen's Mode $7.$

And that's all there are, $16$ modes of limited transposition in total.
There may be additional properties a MOLT should have that the other nine do not; I just don't know specifically what they are.
And it may be, as suggested in your source, that Messaien was simply wrong about the mathematical impossibility of finding a MOLT he did not count, even if his additional criteria are applied.

Note that the classification of modes of limited transposition above does not distinguish $(0,1,3,4,6,7,9,10)$ from $(0,2,3,5,6,8,9,11).$
I consider both of these to be Messaien's Mode $2,$ because if you simply transpose $(0,1,3,4,6,7,9,10)$ up by a whole step ($x\mapsto x+2$)
and start playing at pitch $0$ of the new collection of pitches,
you play $(0,2,3,5,6,8,9,11).$
The reason I think these are not distinct modes of limited transposition are because a MOLT is supposed not to have a tonic (so it should not matter which note of the scale is played first) and because Messaien never identifies two distinct "modes of limited transposition" that can be mapped one to the other in this way.

But if we distinguish a different mode within a MOLT for each transposition that gives a different set of pitch classes, with the restriction that one of those pitch classes must be $0$, then we get the following additional modes:

From $(0,1),\ x\mapsto x+3$, the additional mode $(0,2),\ x\mapsto x+3$.

From $(0,1),\ x\mapsto x+4$, the additional mode $(0,3),\ x\mapsto x+4$.

From $(0,1,2),\ x\mapsto x+4$, the additional modes $(0,1,3),\ x\mapsto x+4$
and $(0,2,3),\ x\mapsto x+4$.

From $(0,2),\ x\mapsto x+6$, the additional mode $(0,4),\ x\mapsto x+6$.

From $(0,1),\ x\mapsto x+6$, the additional mode $(0,5),\ x\mapsto x+6$.

From $(0,1,3),\ x\mapsto x+6$, the additional modes $(0,2,5),\ x\mapsto x+6$
and $(0,3,4),\ x\mapsto x+6$).

From $(0,1,4),\ x\mapsto x+6$, the additional modes $(0,3,5),\ x\mapsto x+6$
and $(0,2,3),\ x\mapsto x+6$.

From $(0,1,2),\ x\mapsto x+6$, the additional modes $(0,1,5),\ x\mapsto x+6$
and $(0,4,5),\ x\mapsto x+6$.

From $(0,1,2,4),\ x\mapsto x+6$,
the additional modes $(0,1,3,5),\ x\mapsto x+6$
and $(0,2,4,5),\ x\mapsto x+6$
and $(0,2,3,4),\ x\mapsto x+6$.

From $(0,1,2,3),\ x\mapsto x+6$,
the additional modes $(0,1,2,5),\ x\mapsto x+6$
and $(0,1,4,5),\ x\mapsto x+6$
and $(0,3,4,5),\ x\mapsto x+6$.

From $(0,1,2,3,4),\ x\mapsto x+6$,
the additional modes $(0,1,2,3,5),\ x\mapsto x+6$
and $(0,1,2,4,5),\ x\mapsto x+6$
and $(0,1,3,4,5),\ x\mapsto x+6$
and $(0,2,3,4,5),\ x\mapsto x+6$.

I count $22$ additional modes this way.
That is, there are a total of $38$ distinct modes (in the sense of "starting on a different degree of the scale") among the modes of limited transposition.


问:
Integral of $e^{-at}\operatorname{erf}(t)$ and $e^{-at} \operatorname{erf}^2(t)$
Can anyone help me with any suggestions for the following triple integral?

$$
\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^t dt_1 e^{-(\gamma_1+\gamma_2)t_1} \int_{-\infty}^{t_1} dt_2 e^{\frac{\gamma_1+\gamma_2}{2}t_2} e^{-\frac{t_2^2}{4\sigma^2}} \int_{-\infty}^{t_2} dt_3 e^{\frac{\gamma_1+\gamma_2}{2}t_3} e^{-\frac{t_3^2}{4\sigma^2}}
$$
It can be put into the equivalent form,
$$
\frac{1}{2\sqrt{2\pi \sigma^2}}\int_{-\infty}^t dt_1 \left[ \left[ e^{-\frac{\gamma_1+\gamma_2}{2}t_1} \Theta(t_1) \right] * e^{-\frac{t_1^2}{4\sigma^2}} \right]^2
$$
where $\Theta(t_1)$ is the Heaviside unit step function and * denotes convolution.

This second integrand can be written in terms of he error function, $\operatorname{erf}()$,

$$
e^{\frac{(\gamma_1+\gamma_2)^2 \sigma^2}{2}}  \frac{1}{2\sqrt{2\pi \sigma^2}}\int_{-\infty}^t dt_1 e^{-(\gamma_1+\gamma_2)t_1} \left[ \frac{1}{2} + \frac{1}{2}\operatorname{erf}\left(\frac{t_1-(\gamma_1+\gamma_2) \sigma_1^2}{2\sigma} \right)  \right]^2
$$
The definition of $\operatorname{erf}()$ being
$$
\operatorname{erf}\left(\frac{t-t_0}{\sigma}\right) = \int_0^{t-t_0} dt_1 \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{t_1^2}{2\sigma^2}}
$$

Mathematica fails with the computation, even numerically, in the last part.

答:
As asked in title
$$I=\int e^{-a t}\, \text{erf}(t)\,dt$$ is simple (just one integration by parts and completing the square)
$$I=\frac{e^{\frac{a^2}{4}}\,
   \text{erf}\left(t+\frac{a}{2}\right)-e^{-a t}\, \text{erf}(t)}{a}$$

But the second one
$$J=\int e^{-a t}\, \big[\text{erf}(t)\big]^2\,dt$$ seems to be much more problematic.


问:
General question about finding the minimal polynomial
Suppose you're given a root $\alpha$ and asked to find the minimal polynomial of $\alpha$ over $F$, where $F$ is some field. The usual strategy I've seen for solving these problems is to set $x=\alpha$ and then manipulate this expression to get a polynomial $f(x)$ for which $\alpha$ is a zero. Then, it suffices to show that $f(x)$ is irreducible.

My question is whether there are cases where you will follow this procedure and get a polynomial that is not irreducible. Are there any obvious examples?

答:
It might depend on how you "manipulate this expression".
For example, if $F = \mathbb Q$ and $\alpha = \exp(2 \pi i/n)$, you might well come up with the polynomial $x^n - 1$.  But this is not irreducible: the polynomial you want is the $n$'th cyclotomic polynomial.


问:
A closed form expression for the finite-difference Euler's method
Consider the ODE

$$f(u) \equiv u' = au+b, \>\>\> u(t_0 = 0)\equiv u^0 = u_0$$

For a timestep $k$, the forward-difference approximator for $u'$ gives the Euler update as

$$
u^{n+1} = u^n + kf(u^n) = u^n + k(au^n + b)
$$

I am trying to come up with a closed form expression for $u^n$, that is, one that involves only the constants $a, b, u_0, k$.

I've tried to do this by first writing $u^{n}$ as a cumulative sum

$$
u^n = u^n + k(au^n + b) = u_0 + \sum_{i=0}^{n-1}k(au^i+b)
$$

I then tried to expand the sum on the right and see if there was any easily identifiable pattern to write more generally, or if it would perhaps resemble an expansion of some other known function.After some ugly algebra, this is something like

$$
\begin{align}
u^n &= u_0 + k(au_0 + b) + k(au_1 + b) + k(au_2 + b) + ...\\
&= u_0 + k(au_0 + b) + (k^2a + k)(au_0 + b) + (k^3a^2 + 2k^2a + k)(au_0 + b) + ...\\
&= u_0 + (k^3a^2 + 3k^2a + 3k + ...)(au_0+b)
\end{align}
$$

Nothing jumps out to me, and I'm not sure this approach is promising. Any ideas?

I've massaged it some more and found that it can be written as

$$u^n = (1+ka)^nu_0 + C_n$$

Where C is a function of $a,b,k,n$, which I've yet to deduce...

答:
From the same principles that you would solve a linear DE with, you get for the linear recursion a complementary part and a particular solution that can be found via undetermined coefficients
$$
u^n=c(1+ka)^n-\frac{b}{a}.
$$
To determine the constant of the homogeneous solution, insert into the initial condition
$$
u^0=c-\frac{b}{ka}\implies u^n=(1+ka)^nu^0+\frac{(1+ka)^n-1}{a}b
$$

(edit/verification) Transformation of the equation to a geometric progression
$$
u^{n+1}=(1+ka)u^n+kb=(1+ka)u^n+\frac{b}{a}((1+ka)-1)
\\~\\
\left(u^{n+1}+\frac{b}{a}\right)=(1+ka)\left(u^n+\frac{b}{a}\right)
\\~\\
u^n=(1+ka)^n\left(u^0+\frac{b}{a}\right)-\frac{b}{a}
=(1+ka)^nu^0+\frac{b}{a}\left((1+ka)^n-1\right)
$$


问:
Find the last two digit of $5^{121}*3^{312}$
The answers key says by using
$$5^k\equiv 25 \pmod{100}, k>=2$$

and $$3^{40}\equiv 1 \pmod{100}$$
can have $$ 5^2 *3^4 \equiv25 \pmod{100},$$
and it follows that the last two digits of  $5^{143}*3^{312}$ are $25$.

Don't know why  $ 5^2 *3^4 \equiv25 \pmod{100}$ can applies that the last two digits of $5^{143}*3^{312}$ are $25$.

Also, the question is asking about $5^{121}$, why the last step relates to $5^{143}$

答:
I have no idea what the book is trying to do.

I guess Id do it this way.

$5^2\cdot 3^4 = 25 *81 = 2025 \equiv 25\equiv 5^2 \pmod {100}$.

So $(5^2 \cdot 3^4)^{78} \equiv 25^{78} \pmod {100}$.

SO $5^{156}\cdot 3^{312} \equiv 5^{156} \pmod {100}$.

Now $5^2 \equiv 25 \pmod {100}$ so by induction if $5^k \equiv 25 \pmod {100}$ then $5^{k+1} \equiv 5\cdot 25 =125\equiv 25 \pmod {100}$.

So for any $k, m \ge 2$ you have $5^k \equiv 5^m \equiv 25 \pmod {100}$.

So $5^{156} \equiv 5^{121} \pmod {100}$ and

$5^{121}\cdot 3^{312} \equiv 5^{156}\cdot 3^{312} \equiv 5^{156}\equiv 25 \pmod{100}$.

So the last two digits are $25$.

......

Actually, that's me trying to guess what the book is trying to do.

I'd do:

So $5^{k} \equiv 25$ for all $k \ge 2$.

Then for all $k \ge 2$ then $5^k \cdot 3 \equiv 25 \cdot 3 \equiv 75 \pmod {100}$.

And $5^k 3^2 \equiv 75 \cdot 3 = 225 \equiv 25 \pmod {100}$.

So by induction $5^k 3^m \equiv \begin{cases} 75 & m\ odd\\25 & m\ even\end{cases}$.

So $5^{121} 3^{312} \equiv 25 \pmod {100}$.


问:
Squarefree totient sum
Does anybody have a reference/proof for the asymptotic growth rate of
$$A(x) = \!\!\!\!\!\!\sum_{\substack{n \leqslant x \\ n \ \text{squarefree}}} \!\!\!\!\!\! \varphi(n)$$
as $x \to \infty$? Here $\varphi(n)$ denotes Euler's totient function. Numerical evidence suggests it grows like $cx^2$ where $c \approx 0.21$.

I tried using the identity
$$ \varphi(n) = n \sum_{d \mid n} \frac{\mu(d)}{d}$$
along with the fact that $\mu(n)^2 = 1$ iff $n$ is squarefree, and I got the expression
$$A(x) = \sum_{d \le x} \mu(d) \sum_{m \le x/d} m \mu(md)^2.$$
I realize that $md$ is squarefree iff $m$ and $d$ are both squarefree and coprime, but it feels like that just makes the problem harder. Any help would be appreciated!

答:
This answer gives a Tauberian argument:

As the Greg Martin suggests in the comment section, the Dirichlet series

$$
F(s)=\sum_{n=1}^\infty{\mu^2(n)\varphi(n)\over n^s}=s\int_1^\infty{A(x)\over x^{s+1}}\mathrm dx\tag1
$$

has a simple pole at $s=2$ with residue

$$
K=\prod_p\left(1+{p-1\over p^2}\right)\left(1-\frac1p\right)
$$

which implies

$$
G(s)={F(s)\over s}-{K/2\over s-2}
$$

is analytic in a neighborhood containing $s=2$. Plugging in (1), we have

$$
\begin{aligned}
G(s)
&=\int_1^\infty{A(x)-Kx^2/2\over x^{s+1}}\mathrm dx \\
&=\int_1^\infty{A(\sqrt t)-Kt/2\over2t^{s/2+1}}\mathrm dt
\end{aligned}
$$

Now, Wiener-Ikehara Tauberian theorem guarantees that the RHS converges at $s=2$, so we know that

$$
\int_1^\infty{2A(\sqrt t)/K-t\over t^2}\mathrm dt
$$

converges. Now, for convenience, we set $B(x)=2A(\sqrt x)/K$, so the above formula becomes

$$
\int_1^\infty{B(t)-t\over t^2}\mathrm dt\tag2
$$

from this, we will show that $B(t)\sim t$ therefore $A(x)\sim Kx^2/2$:

If $B(t)\nsim t$ then there exists a constant $\lambda>1$ such that there exists a strictly increasing sequence $t_n$ such that $B(t_n)\ge\lambda t_n$ for all $n$, meaning

$$
\begin{aligned}
\int_{t_n}^{\lambda t_n}{B(t)-t\over t^2}\mathrm dt
&\ge\int_{t_n}^{\lambda t_n}{\lambda t_n-t\over t^2}\mathrm dt
=\int_{t_n}^{\lambda t_n}{\lambda-t/t_n\over(t/t_n)^2}\mathrm d(t/t_n) \\
&=\int_1^\lambda{\lambda-u\over u^2}\mathrm du>0
\end{aligned}
$$

which violates Cauchy's criterion for convergence of (2), so we have

$$
\limsup_{t\to\infty}{B(t)\over t}\le1
$$

Similar argument can be established for $\liminf_{t\to\infty}B(t)/t\ge1$. Consequently, we see that $B(t)\sim t$, indicating

$$
\lim_{x\to\infty}{A(x)\over x^2}=\frac K2=\frac12\prod_p\left(1+{p-1\over p^2}\right)\left(1-\frac1p\right)
$$

Title: Squarefree totient sum


问:
For given line, find the locus of foot of perpendicular from origin

for a variable line $\frac xa +\frac yb =1$, find the locus of foot of perpendicular drawn from origin to the line under the condition that $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$

The line perpendicular to this line and passing through origin is $y=\frac ab x$

The point of interaction of this line is
$$x=\frac{ab^2}{a^2+b^2}$$ and $$y=\frac{a^2b}{a^2+b^2}$$

While $c^2 =\frac{a^2b^2}{a^2+b^2}$

So $ax=c^2$ and $by=c^2$

How should I proceed from here?

答:
Notice that for $x = \dfrac {ab^2}{a^2+b^2}$, $y = \dfrac {a^2b}{a^2+b^2}$, we have:

$$x^2 + y^2 = \frac {(ab^2)^2 + (a^2b)^2}{(a^2+b^2)^2} = \frac {a^2b^2(a^2+b^2)}{(a^2+b^2)^2} = \frac {a^2b^2}{a^2+b^2} = \left(\frac1{a^2} + \frac1{b^2}\right)^{-1} = c^2$$

What do you think this shape is?

Alternatively, the line $\dfrac xa + \dfrac yb = 1$ passes through the points $(a,0)$ and $(0,b)$. The length of the perpendicular from the origin to this line can be calculated by considering the area of the triangle with $(0,0), (a,0), (0,b)$ as vertices. We would therefore have:

$$\frac{ab}2 = \frac {h\sqrt{a^2+b^2}}2$$

giving $h = \dfrac{ab}{\sqrt{a^2+b^2}}$, or $h^2 = \dfrac {a^2b^2}{a^2+b^2} = c^2 \leadsto h = |c|$.


问:
A $56$-digit number divisible by $13$
Let $N$ be a $56$-digit number with all the digits same except the $32$nd digit from the right which is a different one. If $N$ is divisible by $13$, then which digits can never be the units digit of $N$?

I tried to approach this question with the help of divisibility rules available for $13$ but I was not able to figure out the solution for this. Can someone please explain the way to solve this?

Update : As many of you guys are helping with the solution and thanks for that...I am just putting down the answer over here... the correct answer is 4 and 7.

答:
Let $N$ have digit $a$ $55$ times and digit $b$ once at $32$nd place from right.

Now $13$ divides any block of six same consecutive digits $$13 \mid a\cdot 111111 \cdot 10^m \quad , \quad m \in \{0,1,2,\ldots\} $$

There are five blocks of $aaaaaa$ before $31$st digit and four blocks of $aaaaaa$ after $32$nd digit. So $13$ divides $N$ iff $13$ divides the remaining two digit number formed by $32$nd and $31$st digits.
$$13 \mid ba$$

Looking at two-digit multiples of $13$, $a$ cannot be $0,4,7$.

Title: A $56$-digit number divisible by $13$


问:
Question in the solution of an exercise in Van Dalen's "Logic and Structure"
This is exercise 5 in Van Dalen's "Logic and Structure", and the question is to show:

$\Gamma \vdash \phi \implies \Gamma \cup \Delta \vdash \phi$

$\Gamma \vdash \phi ;  \Delta, \phi \vdash \psi \implies \Gamma \cup\Delta\vdash\psi$

I have two questions:

What is this asking...? Can someone help me understand exactly what's going on in the question here?

There is a solution given by Dr. Kevin T. Kelly's Logic and Computation course, in which he writes the following here:

I have a few questions on how this is written. Firstly, what is the set $\text{der}$ exactly? This is not defined in Van Dalen. Secondly, what does he mean when he writes "clamp $\cal{D'}$ onto $\phi$ in $\cal{D}'$"? Is this like the standard clamp function in analysis?

Cheers

答:
The two results above express the basic property of the derivability relation $\vdash$.

The symbol $\Gamma \vdash \phi$ means that there is a derivation $\mathcal D$ (in the Natural Deduction proof system) of conclusion $\phi$ from the set $\Gamma$ of assumptions (or premises).

In the linked comment, the existence of such a derivation is symbolized with: $\mathcal D \in \text {der}$.

Having said that, the first result says that: having a derivation $\mathcal D$ of formula $\phi$ from the set of assumptions $\Gamma$, we can add further assumption (the new set of formulas $\Delta$) and the derivation still holds.

This property is called Monotonicity of entailment.

The second result expresses the Transitivity of entailment.

About "to clamp"...?

The idea is simply this one: starting from derivation $\mathcal D$ of $\phi$ from $\Gamma$, add to the premises the new set $\Delta$ and copy-paste the derivation $\mathcal D'$ of $\psi$ from $\Delta$ and $\phi$ (that we have already as conclusion of $\mathcal D$).

The result is a new derivation $\mathcal D''$ of $\psi$ with the new set $\Gamma \cup \Delta$ of premises.

This property of the $\vdash$ relation formalizes the usual mathematical practice  of proving some preliminary Lemma to be used in the proof of a main Theorem.


问:
Algebraic Tangent Space and Vector - an intuitive understanding?
In this question I am looking for help in understanding the Algebraic Tangent vector and what the difference is between it and the "regular" Tangent vector.

A "differentiable function" near p is a pair $(f,U)$ where $U \subset$ is open with $p \in U$, and $f:U \rightarrow \mathbb{R}$ is a differentiable function.   We say that two such pairs $(f,U), (g,W)$ are "equivalent" - the notation is $(f,U) \sim (g,W)$, if and only if there exists an open $V \subset U \cap W$ with $p \in V$ such that $f|_{V}$ and $g|_{V}$.  So f and g coincide in a small neighborhood of p. $\sim$ is an equivalence relation.

Note:  I am thinking that this is an equivalence relation (intuitively) because we are just looking at some neighborhood of point p - that which is  $V \subset U \cap W$ and so other mapping $h: Y \rightarrow \mathbb{R}$ where $Z \subset U \cap W$ is also in the equivalence relation defined by:

$$E_{p} \equiv \left\{(f,U)|(f,U) \text{ is a differential function near } p \right\}/\sim$$

Now the equivalence class of a pair $(f,U)$ is denoted by $[f,U]$ and is called the "germ" of a differentiable function at  $p$.

Note: From my understanding $[f,U]$, the germ, is showing that locally these functions $\in [f,U]$ are differentiable.  I am guessing that is just scratching the surface - but furthermore other than knowing something has to be differentiable to find the tangent (at least in the sense of being $C^{1}$).

While I am trying to wrap my head around the above, it seems to get worse..

An element $v$ of the dual space of $E_{p}, v \in Hom(E_{p},\mathbb{R}) =\left\{v:E_{p} \rightarrow \mathbb{R} | \text{ is linear} \right\}$ is called a "derivation" if $\forall [f,U],[g,W] \in E_{p}$ the product rule $$v([f,U] \cdot [g,W]) = v([f,U]) \cdot [g,W](p) + [f,U] \cdot v([g,W])$$ holds

Defn:  The Algebraic Tangent space of a space X at p is: $T_{p}^{alg}X  \equiv \left\{v \in E_{p}^{*} | v \text{ a derivation}\right\}$

An element of $T_{p}^{alg}X$ is called an algebraic tangent vector of X at p.

So what is the point of the Algebraic Tangent vector and space?

Thanks much for any thoughts,

Brian

答:
I don't think I can answer this coherently, so I'll break it up in different parts.

1. Germs.

What you wrote about germs seems right, as it is. Since looking at tangent spaces for non-differentiable functions doesn't make sense, differentiablitiy (at the point where you're looking at the tangent space) is a necessary condition. At places where I encountered the algebraic construction of tangent spaces, the (indeed!) equivalence relation $\sim$ was always defined on the set of differentiable maps.

I consider the next question to be a useful one wrt this subject: Show that the set of germs around 0 of functions from $\mathbb{R}^n$ to $\mathbb{R}$ has zero divisors.

2. Derivations.

The notation you use unfortunately makes things more complicated. The best way to think about derivations, I think, is as derivatives. I will assume you've written $f$ and $g$ instead of $f$ and $f'$ and reserve the prime for standard derivatives. If you have two maps $f,g$ and let $v$ be the standard derivative evaluated at $p$, then $v(\lambda f+g) = (\lambda f+g)'(p)= \lambda f'(p)+g'(p)$. This equailty still holds if you pass to $E_p$, justifying that you look at linear maps. The extra condition then corresponds to the product rule: $v(fg) = (fg)'(p) = (f'g)(p)+(fg')(p)$.

3. What's the point?

If it isn't clear why you should look at algebraic tangent vectors, this is a very good question that most certainly needs addressing. There are three different possible definitions of tangent spaces that can be used: the 'regular', or geometric, tangent space $T_pM$, the algebraic tangent space $T_p^{\text{alg}}M$ and the physicist's tangent space $T_p^{\text{ph}}M$. The main point in having these different definitions, is that the notions are equivalent and indeed, in differential geometry or differential topology, they are used interchangeably. Their main use is that in certain proofs one of the different notions is a lot easier to deal with than the others. For definitions of the notions as well as proofs of their being equivalent, I'd like to refer to the second chapter of Bröcker, Jänich, Introduction to differential topology.

ContentLicense: CC BY-SA 3.0


问:
principal eigenvalue in infinite dimensions
For $A$ a symmetric, positive let $\lambda_1$ be its principal eigenvector. I want to show that
$$\lambda_1=max_{v\neq 0}\frac{<Av,v>}{||v||^2}$$

I found this link: Prove that a $2\times2$ real matrix has an eigenvalue $\lambda$ with $\vert \lambda \vert = r(A)= \max_{x \ne 0} \frac{\|Ax\|}{\|x\|}$

one of the comments shows how to do this in finite dimensions but what does the proof look like in infinite dimension (Hilbert Space)

答:
It's a little involved, here are the steps, for special classes of bounded linear operators $T$ on an infinite dimensional Hilbert space:

prove that, for compact operators, the spectrum is $\{0\}\cup \{$ eigenvalues $\}$ (eigenvalues need not to exist)

if the ground field are the complex numbers, Gelfand's theorem ensures that the spectral radius $r(T)$ of $T$ is the biggest (in absolute value) number contained in the spectrum of $T$. Now, for normal operators, the spectral radius is just the norm, and if that operator is nonzero, you obtain that for compact normal operators the biggest eigenvalue (in absolute value), is exactly $||T||=\sup_x ||Tx||/||x|| $

if the ground field are the reals, you need the operator to be compact and symmetric in order to show the existence of an eigenvalue whose absolute value is the norm of the operator. The eigenvalue in question is $\lim_n (Tx_n,x_n)$, where that sequence is chosen in such a way that its absolute value converges to $||T||$

Long story short, your proposition holds, for instance, for compact normal operators (on the complex numbers), and for compact symmetric operators on the reals.

Do you want further details?


答:
Hint

WLOG $a=c\cos t,b=c\sin t$

$$x/c\cos t+y/c\sin t=1$$

so, the equation of the perpendicular will be $$x/\sin t-y/\cos t=0$$

Solve for $x,y$

and eliminate$t$

Title: For given line, find the locus of foot of perpendicular from origin

Body:
for a variable line $\frac xa +\frac yb =1$, find the locus of foot of perpendicular drawn from origin to the line under the condition that $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$

The line perpendicular to this line and passing through origin is $y=\frac ab x$

The point of interaction of this line is
$$x=\frac{ab^2}{a^2+b^2}$$ and $$y=\frac{a^2b}{a^2+b^2}$$

While $c^2 =\frac{a^2b^2}{a^2+b^2}$

So $ax=c^2$ and $by=c^2$

How should I proceed from here?


问:
Is this set a basis of the vector space $H$?
When $H=\{(x_1,x_2,x_3)^T\in\Bbb{R^3}:3x_1-2x_2+6x_3=0\}.$ I claim that the set $V=\{v_1 = (2,0,3), v_2 = (2,3,0), v_3 = (0,3,1)\}$ is a basis. My logic is this. This set is linearly independent, and it spans the entirety of $\Bbb{R^3}$. Thus it must also span each subspace, including $H$. Is this enough to $V$ is a basis? Are there other methods we could use?

答:
No. The reason being the first vector of your claimed basis i.e. $v_1$ doesn't belong to the space $H$.

$3(2)-2(0)+6(3)=24\ne 0$


答:
Notice that $\frac{10^{56}-1}9$ in decimal is a string of 56 ones, so
$$N=a\frac{10^{56}-1}9+(b-a)10^{31}$$
where $1\le a\le9$ is the repeated digit and $0\le b\le9$ is the different digit.

Looking mod 13, we have $9^{-1}=3$, and using that $10^6\equiv1$, we have
$$
10^{56}=(10^6)^910^2\equiv1^910^2\equiv9\\
10^{31}=(10^6)^510^1\equiv1^510^1=10.
$$

Therefore, the statement that $N$ is divisible by 13 is equivalent to the (mod 13) congruence
$$
0\equiv a(9-1)3+(b-a)10\\
0\equiv 14a+10b\\
3b\equiv a.
$$
This means we can use any digit $a$ as long as there is a digit $b$ such that $a\equiv3b$ (mod 13). The first ten multiples of 3, corresponding to $b=0,\dots,9$, are $a=0,3,6,9,12,2,5,8,11,1$. (The next values $a=4,7,10$ correspond to $b=10,11,12$.) So $a$ can't be 4 or 7.

Title: A $56$-digit number divisible by $13$

Body:
Let $N$ be a $56$-digit number with all the digits same except the $32$nd digit from the right which is a different one. If $N$ is divisible by $13$, then which digits can never be the units digit of $N$?

I tried to approach this question with the help of divisibility rules available for $13$ but I was not able to figure out the solution for this. Can someone please explain the way to solve this?

Update : As many of you guys are helping with the solution and thanks for that...I am just putting down the answer over here... the correct answer is 4 and 7.


问:
Showing that a group has infinite order
Let $G$ be an abelian group such that for all $g \in G$ and for all positive integers $k$, there exists some $h \in G$ such that $h^k = g$.

Now, I would like to show that the group is either the trivial group, or of infinite size. The trivial case is straightforward, but I am having trouble proving the infinite cardinality.

I start by supposing that $G$ has finite cardinality with $|G| = m>1$ where $m$ is some integer. I now pick an abitrary $g$ and $k=m+1$, and the corresponding $h$ so that we have $h^k=g$.

If I am on the right track, I'm lost as to where to go to from here. If I am not on the right track, well then I am quite lost!

Thank you for any help!

答: If $G$ is not the trivial group, take $g\in G\setminus\{e_G\}$. There is some $h_1\in G$ such that $h_1^{\,2}=g$ and there is some $h_2\in G$ such that $h_2^{\,3}=g$. But then $h_1\ne h_2$, since this would imply that $h_1^{\,2}=h_1^{\,3}$, which in turn would imply that $h_1=e_G$, which is impossible, since $g\ne e_G$. Now, take $h_3\in G$ such that $h_3^{\,7}=g$. Then $7\equiv1\pmod2$ and $7\equiv1\pmod3$. So, by the same argument as before, $h_3\ne h_1$ and $h_3\ne h_2$. And so on…

Title: Showing that a group has infinite order


问:
How to find the Image of the linear transformation?
I tried to find the image of the linear transformation: $T:\mathbb{R}^3\to\mathbb{P}_3$ defined:

$T\left[\begin{array}{c}a\\ b\\ c\end{array}\right] = (a+b)x^3+(-b+c)x^2+(a+c)x+(2a+b)$

I define a function of degree 3 as $mx^3+nx^2+px+q$ and equal it to $(a+b)x^3+(-b+c)x^2+(a+c)x+(2a+b)$ , but i get confused. Please, help me.

答:
Choose standard basis of $\mathbb R^3$ s.t. $B=\{(1,0,0),(0,1,0),(0,0,1)\}$

$T(1,0,0)=x^3+x+2$

$T(0,1,0)=x^3-x^2-1$

$T(0,0,1)=x^2+x$

The image vectors of basis elements also spans the image space so $span\{Im(T)\}=\{x^3+x+2,x^3-x^2-1,x^2+x\}$. Can you investigate the linear independence of the vectors in above set?

If they are LI then basis of $Im(T)$ is same as $span\{Im(T)\}$ otherwise, a basis must be a $2-$element subset like $\{x^3+x+2, x^2+x\}$ (why?).


问:
Prove that $f(z)=g(z)$ on $D$ considering Taylor expansions
Let $f$ and $g$ be differentiable on the strip
$D = \{z \in \mathbb{C}: -2 < \operatorname{Im}z <2$. Suppose that $ f(z) =g(z)$ for all $z$ such that $|z|<0.01$. By considering Taylor expansions first about $0$, then $1$, and so on by induction, prove that $f(z)=g(z)$ on $D$.

Is it safe simply to state the Identity Theorem? If not, I am lost and need some assistance

Identity Theorem:  If $f$ and $g$ are differentiable in a domain $D$ and $f(z)=g(z)$ for all $z \in S \subseteq D$ where $S$ has a limit point in $D$, then $f=g$ throughout $D$.

答: The fact that $f=g$ for $|z| <0.01$ implies that the power series expansions of $f$ and $g$ in the disk $D(0,2)$ have the same coefficients: the coefficients are simply $\frac  {f^{(n)} (0)} {n!}$ and  $\frac  {g^{(n)} (0)} {n!}$. This proves that $f=g$ in $D(0,2)$. Now we already know that $f=g$ is some disk around $1$ so we can repeat the argument (using power series around the point $1$) to conlude that $f=g$ in $D(1,2)$, and so on. Thus, $f=g$ in $D(n,2)$ for each $n$ and hence $f=g$ on the part of the domain where $\Re z \geq 0$. Similar argument holds for the part $\Re z \leq 0$.


问:
Diameter metric space
Suppose $(A, d)$ is a metric space with a non-empty subset $E \subseteq A$. Let $S = \{d(a, b) : a, b \in E\}$. Define $\phi (E)$ be the diameter of $E$ so that $\phi (E) = \sup S$, and $\phi(E) = \infty$ if $S$ is unbounded. Prove the following:

(a) If $E \subseteq A$ is non-empty then $\phi(E) = \phi(\overline{E})$ (the diameter of $E$ is equal to that of the closure of $E$).

(b) If $\phi (E) < \epsilon$ for some $\epsilon > 0$ and $E \cap B_{\epsilon}(x) \neq \emptyset$ for some $x$, then $E \subseteq B_{2\epsilon}(x)$.

For (a) I came up with a proof by contradiction but I am curious as to whether there are other proofs. After much time spent, I still cannot figure out a proof for (b). Any assistance or alternatively proofs are appreciated.

答:
First we show that if $E \subset B$ then $\phi(E) \leq \phi(B)$. This follows directly:

$$\phi(E) = \sup_{x,y \in E} d(x,y) \leq \sup_{x,y \in B}d(x,y) = \phi(B).$$

Hence, $\phi(E) \leq \phi(\overline{E})$ follows immediately.
Going the other way (i.e., showing $\phi(E) \geq \phi(\overline{E})$): Cover every point in $E$ by balls of some fixed radius $\epsilon > 0$:

$$E = \bigcup_{x \in E} \{x\} \subset \bigcup_{x\in E} B_\epsilon(x)$$

Next note that $\overline{E} \subset \bigcup_{x\in E} \overline{ B_{\epsilon/2}(x) }\subset \bigcup_{x\in E} B_\epsilon(x) $ and by the triangle inequality, $\phi\left( \bigcup_{x\in E} B_\epsilon(x) \right) \leq \phi(E) + 2\epsilon$.

So by applying the first result (with $B \equiv \bigcup_{x\in E} B_\epsilon(x)$),
$$ \phi(\overline{E}) \leq \phi\left( \bigcup_{x\in E} B_\epsilon(x) \right) \leq \phi(E) + 2\epsilon $$

as this holds for arbitrary $\epsilon > 0$, let $\epsilon \to 0^+$ and hence $\phi(\overline{E}) \leq \phi(E)$.

Title: Diameter metric space


问:
Reducing sum of squared differences to closed form expression
I have the following expression:

$$\sum_{i=0}^{n-1}[\rho^i-\frac{1}{n}(\frac{\rho^{i+1}-1}{\rho-1})]^2 $$

I would like to reduce this to a closed form expression. What I have tried so far:

Remove the square by multiplying out. This give an inner expression of:
$$\rho^{2i}-2\frac{\rho^i}{n}(\frac{\rho^{i+1}-1}{\rho-1})+\frac{1}{n^2}\frac{(\rho^{i+1}-1)^2}{(\rho-1)^2}$$

Split out the sum on each term
$$\sum_{i=0}^{n-1}\rho^{2i}-\sum_{i=0}^{n-1}2\frac{\rho^i}{n}(\frac{\rho^{i+1}-1}{\rho-1})+\sum_{i=0}^{n-1}\frac{1}{n^2}\frac{(\rho^{i+1}-1)^2}{(\rho-1)^2}$$

Hope for some nice expressions to arise e.g. using geometric series.

I have only done some simple manipulations with geometric series before and can't get further than the above. All help is appreciated!

答:
There is further expansion possible:

$$\sum_{i=0}^{n-1} \rho^i (\rho^{i+1} - 1) = \rho \sum_{i=0}^{n-1} (\rho^2)^i - \sum_{i=0}^{n-1} \rho^i,$$

and

$$\sum_{i=0}^{n-1} (\rho^{i+1}-1)^2 = \rho^2 \sum_{i=0}^{n-1} (\rho^2)^i - 2\rho \sum_{i=0}^{n-1} \rho^i + \sum_{i=0}^{n-1} 1.$$

So if we let $$\sum_{i=0}^{n-1} \rho^i = S_1, \quad \sum_{i=0}^{n-1} (\rho^2)^i = S_2,$$ then the given sum equals

$$S_2 - \frac{2}{n(\rho-1)} \left(\rho S_2 - S_1\right) + \frac{1}{n^2(\rho-1)^2} \left(\rho^2 S_2 - 2\rho S_1 + n\right).$$  Now put everything over a common denominator and simplify:

$$\frac{n^2(\rho-1)^2 S_2 - 2n(\rho-1)(\rho S_2 - S_1) + \rho^2 S_2 - 2\rho S_1 + n}{n^2 (\rho-1)^2} = \frac{(n+\rho-n\rho)^2 S_2 - 2(n + \rho - n\rho)S_1 + n}{n^2 (\rho-1)^2}.$$

Now because $$S_1 = \frac{\rho^n - 1}{\rho - 1}, \quad S_2 = \frac{\rho^{2n}-1}{\rho^2-1}, $$ we can substitute these to get the final result, but since it is a rather lengthy expression, I have omitted it.


问:
Piecewise function same as multi part function?
As shown in attached graph,we have a multi part function. Can we also call it piece wise function?

答: The graph is that of piecewise function. Said function also appears to be piecewise linear. I'm unfamiliar with the term "multi part function", but Google seems to suggest that it's synonymous with piecewise function.</p>


问:
Essential singularities of $\frac 1{e^z-1}$
How do I show that $\frac 1{e^z-1}$ has essential singularities (instead of say, poles) at $z=2n\pi i(n\in \mathbb Z)$?

I can't figure out how to show that the function does not go to infinity near $0$, or that it assumes every possible value near $0$. Exhibiting the laurent series around $0$ isn't general enough to show that essential singularities occur at all the stated points.

答:
An idea: using, for example, l'Hospital, you could show that

$$\forall\,n\in\Bbb N\;,\;\;\lim_{z\to 2n\pi i}\frac{z-2n\pi i}{e^z-1}\stackrel{\text{l'H}}=\lim_{z\to 2n\pi i}\frac1{e^z}=1$$

so the singularity $\,z=2n\pi i\;$ is,  in fact, a pole: a simple one with residue $\,1\,$


答:
The division rule of $13$ is that if you take the $k$ and $k+3$ digit of any number, and increase or decrease them both by the same amount, then the divisibility by $13$ of the result will be unchanged.

(This is because increasing of decreasing those digits is the same as adding or subtracting a multiple of $1001$ and $1001$ is a multiple of $13$.)

We can use this method to completely eliminating the smaller digit while subtracting the smaller digit from the larger digit.  Also, if its convenient to do so, we can do this to three pairs of digits at a time and increase any six consecutive digits a time.

(We can think of this as adding or subtracting a multiple of $111111 = 111\cdot 1001$ which is a multiple of $13$.

Sooooo...

$N=\underbrace{\underbrace{aaa.....a}_{24}\color{red}b\underbrace{aaa....}_{31}}_{56}=$

$\underbrace{\underbrace{aaaaaa}.....\underbrace{aaaaaa}}_{4\text{ groups of }6}ba\underbrace{\underbrace{aaaaaa}.....\underbrace{aaaaaa}}_{5\text{ groups of }6}\equiv $

$\underbrace{\underbrace{000000}.....\underbrace{000000}}_{4\text{ groups of }6}ba\underbrace{\underbrace{000000}.....\underbrace{000000}}_{5\text{ groups of }6}\pmod {13}\equiv$

$ba\times 10^{30}=$

$ba\times 10^{30} + ba\times 10^{27} - ba\times 10^{27} =$

$ba\times 1001\times 10^{27} - ba\times 10^{27}\equiv $

$-ba \times 10^{27} \pmod{13}\equiv $

$-ba \times 10^{27} - ba \times 10^{24} + ba\times 10^{24} \equiv $

$-ba\times 1001\times 10^{24} +  ba\times 10^{24}\equiv$

$ba \times 10^{24} \pmod {13}\equiv$

$....$

$ba \times 10^{18} \pmod {13}\equiv$

$....$

$....$

$ba \times 10^0 \equiv ba\equiv 10b + a\pmod{13}$

So if $N$ is divisible by $13$ then $ba$ is divisible by $13$.  Assuming $a$ and $b$ are not both $0$ there are $7$ two digit numbers divisible by $13$ and of the possible values of $a$ only $4$ and $7$ never occur. (The seven two digit multiples of $13$ are $13, 26,39,52, 65, 78,91$.)

........

A more sophisticated way but the exact same idea would be to note $1001 \equiv 0 \pmod{13}$ so $10^3 \equiv -1 \pmod {13}$ and by induction.

$10^{6k}\equiv 1 \pmod {13}$ and $10^{6k + 3}\equiv -1 \pmod {13}$.

This also means $10^{6k +i} a \equiv - 10^{6k + i + 3}a\pmod {13}$ and $10^{6k+i}a + 10^{6k+i+3} \equiv 0 \pmod{13}$

Therefore:

$N = \sum_{i=0; i\ne 31}^{55} 10^ka+ b10^{31}=$

$\sum_{i=0}^{55} 10^k a + (b-a)10^{31}\equiv$

$\sum_{i=0}^1 10^ka + \sum_{j=0}^{53}10^{k+2}a + (b-a)\cdot 10=$

$11a  + \sum_{m=0}^8(\sum_{n=0}^n (10^{6m + n+2}a + 10^{6m + n + 2 + 3}a) + (b-a)\cdot 10 \equiv $

$11a + 0 + (b-a)10 \equiv 10b + a \pmod{13}$

Again we can look at all $7$ of $10b + a\equiv 0 \pmod {13}$ or we can still be more sophisticated.

$10b + a \equiv a -3b \pmod{13}$ so if $N \equiv 0 \pmod{13}$ then

$a \equiv 3b \pmod{13}$.

$13$ is prime so multiplication and addition is a field $\pmod {13}$ and we will $a\equiv 3b \equiv 3\times 10\equiv 30\equiv 4 \pmod{13}$ if and only if $b \equiv 10 \pmod {13}$ and likewise $a \equiv 3b \equiv 3\times 11\equiv 33\equiv 7\pmod {13}$ if and only if $b \equiv 11 \pmod{13}$ and $a \equiv 3b \equiv 3\times 12 \equiv 10\pmod {13}$ if and only if $b \equiv 12 \pmod {13}$.

And as $0 \le b < 10$ we can not have any of those cases, so we can not have $a \equiv 4$ or $7\pmod {13}$. so $a$ can not be $4$ or $7$.

Title: A $56$-digit number divisible by $13$


问:
CDF and the floor!
I am working with Cumulative Distribution Function and trying to understand how to approach a few questions.

Problem:
$F(x)= \left\{ \begin{array} \\ k-\frac{1}{\lfloor x\rfloor}, x\ge 1,\\ 0, x\lt 1,\end{array} \right.$ is a cummulative distibution function (cdf) for some fixed number k.

Find:

$k$,

$x_{min}$ (the smallest number with non-zero probability)

$P(X=4)$

$P(2\lt X\le5)$

QUESTION 1: Somehow I was able to understand that the answer is $1$.

QUESTION 2: I assumed that the smallest number that $x$ can be is $1$ because the statement clarifies that for $k-\frac{1}{\lfloor x\rfloor}$, $x\ge1$. But when I answered $1$ it was wrong.  What am I missing here?

QUESTION 3: When I enter in $4$ for $x$ it then becomes: $1-\frac{1}{4}$ since $\lfloor 4\rfloor$ should equal $4$. So reducing the equation I got $\frac{3}{4}$ and that was wrong.  What am I missing here?

QUESTION 4: Since I am not getting the above questions right I am not sure how to approach this.  I would have thought that I would replace $x$ with the values that are $X$ in $(2\lt X \le5)$ would be $\{3,4,5\}$.

答:
$k=1$ is right: It comes from the fact that $F(x) \to 1$ as $x \to \infty$.

$F(x)=0$ for $x <2$ since $[x]=1$ anf $F(x)=0$ if $1 \leq x <2$ Hence $x_{min} =2$.

$P(X=4)=P(X\leq 4) -P(X<4)=(1-\frac  1 4) -(1-\frac 1  3)$ since $F(x)=1-\frac 1  3$ in $(3,4)$. So $P(X=4)=\frac 1 {12}$ [$P(X<x)$ is the limit of $P(X \leq y) \equiv F(y)$ as $y $ increases to $x$].

For the last part use: $P(2<X \leq 5)=F(5)-F(2)$.

Title: CDF and the floor!


问:
If a forgetful functor between varieties preserves coproducts, does it have a right adjoint?
Suppose that $C,D$ are two varieties (equational classes) and that $U:C\rightarrow D$ is the forgetful functor. Is it true that if this functor preserves coproducts and initial objects then it has a right adjoint?

I remember someone once told me this fact but I do not remember whom it was to ask for details. I just remember that I have been told that it follows from Freyd's criterion and some properties of varieties, but whenever I look at Freyd's criterion, it only talks about the existence of left adjoints and not right adjoints. Maybe this is easy, but I did not study the subject properly but need to use this result but I want to make sure it is true and understand some of its backgrounds.

I would appreciate any clarification/reference where I can find proof for this fact.

答:
Yes, this is true.  I claim first that if the forgetful functor $U:C\to D$ preserves coproducts, then it preserves coequalizers.  Let $f,g:X\to Y$ be $C$-morphisms, let $\sim_C$ be the $C$-congruence relation they generate on $Y$ and let $\sim_D$ be the $D$-congruence relation they generate on $Y$.  To show $U$ preserves coequalizers, we must show $\sim_C$ and $\sim_D$ are the same.

Let us denote the free $C$-algebra on $n$ generators $t_0,\dots,t_{n-1}$ by $F_n$.  Note that $\sim_C$ is just the equivalence relation generated by saying that for any $\alpha(t_0,\dots,t_n)\in F_{n+1}$, any $x\in X$, and any $y_1,\dots,y_n\in Y$, $\alpha(f(x),y_1,\dots,y_n)$ is equivalent to $\alpha(g(x),y_1,\dots,y_n)$.  So, it suffices to show that $\alpha(f(x),y_1,\dots,y_n)\sim_D \alpha(g(x),y_1,\dots,y_n)$ in this situation.

Let $i:F_{n+1}\to Y$ be the $C$-morphism sending the generators to $f(x),y_1,\dots,y_n$ and $j:F_{n+1}\to Y$ be the $C$-morphism sending the generators to $g(x),y_1,\dots,y_n$; it suffices to show that $hi=hj$ for any $D$-morphism $h:Y\to Z$ that coequalizes $f$ and $g$.  Note that $F_{n+1}$ is a coproduct in $C$ of $n+1$ copies of $F_1$, and hence also in $D$ since $U$ preserves coproducts.  So to show that $hi=hj$, it suffices to show their compositions with each coproduct inclusion $F_1\to F_{n+1}$ are the same.  This is trivial for all the inclusions except the first one.  For the first coproduct inclusion, the two maps we are comparing are just $hfk$ and $hgk$, where $k:F_1\to X$ is the $C$-morphism that sends the generator to $x$.  These are equal since $h$ coequalizes $f$ and $g$.

Thus $U$ preserves coequalizers, and hence preserves all colimits.  To show $U$ has a right adjoint, we now just have to verify the solution set condition of the general adjoint functor theorem (you mention being familiar with this theorem giving the existence of left adjoints, but you can get a criterion for the existence of right adjoints by just dualizing).  Concretely, in this case, that just means we need the following: given a $D$-algebra $X$, there exists a cardinal $\kappa$ such that if $Y$ is any $C$-algebra and $f:Y\to X$ is a $D$-morphism, there is a $C$-algebra $Z$ of cardinality at most $\kappa$, a $C$-morphism $g:Y\to Z$, and a $D$-morphism $h:Z\to X$ such that $hg=f$.  (Or more briefly, every $D$-morphism from a $C$-algebra to $X$ can be factored through a $C$-algebra of bounded cardinality.)

To prove this, suppose $Y$ is a $C$-algebra and $f:Y\to X$ is a $D$-morphism.  Let $S$ be the set of $D$-morphisms $F_1\to X$.  Then $f$ induces a function $\varphi:Y\to S$ as follows: given an element $y\in Y$, we get an associated $D$-morphism $F_1\to Y$ which we can then compose with $f$ to get an element of $S$.

Now consider the following diagram in $C$.  The objects are $Y$, and a copy of $F_1$ for each element of $S$ that is in the image of $\varphi$.  For each $y\in Y$, we have a morphism in the diagram from the copy of $F_1$ corresponding to $\varphi(y)$ to $Y$ which sends the generator to $y$.  Let $g:Y\to Z$ be the colimit of this diagram in $C$ (which is also the colimit in $D$ since $U$ preserves colimits).  That is, $g$ is the universal morphism that coequalizes the inclusions from $F_1$ of all pairs of elements $y,y'\in Y$ such that $\varphi(y)=\varphi(y')$.  By definition of $\varphi$, $f$ coequalizes all such pairs of inclusions, so $f$ factors through $g$ (via a $D$-morphism $h:Z\to X$).  On the other hand, note that $|Z|\leq |S|$, since pairs of elements of $Y$ that map to the same element of $S$ under $\varphi$ become equal in $Z$.  Thus, taking $\kappa=|S|$, we have the desired factorization of $f$.

(Incidentally, there really is something nontrivial going on in verifying the solution set condition here--if you drop the assumption that $U$ preserves coproducts, then it may not satisfy the solution set condition!  For instance, the forgetful functor from groups to sets does not, since there exist simple groups of arbitrarily large cardinality and nonconstant maps out of them cannot factor through any smaller group.  Here's the step of the argument that would break down in that case: in order to factor $f$ through $g$, you need to know that $g$ is a colimit not just in $C$ but also in $D$, so you need to know that $U$ preserves colimits.)

Let me close with an explicit construction of the right adjoint, starting with an illustrative example.  Let $C$ be the variety of sets with a unary operation and let $D$ be the variety of sets.  Then the right adjoint $G:D\to C$ sends a set $X$ to the set $X^\mathbb{N}$ of sequences of elements of $X$, with unary operation given by the left shift operator (sending a sequence $(x_0,x_1,x_2,\dots)$ to $(x_1,x_2,x_3,\dots)$).  (To relate this to the construction above, note that $X^\mathbb{N}$ is just the set of $D$-morphisms from the free $C$-algebra on one generator (i.e., $\mathbb{N}$) to $X$.  Indeed, with some work you can extract this description of the right adjoint from the proof of the general adjoint functor theorem using the solution set given above.)

Why is this right adjoint to the forgetful functor?  Well, suppose $Y$ is a set with unary operation $s$ and $f:Y\to X$ is a function.  Then $f$ induces a map $g:Y\to X^{\mathbb{N}}$ sending each $y\in Y$ to the sequence $(f(y),f(s(y)),f(s^2(y)),\dots)$, and $g(s(y))$ is just the left-shift of $g(y)$.  Conversely, any $C$-morphism $g:Y\to X^{\mathbb{N}}$ must have this form for a unique function $f$: if we define $f(y)$ to be the first term of the sequence $g(y)$, then the second term must be $f(s(y))$, the third term must be $f(s^2(y))$, and so on, because $g(s(y))$ is the left-shift of $g(y)$.

More generally, the right adjoint $G:D\to C$ can always be constructed explicitly by defining $G(X)$ as the set $S$ of $D$-morphisms $F_1\to X$ equipped with a $C$-algebra structure defined as follows.  An $n$-ary operation of $C$ can be thought of as a $C$-morphism $F_1\to F_n$.  Given $n$ elements of $S$, we get $n$ morphisms $F_1\to X$, which give a morphism $F_n\to X$ (since $F_n$ is a coproduct of $F_1$s in $D$, not just in $C$).  We can then compose this with our morphism $F_1\to F_n$ to get a morphism $F_1\to X$, i.e. another element of $S$.  Then, given a $D$-morphism $Y\to X$ where $Y$ is a $C$-algebra, the function $\varphi:Y\to S$ described in the proof above is exactly the adjoint $C$-morphism.


问:
Probability calculation when rolling a 6 sided dice
this is a newbie question but I can't seem to get to the answer right.

The scenario is the following - it's said (in this book I'm reading) that the probability to roll a 6-sided dice 30 times and have the number 6 as an outcome 21 of those 30 times is 0.000297.

How do they calculate this result? My calculation is obviously off, since I thought the probability must be $\frac{1}{6}^{21} * \frac{5}{6}^9$ but it obviously isn't because I'm not getting the correct number.

Thanks in advance!

答:
Your working assume that the first $21$ outcome is $6$ and the rest are not.

You need to multiply it by $\binom{30}{21}$.

$$\binom{30}{21}\left( \frac16\right)^{21} \left( \frac56\right)^9$$

Check out binomial distribution.


问:
Characteristic polynomial of a matrix of $1$'s
I am trying to calculate the characteristic polynomial of the $n \times n$ matrix $A = \{ a_{ij} = 1 \}$.

Case $n=2$: I obtained $p(\lambda)=\lambda^2-2\lambda$ .

Case $n=3$: I obtained $p(\lambda)=-\lambda^3+3\lambda^2$.

Case $n=4$: I obtained $p(\lambda)=\lambda^4 - 4\lambda^3$.

I guess that for the general case, we have

$$p(\lambda)=(-1)^n\lambda^{n}+(-1)^{n-1}n\lambda^{n-1}$$

I tried to use induction, but it didn't work, unless I've done wrong. Can somebody help me? Or give me a hint?

答:
We wish to compute the determinant of the $n\times n$ matrix
$$
M=\begin{bmatrix}
1-\lambda&1&1&\cdots&1\\
1&1-\lambda&1&\cdots&1\\
1&1&1-\lambda&\cdots&1\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&1&1&\cdots&1-\lambda
\end{bmatrix}\tag1
$$
Here are two approaches, the second of which uses a generalization of the lemma cited in Rodrigo de Azevedo's answer.

Finding a Similar Matrix

Note that the $n-1$ dimensional subspace orthogonal to $\begin{bmatrix}1&1&1&\cdots&1\end{bmatrix}^\text{T}$ is multiplied by $-\lambda$. (If the subspace is orthogonal to a given vector, we can subtract that vector from each row of the matrix when operating on that subspace.)

$\begin{bmatrix}1&1&1&\cdots&1\end{bmatrix}^\text{T}$ is multiplied by $n-\lambda$. (Just compute it.)

Thus, $M$ is similar to
$$
\begin{bmatrix}
-\lambda&0&0&\cdots&0\\
0&-\lambda&0&\cdots&0\\
0&0&-\lambda&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&\cdots&n-\lambda\\
\end{bmatrix}\tag2
$$
Therefore,
$$
\det(M)=(-\lambda)^{n-1}(n-\lambda)\tag3
$$

$\boldsymbol{\det(\lambda I_n-AB)=\lambda^{n-m}\det(\lambda I_m-BA)}$

As shown in this answer,

$$
\det(\lambda I_n-AB)=\lambda^{n-m}\det(\lambda I_m-BA)\tag4
$$

Let $A=\begin{bmatrix}1&1&1&\cdots&1\end{bmatrix}^\text{T}$ and $B=A^\text{T}$, then $M=AB-\lambda I_n$.

Furthermore, $m=1$ and $BA=\begin{bmatrix}n\end{bmatrix}$.

$(4)$ then says that
$$
\begin{align}
\det(-M)
&=\det(\lambda I_n-AB)\tag{5a}\\[2pt]
&=\lambda^{n-1}\det(\lambda I_1-BA)\tag{5b}\\[2pt]
&=\lambda^{n-1}\det(\lambda I_1-\begin{bmatrix}n\end{bmatrix})\tag{5c}\\
&=\lambda^{n-1}(\lambda-n)\tag{5d}
\end{align}
$$
which, since $\det(-M)=(-1)^n\det(M)$, becomes
$$
\det(M)=(-\lambda)^{n-1}(n-\lambda)\tag6
$$


问:
Partial derivatives of $|Ax - b|^2$?
I'm trying to work out the partial derivatives of a function $L$ in terms of $x_i$:

$$ A \in \mathbb{R}^{m x n} \quad b \in \mathbb{R}^m \quad x \in \mathbb{R}^n $$

$$\begin{aligned} L(x) &= \left\|{Ax - b }\right\|^2 \\&= (Ax-b)^T(Ax-b) \\ &= x^TA^TAx - b^TAx - x^TA^Tb + b^Tb\end{aligned}$$

All four of these terms are scalars so I think I can transpose the third term to get:

$$\begin{aligned} &= x^TA^TAx - 2b^TAx + b^Tb\end{aligned}$$

I'm a little stuck on how to transform $x^TA^TAx$ further.

Calculating the partial derivatives, the final term $b^Tb$ is constant so goes to zero, the second term has coefficient vector $2b^TA$ so we just drop the $x$, its again the first term I'm stuck on:

$$\frac{\partial{L(x)}}{\partial x} = ??? -2b^TA$$

What's the derivative of $x^TA^TAx$ in terms of $x$?  How you work it out?

Update:

After looking at the potential duplicate, I think I'm mostly covered - however it's not immediately obvious to me why:

$$\frac {\partial(x^TMx)} {\partial x}=(M+M^T)x$$

How is this rule derived?

答:
I answer to your last question for the term $f(x)=x^\top A^\top A x$.

You can easily find it in the following way:

$$f(x+dx)= ( x + dx)^\top A^\top A (x+dx) = x^\top A^\top A x + 2x^\top A^\top A dx + dx^\top A^\top A dx\\= f(x) + 2x^\top A^\top A dx + O(dx^2)  $$

As you can see, the linear term in $dx$ is your gradient (or better, the transpose of it, since what you see there is gradient transpose).

So $$\frac{\partial (x^\top A^\top A x)}{\partial x}= 2 A^\top A x$$

Title: Partial derivatives of $|Ax - b|^2$?


问:
a proof in variable separation method for 2-D heat equation
when solving the PDF of 2D heat equation
$U_t= C (U_{xx}+U_{yy})$, via classic separation method to set $U_{(x,y,t)}=X_{(x)}Y_{(y)}T_{(t)}$, then by plugging into the PDE, we may have the following expression: $$\frac{T'}{CT}=\frac{X''}{X}+\frac{Y''}{Y}=constant -(1)$$

I saw the material I have starts to have two ODE:$$X''-AX=0-(2)$$ $$Y''-BY=0-(3)$$
Here comes my question: is there a way to prove $\frac{X''}{X}$ and $\frac{Y''}{Y}$ has to be constants because of (1) so that the (2),(3) is therefore derived?

答:
It's just a variation of the usual argument: you have
$$
\frac{X''(x)}{X(x)} = \lambda - \frac{Y''(y)}{Y(y)}
$$
where the left-hand side is independent of $y$ and the right-hand side is independent of $x$, so that both sides must be constant, say equal to $A$. Then you get the ODEs that you wrote, if you let $B=\lambda-A$.


答:
You don't!

The function in question has simple poles. The easiest way to see this is to note that $\exp(2n\pi i)=1$, so by continuity $e^z-1$ tends to zero as $z \rightarrow 2n\pi i$ and thus the reciprocal tends to infinity.

To show that the poles are simple, note that $e^z-1$ has a zero of degree $1$ at $2n\pi i$ so the reciprocal has a simple pole and so the residue is:

$$\frac{1}{\frac{d}{dz}(e^z-1)}=\frac{1}{e^{2n\pi i}}=1$$

Title: Essential singularities of $\frac 1{e^z-1}$

Body:
How do I show that $\frac 1{e^z-1}$ has essential singularities (instead of say, poles) at $z=2n\pi i(n\in \mathbb Z)$?

I can't figure out how to show that the function does not go to infinity near $0$, or that it assumes every possible value near $0$. Exhibiting the laurent series around $0$ isn't general enough to show that essential singularities occur at all the stated points.


问:
Show that $a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}$ converges to zero
According to WolframAlpha, the following sequence
\begin{align*}
a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}
\end{align*}
seems to converges to zero. However, how can I prove this ? The difficulty I encounter is this "one quarter of an integer" in the Gamma functions.

答:
$$a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}\implies \log(a_n)=\log \left(\Gamma \left(\frac{n}{2}+\frac{1}{4}\right)\right)-\log \left(\Gamma \left(\frac{n}{2}+\frac{3}{4}\right)\right)$$ Now, using Stirling approximation
$$\log (\Gamma (p))=p (\log (p)-1)+\frac{1}{2} \log \left(\frac{2 \pi }{p}\right)+\frac{1}{12
   p}-\frac{1}{360 p^3}+O\left(\frac{1}{p^5}\right)$$ apply it twice and continue with Taylor expansion to obtain
$$\log(a_n)=-\frac{1}{2} \log \left(\frac{n}{2}\right)-\frac{1}{16 n^2}+\frac{5}{128
   n^4}+O\left(\frac{1}{n^5}\right)$$ Continuing with Taylor
$$a_n=e^{\log(a_n)}=\sqrt{\frac{2}{n}} \exp\Big[-\frac{1}{16 n^2}+\frac{5}{128
   n^4} \Big]$$


问:
Uniform distribution of points on a sphere
I'm reading paper by Arnol'd and Krylov ( UNIFORM DISTRIBUTION OF POINTS ON A SPHERE AND SOME ERGODIC PROPERTIES OF SOLUTIONS OF LINEAR ORDINARY DIFFERENTIAL EQUATIONS IN A COMPLEX REGION  ).

I need help in understanding the proof of the following theorem:

Theorem 1. Let A and B be two rotations of the sphere $S^2$ and let $x$ be a point of the sphere. If the sequence of points $$x, Ax, Bx, A^2x, ABx, BAx, B^2x, \ldots \ \ \ \ \ \ (1) $$ is dense on the sphere, then it is uniformly distributed.

Where uniformly distributed means that: if we consider the $2^n$ points produced with exactly $n$ iterations of the rotations,

$$
A^nx , A^{n-1}Bx , A^{n-2}BAx , \ldots , B^nx \ \ \ \ \ \ \ \ \ \ \ \ (2) $$ and we count the proportion of them that lie in a region of the sphere $\Delta$ which is bounded by a piecewise smooth curve, then these ratios converge to the measure of $\Delta$;
$$ \lim_{n\to \infty} \frac{\mbox{Number of points among } (2) \mbox{ lie in } \Delta}{2^n} = \frac{mes \ \Delta}{mes \  S^2}. \ \ \ \  (3)$$

The proof of the theorem as the following:

If any one can clarify to me why $(7)$ is impossible, how by density of the points in $(1)$ we get the commutativity on $R_{l}$ ?

Thanks in advance!

答:
If $x,Ax,Bx,\ldots$ were dense in $S^2$, then their closure gives $S^2$. So any rotation $R$ can be approximated by products $R=\lim_{n\to\infty}C_n$ where $C_n$ is a product of $A$s and $B$s.$\dagger$

$A,B$ commute on $R_l$, since any $g\in R_l$ is $Rf$ for some rotation $R$; hence using (7) $$ABg=ABRf=\lim_{n\to\infty}ABC_nf=\lim_{n\to\infty}BAC_nf=BAg$$
But then, any two rotations $R_1$, $R_2$, commute on $R_l$ since for any $g\in R_l$, $$R_1R_2g=\lim_{n\to\infty}C_nD_ng=\lim_{n\to\infty}D_nC_ng=R_2R_1g$$ This is a contradiction since, in fact, the group of rotations is not commutative on $R_l$ for $l>0$.

$\dagger$ (I don't fully understand this part yet, but the paper says it is easy, so it must follow trivially from some other result; I think one can show this by approximating $R\approx S^m$, where $S$ is a rotation by a small angle; then $Sx\approx C_nx$, and show $R\approx C_n^m$. Perhaps you can ask another question on this point.)


问:
Unions of closed sets
Let $\bar{A}$ be the closure of $A$. Let $B_n = \bigcup_{i=1}^{n}A_i$ for positive integer $n$. I want to show that $\bar{B}_n = \bigcup_{i=1}^{n}\bar{A}_i$.

I am thinking if $x$ is a limit point of $B_n$, if $x \in \bigcup_{i=1}^{n}A_i$ then clearly $x \in \bigcup_{i=1}^{n}\bar{A}_i$. If $x \notin \bigcup_{i=1}^{n}A_i$ then suppose $x$ is not a limit point of any $A_i$. This contradicts that $x$ is a limit point of $B_n$ so then $x \in \bigcup_{i=1}^{n}\bar{A}_i$. So $B_n \subset  \bigcup_{i=1}^{n}\bar{A}_i$. I think the reverse inclusion is straightforward.

For an infinite union $\bigcup_{i=1}^{\infty}\bar{A}_i$ does not contain $\bar{B}_n$ because a limit point $x \in B_n$ may for each neighborhood $N_r(x)$ intersect a different $A_i$ and there are infinite $A_i$ so none of them necessarily have $x$ as a limit point.

答:
If $x$ is a limit point of $B_n = \bigcup_{i=1}^n A_i$, then for some $1\le i \le n$, $x$ is a limit point of $A_i$, for if this were not the case, for every $1 \le i \le n$ we have a ball $N_{r_i}(x)$ for some $r_i>0$, such that $N_{r_i} \cap A_i \subseteq \{x\}$, define $r=\displaystyle\min_{1 \le i \le n} r_i >0$ and note that $$N_r(x) \cap B_n = \bigcap_{i=1}^n N_{r_i}(x)  \cap B_n \subseteq \bigcup_{i=1}^n (N_{r_i}(x) \cap A_i) \subseteq \{x\}$$

which contradicts that $X$ is a limit point of $B_n$, so the assumption was wrong.

Denoting the set of limit points of a set $A$ by $A'$ and using that $\overline{A}= A \cup A'$ we conclude:

$$\overline{B_n} = B_n \cup B_n' = \bigcup_{i=1}^n A_i \cup \left( \bigcup_{i=1}^n A_i \right)' = \bigcup_{i=1}^n A_i \cup \bigcup_{i=1}^n A'_i = \bigcup_{i=1}^n (A_i \cup A_i') = \bigcup_{i=1}^n \overline{A_i}$$

And we're done.

Title: Unions of closed sets


问:
Diagonalizing a matrix where the pairwise difference of eigenvalues are units
Let $R$ be a commutative ring (with $1$), and $M$ be an $n\times n$ matrix over $R$ whose characteristic polynomial is
$$\det(tI - M) = (t-\alpha_1)\cdots(t-\alpha_n)$$
with $\alpha_i \in R$ for all $i$. Assume that $\alpha_i - \alpha_j \in R^\times$ whenever $i\ne j$. Is it true that there is some invertible $P \in GL_n(R)$ with
$$M = P\begin{pmatrix}\alpha_1 & & & \\ & \alpha_2 & & \\ & & \ddots & \\ & & & \alpha_n\end{pmatrix}P^{-1}?$$

This is true if $R$ is a field. The proof here shows that we must have $R^n = \oplus_{i=1}^n \ker(M - \alpha_i I_n)$, at least in the case when $R$ is a local ring (but also it seems to work for general $R$). But would it follow that we can write $M$ in the form above? It is not clear to me that each $\ker(M - \alpha_i I_n)$ is free (if $R$ is not local), so I am not sure if we can necessarily find a basis of $R^n$ consisting of eigenvectors of $M$.

If the above is false, might there be some conditions on $R$ that would make it true, e.g. if we assume $R$ is an algebra over a field?

答:
Take $R = \mathbb{Z}[\sqrt{-5}]$ and $M = \begin{bmatrix}3 & 1+\sqrt{-5} \\ -1+\sqrt{-5} & -2\end{bmatrix}$. This has characteristic polynomial $t(t-1)$.

I claim that we cannot diagonalize $M$ over $R$. If we could, then the $0$-eigenspace and $1$-eigenspace must both be free of rank $1$ over $R$. Say $\begin{bmatrix} x \\ y\end{bmatrix} \in R^2$ generates the $0$-eigenspace. The $0$-eigenspace contains both $\begin{bmatrix} 1+\sqrt{-5} \\ -3\end{bmatrix}$ and $\begin{bmatrix} 2 \\ -1+\sqrt{-5}\end{bmatrix}$, so there must be some $r,s \in R$ such that
$$\begin{bmatrix} 1+\sqrt{-5} \\ -3\end{bmatrix} = r\begin{bmatrix} x \\ y\end{bmatrix}, \quad \begin{bmatrix} 2 \\ -1+\sqrt{-5}\end{bmatrix} = s\begin{bmatrix} x \\ y\end{bmatrix}.$$
In particular, $ry = -3$ and $sx = 2$. Since $2$ and $3$ are irreducible in $R$ (and $\pm 1$ are the units of $R$), then $x,y$ must lie in $\mathbb{Z}$. But
$$\begin{bmatrix} 0 \\ 0\end{bmatrix} = \begin{bmatrix}3 & 1+\sqrt{-5} \\ -1+\sqrt{-5} & -2\end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 3x + (1+\sqrt{-5})y \\ (-1+\sqrt{-5})x - 2y\end{bmatrix}$$
has no integer solutions for $(x,y)$ other than $(0,0)$ (because $\sqrt{-5}$ is irrational). Contradiction.


问:
Is learning (theoretical) physics useful/important for a mathematician?
I'm starting to read The Princeton Companion to Mathematics, at the beginning it says:

A proper appreciation of
  pure mathematics requires some knowledge of applied
  mathematics and theoretical physics.

Some of my professors have told me that modern Mathematics require some knowledge about Quantum Mechanics and theoretical Physics.

I attended the second and third day of the José Adem Memorial Lecture Series by Matilde Marcolli at the beginning of this year, they were about Number theory, Quantum statistical Mechanics and Quantum Field theory. I did not understand a single word but somehow the few things I understood had a big impression on me.

Please give me some examples of pure Mathematics that require/use Physics.

答:
I suggest to read
Missed Opportunities by Freeman Dyson an essay that describes missed opportunities in mathematics due to the fact that mathematicians were not interested in new developments in physics (e.g. investigating the symmetries of the new Maxwell equations).

It is certainly wrong that a knowledge of physics is required for mathematics, but a widespread ignorance of developments in areas that use mathematics leads to missed opportunities and missed intuition.

Edited to add: It might be worthwhile to note that the quote does not say that mathematics or doing mathematics requires knowledge of physics, but the proper appreciation of mathematics requires it. The book sets out to give an overview of mathematics and its development and requires some physics knowledge to do so. The fact that it is perfectly possible to work happily and productively in many branches of pure mathematics without getting your neurons dirty with physical intuition is not in contradiction with this.


问:
Why is $f(x)=x^2$ not equal to zero a.e. in regards to Lebesgue measure?
Question as above.

In our measure theory script (and in further literature also), to illustrate the term "almost everywhere" it's stated that the function $x^2$ is not equal to zero a.e. considering Lebesgue measure while it is equal to zero considering dirac measure in 0. I have difficulties in understanding this.

答:
$x^2$ is equal to $0 $ only at $x=0$ and one point has zero Lebesgue measure. However, this Dirac measure in $0$ of this one-point is $1$. This Dirac measure is defined as
$$
\delta_0(A) = \begin{cases}
0, & x\notin A \\
1, & x\in A \end{cases}
$$
for any measurable set $A$.
So, $\mu(\{x:x^2=0\}) = 0$ and $\mu(\{x:x^2\neq0\}) > 0$, where $\mu$ is Lebesgue measure, and $\delta_0(\{x:x^2=0\}) = 1$ and $\delta_0(\{x:x^2\neq0\}) = 0$. So, $x^2 \neq 0$ a.e. for Lebesgue measure, but not for Dirac measure.


问:
Definite integral of $e^{x/2}$ using Maclaurin polynomial
My professor asked us to find the 3rd degree Maclaurin polynomial of $e^{x/2}$ which I found to be
$$1 +  \frac{x}{2}  +  \frac{x^2}{8}  +  \frac{x^3}{48}$$

I do know that that the series for $f(x)=e^{\frac{x}{2}}$ is

$$\sum_{n=0}^\infty \frac{x^n}{n! 2^n}$$

But then he said to approximate $\int_0^2 e^{\frac{x}{2}} dx$ using part a (the Maclaurin polynomial)

How do I do that? Would I just integrate the polynomial and not the series?

答:
Your professor's question isn't a "realistic" application of this method. It's a highly simplified problem to test you on whether you understand how the method works. Often teachers will ask you to solve problems using a method to demonstrate your ability even if that isn't the best way to solve the problem.

In practice we would immediately integrate this, we wouldn't involve either the the power series or the polynomial approximation, i.e. we would just say that

$\int^2_0 e^\frac{x}{2}dx=[2 e^\frac{x}{2}]^2_0=e^2-1$

converting $e^\frac{x}{2}$ into a Maclaurin just makes the process of finding the integral more complicated.

We use the method of using the Maclaurin series when it's hard to find the integral. For example $\int e^{-x^2}dx$ is not an integral we can find in terms of basic functions. So if we want to find $\int^1_0 e^{-x^2}dx$ then one way we might approach it is using the method your professor is testing you on.

The Maclaurin series of $e^{-x^2}$ is quite straightforward.

$e^{-x^2}=\sum^\limits\infty_{n=0}\frac{(-x^2)^n}{n!}=\sum^\limits\infty_{n=0}\frac{(-1)^n x^{2n}}{n!}$

So we could easily integrate the whole power series. So your question still remains, why would we stop at the $x^3$ term?

Some possible reasons:

we are working with a function where we can't so easily find a formula for the coefficient of $x^n$ in the Maclaurin series,

we only care about getting a certain level of accuracy, so summing to infinity isn't worthwhile,

we don't want to spend too much computing power on this.


问:
Assigning Tiles on a Hyperbolic Grid with Unique Coordinates
I am creating a location in an RPG campaign that deals with non-euclidean space, and I'm currently toying with the idea of a forest with a finite border that takes up infinite space.

The idea is that from the outside, the forest appears to be six trees arranged in a hexagon, but depending on which pair of trees you enter through, you find yourself in a different 'hexagon' of space. The way that space has been expanded inside the forest is based on the idea that the forest contains a hexagonal grid of trees, and you need to walk 720 degrees around a tree to end up where you started. Hence, while you enter a different space depending on which entrance you take, spinning 360 degrees around either of the trees at the entrance to that space will land you at one of the other entrances.

This pattern - with six hexagons surrounding a given point - can be represented slightly more intuitively on a hyperbolic plane using a tiling with six hexagons adjacent to each vertex. I've provided a labelled projection of such a plane on the Poincare disc to (hopefully) help illustrate my idea. While only five are visible on the diagram, it is clear that each entrance hex is repeated infinitely many times on this grid.

My goal is to find a coordinate system which I can use to uniquely label every hex in the forest so that I can easily identify if one hex is equivalent to another hex. Using the green hex on the diagram as a reference, notice how there's two separate paths someone could take to get there depending on which entrance they used? I want to have a way of labelling hexes which lets me say "Entering from here and going NE then SE will get you to the same place as entering from there and going S, NW, N, SW, SE.". It would also be nice to have some clear way of navigating a shortest path from hex A to hex B in this system, if possible.

If the non-euclidean nature of the grid is making this difficult to comprehend, imagine a very basic infinite square grid. I could label a repeating 8x8 section of that grid with the coordinates (0,0) through to (7,7), then apply a modulus to the coordinates whenever I moved so that instead of going from (7,7) to (8,7) I go to (0,7). Equally, to uniquely label every space on an infinite grid, I could represent every square as the movements it took to get there from some origin square (3 norths, 2 easts, and so on), and then whenever I moved I could simplify the expression to ensure that I never give the same square two different names (3 easts, 2 norths and 4 souths would be simplified to 2 souths, 3 easts).

Does such an approach to naming coordinates exist for such a non-euclidean space? Is there a way to unambiguously name each coordinate so that I can tell if two paths reach the same destination?

Obligatory disclaimer: I am aware how rapidly space expands in this sort of system, and I may end up handwaving navigation between landmarks and just let players travel between any two destinations they have directions for. This post is, at least partially, a matter of curiosity and problem solving, not just an attempt to make something for an RPG campaign easy to represent.

答:
I found a solution that involves modelling the overall structure like a tree with certain properties, then assigning coordinates based on the hierarchy of that tree. Notably, this coordinate system works for a significant number of other structures of this form. I will discuss the specifics of this generalisation later.

Solution

We will first describe all hexes who share a vertex with the outside of the structure as hexes in 'Ring 1'. Every other hex that shares a vertex with a hex in Ring 1 is in Ring 2, and so on. Note that every Ring forms a cycle of hexes - every hex is adjacent to exactly two hexes from its Ring.

We can now uniquely identify any given hex by its Ring's number and the hex's position within its Ring. To describe its position within the Ring, we will identify its lineage of parents in previous Rings. Unfortunately, since not every hex is connected to a hex in the previous Ring, we will choose an arbitrary direction like counter-clockwise, and say that parent-less hexes are considered to have the parent that the nearest non-parent-less hex has. Now that every hex can have its ancestry uniquely traced, it can be uniquely identified.

We will denote the most counter-clockwise child of a hex as the child with 'index 0' (numbering the remaining children with consecutive integers, clockwise), and the total number of children belonging to a single parent as that group of children's 'limit' (for consistency, this will also be 'zero-indexed', so a limit of 2 means there are 3 unique indexes in this group). We will use the notation / to indicate a hex's position within its parent's set of children.

We may also note that every hex has either 2 or 3 connections to hexes within its Ring or in the previous Ring. Because every hex is connected to 6 hexes, this means they each connect to a further 4 or 3 hexes on the next Ring. Additionally, in between every hex connected to the previous Ring will be 2 or 3 hexes which exist to enforce the fact that every vertex is surrounded by six hexes. We can actually tell how many children a hex will have based on its index (in this case the limit is irrelevant), specifically any hex with an index divisible by 4 will connect to 3 hexes on the next Ring (and have 8 other children, totalling 11), and other hexes will connect to 4 hexes (and have 11 other children, totalling 15). This will be useful later.

To deal with some base cases we will say the limit of the outside of the structure is 5 (representing the six entrances) and the limit of each group of children on the outermost Ring is 2. So the hexes you enter once you've walked into the forest would be notated 0/5 - 0/2, 1/5 - 0/2, 2/5 - 0/2, and so on.

Finally, to navigate from one hex to another:

When moving into a deeper Ring - Label the paths from your hex to hexes in the next Ring clockwise with the numbers 0 through 3 or 0 through 2 as appropriate. Multiply this number by 4, this is the new index. Figure out the limit of this hex (11 or 15, as stated earlier), then add these new components to the end of your current coordinates.

When moving into an outer Ring - Remove the last component of the coordinate (note that this movement is only possible when your current index is divisible by 4).

When moving across a Ring - Add 1 to the index of your Ring. If that makes it larger than the limit, increase your parent's index, recalculate your limit and set your index to '0'. If your parent's index is now greater than its limit, repeat this process (and so on for its parent, and its parent, until you reach the outside).

Examples

You start by entering 0/5 - 0/2. By continually moving clockwise across this outermost Ring, you move to 0/5 - 1/2, 0/5 - 2/2, 1/5 - 0/2, 1/5 - 1/2 and so on until you get back to 0/5 - 0/2. This is demonstrates the required property of ending up at a new entrance if you swing 360 degrees around one of the outermost trees upon entering.

You start by entering 0/5 - 0/2. From here you can go into the next Ring through 0/5 - 0/2 - 0/10, 0/5 - 0/2 - 4/10 and 0/5 - 0/2 - 8/10.

You're at 3/5 - 2/2 - 14/14 - 13/14. You move clockwise across your Ring, bringing you to 3/5 - 2/2 - 14/14 - 14/14. You move clockwise again, bringing you to 4/5 - 0/2 - 0/10 - 0/10.

Conclusion and Generalisability

We now have a coordinate system which uniquely identifies each hex in the forest, and a means by which to navigate between them.

That said, could a similar technique be used for other impossible geometries? Yes! While deciding what the entrance(s) to your very own non-euclidean forest may look like, you can replicate this coordinate system for any non-euclidean space which can be modelled as a tree (or repetition thereof), so long as all necessary details about a child can be identified by its parents. It will be easiest to do this if every space in the forest has the same number of sides and every tree the same number of vertices connecting to it, but this is far from mandatory.

For example, why not make an identical forest on a square grid? 8 squares around one tree, each square connects to 4 other squares. Or what about a hexagonal forest with 5 hexes around each tree instead of 6? If you make the number of spaces connected a single tree infinite, you could make a tree that you could keep running around forever without getting back to where you started (at this point, you can stop keeping track of limits - every child connected to its parent has an infinite number of spaces on its Ring between it and the next connected sibling, and at that point you may as well start using negative indexes as well as positive ones).

Hope this was helpful to anyone who ends up with the same very specific problem to solve as I!


问:
Are there any real-world applications for the logistic map (as defined in chaos theory)?
Is the logistic map only useful for theoretically exploring nonlinear dynamical systems, or can it be applied to real-world scenarios in any practical way?</p>

答:
Not sure if this answers your question and you must have seen this before. If you have read the Wikipedia article for this closely

This nonlinear difference equation is intended to capture two effects:

reproduction where the population will increase at a rate proportional to the current population when the population size is small.

starvation (density-dependent mortality) where the growth rate will decrease at a rate proportional to the value obtained by taking the theoretical "carrying capacity" of the environment less the current population.

This quality of unpredictability and apparent randomness led the logistic map equation to be used as a pseudo-random number generator in early computers.

This article is very helpful.

If you only want the practical applications then skip ahead to the Butterfly Effect in this article.

PS. The notes after each horizontal rule are not continuations of each other but rather excerpts from mentioned articles.


问:
When I graph $y=x^{1/2}$ why does it only show the positive y values.
I understand the reason for $y=\sqrt{x}$. I've been told that the radical symbol gives out the positive answer. But $y=x^{1/2}$ doesn't use a radical symbol, and it still only shows the positive y values.

答:
It would help to know what software package you are using here.

I tried it myself using GeoGebra, and I had the same as you: entering the formula "y=x^(1/2)" in the formula entry field did indeed yield a graph with positive $y$ only.

However, when I entered "y^2 = x" in that field, it generated both branches of the relation.

There is considerable confusion and individual taste as to exactly what is meant by $x^{1/2}$.

When in the field of complex analysis, the understanding is:
$$z^{1/n} = \{r^{1/n} (\cos(\theta + 2 \pi k / n) + i \sin (\theta + 2 \pi  k / n)): k \in \{0,1,2,…,n−1\}\}$$

from which, restricting to the real number line, you would expect $x^{1/2} = \pm \sqrt x$.

But it seems that this may not be the generally understood case. I don't know why this is and I disagree with it, as a philosophical position. But don't take my word for it, as my opinions and ideas differ from those of mainstream mathematics in plenty of ways.


问:
What is the definition of a "Circular Wedge"?
In Ahlfors' Complex Analysis, chapter 3, section 4, the author claims that a region whose boundary consists of two circular arcs with common end points is either a "circular wedge" or its complement, he also claims that "its angle may be greater than $\pi$". Later on he refers to the complex plane with the interval $[-1,1]$ removed a "wedge with the angle $2\pi$"

My question is, what is exactly a circular wedge (or an non-circular wedge)? and how do you measure its angle?

答:
A circular wedge is a domain whose boundary consists of two arcs of line/circles (the term that can mean either a circle or a line) with common endpoints $a$ and $b$, called vertices. The arcs are allowed to coincide.

The fractional linear transformation $z\mapsto \dfrac{z-a}{z-b}$ transforms the boundary into the union of two half-lines that go from $0$ to $\infty$. The domain itself becomes a sector of opening angle $\alpha\in (0,2\pi]$. Observe that the angle is the same at $0$ and at $\infty$. Since conformal maps preserve angles, the original wedge domain also had angle $\alpha$ at each vertex.

An example of conformal transformation of a circular wedge domain can be found in https://math.stackexchange.com/q/403075/79365.

I don't think there is a formal notion of a "non-circular wedge".


问:
Finding Nth term using a generating function
Good day everyone

I am in need of help, I was using an online program to solve a sequence, the name of the program is WolframAlpha and using it I was able to get the actual nth term that I was looking for several times. But in order to get that nth term the program used the following generating function

$$\frac{ z(-34z  ^ { 2  }  +77z-41)  }{ (z-1)  ^ { 3  }    }  +25$$

Note my sequence was 25,32,36,39,41 and gave me nth term 42 which is what I actually needed.
Now I need to find out how did the program get to find the nth using this generating function, I don't understand how -34z and 77z came to be.

Thank you while I wait for the answer

答:
I think perhaps the fastest way to answer this question is to note that
\begin{align*}
f(z) = \frac{z(-34z^2 + 77z - 41)}{(z-1)^3} + 25 = -9 + \frac{25}{1-z} + \frac{11}{(1-z)^2} - \frac{2}{(1-z)^3}
\end{align*}

The coefficient of $z^n$ in the series expansion of  $(1-z)^{-1}, (1-z)^{-2}, (1-z)^{-3}$ are respectively $1, n+1, \frac{(n+1)(n+2)}{2}$. Therefore, the coefficient of $z^n$ in $f(z)$ is
\begin{align*}
-9\cdot\mathbb{I}_{\{n=0\}} + 25+ 11(n+1) - 2\frac{(n+1)(n+2)}{2}
\end{align*}
where $\mathbb{I}$ is the indicator function.


问:
How do you solve $5\sin^2(x)+8\sin x \cos x-3=0$?
I have tried using compound angle, sum to product, pythag identities but nothing seems to work. I tried turning it into $\sin 2x$ but then I have $\sin^2(x)$ and $\sin 2 x$ together.

Body:
Hint:

$${5\sin^2(x)\over\cos^2(x)}+{8\sin(x)\over\cos(x)}-{3\sin^2(x)+3\cos^2(x)\over \cos^2(x)}=0$$

$$2\tan^2(x)+8\tan(x)-3=0$$


问:
Prove $|f(x)-f(y)|<\varepsilon \implies \sup f - \inf f \leq \varepsilon$
$f$ is continuous on $[a,b]$ and $|f(x)-f(y)|<\varepsilon$. Show that $\sup f - \inf f \leq \varepsilon$

Suppose that $\sup f - \inf f > \varepsilon$.

$|f(x)|-|f(y)| \leq |f(x)-f(y)| < \varepsilon < \sup f - \inf f \Rightarrow |f(x)|-|f(y)| < \sup f - \inf f$

It follows that $\begin{cases}|f(x)|<\sup f\\-|f(y)|<-\inf f \end{cases} \Rightarrow \begin{cases}|f(x)|<\sup f\\|f(y)|>\inf f \end{cases}$

In particular $f(x)<\sup f$ for all $x \in [a,b]$. But $f(x)$ has a maximum on $[a,b]$ since it's continuous and $\max(f)<\sup f$, hence $\sup f$ is not a least upper bound. I've derived a contradiction, so $\sup f - \inf f \leq \varepsilon$.

Not sure if this proof is correct. Is there a better one?

答: A bettter proof: $f(x)-\epsilon <f(y)$ . Take sup over $x$ to get $\sup f -\epsilon \leq f(y)$. This is true for all $y$. Take inf over $y$ to get $\sup f \leq \inf f+\epsilon$.


问:
Constructing Very "Flat" Functions
Consider the function $f(x)=\tan(\sin(x))-\sin(\tan(x))$. It has (what I think is) a remarkable property. The Taylor series expansion of $f(x)$ around $x=0$ is
$$\frac{1}{30}x^7+\frac{29}{756}x^9+\frac{1913}{75600}x^{11}+o(x^{13}).$$

Since $\tan(x)$ and $\sin(x)$ are odd functions, $f(x)$ is also an odd function. But its first nonzero term is $x^7$. This tells us that that $f(x)$ is very "flat" near the origin, which it is indeed.

I haven't calculated an explicit closed form for the series coefficients but looking at the wild oscillations around $x=\pm \pi/2$ (and confirming it by looking at it in the complex plane), the radius of convergence is $\pi/2$. This is simply because of the tangent function.

I have the following questions.

Is this just a coincidence or is there something deeper going on here? Is there any way to build/find more such examples? How can we find two real functions $g_1(x)$ and $g_2(x)$ such that

they are both distinct,

they are both elementary and can be explicitly written in closed form,

and $g_1(g_2(x))-g_2(g_1(x))$ is $o(x^n)$ for some $n>1$ with a strictly positive radius of convergence.

Neither should be a polynomial because this is trivial to do with polynomials. So the Taylor expansion of $g_1(x)$, $g_2(x)$, and $g_1(g_2(x))-g_2(g_1(x))$ should be infinite.

Addressing Calvin Khor's comment

Is it really trivial with polynomials? The first few terms of the Taylor series only depend on the first few derivatives, so it seems to me that understanding how to do it for polynomials should help understand what's happening in your example

Yes, it is trivial with polynomials. Consider
$$g_1(x)=x^{11}+x^{13}$$
$$g_2(x)=x^{15}+x^{17}$$
then
$$g_2(g_1(x))-g_1(g_2(x))=4 x^{167}+50x^{169}+290x^{171}+\cdots$$
which is indeed a finite series with an infinite radius of convergence, because it is only a composition of polynomials. So it is trivial to build such a (finite series) example for any desired $o(x^n)$.

I was looking for a general construction for any given $n$, so that the series is infinite and can be written in explicit closed form using elementary functions, or a proof that it is not possible.

答:
Summary

$O(x^7)$ is only one power more surprising than most $O(x^3)$ analytic perturbations of the trivial pair $g_1(x)=g_2(x)=x$, which give $O(x^6)$ errors. Some distinct pairs $g_1,g_2$ give exact cancellation. The extra power is explained by the fact that $\sin$ and $\tan$ are odd functions (implying the vanishing of the $x^2$ and $x^4$ terms in their expansion.)

Perfect pairs

There is a pair of distinct elementary functions for which the diffference of the two compositions exactly vanishes:
\begin{align} g_1(x)&=\log(1+x),\\ g_2(x) &= e^x-1.\end{align}
as then $$g_1\circ g_2 = \log e^x = x = e^{\log(1+x)}-1 = g_2\circ g_1.$$
Of course, any function-inverse pair will work, like $2\sin(x)$ and $\arcsin(\frac12 x)$. The factor of $2$ shows that you don't need to be close to the identity map itself as guessed in Yves' answer, but this of course still requires the existence of an inverse.

Analysis of a general case

Aside from these perfect pairs,
the following basic calculation gives examples of arbitrary rates of vanishing at $x=0$.

Let $h_1,h_2$ be distinct (elementary, if you want) analytic functions around $0$ whose first $N$ taylor coefficients at $0$ vanish. Then let's consider $$g_1(x) = x+h_1(x), \\ g_2(x) = x+h_2(x).$$
For instance, $h_i$ could be $\tan$ and $\sin$ minus their respective $N$th Taylor polynomials. The tan-sin example you give is of this form with $N=2$ (since their $x^2$ coefficient is zero).

Their compositions are
$$ g_1(g_2(x)) = x+h_2(x) +h_1(x+h_2(x)),$$
$$ g_2(g_1(x)) = x+h_1(x) +h_2(x+h_1(x)).$$
and the subtraction is
$$\Delta:=g_1\circ g_2(x) - g_2\circ g_1(x) = \bigg(  h_1(x+h_2(x))- h_1(x) \bigg) - \bigg(h_2(x+h_1(x))- h_2(x)\bigg). $$
Taylor's theorem gives on a sufficiently small ball around $0$,
$$ h_i(x+h_j(x)) = h_i(x)+h_i'(x)h_j(x)+O(h_j(x)^2),$$
so
$$\Delta =h_1'(x)h_2(x) - h_2'(x)h_1(x) + O(h_1(x)^2+h_2(x)^2).$$
Since these terms vanish to high order （specifically, to order $O(x^{2N+1})$ the vanishing to high order is proven.

Optimal degree of vanishing

For the tan-sin example, this only proves $O(x^5)$, which is worse than the observed $O(x^7)$. In general, the order of vanishing can be improved by a small calculation to $O(x^6)$ (i.e. $O(x^{2N+2})$ in general) but no better. If you further write
$$ h_i(x) = a_i x^{N+1}+ O(x^{N+2})$$
then observe that $$ h_1'(x)h_2(x) = (N+1)a_1 a_2 x^{2N+1} + O(x^{2N+2}) $$
which has the same lowest order term as $h_2'(x)h_1(x)$. So these perfectly cancel, leaving you with an $O(x^{2N+2})$ error term as promised.

This is sharp in the class of functions I considered above, as can be seen by the example
$$ g_1(x)=e^x-1-\frac{x^2}2, \quad g_2(x) = \sin x + x^4, \quad \Delta = \frac{25}{144}x^6 + O(x^7).$$

Proving the $O(x^7)$ bound

Just a little more elbow grease explains the $O(x^7)$ term.

If you used the higher order Taylor expansion
$$ h_i(x+h_j(x)) = h_i(x)+h_i'(x)h_j(x)+h_i''(x)h_j(x)^2/2+O(h_j(x)^3),$$
we see that the terms $h_i''(x)h_j(x)^2/2+O(h_j(x)^3)$ only contribute to high order error terms $O(x^{2N+3})$. So the only term that could be only $O(x^{2N+2})$ is the term $h_i'(x)h_j(x)$. Expanding this to one more term than before:
$$ h_i(x) = a_i x^{N+1}+ b_ix^{N+2} +O(x^{N+3}),$$
we see
$$h_1'h_2(x)= (N+1)a_1a_2 x^{2N+1} + ((N+1)a_1b_2 + (N+2)a_2b_1)x^{2N+2} + O(x^{2N+3}).$$
Note that the second coefficient $(N+1)a_1b_2 + (N+2)a_2b_1$ is not cancelled when we subtract $h_2'(x)h_1(x)$, in contrast with the lower order term. So disregarding miracles like the perfect inverse pairs, we basically need $b_1,b_2=0$ to remove the  $x^{2N+2}$ term. This is precisely the case for the sin-tan example, as (recall $N=2$) there are no $bx^{N+2}=bx^4$ terms in the expansions of $\sin$ and $\tan$.

Title: Constructing Very "Flat" Functions


问:
Upper bound on chromatic number characterization with longest directed path
"Show that $\chi (G)\leq k$ if and only if $G$ admits an orientation such that the longest path has length at most $k-1$."

The greedy algorithm may be helpful for proving one direction of the equivalency. I would like to adapt Brooks' theorem, but so far I have not come up with anything relevant.

答:
Suppose $D=(V,E)$ is a directed graph containing no (simple) directed path of length $k$; I will show that $\chi(D)\le k$.

Let $D_0=(V,E_0)$ be a maximal acyclic spanning subgraph of $D$.

Define $f:V\to\{0,\dots,k-1\}$ by letting $f(v)$ be the maximum length of a directed path in $D_0$ with initial vertex $v$. I claim that $f$ is a proper coloring of $D$, i.e., that $uv\in E\implies f(u)\ne f(v)$.

First note that, if $uv\in E_0$, then $f(u)\gt f(v)$, since $D_0$ is acyclic.

Now suppose $uv\in E\setminus E_0$. Because of the maximality of $D_0$, there is a path from $v$ to $u$ in $D_0$, say $v=v_0,v_1,v_2,\dots,v_n=u$. Then we have $f(v)=f(v_0)\gt f(v_1)\gt f(v_2)\gt\cdots\gt f(v_n)=f(u)$, i.e., $f(u)\lt f(v)$.