问:
How to find parametric curve from differential equations
How would you find the parametric curve for this set of parametric differential equations: $$x'=\frac{1}{y},y'=2xy$$
I tried dividing the second equation by the first and treating it as a normal differential equation. However, this results in a normal cartesian equation which can have multiple parametric representations, some of which don't satisfy the equations. How can I solve these equations while preserving their parametric nature?

Body:
$$x''=-\dfrac {y'}{y^2}=-\dfrac {2x}y$$
$$x''=-2xx'$$
$$x''=-(x^2)'$$
Integrate.
$$x'=-x^2+C$$
This is separable.


问:
How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$?
The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate  the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$

答:
In polar form,
$$x = 35+18 i \sqrt{3} = \sqrt{13}^3 \cdot \exp\left( i \cdot \rm{atan} \frac{18\sqrt{3}}{35}\right)$$
so that

$$\sqrt[3]{x} + \sqrt[3]{\overline{x}} = 2 \sqrt{13} \; \cos\left(\frac{1}{3} \rm{atan} \frac{18\sqrt{3}}{35} \right)$$

which magically evaluates to $7$ in a calculator.

(I know, I know: It's not appropriate to call any one value the cube root of another. Be that as it may ...)

To verify that the calculation is exact, we'll seek $T = \tan\theta$ such that $\tan 3\theta = 18\sqrt{3}/35$.

Invoking the triple-angle tangent formula, we have

$$\tan3\theta = \frac{T ( 3 - T^2 )}{1 - 3 T^2} = \frac{18\sqrt{3}}{35}$$

If I were convinced that $T$ is a relatively nice number, I'd be willing to believe that the $T$ in the left-hand numerator accounts for the $\sqrt{3}$ in the right-hand numerator. Write $T := S \sqrt{3}$, so that:

$$\frac{S \; ( 1 - S^2 )}{1 - 9 S^2 } = \frac{6}{35}$$

Now, note that (negative) $35$ is just one away from (negative) $36$, a perfect square (and multiple of $9$, to boot), so that comparing denominators suggests $S = \pm 2$. By happy coincidence, $S=2$ gives precisely the numerator we'd need.

Therefore, $\tan\theta = T = S\sqrt{3} = 2 \sqrt{3}$, so that $\cos\theta = \frac{1}{\sqrt{1 + T^2}} = \frac{1}{\sqrt{13}}$. This gives

$$\sqrt[3]{x} + \sqrt[3]{\overline{x}} = 2 \sqrt{13} \cos\theta = 2$$

and we're don--- hey, waydaminnit ... $2\neq 7$! What happened?

Well, notice that we actually solved
$$\frac{S(1-S^2)}{1-9S^2} = \frac{-6}{-35} = \tan(3\theta+\pi)$$
("Wait, what's the difference?" "Just keep reading.") so that we actually got
$$T = \tan(\theta+\pi/3) = 2\sqrt{3}$$

That is, what we took for $\theta$ was in fact $\theta+\pi/3$. We can get at the $\theta$ we wanted thusly:

$$\tan\theta = \tan\left((\theta+\pi/3)-\pi/3\right) = \frac{T-\tan\frac{\pi}{3}}{1+T\tan\frac{\pi}{3}} = \frac{2\sqrt{3}-\sqrt{3}}{1+2\sqrt{3}\cdot\sqrt{3}} = \frac{\sqrt{3}}{7}$$
so that $\cos\theta = \frac{7}{2\sqrt{13}}$ and therefore $\sqrt[3]{x} + \sqrt[3]{\overline{x}} = 2 \sqrt{13} \cos\theta = 7$, as expected.

So, why the $2$ before?

When we solved using $\frac{-6}{-35}$ instead of $\frac{6}{35}$, we implicitly replaced $x$ ---a complex number having positive real and imaginary components--- with $-x$. (Recall that $\tan 3\theta$ is the ratio of these components. It can't distinguish $x$ from $-x$.) Consequently, the $T$ we first obtained corresponds to the relation:
$$\sqrt[3]{-x} = \sqrt{13} \left( \frac{1}{\sqrt{13}} + i \frac{2\sqrt{3}} {\sqrt{13}} \right) = 1 + 2 i \sqrt{3}$$

which is to say: $(1+2i\sqrt{3})^3 = - \left( 35 + 18 i \sqrt{3}\right)$.

Then, since the arguments of $x$ and $-x$ differ by $\pi$, the arguments of the calculated cube roots must differ by $\pi/3$, and our second computation gives $\theta$ corresponding to this result:
$$\sqrt[3]{x} = \sqrt{13} \left( \frac{7}{2\sqrt{13}} + i \frac{\sqrt{3}}{2\sqrt{13}}\right) = \frac{1}{2}\left( 7 + i \sqrt{3} \right)$$

ContentLicense: CC BY-SA 4.0

ContentLicense: CC BY-SA 3.0


问:
Why does the Discrete Fourier transform involve the $nth$ data in the transform?
Currently, I'm a 3rd year university student. I don't have a strong background in real analysis or differential equations so please bear with me. I'm trying to learn about the Discrete Fourier transform (DFT) and I came across this video.

I know that a Discrete Fourier transform is essentially decomposing a function $f(x)$ into $N$ sinusoidals where each one has a frequency of $k\omega$ where $\omega$ is some fundamental frequency and $k$ is an integer multiple.

When we're looking at the discrete Fourrier transform with $N$ data points, the $kth$ Fourrier coefficent is computed using the formula

$$\hat{f_k} = \sum_{n=0}^{N-1}f_ne^{-2\pi i k n/N}$$

where $0 \leq k \leq N-1$.

From my understanding $\hat{f_k}$ is a complex value where the magnitude of the complex number gives me the amplitude of the sinusoidal having frequency $k\omega$ and the angle of the complex number is the phase shift of the sinusoidal used in the decomposition of $f(x)$.

The things that confuses me about the formula $\hat{f_k} = \sum_{n=0}^{N-1}f_ne^{-2\pi i k n/N}$ is the fact that $n$ is involved in the exponential. By Euler's formula we know that $$e^{-2\pi i k n/N} = \cos(-2\pi k n/N) + i\sin(-2\pi k n/N)\\ = \cos(2\pi k n/N) - i\sin(2\pi k n/N).$$ Then from this we see that that when we are computing the $nth$ term of $\hat{f_k}$, the frequency inside the sinusoidal is also changing. I wouldv'e expected that the frequency remains at $2\pi k/N$ over all summations instead of $2\pi kn/N$. If the frequency is changing at each term of the summation, how does this formula give the total contribution for frequency $k\omega$.

答:
Your intuition about the discrete Fourier transform seems to be
confusing you. I suggest the following intuition about DFT. Fix
the positive integer $N$. Define the roots of unity
$$ \zeta_k = \zeta_{k,N} := e^{2\pi i k/N} = \zeta_1^k.\tag{1} $$ Define the sequence of functions
$$ z_k(n) := \zeta_k^n \quad\text{ where }\quad 0\le k<N. \tag{2} $$
Check that each $\,z_k(n)\,$ function for $k>0$ is sinusoidal with
period $N$ and frequency $\,k\,\omega=k/N.\,$

Check that the average value of $\,z_k(n)\,$ is $0$ if $\,k>0\,$
while $\,z_0(n)\,$ is the constant value $1$.
Expressed as an equation this is
$$ \delta_{0,k} = \frac1N\sum_{n=0}^{N-1} z_k(n)\quad \forall k. \tag{3}$$
Check that the $\,z(.)\,$ functions satisfy
$$ z_{j+k}(n) = z_j(n)z_k(n)\quad \forall j,k. \tag{4}$$
Assume that we have a sequence that is a sum of sinusoids
$$ f_n := \sum_{k=0}^{N-1} c_k\,z_k(n). \tag{5} $$
Define the DFT of $\,f\,$ with
$$ \hat{f_k} := \frac1N \sum_{n=0}^{N-1}f_nz_{-k}(n) \\
= \frac1N \sum_{n=0}^{N-1}\left( \sum_{j=0}^{N-1} c_j\,z_j(n)\right)z_{-k}(n)\\
 =  \sum_{j=0}^{N-1}c_j \left(\frac1N \sum_{n=0}^{N-1}\,z_{j-k}(n) \right) = c_k. \tag{6}$$
Thus the coefficients $\,c_k\,$ in equation $(5)$ are exactly
the values of the DFT $\,\hat{f_k}\,$ defined in equation $(6)$.


问:
kinetic energy formula
I cannot understand the derivation of the formula for potential energy ($\frac{1}{2}mv^2$).
It is my understanding that mathematically the work done by a force $F$ should be the line integral of $F$ over a curve $\gamma$. So the force must be a funtion $F: \gamma \rightarrow \mathbb{R}^3$. Instead, my book and the wikipedia page treat $F$ as a function of time, which does not make sense to me.
(I read the whole Analysis on manifolds by Munkres, so feel free to explain by means of manifolds and differential forms)

答: Treating $F$ as a function of time is simply defining $F$ in terms of 'when' instead of 'where', which is what you have suggested. This is better since $\gamma$ might cross itself, i.e. be at the same place at two different times, and you might experience two different forces at this position.

Title: kinetic energy formula


答:
Let $F$ be the force vector and let $\gamma$ be the motion of the particle from time $t=a$ to $t=b$. The total work is
$$
W=\int_a^b F(\gamma(t))\cdot \dot{\gamma(t)} dt
$$
Force is the old mass time acceleration so
$$
F(\gamma(t))=m\ddot{\gamma}
$$
Plugging this into the above formula gives
$$
W=\int_a^b m\ddot{\gamma}\cdot \dot{\gamma(t)} dt
$$
Doing an integration by parts gives
$$
\begin{align}
\int_a^b m\ddot{\gamma}\cdot \dot{\gamma(t)} dt=(m|\dot{\gamma}|^2)\big|_a^b-\int_a^b m\ddot{\gamma}\cdot \dot{\gamma(t)} dt
\end{align}
$$
The same integral (the total work) appears on both sides of the equation so
$$
2\int_a^b m\ddot{\gamma}\cdot \dot{\gamma(t)} dt=(m|\dot{\gamma}|^2)\big|_a^b
$$
Substituting the speed $v=|\dot{\gamma}|$ and dividing by 2 gives
$$
W=\int_a^b m\ddot{\gamma}\cdot \dot{\gamma(t)} dt=\frac{1}{2}mv\big|_a^b=\frac{1}{2}m(v(b)^2-v(a)^2)
$$
If we assume that the initial speed of the object was zero ($v(a)=0$) and let $V=v(b)$, we have that the total work from time $t=a$ to $t=b$ is
$$
W=\frac{1}{2}mV^2.
$$

Title: kinetic energy formula

Body:
I cannot understand the derivation of the formula for potential energy ($\frac{1}{2}mv^2$).
It is my understanding that mathematically the work done by a force $F$ should be the line integral of $F$ over a curve $\gamma$. So the force must be a funtion $F: \gamma \rightarrow \mathbb{R}^3$. Instead, my book and the wikipedia page treat $F$ as a function of time, which does not make sense to me.
(I read the whole Analysis on manifolds by Munkres, so feel free to explain by means of manifolds and differential forms)


问:
Show that $A=\{0^k 1^m 0^n |\text{ }n=k+m\}$ has infinitely many equivalence classes.
Show that $A=\{0^k 1^m 0^n |\text{ }n=k+m\}$ has infinitely many equivalence classes.

I'm trying to prove that $A$ is a non-regular language. Alternatively, I could use closure properties - but have been struggling.

答:
Consider the string $S_k = 0^k$. Consider the string $T_k = 1^k$.

Note that for all positive integers $j, k$, we have $S_k T_j \in A$ iff $k = j$.

Therefore, each $S_k$ is in a different equivalence class (since if $k \neq j$, then $S_k T_k \in A$ but $S_j T_k \notin A$).

But a regular language can only have finitely many equivalence classes. So $A$ is not a regular language.


问:
Compute $E[\alpha]$ where $\alpha = \inf\{i > 0 : X_i > X_0\}$ and $\{X_i\}_{i=0}^{\infty}$ are IID.
If $\{X_i\}_0^{\infty}$ are IID discrete, then I'm quite sure $E[\alpha] = \infty$ because
$$
E[\alpha] = \sum_{\{x_j \}_{j=0}^{N}} E[\alpha | X_0 = x_j] P(X_0 = x_j) = \infty = E[\alpha | X_0 = \sup\{x_j \}_{j=0}^{N}]\\
$$

In this set of values, you will eventually condition on $X_0$ being the maximum value, and as long as the probability of obtaining that maximum value is non-zero, then $E[\alpha | X_0 = \sup\{x_j \}_{j=0}^{N}]$ is necessarily $\infty$ because you will never find another $X_i$ such that $i > 0$ and $X_i > X_0$.

Now what if $\{X_i\}_0^{\infty}$ are IID continuous. I feel the same logic should apply, but for continuous distributions, there should be a probability of zero that $X_0$ realizes the maximum value (or any value), which would seem to mean $E[\alpha]$ is finite.

Is this where the "almost surely" comes in? Which would mean we could assume that there is a non-zero probability that $X_0$ realizes the maximum value of the continuous distribution, in which case the logic applied in the discrete case will still apply here.

答:
Assume $X_i$ are IID. Then
$$P(\alpha \ge k) =P \bigl(X_0=\max\{X_0,X_1, \ldots,X_{k-1}\} \bigr) \ge 1/k$$ by symmetry. (If the variables are continuous, then the rightmost inequality is an equality.)
Therefore,$$E(\alpha)=E \left[\sum_{k=1}^\infty {\mathbf 1}_{\{k \le \alpha\}} \right]=
\sum_{k=1}^\infty  P(\alpha \ge k) \ge \sum_{k=1}^\infty 1/k =\infty \,.$$


问:
If $\mathfrak{g} := Lie(G) = V \oplus W$, is it true that some neighborhood of the identity in $G$ is contained in $\exp(V) \exp(W)$?
I apologize if this has been asked before, and would appreciate a reference.

Let $G$ be a real Lie group (if it matters, I'm satisfied with $G$ being a matrix Lie group, say $SL_n(\Bbb{R})$). Suppose that the Lie algebra of $G$ can be written as a direct sum $\mathfrak{g} = V \oplus W$ of vector subspaces. Is it true that there exists an open neighborhood of the identity that is contained in $\exp(V)\exp(W)$? In other words, can I always find an open ball $B_r(e) \subset G$ of identity so that every element in the ball can be written as $\exp(v) \exp(w)$ for some $v \in V$ and $w \in W$?

If not, is it true if in addition, $W$ is a Lie subalgebra of $\mathfrak{g}$? If not, is it true if in addition, $W$ is a nilpotent Lie subalgebra of $\mathfrak{g}$?

Again I'm okay with a solution for $G = SL_n(\Bbb{R})$, though a proof for the general case would be wonderful. Thanks a lot.

答:
Define $\psi: V \oplus W\cong V \times W\to G$ by $\psi(X,Y)=\exp(X)\exp(Y)=\rho(\exp(X),\exp(Y))=\rho(\eta(X,Y))$, where $\eta(X,Y)=(\exp(X),\exp(Y))$.  Then, $\psi$ is differentiable and
\begin{eqnarray}
d(\psi)_{(0,0)}&=&[d(R_{\exp(0)})_{\exp(0)}  \quad d(L_{\exp(0)})_{\exp(0)}]=\operatorname{Id}.
\end{eqnarray}

Therefore, from the Inverse Function Theorem, there exists open nbhd's $U_1 \subset V$, $U_2 \subset W$ and $G'\subset G$ a neighborhood of $1 \in G$ such that $\psi:U_1 \times U_2 \to G'$ is a diffeomorphism.


问:
Topological spaces similar to but not actual manifolds
I'm looking for topological spaces that are similar to but not actual manifolds.
Is there a way to construct topo. spaces that are locally homeomorphic to, say, $S^n$ (rather than $\mathbb R^n$)? By this I mean that every point in my space $X$ has an open neighborhood which is homeomorphic to (in this example) $S^n$.

For example, I can think of gluing circles into a torus where the open sets are the circles that get glued together.
In this case, every point has an open neighborhood (which is $S^1$ by construction), which is not homeomorphic to $\mathbb R$. Note that in this topology, small "patches" are not open sets, only combinations of "cylindrical slices" of the torus. Therefore, there is no open set in my $T^2$ that is homeomorphic to $\mathbb R^n$.

However, this space is not Hausdorff (not even $T_0$), which is a property that I would like to preserve.

Is it possible to construct such Hausdorff spaces?
How similar would they be to typical manifolds?

答:
Every such space is homeomorphic to the product of $S^n$ and a discrete space, i.e. a disjoint union of spheres, in particular, a manifold in the common sense.

The moral: Use noncompact models for your open subsets (if you want to work with Hausdorff spaces).

But, there is an interesting variation on the theme of manifolds where local models are compact (say, spheres of a fixed dimension): Replace open subsets by closed subsets. Thus, you require your space to be a union of (overlapping) topological spheres. To make things more interesting, you require the transition maps to belong to some small subgroup $G$ of the group  of homeomorphisms of the sphere. There is a remarkable class of spaces obtained this way if $G$ is a finite reflection group. The spaces are called buildings. They first appeared in the classical projective geometry and the general theory was developed by Jacques Tits.


答:
There's always going to be an $n$ in the exponent of the summation. Conceptually, the Fourier transform evaluates the inner product of $\{f_n\}_{n=0}^{N-1}$ with the complex exponential $\{e^{-i \omega n}\}_{n=0}^{N-1}$, where $\omega$ is the frequency of interest:
$$\tilde{f}(\omega) = \sum_{n=0}^{N-1} f_n e^{-i\omega n}$$
where $\tilde{f}$ denotes the continuous-frequency version of the Fourier transform. This is a $2\pi$-periodic function of the real-valued frequency $\omega$.

If there were no $n$ in the exponent, then you would have $e^{-i\omega}$, which is just a constant that can be pulled out of the summation, and you would end up with
$$e^{-i\omega}\sum_{n=0}^{N-1}f_n$$
which isn't very interesting!

The discrete Fourier transform evaluates $\tilde{f}$ at specific values of $\omega$, namely $\omega = 2\pi k/N$ for $k=0,1,\ldots N-1$. This is just a uniform sampling of the interval $[0,2\pi)$.
$$\hat{f}_k = \tilde{f}(2\pi k / N) = \sum_{n=0}^{N-1} f_n e^{-i(2\pi k/N) n} = \sum_{n=0}^{N-1} f_n e^{-i2\pi k n / N}$$

Title: Why does the Discrete Fourier transform involve the $nth$ data in the transform?

Body:
Currently, I'm a 3rd year university student. I don't have a strong background in real analysis or differential equations so please bear with me. I'm trying to learn about the Discrete Fourier transform (DFT) and I came across this video.

I know that a Discrete Fourier transform is essentially decomposing a function $f(x)$ into $N$ sinusoidals where each one has a frequency of $k\omega$ where $\omega$ is some fundamental frequency and $k$ is an integer multiple.

When we're looking at the discrete Fourrier transform with $N$ data points, the $kth$ Fourrier coefficent is computed using the formula

$$\hat{f_k} = \sum_{n=0}^{N-1}f_ne^{-2\pi i k n/N}$$

where $0 \leq k \leq N-1$.

From my understanding $\hat{f_k}$ is a complex value where the magnitude of the complex number gives me the amplitude of the sinusoidal having frequency $k\omega$ and the angle of the complex number is the phase shift of the sinusoidal used in the decomposition of $f(x)$.

The things that confuses me about the formula $\hat{f_k} = \sum_{n=0}^{N-1}f_ne^{-2\pi i k n/N}$ is the fact that $n$ is involved in the exponential. By Euler's formula we know that $$e^{-2\pi i k n/N} = \cos(-2\pi k n/N) + i\sin(-2\pi k n/N)\\ = \cos(2\pi k n/N) - i\sin(2\pi k n/N).$$ Then from this we see that that when we are computing the $nth$ term of $\hat{f_k}$, the frequency inside the sinusoidal is also changing. I wouldv'e expected that the frequency remains at $2\pi k/N$ over all summations instead of $2\pi kn/N$. If the frequency is changing at each term of the summation, how does this formula give the total contribution for frequency $k\omega$.


问:
How Many Ways to Choose Dance Partners if Choosing Both Men and Women?
I was working through supplementary exercises in Grimaldi, and I came across a question about dance partners--it asked how to choose a clean-up team of 8 from a group of 12 men, and then asked how many different ways to pair up 8 women from those 12 men.  The answer was simple enough: find the number of permutations of size 8 from a set of 12 men (i.e., $\frac{12!}{4!})$.  But that got me wondering, if we want to choose both the men and the women from a set of 12 each, how does the answer change?  Since I'm completely new to self-studying math, I thought I'd pose the question to see if my approach to answering the question is valid (I thought it might still have something to do with permutations, but I wasn't sure).

So let's say there are 12 men and 12 women at a dance.  How many ways are there to pair off 8 men and 8 women as dance partners?

答:
Let us simplify the question.  Imagine you have to choose 2 pairs from 3 pairs of men  and woman (M1-M3,W1-W3) to form partners.

Step 1:

Chose 2 from  3 men,  $\binom{3}{2}$ =3 choices.

Step 2:

Chose 2 from  3 women, $\binom{3}{2}$ =3 choices.

Step 3:
Then to pair them $\to$

```<code>If no need  to line them up, 3x3 =  9

If need line them up, 3x3x2! =18
</code>```

Now, we can answer the  original question.

Body:
Let us simplify the question.  Imagine you have to choose 2 pairs from 3 pairs of men  and woman (M1-M3,W1-W3) to form partners.

Step 1:

Chose 2 from  3 men,  $\binom{3}{2}$ =3 choices.

Step 2:

Chose 2 from  3 women, $\binom{3}{2}$ =3 choices.

Step 3:
Then to pair them $\to$

If no need  to line them up, 3x3 =  9

If need line them up, 3x3x2! =18

Now, we can answer the  original question.


问:
I don't get how the answer to this problem works, its a practice mathcon problem.
The answer to this problem is c. I don't get how that is the answer.
This is the problem

答:
Ross Millikan posted an answer long before I finished tinkering with that picture, but here is my version.

The volume in question is enclosed by the ellipse with large diameter EF, and the ellipse with large diameter GD (which is just a copy of the first ellipse, shifted up to the point when water overflows). By Cavalieri's principle, this is the same as the volume of a cylinder where the base is the ellipse with large diameter EF, and the height is $h=$ the distance from point F to the top ellipse (with large diameter GD). So, all this is less elegant, but if $t$ is the acute angle between the horizontal and AF, then the large diameter of these ellipses is the length of AF which is $\dfrac6{\sin(t)}$. The small and large radii are respectively $a=\dfrac62=3$ and $b=\dfrac6{2\sin(t)}=\dfrac3{\sin(t)}.$ The area of the ellipse is
$S=\pi ab=\pi\cdot3\cdot\dfrac3{\sin(t)}=\dfrac{9\pi}{\sin(t)}.$ The height $h$ (see picture) is $h=2\sin(t)$, so the volume in question is product of the area $S$ and the height $h$, so it is $\dfrac{9\pi}{\sin(t)}\cdot2\sin(t)=18\pi.$


问:
order of parameters in differential forms
I am completely mystified about parameters in differential forms and wherever I looked it's just glossed over.

Take for example the $2-$form $\omega = dx\land dy$ in $\mathbb{R}^n$.

Usually it's just defined like that, but given two vectors $(x,v_1), (x,v_2)$ in the tangent space at $x$, it shold be that $dx \land dy = dy \land dx$, since
$$(dx \land dy) (x)((x,v_1), (x,v_2)) = $$
$$- (dy \land dx)(x)((x,v_2)(x,v_1))=$$
$$ (dy \land dx)(x)((x,v_1)(x,v_2))=$$

According to my book, tough, $dx \land dy = - dy \land dx$.
Where am i wrong?

答:
Answering my own question.

It's a misunderstanding of the anticommutativity of the wedge product of alternating tensors. I previously thought for some reason that, given $f,g$ alternating k-tensor and m-tensor respectively on $V$, with $V$ vector space over $\mathbb{R}$,
$$(f \land g) (x_1, ..., x_k, ..., x_{k+m}) = (-1)^{km} (g \land f) (x_{k+1}, ..., x_{k+m}, x_1, ..., x_k)$$
while of course it is:
$$(f \land g) (x_1, ..., x_k, ..., x_{k+m}) = (-1)^{km} (g \land f) (x_1, ..., x_k, ..., x_{k+m})$$
so the error is in the first equality.


问:
Logarithm algebra question
The question says: Given $\log_n(9)=x$ and $\log_n(4)=y$, find $\log_n(12)$. I'm just really stuck on how to work this out.

答:
remember $log_n(ab) = log_n(a) + log_n(b) $

also $log_n(a^{c})= c\cdot log_n(a)$

Notice: $12 = 3 \cdot 4$

$$log_n(12) = log_n(4) + log_n(3)$$
We have $log_n(4)$ but not $log_n(3)$

But $3^{2} = 9$

So, $log_n(9) = log_n(3^{2}) = 2log_n(3) = x$ ,therefore $log_n(3) = \frac{x}{2}$

Therefore $$log_n(12) = log_n(4) + log_n(3) = y + \frac{x}{2}$$

Title: Logarithm algebra question


问:
Finding the remainder of $2^{2021}$ divided by $2700$
Finding the remainder of $2^{2021}$ divided by $2700.$ I can't solve this because $2700= 3\cdot 9\cdot 100$ and we cannot use Little Fermat.

答:
We have $2700 = 3^3 2^2 5^2$. We will evaluate $2^{2021}$ modulo each of $3^3$, $2^2$, and $5^2$ and use the Chinese remainder theorem to piece the result back together.

Note that $totient(3^3) = 3^2 \cdot 2 = 18$ and $2021 \equiv 5$ modulo 18, so $2^{2021} \equiv 2^5 = 32 \equiv 5$ modulo 27.

Note that $2^2 | 2^{2021}$, so $2^{2021} \equiv 0$ modulo $4$.

Note that $totient(5^2) = 5 \cdot 4 = 20$ and $2021 \equiv 1$ modulo $20$, so $2^{2021} \equiv 2^1 = 2$ modulo 25.

Now, we must piece these puzzle pieces together. Using the Euclidean algorithm, we compute that

$7 \cdot 4 - 27 = 1$

$13 \cdot 25 - 12 \cdot 27 = 1$

$25 - 6 \cdot 4 = 1$

So to compute a number which is $5$ mod 27 and 0 mod 4 and 2 mod 25, we take $5 \cdot (1 + 27) \cdot (1 + 12 \cdot 27) + (1 - 25) \cdot (1 - 13 \cdot 25)$, which evaluates to $47552 = 1652$ mod $2700$.

The final answer is therefore 1652.


问:
Probability of infection by staphylococcus aureus
Please forgive my innumeracy, but I have a question with which I am hoping someone might be able to help me.

Suppose the following be true. The chance of a prosthetic hip joint becoming infected by staphylococcus aureus is one per cent. The chance of a natural hip joint becoming infected by staphylococcus aureus is 0.1%. In other words (in case I am misusing the word 'chance'), one in one hundred people with prosthetic hip joints will become infected by staphylococcus aureus, whereas only one in one thousand people with natural hip joints will become so infected.

Now suppose that X has a prosthetic hip joint and that X's hip joint becomes infected by staphylococcus aureus.

Given only the information provided here, is it correct to say that X's hip joint probably would not have become infected but for the fact that X has a prosthetic hip joint (instead of a natural hip joint)? In other words (to make it clear what I mean by 'probably'), is it correct to say that there is a greater than 50 per cent chance that X's hip joint would not have become infected but for the fact that X has a prosthetic hip joint? Why or why not?

答:
It is tempting to think of the cases where infection and prosthetic hip joints occur as a superset of the cases where natural hip joints contract the infection. This way we could reach a result that "Only 10% of prosthetic hip joint infections would have been infected with a natural hip joint" but this is not a valid conclusion. The two groups aren't directly comparable. We'd need some data about the correlation of these percentages, which is impossible to acquire because we can't simulate alternate universes where person X who has a prosthetic hip joint doesn't have the prosthetic hip joint.

Even if we could correlate the percentages of people that get a staphylococcus aureus infection in either group, the statement "X's hip joint probably would not have become infected but for the fact that X has a prosthetic hip joint" implies a causal link. We know that correlation doesn't imply causation.

In fact, given that X was any other person with a prosthetic hip joint, they would probably not have acquired the hip joint infection. The probabilities in this case work out similarly.


问:
Recursively enumerable in a one-to-one manner
I've encountered a statement like this (I hope it's correct, if not, let me know)

The sets that can be recursively enumerated in a one-to-one manner are the infinite recursively enumerated sets.

But I don't really understand what the infiniteness condition has to do with one-to-one enumeration. First, to enumerate a set $A$ in a one-to-one manner amounts to (from what I understand) having an enumerator that outputs all the elements of $A$ (and only them) exactly once. Now, if $A$ is recursively enumerable, we have an access to the code of its enumerator. Can't we just modify that code by saying "before outputing something, check whether this 'something' had already been spit out, and if so, don't output it for a second time"? I don't see how infiniteness is used here.

答:
Our "enumerator" is assumed to have domain $\mathbb{N}$. Since there are no injections of any complexity from $\mathbb{N}$ to a finite set, the infinitude requirement is essential.

The precise theorem is:

Suppose $A$ is an infinite r.e. set. Then there is a total computable injective function $f: \mathbb{N}\rightarrow\mathbb{N}$ with $ran(f)=A$.


问:
Show that $\{\cup_{n\in K} (n, n+1]: K \subset \mathbb{Z}\}$ is a $\sigma$-algebra on $\mathbb{R}$
I am trying to prove that $\mathcal{S} = \{\cup_{n\in K} (n, n+1]: K \subset \mathbb{Z}\}$ is a $\sigma$-algebra on $\mathbb{R}$ using the definition.

I want to show two things: (1) $E \in \mathcal{S} \implies E^c \in \mathcal{S}$ and (2) If $E_1, E_2, \dots$ is a sequence of members of $\mathcal{S}$, then $\bigcup_{i=1}^\infty E_i\in\mathcal{S}$.

My attempt: Fix $K \subset \mathbb{Z}$. Let $E\in\mathcal{S}$. Then, $E=\bigcup_{n\in K}(n, n+1]$. So, $E^c = \bigcap_{n\in K} (-\infty, n]\cup (n+1, \infty)$. Now, how can I show that $E^c\in\mathcal{S}$? It seems that as $n$ increases, the intervals that constitute $E^c$ get smaller and smaller, so the intersection of all these intervals would get smaller and smaller. Also, how can (2) be proven? Would be have $2$ unions in this part?

答:
Note that $\mathbb{R} = \bigcup\limits_{n \in \mathbb{Z}} (n, n + 1]$, so the purported $\sigma$-algebra has the whole space as an element. In fact, note that every element of $\mathbb{R}$ is in exactly one of the sets of the form $(n, n + 1]$ for $n \in \mathbb{Z}$.

It follows that the complement of the set $\bigcup\limits_{k \in K} (k, k + 1]$ is the set $\bigcup\limits_{k \in \mathbb{Z} - K} (k, k + 1]$. Thus, the purported $\sigma$-algebra is closed under complements.

Finally, if we take $\bigcup\limits_{i = 1}^\infty \bigcup\limits_{k \in K_i} (k, k + 1]$, this is clearly equal to $\bigcup\limits_{k \in K} (k, k + 1]$ where we define $K = \bigcup\limits_{i = 1}^\infty K_i$. So the purported $\sigma$-algebra is closed under countable union.

Thus, the purported $\sigma$-algebra actually is a $\sigma$-algebra.


问:
Reparametrization of regular curves
I'm studying curves in $\mathbb{R}^3$, and I came up with a question that I don't manage to answer on my own, so I would like your help. Since the definitions can vary from book to book, I'll first give some definitions and conventions that my book uses:

Regular curve

Let $(\gamma,\mathbf{r})$ be a curve

$\mathbf{r}:I\subseteq \mathbb{R} \to \mathbb{R}^3 , I \text{  interval}$

$(\gamma,\mathbf{r})$ is regular iff $\mathbf{r} \in C^1(I)$ and $\mathbf{r}'(t)\neq 0 \ \forall t\in I$.

Intuitively(if I didn't misunderstood) a curve is regular if and only if $\mathbf{r}$ "walks on $\gamma$" in a continous way withouth jumping or coming back or stopping or zig-zagging, etc.
Then there is this definition:

Reparametrization

Let $(\gamma,\mathbf{r}_1)$ and $(\gamma,{\mathbf{r}_2})$ be two regular curves.

$\mathbf{r}_1:I_1\subseteq \mathbb{R} \to \mathbb{R}^3 $

$\mathbf{r}_2:I_2\subseteq \mathbb{R} \to \mathbb{R}^3 $

$(\gamma,{\mathbf{r}_2})$ is a reparametrization of  $(\gamma,{\mathbf{r}_1})$ if there exists a bijection:

$\varphi:I_2 \to I_1$

$\varphi \in C^1(I_2)$

$\varphi'(t)\neq 0 \ \forall t \in I_2$

Such that $\mathbf{r}_2=\mathbf{r}_1 \circ \varphi$

Clearly the properties of $\varphi$ are imposed by the fact that $\mathbf{r}_2$ has to be regular. Now I can start to come to my question.Let's proceed step by step.
I wondered if two regular curves $(\gamma,\mathbf{r}_1)$, $(\gamma,\mathbf{r}_2)$ are always one the reparametrization of the other. And I come up to a negative answer. In fact I can take:
$$\mathbf{r}_1(t)=(\cos(t),\sin(t)) \ \ t\in [0,2\pi]$$
$$\mathbf{r}_2(t)=(\cos(t),\sin(t)) \ \ t\in [0,4\pi]$$
And this curves are not one the reparametrization of the others, because if they were they should pass the same number of times on each point of $\gamma$ in fact if $\mathbf{r}_2=\mathbf{r}_1 \circ \varphi$, let $P$ be a point of $\gamma$:
$$|\mathbf{r}_2^{-1}(P)|=|(\mathbf{r}_1 \circ \varphi)^{-1}(P)|=|\varphi^{-1}(\mathbf{r}_1^{-1}(P))|=|\mathbf{r}_1^{-1}(P)|$$
Where the last step is justified because $\varphi^{-1}$ is a bijection so it preserves cardinalities.
So I tried to solve this problem by imposing another hyphothesis.

Conjecture

Let $(\gamma,\mathbf{r}_1)$ and $(\gamma,{\mathbf{r}_2})$ be two regular curves.

$|\mathbf{r}_1^{-1}(P)|=|\mathbf{r}_2^{-1}(P)|<\aleph_0 \ \forall P \in \gamma$.

$\Rightarrow $ $(\gamma,\mathbf{r}_1)$, $(\gamma,\mathbf{r}_2)$ are one the reparametrization of the other

However I think that I found a counterexample also to this one, even though it's just a sketch:

(The arrows indicate how each parametrizations walks on $\gamma$, the red point is the start point the coincides with the end point, and just to make it clear $\mathbf{r}_1$ and $\mathbf{r}_2$ are injective(except for the start end point))

This is a countexample because if $\mathbf{r}_1$ and $\mathbf{r}_2$ were one the reparametrization of the other the tangent versors should be equal in every point or opposite in every point.
When I come to this point I thought that the problem was that the way that the self-intersection was constructed(because not all self-intersection give problems), but I soonely found out that self-intersections can get pretty nasty and sneaky,so I gave up and I came up with this weaker conjecture:

Final Conjecture

Let $(\gamma,\mathbf{r}_1)$ and $(\gamma,{\mathbf{r}_2})$ be two regular simple curves(if they are close suppose in addiction that the starting\end point of the two parametrizations coincide). Then one is the reparametrization of the other.

I would like to know if this Final Conjecture is true, and if possible I would like some insight about the self-intersections problem and if there is some result about this. I'm sorry if it's a long post, but I'm really interested in this topic so I get carried away.

Thank you in advance.

:)

答:
If I understand your hypotheses correctly, I think the answer to your final conjecture is yes. By your notation $(\gamma, r)$, I take it to mean that $\gamma = r(I)$ is the image of $r$. Then a simple curve presumably means one such that $r: I\to \gamma$ is one-to-one. The key point is that if $r$ is regular at $p\in I$ (meaning $r'(p) \neq 0$), then there is are open neighborhoods $U\subset I$ of $p$ and $V\subset \mathbb R^3$ of $r(p)$, and a $C^1$ function $g: U\to V$, such that $g\circ r = id_U$. In other words, near a regular point $p$, $r$ is a local diffeomorphism to its image. This is a consequence of the inverse function theorem; note that assuming that $r$ is simple means there is an inverse function defined on $\gamma$, but regularity means that the inverse is (locally) the restriction of a differentiable function. Equivalently, if the language is familiar, $\gamma(U)$ is an embedded submanifold of $\mathbb R^3$.

Now if $r_1$, $r_2$ are as in your final conjecture, let $h: \gamma\to I_2$ be the inverse of $r_2$. This exists as a set function since $r_2$ is one-to-one. Moreover, the composition $h\circ r_1: I_1 \to I_2$ is $C^1$: to see this it's enough to check at a given point $p\in I_1$, and since $h$ is the restriction of a $C^1$ function $g$ as above, defined near $r_1(p)$, we get that, near $p$, $h\circ r_1 = g\circ r_1$ is the composition of $C^1$ functions, hence is $C^1$. In fact, $\varphi = h\circ r_1$ is a $C^1$ bijection, since it has an inverse of the form $\tilde h\circ r_2$ where $\tilde h$ is the inverse of $r_1$. Moreover $\varphi$ has nonzero derivative everywhere (from the chain rule and using regularity of $r_1$ and $r_2$), and we have $r_2\circ\varphi = r_1$ since $r_2\circ h = id$ on $\gamma$. So $\varphi$ is the desired reparametrization.

Note that in general for a simple regular curve $r$ there need not be a differentiable function $g$ defined on an open neighborhood of all of $\gamma$ such that $g\circ r = id_{I}$. However, such a $g$ does exist if $I$ is compact.

Title: Reparametrization of regular curves


问:
Is the graph of $y=y$ everything, nothing, or simply an invalid question?
So here's my reasoning:

For any $x$, $y=0x+y$ will simply yield $y$, which would lead me to believe that $y=y$ would produce a horizontal line at, well, $y$... except, since for any $y$ the horizontal line will take at place at said $y$, I'd figure $y=y$ would produce an infinite set of horizontal lines stacked on top of each other.

Hence, all (x, y) pairs are valid solutions to $y=y$. I would assume this also extends to $x=x$ too if it is true.

Now, it is undeniable that both $y=y$ and $x=x$ are true statements, but everyone whom I asked this question either replied with saying this would produce nothing, or simply does not make sense to ask. Desmos seems to agree with the nothing result too.

What am I missing? Why would asking to plot $y=y$ be invalid if $y=c$ where $c$ is some constant or $y=x$ is valid? And if I am right, how could I prove it to those who disagree?

答:
Desmos needs to find how many variables you're using, ie which is $x$ and which is $y$. It does this implicitly, but when you put an equation $y = y$ it evaluates that as $y-y = 0$, ie just a constant. Try putting any constant in Desmos with no variables and you'll see that it will not graph anything.

For $y = y$, there are infinite solutions. In the x,y plane, $x$ is not restricted so it can be any value. So yes, it is the whole x,y plane.

However this equation doesn't really convey any information. The 'solution' is trivial, so plotting it is unnecessary.


问:
The fundamental theorem of calculus and function composition
I'm reading Gong Sheng's Concise Complex Analysis, where in Chapter 1 reviewing calculus, he says

The fundamental theoren1 of calculus plays the most important role in
calculus. There are two equivalent forms of this theorem:

Theorem 1.1 (The Fundo,mental Theorem of Calculus) (differ-form)
Suppose the function $f(x)$ is continuous on $[a, b]$ and
$x\in [a,b]$, Let  $$\Phi(x)=\int_a^x f(t)dt$$ , $a\le x \le b$. Then
$\Phi(x)$ is differentiable in $[a, b]$ and $\Phi'(x) = f(x)$,
$d\Phi(x) = f(x)dx$. In  other words, if the integral of $f(x)$ is
$\Phi(x)$, then the differential of $\Phi(x)$ is  $f(x)$.

Theorem 1.2 (The Fundamental Theorem of Calculus) (integral  form)
Suppose $\Phi(x)$ is differentiable in $[a, b]$; and $d\Phi(x)/dx$ is
equal to  a continuous function $f(x)$. Then
$$\int_a^x f(t) dt = \Phi(x) - \Phi(a)$$
, $(a\le x \le b)$ holds. In other words: If the
differential of $\Phi(x)$ is $f(x)$, then the integral  of $f(x)$ is
$\Phi(x)$.

By this theorem, finding the integral is the inverse operation of
finding  the differential of a function, and properties of
differentials correspond to  properties of integrals. For instance:

$$\frac{d(f(x)\pm g(x))}{dx}=\frac{df(x)}{dx}\pm\frac{d  g(x)}{dx}$$

corresponds to

$$\int (f(x)\pm g(x))dx = \int f(x) dx \pm \int g(x) dx$$ ; and

$$\frac{d}{dx}(fg) = f \frac{dg}{dx}+\frac{df}{dx} g$$

corresponds to  $$\int fg' dx = fg - \int gf' dx$$

So far so good... then he mentions that given $y=g(x)$

$$\frac{d f(g(x))}{dx} = \frac{df}{dy} \cdot \frac{dy}{dx}$$

corresponds to  $$\int f(g(x))g'(x) dx = \int f(y) dy$$

I couldn't figure out, by the Fundamental Theorem of Calculus, and the differential form $\frac{d f(g(x))}{dx} = \frac{df}{dy} \cdot \frac{dy}{dx}$, how could one derive the integral form $\int f(g(x))g'(x) dx = \int f(y) dy$, or vice versa?

答:
Suppose that $F$ is an anti-derivative of $f$. Then the function $F \circ g$ will have the anti-derivative $(F' \circ g) \cdot g' = (f \circ g) \cdot g'$ by the chain rule. That is, the function $h(x) = F(g(x))$ will have derivative $h'(x) = f(g(x)) g'(x)$.

Thus, if we can find an anti-derivative of $f$, we can use this anti-derivative to find an anti-derivative of $f(g(x)) g'(x)$. This is what $u$-substitution is all about.


问:
Find the derived set of $S =\{(2/p)+(3/q) : p,q\in \mathbb N \}$ and check if that set is open.
Find the derived set of $S =\{(2/p)+(3/q) : p,q\in \mathbb N \}$ and check if that set is open.

I guess its derived set should be $D:=\{0\}\cup\{2/p: p \in \mathbb{N}\}\cup\{3/q : q \in \mathbb{N}\}$ but I'm not sure about it.

答: The question is not about the set $S$ itself, but about the derived set $D$ (which is defined as the set of all limit points of $S$ (where, in turn, $x$ is a limit point of $S$ if $x$ is in the topological closure of $S\setminus\{x\}$)). Certainly $0$ is a limit point of $S$ (when $p,q$ both go to $\infty$), so $0$ belongs to $D$. If $p$ is fixed and $q\to\infty$ we see that $2/p\in D$, similarly $3/q\in D.$ The question now is whether there are any other points in $D$, and the answer to that is no (ether both $p,q$ go to $\infty$ in which case the limit point is $0$, or one of $p$ or $q$ is fixed and the other goes to $\infty$, and then the limit points are of the type already described above). So the OP guess is correct, $$D:=\{0\}\cup\{2/p: p \in \mathbb{N}\}\cup\{3/q : q \in \mathbb{N}\}.$$ Clearly the derived set is not open.


答:
Since $\frac{d}{dx}\Phi(g(x))=f(g(x))g'(x)$, we have
$$\int_a^bf(g(x))g'(x)dx=\Phi(g(b))-\Phi(g(a))=\int_{g(a)}^{g(b)}f(x)dx$$

Title: The fundamental theorem of calculus and function composition

Body:
I'm reading Gong Sheng's Concise Complex Analysis, where in Chapter 1 reviewing calculus, he says

The fundamental theoren1 of calculus plays the most important role in
calculus. There are two equivalent forms of this theorem:

Theorem 1.1 (The Fundo,mental Theorem of Calculus) (differ-form)
Suppose the function $f(x)$ is continuous on $[a, b]$ and
$x\in [a,b]$, Let  $$\Phi(x)=\int_a^x f(t)dt$$ , $a\le x \le b$. Then
$\Phi(x)$ is differentiable in $[a, b]$ and $\Phi'(x) = f(x)$,
$d\Phi(x) = f(x)dx$. In  other words, if the integral of $f(x)$ is
$\Phi(x)$, then the differential of $\Phi(x)$ is  $f(x)$.

Theorem 1.2 (The Fundamental Theorem of Calculus) (integral  form)
Suppose $\Phi(x)$ is differentiable in $[a, b]$; and $d\Phi(x)/dx$ is
equal to  a continuous function $f(x)$. Then
$$\int_a^x f(t) dt = \Phi(x) - \Phi(a)$$
, $(a\le x \le b)$ holds. In other words: If the
differential of $\Phi(x)$ is $f(x)$, then the integral  of $f(x)$ is
$\Phi(x)$.

By this theorem, finding the integral is the inverse operation of
finding  the differential of a function, and properties of
differentials correspond to  properties of integrals. For instance:

$$\frac{d(f(x)\pm g(x))}{dx}=\frac{df(x)}{dx}\pm\frac{d  g(x)}{dx}$$

corresponds to

$$\int (f(x)\pm g(x))dx = \int f(x) dx \pm \int g(x) dx$$ ; and

$$\frac{d}{dx}(fg) = f \frac{dg}{dx}+\frac{df}{dx} g$$

corresponds to  $$\int fg' dx = fg - \int gf' dx$$

So far so good... then he mentions that given $y=g(x)$

$$\frac{d f(g(x))}{dx} = \frac{df}{dy} \cdot \frac{dy}{dx}$$

corresponds to  $$\int f(g(x))g'(x) dx = \int f(y) dy$$

I couldn't figure out, by the Fundamental Theorem of Calculus, and the differential form $\frac{d f(g(x))}{dx} = \frac{df}{dy} \cdot \frac{dy}{dx}$, how could one derive the integral form $\int f(g(x))g'(x) dx = \int f(y) dy$, or vice versa?


问:
Proposition 4.11 in Eisenbud's Commutative Algebra
(Proposition 4.11 and its proof are attached below.) In the proof, I'm not sure why $S[\alpha_1]=S[x]/(g)$. As I understand it, we can define the surjective map $S[x]\to S[\alpha_1]$ by sending $x$ to $\alpha_1$. But couldn't the kernel of this map be much larger than $(g)$?

Proposition 4.11. Let $R\subset S$ be rings, and suppose that $f\in R[x]$ is a monic polynomial. If $f$ factors in $S[x]$ as $f=gh$, with $g$ and $h$ monic, then the coefficients of $g$ and $h$ are integral over $R$.

Proof. Adjoining a root $\alpha_1$ of $g$ to $S$ and using long division in the ring $S[\alpha_1]=S[x]/(g)$, we see that $g$ factors as $(x-\alpha_1)g_1$, where the degree of $g_1$ is one less than the degree of $g$. Repeating this process inductively, we may find an extension ring $T$ of $S$ and elements $\alpha_i$, and $\beta_j$ of $T$ such that $g=\prod(x-\alpha_i)$, $h=\prod(x-\beta_j)$ in $T[x]$. Since each $\alpha_i$ and $\beta_j$ is a root of the monic polynomial $f$, the subring $T'$ of $T$ generated as an $R$-algebra by the $\alpha_i$, and $\beta_j$ is integral over $R$. Since the coefficients of $g$ and $h$ are the elementary symmetric functions in the $\alpha_i$, and the $\beta_j$, respectively, they too are integral over $R$. $\qquad\square$

答:
They are defining the notation $S[\alpha_1]$ to mean $S[x]/(g)$, i.e. they are formally adjoining a root of $g$ which is $x+(g)$. This is not taking place in some pre-given bigger ring.

For example, take the polynomial $x^3 - 2$ over $\Bbb R$ and adjoin a formal root $\alpha_1$ by considering the ring $\Bbb R[x]/(x^3-2)$ (which is not an integral domain or a field), and say that $\alpha_1 := x + (x^3-2)$. Note that $\alpha_1 \ne \sqrt[3]2$, and the minimal polynomial of $\alpha_1$ is $x^3-2$, not $x-\sqrt[3]2$. In the same scenario, we would have $x^3-2 = (x - \alpha_1)(x^2 + \alpha_1 x + \alpha_1^2)$.


问:
Find constant $c$ such that all intersection points of two spheres have perpendicular tangent planes
I have come to a problem in a multivariable calculus book that I'm having trouble with.

The problem statement is :

"Find a constant $c$ such that for any point of intersection of the two spheres
$(x-c)^{2} + y^{2} + z^{2} = 3$ and $x^{2} + (y-1)^{2} + z^{2} = 1$, the corresponding
tangent planes will be perpendicular to each other."

So I define two functions :
\begin{align}
f(x,y,z) & = (x-c)^{2} + y^{2} + z^{2} - 3 \\
g(x,y,z) & = x^{2} + (y-1)^{2} + z^{2} - 1
\end{align}
We denote the partial derivative with respect to $x$ as $f_{x}$ and so on...
We see :
\begin{align}
f_{x}(x,y,z) & = 2(x-c) = 2x - 2c\\
f_{y}(x,y,z) & = 2y\\
f_{z}(x,y,z) & = 2z
\end{align}
and :
\begin{align}
g_{x}(x,y,z) & = 2x \\
g_{y}(x,y,z) & = 2(y-1) = 2y - 2 \\
g_{z}(x,y,z) & = 2z
\end{align}
I assume that if the tangent planes of the two spheres are perpendicular at an intersection point, then the normals are also perpendicular. So for every intersection point (x,y,z)
we have :
\begin{equation}
\bigtriangledown f(x,y,z) \cdot \bigtriangledown g(x,y,z) = 0
\end{equation}
So :
\begin{align}
\require{cancel}
(2x - 2c, 2y, 2z) \cdot (2x, 2y-2, 2z) & = 0\\
(2x-2c)2x + (2y)(2y) - 2(2y) + 4z^{2} & = 0\\
4x^{2} - 4xc + 4y^{2} - 4y + 4z^{2} & = 0\\
4(x^{2} + y^{2} + z^{2}) - 4(xc + y) & = 0\\
\cancel{4} \left[ (x^{2}+y^{2}+z^{2}) - (xc + y) \right] & = 0\\
(x^{2}+y^{2}+z^{2}) - (xc + y) & = 0 \\
x^{2} + y^{2} + z^{2} & = xc + y
\end{align}
We also see :
\begin{align}
\require{cancel}
f(x,y,z) & = x^{2} - 2xc + c^{2} + y^{2} + z^{2} - 3 \\
         & = (x^{2} + y^{2} + z^{2}) + (c^{2} - 2xc - 3) \\
g(x,y,z) & = x^{2} + y^{2} - 2y + \cancel{1} + z^{2} - \cancel{1} \\
         & = (x^{2}+y^{2}+z^{2}) - 2y
\end{align}
and :
\begin{align}
f(x,y,z) = 0 & \Rightarrow (x^{2}+y^{2}+z^{2}) = -(c^{2}-2xc-3) = -c^{2} + 2xc + 3\\
g(x,y,z) = 0 & \Rightarrow (x^{2}+y^{2}+z^{2}) = 2y
\end{align}
So we have :
\begin{equation}
xc + y = -c^{2} + 2xc + 3 = 2y
\end{equation}
It is here that I am stuck. It seems that the goal here would be to obtain an expression for $c$ that doesn't include $x$ or $y$, but I do not know how to obtain it.

Can someone help with this ?

答:
Because of the symmetry of spheres with respect to rotation, the statement “for any point of intersection between two spheres $A$ and $B$, the corresponding tangent planes are perpendicular” is equivalent to “between two spheres $A$ and $B$, there is a point of intersection at which the corresponding tangent planes are perpendicular.”

This means that we can reduce this problem to two dimensions by looking at the intersection of the spheres with a plane that passes through the center of both. Doing so we get two intersecting circles, one with radius $1$, the other with radius $\sqrt{3}$.

Rather than look at the two circles actually created by this, for now we will look at the semicircles $y=\sqrt{1-x^2}$ and $y=\sqrt{3-(x-d)^2}$. These intersect at $x=\frac{d^2-2}{2d}$, and for their tangent lines to be perpendicular, their derivatives must be perpindicular, so we have
$$0=\frac{\sqrt{3-(x-d)^2}}{x-d}+\frac{x}{\sqrt{1-x^2}}$$

Substituting for $x$ and solving, we get $d=\pm 2$. Because of this, we know that the centers of the spheres in the problem must be a distance of 2 apart, which is equivalent to $c^2+1^2=2^2 \to c=\pm \sqrt{3}$.


问:
How to find the next numbers in the sequence related to regular polygons

How can I find the next polygons, line segments(in blue) and related numbers (in red) given the sequence in this image?

The question asks up to 7th element, but I don't know the related number or line segment for the nth element other than the polygon will have n+2 sides, and that the number in red is equal to number of polygons "inside" the regular polygon.

@BrianM.Scott Suggestion
possible 5th, 6th and 7th

答:
Welcome to MSE!

draw the blue edges according to the algorithm given by Brian M. Scott,

the sequence is the number of distinct areas created inside the polygon by the interior edges!!!!

inside the triangle there are no interior edges, so there is namely only one area: $\mathbf{\color{red}{1}}$

inside the square there are $4$ triangles: $\mathbf{\color{red}{4}}$

inside the pentagon there are $10$ triangles and a central pentagon: $\color{red}{10+1=\mathbf{11}}$

inside the hexagon there are $12$ triangles and a central hexagon: $\color{red}{12+1=\mathbf{13}}$

Euler’s formula works for the first two graphs since they are planar.


问:
Can $\frac{n!}{(n-r)!r!}$ be simplified?
I'm trying to calculate in a program the number of possible unique subsets of a set of unique numbers, given the subset size, using the following formula:

$\dfrac{n!}{(n-r)!r!}$

The trouble is, on the face of it, you may need an enormous structure to hold the dividend (at least).  Is there a way of simplifying this calculation, so that it can be calculated using something smaller like $64$-bit integers, or are you really going to have to store numbers like $60!$?

Alternatively, is there a formula that is more suited to computing the aforementioned value?

答:
To calculate $\binom{n}{r}$ with

the least likelihood of overflow

keeping any intermediate calculations as small as possible

using only integers (no chance of involving even rationals),

use the following identity:

$$
\begin{array}{rcl}
\binom{n}{r+1}&=&\frac{n-r}{r+1}\binom{n}{r-1}
\end{array}
$$

What this translates to is, in Pascal's triangle, moving along a row, starting from 1.

For example, to compute $$\binom{14}{4} = \frac{14\cdot13\cdot12\cdot11}{1\cdot2\cdot3\cdot4},$$ it is

$$
\begin{array}{ccc}
14*13&=&182\\
     &&/2&=&91\\
     &&&&\cdot12&=&1092\\
     &&&&&&/3&=&364\\
     &&&&&&&&\cdot11&=&4004\\
     &&&&&&&&&&/4&=&1001\\
\end{array}
$$

Since you're moving a long the row, you know that you'll be getting integers every time (you can always divide evenly). Also, you've computed other Pascal numbers along the way so you'll be more likely to remember them next time. What's $14 \choose 5$? Easy, multiply by $10/5$ to get $1001\cdot2$ or 2002.

This is also a manageable method for doing it in your head: easy sequence, low calculation overhead, fewer digits to remember.


问:
Subspace of $ P_3(\mathbb{R})$
Let $p_0,p_1\in P_3(\mathbb{R})$ such that $p_0(x)=5, p_1(x)=1+x$. Suppose that $U=\operatorname{span}(p_0,p_1)$. Find a subspace $W$ of $P_3(\mathbb{R})$ such that
$$ P_3(\mathbb{R}) = U \oplus W,$$ here $\oplus$ denotes direct sum of subspaces and $P_3(\mathbb{R})$ denotes Polynomials with degree at most $3.$

What I have done:

By $U=\operatorname{span}(p_0,p_1)$, we get $U=\{5c_1  + c_2 (1+x) | c_1,c_2 \in \mathbb{R}\}.$ Then I think we need such $W$ subspace of $P_3(\mathbb{R})$
$ P_3(\mathbb{R}) = U \oplus W.$
Now to show the above statement I think we need to show two things

$ P_3(\mathbb{R}) = U \oplus W$

$U$ and $W$ are linearly independent.

But what appropriate $W$ can I use here. Please help. Thank you.

答:
In this case you can take a look at completing the basis of your original vector space. As you know, the direct sum of $U$ and $W$ gives you the vector space $P_3(\mathbb{R})$, and looking at dimensions you can say that $dim_{\mathbb{R}}P_3(\mathbb{R}) = dim_{\mathbb{R}}U +dim_{\mathbb{R}}W$. If you think about it the basis that you are looking for, is actualy the set that contains an $dim_{\mathbb{R}}P_3(\mathbb{R}) - dim_{\mathbb{R}}U$ number of vectors that are linearly independent from the ones in the $U$ basis. In a practical way, you can do the following:

1º- You put the coordinates of the basis of $U$ in the lines of a matrix. Let's call it A:
$$ A = \begin{pmatrix}
5 & 0 & 0 & 0\\
1 & 1 & 0 & 0
\end{pmatrix}
$$
2º - In this step you should do a kinda like a Gaussian elimination (don't know a better name for that in english, sorry), then you'll get a matrix, lets say $A'$  like this:
$$ A' = \begin{pmatrix}
5 & 0 & 0 & 0\\
0 & 1 & 0 & 0
\end{pmatrix}
$$
3º - Now what you need to do is just complete a $dim_{\mathbb{R}}P_3(\mathbb{R})\times dim_{\mathbb{R}}P_3(\mathbb{R})$ matrix, in this case $4\times 4$, in such a way that you place a non-zero number in the $k$-$th$ column of the following rows, were $k$ should be the first column were a number is "missing". Let's call it $M$:
$$ M = \begin{pmatrix}
5 & 0 & 0 & 0\\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix}
$$

If you take the rows that you have added you will get the set $B = \{x^2, x^3\}$ (by the vectors of the third and fourth rows $v_3 = (0\ 0\ 1\ 0)$ and $v_4 = (0\ 0\ 0\ 1)$ ), that is in fact a basis for the $W$ vector sub-space. So your answer could be:
$$ W = \{c_3 x^2 + c_4 x^3\ |\ c_3, c_4 \in \mathbb{R}\}$$

Sorry for the big answer and I hope it helps you. The process may seem silly because we had your basis vectors for $U$ realy simple, but when it isn't that clear it realy helps.

Title: Subspace of $ P_3(\mathbb{R})$


问:
Solving $ x - a = \sqrt{(x-a)^2 + b^2} $ for $x$
Given the following equation:

$$
x - a = \sqrt{(x-a)^2 + b^2}
$$

Is there any way to move $x$ to one side and everything else to the other side?

答:
Apply this rule :

If $A=\sqrt B$, then

$$\begin{cases} A^2=B \\B≥0\\A≥0\end{cases} $$


答:
Square both sides and you would only have $b^{2} = 0$, so a solution exists if $b = 0$

As the square root must be greater than zero, then the LHS $\geq 0$ so, $x-a \geq 0$

So $x \geq a$ and your solution set would then be $\left \{(x,a,b) | x \geq a, b = 0 \right\}$

Title: Solving $ x - a = \sqrt{(x-a)^2 + b^2} $ for $x$

Body:
Given the following equation:

$$
x - a = \sqrt{(x-a)^2 + b^2}
$$

Is there any way to move $x$ to one side and everything else to the other side?


问:
Sum of Random Variables...
Imagine we repeat the following loop some thousands of times:

$$
\begin{align}
& \text{array} = []\\
& \text{for n} = 1: 10 000 \\
& k = 0 \\
& \text{while unifrnd}(0,1) < 0.3 \\
& k = k + 1 \\
& \text{end} \\
& \text{if k} \neq 0 \\
& \text{array} = [\text{array,k}] \\
& \text{end} \\
\end{align}
$$

whereas "unifrnd(0,1)" means a random number drawn from the uniform distribution on the unit interval.

My question is then: What is then the value of k, which is the most often observed - except for k = 0? And is that the expectation of k?

Thanks very much

答:
Since the events "unifrand(0,1) > 0.3" (which lead to a repetition of the same value of $k$) can come in any order and all these orders are equally likely, there is no special value of $k$ that occurs most often.

Edit: With the code as it reads now, the value $k = 1$ will be the most frequent one, and $k$ has indeed a geometric distribution.

Title: Sum of Random Variables...


问:
Simplify a log of a log and series
I'm having a bit trouble trying to write this pattern out

if
$$
f_k(x) =
x \cdot \log(x) \cdot \log(\log(x)) \cdot \ldots \cdot
\log(\ldots(\log(x))).
$$
where the final term has k logs in it
(So $f_0(x) = x$,
$f_1(x) = x \log(x)$,
$f_2(x) = x \log(x) \log(\log(x))$, and so on).
then how would I write the final term?

Is there a way to simplify writing a log of a log?

答:
Welcome to MSE!

If you want to indicate you're composing a function $n$ times, it's standard to write

$$f^{\circ n}(x).$$

That is

$f^{\circ 0}(x) = x$

$f^{\circ 1}(x) = f(x)$

$f^{\circ 2}(x) = f(f(x))$

$f^{\circ 3}(x) = f(f(f(x)))$

etc.

So your term is the product of $\log^{\circ j}$ for $j \leq k$. Or, in symbols:

$$
f_k(x) = \prod_{j = 0}^k \log^{\circ j}(x)
$$

If you aren't familiar with $\prod$ as a product, you can see more here.

I hope this helps ^_^

Title: Simplify a log of a log and series


问:
Graph embeddings on nonorientable surfaces
If a finite graph $G$ can be embedded on an orientable surface of genus $n$, does this mean that it can be embedded on a nonorientable surface of genus $n$? Is the converse of this statement true?

答:
As Hugo Manet's comment suggests, nonorientable and orientable genus don't really line up like that. Euler's formula states that

$$|V|-|E|+|F|=\chi(S),$$

where $\chi(S)$ denotes the Euler characteristic of the surface. The Euler characteristic of the sphere with $g$ handles is $2-2g$, but the Euler characteristic of the sphere with $k$ crosscaps is $2-k$. As a result, maybe the smallest case is really about comparing the Klein bottle and the torus, two surfaces with the same Euler characteristic. Both statements have counterexamples (see Ringel, "Non-existence of graph embeddings"):

$K_7$, the complete graph on 7 vertices, embeds in the torus, but not the Klein bottle.

$K_8-C_4$, the complete graph on 8 vertices with edges of a 4-cycle deleted, embeds in the Klein bottle, but not the torus.


问:
How do I compute $\Delta_x\frac{1}{|x-y|}$ when $x$ has arbitrary $n$ components?
How do I compute $\Delta_x\frac{1}{|x|}$ when $x$ has arbitrary $n$ components?

Is there any good way to simply this?

答:
This is just an exercise in computing partial derivatives. All you need to know is the chain rule and the product rule. Remember in $\mathbb{R}^n$, $$ \vert x-y \vert = \bigg ( \sum_{k=1}^n (x_k-y_k)^2 \bigg )^{\frac 1 2}$$ where $x = (x_1, \dots , x_n)$ and $y = (y_1, \dots , y_n)$. Fix $y \in \mathbb{R}^n$ and let $$u(x) = \vert x - y \vert^{-1} = \bigg ( \sum_{k=1}^n (x_k-y_k)^2 \bigg )^{-\frac 1 2}.$$ Via the chain rule, \begin{align*}
\frac \partial {\partial x_i} \vert x-y \vert &=\frac 1 2  \bigg ( \sum_{k=1}^n (x_k-y_k)^2 \bigg )^{-\frac 1 2} \cdot 2(x_i-y_i) \\
&= \frac{x_i-y_i}{\vert x-y \vert }.
\end{align*} Hence, $$ \frac{\partial u}{\partial x_i} = - \frac{1}{\vert x-y \vert^2} \cdot \frac \partial {\partial x_i} \vert x-y \vert = - \frac{x_i-y_i}{\vert x -y \vert^3}. $$ Now by the product rule, \begin{align*}
\frac{\partial u}{\partial x_i^2} &= - \frac 1 {\vert x -y \vert^3}+3 \frac{x_i-y_i}{\vert x -y \vert^4}\frac \partial {\partial x_i} \vert x-y \vert \\
&= - \frac 1 {\vert x -y \vert^3}+3 \frac{(x_i-y_i)^2}{\vert x -y \vert^5}.
\end{align*} Finally, \begin{align*}
\Delta u &= \sum_{i=1}^n\frac{\partial u}{\partial x_i^2} \\
&= - \sum_{i=1}^n\frac 1 {\vert x -y \vert^3}+3 \sum_{i=1}^n\frac{(x_i-y_i)^2}{\vert x -y \vert^5} \\
&= -\frac n {\vert x -y \vert^3} +  \frac 3 {\vert x -y \vert^3} \\
&= - \frac {n-3} {\vert x -y \vert^3}.
\end{align*}


问:
Why coorientable is a dual notion of orientable?
When I started to study contact geometry, I immediately met some obstacle, the notion of 'coorientable'. In Geiges' "an introduction to contact topology", he defines that the distribution $\xi$ on $M$ with codimension 1 is coorientable if $TM/\xi$ is trivial. I think it is clear, however, it is hard to understand why this definition has the name of 'coorientable'. I hope to find the dual notion of orientable in some sense and understand the connection between coorientable and orientable. Is there some intuitive way to understand the meaning of coorientable in terms of orientable?

答:
$TM/\xi$ is the normal bundle to the distribution. If $M$ has a metric, you're asking to be able to put a consistent orientation on the set of vectors perpendicular to $\xi$.

In an inner product space $V$ it is reasonable to assert that $W \mapsto W^\perp$ is a duality on the subspaces of $V$, as $(W^\perp)^\perp = W$. Whence the "co".

Notice that one has $TM \cong \xi \oplus \xi^\perp \cong \xi \oplus (TM/\xi)$.

Hence if $M$ is orientable, a distribution is orientable if and only if it is co-orientable. More specifically, if $M$ is oriented, an orientation on $\xi$ determines one on $\xi^\perp$ and vice versa (by the demand that $TM \cong \xi \oplus \xi^\perp$ is an isomorphism of oriented vector bundles).

When $M$ is not orientable, the distribution $\xi$ cannot be both orientable and co-orientable. A standard example of a co-orientable distribution which is not orientable is the distribution on $\Bbb{RP}^2 \times S^1$ of tangent planes to the fibers $\Bbb{RP}^2 \times \{\theta\}$. Of course, this is not a contact structure.

There are a number of unproved claims in this answer (for instance, in the line starting "Hence"). I suggest that if they are not obvious to you, it will be an instructive exercise in the notion of orientability of vector bundles to prove it. Whenever we use splittings as above or pass between $TM/\xi$ and $\xi^\perp$ we are using an inner product on $TM$.


问:
Positive integer solution such that $7^m - 3^m$ is prime
Find all $m \in \mathbb{Z}$, for which the expression $7^m - 3^m$ a prime. What would be the general strategy for solving this? I have no clue where to begin. I tried picking $m$ and checking some values but I don't see any relation between them. This question can be solved using modular arithmetic as well but im not sure how. What I do know is that $7^m - 3^m > 1$.

答:
But $7^m - 3^m$ for positive $m$ is not only even but is a multiple of $4$ which is not prime [ETA as noted in comments already]. So definitely not prime.

In general, if $a$, $b$, and $c$ are positive integers satisfying $a-b = c$, then working mod $c$, one can see that $a^m-b^m$ is always a multiple of $c$.


问:
Why does the term $A\mathbf{h}/\|\mathbf{h}\|$ appear in the definition of the derivative?
My analysis course provides the following definition of the derivative of $f$, $Df(\mathbf{a})$:

Definition: A function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ is differentiable at $\mathbf{a} \in \mathbb{R}^{n}$ if there exists a linear function $\mathbf{x} \mapsto A \mathbf{x}$ (where $A$ is an $m \times n$ matrix) from $\mathbb{R}^{n}$ to $\mathbb{R}^{m}$ such that
$$
\frac{f(\mathbf{a}+\mathbf{h})-f(\mathbf{a})-A \mathbf{h}}{\|\mathbf{h}\|} \rightarrow \mathbf{0}.
$$

Now, if we unpack this and go back to the limit definition we get that $Df(\mathbf{a})$ is
$$
\lim_{\|\mathbf{h}\| \rightarrow  0 } \frac{f(\mathbf{a}+\mathbf{h})-f(\mathbf{a})}{\|\mathbf{h}\|} - \frac{A \mathbf{h}}{\|\mathbf{h}\|}.
$$

What I don't understand is the term $\frac{A \mathbf{h}}{\|\mathbf{h}\|}$. Why is the product $A \mathbf{h}$ scaled by the norm of $\|\mathbf{h}\|$ as well?

Would appreciate further insight into this definition.

Additionally, based on this definition of the derivative, it is possible for directional derivatives to exist even the function is not differentiable at a particular vector $\mathbf{a}$. Why is this the case ?

Defining the directional derivative as:

$$D_{\mathbf{v}} f(\mathbf{a})=\lim _{h \rightarrow 0} \frac{f(\mathbf{a}+h \mathbf{v})-f(\mathbf{a})}{h}.$$

答:
First off, minor nitpick:  $A \mathbf{h}$ is not a dot product.  The dot product takes two $n$-dimensional real (or complex) vectors as input, and gives a real number as output.  $A$ is a matrix (not a vector) which represents a linear transformation $\mathbb{R}^m \to \mathbb{R}^n$ which acts on $\mathbf{h}$ (a vector in $\mathbb{R}^m$) in order to produce a vector in $\mathbb{R}^n$.

The Derivative

Addressing the question:  in one dimension, the derivative of $f$ at a point $a$ is generally defined as the limit of some kind of difference quotient, e.g.
$$ f'(a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h},$$
assuming that this limit exists.  This definition is usually justified geometrically:  we think about approximating a tangent line to a curve via secant lines defined by points which are closer and closer to each other on the curve.  In that sense, the definition of the derivative in one dimension is natural.

Manipulate this identity to get
\begin{align}
0
&= \left( \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}\right) - f'(a) \\
&= \lim_{h\to 0} \left( \frac{f(a+h) - f(a)}{h} - f'(a) \right) && (\text{$f'(a)$ is a constant)} \\
&= \lim_{h\to 0} \frac{ f(a+h) - f(a) - hf'(a)}{h}.
\end{align}
With a little more manipulation, the definition of the derivative can be rewritten as

Definition: A function $f : \mathbb{R} \to \mathbb{R}$ is differentiable at a point $a \in \mathbb{R}$ if there exists a linear function $x \mapsto Ax$ from $\mathbb{R}$ to $\mathbb{R}$ (where $A$ is a $1\times 1$ matrix) such that
$$ \frac{f(a+h) - f(a) - Ah}{|h|} \to 0. $$

This version of the definition follows from the "usual" geometric definition, but encapsulates a slightly different idea.  In this case, we think of the derivative as the linear operator which best approximates how the function is changed by small perturbations around $a$.  That is, if we perturb $a$ by $h$, what will be the corresponding change in $f$, assuming that $f$ is (at least locally) linear?

The actual perturbation is the difference between $f(a+h)$ and $f(a)$, while the linear approximation of the perturbation is $Ah$.  The derivative $A$ is the best linear approximation in the sense that the relative error between actual perturbation and the linear approximation can be made as small as possible.

In higher dimensions, the same idea applies: if we perturb $\mathbf{x}$ by just a little, this is going to perturb $f$ by just a little.  We then want to find the linear function which best approximates this perturbation.  We assess this by looking at the relative error, i.e the difference between the actually perturbation and the linear approximation, scaled by the size of the perturbation in $\mathbf{x}$.  The result is the definition of the derivative given in the question.

Directional Derivatives

Regarding the existence of directional derivatives even if the derivative does not exist:  a directional derivative is the best linear approximation of a function in a particular direction, whereas the derivative is the best approximation in any direction.  It is possible for a function to be approximable in one direction, but not in every direction.  As a simple example, consider
$$ f(x,y) = |x|. $$
If you are moving parallel to the $y$-axis, the function is constant, so
$$ D_{\langle 0, 1\rangle} f(x,y) = \lim_{h \to 0} \frac{f(x,y+h) - f(x,y)}{h} = \lim_{h\to 0} \frac{ |x| - |x| }{h} = 0. $$
Here, the directional derivative in the direction of $\langle 0, 1 \rangle$ exists (and is zero).  However, in any other direction, the derivative does not exist wherever $x=0$.  For example
$$ D_{\langle 1,0 \rangle} f(0,0) = \lim_{h \to 0} \frac{ f(0+h, 0) - f(0,0) }{ h }
= \lim_{h \to 0} \frac{ |0+h| - |0| } {h} = \lim_{h\to 0} \frac{|h|}{h}, $$
which does not exist.


问:
Is possible exists a injective function $f\colon \mathbb{R}\to\mathbb{R}$ such that $f(x^2)-f(f(x))\geq 1/4$?

Is possible exists a injective function $f\colon \mathbb{R}\to\mathbb{R}$ such that $f(x^2)-f(f(x))\geq 1/4$?

Anyone can give a tip? It's an exercise about functional equations.

Body:
This is possible. Consider $f(x) = x - 1$. Then

$f(x^2) - f(f(x)) = (x^2 - 1) - ((x - 1) - 1) = x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4} \geq \frac{3}{4} \geq \frac{1}{4}$.


答:
Let's compare this to the ordinary definition of the derivative in 1 variable. Consider a function $f : \mathbb R \to \mathbb R$ and $a \in \mathbb R$. To say that $f$ is differentiable at $a$ means that the following limit exists:
$$\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}
$$
This limit is usually given its own notation:
$$f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}
$$
Now let's manipulate this around a bit:
$$\left( \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \right) - f'(a) = 0
$$
Holding $a$ fixed, as $h$ varies we can treat $f'(a)$ as a constant, and then use the formula for limit of a constant:
$$\left( \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \right) - \lim_{h \to 0} f'(a) = 0
$$
and now we can use the rule for limit of a difference:
$$\lim_{h \to 0} \left( \frac{f(a+h)-f(a)}{h} - f'(a) \right) = 0
$$
Combining into one fraction we obtain
$$\lim_{h \to 0} \frac{f(a+h)-f(a) - f'(a) h}{h} = 0
$$
I'm sure you can see that this is equivalent to
$$\lim_{h \to 0} \frac{f(a+h)-f(a)-f'(a) h}{|h|}  = 0
$$
Good so far.

Now let's compare this with the new definition of the multivariable derivative that you are learning, specialized to the case $m=n=1$. To say that $f : \mathbb R \to \mathbb R$ is differentiable at $a$ means that there exists a linear function $x \mapsto Ax$ (where $A$ is a $1 \times 1$ matrix) such that
$$\lim_{h \to 0} \frac{f(a+h)-f(a)- Ah}{|h|} = 0
$$
Well, these two definitions are equivalent, using the $1 \times 1$ matrix
$$A = (f'(a))
$$
To summarize, one reason the multivariable definition of differentiability is written the way it is so as to generalize the 1-variable definition.

As for our second question, one key feature of a differentiable real valued function $f : \mathbb R^n \to \mathbb R$ is that the directional derivatives at $a$, and in particular partial derivatives which are a special case of directional derivatives, may be evaluated by a useful formula expressed in terms of the $m \times 1$ matrix $A$, assuming that $f$ is differentiable: for any unit vector $\vec v$ one has
$$D_v f(a) = A(v)
$$
So for example if $a = \vec e_i$ is the unit vector in the $i$th coordinate direction, on the left hand side one gets the usual $\frac{\partial f}{\partial x_i}(a)$ and on the right hand side one gets the $i$th coordinate of the matrix $A$. Thus the matrix is the same as the gradient vector of $f$.

But if $f$ is not differentiable at $a$ then all those nice relationships between partial derivatives and the matrix $A$ are out the window, because of course $A$ does not exist. You should definitely crack open your old multivariable calculus book and look for an example of this phenomenon, i.e. an example of a function $f : \mathbb R \to \mathbb R$ and a point $a \in \mathbb R$ such that all partial derivatives $\frac{\partial f}{\partial x_i}(a)$ exist, and yet $f$ is not differentiable at $a$.

On the other hand, if all of the first order partial derivatives of $f : \mathbb R^m \to \mathbb R$ exist, and if each of the partial derivative functions $\frac{\partial f}{\partial x_i} : \mathbb R^m \to \mathbb R$ is continuous, then a miracle occurs: $f$ is differentiable! That's one of the basic theorems that you'll find in an advanced multivariable calculus book.

So, to answer your final question:  It is possible for directional derivatives to exist even when the function is not differentiable at a particular vector? Yes, it is possible, but only when at least one of the directional derivative functions is not continuous.

Title: Why does the term $A\mathbf{h}/\|\mathbf{h}\|$ appear in the definition of the derivative?

Body:
My analysis course provides the following definition of the derivative of $f$, $Df(\mathbf{a})$:

Definition: A function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ is differentiable at $\mathbf{a} \in \mathbb{R}^{n}$ if there exists a linear function $\mathbf{x} \mapsto A \mathbf{x}$ (where $A$ is an $m \times n$ matrix) from $\mathbb{R}^{n}$ to $\mathbb{R}^{m}$ such that
$$
\frac{f(\mathbf{a}+\mathbf{h})-f(\mathbf{a})-A \mathbf{h}}{\|\mathbf{h}\|} \rightarrow \mathbf{0}.
$$

Now, if we unpack this and go back to the limit definition we get that $Df(\mathbf{a})$ is
$$
\lim_{\|\mathbf{h}\| \rightarrow  0 } \frac{f(\mathbf{a}+\mathbf{h})-f(\mathbf{a})}{\|\mathbf{h}\|} - \frac{A \mathbf{h}}{\|\mathbf{h}\|}.
$$

What I don't understand is the term $\frac{A \mathbf{h}}{\|\mathbf{h}\|}$. Why is the product $A \mathbf{h}$ scaled by the norm of $\|\mathbf{h}\|$ as well?

Would appreciate further insight into this definition.

Additionally, based on this definition of the derivative, it is possible for directional derivatives to exist even the function is not differentiable at a particular vector $\mathbf{a}$. Why is this the case ?

Defining the directional derivative as:

$$D_{\mathbf{v}} f(\mathbf{a})=\lim _{h \rightarrow 0} \frac{f(\mathbf{a}+h \mathbf{v})-f(\mathbf{a})}{h}.$$


问:
Is $A - B$ positive definite?

If $A$ is positive definite and $B$ is negative semidefinite, then $A−B$ is positive definite.

If $A$ is indefinite and $B$ is positive definite, then $A+B$ is indefinite.

The definition I am using is if $(Ax,x) > 0$ for all $x$, then $A$ is positive definite. I think (1) is false because $B$ is negative semidefinite so there may be some $x$ for which the negative outweighs the positive from $A$.

For (2), I think the answer is again false. The positive definiteness of $B$ may overweight $A$ so it is possible that $(A+bx,x)$ is positive for all $x$.

答:
$x^T(A-B)x = x^TAx - x^TBx$. If $x \ne 0$, then $x^TAx>0$ and $-x^TBx>0$, we have $x^TAx - x^TBx > 0$. $A-B$ is positive definite.

Counterexample: If $A = \begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$, we can have $B=I_2$ and $A+B$ is positive semidefinite.

Title: Is $A - B$ positive definite?


Title: Positive integer solution such that $7^m - 3^m$ is prime

Body: Find all $m \in \mathbb{Z}$, for which the expression $7^m - 3^m$ a prime. What would be the general strategy for solving this? I have no clue where to begin. I tried picking $m$ and checking some values but I don't see any relation between them. This question can be solved using modular arithmetic as well but im not sure how. What I do know is that $7^m - 3^m > 1$.

Body:
We have $7^{m}-3^{m}$ and m is positive. So,
$$7^{m}-3^{m}=(7-3)(7^{m-1}+7^{m-2}.3+7^{m-3}.3^{2}+......7^{2}.3^{m-3}+7.3^{m-2}+3^{m-1})$$
$$7^{m}-3^{m}=\mathbf{4}.(7^{m-1}+7^{m-2}.3+7^{m-3}.3^{2}+......7^{2}.3^{m-3}+7.3^{m-2}+3^{m-1})$$
$$7^{m}-3^{m}=\mathbf{4}.g(m)$$
So 4 is a root of this for m ($m\in \mathbb{Z^{+}}$) positive. So $7^{m}-3^{m}$ is not a prime.


问:
How Many Ways to Choose Dance Partners if Choosing Both Men and Women?
I was working through supplementary exercises in Grimaldi, and I came across a question about dance partners--it asked how to choose a clean-up team of 8 from a group of 12 men, and then asked how many different ways to pair up 8 women from those 12 men.  The answer was simple enough: find the number of permutations of size 8 from a set of 12 men (i.e., $\frac{12!}{4!})$.  But that got me wondering, if we want to choose both the men and the women from a set of 12 each, how does the answer change?  Since I'm completely new to self-studying math, I thought I'd pose the question to see if my approach to answering the question is valid (I thought it might still have something to do with permutations, but I wasn't sure).

So let's say there are 12 men and 12 women at a dance.  How many ways are there to pair off 8 men and 8 women as dance partners?

答:
There are $\binom{12}8$ ways to choose the men, and $\binom{12}8$ ways to choose the women.  Once the people have been chosen there are $8!$ ways to pair them off.  (Stand the women in a line.  There are $8!$ ways to shuffle the men, then we pair the first man with the first woman, the second man with the second woman, and so on.)

The answer is $$\binom{12}{8}^2\cdot8!=9,879,408,000
$$


问:
Using Nakayama's Lemma to prove isomorphism theorem for finitely generated free modules
Suppose $R \neq 0$ is a commutative ring with $1$. The following is well known:

(Isomorphism Theorem for Finitely Generated Free Modules) [FGFM] $R^{n}\cong R^{m}$ as
  $R$-modules if and only if $n=m$.

One proof of this result (if I recall correctly) was somehow based on the idea that we can quotient out by a maximal ideal of $R$ (whose existence is guaranteed by Zorn's Lemma), and reduce the situation to that of vector spaces. However, I am interested in different approach. Recall

(Nakayama's Lemma) Suppose $M$ is a finitely generated $R$-module, and
  $M=IM$ where $I$ is an ideal contained in Jacobson radical of $R$. Then, $M=0$.

My question is:

Can we prove FGFM using Nakayama's Lemma?

My main motivation in asking this question is two-fold:

1) to see the power and usefulness of Nakayama's Lemma, and

2) to see a nice and short proof of FGFM that does not involve reducing the problem to a result from linear algebra.

I appreciate any input :)

答:
Here's one idea. Use the following consequence of Nakayama:

If $M$ is a finitely generated $R$-module and $f\colon M\to M$ is a surjective module homomorphism, then $f$ is an isomorphism.

Proof: View $M$ as an $R[x]$-module, where the action of $x$ on $M$ is given by $f$. By assumption $xM = M$. The proof of Nakayama's lemma then gives that there is some element $P(x)\in R[x]$ such that $(1 - P(x)x)M = 0$. Then for any $m\in M$, $$0 = (1 - P(x)x)m = m - P(f)(f(m))\implies m = P(f)(f(m)).$$ This shows that $P(f)$ is an inverse of $f$.

To use this to prove FGFM, do the following. Suppose $n\geq m$, and that $R^n\cong R^m$. First, fix a basis $e_1,\ldots, e_n$ of $R^n$, and identify $R^m$ with the submodule of $R^n$ generated by $e_1,\ldots, e_m$, so $R^m\subseteq R^n$. Let $\pi\colon R^n\to R^m$ be the projection map. Since by assumption $R^n\cong R^m$, there is some isomorphism $f\colon R^m\to R^n$. But then $\pi\circ f\colon R^m\to R^m$ is surjective, and hence by the result above, is also an isomorphism. However, since $f$ is surjective, the only way that $\pi\circ f$ can be an isomorphism is if $\pi$ is injective. This of course only happens when $n = m$.


问:
Evaluating $\int_0^z(z-x)^{2m-1}x^{2m-1}e^{-\frac{m}{\Omega}x^2}\times K_0(\frac{2(z-x)}{\alpha}\frac{m}{\Omega}){\rm d}x$
I am trying to proceed with this integration but unable to solve completely
$$f_{Z}(z)=\int_0^z(z-x)^{2m-1}x^{2m-1}e^{-\frac{m}{\Omega}x^2}\times K_0\left(\frac{2(z-x)}{\alpha}\frac{m}{\Omega}\right){\rm d}x$$
where $K_0$ is zeroth order bessel function.

Any help in this regard is highly appreciated.

答:
so the zeroth order Besselfunction of the first kind is going with the series of ~$\sum_{m=0}^{\infty}(z-x)^{2m}$ and then some constant factors... i leave them out, u have to figure them out on ur own.

note that $\partial_x e^{x^2}$ equals $2x*e^{x^2}$ and the integral of $e^{-x^2}$ is the error function. Also note that $\partial_a e^{ax^2}|_{a=1}=x^2e^{x^2}$
so every x powered with a multiple of 2 can be replaced with a $\partial_a$ if u put an $a=1$ into the exponent and then it can be dragged out of the integral because $\partial_a$ and $\int dx$ commute. Dont forget to set a=1 after u are done.

I wont calculate this integral fully for u, but this could make it considerably easier. It is basically integration by differentiation. U just have to keep track of the constants that u get with every derivative.

consider also, that u have very simple integral limits... so when u have an uneven power of x as factor, u get a $e^{-ax^2}/a$ solution which is in this limits equal to $1/2a$ which u then just have to differentiate multiple times by $a$

one example: $\int dx (x^2+x^3)e^{-x^2}=\int dx (-\partial_a-\partial_a *x)e^{-ax^2}|_{a=1}=-\partial_a\int dx e^{-ax^2}-\partial_a \int dx$ $ x*e^{-ax^2}|_{a=1}=-\partial_a \sqrt\pi/(2\sqrt a)-\partial_a(1/2a)|_{a=1}=\sqrt \pi/4+1/2 $ in the limits of $0$ and $\infty$


问:
Integer Solutions of $x^2+y^2=100,003$ Using Modulo Arithmetic
Problem: Prove whether or not $x^2+y^2=100,003$ has any integer solutions using modulo arithmetic.

My professor gave the following solution: $x^2+y^2=100,003 \equiv 3 \pmod 4$. By the previous proof verifying that no integer of the form $4k+3$ with integer $k$ is the sum of two squares. Thus $x^2+y^2=100,000$ has no integer solutions.

Why did my professor know to use modulo 4 with remainder 3 looking at the problem? I can't think of any reason to do that.

答:
If $x=2m, y=2n$, then $x^2+y^2$ is always even. Therefore, the equality is not possible.

If $x=2m, y=2n-1$ or $x=2m-1, y=2n$, then
$$x^2+y^2=4m^2+4n^2-4n+1=4(m^2+n^2-n)+1=100003$$
which is impossible.

There is no need to check $x=2m-1$ and $y=2n$.

If $x=2m-1,y=2n-1$  then $x^2+y^2$ is always even.
So, it is impossible again.


问:
How do I use polar coördinates in order to integrate along a curved axis?
Currently I'm struggeling with a mechanics problem. My issue is not the mechanics, but the mathematics. Therefore I see it fit to post this question here.

I need to integrate along the axis of the beam and I need to do this in polor coordinates. This is where I get stuck. I know from calculus that I need to express the x and y value in an angle in combination with a geometric function.

It looks like this beam is the quarter of a circle. I think this piece of data is usefull, but I don't know how to proceed.

Question: How do I integrate along the axis of this beam in polor coordinates?

答:
How do I use polar coördinates in order to integrate along a curved axis?

I've flipped your sketch left to right.
Let $A=(a,0)$ and $B=(0,a).$ In polar coordinates,
$x=a\cos \theta,$ $y=a\sin \theta$ and $\theta$ goes from $0$ to $\pi\over 2$ radians. For example,
$$\int_0^{\pi/2} a\, d\theta$$
gives the length of your beam.


问:
What it Means for a Moment Generating Function when E(X) is undefined or infinity in a Distribution Function
The other day I was solving some problems relating to distribution functions and expected values.
There was a question asking about the "meaning" of the moment generating function depending on a random variable.
There was a random variable A, which I solved and the $E(X^\alpha) = $undefined and $E(X^\beta) = ∞$.

Does this fact tell us anything about the moment generating function?

I have come across the fact that when $E(X^\alpha)$ is undefined, then the "$\alpha$"th moment of X does not exist. However I do not know what is means for the moment generating function of X when $E(X^\beta)$ = ∞

It would be nice if I can get an explanation about this.

答:
Note that the MGF is defined as $\int_{-\infty}^{\infty} e^{tw}f_W(w)dw$ where $f_W$ is the PDF of the function and is valid for all values of $t$ for which the expected value exists.  To get an i'th moment, you take i derivatives of this integral wrt t, and plug in t = 0.

If you're not getting any valid number, then that i'th derivative does not exist, and so the i'th moment does not exist.


问:
Is there a closed form to this series?
Here is the series in question,

```<code>1 + 2 + 1 + 4 +
1 + 2 + 1 + 8 +
1 + 2 + 1 + 4 +
1 + 2 + 1 + 16 +
1 + 2 + 1 + 4 +
1 + 2 + 1 + 8 +
1 + 2 + 1 + 4 +
1 + 2 + 1 + 32 + ...
</code>```

so for example, given n = 4, the series sum is 1 + 2 + 1 + 4 = 8

I created this series inspired by the game 2048,
and I've been trying to find the closed-form with respect to n for a few hours to no avail.
Does there exist a closed-form? What's the name of this type of series?

答:
The $k$-th term is $\gcd(k,2^k)$, so the sum up to $n$ is $$\sum_{k=1}^n \gcd(k,2^k)$$

It can also be calculated as $$n+\sum _{k=1}^n 2^{k-1} \left\lfloor \frac{n}{2^k}\right\rfloor$$

No closed form is known for the sum, i think probably there isn't one.

Title: Is there a closed form to this series?


问:
determine whether an integral is positive
Given a standardized normal variable $X\sim N\left(0,1\right)$, and constants $ \kappa \in \left[0,1\right)$  and  $\tau \in \mathbb{R}$, I want to sign the following expression:

\begin{equation}
\int_{-\infty}^\tau \left(X-\kappa \tau \right) \phi(X)\text{d}X
\end{equation}

where $\phi$ is the PDF of $X$.  Any comment will be appreciated. I would at least want to know if the sign of the expression can be determined given the information, or whether it hinges on the value of $\tau$.

答:
Let the function be $g$. Evaluating $g(\kappa =1)$, we to obtain

\begin{equation*}
g=\int_{-\infty }^{\tau }\left( X-\tau
\right) \phi \left( X\right) \text{d}X \leq 0
\end{equation*}

the inequality follows since $X-\tau \leq 0$ for all $X \in (-\infty,\tau)$. Similarly, for $\kappa =0$,

\begin{equation*}
g=\int_{-\infty }^{\tau }X\phi \left(X\right) \text{d}X \leq 0
\end{equation*}

since
\begin{equation*}
\int_{-\infty }^{\tau }X\phi \left( X\right) dX<\int_{-\infty }^{\infty
}X\phi \left( X\right) \text{d}X=0
\end{equation*}
and finally taking the derivative of $g$ with respect to $\kappa$

\begin{equation*}
\frac{\partial g}{\partial \kappa }=-\tau \int_{-\infty }^{\tau }\phi \left(
X\right) \text{d}X \leq 0
\end{equation*}

since, if it depends on $\tau$, then we will take $g\left( \kappa
=1\right) $ to find that it is negative. Therefore,

\begin{equation*}
g = \int_{-\infty }^{\tau } \left( X-\kappa \tau \right) \phi \left(X\right) \text{d}X \leq 0
\end{equation*}
and most importantly, this inequality does not depend on $\tau$


问:
Asking if an infinite series will converge or diverge
I have two questions regarding the convergence of infinite series. One, if $\sum_{n=1}^{\infty}a_n$ converges to L, where L is defined, does $a_n < L$ for each n?

Two, if $a_n$ is defined for each n, and $\sum_{n=j}^{\infty}a_n$ converges for some integer j, then does $\sum_{n=j}^{\infty}a_n$ converge for all j?

My response: for one, this is a no for me, since you can make $a_n=0$, which will promptly converge to 0, but $0 \not < 0$. For two, I think this is based right off of one of the properties of series, where if a portion of the series converges, then the entire series converges, right? Correct me if I'm wrong (:

答:
You are more or less correct for both questions, but we can expand on this a little bit.

Note that if $\sum_{n=1}^{\infty}a_{n}$ converges, then $\lim_{n\to\infty}a_{n}=0$. Now, consider $\sum_{n=1}^{\infty}\frac{-1}{n^{2}}=-\frac{\pi^{2}}{6}=L$. It follows that $a_{n}>L$ for sufficiently large $n$ since $L$ is negative and $a_{n}$ must eventually be getting very close to zero.

Your intuition is essentially right here. For example, if $\sum_{j=1}^{\infty}a_{j}$ converges and you ignore finitely-many terms of the series, then the remaining sum still converges.


答:
The discriminant of a quadratic form $$A x^2 + Bxy + C y^2$$  is
$$ \Delta = B^2-4AC  $$
When $\Delta$ is a (positive!) square, the form factors and some things go wrong.

When  $\Delta$ is not a square, as your $-4,$ the form represents certain values $\pmod{ |\Delta|}.$  That is, your professor knew to work $\pmod 4$

Here's an example: show that there is no integer solution to
$$ x^2 +xy+2y^2 = 70000000003  $$

For forms with positive discriminant (so indefinite),  care is needed in treating negative values differently from positive. An example is $x^2-3y^2$   which represents $p$ when prime $p \equiv 1 \pmod {12},$  but instead represents  $-p$ when prime $p \equiv 11 \pmod {12}$

Title: Integer Solutions of $x^2+y^2=100,003$ Using Modulo Arithmetic

Body:
Problem: Prove whether or not $x^2+y^2=100,003$ has any integer solutions using modulo arithmetic.

My professor gave the following solution: $x^2+y^2=100,003 \equiv 3 \pmod 4$. By the previous proof verifying that no integer of the form $4k+3$ with integer $k$ is the sum of two squares. Thus $x^2+y^2=100,000$ has no integer solutions.

Why did my professor know to use modulo 4 with remainder 3 looking at the problem? I can't think of any reason to do that.


答:
Why did my professor know to use modulo 4 with remainder 3 looking at
the problem? I can't think of any reason to do that.

I'm not sure I understand your question properly. You mentioned that:

By the previous proof verifying that no integer of the form 4k+3 with
integer k is the sum of two squares

That was already proven, so you simply need to verify if the two squares are in the form, so by taking mod 4, you'll be left with the remainder. ie $4k + r \equiv r$ (mod 4) so you want $r = 3$ if you want to prove no solution exists.


问:
$\sum_{n=1}^{\infty}a_{2n}$ converges? or no
So, a series of $\sum_{n=1}^{\infty}a_n$, but $a_n=2n^{-1}$ if n is even, but $a_n=2^{-n}$ if $n$ is odd. Does $\sum_{n=1}^{\infty}a_{2n}$ converges? $2n$ is all even numbers, so it's $a_n=2(2n)^{-1}$. This is turned into $\frac{1}{n}$, which diverges, this is correct, yes?

答: The (sub)series for even terms is twice the sum of reciprocals of even natural numbers, i.e., the sum of reciprocals, which diverges. Thus the whole sum diverges.</p>


问:
Does this logic have the downward Skolem-Löwenheim theorem?
Let $\mathcal L_Q$ denote the logic obtained from adding the quantifier $\newcommand{\almost}{\forall^\infty}\almost$ to the usual first-order logic, where the semantic interpretation of $\almost x\varphi$ is "All but finitely many $x$ satisfy $\varphi$", or formally:  $$M\models\almost x\varphi(x)\iff\Big|\{m\in M\mid M\not\models\varphi[m]\}\Big|\text{ is finite}.$$

It's not very hard to show that this logic is not compact$^*$, and does not satisfy the upward Skolem-Löwenheim theorem (e.g. the order $(\Bbb N,\leq)$ has a categorical axiomatization). But what about its downward counterpart?

According to Lindström theorem either compactness fails, or the downward Skolem-Löwenheim theorem should fail. One fails, what about the other?

(*) Please don't discuss the failure of the compactness theorem for $\mathcal L_Q$ here before June 17th, 2013. I gave that part as a homework assignment to my students - some of whom are reading this site.

答:
There's a cute nuke which works here:

Given a structure $\mathfrak{A}$ of arbitrary cardinality in a language $\Sigma$, fix $\theta$ large enough that $\Sigma,\mathfrak{A}\in H_\theta$ and $H_\theta$ satisfies "enough set theory." Now by downward Lowenheim-Skolem for first-order logic, let $M$ be an elementary submodel of $H_\theta$ with $\mathfrak{A},\Sigma\in M$, $\Sigma\subseteq M$, and $\vert M\vert=\aleph_0+\vert\Sigma\vert$. Set $\mathfrak{B}=\mathfrak{A}\cap M$; since finiteness is absolute to $M$, it's not hard to show that $\mathfrak{B}\preccurlyeq_{\mathsf{FOL\forall^\infty}}\mathfrak{A}$ as desired.

And of course this functions as a general recipe for proving appropriate downward Lowenheim-Skolem properties for "concrete" logics.


问:
Tricky Markov Chain Reversibility

I'm very confused about how to do part (a) of the question. I've set up a systems of equations for the stationary distribution based on the transition matrix but thats too messy. I know there's a more elegant way of doing it. I think the proof of reversibility has something to do with irreducibility of the MC. But I'm not sure how to wrangle it out. After the proof, I think there will be some kind of substitution based on the conditional definition of a SD and the detailed balance equations.

答:
Let $\pi_{ijk} = \alpha p_i^2 p_j$,
where $\alpha$ is some normalization constant that ensures that the probabilities add up to one. To establish reversibility, we must show that for all states $ijk$ and $abc$:
$$
\pi_{ijk} P_{ijk,abc} = \pi_{abc} P_{abc,ijk}
$$

To this end, notice that there are three types of nonzero transition probabilities:
\begin{align*}
P_{ijk, ijk} &= p_i \\
P_{ijk, jik} &= p_j \\
P_{ijk, ikj} &= p_k
\end{align*}

It suffices to verify the reversibility equation for the last two types of probabilities. Indeed, observe that:
\begin{align*}
\pi_{ijk} P_{ijk,jik}
&= (\alpha p_i^2 p_j)(p_j)
= \alpha p_i^2p_j^2
= (\alpha p_j^2 p_i)(p_i)
= \pi_{jik}P_{jik,ijk} \\
\pi_{ijk} P_{ijk,ikj}
&= (\alpha p_i^2 p_j)(p_k)
= \alpha p_i^2p_j p_k
= (\alpha p_i^2 p_k)(p_j)
= \pi_{ikj}P_{ikj,ijk} \\
\end{align*}
as desired. $~~\blacksquare$

Title: Tricky Markov Chain Reversibility


问:
Can someone help me solve this misunderstanding please.
Is it a good way to determine isomorphic of two graphs by comparing their adjacent vertices? If their is a different then they are not isomorphic? If they have the same then they are isomorphic? Since they have same number of vertices and each vertex have the same degree. If not, then give an appropriate way of finding their isomorphic please.</p>

答:
It is not enough that the graphs have the same number of vertices of the same degree.  For example, two disjoint triangles and the hexagon, each have six vertices of degree 2.  You are describing the graph isomorphism problem, which is known to be quite hard in general.

To elaborate a little, two graphs are isomorphic if you can pair off the vertices of one with the vertices of the other, and the resulting adjacency matrices are identical.  In the example above, $$\left(\begin{matrix}
1 & 1 & 0 & 0 & 0 & 0\\
0 & 1 & 1 & 0 & 0 & 0\\
1 & 0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 0\\
0 & 0 & 0 & 1 & 0 & 1\\
0 & 0 & 0 & 0 & 1 & 1\end{matrix}\right)$$
and
 $$\left(\begin{matrix}
1 & 1 & 0 & 0 & 0 & 0\\
0 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 0\\
0 & 0 & 0 & 0 & 1 & 1\\
1 & 0 & 0 & 0 & 0 & 1\end{matrix}\right)$$
there is no way to rearrange the vertices of either graph (which will rearrange rows/columns of the matrix), to make the matrices agree.


问:
How do I interpret this with logic notation? "not A doesn't mean not B"
$$
\neg A \rightarrow \neg B \equiv B \rightarrow A
$$

Is this correct?

答:
As negation have more high priority, then implication, we have
$$\neg A \Rightarrow \neg B \equiv \\
(\neg A) \Rightarrow (\neg B) \equiv \\
\neg(\neg A) \lor (\neg B) \equiv\\
A \lor (\neg B) \equiv\\
B \Rightarrow A$$
So, yes, you are correct.


答:
This is an implication so if A is true and b is true it is true
if not A implies not B then A is False implies B is False so the implication is still true.
A implies B is the converse of the statement and the truth of a statement does not imply the truth of the converse</p>

Title: How do I interpret this with logic notation? "not A doesn't mean not B"

Body:
$$
\neg A \rightarrow \neg B \equiv B \rightarrow A
$$

Is this correct?


问:
Examples of isomorphic group presentations with homotopy NONequivalent complexes

(Wikipedia) A presentation complex is a 2-dimensional cell complex associated to any presentation of a group $G$. The complex has a single vertex, and one loop at the vertex for each generator of $G$. There is one 2-cell for each relation in the presentation, with the boundary of the 2-cell attached along the appropriate word.

A group $G$ can be presented in different ways. These presentations need not have homotopy equivalent presentation complexes. I would like to collect as many examples of presentations of a group with presentation complexes that are not homotopy equivalent as possible. I would take the following as sufficient answers to this post:

Referrals to references which have such examples, or discuss methods to construct such presentations;

Examples of such presentations, preferably a large number of them, and preferably not all sharing many common threads (in, say, how they are constructed, dimension, etc.).

The motivation for my question is that I am interested in so-called exotic group presentations. See here for some info on them.

答:
Take
$$G=\langle x,y\mid x^2=y^3\rangle.$$
This is the standard presentation for the fundamental group of the trefoil knot. Of note is the fact that this group is Hopfian and non polycyclic-by-finite.

Now for each $n\in\mathbb{N}$ put
$$G_n=\langle x,y,\overline x,\overline y\mid x^2=y^3,\overline x^2=\overline y^3,x^{2n+1}=\overline x^{2n+1},y^{3n+1}=\overline y^{3n+1}\rangle.$$
The result in each case is a different presentation of $G$. Write $K_n$ for the presentation complex corresponding to $G_n$. Using the fact that the commutator group of $\pi_1K_n\cong G_n\cong G$ is free of rank $2$, and that $G/[G,G]\cong\mathbb{Z}$ you can show without too much difficulty that that $\chi(K_n)=1$ for each $n$.

There are infinitely many distinct homotopy types amongst the $K_n$.

This may be established by studying the homologies of the universal covering spaces $\widetilde K_n$. It can be shown that there are infinitely many integers $m,n$ such that $H_2(\widetilde K_m;\mathbb{Z})$ and $H_2(\widetilde K_n;\mathbb{Z})$ are non-isomorphic $\mathbb{Z}[G]$-modules. An interesting fact is that $K_m\vee S^2\simeq K_n\vee S^2$ although $K_m\not\simeq K_n$.

The calculations for all this are contained in Exotic relation modules and homotopy types for certain 1–relator groups, Algr. Geo. Top. 6, 2163-2173, (2006), by J. Harlander, J. Jensen. They make use of ealier work by P. Berridge and M. Dunwoody in the paper Non‐Free Projective Modules for Torsion‐Free Groups J. London Math. Soc. 2, 433–436, (1979).