问:
Let $f$ be a complex function defined on an open ball $U$. If $\sup |f(z)|=|f(z_0)|$ and if $f'(z_0)$ exists, is $f'(z_0)=0$?
I'm reading Rudin's Principles of Mathematical Analysis and I'm wondering if the following theorem can be generalized to complex functions:

Theorem: Let $f$ be a real function on $[a,b]$. If $f$ has a local maximum (or minimum) at $x_0$, and if $f'(x_0)$ exists, then $f'(x_0)=0$.

I've read about the Maximum modulus principle, and even though I admittedly don't know much about complex functions, this principle doesn't seem to resolve my question since I'm only requiring that $f'$ be defined on $z_0$, and thus $f$ need not be holomorphic on any set containing $z_0$.

答:
While the above is kind of weird since the function cannot be differentiable in a neighborhood of $z_0$ (without being in the trivial constant case), the answer is still yes as follows. Wlog we can assume $z_0=0, |f(0)| \ne 0$ (consider $g(z)=f(z+z_0)$) and then by considering $\alpha f$ (doesn't change hypothesis/conclusion) for some $|\alpha|=1$ we can assume $f(0) >0$ and by scaling $f(0)=1$ so the problem reduces to showing that:

if $\frac{f(z)-1}{z} \to A+iB, z \to 0, |f(z)| \le 1, |z| <r$ then $A=B=0$

But now take $z=x \in (-r,r)$ and notice that the hypothesis implies $\Re \frac{f(x)-1}{x} \to A, x \to 0$ or equivalently $\frac{\Re f(x)-1}{x} \to A, x \to 0, x \in (-r,r)$

Since $\Re f(x) \le |f(x)| \le 1$ the numerator is non-positive, while the denominator takes both plus and minus signs, which clearly forces $A=0$

Similarly with $z=ix$, $\frac{f(ix)-1}{ix} \to iB, x \to 0, x \in (-r,r)$ so  $\frac{f(ix)-1}{x} \to -B, x \to 0, x \in (-r,r)$ and taking real parts and repeating the reasoning forces $B=0$ Done!


问:
Pointwise convergence and differentiability implies uniform convergence
The problem:

True or False:
If $-\infty<a<b<\infty$ and $f_n\in C^1([a,b])$ and $f_n$ weakly converges in $(C([a,b]), \mid \mid \cdot \mid \mid_{\infty})$ to $f\in C([a,b])$, then $f_n$ converges uniformly to $f$; in other words $\mid \mid f_n-f_\infty\mid \mid_{\infty} \to 0$ as $n\to \infty$.

I tried contradiction but it didn't lead anywhere $\ldots$ I looked at Dini's Theorem and I found another theorem which has this conclusion but assumes uniform convergence of $(f_n ')_n$, but I don't have that.

答:
The answer is no and here is a counter example: let $f_n$ be the continuous function on $[0,1]$ whose graph is a poligonal line with vertices at $(0,0)$, $(1/n,1)$, $(2/n,0)$ and $(1,0)$.  Then it is easy to see that $f_n\to0$ pointwise, but not uniformly.  By Riesz, any continuous functional $\phi$  on $C[0,1]$ is given by the integration against a complex measure with finite total variation, so $\phi(f_n)\to0$ by Lebesgue's dominated convergence Theorem.

If you want a sequence of smooth functions, it is enough to take smooth approximations of the above $f_n$.

ContentLicense: CC BY-SA 4.0


问:
Integration of $\int_{0}^{1} \frac{x^k}{1-x} dx$
Solving this using integration by parts and by letting $u = x^k$ and $dv = (1-x)^{-1}$, it follows:
$$\int_{0}^{1}\frac{x^k}{1-x} dx = -x^{k}\ln(1-x) + k\int_{0}^{1} \ln(1-x)x^{k-1} dx$$
It is known that
$$\ln(1-x) = (1-x) - (1-x)\ln(1-x)$$
But doesn't this make the integration more complicated. Can I obtain a recursive formula for such an integration?

答:
Assuming that $k$ is an integer, you could write
$$\frac{x^k}{1-x}=\frac{x^k-1+1}{1-x}=\frac 1{1-x}-\frac{x^k-1}{x-1}=\frac 1{1-x}-\sum_{n=0}^{k-1}x^n$$
and integrate termwise.


问:
Proving that a function is nowhere differentiable so it cannot be analytically continued.
Here is a problem from Stein and Shakarchi

Where the definiton of analytic continuity is given here

This is what I have so far. At the unit circle, $g(z)=\sum_{n=0}^{\infty} 2^{-n \alpha} e^{i 2^n}$.
If g(z) is analytic, then $\forall z \in D$ the coefficients are given by $2^{-n \alpha} = \frac{g^(n)(z_0)}{n!}$.Can I somehow show that these $g^(n)$ don't exist.

答:
The easy way to do this is to use the fact that if an analytic function $h$ is analytically continuable at a point $|\alpha|=1$ then all its derivatives are (why?).

Picking an integer $k >\alpha$ and taking $g_1=zg', g_2=zg_1',...g_k=zg_{k-1}'$ one gets that the function $g_k(z)=\sum 2^{(k-\alpha)n} z^{2^n}$ is continuable at $\alpha$, while it is very easy to show that for any $\beta$ root of unity of order some power of $2$, $|g_k(r\beta)| \to \infty, r \to 1$ and such are dense on the unit circle contradicting that $g_k$ is analytic (hence continuous) on a neighborhood of $\alpha$, so at some such root of unity


问:
How do you find the number of contestants in a tv show based on their eye colors?
The problem is as follows:

To participate in an edition of The Price is Right $60$ people
showed up. The producer of the show collected the following
information:

$10$ women have black eyes $16$ women don't have black eyes $14$ women
do not have brown eyes $10$ men do not have brown or black eyes

Based on information from the show's producer. How many men have brown
or black eyes?

The alternatives in my book are as follows:

$\begin{array}{ll}
1.&\textrm{24 men}\\
2.&\textrm{22 men}\\
3.&\textrm{18 men}\\
4.&\textrm{28 men}\\
5.&\textrm{20 men}\\
\end{array}$

What I attempted to do is summarized in the table from below which showcases my findings. I must indicate that I arranged the characters of the story in two groups. One for women and the other for men. One column indicates $F$ for false of those who do not have a certain characteristic, i.e. A man who does not have black eyes would be in the F column below black eyes, a woman who has black eyes will lie below T in the black eyes column.

Hence this is represented as:

Where $x+y=10$

Adding all of these yields:

$14+10+16 +x+y+a+b=60$

Since it mentions $x+y=10$ and they want to know what is a+b then this would mean:

$a+b=60-50=10$

However this is not within the alternatives.

In my effort to better represent this situation and avoid counting twice I made the diagram from below:

This different representation using an Euler diagram shows what I thought, the set of those who do not have brown eyes must account for those who have black eyes.

There is no interesection in any of these sets as there cannot be someone who has brown and black eyes at the same time. Thus I concluded what they want is the number of men who have brown eyes and those who have black eyes.

This part I require clarification, what is the meaning of OR translated in set theory language?. Does it mean is it a Union?. Is it $\cup$?. Can someone help me with this part?.

Returning to my attempt I then thought that the number $14$ of those who do not have brown eyes will include those who have black eyes therefore.

Adding those who have black eyes with those who have brown eyes with those who dont have brown neither black in the women side. With those on the men side who dont have neither black or brown with those who have brown and those who have black will equate to $60$. This translated into an equation would be as this:

14+16+10+u=60

Assuming $u=\textrm{unknown males with brown with those with black eyes}$

This would become into:

$u=20\,men$

But this is not the right answer. Why?.

Did I made a mistake or something?. Because of this reason I need a very detailed help with this.

Please an answer must include a Venn or Euler diagram strategy or using a table. To better aid the understanding of this problem. I have tried all sorts, but I don't seem to get the right answer.

Can someone include also the representation of set notation using algebraic notation accompanied with an Euler or Table diagram?.

答:
There are 60 contestants

10 women have black eyes 16 women don't have black eyes

10 men do not have brown or black eyes

I will try to give as best an explanation as I can possibly give:

Notice, statement 2 will represent all women, since they have black eyes or don't. Just like if I were to say there are 2 guys with a shirt and 3 without: there are 5 guys. So back to statement 2, $10+16=26$ so there are $\boxed{26 women}$

Since there are 26 women, this means $60-26=34$ so there are $\boxed{34 men}$ since the contestants are women or they aren't just like above, the women either have black eyes or they don't

Statement 3 suggests there are 10 men without brown or black eyes, the opposite of that would be men with brown or black eyes. And as I've done through the whole answer, we have $34-10=24$ $\boxed{24}$ men with brown or black eyes

The logic of this whole answer is basically: There are $n$ number of things with the attribute(s) $X$ and $s$ number of things without attribute(s) $X$, so there are $n+s$ number of things total.


问:
Concrete description of the tensor product of suplattices?
I know the universal property of the tensor product of suplattices: sup-preserving maps $A \otimes B \to C$ represent bi-sup-preserving maps $A \times B \to C$.

But I want to know how to build $A \otimes B$ concretely. For example, if $A$ and $B$ are finite powersets what is $A \otimes B$?

Thank you.

答:
Well, here's a slick way to compute the tensor product of two power sets.  First, $A\otimes \{0,1\}\cong A$: bi-sup-preserving maps $f:A\times \{0,1\}\to C$ are in natural bijection with sup-preserving maps $g:A\to C$ by sending $g$ to the map $f$ defined by $f(a,0)=0$ and $f(a,1)=g(a)$.  Second, tensor products distribute over coproducts (which are the same as products): bi-sup-preserving maps $g:A\times \prod B_i\to C$ are in natural bijection with families of bi-sup-preserving maps $f_i:A\times B_i\to C$, by defining $g(a,(b_i))=\bigvee_i f_i(a,b_i)$.

Putting these together, a power set is just a product of copies of $\{0,1\}$, so $$P(X)\otimes P(Y)\cong\{0,1\}^X\otimes \{0,1\}^Y\cong (\{0,1\}\otimes \{0,1\})^{X\times Y}\cong \{0,1\}^{X\times Y}\cong P(X\times Y).$$  Concretely, the universal bi-sup-preserving map $P(X)\times P(Y)\to P(X\times Y)$ is just $(a,b)\mapsto a\times b$.

Here is a general explicit construction of $A\otimes B$, though I imagine in practice it is often not so easy to use.  If $S\subseteq A\times B$ and $(a,b)\in A\times B$, say that $b\in B$ is the $S$-partner of $a$ if for all $b'\in B$, $(a,b')\in S$ iff $b'\leq b$.  Similarly, say that $a$ is the $S$-partner of $b$ if for all $a'\in A$, $(a',b)\in S$ iff $a'\leq a$.  Say that $S$ is good if every element of $A$ and $B$ has an $S$-partner.  (Equivalently, $S$ is good if for each $a\in A$, $\{b\in B:(a,b)\in S\}$ is a complete ideal in $B$, and for each $b\in B$, $\{a\in A:(a,b)\in S\}$ is a complete ideal in $A$.)   Let $G$ be the set of all good subsets of $A\times B$, ordered by inclusion; this poset will be our tensor product of $A$ and $B$.  The intuition here is that a good subset $S$ represents the formal join of the elements $a\otimes b$ over all $(a,b)\in S$.

I claim first that $G$ is a suplattice.  Joins in $G$ are complicated to describe; it is easier to instead prove the equivalent statement that $G$ has all meets.  Indeed, if $(S_i)$ is a family of good sets, then $\bigcap S_i$ is a good set: if $b_i$ is the $S_i$-partner of $a$ for each $i$, and then $\bigwedge b_i$ is the $\bigcap S_i$-partner of $a$.

Next, I claim there is a bi-sup-preserving map $\mu:A\times B\to G$ defined by $$\mu(a,b)=\{(a',b'):a'\leq a,b'\leq b\}\cup A\times\{0\}\cup\{0\}\times B.$$  First, this set $\mu(a,b)$ is always good: if $a'\leq a$, then the $\mu(a,b)$-partner of $a'$ is $b$, and otherwise it is $0$, and similarly for elements of $B$.  To see that $\mu$ preserves sups in the first variable, let $(a_i)$ be a family of elements of $A$ and $b\in B$.  It is clear that $\mu(\bigvee a_i,b)\supseteq \mu(a_i,b)$ for each $i$.  To prove that $\mu(\bigvee a_i,b)=\bigvee\mu(a_i,b)$, suppose $S\in G$ contains $\mu(a_i,b)$ for each $i$; we must show that $\mu(\bigvee a_i,b)\subseteq S$.  Then for any $b'\leq b$, $(a_i,b')\in S$ for each $i$.  If $a$ is the $S$-partner of $b'$, this means that $a\geq a_i$ for all $i$ so $a\geq\bigvee a_i$.  Thus $(a',b')\in S$ for all $a'\leq \bigvee a_i$, and so $\mu(\bigvee a_i,b)\subseteq S$.  Thus $\mu$ preserves sups in the first variable, and by symmetry also preserves sups in the second variable.

Finally, I claim this $\mu$ is the universal bi-sup-preserving map out of $A\times B$.  To prove this, let $f:A\times B\to C$ be any bi-sup-preserving map.  Note that if $S\in G$ and $(a,b)\in S$, then $\mu(a,b)\subseteq S$.  Thus every element of $G$ is a union of sets of the form $\mu(a,b)$, so the image of $\mu$ generates $G$ as a suplattice.  So it suffices to show that $f$ factors through $\mu$ (uniqueness is automatic).  To do so, define $g:G\to C$ by $$g(S)=\bigvee_{(a,b)\in S}f(a,b).$$  It is easy to see that $g(\mu(a,b))=f(a,b)$, so it suffices to show that $g$ is sup-preserving.  So, suppose $(S_i)$ is a family of elements of $G$; it is clear that $g$ is order-preserving so $g(\bigvee S_i)\geq \bigvee g(S_i)$.  To prove the reverse inequality, suppose $c\in C$ is such that $c\geq g(S_i)$ for all $i$; we must show that $c\geq g(\bigvee S_i)$.  Let $$S=\{(a,b)\in A\times B:f(a,b)\leq c\}.$$  Since $f$ is bi-sup-preserving, $S$ is good: if $a\in A$, the $S$-partner of $a$ is just the join of all $b$ such that $f(a,b)\leq c$, and similarly for elements of $B$.  So $S\in G$, and $S\supseteq S_i$ for each $i$ since $c\geq g(S_i)$ for each $i$.  Thus $S\supseteq \bigvee S_i$, and $g(S)\geq g(\bigvee S_i)$.  But we also have $c\geq g(S)$ by definition of $S$, and hence $c\geq g(\bigvee S_i)$, as desired.

To connect this construction to the power set example above, if $A=P(X)$ and $B=P(Y)$, then the canonical isomorphism $P(X\times Y)\to G$ sends a subset $Z$ of $X\times Y$ to the set of all $(a,b)$ such that $Z$ contains the rectangle $a\times b$.  The good subsets of $P(X)\times P(Y)$ are exactly the collections of rectangles that are "saturated", in that every rectangle contained in their union is in the collection.  In general, you can think of the tensor $a\otimes b\in A\otimes B$ as representing the "rectangle $a\times b$", and a general element of the tensor product is a formal union of such rectangles, with two such formal unions being equal if they generate the same good subset of $A\times B$.

Finally, here is another neat way to think about this construction.  We can identify a good set $S\subseteq A\times B$ with the function $f:A\to B$ sending an element of $A$ to its $S$-partner.  Conversely, given a function $f:A\to B$, we can define $S=\{(a,b)\in A\times B:b\leq f(a)\}$.  For this set to be good, we additionally need every element of $B$ to have an $S$-partner, and this will be true iff $f$ turns joins into meets, i.e. iff $f$ is a sup-preserving map $A\to B^{op}$.  The inclusion order on $S$ corresponds to the pointwise order on such functions $f$, or the opposite pointwise order when we think of the codomain as $B^{op}$.  So, $A\otimes B$ can be identified with $\operatorname{Hom}(A,B^{op})^{op}$.

You can also derive this description via some abstract nonsense.  Recall that the category of suplattices is self-dual, by the contravariant functor taking a suplattice to its opposite and a sup-preserving map to its right adjoint (which is inf-preserving), and this gives not just a bijection but an order-isomorphism $\operatorname{Hom}(A,B)\cong \operatorname{Hom}(B^{op},A^{op})$.  Now, morphisms $A\otimes B\to C$ are by definition in natural bijection with morphisms $A\to \operatorname{Hom}(B,C)\cong \operatorname{Hom}(C^{op},B^{op})$.  These are then in natural bijection with morphisms $A\otimes C^{op}\to B^{op}$, which are also in natural bijection with morphisms $C^{op}\to\operatorname{Hom}(A,B^{op})$.  Finally, by the self-duality, these are in natural bijection with morphisms $\operatorname{Hom}(A,B^{op})^{op}\to C$.  So, by Yoneda, this identifies $A\otimes B$ with $\operatorname{Hom}(A,B^{op})^{op}$.


问:
How is cdf input converted from negative to positive?
Suppose X has normal $(μ,σ^{2})$ distribution, and $P(X \leq 0)=1/3, P(X \leq 1)=2/3$

What are the values of μ and σ?

$\frac{1}{3}=P(X \leq 0)=P(\frac{X-μ}{σ} \leq \frac{0-μ}{σ})=φ(\frac{-μ}{σ})$

$φ(\frac{-μ}{σ})=\frac{1}{3} → φ(\frac{μ}{σ})=\frac{2}{3} → \frac{μ}{σ}=φ^{-1}(\frac{2}{3})=0.43$

I don't understand how this $φ(\frac{-μ}{σ})=\frac{1}{3} → φ(\frac{μ}{σ})=\frac{2}{3}$ happened.

答: You will have to make use of a special property of the CDF of standard normal distributions, namely $\Phi(x)+\Phi(-x)=1\forall x\in\Bbb R$. It is easy to see why this is true:$$\begin{align*}\Phi(x)+\Phi(-x)&=\int_{-\infty}^x f(x)dx+\int_{-\infty}^{-x}f(x)dx\\&=-\int_{-\infty}^x f(-(-x))d(-x)+\int_{-\infty}^{-x}f(x)dx\\&=-\int_\infty^{-x} f(-y)dy+\int_{-\infty}^{-x}f(x)dx\end{align*}$$ where $y=-x$ and $f(x)$ is the PDF of standard normal distribution. $f(x)$ is even so $f(-y)=f(y)$, giving$$\int^\infty_{-x} f(y)dy+\int_{-\infty}^{-x}f(x)dx=1$$You can extend this concept to general normal distributions:$$F(\mu-x)+F(\mu+x)=1$$


问:
Determining the optimal solution for a given function
I have a function that I aim to minimize, however, I could not figure out the rule to reach the optimal. I am first sharing the function below and then providing a small example.

$\min_{\alpha \geq 0} f(\alpha) = [  z \alpha + \sum_{i=1}^{n} c_i \max\{ d_i - \alpha, 0 \}] $ where $z, c _i,  d_i $ are non-negative constant numbers.

For instance, if I try to find the optimal solution for the example shown below,

$\min_{ \alpha \geq 0} f(\alpha) = 0.95\alpha + 0.9 \max\{ 0.6 - \alpha, 0 \} + 0.4 \max\{ 0.11 - \alpha, 0 \} + 0.5 \max\{ 0.25 - \alpha, 0 \} $

The optimal solution is obtained when $\alpha = 0.25$. I am 100% sure that the optimal is attained whenever $\alpha$ is equal to one of given $d_i$, however, I sort of got stuck how to determine the index $i$. I was wondering if someone could provide me a hint.

Also, please note that I do no want to use a solver to reach the optimal.

答:
You can solve the problem via linear programming (LP) as follows.  Let $y_i$ represent $\max\{ d_i - \alpha, 0 \}$.  The problem is to minimize
$$z \alpha + \sum_{i=1}^n c_i y_i$$
subject to
\begin{align}
\alpha + y_i &\ge d_i &&\text{for all $i$} &&(\pi_i \ge 0)\\
\alpha &\ge 0 \\
y_i &\ge 0 &&\text{for all $i$} \\
\end{align}
The dual LP is to maximize
$$\sum_{i=1}^n d_i \pi_i$$
subject to
\begin{align}
\sum_i \pi_i &\le z &&&&(\alpha \ge 0)\\
\pi_i &\le c_i &&\text{for all $i$} &&(y_i \ge 0)\\
\pi_i &\ge 0 &&\text{for all $i$} \\
\end{align}
Without loss of generality, assume $d_1 \ge d_2 \ge \dots \ge d_n$ and $c_i>0$ for all $i$. Then an optimal dual solution is to take
\begin{align}
\pi_1^* &= \min\{z, c_1\}\\
\pi_2^* &= \min\{z - \pi_1^*, c_2\}\\
\pi_3^* &= \min\{z - \pi_1^* - \pi_2^*, c_3\}\\
\vdots \\
\pi_n^* &= \min\left\{z - \sum_{i=1}^{n-1} \pi_i^*, c_n\right\}
\end{align}
In particular, note that $\pi_i^* = 0$ implies $\pi_{i+1}^* = 0$.  Let
$$i^* = \min\left\{i: \sum_{j=1}^i c_j > z\right\}$$
be the smallest index for which $\pi_i^* = 0$.  Because $c_i > 0 = \pi_i^*$ for $i \ge i^*$, complementary slackness implies that $y_i^*=0$ for $i \ge i^*$, which suggests taking $\alpha^* = d_{i^*}$ and $y_i^* = \max\{ d_i - \alpha^*, 0 \}$, which is clearly primal feasible.

Now to certify optimality of this primal-dual pair $(\alpha^*,y^*)$ and $\pi^*$, compare the objective values.

For your example, $z=0.95$ and
\begin{matrix}
i & c_i & d_i & \pi_i^* & y_i^* \\
\hline
1 & 0.9 & 0.6  & 0.9  & 0.35 \\
2 & 0.5 & 0.25 & 0.05 & 0 \\
3 & 0.4 & 0.11 & 0    & 0 \\
\end{matrix}
So we have $i^* = 2$, with $\alpha^* = d_{i^*} = d_2 = 0.25$, as you claimed.


问:
How to prove limit is equal to infinity?
Suppose $f$ and $g$ continuous functions in $[a,b]$, differentiable in $(a,b)$, $c\in [a,b]$ and $g(x)>0$ for all $x\in [a,b]$, $x\neq c$. Let $A=\lim_{x\to c} f(x)$ and $B=\lim_{x\to c} g(x)$. If $A>0$ and $B=0$ prove that $\lim_{x\to c} f(x)/g(x)=+\infty$.

My attempt:
Let $M>0, \exists \delta>0$ such that $0<|x-c|< \delta$, then $f(x)/g(x)=+ \infty$

But I don't know if this is enough to prove the limit.

答:
Hint: for $f$ you need to find neighbourhood of $c$ where $f(x)>A-\varepsilon_1>0$, for $g$ neighbourhood of $c$ where $0<g(x)<\varepsilon_2 \Leftrightarrow \frac{1}{g(x)}> \frac{1}{\varepsilon_2}>0$ and then you'll have $\frac{f(x)}{g(x)}>\frac{A-\varepsilon_1}{\varepsilon_2}$

Addition:

Let's look at definition of $\lim\limits_{x \to c}f(x)=A>0$. As $A>0$, then we can choose  $\varepsilon_1>0$ such that $A-\varepsilon_1>0$. For this $\varepsilon_1$ there $\exists \delta_1>0$ such that in $\delta_1$ neighbourhood of $c$ holds $|f(x)-A|<\varepsilon_1$ so $f(x)>A-\varepsilon_1>0$.

From $\lim\limits_{x \to c}g(x)=0$ and $g(x)>0$ we have that for $\forall \varepsilon_2>0$ $\exists \delta_2>0$ such that in $\delta_2$ neighbourhood of $c$ holds $|g(x)|<\varepsilon_2 \Leftrightarrow 0<g(x)<\varepsilon_2$. This is equivalent to $\frac{1}{g(x)}> \frac{1}{\varepsilon_2}>0$


问:
Calculating population movement between countries... unsure what to tag this as
I apologize if this is a slightly vague question.

Let's say that I have yearly data containing the amount of births, amount of deaths, and the total population in each of the 6 regions throughout the world. Let's assume that this data is accurate.

Using only this information, is there a way that I could calculate the number of people moving from one region to another in a given year or would I need more information?

Here is a simplified example with 3 regions rather than 6.

Pop_1_2000= 1000
Pop_2_2000 = 2000
Pop_3_2000 = 1500
Birth_1_2000 = 50
Birth_2_2000 = 75
Birth_3_2000 = 105
Death_1_2000 = 75
Death_2_2000 = 80
Death_3_2000 = 100

So total population in this "world" would start at 4500(add Pop_i_2000 up) and then be 4,475 after this first year (add starting populations and then subtract the births and deaths)
.
The populations in each of the three countries shift to this though.
Pop_1_2001 = 1200
Pop_2_2001 = 1800
Pop_3_2001 = 1475

Is there any way that I could (only using this information) find out how many people moved from country 1 to country 2, country 2 to country 1, country 3 to country 1, etc.?

答:
You would need more information.  From the starting populations, births, and deaths you can compute what the population would be in each region for each year in the absence of migration.  The difference between this calculation and the actual population is the effect of migration.  With six regions there are fifteen net migration routes.  You have six equations, one from the population of each region, and fifteen unknowns, the net migration between each pair of regions.  It is even worse because there is at least one dependence between the equations.

A simple example would be three regions with no births, deaths, or change in population.  It is consistent with the data that there is no migration at all.  It is also consistent that a given number of people move from 1 to 2, from 2 to 3, and from 3 to 1.


问:
How to find area covered by a car windshield wiper when it swaps a certain angle?
The problem is as follows:

A buggy is set to cross over a snowy terrain. The driver seat has a
rectangular window featured in the diagram from below. When the driver
activates the mechanism for cleaning the window from the snow, the
wiper spins $120^{\circ}$ using the point $M$ as a pivot. Assuming the
rotation is clockwise. Find the area covered by the windshield wiper
in the buggy. Assume the wiper has a T shape and the rectangular
window is $ABCD$ where $BC=2AB=2AE=2AM=1\,m$. Also assume AE is
perpendicular to $MN$ and $N$ is midpoint between $AE$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\left(\frac{\pi}{6}-\frac{\sqrt{3}}{16}\right)\,m^{2}\\
2.&\left(\frac{\pi}{16}-\frac{\sqrt{3}}{6}\right)\,m^{2}\\
3.&\left(\frac{\pi}{4}-\frac{\sqrt{3}}{8}\right)\,m^{2}\\
4.&\left(\frac{\pi}{16}-\frac{\sqrt{3}}{16}\right)\,m^{2}\\
\end{array}$

So far the only thing which I could come up with was the diagram from below:

However I don't know how to find such area. I remember how to calculate the area of a half circle sector by means of the equation:

$A=\frac{\alpha}{360}\pi r^2$

Assuming $\alpha$ is an angle given in sexagesimal units. But in this case the figure doesn't help much. How exactly should I find that weird surface.

Can someone help me with a method relying euclidean geometry or something along precalculus?. I think integrals can be used but I am not looking for such sort of answer.

Please include a drawing in the answer because for me it is not easy to spot how to find such problematic area. Does it require some sort of construction?.

答:
Based on Student1058's  answer I'm posting a revisited version of my attempt solved properly with some annotations which may help somebody in a similar situation.

Typically if you are inside a car and look at the area swept by a windshield wiper you'll notice it generates an area of a semicircle. This happens because the swept is generated by the rotation of the arm which is covered by a cleaning foam or a rubber blade which doesn't cover all the arm of the wiper. This is better understood in the graphic from below:

However in this sort of problem it seems a bit challenging and at the same part confusing if you rush to believe that the segment perpendicular forming the $T$ (meaning $MN \perp AE$) also acts as a normal wiper. But this is not meant as the correct interpretation of the problem as it indicates, only the area which is swept by the arm $AE$. (See images below)

This jumps into the question, how do you find such area?. The best method is to build a tangential circle enclosing the inner part of the wiper. In geometrical terms, this is translated as drawing a circle tangential to point $N$.

The next thing we want to do is to make another circle but this time, one which touches the points $A$ and $E$ (yes similar to the tv channel).

In this case. This happens because the angle on $\angle NAM = 30^{\circ}$.

Because:

Since $AM=\frac{1}{2}\,m$ and $N$ is midpoint of $AM$ then $AN=NE=\frac{1}{4}\,m$

Since $\triangle ANM$ is a right triangle whose sides are:

$NA=\frac{1}{4}$

$\textrm{hypotenuse}=\frac{1}{2}$

Then this proportion lead us to indicate it is a $30-60-90$ Pythagorean triangle, more properly is a special right triangle.

Where the sides are in a proportion of:

$\textrm{hypotenuse}= 2k$

$\textrm{side opposing 30 degrees}= k$

$\textrm{side opposing 60 degrees}= k\sqrt 3$

To obtain $NM=\frac{\sqrt{3}}{4}$ and since $\triangle ANM \cong$ to $\triangle ENM$ this led us to indicate $\angle\,ENM= 30^{\circ}$ and $NM=\textrm{is bisector and median hence a mediatrix}$.

Therefore the $\angle AEM = \angle EMA = 60^{\circ}$ thus making our $\triangle\,AEM$ isosceles into an equilateral triangle. Thus perfectly fitting one-third of the semicircle. Referring to the semicircle between points $AD$.

Had this angle not occured then it would follow the same strategy, you can always make two circles, both tangential to the inner part from where is about to be rotated and the other one touching both ends of the segment, the only difference will lie on the slope of the segment.

Since it is not easy to spot how the arm generates the area. I made succesive stops along the way as the wiper moves touching both cirles. The image from below showcases this motion.

As now it is proved that we must only focus in the areas covered first by the arc between $AN$ and the ring between $NN$ and finally from the arc between $NE$.

From a visual inspection we can notice that since arc $AN$ equals to the arc $NE$ their swept areas will be the same.

Thus we proceed to calculate such areas:

The easiest part is the ring:

This comes from:

$A=\frac{r^2\theta}{2}$

where $\theta=\textrm{given in radian units}$

therefore: $As \frac{120^{\circ}}{180}\pi=\frac{2\pi}{3}$

$A_{NN}=\frac{(r_1^2-r_2^2)\frac{2\pi}{3}}{2}=\frac{(\frac{1}{2}^2-\frac{\sqrt{3}}{4}^2)\frac{2\pi}{3}}{2}$

$A_{NN}=\frac{\pi}{48}$

Now we will focus in the other two arcs:

This can be obtained by subtracting our right triangle from the circle sector on $NAM$

Hence:

$A_{AN}=\left[\frac{r_{AM}^2\theta_{\angle\,NMA}}{2}\right]-A_{\triangle\,NAM}$

since: $\theta_{\angle\,NMA}=\frac{30\pi}{180}=\frac{\pi}{6}$

This means:

$A_{AN}=\left[\frac{\frac{1}{2}^2}{2}\cdot\frac{\pi}{6}\right]-\frac{\frac{1}{4}\cdot\frac{\sqrt{3}}{4}}{2}$

$A_{AN}=\frac{\pi}{48}-\frac{\sqrt{3}}{32}$

Since $A_{AE}=A_{AN}$

Notice here we're referring as $AN$ and $NE$ as the arcs.

The total area would be:

$A_{\textrm{swept area}}=A_{NN}+2A_{AN}=\frac{\pi}{48}+2\left(\frac{\pi}{48}-\frac{\sqrt{3}}{32}\right)$

Finally becoming into:

$A_{\textrm{swept area}}=\frac{\pi}{16}-\frac{\sqrt{3}}{16}\,m$

Which corresponds to the fourth option.

This explanation is the most complete I could write. Which hope may help in any similar cases.


问:
Inequality with integers
I am trying to find an analytical way to prove that for integers $n\ge 6$,

$$\sqrt[n]{n+1}\le \frac{1}{n}+\frac{1}{\sqrt[n]{n+1}}+\frac12$$

答:
Denote $x = \sqrt[n]{n+1}$. The desired inequality is written as
$\frac{-2nx^2 + (n+2)x + 2n}{2nx} \ge 0$.
It suffices to prove that $-2nx^2 + (n+2)x + 2n \ge 0$ which is written as
$$2n(x - x_1)(x - x_2)\le 0$$
where
$$x_1 = \frac{n+2 + \sqrt{17n^2 + 4n + 4}}{4n},
\quad x_2 = \frac{n+2 - \sqrt{17n^2 + 4n + 4}}{4n}.$$
Since $x > 0$ and $x_2 < 0$, it suffices to prove that $x \le x_1$, namely,
$$\sqrt[n]{n+1} \le \frac{n+2 + \sqrt{17n^2 + 4n + 4}}{4n}. \tag{1}$$

For $n = 6, 7, \cdots, 11$, (1) is verified directly.

For $n \ge 12$, since $\sqrt{17n^2 + 4n + 4} \ge 4n + 1$, it suffices to prove that
$$\sqrt[n]{n+1} \le \frac{n+2 + 4n + 1}{4n}$$
or
$$n+1 \le (1 + \tfrac{1}{4})^n (1 + \tfrac{3}{5n})^n.$$
By the binomial theorem, we have $(1 + \tfrac{1}{4})^n \ge 1 + \frac{1}{4}n + (\frac{1}{4})^2\frac{n(n-1)}{2}$
and
$(1 + \tfrac{3}{5n})^n \ge 1 + \frac{3}{5n}\cdot n = 1 + \tfrac{3}{5}$.
It suffices to prove that
$$n + 1 \le [1 + \tfrac{1}{4}n + (\tfrac{1}{4})^2\tfrac{n(n-1)}{2}] (1 + \tfrac{3}{5})$$
or
$$\frac{1}{20}(n-1)(n-12)\ge 0$$
which is true.

We are done.

Title: Inequality with integers


问:
Dummit and Foote 10.3.18 (Decomposition theorem for modules)
Let $R$ be PID and $M$ a $R$-Module. Suppose ${\rm Ann}(M)=(a)$. Let $a=p_1^{e_1}p_2^{e_2}...p_k^{e_k}$ be a factorization of $a$ into primes. Define $M_i=\{m \in M|p_i^{e_i}m=0\}.$ Show $M=\bigoplus M_i$ ($i$ runs 1 to $k$).

So I did the most obiovs thing which was to use C.R.T. to get

$M \cong M\diagup(a)M=M \diagup \prod p_i^{e_i}M\cong M \diagup p_1^{e_1}M \times ...M \diagup p_k^{e_k}M$

After this I'm stuck. If I can show $M_i \cong \faktor{M}{p_i^{e_i}M}$ I think I will be close (I think then I will need only show the uniqueness of the sum), but I have no intuitive feeling for why this should be true.

答:
We proceed by induction on $k$. For the case when $k=2$, write $a=p_1^{\alpha}p_2^{\alpha_2}$ where $p_1,p_2$ are distinct primes in $R$. This means $(p_1)$ and $(p_2)$ are comaximal and so (exercise) $(p_1^{\alpha_1})$ and $(p_2^{\alpha_2})$ are comaximal. Then there exists $x,y\in R$ such that $p_1^{\alpha_1}x+p_2^{\alpha_2}y=1$. Since $a$ annihilates $M$, then $am=0$ for all $m\in M$. This means that
\begin{equation*}
m=p_1^{\alpha_1}xm+p_2^{\alpha_2}ym
\end{equation*}
and multiplying throughout by $p_1^{\alpha_1}$, we see that $p_1^{\alpha_1}\left(m-p_1^{\alpha_1}xm\right)=p_1^{\alpha_1}p_2^{\alpha_2}ym=0$. This means that $p_2^{\alpha}ym\in M_{2}$, and by a similar argument, $p^{\alpha}xm\in M_{1}$. This shows that $M=M_{1}+M_{2}$. Now we show the general case. Write $a=p_{1}^{\alpha_{1}}\cdots p_{k}^{\alpha_{k}}$. Let $\gamma=p_{1}^{\alpha_{1}}\cdots p_{k-1}^{\alpha_{k-1}}$, so that $\gamma$ and $p_{k}^{\alpha_{k}}$ are pairwise coprime and hence the principal ideals generated by them are comaximal. Repeating the argument above, we get the desired result. Thus, we have shown that $M=M_{1}+\dots+M_{k}$.

We now show that the sum is indeed direct. We keep the notation as above. Since $\gamma$ and $p_{1}^{\alpha_{1}}$ are coprime in $R$, then we can write $1=\gamma s+p_{1}^{\alpha_{1}}t$ for some $s,t\in R$. Let
\begin{equation*}
n\in M_{i}\cap\sum_{\substack{j=1\\ j\neq i}}^{k}M_{j}
\end{equation*}
so that $\gamma n=0$ and $p_{1}^{\alpha_{1}}n=0$. Using $1=\gamma s+p_{1}^{\alpha_{1}}t$, then $n=\gamma s n+p_{1}^{\alpha_{1}}tn=0$, and hence the sum is direct.


问:
Any books on mathematical logic that can be serve as supplementary reading for Roger Godement's Algebra Chapter 0?
I am from a non-mathematics background but the course that I am taking in probability class is built on measure theory. I found many theorems stated in the book hard to comprehend and I think that's partly due to my insufficient maths background. I am self-learning real analysis currently and think it may be equally important to systematically learn abstract algebra if I really want to know the logic behind the probability class I am currently taking.

I picked Algebra by Roger Godement but find even the first chapter very hard to comprehend. For example, I find myself completely lost proceeding to the following section of the text.

Let $\mathrm{R}$ be a relation, $\mathrm{A}$ a mathematical object, and $x$ a letter (i.e., a "totally indeterminate" mathematical object). In the assembly of letters and fundamental signs which constitutes the relation $\mathrm{R},$ replace the letter $x$ wherever it occurs by the assembly A. One of the criteria for forming relations is that the assembly so obtained is again a relation, which is denoted $(*)$ by the notation
$$
(\mathrm{A} \mid x) \mathrm{R}
$$
and is called the relation obtained by substituting A for $x$ in $\mathrm{R},$ or by giving $x$ the value $\mathrm{A}$ in $\mathrm{R}$. The mathematical object $\mathrm{A}$ is said to satisfy the relation $\mathrm{R}$ if the relation $(\mathrm{A} \mid x) \mathrm{R}$ is true. It goes without saying that if the letter $x$ does not appear at all in the assembly $\mathrm{R},$ then the relation $(\mathrm{A} \mid x) \mathrm{R}$ is just $\mathrm{R},$ and in this case to say that $\mathrm{A}$ satisfies R means that $R$ is true.

However, I do appreciate the textbook that is self-contained and appreciate the author devoted to mathematical reasoning so rigorously at the beginning of the chapter. I tried to find some textbook about mathematical logic but they are either too abstract or not thorough enough that seems to start from the most fundamental (i.e. from axiom and the most basic rule).

I have read relevant posts on the subject I am asking but can't decide the material right for me. I am wondering if there are any materials or textbooks that introduce mathematical logic rigorously and serve as a supplementary text for me to understand the first chapter of the book? If there really isn't any textbook that is not too abstract but rigorous enough, I am wondering if there are any other textbooks on abstract algebra that start from mathematical logic and build the whole system from the scratch?

答: I liked “A Transition to Advanced Mathematics” by Douglas Smith. It covers a lot of transitional knowledge like predicate calculus, sets, relations, functions, cardinality, and abstract algebra. I found it to be very easy to understand in transition from calculus to upper level maths.</p>


问:
Ellipse application problem. I am not sure where to begin with this problem. Is there an easier way of doing this problem?

An express way is built in the shape of a semielliptical arch and is to have a span of one hundred feet.  The height of the arch, at a distance of forty feet from center, is ten feet.  Find the height of arch at the center.

答:
Let $|AC|=L=100$, $|OA|=|OC|=\tfrac12L$,
$|OE|=d=40$, $|BE|=h=10$.

Consider a circle $\mathcal C$ through the endpoints of $AC$ centered at $O$,
with the radius $\tfrac12L$.
The extension of the line $EB$ intersects with the circle $\mathcal C$
in $D$. Let $|BD|=g$, then $k=\frac h{h+g}$
is an $y$-scaling factor that transforms
the circle $\mathcal C$ into the sought ellipse through $A,B,C$,
and its height $H$ at the center is then $k\cdot\tfrac12L$.

So, first, we need to find $g$.
By the Stewart’s theorem,

\begin{align}
\triangle DOE:\quad
|OE|^2\cdot|BD|+|OD|^2\cdot|BE|
&=|DE|\cdot(|OB|^2+|BD|\cdot|BE|)
\tag{1}\label{1}
,\\
\triangle BOE:\quad
|OB|^2&=d^2+h^2
\tag{2}\label{2}
,
\end{align}
and it follows that
\begin{align}
g&=\tfrac12\,\sqrt{L^2-4d^2}-h
=20
\tag{3}\label{3}
,\\
k&=\tfrac13
\tag{4}\label{4}
,\\
H&=\tfrac{50}3
\tag{5}\label{5}
.
\end{align}


问:
Vitali-type set with given outer measure
Is it possible to construct a non-measurable set in $[0,1]$ of a given outer measure $x \in [0,1]$? This will probably require the axiom of choice. Does anyone have a suggestion?

Edit: I forgot to mention that the set has to be in $[0,1]$ which seemed "clear" in my head, but which obviously isn't.

Edit 2: Now it is clear that I just need a non-measurable set with outer measure 1, from that I can make any other one. I have once seen somewhere that they do this with transfinite recursion (if I recall correctly), but I don't remember where and how because I didn't know that transfinite recursion was back then.

答:
Let $V$ be a $\mathbb{Q}$ vector subspace of $\mathbb{R}$ of codimension at most countable. Then $\mu^*(V\cap I) = \mu(I)$ for every $I$ interval.

Proof:

We may assumed wlog $1\in V$ (just divide by some non-zero $t\in V$).
Conclude  $a V + b \subset V$ for all $a$, $b$ in $\mathbb{Q}$

Since a countable union of translates of $V$ covers $\mathbb{R}$ we have $\mu^*(V)>0$. So there exists an interval $I$ with rational ends such that $\mu^*(V\cap I) >(1-\epsilon) \mu(I)$.  Using the rational invariance of $V$ we conclude
$$\mu^*(V\cap I) >(1-\epsilon) \mu(I)$$
for every interval $I$ with rational ends. Therefore
$$\mu^*(V\cap I) =\mu(I)$$
for every interval with rational ends.
Since every finite interval  is contained in one with rational ends, we have the above equality for every finite interval $I$.

Now, take $V$ a complement of $\mathbb{Q}$ in $\mathbb{R}$, that is $V\oplus \mathbb{Q} = \mathbb{R}$. Then $V$ is a VItali set.   Consider some $x>0$.  We have $\mu^*(V\cap [0,x])=x$.  Complete $V\cap[0,x]$ to a Vitali subset $W$ of $[0,x]$ with $\mu^*(W)=x$.

ContentLicense: CC BY-SA 2.5


问:
Evaluating $\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})} x$
I don't know how can this be solved:
$$
\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})} x
$$
I've tried to multiply by $\pi\sqrt{\cos x}$:
$$
\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})}{x\pi\sqrt{\cos x}}
$$

but we have here a hard problem $\lim_{x\to0}(1/x)$.

You think about the limit on the right and on the left but the exercise supposed to be solved by the limit in $0$ and not on the right or the left of it.

答:
Since you already received good answers for the limit iteself.

You can go beyond the limit if you compose Taylor series (that is to say working one piece at the time).

$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$
$$\sqrt{\cos(x)}=1-\frac{x^2}{4}-\frac{x^4}{96}+O\left(x^6\right)$$
$$\pi \sqrt{\cos(x)}=\pi -\frac{\pi  x^2}{4}-\frac{\pi  x^4}{96}+O\left(x^6\right)$$
$$\sin \left(\pi  \sqrt{\cos (x)}\right)= \sin\left(\frac{\pi  x^2}{4}+\frac{\pi  x^4}{96}+O\left(x^6\right) \right)=\frac{\pi  x^2}{4}+\frac{\pi  x^4}{96}+O\left(x^6\right)$$
$$\frac{\sin(\pi\sqrt{\cos (x)})} x=\frac{\pi  x}{4}+\frac{\pi  x^3}{96}+O\left(x^5\right)$$ which shows the limit and how it is approached.

Moreover, this gives a shorcut evaluation of a not so pleasant expression. Suppose $x=\frac \pi{12}$ (this is quite far away from $0$.
The exact expression would be
$$\frac{\sin\left( \pi\sqrt{\frac {\sqrt 6+\sqrt 2} 4}\right)}{\frac \pi {12}} \approx 0.20612 $$ while the above truncated series gives
$$\frac{\pi ^2 \left(3456+\pi ^2\right)}{165888}\approx 0.20620$$


问:
Optimize the following subject to their respective constraints:
$f(x,y)=xy^2$; constraint $x^2/a^2 + y^2/b^2 = 1$

I took the gradients of $f(x,y)$ and $g(x,y)$. Used substitution method to find $x = \lambda(1/b^2)$ and $y = (\sqrt2\lambda/ab)$. However, I have got lost after this and would really appreciate any kind of help.

答:
You are almost there. Plug in your $x$ and $y$ in the equation for the constraint.$$\frac{\lambda^2}{a^2b^4}+\frac{2\lambda^2}{a^2b^4}=1$$
You get $$\lambda^2=\frac13a^2b^4$$
Now find $x$ and $y$ from here. You should get$$x=\pm\frac1{\sqrt 3}a$$ and $$y=\pm\sqrt{\frac23}b$$
You need to calculate $f(x,y)$ at these four points:$$f(x,y)=\pm\frac2{3\sqrt3}ab^2$$


问:
How many positive semidefinite submatrices can an indefinite matrix with a positive eigenvector/positive eigenvalue have?
Let $A\in\mathbb R^{n\times n}$ be a symmetric matrix such that it has $k<n$ non-negative eigenvalues and $n-k$ negative eigenvalues. I am interested in the following question: can all $k\times k$ principal submatrices of $A$ be positive semidefinite? If not, how many of them can be positive semidefinite?

I was experimenting with Matlab for the case $n=4$, $k=2$, and I couldn't find a way to make $A$ so that all $2\times 2$ matrices were positive semidefinite (somehow $A$ always would get more than two non-negative eigenvalues). Hence the question. I can't figure it out myself.

Edit: there are two cases I already understand: $k=1$ and $k=n-1$.

In the first case it is easy to find a matrix with one positive eigenvalue such that the diagonal is non-negative: say, $A=\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}$ (the ones on the off-diagonals may be any positive numbers). For the second case, consider $-A^{-1} = \begin{bmatrix}1&-1&-1\\-1&1&-1\\-1&-1&1\end{bmatrix}/2$. But what if $n>3,1<k<n-1$?

Another edit: one user suggested to use a rectangular $(n\times m)$ (with $m<n$) matrix $X$ such that every set of $k$ rows of $X$ has full rank (like a Vandermonde matrix) and look at $A=cXX^* - I_n$ for some $c>0$ chosen such that $A$ has $k$ non-negative eigenvalues and $k\times k$ PD submatrices. Indeed, this method was successful. So the question is answered, technically, but there is more: the matrix always seems to have a negative eigenvalue with a positive vector.

This leads to the same question, modified by the assumption: suppose that the subspace $V$ spanned by the eigenvectors corresponding to positive eigenvalues contains a positive (entrywise $>0$) vector.

答:
Let $X\in M_{n,k}(\mathbb R)$ be a rectangular Vandemonde
matrix such that $X_{ij}=x_i^{j-1}$ for some $n$ distinct positive real numbers $x_1,x_2,\ldots,x_n$. By construction, each $k\times k$ submatrix $Y$ of $X$ is nonsingular and entrywise positive.

Let $A=cXX^T-I_n$ for some sufficiently large $c>0$. Then $A$ has $k$ positive eigenvalues (namely, $c\sigma_i(X)^2-1$ for $i=1,2,\ldots,k$) and an eigenvalue $-1$ of multiplicity $n-k$. Moreover, all $k\times k$ principal submatrices of $A$ are in the form of $cYY^T-I_k$. Since $c\sigma_\min(Y)^2-1>0$ when $c$ is large, these principal submatrices are positive definite.

Also, when $c$ is large, $A$ and its submatrices are entrywise positive. Therefore, by Perron-Frobenius theorem, each of its principal submatrices (including $A$ itself) contains a positive eigenvector corresponding to a positive eigenvalue (the spectral radius of the submatrix).


问:
Why I cannot get the same answer if I do substitution with $x=a \cos\theta$ for $\int \sqrt{a^2-x^2}dx$ compared with substitution $x=a \sin\theta$?
Why I cannot get the same answer if I do substitution with $x=a \cos\theta$ for $\int \sqrt{a^2-x^2}dx$ compared with substitution $x=a \sin\theta$?

$\int \sqrt{a^2-x^2}dx = \int \sqrt{a^2 - a^2 \cos^2 \theta} d\theta = \int a \sin\theta d\theta = \int a \sin\theta(-a \sin\theta)d\theta = -a^2 \int sin^2 \theta = -a^2[\frac{\theta}{2} - \frac{\sin(2\theta)}{4}]+C$

However, the correct answer seems to be $a^2(\frac{\theta}{2} + \frac{\sin(2\theta)}{4}) + C$

答:
Method$\#1:$

If $\theta=\arccos\dfrac xa=\dfrac\pi2-\arcsin\dfrac xa,$

$\cos\theta=\dfrac xa,0\le\theta\le\pi$

$$dx=-a\sin\theta\ d\theta\text{ and }\sqrt{a^2-x^2}=a\sin\theta$$

$$\int\sqrt{a^2-x^2}\ dx=-\int a^2\sin^2\theta\ d\theta=\dfrac{a^2}2\int(\cos2\theta-1)\ d\theta=\dfrac{a^2\sin2\theta}4-\dfrac{a^2\theta}2$$

Now $\sin2\theta=2\sin\theta\cos\theta=?$

Method$\#2:$

If $\theta=\arcsin\dfrac xa,$

$\sin\theta=\dfrac xa,-\dfrac\pi2\le\theta\le\dfrac\pi2$

$$dx=a\cos\theta\ d\theta\text{ and }\sqrt{a^2-x^2}=a\cos\theta$$

$$\int\sqrt{a^2-x^2}\ dx=\int a^2\cos^2\theta\ d\theta=\dfrac{a^2}2\int(\cos2\theta+1)\ d\theta=\dfrac{a^2\sin2\theta}4+\dfrac{a^2\theta}2$$

Now $\sin2\theta=2\sin\theta\cos\theta=?$


问:
if the real part of f is a composition of the imaginary part and other function then f is a constant.
I have the next problem:

Let f = u + iv be holomorphic in the region G and satisfy $u = h\circ v$ for some differentiable function $h : \mathbb{R} \longrightarrow \mathbb{R}$. Show that f is constant.

I did the following calculations:

If $f$ is holomorphic then: $u_x=v_y$ and $v_x=-u_y$. Then, $u_x=(h\circ v)_x=h_x(v)v_x$ and $u_y=(h\circ v)_y=h_y(v)v_y$. So, $v_x=-h_y(v)v_y=-h_yh_x(v)v_x$.

I think that $h_y=0$ because $h$ is real. But I'm not sure. I appreciate the help you can give me.

答:
Partial answer: I will assume that $h$ is continuously differentiable.

$h$ is a function from $\mathbb R$ into itself so it has no partial derivatives.

We have $u_x=h'(v)v_x$ so $h'(v) v_x=v_y$. Also $u_y=h'(v)v_y$ so $h'(v)v_y=-v_x$. Consider the subset of $G$where $h'(v) \neq 0$. Suppose this is non-empty.  This an open set. On this set we get $v_x=v_y=0$ so $v$ is a constant. Now C-R equation show that $u$ is also a constant so $f$ is constant. By the Identity Theorem this implies that $f$ is a constant throughout $G$. If our subset is empty the $h$ is a constant so $u$ is a constant an this makes $f$ a  constant.


问:
Prove that the recursively defined sequence is Cauchy.
Define a sequence recursively by $x_1=1$, $x_{n+1}=\frac{5+5x_n}{5+x_n}$. Prove that this sequence is Cauchy. Then find the limit.

We call a sequence $\{a_n\}$ Cauchy, if for any $\epsilon>0$, there exists $N\in\mathbb{N}$ such that for all $n,m\geq M$, $|a_n-a_m|<\epsilon$.

答:
There is another shortcut solution, if you are allowed to use the Banach fixpoint theorem considering the function

$$f(x) = \frac{5+5x}{5+x} = 1+\frac{4x}{5+x}$$

Now, you see immediately that
$$x>0 \Rightarrow f(x) > 1 \text{ and } f(x) < 5$$

It follows that

$$f: [1,5] \rightarrow [1,5]$$

Since

$$f'(x) = \frac{20}{(5+x)^2}\Rightarrow |f'(x)| \leq \frac{20}{36}=\frac 59 < 1 \text{ on } [1,5]$$

it follows, that $f$ is a contraction on $[1,5]$ and hence $f$ has a uniquely determined fixpoint $f(x^{\star}) = x^{\star}$ and any recursion

$$x_{n+1} = f(x_n) \text{ with } x_1 \in [1,5]$$

will converge to this fixpoint.

So, for any starting point $x_1 \in [1,5]$ the recursion is a Cauchy sequence.


问:
Derivative of trace squared and trace of a matrix squared?
Going through the matrix cookbook and other resources I can't find an algebraic proof of the following, if $f:\mathbb{R}^{n\times n} \to \mathbb{R}$ and $F:\mathbb{R}^{n\times n} \to \mathbb{R}$ and $H\in\mathbb{R}^{n\times n}$such that
$$
f(A) :=\left( \text{tr}[A]^2 \right) \text{ and } F(A):= \left( \text{tr}[A^2]\right).
$$
It then asks to calculate $\partial_H f(A)$ and $\partial_H F(A)$ but I can't figure the algebra out and I'm stuck on both.

答:
Most of the calculation relies on linearity of trace.

For the first one,
\begin{align*}
(D_Hf)(A) &= \lim_{t \to 0} \frac{\operatorname{tr}[A + tH]^2 - \operatorname{tr}[A]^2}{t}\\
&= \lim_{t \to 0} \frac{\left(\operatorname{tr}[A] + \operatorname{tr}[tH]\right)^2 - \operatorname{tr}[A^2]}{t}\\
&= \lim_{t \to 0} \frac{\operatorname{tr}[A]^2 + 2 \operatorname{tr}[A]\operatorname{tr}[tH] + \operatorname{tr}[tH]^2 - \operatorname{tr}[A^2]}{t}\\
&= \lim_{t \to 0} \frac{2 t \operatorname{tr}[A]\operatorname{tr}[H] + t^2\operatorname{tr}[H]^2 }{t}\\
&= 2\operatorname{tr}[A]\operatorname{tr}[H].
\end{align*}

For the second one,
\begin{align*}
(D_H F)(A) &= \lim_{t \to 0} \frac{\operatorname{tr}[(A + tH)^2] - \operatorname{tr}[A^2]}{t}\\
&= \lim_{t \to 0} \frac{\operatorname{tr}[A^2 + t(AH + HA) + t^2H^2] - \operatorname{tr}[A^2]}{t}\\
&= \lim_{t \to 0} \frac{\operatorname{tr}[A^2] + t\operatorname{tr}[(AH + HA)] + t^2\operatorname{tr}[H^2] - \operatorname{tr}[A^2]}{t}\\
&= \lim_{t \to 0} \frac{t\operatorname{tr}[(AH + HA)] + t^2\operatorname{tr}[H^2]}{t}\\
&=\operatorname{tr}[AH + HA]\\
&=2\operatorname{tr}[AH].
\end{align*}


问:
Spectrum of bounded linear operator
I'm working on the following problem.

Let $(e_n)$ be a complete orthonormal sequence in a Hilbert space $H$, and let $(\lambda_n)$ be a bounded complex sequence. Let $T$ be a bounded linear operator on $H$ such that $Te_n=\lambda_ne_n$ for each $n$. Find $\sigma(T)$, the spectrum of $T$.

What I've done so far: I know that if $\lambda\in\sigma(T)$ then $T-\lambda I$ is not invertible.

I think that $\{\lambda_n\}_n\subset\sigma(T)$, because $(T-\lambda_nI)e_n=Te_n-\lambda e_n=0$, which means $e_n\in\ker(T-\lambda_n I)$, so $T-\lambda_n I$ is not injective (which means it is not invertible.)

But I don't know if there are other members of $\sigma(T)$. Any help would be appreciated!

答:
Since $\sigma (T)$ is closed we see that the closure of $\{\lambda_n
: n \geq 1\}$ is contained in it. If $\lambda $ does not belong to this closure there is a positive distance $r$ from $\lambda$ to this set. Thus $|\lambda-\lambda_n| \geq r$ for all $n$. I will let you check tat $T-\lambda I$ has  a bounded inverse in this case so $\lambda$ cannot be in the spectrum. [Note that there is an obviosu formula for the inverse of $T-\lambda I$]

Title: Spectrum of bounded linear operator


问:
Showing continuity at a point and solving integral.
I know that, in the problem below, $\int\limits_{-1}^{1}fd\alpha = f(0)$. How can infer this from the proof?

Problem. Let $f:[-1,1]\to R$ be a bounded function, $a,b,c\in R$ be such that $a<b<c$, and let $\alpha : R\to R$ be defined by
$$
 \alpha(x)=\left\{\begin{array}{cc} a & x<0 \\ b & x=0 \\ c & x>0 \end{array}\right. .
$$ Show that if $f\in{R(\alpha)}$ on $[-1,1]$ then $f$ is continuous at $x=0$ and, in this case find the value of $\int\limits_{-1}^{1}fd\alpha$.

Proof. Suppose $f\in R(\alpha)$. Let $\varepsilon>0$. There exists a partition, say $P$, of $[-1,1]$ such that
$$
 U(P,f)-L(P,f)<\varepsilon\cdot\min(b-a,c-b).
$$
Let $P=\{x_{0},x_{1},...,x_{k},0,y_{1},...,y_{kr}\}$, where $-1=x_{0}<x_{1}<...<x_{k}<0<y_{1}<...<y_{r}=1$. Then the only terms in the difference of the sum $U(P,f)-L(P,f)$ that survive are precisely those corresponding to the subintervals $[x_{k},0]$ and $[0,y_{1}]$. Therefore
\begin{align*}
U(P,f)-L(P,f)&=\left(\sup_{y\in[x_{k},0]}f(y)-\inf_{y\in[x_{k},0]}f(y)\right)\left(b-a\right)\\
&\qquad\qquad\qquad+\left(\sup_{y\in[0,y_{1}]}f(y)-\inf_{y\in[0,y_{1}]}f(y)\right)\left(c-b\right)\\
\quad&<\varepsilon\cdot\min\left(b-a,c-b\right).
\end{align*}
Thus, in particular,
$$
\left(\sup_{y\in[x_{k},0]}f(y)-\inf_{y\in[x_{k},0]}f(y)\right)\left(b-a\right)<\varepsilon\cdot\min\left(b-a,c-b\right)
.$$
Hence $$\sup_{y\in[x_{k},0]}f(y)-\inf_{y\in[x_{k},0]}f(y)<\varepsilon\quad\text{and similarly}\quad\sup_{y\in[0,y_{1}]}f(y)-\inf_{y\in[0,y_{1}]}f(y)<\varepsilon.$$
Now $0\in (x_{k},y_{1})$, the latter is an open neighborhood of $0$ and, if $a\in(x_{k},y_{1})$, then
$$
|f(a)-f(0)|\le\max\left(\sup_{y\in[x_{k},0]}f(y)-\inf_{y\in[x_{k},0]}f(y),\sup_{y\in[0,y_{1}]}f(y)-\inf_{y\in[0,y_{1}]}f(y)\right)<\varepsilon.
$$
Taking $\delta=\min(x_{k},y_{1})$, we see that if $|a|<\delta$ then $|f(a)-f(0)|<\varepsilon.$ Therefore, since epsilon was arbitrarily chosen, $f$ is continuous at $0$.

答:
Consider the partition $P=\{-1,x_1,0,x_2,1\}.$ Then, $U(f,P,\alpha)=M_1(b-a)+M_2(c-b)$ and $L(f,P,\alpha)=m_1(b-a)+m_2(c-b)$ where $M_i,m_i:1\le i\le2$ are the maxima (resp. minima) of $f$ on the corresponding subintervals determined by $P.$ Then, $U(f,P,\alpha)-L(f,P,\alpha)=b(M_1-m_1)+a(M_2-m_2).$ Now as $x_1\to 0^-$ and $x_2\to 0^+$ continuity of $f$ at $0$ implies that $M_i-m_i\to 0.$ It follows that $f$ is integrable.

Now, let $\epsilon>0$. Choose $\delta>0$ such that $|x-0|<\delta\Rightarrow |f(x)-f(0)|<\epsilon.$ Then, if $P$ is as above and $|x_i|<\delta: 1\le i\le 2$, we have

$U(P,f,\alpha)-f(0)(c-a)=M_1(b-a)+M_2(c-b)-f(0)(c-b+b-a)=$

$(M_1-f(0))(b-a)+(M_2-f(0))(c-b)=\epsilon(c-a).$ It follows that $\int_{-1}^1fd\alpha=f(0)(c-a).$


问:
uniformizer is generator of maximal ideal
I cannot understand why uniformizer is a generator for the maximal ideal in "Joseph H. Silverman, The Arithmetic of Elliptic Curves", which is as follows:

$$\text{ord}_p: \bar K[C]_p \to \{ 0,1,\dots \} \cup \{ \infty \}$$

where $C$ is a curve and $$\text{ord}_p(f) = \sup \{ d \in \mathbb N : f \in M^{d} _p \} , \\ M_p = \{ g \in \bar K [C] : g(p) = 0 \} .$$

A uniformizer for $C$ at $P$ is any function  $ t \in \bar K [C]$ with $\text{ord}_p(t)＝1$

Silverman says uniformizer is a generator for the ideal $M_p$.

I want to understand the proof of this, that is, the uniformizer is a generator.
I firstly tried to show $t$ is a generator of $M_p/M_p^2$ and use Nakayama's lemma. Maybe this is overthinking.I believe this statement has quite easy proof.

Thank you in advance.

答:
Given $g \in M_p$, then $g(p) = 0$ so $\DeclareMathOperator{\ord}{ord} \ord_p(g) \geq 1$. Since $\ord_p(t) = 1$, then
$$
\ord_p(g/t) = \ord_P(g) - \ord_p(t) = \ord_P(g) - 1 \geq 0
$$
so $g/t \in \overline{K}[C]_p$ since $\overline{K}[C]_p$ is the valuation ring of $\ord_p$. Then there exists a function $h \in \overline{K}[C]_p$ such that $g = ht$, so $g \in (t)$. Thus $M_p \subseteq (t)$ and the other inclusion is immediate.

(You certainly could use Nakayama's lemma to prove this. Since $M_p/M_p^2$ is $1$-dimensional and $t \notin M_p^2$, then $\{t\}$ is a basis for $M_p/M_p^2$.)


问:
Prove that Gauss map on M is surjective
Let $M$ be a closed, orientable, and bounded surface in $\mathbb{R}^3$.

(a) Prove that the Gauss map on $M$ is surjective.

(b) Let $K_+(p) = \max \{0, K(p)\}$.  Show that
$$
\int K_+dA \ge 4\pi.
$$
in the area of $M$.  Do not use the Gauss-Bonnet Theorem to prove this.

答:
For a) there is a simple argument (due to Hilbert IIRC). Pick any unit vector $v$ and imagine a plane with normal vector $v$ with distance $t$ from the origin (imagine $t$ large enough so it is further away from the origin than $M$).

Reduce $t$ until the plane first touches the surface. The point on $M$ where it first touches is a point on $M$ whose image under the Gauss map is $v.$ To make this idea rigourous, you will want to formalize the notion of distance between $M$ and the plane. The compactness of $M$ will ensure the infimum is a minimum.

ContentLicense: CC BY-SA 3.0

Title: Prove that Gauss map on M is surjective


问:
To prove that a set exists, do I need to demonstrate that the set has a superset?
As I read various examples of how to prove that sets with particular properties exist, I notice that a requirement seems to be that the set-to-be-constructed must be a subset of another set that I have already demonstrated to exist. Why must this be so?

To provide the most recent example I came across...I made a post asking about the formal statement for the construction of the cartesian product $S \times T$.

Rather than: $\exists C \forall z [z \in C \leftrightarrow \exists x \exists y (x \in S \land y \in T \land z=(x,y))]$, the correct answer is apparently:

$\exists C \forall z [z \in C \leftrightarrow \color{red}{z \in A} \land \exists x \exists y (x \in S \land y \in T \land z=(x,y))]$, where $A$ is a set that has been previously demonstrated to exist.

I interpreted this as saying "$C$ must be a subset of $A$ in order for $C$ to exist".

From the Axiom Schema of Separation/Comprehension, I know that every set has a subset. But the above example (and a few others I have come across) seems to say that every set must be the subset of another set. At first, I was tempted to say that "If all sets have subsets, then all sets are subsets", but I don't think that is a true implication.

What issues do we run into if we construct a set for which we cannot demonstrate has a superset? I assume that there must be some sort of contradiction that we risk encountering (perhaps related to Russell’s  paradox?) but I am uncertain. Any insight is greatly appreciated. Thanks!

答:
The condition of being a subset of some previously defined set is just important when you are trying to define a set using the axiom of separation. Naively, a set is a collection of all sets satisfying some given property, but we know from Russell's paradox that this is untenable. ZF fixes this issue by requiring that when we define a set by a comprehension that it is a subset of some other set.

But separation is not the only way to define new sets in ZF. And if we only had separation, we couldn't actually define any set, since there would be no supersets to start out with. So say we add an axiom that says the empty set exists. Then we're stuck again with just the empty set. But if we add some combination of pairing, union, power set and replacement, then we can define more and more sets. And then once we add infinity, we get an infinite set and it's off to the races.

Still, these construction axioms (at least if we don't include replacement), are pretty crude. They get us bigger and bigger sets, but don't let us slice and dice them logically... that's what separation is for. And so a large amount of definitions are ultimately definitions from separation, which is why you are seeing many definitions that require a stipulation that the set you are defining is a subset of some pre-defined set.


问:
Equivalence between elements of different sets - how to formally define the "equivalence classes"?
What would be a good formal definition for the set $\mathcal T$ constructed below? I will try and give a simple example.

Consider the set $[\;n\;]=\{1,2,\ldots,n\}$, as well the family $\mathcal{C}$ of all subsets of cardinality 3 of $[\;n\;]$: $$\mathcal{C}=\{\{1,2,3\}, \{1,2,4\}, \ldots, \{n-2,n-1,n\}\}.$$
Obviously, every set $C \in \mathcal{C}$ has the same structure - it consists of three elements, $C=\{c^{C}_{1},c^{C}_{2},c^{C}_{3}\}$ with $c^{C}_{1}<c^{C}_{2}<c^{C}_{3}$ - or in other words, every set $C \in \mathcal{C}$ has a smallest, a second-smallest and a largest element.

I would now like to define a set $\mathcal{T}=\{t_{1},t_{2},t_{3}\}$ which generalizes the sets $C\in \mathcal{C}$ in the following sense: $t_{1}$ is a symbol representing the smallest element of a three-element set,  $t_{2}$ represents the second-smallest element, and $t_{3}$ represents the largest element. (The intention is to view $\mathcal{T}$ as a poset and use its properties to draw conclusions about $\mathcal{C}$, but that is irrelevant for the current question.)

One possible way of defining $\mathcal{T}$ might be to construct its elements as equivalence classes in the following way:

Definition Ver. 1: Consider a multiset $$D=\bigcup_{C\in\mathcal{C}}\bigcup_{j=1}^{3}\left\{ c_{j}^{C}\right\}$$
and the following equivalence relation $\sim$ defined on $D$:
$$c_{j_{1}}^{C_{1}}\sim c_{j_{2}}^{C_{2}}\Longleftrightarrow j_{1}=j_{2}.$$
We define the set $\mathcal{T}$ as $\mathcal{T}=D/\sim$ and its elements as the equivalence classes $t_{j}=\left[c_{j}^{C}\right]$.

What I like about this version is the idea of defining each $t_{j}$ as an equivalence class, which gets rid of this weird "$t_{j}$ is a symbol representing the $j^\text{th}$-smallest element of a set" formulation. On the other hand - and this is my biggest issue here - is this equivalence relation well-defined? The statement $j_{1}=j_{2}$ makes sense while we're talking about the variables $c_{j_{1}}^{C_{1}}$ and $c_{j_{2}}^{C_{2}}$; but I'm somewhat inclined towards looking at them as their values, i.e. the actual elements of the set $[\;n\;]$, at which point they lose their information about which index $j$ and subset $C$ they are associated with... Which interpretation is the correct one here?

Here's an alternative version for this definition:

Definition Ver. 2: Consider the set $$\mathcal{T} = \bigcup_{j=1}^{3}\left\{ t_{j}\right\}.$$
For a given set $C \in \mathcal{C}$, identify every element $c^{C}_{j}$ of $C$ with the element $t_{j}$ of $\mathcal{T}$ via the isomorphism $\tau_{C}:\mathcal{T}\rightarrow C$, $\tau_{C}:t_{j}\mapsto c^{C}_{j}.$

Using isomorphisms kind of makes more sense to me. On the other hand, now I'm stuck with my elements $t_{j}$ of $\mathcal{T}$ just being symbols without any further meaning to begin with. Furthermore, it seems like a strange construction to identify elements $t_{j}$ and $c^{C}_{j}$ by matching their indices $j$ - is it even clear when writing it like this? And is it legitimate to define $\mathcal{T}$ and $\tau_{C}$ together in one breath like that? (I could separate the definitions, but that makes the introduction of the object $\mathcal{T}$ even less understandable...)

Any help to clarify this will be greatly appreciated!

答:
I find your proposed definitions a bit difficult to parse. I think, however, that the following may be close to what you're looking for:

We start by strengthening your initial observation, that each of the elements of $\mathcal{C}$ (thought of as linear orders) are "the same." In fact, they're all the same in a unique way. Precisely, for any two three-element linear orderings $L_1,L_2$ there is a unique isomorphism $L_1\cong L_2$. This uniqueness is crucial: it lets us unambiguously talk about "$L_1$'s version of $x$" when $x$ is in $L_2$.

Thinking in terms of the "$L_1$'s version of $x$"-language, and generalizing to arbitrary structures since we don't really need to talk about linear orderings specifically (for your specific example, see below), this sets up the following idea. Suppose we have a set $\mathcal{X}$ of structures such that for any $A,B\in\mathcal{X}$ there is a unique isomorphism $A\cong B$. Let $$\mathfrak{X}=\{\langle A,a\rangle: A\in \mathcal{X}, a\in A\}$$ be the set of "labelled elements" of elements of $\mathcal{X}$. We get an equivalence relation $\sim$ on $\mathfrak{X}$ given by $$\langle A,a\rangle\sim \langle B,b\rangle\iff f_{A,B}(a)=b$$ where $f_{A,B}$ is the unique isomorphism $A\cong B$. We can then naturally view $\mathfrak{X}/\sim$ as a structure of the same type as the elements of $\mathcal{X}$ - and in fact we'll have $\mathfrak{X}/\sim$ be uniquely isomorphic to each element of $\mathcal{X}$ as expected.

In your specific example, here's what the above approach amounts to:

$\mathfrak{X}$ has $3\cdot{n\choose 3}$ elements, including $\langle\{1,2,3\}, 2\rangle$ and $\langle \{2,3,4\},2\rangle$.

The equivalence relation $\sim$ identifies $\langle A,a\rangle$ and $\langle B,b\rangle$ when $a$ occupies the same "place" in $A$ that $b$ does in $B$. So for example we have $$\langle \{1,2,3\},2\rangle\sim \langle \{1,2,4\},2\rangle$$ but $$\langle \{1,2,3\},2\rangle\not\sim \langle \{2,3,4\},2\rangle$$ (in the latter case, "$2$" is the second element of $\{1,2,3\}$ but the first element of $\{2,3,4\}$: the unique isomorphism between those two three-element linear orders sends $1$ to $2$, $2$ to $3$, and $3$ to $4$).

There are, as desired, exactly three $\sim$-classes of elements of $\mathfrak{X}$. So $\mathfrak{X}/\sim$ is a set with three elements, each of which is itself a set of $n\choose 3$ elements, each of which is itself an ordered pair, the first coordinate of which is a three element subset of $[n]$ and the second coordinate of which is an element of that subset. Whew!

The set $\mathfrak{X}/\sim$ can be turned into a linear order in a natural way: we set $$[\langle A,a\rangle]_\sim \le [\langle B,b\rangle]_\sim$$ iff we have $f_{A,B}(a)\le_Bb$, where

$f_{A,B}$ is the unique isomorphism from $A$ to $B$, and

"$\le_B$" means "$\le$ in the sense of $B$" (which here is really just "$\le$," I'm using the more complicated notation to emphasize that we could be playing with very different $A$s and $B$s but things would still work).

Of course we need to show that this $(i)$ is well-defined and $(ii)$ actually satisfies the linear order axioms, but this isn't hard.

Let me end by saying a bit about the importance of the uniqueness which I mentioned in the second paragraph and used crucially in defining $\sim$.

Suppose I have two structures $A$ and $B$ which are isomorphic but not uniquely isomorphic. For example, maybe $A$ is the linear ordering consisting of the rationals and $B$ is the linear ordering consisting of the dyadic rationals. It turns out that these are isomorphic, although that's not at all obvious. Now my question is:

What's $B$'s version of $1\over 3$?

The point is that there are lots of ways to define an isomorphism between $A$ and $B$. In fact, there are as many as possible in a precise sense: a countable dense linear order is homogeneous. This prevents us from translating from one structure to the other in an unambiguous way.

We may still have some clever way of choosing a particular isomorphism between two structures according to some other special criteria, but if there are multiple isomorphisms to choose from this will probably be difficult. So if you have a huge collection $\mathcal{X}$ of non-uniquely-isomorphic structures, your best bet is probably to just pick a specific element of $\mathcal{X}$ to work with rather than try to whip up an "unbiased" version.


问:
Gauss-Bonnet-Chern theorem
Good morning/day/evening/night,

I was presented to the generalized Gauss-Bonnet-Chern theorem for hypersurfaces in Euclidean space;

For a closed, even dimensional manifold $M$ with dimension $n$ embedded in $\mathbb{R}^{n+1}$ we have
$$\int_M K\mathbb dV = \text{Volume }\mathbb{S}^n\cdot\frac{\chi(M)}{2}.$$

I wonder,

How does this theorem look for compact manifolds with boundary?

Can this theorem be generalized to hold for closed odd dimensional manifolds? (I assume there will be a problem here since the Euler characteristic vanishes.)

Both questions are for hypersurfaces in Euclidean space.

Any reference will be greatly appreciated.

答:
The Gauss Bonnet Chern theorem is a special case of the Atiyah-Singer Index theorem, applied to calculated to the index of $d + d^*$ on differential forms, considered as mapping from even-degree forms to odd degree forms, which is exactly the Euler characteristic.

For a general (oriented) closed manifold, this comes down to
$$ \chi(M) = \int_M \mathrm{Eul}(TM),$$
that is, the Euler Characteristic is the integral over the Euler class of the tangent bundle of the manifold, which can be locally computed in terms of curvature. In two dimensions, the Euler class is just a multiple of the Gauss curvature times the volume form.

This holds for odd-dimensional manifolds as well (and simply states that the Euler genus is zero), and the orientability can be discussed away by either going to the oriented double cover of $M$ or by noticing that the integrand is still a well-defined volume density on $M$ even though it is not a proper differential form.

For the case with boundary, the formula is similar, but involves another curvature term, which is integrated over the boundary of $M$. See here, Thm. 1 for example. In the two-dimensional case (i.e. one-dimensional boundary), this integrand is the geodesic curvature of the boundary.

Title: Gauss-Bonnet-Chern theorem


问:
Show that $\sum_{j=1}^k a^2_j\geq 1$.
Suppose $1 \leq k \leq n$ and $x_1, x_2, \ldots, x_k$ are $k$ vectors in $\Bbb R^n$ satisfying for any $1 \leq i, j \leq k$,
$\langle x_i, x_j\rangle = \delta_{ij}$. For each $1 \leq j \leq k$, let $a_j$ be the 1st component of $x_j$ . Show that
$\sum_{j=1}^k a^2_j\geq 1$. I am kind of getting a feel that I have to use least square method but unable to figure it out. Please help.

Edit:
The above inequality will be reversed as Kavi Rama Murty observed. I actually made a typo but don't want to change the question drastically as suggested by the math people. So I am adding this that we have to prove $\sum_{j=1}^k a^2_j\geq 1$.

答:
This is false. Take $n=3, x_1=(0,1,0), x_2=(0,0,1)$.

EDIT: The reverse inequality follows by Parseval's identity Extend $\{x_1,x_2,...,x_k\}$ to an orthonomal basis $\{x_1,x_2,...,x_n\}$ and conclude that $1=\|e_1\|^{2}=\sum\limits_{i=1}^{n} |\langle e_1, x_i \rangle|^{2} \geq \sum\limits_{i=1}^{k} |\langle e_1, x_i \rangle|^{2}$ where $e_1=(1,0,0,,,0)$.

Title: Show that $\sum_{j=1}^k a^2_j\geq 1$.


问:
Proving that If $A$ and $B$ are compact then $\operatorname{dist}(A,B) > 0$
I have seen this question on here before, but I have a different attempt at the proof for it. We are given that $A$ and $B$ are compact and that $\operatorname{dist}(A,B) = \inf\{\Vert x-y \Vert: x \in A, y \in B\}.$

Here is my attempt at it:

So since we have that $A$ and $B$ are compact, we have that by the Heine-Borel theorem that both $A$ and $B$ are closed and bounded. There is a theorem in my text that states that for $E \subseteq \mathbb{R^n}$ and $E$ closed, then $E$ contains all of its limit points. So from this we have that there exists $x_k \in A$ which converges to $x$, and some $y_j \in B$ that converges to $y$. Now, by the Bolzano Weierstrass theorem, each convergent sequence, has a convergent subsequence, namely we have that $y_{j_t}$ converges to $y$ as $t$ goes to infinity, and similarly for $x_{k_p}$ converges to $x$ as $p$ goes to infinity. But from here I'm unsure where to finish. I feel as if the approximation theorem for infimum can be used here, but maybe not.

答:
You don't say it, but I assume we also know $A \cap B = \emptyset$.  Assume $A, B \subseteq X$.

Define $U_n= \{x \in X \mid d(x, A) \gt \frac 1n \}$.  Each $U_n$ is open and since $A$ is closed (we don't need $A$ compact), $\{ U_n \mid n \in \Bbb N \}$ is an open cover of $X \setminus A$.

Since $A \cap B = \emptyset, \{U_n \cap B \mid n \in \Bbb N \}$ is an open cover of $B$.

$B$ is compact so this open cover must have a finite subcover, so for some $N \in \Bbb N, B \subseteq U_N$ and $\operatorname{dist}(A, B) \ge \frac 1N \gt 0$.


问:
SCH part of Jech's book
I do not understand the purpose of the end of the last but one line and the last line in the following snippet from Jech's book:
if $2^{\operatorname{cf}\kappa}\geq \kappa$ then $\kappa^{\operatorname{cf}\kappa}=2^{\operatorname{cf}\kappa}.$ Namely, why the first condition with $\geq$ is assumed ?

答:
I'll turn my comments into an answer to get this off the unanswered list...

The results of all cardinal exponentiation calculations are determined from the $\gimel$ function ($\kappa\mapsto \kappa^{\operatorname{cf}\kappa}$) on singular cardinals, and the continuum function ($\kappa\mapsto 2^\kappa$) on regular cardinals. Jech demonstrates this in corollary 5.21, before the section on SCH. (Note that the gimel function and the continuum function give the same thing on regular cardinals, so we can really say that cardinal exponentiation is determined by the gimel function alone, which is how Jech puts it.)

What Jech is getting at in this confusingly terse passage is that SCH allows us to reduce the gimel function on singular cardinals to the continuum function on regular cardinals. He splits it up into two cases: The first case is: $2^{\operatorname{cf}\kappa}\ge \kappa,$ then (in ZFC alone), we have $\kappa^{\operatorname{cf}\kappa}= 2^{\operatorname{cf}\kappa},$ which is the value of the continuum function on the regular cardinal $\operatorname{cf}\kappa.$ The second case is $2^{\operatorname{cf}\kappa}< \kappa,$ in which the SCH says $\kappa^{\operatorname{cf}\kappa}$ takes its minimum possible value $\kappa^{\operatorname{cf}\kappa}=\kappa^+.$

So, since we can get gimel in terms of values of the continuum function on regular cardinals we should be able to reduce any computation $\kappa^\lambda$ to that as well. Jech puts this together in theorem 5.22, just after the passage you quoted.

Title: SCH part of Jech's book


答:
$I(x)$ integral done by two different methods yields $I_1(x), I_2(x)$. These two differ only  by a constant independent of $x$: $I_1(x)-I_2(x)=$Cosnstant, eventually.
In @lab Bhattacharjee 's answer above
$$I_2(x)=\frac{x}{2}\sqrt{a^2-x^2}-\frac{a^2}{2} \cos^{-1}(x/a)$$
and$$I_2(x)=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2} \sin^{-1}(x/a)$$
So $$I_1(x)-I_2(x)=\frac{a^2}{2}[-\cos^{-1}(x/a)-\sin^{-1}(x/a)]=-\frac{a^2\pi}{4}$$

Body:
Why I cannot get the same answer if I do substitution with $x=a \cos\theta$ for $\int \sqrt{a^2-x^2}dx$ compared with substitution $x=a \sin\theta$?

$\int \sqrt{a^2-x^2}dx = \int \sqrt{a^2 - a^2 \cos^2 \theta} d\theta = \int a \sin\theta d\theta = \int a \sin\theta(-a \sin\theta)d\theta = -a^2 \int sin^2 \theta = -a^2[\frac{\theta}{2} - \frac{\sin(2\theta)}{4}]+C$

However, the correct answer seems to be $a^2(\frac{\theta}{2} + \frac{\sin(2\theta)}{4}) + C$


问:
Riesz Representation Theorem, true for pre-Hilbert spaces and any functional?
I have some doubts about the Riesz theorem.
Firstly can you check my proof?

Fa = for all

Fa (H,<,>) a pre Hilbert space

Fa x in H different from the zero vector

Fa F a functional on H

Fa T: (H->R) -> H : T(F) = F(x) x/||x||^2 a map from a functional on H to H

Then F(x) = F(x) < x, x > / ||x||^2 = < F(x) x/||x||^2, x > = < T(F), x >

End of proof, where F(x) = < T(F), x > for any vector x in a pre Hilbert space.

If my proof is correct, it implies that H doesn’t have to be an Hilbert space but only a pre Hilbert one and that the functional doesn’t have to be neither linear and continuous, but only a functional.

Is it true or I’m missing something?

答:
The Riesz Representation theorem is:

$$\forall F \in H^*, \exists y \in H, \forall x \in H, F(x) = \langle y, x\rangle$$

What you have shown is

$$\forall F \in H^*, \forall x \in H, \exists y \in H, F(x) = \langle y, x\rangle$$

$y$ or $T(F)$ in the Riesz Representation theorem is picked knowing only $F$, not $x$. This single value of $y$ must work for every $x$. In your result, you pick $x$ first, and then define $T(F)$ in terms of it. Thus your $y$ changes from one value of $x$ to another.


问:
Does $\sum\frac{4^n(n!)^2}{(2n)!}$ converge?
Does $\sum\frac{4^n(n!)^2}{(2n)!}$ converge? The ratio test is inconclusive. I know that $\sum\frac{2^n(n!)^2}{(2n)!}$ converges, if that's of any help. </p>

答:
Try

$$
  a_n = \frac{4^n (n!)^2}{(2n)!}
$$
$$
  \frac{a_{n+1}}{a_{n}} = 4 \frac{(n+1)^2}{(2n+1)(2n+2)} > 1
$$

conclude that $a_n$ is monotonically increasing and also $a_n > 0$ so
$$
    \lim \limits_{n \to \infty} a_n \ne 0
$$
So
$\sum a_n $ diverge.


问:
Characters of irreducible representations of odd-order groups
I want to prove that if $G$ is a group and the order of $G$ is odd, and $\chi$ is a real-valued irreducible character of $G$, then $\chi$ must be the trivial representation, $\chi = \epsilon$.

So far, I know that $|G| = \sum_{i = 1}^{n} n_i\chi_i^2$, and that $|G|$ odd $\implies \exists j \in \mathbb{N}: |G| = 2j+1$. Therefore:

$2j+1 = n_1 \chi^2 + \sum_{i=2}^n n_i \chi_i^2$, and therefore $n_1\chi^2$ must be even and $\sum_{i=2}^n n_i \chi_i^2$ even, or both of them must be odd. Since I have no further information about the characters of the other irreducible representations of this group, I am not sure how to proceed. Any tips?

答:
If you know the Frobenius–Schur indicator, you can argue as follows:

Suppose not, i.e., $\chi$ is not the principal character, then by the orthogonality relationship, we have $\frac{1}{|G|}\sum_{g\in G}\chi(g)=0$.

Note that the Frobenius-Schur indicator can be computed as:
\begin{align}
\nu_2(\chi)&=\frac{1}{|G|}\sum_{g\in G}\chi(g^2).
\end{align}
But $|G|\equiv 1\pmod 2$ implies $g\mapsto g^2$ is a bijection of $G$. Hence
$$\nu_2(\chi)=\frac{1}{|G|}\sum_{g\in G}\chi(g^2)=\frac{1}{|G|}\sum_{g\in G}\chi(g)=0.$$

The Frobenius-Schur indicator equals $0$ can only happen when $\chi\ne\overline\chi$ which is a contradiction.


问:
Counting The Number of Equivalence Relations on a Set
I am trying to solve the following problem from a textbook:

Let $S$ be a set with cardinality $2m$. Give a closed-form expression for the number of equivalence relations on $S$ where every equivalence class has two elements.

My approach thus far: Every equivalence relation partitions $S$ into $m$ different subsets/classes. Thus, counting the number of equivalence relations is equivalent to counting the number of ways of choosing $m$ different objects in a $2m$-element set. And this is given by
$\binom{2m}{m}$.

Is my approach correct? If not, how can I solve this problem? I would appreciate any help on this!

答:
Unfortunately your approach chooses $m$ elements from a set of $2m$ elements, but it doesn't specify who their pairs would be.

To solve this: Choose the first class of two elements. the order of choice doesn't matter so divide by two. Choose the second class of two elements from what is left. Again divide by two. Continue this way until there are no more elements. Then you have to divide by the order of the pairs, which also didn't matter.

This way you get $[2m\cdot(2m-1)/2]\cdot[(2m-2)\cdot(2m-3)/2]...[2\cdot1/2]/m!$ so this is given exactly by $$\frac{(2m)!}{2^mm!}= {2m\choose{2,2,2,...,2}}/m!$$


问:
Newton-Raphson Method - need help understanding an example
I am currently trying to finish an assignment regarding Newton-Raphson method. Anyone able to explain an example to me? The function $f(x) = \sin(x) + \sin(\frac{10x}{3})$ is shown on the picture. The red dot is the starting point and the cross is a local extrema found by using the NR method. My question is why isn't the cross on the local extremas near the starting point, specifically where $x \approx 1.5$ and $x \approx 0.5$ which is visible on the picture. Why did the point end up where it did and passed two extremas on its way? Could this be an implementation issue, or is this how it's supposed to be ?

答:
If I poperly understood, you are looking for the minimum of
$$f(x)= \sin(x) + \sin(\frac{10}{3}x)$$ starting with $x_0=1$.

This means that you are looking for the zero of function
$$g(x)=f'(x)=\cos (x)+\frac{10}{3} \cos \left(\frac{10 }{3}x\right)$$ which, repeating your calculations gives the following iterates
$$\left(
\begin{array}{cc}
 0 & 1.00000 \\
 1 & 3.14110 \\
 2 & 3.41900 \\
 3 & 3.38666 \\
 4 & 3.38725
\end{array}
\right)$$

The problem is that
$$g''(x)=-\cos (x)-\frac{1000}{27} \cos \left(\frac{10 }{3}x\right)$$

At the starting point we have
$$g(1)=\cos (1)+\frac{10}{3} \cos \left(\frac{10}{3}\right)\approx -2.73194 $$
$$g''(1)=-\cos (1)-\frac{1000}{27} \cos \left(\frac{10}{3}\right)\approx 35.818$$ that is to say $g(1)\times g''(1) <0$.

By Darboux theorem, because of this result, you will have an overshoot of the solution.


问:
One root common to $ax^2+2bx+c=0$ and $dx^2+2ex+f=0$

If three distinct numbers $a,b,c$ are in GP, and the equations $ax^2+2bx+c=0$ and $dx^2+2ex+f=0$ have a common root, then which of the following statements is correct?
 $1.$ $d,e,f$ are in GP.
 $2.$ $d,e,f$ are in AP.
 $3.$ $\frac da,\frac eb, \frac fc$ are in GP.
 $4.$ $\frac da,\frac eb, \frac fc$ are in AP.

My attempt:

Let $r$ be the common ratio in $a,b,c$. So, $b=ar, c=ar^2$.

So, the first equation becomes $ax^2+2arx+ar^2=0\implies x^2+2rx+r^2=0$.

Let $\alpha$ be the common root. So, $\alpha^2+2r\alpha+r^2=0$. Also, $d\alpha^2+2e\alpha+f=0\implies \alpha^2+2\frac ed\alpha+\frac fd=0$.

On comparing, I get $r=\frac ed, r^2=\frac fd\implies(\frac ed)^2=\frac fd\implies e^2=fd$.

So, I am getting option $1$ as correct. But the answer is given as $4$. What's my mistake?

While this post indeed has a lot of good answers, my question was about my mistake in the method I followed. Scilife has answered that in the comments below.

答:
Following your notation, let us say that the common ratio of the geometric progression is $r$. The equation, as you currently noted, becomes:
$$ x^2 + 2rx + r^2 = 0 $$
It can easily be seen that the above equation has a double root, which is $-r$. Hence, this must be the root in common with the second quadratic equation as well.

Substituting the root into the second equation, we get:
$$dr^2 - 2er + f = 0$$
Using the fact that $r^2 = \frac{c}{a}$ and $r = \frac{c}{b}$ and dividing both sides of the equation by $c$, we see that:
$$\frac{d}{a} - 2\frac{e}{b} + \frac{f}{c} = 0 $$
Which is the condition for 3 numbers to be in arithmetic progression.


问:
Expected value inequality for non-negative random variable
Let $X$ be a non-negative random variable. Using the fact the next inequality holds:
$$n\ \mathbb{I}_{n\leq X < n+1} \leq X\ \mathbb{I}_{n\leq X < n+1} \leq (n+1)\ \mathbb{I}_{n\leq X < n+1}.$$

I need to prove the following inequality:

$$\sum_{n=1}^{\infty} \mathbb{P}\left(X \geq n\right) \leq \mathbb{E}(X) \leq 1 + \sum_{n=1}^{\infty} \mathbb{P}\left(X\geq n\right).$$

My attempt:

Let $n \in \mathbb{N}$, then, using the first inequality and applying $\mathbb{E}$, then:
$$\mathbb{E}[n\ \mathbb{I}_{n\leq X < n+1}] \leq \mathbb{E}[X\ \mathbb{I}_{n\leq X < n+1}] \leq \mathbb{E}[(n+1)\ \mathbb{I}_{n\leq X < n+1}].$$

Then, summing over $n$ then
$$\sum_{n=0}^{\infty}\mathbb{E}[n\ \mathbb{I}_{n\leq X < n+1}] \leq \sum_{n=0}^{\infty}\mathbb{E}[X\ \mathbb{I}_{n\leq X < n+1}] \leq \sum_{n=0}^{\infty}\mathbb{E}[(n+1)\ \mathbb{I}_{n\leq X < n+1}]$$

Using expected value properties, it holds
$$\sum_{n=0}^{\infty}n\ \mathbb{P}[n\leq X < n+1] \leq \sum_{n=0}^{\infty}\mathbb{E}[X\ \mathbb{I}_{n\leq X < n+1}] \leq \sum_{n=0}^{\infty}(n+1)\ \mathbb{P}[n\leq X < n+1]$$

Then
$$\sum_{n=0}^{\infty}n\ \mathbb{P}[n\leq X < n+1] \leq \sum_{n=0}^{\infty}\mathbb{E}[X\ \mathbb{I}_{n\leq X < n+1}] \leq \sum_{n=0}^{\infty}n\ \mathbb{P}[n\leq X < n+1] + \sum_{n=0}^{\infty} \mathbb{P}[n\leq X < n+1].$$

Here's the deal:

For $\sum_{n=0}^{\infty}\mathbb{E}[X\ \mathbb{I}_{n\leq X < n+1}]$ (my) intuition says that because  we're summing over all possible values of $n$, and the indicator function is considering the set of $\{n\leq X < n+1\}$, then we're considering all the half-open intervals $[n,n+1)$, since they're disjoint, the union of all of them gives us $[0,\infty)$ and so, since $X$ is a non-negative r.v., then $\sum_{n=0}^{\infty}\mathbb{E}[X\ \mathbb{I}_{n\leq X < n+1}] = \mathbb{E}[X]$.

For $\sum_{n=0}^{\infty} \mathbb{P}[n\leq X < n+1]$, again, since we're considering all the posible values of $n$ and we're looking for $n\leq X < n+1$, then essentially we're looking for the probability of $X \in [0,\infty)$, since $X$ non-negative, then $\sum_{n=0}^{\infty} \mathbb{P}[n\leq X < n+1] = 1$.

Up to this point, if my intuition in the last two arguments is correct, all that remains is to prove
$$\sum_{n=0}^{\infty}n\ \mathbb{P}[n\leq X < n+1] = \sum_{n=1}^{\infty}\mathbb{P}[X\geq n]$$
but here's where I'm having trouble. I know that, if we're talking about discrete r.v. then:
$$\sum_{n=1}^{\infty}\mathbb{P}[X\geq n] = \mathbb{P}[X=1] + 2\mathbb{P}[X=2] + \ldots = \sum_{n=1}^{\infty}n\ \mathbb{P}[X=n]$$
and in this case $\{n\leq X < n+1\}$ reduces to $\{X=n\}$, and so,
$$\sum_{n=0}^{\infty}n\ \mathbb{P}[n\leq X < n+1] = \sum_{n=1}^{\infty}n\ \mathbb{P}[X= n] = \sum_{n=1}^{\infty}\mathbb{P}[X\geq n]$$
and the proof will be done. But the problem is that, the only hypothesis I have for $X$ is that it is non-negative, so I would make a mistake considering $X$ discrete r.v.

I'm lost by this point to prove it in general, and I don't know if I'm missing any other property of expected value that could be useful. Any help would be appreciated.

Body: Let $b_n=P(X \geq n)$. Then  $ \sum\limits_{n=0}^{\infty} nP(n \leq X <n+1)=\sum\limits_{n=0}^{\infty} n (b_n-b_{n+1})=1(b_1-b_2)+2(b_2-b_3)+...=b_1+b_2+b_3+...= \sum\limits_{n=1}^{\infty} P(X \geq n)$.


问:
If $f:M\to \Bbb R$ is smooth and $f^{-1}(0)$ is a compact regular level set, then each $t\in (-\epsilon,\epsilon)$ is a regular value of $f$
Let $M$ be a smooth manifold and $f:M\to \Bbb R$ a smooth function. Suppose $0\in \Bbb R$ is a regular value of $f$, so that $f^{-1}(0)$ is an embedded submanifold of $M$. Also suppose that $f^{-1}(0)$ is compact. My goal is to show that $t$ is a regular value of $f$ for sufficiently small $|t|$.

My attempt: For each point $p\in f^{-1}(0)$, $f$ is a submersion at $p$, so $f$ is a submersion in a neighborhood $U_p\subset M$ of $p$. Now $\{U_p\}_{p\in f^{-1}(0)}$ is an open cover of the compact set $f^{-1}(0)$, so there must exist finitely many points $p_1,\dots,p_n\in f^{-1}(0)$ so that $f^{-1}(0)\subset U_{p_1}\cup \cdots \cup U_{p_n}=:U$. Also $f$ is a submersion in $U$. I would be done if I can show that $U$ contains $f^{-1}(-\epsilon,\epsilon)$ for some $\epsilon>0$, but I can't show this. Should I need a different approach? Thanks in advance.

答:
The statement is false. Let $M$ be the real line. Take any smooth map $f: \mathbb{R} \to \mathbb{R}$ such that $f$ vanishes only at zero but $f(x)$ goes to zero when $x$ goes to the infinity. Assume there exists a sequence $\{y_n\}$ such that $y_n \to \infty$ and $f'(y_n) = 0$. Then $f^{-1}(f(y_n))$ is not a regular fibre but $f(y_n)$ can be as close as we want to $0$. It is easy to construct such functions using smooth bump functions. Indeed, consider $f : \mathbb{R}^{\star} \to \mathbb{R}$ defined by
$$ f(x) = \frac{1}{\int_0^{x^2} \sin^2(2\pi y)\ dy}.$$
Then $f'(k) = 0$ for any $k \in \mathbb{Z}$ and $f > 0$ on $\mathbb{R}^{\star}$ and $f(x) \to 0$ when $|x| \to \infty$. Now, using a bump function, you can modify the function $f$ at the origin in such way $f(0) = 0$, $f$ is till positive outside the origin and $f'(0) = 1$ for instance.

If there exists a compact neighbourhood $K$ of $0$ such that $f^{-1}(K)$ is compact then all the values of $f$ in a small neighbourhood of $f$ have to be regular. Indeed, arguing by contradiction, assume there exist two sequences $\{x_n\}$ and $\{y_n\}$ such that $x_n$ tends towards $0$ and $f(y_n) = x_n$ and the Jacobian matrices $\mathrm{J}_f(y_n)$ have not full rank. We can assume that $x_n \in K$ and $\{y_n\}$ converges to some $y$ since $f^{-1}(K)$ is compact. By continuity, we should have $f(y) = 0$. Thus, $\mathrm{J}_f(y)$ has full rank, which is a contradiction.

In particular, the statement holds when $f$ is proper.


问:
How to solve an equation with variable exponents?
I have an equation of the form

$$x^a+(1-x)^b = 1$$

where $a$ and $b$ are constants.

Is there some way to manipulate this equation to solve for $x$?

Below is some more information about the problem for some context, although I don't think it affects the mathematics from the above equation.

I am looking at mixing relations for the electrical resistivity in porous media. Usually, we use something called Archie's Law:

$$\rho_b = \rho_f \phi ^{-m}$$

where $\rho_b$ is the bulk resistivity, $\rho_f$ is the fluid resistivity, $\phi$ is the porosity, and $m>0$ is a cementation exponent. In my case, the porosity, $\phi$, is unknown. For the above equation, it is easy to manipulate to solve for $\phi$. However, this simplistic equation is not sufficient for all problems and there is an alternative known as Modified Archie's Law which takes the form:

$$\rho_b\rho_m\phi^m + \rho_b\rho_f(1-\phi)^p = \rho_f\rho_m$$

This form incorporates the surrounding rock resistivity (i.e. the matrix resistivity, $\rho_m$) and uses an additional exponent, $p$. Now, if I want to solve for $\phi$, it is not clear how I can manipulate the equation.

Any help is appreciated.

答:
You may not always be able have other solutions besides those mentioned by lulu in their comments (when $a,b>0$, of course).

In the sequel, I presume, for well-definedness, that $x\in(0,1)$—allowing for $x\in[0,1]$ when $a,b>0$. For our particular demonstration, we consider the general equation $$f(x)=mx^a+n(1-x)^b\,,$$ where $m,n$ are of the same sign—we’d assume positive— and $a,b\notin\{0,1\}$. For now, assume $a=b$. Differentiating $f$ gives $$f’(x)=a(mx^{a-1}-n(1-x)^{a-1})$$ and $$f’’(x)=a(a-1)(mx^{a-2}+n(1-x)^{a-2})\,.$$ This shows that $x=\frac{c}{1+c}$, where $c:=\left(\frac{n}{m}\right)^{1/(a-1)}$ is a global minimum (respectively, maximum) of $f$ if $|a-\frac{1}{2}|> \frac{1}{2} $ (respectively, $a\in(0,1)$), which in either case is simply $$f(\frac{c}{1+c})=\frac{mc^a+n}{(1+c)^a}\,.$$ For $a<0$, in which case it is a minimum, it follows that $f(x)\ge 2$ when $m=n=1$ and there’s no solution to your equation (more generally, there’s no solution when the minimum is $>1$). On the other hand, it is easy to observe that before and beyond $x= \frac{c}{1+c}$, then $f$ is strictly monotonic, which implies that besides $x\in\{0,1\}$ —when $m=n=1$—we cannot have that $f(x)=1$ (and this is generally the case if the maximum is $>1$).

From the above, owing to continuity — and as probably hinted in G Cabs comment— it follows that a small perturbation of $b$ from $a$ would not alter the solution set as obtained above; in particular, there’d exist $\epsilon$ (may or may not depend on $a,b$) such that whenever $|a-b|\le\epsilon$, then $f(x)=1$ has no solutions when the case with $a=b$ also has no solutions.


问:
Topology of $P(\omega)$ as $2^\omega$ (ultrafilters are open?)
I'm looking for a sanity check more than anything else. We may identify $P(\omega)$ with $2^{\omega}$ in the obvious manner.

Let $U$ be a principal ultrafilter on $\omega$, so $U \subseteq 2^{\omega}$ (identifying $P(\omega)$ with $2^{\omega}$).

Is it true that $U$ is an open set with the cantor space topology on $2^{\omega}$? This is the topology generated by the basic open sets $N_s = \{ \alpha \in 2^{\omega} : s$ is an initial segment of $\alpha\}$ for every $s \in 2^{<\omega}$.

Since $U$ is principal, there is an $n \in \omega$ such that $U = \{A \subseteq \omega : n \in A \}$ so that for every string $\alpha \in 2^{\omega}$ with $\alpha(n) = 1$, $\alpha \in U$. But then there are $2^n$ many finite binary strings $s_i$ of length $n+1$ such that $s_i(n) = 1$, and then $U = \cup_{i \leq 2^n}N_{s_i}$ (if $\alpha \in N_{s_i}$ then it's initial segment is $s_i$ and thus $\alpha(n) = s_i(n) =1$, so $\alpha \in U$, and the other direction is similar.

So every principal ultrafilter is an open set in the cantor space? I'm pretty sure this is not true, but I also don't see the mistake either (I am quite sleep deprived right now) so I'm posting this here as a sanity check hoping that someone would correct me. Sorry if I'm missing something blaringly obvious.

答:
It is true. That topology is just the product topology when $2$ is given the discrete topology. It has as subbase the sets

$$\left\{\alpha\in 2^\omega:\alpha(n)=0\right\}$$

and

$$\left\{\alpha\in 2^\omega:\alpha(n)=1\right\}$$

as $n$ ranges over $\omega$. Thus, the principal ultrafilter over $n$ is not just open: it’s a member of this subbase for the topology.


问:
Inner Measure not a Measure on Power Set, and Equals Outer Measure for Lebesgue Measurable set
Is there any advice on this problem:

For $A \subset \mathbf{R},$ the quantity
$\sup \{|E|: E$ is a closed bounded subset of $\mathbf{R}$ and $E \subset A\}$
is called the inner measure of $A .$

(a) Show that if $A$ is a Lebesgue measurable subset of $\mathbf{R}$, then the inner measure of $A$ equals the outer measure of $A .$

(b) Show that inner measure is not a measure on the $\sigma$ -algebra of all subsets of $\mathbf{R}$

For (b) would the fact that $\sup(\varnothing)=-\infty$ be of use?

答: The reason why this holds is because there is no "perfect" measure. In Axler's text, this cannot be a measure since it is not a measure on all sets of $\mathbb{R}$


问:
Minimum variance for investment portfolio problem

Suppose you wish to invest a fixed sum of money in to assets that yield returns of $X$ and $Y$, where $X$, and $Y$ are random quantities. You will invest a fraction $a$ in $X$ and the remaining in $Y$. Because there is variability associated with the returns on these two assets we wish to choose $a$ such that the total risk or variance is minimized. We want to minimize $\mathbb{Var}(aX+(1-a)Y)$.

Find $a$ such that the risk is minimized.

Here's what I have gotten till so far:
$$
a^2 \mathbb{Var}[X]+(1-a)^2 \mathbb{Var}[Y]+a(1-a)\mathbb{E}{[X-\mathbb{E}(X)][Y-\mathbb{E}(Y)]}
$$

What's the next step for finding the minimum variance? Which terms should be expanded and how can they be rearranged together?

答:
The variance is
$$\text{Var}(aX+(1-a)Y) = a^2 \text{Var}(X) + (1-a)^2 \text{Var}(Y) + 2a(1-a) \text{Cov}(X,Y).$$
This is a quadratic of the form
$$c_2 a^2 + c_1 a + c_0$$
with
\begin{align}
c_2 &= \text{Var}(X) + \text{Var}(Y) - 2 \text{Cov}(X,Y)
\\
c_1 &= -2 \text{Var}(Y)  + 2\text{Cov}(X,Y)
\\
c_0 &= \text{Var}(Y)
\end{align}

If $c_2 < 0$, then this is a $\cap$-shaped quadratic, which will be minimized either at $a=0$ or $a=1$, depending on which of $\text{Var}(X)$ or $\text{Var}(Y)$ is smaller.

If $c_2 = 0$, then this is a line, will be minimized either at $a=0$ or $a=1$.

If $c_2 > 0$, then this is a $\cup$-shaped quadratic, which is minimized at $-\frac{c_1}{2c_2} = \frac{\text{Var}(Y) - \text{Cov}(X,Y)}{\text{Var}(X) + \text{Var}(Y) - 2 \text{Cov}(X,Y)}$ if this quantity is between $0$ and $1$; otherwise it is again minimized either at $a=0$ or $a=1$.


问:
Area of a graph above a region given particular values.
Suppose $S$ is a graph above the region $\Omega$ of the $(xy)$-plane defined by $x^2+y^2\leq a^2$ for a constant $a$, and we know that $|\frac{\partial z}{\partial x}|\leq 2$ and $|\frac{\partial z}{\partial y}|\leq 2$. What is the area of the surface $S$?

I think that the formula I need to use is:
$$
\iint_\Omega f(x,y,z(x,y))\sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2} \,dx\,dy
$$
So I think the area after calculation is $\frac{1}{2}a^2\leq Area(S) \leq 3a^2$

But I may have calculated wrong or maybe I'm missing something... Would anyone be able to help?

答:
$|r_x \times r_y| = \sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2} = \sqrt{1 + 2^2 + 2^2} = 3$

Surface Area of $S = \displaystyle \int_{-a}^{a}\int_{-\sqrt{a^2-y^2}}^{\sqrt{a^2-y^2}} 3 \,dx\,dy = 3 \pi a^2$

Or write it in polar coordinates for easier integration

$S = \displaystyle \int_{0}^{2 \pi}\int_{0}^{a} 3 r dr d\theta = 3\pi a^2$


问:
Convergence of $\sum_{k=1}^\infty ke^{-\sqrt{k}}$
So I've tried ratio test and it was inconclusive. I think comparison test would work with $\frac {k}{e^\sqrt{k}} \le \frac {k}{k^3}=\frac {1}{k^2}$ since $e^\sqrt k \ge k^3$ for large enough k. But I couldn't prove that inequality with induction. Can you help me prove that inequality or give me another way to show the convergence?

Thanks for your effort and time.

答:
It suffices to show that the inequality $e^\sqrt k \ge k^3$, that is $\frac{k^3}{e^\sqrt{k}}\leq 1$, holds eventually. This follows from the fact that
$$\lim_{k\to \infty}\frac{k^3}{e^\sqrt{k}}=0<1.$$
Actually $\lim_{k\to \infty}\frac{k^a}{e^{k^b}}=0$ for any positive numbers $a$ and $b$.


问:
What does $\leq ^{-1}$ mean?
Just came across a question on my homework assignment. The only thing it says is:

What is $\leq ^{-1}$?

It is on chapter for relations.

I'm guessing it means:

$R \leq R^{-1}$, for $R$ being a relations, then having $R$ factored out on each sides??

If it helps, the book I'm using is Mathematics: A Discrete Introduction, Chapter 14, exercise 14.12

答: Given a relation $R$, the relation $R^{-1}$ is another relation:$$a\mathrel{R^{-1}}b\iff b\mathrel Ra.$$

Title: What does $\leq ^{-1}$ mean?


问:
Simple gradient/line intersect question
Very, very basic question here: Given an x,y coordinate and a gradient (but no equation), how can I find the x and y axis intercepts? (assuming the line is linear)</p>

答:
You are given a point $(x_l,y_l)$ on a line with gradient $m$ and want to find out what value of $t$ such that $(t,0)$ lines on the line (and the same problem for $y$),

so when are both $(x_l,y_l)$ and $(t,0)$ on a line with gradient $m$? It is exactly when $\frac{x_l-t}{y_l-0} = m$, by algebra this gives $t = x_l - m y_l$.

Title: Simple gradient/line intersect question


问:
Find the inverse of the matrix
Find the inverse of the matrix

```<code>B=    1 0 2
      0 -1 1
      1 1 0

Thats easy -1  2  2
            1 -2 -1
            1 -1 -1
</code>```

But how do i show this

```<code>Hence solve BX = -3   where X = x1
                  4             x2
                  5             x3
</code>```

答: Let $v$ be the given vector (that is, the column vector $(-3,4,5)$). You have the equation $BX=v$. Multiplying on both sides by $B^{-1}$ gives $B^{-1}BX=B^{-1}v$, or $X=B^{-1}v$. You can easily compute the right hand side to find $X$. </p>

Title: Find the inverse of the matrix


Title: Convergence of $\sum_{k=1}^\infty ke^{-\sqrt{k}}$

Body:
So I've tried ratio test and it was inconclusive. I think comparison test would work with $\frac {k}{e^\sqrt{k}} \le \frac {k}{k^3}=\frac {1}{k^2}$ since $e^\sqrt k \ge k^3$ for large enough k. But I couldn't prove that inequality with induction. Can you help me prove that inequality or give me another way to show the convergence?

Thanks for your effort and time.

Body: $e^{\sqrt k} \geq \frac {(\sqrt k)^{4}} {4!}$. Hence $\frac k {e^{\sqrt k}} \leq \frac {24} {k^{2}}$.


问:
Differentiability at point of inflection
Point of inflection literally means that the slope does not change at that point so shouldn't all points of inflection be compulsarily differentiable?
I don't quite understand point of inflection. It's a point where the concavity of graph changes, but that seems too vague. The slope is constant at that point, but how can a slope be constant at a 'POINT'?? Don't we need nearby points for the slope? What exactly is so special about inflection point when all the points that are differentiable already have same left hand and right hand derivative? When x is tending to 0, don't all differentiable points have equal slope on left and right side? So why is inflection point any different?

Hope you got my question!

答:
As also illustrated in this answer, it is incorrect to assume that the condition $f''(x) = 0$ is the sole criteria to determine whether a point is a point of infection. I will quote the same example just to make this point.

Consider the function:
$$f(x)=\cases{ -x^2 & $x\le 0$ \\ x^2 & $x>0$ }$$

If you sketch the graph of the function, you can note quite clearly that the concavity of the above function changes at $x = 0$. However, the second derivative is not continuous at that point.

Therefore, your argument that a point of inflection implies that the slope "does not change" is also quite incorrect. While it is true that if the second derivative is continuous at a point $x = a$ and the second derivative exists at the point, then $f''(a) = 0$, we cannot simply conclude a point is a point of inflexion by noting the double derivative vanishes. You can also consider the graphs of the functions $y = x^{2n}$ for $2 \le n$ where the double derivative vanishes at $x=0$ but the concavity does not change.