TITLE: Why derivative of analytic polynomial equals ${\partial \over \partial x}$
QUESTION [2 upvotes]: I solved the following exercise:
Find the derivative $P'(z)$ of
$$ P(x,y) = x^3 - 3xy^2 - x + i(3x^2 y - y^3 -y)$$
and
$$ P(x,y) = 2xy + i(y^2 -x^2)$$
Show that in each case $P'(z) = P_x$. Explain.
The only part I didn't manage was the "explain" part. For an analytic polynomial we have $P_y = i P_x$. And Since the derivative from any direction has to be equal I figured that we'd have to have
$$ P'(z) = i P_x$$
Instead, we have $P'(z) = P_x$.

What am I missing? Why is $P'(z) = P_x$ and not $P'(z) = iP_x$ as I
  expected?

REPLY [1 votes]: This is taken, I guess, from the book by Bak and Newman. In the corresponding chapter the authors show that a polynomial $P=P(x,y)$ is analytic is and only if $P(x,y)=P(x+iy,0)$ identically. If you give an increment to $z=x+iy$, you will just vary the first argument of $P(x,y)$. Hence it is really reasonable that $P'(z)=P_x$.


TITLE: Is it possible to improve the efficacy of light bulbs with a semipermeable mirror? (The greenhouse bulb)
QUESTION [3 upvotes]: Let us imagine that there exists a material that reflects infrared radiation, but is transparent for visible light. Could we take an incandescent bulb and add this material to the inner surface of the glass hull? Then most of the infrared radiation isn't lost, but reflected onto the wire again. So we would need much less energy to heat up the wire to a sufficient temperature. The result would be an efficient light bulb with a broad, gapless spectrum in the visible range.
Is this possible like that? Have I done a fundamental mistake? Or do the required materials not exist?

REPLY [2 votes]: Many light bulbs already do this.  See for example this article.  I was in the lighting technology business at one point.  At that time it was done in some tungsten-halogen incandescents.  I don't know current state of things.


TITLE: How to show the existence of the limit $\lim_{n\to \infty}\frac{x_n}{n}$ if $x_n$ satisfy $x^{-n}=\sum_{k=1}^\infty (x+k)^{-n}$?
QUESTION [6 upvotes]: Suppose $x_n$ is the only positive solution to the equation  $x^{-n}=\sum\limits_{k=1}^\infty (x+k)^{-n}$,how to show the existence of the limit $\lim_{n\to \infty}\frac{x_n}{n}$?
It is easy to see that $\{x_n\}$ is increasing.In fact,  the given euation equals
$$1=\sum_{k=1}^\infty(1+\frac{k}{x})^{-n} \tag{*}$$
If $x_n\ge x_{n+1}$,then notice that for any fixed$ k$,$(1+\frac{k}{x})^{-n}$ is increasing,thus we can get
$$\frac{1}{(1+\frac{k}{x_n})^n}\ge \frac{1}{(1+\frac{k}{x_{n+1}})^n}>\frac{1}{(1+\frac{k}{x_{n+1}})^{n+1}}$$
By summing  up all k's from 1 to $\infty$,we can see
$$\sum_{k=1}^\infty\frac{1}{(1+\frac{k}{x_n})^n}>\sum_{k=1}^\infty\frac{1}{(1+\frac{k}{x_{n+1}})^{n+1}}$$
then from $(*)$ we see that the two series in the above equality are all equals to $1$,witch is a contradiction!
But it seems hard for us to show the existence of  $\lim_{n\to \infty}\frac{x_n}{n}$.What I can see  by the area's principle is
$$\Big|\sum_{k=1}^\infty\frac{1}{(1+\frac{k}{x_n})^n}-\int_1^\infty \frac{1}{(1+\frac{x}{x_n})}dx\Big|<\frac{1}{(1+\frac1{x_n})^n}$$
or
$$\Big|1-\frac{x_n}{n-1}(1+\frac{1}{x_n})^{1-n}\Big|<\frac{1}{(1+\frac1{x_n})^n}$$

REPLY [0 votes]: We can rewrite  $$x^{-n} = \sum_{k=1}^\infty (x+k)^{-n}$$
as
$$1= \sum_{k=1}^\infty e^{- n\ln (1+ k/x_n)}.$$
Now
$\ln (1+k/x_n) \le k/x_n$, therefore
$$1 \le \sum_{k=1}^\infty e^{-\frac{n}{x_n}k} = \frac{1}{e^{n/x_n}-1}.$$
From this it follows that
$$ (*) \quad  n /x_n \ge \ln 2.$$
Suppose now that $\limsup_{n\to\infty} n/x_n=M>c$. Then for all $n$ large, we have  $n/x_n>c$ and
\begin{align*} 1 &= \sum_{k=1}^\infty e^{-n \ln (1+\frac{k}{n} \times \frac{n}{x_n})}\\
& \le \sum_{k=1}^\infty e^{-n \ln (1+ \frac{k}{n} c)}\\
& = \sum_{k=1}^\infty (1+\frac{k}{n}c)^{-n} \\
& \to \sum_{k=1}^\infty e^{-kc}=\frac{1}{e^c-1}.
\end{align*}
by dominated convergence (note: $(1+\frac{k}{n}c)^{-n} \le (1+\frac{kc}{2})^{-2}$).
Thus, $e^c-1 \le 1$, or $c \le \ln 2$. It follows that
$$(**) \quad \limsup n/x_n \le \ln 2.$$
Now $(*)$ and $(**)$ give
$$\lim_{n\to\infty} \frac{x_n}{n} = \sup_{n} \frac{x_n}{n} = \frac{1}{\ln 2}.$$


TITLE: A Hilbert-space completion of a Hilbert $ C^{*} $-module over a separable $ C^{*} $-algebra
QUESTION [4 upvotes]: Let $ B $ be a separable $ C^{*} $-algebra and $ \mathcal{E} $ a Hilbert $ B $-module. We know that $ B $ has a faithful state $ \phi $. Using $ \phi $, we can construct a $ \mathbb{C} $-valued pre-inner product $ [\cdot,\cdot] $ on $ B $ by
$$
\forall \xi,\eta \in \mathcal{E}: \quad
[\xi,\eta] \stackrel{\text{df}}{=} \phi(\langle \xi,\eta \rangle_{B}),
$$
where $ \langle \cdot,\cdot \rangle_{B} $ denotes the $ B $-valued inner product on $ \mathcal{E} $.

Question: Is it necessarily true that $ \mathcal{E} $ is already complete with respect to the metric induced by $ [\cdot,\cdot] $? If the answer is ‘no’, then is there a judicious choice of a faithful state $ \phi $ on $ B $ that would make this true?

Thank you very much.

REPLY [5 votes]: One example of such a bimodule $\mathcal E$ is $B$ itself, with the inner product $\langle b,b'\rangle_B = b^* b'$.  Choose $B\supset B_0 = C(X)$ a unital abelian $*$-subalgebra, which can be identified with the algebra of bounded continuous functions on a compact space $X$.  If what you are asking were true, you would get that $C(X)\subset B$ is the same as its $L^2$-completion $L^2(X,\mu)\subset L^2(B,\phi)$ for some measure $\mu$ which assigns non-zero values to each non-empty open subset of $X$.  From this  you see that with the choice $\mathcal E=B$, the state $\phi$ with the property you want exists iff $B$ is finite-dimensional.


TITLE: Where completeness and separability are used in a proof of this lemma?
QUESTION [1 upvotes]: I'm reading Lemma 2.30. from this note.

Lemma 2.30. Let $\mathcal{V}$ and $\mathcal{D}$ be two algebras of subsets of a separable complete metric space $T$. Assume $\mathcal{V} \subseteq \mathcal{D}$ and $\mu: \mathcal{D} \rightarrow[0, \infty)$ is (finitely) additive and for every $B \in \mathcal{V}$ we have
$$
\mu(B)=\sup \{\mu(K): K \subseteq B, K \in \mathcal{D}, K \text { compact }\}
$$
then $\mu$ is $\sigma$-additive on $\mathcal{V}$.

Below is the proof by the author.
Proof. Suppose not: there are $B_1, B_2, \ldots$ in $\mathcal{V}$ disjoint and $B \in \mathcal{V}$ such that $B=\bigcup_{i=1}^{\infty} B_i$ and $\delta:=\mu(B)-\sum_{i=1}^{\infty} \mu\left(B_i\right)>0$. Let $C_k:=B \backslash \bigcup_{j=1}^k B_j$. Then $C_k \in \mathcal{V}, C_k \supseteq C_{k+1}$, $\bigcap_{k=1}^{\infty} C_k=\emptyset$, and $\mu\left(C_k\right) \geq \delta$ for all $k$. Take for each $k$ a compact set $K_k \in \mathcal{D}$ with $K_k \subseteq C_k$ such that $\mu\left(C_k \backslash K_k\right)<2^{-k} \delta / 2$. Then
$$
\begin{aligned}
\mu\left(K_1 \cap \cdots \cap K_n\right) &=\mu\left(\left(C_1 \cap \cdots \cap C_n\right) \backslash \bigcup_{k=1}^n\left(C_k \backslash K_k\right)\right) \\
& \geq \mu\left(C_n\right)-\sum_{k=1}^n \mu\left(C_k \backslash K_k\right) \\
& \geq \mu\left(C_n\right)-\delta / 2 \geq \delta / 2,
\end{aligned}
$$
so $K_1 \cap \cdots \cap K_n \neq \emptyset$ for all $n$. Since the sets $K_k$ are compact it follows that $\bigcap_{k=1}^{\infty} K_k \neq \emptyset$. This contradicts $C_k \supseteq K_k$ and $\bigcap_{k=1}^{\infty} C_k=\emptyset$.

My understanding:

Because $\mathcal{V}$ is an algebra and $B, B_1, \ldots, B_k \in \mathcal{V}$, we get $C_k:=B \backslash \bigcup_{j=1}^k B_j \in \mathcal{V}$.

Because $\mathcal{D}$ is an algebra and $C_k, K_k \in \mathcal{D}$, we get $C_k \setminus K_k \in \mathcal{D}$.

The fact that $\mu$ is tight on every set in $\mathcal{V}$ justifies the existence of $K_k$'s. By finite additivity of $\mu$, we have
$$
\mu\left(\left(C_1 \cap \cdots \cap C_n\right) \backslash \bigcup_{k=1}^n\left(C_k \backslash K_k\right)\right)  \geq \mu\left(C_n\right)-\sum_{k=1}^n \mu\left(C_k \backslash K_k\right).
$$

Because compact sets in a metric space are closed, every $K_k$ is closed. Then we apply the finite-intersection characterization of compactness to get $\bigcap_{k=1}^{\infty} K_k \neq \emptyset$.

My question: Where the assumption that $T$ is complete separable, is used? It seems to me it's enough that $T$ is Hausdorff.

REPLY [0 votes]: In fact, it's enough that $T$ is Hausdorff. This lemma is taken from Dudley's Real Analysis and Probability, i.e.,

Now given any two algebras $\mathcal{V} \subset \mathcal{D}$ of subsets of a set $S$, where $(S, \mathcal{T})$ is a Hausdorff topological space, and a finite, nonnegative, finitely additive function $\mu$ defined on $\mathcal{D}$, it will be said that $\mu$ is regular on $\mathcal{V}$ for $\mathcal{D}$ iff for every $B \in \mathcal{V}$,
$$
\mu(B)=\sup \{\mu(K): K \subset B, K \in \mathcal{D}, K \text { compact }\} .
$$
Lemma 10.2.4 If $\mathcal{V} \subset \mathcal{D}$ are two algebras, $\mu$ is finite and finitely additive on $\mathcal{D}$, and $\mu$ is regular on $\mathcal{V}$ for $\mathcal{D}$, then $\mu$ is countably additive on $\mathcal{V}$.


TITLE: How to prove this to be an Irrational number?
QUESTION [5 upvotes]: $$ \int^1_0 e^{-x^2} \, \mathrm{d} x $$
It seems that needs more than 30 word to make a discription of this problem,but actually that all included in the title. Thank you for your answer.

REPLY [3 votes]: EDIT: the proof isn't correct, the factors $(2b-1), (2b-3),\cdots$ spoil it. I will try to fix it.
$$
I=\int_0^1e^{-x^2}\,dx=
\int_0^1\sum_{n=0}^\infty{(-x^2)^n\over n!}\,dx=
\int_0^1\sum_{n=0}^\infty{(-1)^nx^{2n}\over n!}\,dx = \\ =
\sum_{n=0}^\infty\int_0^1{(-1)^nx^{2n}\over n!}\,dx=
\sum_{n=0}^\infty{(-1)^n\over(2n+1)n!}.
$$
If $I={a\over b}\in\Bbb Q$, consider
$$
X=(2b+1)b!\left(I-\sum_{n=0}^b{(-1)^n\over(2n+1)n!}\right)= \\
= a(2b+1)(b-1)!-\sum_{n=0}^b{(-1)^n(2b+1)b!\over(2n+1)n!}=
\sum_{n=b+1}^\infty{(-1)^n(2b+1)b!\over(2n+1)n!},
$$
and the $|\ |$ of the last series will be $<1$ (is alternating).


TITLE: Why do the characters not determine the group?
QUESTION [10 upvotes]: Motivation/Context
Looking at finite groups $G$. Of course the character table is (up to permutation of rows and columns) determined by $G$ up to isomorphism. I thought about why the converse is not true (question 1)?
Question 2
Given a complete set of characters of a finite group $G$, but not the group table (or the generators). What is exactly the minimum amount of information that is missing, necessary to determine the group $G$ (i.e. the group table) up to isomorphism, uniquely?
Own efforts
I have been looking at the famous example of the quaternion group $Q$ and the dihedral group $D_4$. They have up to permutation of rows and elements the same character table. However, they disagree in the order. I understand that a large part of the information necessary to determine the group table up to isomorphism must be contained in the character table, but I fail to pinpoint what is exactly the missing information in the general case.

REPLY [5 votes]: One possible answer is given by the following paper.

Hoehnke, H.-J.; Johnson, K. W. The 1-, 2-, and 3-characters determine a group.
  Bull. Amer. Math. Soc. (N.S.) 27 (1992), no. 2, 243–245.

Here is a brief explanation. For a character $\chi$ of a finite group $G$, define the corresponding $2$-character $\chi^{(2)} : G \times G \rightarrow \mathbb{C}$ by $$\chi^{(2)}(x,y) = \chi(x)\chi(y) - \chi(xy)$$ for all $x, y \in G$. We also define the corresponding $3$-character $\chi^{(3)}: G \times G \times G \rightarrow \mathbb{C}$ by $$\chi^{(3)}(x,y,z)=\chi(x)\chi(y)\chi(z)−\chi(x)\chi(yz)−\chi(y)\chi(xz)−\chi(z)\chi(xy)+\chi(xyz)+\chi(xzy)$$ for all $x, y, z \in G$.
There are similar definitions of $k$-characters $\chi^{(k)}$ and these go back to Frobenius. A paper of Formanek and Sibley from 1991 shows that the group determinant determines $G$. One consequence of this is that $G$ is determined by its irreducible characters $\chi$ and their $k$-characters $\chi^{(k)}$.
In their paper, Hoehnke and Johnson improved this by showing that the irreducible characters $\chi$ along with $\chi^{(2)}$ and $\chi^{(3)}$ suffice to determine $G$:

Theorem. Let $G$ be a finite group with complex irreducible characters $\chi_1$, $\ldots$, $\chi_t$. Then $G$ is determined up to isomorphism by the $\chi_i$, $\chi_i^{(2)}$, $\chi_i^{(3)}$, $1 \leq i \leq t$.

In a sense this result is optimal, since in the following paper Johnson and Sehgal show that the knowledge of $\chi$ and $\chi^{(2)}$ does not determine $G$ in general.

Johnson, Kenneth W.; Sehgal, Surinder K. The 2-character table does not determine a group. Proc. Amer. Math. Soc. 119 (1993), no. 4, 1021–1027.


TITLE: Application of mean value theorem (bounded derivative)
QUESTION [2 upvotes]: Im asked to show $|f(x)-f(y)|\le M|x-y|$ if $|f'(x)|\le M$ $x \in I$ and $f$ is continuous on $I$. I doubt this right as I'm not quite sure what it's asking.
Mean Value Theorem gives
$$
f(x)-f(y)=f'(c)(x-y) \implies |f(x)-f(y)|=|f'(c)(x-y)| = |f'(c)||x-y|
$$
Since $c\in I \implies |f'(c)|\le M$
$$
|f(x)-f(y)|=|f'(c)||x-y| \le M|x-y|
$$
Which doesn't seem difficult enough so I think I've misunderstood the question. Any input is appreciated.

REPLY [2 votes]: You are right with that.
I prefer a more direct approach:
If $x, y \in I$ and if $x < y$, then
$$
|f(x) - f(y)| \leq |x-y|\sup_{x < t < y}|f'(t)| \leq |x-y|M
$$
by assumption.


TITLE: What are the application of Scott-Topology in theoretical computer science?
QUESTION [7 upvotes]: During a work I came across the Scott-Topology and I see that Scott-continuous functions show up in the study of models for lambda calculi. What I cannot understand is how this enrich the lambda-calculus as we know.
I'm searching for paper that give -maybe- some application of Scott-topology in the computability field, as I have not find anything related.
Hoping for help from this great community

REPLY [3 votes]: Scott-continuity emerged when Dana Scott build the first model of untyped λ-calculus, while trying to prove that no such model can exist (since  any such model $D$ needs to be, simplifying a bit, isomorphic to the function space $D \rightarrow D$ which is not possible set-theoretically, but turns out to be possible when you restrict your attention to computable functions).
Scott-continuity can be understood as a mathematically well-behaved approximation to computability.
[1] is a gentle introduction to the general area of order theory that Scott continuity emerged out of, and [2] is a reference article. [3] has a bit on domain-theory and Scott-continuity and might be the easiest  introduction for computer scientists.

B. A. Davey, H. A. Priestley, Introduction to Lattices and Order.

S. Abramsky, A. Jung, Domain theory, https://www.cs.bham.ac.uk/~axj/pub/papers/handy1.pdf

G. Winskel, The Formal Semantics of Programming Languages: An Introduction.


TITLE: Group with a property that $x^{-1}=x$
QUESTION [0 upvotes]: Suppose that we have some group G in which the following holds for all $x $ belongs to G: $x^{-1} =x$. Prove that G is abelian.
How would you guys solve this one?

REPLY [0 votes]: Abelian or not: $(ab)^{-1} = b^{-1}a^{-1}$.
(Because 1) $(ab)(b^{-1}a^{-1}) = a(bb^{-1})a^{-1} = a*e*a^{-1}= aa^{-1} =e$ and 2) inverses are unique (because $fa = af =e$ and $ga=ag = e$ means $fag = fe =f$ and $fag =eg=g$ so $f=g$))
But we are told $x^{-1} = x$ so $(ab)^{-1}=ab$ and $b^{-1} = b$ and $a^{-1} = a$ so....


TITLE: Fleming's "right-hand rule" and cross-product of two vectors
QUESTION [2 upvotes]: I have been throwing around hand gestures for the past hour in a feeble attempt at trying to solve this question involving a cross product of two vectors $a$ x $b$. So far, I haven't found any satisfactory explanation on the internet as to how this rule is supposed to help me find out whether the components of $a$ and $b$ are positive, negative or zero. E.g., James Stewart explains it as though it is obvious (it isn't):

If the fingers of your right hand curl through the angle $\theta$ from $a$ to $b$, then your thumb points in the direction of $n$

What does he mean by "curl"? Sorry if this is a dumb question but I am getting really frustrated trying to use this "rule" to solve what is seemingly a really elementary question. I imagine it would be more intuitive if someone could just "show" me how to do it, but since that is not possible, if anyone could offer an explanation that is more helpful than the above, I would appreciate it.
(This question is specifically in reference to Exercise #4, Chapter 9.4 in Stewart's Single-variable calculus book, 4th edition.)

The figure shows a vector a in the xy-plane and a vector $b$ in the direction of $k$. Their lengths are $||a|| = 3$ and $||b|| = 2$.
     a) find $||a$ x $b||$
     b) Use the right-hand rule to decide whether the components of a x b are positive,         negative or zero

I am stuck on part b).

REPLY [4 votes]: You're supposed to curl your hand like so:
$\hskip 1.3in$
Your knuckles are supposed to form an angle while your fingers are otherwise straight, and your knuckles themselves point in the direction of $a$ while your fingertips point in the direction of $b$.
Using this, part (b) presumably wants you to use RHR to figure out which of the 8 octants $\vec{n}$ lies in (or between!), which would determine the sign of the components. I can't say much without seeing the picture of $a$ and $b$ though...

Update (sorry, MSPaint...):
$\hskip 1in$
I assume $a$ is on the $xy$ plane, so your thumb is too. This implies $z=0$. Moreover, the thumb is still close enough to the positively oriented $x$-ray (haha, pun) that it has positive $x$ component, but far enough from the positively oriented $y$-ray to have negative $y$ component.


TITLE: Product of two random variables should belong to additively closed family of distributions
QUESTION [0 upvotes]: Suppose $X$ and $Y$ are two independent random variables in $\mathcal{D_1}$ and $\mathcal{D_2}$ respectively(two families of distributions).
Can you suggest some pairs of $\mathcal{D_1}$ and $\mathcal{D_2}$ such that the product $XY$ belongs to a family $\mathcal{D_0}$ which is additively closed?
By additively closed I mean given $a\in \mathcal{D_0}$ and $b\in \mathcal{D_0}$, $(a+b)\in \mathcal{D_0}$.

REPLY [0 votes]: Well, if I understand the question properly, Normal distributions are additively closed so if we take $D_0=N$ and $D_1=N$ and $D_2=IG$ (inverse gamma) then $D_1 *D_2 \in D_0$ which is additively closed.


TITLE: Do ultrasonic motion detectors (like the Arduino sensor) detect small particles like dust when receiving waves? If they don't why is that the case?
QUESTION [0 upvotes]: This may be a bit of an elementary question but I've recently started studying physics for fun so this has confused me. Do ultrasonic sensors have the capability to detect very small particles like dust, or is there some limit to the size of the objects that the waves can bounce off of? I assume that if the area was void of other objects, the sensor could pick up the particles of dust since the waves would rebound off of the particles from the transmitter into the receiver. I know this falls in the area of electrical engineering I'm pretty sure, so sorry if its not fully physics.

REPLY [0 votes]: Waves of any kind tend not to interact with particles or obstacles that are much smaller than their wavelength. I don't know the frequency/wavelength of "the Arduino sensor," But let's just guess that it's 40kHz. That is a commonly used frequency for open-air ultrasonic applications. At 40kHz, under standard conditions, the wavelength in air would be roughly eight millimeters.


TITLE: Lost on an induction proof my colleague called basic.
QUESTION [0 upvotes]: I just have no idear what is being asked (like what is set J) and if it is so basic why is it so general (I'm probably an idiot) .
Prove by induction on n that for any finite set I with cardinality n with $\forall i \in I$ and $\forall a_i \in \mathbb{R}$ $$\prod _{i\in I} (1 + a_i) = \sum _{j \in \mathcal{P}(I)} \left ( \prod_{j \in J} a_j \right )$$
I do not even know what $S_0$ would say. I think I am just so unfamiliar with indexed sets in pi notation that I am confusing myself.

REPLY [1 votes]: The theorem is just stating how to expand $(1+a_1)(1+a_2)...(1+a_n)$
P(I) is the power set of I, i.e. the set of all subsets of I.
j is an element of P(I).
Also, the pi notation is similar to the $\Sigma$ notation. Just replace summation by multiplication.
Hope this helps.


TITLE: linear algebra from beginner to intermediate level
QUESTION [3 upvotes]: I was wondering if there is a way to learn linear algebra from beginner to advanced level by studying it myself. I want to collect a number of books, video lectures, tutorials and other resources to learn it. Stackoverflow has a similar question on learning C++, so I thought this would be a good place to ask this question.
Two things:
1. Open video lectures, if any, will be really helpful.
2. As I am trying to self study, problem oriented books (possibly with hints and solutions) will be very helpful.

REPLY [2 votes]: You could try this link. It covers the basics and you could use it also for organizing your learning. I mean what to start with and what to learn next.
http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/