问:
Is there such a thing as partial integration?
Recently in my mathematics courses I was taught partial derivatives, and I wondered if the reverse exists for integrals.

This may sound like a stupid question, and it probably is, but let me explain:

By the fundamental theorem of calculus:
$$ \int \frac{d}{dx}f(x)~dx = f(x) $$
So is there an operator such that:
$$ \int \frac{\partial}{\partial x}f(x,y)~\partial x = f(x,y) $$
or is this complete bogus?

What I am saying is that if the partial derivatives of a multi-variable equation is the slope of the line along the derivative axis, is there an operator, a "partial integral" that is the area under the curve along the integrated axis?

Or is that just $\int f(x,y) ~dx$, since doing it over more than one axis requires $\iint f(x,y) dx dy$?

Physically, this could be related to finding the x-component of the velocity given the acceleration function.

Also, how stupid of a question is this?

答:
Seeing the other answers above, I wanted to make a quick clarification. When you take the derivative or the integral of some function, you do it with respect to a specific variable.

Consider $f: \mathbb{R}^2 \to \mathbb{R}$ with such that $f(x,y) = z$. There are two input variables, so there are two derivatives given by
\begin{equation}
  {f_y}'(x) = \frac{\partial f}{\partial x} \mathrm{~~~and~~~} {f_x}'(y) = \frac{\partial f}{\partial y}
\end{equation}

The gradient is, by definition, a vector which yields the direction and rate of the greatest increase. Here, the gradient is given by a two-dimensional vector along both the $x$-axis and the $y$-axis with
\begin{equation}
  \nabla f(x,y) = \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{bmatrix}
\end{equation}

Consider the curve $\mathcal{C}$ in the closed interval $[a,b]$ given by
\begin{equation}
  \mathbf{r}(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} \mathrm{~~~with~} t \in \mathbb{R}
\end{equation}

We then evaluate the line integral along $\mathcal{C}$ with
\begin{align}
  f(\mathbf{r}(b)) - f(\mathbf{r}(a)) & = \int_a^b \left[f(\mathbf{r}(t))\right]' \; dt                                                                                \\
                                      & = \int_a^b \left[{f_y}'(x(t)) \cdot x'(t) + {f_x}'(y(t)) \cdot y'(t)\right] \; dt                                                   \\
                                      & = \int_a^b \left[\frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt}\right] \; dt \\
                                      & = \int_a^b \nabla f(\mathbf{r}(t)) \bullet {\mathbf{r}'}(t) \; dt                                                                   \\
                                      & = \int_\mathcal{C} \nabla f(\mathbf{u}) \bullet \mathbf{du}
\end{align}

And with $\mathbf{p} = \mathbf{r}(a)$ and $\mathbf{q} = \mathbf{r}(b)$ the gradient theorem is given by
\begin{equation}
  \int_\mathcal{C} \nabla f(\mathbf{u}) \bullet \mathbf{du} = f(\mathbf{p}) - f(\mathbf{q})
\end{equation}

The total differential of $f(x,y)$ is given by
\begin{equation}
  df = \frac{\partial f}{\partial x} \; dx + \frac{\partial f}{\partial y} \; dy
\end{equation}

Note that integrating with respect to multiple variables has no meaning, consider the following
\begin{equation}
  \int df = \int \frac{\partial f}{\partial x} \; dx + \int \frac{\partial f}{\partial y} \; dy
\end{equation}

This is clearly invalid. The integral sign has no meaning on its own and must always be coupled with a dummy integration variable.

Instead, you may want to integrate the gradient along a linear path in $\mathbb{R}^2$ as shown in the gradient theorem above.

Consider a linear path $\gamma$ from $(0,0)$ to $(x_0,y_0)$ given by
\begin{equation}
  \gamma(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = \begin{bmatrix} tx_0 \\ ty_0 \end{bmatrix} \mathrm{~~~with~} t \in [0,1] \mathrm{~and~} (x_0,y_0) \in \mathbb{R}^2
\end{equation}
and clearly we have $f(\gamma(0)) = f(0)$ and $f(\gamma(1)) = f(x_0,y_0)$.

Thus per the gradient theorem above we have
\begin{equation}
  f(x_0,y_0) = \int_{\gamma[0,1]} \nabla f(\mathbf{u}) \bullet \mathbf{du}
\end{equation}

The gradient theorem works in this case because we integrate the gradient along both the $x$-axis and the $y$-axis in one straight linear path in $\mathbb{R}^2$ with the parametrization given above.

There is also another method which relies on solving for remaining parts after integrating the partial derivatives.

Let us take a concrete example with $f(x,y) = xy + y$ so that we have
\begin{equation}
  {f_y}'(x) = \frac{\partial f}{\partial x} = y \mathrm{~~~and~~~} {f_x}'(y) = \frac{\partial f}{\partial y} = x + 1
\end{equation}

In fact, we easily see that
\begin{align}
  f(x,y) & \neq \int \frac{\partial f}{\partial x} \; dx + \int \frac{\partial f}{\partial y} \; dy \\
         & \neq 2xy + y
\end{align}

Because we integrate along both the $x$-axis and the $y$-axis, we obtain twice the result for variables that are bound together.

The correct approach is either to use the gradient theorem as shown above or to integrate with respect to one chosen variable instead, and then solve for the remaining parts.

When integrating with respect to $x$, we have
\begin{align}
  f(x,y) & = \int \frac{\partial f}{\partial x} \; dx + \psi(y) + K_1 \\
         & = xy + \psi(y) + K_1
\end{align}
with $\psi(y)$ is a function of $y$ only and $K_1$ is a constant.

Note carefully that the constant of integration here is any differentiable function of $y$ denoted by $\psi(y)$ since any such function would vanish upon partial differentiation with respect to $x$ (just as any pure constant $C$ would vanish upon ordinary differentiation).

On the other hand, when integrating with respect to $y$, we have
\begin{align}
  f(x,y) & = \int \frac{\partial f}{\partial y} \; dy + \phi(x) + K_2 \\
         & = xy + y + \phi(x) + K_2
\end{align}
with $\phi(x)$ is a function of $x$ only and $K_2$ is a constant.

Equating the two results we obtain
\begin{align}
  xy + \psi(y) + K_1 & = xy + y + \phi(x) + K_2                                   \\
  \psi(y) + K_1      & = y + \phi(x) + K_2                                        \\
  \psi(y)            & = y + \underbrace{\phi(x) + K_2 - K_1}_{\mathrm{constant}}
\end{align}
and because $\phi(x) + K_2 - K_1$ evaluates to a constant in function with respect to $y$ we clearly see that $\psi(y) = y$.

The same result is obtained by differentiating with respect to $y$ the following expression
\begin{equation}
  \frac{\partial f}{\partial y} \psi(y) = \frac{\partial f}{\partial y} \left[f(x,y) - \int \frac{\partial f}{\partial x} \; dx + K_1\right]
\end{equation}
so that we have
\begin{align}
  \psi'(y) & = \frac{\partial f}{\partial y} - \frac{\partial}{\partial y} \int \frac{\partial f}{\partial x} \; dx \\
           & = 1
\end{align}
and integrating with respect to $y$ we obtain $\psi(y)$ with
\begin{align}
  \int \psi'(y) \; dy & = \int dy \\
                      & = y + C
\end{align}


问:
Finding a function which is distance $1$ from $x^2$ along its normals
So I recently thought of a problem, which I have tried to solve many ways but couldn't. It's very simple to describe...

Find the function which is distance $1$ from parabola $y=x^2$ along its normals. Visual of what I mean.

This one of the things I've tried...

This function would get the corresponding $x$ value on the function we want to find from an $x$ value on curve $y=x^2$. The inverse would give the corresponding $x $ value on $x^2$ for a given $x$ value which would make it trivial to determine the $y$ value.
$$
f(x)=x+\sin(\arctan(2x))
$$
This can be resolved to this...
$$
f(x)=x+\frac{2x\sqrt{1+4x^2}}{1+4x^2}
$$
However I don't know how to get the inverse of this function.

答:
We can modify by offsetting from standard parametrization. Used $ f=1, r= (-0.2,0,+0.2)$ in the plot ( used offset 0.2 in place of 1.0 for graph clarity ).

```<code>f = 1; r = 0;
g1 = ParametricPlot[{2 f t, f t^2} +
   r {-t/Sqrt[1 + t^2], 1/Sqrt[1 + t^2]}, {t, -1, 1},
  GridLines -> Automatic]
 r = 0.2;
g2 = ParametricPlot[{2 f t, f t^2} +
   r {-t/Sqrt[1 + t^2], 1/Sqrt[1 + t^2]}, {t, -1, 1},
  PlotStyle -> {Thick, Blue}]
r = -0.2;
g3 = ParametricPlot[{2 f t, f t^2} +
   r {-t/Sqrt[1 + t^2], 1/Sqrt[1 + t^2]}, {t, -1, 1},
  GridLines -> Automatic, PlotStyle -> {Thick, Red}]
Show[{g1, g2, g3}, PlotRange -> All]
</code>```

We add or remove distances along normal and the tangent with offset $r$
$$ (x,y)= ( 2 f t,f t^2 ) ;\; t = \tan \phi $$

$$ x_1= x - r \sin \phi, y_1= y+ r \cos \phi $$

ContentLicense: CC BY-SA 4.0

Body:
We can modify by offsetting from standard parametrization. Used $ f=1, r= (-0.2,0,+0.2)$ in the plot ( used offset 0.2 in place of 1.0 for graph clarity ).

f = 1; r = 0;
g1 = ParametricPlot[{2 f t, f t^2} +
   r {-t/Sqrt[1 + t^2], 1/Sqrt[1 + t^2]}, {t, -1, 1},
  GridLines -> Automatic]
 r = 0.2;
g2 = ParametricPlot[{2 f t, f t^2} +
   r {-t/Sqrt[1 + t^2], 1/Sqrt[1 + t^2]}, {t, -1, 1},
  PlotStyle -> {Thick, Blue}]
r = -0.2;
g3 = ParametricPlot[{2 f t, f t^2} +
   r {-t/Sqrt[1 + t^2], 1/Sqrt[1 + t^2]}, {t, -1, 1},
  GridLines -> Automatic, PlotStyle -> {Thick, Red}]
Show[{g1, g2, g3}, PlotRange -> All]

We add or remove distances along normal and the tangent with offset $r$
$$ (x,y)= ( 2 f t,f t^2 ) ;\; t = \tan \phi $$

$$ x_1= x - r \sin \phi, y_1= y+ r \cos \phi $$


问:
Proof of: If $x_0\in \mathbb R^n$ is a local minimum of $f$, then $\nabla f(x_0) = 0$.

Let $f \colon \mathbb R^n\to\mathbb R$ be a differentiable function.
If $x_0\in \mathbb R^n$ is a local minimum of $f$, then $\nabla f(x_0) = 0$.

Where can I find a proof for this theorem? This is a theorem for max/min in calculus of several variables.

Here is my attempt:

Let $x_0$ = $[x_1,x_2,\ldots, x_n]$
Let $g_i(x) = f(x_0+(x-x_i)e_i)$ where $e_i$ is the $i$-th standard basis vector of dimension $n$.

Since $f$ has local min at $x_0$, then $g_i$ has local minimum at $x_i$.
So by Fermat's theorem, $g'(x_i)= 0$ which is equal to $f_{x_i}(x_0)$.
Therefore $f_{x_i}(x_0) = 0$ which is what you wanted to show.
Is this right?

答:
For small $\varepsilon$, we can define a curve $\gamma:(-\varepsilon,\varepsilon)\to\mathbb{R}^n$ by $\gamma(t)=x_0+t\nabla f(x_0)$. Then $f \circ \gamma$ has a local minimum at $0$, so $(f\circ\gamma)'(0)=0$. By the chain rule,
$$0=(f\circ\gamma)'(0)=\nabla f(x_0) \cdot \nabla f(x_0),$$ and $\nabla f(x_0)=0$.</p>

ContentLicense: CC BY-SA 3.0


问:
(Topologically) different tangent bundles on the same manifold
Is there an example of a topological manifold in which different smooth structures give rise to tangent bundles which are not isomorphic as topological vector bundles?

My (our) own attempts or remarks, mostly obtained from discussing this question with other people:

By the Wu formula, the two tangent bundles would have in any case the same Stiefel--Whitney classes. This is in fact what originally motivated my question.

Exotic 7-spheres are no good to produce such examples because they all have trivial tangent bundles.

"Different smooth structures correspond to different (stable) linear structures on the tangent microbundle."

答:
Yes, there are. In fact, there are examples where the tangent bundles aren't even stably equivalent.

In section 9 of Microbundles: Part I, Milnor constructs an open set $U \subset \mathbb{R}^m$. With its standard smooth structure, the (stable) tangent bundle of  $U\times\mathbb{R}^k \subset \mathbb{R}^{m+k}$ is trivial, while in Corollary 9.3, Milnor shows that it admits a smooth structure for which the tangent bundle has a non-zero Pontryagin class. As Pontryagin classes are stable, the stable tangent bundle of the latter manifold is not trivial, and hence not isomorphic to the stable tangent bundle of $U\times\mathbb{R}^k$ with its standard smooth structure.

Milnor, John W., Microbundles: Part I, Topology 3, Suppl. 1, 53-80 (1964). ZBL0124.38404.

A closed example is constructed in this answer.


问:
Plotting parametric curve with implicit parameter function
I need to plot a Smale diagram as a parametric curve $(p(x), h(x))$ where $h$ is amended potential and $p = \omega^2$ is parameter of bifurcation:
$$h:=V_\omega(x,y)=-2\cos(x)-\cos(y)-\frac{1}{2}\omega^2\bigl(\sin^2(x)+(\sin(x)+\sin(y))^2\bigr)$$
There is system of equilibrium equations (partial derivatives of amended potential are zero):
$$h_x'=2\sin(x)-\omega^2\cos(x)(2\sin(x)+\sin(y))=0$$
$$h_y'=\sin(y)-\omega^2\cos(y)(\sin(x)+\sin(y))=0$$

I am trying to get $y = y(x)$ and $p = p(x) = p(x, y(x))$ from this system. For example, isolating $\sin(y)$ and $\cos(y)$, and using $\sin^2(y)+\cos^2(y)=1$, it can be turned into:
$$4\bigl(1-\omega^2\cos(x)\bigr)^2\Bigl(\cos^2(x)+\bigl(2-\omega^2\cos(x)\bigr)^2\sin^2(x)\Bigr)=\omega^4\cos^2(x)(2-\omega^2\cos(x))^2$$
Then, substituting $\cos(y)$ and $\sin(y)$ in $h$:
$$h=-2\cos(x)-\frac{2\bigl(1-\omega^2\cos(x)\bigr)}{\omega^2\bigl(2-\omega^2\cos(x)\bigr)}-\frac{1}{2}\omega^2\sin^2(x)\biggl(1+\Bigl(1+\frac{2\bigl(1-\omega^2\cos(x)\bigr)}{\omega^2\cos(x)}\Bigr)^2\biggr)$$

So I got something like parametrization for $h$ and $p$ with the parameter $x$, but for $p(x)$ it is implicit. I had attempted to plot this in Maple 13, but unsuccessfully.

(it's described here)

What I want to know - is there a way to plot this parametric curve on $(h,\omega^2)$ coordinates with computational methods, even numerically (Wolfram Mathematica or Maple would be the best)?

Particularly, for one curve $h_-(\omega^2)$, separating from point $(x,y,\omega^2)=(0,0,2-\sqrt{2})$ with the following conditions:
$$\omega^2>2-\sqrt{2}$$
$$h(2-\sqrt{2})=-3$$
$$\sin(x)\sin(y)>0, (x,y)\in O_\varepsilon (0)$$

答:
From $h'_y=0$, you have
$$\sin (x)+\sin (y)=\frac{\tan (y)}{\omega ^2}$$
Replacing in $h'_x=0$, then
$$y=\tan ^{-1}\left(2 \tan (x)-\omega ^2 \sin (x)\right)$$


问:
Fatou: Reverse?
Attention

The usual problems are about absolute convergence:
$$\int|g_n|\mathrm{d}\mu\quad(g_n=f_n,f-f_n,s_m-s_n,\ldots)$$

(There Fatou may help out!)

But as proceeding with Fatou one encounters that one can't distort to the limessuperior:
$$\int\limsup_n|g_n|\mathrm{d}\mu\nleq\int\liminf_n|g_n|\mathrm{d}\mu\leq\liminf_n\int|g_n|\mathrm{d}\mu\leq\limsup\int|g_n|\mathrm{d}\mu$$

So the real question is about the analogue for limessuperior!!

(And not the analogue for negative functions...)

Problem

Given a measure space $\Omega$.

The lemma of Fatou states:
$$f_n\geq0:\quad\int\liminf_nf_n\mathrm{d}\mu\leq\liminf_n\int f_n\mathrm{d}\mu$$
Does the reverse hold true:
$$f_n\geq0:\quad\int\limsup_nf_n\mathrm{d}\mu\leq\limsup_n\int f_n\mathrm{d}\mu$$
Certainly, for convergent examples this holds true:
$$f_n\geq0:\quad\int\limsup_nf_n\mathrm{d}\mu=\int\lim_nf_n\mathrm{d}\mu\leq\lim_n\int f_n\mathrm{d}\mu=\limsup_n\int f_n\mathrm{d}\mu$$
So one needs to dig deeper to find an honest counterexample!!!!

(I intend to answer my own question!)

(I have to admit that my earlier answer was lame!)

答:
Here's a simpler counterexample.

For $x \in [0,1]$, $f_{n}(x) = 1$ if $n$ is odd, and $0$ otherwise.

For $x \in (1,2]$, $f_{n}(x)=1$ if $n$ is even, and $0$ otherwise.

For any other $x$, $f_{n}(x) = 0$.

In this case
$$
\int \limsup_{n} f_{n}(x) dx = 2,\\
\limsup_{n} \int f_{n}(x) dx = 1.
$$

Title: Fatou: Reverse?


问:
What is a resolution in logic.
(In "Introduction to mathematical logic" by mendelson , page 32)

I found this concept that I seemingly not been able to grasp , a resolution. It is represented by the function $Res(\beta)$ (Here $\beta$ is a cnf) . Now , I couldn't really understood this from the textbook itself. And , I tried to find this on the internet , but I only find "resolution proof" , which seems to be not related to what I wanted(I might be wrong). It is said in the book that $Res(\beta)$ is like a repetitive algorithm . But the algorithm itself seems to be unclear to me in many ways.

Can someone tell me if there is any books where they have more info about this algorithm for resolution?

答:
The resolution rule in simplified notation is

$$(\ldots_1 \lor A \lor \ldots_2) \land \ldots \land (\ldots_3 \lor \neg A \lor \ldots_4)\\
\rightsquigarrow_\mathcal{R} (\ldots_1 \lor \ldots_2 \lor \ldots_3 \lor \ldots_4)$$

where the $\ldots$s are (possibly empty, but see below) disjunctions of literals.

This step is called resolution on $A$, and the conclusion of the rule is called the resolvent.

The resolution algorithm consists of simply repeating the resolution rule on the conjoined output of the previous steps until there are no more occurrences of literals $A, \neg A$ to resolve. The result of this is called $\mathcal{Res}(\mathcal{B})$: $\mathcal{Res}(\mathcal{B}) = \mathcal{B} \land \mathcal{R}_1 \land \ldots \land \mathcal{R}_n$, where each resolvent $\mathcal{R}_i$ is obtained by applying one resolution step on the previous result $\mathcal{B} \land \ldots \land \mathcal{R}_{n-1}$.

Resolution does have to do with resolution proof.

Mendelson requires the disjunctions formed from resolution to be non-empty, and calls the limit case of disjunctions consisting of only one resolvable literal "blatant contradictions". Normally one would define resolution also for this limit case, when the two disjunctions consist of only one literal before the resolution step and of zero literals afterwards,

$$(A) \land (\neg A)\\
\rightsquigarrow_\mathcal{R} \Box$$

$\Box$ is called the empty clause and is unsatisfiable; the idea being that a disjunction is true iff at least one of its disjuncts is true, and if the disjunction is empty, there is nothing to satisfy it, so it is contradictory.

A resolution proof of a formula $\mathcal{B}$ consisits of negating $\mathcal{B}$, forming its CNF, and applying the resolution iteration on it, i.e. computing $Res(CNF(\neg \mathcal{B})))$. If and only if the resolution procedure yields the empty clause, the negation of $\mathcal{B}$ is unsatisfiable, so $\mathcal{B}$ must be a tautology, and the derivation of the empty clause is a resolution proof of $\mathcal{B}$: $$\vDash \mathcal{B} \text{ iff } Res(CNF(\neg \mathcal{B}))) = \Box$$

This can be generalized to resolution proofs of inferences by performing the procedure on the conjunction between the unnegated premsies and the negated conclusion: $$\mathcal{A}_1, \ldots, A_n \vDash \mathcal{B} \text{ iff } Res(CNF(\mathcal{A_1} \land \ldots \land \mathcal{A}_n \land \neg \mathcal{B})) = \Box$$

The keyword for finding resources covering resolution is "logic programming"; one I have lying around is Kees Doets (1994), "From logic to logic programming". The style of presentation varies between different textbooks; Doets uses set notation instead of the $\lor$ connective for disjunctions and a tree-style presentation of derivations, others use a sequent notation $\vdash$ for the clauses, but the idea is all the same.

Title: What is a resolution in logic.


问:
Inverting $3 \times 3$ matrix via Gauss-Jordan, cannot find my mistake
Can't find my mistake trying to find the inverse:

答: When you do $R_2 \to R_2-2R_3$ on the 2nd page, you get in the 1st column $\frac23-2\times\frac19=\frac13$, while it should be $\frac49$.


问:
How did the answer come about in this thermometer reading Differential Equation problem?

A  thermometer  reading  70oF  is  taken  outside.    Five  minutes  later  the thermometer reads 40 oF.  Thirty-five minutes after being taken outside, the  thermometer  reading  is  within  one-half  a  degree  of  the  outside temperature.    Find  the  value  of  the  outside  temperature  assuming  it  is constant.

Ans.  9.5oF

答:
HINT

Determine two constants in the Newton's Law of cooling:

$$( T-T1) = e^{- k t}$$


问:
Polish notation, rooted tree
Hi im suppost to write the expression

(w+x - y)/(pi * z^3)
in Polish notation.

My answer was / - + w x y * pi ^ z 3

But the solution is / + w - x y * pi ^ z 3

Is my answer the same as the solution and if not what am I doing wrong?

答: You've translated the potentially ambiguous formula $w + x - y$ that appears as the numerator of the fraction as if it read $(w + x) - y$, the other solution has taken it as $w + (x - y)$. The two readings are algebraically equivalent, but I think your reading agrees with the BODMAS rules (by working left to right given a sequence of additions and subtractions) - so I would challenge the other solution if I were you. (Note that if it were $w - x + y$, then reading it as $w - (x + y)$ rather than $(w - x) + y$ is definitely wrong.)

Title: Polish notation, rooted tree


问:
Show if $u$ and $v$ (non zero vectors) are perpendicular, then they are linearly independent
Let u and v be non zero vectors in Rˆn. Show if u and v are perpendicular, then {u,v} is linearly independent.

I started my proof as following:

If $u$ and $v$ are perpendicular, then $\langle u,v\rangle = 0$

So we can write $u_1 v_1 + \cdots + u_n v_n = 0$.

As $u$ and $v$ are non zero vectors there may exist a scalar $k$ that satisfies the identity.

Thus, $k = 0$ and we have a linearly independent combination of $u$ and $v$.

Is it right? wrong? How can I improve it?

答:
Your proof is not clear.

Let $t,s$ be scalars such that $tu+sv=0.$ Then

$$0==t+s=t||u||^2.$$

This shows that $t=0$, since $||u||^2 \ne 0.$ Hence $sv=0$. Since $v \ne 0$, we get $s=0.$

Body:
Your proof is not clear.

Let $t,s$ be scalars such that $tu+sv=0.$ Then

$$0=<u,tu+sv>=t<u,u>+s<u,v>=t||u||^2.$$

This shows that $t=0$, since $||u||^2 \ne 0.$ Hence $sv=0$. Since $v \ne 0$, we get $s=0.$


问:
How does integration of forms make sense?
I don't understand how integration of a $k$-form on a manifold makes sense. For simplicity let's assume $M$ is a smooth $k$-manifold, $\gamma: [a,b] \rightarrow M$ is a smooth curve on $M$ and that $\omega \in \Omega^1(M)$. Then by definition, $$\int_{\gamma} \omega = \int_{[a,b]} \gamma^* \omega$$ In coordinates $(x^i)$ we can write $\omega = \omega_i dx^i$ so that we then have $$\int_{[a,b]} \gamma^* \omega = \int_a^b (\omega_i \circ \gamma)d\gamma^i = \int_a^b (\omega_i \circ \gamma) \frac{d\gamma^i}{dt}dt$$

From an elementary calculus point of view, when integrating, we take the value of $(\omega_i \circ \gamma) \frac{d\gamma^i}{dt}$ at a point on some subrectangle of some partition, which gives us a real number. We then multiply it by the length of the subrectangle (which is approximated by $dt$), which is also a real number and we add these products up, then let the mesh of the partition tend to $0$.

From a differential forms point of view, $(\omega_i \circ \gamma) \frac{d\gamma^i}{dt}$ is a function that is being multiplied by the $1$-form $dt$. The value of $(\omega_i \circ \gamma) \frac{d\gamma^i}{dt}$ at a point of a subrectangle is a real number, but the value of $dt$, since it is a covector field, is a covector. Thus the value of $(\omega_i \circ \gamma) \frac{d\gamma^i}{dt} dt$ at a point of a subrectangle is a covector as well. How does it make sense to sum up a bunch of covectors to obtain a real number (the value of the integral)?

答:
How does it make sense to sum up a bunch of covectors to obtain a real number (the value of the integral)?

A differential n-form measures oriented n-volumes of supplied n vectors. Those measurements are real numbers and can be summed up.

Integration operator feeds $dx$ with (infinitesimal) tangent vectors at (continuum) consecutive points along a curve: $\int_C dx=\lim\sum dx_{p_i}(v_i)$.


问:
Calculate the following infinite sum in a closed form $\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$?
Is it possible to calculate the following infinite sum in a closed form? If yes, please point me to the right direction.
$$\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$$</p>

Body:
We have
$$\text{arccot}(x) = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)x^{2k+1}}\implies x\text{arccot}(x)-1 = \sum_{k=1}^{\infty} \dfrac{(-1)^k}{(2k+1)x^{2k}}$$
Hence,
$$\sum_{n=1}^{\infty}\left(n\text{arccot}(n)-1\right) = \sum_{n=1}^{\infty}\sum_{k=1}^{\infty} \dfrac{(-1)^k}{(2k+1)n^{2k}} = \sum_{k=1}^{\infty} \dfrac{(-1)^k \zeta(2k)}{2k+1}$$


问:
Proof of Fatou-Lebesgue Theorem
Good evening everyone,

how can I prove the following inequality?

Let $f,f_n\in L(X,\mu)$ on a measure space with $0\leq f_n(x)\leq f(x)$. If $f,f_n$ are $\mu$-almost everywhere in $X$, then is

$$\int_X\liminf_{n\to\infty} f_n d\mu \leq \liminf_{n\to\infty}\int_X f_n d\mu\leq\limsup_{n\to\infty}\int_X f_n d\mu\leq \int_X\limsup_{n\to\infty} f_nd\mu.$$

How can I prove it?

Firstable, I know that

$$\liminf_{n\to\infty} \int_X f_n d\mu \leq \limsup_{n\to\infty}\int_X f_n d\mu.$$

But I don't know how I can prove the inequality. Could someone help me please?

答:
The inequality you're asking about has nothing to do with sequences of integrals (as pointed out in the first comment). In general $$\liminf_{n} x_{n} \leq \limsup_{n} x_{n}$$ for any sequence $\{x_{n}\}$.

If you really need to prove the inequality, you should do it for a sequence of real numbers (or extended real numbers). It's a matter of checking the definitions of $\liminf$ and $\limsup$.

Title: Proof of Fatou-Lebesgue Theorem


问:
Integral of a sequence of function not uniformly convergent
Let $$f_n(x)=(\ln x)^n\quad\forall n\in\mathbb{N}$$Prove that $$\lim_{n\to+\infty}{\int_1^e{f_n(x)}dx=0}$$
I studied the convergence of the sequence of functions and I found that, point wise, $$f_n(x) \to f(x)$$ with $$f(x)=\left\{\begin{array}{ccl}0&\text{for}& e^{-1}<x<e\\
1&\text{for}& x=e\end{array}\right.$$
So there is not uniform convergence on $[1,e]$ because $f(x)$ is discontinuous while $f_n(x)$ is continuous $\forall n\in\mathbb{N}$ and
this prevent me from exchanging the limit with the integral.

My idea is to find a function that is greater than $f_n(x)$  over $[1,e]$ and whose integral tends to $0$ when $n\to+{\infty}$ but I have not been able to find it yet. Any Idea?

By the way, we are doing only Riemann-integrals, not Lebesgue-integrals.

答:
You don't need uniform convergence. Since $0 \le \mbox{ln}(x)^n \le 1$ on $[1, e]$, you have that your $f_n$ are all bounded between $0$ and $1$.

Then, from the dominated convergence theorem we know that $\displaystyle{\int_1^e} f_n(x)dx \underset{n \to \infty}{\longrightarrow} \displaystyle{\int_1^e} f(x)dx = 0$.

Other solution: since $\mbox{ln}$ is concave, $\mbox{ln}$ is under its tangent at $x=e$, which equation is $y = \frac{x}{e}$. Thus $$\displaystyle{\int_1^e} f_n(x)dx \le \displaystyle{\int_1^e} \big(\frac{x}{e}\big)^n dx \le \frac{1}{e^n} \Big[ \frac{x^{n+1}}{n+1}\Big]_1^e \le \frac{e}{n+1}$$

so we get the same conclusion.


问:
Cartesian Coordinates vs Spherical Coordinates vs Spherical Basis
I would like to properly understand spherical coordinates once and for all:

In the years of innocence and youth we are all introduced to cartesian coordinates.

This are pretty simple to understand: we define three fixed, mutually orthogonal, direction in space, represented by three versors $\hat{x},\hat{y},\hat{z}$; and then to represent a point in 3D space1 we can simply write:
$$\vec{r}=x\hat{x}+y\hat{y}+z\hat{z} \ \ \ \ \ \ (1)$$
in fact the instruction to find the point can be simply derived from (1): start from the origin of the coordinate system, take $x$ steps in the $\hat{x}$ direction, then $y$ steps in the $\hat{y}$ direction and the same again for $z$, and then you are done! You have found the point.2

Then soon after we are introduced to spherical coordinates:

Here we fix only two orthogonal direction in space, and we use those to measure the angles $\theta$ and $\phi$, then we can find a point by giving the two angles and the distance from the origin a.k.a the radius $r$. Intuitively this is completely fine, the value of $r$ locks the point into a sphere and then we can use the angles to find the point, but then when we want to explicitly write down the formula for $\vec{r}$ how should we do it? Is pretty common to find the line element written like this:
$$d\vec{r}=dr\hat{r}+rd\theta\hat{\theta}+r\sin{\theta}d\phi\hat{\phi}$$
So are we allowed to write $\vec{r}$ as
$$\vec{r}=r\hat{r}+r\theta\hat{\theta}+r\sin{\theta}\phi\hat{\phi} \ ? \ \ \ \ \ \ (2)$$
I have my doubts, since $\hat{\theta},\hat{\phi}$ are not constant versors, they kinda rotate around and change direction.3 So how should we write $\vec{r}$ in this coordinate system? Are we forced to write $\vec{r}$ by integrating the line element? Seems pretty inconvenient to me!

And at last, when our innocence is completely gone for sure, we are introduced to the spherical basis.

To be honest when I first heared the term I thought it was a synonymous to spherical coordinates, oh was I wrong! Turns out that the spherical basis is a kind of coordinate system of 3D real space involving complex numbers. On the bright side this coordinate system allows us to write a vector in a simple way reminiscent of the cartesian coordinate system:
$$\vec{r}=A_+\hat{e}_++A_-\hat{e}_-+A_0\hat{e}_0 \ \ \ \ \ \ (3)$$
At a first glance this seems just a relabeling of the cartesian basis, no real change whatsoever, but this is not the case! From what I currently understand $\hat{e}_+,\hat{e}_-\hat{e}_0$ are complex vectors and $A_+,A_-,A_0$ are complex numbers! Now I really struggle to understand how this coordinate system works, I mean, I struggle to build an intuition, a picture in my head on how we can represent 3d space using this one. I suppose this is analogous to represent 2D real space using a single complex number, but this supposition seems inadeguate: when we represent 2D real space in this way there are no versors, only one complex number4
$$z=re^{i\theta}.$$
Furthermore I would expect something like (3) to represent a 6D real space, not a 3D one. On top of all of this it turns out that this basis is really important to understand fundamental concepts in Quantum Mechanics, such as the Spherical Harmonics, so it is not something that we can brush off as marginal; we really need to build a deep understanding of this one.

So I guess that my question regarding this coordinate system/base is: How does it work? How can it represent 3D real space? How can we build an intuition about this?

(1): Or to represent a vector in 3D space, since we can think of putting the tip of the vector onto the point and the tail fixed in the origin.

(2): Of course there is the hole question of: How big is your step? Since the versors have module equal of one the answer is one; but from a physical point of view we have to chose an unit of measurement for lenght, and we can do that, but it is beside the point here. So no problem.

(3): On the other hand the fact that $\hat{\theta},\hat{\phi}$ have no constant direction is not a problem when writing the line element, since we are dealing with an infinitesimal vector. In fact in this case we can think of $\hat{\theta},\hat{\phi}$ as constant since we perform the sum in an infinitesimal region of space.

(4):We then take the imaginary part to be $y$ and the real part to be $x$, or alternatively we take $r,\theta$ in polar coordinates.

答:
$\hat{r}$ is defined for $r\ne0$, as $\vec{r}/r$ so $\vec{r}=r\hat{r}$. But $d\vec{r}$ isn't in general parallel to $\hat{r}$, as $\vec{r}$ can evolve in an arbitrary direction, not in general parallel to $\vec{r}$, to $\vec{r}+d\vec{r}$.

Usine Eq. (3B) here, $\vec{r}=(x(\vec{e}_--\vec{e}_+)+iy(\vec{e}_-+\vec{e}_+))/\sqrt{2}+z\hat{z}$, which indeed needs complex coefficients. Note in particular that $r^2$, defined as the unique element of the $1\times1$ matrix $r^\dagger r$, is $\frac{|x+iy|^2+|x-iy|^2}{2}+z^2=x^2+y^2+z^2$.


问:
A subgroup such that every left coset is contained in a right coset.
Let $G$ be any group, and $H \leq G$ a subgroup.

Suppose that for each $x \in G$, there exists a $y \in G$ such that $xH \subseteq Hy$. In other words, every left coset of $H$ is contained inside some right coset of $H$.

Question: what can we say about $H$? In particular, does this imply that $H$ is normal?

I know that if $G$ is finite, then $H$ must be normal, because then $xH \subseteq Hy$ implies $xH = Hy$, since $|xH| = |Hy|$.

答:
Theorem 1 Let $H,K \leq G$ and assume that a left coset of $H$ is contained in a left coset of $K$. Then $H \subseteq K$.

Proof Assume first that $x \in G$ with $xH \subseteq K$. Then $x=x \cdot 1 \in K$. Pick an $h \in H$, then $xh=k$ for some $k \in K$, so $h=x^{-1}k \in K$, and it follows that $H \subseteq K$.

Now suppose $xH \subseteq yK$ for some $x,y \in G$. Then $y^{-1}xH \subseteq K$ and from the previous it now follows that $H \subseteq K$.

Corollary 1 Let $H,K \leq G$ and assume that a right coset of $H$ is contained in a right coset of $K$. Then $H \subseteq K$.

Proof One can mimic the proof above, or, in general, if $Hx \subset Ky$ for some $x,y \in G$, this is equivalent to $x^{-1}H \subseteq Ky^{-1}$.

So what happens if we mix left and right.

Theorem 2 Let $H,K \leq G$, $x,y \in G$ and assume $xH \subseteq Ky$. Then $Kx=Ky$.

Proof Since $xH \subseteq Ky$, $H^{x^{-1}} \subseteq Kyx^{-1}$, so Corollary 1 yields $H^{x^{-1}} \subseteq K$, that is, $H \subseteq K^x$ and equivalently, $xH \subseteq Kx$. Hence, $\emptyset \neq xH \subseteq Ky \cap Kx$, implying $Kx=Ky$.

Corollary 2 Let $H \leq G$ and $x,y \in G$ with $xH \subseteq Hy$. Then $Hx=Hy$. In particular if $H$ is finite then $xH=Hx=Hy$.

Proof Take $H=K$ in Theorem 2, yielding $Hx=Hy$. If $H$ is finite then $xH \subseteq Hy$ implies $xH=Hy$ and the result follows.


问:
Compute the norm of $T$.
I need some help with this problem:

Let be $X=C([0,1])$ with te uniform norm (the max-norm). We define $T:X \longrightarrow X$ as $T(f)(x)=\displaystyle \int_{0}^{x} f(t^2)dt$. Compute $||T||=\sup\{||Tx||:||x||\leq 1\}$.

I proved that $||T|| \leq 1$, but I'm trying to show that $||T|| \geq 1$ in order to prove that $||T||=1$, but I don't arrive to the solution. Any help, please?

答: Take $f(x)$ to be the constant function $1$ on $[0,1]$. Then $||f||=||T(f)||=1$, and so $||T||\geq 1$.

Title: Compute the norm of $T$.


答:
Let $f$ be the constant function $1$. Then

$$T(f)(x)=x$$

for all $x \in [0,1]$

This gives

$$||T(f)||=1= ||f||.$$

Title: Compute the norm of $T$.

Body:
I need some help with this problem:

Let be $X=C([0,1])$ with te uniform norm (the max-norm). We define $T:X \longrightarrow X$ as $T(f)(x)=\displaystyle \int_{0}^{x} f(t^2)dt$. Compute $||T||=\sup\{||Tx||:||x||\leq 1\}$.

I proved that $||T|| \leq 1$, but I'm trying to show that $||T|| \geq 1$ in order to prove that $||T||=1$, but I don't arrive to the solution. Any help, please?


问:
Find the ratio of area of $K$ to that of $M$.

Circle $K$ has diameter $AB$. Circle $L$ is tangent to circle $K$ and to $AB$ at the center of circle $K$. Circle $M$ is tangent to circle $K$, to circle $L$ and to $AB$. Find the ratio of area of $K$ to that of $M$.

What I Tried: Here is a picture :-

Mark all the points, do the labelling and draw perpendiculars and all, and in the end doing all these in Geogebra I found that $PN$ is almost equal to $\frac{1}{4}$th of $KB$'s length, where $KB$ is the radius of the large circle. That is possible only when $\Delta OQP \cong \Delta KQP$ , but I am really having a hard time showing this, and I got no idea how to proceed. I tried joining more lines and even doing Pythagorean Theorem but it had no use.

Can anyone help me? Thank You.

Source:- $1976$ AHSME Problem $24$ (Check here :- https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems)

答:
Let $K(R)$, $L(R/2)$, $M(r)$.

$$PQ^2= OP^2 - OQ^2 = \color{blue}{PK^2} - QK^2$$

$$\Rightarrow (R/2+r)^2 - (R/2-r)^2 = \color{blue}{(R-r)^2} - r^2$$

$$\therefore \boxed{r=R/4}$$

Thinking method : Centers of two tangent circles are collinear with the tangency point. You did not use this condition for circles $K$ and $M$.


问:
Sketch the function $y = {1 \over {{x^2}}}\ln x$
I don't know where to begin with this, the ${1 \over {{x^2}}}$ part of the function throws me off, how to do I go about this? How does one generally approach a question like this?</p>

答:
If you aren't supposed to be using a calculator/comptuer: plug in lots of values of $x$ for which you'll be able to easily approximate the value of $f(x)$, plot the resulting points $(x,f(x))$, and connect them with a reasonable-looking curve.

For example, for the function $f(x)=\frac{1}{x^2}\ln(x)$, we'll want to plug in values of $x$ for which we know the approximate value of $\ln(x)$. Some such $x$'s are powers of $e$. For example, let's say that $e\approx 2.7$. Then we can approximate
$$f(2.7)=\frac{1}{(2.7)^2}\ln(2.7)\approx \frac{1}{(2.7)^2}\cdot 1=\frac{1}{7.29}$$

If you are allowed to use a calculator/comptuer, just tell it to do it :)

Entering this code in Mathematica

```<code>Plot[(1/x^2) Log[x], {x, 0, 5}, PlotRange -> {-1, 1}]
</code>```

produces

Body:
If you aren't supposed to be using a calculator/comptuer: plug in lots of values of $x$ for which you'll be able to easily approximate the value of $f(x)$, plot the resulting points $(x,f(x))$, and connect them with a reasonable-looking curve.

For example, for the function $f(x)=\frac{1}{x^2}\ln(x)$, we'll want to plug in values of $x$ for which we know the approximate value of $\ln(x)$. Some such $x$'s are powers of $e$. For example, let's say that $e\approx 2.7$. Then we can approximate
$$f(2.7)=\frac{1}{(2.7)^2}\ln(2.7)\approx \frac{1}{(2.7)^2}\cdot 1=\frac{1}{7.29}$$

If you are allowed to use a calculator/comptuer, just tell it to do it :)

Entering this code in Mathematica

Plot[(1/x^2) Log[x], {x, 0, 5}, PlotRange -> {-1, 1}]

produces


问:
General Solution of Linear Homogenous Difference Equations
I'm currently learning about linear inhomogenous difference equations and I completely understand why knowing a particular solution and all solutions of the appropriate homogenous equation is enough. However, I don't quite understand how we know that we have captured all the solutions of the homogenous equation. For example, consider the general form of second order linear homogenous difference equations:

$$u_{n+1} = au_n + bu_{n-1}$$

So, the auxiliary equation is $$\lambda^2 - a\lambda - b = 0$$ with roots $\lambda_1$, $\lambda_2$. Now I see that every function of the form $A\lambda_1^n + B\lambda_2^n$ is a solution to the given homogenous difference equation, assuming that $\lambda_1 \neq \lambda_2$. However, why is every solution of that form?

I know that the solutions form a vector space over the real numbers, but why is the dimension of the vector space equal to 2?

答:
If we can solve these equations directly we will have a better understanding of the roles of the underlying characteristic equation.

By using the forward shift operator $\mathbb{E}$ such that $\mathbb{E} a_n=a_{n+1}, \mathbb{E}^2 a_n=a_{n+2}$, etc., a linear difference equation can then be written as $f(\mathbb{E})a_n=0$, where $f$ is a polynomial.

First order LDE: $(\mathbb{E} - r)a_n=0$. The solution is simply $a_n = r^n a_0.$

Second order LDE: $f(\mathbb{E})a_n=(\mathbb{E}-r)(\mathbb{E} - s)a_n=0$. From the first order case we have $$(\mathbb{E}-s)a_n = r^n (\mathbb{E}-s)a_0 \tag1$$ or equivalently

$$a_{n+1} - s a_n = r^n (a_1-s a_0)\tag 2$$
You can then solve this recursively (denote $C_1=a_1-sa_0$)
$$a_{n}=C_1 r^{n-1}+sa_{n-1} = C_1 r^{n-1} + s(C_1 r^{n-2}+sa_{n-2}) = C_1 (r^{n-1}+sr^{n-2}) +s^2 a_{n-2} = \cdots \tag{A}$$
And that's why we see the $r^n$ and $s^n$ terms in our final solution. This can be tedious but fortunately we have a shortcut method to solve it much more quickly:

By symmetry
$$(\mathbb{E}-r)a_n = s^n (\mathbb{E}-r)a_0 \tag3$$
Or equivalently
$$a_{n+1} - r a_n = s^n (a_1-ra_0) \tag4$$

Subtract (4) from (2) you get
$$a_n=\frac{a_1-s a_0}{r-s} r^n  - \frac{a_1-r a_0}{r-s} s^n$$

You now see clearly that from first order to second order, a new geometric sequence component via the second root (assuming distinct roots) is "added" to the geometric sequence solution to the first order, regardless of the initial values. They always have the same form, the only differences are the coefficients which can be determined from the initial values.

We can apply this factoring technique repeatedly to a higher order LDE. There won't be a shortcut method like in the second order case but you can solve recursively as in $(A)$.

Another clean way to solve is via linear algebra. It's easy to show that big transition matrix has the same rank as the order of the LDE.


问:
Variance of present value of annuity
Let $Y$ be a present value of whole life annuity that pays $1$ for year of each $x$-years old. We know that $\ddot{a}_x=10$ for $i'=1/24=e^\delta-1$ and $\ddot{a}_x=6$ for $i=e^{2\delta}-1$. I want to calculate the variance of $Y$. To be honest I don't know where to start.

I want to use the equation:

$$VarY=\frac{^2A_x-(A_x)^2}{d^2}.$$
Where: $$A_x=\sum_{j=0}^\infty v^{j+1} {}_j| q_{x};$$
$$^2 A_x=\sum_{j=0}^\infty (v^2)^{j+1} {}_j| q_{x}$$
Also we have: $\ddot{a}_x=\frac{1-A_x}{d}$ where $d=\frac{i}{1+i}$

The definition of $\ddot{a}_x=\sum_{j=0}^\infty v^j\  _{j}p_x$ and $v=\frac{1}{1+i}$.

All I can think of is to calculate $\delta$ and other constants as $v$ and $d$  for $i'$ and $i$ but I don't really know where it should go from there.

答: You can solve for $A_x$ and ${}^2 A_x$ using the formula $\ddot{a}_x = \frac{1-A_x}{d}$ with the two given values for $\ddot{a}_x$.  Then you can solve for $\delta$ using the given $e^\delta - 1 = 1/24$ and use that to solve for $d$.

Title: Variance of present value of annuity


问:
Sequence of sequences with sup-norm goes to 0 but 2-norm stay positive
I am trying to show sup-norm is strictly weaker than 2-norm in sequence space. In $\Bbb R^n$, we have sup-norm equivalent to 2-norm since we are doing finite sum, $\sqrt{\sum^Na_n^2}\le\sqrt{\sum^N||a_n||_\infty^2}=\sqrt N||a_n||_\infty$. But, in sequence space, we have infinitely many terms, I think it is possible to keep the sup-norm very low but the 2-norm say positive. Therefore, we can find a sequence of sequences converges to 0 in supnorm but stay positive in 2-norm. But I got stuck in finding such an example.

Body:
Let $(x_n)_{n=1}^\infty$ be a sequence of sequences defined as follows
\begin{align*}
x_1 & = (1,0,0\cdots)\\
x_2 & = (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,\cdots)\\
x_3 & = (\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},0,\cdots)\\
&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\vdots\\
x_n & = (\underbrace{\frac{1}{\sqrt n},\frac{1}{\sqrt n},\cdots,\frac{1}{\sqrt n}}_{\text{first }n \:\text{terms}},0,0\cdots)\\
&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\vdots\\
\lvert\lvert x_n\rvert\rvert_\infty&=\frac{1}{\sqrt{n}} \:\:\text{but}\:\:\lvert\lvert x_n\rvert\rvert_2 =1
\end{align*}


问:
Haar measure from axis-angle representation of $SO(3)$
My aim is to study some special representations of $SO(3)$ using characters, and to do this I need an explicit expression of the Haar measure on $SO(3)$. I've found some "versions" of the Haar measure on $SO(3)$, and the most common one seem to be the acquired by parameterizing $SO(3)$ using Euler angles (can be found here).

Since the characters of $SO(3)$ only depend on the angle from the axis-angle representation, it seems as if it would be suitable to find a Haar measure based on this parameterization. However, I don't really know where to start, and I've searched the web like a maniac.

Could someone provide some help?

Thanks in advance!

答:
I will add an extension to the answer by Brightsun, since the reply posted above does not explicitly answer the original question.

I understand "angle-axis" representation as a parametrisation of the group manifold of $SO(3)$ in terms of an angle of rotation $\theta$ and an axis of rotation $\hat{n}$ of unit length. Explicitly, the group manifold of $SO(3)$ in terms of $\theta$ and $\hat{n}(\beta_1,\beta_2)$ is parametrised by coordinates $(\theta,\beta_1,\beta_2)$. In these coordinates, a general group element $R(\theta,\beta_1,\beta_2) \in SO(3)$ is given by
$$R(\theta,\beta_1,\beta_2)=\mathbb{I}+\sin (\theta)K(\beta_1,\beta_2)+(1-\cos (\theta))K^2,$$
where $\mathbb{I}$ is the unit matrix in three dimenions. The matrix $K(\beta_1,\beta_2)$ is given in terms of the axis of rotation $\hat{n}=(\hat{n}_1,\hat{n}_2,\hat{n}_3)$ of unit length,
$$K(\beta_1,\beta_2)=\left( \begin{matrix}
0&-\hat{n}_3&\hat{n}_2\\
\hat{n}_3&0&-\hat{n}_1\\
-\hat{n}_2&\hat{n}_1&0
\end{matrix} \right).$$
The axis of rotation $\hat{n}(\beta_1,\beta_2)$ is explicitly given by
$$\hat{n}(\beta_1,\beta_2)=\left(
\begin{matrix}
\sin(\beta_1)\cos(\beta_2)\\
\sin(\beta_1)\sin(\beta_2)\\
\cos(\beta_1)
\end{matrix} \right).$$
The rotation matrix $R(\theta,\beta_1,\beta_2)$ corresponds to a rotation of an angle $\theta$ around the axis $\hat{n}(\beta_1,\beta_2)$.

With the procedure described by Brightsun, using $R(t)\equiv R(\theta(t),\beta_1(t),\beta_2(t))$, the Haar measure $\omega$ of $SO(3)$ in the axis-angle representation is
$$\omega=4\sin(\frac{\theta}{2})^2\sin(\beta_1)d\theta \ \wedge\beta_1\wedge\beta_2.$$
Note that the determinant of the Killing metric can be read of from $\omega$.


问:
What are some examples of "visually beautiful" math texts?
Are there good examples of math textbooks that seamlessly combines art, cartoon drawing or computer graphics with the texts? I know that there are a few books on the market, which are not text-book per se, that are geared towards general audience, where there are plenty of artistic figures naturally generated by the phenomena under study, usually things of geometric nature, Escher tiling, or things like chaos.

Wolfram's a New Kind of Science comes to mind. What are some other examples of "visually beautifu" or "artistic" math textbooks or texts.

答: Here's a very old suggestion: Euclid's Elements. Specifically, Oliver Byrne's 1847 rendition of the text with its elegant diagrams and design aesthetic, and Nicholas Rougeux's online conversion of the text into a digital edition, available for free here.


问:
Every curve is a geodesic??
I've been reading up on how isometries send geodesics to geodesics. I recently saw a proof of another theorem that used the fact:

The set of fixed points of an isometry is a geodesic.

But isnt the Identity always an isometry, which would then imply every curve, in say the Poincare half plane, is a geodesic. Whats wrong with my reasoning?

Thanks

答:
You're probably recalling or interpreting the fact they used incorrectly. The relevant fact is the following:

If $M$ is a Riemannian manifold and $f: M \longrightarrow M$ an isometry, then each connected component of $\mathrm{Fix}(f)$ is a closed totally geodesic submanifold of $M$.

Note that a submanifold $N \subset M$ being totally geodesic doesn't mean that $N$ is a geodesic in $M$, it means that every geodesic in $N$ is a geodesic in $M$ as well.

Title: Every curve is a geodesic??


问:
Are the two spans equal?
Let $x$,$y$ and $z$ be three linearly independent vectors. Explain, with justification, whether or not
$\operatorname{span}\{x,\,y,\,z\}=\operatorname{span}\{x+y,\,y+z,\,z+x\}$.

Title: Are the two spans equal?

Body:
$x = \frac{1}{2}(x+y)-\frac{1}{2}(y+z)+\frac{1}{2}(z+x)$

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

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

So $x,y,z$ can be written as a linear combination of $(x+y),(z+x),(y+z)$ , hence the spans are equal.


答:
$$\begin{cases}
x^2+y^2=(2-y)^2&\text{Pyth. th. in } \Delta KNP\\
x^2+(1-y)^2=(y+1)^2&\text{Pyth. th. in } \Delta POQ\\
\end{cases}
$$
Solution is $x= \sqrt{2},y= \frac{1}{2}$

Body:
Circle $K$ has diameter $AB$. Circle $L$ is tangent to circle $K$ and to $AB$ at the center of circle $K$. Circle $M$ is tangent to circle $K$, to circle $L$ and to $AB$. Find the ratio of area of $K$ to that of $M$.

What I Tried: Here is a picture :-

Mark all the points, do the labelling and draw perpendiculars and all, and in the end doing all these in Geogebra I found that $PN$ is almost equal to $\frac{1}{4}$th of $KB$'s length, where $KB$ is the radius of the large circle. That is possible only when $\Delta OQP \cong \Delta KQP$ , but I am really having a hard time showing this, and I got no idea how to proceed. I tried joining more lines and even doing Pythagorean Theorem but it had no use.

Can anyone help me? Thank You.

Source:- $1976$ AHSME Problem $24$ (Check here :- https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems)


问:
Construnction of concrete example about the normal scheme
According to the Hartshorne textbook, $X=\operatorname{Spec}A $ is normal if $A$ is UFD. (such context is proof of why the divisor class of the group of $X$ is zero if $A$ is UFD) I have a curiosity why such a fact holds. However, I want to make a concrete example rather than to prove the given fact.

For example, take $A=\mathbb{Z}$ (It is undoubtly UFD!). Then, the rational number,  $\mathbb{Q}$, is  a ring extension of $\mathbb{Z}$.(or, such extension is viewed as a field of fraction).  $X=\operatorname{Spec}A = \left \{ (0) \right \}U \left \{ (p) | ~p:prime~number \right \} $. (I deonte $(p)$ by $\mathfrak{p}$). Next, when considering the defintion of normal scheme, all of local rings are integrally closed domain, our claim is

Claim : The local ring $\mathcal{O}_{X,\mathfrak{p}}$ is a integrally
closed domain. i.e (1) $\mathcal{O}_{X,\mathfrak{p}}$ is an integral domain (2) its integral closure in its field of fractions is $\mathcal{O}_{X,\mathfrak{p}}$ for all $\mathfrak{p} \in \operatorname{Spec}A$.(*Now, I do not want to foucs on the generic point.. )

When obseriving $\mathcal{O}_{X,\mathfrak{p}}$, $\mathcal{O}_{X,\mathfrak{p}}$ is isomorphic to $\mathbb{Z}_{\mathfrak{p}}$ by prop2.2 in Hartshorne. And $\mathbb{Z}_{\mathfrak{p}}=\left \{ \frac{a}{b}| a \in \mathbb{Z}, p \not | b \right \} \subset \mathbb{Q} $ (Here is local ring at prime ideal $\mathfrak{p}$. And it is clearly integral domain. However, (2) is not easy for me ; I do not know how to collect integral closure $I$. Clearly, $\mathbb{Z} \subset \mathbb{Z}_{\mathfrak{p}} \subset \mathbb{Q} $ holds. Then is $I\overset{??}=\left \{ q \in \mathbb{Q} : q ~ is~integrable~over~ \mathbb{Z}_{\mathfrak{p}} \right \}$ the collection of integral closure? I do not see which elements are integral closure. Anyway, the proof will finish just by showing $I= \mathbb{Z}_{\mathfrak{p}} $ , but I do not imagine the collection of integral closure.

答:
Assuming $\frac{a}{b} \in \mathbb{Q}$ is integral over $\mathbb{Z}_\mathfrak{p}$, it means
that there exists $c_1,\ldots,c_n \in \mathbb{Z}_\mathfrak{p}$ satisfied
$$(\frac{a}{b})^n+c_1(\frac{a}{b})^{n-1}+\ldots+c_n=0.$$
Since $c_i=\frac{x_i}{y_i}$ where $y_i \nmid p$, after multiplying $b^ny_1y_2\ldots y_n$ one gets
$$a^n(y_1\ldots y_n)+x_1ba^{n-1}(y_2\ldots y_n)+\ldots+x_nb^n(y_1\ldots y_{n-1})=0.$$
If $\frac{a}{b} \notin \mathbb{Z}_\mathfrak{p}$, one can assume $p \mid b$ and $p \nmid a$. But it's clearly wrong since the first term is not divisible by $p$ but the others are not. It means $\frac{a}{b} \in \mathbb{Z}_\mathfrak{p}$.


问:
How can I prove that the cyan triangle in the picture is isosceles?
There is a circumference $\gamma$ which has a square $ABCD$ inscribed.
There is also a semi-circumference $\gamma_1$ centered in $H$ with radius $\overline{OH}=\overline{AH}=\overline{HB}$.
Then there is a half line having origin in $A$ and intersecting $\gamma_1$ and $\gamma$ in $E$ and $F$ respectively.
I need to prove that the cyan triangle $BEF$ is right-angled and isosceles using only euclidean geometry theorems (except triangle similitude criteria). I managed to prove that the angle $B\hat{E}F$ is right, then I tried to extend the segment $\overline{BE}$ intersecting $\gamma$ in G and tried to prove that $GEF \equiv ABE$ with one of the congruence criteria, but no luck. I then tried a lot of constructions, for example drawing the semi-circumference intersecting the points $E$, $F$, and $B$, but I still get stuck on proving that there is some length on one side which is equal to some other length on the other side.

答: Hint: $\angle ACB = \angle AFB = 45^\circ$ follows from angles in the same segment.


答: First, note that $\angle ACB$ and $\angle AFB$ subtend the same arc. Then, since $\angle ACB = 45^{\circ}$, $\angle AFB = 45^{\circ}$. Also, $\angle AEB$ is inscribed in a semicircle, so it is a right triangle. Then, $\angle FEB = 90^{\circ}$ and $\angle EFB = 45^{\circ}$, so $\angle FBE = \angle EFB = 45^{\circ}$, and $\triangle FBE$ is isosceles. $\blacksquare$

Title: How can I prove that the cyan triangle in the picture is isosceles?

Body:
There is a circumference $\gamma$ which has a square $ABCD$ inscribed.
There is also a semi-circumference $\gamma_1$ centered in $H$ with radius $\overline{OH}=\overline{AH}=\overline{HB}$.
Then there is a half line having origin in $A$ and intersecting $\gamma_1$ and $\gamma$ in $E$ and $F$ respectively.
I need to prove that the cyan triangle $BEF$ is right-angled and isosceles using only euclidean geometry theorems (except triangle similitude criteria). I managed to prove that the angle $B\hat{E}F$ is right, then I tried to extend the segment $\overline{BE}$ intersecting $\gamma$ in G and tried to prove that $GEF \equiv ABE$ with one of the congruence criteria, but no luck. I then tried a lot of constructions, for example drawing the semi-circumference intersecting the points $E$, $F$, and $B$, but I still get stuck on proving that there is some length on one side which is equal to some other length on the other side.


答:
Note that

$$2\angle AFB= \angle AOB =\angle AEB  =90^\circ$$
which leads to
$\angle AFB = \angle FBE =45^\circ$.


问:
Is formula such as $\forall x\exists x( x\neq 0)$ allowed?

In the book I am reading, it seems that $\forall x\exists x (x\neq 0)$ is also allowed. Why does it not restrict the times of occurrences of quantification of the same variable to be $\leq 1$ ? What does $\forall x\exists x( x\neq 0)$ mean here?

It's Raymond's First-order logic.

答:
It depends on which textbook you're using. Some forbid vacuous quantification by defining

If $x$ is a variable and $\phi$ is a formula in which $x$ occurs free, then $\forall x \phi$ and $\exists x \phi$ are formualas.

This would rule out prepending $\forall x$ to $\exists x (x \neq 0)$ because in $x$ occurs only bound, and also rule out applying $\forall x$ to $(1 \neq 0)$ because $x$ does not occur at all, whereas $\forall x (x \neq 0)$ would be allowed.

Usually though the restriction in bold is dropped, and that's what your textbook does too, so $\forall x \exists x (x \neq 0)$, and likewise a formula like $\forall x (1 \neq 0)$, is allowed. While vacuous quantification isn't very useful, it also doesn't hurt to have it, and additional restrictions just unnecessarily complicate the syntax of FOL.

It follows from the semantics of FOL that the first quantifier will be vacuous and only the inner one actually takes effect on the formula: With every quantifier the evaluation of the following formula happens under the modified assignment which maps the variable bound by the quantifier to whichever element is currently being iterated over, thereby "overriding" the variable assignment on the variable both of the initial variable assignment for free variables and quantifiers further out:

$\newcommand{\M}{\mathcal{M}}\newcommand{\D}{\mathcal{D}}\newcommand{\I}{\mathcal{I}}$

$\M \models \forall x \exists y (x \neq y)$

$\Longleftrightarrow \text{for all } v: \M, v \models \forall x \exists x (x \neq 0)$

$\Longleftrightarrow \text{for all } a \in \D: \M, v^{[x \mapsto a]} \models \exists y (x \neq 0)$

$\Longleftrightarrow \text{for all } a \in \D: \text{there exists } b \in \D: \M, v^{[x \mapsto a][x \mapsto b]} \models x \neq 0$
$\Longleftrightarrow \text{for all } a \in \D: \text{there exists } b \in D: v^{[x \mapsto a][x \mapsto b]}(x) \neq \I(0)$

By the definition of $v^{[x \mapsto d]}$:

$$v^{[x \mapsto d]}(u) = \begin{cases}d & \text{if } u = x\\v(u) & \text{else}\end{cases}$$

we get that $v^{[x \mapsto a][x \mapsto b]} = v^{[x \mapsto b]}$, so at any evaluation of the subformula $x \neq 0$, the value of $x$ will be that assigned by the existential quantifier, whereas the universal quantifier dose not actively take effect on anything, and the formula "means" in effect $\exists x (x \neq 0)$.

Edit: In the notation used in Smullyan, $v \models A_a^x$ instead of $v^{[x \mapsto a]} \models A$; but the mechanism is exactly the same.


问:
Does 'incomplete' mean the same thing in formal arithmetic compared to Godel's work?
I am struggling to understand what is different from 'Godel incompleteness' (GI) and call it 'ordinary completeness' (OC). Godel showed FOL to be complete. Yet, Godel's theorems establish that the language Q (an axiomatic extension of FOL) is incomplete.

What distinguishes GI from OC in terms of FOL?

答: The completeness of first-order logic says that any sentence that is true in all models is provable. This is still true in a system like Robinson's system $Q$ that can model enough of natural number arithmetic for Gödel's incompleteness theorems to apply. The incompleteness theorems imply that there are sentences that are true in the standard model of arithmetic that cannot be proved: these sentences, e.g., a sentence $\mathrm{Con}(Q)$ that asserts the consistency of $Q$, will be false in certain non-standard models. Non-standard models include non-standard "numbers" that are greater than any number that can be obtained from $0$ by adding $1$ a finite number of times. A non-standard model can satisfy the axioms of $Q$, while admitting non-standard numbers that behave as if they were the Gödel number of a proof of the inconsistency of $Q$.


问:
Use of the Well Ordering Principle in the proof of the Fundamental Theorem of Arithmetic
Unlike a previous question that was closed on accounts of ambiguity, I perfectly understand the 'need' of the Well Ordering Principle in the proof of the Fundamental Theorem of Arithmetic. What I do not understand is which is the set on which the well ordering principle is applied? Is it the set of numbers that cannot be written as a product of primes? Is the step intuitive?
The link contains the PDF that can be referred to for details on the proof:
http://math.hawaii.edu/~lee/courses/fundamental.pdf</p>

答:
The set of natural numbers which do not have a unique prime factorisation is either empty or it isn't. If it isn't empty, the WOP applies, and there is a smallest such number, $N$.

Now, it's not a question of having or not having a prime factorisation. It's easy to prove that a number has a prime factorisation. Any simple, naive factorisation algorithm is entirely deterministic and conclusive.

The issue is whether there can be more than one factorisation. And assuming $N$ as defined above exists, with multiple distinct prime factorisations, we can use Euclid's lemma to find a strictly smaller such number, contradicting the definition of $N$.


问:
Show that $\Phi \vDash \psi \Longleftrightarrow \Phi \cup \{\lnot \psi\} \ \text{is not satisfiable}$
The main algorithmic problems in logic (in my bachelors CS course) are: $0)$ evaluation problem "Auswertungsproblem", $1)$ equivalence problem, $2)$ logical consequence, $3)$ satisfiability problem and $4)$ validity problem. All of these problems can be reduced to the satisfiability problem or the evaluation problem.

The equivalence problem states: "Given $\varphi, \psi \in AL \ \text{is} \ \varphi \equiv \psi$?". I can reduce this problem to the logical consequence problem because if $\varphi \equiv \psi \Longleftrightarrow \varphi \vDash \psi \land \psi \vDash \varphi$. If we can determine such a consequence then $\varphi$ and $\psi$ are equivalent.

The logical consequence problem states that: "$\text{Given} \ \Phi \subseteq AL \ \text{and } \ \psi \in AL, \ \text{is} \ \Phi \vDash \psi \ \text{valid}$?" We can reduce this problem to the satisfiability problem because we know that $\Phi \vDash \psi \Longleftrightarrow \Phi \cup \{\lnot \psi\} \ \text{is not satisfiable}$.

I'm having a problem with the very last section. Why is "$\Phi \vDash \psi \Longleftrightarrow \Phi \cup \{\lnot \psi\} \ \text{is not satisfiable}$" true? I can't seem to understand how to go about showing this.

I mean the idea for the whole exercise is to understand that instead of solving $5$ different problems, it is more efficient to reduce them to ones that're "easier" to solve.

答:
Regarding the title question:

(i) if $\Phi \vDash \psi$, then $\Phi \cup \{ ¬ \psi \}$ is unsatisfiable.

Assume not, i.e. $\Phi \cup \{ ¬ \psi \}$ is satisfiable; this means that there is some valuation such that all of $\Phi$'s are TRUE and also $\lnot \psi$ is, i.e. $\psi$ is FALSE, contradicting the fact that $\Phi \vDash \psi$ means: in every valuation where all of $\Phi$'s are TRUE, $\psi$ must be TRUE also.

(ii) if $\Phi \cup \{ ¬ \psi \}$ is unsatisfiable, then $\Phi \vDash \psi$.

The proof is similar.

The body of the question is not very clear, but yes, the various concept are strictly linked.

With the previous result we have showed how to reduce the problem of logical consequence [locical implication] to that of satisfiability.

Also validity can be so reduced: a formula $\varphi$ is valid iff $\lnot \varphi$ is unsatisfiable.

And also equivalence is: two formulas are logically equivalent iff they imply each other.


问:
Are rotations of $(0,1)$ by $n \arccos(\frac{1}{3})$ dense in the unit circle?
Under which conditions will successive rotations of $(0, 1)$ by an angle $\theta$ guarantee that given $\delta > 0$ and some point $p$ on the unit circle, there exists some $n$ such that rotating $(0, 1)$ by $n\theta$ yields a point within $\delta$ radians of $p$?  That is, are there certain angles that will guarantee that the set generated by rotating $(0, 1)$ by integral multiples of $\theta$ is dense in the unit circle?

More specifically, I am working with the angle $\cos^{-1}\left(\frac{1}{3}\right)$ and wish to show that this property is true, but I'm not sure how to go about proving this.  It seems likely I'm missing something simple here...  Does it hold for any angle that is an irrational multiple of $\pi$?  If so, why?

答: It's dense if and only if $\frac{\theta}{\pi}$ is irrational</p>


问:
Let $f$ be defined on $[a,b]$, Prove that if f has a local maximum at a point $x \in (a,b)$, and if $f'(x)$ exists, then $f'(x)=0$
Is this proof correct:

Let's choose a $\delta$ to that

$a < x - \delta < x < x + \delta < b$

If $ x - \delta < t < x$

then $\frac {f(t) - f(x)} {t-x} \geq 0$

Letting $t \rightarrow x$, we see that $f'(x) \geq 0$

If $ x < t < x + \delta$,

then $\frac {f(t) - f(x)} {t-x} \leq 0$

Which shows that $f'(x) \leq 0$. Hence $f'(x) = 0$

答: Yes, it is. Maybe you want to say $t\to x^+$ and $t\to x^-$ explicitly.</p>


答:
Even assuming that the expression is "syntactically" valid, you can't assign a truth value to

$$\forall x:\exists x:x\ne0$$ because

$$\forall x':\exists x'':x'\ne0$$ is false and $$\forall x':\exists x'':x''\ne0$$ is true.

Title: Is formula such as $\forall x\exists x( x\neq 0)$ allowed?

Body:
In the book I am reading, it seems that $\forall x\exists x (x\neq 0)$ is also allowed. Why does it not restrict the times of occurrences of quantification of the same variable to be $\leq 1$ ? What does $\forall x\exists x( x\neq 0)$ mean here?

It's Raymond's First-order logic.


问:
Proving a specific eigenvalue of a 6x6 Matrix.
Currently, I am looking at Markov diagrams and their associated transition matrices. I am trying to prove that the transition matrix has eigenvalue $\lambda=1$. I am aware that to find the eigenvalues of a matrix we use:

$$\det (A - \lambda I_{6}) = 0$$

and then each $\lambda$ gives an eigenvalue of the matrix. But this seems like a very lengthy process for a $6 \times 6$ matrix.

I'm also aware that every stochastic matrix has the eigenvalue of $\lambda=1$ but I would like to learn how to work this out for this specific matrix.

$$
\begin{bmatrix}
    0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & 0 \\
    \frac{1}{2} & 0 & 0 & \frac{1}{3} & 0 & 0 \\
    \frac{1}{2} & 0 & 0 & \frac{1}{3}& \frac{1}{2}& 0 \\
    0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & \frac{1}{2} \\
    0 & 0 & \frac{1}{3}& 0 & 0 & \frac{1}{2} \\
    0 & 0 & 0 & \frac{1}{3} & \frac{1}{2} & 0 \\
\end{bmatrix}
$$

Any help would be greately appreciated.

edit

Having using the eigenvector of $\large[1\ 1\ 1\ 1\ 1\ 1\ \large]$
to show that $\lambda =1$ is an eigenvalue I now want to find a stable distribution.

So I am going to labe my horizontal and vertical entries M, E, C, S, W, N respectively. and will set $\large[M\ E\ C\ S\ W\ N \large]$ to be a vector to multiply my transition matrix by.

So i get:

$1/2E+1/3C = M$

$1/2M+1/3S=E$

$1/2M+1/3S+1/2W=C$

$1/2E+1/3C+1/2N=S$

$1/3C+1/2N=W$

$1/3S+1/2W=N$

So Im wondering if what I'm doing is right and whether my next steps should be to try and work out what the values of M E C S W N are?

答:
HINT

If you are looking for a specific eigenvalue, compute the matrix $B = A - \lambda I$, and show that $\det(B) = 0$.

If you can guess the corresponding eigenvector $\vec{x}$, it becomes even easier. Multiply matrices to show that
$$
A \vec{x} = \lambda \vec{x}
$$
and you are done.

In your case, however, it is even easier. The sum of all entries in each column is $1$, which makes you matrix stochastic. For such a matrix, the vector of all $1$'s is a left eigenvector, in other words,
$$
\begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix}
\begin{bmatrix}
    0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & 0 \\
    \frac{1}{2} & 0 & 0 & \frac{1}{3} & 0 & 0 \\
    \frac{1}{2} & 0 & 0 & \frac{1}{3}& \frac{1}{2}& 0 \\
    0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & \frac{1}{2} \\
    0 & 0 & \frac{1}{3}& 0 & 0 & \frac{1}{2} \\
    0 & 0 & 0 & \frac{1}{3} & \frac{1}{2} & 0 \\
\end{bmatrix}
= \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix}
$$

UPDATE

Typically to find the eigenvectors and eigenvalues of a matrix $A$, first solve $\det(A-\lambda I)=0$ and then when you get the eigenvalues, plug them into $(A - \lambda I) \vec{x}=\vec{0}$ and solve for each $\vec{x}$ separately.


问:
Real Spectral Theorem Induction Proof
I am currently reading Linear Algebra Done right and was having trouble with the proof of the Real Spectral Theorem.

Statement:

7.13 Real Spectral Theorem: Suppose that V is a real inner-product space and T ∈ L(V). Then V has an orthonormal basis consisting of
eigenvectors of T if and only if T is self-adjoint.

The theorem he is refering to in the proof is:

7.12 Lemma: Suppose T ∈ L(V) is self-adjoint. Then T has an eigenvalue.

I am only interested in the proof from one side. Here it is:

“To prove the other direction, now suppose that $T$ is self-adjoint. We will prove that $V$ has an orthonormal basis consisting of eigenvectors of $T$ by induction on the dimension of $V$. To get started, note that our desired result clearly holds if $\operatorname{dim} V=1$. Now assume that $\operatorname{dim} V>1$ and that the desired result holds on vector spaces of smaller dimension.

The idea of the proof is to take any eigenvector $u$ of $T$ with norm 1 then adjoin to it an orthonormal basis of eigenvectors of $\left.T\right|_{\{u\}^{\perp}}$. Now
for the details, the most important of which is verifying that $\left.T\right|_{\{u\}^{\perp}}$ is self-adjoint (this allows us to apply our induction hypothesis).

Let $\lambda$ be any eigenvalue of $T$ (because $T$ is self-adjoint, we know from the previous lemma that it has an eigenvalue) and let $u \in V$ denote a corresponding eigenvector with $\|u\|=1$. Let $U$ denote the one-dimensional subspace of $v$ consisting of all scalar multiples of $u$. Note that a vector $v \in V$ is in $U^{\perp}$ if and only if $\langle u, v\rangle=0$.
Suppose $v \in U^{\perp}$. Then because $T$ is self-adjoint, we have
$$
\langle u, T v\rangle=\langle T u, v\rangle=\langle\lambda u, v\rangle=\lambda\langle u, v\rangle=0
$$
and hence $T v \in U^{\perp}$. Thus $Tv \in U^{\perp}$ whenever $v \in U^{\perp}$. In other words, $U^{\perp}$ is invariant under $T$. Thus we can define an operator $S \in \mathcal{L}\left(U^{\perp}\right)$ by $S=\left.T\right|_{U^{\perp}}$. If $v, w \in U^{\perp}$, then
$$
\langle S v, w\rangle=\langle T v, w\rangle=\langle v, T w\rangle=\langle v, S w\rangle
$$
which shows that $S$ is self-adjoint (note that in the middle equality above we used the self-adjointness of $T$ ). Thus, by our induction hypothesis, there is an orthonormal basis of $U^{\perp}$ consisting of eigenvectors of $S$. Clearly every eigenvector of $S$ is an eigenvector of $T$ (because $S v=T v$ for every $v \in U^{\perp}$ ). Thus adjoining $u$ to an orthonormal basis of $U^{\perp}$ consisting of eigenvectors of $S$ gives an orthonormal basis of $V$ consisting of eigenvectors of $T$, as desired.”

It isn't fully clicking to me. He defined an artificial operator himself and is saying this must have an orthogonal basis even though he never showed it for matrices greater than dimension 1. Dimension 1 is such an unrealistic case for matrices. Can anyone clear out his thinking or approach for me?

答:
@Chris Eagle gave a very detailed explanation. I want to make a brief summary of his answer.

The logic of the proof in Linear Algebra Done Right is as follows:

it's true for $dim V = 1$.

let $dim V = n+1$. we assume it's true for real inner product spaces with dimensions up to $n$, in this proof, this real inner product space is $U^\perp$.

$dim U = 1$, $dim U^\perp = n$, we can find an orthonormal basis of $V$.

Title: Real Spectral Theorem Induction Proof


答:
In order to prove that 1 is indeed an eigenvalue, you just need to prove that $\det(A-I_6)=0$. You can prove that by finding a linear combination of the rows or columns that are $0$.

Here, you can use that the matrix is stochastic, so the sum of each row is 1, when you substract $I_6$, you get that the sum of all the rows of $A-I_6 $ is 0, so 1 is an eigenvalue of A

Title: Proving a specific eigenvalue of a 6x6 Matrix.

Body:
Currently, I am looking at Markov diagrams and their associated transition matrices. I am trying to prove that the transition matrix has eigenvalue $\lambda=1$. I am aware that to find the eigenvalues of a matrix we use:

$$\det (A - \lambda I_{6}) = 0$$

and then each $\lambda$ gives an eigenvalue of the matrix. But this seems like a very lengthy process for a $6 \times 6$ matrix.

I'm also aware that every stochastic matrix has the eigenvalue of $\lambda=1$ but I would like to learn how to work this out for this specific matrix.

$$
\begin{bmatrix}
    0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & 0 \\
    \frac{1}{2} & 0 & 0 & \frac{1}{3} & 0 & 0 \\
    \frac{1}{2} & 0 & 0 & \frac{1}{3}& \frac{1}{2}& 0 \\
    0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & \frac{1}{2} \\
    0 & 0 & \frac{1}{3}& 0 & 0 & \frac{1}{2} \\
    0 & 0 & 0 & \frac{1}{3} & \frac{1}{2} & 0 \\
\end{bmatrix}
$$

Any help would be greately appreciated.

edit

Having using the eigenvector of $\large[1\ 1\ 1\ 1\ 1\ 1\ \large]$
to show that $\lambda =1$ is an eigenvalue I now want to find a stable distribution.

So I am going to labe my horizontal and vertical entries M, E, C, S, W, N respectively. and will set $\large[M\ E\ C\ S\ W\ N \large]$ to be a vector to multiply my transition matrix by.

So i get:

$1/2E+1/3C = M$

$1/2M+1/3S=E$

$1/2M+1/3S+1/2W=C$

$1/2E+1/3C+1/2N=S$

$1/3C+1/2N=W$

$1/3S+1/2W=N$

So Im wondering if what I'm doing is right and whether my next steps should be to try and work out what the values of M E C S W N are?


问:
determine with proof, all the positive integers $n$ for which $n$ is not the square of any integer and $[\sqrt{n}]^3$ divides $n^2$
determine with proof, all the positive integers $n$ for which:

$n$ is not the square of any integer and,

$[\sqrt{n}]^3$ divides $n^2$. where $[x]$ is the greatest integer less than or equal to $x$.

My Approach:

Well I tried a few test cases and I concluded that no prime number greater than 4 holds true to this statement. Except that, I couldn't recognize a pattern or make a solution on my own so please help me out.

答:
Hint: $\,[\sqrt n]\! =\! k < \sqrt n < k\!+\!1 \!\iff\! k^2 < n < (k\!+\!1)^2 \!\!\iff\! 0 < \overbrace{n\!-\!k^2}^{\color{#0a0}{\textstyle j}} \color{#0a0}{< 2k+1}$

Thus $\, k^3\mid (k^2\!+\!j)^2 \Rightarrow  k^2\mid j^2 \Rightarrow k\mid j\, $ via Rational Root Test, so cancelling $\,k^2$

yields $\ k\mid (k\!+\!j/k)^2 \Rightarrow \,\bbox[5px,border:1px solid #c00]{\color{#c00}k\mid (j/k)^2\,\color{#c00}{\mid\, 4}}\ $ by $\,\color{#0a0}{j/k \le 2}$


问:
Number of solution of three equation in term of $x,y,z,\theta$
Find the number of all possible values of $\theta$, where $0<\theta<\pi$, for which the system of equations

$\left( {y + z} \right)\cos 3\theta  = xyz\sin 3\theta $

$x\sin 3\theta  = \frac{{2\cos 3\theta }}{y} + \frac{{2\sin 3\theta }}{z}$

$xyz\sin 3\theta  = \left( {y + 2z} \right)\cos 3\theta  + y\sin 3\theta $
have a solution $x_0,y_0,z_0$ with $y_0z_0\ne 0$ .

My approach is as follow

$\left( {y + z} \right)\cos 3\theta  = xyz\sin 3\theta  \Rightarrow \frac{{y + z}}{{xyz}} = \tan 3\theta $

$x\sin 3\theta  = \frac{{2\cos 3\theta }}{y} + \frac{{2\sin 3\theta }}{z} \Rightarrow \sin 3\theta \left( {x - \frac{2}{z}} \right) = \frac{{2\cos 3\theta }}{y} \Rightarrow \tan 3\theta  = \frac{2}{{y\left( {x - \frac{2}{z}} \right)}} = \frac{{2z}}{{y\left( {xz - 2} \right)}} = \frac{{2z}}{{\left( {xyz - 2y} \right)}}$

$xyz\sin 3\theta  = \left( {y + 2z} \right)\cos 3\theta  + y\sin 3\theta  \Rightarrow \sin 3\theta \left( {xyz - y} \right) = \cos 3\theta \left( {y + 2z} \right) \Rightarrow \tan 3\theta  = \frac{{y + 2z}}{{\left( {xyz - y} \right)}}$

$\tan 3\theta  = \frac{{y + z}}{{xyz}} = \frac{{y + 2z}}{{\left( {xyz - y} \right)}} = \frac{{2z}}{{\left( {xyz - 2y} \right)}}$

$xyz = t \Rightarrow \tan 3\theta  = \frac{{y + z}}{t} = \frac{{y + 2z}}{{t - y}} = \frac{{2z}}{{t - 2y}}$.

How do we proceed from here.

答:
$3\theta=t$

By first and third condition $$z\cos t+y\sin t=0$$ By mutiplying $yz$ second condition $$xyz \sin t=2z\cos t+2y\sin t=0$$ Hence  either $x=0$ or $\sin t=\sin 3\theta=0$

It remains to check which values of $\theta$ for which $x=0$ and whether $\sin 3\theta=0$ satisfies the 3 conditions and how many solutions it has .


问:
Prove for any natural number $n$, ${a^n}$ + ${b^n}$ is a natural number not divisible by $5$.

Let $a = 3 + 2\sqrt{2}$ and $b = 3 - 2\sqrt{2}$. Prove that for every natural number $n$, ${a^n}$ + ${b^n}$ is a natural number, not divisible by $5$.

It's trivial that this number is natural by using binomial formula, but I can't figure out how to prove that it's not divisible by $5$.

答:
I am well aware that what follows is not the simplest way to answer to this question, but I'd like to put the problem in a more structured context.

Consider the field
$$
\Bbb Q(\sqrt 2)=\{a+b\sqrt2\,\mid\,a, b\in\Bbb Q\}.
$$
The field is endowed with an automorphism $\sigma$ defined as $\sigma(a+b\sqrt2)=a-b\sqrt2$ which has the property that $\Bbb Q$ is (the subfield) fixed.

Note that the two elements given are
$$
x=3+2\sqrt2\qquad\text{and}\qquad
x_\sigma=\sigma(x)=3-2\sqrt2
$$
This tells immediately that $x^n+x_\sigma^n\in\Bbb Q$ since $\sigma(x^n+x_\sigma^n)=x^n+x_\sigma^n$. Then it is pretty clear that it is actually in $\Bbb Z$ since taking positive powers never introduces denominators.

Next observe that $xx^\sigma=1$ so that in fact
$$
x^n+x^n_\sigma=\frac{1+x^{2n}}{x^n}\qquad\qquad(\ast)
$$
We now want to study divisibility by $5$. Thus we shall work modulo $5$ in $\Bbb Z[\sqrt 2]$. One has
$$
\frac{\Bbb Z[\sqrt 2]}{(5)}\simeq\frac{\Bbb Z[T]}{(5,T^2-2)}\simeq
\frac{\Bbb F_5[T]}{(T^2-2)}\simeq\Bbb F_{25},\qquad(\ast\ast)
$$
the field with $5^2$ elements, since $2$ is not a square modulo $5$.
Denote $\bar z$ the class of $z\in\Bbb Z[\sqrt2]$ in this quotient.

Note that $\bar x\neq0$ so that $\bar x^n\neq 0$ for all $n\in\Bbb Z$ so that after $(\ast)$ we need to prove that $\bar x^{2n}\neq-1$.

Now if $u=\bar T$ in $(\ast\ast)$ we can write $\bar x=3+2u$ and a simple modulo $5$ computation shows that $\bar x^2=-\bar x_\sigma$, i.e.
$$
\bar x^3=-1.
$$
Thus $\bar x^6=1$ and therefore
$$
\bar x^k=-1\qquad\Longleftrightarrow\qquad k\equiv3\bmod 6
$$
showing that the numerator of $(\ast)$ is never $0$ modulo $5$.


问:
Sketch the function $y = {1 \over {{x^2}}}\ln x$
I don't know where to begin with this, the ${1 \over {{x^2}}}$ part of the function throws me off, how to do I go about this? How does one generally approach a question like this?</p>

答:
It is indeed an interesting function. You can try finding a lot of points: select a range of $x$ values, and evaluate $$f(x) = \dfrac{\ln x}{x^2}$$ for each x.  Given enough data points, plotting them as you go along, you can try to "connect the dots" to see what you get.

You can also check for asymptotes, critical values, if any (differentiate!!) to find any extrema, if they exist, etc: that information will help reveal if and where the function peaks and/or bottoms out, where it is increasing, decreasing, etc.

For example, if we take the limit of $f(x)$ as $x\to +\infty$,  $x \to 0^+$, e.g., and we will see some interesting things.

Computing the derivative gives us $$f'(x) = \frac{1-2\ln x}{x^3}$$ So the derivative $f'(x) = 0$ at $x=e^{1/2}$, is positive for $x\lt e^{1/2}$ (hence increasing) and is negative, hence decreasing, when $x\gt e^{1/2}$.

Maximum: $x=e^{1/2}$.

You can also "take a peak" to see what you're trying to approximate: [Disclaimer Wolfram Alpha's graph is taking, I assume, $\ln x$ in the numerator to be $\ln|x|$, in which case you'd have a symmetrical graph (the left portion mirroring the right-hand portion).


问:
Why Jacobson radical is 'radical'?
Nilradical of ring R, $\text{nil}(R)$ is intersection of prime ideals, is radical of $(0)$.
So I feel the name  'radical' is natural.

But Jacobson radical, that is an intersection of maximal ideals, seem not to be radical of a certain ideal.
Why Jacobson radical is nevertheless called 'radical'? Thank you.

答:
To add another viewpoint: the "bad element" description of radicals is somewhat useful, but I also feel like it might be someone's atttempt at a retrospective description that just caught on.

I tried tracking down the original usage in ring-theory prehistory just now, but I didn't get very far. A biographical article on Wedderburn hints that it may lie in the work of Cartan, Molien and Frobenius.  (It would be interesting to know where to find that but I reached the end of my timebox on the subject. )

Really you should probably interpret "radical" as a cousin of "$\sqrt{\cdot}$" descended from the same latin root meaning... "root."

A little caveat: The following explanation is speculative. I don't know if this was the original intention or if I am just the next person with a clever retrospective explanation.

The following description applies pretty well:

A root is something important that supports a structure but is itself largely hidden.  When viewing the elements through a certain lens, the "radical" for that lens is the set of elements which you can't see through the lens.

When your lens is representation theory, the elements that are invisible to the semisimple representations are the Jacobson radical elements. They don't act at all on any semisimple representations, so they are "hidden."

When your lens is classical algebraic geometry, the nilpotent elements are invisible to the Zariski topology.

I think this is essentially what is captured by the definition of a radical class and that the corresponding ideal is essentially what is invisible to the respective torsion theory.  I am not an expert in this subject, though, so take it with a grain of salt.

The elements themselves are not always obviously "bad" in the sense of "badly behaved." Are the elements of the maximal ideal of a local domain really 'bad'?  Not with respect to the ring operation, I don't think. But in terms of shedding light on semisimple modules for the ring, they don't contribute.

Title: Why Jacobson radical is 'radical'?


问:
Is the winding number of a map $\Omega: \; S^n \mapsto S^n$ dependent on the radius of both spheres?
I have 2 questions regarding homotopy groups.

My first questions comes from the fact that I've read different definitions of homotopy groups. The book by Manton and Sutcliffe defines them as:

''The set of homotopy classes of based maps $\Psi: S^n \mapsto Y$'', where $S^n$ is a n-dimensional unit sphere. But most other sources that I've read about about this topic do not define $S^n$ necessary as a unit sphere, they don't define the radius of the sphere at all, therefore I'm not sure what is right or wrong.

My second question is for the situation where we consider a mapping from $\Psi: S^n \mapsto S^n$, i.e. $\pi_n(S^n)=\mathbb{Z} \; \forall \; n$. Now I've read in a lot of sources (for instance in Magnetic Monopoles by John Preskill) that say the winding number is a conserved quantity. I understand that the winding number is "preserved by continuous deformations''.
However, I wonder if the winding number is also conserved if the radius of either n-spheres is changing with time. For instance, there exist a situation where $\pi_1(S^1)=2$, but there is surely also a situation where $\pi_1(S^1)=3$. I.e. they have different winding numbers, in my mind (I'm most likely wrong) this is because the radius of either spheres have changed. I don't, intuitively, understand how else this is possible.

Just in case my background is important: I'm physics student, so have no mathematical skills regarding topology. I'm not very good in mathematical proofs, but I would really like to intuitively understand this topic.

Thanks in advance!

答:
$\pi_1(S^1) \simeq \mathbb{Z}$ means that the fundamental group of the circle is isomorphic to the entire group of the integers, so it does not make sense to write $\pi_1(S^1) = 2$ or $\pi_1(S^1) = 3$.

Perhaps you mean that there could be two maps $f$ and $g: S^1 \to S^1$ such that the winding number of $f$ is $2$ and the winding number of $g$ is $3$.  This can definitely happen.  Think of the unit circle as being parametrized by radians from $0$ to $2\pi$ with $0 = 2\pi$.  (In topology, we'd say that the circle is homeomorphic to the interval $[0,2\pi]$ with its ends identified.)  The map $\theta \mapsto 2\theta$ has winding number $2$, while the map $\theta \mapsto 3\theta$ has winding number $3$.  If you prefer to think of the unit circle as sitting in $\mathbb{R}^2$, think about the maps $\theta \mapsto e^{2i\theta}$ and $\theta \mapsto e^{3i\theta}$.

So radius doesn't play a role here.

One more thing: algebraic topologists generally don't think of $S^n$ as sitting inside $\mathbb{R}^{n+1}$ -- they prefer a perspective like "$[0,2\pi]$ with the ends identified."  Because of that, they call tend to call winding number "degree."  I tell you this only because it might be helpful in finding out more from mathematicians -- I defer to your judgment about the correct term to use in physics!


问:
Are all these 3 graphs correct f(x)=x^2
My sir said all these 3 graphs are correct for $f(x)=x^2$ where x belongs to [-1,2].

He said that we just prefer using 1st one.Is that right ?

答:
I can imagine the way how the graphs $2$ and $3$ can be correct - but this requires the marks on the $y$ axis to not be equidistant, and actually to be very precisely tailored to the picture you want to show.

For example, if on $y$ axis the $1/4$ mark was half-way through (rather than a quarter of the way through) between the $0$ and $1$ marks, then the point $(1/2, 1/4)$ would appear on the straight line on the graph $2$.

Having said all that, unless making up a trick question - why would anyone do that? People expect equidistant marks on both $x$ and $y$ axis, and if that is the case, only graph $1$ is correct. Any other setup of the axes (including the case when the scale is logarithmic on one or both axes) needs to be specifically noted under the graph, so that people would know what they are looking at.

Title: Are all these 3 graphs correct f(x)=x^2


问:
Evaluate $\lim_{n\to \infty} (n+1)^{\frac 23} -(n-1)^{\frac 23}$
Clearly, $\infty -\infty \not =0$

I have a feeling the squeeze theorem can be applied here, but I am not sure how to write the required terms separately. Can I get a hint?

Body:
Your expression almost looks like the symetric derivative of $f(x)=x^{2/3} $, so you have :

$$(n+1)^{2/3}-(n-1)^{2/3}\sim_{\infty}2\times\frac{2}{3}n^{-1/3} $$

Therefore, $\lim_{n\to\infty}(n+1)^{2/3}-(n-1)^{2/3}=0 $

edit : $$(n+1)^{2/3}-(n-1)^{2/3}= n^{2/3}\left(\left(1+\frac{1}{n}\right)^{2/3} -\left(1-\frac{1}{n}\right)^{2/3}\right)\\
=n^{2/3}\left(1+\frac{2}{3n}+o\left(\frac{1}{n}\right) -1+\frac{2}{3n}+o\left(\frac{1}{n}\right)\right)\\
=\frac{4}{3}n^{-1/3}+o\left(n^{-1/3}\right) $$


问:
existence of faithful state in the predual of a von Neumann algebra
Let $M$ be a von Neumann algebra. $M_{*}$ is the predual of $M$.

Does there must exist a faithful state $\psi\in M_{*}$? If such faithful state exists, how to construct it ?

答:
If $M_*$ is separable, just take $\psi=\sum_n2^{-n}\phi_n$, where the $\phi_n$ form a dense subset of the normal state space.

Otherwise such a state might not exist.


问:
How could I find $x$ in this equation $x^2-x+6 \equiv 0 \pmod {9}$
This equation is like this:
$$x^2-x+6 \equiv 0 \pmod{9}$$

I want to find x by using modular arithmetic. How I could do that?

My solution was:

$$\begin{align*}
 x(x-1) &\equiv -6 \pmod{9}\\
x(x-1) &\equiv 3 \pmod{9}
\end{align*}$$

So $x$ is either:
  $$ x \equiv 3\pmod{9}$$

or

$$x - 1 \equiv 3 \pmod{9}$$

Which is $x \equiv 4 \pmod{9}$

I know what I have done is very dumb. So what is the correct solution? Give me the instructions also please.

答:
Update: I just noticed that this answer is in the same vein as the answer given by JMP, but the step-by-step instructions(method) follows a predetermined pedagogical flow.

Solve

$\tag 1 x^2 - x + 6 \equiv 0 \pmod9$

Let $T = \{0,1,2\}$ and recall that every integer $n$ satisfying $0 \le n \lt 9$ has a representation

$\tag 2  n = 3q + r \quad \text{where } r,q \in T$

Using simple algebra,

$\tag 3 n^2 -n + 6 = 9 q^2 + 6 q r - 3 q + r^2 - r + 6$

Setting $r = 0$ on the rhs of $\text{(3)}$,

$\quad 9 q^2 - 3 q + 6 \equiv 0 \pmod9 \; \text{ implies } \; 3q \equiv 6 \pmod9$

and we see that $[x] = [3\cdot2 + 0] = [6]$ is a solution to $\text{(1)}$

Setting $r = 1 $on the rhs of $\text{(3)}$,

$\quad  9q^2 + 3q + 6 \equiv 0 \pmod9 \; \text{ implies } \; 3q + 6 \equiv 0 \pmod9$

and we see that $[x] = [3\cdot1 + 1] = [4]$ is a solution to $\text{(1)}$

Setting $r = 2$ on the rhs of $\text{(3)}$,

$\quad  9 q^2 + 9 q + 8 \equiv 0 \pmod9 \; \text{ implies } \; 8 \equiv 0 \pmod9$

We conclude that $[4]$ and $[6]$ are all the solutions to $\text{(1)}$.


Title: Evaluate $\lim_{n\to \infty} (n+1)^{\frac 23} -(n-1)^{\frac 23}$

Body:
Clearly, $\infty -\infty \not =0$

I have a feeling the squeeze theorem can be applied here, but I am not sure how to write the required terms separately. Can I get a hint?

Body: $n=\frac{1}{t}$ $$\lim_{t\to 0}\frac{{(1+t)}^{2/3}-{(1-t)}^{2/3}}{t^{2/3}}$$  Now apply L-Hospitals rule.. to get $$\lim_{t\to 0}\left(\frac{t}{\sqrt[3]{1+t}}+\frac{t}{\sqrt[3]{1-t}}\right)=?$$


Body:
Let $f:[n-1,n+1]\to\mathbb{R}$ and $f(x)=x^{\frac{2}{3}}$. Then from Mean value theorem (Lagrange’a) exist $\xi\in (n-1,n+1)$ such that:

$$\frac{f(n+1)-f(n-1)}{n+1-(n-1)}=f'(\xi)$$

but (check it out)
$$\frac{2}{3 \sqrt[3]{n+1} }\le f'(\xi)\le \frac{2}{3 \sqrt[3]{n-1} } $$

so

$$\frac{4}{3 \sqrt[3]{n+1} }\le f(n+1)-f(n-1 )\le \frac{4}{3 \sqrt[3]{n-1} } $$

$$\frac{4}{3 \sqrt[3]{n+1} }\le (n+1)^{\frac{2}{3}}-(n-1 )^{\frac{2}{3}}\le \frac{4}{3 \sqrt[3]{n-1} } $$

now You can aplay Squeeze theorem.


Body:
$$(n+1)^{2/3} -(n-1)^{2/3}=\frac{(n+1)^2-(n-1)^2}{(n+1)^{4/3}+(n+1)^{2/3}(n-1)^{2/3}+(n-1)^{4/3}}
\\=\frac{n}{n^{4/3}}\frac4{(1+n^{-1})^{4/3}+(1-n^{-2})^{2/3}+(1-n^{-1})^{4/3}}\to 0$$


Body:
Set $\dfrac1n=y\implies y\to0^+$ to find

$$F=\lim_{y\to0^+}\dfrac{(1+y)^{2/3}-(1-y)^{2/3}}y$$

Now rationalize the numerator using $a^3-b^3=(a-b)\cdot (?)$
to find

$$F=\lim_{y\to0^+}\dfrac{(1+y)^2-(1-y)^2}y\cdot\lim_{y\to0^+}\dfrac1{(1+y)^{4/3}+(1+y)^{2/3}(1-y)^{2/3}+(1+y)^{4/3}}=?$$


答: $x^{-2}>1$ when $|x|<1$ and if you know the graph of $lnx$ then the guessing part should go easy. $lnx$ goes off to $-\infty$ as $x$ approaches $0$. Because of $x^{-2}$ in that region, it will shoot off to $-\infty$ at a quicker rate. Whereas for $x>1$ region, it should be noted that $x^{-2}$ has a growth rate larger than $lnx$ and hence $x^{-2}$ would manage to calm $lnx$ down. This knowledge coupled with the knowledge of maxima should be enough to give a good estimate for all $x>0$.</p>

Title: Sketch the function $y = {1 \over {{x^2}}}\ln x$

Body: I don't know where to begin with this, the ${1 \over {{x^2}}}$ part of the function throws me off, how to do I go about this? How does one generally approach a question like this?</p>


答:
Starting from the hint in the great @lulu comment we have that
$$\left(3-2 \sqrt{2}\right)^n+ \left(2 \sqrt{2}+3\right)^n$$
is the solution of the recurrence having characteristic polynomial
$$\lambda^2-6\lambda+1$$
generated by the recurrence
$$s_n=6s_{n-1}-s_{n-2};\;s_0=2;\;s_1=6$$
If we rewrite the recurrence $\text{mod }5$ we get
$$s'_n=s'_{n-1}-s'_{n-2};\;s'_0=2;\;s'_1=1$$
which gives the values $\text{mod }5$ of the recurrence above.

The solution can be easily found by the characteristic equation
$$x^2-x+1=0\to x_1=\frac{1-i \sqrt{3}}{2};\;x_2=\frac{1+i \sqrt{3}}{2}$$
so we have
$$s'_n=a \left(\frac{1-i \sqrt{3}}{2}\right)^n + b\left(\frac{1+i \sqrt{3}}{2}\right)^n$$
Plugging the initial values $s'_0=2;\;s'_1=1$ we get
$$\begin{cases}
a+b=2\\
\left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right) a+\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right) b=1\\
\end{cases}
$$
Solution is $a=1;\;b=1$ therefore the solution of the second recurrence is
$$s'_n= \left(\frac{1-i \sqrt{3}}{2}\right)^n + \left(\frac{1+i \sqrt{3}}{2}\right)^n$$
These are the numbers of the set $a^n+b^n$ modulus $5$ and can be easily seen that they are never $0$.

Finally we can say that $a^n+b^n$ is never divisible by $5$.

Title: Prove for any natural number $n$, ${a^n}$ + ${b^n}$ is a natural number not divisible by $5$.

Body:
Let $a = 3 + 2\sqrt{2}$ and $b = 3 - 2\sqrt{2}$. Prove that for every natural number $n$, ${a^n}$ + ${b^n}$ is a natural number, not divisible by $5$.

It's trivial that this number is natural by using binomial formula, but I can't figure out how to prove that it's not divisible by $5$.


问:
Weierstrass nowhere differentiable function is Lip$_\alpha$ for $\alpha < 1$
Let $f(x) = \sum_{n=1}^{\infty}{(\frac{3}{4})^{n} \phi(4^nx) }$, where $\phi(x)$ is defined as follows.

$x-n$ if $n \leq x <n+1$ and $n$ is even.

$n + 1 - x$ if $ n \leq x < n+1 $ and n is odd.

Then it can be proved that $f$ is continuous(actually, uniformly continuous) but nowhere differentiable. How can I prove that $f$ is in Lip$_\alpha$ for any $\alpha < 1$? (It means there is a constant $C > 0$ such that for any $s, t$, it holds that $|f(s) - f(t)| \leq C|s - t|^\alpha$. I was able to prove that $f$ is not Lip$_1$.)

答:
Actually it's not true  that $f\in Lip_\alpha$ for every $\alpha<1$. It's true for some $\alpha$:

$\exists\alpha f\in Lip_\alpha$:

First note two things about $\phi$, namely that

(i) $|\phi(x)|\le1$

and

(ii) $|\phi(x)-\phi(y)|\le|x-y|$.

Write $$f(x)-f(y)=\sum\left(\frac34\right)^n(\phi(4^n x)-\phi(4^ny)).$$
Choose $N$ so $$\frac14\le4^N|x-y|\le1,$$and write $\sum=\sum_1^N+\sum_{N+1}^\infty=I+II$ above. Choose $\alpha$ so $4^{(-\alpha+1)}=3$, and use (ii) to estimate I: $$|I|\le\sum_1^N\left(\frac34\right)^n4^n|x-y|\sim 3^N|x-y|\sim4^{(-\alpha+1)N}4^{-N}=4^{-N\alpha}\sim|x-y|^\alpha.$$Similarly (i) shows that $|II|\le c|x-y|^\alpha$.

$f\notin\bigcap_{\beta<1}Lip_\beta$:

Since $\phi\ge0$ and $\phi(x)=x$ for $0<x<1$ we have $$f(4^{-N})-f(0)
=f(4^{-N})\ge\sum_{n=1}^N\left(\frac34\right)^n4^{n-N}=c\left(\frac34\right)^N=c(4^{-N})^\alpha,$$where $\alpha$ is as above. Hence $f\notin Lip_\beta$ for $\beta>\alpha$.

Exercise. Let $g(x)=\sum4^{-n}\phi(4^nx)$. Then $g\in Lip_\alpha$ for every $\alpha\in(0,1)$.


问:
Prove it by using that $f^{-1}$ of an open set is open
If $f$ is continuous from a metric space $X$ to a metric space $Y$ prove that for any $A\subseteq X$, $f(\overline{A})\subseteq\overline{f(A)}$.

My attempt: I have already proved it in two different ways, one using convergent sequence and another using the fact that if $f$ is continuous then the inverse image of a closed set is closed. But I am not satisfied with these two. I am trying to prove it by using inverse image of open set is open. Can somebody give me a hint?

答: Assume $f$ is continuous. We show that $x\in \overline{A}$, then $f(x)\in\overline{f(A)}$.Let $A$ be a subset of $X$ Let $N$ be a neighborhood of $f(x)$. Then $f^{-1}(N)$ is an open set of $X$ containing $x$; it must intersect in some point $y$. Then $N$ intersects $f(A)$ in the point $f(y)$, so that $f(x)\in \overline {f(A)}$.


问:
Is $\sin(x^2)$ uniformly continuous for all $x \in (0, +∞)$? Where is my error?
The answer is that is not uniformly continuous. But I get the contrary. Here is my solution.
Let $f(x) = \sin(x^2)$, if $x ≠ 0$ and 0 if $x = 0$. In this case $f \in \Bbb{R}_{≥0}$ and $f(x)$ is continuous on $\Bbb{R}_{≥0}$.
According to Heine–Cantor's theorem, if a function is continuous on $[a, b]$ then it is uniformly continuous. Let $a = 0$, and $b_n \to +\infty$ is the sequence of all numbers that belong to $\Bbb{R}_{≥0}$. Then $f(x)$ is uniformly continuous on $[0, b_n]$. And so it is uniformly continuous on $\Bbb{R}_{≥0}$. Now, I think that we can claim that $\sin(x^2)$ is uniformly continuous for any real $x$.
Where is my error?

答: $f$ is uniformly continuous on $[0,b_n]$ for each $n$. But this does not imply that $f$ is uniformly continuous on $\Bbb{R}_{≥0}.$


问:
Example of a basis which is not a Riesz basis?
I'm looking for an example of a countable "basis" $B=(\phi_i)_{i\in I}$ in a real Hilbert space $\mathcal{X}$ which is not a Riesz basis. So, we only require that the closure of the span of $B$ is $\mathcal{X}$ and that, for every $N\in\mathbb{N}$ and every cardinality-$N$ subset $\{i_1,\ldots,i_N\}$ of $I$,
\begin{equation}
(\forall (\alpha_{i_1},\ldots,\alpha_{i_N})\in\mathbb{R}^N)\quad
\alpha_{i_1}\phi_{i_1} + \alpha_{i_2}\phi_{i_2}+\cdots+\alpha_{i_N}\phi_{i_N}=0\Rightarrow (\alpha_{i_j})_{1\leq j\leq N}\equiv 0.
\end{equation}
(note that, by "span" I mean finite linear combinations.)

A few observations / questions

This must occur in infinite-dimensional $\mathcal{X}$.

I think this would mean that the basis decomposition operator $L\colon\mathcal{X}\to\ell_2\colon x\mapsto(\langle x\,|\,\phi_i\rangle)_{i\in I}$ is unbounded?

答:
Start with an orthonormal basis $ \{e_n\}_n$ and put
$\phi_n=e_n/n$, or $\phi_n=ne_n$.  A less trivial example would be
$$
\phi_n =e_n/n + \sum_{i=1}^{n-1}e_i.
$$

EDIT. Here is a concrete criteria for identifying, or debunking,  Riesz bases.

Recall that a sequence $\{\phi_n\}_n$  in a Hilbert space $H$ is called a Riesz basis if it spans a dense subspace
of $H$, and there are positive  constants
$c$ and $C$  such that
$$
  c\left(\sum_n|x_n|^{2}\right)\leq
  \left\Vert \sum_nx_n\phi_n\right\Vert ^{2}\leq
  C \left(\sum _n|x_n|^{2}\right),
  \tag 1
  $$
for every finitely supported sequence $\{x_n\}_n$ of scalars.

This is obviously equivalent to the fact that the correspondence
$$
  T : (x_n)_n \in  \ell^2 \mapsto \sum_nx_n\phi_n\in  H
  $$
defines a (not necessarily isometric) isomorphism from  $\ell^2$  onto $H$.

Theorem.  Let $\{\phi_n\}_n$ be a sequence  in  $H$ spanning  a dense subspace.  Then $\{\phi_n\}_n$ is a Riesz
basis iff the matrix
$$
  A=\{\langle \phi_j,\phi_i\rangle \}_{i, j}
  $$
represents an invertible operator on $\ell^2$.

Proof.  Assuming that $\{\phi_n\}_n$ is a Riesz basis, let $T$ be the operator defined above.
Denoting by
$\{e_k\}_k$ the standard orthonormal basis of $\ell^2$, notice that the matrix $A=\{a_{i, j}\}_{i, j}$ representing the
operator $T^*T$ is given by
$$
  a_{i,j}=  \langle T^*T(e_j),e_i\rangle  = \langle T(e_j),T(e_i)\rangle  = \langle \phi_j,\phi_i\rangle .
  $$
Since $T$ is invertible, if follows that $T^*T$ is also invertible, so this concludes the proof of the "only if" part.

Conversely, suppose that $A=\{\langle \phi_j,\phi_i\rangle \}_{i, j}$
represents an invertible operator on $\ell^2$.  Then, for every finitely supported sequence
$x=\{x_n\}_n$  of scalars we have that
$$
  \langle Ax,x\rangle  =
  \sum_{i,j} \langle \phi_j,\phi_i\rangle x_j\overline{x_i} =
  \sum_{i,j} \langle x_j\phi_j,x_i\phi_i\rangle  =
  \left\|\sum_ix_i\phi_i\right\|^2.
  $$
This shows that $A$ is a positive operator and then $B:=A^{1/2}$ is an invertible  self-adjoint operator  satisfying
$$
  \|Bx\|^2 =
  \langle Bx,Bx\rangle  =
  \langle B^2x,x\rangle  =
  \langle Ax,x\rangle  =
  \left\|\sum_ix_i\phi_i\right\|^2.
  $$
Combining this with the fact that
$$
  \|B^{-1}\|^{-1}\|x\| \leq      \|Bx\|\leq     \|B\|\|x\|,
  $$
we deduce that
$\{\phi_n\}_n$ satisfies (1) and hence is  a Riesz basis.  QED

This said, it is very easy to build examples of linearly independent sets spanning a dense subspace which are not  Riesz
bases.  A typical obstruction for this would   be when the matrix  $A$ above has rows or columns which are not square summable.


问:
Prove that if $x, y ∈$ B̅$(z,r)$ such that $||x - y|| ≥ εr$, then $||z - {{x+y} \over 2}|| ≤ r \sqrt{1 - {{ε^2} \over 4}}$.
Let $z ∈ ℝ^n$, $r > 0$, and $ε ∈ (0, 2]$. Prove that if $x, y ∈$ B̅$(z,r)$ such that $||x - y|| ≥ εr$, then $||z - {{x+y} \over 2}|| ≤ r \sqrt{1 - {{ε^2} \over 4}}$.

my take on this exercise (not much, but I tried):

Knowing that B̅$(z,r)$ is a closed ball, we can say:

$||x-z|| ≤ r$ and $||y-z|| ≤ r$

From the triangle inequality we also know:

$| ||x|| - ||y|| ≤ ||x-y||$

How to continue to get the conclusion?

答:
Let $\alpha=z-x$, $\beta=z-y$ and $\gamma=y-x=\alpha-\beta$, where
$$|\alpha|^2,|\beta|^2\leq r^2,\qquad |\gamma|^2\geq \varepsilon^2 r^2.$$
We only need to prove
$$(\alpha+\beta,\alpha+\beta)\leq r^2(4-\varepsilon^2).$$
Note that
$$\begin{aligned}
(\alpha+\beta,\alpha+\beta)&=|\alpha|^2+|\beta|^2+2(\alpha,\beta),\\
\varepsilon^2 r^2\leq |\gamma|^2=(\alpha-\beta,\alpha-\beta)&=|\alpha|^2+|\beta|^2-2(\alpha,\beta).
\end{aligned}$$
It follows that
$$\begin{aligned}
|\alpha+\beta|^2&\leq 2(|\alpha|^2+|\beta|^2)-\varepsilon^2 r^2\\
&\leq (4-\varepsilon^2)r^2.
\end{aligned}$$