问:
Derivation of a log of an integral
I've come across the following derivation

$$
\frac{\partial}{\partial \alpha}  \log_2 \left(\int p(x)^\alpha \, \mathrm{d}x \right)
$$

but cannot figure out, what tricks are required to solve it nicely. Definitely some chain rule but the log of the integral looks scary.

($p(x)$ is probability density; an alternative version with probability mass function, with sum instead of integral, was easy)

答:
$$\frac{\partial \log _2\left(\int p(x)^{\alpha } \, dx\right)}{\partial \alpha }=\frac{\partial \frac{\log \left(\int p(x)^{\alpha } \, dx\right)}{\log 2}}{\partial \alpha }=\frac{1}{\log 2}\frac{\partial \log \left(\int p(x)^{\alpha } \, dx\right)}{\partial \alpha }=$$
$$=\frac{1}{\log 2}\frac{\int p(x)^{\alpha } \log (p(x)) \, dx}{\int p(x)^{\alpha } \, dx}$$

Example

$$\log _2\left(\int x^{\alpha } \, dx\right)=\frac{\log \left(\frac{x^{\alpha +1}}{\alpha +1}\right)}{\log 2}+C$$
$$\frac{\partial}{\partial\alpha}\log _2\left(\int x^{\alpha } \, dx\right)=\frac{\partial}{\partial\alpha}\frac{\log \left(\frac{x^{\alpha +1}}{\alpha +1}\right)}{\log 2}=\frac{1}{\log 2}\frac{\partial}{\partial\alpha}\left((\alpha+1)\log x-\log(\alpha+1)\right)=\frac{1}{\log 2}\left(\log x-\frac{1}{\alpha+1}\right)$$

Using the formula above

$$\frac{\partial}{\partial\alpha}\log _2\left(\int x^{\alpha } \, dx\right)=\frac{1}{\log 2}\frac{\int \left(x^{\alpha}\log x\right)\,dx}{\int x^{\alpha}\,dx}=$$
$$=\frac{1}{\log 2}\frac{\frac{x^{\alpha +1} ((\alpha +1) \log x-1)}{(\alpha +1)^2}}{\frac{x^{\alpha +1}}{\alpha +1}}=\frac{1}{\log 2}\frac{x^{\alpha +1} ((\alpha +1) \log x-1)}{(\alpha +1)^2}\cdot \frac{\alpha +1}{x^{\alpha +1}}=\frac{1}{\log 2}\frac{(\alpha +1) \log (x)-1}{\alpha +1}$$

Title: Derivation of a log of an integral


问:
Which topics of mathematics should I study?
I'm a first year econometrics student with a great interest in mathematics. I very much enjoy my study, but still I am interested to learn about more topics in mathematics which are not part of my study. Some of the main topics which I am already familar with or which will soon be covered in my study are: calculus, linear algebra, optimization, statistics/probability/combinatorics.

Which topics in mathematics would you advise me to study still (that are not part of this list)? I have a general interest in mathematics, so any advice on interesting and or essential topics in mathematics that are worth studying is appreciated.

If possible, could you also give me advice on books/references which I should study from?

Thanks in advance.

Edit: To be more specific, I am looking for topics on which the general consensus is that they are essential to know of for any mathematician or very interesting to study.

答:
This is a link to the Mathematics Programs offered at the University of Toronto (St. George):

http://www.artsandscience.utoronto.ca/ofr/archived/1213calendar/crs_mat.htm

If you scroll down you'll find the course requirements for "Mathematical Applications in Economics and Finance Specialist Program" which includes subjects like Real and Complex Analysis and PDE's which aren't on your list. However, if you'd like to follow the Mathematics Specialist program I could tell you which texts they use/have used for quite a few of them. A course number with a Y indicates a full year course (72 hrs of lecture) and a course number with H indicates a half year course (36 hrs of lecture):

First Year

MAT157Y1 - Analysis I
Text: Calculus by Spivak.
Used in the past: Principles of Mathematical Analysis by Rudin.

If you have never been exposed to abstract mathematics Spivak is probably better to go with. UofT has been teaching from Spivak's for awhile now.

MAT240H1 & Mat247H1: Linear Algebra I & II
Text: Linear Algebra by Friedberg et al.
Used in the past: Linear Algebra Done Right by Axler.

Second Year

MAT257Y1 - Analysis II

Text - Analysis on Manifolds by Munkres
Used in th past: Calculus on Manifolds by Spivak

Go with Munkres on this one. Spivak is barely a little over 100 pages in length! So you can imagine how terse it is.

MAT267H1 - Advanced Ordinary Differential Equations
Text - Differential Equations, Dynamical Systems, & Introduction to Chaos by Hirsch et al. & Elementary Differential Equations by Boyce and DiPrima

Third Year

MAT347Y1 - Groups, Rings, & Fields
Text: Abstract Algebra by Dummit and Foote

MAT354H1 - Complex Analysis I
Text: Complex Analysis by Stein & Shakarchi.
Used in the past: Real and Complex Analysis by Rudin

MAT315H1 - Introduction to Number Theory
Text: An Introduction to the Theory of Numbers by Niven.
Used in the past: A Friendly Introduction to Number Theory by Silverman.

MAT344H1 - Introduction to Combinatorics
Text: Applied Combinatorics by Tucker

MAT327H1 - Introduction to Topology
Text: Topology by Munkres.

MAT357H1 - Real Analysis I
Text: Real Mathematical Analysis by Pugh.
Used in the past: Real and Complex Analysis by Rudin.

MAT363H1 - Introduction to Differential Geometry
Text: Elementary Differential Geometry by Pressley.

Fourth Year

A lot of these courses are cross listed so they're actually graduate courses. Check here for texts and references:

http://www.math.toronto.edu/cms/tentative-2012-2013-graduate-courses-descriptions/

Hope this helps!

ContentLicense: CC BY-SA 3.0


问:
The meaning of geodesic curvature, for a geodesic curve
If $\alpha : [0,1] \to M$  is a geodesic curve in a surface $M\subseteq \mathbb R^3$, then is the "geodesic curvature of $\alpha$" equal to the "curvature of $\alpha$" (defined using the Frenet apparatus)?

Or is "geodesic curvature" an unrelated thing from "curvature of a curve" (especially for geodesic) but only their names are chosen to be very similar?

Thank you in advance.

答:
The total (Frenet), normal and geodesic curvatures satisfy $\kappa^2=\kappa_n^2+\kappa_g^2$. For a geodesic the geodesic curvature vanishes $\kappa_g=0$.

So, they are not completely unrelated, but they are indeed different things. You can find this in many books about differential geometry of curves and surfaces.

ContentLicense: CC BY-SA 4.0


问:
Notation for indicating closure.
I'm trying to record in my notes a general definition for what has been introduced to me as the "generating class argument." A form of such argument is commonly seen in elementary measure theory, where one might argue that since $A \subset B$, it must be that $\sigma(A) \subset \sigma(B)$, where $\sigma()$ here indicates something like a closure operator which sends generators to sigma algebras/fields. I have not studied Category Theory yet, but would like to extend the notion a bit, so that I could say something like $$ A \subset B \Rightarrow \operatorname{cl}_{Y}(A) \subset \operatorname{cl}_{Y}(B)$$
where $Y$ is some notion that is idempotent,extensive,monotone (but doesn't necessarily send $\mathcal{P}(\mathcal{X}) \to \mathcal{P}(\mathcal{X})$). E.g. topological closedness. Is this intelligible/ is there such a notion of closure under $Y$? Apologies if the question is flawed, I'm a bit out of my depth here.

答:
I'm not aware of any specific notation, we just say that $C:P(X)\to P(X)$ is a closure operator if your conditions hold.

More generally you can define closure operators over arbitrary posets just the way you did:

$C:P\to P$ is a closure operator if $(1)\ x\le C(x),\ (2)\ x\le y\implies C(x)\le C(y)$ and $(3)\ C(C(x))=C(x)$.

Remarkably, this is a special case of monads where arbitrary categories replace posets.

A monad on a category $\mathcal A$ is a functor $C:\mathcal A\to\mathcal A$ (corresponding to $(2)$) together with natural transformations $\eta:{\rm id}_{\mathcal A}\to C$ (implying the existence of arrows $a\to C(a)$ hence corresponding to $(1)$) and $\mu:C^2\to C$ (corresponding to $(3)$ in the presence of $(1)$) that satisfy certain commutativity conditions which all are automatic in a posetal category since every pair of parallel arrows commute there.

Title: Notation for indicating closure.


问:
Let $N$ be Poisson with parameter $m$. Find $\frac{E[ \log (N+\frac{1}{m} )]}{\log(m)}$
Let $N$ a Poisson random variable with parameter $m$.  How to find the following limit:
\begin{align}
\lim_{m \to \infty} \frac{E[\log(N+\frac{1}{m})]}{\log(m)} ?
\end{align}

Using Jensen's inequality we have that
\begin{align}
\lim_{m \to \infty} \frac{E[\log(N+\frac{1}{m})]}{\log(m)} \le\lim_{m \to \infty} \frac{\log(E[ N+\frac{1}{m}])}{\log(m)}= \lim_{m \to \infty} \frac{\log(m+\frac{1}{m})}{\log(m)}=1.
\end{align}

I am wondering if this is the correct behavior.

答:
Indeed it is, note that $$ \ln(N+\frac{1}{m})  = \ln(\frac{N+\frac{1}{m}}{m}) + \ln(m)$$ so $$ \mathbb E[\frac{\ln(N+\frac{1}{m})}{\ln(m)}] =  = \frac{1}{\ln(m)}\mathbb E[\ln(\frac{N+\frac{1}{m}}{m})] + 1 $$

We need to prove that $$ \frac{1}{\ln(m)}\mathbb E[\ln(\frac{N+\frac{1}{m}}{m})]$$ converges to $0$. As $N \sim \mathcal Po(m)$ we get (second one we bounded by $ln(x) \le x$ for $x>0$) $$ |\mathbb E[\ln(\frac{N+\frac{1}{m}}{m})]| \le |e^{-m}\sum_{k=0}^m \frac{m^k}{k!}\ln(\frac{k+\frac{1}{m}}{m})| +  e^{-m} \sum_{k=m}^\infty \frac{m^{k-1}}{k!}(k+\frac{1}{m}) $$
We can split the second series into $2$, namely $$ e^{-m}\sum_{k=m}^\infty \frac{m^{k-1}}{(k-1)!} + e^{-m} \frac{1}{m} \sum_{k=m}^\infty \frac{m^{k-1}}{k!} $$
And it is not hard to see that both series are bounded by $e^m$, hence the whole series is bounded by something convering to $1$ and by multiplying it by $\frac{1}{\ln(m)}$ we have convergence to $0$.

The first $m$ terms are harder. Split the sum into $k \in \{0,...,\frac{m}{2}\}$ and $k \in \{\frac{m}{2} , ... , m\}$. The second one can be bounded by $$ | \sum_{k=\frac{m}{2}}^m e^{-m}\frac{m^k}{k!} \ln(\frac{1}{2} + \frac{1}{m^2})| \le |\ln(\frac{1}{2} + \frac{1}{m^2}|$$ which converges to $0$ if we multiply it by $\frac{1}{\ln(m)}$.

While the second one requires either analitic computation or central limit theorem. Indeed, note $$ \frac{1}{\ln(m)}|\sum_{k=0}^{\frac{m}{2}} e^{-m}\frac{m^k}{k!}\ln(\frac{k}{m} + \frac{1}{m^2})| \le \frac{1}{\ln(m)}|\ln(\frac{1}{m^2}) | \sum_{k=0}^{\frac{m}{2}} e^{-m}\frac{m^k}{k!}  = 2 \mathbb P(N \le \frac{m}{2})$$

It remains to show that $\mathbb P(N \le \frac{m}{2})$ converges to $0$, but as $N$ is sum of $n$ independent $\mathcal Po(1)$ random variables (call them $X_k$) we have due to central limit theorem $$ \mathbb P(N \le \frac{m}{2}) = \mathbb P( \sum_{k=1}^m X_k - m \le -\frac{m}{2}) = \mathbb P( \frac{\sum_{k=1}^m (X_k-\mathbb E[X_k])}{\sqrt{m}} \le -\frac{\sqrt{m}}{2}) \to \mathbb P(\mathcal N(0,1) \le -\infty) = 0$$

W which proves our claim.


问:
Equivalent definitions of subharmonicity (Mean Value Inequality vs Viscosity vs Classic)
Consider the following problem:

Let $\Omega \subset \mathbb R^N$ be a domain and $u \in C(\Omega)$. Show that $u$ is subharmonic in the Mean Value Inequality sense, that is,
$$
u(x) \leq \frac{1}{\omega_N r^N} \int_{B_r(x)} u (y) \ dy, \quad \forall x \in \Omega, \forall 0 < r < d(x, \partial \Omega).
$$
if and only if for every $x_0 \in \Omega$ and every $\phi \in C^2(\Omega)$ such that $u - \phi$ attains a local maximum at $x_0$ it holds that
$$
- \Delta \phi(x_0) \leq 0,
$$
that is, $u$ is a viscosity subsolution of the Laplace equation.

For the $\Rightarrow$ implication, I managed to show that $\phi$ is subharmonic in the sense of the (volumetric) Mean Value Inequality. Since it is $C^2$, it would suffice to show that it implies that $\phi$ is subharmonic in the classic sense. How to do this, if true?

For the $\Leftarrow$ implication I have no ideas, and any hints are welcome.

Thanks in advance.

EDIT

For the $\Rightarrow$ implication, it almost suffices to show that $\phi$ is subharmonic in the Mean Value Property sense. Indeed, suppose that $- \Delta \phi(x_0) > 0$. Then we would have the opposite inequality.

答: There are lots of ways to prove this, depending on what you know about PDEs and viscosity solutions. Here is one hint: Fix your ball $B_r(x)$ and solve Laplace's equation $\Delta v = 0$ on $B_r(x)$ with boundary conditions $v=u$ on $\partial B_r(x)$ (I assume you know how to construct the solution with Poisson's Integral Formula). Then $v$ is $C^2$, and the viscosity subsolution property for $u$ can be used to show that $u\leq v$ on $B_r(x)$ (this does not require the full comparison principle machinery of viscosity solutions, and is easy to prove by hand; I can add a comment on this if needed). Then since $v$ is harmonic and smooth, use the mean value property for $v$.


问:
MGF for chi-square distribution
I am trying to get the mgf for the chi-squared distribution but I keep getting $(1-2t)^{1/2}$ instead of $(1-2t)^{-\frac{1}{2}}$.

My method was:

$E(e^{tZ}) = \int_{-\infty}^{\infty}{e^{tz}\frac{}{\sqrt z\sqrt {2\pi}}{e^{-z/2}}} dz$

Then multiplying in I get: $\int_{-\infty}^{\infty}\frac{e^{\frac{-z(1-2t)}{2}}}{\sqrt z \sqrt {2\pi}} dz$

Now I want to force a $\sqrt {1-2t}$ into the denominator and cancel the integral to $1$. But then I have to multiply by $\frac{\sqrt{1-2t}}{\sqrt{1-2t}}$ and then the denominator of the integral absorbs the denominator of the fraction leaving me with $\sqrt{1-2t}$ as the M.G.F. Where am I going wrong???

答: Hint: use the substitution $u = \frac{1-2t}{2}x$. Then $du= \frac{1-2t}{2}dx$. Also, remember that you're only integrating over $(0, \infty)$, not $(-\infty, \infty)$.

Title: MGF for chi-square distribution


问:
If $L/K$ is a finite field extension and $L$ is perfect, then $K$ is perfect.
I'm aware that this question has been answered here. However, I wanna try and prove this directly:

Let $L/K$ be a finite field extension, suppose $L$ is perfect and $\text{char}(K) = p > 0$. Show that $L/K$ is separable. (Hint: Show that there exists an $n \in \mathbb{N}$ s.t $x^{p^n} \in K_s = \{a \in L\; |\; a\; \text{is separable over}\; K\}$ for every $x \in L$.)

Since $L$ is perfect, every $x \in L$ has a $p$-th root in $L$. I don't see how to prove the hint though. Any help is greatly appreciated!

答:
For $K(a)/K$ a finite extension in characteristic $p$ let $$f(x)=\prod_j (x-a_j)^{p^{e_j}}\in K[x]$$ be its minimal polynomial and $F$ the splitting field. There is an automorphism $\sigma_j \in Aut(F/K)$ sending $a_1\to a_j$ thus $e_1=e_j$ so that $$f(x)=g(x)^{p^{e_1}}, \quad g(x)=\prod_j (x-a_j)=\sum_{n=0}^d c_n x^n$$
$$f(x)=\sum_{n=0}^d c_n^{p^{e_1}} x^{p^{e_1}n}, \qquad \sum_{n=0}^d c_n^{p^{e_1}} x^{n}=\prod_j (x-a_j^{p^{e_1}}) \in K[x]$$
Whence $K(a^{p^{e_1}})/K$ is separable.

Going back to your extension, if $n= [L:K]$ then $K(L^{p^n})/K$ is separable.

But $L=L^{p^n}$ so $K(L^{p^n})=L$ and $L/K$ is separable.


问:
List of Local to Global principles
What are some of the local to global principles in different areas of mathematics?</p>

答:
Analysis

$$\prod_{p}|x|_p = |x|^{-1}$$ gives a way to piece together all local norms to find the value of a global norm.

Title: List of Local to Global principles


问:
possible combinations of 3-digit
How many possible combinations can a 3-digit safe code have?

Because there are 10 digits and we have to choice 3 digits from this,
 then we may get $10^P3$ but A author used the formula $n^r$, why is that. What the problem in my calculations?

答:
The error is exactly the one that I suspected in my comment. $_{10}P_3=\frac{10!}{3!}=10\cdot9\cdot8$. This is the number of ways of choosing a $3$-digit combination in which the three digits are all distinct: there are $10$ ways to choose the first digit, but then only $9$ choices remain for the second digit, and once those have been chosen only $8$ choices remain for the last digit.

The problem, however, does not require the digits to be distinct. Thus, there are $10$ choices for the first digit, and since you’re allowed to repeat digits, there are still $10$ choices available for the second digit, and again $10$ for the third. These choices can be made in $10\cdot10\cdot10=10^3$ different ways, so that’s the number of $3$-digit combinations.

As an independent check, notice that the $3$-digit combinations are just the integers from $0$ through $999$, padded on the left with zeros to bring them up to $3$ digits. There are $999$ integers from $1$ through $999$, and $000$ makes the thousandth, so there are $1000=10^3$ such integers.

Title: possible combinations of 3-digit


问:
Why is $(\mathbb{Z}\oplus\mathbb{Z})/\langle(3,6)\rangle$ not isomorphic to $\mathbb{Z}$?

Use first isomorphism theorem to show that $(\mathbb{Z}\oplus\mathbb{Z})/\langle(1,3)\rangle$ is isomorphic to $\mathbb{Z}$.

Explain why $(\mathbb{Z}\oplus\mathbb{Z})/\langle(3,6)\rangle$ is not isomorphic to $\mathbb{Z}$.

For the first part I think I have to take the mapping $\mathbb{Z}\oplus\mathbb{Z}\rightarrow\mathbb{Z}$ defined by $(a,b)\mapsto (3a - b)$ then kernel will be $\langle (1,3)\rangle$.

For the second part I have no idea how do I show that it is not a isomorphism. What I think is since here $\gcd(3,6)= 3$ but in the first case $1$ and $3$ where coprime (this is somewhere related with this kind of isomorphism). May be i am wrong. Any hint will be enough. Thank you.

答:
Hint: Unlike $\Bbb Z$, the group $\Bbb Z \oplus \Bbb Z/ \langle (3,6) \rangle$ contains a non-identity element of finite order.

In other words, its torsion subgroup is non-trivial.


问:
Find out if a point belongs to the shape (or lies on its border).
There is a figure shown in the image.

$a$ is the radius of the circle (the circle lies at point $(0, 0)$), the equation of the line is defined as $y=kx$.

There is also a set of random points.
For all points, I need to determine whether they lie in the given shape (or on its border, or outside the shape).
Please help me to derive the formula. I will be very grateful.

答: Any point $(\alpha,\beta)$ will be inside the circle iff $$\alpha^2+\beta^2\lt a^2$$ and below the line iff $$ \beta \lt k\alpha$$ so you need both of these conditions to be true for a point to lie inside the region you shaded. Reverse the inequality signs for outside points. For the point to lie on the boundary, you need either $$\alpha^2+\beta^2=a^2 \ \text{and} \ \beta \le k\alpha$$ or $$\beta=k\alpha \ \text{and} \  \alpha^2+\beta^2 \le a^2$$


问:
Is T discrete topology on X or not?
I want to verify my argument  from  question of my assignments in topology.

For each n$\in  \mathbb{N}$ , let $X_n =\mathbb{R}$ and let $T_n$ be discrete topology on $X_n$. Let T be product topology on X=$\prod_{n \in \mathbb{N}} X_n$ . Is T discrete topology on X or not?

I think It will be discrete topology as every element of  $P(\prod_{\mathbb{N}} X_n)$, where P(X) means power set of X are in X.

Am I right?

答:
No, it is not the discrete topology. The set $\{(0, 0, 0, \dots,)\}$ is not open, as infinitely many of the pre-images under the projection maps are not the entire space.

To provide more detail: for $T$ to be the discrete topology, every subset of $X$ must be open. It will therefore suffice to show that a single subset of $X$ is not open. This set will be the singleton $A = \{(0, 0, \dots)\}$.

Let $\pi_n$ be the canonical projection function from $X_n$ to $X$. As $X$ has the product topology, a subset $V$ of $X$ is open iff every point of $V$ has a neighbourhood $U$ such that $\pi_n^{-1}(U) = X_n$ for all but finitely many $n$. However, $\pi_n^{-1}(A) = \{0\}$ for all $n$. Infinitely many preimages fail to be the entire space. The defining characteristic of open sets in $X$ has failed, and so, $A$ is not open. Thus, the topology of $T$ is different from the discrete topology.

Title: Is T discrete topology on X or not?


答:
In essence, this results from the fact that, over a field, $F$, of characteristic $p>0$, and any element, $\alpha$, that is algebraic over $F$, we have that $\alpha$ is separable over $F$ iff $F(\alpha)=F(\alpha^p)$. Put another way (as we trivially have $F(\alpha^p)\leq F(\alpha)$ regardless of whether or not $\alpha$ is separable and $F(\alpha)$ is obviously a finite extension of $F$), we have that an element $\alpha$ that is algebraic over $F$ is separable over $F$ iff $[F(\alpha):F]=[F(\alpha^p):F]$. The contrapositive of this is that an element, $\alpha$, that is algebraic over $F$ is inseparable over $F$ iff $[F(\alpha):F]>[F(\alpha^p):F]$. One more minor fact we need is that if $\alpha$ is separable so is $\alpha^p$ (this follows from, e.g., the fact that the separable closure is a field, or the fact that a simple extension by a separable element is separable).

Now addressing your question directly: as $L/K$ is finite, we may choose some integer $N\geq[L:K]-1$. I claim this is our desired integer. For suppose by way of contradiction that there exists some $y\in L$ such that $y^{p^N}$ is inseparable over $K$. By the minor fact stated above, this means $y^{p^i}$ is inseparable over $K$ $\forall\; 1\leq i\leq N$. So for each $1\leq i\leq N$, we have $[K(\alpha^{p^i}):K]>[K(\alpha^{p^{i+1}}):K]$ giving us the chain of inequalities:

$[K(\alpha):K]>[K(\alpha^p):K]>[K(\alpha^{p^2}):K]>\dots>[K(\alpha^{p^N}):K]>[K(\alpha^{p^{N+1}}):K]$

This is a sequence of $N+2\geq [L:K]+1$ integers. At the same time, as $L\geq K(\alpha)$ and $K(\alpha^{p^{N+1}})\geq K$, we have $[L:K]\geq [K(\alpha):K]$ and $[K(\alpha^{p^{N+1}}):K]\geq [K:K]=1$. So this is in fact a sequence of at least $[L:K]+1$ integers between $1$ and $[L:K]$, a contradiction by the pigeonhole principle. It follows that $x^{p^N}$ is separable over $K$ $\forall\; x\in L$.

There is another argument that proves the above that uses minimal polynomials of inseparable elements instead. That argument allows one to do some optimization and show that in fact any $N$ such that $p^N\geq [L:K]$ will work (and this is the best possible lower bound, I think).

Title: If $L/K$ is a finite field extension and $L$ is perfect, then $K$ is perfect.

Body:
I'm aware that this question has been answered here. However, I wanna try and prove this directly:

Let $L/K$ be a finite field extension, suppose $L$ is perfect and $\text{char}(K) = p > 0$. Show that $L/K$ is separable. (Hint: Show that there exists an $n \in \mathbb{N}$ s.t $x^{p^n} \in K_s = \{a \in L\; |\; a\; \text{is separable over}\; K\}$ for every $x \in L$.)

Since $L$ is perfect, every $x \in L$ has a $p$-th root in $L$. I don't see how to prove the hint though. Any help is greatly appreciated!


问:
Can a function be of class C1 even if its partial derivatives are not continuous?
I know that if all partial derivatives of a function f exist and are continuous then the function is said to be of class C1 (continuously differentiable).

However, I was not able to find whether this is a necessary or a sufficient condition. What I mean is: Could a function be of class C1 even though its partial derivatives are not continuous?

PS. Can a function be continuous/differentiable even if its partial derivatives are not continuous?

Thank you!

答:
I was looking for a specific question when I jumped into this old one, and I think that I can share some thoughts about it.

Here's the definition of $C^1$ function:

Let $A$ be an open set of $\mathbb{R}^n$. The set $C^1(A;\mathbb{R}^m)$, $m\geq 1$, is defined to be the set of all the functions $f:A\rightarrow\mathbb{R}^m$ such that all the partial derivatives $$\frac{\partial f_j}{\partial x_i},\quad i=1,\dots,n\quad j=1,\dots,m$$ exist and are continuous.

Hence you can see that by definition a $C^1$ function cannot have discontinuous partial derivatives. For what regards the second part of the question, a function can be continuous without being even differentiable, let alone being $C^1$. An easy example is the absolute value function, which is continuous but not differentiable at $0$.

The very interesting part is whether there exists or not a function which is differentiable but also has discontinuous partial derivatives. Of course there is a theorem that says that if a function is $C^1$ then it is differentiable, but does the converse hold? The answer is no, and a good counterexample is the function
$$
f(x,y)=
\begin{cases}
(x^2+y^2)\sin\bigg(\frac{1}{\sqrt{x^2+y^2}}\bigg),\,\, \text{if}\,\, (x,y)\ne (0,0)\\
0,\,\,\text{if}\,\, (x,y)=(0,0)
\end{cases}
$$
whose differentiable at the origin with differential equal to the zero function, and whose partial derivatives are such that:
$$
\frac{\partial f}{\partial x}(0,0)=0\qquad\frac{\partial f}{\partial y}(0,0)=0
$$
but they are not continuous at $(0,0)$, as you can verify by computing the generic partial derivatives and the limit for $(x,y)\rightarrow (0,0)$.


问:
Limit: $\lim_{x \to 0}{[1+\ln{(1+x)}+\ln(1+2x)+\ldots+\ln(1+nx)]}^{1/x}.$
How can I find the following limit?

$$\lim_{x \to 0}{[1+\ln{(1+x)}+\ln(1+2x)+\ldots+\ln(1+nx)]}^{1/x}.$$

It's a limit of type $\displaystyle 1^{\infty}$ and if I note with $\displaystyle f(x)=\ln{(1+x)}+\ln(1+2x)+\ldots+\ln(1+nx)$ then I must do:

$$\displaystyle \lim_{x \to 0}{(1+f(x))}^{1/x}=\lim_{x \to 0}[(1+f(x))^{\frac{1}{f(x)}}]^{\frac{f(x)}{x}}. $$

Is ok? Thanks :)

答:
You can also argue $$\lim_{x\to 0}\left[1+\sum_{k=1}^n\log(1+kx)\right]^{\frac{1}{x}}=e^{\lim_{x\to0}\frac{1}{x}\log\left[1+\sum_{k=1}^n\log(1+kx))\right]}$$
and then use L'Hopital's Rule to compute the remaining limit. (To spell this out rigorously, you're using the fact that you can pass limits through continuous functions. In other words, $$\lim_{x\to a}f(g(x))=f(\lim_{x\to a} g(x))$$ when $f$ is continuous at $\lim_{x\to a}g(x)$.) In fact, the most common trick for dealing with indeterminate forms of the form $1^\infty$ is to rewrite $\lim f(x)$ as $\lim e^{\log f(x)}$ and then use the fact I just mentioned.</p>


问:
How to intuitively understand parallel transport
In the article I've referenced below, and many other articles for that matter, the notion of parallel transport along a line of latitude $\theta=\theta_0$ on the unit 2-sphere is spoken about. What I can't understand is, intuitively, how there is any "rotation" after circuiting the full line-of-latitude path.

The above illustration is my understanding of what is meant by "parallel transport of a vector along the $\theta$ or $\phi$ direction". Clearly, after completing a full path, the rotation of the vectors is always exactly $2\pi$, and hence the parallely transported vectors are left unchanged. What have I misunderstood? My understanding doesn't seem to match with the results derived in the article below.

Reference: http://www.physics.usu.edu/Wheeler/GenRel2013/Notes/Geodesics.pdf

答: Hilbert and Cohn-Vossen in their “Geometry and Imagination” give the example of driving a small car on the globe. You never have to turn the front wheel from the straight position only if you’re driving on a great circle of the globe. The blue vectors need to turn to compensate for the wheel turning required on the latitudinal drive.</p>


问:
Does norm equivalence preserve the reflexivity of Banach Spaces?
I am working on a problem right where I have a sequence $(x_n)$ which converges weakly to some $x$ in a Hilbert space $(H, \left<\cdot, \cdot\right>)$. Denote by $||\cdot||$ the norm induced by the inner-product. Since an induced norm is (sequentially) weakly lower-semicontinuous, we have that $||x|| \le \liminf_{n \to \infty} ||x_n||$.

Now consider the Banach space $(H, ||\cdot||_0)$ and suppose that $||\cdot||$ and $||\cdot||_0$ are bi-Lipchitz equivalent. That is, $\exists \alpha, \beta \in \mathbb{R}$ such that $\alpha ||v|| \le ||v||_0 \le \beta ||v||$ for all $v \in H.$ Can I similarly conclude that $||x||_0 \le \liminf_{n \to \infty} ||x_n||_0$?

It's my understanding that, since the norms are equivalent, $(H, ||\cdot||)$ and $(H, ||\cdot||_0)$ must have the same continuous dual space, and hence the same weak topology. Therefore, $x_n \rightharpoonup x$ with respect to the weak topology of $(H, ||\cdot||_0)$. I would also imagine that the reflexivity of $(H, \left<\cdot, \cdot\right>)$ is carried over to $(H, ||\cdot||_0)$ for similar reasons, after which I should have that $||\cdot||_0$ is (sequentially) weakly lower-semicontinuous as well. Does this hold?

答:
Hm, perhaps, you have thought too much?

For all normed vector space $E$, if $x_n$ converges weakly to $x$ ($x,x_1,x_2,... \in E$ ), we have $|x| \le \liminf |x_n|$

Now, if you have already the weak convergence of $x_n$ to $x$ in the new norm, then clearly, you have that norm inequality.


问:
Prove that there is exactly one identity relation $I_A$ for each set $A$
Originally, to solve this question. I found the 2 relations and checked if they were the subset of each other to prove that they were the same thing. But the question had a hint which stated,

"As a hint, you can write a rigorous proof of this result without referencing what it means for one relation to be a subset of another. Look at the definition of an identity relation, which gives a series of equalities involving an identity relation. Can you make use of those equalities here?"

I came up with a proof but that used equalities but i'm not really sure if i've done it right. I was wondering if someone can check the proof and critique it. Also if you have another way of proving, that would be very helpful too.

My proof:

Let's assume $A$ is an arbitrary set. We will prove that $A$ has exactly one identity relation.

Let $I_1, I_2$ be identity relations where for every $x$ and every $y$, $x=y$ and $(x,y) \in$ $A$x$A$.

To show $I_1 = I_2$, we can say that if every $x$ and $y$ in $A$x$A$ such that $x = y$ then $(x,y) \in I_1 = (x,y) \in I_2$.

Therefore we can conclude that there is only one identity relation for each set A. $\blacksquare$

答:
This is my understanding of the hint you're interested in.

Look carefully at the definition of the identity relation, it says:

Given a set $A$, a binary relation $I_A$ is called an identity relation over $A$ if the following statement is true for every binary relation $R$ over $A$:
$I_A \circ R = R \land R \circ I_A = R$.

So, pay close attention to the phrase for every binary relation $R$. So like we can use any binary relation instead of $R$ and still get that $I_A \circ R = R \land R \circ I_A = R$ (of course if $I_A$ is identity relation over $A$.

So assuming that $I_{A1}$ and $I_{A2}$ are identity relations over $A$, we sure know that they are also binary relations over $A$. So, consider $I_{A1} \circ R = R \land R \circ I_{A1} = R$ and let $R$ = $I_{A2}$, then
$$I_{A1} \circ I_{A2} = I_{A2} \land I_{A2} \circ I_{A1} = I_{A2}$$
Next, consider $I_{A2} \circ R = R \land R \circ I_{A2} = R$ and let $R$ = $I_{A1}$, then
$$I_{A2} \circ I_{A1} = I_{A1} \land I_{A1} \circ I_{A2} = I_{A1}$$

Note, that $I_{A1} \circ I_{A2} = I_{A1}$ and $I_{A1} \circ I_{A2} = I_{A2}$. Thus, $I_{A1} = I_{A2}$.

You can argue why I assume that $R = I_{A1}$ and $R = I_{A2}$. In fact, I just picked any of all the binary relations over $A$. In the first case I picked $I_{A2}$ and in the second case I picked $I_{A1}$.


问:
evaluate $\lim_{x,y \to 0,0} \frac{\sin(x+y)-(x+y)}{\sqrt{x^2 +y^2}}$
I am trying to evalute the following limit:

$\lim_{x,y \to 0,0} \frac{\sin(x+y)-(x+y)}{\sqrt{x^2 +y^2}}$

However , i have a problem. The answerkey of my textbook says that its limit is zero ,but wolfram-alpha says that limit does not exists.

Which answer is correct ? Can you help me to solve this limit.

答:
HINT

Show that $\dfrac{\sin(x+y)-(x+y)}{x+y}\to0$.

Show that $0\le \left|\dfrac{x+y}{\sqrt{x^2+y^2}}\right|\le2$


问:
Angle of triangle given three altitude vectors adding to $\mathbf{0}$

Let $\overline{AD},$ $\overline{BE},$ $\overline{CF}$ be the altitudes of acute triangle $ABC.$ If
$9 \overrightarrow{AD} + 4 \overrightarrow{BE} + 7 \overrightarrow{CF} = \mathbf{0},$ compute $\angle ACB,$ in degrees.

I know that, in general, $\overrightarrow{XY} = \overrightarrow{Y}-\overrightarrow{X}$, so then I changed the first equation to $$9\overrightarrow{D}+4\overrightarrow{E}+7\overrightarrow{F}-9\overrightarrow{A}-4\overrightarrow{B}-7\overrightarrow{C} = \mathbf{0}.$$

Additionally, $\overline{AD}$ and $\overline{BC}$ are orthogonal, and same with the other two altitudes. So that means $$\overrightarrow{AD}\cdot \overrightarrow{BC} = \mathbf{0}$$$$(\overrightarrow{D}-\overrightarrow{A})(\overrightarrow{C}-\overrightarrow{B})=\mathbf{0}$$$$\overrightarrow{DC}-\overrightarrow{DB}-\overrightarrow{AC}+\overrightarrow{AB}=\mathbf{0}.$$

However, I can't seem to relate this equation to the original one. I also know that $\cos{\angle{ACB}} = \frac{\overrightarrow{AC}\cdot \overrightarrow{BC}}{||AC||\text{ }||BC||}$, but I'm not sure how to find the magnitudes of $\overrightarrow{AC}$ and $\overrightarrow{BC}$, since there are no concrete side lengths. I could also find the $\sin$ of the angle using cross product, but I don't think that will come up either.

Help is greatly appreciated, thanks!

答:
Let $\triangle ABC$ have sides ${a,b,c}$. Let the corresponding unit altitude vectors ${\hat h_a, \hat h_b, \hat h_c }$. Also let area of $\triangle ABC = \Delta$.

We use the following result :
$$a\hat h_a+ b\hat h_b+ c\hat h_c =0 \tag{1}$$

(To prove this bring these three vectors with tails at a common point, mark the angles and compare the components. One should obtain the familiar sine-rule and projection-rule.)

Given is $$9\cdot \frac{2\Delta}{a}  \, \hat h_a+ 4\cdot \frac{2\Delta}{b}  \, \hat h_b+ 7\cdot \frac{2\Delta}{c}  \, \hat h_c =0 $$

Comparing this with (1), we get $$a:b:c = 3:2:\sqrt 7$$

Now use cosine-rule to obtain $$\angle C=60^{\circ}$$


问:
Derivative of a smooth function as an equivalence class
In my differential manifolds class, the derivative of a function $f \in C^\infty (M)$, where $M$ is a manifold, at $p \in M$ was defined as the image of the linear map
$$(df)_p := C^\infty (M) \mapsto C^\infty (M)/Z_p$$

where $Z_p$ is the set of all smooth functions that have zero derivative at $p$. In other words, the derivative is an equivalence class of $C^\infty (M)$ defined by the relation that two functions are related if their derivative vanishes at $p$, the derivative of a function at a point is an equivalence class.

I'm struggling to wrap my head around this. How can a derivative become an equivalence class? Does this hold in elementary calculus too?

答: Maybe you are over-interpreting it a bit. It just says that if $f=g+h$ with $\frac{\partial h}{\partial {x}^i}=0$ for any chart, you have $f\in [g]$. In other words, $[f]=[g]$ iff $\frac{\partial f}{\partial {x}^i}=\frac{\partial g}{\partial {x}^i}$. It's just a way to express the idea of $df=\frac{\partial f}{\partial {x}^i} dx^i$ without coordinate frames.


问:
Question on integral for area in polar coordinates
My book defines this:

$a(S) = \frac 1 2 \sum_k^n (s_k)^2 (\theta_k - \theta_{k-1}) = \frac 1 2 \int_a^b s^2(\theta) d\theta$

where $a(S)$ is area of a sector, $s_k$ is a partition index $s(\theta) = s_k$.

The following picture describes this.  My question is why is $s^2$ vs $s$?  Visually it seems like the area of a sector is $\frac 1 2 b*h$ of a triangle, where $b,h$ are interchanged with $s,\theta$.

答:
The area of a circle of radius $r$ is $\pi \, r^2$. The area of a sector of a circle with angle $\theta$ is $\left(\frac{\theta}{2\pi}\right) \pi \, r^2$, since the ratio $\theta/2\pi$ is the fraction of the whole circle that the sector takes up.

In each sector, using $s_k$ as the radius, and $\theta_{k+1}-\theta_k$ in place of what I called just $\theta$ above, you get the formula:

$$ \mathrm{area} = \frac{\theta_{k+1}-\theta_k}{2\pi} \cdot \pi s_k^2 = \frac{\theta_{k+1}-\theta_k}{2} \, s_k^2 $$

The $\pi$'s cancel out, which is why they don't appear in the formula.


答:
$$
\begin{gathered}
  \Delta s = \frac{1}{2}rr'\sin \Delta \theta  \hfill \\
  \Delta \theta  \to 0;r' \to r;\sin \Delta \theta  \to d\theta  \hfill \\
  ds = \frac{1}{2}r^2 d\theta  \hfill \\
   S=\int_{\alpha}^{\beta}\frac12 r^2(\theta)\,d\theta\hfill\\
\end{gathered}
$$

Title: Question on integral for area in polar coordinates

Body:
My book defines this:

$a(S) = \frac 1 2 \sum_k^n (s_k)^2 (\theta_k - \theta_{k-1}) = \frac 1 2 \int_a^b s^2(\theta) d\theta$

where $a(S)$ is area of a sector, $s_k$ is a partition index $s(\theta) = s_k$.

The following picture describes this.  My question is why is $s^2$ vs $s$?  Visually it seems like the area of a sector is $\frac 1 2 b*h$ of a triangle, where $b,h$ are interchanged with $s,\theta$.


答:
Using
$$ \sin t=t+\frac{1}{2}\cos(\xi)t^2, |a|\le\sqrt{a^2+b^2}, $$
where $\xi$ is between $0$ and $t$, one has
$$\bigg|\frac{\sin(x+y)-(x+y)}{\sqrt{x^2 +y^2}}\bigg|=\frac12\bigg|\frac{\cos(\xi)(x+y)^2}{\sqrt{x^2 +y^2}}\bigg|\le\frac12\frac{|x|+|y|}{\sqrt{x^2+y^2}}(|x|+|y|)\le|x|+|y|$$
from which one has
$$\lim_{x,y \to 0,0} \frac{\sin(x+y)-(x+y)}{\sqrt{x^2 +y^2}}=0$$

Title: evaluate $\lim_{x,y \to 0,0} \frac{\sin(x+y)-(x+y)}{\sqrt{x^2 +y^2}}$

Body:
I am trying to evalute the following limit:

$\lim_{x,y \to 0,0} \frac{\sin(x+y)-(x+y)}{\sqrt{x^2 +y^2}}$

However , i have a problem. The answerkey of my textbook says that its limit is zero ,but wolfram-alpha says that limit does not exists.

Which answer is correct ? Can you help me to solve this limit.


问:
Parameterfree heir-coheir relation in stable theories
Let $M\models T$, a stable theory. Let $a,b\in U\succ M$.
Are the following equivalent?

$\varphi(M,b)\neq\varnothing$ for every $\varphi(x,y)\in \color{red}{L(M)}$ such that $\varphi(a,b)$

$\varphi(M,b)\neq\varnothing$ for every $\varphi(x,y)\in \color{red}{L}$ such that $\varphi(a,b)$

答:
No, parameters from $M$ can be important for witnessing dependence over $M$. Here's an example.

Let $M$ be an infinite $k$-vector space and $U$ a proper elementary extension. Let $v\in M$ with $v\neq 0$, let $b\in U\setminus M$, and let $a = b+v$. Let's observe two facts:

$a\notin M$. Indeed, if $a\in M$, then $b = a - v\in M$, contradicting the choice of $b$.

$a$ and $b$ are linearly independent. Indeed, since $b\neq 0$, if $a$ and $b$ are linearly dependent, then $a = cb$ for some $c\in k$. Further, $c\neq 1$, since $v\neq 0$ implies $a\neq b$. But then $cb = b+v$, so $v = (1-c)b$, and hence $b = (1-c)^{-1}v$, so $b\in M$, contradicting the choice of $b$.

Let $\varphi(x,y)$ be the $L(M)$ formula $x = y + v$. Then $\varphi(a,b)\in \text{tp}(a/Mb)$, but is not satisfiable in $M$. Indeed, $a$ is its only realization in $U$, and $a\notin M$.

On the other hand, every $L$-formula in $\text{tp}(a/b)$ is satisfiable in $M$. In fact, every non-zero element of $M$ realizes the whole type! Indeed, for any non-zero $a'\in M$, $a'$ and $b$ are linearly independent (since $b\notin M$), so there is an automorphism of $U$ fixing $b$ and moving $a$ to $a'$.


问:
Random walk given two random variables
Each step of a random walk, which starts at $X_0=0$, has a probability$P(X_n=X_{n-1}+1)=P(X_n=X_{n-1}-1)=1/2$. Let Y be another r.v. as the unique positive integer s.t. $X_Y = 0\ \forall \ 0<i<Y$ (it returns to the origin for the first time). Define another random variable $M_k$ as the number of times this random walk visits k before step $Y$ (return to zero). In other words, $M_k=\left| \left\{ i:0\lt i\lt Y, X_i=k \right\} \right|$

Find $P(M_k\ge 1)$ and $P(M_k=n), \forall n>0$

Thank you, any help would be appreciated.

答:
Let $\mathsf{P}_v$ be the law of the random walk $X_n$ starting at $v$ (i.e., $X_0=v$) and let $T_a:=\inf\{n\ge 1:X_n=a\}$. We will use the fact that for $a<0<b$,
$$
\mathsf{P}_0(T_a<T_b)=\frac{b}{b-a}.
$$

It is clear that $\mathsf{P}_0(M_1\ge 1)=1/2$. For $k>1$,
\begin{align}
\mathsf{P}_0(M_k\ge 1)&=\frac{1}{2}\mathsf{P}_0(M_k\ge 1\mid X_1=1)+\frac{1}{2}\mathsf{P}_0(M_k\ge 1\mid X_1=-1) \\
&=\frac{1}{2}\mathsf{P}_1(M_k\ge 1)+0=\frac{1}{2}\mathsf{P}_0(T_{k-1}<T_{-1})=\frac{1}{2k}.
\end{align}

Using the strong Markov property,
\begin{align}
\mathsf{P}_0(M_k\ge 2)&=\mathsf{P}_0(M_k\ge 2,M_k\ge 1)=\mathsf{E}_0[\mathsf{P}_k(M_k\ge 1)1\{M_k\ge 1\}] \\
&=(1-\mathsf{P}_0(M_k\ge 1))\mathsf{P}_0(M_k\ge 1).
\end{align}

Finally, using induction, for $n\ge 1$,
$$
\mathsf{P}_0(M_k\ge n)=(1-\mathsf{P}_0(M_k\ge 1))^{n-1}\mathsf{P}_0(M_k\ge 1).
$$
Therefore,
$$
\mathsf{P}_0(M_k=n)=(1-\mathsf{P}_0(M_k\ge 1))^{n-1}\mathsf{P}_0(M_k\ge 1)^2.
$$

An interesting implication of this result is that
$$
\mathsf{E}_0 M_k=\sum_{n\ge 1}^{\infty}\mathsf{P}_0(M_k\ge n)=1.
$$

Title: Random walk given two random variables


问:
Solve the recursion, $a_n = 3a_{n-1}-3a_{n-2}+a_{n-3}+8$
Bring the following recursion relation to an explicit expression:

$$a_n = 3a_{n-1}-3a_{n-2}+a_{n-3}+8$$
$a_{0} = 0$, $a_1 = 1$, $a_2 = 2$

All the examples I have seen were with maximum 2 steps back ($a_{n-2}$) and I thought I know how to solve those but I'm having a hard time both with the Generating function and the Characteristic polynomial methods.

The Generating function should start from $n = 3$ but what would happen to the defined values?

For the Characteristic polynomial, Does it mean I'll have to find the roots of a polynomial from a 3rd degree?

答:
Rewrite the expression for simplicity as
$$
a_{k+3}=3 a_{k+2} -3 a_{k+1} +a_k +8
$$
Multiply by $z^k$ and sum over $k$ to get the generating function $G(z)=\sum_{k \geq 0}a_k z^k$. On LHS you'll get
$$
\frac{1}{z^3}\bigg(G(z)-a_0 -a_1 z -a_2 z^2\bigg)
$$
On RHS you get
$$
\frac{3}{z}\bigg(G(z)-a_0\bigg)-\frac{3}{z^2}\bigg(G(z)-a_0-a_1 z\bigg)+G(z)+\frac{8}{1-z}
$$
Now do the algebra carefully to get on LHS $G(z)$ and on RHS some expression of the form $\sum_{k=0}^{\infty}\varphi_k z^k$. Can you handle from here?</p>


问:
Uniform convergence of $f_{n}=\frac{e^{n^2 x^2}}{n}$ in $\mathbb{R}$
Let $\displaystyle f_{n}(x)=\frac{e^{n^2 x^2}}{n}$ for $x \in \mathbb{R}$, $n\geq 1$. Show that $f_{n}\rightarrow 0$ uniformly in $\mathbb{R}$ and $f_{n}' \rightarrow 0$ pointwise in $\mathbb{R}$ but the sequence $(f_{n}')$ does not converge uniformly in a interval which contains the origin.

My first approximation to this was with $M_{n}=\displaystyle\sup_{x\in \mathbb{R}}|f_{n}(x)-f(x)|=\sup_{x \in \mathbb{R}}\frac{e^{n^2 x^2}}{n}$ and find critical points but I only find a minimum, other approximation was with the theorem of differentiation and uniform convergence but this fail when I take $x=0$ and I don't know how I can bounded this:
\begin{eqnarray*}
|f_{n}'(x)-f'(x)|=2nxe^{n^2x^2}
\end{eqnarray*}
How can I bound this? or What way a I have to take in this problem?. Thank you. I will greatly appreciate your advice.

答:
It doesn't:

$$\frac{1}{n}e^{n^2}>\frac{1}{n}\left(1+n^2\right)=\frac{1}{n}+n>n\to\infty$$

If on the other hand

$$f_n(x)=\frac{1}{n}e^{-n^2x^2}$$

then the results follows. First, for all $x\in\mathbb{R}$ and $n\geq m\geq \epsilon^{-1}$ we have

$$|f_n-f_m|=\left|\frac{1}{n}e^{-n^2 x^2}-\frac{1}{m}e^{-m^2 x^2}\right|$$

Since $e^{-x}$ and $\frac{1}{x}$ are decreasing functions, this simplifies to

$$=\frac{1}{m}e^{-m^2 x^2}-\frac{1}{n}e^{-n^2 x^2}<\frac{1}{m}e^{-m^2 x^2}\leq\frac{1}{m}e^{0}=\frac{1}{m}\leq \epsilon$$

Thus, $f_n(x)$ converges uniformly to $0$ for all $x$. For the second portion, we have that

$$f_n^{'}(x)=-2nxe^{-n^2 x^2}$$

For any $x\in\mathbb{R}$ ($x\neq 0$) we have

$$|-2nxe^{-n^2 x^2}|=2n|x|e^{-(n|x|)^2}$$

Substituting $y=n|x|$ (note that $y\to\infty $ as $n\to\infty$ since $x\neq 0$) we see that

$$\lim_{y\to\infty}\frac{e^{y^2}}{2y}\geq  \lim_{y\to\infty}\frac{1+y^2}{2y}=\lim_{y\to\infty}\left(\frac{1}{2y}+\frac{y}{2}\right)\geq \lim_{y\to\infty}\frac{y}{2}=\infty$$

Thus

$$\lim_{y\to\infty}2ye^{-y^2}=0$$

For $x=0$ it is obvious that $f_n^{'}(0)=0$. We conclude that $f_n^{'}(x)$ converges pointwise to $0$ for all $x\in\mathbb{R}$. For the final portion, consider the sequence $x_n=\frac{1}{n}$ (if the interval is only less than zero, take $x_n=-\frac{1}{n}$). This is

$$|f_n^{'}(x_n)|=2e^{-1}>\frac{1}{2}$$

Since this is constant, we have found a sequence of $x_n\to 0$ such that $f_n^{'}(x_n)>\frac{1}{2}$ for all $n\in\mathbb{N}$. We conclude $f_n^{'}(x)$ does not converge uniformly around $x=0$.


问:
Sampling 3 out of 7 items; with and without replacement
A set of numbers (1, 2, 3, 4, 5, 6, 7) is given. What is the probability of getting 3 odd numbers when choosing 3 numbers among the set of numbers at random?

I'm not sure how to solve a question like this. I'm also not sure if the sets of 3 numbers must be sets of 3 unique numbers or if a number can be repeated more than once in a single set. Please try to explain the procedures of solving the question for both cases (unique and non-unique).

答:
Because you have not shown any attempt to solve these
problems, there is no way for me to know which of several
approaches you are expected to take. I will show a couple
of approaches along with the numerical answers obtained.

That way, at least you will know the correct answers if
you follow @JMoravits's extensive clues toward a solution
via combinatorial methods. [By whatever method, my goal is that you will be able to solve such problems on your own. Like, say, on an exam.]

Binomial distribution "with replacement". If sampling with replacement (repetitions allowed), then the number $X$ of odd
items chosen from among $\{1,2,3,4,5,6,7\}$ is
distributed according to a binomial distribution with
$n = 3$ trials and success probability $p = 4/7.$
The exact answer from R statistical software, where dbinom is a
binomial PDF, is shown below. Look at the binomial PDF formula in your text.

```<code>dbinom(3,3,4/7)
[1] 0.1865889
(4/7)^3
[1] 0.1865889
</code>```

Hypergeometric distribution "without replacement". If repetitions are not allowed the number of odd items selected
follows as hypergeometric distribution with four favorable (odd)
items and three unfavorable items, where three items are selected
at random without replacement. Again here, you can check the exact answer against the appropriate hypergeometric PDF in your text.

```<code>dhyper(3, 4,3, 3)
[1] 0.1142857
24/210
[1] 0.1142857
choose(4,3)/choose(7,3)
[1] 0.1142857
</code>```

Simulation of many draws of three items from among seven, of which four are odd, can provide useful approximate answers.
Let the population consist of the vector c(1,1,1,1,2,2,2), in which 1 stands for 'Odd' and 2 stands for 'Even'. In R, the sample procedure samples without replacement unless sampling
with replacement is specifically indicated. With 100,000 iterations it is reasonable to expect two place accuracy.

```<code>set.seed(1312)
pop = c(1,1,1,1,2,2,2)
nr.wo = replicate(10^5, sum(sample(pop,3)==1))
mean(nr.wo==3)
[1] 0.11328  # aprx Answ without replacement 0.1142857
</code>```

Now, continuing from above, we specify sampling with replacement, by using parameter rep=T:

```<code>nr.w = replicate(10^5, sum(sample(pop,3,rep=T)==1))
mean(nr.w == 3)
[1] 0.18699 # aprx Answ with replacement 0.1865889
</code>```

Note: After the first simulation, the vector nr.wo contains
100,000 counts between 0 and 3 for the numbers of odd items chosen. Then the logical vector nr.w0 == 3 has 100,000 TRUEs and
FALSEs. The mean of this vector is its proportion of TRUEs. (Similarly, for the second simulation.)

Body:
Because you have not shown any attempt to solve these
problems, there is no way for me to know which of several
approaches you are expected to take. I will show a couple
of approaches along with the numerical answers obtained.

That way, at least you will know the correct answers if
you follow @JMoravits's extensive clues toward a solution
via combinatorial methods. [By whatever method, my goal is that you will be able to solve such problems on your own. Like, say, on an exam.]

Binomial distribution "with replacement". If sampling with replacement (repetitions allowed), then the number $X$ of odd
items chosen from among $\{1,2,3,4,5,6,7\}$ is
distributed according to a binomial distribution with
$n = 3$ trials and success probability $p = 4/7.$
The exact answer from R statistical software, where dbinom is a
binomial PDF, is shown below. Look at the binomial PDF formula in your text.

dbinom(3,3,4/7)
[1] 0.1865889
(4/7)^3
[1] 0.1865889

Hypergeometric distribution "without replacement". If repetitions are not allowed the number of odd items selected
follows as hypergeometric distribution with four favorable (odd)
items and three unfavorable items, where three items are selected
at random without replacement. Again here, you can check the exact answer against the appropriate hypergeometric PDF in your text.

dhyper(3, 4,3, 3)
[1] 0.1142857
24/210
[1] 0.1142857
choose(4,3)/choose(7,3)
[1] 0.1142857

Simulation of many draws of three items from among seven, of which four are odd, can provide useful approximate answers.
Let the population consist of the vector c(1,1,1,1,2,2,2), in which 1 stands for 'Odd' and 2 stands for 'Even'. In R, the sample procedure samples without replacement unless sampling
with replacement is specifically indicated. With 100,000 iterations it is reasonable to expect two place accuracy.

set.seed(1312)
pop = c(1,1,1,1,2,2,2)
nr.wo = replicate(10^5, sum(sample(pop,3)==1))
mean(nr.wo==3)
[1] 0.11328  # aprx Answ without replacement 0.1142857

Now, continuing from above, we specify sampling with replacement, by using parameter rep=T:

nr.w = replicate(10^5, sum(sample(pop,3,rep=T)==1))
mean(nr.w == 3)
[1] 0.18699 # aprx Answ with replacement 0.1865889

Note: After the first simulation, the vector nr.wo contains
100,000 counts between 0 and 3 for the numbers of odd items chosen. Then the logical vector nr.w0 == 3 has 100,000 TRUEs and
FALSEs. The mean of this vector is its proportion of TRUEs. (Similarly, for the second simulation.)


问:
Itô differential coincides with classical differential for classically differentiable process
While studying Itô calculus, I came across the following question. Suppose we have the canonical $d$-dimensional Wiener process $(B_t)_{t \geq 0}$, then for sufficiently smooth $h = h(x,y)$, if we put $Y_t := h(t,B_t)$, the Itô formula
$$Y_t - Y_0 = \int_0^t \partial_xh(t,B_t) + \frac{1}{2} \Delta_y^2h(t,B_t) \, dt + \int_0^t \nabla _yh(t,B_t) \cdot dB_t$$
holds. We can write this in formal "differential" notation:
$$dY_t = \left (\partial_x h(t,B_t) +  \frac{1}{2} \Delta_y^2h(t,B_t) \right) dt + \nabla _yh(t,B_t) \cdot dB_t .$$
Somewhere in my book the process $Z_t := \exp(-\int_0^tf(B_s) \, ds)$ appeared, where $f$ is some smooth function. The author claims that $dZ_t = -fZ_t \, dt$, but meant in the sense of Itô's formula above. This confused me since I was not able to write $Z_t$ as some function of $B_t$ and apply Itô's formula (it is rather a functional of $(B_t)_{t\geq 0}$).

However, $-fZ_t $ is of course the classical derivative of the differentiable function $t\mapsto Z_t$. Is it true that these two notions coincide in this case and if yes where can I find a proof of this fact?

答: Remember that Ito's formula is really only used to handle the fact that $t \rightarrow B_t$ is not differentiable, or even finite variation.  Here, the function we're differentiating is $Z_t := \exp(-F_t)$ where $F_t := \int_0^t f(B_s)ds$ and $t \rightarrow F_t$ is differentiable even though $B_t$ is not.  It's similar to the fact that if $g: \mathbb{R} \rightarrow \mathbb{R}$ is any integrable function then $F(t):=\int_0^t f(g(s))ds$ is differentiable and $F'(t) = f(g(t))$ no matter what $g$ is, even if $g$ is not differentiable or even continuous!  Here we just have that $g(t) = B_t$, so the same reasoning applies.  Then $dZ_t = -f(B_t)Z_t dt$ just follows from the chain rule.


问:
Is common write $\lim f(2)$ instead of $\underset{x \to 2}{\lim} f(x)$
I didn't saw anyone writing like it, but is it understandable? If yes why anyone write $\lim f(2)$?

答:
No. One might say that $f(n)=\lim_{x\to n} f_1(x)$, where $f_1: \mathcal D(f) / \{n\}\mapsto \mathcal R(f)$ s.t. $f_1(x)=f(x)$ in order to extend a function $f_1$ on a domain $\mathcal D(f_1)$ piecewise across a larger domain $\mathcal D(f)$: $$f: \mathcal D(f)\mapsto \mathcal R(f) ,  f(x)=\begin{cases} f_1(x)  & x\in \mathcal D(f_1) & \\  \lim_{t\to x} f_1(t) & x\notin\mathcal D(f_1) \end{cases}$$

However, $\lim f(n)$ has no easily understandable meaning. It could be what you wish to convey, but, as the comments point out, it could also be $\lim_{n\to\infty} f(n)$.

Where notation is ambiguous, it's best to avoid it even at the cost of brevity and stick to more unambiguous notation.


问:
How to prove $\sup(f(x)^2) = (\sup(f(x)))^2$
I started the proof with this:

$$\sup(f(x)^2) = \sup(f(x)*f(x)) = \sup(f(x)) * \sup(f(x) = (\sup(f(x)))^2$$

But it doesn't feel right. Any ideas or comments?

答: Let $f:[-10,1] \to \mathbb{R}$ such that $f(x)=x$. Then $sup(f(x)^2)=100$ and $(sup(f(x)))^2=1$. They are not equal.


问:
Drawing balls with replacement, until I have one of each.
A urn has (n+1) types of balls, n of unique colors and the rest black. When picking a ball randomly from the urn, a colored (non black) ball has a probability of p of being picked. Each ball of color has equal probability of being picked, ie each has a (p/n) chance of being picked. The urn has infinite balls/we are picking with replacement.

Let X be the number of picks until we get at least 1 of each n colored balls.

What is the Expected value of X ? What is the probability distribution of X ?

答: Hint:  you have to pick $\frac 1{p}$ balls to get a colored one, so multiply that by the solution to the Coupon collector's problem</p>


问:
In the definition of a Stopping Time, how important are the conditions on the Filtration being complete and right-continuous?
I'm reading a book and, in its section on the definition of a stopping time(continuous), the author declares at the start that for the whole section every filtration will be complete and right-continuous.

So, in the definition of a Stopping Time, how important are these conditions? Why would they matter?

答:
user6247850 explains nicely the role of right continuity of a filtraton. In that discussion, "completeness" of the filtration plays no role. The main role of completeness is in showing that certain hitting times are in fact stopping times. For example, if the real-valued process $(X_t)_{t\ge 0}$ is adapted to $(\mathcal F_t)$ and has right-continuous paths, and if $A$ is a Borel subset of the real line, then
$$
D_A:=\inf\{t\ge 0: X_t\in A\}
$$
can be shown to satisfy
$$
\{D_A<t\}\in\mathcal F_t,\qquad\forall t>0,
$$
provided $(\Omega,\mathcal F,P)$ is complete and each $\mathcal F_t$ contains all of the elements of $\mathcal F$ of measure $0$. If also $(\mathcal F_t)$ is right continuous, then (referring back to the answer of user6247850) we know that
$$
\{D_A\le t\}\in \mathcal F_t,\qquad\forall t\ge 0.
$$


问:
Show that $x^\frac{1}{n}$ is continuous at all $a \in [0,\infty)$
I have to show $x^\frac{1}{n}$ is continuous using:
$|x-a| = |(x^\frac{1}{n})^n - (a^\frac{1}{n})^n| = |x^\frac{1}{n} - a^\frac{1}{n}||\displaystyle \sum_{k=1}^{n-1} (x^\frac{1}{n})^{n-1-k} + (a^\frac{1}{n})^{k} |$

Here's what I did:

$a=0$:

$|x^\frac{1}{n}| < \epsilon => |x| < \epsilon^n$
Pick $\delta = \epsilon^n$, and we have $f(x)$ is continuous at $x=0$

$a>0$:

$|x-a| = |x^\frac{1}{n} - a^\frac{1}{n}||\displaystyle\sum_{k=1}^{n-1} (x^\frac{1}{n})^{n-1-k} + (a^\frac{1}{n})^{k}| < \epsilon \cdot |\displaystyle\sum_{k=1}^{n-1} (x^\frac{1}{n})^{n-1-k} + (a^\frac{1}{n})^{k}|$

Since this sum evaluates to a real number $R > 0$, we can pick $\delta = R\epsilon $  and we get:

$0<|x-a| < \delta => |x^\frac{1}{n} - a^\frac{1}{n}| < \epsilon$

Is this correct or is there anything else I need to show?
Also, this is my first time formatting with MathJax, so if there are any errors please let me know!

答:
First you need the correct factorization

$$x- a = (x^{\frac{1}{n}} - a^{\frac{1}{n}})\sum_{k=0}^{n-1}x^{\frac{n-1-k}{n}} a^{\frac{k}{n}}$$

Whence,

$$|x^{\frac{1}{n}} - a^{\frac{1}{n}}|= \frac{|x-a|}{\left|\sum_{k=0}^{n-1}x^{\frac{n-1-k}{n}} a^{\frac{k}{n}} \right|} $$

Note that by the reverse triangle inequality $|a|- |x| \leqslant |x-a|$ which implies that $|x| \geqslant |a| - |x-a|$.

If $a > 0$ and $|x-a| \leqslant \frac{a}{2}$, then we have $x = |x| \geqslant |a| - |x-a| \geqslant a - \frac{a}{2} = \frac{a}{2}.$

Hence, $\displaystyle x^{\frac{n-1-k}{n}} a^{\frac{k}{n}} \geqslant 2^{\frac{k}{n}}\left(\frac{a}{2}\right)^{1 - \frac{1}{n}},$ and

$$\left|\sum_{k=0}^{n-1}x^{\frac{n-1-k}{n}} a^{\frac{k}{n}} \right|\geqslant  \left(\frac{a}{2}\right)^{1 - \frac{1}{n}}\sum_{k=0}^{n-1}2 ^{\frac{k}{n}}  = \left(\frac{a}{2}\right)^{1 - \frac{1}{n}}\frac{1}{2^{\frac{1}{n}} -1} := C_n$$

It follows that if  $|x-a| < \delta = \min\left(\epsilon C_n, \frac{a}{2}\right)$, then

$$|x^{\frac{1}{n}} - a^{\frac{1}{n}}|\leqslant \frac{|x-a|}{C_n} < \epsilon,$$

and $x \mapsto x^{\frac{1}{n}}$
is continuous at $a > 0$.


问:
Simple doubt about cumulative distribution function property proof
Let's fix a probability space $(\Omega,P)$ where $|\Omega| \le \aleph_0$, and a random discrete real variable $X\colon \Omega \to \mathbb{R}$.

We define the cumulative distribution function $F_X\colon \mathbb{R} \to [0,1] \mid F_X(x)=P(X \le x)$.

I want to prove that $\lim_{x \to +\infty} F_X(x)=1$.

My textbook proof goes as follows:

$ $

Let's fix a sequence $(x_n)_{n \in \mathbb{N}} \subseteq \mathbb{R} \mid x_n \to +\infty$.

We obviously have $\Omega=\bigcup_{n \in \mathbb{N}} \{X \le x_n\}$, where $\{X \le x_n\} \equiv \{\omega \in \Omega \mid X(\omega) \le x_n\}$.

Then $1=P(\Omega)=P(\bigcup_{n \in \mathbb{N}} \{X \le x_n\})=\lim_{n \to +\infty} P(X \le x_n)=\lim_{n \to +\infty} F_X(x_n)$.

Then $\lim_{x \to +\infty} F_X(x)=1$.

This last line follows from the following theorem:

Let's give a function $f\colon D \subseteq \mathbb{R} \to \mathbb{R}$.

Then $\lim_{x \to +\infty} f(x)=l \iff \forall \,(x_n)_{n \in \mathbb{N}} \subseteq D \mid x_n \to +\infty \quad \lim_{n \to +\infty} f(x_n)=l$.

$ $

All of this is pretty clear to me.

BUT I have a doubt. We have used in this proof the identity:  $P(\bigcup_{n \in \mathbb{N}} \{X \le x_n\})=\lim_{n \to +\infty} P(X \le x_n)$, but this is true only if the sequence of events $E_n=\{X \le x_n\}$ is increasing (namely $E_n \subseteq E_{n+1}$), but this is not necessarily the case, since $(x_n)_{n \in \mathbb{N}}$ is not in general increasing.

Thank you very much for any help!

答:
Let $g$ be a nondecreasing function with $g\leq1$.
Suppose there exists a sequence $(x_{n})_{n}$ such that $x_{n}\rightarrow\infty$ and $\lim_{n\rightarrow\infty}g(x_{n})=1$.
Now, let $\epsilon>0$ and pick $N(\epsilon)$ such that $1-g(x_{n})<\epsilon$ whenever $n\geq N(\epsilon)$.
Since $g$ is nondecreasing, it follows that $1-g(x)<\epsilon$ whenever $x\geq x_{N(\epsilon)}$.
Therefore, by definition, $\lim_{x\rightarrow\infty}g(x)=1.$

Returning to your original question, the above suggests that it is sufficient to prove your claim by considering a nondecreasing sequence $(x_{n})_{n}$ such that $x_{n}\rightarrow\infty$.
Letting $E_n = \{X\leq x_{n}\}$, the containment $E_1 \subset E_2 \subset \cdots$ holds true and you can conclude, by continuity of measure,
$$
1=\mathbb{P}(\Omega)=\mathbb{P}(\cup_{n}E_n)=\lim_{n}\mathbb{P}(E_{n})=\lim_{n}F(x_{n}).
$$


问:
Normality result using Schur form

Let $A\in\mathbb{C}^{n\times n}$ have eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$ Using Schur Form so that
\begin{equation}\sum_{i,j=1}^{n}|a_{ij}|^2=\sum_{i=1}^{n}|\lambda_i|^2\implies A\text{ normal}\end{equation}
Using this result; Prove that 2 unitarily similar matrices $A=(a_{ij})$ and $B=(b_{ij})$ satisfy the condition
\begin{equation}\sum_{i,j=1}^n|a_{ij}|^2=\sum_{i,j=1}^n|b_{ij}|^2\end{equation}
and therefore show the converse of the first statement is true, that is $A$ normal $\implies \sum_{i,j=1}^n|a_{ij}|^2=\sum_{i}^n|\lambda_i|^2.$

I have carried out the following working however I am not sure I am using the correct approach to this problem and have somewhat hit a wall on what to do.

We know by Schur form that we can have $A=UTU^*$ for some unitary $U$ and a triangular $T$ with $t_{ii}=\lambda_i.$ Notice that $trace(A^*A)=\sum_{i,j=1}^n|a_{ij}|^2$, and $A^*A=UT^*TU^*,$ at this point i have $trace(UT^*TU^*)=trace(U^*UT^*T)=trace(T^*T)=\sum_{i,j=1}^n|t_{ij}|^2,$ but i cant see how to get this to be the eigenvalues given it isnt the diagonal entries in the summation, or how to get this implying that $A$ is normal.

I think the above working might be able to be applied to the proof regarding the condition for unitary similar matrices, but I am unsure on how to tackle the rest of the problem.

答:
You are on somewhat of a right track. The point is that for a triangular matrix the eigenvalues are on the diagonal, so that the sum of squares of the eigenvalues is equal to the sum of squares of all entries if and only if there are no other entries, i.e. if the matrix is diagonal. So now we can apply this:

If $\sum_{i} \lambda_i^2=\sum_{i,j} |a_{ij}|^2$  then we get

$$\sum_i t_{ii}^2=\sum_{i} \lambda_i^2=\sum_{i,j} |a_{ij}|^2= \sum_{i,j} |t_{ij}|^2$$

so $T$ is diagonal and $A=UDU^*$, is normal.

Conversely, if $A$ is normal then (by diagonalizations theorem for normal matrices) $A=UDU^*$ and

$$\sum_{i,j} |a_{ij}|^2=\sum_{i,j} |d_{ij}|^2=\sum_{i} d_{i,i}^2=\sum_{i} \lambda_i^2$$

Title: Normality result using Schur form


问:
Is $f$ is a contraction, could be $F$ the empty set?
Suposse we have a function $f: H \rightarrow H$ where $H$ is a hilbert space and $f$ is a contraction.

Is it possible that the set $F=\{x \in H : \exists \alpha \in (0,1) / x=\alpha f(x)\}$ is empty and $f$ have a fixed point?

答:
Note that for any $\alpha \in (0, 1)$, the function $\alpha f$ is a contraction with constant $L \alpha$, where $L$ is the contraction constant of $f$. So according to Banach's fixed point theorem, for every $\alpha \in (0, 1)$ there is exactly one $x_\alpha \in H$ such that $x_\alpha = \alpha f(x_\alpha)$. So the set $F$ is never empty but consists of all the $x_\alpha$.

But $f$ has a fixed points according to Banach's fixed point theorem. Hence: No it is not possible.


问:
$xy''-(2x^2+1)y'=x^5e^{x^2}$
$xy''-(2x^2+1)y'=x^5e^{x^2}$

Hi there,
needs help with the second order ODE.
I've tried (for the homogenous part) to substitute $\frac{dy}{dx} =u$ to get $ u'-(2x+\frac{1}{x})u=0$

solving for the first non trivial solution

$y_1=C_1 e^{x^2}$
however with $y_2$ and the particular solution i stuck.
asks for help.

答: Set $z:=y^\prime$ so $z^\prime+Pz=Q$ with $P=-2x-1/x,\,Q=x^4e^{x^2}$. We may take the integration factor as $R=\exp(-x^2)/x$ so $(\ln R)^\prime=P$. Thegeneral solution is$$z=x\exp(x^2)\int x^3dx=(\tfrac14x^5+Ax)\exp(x^2),$$i.e.$$y=(\tfrac18x^4-\tfrac14x^2+\tfrac12A+\tfrac14)\exp(x^2)+B.$$

Title: $xy''-(2x^2+1)y'=x^5e^{x^2}$


答:
If you use the characteristic polynomial, it will indeed be a cubic, but it’s a cubic that factors very easily: it’s

$$x^3-3x_2+3x-1=(x-1)^3\;.$$

If you use generating functions, note that you can write the recurrence as

$$a_n=3a_{n-1}-3a_{n-2}+a_{n-3}+8-8[n=0]-7[n=1]-9[n=2]\;,\tag{1}$$

valid for all $n\in\Bbb Z$ if you make the blanket assumption that $a_n=0$ for all $n<0$. The last three terms are Iverson brackets and are there to make the recurrence yield the correct initial values.

Now multiply $(1)$ by $x^n$ and sum over $n\ge 0$:

$$\sum_{n\ge 0}a_nx^n=3\sum_{n\ge 0}a_{n-1}x^n-3\sum_{n\ge 0}a_{n-2}x^n+\sum_{n\ge 0}a_{n-3}x^n+8\sum_{n\ge 0}x^n-8-7x-9x^2\;.\tag{2}$$

If the generating function is $A(x)$, $(2)$ can be rewritten as

$$A(x)=3xA(x)-3x^2A(x)+x^3A(x)+\frac8{1-x}-8-7x-9x^2\;,$$

which you can then solve for $A(x)$ and resolve into partial fractions.

Title: Solve the recursion, $a_n = 3a_{n-1}-3a_{n-2}+a_{n-3}+8$

Body:
Bring the following recursion relation to an explicit expression:

$$a_n = 3a_{n-1}-3a_{n-2}+a_{n-3}+8$$
$a_{0} = 0$, $a_1 = 1$, $a_2 = 2$

All the examples I have seen were with maximum 2 steps back ($a_{n-2}$) and I thought I know how to solve those but I'm having a hard time both with the Generating function and the Characteristic polynomial methods.

The Generating function should start from $n = 3$ but what would happen to the defined values?

For the Characteristic polynomial, Does it mean I'll have to find the roots of a polynomial from a 3rd degree?


问:
The meaning of geodesic curvature, for a geodesic curve
If $\alpha : [0,1] \to M$  is a geodesic curve in a surface $M\subseteq \mathbb R^3$, then is the "geodesic curvature of $\alpha$" equal to the "curvature of $\alpha$" (defined using the Frenet apparatus)?

Or is "geodesic curvature" an unrelated thing from "curvature of a curve" (especially for geodesic) but only their names are chosen to be very similar?

Thank you in advance.

答:
It has no meaning for curves in space defined on a single parameter.

On a surface 'geodesic curvature' symbol $k_g$ is the same thing  as " tangential curvature of a curve ". It is the deviation from straightness. Zero for straight, positive for right turn , negative for left turn in the tangential plane. Curvature has two components resolved along the tangent and normal respectively as $(k_g, k_n)$

$k_g$ is an isometric invariant never changing during bending along with Gauss curvature, integral curvature etc. other first fundamental form (FFF) definable scalars.

I also at first felt unhappy about the nomenclature. A geodesic or straight line has no curvature. Sounds as a contradiction in its own terms. ( terms introduced by Bonnet?) Imho should be referred to as curvature departure/ difference etc. On the other hand geodesic torsion τg makes full sense ( e.g., for helices on a cylinder). But be cautioned this may be an opinion only. Please read what is in the text books of differential geometry and stick to its definition by Liouville etc. on surfaces using coefficients/derivatives of FFF.


问:
Polar to cartesian form of $r=\cos(4θ)$?

Consider an equation in polar coordinates, $r = \cos(4θ)$. Find the equation of the curve in the first quadrant in Cartesian coordinates.

This is for an assignment and this is what help I have received so far from user170231-

$$\begin{align}
r(\theta)&=\cos(4\theta)\\[1ex]
&=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\\[1ex]
&=\frac{r^4\left(\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\right)}{r^4}\\[1ex]
&=\frac{(r\cos \theta)^4-6(r\cos \theta)^2(r\sin \theta)^2+(r\sin \theta)^4}{\left(r^2\right)^2}
\end{align}$$

He told me to see if I could solve it now, and here's what I could come up with:

$$\sqrt{x^2+y^2} = \frac{x^4-6x^2y^2+y^4}{(x^2+y^2)^2}$$

Is this step correct? If so, am I done with the transformation of $r=\cos(4θ)$ to cartesian for the first quadrant only? If not, where should I go from here? Any help is appreciated.

答: Welcome to MSE! Here you can see how $r=cos(4\theta)$ is negative and positive and the importance of solving for both $\pm \sqrt{x^2+y^2}$


问:
Are functions: $1$; $\cos(x)$; $\cos^2(\frac{x}{2})$ linearly independent?
I used the definition.

$1$; $\cos(x)$; $\cos^2(\frac{x}{2})$

$c_1\cdot1+c_2\cdot\cos(x)+c_3\cdot\cos^2(\frac{x}{2}) = 0$

I tried converting $\cos^2(\frac{x}{2})$ into something better: $\frac{1+\cos(x)}{2}$

$c_1+c_2\cdot\cos(x)+c_3\cdot\frac{1+\cos(x)}{2} = 0$

$c_1+\cos(x)\cdot\left(c_2+c_3\cdot\frac{1+\cos(x)}{2\cos(x)}\right) = 0+0\cdot\cos(x)$

$c_1=0$

$c_2+c_3\cdot\frac{1+\cos(x)}{2\cos(x)}=0 \to c_2=-c_3\cdot\frac{1+\cos(x)}{2\cos(x)} \to \frac{c_2}{c_3}=\frac{-1-\cos(x)}{2\cos(x)}$

$c_2=-1-\cos(x)$ and $c_3=2\cos(x)$

There exist $c_i$-s outside of $c_i=0$ which can be solutions. So these functions are linearly dependent.

Is this correct?

Thanks.

答: From $\cos^2(\tfrac{x}{2}) = \tfrac12+\tfrac12\cos(x)$ we get $$\tfrac12 + \tfrac12\cos(x) - \cos^2(\tfrac{x}{2}) = 0$$ that is, $c_1\cdot1+c_2\cdot\cos(x)+c_3\cdot\cos^2(\frac{x}{2}) = 0$ where $c_1=c_2=\tfrac12$ and $c_3=-1$.


答: Once you know that $$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2},$$ you already know that $$(1/2, 1/2, -1) \cdot (1, \cos x, \cos^2 \tfrac{x}{2}) = 0.$$

Title: Are functions: $1$; $\cos(x)$; $\cos^2(\frac{x}{2})$ linearly independent?

Body:
I used the definition.

$1$; $\cos(x)$; $\cos^2(\frac{x}{2})$

$c_1\cdot1+c_2\cdot\cos(x)+c_3\cdot\cos^2(\frac{x}{2}) = 0$

I tried converting $\cos^2(\frac{x}{2})$ into something better: $\frac{1+\cos(x)}{2}$

$c_1+c_2\cdot\cos(x)+c_3\cdot\frac{1+\cos(x)}{2} = 0$

$c_1+\cos(x)\cdot\left(c_2+c_3\cdot\frac{1+\cos(x)}{2\cos(x)}\right) = 0+0\cdot\cos(x)$

$c_1=0$

$c_2+c_3\cdot\frac{1+\cos(x)}{2\cos(x)}=0 \to c_2=-c_3\cdot\frac{1+\cos(x)}{2\cos(x)} \to \frac{c_2}{c_3}=\frac{-1-\cos(x)}{2\cos(x)}$

$c_2=-1-\cos(x)$ and $c_3=2\cos(x)$

There exist $c_i$-s outside of $c_i=0$ which can be solutions. So these functions are linearly dependent.

Is this correct?

Thanks.


问:
Why is the cone measurable?
Problem

Denote by $\lambda_n$ the Lebesgue measure on $\mathbb {R}^n$.
Let $h \in \mathbb {R}_{>0}$. Let $A \subset \mathbb {R}^{n-1}$ be Lebesgue-measurable with finite measure.
Define the cone $C(A,h)$ as the union of all straight lines connecting a point in $A \times\{0\}$ with the point $(0, h) \in \mathbb {R}^{n-1} \times \mathbb{R}$. Prove that $\lambda_n (C(A,h))=\frac{h}{n}\cdot \lambda_{n-1}(A)$.

Where I struggle

Assuming that $C(A,h)$ is (Lebesgue-)measurable, this is a simple application of Fubini. How Fubini is used is all clear to me.

However, why is $C(A,h)$ measurable?
It seems one could roughly argue as follows: As $A$ and $[0,h]$  are measurable so is $A \times [0,h]$. As $A \times [0,h]$ is measurable it is the union of a null set and a Borel set. $C(A,h)$ is the image of $A \times [0,h]$ under a certain diffeomorphism (c.f. Deducing a formula for the volume of a cone over a $J$-measurable set.). Diffeomorphisms map null sets to null sets, and Borel sets to Borel sets. Hence, $C(A,h)$ is (as a finite union of measurable sets) measurable.

Unfortunately, I am not allowed to use the above properties of diffeomorphisms. What are other ways to prove the claim? Please only give a hint to set me thinking.

答:
One can reason as follows: $E\subset \mathbb{R}$ is Lebesgue-measurable if and only if for every $\varepsilon >0$ there exist an open set $G$ and a closed set $F$ such that $ F\subset E\subset G$
and $ \lambda (G\setminus F)<\varepsilon$.

I'll say that $F$ and $G$ are "squeeing" for $E$ with tolerance $\varepsilon$.

Now fix $\epsilon>0$ and let $\mu>$ be such that $2\frac{h}{n}\mu< \epsilon$. Then, since $A$ is Lebsgue-measurable, there are $F$ and $G$ in $\mathbb{R}^n$ squeezing $A$ with tolerance $\mu$. The closed cone (i.e. including endpoints) on $F$ and open cone on $G$ (without endpoints) unioned with some "padding" - a small ball around the cone point and small open neighborhood of $A$ (of total additional  measure at most $\epsilon/2$) - provide squeezing for (the closed cone) $C(A)$ with tolerance $\epsilon$ (by the Fubini computation for $C(F)$ and $C(G)$ plus extra for the "padding" of $C(G)$). Thus $C(A)$ is measurable.

Side note: I think one could also do the something similar with  $G_\delta$ and $F_\sigma$ squeezing.

Title: Why is the cone measurable?


问:
Derivative of a smooth function as an equivalence class
In my differential manifolds class, the derivative of a function $f \in C^\infty (M)$, where $M$ is a manifold, at $p \in M$ was defined as the image of the linear map
$$(df)_p := C^\infty (M) \mapsto C^\infty (M)/Z_p$$

where $Z_p$ is the set of all smooth functions that have zero derivative at $p$. In other words, the derivative is an equivalence class of $C^\infty (M)$ defined by the relation that two functions are related if their derivative vanishes at $p$, the derivative of a function at a point is an equivalence class.

I'm struggling to wrap my head around this. How can a derivative become an equivalence class? Does this hold in elementary calculus too?

答: Adam Zalcman's answer is already really great so I don't have much to add, except for an explanation of why the differential has this form; which I found in Introduction to Smooth Manifolds by John Lee (chapter 11) and I found quite nice:


问:
$[0,1]$ and $(0,1)$ both endowed with the discrete topology are homeomorphic?
I've started reading 'Introduction to Topology' by Vassiliev and I stumbled upon this problem in the first chapter - prove that $[0,1]$ and $(0,1)$ both endowed with the discrete topology are homeomorphic, but I can't think of a bijective map from the open sets of $[0,1]$ with the discrete topology (all points are open) to $(0,1)$ - where would the end points from the closed interval get mapped to in the open interval? Does anyone have a suggestion of how to start? Thank you.

答: The two sets have the same cardinality.  Thus there is a bijection between them.  Any such bijection is a homeomorphism, since both it and its inverse map from spaces with the discrete topology.</p>


问:
Determinant and inverse matrix calculation of a special matrix
Is there any smart way of calculating the determinant of this kind matrix?

\begin{pmatrix}
1 & 2 & 3 & \cdots &  n \\
2 & 1 & 2 & \cdots & n-1 \\
3 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & \cdots & 1 \end{pmatrix}

I encountered this in a problem for the case $n=4$. I need to find the inverse matrix.

I doubt the idea of this problem is to calculate all the cofactors and then the inverse matrix in the usual way.

I see the pattern here and some recursive relations... but I am not sure if this helps for calculating the determinant of the existing matrix.

答:
$$\begin{vmatrix}
1 & 2 & 3 & \cdots &  n \\
2 & 1 & 2 & \cdots & n-1 \\
3 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & \cdots & 1 \end{vmatrix}$$
Notice the sum of an element from the first column and an element from the last column (with the same row) will always be equal to $n+1$,thus we can do $(C_1 = C_1+C_n)$
$$ = \begin{vmatrix}
n+1 & 2 & 3 & \cdots &  n \\
n+1 & 1 & 2 & \cdots & n-1 \\
n+1 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n+1 & n-1 & n-2 & \cdots & 1 \end{vmatrix} = (n+1)
\begin{vmatrix}
1 & 2 & 3 & \cdots &  n \\
1 & 1 & 2 & \cdots & n-1 \\
1 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & n-1 & n-2 & \cdots & 1 \end{vmatrix}$$
From here, we can do as follows:

go from the last row towards the first and decrease each row's value with the one above it (for any row but the first one).
($\forall i \neq 1, R_i=R_i-R_{i-1}$, Starting with $i=n$ then $i=n-1 ... i=2$).
$$ = (n+1)\begin{vmatrix}
1 & 2 & 3 & \cdots &  n \\
0 & -1 & -1 & \cdots & -1 \\
0 & 1 & -1 & \cdots & -1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 1 & 1 & \cdots & -1 \end{vmatrix}$$
Expand $C_1$:
$$ = (n+1)\begin{vmatrix}
-1 & -1 & -1 & \cdots &  -1 \\
1 & -1 & -1 & \cdots & -1 \\
1 & 1 & -1 & \cdots & -1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & -1 \end{vmatrix}$$
Now Add the first row to all of the other rows ($\forall i \neq 1, R_i = R_i + R_1$).
$$ = (n+1)\begin{vmatrix}
-1 & -1 & -1 & \cdots &  -1 \\
0 & -2 & -2 & \cdots & -2 \\
0 & 0 & -2 & \cdots & -2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & -2 \end{vmatrix} = (n+1)[-(-2)^{n-2}]$$


问:
Show that the dimension of the set of all linear transformations from $V$ to $W$ has dimension equal to $mn$.
Let $V, W$ be vector spaces. The set $L(V,W)$ is the set of all linear transformations $T: V \to W$, with the operations:

$(T_1 + T_2)(v) = T_1(v)+T_2(v)$

$(\alpha T)(v) = \alpha T(v)$

for $v\in V$.

Prove that the dimension of the set $L(V,W)$ is $\mathrm{dim}(L(V,W)) = \mathrm{dim}(V) \cdot \mathrm{dim}(W)$

I know I have to (or can) use the bases of the vector spaces to prove this but I cannot come up with a solution.

答:
This explanation helps if you know that two finite-dimensional vector spaces over the same field, $F$, are isomorphic if and only if they have equal dimensions.

If you're using Sheldon Axler's Linear Algebra Done Right (3rd. Edition), you might have seen $\mathcal{L}(V,W)$ as a domain in the context of using the linear map $\mathcal{M}$, which maps $\mathcal{L}(V,W)$ to $F^{mn}$, where $F^{mn}$ is the set of all m-by-n matrices, where each entry is in $F$.

Formally:

$\mathcal{M} : \mathcal{L}(V,W) \rightarrow F^{mn}$

So using the Fundamental Theorem of Linear Maps, we know that:

dim $\mathcal{L}(V,W) =$ dim null $\mathcal{M}$ $+$ dim range $\mathcal{M}$

So our goal now is to prove that: dim $\mathcal{L}(V,W) =$ (dim $V$)(dim $W$)

Let's start with a basis $v_1, ..., v_n$ for $V$ and a basis $w_1, ..., w_m$ for a basis of W. This means that dim $V = n$ and dim $W = m$. And this also makes sense because $\mathcal{M}(T)$, where $T \in \mathcal{L}(V,W)$, represents an m-by-n matrix that holds the coefficients for the basis of $W$ to make $Tv_k$:

$Tv_k = A_{1,k}w_1 + A_{2,k}w_2 + ... + A_{m,k}w_m$, where $A_{j,k}$ are entires of the k-th column of $\mathcal{M}(T)$ for $1 \leq j \leq m$

If you need help convincing yourself that dim $F^{mn} = mn =$ (dim $V$)(dim $W$), just imagine the basis of $F^{mn}$ to be a list of m-by-n matrices where each matrix has entries of all $0$'s except one where there is a $1$:

$\begin{equation}
\begin{bmatrix}
1 & 0 & \ldots & 0\\
0 & 0 & \ldots & 0\\
\vdots & \ldots & \ldots & 0\\
0 & \ldots & \ldots & 0\\
\end{bmatrix},
\
\begin{bmatrix}
0 & 1 & \ldots & 0\\
0 & 0 & \ldots & 0\\
\vdots & \ldots & \ldots & 0\\
0 & \ldots & \ldots & 0\\
\end{bmatrix},
\
\ldots , \
\
\begin{bmatrix}
0 & 0 & \ldots & 0\\
0 & 0 & \ldots & 0\\
\vdots & \ldots & \ldots & 0\\
0 & \ldots & \ldots & 1\\
\end{bmatrix}
\end{equation}$

We see that there are a total of $mn$ matrices in the basis of $F^{mn}$.

Now, in order for dim $\mathcal{L}(V,W) = mn =$ (dim $V$)(dim $W$), we need to show that dim $\mathcal{L}(V,W) = $ dim $F^{mn}$. So that means that we need to show that dim null $\mathcal{M} = 0$ and dim range $\mathcal{M} = $ dim $F^{mn}$ (in other words, show that $\mathcal{M}$ is injective and surjective).

To show that $\mathcal{M}$ is injective, we need to show that $\mathcal{M}(T) = 0$, where $T = 0$. In order to do so, let's suppose that $\mathcal{M}(T) = 0$, where $0$ is the m-by-n matrix that has all entries as $0$. What this means is that: $Tv_k = 0$ because:

$Tv_k = 0w_1 + 0w_2 + ... + 0w_m$

Since $v_k$ is an element from the basis of $V$, $v_k \neq{0}$, therefore $T = 0$. Hence, null $\mathcal{M} = \{0\}$ and thus dim null $\mathcal{M} = 0$.

So what we have now is dim $\mathcal{L}(V,W) = $ dim range $\mathcal{M}$, where we need to show that range $\mathcal{M} = F^{mn}$.

Let's suppose that $A \in F^{mn}$. Remember that:

$Tv_k = A_{1,k}w_1 + A_{2,k}w_2 + ... + A_{m,k}w_m$, where $A_{j,k}$ are entires of the k-th column of $\mathcal{M}(T)$ for $1 \leq j \leq m$

So, we see that $\mathcal{M}(T) = A$. By using the element-method, we see that range $\mathcal{M} \subseteq F^{mn}$ and $F^{mn} \subseteq$ range $\mathcal{M}$, thus range $\mathcal{M} = F^{mn}$.

Therefore, dim $\mathcal{L}(V,W) = $ dim $F^{mn} = mn = $ (dim $V$)(dim $W$).

As we see, we showed that because these two vector spaces are isomorphic, they possess equal dimensions.


问:
Tensor power- Notation question
I have a notational question, which is written usually in papers, but I can not figure it out what could be. Let $M$ be an $A$-module. I have seen this notation
$$M^{\otimes -n}$$

I think this would mean $Hom_A(M^{\otimes n},A)$, but in other way this usually is denoted by $(M^{\otimes n})^*$. Am I right?

答: Yes, you are right. If you're curious, the reason for that notation is that you can view the set of isomorphism classes of A-modules as a group, with the operation of tensor product. Since $M \otimes M^*$ is isomorphic to $A$, we write $M^{-1} = M^*$, and similarly for products.</p>

Title: Tensor power- Notation question


问:
Square-Trangular Numbers Checking Answer
Problem: The first 2 numbers that are both squares and triangles are 1 and 36. Find the next one and if possible, the one after that.

Answer: 1225, 41616

Problem: Can you figure out an efficient way to find triangular-square numbers?

Answer: $s^2 = t(t+1)/2$, where $s,t \epsilon{} \textbf{Z}$.

Problem: Do you think that there are infinitely many?

Answer: Yes. (Can someone give me an justification for my answer?)

答: Hint:  rearrange your equation as a quadratic in $t$.  You will find you need $\sqrt {1+8s^2}$ to be a perfect square to make $t$ an integer.  Then look up Pell's equation.</p>


问:
Integration of Gaussian's
I need to find the integration of
$$
\operatorname{G}\left(x\right) =
\int_{0}^{x}\mathrm{d}y\,
\exp\left(\frac{\left[1 - y^{2}\right]^2}{4a}\right) \int_{y}^{\infty}\mathrm{d}z\,
\exp\left(-\,\frac{\left[1 - z^{2}\right]^{2}}{4a}\right)
$$
for $a > 0$.

I have tried to do it a lot of times but could not get it.

答:
we have $z\in[y,\infty),y\in[0,x]$ which is the same as:
$$0\le z<\infty,0<y<x$$
minus a triangular region at the bottom i.e.
$$G(x)=\int_0^x\int_0^\infty e^{\frac{(1-y^2)^2-(1-z^2)^2}{4a}}dzdy-\int_0^x\int_0^ye^{\frac{(1-y^2)^2-(1-z^2)^2}{4a}}dzdy$$

if you were to try something like converting to polar coordinates with:
$$z=r\cos\theta,y=r\sin\theta$$
then you can try some manipulations:
$$(1-y^2)^2-(1-z^2)^2$$
$$=(1-r^2\sin^2\theta)^2-(1-r^2\cos^2\theta)^2$$
$$=\left(1-2r^2\sin^2\theta+r^4\sin^4\theta\right)-\left(1-2r^2\cos^2\theta+r^4\cos^4\theta\right)$$
$$=2r^2\left(\cos^2\theta-\sin^2\theta\right)+r^4\left(\sin^4\theta-\cos^4\theta\right)$$
now we need to try and use some manipulations:
$$\cos^2\theta-\sin^2\theta=\cos2\theta$$
$$\sin^4\theta-\cos^4\theta=\left(\sin^2\theta+\cos^2\theta\right)\left(\sin^2\theta-\cos^2\theta\right)=-\cos2\theta$$
and so:
$$(1-y^2)^2-(1-z^2)^2=2r^2\cos2\theta-r^4\cos2\theta=r^2(2-r^2)\cos2\theta$$
Although neither of these give nice integrals and so i'd assume that if changing the order of integration doesn't work then change of variables will be the only way

Title: Integration of Gaussian's


Title: $xy''-(2x^2+1)y'=x^5e^{x^2}$

Body:
$xy''-(2x^2+1)y'=x^5e^{x^2}$

Hi there,
needs help with the second order ODE.
I've tried (for the homogenous part) to substitute $\frac{dy}{dx} =u$ to get $ u'-(2x+\frac{1}{x})u=0$

solving for the first non trivial solution

$y_1=C_1 e^{x^2}$
however with $y_2$ and the particular solution i stuck.
asks for help.

Body:
$$y'(x)=u(x)$$
$$x u'(x)-\left(2 x^2+1\right) u(x)=0$$
$$u(x)=c_1 e^{x^2} x$$
$$y(x)=\int c_1 e^{x^2} x\,dx=\frac{c_1 e^{x^2}}{2}+c_2$$
To solve the non homogeneous equation $$xy''-(2x^2+1)y'=x^5e^{x^2}$$
we guess the particular solution as $y_1(x)=(a x^4+b x^3+c x^2+d x+e)e^{x^2}$
$$y_1'(x)=2 e^{x^2} x \left(a x^4+b x^3+c x^2+d x+e\right)+e^{x^2} \left(4 a x^3+3 b x^2+2 c x+d\right)=\\=e^{x^2} \left(2 a x^5+4 a x^3+2 b x^4+3 b x^2+2 c x^3+2 c x+2 d x^2+d+2 e x\right)$$
$$y_1''(x)=2 e^{x^2} \left(2 a x^6+9 a x^4+6 a x^2+2 b x^5+7 b x^3+3 b x+2 c x^4+5 c x^2+c+2 d x^3+3 d x+2 e x^2+e\right)$$
So plugging this in the equation we get
$$8 a x^5+6 b x^4+ x^3 (8 a+4 c)+x^2 (3 b+2 d)-d\equiv x^5e^{x^2}$$
$a=\frac18;\;c=-\frac{1}{4}$ so the particular solution is

$$y_1=\frac{1}{8} e^{x^2} x^2 \left(x^2-2\right)$$
The general solution of the DE is
$$y=\frac{c_1 e^{x^2}}{2}+c_2+\frac{1}{8} e^{x^2} x^2 \left(x^2-2\right)$$
$$y=\frac{1}{8} e^{x^2} \left(x^4-2 x^2+4 c_1\right)+c_2$$


问:
Prove the property of complex roots
Let $n$ be a positive integer. Prove that if $z_1$ is any $n$-root of $z$, that is, $z_1^n=z$, then, $z_1,z_1\omega, \dotsc, z_1\omega^{n-1}$ are the $n$-roots $n$-th of $z$ where $\omega = \cos(2\pi/n)+i\sin(2\pi/n)$.

Title: Prove the property of complex roots

Body: Note that $\omega = e^{2\pi i/n}$, and $\omega^n=(e^{2\pi i/n})^n=e^{2\pi i}=\cos(2\pi)+i\sin(2\pi)=1$. If $z_1\omega^{n-1}$ is the n-root n-th of $z$ then $$(z_1\omega^{n-1})^n=z_1^n(\omega^{n-1})^n=z(\omega^n)^n(\omega^{-1})^n=z(1)^n(\omega^n)^{-1}=z$$


问:
Pigeonhole for threshold on sum of non-negative random variables
Let $X = X_1 + X_2 + \cdots + X_n$ , where $X_i$ ($1 \le i \le n$) are non negative random variables ($X_i$ are not independent). Then using pigeonhole principle would it be correct to state the following two

$$
\Pr[X \ge t] \le \Pr\bigg[ \bigcup_{i=1}^n\big(X_i \ge \frac{t}{n}\big) \bigg]
$$

$$
\Pr[X \le t] \le \Pr\bigg[ \bigcup_{i=1}^n\big(X_i \le \frac{t}{n}\big) \bigg]
$$

where $t \ge 0$.

If $X \ge t$, then by pigeonhole principle there will be an $i$ s.t. $X_i \ge \frac{t}{n}$. But the implication is not true in the other direction.

Similarly for when $X \le t$, there would be an $i$ s.t. $X_i \le \frac{t}{n}$.

If the claim above are correct then I can write

$$
\Pr[X = t] \le \Pr\bigg[ \bigcup_{i=1}^n\big(X_i \ge \frac{t}{n}\big) \bigg]
$$

$$
\Pr[X = t] \le \Pr\bigg[ \bigcup_{i=1}^n\big(X_i \le \frac{t}{n}\big) \bigg]
$$
Which seems fine to me, but bit weird that I can break $\Pr[X=t]$ into two forms one where $X_i \ge \frac{t}{n}$ are involved and another where $X_i \le \frac{t}{n}$.

答: Claim and the given argument is correct; a reason why this should still be conceptually OK is because those bounds, in general, are going to be quite loose (especially the final step of going from $P(X \geq t) \geq P(X = t)$ is extremely loose for $X$ continuous, say).


问:
Expectation and variance of the maximum of k discrete, uniform random variables
Can anyone help me with this? I need to know if I'm on the right track.

I have k discrete, uniformly, identically distributed, independent random variables on the interval [1,...,n]. I want to learn the variance and expectation of the M - the maximum of the k variables.

I first calculated the CDF, then the PMF with the difference between 2 CDFs and now I need to know if I will get somewhere in regards to variance and expected value this way, by using the definition. My main problem is expressing $E[M]$ as a fixed term expression. Given that, I can calculate $Var(M)$ using $Var(M) = E[M^2]-(E[M])^2$. I'm also hoping I can approach calculating the $E[M^2]$ the same way I would for $E[M]$.

More formally here's the problem statement:

"You observe  $k$  i.i.d. copies of the discrete uniform random variable  $X_i$ , which takes values  $1$  through  $n$  with equal probability.

For  $n=2$ , find  $E[M]$  and  $Var(M)$  as a function of  $k$. The random variable  M  as the maximum of these random variables,  $M=\max_i(Xi)$."

Any hints would be appreciated. Thank you!

答:
Finding P.M.F of maximum ordered statistic of discrete uniform distribution.

check this thread, you get the pdf, use the pdf to calculate E[M] and E[M^2].
for n = 2, E[M] = 1*(1/2)^k + 2*(1-(1/2)^k), E[M^2] = (1^2)(1/2)^k + (2^2)(1-(1/2)^k)

You can easily derive the expectation and the variance. If you still need help, I think this can help you as well:
https://www.jiskha.com/questions/1843175/you-observe-k-i-i-d-copies-of-the-discrete-uniform-random-variable-xi-which-takes


问:
Complete the square for $f(x) = 2x^2 + 4x - 6$
I'm studying for a math test. This is the question:

$f(x) = 2x^2 + 4x - 6$. complete the square.

This is how much I get out of the question:
$$2x^2 + 4x - 6$$
$$2(x^2 + 2x - 3)$$
$$2(x^2 + 2x + 1^2 - 1^2 - 3)$$
$$2((x + 1)^2 - 4)$$

But I get stuck here. Can someone complete it and explain?

I'm sorry i forgot a important piece off the question
it says: Complete the square and derived to get the minimum value off $f(x)$

答:
Well, I'll first give an example to motivate the method. Think for example in $(x+1)^2$ if you expand this normally you'll get $x^2+2x+1$, notice that this is square of the first term plus twice the first term times the second term plus the second term squared. So if we have an arbitrary expression like your's it'll be possible to extract from it one expression that is the square of the sum of two numbers plus another number.

You have $f(x) = 2x^2+4x-6$, so divide this by $2$ to get $f(x)/2 = x^2+2x-3$, the reason I did this will soon appear. Now look just on the RHS, if we want to write this as $(x+\text{something})^2$ we know that it'll have before anything else the first term squared, that is, $x^2$, but look there! You already have $x^2$ in your expression (and it's exactly the first term squared because we divided by $2$ so that only $x^2$ remains).

Now, you have $2x$, remember tha the second term appearing after expanding the square should be twice the first term times the second term, now you have there the second term $2x$, and you know this is the second term because it has the first term times something. Note that in the expansion, only in this place appears the first term times something without being squared. Now, since it is twice the first term times the second term, and sice $x$ is the first term, if we divide this by $2\cdot x$ (twice the first term) we'll get the second term! Now, doing this we have that the second term must by $1$.

Now, we have there a minus $3$, not the square of the second term as expected. Now we fix that completing the square, we simply sum and subtract the second term squared (and it won't alter anything since summing and subtracting together aims to summing zero), so we write:

$$f(x)/2 = x^2+2x +1 - 1 - 3$$

Now, the thing $(x^2+2x+1)$ will be just $(x+1)^2$ and $-1-3 = -4$ so that we have in the end $f(x)/2 = (x+1)^2 - 4$ and so $f(x) = 2(x+1)^2 - 8$. You can easily check that's the same thing you had before:

$$f(x) = 2(x^2+2x+1)-8 = 2x^2+4x+2 - 8 = 2x^2 + 4x -6$$

I hope it helps you out. My suggestion is that you try to repeat this argument to another problem so that you get better practice.


问:
For which a the matrix is positive-definite
We have $A=\begin{pmatrix}a&2\\ 2&a+1\end{pmatrix}$

And I want to know for which a the matrix is positive-definite.

So our requirement is $det(A)\ge0$

$(a*(a+1))-4\ge0$

$a^2+a-4\ge0$

I understand that in the points $\frac{-1+\sqrt{17}}{2},\frac{-1-\sqrt{17}}{2}$ the function are equal to zero.

Is it true to say that because the function is equal to zero in the points so $ ∀a>\frac{-1+\sqrt{17}}{2},\frac{-1-\sqrt{17}}{2}$

$a^2+a-4\ge0$?

答: The last step is to note that since $a^2+a-4=(a-(-1+\sqrt{17})/2)(a-(-1-\sqrt{17})/2)$, we have that it is positive precisely on $(-\infty,(-1-\sqrt{17})/2)\cup((-1+\sqrt{17})/2,\infty)$.


问:
Simultaneous semisimplicity of commuting endomorphisms
We say that an endomorphism $f$ of a $\mathbb{k}$-vector space $V$ is semisimple if the resulting $\mathbb{k}[x]$-module structure on $V$ is semisimple.

Question. Let $V$ be a $\mathbb{k}$-vector space and let $f_1, \dotsc, f_n$ be semisimple endomorphisms of $V$ such that $f_i$ commutes with $f_j$ for all $i, j = 1, \dotsc, n$. Is the resulting $\mathbb{k}[x_1, \dotsc, x_n]$-module structure on $V$ semisimple?

The question is motivated by the following special case:
Suppose that the field $\mathbb{k}$ is algebraically closed and that the vector space $V$ is finite-dimensional. Under this assumptions, semisimplicity is equivalent to (simultaneous) diagonalizability.
The question is then equivalent to the well-known theorem that finitely many diagonalizable, commuting endomorphisms are simultaneously diagonalizable.

答:
Here is a counterexample.  Let $k$ be a field of characteristic $p$ that is not perfect and let $a\in k$ have no $p$th root.  Let $K=k[x]/(x^p-a)$, let $V=K\otimes_k K$ and let $f_1,f_2:V\to V$ be multiplication by $x\otimes 1$ and $1\otimes x$, respectively.  Each of $f_1$ and $f_2$ is semisimple over $k$, since their minimal polynomial $x^p-a$ is irreducible over $k$.  However, as a $k[x_1,x_2]$-module, $V$ is the quotient $k[x_1,x_2]/(x_1^p-a,x_2^p-a)\cong K[x_2]/(x_2^p-a)$ which is not semisimple since $x_2^p-a$ factors as $(x_2-x)^p$ over $K$.

However, this is essentially the only sort of counterexample.  In particular, let us suppose not just that $V$ is semisimple as a $k[x]$-module with respect to each of $f_1,\dots,f_n$, but that it is a sum of simple $k[x]$-modules which are separable extensions of $k$ (in particular, this is automatic if $k$ is perfect).  Equivalently, $V$ is a torsion $k[x]$-module with respect to each $f_i$ and the minimal polynomial of $f_i$ on any finite-dimensional invariant subspace of $V$ is separable.  Then I claim that $V$ is semisimple over $k[x_1,\dots,x_n]$.

By induction on $n$, we may assume $V$ is semisimple over $k[x_1,\dots,x_{n-1}]$.  So, we can decompose $V$ as a direct sum $\bigoplus_m V_m$, where $m$ ranges over the set of maximal ideals of $k[x_1,\dots,x_{n-1}]$ and each $V_m$ is a direct sum of copies of $K_m=k[x_1,\dots,x_{n-1}]/m$.  Since $f_n$ commutes with $f_1,\dots,f_{n-1}$, it maps each $V_m$ to itself, so it suffices to show each $V_m$ is semisimple over $k[x_1,\dots,x_n]$.  Since $m$ acts trivially on $V_m$, we can think of its $k[x_1,\dots,x_n]$-module structure as a $K_m[x_n]$-module structure.

So, here's the situation.  We have a vector space $V_m$ over a field $K_m$ which is an extension of $k$, and an endomorphism $f_n$ of $V_m$ which is semisimple and separable (in the above sense) over $k$, and we wish to show $f_n$ is also semisimple over $K_m$.  If $f_n$ were not semisimple over $K_m$, then there would be a finite-dimensional $f_n$-invariant $K_m$-subspace $W\subseteq V_m$ such that the minimal polynomial $p\in K_m[t]$ of $f_n$ on $W$ over $K_m$ is not squarefree.  By hypothesis, the minimal polynomial $q\in k[t]$ of $f_n$ on $W$ over $k$ is separable.  In particular, this means $q$ is squarefree as an element of $K_m[t]$.
But $p$ is a factor of $q$ and is not squarefree, so this is a contradiction.


问:
Find the limit as $n \rightarrow \infty$ of $\sum_{k=1}^{n} \frac{k^3}{n^4}$
I would appreciate suggestions or hints for this homework question.

I'm asked to find the limit as $n \rightarrow \infty$ of the above series. The question is in a chapter on Riemann integration, so I assume that I need to formulate it as a Riemann sum (although I could be wrong about that).

So far, I have expanded the sum out to the partial, $S_{N} = \frac{1}{N^4} (1^3 + 2^3 + ... + (N-1)^3 + N^3)$.

But I'm not sure where to go from here.

Body: a huge hint is that $1^3+2^3+...+N^3=\frac{(N(N+1))^2}{4}$


Title: Find the limit as $n \rightarrow \infty$ of $\sum_{k=1}^{n} \frac{k^3}{n^4}$

Body:
I would appreciate suggestions or hints for this homework question.

I'm asked to find the limit as $n \rightarrow \infty$ of the above series. The question is in a chapter on Riemann integration, so I assume that I need to formulate it as a Riemann sum (although I could be wrong about that).

So far, I have expanded the sum out to the partial, $S_{N} = \frac{1}{N^4} (1^3 + 2^3 + ... + (N-1)^3 + N^3)$.

But I'm not sure where to go from here.

Body:
$$S_n=\frac 1n\Bigl((\frac 1n)^3+(\frac 2n)^3+...(\frac nn)^3\Bigr)$$

$$=\frac{1-0}{n}\sum_{k=1}^n(0+k\frac{1-0}{n})^3$$
$$=\frac{1-0}{n}\sum_{k=1}^nf(0+k\frac{1-0}{n})$$

with $ f(x)=x^3$.

$ f $ is continuous and integrable at $ [0,1] $, so
$$\lim_{n\to+\infty}S_n=\int_0^1x^3dx$$
$$=\Bigl[\frac{x^4}{4}\Bigr]_0^1=\frac 14$$.


问:
How to develop patience in mathematics?
How  to develop the patience in mathematics?

When  you read the books like Rudin, Conway, etc., it takes a lot of time for one problem; sometimes $10$ hours each.

I usually  devote one  problem for $30$ minutes. If I can't solve the problem within $30$ minutes, then I post the problem on this site for a solution.

Actually, I have no patience and can't wait for long hours for thinking about the problem.

How  can I improve myself and develop the patience?

答: Whenever I need to study something new, I study with friends. When you do that, as everyone is thinking in the problem, you solve problems faster, and it is easier to keep yourself concentrated.</p>

Title: How to develop patience in mathematics?


答:
You can prove a matrix is positive definite by checking if the determinant of all of its leading minors are greater then 0.
$$\begin{vmatrix} a\end{vmatrix}>0 \iff a>0 \space\space\space(1)$$

$$\begin{vmatrix}
a & 2 \\
2 & a+1
\end{vmatrix} = a^2+a-4$$
$$a^2+a-4>0 \iff a>\frac{-1+\sqrt{17}}{2} \space\space\space OR \space\space\space\space a<\frac{-1-\sqrt{17}}{2}\space\space\space\space\space(2)$$
From $(1)$ and $(2)$ we can get that A is positive definite for any $a>\frac{-1+\sqrt{17}}{2}$

Title: For which a the matrix is positive-definite

Body:
We have $A=\begin{pmatrix}a&2\\ 2&a+1\end{pmatrix}$

And I want to know for which a the matrix is positive-definite.

So our requirement is $det(A)\ge0$

$(a*(a+1))-4\ge0$

$a^2+a-4\ge0$

I understand that in the points $\frac{-1+\sqrt{17}}{2},\frac{-1-\sqrt{17}}{2}$ the function are equal to zero.

Is it true to say that because the function is equal to zero in the points so $ ∀a>\frac{-1+\sqrt{17}}{2},\frac{-1-\sqrt{17}}{2}$

$a^2+a-4\ge0$?


Title: Prove the property of complex roots

Body: Let $n$ be a positive integer. Prove that if $z_1$ is any $n$-root of $z$, that is, $z_1^n=z$, then, $z_1,z_1\omega, \dotsc, z_1\omega^{n-1}$ are the $n$-roots $n$-th of $z$ where $\omega = \cos(2\pi/n)+i\sin(2\pi/n)$.

Body: $$(z_1\omega^k)^n=z_1^n\omega^{kn}=z\left(\cos\left(\frac{2\pi}nkn\right)+i\sin\left(\frac{2\pi}nkn\right)\right)=z.$$


问:
Prove: If two finite-dimensional vector spaces are isomorphic, then they have the same dimension.
Please comment on the validity of my proof. Any tips and suggestions are appreciated.

Prove: If two finite-dimensional vector spaces are isomorphic, then they have the same dimension.

Let $B=${$b_1,...,b_n$} be a basis for a finite-dimensional vector space $V$. Let $W$ be a finite-dimensional vector space, and define $T: V→W$ as an isomorphism between them.

Let $x∈V$, then $T(x)=T(c_1b_1+...+c_nb_n)$ for some real scalars $c_1...c_n$ as $B$ is a basis for $V$. Also, because $T$ is an isomorphism, $T$ is a linear transformation, onto, and one-to-one.

Because $T$ is onto, for any $y∈W$, $y=T(x)=T(c_1b_1+...+c_nb_n)=c_1T(b_1)+...+c_nT(b_n)$. Thereofore, $span${$T(b_1),...T(b_n)$}=$W$.

Let $c_1T(b_1)+...+c_nT(b_n)=0$. Because $T$ is one-to-one and linear $c_1T(b_1)+...+c_nT(b_n)=T(c_1b_1+...+c_nb_n)=0$ implies that $c_1b_1+...+c_nb_n=0$. But because $b_1...b_n$ is a basis, it is linearly independent and so $c_1=...=c_n=0$. Therefore {$T(b_1),...T(b_n)$} is a linearly independent set.

So, {$T(b_1),...T(b_n)$} is a basis for $W$ and is clearly $n$ dimensional. So, $dim(V)=dim(W)$.

答:
Your proof is correct. It is also written down very nicely and formally clean. I like it and do not have anything to add. I honestly think there isn't much to add.

Only suggestion I have is to replace onto and one to one with surjective and injective, as I think those are more widely understood. But then again, I'm not from a english speaking country so maybe I'm wrong about that and it's probably personal taste.