问:
How to Find $E(X^2+Y^2)$
I have a question consisting of 5 parts. It is an old practice midterm question I wanted to explain it to my friend. But I seem to have forgotten how to do it.  I could do the easy parts A and B not the rest. I think D is $0$ since they are independent variables but I want to double check with you guys. BUT I need help with the rest.

QUESTION:
Given: $X$ and $Y$ are independent random variables. $X$ has mean $4$ and variance $14$. $X+Y$ has mean $6$ and variance $20$.

Solve:

A-)mean $Y \Rightarrow E(X+Y)=E(X)+E(Y)=6 \Rightarrow 4+E(Y)=6 \Rightarrow$ ANSWER: $E(Y)=2$

B-)variance $y \Rightarrow Var(X+Y)=Var(X)+Var(Y)=20 \Rightarrow 14+Var(Y)=20 \Rightarrow$ ANSWER: $Var(Y)=6$

C-) $E(X^2+Y^2)$: I can't figure this out.

D-)Correlation$(X,Y)$: Should be $0$ since they are independent right?

E-)$COV(XY,X+Y)$: $\Rightarrow COV(XY,X)+COV(XY,Y)$: Can't remember what is next.

Can anyone help me?

答:
$\newcommand{\Cov}{\mathrm{Cov}}$
For c)

$E[X^2+Y^2] = E[X^2]+E[Y^2] = \bigg( Var[X] + E^2[X] \bigg) + \bigg( Var[Y] + E^2[Y] \bigg)$

We used linearity of expectation and definition of variance here

For d) yes independence gives us that the correlation coefficient is zero. Note that the converse is not true in general. Cf. Correlation and Dependence

For e)

$\Cov[XY, X+Y] = \Cov[XY,X] + \Cov[XY,Y] =E\bigg[(XY-E[XY])(X-E[X]) \bigg]+E\bigg[(XY-E[XY])(Y-E[Y]) \bigg] $

Title: How to Find $E(X^2+Y^2)$


问:
Prove that $\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$
$a$ and $b$ are both positive Real Numbers .

I saw this in a math olympiad, and within two days i couldn't solve it.

There is a simple case where:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
We can solve it like this:
$$(a-1)^{2}\geq 0$$
$$\Leftrightarrow a^{2}-2a+1 \geq0$$
$$\Leftrightarrow a^{2}+1 \geq2a$$
$$\Leftrightarrow \frac{a^{2}+1}{a} \geq2$$
With the same method:
$$\frac{b^{2}+1}{b} \geq2 $$
And when we add the two inequalities:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
But in case where:
$$\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$$
I can’t solve it,
Thank you for your help.

答:
Hint: 1) Use AM-GM : $x^2+ 1\ge 2x$
2) UseAM-GM : $ t+\dfrac{1}{t} \ge 2$

ContentLicense: CC BY-SA 4.0


问:
how to understand the (f+g)(x) and (αf)(x) notation
In the following notations:
$$ (f+g)(x)=f(x)+g(x) $$
$$ (αf+βg)(x)=αf(x)+βg(x) $$

can anyone help to understand:

$$(f+g)(x) $$
$$ (αf+βg)(x) $$

?

It's no problem if $f$ and $g$ are linear transformation matrices $A$ and $B$ and $x$ is a vector variable $X$ because I can compute and compare $(A+B)X$ and $AX+BX$.

Otherwise, when it's non-linear, the notation is hard to understand.

for example,

$$ f(x)=\cos x+1 $$
$$ g(x)=e^x-2 $$

I can write:
$$ f(x)+g(x)=\cos x+1 + e^x-2 $$
$$ αf(x)+βg(x) = α\cdot \cos x+α + β\cdot e^x -2β$$

In this case, what are the definition of (and how to evaluate):

$$(f+g)(x) $$
$$ (αf+βg)(x) $$

?

I'm asking this in order to understand the definition of dual space (en.wikipedia.org/wiki/Dual_space) which suggests $(f+g)(x)=f(x)+g(x)$ and $(αf+βg)(x)=αf(x)+βg(x)$ are to be proved/verified (note the word "satisfying" the conditions)

答:
$f+g$ is the function defined by
$$
(f+g)(x) = f(x) + g(x)
$$
for all $x$. So '$f+g$' is just a name we give to a function—but it is a meaningful name, since when people talk of the function '$f+g$' I immediately know what function they are talking about. But if you don't like this notation, then you could just call this function something else instead. E.g. you could let $h$ be the function defined by
$$
h(x)=f(x)+g(x) \, .
$$
It's important to distinguish between a function and its values: $f$ is the function, whereas $f(x)$ is the function evaluated at an arbitrary point $x$. When people speak of 'the function $f(x)$', you should keep in mind that this is just an imprecise yet convenient shorthand. It's the same story with the function $f+g$. The function $f+g$ has the property that when it is evaluated at an arbitrary point $x$,
$$
(f+g)(x)=f(x)+g(x) \, .
$$
Hopefully from this you can guess the meaning of $f \cdot g$ as well: it's the function with the property that
$$
(f \cdot g)(x) = f(x) \cdot g(x) \, .
$$
Sometimes we write simply $fg$ to mean $f \cdot g$, but this notation can be ambiguous since $fg$ can also denote the composite function $f \circ g$.


问:
Taylor series expansion and approximation
I found this amazing question in the last calculus exam, but I don't know how to answer.

Let $T(x) = \ln(1+a) + \frac{1}{1+a}(x-a) - \frac{1}{2(1+a)^2}(x-a)^2 +...+ \frac{(-1)^{n+1}}{n(1+a)^n}(x-a)^n...$

be the expansion of Taylor series $T(x)$ of the function $f(x)=\ln(1+x)$ centered at $a>-1$

and $-1<x\le1+2a$ the interval of convergence of $T(x)$. In that interval $T(x)=f(x)$

1) How can I find all the values of $a$ for calculating $\ln(2)$ with $T(x)$?

2) Furthermore, for $a=0$, how I can determine the degree $n$ of the $T_{n}(x)$ for estimating $\ln(2)$ with an error less than or equal to $1/4$? And give the estimate value of $\ln(2)$.

Any ideas where I can start?

Thanks in advance

答:
The natural choice for $a$ is $a=0$, since that makes all the coefficients easy to compute.  Another plausible choice is $a=e-1$; this is closer to $2$ so the approximation will be better, but it's no longer possible to meaningfully calculate by hand.  If for some reason you have a calculator with an "e" button but a broken "log" button, this is the way to go.

As for estimating error, with $a=0$ and $x=1$, you have an alternating series, so you can use the alternating series error estimate.

ContentLicense: CC BY-SA 3.0


答:
One can actually just use AM-GM directly:
$$\frac{a^2}b+\frac1b+\frac{b^2}a+\frac1a\geq 4\sqrt[4]{\frac{a^2}{b}\cdot\frac{1}{b}\cdot\frac{b^2}a\cdot\frac1a}=4\sqrt[4]{1}=4.$$

Title: Prove that $\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$

Body:
$a$ and $b$ are both positive Real Numbers .

I saw this in a math olympiad, and within two days i couldn't solve it.

There is a simple case where:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
We can solve it like this:
$$(a-1)^{2}\geq 0$$
$$\Leftrightarrow a^{2}-2a+1 \geq0$$
$$\Leftrightarrow a^{2}+1 \geq2a$$
$$\Leftrightarrow \frac{a^{2}+1}{a} \geq2$$
With the same method:
$$\frac{b^{2}+1}{b} \geq2 $$
And when we add the two inequalities:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
But in case where:
$$\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$$
I can’t solve it,
Thank you for your help.


问:
Solving an ODE where the function is evaluated on different points
How does one solve a differential equation of the form
$$f'(x) = f(x) - f(x-1)? $$
Is there any analytic way to obtain a solution?

答:
I haven't found any good general ways to analytically construct solutions to DDEs (I have asked about it before here, not to much avail). In this case, we can do kind-of a guess-and-check type approach by looking for solutions of the form $f(x)=e^{rx}$. Linear combinations of these solutions will give more general solutions to the DDE. We obtain the equation $$
re^{rx} = e^{rx} e^{-r} - e^{rx}
$$
hence $$
r + 1 = e^{-r}
$$
This has a unique real solution $r = 0$, corresponding to the solution $f(x) = 1$. Not very exciting, but fear not, the non-real solutions to $r+1 = e^{-r}$ can give us additional real-analytic solutions. The equation solved gives us $r = W_n(e) - 1$, where $W_n$ is the $n$th branch of the Lambert-W function. Let $r_n = W_n(e) - 1$. The conjugate-symmetry of the branches of the W-function implies that $r_n = \overline{r_{-n}}$, hence $e^{r_n x} + e^{r_{-n} x}$ is a real-valued function, which also solves the equation, as is $i\cdot\left(e^{r_n x} - e^{r_{-n} x}\right)$. We will have the general solutions to the DDE should have the Fourier-series-like form $$
f(x) = \sum_{n=-\infty}^{\infty} a_n e^{r_n x}
$$
where $a_n$ is some sequence of constants such that the series converges. If $a_{-n} = \overline{a_n}$, then the series will be something real-valued. If $a_{n}$ decays fast enough, then it will be analytic.


问:
Hamilton equations by flow of Hamiltonian vector field
I am working on an exercise for my differential geometry class and would like to know if my solution to a problem is correct. The problem is as follows:

A symplectic manifold $S$ is - by definition - endowed with a closed non-degenerate 2-form $\omega$. The map

$$T(S) \longrightarrow T^{\ast}(S) : \, v \longmapsto -i_v \, \omega$$

is an isomorphism (by non-degeneracy of $\omega$), so it has an inverse
$$I: T^{\ast}(S) \longrightarrow T(S) \quad .$$

This can be used to assign to any function $H \in C^{\infty}(S)$ ("Hamiltonian") the vector field $I dH$ ("Hamiltonian vector field"). Let $S$ be described locally by $2n$ coordinates

$$\lbrace x^j \rbrace_{j=1,...,2n} \enspace = \enspace \lbrace p_i, q^i \rbrace_{i=1,...,n} \qquad \text{with} \quad \omega \enspace = \enspace \sum_{i=1}^n dp_i \land dq^i \quad .$$

Show that the flow equation determined by the Hamiltonian vector field corresponds to Hamilton's equations, i.e.

$$\frac{dx^j}{dt} \, \frac{\partial}{\partial x^j} \enspace = \enspace I \, dH \qquad \Longleftrightarrow \qquad \frac{dp_i}{dt} \enspace = \enspace - \frac{\partial H}{\partial q^i} \quad , \qquad \frac{dq^i}{dt} \enspace = \enspace \frac{\partial H}{\partial p_i} \quad .$$

My solution is provided as an answer. Please let me know if it is correct and if not, where my errors are.

答:
The equation

$$\frac{dx^j}{dt} \, \frac{\partial}{\partial x^j} \enspace = \enspace I \, dH$$

can be rewritten as

\begin{align*}
\frac{dx^j}{dt} \, \frac{\partial}{\partial x^j}\enspace &= \enspace I \, dH \\
&= \enspace I \Big( \frac{\partial H}{\partial x^j} \, dx^j \Big) \\
&= \enspace I \Big( \frac{\partial H}{\partial q^i} \, dq^i + \frac{\partial H}{\partial p_i} \, dp_i \Big) \quad .
\end{align*}

Now consider the LHS of this equation. We find

\begin{align*}
\text{LHS} \enspace &= \enspace \frac{dx^j}{dt} \, \frac{\partial}{\partial x^j} \\
&= \enspace \frac{dq^i}{dt} \, \frac{\partial}{\partial q^i} + \frac{dp_i}{dt} \, \frac{\partial}{\partial p_i} \\
&= \enspace \dot{q}^i \, \frac{\partial}{\partial q^i} + \dot{p}_i \, \frac{\partial}{\partial p_i}
\end{align*}

Multiplying both the LHS and the RHS with $dq^k$ yields for the LHS:

\begin{align*}
dq^k \cdot \text{LHS} \enspace &= \enspace \dot{q}^i \, dq^k \cdot \frac{\partial}{\partial q^i} + \dot{p}_i \, dq^k \cdot \frac{\partial}{\partial p_i} \\
&= \enspace \dot{q}^i \, \delta^k_i + \dot{p}_i \cdot 0 \\
&= \enspace \dot{q}^k
\end{align*}

Since $I$ is an isomoprhism, the RHS becomes under this multiplication:

\begin{align*}
dq^k \cdot \text{RHS} \enspace &= \enspace dq^k \cdot I \Big( \frac{\partial H}{\partial q^i} \, dq^i + \frac{\partial H}{\partial p_i} \, dp_i \Big) \\
&= \enspace dq^k \cdot I \Big(  dq^i \Big) \, \frac{\partial H}{\partial q^i} + dq^k \cdot I \Big( dp_i \Big) \, \frac{\partial H}{\partial p_i}
\end{align*}

In order to get (one of) the desired Hamilton equations, we need to show that $dq^k \cdot I \Big(  dq^i \Big)$ equals $0$ and that $dq^k \cdot I \Big( dp_i \Big)$ is equal to $\delta^k_i$. This is done in the following way: Let $\partial_i \equiv \frac{\partial}{\partial q^i}$. Then

\begin{align*}
I^{-1} \big( \partial_i \big) \enspace &= \enspace - i_{\partial_i} \omega \\
&= \enspace - i_{\partial_i} \sum_{\ell = 1}^n dp_{\ell} \land dq^{\ell} \\
&= \enspace - \sum_{\ell = 1}^n \big( i_{\partial_i} dp_{\ell} \big) \land dq^{\ell} - dp_{\ell} \land \big( i_{\partial_i} dq^{\ell} \big) \\
&= \enspace - \sum_{\ell = 1}^n 0 \land dq^{\ell} - dp_{\ell} \land \delta^{\ell}_i \\
&= \enspace dp_{i} \land 1 \\
&= \enspace dp_{i}
\end{align*}

We therefore know that $I^{-1} \Big( \frac{\partial}{\partial q^i} \Big) = dp_i$, meaning that $I(dp_i) = \frac{\partial}{\partial q^i}$. Analogously we get $I(dq^i) = - \frac{\partial}{\partial p_i}$. Inserting this in the above equation for the RHS we get:

\begin{align*}
dq^k \cdot \text{RHS} \enspace &= \enspace dq^k \cdot I \Big( \frac{\partial H}{\partial q^i} \, dq^i + \frac{\partial H}{\partial p_i} \, dp_i \Big) \\
&= \enspace - dq^k \cdot \frac{\partial}{\partial p_i} \, \frac{\partial H}{\partial q^i} + dq^k \cdot \frac{\partial}{\partial q^i} \, \frac{\partial H}{\partial p_i} \\
&= \enspace \delta^k_i \, \frac{\partial H}{\partial p_i} \\
&= \enspace \frac{\partial H}{\partial p_k} \quad .
\end{align*}

Putting all together we find

\begin{align*}
dq^k \cdot \text{LHS} \enspace = \enspace \underbrace{\dot{q}^k \enspace = \enspace \frac{\partial H}{\partial p_k}}_{\text{Hamilton equation of motion}} \enspace = \enspace dq^k \cdot \text{RHS}
\end{align*}

In a similar approach we get the other Hamilton equation of motion.


问:
Mistake in paper? Subgroup of order $4$ in group of order $16$.
I'm reading this proof from this article and I don't see why one argument works. In the Lemma, $n_4(G)$ is the number of elements of order $4$ in $G$.

In the first step of the induction we have a group $G$ with $|G|=16$ and $exp(G)=4$, three maximal subgroups $A,B,C \leq G$, s.t. $A \cap B \cap C =H$ with $|H|=4$.

Now the author says, that if $C_G(H)=G$, then $G$ is abelian.

I don't see, why this is true and I think I found a counterexample:

I looked up $SmallGroup(16,3)=C_2^2 \rtimes C_4$ in which exist three maximal subgroups with intersection $H=Z(G) \cong C_2^2$. Clearly $C_G(H)=G$, but $G$ is not abelian. What am I missing?

SmallGroup(16,3)

Edit:
I think the next argument is wrong aswell. If $C_G(H)\not = G$, then $C_G(H)=A$ is indeed abelian, but its not true, that $B\cong C$. See $SmallGroup(16,13)$ and chose $Q_8$ as a maximal subgroup. It will always lead to $(A,B,C)=(C_4 \times C_2,Q_8,D_8)$ (up to ordering). Is there any quick way to see that all groups of order $16$ contain a number of elements of order 4, that is divisible by $4$?

答:
I think the parts of the proof which you rightly question aren't actually necessary.

We have $n_4(G)\equiv n_4(A)+n_4(B)+n_4(C)\pmod 4$. Since $C_G(H)\ne H$ we can suppose $A$ is abelian and therefore $n_4(A)\equiv 0\pmod 4$.

If both $B$ and $C$ are non-abelian

Then  $n_4(B)\equiv n_4(C)\equiv 2\pmod 4$ and we are finished.

If, say, $B$ is abelian

Then $G=\leq C(H)$ and so $C$ is also abelian. Then  $n_4(B)\equiv n_4(C)\equiv 0\pmod 4$ and we are finished.

Body:
I think the parts of the proof which you rightly question aren't actually necessary.

We have $n_4(G)\equiv n_4(A)+n_4(B)+n_4(C)\pmod 4$. Since $C_G(H)\ne H$ we can suppose $A$ is abelian and therefore $n_4(A)\equiv 0\pmod 4$.

If both $B$ and $C$ are non-abelian

Then  $n_4(B)\equiv n_4(C)\equiv 2\pmod 4$ and we are finished.

If, say, $B$ is abelian

Then $G=<A,B>\leq C(H)$ and so $C$ is also abelian. Then  $n_4(B)\equiv n_4(C)\equiv 0\pmod 4$ and we are finished.


问:
convergence of series with $k!$
check if the following series converges:

$\sum\limits_{k=1}^{\infty} (-1)^k \dfrac{(k-1)!!}{k!!}$

where $k!!=k(k-2)(k-4)(k-6)...$

I came across this exercise while going trough some old exams. I'm pretty sure we have to bound the sequence and apply Leibniz-criteria but after a while i gave up. If you have a little Hint for me just to get me strted , that would be great.

答:
Set $a_k = (k-1)!!/k!!$, so that we're looking at the series $\sum_{k=1}^\infty (-1)^k a_k$. Notice that the odd terms are exactly the normalized central binomial coefficients:
$$
a_{2n} = \frac1{4^n} \binom{2n}n.
$$
By known asymptotics, $a_{2n} \sim 1/\sqrt{\pi n}$ as $n$ grows large. Similarly,
$$
a_{2n-1} = \frac1{2na_{2n}} \sim \frac{\sqrt{\pi n}}{2n} = \frac{\sqrt\pi}{2\sqrt n}
$$
as $n$ grows large. Therefore
\begin{align*}
\sum_{k=1}^{2N} (-1)^k a_k &= \sum_{n=1}^N ( -a_{2n-1}+a_{2n} ) \\
&\approx  \sum_{n=1}^N \bigg( {-}\frac{\sqrt\pi}{2\sqrt n}+\frac1{\sqrt{\pi n}} \bigg) \sim c\sqrt N
\end{align*}
where $c = -\sqrt\pi+2/\sqrt\pi < 0$. So the sum diverges to $-\infty$.</p>

Title: convergence of series with $k!$


问:
Interpretation of transport equation for a two-form
The vorticity equation arising from the Euler equation for a fluid with velocity $\bf u$ reads:

$$
\partial_t {\bf{w}} + \nabla \times( {\bf{w}} \times  {\bf{u}}) =0
$$

where the vorticity is  ${\bf{w}} =\nabla \times  {\bf{u}}$.

Now, assume that we have a 1-form $u$ and its associated "vorticity" 2-form $w=du$. Is it possible to interpret the above vorticity equation in terms of Lie derivative $L_u$ (or other kind of derivatives) and, possibly, Hodge duality? This would give a geometric meaning to the second term in the equation. Maybe something like

$$
\partial_t {w} + L_u w =0 \, ?
$$

In this way, the vector version of the vorticity equation would simply tell us that the field $\bf{w}$ is transported by $\bf{u}$ in a way to satisfy the Reynolds transport theorem.

答:
All you need is the exterior derivative and the Hodge star; you can already anticipate this because the expression you're looking at is $\nabla \times (\mathbf{w}\times \mathbf{u})$, and the curl is really a manifestation of $d$, the metric tensor and the Hodge star, while the cross product is "essentially" the wedge product of two 1-forms once you make the relevant identifications I describe below. You shouldn't expect any Lie derivatives $L_\mathbf{u}$ because they measure the rate of change of tensor fields as you flow them along $\mathbf{u}$, so in the classical vector calculus this would somehow manifest as a dot product with $\mathbf{u}$.

Just so we're on the same page, we equip $\Bbb{R}^3$ with its usual metric tensor $g$, the standard orientation, and let $\star$ be the associated Hodge star. Also, let $g^{\flat}$ and $g^{\sharp}$ be the associated musical isomorphisms. Given any vector fields $\mathbf{w},\mathbf{u}, \mathbf{v}$, we have the following definitions:

The cross product is $\mathbf{w}\times \mathbf{u}:= g^{\sharp}[\star (g^{\flat}(\mathbf{w})\wedge g^{\flat}(\mathbf{u}))]$. i.e given our vector fields, we convert them into $1$-forms, and take their wedge product. This yields a $2$-form, so we use $\star$ to convert this into a $1$-form, then finally use $g^{\sharp}$ to convert this into a vector field again.

The curl is $\nabla \times \mathbf{v}:= g^{\sharp}[\star d g^{\flat}(\mathbf{v})]$, i.e we convert the vector field into a $1$-form, take the exterior derivative, which gives a $2$-form, then use $\star$ to make this a $1$-form, then use the metric to get a vector field again.

If you denote $\alpha := g^{\flat}(\mathbf{u})$, and $\beta=d\alpha$, then you should find that $g^{\flat}(\mathbf{w})=g^{\flat}(\nabla \times \mathbf{u}) = \star d\alpha = \star \beta$, by directly plugging into the above definitions (and recalling that $g^{\flat}$ and $g^{\sharp}$ are inverses).

So, if we start with the expression $\partial_t {\bf{w}} + \nabla \times( {\bf{w}} \times  {\bf{u}})$, and apply $g^{\flat}$ to make this a $1$-form, then you should find by unwinding the definitions that
\begin{align}
g^{\flat}\left(\partial_t {\bf{w}} + \nabla \times( {\bf{w}} \times  {\bf{u}})\right) &= \partial_t(\star d\alpha) + \star d\left((\star d\alpha)\wedge \alpha\right)
\end{align}
If we now apply $\star$ to both sides of this equation, and note that $\star\star = (-1)^{k(n-k)}\text{identity}$, where $k$ is the degree of the form and $n$ is the dimension of the space, it follows that for $n=3$ and $k\in\{0,1,2,3\}$, this is always the identity. Hence,
\begin{align}
\star g^{\flat}\left(\partial_t {\bf{w}} + \nabla \times( {\bf{w}} \times  {\bf{u}})\right) &= \partial_t( d\alpha) +  d\left((\star d\alpha)\wedge \alpha\right)\\
&\equiv \partial_t\beta + d\left((\star \beta)\wedge \alpha\right).
\end{align}
So, this is how you can convert your vector equation into an equation regarding differential forms:
\begin{align}
\partial_t\beta + d\left((\star \beta)\wedge \alpha\right)&= 0, \quad
\text{where $\alpha := g^{\flat}(\mathbf{u})$, $\beta=d\alpha$}.
\end{align}


问:
How would you define $>$ and $<$ on $\mathbb{C}$?
Is there a way to find out whether complex numbers are greater then or less than another complex number?

For numbers like $i$ and $-i$, it is clear than one is bigger than the other. However, what about numbers like $2+3i$ and $5+2i$?

Here's what I think. In the real numbers, you measured how big numbers are with their magnitude and direction. If two real numbers pointed in the same direction, the one with the bigger magnitude is bigger, and if two real numbers were pointed in opposite directions, the one facing the positive direction is bigger.

So, I tried the same thing for complex numbers. For a complex number $a+bi$, the positive directions are along the north of $a$ and east of $b$ axes. So, $2+3i \gt -2-3i$ because $2+3i$ goes in both positive directions, while $-2-3i$ goes in both negative directions. Also,  $2+3i \gt -2+3i$ because $2+3i$ goes in both positive directions, while $-2-3i$ goes positive in only one direction. You have to compare numbers like $3-5i$ and $-3+6i$ using their magnitudes because they both go in a positive and negative direction.

Also, for numbers lying on the axes themselves, you can't compare them unless they are on the same axis.

So, that is my suggestion, does it hold up?

答:
It's not at all clear that one of $\mathrm i$ or $-\mathrm i$ is bigger than the other! What makes you think so?

That's a rhetorical question. I know that you think so because of the sign. But there's a big difference between imaginary and real numbers when it comes to signs: $-1$ behaves in a way that is distinctly different from $1$. For instance, if we square $1$, we get back the same number, but if we square $-1$, we obtain a different number. So if I give you a mystery number that is either $1$ or $-1$, you can find out which one it is by asking me what its square is. If you get back the same number it's $1$, otherwise it's $-1$. Such things don't work to differentiate $\mathrm i$ from $-\mathrm i$. If I give you a mystery number that is either $\mathrm i$ or $-\mathrm i$, it will always remain a mystery which one it is, no matter how much you know about its behavior. For instance, if you square the mystery number, you're guaranteed to get $-1$, no matter which of the two it is. If you take the mystery number to the power of $3$, you'll get minus the mystery number, no matter which number it is. Specifically, there is no algebraic way (that is, using only multiplication and addition) to tell the difference. For this reason, there is no algebraic way to tell which of the two should be larger. And when ordering fields, we always want to be able to do so algebraically, since algebra is the only thing we can natively do with fields.


问:
Find the reaction of the table

A uniform rod of mass $m$ and length $2a$ is held inclined to the vertical at an angle $\alpha$ with its lower end in contact with a smooth horizontal table and is released from rest. Prove that when the inclination of the rod to the vertical is $\theta$, the reaction of the table is
$$\frac{mg(4-6\cos{\alpha}\cos{\theta}+3\sin^2{\theta})}{(1+3\sin^2{\theta})^2} $$

I got reaction of the table as $\dfrac{mg(4-6\cos{\alpha}\cos{\theta}+3\cos^2{\theta})}{(1+3\sin^2{\theta})^2} $ which is not what we need to prove but I think my answer is correct, If $\alpha=0$ and $\theta=0$ then rod is vertical so reaction must be $mg$ which is satisfy my answer but $\dfrac{mg(4-6\cos{\alpha}\cos{\theta}+3\sin^2{\theta})}{(1+3\sin^2{\theta})^2} $ is not satisfied

Can anyone help me for this? Am I correct?

答:
I got the same result as you.

The motion of the rod can be understood as translation of the center od mass (midpoint of rod) and rotation around the center of mass. As there are no horizontal forces acting on the center of mass, the center of mass only moves vertically. If we denote its height by $y$, the conservation of energy gives
$$\frac12 m\dot{y} + \frac12 I\omega^2 + mgy = mgy(0).$$
where the moment of inertia is $I = \frac{ma^2}3$ and $\omega = \dot{\theta}$ is angular rotation around the center of mass. We can note that $y = a\cos\theta$ so $\dot{y} = -a\omega\sin\theta$. Plugging this in we get
$$\frac12ma^2\omega^2\sin^2\theta + \frac16ma^2\omega^2 + mga\cos\theta=mga\cos\alpha $$
which implies
$$\omega = \sqrt{\frac{6g(\cos\alpha - \cos\theta)}{a(1+3\sin^2\theta)}}.$$

Now, since the table is smooth, the reaction force $\vec{N}$ acts only upwards. Its projection orthogonally to the rod is $N\sin\theta$. The resulting torque $Na\sin\theta$ drives the rotation of the rod. The rotational equation of motion is
$$N\sin\theta = I\dot{\omega}.$$
After some calculation, we get
$$\dot\omega = \sqrt{\frac{6g}{a}} \frac1{2\sqrt{\frac{\cos\alpha - \cos\theta}{1+3\sin^2\theta}}}\frac{(4-6\cos\alpha\cos\theta +3\cos^2\theta)\sin\theta}{2(1+3\sin^2\theta)^2} \omega = \frac{3g(4-6\cos\alpha\cos\theta +3\cos^2\theta)\sin\theta}{2a(1+3\sin^2\theta)^2}$$
so
$$N = I\frac{\dot\omega}{\sin\theta} = \frac{ma^2}3\frac{3g(4-6\cos\alpha\cos\theta +3\cos^2\theta)}{2a(1+3\sin^2\theta)^2} = \frac{mg(4-6\cos\alpha\cos\theta +3\cos^2\theta)}{(1+3\sin^2\theta)^2}.$$

Title: Find the reaction of the table


问:
Restrictions on an array of numbers
We have an array of numbers. The mean of these numbers is $200$. $X$ is the largest element and $Y$ is the smallest one. We have to satisfy $X=3+2Y$. There's no limit on the size of the array (the amount of elements it contains).

I want to find out which numbers couldn't appear in the array.

Since the smallest element have to be lesser than $200$ (so the mean can be $200$), than the largest number have to be lesser than $3+2*200=403$.

My question is: are there any other restrictions on the numbers that could appear in the array?

答:
Working over the integers temporarily, you could have $Y = 99$ so that $X = 201$.  The array consisting of 99 (once) and 201 repeated 101 times has mean 200.

At the other extreme, you could have $Y = 199$ so that $X = 401$.  The array consisting of 199 repeated 201 times and a single 401 has mean 200.

Switching to the reals, you choose $Y$ so that $X$ is greater 200 by an arbitrarily small amount and make an array with mean 200.  Similarly, you can choose $Y$ less than 200 by an arbitrarily small amount and make an array with mean 200.

Are there other kinds of restrictions you were hoping for?

Subsequent addition. Following up on the previous comments gives bounds for $X$ and $Y$:

It's clear that no $Y \ge 200$ could work as the smallest number in a list with mean 200 (since a larger $X$ has to be included). Since any $Y < 200$ can work, that gives $X = 3 + 2Y < 403$.

Similarly you cannot have $X \le 200$ as the largest number of a list with mean 200.  Since any $X>200$ can work, the equivalent expression $Y = (X-3)/2$ gives $Y > 197/2 = 98.5$.

Title: Restrictions on an array of numbers


答:
You could define inequalities for complex numbers. For example, we could make the following definition:

We say that $z$ is 'greater than' $w$ if $|z|>|w|$.

The only question that remains is whether this definition is helpful. And the answer to that is: not really. The real numbers are closed under multiplication and addition; that is, if $a$ and $b$ are positive, then $a \cdot b$ is positive, and $a + b$ is positive. It turns out that these two properties are crucially important for defining inequalities in a meaningful way. And neither of these properties hold for complex numbers, meaning that writing $z>w$ just doesn't make that much sense.

To explain why closure under addition and multiplication is relevant, we must first define what we actually mean by $a>b$. One common way to do this is to take the notion of a positive number for granted. Let's denote the collection of positive numbers as $\mathbf{P}$. One of the most important properties of $\mathbf{P}$ is the trichotomy law:

For every number $x$, one and only one of the following holds:

$x=0$.

$x$ is in the collection $\mathbf{P}$.

$-x$ is in the collection $\mathbf{P}$

While this may seem obvious, it is necessary for the following definitions:

$a>b$ if $a-b$ is in $\mathbf{P}$

$a<b$ if $b>a$

$a \geq b$ if $a>b$ or $a=b$

$a \leq b$ if $b \geq a$

If it were possible for $x$ to be in the collection $\mathbf{P}$, and $-x$ also be in the collection, then we could have $a>b$ and $b>a$ ! Hence, the trichotomy law is necessary if we are to maintain our sanity. Another important property is closure under multiplication and addition, as mentioned earlier. If the real numbers weren't closed under multiplication, this would also cause headaches. For example, we wouldn't be able to say that if $a>0$, and $b>0$, then $ab>0$.

Fortunately, the real numbers do contain a collection $\mathbf{P}$ with all these properties, but the complex numbers do not. This can even be proven fairly easily. Suppose that the complex numbers did satisfy the trichotomy law and closure under multiplication and addition. Let $x$ be an arbitrary complex number. If $x$ is positive, then $xx = x^2$ would also be positive. If $x$ is negative, then $-x$ would be positive, and so $(-x)(-x) = (-x)^2=x^2$ would be positive. It follows that for all non-zero $x$, $x^2$ is positive. In particular, if $x=i$, then $x^2 = -1$ would be positive. But $1=1^2$ would also be positive. This violates the trichotomy law: $1$ and $-1$ can't both be positive.

Hence, the idea of positivity is incoherent for complex numbers, and so too is the idea of inequalities. The best we can do is say that two complex numbers are either equal, or unequal.

Title: How would you define $>$ and $<$ on $\mathbb{C}$?

Body:
Is there a way to find out whether complex numbers are greater then or less than another complex number?

For numbers like $i$ and $-i$, it is clear than one is bigger than the other. However, what about numbers like $2+3i$ and $5+2i$?

Here's what I think. In the real numbers, you measured how big numbers are with their magnitude and direction. If two real numbers pointed in the same direction, the one with the bigger magnitude is bigger, and if two real numbers were pointed in opposite directions, the one facing the positive direction is bigger.

So, I tried the same thing for complex numbers. For a complex number $a+bi$, the positive directions are along the north of $a$ and east of $b$ axes. So, $2+3i \gt -2-3i$ because $2+3i$ goes in both positive directions, while $-2-3i$ goes in both negative directions. Also,  $2+3i \gt -2+3i$ because $2+3i$ goes in both positive directions, while $-2-3i$ goes positive in only one direction. You have to compare numbers like $3-5i$ and $-3+6i$ using their magnitudes because they both go in a positive and negative direction.

Also, for numbers lying on the axes themselves, you can't compare them unless they are on the same axis.

So, that is my suggestion, does it hold up?


问:
Distance between points on a polar coordinate system
Imagine a polar coordinate system with radius = 1. The line that connects these two coordinates is a spiral: (0,0) (10π,1).

How can the length of this spiral be calculated? (Points above are just samples for a "multi rotation" situation... But it can be any two points..)

答:
I'll give a hint on how to setup the problem:

You can use the formula for the arc-length in polar coordinates.

$$ds=\sqrt{r^{2}+(\frac{dr}{dθ})^{2}}dθ$$

with $L=\int ds$.

With a standard spiral you have r to be a linear function of theta $r(\theta) = r_{0}+b(\theta - \theta_{0})$.


问:
Metrizability is a topological property?
How could I show that metrizability is a topological property?

Well, this means that if I have a set $X$ that is metrizable and a homeomorphic function $f$ from $X$ to $Y$, then I need to show that $Y$ is metrizabke, correct?

If I let $d$ be a metric in $X$? How do I construct a metric in $Y$ using the bijection $f$?

答: We can use the Nagata- Smirnov metrization theorem.</p>


问:
Multivariate piecewise linear interpolation
Let $f:X\rightarrow E$ be a uniformly continuous function into some Banach space $E$ ($\dim(E)\geq 1$) from a (non-empty) compact subset $X\subseteq \mathbb{R}^n$ where $n\geq 1$.

Why, for every $\epsilon>0$, is it true that there exists a piecewise linear function $\hat{f}$ satisfying
$$
\sup_{x\in X} \|f(x)-\hat{f}(x)\|_E<\epsilon?
$$

I was thinking of doing the following:

Since $K$ is compact then $K\subseteq [-diam(K),diam(K)]^n$

Pick an equally spaced mesh of points $\{k_n\}_{n\leq N}$ such that $\cup_{n}\operatorname{Ball}(k_n,\delta)$ covers $[-diam(K),diam(K)]^n$; which we know must exist by compactness of $[-diam(K),diam(K)]^n$ (for any given $\delta>0$).

Pick $\delta$ so that $\omega_f(\delta)\leq \epsilon$; where $\omega_f$ is the modulus of continuity of $f$; which we can do since $\omega_f$ is right continuous (or we may wlog assume it so),

Now here is where I get stuck:
How do we then prove that there is a function which linearity interpolates the points
$\{f(x_n)\}_n$?

If we can do that; then by the convexity of the norm $\|\cdot\|_E$ we can infer the existence of $\hat{f}$; no?

Follow-up: This problem seems like it must be covered by some literature (it's too natural not too be).  Does anyone know any good references where I can read about this problem (with Banach co-domain)?

答:
I would try a little different strategy: As $X$ is compact and $f$ is continuous it follows that $f$ is uniformly continuous. Let's $\varepsilon >0$. As $f$ is  uniformly continuous there exists $\delta$ such that $\lVert f\left(x\right)-f\left(y\right)\rVert_E\lt\varepsilon/2$ whenever $\lvert x- y\rvert\lt\delta$. Let $T=\left\{t_{i_1,i_2,\dots i_n}\right\}$ be equaly spaced mesh of points in $X$  indexed with "coordinates" $(i_1, i_2,\dots, i_n)\in\mathbb N^n$ such that $\lvert t_{i_1,i_2,\dots,i_n} - t_{i'_1,i'_2,\dots,i'_n}\rvert\lt \delta$ for any $t_{i_1,i_2,\dots,i_n} ,t_{i'_1,i'_2,\dots,i'_n}\in T$ such that $\max\left\{\lvert i_1-i'_1\rvert,\lvert i_2-i'_2\rvert,\cdots,\lvert i_n-i'_n\rvert\right\}\le 1$. (in other words, distance between two "neighborhood" points is less then $\delta$, including "diagonal neighborhood").

Let $C_{i_1,i_2,\dots,i_n}$ be $n$-dimensional cube with vertices $\{t_{i_1+p,i_2+p,\dots,i_n+p}\}$ where $p\in\{0,1\}$

Lets define linear function $f_{i_1,i_2,\dots,i_n}$ in following way $$f_{i_1,i_2,\dots,i_n}(x)=f_{i_1,i_2,\dots,i_n}(x_1,x_2,\dots,x_n)=\begin{cases} \Lambda(x)  & x\in C_{i_1,i_2,\dots,i_n} \\ 0 & x\not\in C_{i_1,i_2,\dots,i_n}\end{cases}$$ where $\Lambda$ is linear function that agrees with $f$ on verices of cube $C_{i_1,i_2,\dots,i_n}$. More on this bellow...

Now set $$\hat f\left(x\right)=\sum_{i_1,i_2,\dots,i_n} f_{i_1,i_2,\dots,i_n}\left(x\right).$$ For every $x\in X$ we have
$$\left\lVert f\left(x\right)-\hat f\left(x\right)\right\rVert_E=\left\lVert f\left(x\right)-f_{i_1,i_2,\dots,i_n}\left(x\right)\right\rVert_E$$ where $(i_1,i_2,\dots,i_n)$ is such that $x\in C_{i_1,i_2,\dots,i_n}$ and $$\left\lVert f\left(x\right)-f_{i_1,i_2,\dots,i_n}\left(x\right)\right\rVert_E\le\left\lVert f\left(x\right)-f\left(x_{i_1,i_2,\dots,i_n}\right)\right\rVert_E+\left\lVert f\left(x_{i_1,i_2,\dots,i_n}\right)-f_{i_1,i_2,\dots,i_n}\left(x\right)\right\rVert_E\lt \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$ That proves that $$ \sup_{x\in X} \|f(x)-\hat{f}(x)\|_E\lt\varepsilon$$

So how can we construct $\Lambda$? Well, it would be very ugly expression if I try to write it down, but you can get it easily with induction on dimension $n$.

Fer example, lets find a linear function $\Lambda\left(x,y\right)$ such that $\Lambda\left(0,0\right)=A_{00}$, $\Lambda\left(1,0\right)=A_{10}$, $\Lambda\left(0,1\right)=A_{01}$ and $\Lambda\left(1,1\right)=A_{11}$. This will work: $$ \Lambda\left(x,y\right) = A_{11}xy+A_{01}\left(1-x\right)y+A_{10}x\left(1-y\right)+A_{00}\left(1-x\right)\left(1-y\right).$$

Lets consturct $\overline{\Lambda}\left(x,y\right)$ such that $\overline\Lambda\left(0,0\right)=B_{00}$, $\overline\Lambda\left(1,0\right)=B_{10}$, $\overline\Lambda\left(0,1\right)=B_{01}$ and $\overline\Lambda\left(1,1\right)=B_{11}$. Now if we set $$\Psi\left(x,y,z\right)=\left(1-z\right)\Lambda\left(x,y\right)+z\overline\Lambda\left(x,y\right)$$ we will get function such that $\Psi\left(0,0,0\right)=A_{00}$, $\Psi\left(0,0,1\right)=B_{00}$, $\Psi\left(1,0,0\right)=A_{10}$, and so on, and so on....


答:
I think that to define an order on a field, you would want a positive cone $\Bbb P$in that field, behaving like the set of positive reals within $\Bbb R$. So you would want, for complex numbers $z$ and $w$, satisfaction of the three conditions
\begin{align}
\Bbb P+(-\Bbb P)=\Bbb C\\
z,w\in\Bbb P&\Longrightarrow z+w\in\Bbb P\\
z,w\in\Bbb P&\Longrightarrow zw\in\Bbb P\,.
\end{align}
The first condition  means that every complex number is to be the sum of a “positive” number and the negative of some positive number. I think you’ll persuade yourself, no matter how you try, that with the first condition in place, the third condition is unsatisfiable.


问:
Number of Paths on a Grid From S to M
I was reading about a the derivation of the formula for the number of paths from one corner to another corner of a H by W grid here and I wondered whether it is possible to apply the result: $\binom{(H-1)(W-1)}{H-1}$ to find the number of paths from a given square on the top row of the grid to another selected square in the bottom row.

For example the number of paths from B to J.
I thought of reducing the grid to just the columns B to D, counting the paths there and then adding the possible paths from every other square outside the reduced grid. However I had trouble in finding a formula for the possible paths from outside of the reduced grid.

In a path you cant repeat a square, and you can move to any adjacent square.

答:
A possible (but involved) approach

You could represent moves by complex numbers:

$1$ is move right, $-1$ is move left, $i$ is move up, $-i$ is move down.

Then a path is a sequence such as $1,-i, -1, ...$

The conditions for a valid path then have simple arithmetical equivalences.

To move from B to J The total sum of the numbers must be $2-5i$.

To stay in the grid For any natural number $n$, the sum $S_n$ of the first $n$ numbers must  satisfy $4\ge \Re (S_n)\ge -1, 0\ge \Im (S_n)\ge -5. $

To not visit any square twice No sum of successive numbers must have sum $0$.

This looks suitable to be programmed if that is of interest.

Title: Number of Paths on a Grid From S to M


问:
Should equality, like any other predicate, involve only bound variables in a valid FOL sentence?
I have found a nice interactive tutorial about sequent calculus. It contains all the inference rules for the first order logic (13 rules in the green box). One can also find these rules on Wikipedia.

For my project I need to extend these rules to the case of FOL with equality. I found this document, which seems to be exactly what I need. In particular, what I have found is that one need to add one more axiom (page 17):

$
\frac{}{\Gamma \implies \Delta, t = t}
$

One also adds the following three derivation rules:

$
\frac{\Gamma, t = t \implies \Delta}{\Gamma \implies \Delta}
$

$
\frac{\Gamma, t_1 = t_2, \phi(t_2) \implies \Delta}{\Gamma, t_1 = t_2, \phi(t_1) \implies \Delta}
$

$
\frac{\Gamma, t_1 = t_2 \implies \Delta, \phi(t_2)}{\Gamma, t_1 = t_2 \implies \Delta, \phi(t_1)}
$

What is not clear to me, is that equality is a binary predicate (predicate that takes two arguments). So, $t_1 = t_2$ is not a valid FOL sentence because a valid sentence should not contain free variables. In other words, if we have a predicate operating on some variables, then these variables should be "removed" by an existential of universal quantifier.

So, my question is: Can equality be applied to unbound variables?

Maybe my problem is that I misinterpret the rule and the rule does not imply that there are unbound variables in the expressions.

ADDED:

To explain better what I mean, an example of a valid FOL sentence is:

$
\forall q_1 : ((\exists q_2 : A(q_1, q_2)) \land (\forall q_2 : B(q_1, q_2)))
$

In the above expression, we have variables $q_1$ and $q_2$ but the whole expression does not "depend" on them because these dependencies are removed by the quantifiers.

Now, we can remove predicates $A$ or $B$ by equality. So, that instead of $A(q_1, q_2)$ we will have $q_1 = q_2$. In this case, equality will be applied to bound variables.

SO, it is not clear to me how equality can be applied to unbound variables. Or, in other words, if we have it, how should we interpret this expression? We say that $q_1 = q_2$ without saying anything about $q_1$ and $q_2$? So, what are $q_1$ and $q_2$ in this case?

答:
If you want to stay in the realm of sentences, then yes, free variables are not allowed at all. However, a sequent calculus doesn't need to restrict itself to sentences; in fact, most presentations I've seen operate on sets of formulas (which personally I find annoying, but that's just my preference), which may or may not include free variables.

This is what's going on here. To quote from page $12$:

A sequent $S:=(\Gamma,\Delta)$ is a pair of formula sets

(emphasis mine). So there's no error either in the text or in your understanding of sentences, the text is simply setting up a different system than you were expecting.

It's a good idea at this point to carefully present the relevant completeness theorem for this sequent calculus. When only sentences are involved there are no difficulties, but once formulas enter the picture we have to be a bit careful. I'll follow the terminology from the text above.

First, the definitions (which differ a bit from most texts):

Signatures and structures are defined in the usual way, except that we demand countability. This is a weird thing to do from a "pure logic" perspective, but in a computation-oriented context it makes sense - and by the downward Lowenheim-Skolem theorem, if we're restricting attention to countable languages we can also restrict attention to countable structures without affecting basic properties like validity, satisfiability, and entailment.

An interpretation for a set of formulas $\Phi$ is then a pair $(\mathcal{S},\mathcal{V})$ where $\mathcal{S}$ is a structure and $\mathcal{V}$ is a variable assignment, that is, a map from the free variables of $\Phi$ (= the variables which are free in some element of $\Phi$) to (the underlying set of) $\mathcal{S}$.

Note that this differs from the usual meaning of "interpretation" in the semantics of first-order logic, where it is used instead to refer to the part of a structure giving meaning to the non-logical symbols.

The notion of an interpretation satisfying a set of sentences is then defined in the usual Tarskian recursive manner. When an interpretation $(\mathcal{S},\mathcal{V})$ satisfies a set of formulas $\Phi$, we write "$(\mathcal{S},\mathcal{V})\models\Phi$" and say that the pair $(\mathcal{S},\mathcal{V})$ is a model of $\Phi$.

There's actually a bit of sloppiness in the text w/r/t the above two bulletpoints. The relevant notions are only introduced (e.g. in Definition $6$) for individual formulas in place of the *set of formulas $\Phi$, but later on (page $12$) everything is generalized to sets of formulas. In the above, I've written things as they should be.

With this in hand, we can state the completeness theorem:

Using the definition of $\vdash$ in the text, for all sequents $(\Gamma,\Delta)$, we have $$\Gamma\vdash\Delta\quad\iff\quad\forall \mbox{ interpretations }I, [\forall \gamma\in\Gamma(I\models\gamma)\implies \exists \delta\in\Delta(I\models\delta)].$$ (Technically the left-to-right direction is soundness, but due to its fairly trivial nature we generally consider it part of the completeness theorem. This is bad practice if we're working in a context where soundness isn't obvious of course, but that isn't happening here.)


问:
What is $\arg(z)$ if $z=\left(1+i\sqrt3\right)^{2i}$?
$z=(1+i\sqrt{3})^{2i}$, what is the respective $\arg(z)$? How do I calculate the exponent $2i$?

答: This exponentiation is multivalued. Since $1+i\sqrt{3}=e^{\ln 2+\pi i/3+2n\pi i}$ for $n\in\Bbb Z$, each such $n$ affords a choice $z=e^{2i\ln 2-2\pi/3-4n\pi}$ corresponding to one branch of $\ln z$.


问:
Notation for spectral decomposition
One of the conditions for the spectral decompositions $P_{\lambda} (\lambda \in Sp(A))$ of diagonalizable operator $A$ acting upon a finite dimensional vectorspace is $P_{\lambda} P_{\mu} = \delta_{\lambda, \mu} P_{\mu}$.

What does $\delta_{\lambda, \mu}$ mean? The notes that I am using do not define it, and I am struggling to find it online.

答: It is a the Kronecker delta, a function of $\lambda$ and $\mu$ --- equal to one when $\lambda=\mu$, and zero otherwise. It is expressing that $P_\mu$ and $P_\nu$ are orthogonal ('perpendicular').

Title: Notation for spectral decomposition


问:
Concern regarding a proof from brilliant
Here in the proof part they deduced that the arc length is $$\lim_{n \to \infty}\sum_{n=1}^{\infty}\sqrt{1+(f'(x_{i}*))^2}\Delta x$$
then said that it was equal to the definite integral. But I didn't understand why. Shouldn't it be $f(x_{i})$ for the limit to be an integral? Since the $x_{i}*$ are not continuous and evenly spaced as $x_i$. Please explain.

答:
Different authors define Riemann integration in different ways.
In the definition that I think I remember from high school,
to integrate $f$ over the interval $[a,b]$ we define an approximate integral
of $n$ steps by setting $x_k = a + k\frac{b-a}{n}$ for $k = 0, 1, 2, \ldots, n.$
This divides the interval $[a,b]$ into $n$ equal-sized intervals between
consecutive subscripted $x$ values, $[x_{k-1}, x_k].$
Then the $n$th approximation of the integral is
$$ \frac{b-a}{n} \sum_{k=1}^n f(x_k). \tag1$$
Or perhaps it was
$$ \frac{b-a}{n} \sum_{k=1}^n f(x_{k-1}). \tag2$$
There actually are distinct names for these sums: $(1)$ is the "right endpoint" sum because you can get it by evaluating $f$ at the right-hand end of each interval
$[x_{k-1}, x_k],$ while $(2)$ is the "left endpoint" sum because you can get it by evaluating $f$ at the left-hand end of each interval $[x_{k-1}, x_k].$

But there are other ways to define the integral. You could evaluate $f$ at an $x$-value midway between the two ends of each interval. That's the "midpoint" sum.
You could average the values of $f$ at the left and right ends of each interval.
That's a "trapezoid" rule.

You could take the "upper" sum, which means you look for the greatest value of $f$ in each interval $[x_{k-1}, x_k].$ This largest value might occur at $x_{k-1}$ (left end), or at $x_k$ (right end), or (in cases where the function has a local maximum in the interval) anywhere between $x_{k-1}$ and $x_k.$

You could take the "lower" sum, which means you look for the least value of $f$ in each interval $[x_{k-1}, x_k].$ Again this might use an $x$-value at the left end, the right end, or somewhere in between.

All of these kinds of sums (except one) are illustrated in an answer to What does it mean for a function to be Riemann integrable?
And as it turns out, for functions that satisfy some reasonable criteria, it really does not matter which of these definitions we use--in every case we are going to to define the value of the integral as the limit of the sequence of sums produced by letting $n$ increase without bound, and the limit always comes out to the same number.

Now we can do one better: do not even make a rule that says which $x$ to pick within each interval. The only constraint on the $k$th term of the sum is it has to use an $x$ from somewhere in the interval $[x_{k-1}, x_k].$
If we say that $\xi_k$ represents a value somewhere in the interval $[x_{k-1}, x_k],$
the $n$th approximate integral looks like this:
$$ \frac{b-a}{n} \sum_{k=1}^n f(\xi_k).$$
Some authors use a different symbol than $\xi_k$.
For example, the symbol might be $x_k^*.$

We can go even further than this, though.
Why does each $x_k$ have to follow the rigid formula $x_k = a + k\frac{b-a}{n}$?
Instead, we merely insist that
$a = x_0 < x_1 < x_2 < \cdots < x_{n-1} < x_n = b.$
Now the sum looks like this:
$$ \sum_{k=1}^n (x_k - x_{k-1}) f(\xi_k).$$
So now even the $x_k$s are not necessarily uniformly spaced.
This means we cannot rely on the number $\frac{b-a}{n}$ to be the "grid size":
instead, we choose the largest of all interval lengths $x_k - x_{k-1}$
as the "grid size". This also means it is not good enough to say $n$ increases without bound; what we really need is to know that the grid size decreases to zero.
That's the definition shown in On definition of Riemann integral.

It appears that the definition of integration that was used in that answer on brilliant.com was different from the one you learned.


答:
You write
$$
z=\exp(2i\log(1+i\sqrt{3}))
$$
where you choose a branch of the logarithm. Since
$$
1+i\sqrt{3}=2e^{i\pi/3+2ki\pi}
$$
you have $\log(1+i\sqrt{3})=\log2+i\alpha$ (where $\alpha=\pi/3+2k\pi$).

Then we have
$$
2i\log(1+i\sqrt{3})=2i\log2-2\alpha
$$
and therefore
$$
\exp(2i\log(1+i\sqrt{3})=\exp(2i\log2-2\alpha)=e^{-2\alpha}e^{2i\log2}
$$
Thus the argument is $2\log2$. The modulus is not determined uniquely and depends on what branch of the logarithm you choose.

Title: What is $\arg(z)$ if $z=\left(1+i\sqrt3\right)^{2i}$?

Body: $z=(1+i\sqrt{3})^{2i}$, what is the respective $\arg(z)$? How do I calculate the exponent $2i$?


问:
Are Differential equations without "$x$" considered separable?
For example:
Do we consider all differential equations of the form $y'(x)=f(y)$ separable?
why/why not?

答:
It would be separable. Check if $f(y) = 0$ is a solution. Then ignore this case to get
$$\frac{y'}{f(y)} = 1 $$
so your solution looks like
$$\int\frac{y'}{f(y)} = x + C $$


答:
I am interpreting what you have written as:

$$y'(x)=f(y(x))$$

Separable equations can be written in the form:

$$A(y(x))y'(x) = B(x)$$

So if you choose $A(y) = \frac{1}{f(y)}$ and $B(x) = 1$. $f(y)$ should not equal zero on your interval of analysis in order for you to be able to make this transformation.

Title: Are Differential equations without "$x$" considered separable?

Body:
For example:
Do we consider all differential equations of the form $y'(x)=f(y)$ separable?
why/why not?


问:
Find the general solution of the next differential equation
Im trying to solve this problem but have no idea how start

$$y''-\frac{2x}{x^{2}+1}y'+\frac{2}{x^{2}+1}y=(x^2+1)^2$$

答:
Hint:
$$y''-\dfrac{2x}{x^{2}+1} y'+\dfrac{2}{x^{2}+1}y=0$$
Note that $y=x$ is a solution. So try to  reduce the order with $y=vx$.


问:
An integral involving Fresnel integrals $\int_0^\infty \left(\left(2\ S(x)-1\right)^2+\left(2\ C(x)-1\right)^2\right)^2 x\ \mathrm dx,$
I need to calculate the following integral:
$$\int_0^\infty \left(\left(2\ S(x)-1\right)^2+\left(2\ C(x)-1\right)^2\right)^2 x\ \mathrm dx,$$
where
$$S(x)=\int_0^x\sin\frac{\pi z^2}{2}\mathrm dz,$$
$$C(x)=\int_0^x\cos\frac{\pi z^2}{2}\mathrm dz$$
are the Fresnel integrals.

Numerical integration gives an approximate result $0.31311841522422385...$ that is close to $\frac{16\log2-8}{\pi^2}$, so it might be the answer.

答:
Following on from Ron...

Make the substitutions: $x=\sqrt{y}$, $p=1/2+1/2\,\sqrt {1+4\,r}$, to get:

$\displaystyle \dfrac{64}{{\pi }^{2}}\,\int _{0}^{\infty }\!x \left( \int _{1}^{\infty }\!{\frac {\sin
 \left( 2\,\pi \,{x}^{2}p \left( p-1 \right)  \right) }{p}}{dp}
 \right) ^{2}{dx}$,

$\displaystyle=\dfrac{32}{{\pi }^{2}}\,\int _{0}^{\infty }\! \left( \int _{0
}^{\infty }\!{\frac {2\,\sin \left( 2\,\pi\, y\, r \right) }{\sqrt {1+4\,
r} \left( 1+\sqrt {1+4\,r} \right) }}{dr} \right) ^{2}{dy}
$,  see Appendix,

$\displaystyle=\dfrac{16}{{\pi }^{2}}\,\int _{-\infty}^{\infty }\! \left( \int _{0
}^{\infty }\!{\frac {2\,\sin \left( 2\,\pi \,y \,r \right) }{\sqrt {1+4\,
r} \left( 1+\sqrt {1+4\,r} \right) }}{dr} \right) ^{2}{dy}
$,

$\displaystyle=\dfrac{16}{{\pi }^{2}}\,\int _{-\infty}^{\infty }\! \left( \int _{-\infty
}^{\infty }\!{\frac {H(r)\,\sin \left( 2\,\pi \,\,y\,r \right) }{\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right) ^{2}{dy}
$ :  $H \left( r \right) =\cases{1&$0\leq x$\cr -1&$x<0$\cr}$.

Note then that:

$\displaystyle\left(\int _{-\infty
}^{\infty }\!{\frac {H(r)\,\sin \left( 2\,\pi \,\,y\,r \right) }{\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right) ^{2}=\displaystyle\left|-\dfrac{i}{2}\int _{-\infty
}^{\infty }\!{\frac {H(r)\,e^{ -i2\pi \,y\,r } }{\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr}+\dfrac{i}{2}\int _{-\infty
}^{\infty }\!{\frac {H(r)\,e^{ i2\pi\, y\,r } }{\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right| ^{2}$

$=\displaystyle\left|\int _{\infty
}^{\infty }\!{\frac {\left(H(r)-H(-r)\right)\,e^{ i2\pi \,y\,r } }{2\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right| ^{2}=\displaystyle\left|\int _{\infty
}^{\infty }\!{\frac {\,e^{ i2\pi \,y\,r } }{\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right| ^{2}.$

We can now use Plancherel's theorem which states:

$\displaystyle \int _{-\infty }^{\infty }\!  \left| f \left( y \right)
 \right|   ^{2}{dy}=\int _{-\infty }^{\infty }\!
 \left| F \left( r \right)  \right|   ^{2}{dr}
$ : $ \displaystyle F \left( r \right) =\int _{-\infty }^{\infty }\!f \left( y \right) {
e^{-i2\pi r y}}{dy}
$,

and thus:

$\displaystyle\dfrac{16}{{\pi }^{2}}\,\int _{-\infty}^{\infty }\! \displaystyle\left|\int _{\infty
}^{\infty }\!{\frac {\,e^{ i2\pi \,y\,r } }{\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }}{dr} \right| ^{2}{dy}$

$=\displaystyle\dfrac{16}{{\pi }^{2}}\,\int _{-\infty}^{\infty }\! \displaystyle\left|{\frac {1 }{\sqrt {1+4\,
|r|} \left( 1+\sqrt {1+4\,|r|} \right) }} \right| ^{2}{dr}$,

$=\dfrac{16}{{\pi }^{2}}\displaystyle\int _{0}^{\infty }\!{\frac {2}{ \left( 1+4\,r \right)  \left( 1+
\sqrt {1+4\,r} \right) ^{2}}}{dr}$.

Now if we undo the substitution made earlier by letting $r = p^2-p$:

$\dfrac{16}{{\pi }^{2}}\displaystyle\int _{0}^{\infty }\!{\frac {2}{ \left( 1+4\,r \right)  \left( 1+
\sqrt {1+4\,r} \right) ^{2}}}{dr}=\dfrac{16}{{\pi }^{2}}\int _{1}^{\infty }\!{
\frac {1}{2{p}^{2} \left( 2\,p-1 \right) }}{dp}$,

and then make one final substitution $p=\dfrac{1}{q+2}$ we get:

$\displaystyle\dfrac{16}{{\pi }^{2}}\int _{1}^{\infty }\!{
\frac {1}{2{p}^{2} \left( 2\,p-1 \right) }}{dp}=-\dfrac{16}{{\pi }^{2}}\int_{-2}^{-1}\dfrac{1}{2}+\dfrac{1}{q}{dq}=\dfrac{16\ln(2)-8}{\pi^2}$.

Appendix:

Note that the integral: $$\int _{0
}^{\infty }\!{\frac {2\,\sin \left( 2\,\pi \,y \,r \right) }{\sqrt {1+4\,
r} \left( 1+\sqrt {1+4\,r} \right) }}{dr} $$

converges by the Chartier-Dirichlet test because: $$f(r)={\frac {2\, }{\sqrt {1+4\,
r} \left( 1+\sqrt {1+4\,r} \right) }}$$

is monotonic and continuous on $ \mathbb R^+ $ and $f(r)\rightarrow0$ as $r\rightarrow \infty$, and because:
$$\left|\int_{0}^{b}\sin\left(2\pi\,y\,r\right){dr}\right|$$
is bounded as $b\rightarrow\infty$.


问:
What is the equivalent to $\tan\left(\frac{7\pi}{4}-\theta\right)$?

What is the equivalent to $\tan\left(\frac{7\pi}{4}-\theta\right)$?

I'm very new to the mathjax formatting and the site in general but when I work the problem out it seems as though the answer should be $-1$.

$$\frac{\left(-1\ -\ \tan\left(\theta\right)\right)}{1+\left(-1\cdot\tan\left(\theta\right)\right)}$$

$$\frac{\left(-1\ -\ \tan\left(\theta\right)\right)}{1-\tan\left(\theta\right)}$$

答: The answer would be $-1$ only if numerator was the negative of denominator. But it isn't.


问:
Lattice Point Geometry
A student of mine asked me the following question:

Three lattice points (points with integer coordinates) are chosen at random with replacement (meaning you can select the same point more than once) in the interior of the square defined by $-99 \le x \le 100$ and also $-99 \le y \le 100$. The probability that the area of the triangle thus formed (which may be degenerate) is an integer can be expressed in the form $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

I know the answer. However, I need an explanation with the answer.

Thank you!

Also I have just a few hours before I go back to teach him.

答:
Let the three lattice points be $(a_1, b_1)$, $(a_2, b_2)$, and $(a_3, b_3)$. By the shoelace formula (as @Jeannette said), the area of the triangle is
$\frac{|a_1b_2 + a_2b_3 + a_3b_1 - b_1a_2 - b_2a_3 - b_3a_1|}{2}.$
Note that whether or not the triangle has integer area depends only on the parity of the points.

From $-99 \leq x \leq 100$ and $-99 \leq y \leq 100$, we see that each coordinate has an equal probability of being even or odd. Hence, all three points are chosen from the four parity combinations $(even,even)$, $(even,odd)$, $(odd,even)$, and $(odd,odd)$, with replacement. It is true that the triangle have non-integer area iff they are all of different parity combinations (think about it).

We can choose any of the four parity combinations for the first point. The second point has a probability of $\frac{3}{4}$ of having a different parity from the first point. The third point has a probability of $\frac{1}{2}$ of having a different parity than the other two points. Hence, the probability that all three points have different parities (and therefore the triangle has non-integer area) is $\frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8}$.

Therefore, the probability that the triangle has integer area is $1 - \frac{3}{8} = \frac{5}{8}$. The final answer is $5 + 8 = \boxed{13}$.

Title: Lattice Point Geometry


问:
How to calculate with $\mathbb{Z}/2\mathbb{Z}$ with an unknown variable?
When calculating with numbers from a $\mathbb{Z}/2\mathbb{Z}$ how do you deal with unknown variables? For example, if I have the following term:

$(a - 1)(a - 1) - (a - 1) - (a - 1) = a^2 + 1$

Or is this incorrect?

答:
$\newcommand{\Z}{\mathbb{Z}}$First note that
\begin{align*}
(a-1)(a-1) - (a-1) - (a-1)
&= a^2 + 1 - a + 1 - a + 1\\
&= a^2 - 2a + 3
\end{align*}
At least under more normal circumstances. In $\Z/2\Z$, though, $2x=0$ for every $x \in \Z / 2 \Z$. This does, in fact, include variables. (After all, the variables are eventually elements from $\Z / 2 \Z$, we just don't know which ones.) Thus,

$$-2a = -(2a) = 0 \qquad 3 = 2 + 1 = 0 + 1 = 1$$

(This might be easier to grasp if you recall that $\Z/2\Z$ is, in reality, sets of equivalence classes rather than strictly numbers: that is, the statement $3=1$ means, rather, "$3$ and $1$ share the same equivalence class in $\Z/2\Z$.")

And thus,

$$(a-1)(a-1) - (a-1) - (a-1) = a^2 + 1$$

as you would expect.


答:
$\mathbb{Z}/2\mathbb{Z}$ is a ring with characteristic $2$, so that means that, for any $x\in\mathbb{Z}/2\mathbb{Z}n$, then $x+x=0$. From this, we know directly that the part $-(a-1)-(a-1)=0$, so we're left with

$$(a-1)(a-1),$$
and calculating it normally we get that
$$(a-1)(a-1)=a^2-2a+1,$$
but $2 \equiv 0$ in $\mathbb{Z}/2\mathbb{Z}$, so we conclude that
$$(a - 1)(a - 1) - (a - 1) - (a - 1) = a^2 + 1$$
and you were right.

Title: How to calculate with $\mathbb{Z}/2\mathbb{Z}$ with an unknown variable?

Body:
When calculating with numbers from a $\mathbb{Z}/2\mathbb{Z}$ how do you deal with unknown variables? For example, if I have the following term:

$(a - 1)(a - 1) - (a - 1) - (a - 1) = a^2 + 1$

Or is this incorrect?


问:
What happens to point at infinity when you complete the square of an elliptic curve?
Suppose we have some elliptic curve (just suppose we are working over $\mathbb{R}$) defined by the equation
$$Y^2Z + a_1XYZ + a_3YZ^2 = X^3 + a_2X^2Z + a_4XZ^2 + a_6Z^3 $$
with a point at infinity $[0,1,0]$. Now we can go to its affine part
$$y^2 + a_1xy + a_3y = x^3 + a_2x^2 + a_4x + a_6 $$
Now I can visualize this curve well, and I can visualize the point at infinity as being where all vertical lines intersect.

Now the typical thing I see (since we aren't in characteristic 2) is complete the square, letting $y \mapsto (y-a_1x - a_3)/2$ our equation becomes
$$y^2 = 4x^3 + b_2x^2 + 2b_4x + b_6 $$
where we have $b_2$, $b_4$, $b_6$ defined in the appropriate way.

Now my question is, how is this transformation affecting our point at infinity? What does this transformation look like in terms of the projective equation, not just the affine equation? Im struggling to see how this works in terms of the projective curve, and why we are allowed to do this and why it stays an elliptic curve (why it still looks like the initial equation in some $X,Y,Z$).

答: This change of coordintes is linear, so it extends to a change of coordinates on $\Bbb P^2$ by $X\mapsto X$, $Y\mapsto (Y-a_1X-a_3Z)/2$, $Z\mapsto Z$. You can see directly that this preserves the point at infinity $[0:1:0]$, and that it gives you the projective equation $Y^2Z=4X^3+b_2X^2Z+2b_4XZ^2+b_6Z^3$.


问:
Is that statement a well-known result in probability?
Here is the statement :

Let $(\{\pm1\}^n,\mathcal{P}(\{\pm1\}^n), \mathbb{P})$ a probability space, $(X_i)_{1\le i\le n}$, $\ n$ random real variables which are mutually independents. For all $i\in\{1,...,n\}$ : $\mathbb{P}(X_i =1)=\frac{1}{2}=\mathbb{P}(X_i = -1)$. Let $(v_i)_{1\le i\le n}$, $\ n$ unit vectors of $E$ (a $\mathbb{R}$-vector space with the scalar product : $E\to \mathbb{R}, (x,y)\mapsto\langle x,y\rangle$).

We define the random variable $U : \{\pm1\}^n \to \mathbb{R},\ \omega \mapsto \Vert \sum\limits_{i=1}^{n}X_i (w)v_i \Vert^2$. Then there exists $(\epsilon_1,...,\epsilon_n) \in  \{\pm1\}^n$ such that : $\Vert\sum\limits_{i=1}^n \epsilon_iv_i\Vert\le \sqrt{n}$. And moreover, if the $(v_i)_{1\le i\le n}$ are not pairwise orthogonals then there exists $(\epsilon_1,...,\epsilon_n) \in  \{\pm1\}^n$ such that : $\Vert\sum\limits_{i=1}^n \epsilon_iv_i\Vert>\sqrt{n}$.

It seems to be linked with Pythagoras theorem for euclidean space. Is this result well-know in probability theory or does this statement have a name ?

Moreover, what would happen if if we took $\{\pm1\}^{\mathbb{N}}$ instead of $\{\pm1\}^n$ ?

Thanks in advance !

答:
A simplified statement of the problem is this:

Given $n$ unit vectors $\{v_1, ..., v_n\}$ on an inner product space with inner product $\langle v, z \rangle$ and norm $||v|| = \sqrt{\langle v,v\rangle}$, prove

a) There is a vector $(w_1, ..., w_n) \in \{-1,1\}^n$ such that $||\sum_{i=1}^n w_iv_i||\leq \sqrt{n}$.

b) If there is some pair of distinct vectors $v_i, v_j$ such that $\langle v_i, v_j\rangle \neq 0$ then there is a vector $(w_1,..., w_n) \in \{-1,1\}^n$ such that $||\sum_{i=1}^n w_i v_i||> \sqrt{n}$.

While this problem has nothing to do with probability, one way to solve it is to define a random vector $(W_1, ..., W_n) \in \{-1,1\}^n$ with i.i.d. entries and use the fact that if $Z$ is a random variable that satisfies $E[Z]\leq 0$ then there is an outcome $\omega$ such that $Z(\omega) \leq 0$.


问:
Historical context of the quadruple $(x,y,z,t)$ where $x^2+y^2=z^2+t^2$
Recently my study is related to an equation $x^2+y^2=z^2+t^2$. I have learned that Euler studied about numbers which can be represented as a sum of two squares in two different way. But I do not know if there is anyone who studied this equation before him.

So I would like to ask if there is a name and the historical context of the quadruple $(x,y,z,t)$ so that $x^2+y^2=z^2+t^2$.

答:
Any Pythagorean triple that has a hypotenuse that factors into $2$ distinct primes, to any power will have two sets of $(A,B)$ pairs where $A^2+B^2=C^2$. For example $65=5\times13$ so
$$33^2+56^2=65^2\quad
63^2+16^2=65^2\\
\implies 33^2+56^2=63^2+16^2
$$

There are infinite numbers of such triple pairs. There will be $2^{n-1}$ primitive triples for any given hypotenuse where $(n)$ is the number of its unique prime factors. Some will have more triples than this but the extras will be non-primitive.

The only historical context I can think of is the development of sums of squares representing integers whether they are squares or not. Some studies on the subject are here and here.


问:
How can I rewrite this series of equations?
I have the following series of equation:

$m_1=m_1$

$m_2=m_1q_1$

$m_3=m_1q_1q_2$

$m_4=m_1q_1q_2q_3$

and so on till some $i=M$.

What I want is to write this in one function such that by plugging in any $i$ it outputs me the corresponding right hand side of the function.

What I found is the following:

$m_i=m_1\prod_{2}^i (q_{i-1})$

However, this only holds for $i>1$.

It it possible to construct such a thing for all $i$>0?

答:
The formula
$$m_i = m_1 \prod_{j=1}^{i-1} q_j$$ holds for all $i>0$, under the convention that an empty product is $1$.


问:
A question regarding pluriharmonic functions
A real-valued function $u$ that is defined on a domain $D$ of $\mathbb{C}^n$ is pluriharmonic if

$u$ is of class $C^2$ and

for all $a\in D$ and $b\in\mathbb{C}^n$ the function  $\lambda\mapsto u(a+\lambda b)$ is harmonic on the set
$$D=\{\lambda\in\mathbb{C}: a+\lambda b\in D\}.$$

My question is: What if the second condition is satisfied only for $b\not=0$? Is $u$ still pluriharmonic? (the first condition is satisfied)

答: The condition is automatically satisfied for $b = 0$ because $\lambda \mapsto u(a)$ is a constant function and therefore harmonic.


问:
Complex Integral : How can I evaluate $\int_{C_R} f(z) dz$?
I have to calculate $I$ ,using complex integral.
\begin{equation}
I:=\displaystyle\int_{-\infty}^{\infty} \dfrac{\text{log}{\sqrt{x^2+a^2}}}{1+x^2}.(a>0)
\end{equation}

Let $f(z)=\dfrac{\text{log}(z+ia)}{1+z^2}$.

$
C_R : z=Re^{i\theta}, \theta : 0 \to \pi.$

$C_1 : z=t, t : -R \to R.$

From Residue Theorem,
\begin{equation}
\displaystyle\int_{C_1} f(z) dz + \displaystyle\int_{C_R} f(z) dz = 2\pi i \text{Res}(f, i).
\end{equation}

If $R \to \infty$, $\displaystyle\int_{C_1} f(z) dz  \to \displaystyle\int_{-\infty}^{\infty} f(x) dx. $

And my mathematics book says that if $R \to \infty$, $\displaystyle\int_{C_R} f(z) dz \to 0.$ I cannot understand why this holds.

My attempt is following :

\begin{align}
\Bigg|\displaystyle\int_{C_R} f(z) dz \Bigg|
&=\Bigg|\displaystyle\int_0^{\pi} f(Re^{i\theta}) i Re^{i\theta} d\theta \Bigg| \\
&\leqq \displaystyle\int_0^{\pi} \Bigg| R \dfrac{\text{log} (Re^{i\theta} + ia)}{1+R^2e^{2i\theta}} \Bigg| d\theta \\
&\leqq \displaystyle\int_0^{\pi} \dfrac{R}{R^2-1} \Bigg| \text{log} (Re^{i\theta}+ia) \Bigg| d\theta \\
&= \displaystyle\int_0^{\pi} \dfrac{R}{R^2-1} \Bigg| \text{log} (R\cos \theta+i(R\sin \theta +a)) \Bigg| d\theta
\end{align}

I expect that
$ \displaystyle\int_0^{\pi} \dfrac{R}{R^2-1} \Bigg| \text{log} (R\cos \theta+i(R\sin \theta +a)) \Bigg| d\theta \to 0$,
but I cannot prove this.

I would like to give me some ideas.

答:
I fact the integral can be evaluated by means of contour integrating in the complex plane.
Let' consider the following closed contour C in the upper half-plane with the cut from $x=ia$ to $x=0$. We also choose the integrand $f(x)=\frac{\log(x^2+a^2)}{1+x^2}$ on the positive part of axis (right side). For the convenience of calculations we take $a<1$.

$x=i$  is a simple pole, and $x=ia$ is a logarithm branch point.

Due to the cut our integrand is a single-valued function in the upper half-plane. We integrate from $x=0$  along the positive axis, then along a big circle (radius R) counter clockwise, then along negative part of axis, from $x=0$ to $x=ia$, clockwise around the point $x=ia$ (circle of radius $r$) and then from $x=ia$ along the second bank of the cut to the initial starting point. It is easy to see that the integral along the big circle $\to0$ as soon as $R\to{\infty}$, and so does the integral around the point $x=ia$ when $r\to0$ . In the left quarter of half-plane (on the negative part of axis) the function $(x^2+a^2)$ will get the factor $\exp(2\pi{i})$ due to the logarithm branch point $x=ia$ (which we go around in the positive direction when integrating along a big circle).

Taking all together we get

$\oint_C\frac{\log(x^2+a^2)}{1+x^2}$=$\int_0^{\infty}\frac{\log(x^2+a^2)}{1+x^2}dx+\int_R+\int_{-\infty}^0\frac{\log[(x^2+a^2)\exp(2\pi{i})]}{1+x^2}dx$+$\int_0^{ia}\frac{\log[(x^2+a^2)\exp(2\pi{i})]}{1+x^2}dx+\int_r + $
$+\int_{ia}^{0}\frac{\log(x^2+a^2)}{1+x^2}dx=$$2\pi{i}Res_{x=i}\frac{\log(x^2+a^2)}{1+x^2}$

Setting $R\to{\infty}$ and $r\to0$, and taking into consideration that there are integrals cancelling each other, we get:

$2I=\int_{-\infty}^{\infty}\frac{\log(x^2+a^2)}{1+x^2}dx=$$-\int_{-\infty}^{0}\frac{2\pi{i}}{1+x^2}dx-\int_0^{ia}\frac{2\pi{i}}{1+x^2}dx+2\pi{i}Res_{x=i}\frac{\log(x^2+a^2)}{1+x^2}$.

When calculating the residual we have to bear in mind that $\log(a-1)=\log[\exp(\pi{i}(1-a)]=\pi{i}\log(1-a)$- due to the choice of logarithm branch (and the choice $a<1$).

Finally we get $2I=-{\pi}^2i+2\pi\int_0^{a}\frac{1}{1-x^2}dx+\pi\log(a^2-1)=\pi\log(\frac{1+a}{1-a})+\pi\log[(1+a)(1-a)]$.

$$I(a)=\int_{-\infty}^{\infty} \dfrac{\text{log}{\sqrt{x^2+a^2}}}{1+x^2}=\frac{1}{2}\int_{-\infty}^{\infty} \dfrac{\text{log}(x^2+a^2)}{1+x^2}=\pi\log(1+a); a<1$$

Calculations for $a>1$ can be done in the same fashion.


问:
Weak-*-Topology and sequences
Let $X$ be a Banach Space , $x\in X$ and $(x^{*}_{n})_{n\in\mathbb{N}}$ a sequence in $X'$ with
a weak-$*$-clusterpoint $x^{*}$. Does this imply that $x^{*}(x)$ is a clusterpoint of $(x^{*}_{n}(x))_{n\in\mathbb{N}}$ (which is a sequence in $\mathbb{R}$)?
I have difficulties with this topology beacause i don't know which norm or metric i have to use (if there is any). How does a neighbourhood in the weak-*-topology look like?

答: Weak* topology is neither normable nor metrizable. A basic neighborhood of $x^{*}$ is of the form $\{y^{*}: |y^{*}(x_i)-x^{*}(x_i)|<r_i \forall i \in \{1,2,...,N\}\}$ for some integer $N$, some positive numbers $r_i$ and some vectors $x_1,x_2,..,x_n \in X$. To answer your question take a neighborhod with $N=1$ and $x_1=x$.

Title: Weak-*-Topology and sequences


问:
Localization of Kähler differential
Let $\phi: R\rightarrow S$ be a ring homomorphism, $\mathfrak{q}$ be a prime of $S$ and $\phi^{-1}(\mathfrak{q})=\mathfrak{p}$.

Then I'd like to know if $\Omega_{S/R}\otimes R_\mathfrak{p}\cong \Omega_{S_\mathfrak{q}/R_\mathfrak{p}}$ is correct. It seems that Akhil Mathew uses this in proposition 1.8 on page 6, but I do not understand the proof:

Proposition 1.8. Let $\phi: R\rightarrow S$ be a morphism of finite type. Then $\phi$ is unramified if and only if for every $\mathfrak{p}\in\operatorname{R}$, we have $k(\mathfrak{p})\rightarrow S\otimes_R k(\mathfrak{p})$ is unramified.

Proof: Suppose $k(\mathfrak{p})\rightarrow S\otimes_R k(\mathfrak{p})$ is unramified for all $\mathfrak{p}\subset R$. We then know that $\Omega_{S/R}\otimes_R k(\mathfrak{p})=\Omega_{S\otimes_R k(\mathfrak{p})/k(\mathfrak{p})}=0$ for all $\mathfrak{p}$. By localization, it follows that $\mathfrak{p}\Omega_{S_\mathfrak{q}/R_{\mathfrak{p}}}=\Omega_{S_{\mathfrak{q}}/R_{\mathfrak{p}}}$ for any $\mathfrak{q}\in\operatorname{S}$ lying over $\mathfrak{p}$.

In fact, he deduces $\mathfrak{p}\Omega_{S_\mathfrak{q}/R_\mathfrak{p}}=\Omega_{S_\mathfrak{q}/R_\mathfrak{p}}$ from $\Omega_{S/R}\otimes \frac{R_\mathfrak{p}}{\mathfrak{p}R_\mathfrak{p}}=0$. So I guess that he uses $\Omega_{S/R}\otimes R_\mathfrak{p}\cong \Omega_{S_\mathfrak{q}/R_\mathfrak{p}}$.

From my point of view, $\Omega_{S/R}\otimes R_\mathfrak{p}\cong\Omega_{S\otimes_RR_\mathfrak{p}/R_\mathfrak{p}}$, this is the tensor property of Kahler differential. But I don't think $S\otimes_RR_\mathfrak{p}\cong S_\mathfrak{q}$ holds, since this should require that $\phi(R-\mathfrak{p})=S-\mathfrak{q}$, but this is not true in general.

Any help is much appreciated.

答:
You are correct that $\Omega_{S/R}\otimes_R R_p = \Omega_{S_p/R_p}$ which is not the same as $\Omega_{S_q/R_p}$ in general (any of the counterexamples at your previous question will suffice here too). What is true is that $\Omega_{S_q/R_p}$ is the localization of $\Omega_{S_p/R_p}$ at the multiplicative set $S\setminus q$.

The way this is used in the proof is that $$\Omega_{S\otimes_R k(p)/k(p)}\cong \Omega_{S/R}\otimes_R (R_p/pR_p) \cong \Omega_{S/R} \otimes_R R_p \otimes_{R_p} R_p/pR_p \cong \Omega_{S_p/R_p} \otimes_{R_p} R_p/pR_p,$$ which is isomorphic to $\Omega_{S_p/R_p}/p\Omega_{S_p/R_p}$, as $M\otimes_R R/I\cong M/IM$ for any ring $R$, ideal $I$, and module $M$. So $p\Omega_{S_p/R_p}=\Omega_{S_p/R_p}$, and localizing at $S\setminus\mathfrak{q}$ and using the fact that $T^{-1}\Omega_{S/R}=\Omega_{T^{-1}S/R}$ we obtain the desired equality.

Title: Localization of Kähler differential


问:
How do I solve for the reaction force at the hinge
The question given is

A uniform rod, of mass $m$ and length $2a$ smoothly hinged to a vertical wall is connected to a point on the wall above the hinge by a light inelastic string. Find the magnitude and direction of the force on the rod from the hinge.

The diagram given is

I have constructed the forces on to the freebody diagram as such

Using the vertical component of net force:
$$F_y=T \sin(20)-mg+R_V=0$$
$$\implies T\sin(20)=mg-R_V$$

Using the horizontal component of net force:
$$F_x=T \cos(20)-R_H=0$$
$$\implies T\cos(20)=R_H$$
$$\implies T=\frac{R_H}{\cos(20)}$$

Finding the moment around the hinge A:
$$\tau=T\sin(20)*2a\sin(50)-mg*a\sin(50)=0$$
$$\implies T\sin(20)=\frac{mg}{2}$$
$$\implies T=\frac{mg}{2\sin(20)}$$

Then, equating the two equations for $T\sin(20)$ I get:
$$\frac{mg}{2}=mg-R_V$$
$$\implies 2R_V=mg$$

Finally equating both equations for $T$:
$$\frac{mg}{2\sin(20)}=\frac{R_H}{\cos(20)}$$
$$\implies \frac{2R_V}{2\sin(20)}=\frac{R_H}{\cos(20)}$$
$$\implies \frac{R_V}{R_H}=\frac{\sin(20)}{\cos(20)}$$

And since $R_V,R_H$ are the vertical and horizontal components of the same force respectively, the $tan^{-1}$ of this ratio is equal to the angle between the resultant and the horizontal hence:

$$\tan^{-1}(\frac{\sin(20)}{\cos(20)})=20°$$

Which is wrong as the answer given in the book is that the angle is $82.7°$ and the reaction force is $0.474mg$, I am aware that this entire method is a hackjob but I have no clue on how to tackle this question properly, are the reaction forces I drew the problem or is it my working onwards?

How would I tackle this problem properly?

答:
HINT for "How would I tackle this problem properly?"

There is a very clever method for tackling this type of problem. The key is not to split the forces up into components as you have done.

The point is that the rod is acted upon by just three forces.

These must add up to $0$ and so can be drawn head-to-tail to form a triangle and trig can then be used e.g. cosine rule/sine rule, ...

Even more usefully the three forces must pass through the same point! In your diagram, the forces of $mg$ and $T$ pass through a point, $P$ say,  on the rope directly over the middle of the rod. Then these two forces have no moment about $P$. But the moment of all the forces about $P$ must be zero and so the reaction must also pass through $P$ in order to have zero moment about $P$.

The entire question is thus reduced to the geometry of triangles.


问:
Prove that $\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$
$a$ and $b$ are both positive Real Numbers .

I saw this in a math olympiad, and within two days i couldn't solve it.

There is a simple case where:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
We can solve it like this:
$$(a-1)^{2}\geq 0$$
$$\Leftrightarrow a^{2}-2a+1 \geq0$$
$$\Leftrightarrow a^{2}+1 \geq2a$$
$$\Leftrightarrow \frac{a^{2}+1}{a} \geq2$$
With the same method:
$$\frac{b^{2}+1}{b} \geq2 $$
And when we add the two inequalities:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
But in case where:
$$\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$$
I can’t solve it,
Thank you for your help.

答:
You can use your method, making $(a-1)^2$ and $(b-1)^2$ appearing was a nice move.

$$\frac{a^2+1}{b}+\frac{b^2+1}{a}-4=\frac 1{\underbrace{ab}_{>0}}(\underbrace{a^3+a+b^3+b-4ab}_E)$$

Now use $2ab\le a^2+b^2$

$E\ge a^3+a+b^3+b-2a^2-2b^2=a(a-1)^2+b(b-1)^2\ge 0$


问:
How would you define $>$ and $<$ on $\mathbb{C}$?
Is there a way to find out whether complex numbers are greater then or less than another complex number?

For numbers like $i$ and $-i$, it is clear than one is bigger than the other. However, what about numbers like $2+3i$ and $5+2i$?

Here's what I think. In the real numbers, you measured how big numbers are with their magnitude and direction. If two real numbers pointed in the same direction, the one with the bigger magnitude is bigger, and if two real numbers were pointed in opposite directions, the one facing the positive direction is bigger.

So, I tried the same thing for complex numbers. For a complex number $a+bi$, the positive directions are along the north of $a$ and east of $b$ axes. So, $2+3i \gt -2-3i$ because $2+3i$ goes in both positive directions, while $-2-3i$ goes in both negative directions. Also,  $2+3i \gt -2+3i$ because $2+3i$ goes in both positive directions, while $-2-3i$ goes positive in only one direction. You have to compare numbers like $3-5i$ and $-3+6i$ using their magnitudes because they both go in a positive and negative direction.

Also, for numbers lying on the axes themselves, you can't compare them unless they are on the same axis.

So, that is my suggestion, does it hold up?

答:
You can define $<$ (nearly) anyway you like.

If wanted two I could define $x < y$ if when you spell them out in english words $x$ occurs lower in alphabetical order than $y$.

so $8 < 18 <805   < 85 < 82< 11< 10 < 3 < 0 < 0.5 < 0.7 < 0.6 < 0.1 < 0.2$ and so on

Because eight < eighte < eighth < eightyf < eightyt < el < te < thr < zero < zero point f < zero point se < zero point si < zero point o < zero point t.

Kinda dumb though.

And we could define an order of compairing $a+bi$ to $c + di$ as $a+bi < c+di$ if $a < c$ or if $a=c$ and $b < d$.  But $a + bi = c+di$ if $a=c$ and $b=d$.  And $a +bi > c+di$ if $a > c$ or if $a=c$ and $b >  d$.

That's called the lexigraphical order and it looks okay by it's actually dumb too.

The thing is if we want to define an order we want it to obey in certain ways.

One thing we want is we want if $a < b$ then $a+c < b+c$ for all $c$

The lexigraphical order does do this but what you, incorrectly, claim is the order on the reals does not.

You claim, incorrectly,  $-5 > -2$ because $|-5| > |-2|$ and $-5, -2$ both point in the same direction.  If this were true we would like $-5 + 10 > -2 + 10$ and that would mean $5 > 8$ which it isn't.

We also want that if $a < b$ and $m > 0$ then $am < bm$.

This fails on the lexigraphical order.  $i = 0 + i > 0 + 0i = 0$.  so $i*i$ should be greater than $0*i = 0$.  but $i*i=-1 < 0$.

And it is because of this requirement we can not have a reasonable order on the complex.  That just isn't possible.

Consider:   Is $0 < 1$ or is $0 > 1$?

If $0 > 1$ then $0 + (-1) > 1 +(-1)$ and $-1 > 0$.  Therefor $(-1)*(-1) > (-1)*0$ so $1 > 0$.  But that contradicts what we assumed.  So we can't have $0 > 1$ so we must have $0 < 1$.

And therefor $0 + (-1) < 1 + (-1)$ and $- 1 < 0$.

[Note, this means also that if $5 > 2$ the $5-5 > 2-5$ and $0 > -3$ and $0 + (-2) > -3+(-2)$ and so $-2 > -5$.  this is why your claim that $-5 > -2$ was wrong.  If they point in the negative direction the one with a bigger magnitude must be less than the one with the smaller magnitude.]

Okay, but what about $i$.  Is $i > 0$.   If so then $i*i > 0*i$ and $-1 > 0$.  But we just showed we had to have $-1 < 0$.

Okay, so $i < 0$.  But that means $i + (-i) < 0 + (-i)$ so $0 < -i$.  So $0*(-i) < (-i)(-i)$.  So $0 < (-i)^2 = (-1)^2(i)^2 = 1*(-1) = -1$.  So $0 < -1$.... but we just said....

So it's impossible.

If we want $a < b \implies a+c < b+c$ and we also want $a 1 \implies am < bm$ then it's just not going to be possible to compare two complex numbers.  It just can not be done.

Body:
You can define $<$ (nearly) anyway you like.

If wanted two I could define $x < y$ if when you spell them out in english words $x$ occurs lower in alphabetical order than $y$.

so $8 < 18 <805   < 85 < 82< 11< 10 < 3 < 0 < 0.5 < 0.7 < 0.6 < 0.1 < 0.2$ and so on

Because eight < eighte < eighth < eightyf < eightyt < el < te < thr < zero < zero point f < zero point se < zero point si < zero point o < zero point t.

Kinda dumb though.

And we could define an order of compairing $a+bi$ to $c + di$ as $a+bi < c+di$ if $a < c$ or if $a=c$ and $b < d$.  But $a + bi = c+di$ if $a=c$ and $b=d$.  And $a +bi > c+di$ if $a > c$ or if $a=c$ and $b >  d$.

That's called the lexigraphical order and it looks okay by it's actually dumb too.

The thing is if we want to define an order we want it to obey in certain ways.

One thing we want is we want if $a < b$ then $a+c < b+c$ for all $c$

The lexigraphical order does do this but what you, incorrectly, claim is the order on the reals does not.

You claim, incorrectly,  $-5 > -2$ because $|-5| > |-2|$ and $-5, -2$ both point in the same direction.  If this were true we would like $-5 + 10 > -2 + 10$ and that would mean $5 > 8$ which it isn't.

We also want that if $a < b$ and $m > 0$ then $am < bm$.

This fails on the lexigraphical order.  $i = 0 + i > 0 + 0i = 0$.  so $i*i$ should be greater than $0*i = 0$.  but $i*i=-1 < 0$.

And it is because of this requirement we can not have a reasonable order on the complex.  That just isn't possible.

Consider:   Is $0 < 1$ or is $0 > 1$?

If $0 > 1$ then $0 + (-1) > 1 +(-1)$ and $-1 > 0$.  Therefor $(-1)*(-1) > (-1)*0$ so $1 > 0$.  But that contradicts what we assumed.  So we can't have $0 > 1$ so we must have $0 < 1$.

And therefor $0 + (-1) < 1 + (-1)$ and $- 1 < 0$.

[Note, this means also that if $5 > 2$ the $5-5 > 2-5$ and $0 > -3$ and $0 + (-2) > -3+(-2)$ and so $-2 > -5$.  this is why your claim that $-5 > -2$ was wrong.  If they point in the negative direction the one with a bigger magnitude must be less than the one with the smaller magnitude.]

Okay, but what about $i$.  Is $i > 0$.   If so then $i*i > 0*i$ and $-1 > 0$.  But we just showed we had to have $-1 < 0$.

Okay, so $i < 0$.  But that means $i + (-i) < 0 + (-i)$ so $0 < -i$.  So $0*(-i) < (-i)(-i)$.  So $0 < (-i)^2 = (-1)^2(i)^2 = 1*(-1) = -1$.  So $0 < -1$.... but we just said....

So it's impossible.

If we want $a < b \implies a+c < b+c$ and we also want $a<b; m > 1 \implies am < bm$ then it's just not going to be possible to compare two complex numbers.  It just can not be done.


问:
Integrating an Infinite exponent tower
$$ \int_0^1 x^{2^{x^{2^{x^{\ldots}}}}} ~~ dx =  ~~?$$

What I've tried so far:

$$ x^{2^{x^{2 \ldots}}} \overbrace{=}^{\text{def}} y \\ x^{2^{y}} = y \\ 2^y \ln (x) = \ln(y)$$

Now applying the chain rule:

$$\begin{align} \\&2^y \cdot \frac{dy}{dx} \cdot \ln(2) \cdot \ln(x) + \frac{1}{x} \cdot 2^y \ln(y) =  \frac{dy}{dx} \cdot \frac1y \\ &\implies \frac{dy}{dx} \left (2^y \cdot \ln(x)  \cdot \ln(2) - \frac1y \right) = - \frac1x 2^y \cdot \ln(y) \\ &\implies \frac{dy}{dx} = \left ( \frac{- \frac1x 2^y \cdot \ln(y)}{2^y \cdot \ln(x)  \cdot \ln(2) - \frac1y} \right) \end{align}$$

The problem that this looks like a pretty harsh equation to solve for $dy$ and then substitute it back to the integral, with the fact that we need to find the value of $x$ in terms of $y$.

Any help would be appreciated!

答:
We can transform this to an integral of an elementary function:
\begin{align}
\int_0^1ydx&=[xy]_0^1-\int_0^1xdy\\
&=1-\int_0^1y^{1/(2^y)}dy.\\
\end{align}
However, I don't think there is a closed-form to this integral since Wolfram Alpha doesn't show anything. I tried to use some series expansion, but that didn't seem to work.

Title: Integrating an Infinite exponent tower


问:
Image of the Norm on a Finite Dimensional Extension of $\mathbb{Q}_p$
I've been trying to see whether following assertion is true in order to give a quick proof of another problem I was doing: if $K$ is a finite dimensional extension of the $p$-adic numbers $\mathbb{Q}_p$, we have the multiplicative field norm $N_{K/\mathbb{Q}_p}: K \rightarrow \mathbb{Q}_p$.  While the norm is not guaranteed to be surjective, is it always possible to find some $k \in K$ such that $N_{K/\mathbb{Q}_p}(k)$ has ($p$-adic) absolute value $1/p$?

One way I would imagine to do this is to show that there exist $\alpha, \beta \in K$ such that $|N(\alpha)| = p^A$ and $|N(\beta)| = p^B$ with $A, B$ relatively prime.  The assertion then follows because there exist integers $x, y$ such that $xA + yB = 1$, whence $\alpha^{-x}\beta^{-y} \in K$ with $|N(\alpha^{-x}\beta^{-y})| = |N(\alpha^{-x})N(\beta^{-y})| = |a|^{-x}|b|^{-y} = p^{-Ax}p^{-By} = 1/p$.

My claim seems to me altogether reasonable since otherwise there exists some prime $q$ such that the absolute value $p^s$ of every $N(k)$ as $k$ runs through all of $K$ will be such that $s$ is always divisible by $q$, which doesn't seem right.

答:
You ask whether, given a finite extension $K/\mathbf Q_p$, the image of the norm $N_K : K^* \to \mathbf Q_p^*$ contains a uniformiser $\pi \in \mathbf Q_p^*$. In other words, does $\nu(N_K(a)) = 1$ have a solution $a \in K$?

In general, it is not true. For simplicity, suppose $K/\mathbf Q$ is Galois of degree $n$. Since all the conjugates of $a$ have the same $p$-adic valuation, $\nu (N_K(a)) = n\cdot \nu(a)$ for every $a$. Recall that $\nu(K^*) =\frac{1}{e}\mathbf Z$, where $e$ is the ramification index of $K/\mathbf Q_p$. Therefore $\nu(N_K(K^*)) = n\cdot \nu(K^*) = (n/e) \mathbf Z$; so a necessary and sufficient condition for a uniformiser to be in the image of the norm, in the case of a Galois extension, is that $n=e$; i.e. that $K/\mathbf Q_p$ be totally ramified.


答:
Notice that $a^2\equiv a$ in $\mathbb Z/2\mathbb Z\quad$ (since $0^2=0$ and $1^2=1$).

So you can also simplify this way

$\require{cancel}\underbrace{(a-1)(a-1)}_{=(a-1)}-(a-1)-(a-1)\equiv \cancel{(a-1)}-\cancel{(a-1)}-(a-1)\equiv 1-a\pmod 2$

Since $2a\equiv 0$ you also have $1-a\equiv 1-a+2a\equiv 1+a\pmod 2$

Which is the same as $\quad 1+a^2\equiv 1+a\pmod 2\quad$ by the way.

We can go further and write this $\tilde a$ since $1+a$ will just invert the parity of $a$.


答:
In general -- for any functions, not just for trig functions --
if $g(\theta) = f(\theta - k)$ for all $\theta,$ where $k$ is a constant,
then the graph of $g$ is the graph of $f$ shifted $k$ units to the right.

If $h(\theta) = g(-\theta),$ then the graph of $h$ is the mirror image of the graph of $h$, flipped left to right and right to left across the $y$ axis.

So the graph of $\tan\left(\frac{7\pi}{4} - \theta\right)$ is the mirror image of
the graph of $\tan\left(\theta - \frac{7\pi}{4}\right),$ which in turn is
the graph of $\tan\left(\theta \right)$ shifted $\frac{7\pi}{4}$ units to the right.

This does not tell you the simplest way to write the function as a formula
(in fact you can't get much simpler than it already is), but it should help you to avoid stumbling into wrong formulas. Whatever formula you come up with has to have a graph that looks a lot like the graph of $\tan(\theta)$, just mirror-imaged and moved somewhat sideways. It is certainly not a flat line at height $-1.$

Body:
What is the equivalent to $\tan\left(\frac{7\pi}{4}-\theta\right)$?

I'm very new to the mathjax formatting and the site in general but when I work the problem out it seems as though the answer should be $-1$.

$$\frac{\left(-1\ -\ \tan\left(\theta\right)\right)}{1+\left(-1\cdot\tan\left(\theta\right)\right)}$$

$$\frac{\left(-1\ -\ \tan\left(\theta\right)\right)}{1-\tan\left(\theta\right)}$$


问:
Help with Conditional Probability Question
I'm trying to solve the following problem:

Susan tosses a fair coin three times. What is the probability that she tosses "all tails" if she tosses at least two tails?

The probability space is:

```<code>HHH
HHT
HTT
TTT
THH
TTH
</code>```

The conditional probability formula states that $P(A|B)=\dfrac{P(A \cap B)}{P(B)}$.

In this case, "$A$" is tossing three tails. I believe the probability of tossing $3$ tails is $\dfrac{1}{6}$. In this case, "$B$" is tossing at least two tails.  I believe the probability of this event is $\dfrac{1}{2}$.

Then, according to the formula above, the conditional probability of tossing $3$ tails given that she tosses at least $2$ tails is:

$$\frac{\left(\frac{1}{6} \cdot \frac{1}{2}\right)}{\frac{1}{2}}=\frac{1}{6}$$

Is my understanding correct?  Thanks in advance!

答:
You have two mistakes, but based on your current work I am confident you can fix them.

You used $P(A\cap B) = \frac16 \cdot \frac12$ for some reason.
Actually, it can be calculated that $A\cap B=A$.
Thus, you should have used
$P(A\cap B)=P(A)=\frac16$, which is different than $\frac16\cdot\frac12$.

Your probability space is wrong.
You forgot

HTH
THT

So, the probability spaces has 8 elements instead of 6.
Unfortunately, this means you have to redo all calculations
(and even my numbers from the first mistake are wrong).


问:
Fibonacci, compositions, history
There are three basic families of restricted compositions (ordered partitions) that are enumerated by the Fibonacci numbers (with offsets):

a) compositions with parts from the set {1,2} (e.g., 2+2 = 2+1+1 = 1+2+1 = 1+1+2 = 1+1+1+1)

b) compositions that do not have 1 as a part (e.g., 6 = 4+2 = 3+3 = 2+4 = 2+2+2)

c) compositions that only have odd parts (e.g., 5 = 3+1+1 = 1+3+1 = 1+1+3 = 1+1+1+1+1)

The connection between (a) & the Fibonacci numbers traces back to the analysis of Vedic poetry in the first millennium C.E., at least (Singh, Hist. Math. 12, 1985).

Cayley made the connection to (b) in 1876 (Messenger of Mathematics).

$\bullet$ Who first established the connection with (c), odd-part compositions? It was known by 1968 (Hoggatt & Lind, J. Comb. Th.), but I suspect it was done before that. Thanks for any assistance, especially with citations.

By the way, it is a nice exercise to give combinatorial proofs of why each family is counted by the Fibonacci numbers, and establish direct connections between each pair of families.

PS: Apologies for cross-posting from MathOverflow, want to see if the audience here has more knowledge of such things.

答:
Given your interest in this matter, you may very well have encountered the OEIS entry:

"F(n) = number of compositions of n into odd parts; e.g. F(6) counts 1+1+1+1+1+1, 1+1+1+3, 1+1+3+1, 1+3+1+1, 1+5, 3+1+1+1, 3+3, 5+1. - Clark Kimberling (ck6(AT)evansville.edu), Jun 22 2004"

Perhaps you can learn more by contacting Clark Kimberling, at email listed above, who may very well have researched the origins of this observation?

Edit:

I did find a scanned copy of an article authored by Hoggatt and Lind ("Combinations and Fibonacci Numbers", The Fibonacci Quarterly, Vol. 7 (3), 1969), which includes a reference to their 1968 publication.  No direct reference is given to their claim and demonstration of the property in question, and it seems to be, as Lind commented in your communication, original.) However, the opening section of this paper discussed the motivation for it, including references to earlier work upon which the authors hope to generalize.

The link I have is a directory with a link to table of contents of the relevant issue, and the article (scanned, "Hoggatt.pdf").

Title: Fibonacci, compositions, history


问:
Introductory/Intuitive Functional Analysis Book
Can you recommend a gentle introduction to the abstract thinking and motivation of functional analysis? I'm looking for a book that holds you by the hand and shows the details of exercises, etc.

Thanks.

答:
Erwin Kreyszig Introductory Functional Analysis
or Rynne & Youngson Linear Functional Analysis.

Both are elementary, well-written and self-contained.


答:
I would do this the following way--no knowledge of AM--GT needed: Assume WLOG $a \ge b$.
Then this inequality is clearly true for $a \ge 4$, as $\frac{a^2+1}{b} > a$ for all $a,b \in \mathbb{N}$ satisfying $a \ge b$. So now we are left to check the cases where both $a,b \le 3$; $a\ge b$. This can be done directly quite easily.


问:
Infimum of an Matrix-induced inner product
Hello this is my first post here!
Does anyone have a proof for this standard fact?

If $H$ is an $n \times n$ positive definite matrix and $\mathbf n \in \Bbb R \setminus \{0\}$ then
$$
\inf_{\mathbf y \in \Bbb R^n, \langle \mathbf n, \mathbf y \rangle=1}
\langle \mathbf y,H \mathbf y \rangle =
\frac{1}{\langle \mathbf n, H^{-1} \mathbf n\rangle}
\quad \text{and} \quad
\underset{\mathbf y \in \Bbb R^n, \langle \mathbf n, \mathbf y \rangle=1}{\operatorname{argmin}} = \frac{H^{-1} \mathbf n}{\langle \mathbf n, H^{-1}\mathbf n \rangle}.
$$

Couldn't find it anywhere. Thank you!

答:
Note: I assume in this context that "positive definite" implies symmetric, so that $H$ is symmetric.

One approach is to use the method of Lagrange multipliers. We want to minimize $f(\mathbf y )= \langle \mathbf y,H \mathbf y\rangle$ subject to the constraint that $g(\mathbf  y) = \langle \mathbf n,\mathbf y \rangle = 1$. We compute
$$
\nabla f = 2H \mathbf y, \quad \nabla g = \mathbf n.
$$
Thus, if $\mathbf y$ for which the minimum is achieved, then there must be a $\lambda$ such that
$$
\nabla f = \lambda \nabla g \implies 2H \mathbf y = \lambda \mathbf n \implies \mathbf y = \frac{\lambda}2 H^{-1}\mathbf n.
$$
On the other hand, because $g(\mathbf y) = 0$, we have
$$
\left \langle \mathbf n,
\frac{\lambda}{2}H^{-1} \mathbf n
\right \rangle = 1 \implies \lambda = \frac{2}{\langle \mathbf n, H^{-1}\mathbf n\rangle}.
$$
Thus, if a minimum is achieved, it must be achieved at
$$
\mathbf y = \frac {\lambda}{2}H^{-1}\mathbf n = \frac{H^{-1} \mathbf n}{\langle \mathbf n,H^{-1} \mathbf n\rangle}
$$
and the minimum must be
$$
f \left( \frac{H^{-1} \mathbf n}{\langle \mathbf n,H^{-1} \mathbf n\rangle} \right) =
\frac {1}{\langle \mathbf n, H^{-1}\mathbf n \rangle}.
$$

An alternative proof: let $M$ be such that $H = M^TM$ (such an $M$ could be found via Cholesky decomposition for instance). Make the substitution $\mathbf x = M\mathbf y$. We can rewrite the minimization as
$$
\inf_{\langle \mathbf n,  M^{-1}\mathbf x \rangle = 1} \langle M^{-1}\mathbf x, H M^{-1}\mathbf x\rangle = \\
\inf_{\langle M^{-T}\mathbf n, \mathbf x \rangle = 1} \langle M^{-1}\mathbf x, M^T(M M^{-1})\mathbf x\rangle = \\
\inf_{\langle M^{-T}\mathbf n, \mathbf x \rangle = 1} \langle \mathbf x, \mathbf x\rangle.
$$


问:
In the $(p + 1)$-dimensional input-output space, $(X, \hat{Y})$ represents a hyperplane?
I am currently studying the textbook The Elements of Statistical Learning, second edition, by Hastie, Tibshirani, and Friedman. Chapter 2.3.1 Linear Models and Least Squares says the following:

The linear model has been a mainstay of statistics for the past 30 years and remains one of our most important tools. Given a vector of inputs $X^T = (X_1, X_2, \dots, X_p)$, we predict the output $Y$ via the model
$$\hat{Y} = \hat{\beta}_0 + \sum_{j = 1}^p X_j \hat{\beta}_j . \tag{2.1}$$
The term $\hat{\beta}_0$ is the intercept, also known as the bias in machine learning. Often it is convenient to include the constant variable $1$ in $X$, include $\hat{\beta}_0$ in the vector of coefficients $\hat{\beta}$, and then write the linear model in vector form as inner product
$$\hat{Y} = X^T \hat{\beta}, \tag{2.2}$$
where $X^T$ denotes vector or matrix transpose ($X$ being a column vector). Here we are modelling a single output, so $\hat{Y}$ is a scalar; in general $\hat{Y}$ can be a $K$-vector, in which case $\beta$ would be a $p \times K$ matrix of coefficients. In the $(p + 1)$-dimensional input-output space, $(X, \hat{Y})$ represents a hyperplane.

I'm struggling to make sense of this. If $X^T$ is a vector or matrix transpose and $\hat{Y}$ is a $K$-vector, then what exactly are the dimensions of $X^T$, $\hat{Y}$, and $\hat{\beta}$ for this all to make mathematical sense? Furthermore, how does the $\beta$ relate to this (this might be a typo by the authors)? And lastly, how does this then show that, in the $(p + 1)$-dimensional input-output space, $(X, \hat{Y})$ represents a hyperplane?

答:
You have to think that $(X, \hat{Y}) $ is a vector with $p+K$ components, given by $ (X_1,\ldots, X_p, \hat{Y}_1, \ldots, \hat{Y}_K ) $.

In general, let $B$ be the $(p+K) \times K$ matrix which consists of two blocks, one over the other:

The upper block is the $p\times K$ matrix $\hat{\beta}$;

The lower block is the $K \times K$ matrix given by $-Id$.

Let me denote by $\hat{\beta}^i$ the $i$-th column of $\hat{\beta}$. Note that the $i$-th column of $B$ is then given by $(\hat{\beta}^i, -e_i) $; here I am using the same convention of $(X, \hat{Y}) $ by concatening two vectors one after another (the first with $p$ components and the second with $K$ components).

Let's see what does the equation $(X, \hat{Y})^T B = 0$ means.  The output is a vector with $K$ components. The $i$-th component is the scalar product of the the input-output with the $i$-th column of $B$, that is

$$ 0= (X, \hat{Y}) \cdot (\hat{\beta}^i, -e_i) = (X, \hat{\beta}^i) + (\hat{Y}, -e_i) $$

Which is equivalent to

$$ \hat{Y}_i =  (X, \hat{\beta}^i) $$

In the same spirit of above, note that the RHS is the i-th component of $X^T \hat{\beta}$! This means our equation is equivalent to

$$ \hat{Y} = X^T \hat{\beta}$$

Which is the original equation! All this stuff was a fancy way to demonstrate you that, whenever you have some variables constrained by linear equations, you can always put your equations into the form $(variables) ^T (matrix) = 0$, and this is what happening in your case.

In the pleasant case in which you have just one equation ($K$=1), the matrix $B$ will have just one column; in other words $B$ is a vector, and the $B$ equation reads as

$$ (X, \hat{Y}) \cdot B =0$$

Such an equation describes exactly the hyperplane perpendicular to the vector $B$! Think about this geometrically: if I ask you what is the locus of points in $\mathbb{R}^3$ perpendicular to the vector $(0, 0,1) $, this will be the $x-y$ hyperplane.