问:
Manipulating a series with big-O-notation
Let's assume we are given a series $\sum\limits_{k=1}^{\infty}a_k$ which we want to check for convergence. Two scenarios:

1.) After some manipulations of $a_k$ we get something like: $a_k= \frac{1}{k}+f(k)=\frac{1}{k}+O\left(\frac{1}{k^2}\right)$. Now, if someone asks you to prove the divergence via comparison test then I would simply conclude (for a $k$ that is large enough):

$$f(k)= O\left(\frac{1}{k^2}\right)\implies |f(k)|\leq C\frac{1}{k^2}\implies -C\frac{1}{k^2}\leq f(k)\leq C\frac{1}{k^2}\\\implies \Big| \frac{1}{k}-C\frac{1}{k^2}\Big|= \frac{1}{k}-C\frac{1}{k^2}\leq\frac{1}{k}+f(k)$$ where $C>0$. Further, we know that $\sum\limits_{k=1}^{\infty} \frac{1}{k}-C\frac{1}{k^2}$ diverges and hence $\sum\limits_{k=1}^{\infty}a_k$ diverges due to comparison test.

Is this right?

2.) After some manipulations of $a_k$ we get something like: $a_k= g(k)=O\left(\frac{1}{k}\right)$. Now, if someone asks you to prove the divergence via comparison test then I would argue that for a $k$ that is large enough we run into trouble, namely:

$$g(k)= O\left(\frac{1}{k}\right)\implies |g(k)|\leq C\frac{1}{k}\implies -C\frac{1}{k}\leq g(k)\leq C\frac{1}{k}\\\implies \color{red}{\Big|?\Big|}\leq g(k)\leq C\frac{1}{k}$$ where $C>0$. So in this case we cannot construct another series which is divergent and could be used to perform the comparison test, right?

Although our series $\sum\limits_{k=1}^{\infty}a_k$ has an $a_k$ which has the same behavior like $\frac{1}{k}$ it doesn't help us to decide if the series is divergent or not, right?

答:
With $$a_k=\frac{\sin(k)}{k}$$
and

$$b_k=\frac{|sin(k)|}{k}$$
we have

$$a_k=O(\frac 1k)\;\; and \;\; b_k=O(\frac 1k)$$
but
$$\sum a_k \;\; converges$$
while
$$\sum b_k \;\; diverges$$

What we know is, if $ C>0 $,

$$ a_k\le \frac{-C}{k}\implies \sum a_k\; \; diverges$$
and

$$\frac{C}{k}\le a_k\implies \sum a_k\;\; diverges$$

but

$$\frac{-C}{k}\le a_k\le \frac{C}{k}$$
does not imply the divergence.


问:
Find the eigenvalues of an identity matrix minus a rank 1 matrix
This problem comes from Linear Algebra and its Applications by David C Lay and Steven R Lay.

Let $c$ be a unit vector in $\mathbb{R}^3$. Consider the matrix $A = I - 2cc^T$

a ) What are the eigenvalues of $A$?

b) Find an easily computable eigenvector of $A$

My attempt:

Firstly we can see that $cc^T$ is a symmetric $rank\ 1$ matrix.

Let $v_i$ be a eigenvector of $A$ for $i = 1, 2, 3$

$$Av_i = \lambda_iv_i = (I - 2cc^T)v_i \\ = Iv_i - 2cc^Tv_i \\ = v_i  - 2cc^Tv_i$$

I'm stumped on where to go next. Is there something I'm missing?

Any pointers would be greatly appreciated :)

答:
First observe that
$$
Ac=(I-cc^T)c=c-cc^Tc=(1-c^Tc)c
$$
Hence, $c$ is an eigenvector, with eigenvalue $\lambda=1-c^Tc$.

Next, if $u$ and $v$ linearly independent and perpendicular to $c$, i.e., $c^Tu=c^Tv=0$, then
$$
Au=(I-cc^T)u=u, \quad Au=(I-cc^T)v=v.
$$
Hence the eigenvalues of $A$ are $\,1,\,1,\,1-c^Tc$.

ContentLicense: CC BY-SA 4.0


问:
Exercise on Injective and surjective functions
Let $ A, B, C $ be non-empty sets and let $ f, g, h $ be functions such as u $ f: A \to B, g: B \to C $ and $ h: B \to C $. If $ f $ is injective and $ g \circ f = h \circ f $, then $ g = h $.

This is false, right? For this to be fulfilled, the $ f $ function has to be surjective, it is not enough to be injective, is this correct?

答: If $f$ is not surjective, $g$ and $h$ could do whatever they feel like doing off the image of $A$ under $f$.


问:
Finding the value of t where tangent line is perpendicular to x axis
For the curve x = t$^2 - 1, y = t^2 - t$, the tangent line is perpendicular to x-axis, where

Options are :

a ) t = 0

b)  $t \to \infty$

c)  $t = \frac{1}{\sqrt{3}}$

d)  $t = \frac{-1}{\sqrt{3}}$

Here we have $\frac{dx}{dt} = 2t$   $\& $  $ \frac{dy}{dt} = 2t-1$

$\therefore, \frac{dy}{dx} = \frac{2t-1}{2t}$ If this tangent is parallel to x axis then $\frac{dy}{dx}=0$ and if this is perpendicular then $\frac{-dx}{dy} =0$ $\Rightarrow  \frac{2t}{2t-1}=0 \Rightarrow 2t = 0 \Rightarrow t =0$

I hope this is the correct approach please suggest.. thanks..

答: Yes,you are absolutely right in your procedure to find first dy/dx and then,as it denotes the slope of the tangent at a point,equating it to -(infinity) is the right way....however,let's see if anyone point out any mistake(if any remains...so far i could not find)....thanking you that you yourself solved it...:)</p>

ContentLicense: CC BY-SA 3.0


问:
Reducing $A^c\cup(A\cap B^c)\cup(A\cap B\cap C^c)\cup(A\cap B\cap C\cap D^c)\cup\cdots$
I would like to know if there is any way I can reduce or know what this whole set of operations between sets is equivalent to
$$A^c\cup(A\cap B^c)\cup(A\cap B\cap C^c)\cup(A\cap B\cap C\cap D^c)\cup\cdots$$

答:
You can try to find the complement:

$$(A^c\cup(A\cap B^c)\cup(A\cap B\cap C^c)...)^c$$

Use de-morgan to get

$$A\cap(A^c\cup B)\cap(A^c\cup B^c\cup C)...$$
Note that: $A\cap(A^c\cup B)=A\cap B$

So now we have:

$$(A\cap B)\cap(A^c\cup B^c\cup C)...$$

Similarly, $(A\cap B)\cap(A^c\cup B^c\cup C)=A\cap B\cap C$

I think that you can see where I am going with it, the complement is:
$$A\cap B\cap C\cap D...$$

Thus, we can use de-morgan again to deduce that the original set is:

$$A^c\cup B^c\cup C^c\cup D^c...$$

In fact, you don't really have to use de-morgan, but the complement turns out to be more comfortable in my opnion...

Moreover, If you want a proof - you can easily prove it by induction.


问:
Identity concerning the Levi-Civita symbol and determinant
I'm asked to prove $R(\vec{u} \times \vec{v}) = (R \vec{u}) \times (R \vec{v})$ for orthogonal matrices (rotations) $R$ in 3-dimensional euclidean space and use the property $\varepsilon_{ijk} \det M = \varepsilon_{i'j'k'} M_{ii} M_{jj'} M_{kk'}$.

I started with:

$$\left[ (R \vec{u}) \times (R\vec{v}) \right]_{i }= \varepsilon_{ijk} R_{j l} u_{l} R_{km}v_{m},$$

where I used $u'_{i} =(R\vec{u})_{i} = R_{ij} u_{j}$ and $(\vec{u} \times \vec{v})_{i} = \varepsilon_{ijk} u_{j} v_{k}$. So

$$\left[ (R \vec{u}) \times (R\vec{v}) \right]_{i } = \varepsilon_{ijk} R_{j l} R_{km} u_{l} v_{m} = \delta_{ir} \varepsilon_{rjk}R_{j l} R_{km} u_{l} v_{m}, $$

since $\delta_{ij} A_{jm} = A_{im}$. Now because $R$ is orthogonal, we have $R_{ik} R_{jk} = R_{ki} R_{kj} = \delta_{ij}$ from $RR^T = R^T R = I$. Then, using $\delta_{ir} = R_{is} R_{rs}$ we have

$$\left[ (R \vec{u}) \times (R\vec{v}) \right]_{i } = \delta_{ir} \varepsilon_{rjk}R_{j l} R_{km} u_{l} v_{m} = \varepsilon_{rjk} R_{is} R_{rs} R_{jl} R_{km} u_{l} v_{m} = \varepsilon_{rjk} R_{rs} R_{jl} R_{km} R_{is} u_{l} v_{m}$$

So I have something pretty close to the identity I need but not quite. If I set $\varepsilon_{rjk} R_{rs} R_{jl} R_{km} = \varepsilon_{slm} \det R$ I can finish the proof. So I have two questions:

How can I put my proof in a form that I use the identity I need, that is, how can I make the following transformation $\varepsilon_{rjk} R_{rs} R_{jl} R_{km} \to \varepsilon_{rjk} R_{sr} R_{lj} R_{mk}$?

Is $\varepsilon_{rjk} R_{rs} R_{jl} R_{km} R_{is} = \varepsilon_{slm} \det R$ ? If yes, how can I prove it?

答:
Actually I just had to notice that $\varepsilon_{rjk} R_{rs} R_{jl} R_{km} R_{is} u_{l} v_{m} = \varepsilon_{rjk} (R^{T})_{sr} (R^{T})_{lj} (R^{T})_{mk} R_{is} u_{l} v_{m}  = \varepsilon_{slm} \det (R^{T})$. That way we have

$$\varepsilon_{rjk} R_{rs} R_{jl} R_{km} R_{is} u_{l} v_{m} = \varepsilon_{rjk} (R^{T})_{sr} (R^{T})_{lj} (R^{T})_{mk} R_{is} u_{l} v_{m}  = \\
= \varepsilon_{slm} \det (R^{T}) R_{is} u_{l} v_{m} =\varepsilon_{slm} \det(R) R_{is} u_{l} v_{m} = \varepsilon_{slm} R_{is} u_{l} v_{m} = \\
= R_{is} \varepsilon_{slm} u_{l} v_{m} = R_{is} \left(\vec{u} \times \vec{v}\right)_{s} = \left[ R(\vec{u} \times \vec{v})\right]_{i}.
$$


问:
Comparison Test for series.
I need to use the comparison test to determine whether the series:

$\sum\frac{3^n+7^n}{3^n+8^n}$

converges or diverges.

So far, I've tried splitting it into $\frac{3^n}{3^n + 8^n} + \frac{7^n}{3^n + 8^n} $ but have gotten nowhere

答:
hint

For the numerator,
$$3^n+7^n\le 2.7^n$$

For the denominator,
$$3^n+8^n\ge 8^n$$

thus

$$0<\frac{3^n+7^n}{3^n+8^n}\le 2\frac{7^n}{8^n}$$

Title: Comparison Test for series.


问:
Reorganising alternating sum of products of binomial coefficients
The summation
$$
\sum_{k=0}^{\lfloor\frac{p}{s}\rfloor}(-1)^k {n\choose k}{p-ks+n-1\choose n-1}\quad;n > 0, s > 0 \text{ and } 0\le p\le ns,
$$
with ${n\choose k}$ denoting the binomial coefficient, has the limitation that implemented fixed sized integers, there are cases where individual terms exceed the integer size even though the result does not. (Even assuming ideal calculation of binomial coefficients.) Is there a way to reorganise that summation to avoid that? The integer size bound can be assumed to be $s^n$, the sum of that result for all $p$. A stronger result would that all terms summed do not exceed their sum in absolute value. A yet stronger result would be that all terms summed are non-negative. But even the weakest case is fine.

答:
If you know that the final answer fits into the fixed integer size, then you don't have to worry about intermediate overflow. Computation with, say, $64$-bit integers is equivalent to computation modulo $2^{64}$. So even if intermediate answers overflow and give you something that's only correct modulo $2^{64}$ it still guarantees that the final answer is correct modulo $2^{64}$, which will be actually correct.

So for instance in $8$-bit integer arithmetic I can compute $$100 + 200 - 50 - 150$$ and get $100$, even though the partial sums don't fit into $8$ bits:

With unsigned $8$-bit integers, $100+200$ would come to $44$, subtracting $50$ would yield $250$, and subtracting $150$ would yield $100$.

With signed $8$-bit integers, the individual terms wouldn't even fit, and we'd be computing $100 + (-56) - 50 - (-106)$. The partial sums would be $100 + (-56) = 44$, $44-50 = -6$, and $-6 - (-106) = 100$.


问:
is $\frac{Γ(r+x)}{Γ(r)Γ(x+1) }$ the same as ${r+x-1 \choose x}$?
is $\frac{Γ(r+x)}{Γ(r)Γ(x+1) }$ the same as ${r+x-1 \choose x}$ ?

I have two pmfs $$P(X=x\mid\theta)={r+x-1 \choose x}(1-\theta)^x\theta^r $$
and $$P(X=x\mid\theta)=\frac{Γ(r+x)}{Γ(r)Γ(x+1) }(1-\theta)^x\theta^r $$

I was wondering if they would yield the same MlE

答:
Assuming that both $x$ and $r$ are integers, yes. Since $\Gamma(n)=(n-1)!$ for all positive integers, then
$$ \frac{\Gamma(r+x)}{\Gamma(r)\Gamma(x+1)} = \frac{(r+x-1)!}{(r-1)!x!} = \binom{r+x-1}{x}. $$


答: You're right, but you should provide an example. A simple one is with $A=\{0\}$, $B=C=\{1,2\}$. Define $f(0)=1$, $g(1)=1$, $g(2)=1$, $h(1)=1$ and $h(2)=2$.

Title: Exercise on Injective and surjective functions

Body:
Let $ A, B, C $ be non-empty sets and let $ f, g, h $ be functions such as u $ f: A \to B, g: B \to C $ and $ h: B \to C $. If $ f $ is injective and $ g \circ f = h \circ f $, then $ g = h $.

This is false, right? For this to be fulfilled, the $ f $ function has to be surjective, it is not enough to be injective, is this correct?


问:
Singular values of Sigma in SVD
I am sorry if this is a dumb question. Today I've come across for the first time with the notion of spectral norm of a matrix. I have read that it is the biggest singular value of a matrix.

In the context of SVD, let $X=U \Sigma V^*$

What is the spectral norm of $\Sigma$? Are the singular values of $X$ (contained in $\Sigma$) also the singular values of $\Sigma$?

答: Yes, the singular values of $X$ are equal to the singular values of $\Sigma$.

Title: Singular values of Sigma in SVD


问:
Axiom of choice in proof of theorem 2.41 in Rudin's principles of real analysis?
I have looked through the many questions about this theorem on this website, but none that I could see adressed exactly what I am wondering about. For completeness, I will state the theorem below and the part of the proof I am wondering about.

2.41 Theorem If a set $E$ in $\mathbb{R}^{k}$ has one of the following three properties, then it has the other two:

$E$ is closed and bounded.

$E$ is compact.

Every infinite subset of $E$ has a limit point in $E$.

The part of the proof I am wondering about concerns proving that (3) implies that $E$ is closed. The proof is set up by proving the contrapositive.

Proof If $E$ is not closed, then there is a point $\mathbf{x_{0}}\in R^{k}$ which is a limit point of $E$ but not a point of $E$. For $n=1,2,3,...,$ there are points $\mathbf{x_{n}}\in E$ such that $|\mathbf{x_{n}} - \mathbf{x_{0}}| < 1/n$. Let $S$ be the set of the points $\mathbf{x_{n}}$.

The proof continues, but the part I am wondering about is the last part, namely "let $S$ be the set of points $\mathbf{x_{n}}$." I understand that because $\mathbf{x_{0}}$ is a limit point, there must exist such points for all $n$. However, is just knowing that such points exist enough to form a set consisting of such points? For each $n\in \mathbb{N}\setminus \{0\}$ there must exist infinitely many such points since $\mathbf{x_{0}}$ is a limit point, must there not? And if so (since such points are not unique), if we want to form the set $S$ described above, must we not either specify some procedure describing these points for each $n$, or appeal to the axiom of (countable?) choice to choose one such point for every $n$?

Unfortunately my instructor has deemed set theory "boring" every time I have tried to talk to him about the potential usage of the axiom of choice. Thank you very much in advance to anyone that took the time to read this post. Any answer is much appreciated.

答: The first two are equivalent without the axiom of choice (see here and here), but the third is not: by Theorem $\bf{3.27}$ of Horst Herrlich, Axiom of Choice, Springer-Verlag, $2006$, the equivalence of $(2)$ and $(3)$ for arbitrary pseudometric spaces is equivalent to the axiom of countable choice. In Disaster $\bf{4.53}$ he also notes that it is consistent with $\mathsf{ZF}$ that $\Bbb R$ have a subset $A$ with an accumulation point $x$ that is not the limit of any sequence in $A$.


问:
Showing $F_{64}/F_2$ is Normal
I need to show that $F_{64}/F_2$ is normal. So far I generated $F_{64}$ by doing $F_2[x]/(x^6+x+1)$. But I'm not sure where to go from here. I think what I need to do is show that $F_{64}$ is a splitting field for $x^6+x+1$? Though I'm not sure why just showing it for this one polynomial is enough to show that the extension is normal. And I've been trying to factor it but I don't see how to go about that?

答:
During my qualifying oral exam, I was asked to construct a field of order $27$; so I wrote one polynomial and invoked a certain result, but was asked to do a more constructive construction that did not invoke the existence of splitting fields. I found an irreducible polynomial over $\mathbb{F}_3$ and did the usual construction. Great. Then I was given a different irreducible polynomial over $\mathbb{F}_3$ and was asked to prove the resulting extension was isomorphic to the one I had initially given. After 5 minutes of flailing to try to construct an explicit isomorphism, I was told to step back and look at the polynomial I had originally written down: $x^{27}-x$.

Basically: since a finite subgroup of the multiplicative group of a field (in fact, of an integral domain) must be cyclic, the multiplicative subgroup of $\mathbb{F}_{p^n}$ is cyclic of order $p^{n}-1$, and therefore every nonzero element satisfies the polynomial $x^{p^n-1}-1$. Therefore, the elements of $\mathbb{F}_{p^n}$ are all roots of $x^{p^n}-x$, and these are all the roots in an algebraic closure of $\mathbb{F}_{p}$. And no strictly smaller field has all the roots, because this polynomial is separable (its derivative is $-1$). Thus, $\mathbb{F}_{p^n}$ is the splitting field of $x^{p^n}-x$ over $\mathbb{F}_p$, and therefore it is normal (and unique up to isomorphism over $\mathbb{F}_p$).

Title: Showing $F_{64}/F_2$ is Normal


问:
Chain rule use in discontinuous generalized function derivative
Let's assume we have a function of the following form $f(x,a):=g(H(a-x))$, where $H$ is the Heaviside step function.
We now would like to look at the derivative $\frac{\partial}{\partial x}\int_0^1 f(x,a)da$. (This form should explain the "generalized functions" in the title.)

The actual question now is if it is allowed to write the following:
$$\frac{\partial}{\partial x}\int_0^1 f(x,a)da = \int_0^1 \frac{\partial g(y)}{\partial y} \cdot \frac{\partial H(a-x)}{\partial x} da,$$
where $y = H(a-x)$.

In a general setting the chain rule can to my understanding not be used under such circumstances, since $H(x-x)=\infty$ and thus the differentiability of $g$ at this point is a little hard to define. Additionally $H(a-a$ neither is differentiable at this point.
Also when doing this we have the weird situation of the multiplication of a step function with a Dirac delta, both having discontinuities at $x=a$.

Does someone know about the rules in such points? Does there exist a chain rule for Heaviside step functions?

答:
Not quite. For simplicity assume $g$ is $C^1$

$g(H) = g(0)+ (g(1)-g(0))H$ whose distributional derivative is $(g(1)-g(0))\delta$.

Taking $\phi \in C^\infty_c,\int \phi=1,\phi_n(x)=n\phi(nx)$ then $H\ast \phi_n \to H$.

$g(H\ast \phi_n)\to g(H)$ in the sense of distributions, and we'll have no fear to say that

$$(g(H\ast \phi_n))'=g'(H\ast \phi_n) (H\ast \phi_n)'=g'(H\ast \phi_n) \phi_n \to (g(H))'$$ in the sense of distributions.

So $\lim_{n\to \infty}g'(H\ast \phi_n) (H\ast \phi_n)'$ is the true distributional meaning of $g'(H) H'$. It is merely saying that the correct value  of $g'(H(0))$, which gives
$$(g(H))'=  g'(H(0)) \delta$$
is $g'(H(0))=\int_0^1 g'(x)dx$.


问:
geodesic on Riemann manifold and tangent space
Knowing almost nothing about differential geometry, I have to understand this statement from the book that i am reading (page 115):

Recall that a geodesic on a Riemannian manifold $M$ is a path $\gamma(t)$, which minimizes the distance
$$L(\gamma) = \sqrt{\int_{t_1}^{t_2} |\dot{g}(t)|^2 dt}$$

among all curves $g:[t_1,t_2]\rightarrow M$ constrained by the boundary conditions $g(t_1)=\gamma(t_1), g(t_2)=\gamma(t_2)$, and this minimization property should hold true, whenever  $t_2$ is close enough to $t_1$. It is equivalent to stating, that the accelleration of the curve $\gamma$, viewed from the tangent space to the manifold vanishes identically.

What does that last sentence mean? My Interpretation is that $\langle \ddot{\gamma}(t), T_{\gamma(t)}M \rangle = 0$, i.e. that the second time derivative of the curve is orthogonal to the tangent space. But intuitively, I don't see, why that should be an equivalent statement to the minimization property. I would rather expect that any curve on $M$ has its acceleration  in the span of the tangent space of $M$.

答:
The definition of geodesics is not exactly what you stated. A curve is a geodesic if it minimizes locally the distance between points, not globally. But this is not so important.

In the case where $M$ is a submanifold of $\mathbb{R}^n$ with the induced metric, that is, $M \subset \mathbb{R}^n$ and if $u,v \in T_pM$, that is $g_p(u,v) = \langle u,v\rangle$ the usual scalar product, then a smooth curve $\gamma : I \mapsto M$ is a geodesic if and only if its acceleration is orthogonal to $M$, that is, for all $t\in I$, $\gamma''(t) \perp T_{\gamma(t)}M$.

If $(M,g)$ is an abstract riemannian manifold, there is no ambiant space to state that the acceleration of a curve is orthogonal to $M$. In fact, there is nothing that can define the acceleration on an abstract manifold! Even if smoothness is well-defined, second derivative of a function is not well-defined as it is closely coordinate-dependant. So there is no possibility to talk about high order differentiation.

The key is that if $M$ is a riemannian manifold, there exists an intrinsic object call the Levi-Civita connection or covariant derivative, denoted by $\nabla$ or $D$ usually, that allows you to talk about acceleration. It is an algebraic objet that allows you to derive vector fields in the direction of vectors fields. In case $M\subset \mathbb{R}^n$ with the induced metric, the Levi-Civita connection is the orthogonal projection onto the tangent space of the derivative: if $v$ is a vector field and $\gamma(t)$ is a curve, the covariant derivative of $v$ along $\gamma$ is
$$ \nabla_{\gamma'(t)}v(t) = {v(\gamma(t))'}^{\perp} $$
In the general setting, one define the acceleration of a curve $\gamma$ to be the vector field along $\gamma$ defined by $\nabla_{\gamma'}\gamma'$. One can show a geodesic is a curve that is solution to the equation of geodesics
$$\nabla_{\gamma'}\gamma'=0$$
Once again, if $M\subset \mathbb{R}$, a curve $\gamma : I \to M$ is a geodesic if and only if its acceleration is orthogonal to $M$, if and only if $M$ is a solution of the equation of geodesics. Thus, saying $\nabla_{\gamma'}\gamma'=0$ is a generalisation of saying the acceleration of $\gamma$ is orthogonal to $M$: if there was an ambiant space, it would be.

If for example $M = \mathbb{S}^n \subset \mathbb{R}^{n+1}$, a geodesic is a great circle of the sphere. It can be parametrized by
$$\gamma(t) = \cos(t\|v\|)p + \sin(t\|v\|)\frac{v}{\|v\|}$$
where $p\in \mathbb{S}^n$ and $v \in T_p\mathbb{S}^n = p^{\perp}$. The usual differentiation of $\gamma$ in the ambiant space shows that
$$ \gamma''(t) = -\|v\|^2\cos(t\|v\|)p  -\|v\|^2\sin(t\|v\|)\frac{v}{\|v\|}=-\|v\|^2\gamma(t) $$
which is colinear to $\gamma(t)$ and thus, orthogonal to $T_{\gamma(t)}\mathbb{S}^n$. Its orthogonal projection onto $T_{\gamma(t)}\mathbb{S}^n$ is then $0$.


答:
It will be simpler to use  the limit comparison test via *asymptotice equivalence of functions:
$$3^n+7^n\sim_\infty 7^n,\quad3^n+8^n\sim_\infty 8^n,\quad\text{ hence }\quad \frac{3^n+7^n}{3^n+8^n}\sim_\infty \frac{7^n}{8^n}=\Bigl(\frac78\Bigr)^n,$$
which converges.

Title: Comparison Test for series.

Body:
I need to use the comparison test to determine whether the series:

$\sum\frac{3^n+7^n}{3^n+8^n}$

converges or diverges.

So far, I've tried splitting it into $\frac{3^n}{3^n + 8^n} + \frac{7^n}{3^n + 8^n} $ but have gotten nowhere


问:
A basic doubt on the definition of a Poisson random variable
What is the significance of "large city" in the definition of the following Poisson variable :

"Number of phone calls placed during a ten second interval in a large city"

I guess either $n \to \infty$ or $p$ very small is not implied by "large city".

答:
I assume the OP is referring to this http://en.wikipedia.org/wiki/Poisson_limit_theorem for $n, p$.

$p$ is probability that a person makes a phone call during a particular 10-second interval, which is very small. $n$ is the number of people in the city. The large city assumption guarantees $n$ is large, so this can be modeled by $Poisson(np)$.


问:
Partial derivatives of matrices?
Assume I have the following equation:

$$\textbf{f}(\textbf{x})=\frac{1}{2}\textbf{x}^T(\textbf{A}^T\textbf{A})\textbf{x} - \textbf{c}^T\log(\textbf{B}\textbf{x})$$

where $\textbf{x}$ is a $m \times 1$ vector of parameters, abd $\textbf{A}$ and $\textbf{B}$ are known $n \times m$ matrices, and $\textbf{c}$ is a known $n \times 1$ vector. I wish to find $\nabla \textbf{f}(\textbf{x})$, where

$$\nabla=\left[\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2},...,\frac{\partial}{\partial x_m}\right]^T$$

hence:

$$\nabla\textbf{f}(\textbf{x})=\frac{1}{2}\nabla\textbf{x}^T(\textbf{A}^T\textbf{A})\textbf{x} - \nabla\textbf{c}^T\log(\textbf{B}\textbf{x})$$

Is it possible to solve this system analytically? I imagine the first term could be simplified, as I believe a simple dot product with $\textbf{x}^T$ would yield the identity matrix ($\nabla\textbf{x}^T=\mathbb{I}$). How would I solve the second term, though? The logarithm in the second term is element-wise.

I would be grateful for any pointers, and ideally references to resources on how to solve such problems in general. ${}$

答:
For convenience, define the variables
$$z = Ax,\quad y = Bx,\quad Y = {\rm Diag}(y)$$
Write the function in terms of these new variables, then calculate the differential and gradient.
$$\eqalign{
f &= \tfrac 12z^Tz - c^T\log(y) \\
df &= z^Tdz - c^T(Y^{-1}dy) \\
   &= z^TA\,dx - c^TY^{-1}B\,dx \\
   &= (A^Tz - B^TY^{-1}c)^T\,dx \\
\frac{\partial f}{\partial x}
   &= A^Tz - B^TY^{-1}c \\
   &= A^TAx - B^T{\rm Diag}(Bx)^{-1}c \\
}$$
In general, if you can manipulate your differential into the form
$$df = g^Tdx$$
then the gradient can be identified as
$\;g=\left(\frac{\partial f}{\partial x}\right)$

Let $f'$ denote the ordinary derivative of the function $f$. When $y$ is a vector, the function is evaluated element-wise, and the vector differential can be written as
$$\eqalign{
df &= f'(y)\odot dy \\&= {\rm Diag}\Big(f'(y)\Big)\,dy \\
}$$
where $\odot$ denotes the elementwise/Hadamard product.

Title: Partial derivatives of matrices?


问:
Functions of the same self-adjoint operator commute on Hilbert space
Let $H$ be a self-adjoint operator on a Hilbert space $\mathcal{H}$, not necessarily bounded. Consider two measurable functions $f$ and $g$ and define the operators $f(H)$ and $g(H)$ symbolically. In what sense are these operators defined? And is it true that $[f(H), g(H)] = 0$?

I have some idea of "functional calculus" and the spectral theorem but its vague and I'd like some help for clarification.

答:
By the spectral Theorem you are allowed to think that $\mathcal H=L^2(X,\mu)$, where $(X,\mu)$ is a $\sigma$-finite measure space, and that your operator (which I am going to call) $T$ is given by
$$
  T(\xi)|_x = \phi(x)\xi(x), \quad \forall \xi\in \mathcal H, \quad \forall x\in X,
  \tag 1
  $$
where $\phi$ is some measurable real valued function on $X$, the domain of $T$ being
$$
  D(T)= \{\xi\in L^2(X,\mu): \phi\xi\in L^2(X,\mu)\}.
  $$

Incidentally it is interesting to notice that, no matter how wildly unbounded $\phi$ is, the domain $D(T)$ will always be dense!

This said, given any Borel measurable function  $f$ on $\mathbb R$, one has that $f(T)$ is defined exactly as in (1), except that $\phi$ is replaced by $f\circ\phi$.  Doing the same for $g$ you may now easily check that the domains of $f(T)g(T)$ and  $g(T)f(T)$ coincide and that $f(T)$ commutes with $g(T)$ on that common domain, simply because pointwise multiplication is commutative!

All of this bonanza takes place because we are basing everything on a single self-adjoint operator.  Should we be interested in working with two different operators (think position and momentum in quantum mechanics) hell would quickly break loose!


问:
Maximum of quasi-convex functions
A function $f$ is quasiconvex if all its sub-level sets are convex (i.e., $\{ x: f(x) \le \alpha\}$ is convex for all $\alpha$.)

For a convex function $f$, it is true that $f$ acheives its maximum over a convex set $S$ at an extreme point of $S$. What is the analogue of this for quasi-convex functions?

答:
I think that assuming that $S$ is compact and that $f$ achieves its supremum over $S$ you are right - see below.

If $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is quasiconvex, then for any $0\leq\theta\leq 1$

$$f(\theta x+(1-\theta)y)\leq\max\{f(x),f(y)\}.$$

Proof: To get a contradiction assume there exists an $y,z$ and $0\leq\theta\leq 1$ such that

$$f(\theta y+(1-\theta)z)>\max(f(y),f(z)).$$

Consider the sub-level set $A:=\{x:f(x)\leq\max(f(y),f(z))\}$. Then $y,z\in A$, but $\theta y+(1-\theta)z\not\in A$ which contradictions quasiconvexity of $f$.

One can iterate the above to get that:

If $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is quasiconvex, then for any $x_1,\dots,x_m\in \mathbb{R}^n$ and $\theta_1\geq 0,\dots,\theta_m\geq0$ such that

$$\sum_{i=1}^m\theta_i=1,$$

then

$$f\left(\sum_{i=1}^m\theta_i x_i\right)\leq \max_{i=1,\dots,m}\{f(x_i)\}.\quad\quad(*)$$

We can use the above to show that:

If $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is quasi-convex, $S$ is a compact and convex subset of $\mathbb{R}^n$, and there exists an $x^*\in S$ such that $f(x^*)\geq f(x)$, then $S$ has an extreme point $y$ such that $f(y)= f(x^*) $.

Proof: By the Krein-Milman Theorem $S$ has at least a single extreme point. In addition, if $K$ denotes the set of all extreme points of $S$, then $S$ is equal to the convex hull of $K$.

Since $x^*\in S$, and using the definition of convex hull, there exists $x_1,\dots,x_m\in K$ and $\theta_1\geq0,\dots,\theta_m\geq0$ with $\sum_{i=1}^m\theta_i=1$, such that $x^*=\sum_{i=1}^m\theta_i x_i$. Applying $(*)$ completes the proof.

Note that if $f$ is continuous, since $S$ is compact, then the existence of $x^*$ in the premise above is guaranteed.

Title: Maximum of quasi-convex functions


问:
Let $X$ be the set of all sequences whose elements lie in $\{0,1\}$. Now consider the metric space $(X,d)$
Let $X$ be the set of all sequences whose elements lie in $\{0,1\}$. Now consider the metric space $(X,d)$  with $d(x,y)=\sum\limits_{i=1}^{\infty}|x_i-y_i|2^{-i}$ where $x=(x_i)$ and $y=(y_i)$ lies in $X$. Now define a map $f:X\rightarrow[0,1]$ by $f(x)=\sum\limits_{i=1}^{\infty}\frac{x_i}{2^i}$ where $x=(x_i)\in X$. Show that $f$ is not an open map.

A map is said to be open if it maps open sets to open sets.

Please given some idea. I know that $f$ is bounded, onto and continuous

答:
Take for example
$$
x=(0,1,0,0,0,\ldots),
\qquad
\text{then}
\qquad
f(x)=\frac{1}{4}.
$$
If $B(x,1/8)=\{y\in X: d(x,y)<1/8\}$, then $B(x,1/8)$ is open, and if
$y=(y_1,y_2,\ldots)\in B(x,1/8)$,
then $y_1=0$ and $y_2=1$. Hence
$$
\frac{1}{4}\in f\big(B(x,1/8)\big)
\quad\text{and}\quad f\big(B(x,1/8)\big)\subset [¼,1],
$$
which means that $f\big(B(x,1/8)\big)$ can not be open.


问:
How can I simplify this complex number?
So I have the complex number $\frac{(7+5i\sqrt{3})^{145}}{(92+48i\sqrt{3})^{72}}$ and I have to simplify it, with the answer being in algebraic form.
I tried using the trigonometric form of the two numbers and the dividing but $|7+5i\sqrt{3}|=\sqrt{124}$ and $\sin(x)=\frac{{5}\sqrt{3}}{\sqrt{124}}$ and $\cos(x)=\frac{{7}}{\sqrt{124}}$. Did the same for the second number but then I would get numbers I cannot easily divide.
So what should I do?

答: A direct computation shows that$$\frac{\left(7+5\sqrt3\,i\right)^2}{92+48\sqrt3\,i}=\frac12+\frac{\sqrt3}2i=\cos\left(\frac\pi3\right)+\sin\left(\frac\pi3\right)i.$$So$$\frac{\left(7+5\sqrt3\,i\right)^{12}}{\left(92+48\sqrt3\,i\right)^6}=1.$$Can you take it from here?

Title: How can I simplify this complex number?


问:
Proving absolute convergence of a series containing parameter
I have the following series:

$$\sum_{n=1}^\infty \frac{\sin(nx)}{n^a}, a>0 $$ $$ 0<x<\pi$$

Now, I have to show for which values of $a$ the series absolutely converges and diverges.

I have figured out that the series converges for $a>1$:

$$\sum_{n=1}^\infty |\frac{\sin(nx)}{n^a}| \le \frac{1}{n^a}$$
which converges for $a>1$ so we proved the absolute convergence of our series but only for $a>1$

However, I have no idea how to prove that the series does not absolutely converge for $0<a\le1$.

I tried finding a lower bound for $|\sin(nx)|$ but all I could come up with is that it is greater than or equal to zero which does not mean anything. Could anyone shed some light?

Body: $\sum | \frac {\sin (\frac {n\pi} 2)} {n^{a}}| \geq \sum | \frac {\sin (\frac {(2n+1)\pi} 2)} {(2n+1)^{a}}| =\sum \frac 1 {(2n+1)^{a}}=\infty$ for any $a \in (0,1]$.


问:
Integrate over the uniform distribution on the simplex
Let $p=(p_1,\ldots,p_n)$ correspond to points in a simplex that add up to one, i.e. $p$ is a discrete probability distribution. I would like to compute an integral of the form $\int dp_1\ldots\int dp_n\sum_{i=1}^np_if(p_i)$ with $p$ uniformly distributed on the $n-1$ dimensional simplex. My question is, how can I parameterize $p_i$ such that the integral covers the simplex uniformly?</p>

答:
If $p$ is uniform on the simplex $\Delta_n=\{(p_i)_{1\leqslant i\leqslant n}\mid p_i\geqslant0,p_1+\cdots+p_n=1\}$, then each (continuous, for every $n\geqslant2$) random variable $p_i$ has density $(n-1)(1-x)^{n-2}\mathbf 1_{0\leqslant x\leqslant 1}$ with respect to the Lebesgue measure $\mathrm dx$. Hence
$$
\int_{\Delta_n}\sum_{i=1}^np_if(p_i)\mathrm d\sigma(p_1,\ldots,p_n)=n(n-1)\int_0^1f(x)x(1-x)^{n-2}\mathrm dx.
$$</p>


问:
If $(X, \|\cdot\|)$ is a normed vector space, then $(X\setminus\{0\}, d)$ is a metric space for $d(x, y) = \frac{\|x-y\|}{\|x\|+\|y\|}$
Consider the following statement: If $(X, \|\cdot\|)$ is a normed vector space, then $(X\setminus\{0\}, d)$ is a metric space for $$d(x, y) = \frac{\|x-y\|}{\|x\|+\|y\|}$$ I'm trying to figure out whether this statement is true (and if it is, whether it's only true in a special setting, e.g. uniformly convex Banach spaces). The context is that I'm trying to define a "similarity metric" between two vectors in a general normed vector space with the invariance $d(x, y) = d(\alpha x, \alpha y)$ for $\alpha\neq 0$, among some other properties. It seems like it could maybe be a textbook problem, but I'm not quite sure if I've seen it somewhere or not. It's pretty clear that $d(x, y) = 0$ implies $x = y$ and $d(x, y) = d(y, x)$, but the triangle inequality is a bit sticky. We can rearrange $d(x, y)\leq d(x, z)+d(z, y)$ as $$\|x-y\|\leq \left(\frac{\|x-z\|}{\|x\|+\|z\|}+\frac{\|y-z\|}{\|y\|+\|z\|}\right)(\|x\|+\|y\|)$$ which feels stronger than the standard triangle inequality on $(X, \|\cdot\|)$. Does anybody have some insights on this? Am I missing a standard trick here? Thanks!

答: Consider the special case of $X=\mathbb R$. Then, $d(x,y)=\frac{|x-y|}{|x|+|y|}$ is not continuous at $(x,y)=(0,0)$. Along $x=y,\ (x,y)\to (0,0)\Rightarrow d(x,y)\to 0$ whereas along $x=0$ we have $d(x,y)\to 1.$ So even if we define $d(0,0)=0$ the triangle inequality can not be satisfied.


问:
Using Lagrangian multiplier method with multiple constraints
So I am trying to find the minimum and maximum of the function $f(x,y,z)= x^2 + y^2 - z^2$ on the curve defined by $y^2 + z^2 =1$ and $x=y.$

My work thus far is the following:

$\text{Proof.}$ Let $g=y^2 + z^2 -1=0, h=x-y=0,$ and taking partials, $$f_x = 2x, f_y = 2y, f_z = -2z$$ $$g_x = 0, g_y = 2y, g_z = 2z,$$ $$h_x = 1, h_y = -1, h_z = 0$$ and by definition of the Lagrange multiplier with multiple constraints, we have $\nabla f = \lambda \nabla g + \mu\nabla h$ corresponding to each parameter. Thus, $$\nabla f_x = \lambda g_x + \mu\nabla h_x \implies x = \frac{\mu}{2}$$ $$\nabla f_y = \lambda g_y + \mu\nabla h_y \implies y = \lambda y - \frac{\mu}{2}$$ $$\nabla f_z = \lambda g_z + \mu\nabla h_z \implies z = -\lambda z$$ Now from here I am having issues because I cannot find a solution for $y,z$ Can someone please help me? Thank you.

答:
Assuming your setup is correct up to this point, basically there is a bunch of case work to do. If you're allowed to divide by whatever you want then you can do this:

$$y=\lambda y - \mu/2 \Rightarrow y=\frac{\mu/2}{\lambda-1} \\
z = -\lambda z \Rightarrow \lambda = -1 \\
\Rightarrow y = \frac{-\mu}{4} \\
\Rightarrow x = \frac{-\mu}{4}$$

However we also know $x=\frac{\mu}{2}$, so $\mu=0$, and so you get $z=\pm 1$ so you have the points $(0,0,1)$ and $(0,0,-1)$. You probably don't actually care but the values of the Lagrange multipliers are $\lambda=-1$ and $\mu=0$.

Now what assumptions did you make along the way? You assumed $\lambda \neq 1$ and $z \neq 0$. What happens if you violate one or both of those assumptions?

Incidentally I would say that an easier approach to this particular problem would be to just substitute $x=y$ and do a one-constraint Lagrange problem. Thus you look at $f(y,z)=2y^2-z^2$ subject to $y^2+z^2-1=0$, so $4y=\lambda 2y$ and $-2z=\lambda 2z$, from which you readily conclude that at least one of $y$ or $z$ must be zero (since $\lambda$ cannot be both $2$ and $-1$ at the same time).


答:
$$
\sum_{n=1}^\infty \frac{\cos nx}{n^a}, \quad x\ne 2k\pi
$$
converges conditionally for every $a>0$.

This is shown using Abel's summation method: Set $s_n(x)=\sum_{k=1}^n \cos kx$, and then observe that
$$
\sum_{k=1}^n \frac{\cos kx}{k^a}=\sum_{k=1}^n \frac{s_k(x)-s_{k-1}(x)}{k^a}=
\sum_{k=1}^n \frac{s_k(x)}{k^a}-\sum_{k=1}^n \frac{s_{k-1}(x)}{k^a} \\
=
\sum_{k=1}^n \frac{s_k(x)}{k^a}-\sum_{k=0}^{n-1} \frac{s_{k}(x)}{(k+1)^a}=\frac{s_n(x)}{n^a}+\sum_{k=1}^{n-1}s_k(x)\left(\frac{1}{k^a}-\frac{1}{(k+1)^a}\right).
$$
Next observe that $s_n(x)$ is bounded for a fixed $x\ne 2k\pi$ and

$$
\frac{1}{k^a}-\frac{1}{(k+1)^a}=\frac{a}{k^{a+1}}+\mathcal O(k^{-a-2}),
$$
and hence the sequence
$$
\frac{s_n(x)}{n^a}+\sum_{k=1}^{n-1}s_k(x)\left(\frac{1}{k^a}-\frac{1}{(k+1)^a}\right).
$$
converges.

Title: Proving absolute convergence of a series containing parameter

Body:
I have the following series:

$$\sum_{n=1}^\infty \frac{\sin(nx)}{n^a}, a>0 $$ $$ 0<x<\pi$$

Now, I have to show for which values of $a$ the series absolutely converges and diverges.

I have figured out that the series converges for $a>1$:

$$\sum_{n=1}^\infty |\frac{\sin(nx)}{n^a}| \le \frac{1}{n^a}$$
which converges for $a>1$ so we proved the absolute convergence of our series but only for $a>1$

However, I have no idea how to prove that the series does not absolutely converge for $0<a\le1$.

I tried finding a lower bound for $|\sin(nx)|$ but all I could come up with is that it is greater than or equal to zero which does not mean anything. Could anyone shed some light?


问:
Finding integer solutions to a system of equations
I am solving a problem and I reached the following system: $$\begin{cases}2y=x+z\\z^2=yt\\x+t=37\\y+z=36\end{cases}.$$
I have to find only the integer solutions, if that helps. I would be very grateful if you could give me some tips on solving it.

答:
I started as so:

We have $$\begin{cases}2y=x+z\\z^2=yt\\x+t=37\\y+z=36\end{cases}.$$ and I labeled equation the equations from 1-4. By equation 1, we have that $z=2y-x$ and by equation 4 we have that $z=36-y$. We can now equate the both z equations and we have that $36-y=2y-x$. Solving this for $y$ we have that $y=12+\frac{x}{3}$. By equation 3 we have that $t=37-x$. As stated earlier we know what Fromm equation 1 that $z=2y-x$, so by equation 2 we have that $z^2 = (2y-x)^2 = (4y^2-4yx+x^2)$. Where $z^2=yt$, where $t=37-x$. Thus we have that $(4y^2-4yx+x^2)=(37-x)(y)$ where $y=12 +\frac{x}{3}$. By plugging this into the equation we have now, we have $\frac{x^2-144x}{9} +576 = \frac{-x^2}{3} + \frac{x}{3} +444$. To get rid of all of the fractions we can multiply both sides by 9 and get $x^2-144x +5184 = -3x^2+3x+3996$, and by simplifying this we have $4x^2-147x+1188=0$ which we can solve with the quadratic formula. We then get that $x=99/4, x=12$ since we only want integer solutions we take the $x=12$ solution and plug it into our $y=12+\frac{x}{3}$ and get that $y=16$. From earlier we had that $t=37-x$ so $t=25$. Lastly since $z=2y-x$ we have that $z=20$. So your integer solutions are $x=12,y=16,t=25,z=20$. I hope this process makes sense!


问:
Find all real polynomials $p(x)$ that satisfy $\sin( p(x) ) = p( \sin(x) )$

Find all real polynomials $p(x)$ that satisfy $\sin( p(x) ) = p( \sin(x) )$.

Is there an easy way to prove this?

答:
Assume that $p\in\Bbb R[x]\space\land\space\deg (p)\ge 2$. May assume that $\lim\limits_{x\to \infty} p(x) = \infty$. Then there exists an interval $[c,\infty)$ such that $p\colon [c, \infty)\to [d, \infty)$ is bijective.

Consider the values $n\pi$ ($n\ge n_0$)  in the interval $[d, \infty)$ and their preimages $x_n \in [c, \infty)$.

Now $p(\sin x_n) = \sin(p(x_n) ) = \sin (n\pi) = 0$

But we have
$$x_n  \to \infty \ \  \textrm{and} \\
 (x_{n+1}-x_n) \to 0$$

(the last limit because $\lim\limits_{x\to \infty} p'(x) = \infty$ ), and so $(\sin x_n)_{n\ge n_0}$ is dense in $[-1,1]$.

We conclude $p\equiv 0$ on $[-1,1]$, contradiction.

Therefore, $p$ is of degree $\le 1$.

Now use $\sin(p(0)) = p(\sin 0) = p(0)$, and this implies $p(0) = 0$.

We are left with $p(x) = a x$, so $\sin a x = a \sin x$. Taking the derivative we get $a \cos a x = a \cos x$. We see from here that the solutions are $a =-1, 0, 1$.


问:
Linear map $T: \mathbb{R^3} \to \mathbb{R^3}$ has a two dimensional invariant subspace
I am supposed to show $T: \mathbb{R^3} \to \mathbb{R^3}$ has both a 1 dimensional and 2 dimensional invariant subspace, using the simplest machinery possible.

My approach: I have considered the 3-degree characteristic polynomial of $T$. If it has 3 real roots, we are done. If it has only one real root, that will be an eigenvalue and hence it's eigenspace is invariant 1D subspace. The natural choice for the 2D invariant subspace seems to be the orthogonal complement of that subspace,  but I am unable to show that the image of that subspace under $T$ is still orthogonal to our 1D invariant subspace. Am I going in the right direction?

答:
Let $\lambda$ be the real eigenvalue, and consider the subspace $W=\text{Range}(T - \lambda I)$.

 Showing it is invariant: if $w \in W$, then it is of the form $w=(T-\lambda I)u$ for some $u \in \mathbb{R}^3$. Then $Tw=T(T-\lambda I) u = (T-\lambda I) T u \in W$.

 Showing it has dimension $2$: use the rank-nullity theorem.

Context: this subspace appears in the inductive proof of the Jordan form theorem. The real Jordan form of your matrix (in the case where the characteristic polynomial has one real root) will have two blocks of size 1 and 2 respectively, where the block of size two is the "real Jordan block" associated with the complex conjugate pair of complex eigenavlues.


问:
Is the Midpoint Rule the average of the lower and upper estimates of an integral?
Does the average of the lower estimate, $L_n$,  and upper estimate, $R_n$, of an integral give the same or very similar approximation of an integral when using the Midpoint Rule or is there no relevant connection?

Where  $f$ is increasing

$$L_n = \sum_{i=1}^n f(x_{i-1}) \Delta x$$

And

$$R_n = \sum_{i=1}^n f(x_{i}) \Delta x$$

Does

$$\sum_{i=1}^n f(\frac{x_{i-1} + x_{i}} {2}) \Delta x  =  \frac{L_n + R_n}{2}$$?

答:
It's not true that $$f\left(\frac{x_{i-1} + x_i}{2}\right) = \frac{f(x_{i-1}) + f(x_i)}{2}$$
unless $f$ is linear. So your last equation is not true:

$$M_n := \sum_{i=1}^n  f\left(\frac{x_{i-1} + x_i}{2}\right) \Delta x_i \neq \frac{L_n + R_n}{2}$$

In general, it's not even true that $M_n$ will be between $L_n$ and $R_n$. If the function is increasing, then you know that $L_n \leq M_n \leq R_n$.

Also, this is not the lower or upper integral. This the left end point and right end point Riemann sum. In general, one cannot say which of $L_n$, $R_n$, or $M_n$ gives a better approximation. However, if $f$ is integrable, then you know that

$$\lim_{n\to \infty} L_n = \lim_{n\to \infty} M_n = \lim_{n\to \infty} R_n = \int_a^b f$$


问:
Ordinals and Cardinals larger than the fixed points of $α↦ω_α$ and $α↦\aleph_α$?
Surely there is no limit to how high we can go, so how do we talk about ordinals and cardinals higher than the fixed points of the functions $α↦ω_α$ and $α↦\aleph_α$? Is the power set of the $\aleph$-fixed point even larger than the $\aleph$-fixed point itself? Or is the $\aleph$-fixed point so mind-bogglingly huge that it can't be rivaled by the same methods we have previously used to reach large cardinals?

答:
Cantor's theorem says that $\mathcal{P}(x)$ is always strictly greater than $x$; it doesn't matter whether $x$ is an $\aleph$-fixed point or not. Similarly, every cardinal has a successor cardinal which can be constructed just as we construct $\omega_1$ from $\omega$.

Note that any strictly increasing operation on cardinals, like the successor operation $\kappa\mapsto\kappa^+$, can get us out of the "fixed point trap." For example, there is no fixed point of the map $\alpha\mapsto\omega_\alpha^+$. So we can quite easily go well beyond the least $\aleph$-fixed point.


问:
Why is it that $a\equiv b \pmod{m} \iff m|(a-b)$
Why is it that $a\equiv b \pmod m \iff m\mid(a-b)$? I know this is a definition, but it's a definition I don't understand.
I know that if $m\mid(a-b)$ then there exists some $q\in\mathbb Z$ such that $mq=a-b$.

But when I take $a\equiv b \bmod m$ I obtain $b=mq+a$, solving for $mq$ I get $mq=b-a$ instead of $mq=a-b$.

Am I getting something wrong about what $a\equiv b\pmod m$ means? My understanding says $b$ is a dividend, $m$ is a divisor, and $a$ is a remainder. The devision algorithm says dividend${}=(\text{divisor}\times\text{quotient})+\text{remainder},$ so that's how I get $b=mq+a$.

答:
Actually, $a\equiv b\pmod m$ doesn't mean that $b$ is a dividend, $m$ is a divisor, and $a$ is a remainder. It means the the divisions of $a$ and $b$ by $m$ have the same remainders.

And if $q$ is such that $mq=a-b$, then $m\times(-q)=b-a$.


问:
How can I prove $\sum_{i=1}^n \sum_{j=1}^n \frac{a_ia_j}{i+j-1}$ is never negative for any set of n real numbers $a_i$
I can't figure this out, can someone help me prove it? I know you guys will come up with an incredibly elegant solution</p>

Body:
Hint: Since $\dfrac{1}{i+j-1} = \displaystyle\int_{0}^{1}x^{i+j-2}\,dx$, we have $$\sum_{i = 1}^{n}\sum_{j = 1}^{n}\dfrac{a_ia_j}{i+j-1} = \sum_{i = 1}^{n}\sum_{j = 1}^{n}\int_{0}^{1}a_ia_jx^{i+j-2}\,dx = \int_{0}^{1}\sum_{i = 1}^{n}a_ix^{i-1} \cdot \sum_{j = 1}^{n}a_jx^{j-1}\,dx$$

Can you see why this is non-negative?


问:
$0 \to G' \xrightarrow{\alpha} G \xrightarrow{\beta} G'' \to 0$ exact sequence, $G$ is torsion group if and only if $G'$ and $G''$ are torsion groups.
Let's work in abelian groups. We say a group $G$ is a torsion group if for every $g \in G$ there is a $n \geq 0$ such as $ng=0$. I want to prove that for an exact sequence

$$0 \to G' \xrightarrow{\alpha} G \xrightarrow{\beta} G'' \to 0,$$

$G$ is a torsion group if and only if $G'$ and $G''$ are torsion groups.

I cant prove this problem so I formulate an easier but similar problem.

Let's take the particular exact sequence

$$0 \to G' \xrightarrow{i} G \xrightarrow{\pi} \frac{G}{G'} \to 0,$$

where $i$ is the inclusion of the subgroup $G'$ of $G$ and $\pi$ is a projection. If $G$ is a torsion  group then it is inmediate $G'$ is also a torsion group. Also if $g+G' \in \frac{G}{G'}$,and as $g \in G$ where $G$ is a torsion group then there is a $n \geq 0$ such $ng=0$. Then $n(g+ G')=ng +G'=0+G$, proving $\frac{G}{G'}$ is a torsion group.

If $G'$ and $\frac{G}{G'}$ are torsion groups and lets take $g \in G$, then as $\pi(g)=g+G' \in \frac{G}{G'}$ so there is a $n \geq 0$ such $n(g+G')=ng+G'=0$, so $ng \in G'$ and $G'$ is torsion group so there is a $n' \geq 0$ such $n'(ng)==(n'n)g=0$. Proving $G$ is a torsion group.

Still cant generalize this problem to te original one. Any help solving the original proof with this or other aproach will be apreciated.

答:
Let $1\to G’\stackrel{\alpha}{\to}G\stackrel{\beta}{\to}G’’\to 1$ be an exact sequence of (abelian) groups. We want to show that $G$ is torsion if and only if both $G’$ and $G’’$ are torsion.

If $G$ is torsion, let $x\in G’$. Then $\alpha(x)$ is torsion, so there exists $n\gt 0$ such that $\alpha(x)^n = e$; then $\alpha(x^n)=e$, so $x^n\in\mathrm{ker}(\alpha)$. Therefore, $x^n=e$. This shows every element of $G’$ is torsion. If $y\in G’’$, then we know there exists $g\in G$ such that $\beta(g)=y$. Since $G$ is torsion, there exists $m\gt 0$ such that $g^m=e$. Then $y^m = \beta(g)^m = \beta(g^m) = \beta(e)= e$. Thus, every element of $G’’$ is torsion.

Now assume both $G’$ and $G’’$ are torsion, and let $x\in G$. Then $\beta(x)$ is torsion, so there exists $n\gt 0$ such that $e=\beta(x)^n = \beta(x^n)$. Thus, $x^n\in\mathrm{ker}(\beta)=\mathrm{Im}(\alpha)$. Hence, there exists $g\in G’$ such that $\alpha(g)=x^n$. Since $G’$ is torsion, there exists $m\gt 0$ such that $g^m=e$. Then
$$x^{nm} = (x^n)^m = (\alpha(g))^m = \alpha(g^m) = \alpha(e) = e.$$
Thus, $x^{nm}=e$, so $x$ is torsion.


问:
How to apply definite integral on $\sqrt{1+\sin x}$
I want to solve the following integral:

$$\int_0^\pi \sqrt{1+\sin x} \, dx$$

I want to do this with the substitution:

$$u=\sin x$$

so that:

$$du=\cos x\,dx=\sqrt{1-\sin^2x}\,dx = \sqrt{1-u^2} \, dx$$

$$dx = \frac{du}{1-u^2}$$

$$u(0) = 0, \quad u(\pi/2)=1, \quad u(\pi)=0$$

$$\int\sqrt{1+\sin x} \, dx = \int\frac{1}{\sqrt{1-u}} \, du = -2\sqrt{1-u}$$

So since $u$ isn't defined for 1 (corresponding $x$ is $\pi/2$), I split the integration to ($0, \pi/2$) and ($\pi/2, \pi$)
However when I assign the substitution I get:

$$\int_0^1 \frac{1}{\sqrt{1-u}} \, du + \int_1^0 \frac{1}{\sqrt{1-u}} \, du = 2-2 = 0$$

But the result should be $4.$ It seems like I should flip the sign here for some reason but I can't figure it out, help would be appreciated.

答:
"I want to specifically know why my substitution fails".  (OP quoted from a comment on a now-deleted answer.  I agree that it would have been nice for OP to actually put his question in the Question.)

$$  \cos x = \sqrt{1 - \sin^2 x}  $$
is only true when $\cos x \geq 0$ (because the radical on the right-hand side is always nonnegative).  This fails for $x \in (\pi/2, \pi]$.  In that interval, you need the other choice of sign of the radical , which gives your missing minus sign.


问:
Sum of squared/cubed combinations
I was wondering if there is a closed formula for sum of cubed combinations. More precisely, I'd like to compute $$\sum_{k=1}^n \left ( \begin{array}{c}n\\k\end{array}\right )^3$$

Obviously, without the "$^3$", the sum is $2^n-1$....

Any idea ?

thanks

答:
These are almost the same as the Franel numbers - just replace the lower limit on $k$ with 0 instead of 1 ($n\choose 0$ is always 1, so this is just a simple shift of the resulting sequence). While there is a closed form expression for the Franel numbers, it requires a generalised hypergeometric function, so it isn't enormously satisfying:

$$
\sum_{k=0}^n {n \choose k}^3 = {}_3F_2 (-n,-n,-n;1,1;-1)
$$

Title: Sum of squared/cubed combinations

OwnerDisplayName: Martin O'Leary


问:
Chang & Keisler Exercise 4.3.9
I'm having a lot of trouble sorting out an exercise from Chang & Keisler's Model Theory:

4.3.9: Let $I$ be an infinite set of power $\alpha$. If $E \subset P(I)$, $|E| \leq \alpha$, and the filter generated by $E$ is uniform, then $E$ can be extended to an $\alpha$-regular ultrafilter $D$ over $I$.

In chapter 4.3, CK proves that for any infinite $\alpha$, there exists an $\alpha$-regular ultrafilter on $\alpha$. Their construction involves working with the set of finite subsets of $\alpha$ and I tried mimicking that argument, but I'm not sure how to employ the given hypotheses. Any and all hints / suggestions are appreciated.

答:
A modification of the proof used in Chang and Keisler can be used to build a regularizing family $E_i$ which is compatible (ie has the finite intersection property) with the uniform filter created by extending $E$ as given in the hypothesis of the problem. Any extension of an $\alpha$-regular filter is $\alpha$-regular so this concludes to problem. Uniformity is used to ensure that the regularizing family is indeed compatible since otherwise the construction would fail. This can also be seen by the fact that every regular filter is uniform. Here is my proof:

Denote the set of finite subsets of $I$ by $P_\omega(I)$. Let $f:I\rightarrow{} S_\omega(I)$ be an arbitrary surjection. For all elements $i\in I$ define the sets
$E_i=\{j\in I: i\in f(j)\}$
Now it is clear that the family of sets $\{E_i\}$ for all $i\in I$ has the finite intersection property since

$E_i\cap E_k=\{j\in I: i\in f(j)\wedge k\in f(j)\}\neq \emptyset$

because there exists a finite sequence in $P_\omega(I)$ which contains any pair $i,k$.

Now if we consider the uniform filter $E'$ generated by $E$ given at the start of the problem, we know that $E'$ has the finite intersection property by way of already being a filter. If we can show that $E'\cup\{E_i\}$ has the finite intersection property then the union can be extended into a regular ultrafilter over $I$ by the ultrafilter lemma.
Since $E'$ and $\{E_i\}$ have the finite intersection property within their respective families it remains to be shown that any intersection between $A\in E'$ and $E_i$ for $i\in I$ is nonempty for arbitrary $A\in E'$ and $i\in I$.

Before showing this intersection is non empty recall that the function $f$ which defined the $E_i$ was arbitrary. We must now define it so that the finite intersection property holds.
Enumerate $A\in E'$ as $A_{\beta}, \beta<\alpha$ as well as every element within $A_\beta$ as $\lambda_\beta<\alpha$.
Likewise enumerate the elements of $P_\omega(I)$ as $\gamma$ with $\gamma<\alpha$. Now we build $f$ inductively by starting with a series of partial functions $g_i$ which are then extended to partial functions $f_i$ which is extended to $f$.

Let $g_{0_0}$ map $0_0\in A_0$ to $\gamma=0$. Let $g_{0_1}$ map $0_1\in A_1$ to $\gamma=0$ if $A_0$ does not contain $\gamma=0$ . Likewise let $g_{0_\beta}$ map $0_\beta\in A_\beta$  to $\gamma=0$ if $_\beta$ does not contain $0$ otherwise do nothing. Let $f_0=\cup g_0$. Now repeat the procedure with $g_{1_\beta}$ taking the union to get $f_1$ and so on to get $f_i$ finally get $f$ which is then defined upon every element in $I$. The reason this can be done is that $E'$ is uniform and that every set $A_\beta$ is of size $\alpha$. This ensures that there is always "room" to extend the function since otherwise would amount to an element in the filter of size less than $\alpha$. $f$ has the special property that every set $A_0$ is surjective onto the domain $P_\omega(I)$. Thus the for any set $A\in E'$ and $E_i$, $A\cap E_i$ is nonempty. Therefore as desired the family of sets $E'\cup \bigcup_{i\in I }E_i$ has the finite intersection property and a regularizing family of size $\alpha$. This ensures that it can be extended to a $\alpha$-regular ultrafilter as desired.

Title: Chang & Keisler Exercise 4.3.9


问:
What positive integer $n$ maximizes the function $f(n) = \sigma_0(n)/n$?
A friend and I were having a discussion on which base would be best. I argued that 12 would be the best because it has the most divisors relative to its size. However, I'm not sure that 12 is actually the number that maximizes this ratio. To investigate, I formalized my observation by claiming that 12 maximizes the function $f(z) = \sigma_0(z)/z$ where $\sigma_0(n) = \sum_{d|n} d^0$ is the function which counts the divisors of $n$. I found some articles and some interesting properties of $\sigma_0$ but nothing which I was able to use to prove this property. I'm not too familiar with this kind of thing so I wasn't sure exactly how to go about it.

Does anyone have any idea as to how one could prove this? Right now, it seems like the formula which would be most useful would be that $$\sigma_o(n) = \Pi_{i = 1}^{\omega(n)}(1 - a_i)$$ where $\omega(n)$ is the number of distinct prime factors of $b$ so that $n = \Pi_{i = 1}^{\omega(n)}p_i^{a_i}$.

Thank you in advance!

EDIT: On thinking about it a little more, it seems like 12 definitely does not maximize this. For example, 6 has 4 divisors whereas 12 has 6 of them. As a commenter also pointed out, 3 has 2 divisors. The best does seem to be 2, though, with two divisors. If $\sigma_0(n) = n$, then for all $m \leq n$, we would have that $m|n$. That would imply that every prime less than $n$ would be included in the prime factorization of $n$. This is a fairly strong property that I suspect only 2 holds.

答:
From this answer, we know that

$$\sigma_0(n)\leq n^{\frac{1.0660186782977...}{\log \log n}}<n^{ \frac{2}{\log \log n}}$$

(with equality at $n=6983776800$). This then implies that

$$\frac{\sigma_0(n)}{n}<n^{ \frac{2}{\log \log n}-1}$$

Now, it is easy to see that for $n\geq 1619$ we have

$$\frac{2}{\log \log n}-1<0$$

Then for $n\geq 1619$ we know

$$\frac{\sigma_0(n)}{n}<n^{ \frac{2}{\log \log n}-1}<n^0=1$$

But

$$\frac{\sigma_0(1)}{1}=\frac{\sigma_0(2)}{2}=1$$

Now, we only have to check all integers $3\leq n\leq 1618$. These are easily checked and we conclude that the function is maximized at $n\in\{1,2\}$.

EDIT: If you wanted the case $n\geq 3$, then in much the same manner we see that for $n\geq 2880$ we have

$$n^{\frac{2}{\log \log n}-1}<\frac{3}{4}$$

Then after checking all integers $5\leq n\leq 2879$ we may conclude that the function is maximized at $n=4$ where

$$\frac{\sigma_0(4)}{4}=\frac{3}{4}$$


问:
What is the sum of:
$\Sigma_{n=0}^{\infty}\Sigma_{m=0}^{\infty} \frac{1}{(m-n)}$

How to treat this double sum?

答:
For all $n$, $$\sum_{m=0,\\ m\ne n}^\infty\frac1{m-n}=\sum_{m=0}^{n-1}\frac1{m-n}+\sum_{j=1}^\infty\frac1j=\infty$$

Therefore the double sum $\sum_{n=0}^\infty\sum_{m=0,\\ m\ne n}^\infty\frac1{m-n}$ is, in the elementary sense, $\infty$.

Title: What is the sum of:


问:
Conditions for level sets to be diffeomorphic to each other
I have a map $\pi:\mathbb R^d\longrightarrow\mathbb R$ which is Lipschitz continuous and $\|\nabla\pi\|>0$ almost everywhere. I know also that the level set $\pi^{-1}(0)$ is compact. Can I conclude that the level sets $\pi^{-1}(c)$ are diffeomorphic for all $c\geq0$?

If not, what are the additional conditions for which I can conclude it?

Thank you

答:
As Ted Shifrin noted, you certainly have to assume there are no criticial values in $[0,\infty)$, because typically the topology of the level sets changes when you cross a critical value. But that's not sufficient.

Here's a counterexample: Consider the function $\pi\colon \mathbb R^2\to\mathbb R$ given by
$$ \pi(x,y) = \frac{ x^2 + y^2 -1}{x^2 + 1}.$$
The only criticial point of $\pi$ is the origin, which is on the level set $\pi^{-1}(-1)$, so there are no critical values in $[0,\infty)$. And $\pi^{-1}(0)$ is the unit circle, which is compact. However, $\pi^{-1}(1)$ is the pair of lines $y=\pm\sqrt{2}$, and the level sets $\pi^{-1}(c)$ for $c>1$ are hyperbolas.

One sufficient condition to guarantee that the level sets are diffeomorphic is that $\pi$ be an exhaustion function, meaning the preimage of every compact interval $[0,b]$ is compact, or equivalently that $\pi(x)\to +\infty$ as $|x|\to \infty$. The proof that the level sets are diffeomorphic in that case is (a slight modification of) a standard argument in Morse theory: Define a vector field $X = \operatorname{grad} \pi/|\operatorname{grad}\pi|^2$, and for each $t\in [0,\infty)$ let $\phi_t\colon \mathbb R^n\to \mathbb R^n$ be the flow of $X$ (wherever it's defined). A computation shows that $\pi(\phi_t(x)) = \pi(x)+t$ whenever $\phi_t(x)$ is defined. Thus as long is $t$ is in a compact interval $[0,b]$, the flow stays in the compact set $\pi^{-1}([0,b])$, which implies that the flow continues to exist for all $t\ge 0$. (This is a consequence of the Escape Lemma, which says roughly that any maximal integral curve that doesn't exist for all positive time must escape from every compact set.)  Therefore $\phi_t$ maps $\pi^{-1}(0)$ to $\pi^{-1}(t)$ diffeomorphically.

Edit: I just realized that to make this argument work, you probably need to assume more regularity than just Lipschitz. In order for the flow of $X$ to be defined, you should assume $\pi$ is locally $C^2$, or at least locally $C^{1,1}$ (continuously differentiable with locally Lipschitz derivatives).


问:
Operator Norm of an augmented matrix vs unaugmented matrix
This problem comes from Linear Algebra and its Applications by David C Lay and Steven R Lay.

Show that $$||[A B]|| \geq ||[A]||$$

Where $[A B]$ is a a matrix $A$ augmented with another matrix $B$
So if A = $$A = \begin{bmatrix}
a & b \\
c & d\\
\end{bmatrix}$$ and $$B = \begin{bmatrix}
e & f \\
g & h\\
\end{bmatrix}$$

Then $$[A B] = \begin{bmatrix}
a & b & e & f \\
c & d & g & h\\
\end{bmatrix}$$

My attempt:

We know that the operator norm of a matrix is the square root of the largest eigenvalue of the matrix. What's really throwing me off here is that what if by augmenting the matrix $B$, the eigenvalues of $[A B]$ are less than that of just $A$.

I think the approach I should be going for starts with an SVD approach(?)
$$[AB] = U_{AB}\Sigma_{AB}V^T_{AB}$$
$$[A] = U_{A}\Sigma_{A}V^T_{A}$$

From there I need to prove that the largest singular value of $[AB]$ is greater than that of $[A]$. Is this correct? Can someone share some insight?

答:
You want to use the actual definition of the operator norm, which is
$$ \|[AB]\| = \sup\{\|[AB]v\| : v\in \mathbb{R}^4 \text{ and } \|v\|=1\}$$
$$ \|A\| = \sup\{\|Av\| : v\in\mathbb{R}^2\text{ and } \|v\|=1\}.$$
Given any vector $(x,y)\in \mathbb{R}^2$, you have the vector $(x,y,0,0)\in \mathbb{R}^4$ with $[AB](x,y,0,0)$ = $A(x,y)$.
From this, the set that $\|[AB]\|$ is the sup of contains the set that $\|A\|$ is the sup of.


问:
Scale Space - Scales and Octaves
So I'm desperately trying to understand scale space for signals, specifically for 2D images...

I'm having trouble with algorithms that discuss creating a pyramid. Specifically, I don't understand how we go between resolutions (octaves). Within in a pyramid, I realize that an octave is a difference in resolution (or subsampled) but I don't understand how it relates to the scale parameter. (I understand that smoothing enough times is equivalent to subsampling).

For example, if we have five signals/images in an octave which have iteratively been smoothed with a Gaussian, how do the images at the edges relate to each other? (Image 5 and Image 6). Is Image 6 subsampled from Image 1 or from Image 5?

What about the scale parameter, discussed often throughout many articles on Scale Space, how does it relate to subsampling?

I've been reading papers all day, my head is swimming and I'm still confused.

From Wikipedia, I'm having intense trouble understanding this, but answers what I'm trying to understand:

An image pyramid is a discrete representation in which a scale space is sampled in both space and scale. For scale invariance, the scale factors should be sampled exponentially, for example as integer powers of 2 or root 2. When properly constructed, the ratio of the sample rates in space and scale are held constant so that the impulse response is identical in all levels of the pyramid. Fast, O(N), algorithms exist for computing a scale invariant image pyramid in which the image or signal is repeatedly smoothed then subsampled. Values for scale space between pyramid samples can easily be estimated using interpolation within and between scales.

答:
I found a great book discussing this topic thanks to a coworker:

Computer Vision: A Modern Approach by D. A. Forsyth and J. Ponce

There's a section labeled:

Linear Filters - Technique: Scale and Image Pyramids

Applying concepts from the book with the wikipedia article explaining the scale parameter helped... The whole goal here is to prevent aliasing.

Title: Scale Space - Scales and Octaves


答:
First note that $\displaystyle \frac{\sigma_0(n)}{n} = \prod_p \frac{\alpha_p+1}{p^{\alpha_p}}$ where $\alpha_p \ge 0$.

But $\displaystyle\frac{\alpha_p+1}{p^{\alpha_p}} < 1$ for all prime $p$ and all $\alpha >0$ with the only exception of $p = 2$ and $\alpha = 1$, which means that the maximum is achieved when all the $\alpha$'s are $0$ ($n = 1$) or when all but $\alpha_2$ are $0$ and $\alpha_2 = 1$ (n = 2).

Title: What positive integer $n$ maximizes the function $f(n) = \sigma_0(n)/n$?

Body:
A friend and I were having a discussion on which base would be best. I argued that 12 would be the best because it has the most divisors relative to its size. However, I'm not sure that 12 is actually the number that maximizes this ratio. To investigate, I formalized my observation by claiming that 12 maximizes the function $f(z) = \sigma_0(z)/z$ where $\sigma_0(n) = \sum_{d|n} d^0$ is the function which counts the divisors of $n$. I found some articles and some interesting properties of $\sigma_0$ but nothing which I was able to use to prove this property. I'm not too familiar with this kind of thing so I wasn't sure exactly how to go about it.

Does anyone have any idea as to how one could prove this? Right now, it seems like the formula which would be most useful would be that $$\sigma_o(n) = \Pi_{i = 1}^{\omega(n)}(1 - a_i)$$ where $\omega(n)$ is the number of distinct prime factors of $b$ so that $n = \Pi_{i = 1}^{\omega(n)}p_i^{a_i}$.

Thank you in advance!

EDIT: On thinking about it a little more, it seems like 12 definitely does not maximize this. For example, 6 has 4 divisors whereas 12 has 6 of them. As a commenter also pointed out, 3 has 2 divisors. The best does seem to be 2, though, with two divisors. If $\sigma_0(n) = n$, then for all $m \leq n$, we would have that $m|n$. That would imply that every prime less than $n$ would be included in the prime factorization of $n$. This is a fairly strong property that I suspect only 2 holds.


问:
prove that $U(\frac{1}{f},P)-L(\frac{1}{f},P)\leq \frac{1}{c^2}(U(f,P)-L(f,P))$ where $c$ is a lower bound for $f$
Let $f: [a,b]\to \mathbb{R}$ bounded below by $c>0$, $U$  denote the upper sum and $L$ the lower sum.
Prove that
$$U(\frac{1}{f},P)-L(\frac{1}{f},P)\leq \frac{1}{c^2}(U(f,P)-L(f,P))$$
for any partition $P$.

答:
HINT:
For any $x,y$,

$$ \left|\frac{1}{f(x)} - \frac{1}{f(y)}\right| =  \frac{|f(x) - f(y)|}{|f(x)f(y)|}
 \leq \frac{|f(x) - f(y)|}{c^2}$$

You should be able to continue now.


问:
Prove $2 < (1 + \frac{1}{n})^n$ for all $n \ge 2$
Prove $2 < (1 + \frac{1}{n})^n$ for all $n \ge 2$.  I believe this will come out by induction.  It looks so much like induction, but I can't seem to find enough algebra to solve it.  The base case: for $n=2$, we get $(1 + \frac{1}{2})^2 = 2 + \frac{1}{4} > 2$. Suppose now there is a $k \ge 2$ such that $(1 + \frac{1}{k})^k > 2$.  I must show $(1 + \frac{1}{k+1})^{k+1} > 2$.  From this, I tried to undo the exponent as in

\begin{align*}
\left(1 + \frac{1}{k+1}\right)^{k+1} &= \left(1 + \frac{1}{k+1}\right)^k \left(1 + \frac{1}{k+1}\right)
\end{align*}

Now I'm looking for a way to use $(1 + \frac{1}{k})^k > 2$, which is true by hypothesis.  But it doesn't seem obvious how to turn a fraction $\frac{1}{k+1}$ into $\frac{1}{k}$.

By the way, I know $(1 + \frac{1}{n})^n$ approaches $e$ as $n$ approaches infinity, but I'm trying not to use any related identities to solve this problem.  Can we do it by induction?  What is the algebra that I'm missing there?  Any hints?  Thank you!

Thanks.  Lots of good answers down there.  I selected what seemed to be the most straightforward one.  Though it looked to me like induction, it seems easier to apply the binomial theorem as @jjagmath kindly pointed out.

Solution. Considering just the first two terms of the binomial theorem expansion, we get

\begin{align*}
   (1 + 1/n)^n &= 1 + n (1/n) + ... + C(n,n) 1/n^n\\
               &= 2 + ... + C(n,n) 1/n^n\\
               &> 2.
\end{align*}

We must restrict $n \ge 2$ because $n=1$ fails with $3/2 > 2$.

答: Hint: It's probably easier to prove the more general fact that $(1+x)^n>1+nx$ for all $n \ge 2$ and all $x>0$. You could do this by induction, or by using the binomial theorem.


问:
Can the map sending a presentation to its group be considered as a functor?
It is well-known that the functor $Grp \to Set$ sending a group $G$ to its underlying set $UG$ has a left adjoint, the functor $Set\to Grp$ sending a set $X$ to the free group $FX$.

I wonder whether one can consider the following modification of the functor $F$: to each presentation $\langle S\mid R\rangle$ we assign the group $FS/N$, where $N$ is the normal subgroup induced by $R$.

Is this a functor?

For this to be true, there need to be a category of presentations. Is there such a category?

Does the functor $\langle S\mid R\rangle\mapsto FS/N$ have a right adjoint (maybe similar to $U$)?

Of course, my questions are not precise, because I haven't specified what the morphisms between presentations are. The question is if one can do all this in a sensible way.

答:
Yes, yes, yes, to all three questions.

And it can be done very generally and very nicely for universal algebra $\ $ [or even for first order structures].

Let's fix an algebraic signature consisting of named finitary operation symbols, and a variety $\mathcal{Alg}$ of algebras of that signature that we're working on. (In your question, the groups.)

As in your question, we use $F:\mathcal{Set}\to\mathcal{Alg}$ for the free functor and $U:\mathcal{Alg}\to\mathcal{Set}$ for the underlying set functor.

A presentation $\langle A,E\rangle$ is a pair where $A$ is a set and $E$ is a subset of equations, i.e. pairs of terms (in other words $E\subseteq F(A)^2$) $\ $ [or, $E$ is a set of atomic formulas].

In the case of groups, an equation can be arranged on one side, leaving the identity on the other side, so then $E\subseteq F(A)$ is enough, as learned.

We can define a morphism of presentations $\langle A,E\rangle\to\langle B,\Sigma\rangle$ to simply be a function $f:A\to B$ such that
$$(\tau,\sigma)\in E\ \implies\  (Ff(\tau),\,Ff(\sigma))\in\Sigma\,.$$

By an interpretation of the presentation $\langle A,E\rangle$ in an algebraic structure $\mathfrak B$ we mean a function $h:A\to U\mathfrak B$ such that all elements of $E$ are evaluated equal $\,$ [all atomic formulas of $E$ are true], i.e. for every $(\tau,\sigma)\in E$ we have
$$\varepsilon_{\mathfrak B}\left( Fh(\tau)\right) \ =\ \varepsilon_{\mathfrak B}\left( Fh(\sigma)\right)\ \text{ that is, }\
(\tau,\sigma)\in\ker(\varepsilon_{\mathfrak B}\circ Fh)$$
where $\varepsilon:FU\to 1_{\mathcal{Alg}}$ is the counit of the free-forgetful adjunction.

Observe that we can compose morphisms of presentations with interpretations as well as with homomorphisms, in an associative way, since all are just functions.

So this actually connects up the category of presentations $\mathcal{Pres}$ with the category of algebras $\mathcal{Alg}$ by (so called hetero-)morphisms between them, yielding to a bigger category, containing both $\mathcal{Pres}$ and $\mathcal{Alg}$ as a full subcategory.

Such an animal is called a profunctor, and it is determined by the restriction of its hom functor to the heteromorphisms, i.e. of a functor of the form $\mathcal{Pres}^{op}\times\mathcal{Alg}\to\mathcal{Set}$.

Now, every presentation $\langle A,E\rangle$ has a reflection in $\mathcal{Alg}$, namely $FA/(E)$ where $(E)$ is the congruence relation generated by $E$.

This defines the free functor on presentations.

Conversely, every algebraic structure $\mathfrak B$ has a coreflection in $\mathcal{Pres}$, namely $\langle U\mathfrak B,\Delta_{\mathfrak B}\rangle$ where $\Delta_{\mathfrak B}$ is the set of all equations that can be written on letters of $U\mathfrak B$ and hold in $\mathfrak B$ $\,$ [the positive diagram of $\mathfrak B$].

This defines the right adjoint.

Note that this is a fully faithful functor.


答: Using Bernoulli's Inequality we can observe that $(1+x)^n>1+nx \implies$ $(1+\frac{1}{n})^n>1+\frac{1}{n}\cdot n=1+1=2$

Body:
Prove $2 < (1 + \frac{1}{n})^n$ for all $n \ge 2$.  I believe this will come out by induction.  It looks so much like induction, but I can't seem to find enough algebra to solve it.  The base case: for $n=2$, we get $(1 + \frac{1}{2})^2 = 2 + \frac{1}{4} > 2$. Suppose now there is a $k \ge 2$ such that $(1 + \frac{1}{k})^k > 2$.  I must show $(1 + \frac{1}{k+1})^{k+1} > 2$.  From this, I tried to undo the exponent as in

\begin{align*}
\left(1 + \frac{1}{k+1}\right)^{k+1} &= \left(1 + \frac{1}{k+1}\right)^k \left(1 + \frac{1}{k+1}\right)
\end{align*}

Now I'm looking for a way to use $(1 + \frac{1}{k})^k > 2$, which is true by hypothesis.  But it doesn't seem obvious how to turn a fraction $\frac{1}{k+1}$ into $\frac{1}{k}$.

By the way, I know $(1 + \frac{1}{n})^n$ approaches $e$ as $n$ approaches infinity, but I'm trying not to use any related identities to solve this problem.  Can we do it by induction?  What is the algebra that I'm missing there?  Any hints?  Thank you!

Thanks.  Lots of good answers down there.  I selected what seemed to be the most straightforward one.  Though it looked to me like induction, it seems easier to apply the binomial theorem as @jjagmath kindly pointed out.

Solution. Considering just the first two terms of the binomial theorem expansion, we get

\begin{align*}
   (1 + 1/n)^n &= 1 + n (1/n) + ... + C(n,n) 1/n^n\\
               &= 2 + ... + C(n,n) 1/n^n\\
               &> 2.
\end{align*}

We must restrict $n \ge 2$ because $n=1$ fails with $3/2 > 2$.


答:
Just use the Newton binomial theorem:
$(1+\frac{1}{n})^n = 1 + \binom{n}{1}\frac{1}{n} + \cdots > 2$

Title: Prove $2 < (1 + \frac{1}{n})^n$ for all $n \ge 2$


答:
$$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+...$$
and so:
$$\left(1+\frac1n\right)^n=1+1+\frac1{2!}\frac{n-1}{n}+...$$
and so you can see that if $n=1$ then the value is $2$ and for $n\ge2$ there are non-zero terms after and so it has to be greater than two


问:
Let $X_k$ be a sequence of IID random variable with $E(X_k)=0$ and finite variance. Show that $\frac{S_n}{n^p} \xrightarrow{a.s.} 0$ for p > 1/2
My idea to this question is
$\frac{S_n}{n^p}=\frac{S_n}{\sigma\sqrt{n}}\cdot\frac{\sigma}{n^{p-1/2}}$

By CLT,
$\frac{S_n}{\sigma\sqrt{n}} \xrightarrow{d} N(0,1)$

Also, it can be shown
$\frac{\sigma}{n^{p-1/2}} \xrightarrow{d} 0$

Therefore,
$\frac{S_n}{\sigma\sqrt{n}}\cdot\frac{\sigma}{n^{p-1/2}}=\frac{S_n}{n^p} \xrightarrow{d} 0$

Since it converges to a constant, it implies
$\frac{S_n}{n^p} \xrightarrow{p} 0$

We know if $\frac{S_n}{n^p} \xrightarrow{p} 0$, then there is a subsequence $\frac{S_{n_{k}}}{n_k^p} \xrightarrow{a.s.} 0$.

My question is how to show such subsequence $\frac{S_{n_{k}}}{n_k^p}\xrightarrow{}  \frac{S_n}{n^p}$ ?

答:
Hmmm, it's very weird that some computational trick doesn't work on this. So let's do it this way.

Demonstration

For simplicity, I'll call out a function $h$ from $\mathbb{N}$ taking values in $\mathbb{N}$ with some specific properties without explicitly specifying it.

Let $h$ be a function from $\mathbb{N}$ to $\mathbb{N}$ such that:

$h$ is strictly increasing, thus $h(n) \ge n$

$ \sum_{n \ge 1} \dfrac{1}{ h(n)^{2p-1}} < +\infty$

$  \dfrac{h(n)}{h(n+1)} > a$ for all $n$ for some $a$ positive

(A candidate for $h$ is $h(n)=n^q$ for $q$ big enough)

Also, for simplicity, WLOG: $\sigma =1$

By Doob's, we have
$$ \mathbb{E}\left( \max_{  m \le h(n)} |S_{m}|^2 \right) \le 4\mathbb{E}( (S_{h(n)})^2 )= 4h(n)$$
Thus,
$$ \sum_{n \ge 1} \mathbb{E}\left( \dfrac{1}{h(n)^{2p}}\left( \max_{  m \le h(n)} |S_{m}|^2 \right) \right) \le \sum_{n \ge 1} \dfrac{4}{h(n)^{2p-1}} < +\infty$$

Thus,
$$ \lim_{n \rightarrow +\infty} \dfrac{1}{h(n)^{p}}\max_{   m \le h(n)} |S_{m}| = 0 \quad a.e \quad (**)$$

As $h(m)> a h(m+1)$, we have that for all $n$:
$$  \dfrac{1}{n^{p}} |S_n| \le \dfrac{1}{h(g(n)-1)^p} \max_{   k \le h(g(n))}  |S_{k}| \le \dfrac{1}{a^p} \dfrac{1}{ h(g(n))^p} \max_{   k \le h(g(n))}  |S_{k}| $$

Where $g(n)$ is defined as the smallest number such that $ h(g(n)-1 ) \le m \le h(g(n))$.

As $h$ increases to infinity, $g$ must also increase to infinity.
Hence, conclusion $\square$

Comments

The same method can be applied to prove that $\dfrac{S_n}{ n^{1/2}\ln(n)^s } \longrightarrow 0 \quad a.e$ for all $s>1/2$

If $(X_i)$ are gaussian, we can even have a tight bound for $\limsup |S_n|$

Many conditions in the statement can be relaxed.

If I remember correctly, the rate when $(X_i)$ are Gaussian is $O( \dfrac{1}{ n^{1/2} \sqrt{ ln(ln(n))} } )$


问:
Kernel and rank of sum of finite linear operators
How do you think this excercise can be done?

Let $U,V$ finite dimensional vector spaces and $A,B:U\to V$ linear transformations. Then $$\mathrm{dim}(\mathrm{Ker}(A+B))\le\mathrm{dim}(\mathrm{Ker}(A)\cap\mathrm{Ker}(B))+\mathrm{dim}(\mathrm{Ran}(A)\cap\mathrm{Ran}(B))$$

I know that $\mathrm{Ker}(A)\cap\mathrm{Ker}(B)\subseteq\mathrm{Ker}(A+B)$, so i was thinking take a basis for $\mathrm{Ker}(A)\cap\mathrm{Ker}(B)$ and extend it to a basis for $\mathrm{Ker}(A+B)$ and extend it to a basis for $U$, but i don't get a good approach.

答:
Note that if $x \in \ker(A + B)$ (i.e. $x$ is such that $(A + B)x = 0$), then one of the following statements must hold:

$Ax = 0$, which means we must also have $Bx = 0$,

$Ax \neq 0$, in which case we observe that $Ax = -Bx \in \operatorname{ran}(B)$.

Per the hint, consider $A|_{\ker(A + B)}$. Let $x_1,\dots,x_j$ be a basis of $\ker(A|_{\ker(A + B)}) = \ker(A) \cap \ker(B)$, and extend this to a basis, $x_1,\dots,x_k$ of $A|_{\ker(A + B)}$. The vectors $Ax_{j+1},\dots,A x_k$ form a linearly independent subset of $\operatorname{ran}(A) \cap \operatorname{ran}(B)$. Thus, it must hold that $k - j \leq \dim \operatorname{ran}(A) \cap \operatorname{ran}(B)$.

With that, we can reach the desired conclusion:
$$
\begin{align}
\dim \ker(A + B) &= k = \dim \ker(A) \cap \ker(B) + (k - j)
\\ & \leq \dim \ker (A) \cap \ker(B) + \dim \operatorname{ran}(A) \cap \operatorname{ran}(B).
\end{align}
$$


问:
Minimizing the free energy
$\DeclareMathOperator{\tr}{tr}$
Let $H$ be a self-adjoint matrix and define the free energy as
$$
F(\Gamma)=\tr(H\Gamma+\Gamma \log \Gamma +(1-\Gamma) \log(1-\Gamma))
$$
where $\Gamma$ satisfies $0\le \Gamma \le 1$. I have seen physics papers suggest that $F(\Gamma)$ is minimized if we take
$$
\Gamma =\frac{1}{e^H +1}
$$
but I can't think of a rigorous proof.

EDIT. The "usual" non-rigrous method of proof would be to use variational method, i.e.,
\begin{align}
\delta F &=\tr\left(H\delta\Gamma+\log\left(\frac{\Gamma}{1-\Gamma}\right)\delta\Gamma \right)\\
0&=H+\log\left(\frac{\Gamma}{1-\Gamma}\right)\\
\Gamma &= \frac{1}{e^H+1}
\end{align}
How would one make this argument rigorous?

EDIT 2. I just realized that $F(\Gamma)$ is convex, so I think that the above calculation is almost rigorous, except that $\delta\Gamma$ may not commute with $\Gamma$, as pointed out by @Sangchul Lee.

EDIT 3. Thank you @Sangchul Lee for the proof. I think I also found another proof, that may be a little more straightforward.

Let $\Gamma_0=1/(e^H+1)$ and let $0\le \Gamma_1 \le 1$. Let $\Gamma(t)=(1-t)\Gamma_0 +t\Gamma_1 =\Gamma_0 +t\Delta$ where $\Delta = \Gamma_1-\Gamma_0$. Let $f(x)=x \log x +(1-x) \log (1-x)$. Hence,
$$
F(\Gamma)-F(\Gamma_0) = \tr(tH\Delta)+\tr (f(\Gamma)-f(\Gamma_0))
$$
Since $f$ is convex, we can apply Klein's inequality and see that
$$
F(\Gamma)-F(\Gamma_0) \ge \tr(t\Delta (H+\log\Gamma_0 -\log (1-\Gamma_0))=0
$$
Also since $f$ is strictly convex, we see that $\Gamma_0$ is the unique global min.

By the way, not sure why this question was voted to close. Please vote to reopen if you think otherwise.

EDIT 4. After further looking into the proof of Klein's inequality, there is a subtle "problem" of taking the derivative of a trace function (which is not fully explained in the wikipedia article). @Sangchul Lee deals with this explicitly for this particular case, but I would like to think that it should be able to be done more generally. Hence, I posted another question here.

答:
In this answer, we will write

$$ D^+_{B}F(A) := \lim_{\epsilon \to 0^+} \frac{F(A+\epsilon B) - F(A)}{\epsilon} $$

whenever the limit exists. Then the following lemma will be useful:

Lemma. We have
$$ D^+_{B}\exp(A) = \int_{0}^{1} e^{sA}Be^{(1-s)A} \, \mathrm{d}s. $$

Proof. We have

$$ D^+_{B}\exp(A)
= \sum_{n=1}^{\infty} \frac{1}{n!} D^+_{B}(A^n)
= \sum_{n=1}^{\infty} \frac{1}{n!} \sum_{k=0}^{n-1} A^k B A^{n-1-k}
= \sum_{k,l\geq 0} \frac{A^k B A^l}{(k+l+1)!}. $$

Then the claim follows from the beta integral $\int_{0}^{1} u^k(1-u)^l \, \mathrm{d}s = \frac{k!l!}{(k+l+1)!} $. $\square$

Now let $\Gamma_0$ and $\Gamma_1$ be self-adjoint matrices such that $0 \leq \Gamma_k \leq 1$ for $k = 0, 1$. Interpolate $\Gamma_0$ and $\Gamma_1$ by letting

$$ \Gamma_t = (1-t)\Gamma_0 + t\Gamma_1, \quad 0 \leq t \leq 1. $$

We also write $\Delta = \frac{\mathrm{d}}{\mathrm{d}t} \Gamma_t = \Gamma_1 - \Gamma_0$ since this will appear frequently. Then by the Frullani's integral and the functional calculus,

\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t} \log \Gamma_{t}
&=  \frac{\mathrm{d}}{\mathrm{d}t} \int_{0}^{\infty} \frac{e^{-x} - e^{-x\Gamma_t}}{x} \, \mathrm{d}x \\
&= \int_{0}^{\infty} (D^+_{\Delta}\exp)(-x\Gamma_t) \, \mathrm{d}x \\
&= \int_{0}^{\infty} \int_{0}^{1} e^{-sx\Gamma_t}\Delta e^{-(1-s)x\Gamma_t} \, \mathrm{d}s \mathrm{d}x.
\end{align*}

From this, we obtain

\begin{align*}
\operatorname{Tr}\left( \Gamma_t \frac{\mathrm{d}}{\mathrm{d}t} \log\Gamma_t \right)
&= \operatorname{Tr}\left( \int_{0}^{\infty} \int_{0}^{1} \Gamma_t e^{-sx\Gamma_t}\Delta e^{-(1-s)x\Gamma_t} \, \mathrm{d}s \mathrm{d}x \right) \\
&= \int_{0}^{\infty} \int_{0}^{1} \operatorname{Tr}\left( \Gamma_t e^{-sx\Gamma_t}\Delta e^{-(1-s)x\Gamma_t} \right) \, \mathrm{d}s \mathrm{d}x \\
&= \int_{0}^{\infty} \int_{0}^{1} \operatorname{Tr}\left( \Gamma_t e^{-x\Gamma_t}\Delta \right) \, \mathrm{d}s \mathrm{d}x \\
&= \operatorname{Tr}(\Delta),
\end{align*}

and similarly

$$ \operatorname{Tr}\left( (1-\Gamma_t) \frac{\mathrm{d}}{\mathrm{d}t} \log(1-\Gamma_t) \right) = -\operatorname{Tr}(\Delta). $$

So we obtain

$$ \frac{\mathrm{d}}{\mathrm{d}t} F(\Gamma_t)
= \operatorname{Tr}\left( H\Delta + \Delta\log(\Gamma_t) - \Delta\log(1-\Gamma_t) \right) \tag{1} $$

Differentiating both sides with respect to $t$ again,

\begin{align*}
\frac{\mathrm{d}^2}{\mathrm{d}t^2} F(\Gamma_t)
&= \operatorname{Tr}\left(\Delta \frac{\mathrm{d}}{\mathrm{d}t}\log(\Gamma_t) - \Delta \frac{\mathrm{d}}{\mathrm{d}t}\log(1-\Gamma_t) \right) \\
&= \operatorname{Tr}\left( \int_{0}^{\infty} \int_{0}^{1} \Delta e^{-sx\Gamma_t}\Delta e^{-(1-s)x\Gamma_t} \, \mathrm{d}s \mathrm{d}x \right) \\
&\qquad + \operatorname{Tr}\left(
 \int_{0}^{\infty} \int_{0}^{1} \Delta e^{-sx(1-\Gamma_t)}\Delta e^{-(1-s)x(1-\Gamma_t)} \, \mathrm{d}s \mathrm{d}x \right) \\
&= \int_{0}^{\infty} \int_{0}^{1} \left\| e^{-\frac{1}{2}sx\Gamma_t}\Delta e^{-\frac{1}{2}(1-s)x\Gamma_t} \right\|^2 \, \mathrm{d}s \mathrm{d}x \\
&\qquad + \int_{0}^{\infty} \int_{0}^{1} \left\| e^{-\frac{1}{2}sx(1-\Gamma_t)}\Delta e^{-\frac{1}{2}(1-s)x(1-\Gamma_t)} \right\|^2 \, \mathrm{d}s \mathrm{d}x \tag{2}
\end{align*}

where $\| X \|^2 = \operatorname{Tr}(X^* X)$ is always a non-negative real number.

Now we are ready to prove the claim.

From $\text{(2)}$, we know that $\frac{\mathrm{d}^2}{\mathrm{d}t^2} F(\Gamma_t) \geq 0$, which in turn implies that $F$ is convex.

$\text{(1)}$ can be used to show that $\Gamma_{\text{m}} = \frac{1}{e^H + 1}$ is a local extremum of $F$. Then by the convexity, this is a local minimum of $F$.

Therefore $\Gamma_{\text{m}}$ minimizes $F$.

Title: Minimizing the free energy


问:
Logic - Reducing a first-order logic involving a material conditional
Suppose that
$$
\forall x \forall y: P(x,y) \implies Q(x)
$$
and
$$
\forall x \exists y: P(x,y)
$$

Then, can I conclude that
$$
\forall x: Q(x)
$$

If it is true, what is the rationale for it?

What I tried:

$$
\begin{align}
&\forall x \forall y: P(x,y) \implies Q(x)\\
&\iff
\forall x \forall y:\lnot P(x,y) \lor Q(x)\\
&\iff
\forall x: (Q(x) \lor \forall y:\lnot P(x,y))\\
&\iff
\forall x: (Q(x) \lor \lnot (\exists y:P(x,y)))
\end{align}
$$
Combining the result with $\forall x \exists y: P(x,y)$, I concluded that $Q(x)$ should be true for all $x$ since $\lnot (\exists y:P(x,y))$ is always false.

$$
\begin{align}
&\forall x: (Q(x) \lor \lnot (\exists y:P(x,y)))
\land \forall x \exists y: P(x,y)\\
&\iff \forall x: ((Q(x) \lor \lnot (\exists y:P(x,y))) \land \exists y: P(x,y))\\
&\iff \forall x: (Q(x) \land \exists y: P(x,y))\\
&\implies \forall x: Q(x)
\end{align}
$$

答:
That's all correct.  for that last step you can actually use the distribution of $\forall$ over $\land$ again:

$$\forall x : (Q(x) \land \exists y : P(x,y))$$

$$\Leftrightarrow$$

$$\forall x : Q(x) \land \forall x \exists y : P(x,y))$$

$$\Rightarrow$$

$$\forall x : Q(x)$$

You haven't learned formal derivations yet?


答:
Yet another method, using the AM-GM inequality:

First, write $1+\frac{1}{n}=\frac{n+1}{n}=\frac{2+\sum_{i=1}^{n-1} 1 }{n}$

Now we get by the AM-GM inequality:

$$
1+\frac{1}{n}=\frac{2+\sum_{i=1}^{n-1} 1}{n}\geq \left(2\cdot\prod_{i=1}^{n-1} 1\right)^{\frac{1}{n}}=2^{\frac{1}{n}}
$$

Since $2\neq 1$ the above inequality is strict for $n\geq 2$ (since for such $n$ we have both $2$ and $1$ in the means), now taking the $n$-th power of both sides gives the desired inequality (since $x\mapsto x^n$ is monotonically increasing):

$$
\left(1+\frac{1}{n}\right)^{n} > 2
$$


问:
Coloring the 1 x n field
I have a field of 1 x n size. I need to color it using: red, orange, green, blue. Also, I can color red only even amount of blocks, and orange only odd amount of blocks.

Finally I need to find a generating function describing how many combinations of different colors I can use.

How can I do it?

答:
If order doesn’t matter, you’re looking at partitions of $n$ into four parts labelled red, orange, green, and blue, where any of the parts except the orange part can be empty, the red part must be even, and the orange part must be odd. Look at the coefficient of $x^n$ in the product

$$\begin{align*}
&\underbrace{(1+x^2+x^4+\ldots)}_{\text{red}}\underbrace{(x+x^3+x^5+\ldots)}_{\text{orange}}\underbrace{(1+x+x^2+\ldots)}_{\text{green}}\underbrace{(1+x+x^2+\ldots)}_{\text{blue}}=\\\\
&\left(\sum_{n\ge 0}x^{2n}\right)\left(\sum_{n\ge 0}x^{2n+1}\right)\left(\sum_{n\ge 0}x^n\right)\left(\sum_{n\ge 0}x^n\right)\;.
\end{align*}$$

All of these summations have simple generating functions.

Title: Coloring the 1 x n field


问:
Extend an integral inequality from $C[0,1]$ to the whole $L^{2}([0,1])$ [related to operator norm].
Consider the following integral defined for $f\in L^{2}[0,1]$ and $x\in [0,1]$, that $$(T^{n}f)(x)=\dfrac{1}{(n-1)!}\int_{0}^{x}(x-r)^{n-1}f(r)dr.$$

This shows that for $f\in C[0,1]$ and for any $x\in [0,1]$, we have $$|(T^{n}f)(x)|\leq \dfrac{1}{(n-1)!}\int_{0}^{x}(x-r)^{n-1}|f(r)|dr\leq \dfrac{\|f\|_{\infty}}{n!}.$$ By definition of operator norm, this implies that the operator norm of $T^{n}$ $$\|T^{n}\|\leq \dfrac{1}{n!},$$ if the operator is on $C[0,1]$.

However, I want to prove the same bound of this operator norm on the whole $L^{2}([0,1])$.  Is there anyway for me to extend this proof to all of $f\in L^{2}([0,1])$? and to conclude $\|T^{n}\|\leq\frac{1}{n!}$

My attempt was to use density of continuous function of compact support in $L^{2}$ to find $g$ such that $f=g+\epsilon$ where $\epsilon>0$ is arbitrarily fixed. Then we bound the integral in the same way, and you will have a summation, the second term in the summation will go to $0$ when $\epsilon\searrow 0$. Therefore, we again have $$|(T^{n}f)(x)|\dfrac{\|g\|_{\infty}}{n!}.$$  But this cannot say anything to the operator norm $\|T\|$.

What should I do? Thank you!

Edit:

The overall purpose is to derive that $\|T\|\leq\frac{1}{n!}$. So if there is any other way to find this, it will also be really good.

Edit2: Proof

Okay, I confused myself in the first place. The operator norm has nothing to do with the sup-norm since $T:L^{2}\longrightarrow L^{2}$, so what we should do is to use Cauchy-Schwarz and $\|f\|_{L^{2}}$. See below my own answer to my post for details.

Also, Ruy gave a general theorem about how to compute the operator norm for Hilbert-Schmidt operator, and my computation will show you the idea about the proof. Basically, the absolute value of operator will be bounded by $\|f\|_{L^{2}}$ multiplication of the $L^{2}-$integral of the kernel, so we should expect such a theorem.

Thank you so much for all the users who helped me!!

答:
Just promoting my comments to an answer because it gives the constant that appears in the question; Ruy gives a (better) method that works for more general kernels, but the kernel here is nice (it is continuous, bounded, $L^1$, ...) so we can apply generalised Minkowski as follows:

\begin{align}\left\| \int_{0}^x K(r) f(x-r)dr\right\|_{L^p([0,1])}
&\le \int_0^1 \|K(r)f(x-r)\mathbf 1_{r<x}\|_{L^p (x\in[0,1])}dr
\\
&= \int_0^1|K(r)| \|f(x-r)\mathbf1_{r<x}\|_{L^p  (x\in[0,1])}dr
\\
&\le \|K\|_{L^1([0,1])}\|f\|_{L^p([0,1])}. \end{align}
So the operator norm is at most
$$\|K\|_{L^1(0,1)}=\frac{\int_0^1r^{n-1}dr}{(n-1)!}=\frac1{n!}.$$


问:
How do you factorise $12^n - 4^n - 3^n +1$?
So I split the $12$ into $(4)(3)$, but I'm not sure what to do next. I think a subtle hint may guide me - I bet its something about "adding zero" or turning the$1$ into something in terms of $4$ or $3$

答:
$12^n-4^n-3^n+1=4^n(3^n-1)-3^n+1=(4^n-1)(3^n-1)$

and then you can use classical formulae to factor $(x^n-1)$.

Looking at the case n=1 you can easily see that this is the best you can do in the largest generality.


问:
Prove isomorphism of rings using Yoneda lemma
I have seen many questions which prove the isomorphism of rings using the Yoneda lemma, for example, this question, this question and this question. The basic idea is that we can prove the two functors (from Ring category to Set category) $\hom(X,*)$ and $\hom(Y,*)$ are isomorphism, then by Yoneda lemma we know $X$ and $Y$ are isomorphic.

For example, the first question proves that $(A/I)\otimes (B/J)\cong (A\otimes B)/(I\otimes 1+1\otimes J)$.

Question: By using category theory, can we see this isomorphism is natural?

What I mean by natural isomorphism is that the isomorphism between $(A/I)\otimes (B/J)$ and $(A\otimes B)/(I\otimes 1+1\otimes J)$ is exactly given by the obvious map $\bar{a}\otimes\bar{b}\rightarrow\overline{a\otimes b}$. I think that using category theory, one can indeed prove that $(A/I)\otimes (B/J)\cong (A\otimes B)/(I\otimes 1+1\otimes J)$, but one can't actually know what the explicit isomorphism is.

I think in commutative algebra and algebraic geometry, the naturalness or canonicalness of an isomorphism is very important, right?

答:
I completely agree knowing the explicit isomorphism is important, and I claim you can recover the isomorphism with the Yoneda lemma proof, you just need to know to know what's going on "inside" the Yoneda lemma. Recall the Yoneda lemma (in one form) says for any object $X$ of a category $\mathscr C$ and a functor $G:\mathscr C\to\mathbf{Set}$, we have a bijection of sets $\operatorname{Nat}(h^X,G)\cong G(X)$ which is natural in $X$. Explicitly this is given as follows: if we have $\varphi:h^X\to G$, then we take $\varphi_X(\mathrm{id}_X)\in G(X)$.

If we are given $X,Y$ in $\mathscr C$ then taking $G=h_Y$, we see that $\operatorname{Nat}(h^X,h^Y)\cong h^Y(X)=\hom(Y,X)$; this is the "fully faithfullness" of the Yoneda embedding, and in particular if we have an isomorphism $h^X\cong h^Y$ then the corresponding morphism $Y\to X$ is our isomorphism, which we see from the above paragraph we can recover as $\varphi_X(\mathcal id_X)$.

Let's apply this to the example you give; let's figure out exactly what is the explicit isomorphism $\hom(A/I\otimes_R B/J,T)\cong \hom((A\otimes_RB)/(I\otimes1+1\otimes J),T)$: I will follow the isomorphisms given in the answer you linked to, so you can compare line by line. Starting with $h\in\hom(A/I\otimes_R B/J,T)$ we have

\begin{align*}
h&\mapsto(\bar a\mapsto h(\bar a\otimes1),\,\bar b\mapsto h(1\otimes\bar b))\\
&\mapsto (a\mapsto h(\bar a\otimes 1),\,b\mapsto h(1\otimes\bar b))\\
&\mapsto (a\otimes b\mapsto h(\bar a\otimes 1)h(1\otimes\bar b)=h((\bar a\otimes 1)(1\otimes\bar b))=h(\bar a\otimes\bar b))\\
&\mapsto (\overline{a\otimes b}\mapsto h(\bar a\otimes\bar b))
\end{align*}

Now taking $h=\mathrm{id}:A/I\otimes_R B/J\to A/I\otimes_R B/J$ we see that the isomorphism $(A\otimes_R B)/(I\otimes 1+1\otimes J)\to A/I\otimes_R B/J$ coming from Yoneda is exactly $\overline{a\otimes b}\mapsto\bar a\otimes\bar b$.


Title: How do you factorise $12^n - 4^n - 3^n +1$?

Body: So I split the $12$ into $(4)(3)$, but I'm not sure what to do next. I think a subtle hint may guide me - I bet its something about "adding zero" or turning the$1$ into something in terms of $4$ or $3$

Body: How about this: $$(3^n-1)(4^n-1)$$


问:
How to apply definite integral on $\sqrt{1+\sin x}$
I want to solve the following integral:

$$\int_0^\pi \sqrt{1+\sin x} \, dx$$

I want to do this with the substitution:

$$u=\sin x$$

so that:

$$du=\cos x\,dx=\sqrt{1-\sin^2x}\,dx = \sqrt{1-u^2} \, dx$$

$$dx = \frac{du}{1-u^2}$$

$$u(0) = 0, \quad u(\pi/2)=1, \quad u(\pi)=0$$

$$\int\sqrt{1+\sin x} \, dx = \int\frac{1}{\sqrt{1-u}} \, du = -2\sqrt{1-u}$$

So since $u$ isn't defined for 1 (corresponding $x$ is $\pi/2$), I split the integration to ($0, \pi/2$) and ($\pi/2, \pi$)
However when I assign the substitution I get:

$$\int_0^1 \frac{1}{\sqrt{1-u}} \, du + \int_1^0 \frac{1}{\sqrt{1-u}} \, du = 2-2 = 0$$

But the result should be $4.$ It seems like I should flip the sign here for some reason but I can't figure it out, help would be appreciated.

Body: $$I=\int \sqrt{1+\sin x} dx= \int (\sin(x/2)+ \cos(x/2)) dx=2[\cos(x/2)-\sin(x/2)]+C$$


问:
adding details to the proof that natural isomorphism between finite dimensional V and $V^{*}$
Assume that V is finite dimensional vector space over $\mathbb{K}$. Let $<-,->$ denote an inner on V.

Let $D : V \rightarrow V^{*}$ defined as
$$v \mapsto D(v)= \ <v,-> \ : V \rightarrow \mathbb{K}$$

It is clear that D is linear. Assume that $D(v) = 0 \implies <v,->$ is the zero map.

In particular $<v,v> = 0 \implies \|v\| = 0 \implies v = 0$

I am having troubles proving that D is surjective.

答:
For a finite dimensional inner product space $V$, we know that $V^*$ has dimension $\dim(V)\dim(K) =\dim(V)$. To show that $D$ is onto, it suffices to exhibit a linearly independent subset of $\mathrm{Im}(D)$ with $\dim(V)$ elements.

To that end, let $\mathbf{v}_1,\ldots,\mathbf{v}_n$ be an orthonormal basis for $V$. Then note that $D(\mathbf{v}_i)(\mathbf{v}_j) = \delta_{ij}$ (Kronecker’s delta). If $\alpha_1D(\mathbf{v}_1)+\cdots+\alpha_nD(\mathbf{v}_n) = \mathbf{0}$, then evaluating at $\mathbf{v}_i$ we conclude that $\alpha_i=0$. Thus, these images are linearly independent, and hence form a basis for $V^*$. This shows $D$ is surjective.

We need to use finite dimensionality, since the algebraic dual of an infinite dimensional vector space is not isomorphic to the space (canonically or otherwise). See here for a proof. The finite dimensionality occurs twice in the above proof: the existence of an orthonormal basis, and the dimension of $V^*$.


问:
Exercise 5.P. - The Elements of Integration and Lebesgue Measure (2011), Bartle's book,
I'm working on the this exercise but I don't know how to prove it. I am new with this subject.
I know that I need to apply the Lebesgue Dominated Convergence Theorem, but I think that there are little diferent because it consider |X| and not X.
I apologize for not present a point of start.

Let $f_n \in L(X,X,\mu)$, and suppose that $(f_n)$ converges to a function $f$. Show that if
$$\lim \int |f_n - f| d\mu = 0$$, then $$\int |f| d\mu = lim \int |f_n| d\mu$$
thanks

答:
Notice that $|f|=|f-f_n+f_n|$. Using the triangle inequality
$$
|f| \leq |f-f_n|+|f_n|.
$$
Hence
$$
\int|f|d\mu \leq \int|f-f_n|+|f_n|d\mu=\int|f-f_n|d\mu+\int|f_n|d\mu
$$
and
$$
\int|f|d\mu-\int|f_n|d\mu \leq \int|f-f_n|d\mu.
$$
Again using the triangle inequality $|f_n|=|f_n-f+f|$
$$
|f_n| \leq |f-f_n|+|f|.
$$
Hence
$$
\int|f_n|d\mu \leq \int|f-f_n|+|f|d\mu=\int|f-f_n|d\mu+\int|f|d\mu
$$
and
$$
\int|f|d\mu-\int|f_n|d\mu \geq -\int|f-f_n|d\mu.
$$
Then it follows that
$$
\int|f-f_n|d\mu \geq \int|f|d\mu-\int|f_n|d\mu \geq -\int|f-f_n|d\mu.
$$
Since $\lim\int |f-f_n| d\mu=0$,
$$
\lim\int|f_n|d\mu=\int|f|d\mu.
$$


问:
Find the center and the radius of convergence for this complex series.
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} (z - \frac{1}{2}\pi)^{2n}$$

How do I know that the ratio test can be used here? Can it? How do I determine the method for solving this problem?

答: For power series, it is usually either the root or ratio test. Factorials work much nicer with ratio test than the root test.</p>


答:
Let $L \in V^{*}$. Choose orthonormal basis {e_1,\ldots,e_n} of V.

$$L(v) = L(x_1e_1 + \ldots + x_ne_n) = x_1L(e_1) + \ldots x_nL(e_n)$$

Let $v^{\prime} = e_1 L(e_1) + \ldots e_n L(e_n)$. It follows that

$$D(v^{\prime})(v) =  D(e_1 L(e_1) + \ldots e_n L(e_n))(v) = \langle e_1 L(e_1) + \ldots e_n L(e_n), x_1e_1 + \ldots x_ne_n> = x_1L(e_1) + \ldots x_nL(e_n) \rangle  = T(v)$$

Title: adding details to the proof that natural isomorphism between finite dimensional V and $V^{*}$

Body:
Assume that V is finite dimensional vector space over $\mathbb{K}$. Let $<-,->$ denote an inner on V.

Let $D : V \rightarrow V^{*}$ defined as
$$v \mapsto D(v)= \ <v,-> \ : V \rightarrow \mathbb{K}$$

It is clear that D is linear. Assume that $D(v) = 0 \implies <v,->$ is the zero map.

In particular $<v,v> = 0 \implies \|v\| = 0 \implies v = 0$

I am having troubles proving that D is surjective.


问:
Proving the quadratic formula (for dummies)
I have looked at this question, and also at this one, but I don't understand how the quadratic formula can change from $ax^2+bx+c=0$ to $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$. I am not particularly good at maths, so can someone prove the quadratic formula in a simple way, with no complicated words? All help appreciated.</p>

答:
There are several ways to prove the quadratic formula.  Using "completion of squares" is the simplest:  rather than write it all out again, I can refer you to a post I wrote as part of a series of proofs.

Actually, though, it seems like a number of the links other people have pointed to have laid out the steps just as well.  What specifically is the part that you don't understand?