How to show every subgroup of a cyclic group is cyclic? I'm teaching a group theory course now, and I wanted to give my students a proof that every subgroup of a cyclic group is cyclic. The easiest way I could think to do this is to say that any cyclic group corresponds to a $\mathbb{Z}/n\mathbb{Z}$ and then (implicitly) use the first isomorphism theorem (which they haven't learned yet) to "reduce" it to the theorem that every subgroup of $\mathbb{Z}$ is of the form $m\mathbb{Z}$. The problem is that this latest statement is not simple at all (every subgroup is an ideal, and $\mathbb{Z}$ is a PID because $\mathbb{Z}$ is Euclidean).

This all seems way too convoluted for such an easy statement. What's the simplest most elementary proof that I can give?

If the original group is generated by $a$, and $d$ is the least positive integer such that $a^d$ is in the subgroup, then show using the division algorithm that $a^d$ generates the subgroup.

To elaborate, the division algorithm says:

  If $m$ and $n$ are integers and $n$ is positive, then there are unique integers $q$ and $r$ such that $m=nq+r$ and $0\leq r\lt n$.

You could prove this by taking $r$ to be the least nonnegative integer expressible in the form $m-np$ as $p$ ranges over the integers, which then determines $q$.  (There are details here which I am omitting.)

Suppose $G$ is a cyclic group (finite or infinite) with generator $a$, i.e., $G=\langle a\rangle$.  Let $H$ be a subgroup of $G$.  Let $d$ be the smallest element of the set of positive integers $n$ such that $a^n$ is in $H$.  (Unless $G$ is infinite and $H$ is trivial, this exists.)  If $b$ is an arbitrary element of $H$, then because $b$ is in $G$, $b=a^m$ for some integer $m$.  Apply the division algorithm to write $m=dq+r$ for integers $q$ and $r$ with $0\leq r\lt d$.  Then $b=a^m=(a^d)^q\cdot a^r$.  Since $b$ is in $H$ and $a^d$ is in $H$, it follows that $a^r=a^m\cdot(a^d)^{-q}$ is in $H$.  If $r$ were positive, this would contradict the choice of $d$, so $r=0$, and $b=a^m=(a^d)^q$.  Since $b$ was arbitrary, this shows that $H\subseteq \langle a^d\rangle$.

How about:

Let G be a group. By definition, any quotient group H of G is realized by a surjective homomorphism $\phi: G \twoheadrightarrow H$.

Now suppose $G = \langle x \rangle$. Then, since $\phi$ is surjective, $H = \langle \phi (x) \rangle$.


How many ways are $3$ letters placed consecutively. ($6$ letters from $8$ letters) Here are the questions:

  $i)$How many ways can you arrange $6$ letters from the letters $a,b,c,d,e,f,g,h$?

  $ii)$How many of those have $b,c$ and $g$ placed consecutively?

I think the solution to the first one is $8\choose 6$, I'm still not sure so please correct me if I'm wrong. I don't have an idea though how to answer the next question. Please help.
Thanks :)

The number of ways to choose 6 letters from the 8 is $\binom 8 6 = \frac{8!}{6! 2!}$

each of these choices can be arranged in $6!$ ways - so the answer to i) is $\frac{8!}{ 2!}$

for ii) - there are $3!$ ways to arrange b,c,g consecutively and  $\binom 5 3 = \frac{5!}{3! 2!}$ choices of the other three letters. For each of these choices there will be $4!$ ways to arrange the 3 letters and the b,c,g block.

$$N=   \frac{3! 4! 5!}{3! 2!} =  \frac{4! 5!}{ 2!}$$

For B , C , G to occur consecutively take them as one letter only . Like X={b,c,g} you are left with 5 letters and select any 5 of them by 5C5 and the 6th would be X . You can arrange the letters inside X by 3! Ways and the six letters(which include X) by 6! Ways hence the answer should be 6! X 3!


Which trigonometric identity is applicable to show $\sin\left(\frac{(2n+1)\pi}{2a}(x + a)\right) = \cos\left(\frac{(2n+1)\pi x}{2a} + n\pi \right)$? I'm confused with regards to the following and want to know which trigonometric identity is applicable here:
$$\sin\left(\frac{(2n+1)\pi}{2a}(x + a)\right) = \cos\left(\frac{(2n+1)\pi x}{2a} + n\pi \right)$$
where $a$ is a constant and $n \in \mathbb{Z}$.
Furthermore, I want to know why the above expression is equal to $$(-1)^n\cos\bigg(\frac{(2n+1)\pi x} {2a}\bigg).$$

Your first equation implies that
$$\sin\left(\phi+\frac{\pi}{2}\right)=\cos(\phi)$$
for any angle $\phi$, which is a defining identity of the trigonometric functions $\sin$ and $\cos$. Furthermore, you can easily see that both hand sides of your last equation are equal to
$$(-1)^n\cos(\phi)\,,$$
since $\cos(a+b)=\cos(a) \cos{b}-\sin(a) \sin{(b)}$.


Uncountable null sets in arbitrary measure spaces Does every measure space $(X,\Sigma,\mu)$ with $X$ and $\Sigma$ uncountable, have an uncountable null set?

The counting measure on an uncountable set does not have an uncountable null set.


Is it true that $a_n \to \infty$ implies that $a_n$ is definitely increasing? Consider a sequence $(a_n)_n$ such that $a_n \to \infty$; is it true that $a_n$ is (at least definitely) increasing?
Intuitively I would say yes, I don't see how is possible that a sequence can go to infinity without increase (at least from a certain point on); I've tried to think to a counterexample but I couldn't find it, so I've tried to prove it but I suspect that I'm doing something wrong.
By hypothesis $a_n \to \infty$, so for all $M\in\mathbb{R}$ there exists $N_M\in\mathbb{N}$ such that $n \geq N_M \implies a_n \geq M$; now (this is the step I'm unsure about) I would like to use the fact that $M\in\mathbb{R}$ is arbitrary to let $M=a_{n-1}$, so I would get that $n \geq N_M \implies a_n \geq a_{n-1}$ and this is the definition of definitely increasing function since this is valid only for $n \geq N_M$.
However I do believe that this proof is not valid, because in the definition of limit $M$ is considered fixed while $a_{n-1}$ can vary because of its dependence on $n$; so I think I'm not allowed to choose $M=a_{n-1}$. Is my suspect correct and this proof is not valid?
Hints of are more welcome that resolutions :) thanks!

The sequence
$$a_n = n + 2 \times (-1)^n$$
tends to $+\infty$ but is not definitely increasing.

A counter-example can be constructed by taking any sequence which tends to $\infty$ and replacing every other term in a way that the pattern is one of:
decrease, increase, decrease, increase... or similar
(actually some of the increases could be decreases as well, but the point is you can build in decreases throughout the whole sequence).  The replacement terms still need to tend $\infty$ as well but there are many ways to ensure that.


Linear structure on the category of formal groups Let $R$ be a commutative ring. If $R$ is a $\mathbb{Q}$-algebra, then the category of formal groups over $R$ (or the category of formal group laws) carries the structure of an $R$-linear category; this is because it is equivalent to the category of Lie algebras over $R$  (which are free of finite rank as $R$-modules) by the $\mathbb{Q}$-Theorem, and the latter category is $R$-linear. If $R$ is not a $\mathbb{Q}$-algebra, then the category of formal groups over $R$ doesn't have to be $R$-linear. Nevertheless, I would like to ask: Is there any way to see the $R$-linear structure on the category of formal groups over $R$ directly, without the classification in terms of Lie algebras, when $R$ is a $\mathbb{Q}$-algebra? For example, why should be expect there to be a map $R \to \mathrm{End}_R(G)$ for every formal group $G$ over $R$?

If $G$ is a formal group over $R$ and $R$ is a $\mathbb Q$-algebra, its logarithm is defined over $R$ (see the explicit formula for the logarithm of a formal group) thus giving an isomorphism from $G$ to the additive formal group $F_a$. It remains to note that ${\rm End}_R(F_a)$ is isomorphic to $R$.


Is a definition like $ g(f(x)) = 0 $ possible? Is it (in any way) possible to define function $g$ like this?

$$ f(x) = x $$
$$ g(f(x)) = 0  $$

Why?

Better write,

$$g:\mathbb R\to\mathbb R:g(x)=0$$

or

$$g=\{x\in\mathbb R:(x,0)\}.$$

$f$ plays no role.

More generally,

$$g:\text{range}(f)\to\mathbb R:g(x)=0$$

or

$$g=\{x\in\text{range}(f):(x,0)\}.$$


Exercise: The measure of the boundary of a level set of a Holder continuous function is positive Let $M$ be a compact manifold in $R^3$ and consider a non negative
function $f\in C^{1,\alpha}(M)$ for some $0<\alpha<1$. I am interested
in showing that $|\partial \{f>0\} |>0$, where $|\cdot|$ stands for
the Hausdorff measure and $\partial A$ denotes
the boundary of $A$.

My Attempt:
By definition, $x^* \in M$ is a boundary point of $\{f>0\}$, if for every $r>0$ it holds that $B_r(x^*) \cap \{f>0\} \neq \emptyset$ and $B_r(x^*) \cap \{f=0\} \neq \emptyset$. In other words, if for every ball with center at $x^*$, I can find not only points of $\{f>0\}$ but also of its complement. Now, it is also true that $M=\{f>0\}\cup \{f=0\} $. Hence, if I could find/construct a dense subset of $\{f>0\}\cup \{f=0\}$, let's say $D$, I could obtain that $B_r(x^*) \cap D \neq \emptyset$ by density. This in particular would yield that $B_r(x^*) \cap (\{f>0\}\cup \{f=0\})=(B_r(x^*) \cap \{f>0\})\cup (B_r(x^*) \cap\{f=0\}) \neq \emptyset$. At this point, two issues arise:

The latter does not imply that both intersections must be non empty, but only at least one between them.
I am not sure if such a dense set exists.

DISCLAIMER: I have seen some similar question here in which the answer is given by a Cantor set. However, since I am not that familiar with topology, I would prefer to prove this (if possible) using the "slightest" tools.
I feel that density is a key part here but I am not able to tackle this problem properly. Any help or hint will be much appreciated!
Thanks in advance.

I do not understand why the dense set would help you in this situation as nonempty sets can still have zero Hausdorff measure. In this answer I will denote the $2$-dimensional Hausdorff measure by $\mathcal{H}^2$ (using the absolute value seems very misleading to me). In the comments I gave some heuristics. Let me back this up by something a bit more rigorous.
As I pointed out in the comments it depends on the desired notion of boundary (either withing the manifold with the subspace topology or in $\mathbb{R}^3$). As you are using $\{ f> 0 \}^c = \{ f=0 \}$, you seem to take the boundary within the manifold. In this case, your claim is not true. Consider
$$\mathbb{T}^2 = \{ (x,y,z) \in \mathbb{R}^3 \ : \ (x^2+y^2-2)^2+z^2 =1 \}$$
(some 2-dimensional torus) embedded in $\mathbb{R}^3$. This is a compact, smooth manifold. Now consider the map $f: \mathbb{T}^2 \rightarrow \mathbb{R}, f(x,y,z)=z^2$. Then $f$ is a smooth, nonnegative map. However, the boundary of the zero locus is equal to two circles (1-dimensional manifolds) and thus, the $2$-dimensional Hausdorff measure will vanish.
If you were to pick the boundary in the topology of $\mathbb{R}^3$, (meaning, the balls should intersect $\{ f>0 \}$ and $\{ f>0 \}^c$), then you get $\{f>0 \} \subseteq \partial \{ f>0 \}$ (that is because any ball in $\mathbb{R}^3$ intersects $M^c$). However, $\{ f>0 \}$ is an open subset of a $2$-dimensional, $C^{1,\alpha}$ manifold and thus is a $2$-dimensional $C^{1,\alpha}$ manifold itself. However, nonempty $2$-dimensional $C^{1,\alpha}$ manifolds have positive $2$-dimensional Hausdorff measure. As $M$ is embedded, for every point $p\in M$ there exists a diffeomorphism $\varphi: B_1(0) \rightarrow \mathbb{R}^3$ such that $$\varphi((-1,1)^3) \cap \{ f>0\} = \{ \varphi(x,y,0) \ : \ (x,y,0)\in (-1,1)^3 \}$$
and such that $\varphi(0)=p$. By the monotonicity of measures, we get
$$ \mathcal{H}^2( \varphi([-0.5, 0.5] \times [-0.5,0.5] \times \{0\})) \vert \leq \mathcal{H}^2( \{ f>0 \} ) \leq \mathcal{H}^2( \partial \{ f>0 \} ). $$
However, as $\varphi$ is $C^1$, we have that its restriction to $[-0.5,0.5]^3$ is bilipschitz (lets say with the inverse having lipschitz constant equal to $L$). By Measure of image of Lipschitz function is bounded? (the $C^{1,\alpha}$ regularity actually would allow you for a simpler proof of this fact, in fact we do not really need $M$ to be more than a topological manifold, see the next paragraph) we have
$$ 0<1 = \mathcal{H}^2([-0.5,0.5] \times [-0.5,0.5]\times \{0\})=\mathcal{H}^2(\varphi^{-1} \varphi([-0.5, 0.5] \times [-0.5,0.5] \times \{0\}))
\leq L^2 \mathcal{H}^2(\varphi([-0.5, 0.5] \times [-0.5,0.5] \times \{0\})). $$
Hence, we get
$$ 0< 1/L^2 \leq \mathcal{H}^2(\partial \{ f>0\}). $$
Added: The $C^{1, \alpha}$ assumption is not really need. Everything works in the topological setting if your manifold is locally given as a graph. Then we can use the projection map. I.e. we have
$$ \mathcal{H}^2((-1,1)^2)=\mathcal{H}^2(P(\{ (x,y, g(x,y)) \in \mathbb{R}^3 \ : \ (x,y)\in (-1,1)^2 \})) \leq \mathcal{H}^2(\{ (x,y, g(x,y)) \in \mathbb{R}^3 \ : \ (x,y)\in (-1,1)^2 \})  $$
where $g$ is any map (we need no regularity, only the fact that we can write it as a graph) and $P: \mathbb{R}^3\rightarrow \mathbb{R}^2, (x,y,z)=(x,y)$ is the projection. So you only need the estimate for the projection which is much easier to show (also because it has all the regularity one can hope for).

I think I figured out the missing keys so I will post the whole argument here as an answer. In case, there are still mistakes, please correct me.
We know that $f \in C^{1,\alpha}(M)$ for some $0<\alpha<1$. Hence, $f \in C^{1,\beta}(M)$ for some $0<\beta<\alpha<1$. We can find a sequence of smooth functions $g_n$ converging to $f$ such that $g_n(x)>0$ for all $x\in B_{\delta_n}(x)$ and $g_n(x)=0$ otherwise (here $\delta_n$ is chosen arbitrarily small). Moreover, since $M$ is a compact manifold, there exists a countable dense subset $\{ x_n, n \in \mathbb N\}$ in $M$. Passing to a subsequence $\hat x_n$ we obtain by density that $\hat x_n \to x^* \in M$.
We can write $M=M \setminus \big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big) \cup \big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big)$. We recall that $\big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big) \neq \emptyset$ since $\{x_n\}$ is dense. At this point, we distinguish cases. If $x^*\in M \setminus \big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big) $ then $f(x^*)=0$. If $x^* \in \big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big)$ then due to $g_n$ and by passing to a subsequence, we obtain that $f(x^*)>0$. This completes the argument.


Does "Doing a thing to both sides of an equation" have a name? A two part question.

1

True or False: when working with an equation or inequality, everything that you do is either:

a substitution, or
an operation performed on each side

Note that algebraic or numerical simplifications are substitutions - $2+2=4$, so we're free to substitute $4$ where $2+2$ is present in an equation.

2

Is there a generic name for 'performing an operation on both sides' of an equation/inequality?

Transformation? Equivalence derivation? I think that I've heard 'transpose' for the case of adding/subtracting something to both sides in order to eliminate it from one side, but I don't believe that this is used for, say, doubling or squaring both sides.

Is there a generic name for 'performing an operation on both sides' of an equation/inequality?

Yes. The name is algebra (maybe "doing algebra" is more appropriate and/or sounds better).

Here there are three sources:

The word "algebra" comes from a book written in 830 by the astronomer Mohammed ibn Musa al-Khowarizmi (c. 825), titled Al-jabr w'al muqâbala. The word al-jabr meant "restoring," in this context, restoring the balance in an  equation by placing on one side of an equation a term that has been removed from the other; thus if $-7$ is removed from $x^2 - 7 = 3$, the balance is restored by writing $x^2 = 7 + 3$. Al' muqâbala meant "simplification," as by combining $3x$ and $4x$ into $7x$ or by subtracting equal terms from both sides of an equation. [...] When
al-Khowarizmi's book was first translated into Latin in the twelfth century,
the title was rendered as Ludus algebrae et almucgrabalaeque, though other titles were also used. The name of the subject was finally shortened to algebra. (Kline's book)
He [al-Khwarizmi] did this in his book al-jabr w al-muqabalah. “Al-jabr” (from which stems our word “algebra”) denotes the moving of a negative term of an equation to the other side so as to make it positive, and “al-muqabalah” refers to cancelling equal (positive) terms on both sides of an equation. These are, of course, basic procedures for solving polynomial equations. Al-Khwarizmi (from whose name the term “algorithm” is derived) applied them to the solution of quadratic equations. (Kleiner's book)
The mathematical sense [of the word algebra] comes from the title of a 9th-century Arabic book ilm al-jabr wa'l-mukabala, ‘the science of restoring what is missing and equating like with like’, written by the mathematician al-Kwarizmi. (Oxford Dictionaries)

EDIT

@BillDubuque pointed out that a correct answer for the question should give a name for the following rule: equalities are preserved by "performing an operation on both sides" (see the comments in this post).

In the Patrick Suppes terminology, the name of this rule can be Consequence of the Rule Governing Identities.

  Rule Governing Identities (RGI): If $S$ is an open formula, from $S$ and $t_1=t_2$, or from $S$ and $t_2=t_1$ we may derive $T$, provided that $T$ results from $S$ by replacing one or more occurences of $t_1$ in $S$ by $t_2$. Moreover, the identity $t=t$ is derivable from the empty set of premises. (Suppes book)

Remark 1: Given an operation $f$, let $S$ be the formula $f(z)=f(z)$. Then, by the RGI,
$$x=y\quad\Longrightarrow\quad f(x)=f(y).$$
In words, "equalities are preserved by performing an operation on both sides".

Remark 2: The RGI also justifies general substitutions in equalities. For example, $x+y=2$ and $x=y+3$ implies $(y+3)+y=2$ (here, $S$ is the formula $x+y=2$). To understand the word "general" see the comments in the ASKASK's answer.

Remark 3: Other names (probably, the usual ones) are Replacement, Substitution Property and Substitution by equality (see other answers and comments).

From a categorial perspective, this might be  called the functoriality of functions.

In particular, observe the similarities:

If $f : X \rightarrow Y$ is a function between sets and $x,x'$ are elements of $X$, then we get an implication like so: $$x = x' \rightarrow f(x) = f(x')$$
If $F : \mathbf{C} \rightarrow \mathbf{D}$ is a functor between categories and $X,X'$ are objects of $\mathbf{C}$, then we get a function like so: $$\mathrm{Hom}(X,X') \rightarrow \mathrm{Hom}(FX,FX')$$


Why use stopping times rather than a deterministic sequence to localise a martingale? I am a beginner on stochastic processes

I am wondering why , to localise a martingale, require the existence of one non-decreasing sequence of stopping times  [$ \tau_1 , \tau_2$,...] such that the corresponding stopped processes $\{M_{\tau_n \land t} \}$ are all martingales, rather than simply require the existence of one increasing infinite partitioning sequence [$t_1 \lt  t_2 \lt ... \lt t_n \lt ... , t_n \rightarrow \infty$] such that the stopped processes $\{ M_{t_i \land t} \}$ are all martingales ?

I mean, what is to be gained by having stopping times rather than a deterministic sequence here ?

Stopping time is more general. Sometimes, you don't know what a particular time will be, e.g. the time the process hits $5$ for the first time is not constant, but is a stopping time.


Finding Maximum "X" Contiguous subsequence I am aware of how to solve maximum contiguous sub sequence given different constraints (increasing, sum, product, etc). However, I'm having problems with trying to satisfy the following constraint.

I want to the sub sequence with the largest $X$ where $X$ is defined as the product of the sub seq length and the largest val in the sub seq.

I tried a small test case with 4 numbers: {1, 10, 10, 1}. The largest sequence would be in the middle with {10, 10} as this gives an $X$ value of 20. To approach this, I'd break the sequence down the middle into two groups. The largest $X$ on each side can be obtained by simply $max(sequence)*length$ on each side. But how do I traverse through the middle efficiently?

I see that you do a divide and conquer method but that’s do not work ,if I understand it properly ,because if you take {10,10,1} ,X will be 30. My solution will be that you traverse the list, find the maximum element and the maximum subsequence will be the subsequence with the last n-1 elements if the max is not the first element and if the max is the first it’s the first n-1 elements . So you will have the maximum X.


linear algebra, dimension of vector of subspace Suppose that $v$, $w$ are linearly independent vectors of $R^3$ then the subspace of $R^3$ generated by the vecors $v$,$w$ and $v-w$ is of dimension

$A)\quad 1$

$B)\quad2$

$C)\quad3$

$D)\quad-1$

I am thinking in this question as $v$, $w$ are linearly independent so $v-w$ is linearly dependent so dimension of linearly dependent vectors is always less then $\dim(R^3)$ so is the answer $2$ ? am I correct or not? Please help me out to solve this.

HINT: $w$ and $v$ are linearly independent, so dimension is at least 2. Why the dimension of a space generated by $v$, $w$, $v-w$ must be less than 3?

Your selection is correct, but your answer doesn't explain, why the dimension is not equal to 1.

Note that $v-w$ is a linear combination of $v$ and $w$ that is

$v-w=1\cdot v-1\cdot w$

or also that in the basis $\{v,w,z\}$, with $z$ a third linearly independent vector, we have

$v=(1,0,0)$
$w=(0,1,0)$
$v-w=(1,-1,0)$


Maximize $\sqrt{2x + 13} + \sqrt[3]{3y+5} + \sqrt[4]{8z+12}$ Given three non-negative (as pointed out by Calvin Lin) real numbers $x+y+z = 3$, find the maximum value of $\sqrt{2x + 13} + \sqrt[3]{3y+5} + \sqrt[4]{8z+12}$.

(Source : Singapore Math Olympiad 2012, Senior section, Round 1, question 29).

I tried using the fact that $2x +13, 8z + 12\ge 0$ to deduce that $y \le 11$, but I couldn't continue from there on. The answer should be an integer, since only integer answers were allowed.

The competition was designed for 15/16 year olds. A simple yet elegant solution would be nice.

For reference of the original problem,

If $y$ is allowed to be negative, then there is no maximum.

Take $ x =k - \frac{13}{2}$, $ y = -k - \frac{5}{3}$, $ z= 3 + \frac{13}{2} + \frac{5}{3}$

The important part of the value looks like $ \sqrt{2k} + \sqrt[3]{-3k} + C$, which tends to infinity as $k\rightarrow \infty$.

Otherwise, if $y$ is positive, see Math110's solution.

math110's AM-GM solution is no doubt (to me at least) what the author's wanted. But a less inspired approach is that with Lagrange Multipliers, you need to solve the system of equations:

$$\left\{\begin{aligned}
x+y+z&=3\\
\frac{1}{\sqrt{2x+13}}&=\lambda\\
\frac{1}{\sqrt[3]{(3y+5)^2}}&=\lambda\\
\frac{2}{\sqrt[4]{(8z+12)^3}}&=\lambda\\
\end{aligned}\right.$$

The second and third imply that $\sqrt{2x+13}=\sqrt[3]{(3y+5)^2}$, so $(2x+13)^3=(3y+5)^4$.

Similarly the second and fourth imply that $2\sqrt{2x+13}=\sqrt[4]{(8z+12)^3}$, so $16(2x+13)^2=(8z+12)^3$.

Now you can forget about $\lambda$ and work with the system
$$\left\{\begin{aligned}
x+y+z&=3\\
(2x+13)^3&=(3y+5)^4\\
16(2x+13)^2&=(8z+12)^3\\
\end{aligned}\right.$$

Eliminating $z$:
$$\left\{\begin{aligned}
(2x+13)^3&=(3y+5)^4\\
(2x+13)^2&=4(9-2x-2y)^3\\
\end{aligned}\right.$$

If there are any solutions where the sides of these equations are integers, we would have to have some integer to the $12$th on both sides of the first equation. Trying something nice and small like $2^{12}$ implies that $x$ would have to be $1.5$ and $y$ would have to be $1$. And this is also a solution to the second equation. So one solution is $(x,y,z)=(1.5,1,0.5)$, which yields a value of $8$. Maybe the relative simplicity of the the curves in this last system can show that there are no other solutions:

It's a little tedious, but we can solve for $y$ in each equation, and the difference of the two functions of $y$ is $$\frac{(2x+13)^{3/4}-5}{3}-\left(\frac{\left(\frac{(2x+13)^2}{4}\right)^{1/3}-9+2x}{2}\right)$$
whose derivative is
$$\frac{1}{2(2x+13)^{1/4}}-\frac{2^{1/3}}{3(2x+13)^{1/3}}-1$$
For positve $x$, this is always negative. So the curves from the last system above can only cross once in the domain where $x$ is positive, and we already found where they do.

And then it remains to show that this yields a local maximum value, not a minimum value or a degenerate solution to the Lagrange multiplier problem.

If there are constraints on $x$, $y$, $z$ being positive, then the boundaries of the planes $x=0$ (subject to $y+z=3$), $y=0$ (subject to $x+z=3$), and $z=0$ (subject to $x+y=3$) need to be separately checked, followed by the boundaries of those boundaries: $(3,0,0)$, $(0,3,0)$, and $(0,0,3)$.


$\sqrt{2^2} = 2$, $\sqrt{2^4} = 2^2 = 4$, $\sqrt{2^6} = 8 \neq 6$, $\ldots$ In general, if $t \in \mathbb{N}$, when is it true that
$$\sqrt{2^{2t}} = 2^t = 2t?$$

I know of course that it is true when $t - 1 = \log_{2}(t)$.  Are there other (more arithmetic) conditions, for which the equation $2^t = 2t$ holds, for $t \in \mathbb{N}$?

It appears to be true when $t = 2^s$, for some $s \in \mathbb{N}$.

The first equality is true: $\sqrt{2^{2t}}=\sqrt{2^{t+t}}=\sqrt{2^t\cdot2^t}=2^t$.

The second holds only for $t=1$ and $t=2$. Indeed, define for $t\in\Bbb R$ $$f(t)=\cfrac{2^{t}}{2t}$$ and $$g(t)=\ln f(t)=t\ln 2-\ln 2-\ln t$$
Now,
$$g'(t)=\ln2-\frac1t$$

We see that $g'$ vanishes only at one point, namely $t=1/\ln 2$. By Rolle's theorem, $g$ has no more than two zeros, and hence the equation $f(t)=1$ can not have more than two solutions.


Why is Euler's Gamma function the "best" extension of the factorial function to the reals? There are lots (an infinitude) of smooth functions that coincide with $f(n)=n!$ on the integers. Is there a simple reason why Euler's Gamma function $\Gamma (z) = \int_0^\infty t^{z-1} e^{-t} dt$ is the "best"?  In particular, I'm looking for reasons that I can explain to first-year calculus students.

The Bohr–Mollerup theorem shows that the gamma function is the only function that satisfies the properties

$f(1)=1$;
$f(x+1)=xf(x)$ for every $x\geq 0$;
$\log f$ is a convex function.

The condition of log-convexity is particularly important when one wants to prove various inequalities for the gamma function.

By the way, the gamma function is not the only meromorphic function satisfying
$$f(z+1)=z f(z),\qquad f(1)=1,$$
with no zeroes and no poles other than the points $z=-n$, $n=0,1,2\dots$. There is a whole family of such functions, which, in general, have the form
$$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{m=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$
where $g(z)$ is an entire function such that
$$g(z+1)-g(z)=\gamma+2k\pi i,\quad k\in\mathbb Z, $$
($\gamma$ is Euler's constant). The gamma function corresponds to the simplest choice
$g(z)=\gamma z$.

Edit: corrected index in the product.

For me another argument is the convincing one.

Consider the log of the factorial resp the gamma-function; this is for the integer arguments a sum of logarithms of the integers. Now for the interpolation of sums to fractional indexes (which is required for the gamma to noninteger arguments) there exists the concept of "indefinite summation", and the operator for that indefinite summation can be expressed by a power series. we find, that the power series for the log of the Eulerian gamma-function matches exactly that of that operator for the indefinite summation of the sum of logarithms.
I've seen this argument elsewhere; I thought it has been here at mse before (by the used "anixx") but may be it is at MO; I'm not aware of a specific literature at the moment, but I've put that heuristic in a small amateurish article on my website; in the essence the representation of that indefinite summation is fairly elementary and should be existent in older mathematical articles. See "uncompleting the gamma" pg 13 if that seems interesting.

Conclusion: the Eulerian gamma-function is "the correct one", because it is coherent with the indefinite summation-formula for the sums of consecutive logarithms.


3D projection onto 2D plane to determine transformation matrix? I'm not sure if there is an actual solution to this problem or not, but thought I would give it a shot here to see if anyone has any ideas. So here goes:

I basically have three vertices of a rigid triangle with known 3D coordinates.  The vertices are projected onto a 2D plane, in which I also know the 2D coordinates.  A transformation matrix is applied to the original three points (can be a combination of rotation and translation) and I now know the new 2D projection coordinates.

Is it possible to obtain either the unknown transformation matrix or the new coordinates?  Any ideas are much appreciated.  Thanks!

all you need to do is measure how the vectors vec3(1,0,0) vec3 (0,1,0) vec3(0,0,1) get changed by any matrix.
vec3(1,0,0)'s transformation is equal to the first  column of the matrix
vec3(0,1,0)'s transformation is equal to the second column of the matrix
vec3(0,0,1)'s transformation is equal to the second column of the matrix (only if it has 3 dimensions)
vec4(0,0,0,1)'s transformation is equal to the second column of the matrix (only if it has 4 dimensions),...

yes, learn how matrices function on basic core level:

in a 2d to 2d matrix transformation, 2 matrix colums are equal to what the 2 base-vectors vec2(1,0) and vec2(0,1) are changed into by the matrix. any scalar of that is also transformed linearily scaled to that, as a matrix transformation is a linear transformation.

in a 3d to 3d matrix transformation, same goes for 3 base-vectors vec3(1,0,0) vec3 (0,1,0) vec3(0,0,1) being changed by 3 matrix columns.

in a 3d to 2d matrix, 3 columns of 2 lines tell how 3 3d basis vectors as in the above are scamed in 3d space, as a sum of three 2d vectors (that get scaled by the matrix).


Is squaring both sides of a differential equations valid? In the equation $ \int \sqrt{y} dy= \sqrt{x} dx $ is it valid to square both sides to obtain $ \int y dy = \int x dx $

In general no, $\int f(y) \,dy = \int g(x) \,dx$ does not imply that $\int f(y)^2 \,dy = \int g(x)^2 \,dx$.

We can see this readily in your example by integrating both equations and observing that we get different solution sets. (I recommend you try carrying this out yourself---I've put the solution sets in a spoiler.)

  The general solution to $\sqrt{y} \,dy = \sqrt{x} \,dx$ is $y = (x^{3 / 2} + C)^{2 / 3}$, but the general solution to $y \,dy = x \,dx$ is $y(x) = \pm \sqrt{x^2 + C}$.

Assuming it is   $$\int \sqrt{y}dy=\int\sqrt{x}dx$$

Yes you can, but this may increase the number of solutions you get upon solving that equation for some initial term, and a specific value of x.

When you do take the square, keep in mind that square will also work upon $dy$ and $dx$, like in your case:
squaring yields $$ \int y \cdot (dy)^2=\int x\cdot(dx)^2$$

This is useless in your question, squaring both sides will just increase its difficulty, without yielding anything useful.

So, for   $$\int \sqrt{y}  dy=\int \sqrt{x} dx$$
simply integrate both the sides, you get $$ \frac{2}{3} y^{\frac{3}{2}}= \frac{2}{3}
 x^{\frac{3}{2}}+c$$
$$ y^{\frac{3}{2}}=x^{\frac{3}{2}}+c'$$ as your equation.


Self studying math, how can I learn the most? I am currently studying Pre-Calculus on my own. I have a few texts I am working with but feel like I could learning a lot more than I am.

When people typically ask these kind of questions the common response for Pre-calculus->Calculus is to watch "Khan Academy" etc.

Is this really a wise approach, especially when confronting future topics? Wouldn't it better to develop more independent, self-sufficient learning strategies beforehand?

I always wonder how mathematicians teach themselves so much knowledge without resources like "Khan Academy", lectures or even teachers. People learned math before the internet, what was the strategy then?

Some other questions:

Which is the better source to learn from, video lectures or reading the text? I feel like video lectures are too passive. Would time be better spent doing more exercises and less video lectures?
If you watch a video lecture, should it be watched before reading the text or after?
Math is quite concise as presented in textbooks, is summarizing the chapter as you read in a notebook effective or unnecessary?
Should I completely read chapters before attempting exercises or try questions first and refer to the text as needed?

Final question:
If a person was attempting to learn a topic in math from scratch with minimal knowledge on the content, what is an approach they could apply to learn it that would be both effective and efficient?

Thanks

Video lectures are actually better for rapidly getting the intuition than studying directly from a textbook. And that's because the lecturer reduces the effort you have to make to understand, especially if his name is Sal.
Yes, textbooks are superior to videos for genuinely learning the material.
It depends on how you summarize your chapter. Try to look up for the Feynman technique of learning, I'm sure you'll appreciate it and it will help.
Well, you can proceed as follows: ${\text{(1)}}$ Read first the chapter. $\text{(2)}$ Do some simple to medium exercises. ${\text{(3)}}$ Read the same chapter again, but this time more intensively and try to think about what you read a bit more. $\text{(4)}$ Do the hard exercises.

  If you are starting a topic from scratch with minimal knowledge on the content, what is an approach that is found to be effective and efficient?

Searching for the best textbook out there and doing the procedure I explained in the last sentences. And for your purpose, i.e. Pre-calculus the best book that I really recommend is this one: Precalculus with unit circle geometry - David Cohen. It discusses lot of aspects from the fundamental theorem of algebra to matrices that you can find in my answer here.

After mastering Precalculus you may have a look at the other undergraduate topics.

Mathematics is somewhat like playing an instrument or programming a computer. Musicians start their study with music in their heads they want to be able to play. Starting programmers may already see that killer game they want to program. Mathematicians want to solve problems or create new mathematics. That is where you start, by working on a problem: your problem. It is essential that you have one. Before anything I would read as much as I can on how to read mathematics, how to study mathematics and on how to solve problems. - About video lectures on internet I would like to say that they lack an essential quality: the possibility for the student to ask questions, you can't really interact and neither you are part of the group. I doubt if they really work.


How to teach mathematical induction? Some students are not convinced that a proof by mathematical induction is a proof. I have given the analogy of dominoes toppling but still some remain unconvinced. Is there very convincing way of introducing mathematical induction? I need something which will have an impact. Are there any real life applications of induction?

I'm surprised everyone seems to be in general agreement that the domino analogy is good. It seems poor to me: what have dominos got to do with proving things? It's a nice metaphor, but it has simply nothing to do with the issue at hand. I don't remember it having been helpful when I came across it before understanding induction, either. Maybe I'm a special case in that regard.

Which brings me to my answer: I would look for theorems that admit a particularly simple inductive proof. I would also relate those proofs to the students in a more conversational language. Ie, rather than:

  First, prove it for 0. Then, prove that if it is true for n, it is true for n + 1. So it must be true for all n.

There are two points here that may be confusing:

The notion of proving an "if then" statement. I suspect many students initially find it hard to see that an "if then" can even be considered a "statement" at all, people are more used to thinking of them as combinations of statements. The idea of proving "A if B", without even considering whether A is true, seems bizarre. Supposing I proved that if there were a unicorn, it would like tea.
The use of a variable $n$ may be an unnecessary distraction for some students with low mathematical ability, many students remain uncomfortable with variables all the way through high school. This is admittedly a less important point.

To beat issue (1), work by example. State a simple theorem and propose to prove that it's true for all numbers (don't say "for all n"! Just point to the number in the statement, written on the board, and say "we'll prove it's true no matter what this is"). Prove it for zero. Then, without proving or stating the inductive step as a separate statement, go right ahead show "well because it's true for 0, it must be true for 1, because...". Then, "but because it's true for 1, we can use the same argument to show it must be true for 2, after all...". Repeat maybe once more. Finish up by pointing out that as the same argument works for any number, you can keep going as far as you need to and the statement is true for all numbers.

Imagine you had never been taught induction, that you were living in an age before abstract mathematics, and remember the principle meaning of the word proof. A proof is what you use to convince your fellow human that something must be true. If you were simply trying to do that, you wouldn't be fussing around with separating out the inductive step as a proposition in its own right, you'd use the much more easily grasped (and much more convincing!) sort of proof by example above. The illustration of why that means it's true for all $n$ is in the language of the proof itself: it's true because we can keep doing the thing we just did two or three times.

A mathematical axiom should be self-evident. It should be unnecessary to explain why something is self-evident.

One should prove mathematical induction based on the self-evident proposition that every set of natural numbers has a least element.

I have found that students immediately understand the truth of that proposition whereas they do not immediately understand the principle of mathematical induction.

Mathematical induction is better approached as a theorem rather than an axiom.


How to determine the map between the singular homology groups of a torus induced by a linear map $f:\mathbb R^2 \to \mathbb R^2$? Let $f:\mathbb R^2 \to \mathbb R^2$ be the linear map whose matrix under the standard basis is $ \begin{bmatrix}
2 & 1 \\
1 & 1
\end{bmatrix} $.

Clearly $f$ induces the map from $\mathbb R^2/ \mathbb Z^2$ to itself. I wonder how to find the induced map between homology groups of the torus.

There are three nontrivial homologies: $H_0(T^2)=\mathbb Z, H_1(T^2)=\mathbb Z^2, H_2(T^1)=\mathbb Z$. Since $f: \mathbb R^2/\mathbb Z^2 \to \mathbb R^2/\mathbb Z^2$ is an isomorphism, by the functorality, so are $f_*: H_*(T^2)\to H_*(T^2)$.

But I can't figure out if $H_i(T^2)\to H_i(T^2)$ are the identity map or $-1$ maps when $i=0,2$. For $i=1$, I have a feeling that the induced homology map $f_*$ sends $(x,y)\in \mathbb Z^2$ to $(2x+y,x+y)$ (natively induced by the matrix), but I don't know how to show it rigorously.

I hope the method is not too advanced (best if it only uses the definition).

The best way to understand the map in $H_0$ is to consider this map as a map of CW-complexes. Than marked point is mapped to itself, so the map is an isomorphism.

If a map in $H_2$ would be multiplication by $-1$ it would mean that the map does not preserve the orientation. But your map does preserve it since it preserves the orientation of a plane(because the determinant is equal to 1). So the map on $H_2$ is also an identity.


Using definite quantification in logic I am looking to translate the following sentence into formal logic:

"Haters hate the hated"
Where Hxy: x hates y

I have translated it to:
∃x((Hxy ∧ ∀y(Hxy → x = y))

However I am not sure if the last part (x=y) is essential as I have already expressed any x hates y and all y is hated by x.

I would appreciate any help clarifying this, thank you.

(I am beginner in logic so please don't judge my attempt too harshly!)

Try and do these translations of complex statements step by step: divide and concur.  And, stick to the division between subject and term (those things about which you say something) and predicate term (what you say about those things), so you can follow the standard forms for translating those.

'Haters hate the hated' is of course 'all haters hate anyone who is hated'

So, subject term: 'haters'

Predicate term: 'hate anyone who is hated'

So this becomes:

$$\forall x (\text{'x is a hater'} \rightarrow \text{'x hates anyone who is hated'})$$

Now, 'x is a hater' translates as:

$$\exists w H(x,w)$$

while 'x hates anyone who is hated' can be paraphrased as 'all those who are hated are hated by x', which has:

Subject term: 'hated'

Predicate term: 'hated by x'

So the form here is:

$$\forall y (\text{'y is hated'} \rightarrow \text{'x hates y'})$$

'y is hated' becomes:

$$\exists z H(z,v)$$

and 'x hates y' is of course:

$$H(x,y)$$

OK, so we can plug in the last two parts:

$$\forall y (\exists z H(z,y) \rightarrow H(x,y))$$

And now we can plug that into the very first form:

$$\forall x (\exists w H(x,w) \rightarrow \forall y (\exists z H(z,y) \rightarrow H(x,y)))$$

I think the sentence should be taken to mean that everyone who is a hater hates everyone who is hated by someone.

Formally this amounts to: $\forall x(Hx \rightarrow \forall y(\exists z Dzy \rightarrow Dxy)) $, where $H, D$ stand for the predicates 'is a hater' and 'hates' respectively.


Understanding two similar definitions: Fréchet-Urysohn space and sequential space Here are the definitions:

Fréchet-Urysohn space: A topological space $ X $ where for every $ A \subseteq X $ and every $ x \in \text{cl}(A) $, there exists a sequence $ (x_{n})_{n \in \mathbb{N}} $ in $ A $ converging to $ x $.
Sequential space: A topological space $ X $ where a set $ A \subseteq X $ is closed iff $ A $ contains the limit points of every sequence contained in it.

As the title explains, I would like to know the difference between them. Thanks for any help.

Consider the following operation on a subset $A$ of a space $X$, defining a new subset of $X$: $$\mbox{s-cl}(A) = \{ x \in X \mid \mbox{ there exists a sequence } (x_n)_n \mbox{ from } A \mbox{ such that } x_n \rightarrow x \}\mbox{.}$$ This set, the sequential closure of $A$, contains $A$ (take constant sequences) and in all spaces $X$ it will be a subset of the $\mbox{cl}(A)$, the closure of $A$ in $X$.

We can define $\mbox{s-cl}^{0}(A) = A$ and for ordinals $\alpha > 0$ we define $\mbox{s-cl}^\alpha(A) = \mbox{s-cl}(\cup_{\beta < \alpha} \mbox{s-cl}^\beta(A))$, the so-called iterated sequential closure.

A space is Fréchet-Urysohn when $\mbox{s-cl}(A) = \mbox{cl}(A)$ for all subsets $A$ of $X$, so the first iteration of the sequential closure is the closure.

A space is sequential if some iteration $\mbox{s-cl}^\alpha(A)$ equals the $\mbox{cl}(A)$, for all subsets $A$.

So basically by taking sequence limits we can reach all points of the closure eventually in a sequential space, but in a Fréchet-Urysohn space we are done after one step already.

For more on the differences and the "canonical" example of a sequential non-Fréchet-Urysohn space (the Arens space), see this nice topology blog, and the links therein.

a space is a Sequential space  space iff every sequencially closed set is closed.But there can be closed sets which are not sequencially closed.
a space is frechet-urysohn if sequencial closure of a set and usual closure coincide


Stopped continuous-time supermartingales I am trying to prove that if I have a right-continuous supermartingale $(S_t,\mathcal{F}_t)_{t\geq0}$ and $\tau <\infty$ a stopping time, that $(S_{\tau \wedge t},\mathcal{F}_{\tau \wedge t})_{t\geq0}$ is also supermartingale.

I proved it for the discrete time case but I don't know how I can take the limit and prove it for the continuous time case.

I tried to take $\tau_n \searrow \tau$ and show uniform integrabillity of $S_{\tau_n \wedge t}$ but I am stuck.

Thank you

Here : Proof that a stopped continuous-time martingale is a martingale.

Optional sampling theorem : If $X = (X_t,\mathcal{F}_t)$ is a supermartingale and $T$ is an arbitrary stopping time, then the stopped process $X^T=(X_{T\wedge t},\mathcal{F}_t)$ is also a supermartingale.

Proof : Here we give a sketch of the proof. Assume $X$ is a supermartingale. First, for a fixed $n\ge 1$, let
$$
D_n = \left\{\frac{k}{2^n}, k= 0,1,2,\ldots\right\}\subset D_{n+1}\subset \cdots
$$
be the set of non-negative dyadic rationals of order no greater than $n$. It follows that
$$
X = (X_t,\mathcal{F}_t; t\in D_n)
$$
is a super martingale (discrete time).

Second, we construct a stopping time $T_n$ such that $T_n\ge T$ and $T_n$ only take values in $D_n$. Indeed, let
$$
T_n(\omega) = \inf\left\{t\in D_n; t\ge T(\omega)\right\}.
$$
Then $T_n\ge T_{n+1} \ge \cdots $ and $T_n$ is a stopping time. Fix $0\le s\le t$, we wish to show that
$$
\mathbb{E}\left[X_{t\wedge T}|\mathcal{F_s}\right]\le X_s
$$
almost surely. Similarly define
$$
t_n = \inf\left\{u\in D_n; u\ge t\right\} \ge t_{n+1}\ge\cdots
$$
and
$$
s_n = \inf\left\{u\in D_n; u\ge s\right\} \ge s_{n+1}\ge\cdots
$$
It follows from the discrete time optional sampling theorem that
$$
\mathbb{E}\left[X_{t_n\wedge T_n}|\mathcal{F}_{s_m}\right] \le X_{s_m \wedge T_n}
$$
for any integers $m\ge n$. Letting $m\to\infty$, we have $s_m \to s$ decreasing and $\mathcal{F}_{s_m}\to\mathcal{F}_s$. By Lévy's Downward theorem and the continuity of process $X$, we have
$$
\mathbb{E}\left[X_{t_n\wedge T_n}|\mathcal{F}_{s}\right] \le X_{s_m \wedge T_n}
$$
Observe that $(X_{t_n\wedge T_n},\mathcal{F}_{t_n\wedge T_n})$ is a backward martingale, whence it is uniformly integrable. Letting $n\to\infty$, we arrive at
$$
\mathbb{E}\left[X_{t\wedge T}|\mathcal{F}_{s}\right] \le X_{s\wedge T}.
$$
This completes the proof.


Let $f:\mathbb R→\mathbb R^3$ be defined by $f(x)=\langle 8x,6x,−8x\rangle$. Is f a linear transformation? I am not sure on what process I should have when dealing with this problem?

I don't know what f(y) is equal to..

To fill in the blanks, the answers must be be written as a formula that returns a vector, which can be done using angle brackets.

For example:<8,6,-8> which is a result with 3 dimensions.
If I put a result that doesn't have 3 dimensions, then I will get an error.

In the "choose" drop-down menu for part a) and part b), it has 2 options that can be selected. The 2 options are "Yes, they are equal" and "No, they are not equal"

For part c), the two options are "f is a linear transformation" and "f is not a linear transformation"

When we write $f(x) = \langle 8x,6x,-8x \rangle$, it means that $f$ assigns to each real number $x$ the 3-dimensional vector $\langle 8x,6x,-8x \rangle$. Thus, for example, $f(1) = \langle 8,-6,-8 \rangle$ and $f(0)$ is the zero vector.

So, $f(y)$ is just the vector $\langle 8y,6y,-8y \rangle$ and $f(x+y)$ is...

It is easy to check that, in fact, $f(x+y) = f(x) + f(y)$.

Please find the solution, let me know if you have questions.


Homeomorphism and Mapping Having some trouble understanding this:

$(S,\circ)$ and $(J,\times)$ be two groups and $f\colon S \to J$ be a homeomorphism. Prove that $f$ maps the identity of $S$ to the identity in $J$.

Any guides on where to step first. Feeling lost.

$f(e)=f(ee)=f(e)f(e)$
But $f(e)=e_Mf(e)$ so we have:
$$f(e)f(e)=e_Mf(e)$$
But recall the cancellation laws, $ax=bx \implies a=b$
We have $f(e)f(e)=e_Mf(e)\implies f(e)=e_M$ as required.
This completes the proof.

$f(g) = f(e\circ g) = f(e)\times f(g)$ and  $f(g) = f(g\circ e) = f(g)\times f(e)$

this is true for all $g\in S$ thus $f(e)$ has to be the identiy of $J$


prove that $\lim\limits_{n\to \infty} 2^{\frac{1}{\sqrt{n}}} =1 $ without using continuity of functions Prove that $$ \lim_{n\to\infty} 2^{\frac{1}{\sqrt{n}}} =1 $$ I tryed to solve this using the sandwich theorem and the fact that $\lim 2^{\frac{1}{{n}}} =1 $. We did not learn continuity of functions so I am not alowed to use this. Any suggestions?

First: $2^{1/n} \to 1.$ Proof: Let $2^{1/n} = 1 + a_n.$ Then $a_n > 0$ and the binomial theorem gives

$$2 = (1+a_n)^n = 1 + na_n + \cdots \ge 1 + na_n.$$

Thus $0< na_n \le 1 \implies a_n \to 0 \implies 2^{1/n}\to 1.$

As for $2^{1/\sqrt n}:$ Let $\epsilon>0.$ From the above there exists $m$ such that $1< 2^{1/m} = 2^{1/\sqrt {m^2}}<1+\epsilon.$ Because $p< q \implies 2^p< 2^q,$ we have $1< 2^{1/\sqrt {n}}<1+\epsilon$ for $n\ge m^2.$ Thus $2^{1/\sqrt n}\to 1.$

Think about this intuitively, as n approaches infinity, what does the value of 1/sqrt(n) become? Well, I think we agree that as n approaches infinity, the square root of n becomes infinity. So, what is 1 divided a really really big number? It is a really small number aka zero; 2 to the zero-th power is one. Therefore, the limit is 1.


Finitely addivite finite regular set function on real Borel $\sigma$ algebra is already a measure. Let $\mu$ be a finitely additive set function on the Borel $\sigma$-algebra $B_{\mathbb{R}}$ with $\mu(\mathbb{R})<\infty$ and $\mu (A) = \sup\{\mu(K) \mid K \subseteq A, \ K \text{ compact} \} $ for every $A \in B_{\mathbb{R}}$. Then $\mu$ is $\sigma$-additive and therefore a measure.

So far I did not find any way to tackle the above problem. Thanks for any inspiration!

We may assume that each $A_i$ is bounded, for $K$ is bounded. Now it suffices to show that for each $i$, we can find open set $O_i \supset A_i$ and having measure not "slightly" greater than that of $A_i$. For this, first enclose $A_i$ by an open interval $I_i$ and find a "big" compact $L_i$ inside $I_i \setminus A_i$. Finally put $O_i = I_i \setminus L_i$.


Derivative of a vector with respect to a matrix let $W$ be a $n\times m$ matrix and $\textbf{x}$ be a $m\times1$ vector. How do we calculate the following then?

$$\frac{dW\textbf{x}}{dW}$$

Thanks in advance.

The quantity in question  is a $3^{rd}$ order tensor.

One approach is to use index notation
$$\eqalign{
  f_i &= W_{ij} x_j \cr\cr
 \frac{\partial f_i}{\partial W_{mn}}
 &= \frac{\partial W_{ij}}{\partial W_{mn}} \,x_j \cr
 &= \delta_{im}\delta_{jn} \,x_j \cr
 &= \delta_{im}\,x_n \cr
}$$
Another approach is vectorization
$$\eqalign{
  f &= W\,x \cr
    &= I\,W\,x \cr
    &= (x^T\otimes I)\,{\rm vec}(W) \cr
    &= (x^T\otimes I)\,w \cr\cr
 \frac{\partial f}{\partial w} &= (x^T\otimes I) \cr
}$$

For independent case:

If $\mathbf{x}$ is independent of $W$, this problem can be calculated as follows.

$$\cfrac{\partial W\mathbf{x}}{\partial W}=
\cfrac{\partial}{\partial W}
\begin{bmatrix}
w_{11} & w_{12} & \cdots & w_{1m} \\
w_{21} & w_{22} & \cdots & w_{2m} \\
\vdots & \vdots & \ddots & \vdots \\
w_{n1} & w_{n2} & \cdots & w_{nm}
\end{bmatrix}
\mathbf{x}
$$

$$
=
\begin{bmatrix}
\cfrac{\partial w_{11}}{\partial w_{11}} & \cfrac{\partial w_{12}}{\partial w_{12}} & \cdots & \cfrac{\partial w_{1m}}{\partial w_{1m}} \\
\cfrac{\partial w_{21}}{\partial w_{21}} & \cfrac{\partial w_{22}}{\partial w_{22}} & \cdots & \cfrac{\partial w_{2m}}{\partial w_{2m}} \\
\vdots & \vdots & \ddots & \vdots \\
\cfrac{\partial w_{n1}}{\partial w_{n1}} & \cfrac{\partial w_{n2}}{\partial w_{n2}} & \cdots & \cfrac{\partial w_{nm}}{\partial w_{nm}}
\end{bmatrix}
\mathbf{x}
$$

Therefore, all elements are $1$. Eventually, the result is below.

$$
\cfrac{\partial W\mathbf{x}}{\partial W}=
(\mathbf{x}^{\text{T}}\mathbf{1_{m}}) \mathbf{1_{n}}
$$

Then $\mathbf{1_{k}} \in \mathbf{R}^{k}$ is

$$\mathbf{1_{k}}=[1 \ 1 \ \cdots 1]^{\text{T}}$$

For dependent case:

If $\mathbf{x}$ is dependent of $W$, it is more difficult than independent case. Likewise,

$$
\cfrac{\partial W\mathbf{x}}{\partial W}=
(\mathbf{x}^{\text{T}}\mathbf{1_{m}}) \mathbf{1_{n}} +
W \cfrac{\partial F(W) }{ \partial W}\mathbf{x}_{0}
$$

Then, $\mathbf{x}$ can be replaced as follows

$$\mathbf{x}=F(W)\mathbf{x}_{0}$$

where, $F(W) \in \mathbf{R}^{m \times n}$ is a matrix function, for which parameters are $W$, and $\mathbf{x}_{0} \in \mathbf{R}^{m}$ is independent of $W$.


Generators of the SU(2) matrix group Let $X_i$ be the generators of $SU(2)$ and let the parameters of the rotation be $\theta, \phi, \delta$ such that the matrix $R = e^{i(\phi X_{1} + \delta X_{2} + \theta X_{3})}$, where $R$ is an element of $SU(2)$.

So, $R = 1 + i(\phi X_{1} + \delta X_{2} + \theta X_{3}) + \frac{1}{2!}[i(\phi X_{1} + \delta X_{2} + \theta X_{3})]^2 + ...$ via Taylor expansion about the zero matrix $\textbf{0}$.

So, $-i \frac{\partial R}{\partial \phi}|_{\theta, \phi, \delta = 0} = -i\ [iX_{1} + (i \phi X_{1}+i \delta X_{2} + i \theta X_{3})(i X_{1}) + ...]|_{\theta, \phi, \delta = 0} = X_{1}$, and similarly $-i \frac{\partial R}{\partial \delta}|_{\theta, \phi, \delta = 0} = X_{2}$, and $-i \frac{\partial R}{\partial \theta}|_{\theta, \phi, \delta = 0} = X_{3}$.

Therefore, $X_{1} = -i \frac{\partial R}{\partial \phi}|_{\theta, \phi, \delta = 0} = -i \frac{\partial}{\partial \phi}\begin{pmatrix}a & b \\ -b^* & a^* \end{pmatrix}|_{\theta, \phi, \delta = 0} = -i \frac{\partial}{\partial \phi}\begin{pmatrix}e^{i\theta}cos \phi & e^{i\delta}sin \phi \\ -e^{-i\delta}sin \phi & e^{-i\theta}cos \phi \end{pmatrix}|_{\theta, \phi, \delta = 0} = -i \begin{pmatrix}-e^{i\theta}sin \phi & e^{i\delta}cos \phi \\ -e^{-i\delta}cos \phi & -e^{-i\theta}sin \phi \end{pmatrix}|_{\theta, \phi, \delta = 0} = -i \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix} =  \begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix}$, which is one of the Pauli matrices.

But, $X_{2} = -i \frac{\partial R}{\partial \delta}|_{\theta, \phi, \delta = 0} = -i \frac{\partial}{\partial \delta}\begin{pmatrix}a & b \\ -b^* & a^* \end{pmatrix}|_{\theta, \phi, \delta = 0} = -i \frac{\partial}{\partial \delta}\begin{pmatrix}e^{i\theta}cos \phi & e^{i\delta}sin \phi \\ -e^{-i\delta}sin \phi & e^{-i\theta}cos \phi \end{pmatrix}|_{\theta, \phi, \delta = 0} = -i \begin{pmatrix}0 & ie^{i\delta}sin \phi \\ ie^{-i\delta}sin \phi & 0 \end{pmatrix}|_{\theta, \phi, \delta = 0} = -i \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix} =  \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$. But, I don't think that the zero matrix is one of the generators of the $SU(2)$ group. In fact, it is not even a Pauli matrix. I was wondering if someone can point out the error in my calculation of $X_{2}$.

So I discussed this today and as it turns out there's a coordinate singularity at the identity in those co-ordinates. So the identity is given by $\theta=\phi=\delta=0$, but equally well the identity is given by $\theta=\phi=0, \delta=\text{anything}$. In fact they're singular whenever $\phi=0$, since the parameter $\delta$ is redundant.

Now we want to compute the lie algebra which is identified with the tangent space at the identity of the group, but one of our co-ordinates $\delta$ isn't good there, so differentiating with respect to that co-ordinate makes no sense, and gives the non-nonsensical result that the zero matrix is a generator, when they should be pauli matrices.

However suppose instead we let $a=x+iy$ and $b=u+iv$, we have the relation that $x^2+y^2+u^2+v^2=1$, so we can let $x=+\sqrt{1-(y^2+u^2+v^2)}$ and our matrix is:

$$\begin{pmatrix}x+iy & u+iv \\ -u+iv & x-iy \end{pmatrix},$$

and we compute derivatives with respect to $y,u,v$ evaluated at $y=u=v=0$.

Then $-i\cdot \frac{\partial}{\partial y}\Big|_{y=u=v=0}$ gives $\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}=\sigma_z$

And $-i\cdot \frac{\partial}{\partial u}\Big|_{y=u=v=0}$ gives $\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix}=\sigma_y$

And finally $-i\cdot \frac{\partial}{\partial v}\Big|_{y=u=v=0}$ gives $\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}=\sigma_x$


Given two relations on a set, I want to see if one is reflexive/transitive/symmetric, and find some relation compositions. As there is limit for title to put the full question, I will put the full question below.

Question: Given $R$ and $S$ are relations on $A = \{1,2,3,4\}$ such that $R = \{(2,2),(2,3),(2,4),(3,2),(3,3),(3,4),(4,4)\}$ and $S = \{(1,1),(1,3),(3,4)\}$...

Is $R$ reflexive? Transitive? Symmetric?
Find $S \circ R$ and $S \circ S$.

Well..Since $1$ from set $A$ is not included in set $R$, can I straightly say that since $1$ is not included in $R$ so $R$ is none of the symmetric, transitive and reflexive?
And also are $S \circ R = \{(2,4),(3,4)\}$ and $S\circ S = \{(1,1),(1,4)\}$?

$
\newcommand{\CC}{\mathcal{C}}
$
Observe that, if we represent your initial relations $R,S$ as a directed graph (where $x$ points to $y$ iff $(x,y) \in R$ or $S$, depending on the relation), then we have these graphs respectively:

In graphs like these, certain properties can be intuitively and visually observed:

Reflexivity appears if every node points to itself.
Symmetry happens if, whenever $x$ points to $y$ (and thus a path exists), then $y$ points to $x$ (and thus a path to "backtrack" exists). Basically, there are no "stray," one-way edges - only closed loops.
Transitivity is visible when, if there exists paths $x$ to $y$ and $y$ to $z$, then there is a path $x$ to $z$. For distinct nodes, then this means that, whenever two sides of a triangle are filled in, the last one is too, though the direction depends on the direction of the other sides in question.
Your equivalence classes are all of the nodes in each of the disjoint graphs you have. (Two graphs are disjoint if, basically, one is entirely separate from another, totally isolated.)

Is $R$ reflexive? Transitive? Symmetric?

$R$ is not reflexive, since $1$ is not pointing to itself, i.e. $(1,1) \not \in R$.
$R$ is not symmetric, since while $2$ and $3$ point to $4$, they do not point back. That is, $(2,4) \in R$ but $(4,2) \not \in R$; similarly, $(3,4) \in R$ but $(4,3) \not \in R$.
$R$ is transitive, since every necessary "triangle" is closed, with the third side always pointing in the right direction.

Note that the three properties above are independent of each other, to some degree. A relation can be any one of these while not the others. For instance

Also note that "$R$ is not reflexive" is not necessarily the same as "$R$ is irreflexive", and similar for symmetry and antisymmetry. A relation can be both, or neither, or one, or the other.
Note in particular: while $(1,1) \not \in R$ shows that $R$ is not reflexive, it does not show symmetry or transitivity (or, rather, the lack thereof), because $(x,1),(1,x) \not \in R$ for any $x \in A$ in the first place, so it's not really a concern.

Find $S \circ R$ and $S \circ S$.

Personally, when possible, I like to approach this sort of problem visually. Write out the elements of the relevant set, in columns -- three for any given composition of relations, but since we're finding multiple compositions we can do this more compactly with four columns to find both compositions.
For convenience, I will also label the columns $\mathcal C_i$, for $i=1,2,3,4$. Then, in the gaps between $\CC_1$ and $\CC_2$, draw arrows between the two, where $x$ in $\CC_1$ points to $y$ in $\CC_2$ if and only if $(x,y) \in R$. Next, draw arrows similarly in the gap for $\CC_2$ and $\CC_3$ representing the relation $S$: $x$ in $\CC_2$ points to $y$ in $\CC_3$ if and only if $(x,y) \in S$. Do this exact same thing for the next column.
In short: the arrows in the first gap represent the first "mapping" for the relation under $R$, and those in the subsequent gaps represent that for $S$. As you might imagine, following from $\CC_1$ to $\CC_3$ gives you $S \circ R$; followings from $\CC_2$ to $\CC_4$ gives you $S \circ S$. (This in particular might be more intuitive if you read $S \circ R$ as "$S$ after $R$," for instance, and think about function composition - in a definition sense, functions are just special relations.)
Anyhow, the diagram you get for your case is this:

You might be guessing what comes next. Essentially, if $x$ in $\CC_1$ points to $y$ in $\CC_2$, and then to $z$ in $\CC_3$, then $(x,z) \in S \circ R$. Or put differently: if there is a path from $x$ in $\CC_1$ to $z$ in $\CC_3$ you can walk along, then $(x,z) \in S \circ R$.
Similarly, the existence of such a path for $x$ in $\CC_2$ to $z$ in $\CC_4$ gives $(x,z) \in S \circ S$.
It will usually be easiest to just start in $\CC_3$, and see if you can trace backwards to $\CC_1$ (and similar for $\CC_4$ and $\CC_2$), but that's up to you. Anyhow, with this in mind, it becomes clear that
$$S \circ R = \big\{ (2,4),(3,4) \big\} \qquad
S \circ S = \big\{ (1,1),(1,3),(1,4) \big\}$$
Thus, your answers in this respect are correct (once we account for your comments in Bram28's answer).

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.

1) The element $1$ not being used in $R$ indeed makes it irreflexive. However, it does not mean that it is not symmetric or transitive. For example, the relation $Q = \{ \}$ (i.e. the empty relation) would be perfectly symmetric and transitive, even if it contains no elements at all of the set it is defined over!  So, you'll need to do some more work on that one

2) Your SoS is missing something ...


Maximum principle of harmonic function on compact manifold Theorem (Maximum Principle).  Let $h$ be a harmonic function on a domain $D$ in $C$.

  (a) If $h$ attains a local maximum in $D$, then $h$ is constant.

  (b) Suppose that $D$ is bounded and $h$ extends continuously to the boundary $\partial D$ of $D$. If $h\leq 0$ on $\partial D$, then $h\leq 0$ on $\overline{D}$.

The above is a very well known result.

Can anyone please suggest a reference of maximal principle for harmonic function on compact Riemannian manifold with boundary? I only need it for dimension 2
manifold.

I found a paper online which may be helpful for your question. The link is: https://projecteuclid.org/download/pdf_1/euclid.ojm/1353054807


How does Schwartz's paradox of surface area effect modelling of 3D objects? Question

I just became aware of Schwartz's paradox of surface area (explanation below for the unfamiliar). How does this effect mathematical modelling of real-life surfaces? For example, suppose I wanted to measure the surface area of a mountain and had the elevation data. I've found approaches that produce a polyhedral approximation (here), but how do we know this polyhedral approximation is actually approaching the surface area of the mountain? Thanks!

PS Maybe this is a better physics question? Also, Mandlebrot's first fractal paper comes to mind as a similar problem.

Schwartz's Paradox Explanation

If I understand correctly, Schwartz's Paradox shows that simply because a polyhedral approximation, $P_n$, of a curved surface $S$ approaches the curved surface as $n \to \infty$, the surface area of the polyhedral approximation, $A(P_n)$, does not approach the geometrically intuitive surface area of the surface, $A(S)$. In summary,

$$\lim_{n\to\infty} P_n = S \not\Rightarrow \lim_{n\to\infty} A(P_n) = A(S) $$

I surmised this from the following paper.

When approximating the cylinder, for example, such an approximation does work, so long as we also have convergence of the normals.

The approximation fails for Schwarz lanterns because the normals of the triangles in the Schwarz lanterns don't converge at all, much less to the normals of the cylinder.

See for example here of how we can use triangulations to approximate surface area: http://arxiv.org/pdf/1404.1823v1.pdf

Also look at the last paragraph of the paper linked to in the original question -- Lebesgue already solved this problem in 1902 by demanding uniform convergence instead of just pointwise convergence. Clearly Schwarz's lanterns converge pointwise but not uniformly -- the reason why the surface area blows up is because the triangular approximations can become arbitrarily jagged (which is also why the normals fail to converge). With uniform convergence, this doesn't happen.

See this Wolfram demo for how it works: http://demonstrations.wolfram.com/CylinderAreaParadox/

If it is reasonable to demand uniform convergence of functions instead of just pointwise convergence, then it should also be reasonable for surface areas.

In short, it doesn't.

I don't have a general proof, but let's take a look at the original paradox. For a cylinder, the area is given by:

$$A(n,m) = 2nr\sin (\pi/n)\sqrt{h^2+r^2m^2(1-\cos^2 (\pi/n))^2}$$

Now, it's true that if you let $m$ go to infinity independently of $n$, you'll end up with nonsense. But in all real life scenarios, you only have finite sampling, and therefore, the ratio $c=m/n$ is bound by some constant. You can then show that if $m=cn$, the limit is indeed $2\pi hr$ irrespective of $c$.


Convergence of sequences in topological spaces Consider R with co-countable topology then for Q'(set of irrational numbers) subset of R ,how can we show that there is no sequence in Q' converges to 0?

If $(x_n)$ is a sequence of irrational numbers converging to $0$ then $\mathbb R \setminus \{x_1,x_2,..\}$ is an open set in this topology containing $0$ so it must contain $x_n$ for $n$ sufficiently large. But it contains no $x_n$!

Any sequence $x_n$ converges to $0$ if for all open $\theta$ with $0\in \theta$ then exist $N$ a natural number with $x_n \in \theta$ for all $n\geq N$.

Let $\theta = \mathbb{R}\setminus \{x_n\}$. Then $\theta$ is an open set, because his complement is a contable set, it's mean, $\{x_n\}$ is contable. For the other hand, for all $n\in\mathbb{N}$, $x_n \not\in\theta$.

A more general result :

If $x$ is a rational number, there isn't any sequence $x_n$ of irrational numbers with $x_n\to a$.


Differential Calculus and Integral Calculus Is differential calculus a prerequisite for integral calculus? Because almost always you see that differential is taught before integral. Does that have a specific reason? Would it be recommended to start with integral calculus? Or is it an absolute non-go

The wonderful Analysis by Its History by Hairer and Wanner presents a first year calculus course by introducing the different topics in chronological order of their discovery. Of course this means that a fair bit of integration comes before some differentiation. The historical route is a great way to motivate the different definitions etc, and to see how mathematical knowledge is constructed/discovered. And its also a great way to learn calculus - it's not in the least pedagogically awful!

In integration, we make use of differentiation, particularly when we are making substitutions.

While learning differentiation, we need not have the knowledge of differentiation.

Further, integration is called the reverse process of differentiation.

Hence, we have to learn differentiation before integration.
Differentiation helps one to understand a whole with micro-level approach to understand whole from inside out in the entire domain of a function, and it's mechanics is pathological analysis in nature. Integration is understanding of coverage of a given function in the Cartesian domain, in any co-ordinate definition system. Approach is reverse of approach to human understanding. Differential calculus is a child while integral calculus is grand parent. One first learns the evolution of child and then understands the old person. Differential coefficient, being tangent of inclination of function, is akin to psychology and behavioral pattern of the child; while integral calculus approach is the tendency of old generation to define it's domain boundaries.

It is, hence, essential to start understanding the child in all its ramifications before undertaking to understand of its higher generation in terms of latter's domain perceptions.


Comparing 2 Point of Estimate Question

  $X_1$ and $X_2$ are random variables that satisfy $\operatorname{E}[X_1] = \operatorname{E}[X_2] = \mu$ and $\operatorname{Var}[X_1] = \operatorname{Var}[X_2] = 1$.  Specify the range for $\mu$ so that the point estimate $\hat \mu_1 = \frac{X_1 + X_2}{3} + 3$ has a lower mean squared error than the point estimate $\hat\mu_2 = \frac{X_1 + X_2}{2}$.

Hi, I was trying to calculate the $u$ that satisfies the stated condition. However,

I found $\frac{1}{2} + \frac{(u-9)^2}{3} < \frac{1}{2} + 0^2$  (By using mse(T) = var(T) + bias(t)^2)

But there is no value for u for this inequality, so I wonder if I have a mistake in some point or not.

Note under the assumption of independence, $$\operatorname{MSE}[\hat \mu_1] = \operatorname{Var}\left[\frac{X_1 + X_2}{3} + 3\right] + \operatorname{E}^2\left[\frac{X_1 + X_2}{3} + 3 - \mu\right] = \frac{2}{9} + \left(3-\frac{\mu}{3}\right)^2$$ whereas $$\operatorname{MSE}[\hat \mu_2] = \operatorname{Var}\left[\frac{X_1 + X_2}{2}\right] + \operatorname{E}^2\left[\frac{X_1 + X_2}{2} - \mu\right] = \frac{1}{4}.$$


The name of matrix operation If I have a matrix:
$$
A =
\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
\end{bmatrix}
$$

and a vector:
$$
B =
\begin{bmatrix}
b_1 \\
b_2 \\
\end{bmatrix}
$$

What is the name of the operation, which produces a vector:

$$
\begin{bmatrix}
b_1 \cdot a_{11} + b_1 \cdot a_{12} + b_1 \cdot a_{13} \\
b_2 \cdot a_{21} + b_2 \cdot a_{22} + b_2 \cdot a_{23} \\
\end{bmatrix}
$$

Since your operation is invariant under row-permutations of matrix $A$ and only the sum matters then given $A$ we have
$$
\tilde{A} = \begin{pmatrix} a_{11} + a_{12} + a_{13} \\ a_{21} + a_{22} + a_{23} \end{pmatrix}
$$
and $b = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$. Now your operation becomes the Hadamard product of $\tilde{A}$ and $b$.

This is a dot product, also called Matrix Multiplication when it involves a vector and a matrix, or two matrices.

In wikipedia article on Matrix Multiplication, in the section Definition, you have the following formula for the elements $c_{ij}$ of a matrix $\bf C=\bf A \bf B$ for $\bf A, \bf B$ and $\bf C$ all $m \times m$ matrices.

$$
c_{ij} = \sum_{k=1}^m a_{ik} b_{kj}
$$

If you treat your $\bf B$ matrix as a vector $\bf b$, with $m$ entries, instead of an $m \times m$ matrix, the operation will produce a vector $\bf c$ with $c_i$ elements given by:

$$
c_i = \sum_{k=1}^m a_{ik} b_k
$$

Which is exactly what you want.


What is a term for two events where the absense of one assures the absense of the other? With two events, there are three types of what i've dubbed 'mutual patterns':

Mutual Inclusion

A    B
1 -> 1 If A then B
1 <- 1 If B then A

A <-> B

Mutual Exclusion

A    B
1 -> 0 If A then NOT B
0 <- 1 If B then NOT A

NOT A <-> B
NOT B <-> A

Mutual erm...   ¯\_(ツ)_/¯

A    B
0 -> 0 If NOT A then NOT B
0 <- 0 If NOT B then NOT A

NOT A <-> NOT B

See, here's where the confusion lies. For 2 of the three patterns, you have words used to describe those two patterns which are antonyms to each other. I.e their opposites. So there is no third term beyond exclusion and inclusion, but the type of pattern where the absence of A assures the absence of the other B and vice versa, still shares this notion of mutuality. In other words, it's like am looking for a third opposite... Which I don't think is possible.

The statement:
$$
(\textrm{Not} \ A) \Leftrightarrow (\textrm{Not} \ B)
$$
is equivalent to the statement:
$$
 A \Leftrightarrow  B
$$
P.S.: You can easily check that the above statements are contrapositives.

I'm not sure I understand this question.

That said, we might write:

1. A is absent of truth, B is absent of truth.
2. A is absent of truth, B is present with truth.
3. A is present with truth, B is absent of truth.
4. A is present with truth, B is present with truth.

Here, the cases where the absence of one proposition assures the absence of the other are cases 1. and case 4.  This accords with the truth table for exclusive disjunction.


Projection of linear combination of real numbers Suppose a real number $\alpha$ is a rational linear combination of two linearly independent real numbers, such as $\pi$ and $\sqrt 3$. For instance, say $$\alpha = \tfrac12\pi + 5\sqrt 3.$$
By a simple linear algebra argument ($\{\pi,\sqrt3\}$ is a $\mathbb Q$-basis), it's clear that the coefficients of $\alpha$ are uniquely determined.

Question: Given $\alpha\in\operatorname{span}_\mathbb Q(\{\pi,\sqrt 3\})$, how do we determine the coefficients of $\pi$ and $\sqrt 3$?

It feels like there is enough information, but at the same time, I imagine that we can find different coefficients which give us numbers arbitrarily close to $\alpha$, so it is important that we have an exact expression for $\alpha$.
(In other words, if we only know that $\alpha\approx 10.23105$, say, I imagine it would be hopeless to expect that we can determine the coefficients, even if we restrict our view to the module $\operatorname{span}_{\mathbb Z}(\{\pi,\sqrt 3\})$.)

The reason I ask this question is because for each $n$, the integral $$I(n)=\int_0^{\pi/3}\sin^{2n}x\,dx$$ is a number of the form $a\pi+b\sqrt 3$. It's easy to obtain the recurrence relation
$$I(n) = \begin{cases} \hfil\frac\pi3\hfil & \text{if $n=0$}\\[3pt]
(1-\tfrac1{2n})\,I(n-1)-\frac1{4n}\big(\tfrac{\sqrt3}2\big)^{2n-1} & \text{otherwise}
\end{cases}$$
for $I(n)$, but I want to be able to compute the coefficients of $\pi$ and $\sqrt 3$ separately (in exact form) using a computer program.

I appreciate any help with solving either of the two problems, I'm assuming a solution to one necessarily sheds light on the other, that's why I haven't posted these as two separate questions.

Here's easy way to determine them:
let $t \in [0..1]$, we define
$$ a_0 = t * \alpha $$
$$ a_1 = (1-t) * \alpha $$
Basically we have $a_0 + a_1 = \alpha$.
Now, after we have $a_0$ and $a_1$, we can calculate co-efficients:
$$c_0 = \frac{a_0}{\pi}$$
$$c_1 = \frac{a_1}{\sqrt{3}}$$.
I have a feeling that the $t$ cannot be calculated from the $\alpha$, given that they are both located in same dimension, basically you need some criteria how $\alpha \in R$ can be divided to two separate dimensions. Thus your end result is large number of possible alternative coefficients. Every choice of $t$ will have different coefficients. Limiting $c_0,c_1 \in R$ to $Q$ sounds kinda difficult.


Proving series convergence I want to show that
$$\prod_{p\in \mathbb{P}}(1-p^{-s})^{-1}$$
converges if Re$(s)>1$, and $\mathbb{P}$ is the set of all prime numbers. In order to show that the infinite product converges, we'll look at
$$-\sum_{p\in \mathbb{P}}\ln(1-p^{-s}).$$
How should I show that it converges?

Apply some basic results: That $\sum_{n\in N} n^{-t}$ converges for $t>1,$ and these two theorems:

Theorem 1. Let $0\leq a_n<1$ for $n\in N.$ Then $$(1)....\quad \prod_{n\in N}(1-a_n)>0 \iff \sum_{n\in N}a_n<\infty.$$ Theorem 2. Let $b_n\geq 0$ for $n\in N.$ Then $$(2)....\quad \prod_{n\in N}(1+b_n)<\infty \iff \sum_{n\in N}b_n<\infty.$$
 Now for $Re(s)>1$ and prime $p$ we have $$(3)....\quad 0<1-p^{-Re(s)}\leq |1-p^{-s}|\leq 1+p^{-Re(s)}.$$ Therefore  $$(4)....\quad \prod_{p\in P}(1-p^{-Re(s)})\leq   \prod_{p\in P}|1-p^{-s}|\leq \prod_{p\in P}(1+p^{-Re (s)}).$$ By Theorem 1  we have $$(5)....\quad 0<\prod_{n\in N}(1-n^{-Re(s)})\leq \prod_{p\in P}(1-p^{-Re(s)}).$$ By Theorem 2 we have $$(6)....\quad \infty>\prod_{n\in N}(1+n^{-Re(s)})\geq \prod_{p\in P}(1+p^{-Re(s)}).$$ So $$(7)....\quad 0<\prod_{p\in P}|1-p^{-s}|<\infty.$$ So, with $s_n=\prod_{p\in P \land p\leq (n+1)}(1-p^{-s} )$, the sequence $S=(s_n)_{n\in N}$  is bounded in modulus by a positive lower bound and a finite upper bound. Therefore $S$ has a convergent subsequence. We can now use the  inequalities of (3), the convergence of the products in (4) & (5), and the basic $\epsilon, \delta$ def'n of convergence to show then that $S$ has a unique non-zero limit.

As it relates to the Reimann zeta function.

Euler showed that:

$\zeta (s) = \sum_\limits{n=1}^{\infty} n^{-s} = \prod_{p\in \mathbb{P}}(1-p^{-s})^{-1}$

https://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function

Now we just have to prove that $\sum_\limits{n=1}^{\infty} n^{-s}$ converges when $s>1.$

The most direct way would be the integral test.

Following the logic above:

$-\sum_{p\in \mathbb{P}}\ln(1-p^{-s})$

$p^{-s}<-\ln (1-p^{-s})<(p^s-1)^{-1}$

and the $n{th}$ prime number is always greater than $n.$

If $\sum_{n\in \mathbb{N}}n^{-s}$ converges, $-\sum_{p\in \mathbb{P}}\ln(1-p^{-s})$ converges by the comparison test.


If $\dim (\operatorname{Ker}f)^{\perp} = 1$ then $f$ is continuous for $f$ linear functional on Hilbert space I need to prove the following result:

  If $\dim (\operatorname{Ker}f)^{\perp} = 1$ then $f$ is continuous for $f$ linear
  functional on Hilbert space.

I have the following ingredients in mind:

Use sequential characterization of continuity knowing that since $\operatorname{Ker}(f)$ is closed, projection theorem gives that $x_n = m_n + u_n \to x = m + u$ is equivalent to $m_n \to m \land u_n \to u$.
Someone suggested me to further prove that $H = \operatorname{Ker}(f) \oplus \operatorname{Ker}(f)^{\perp} = \operatorname{Ker}(f) \oplus\operatorname{Img}(f)$ but I don't see how this is of any help.

Any ideas?

Suppose $\dim(Ker(f)^{\perp})=1$, where $f$ is a linear functional. Then there exists a unit vector $x$ such that $x\perp Ker(f)$. In particular $f(x)\ne 0$. So,
$$
        f\left(y-\frac{f(y)}{f(x)}x\right) = 0,
$$
which forces the following for all $y$:
$$
             \left\langle y-\frac{f(y)}{f(x)}x,x\right\rangle = 0 \\
           \langle y,x\rangle = \frac{f(y)}{f(x)} \\
        f(y) = \langle y,\overline{f(x)}x\rangle,\;\; y\in H.
$$
So $f$ is a continuous linear functional because it has a Riesz representation.

Let $x$ non zero in the orhogonal of $Ker(f)$, the linear function $\langle x, .\rangle$ is continuous, this implies that its kernel $Ker(f)$ is closed and $f$ is continuous.

Closed kernel implies continuous linear functional : Zorn's Lemma


How do I solve $\int \frac{1}{v(1+v^2)}dv$? I'm trying to solve $y'=\frac{2xy}{x^2-y^2}$ and I ended up with the integral $\int \frac{1}{v(1+v^2)}dv$ as part of the solution.

I got a big ugly solution using integration by parts, but I'm hoping there's a simpler way.

Now use partial decomposition:

$$\int\frac{1}{v(1+v^2)}dv = \int[\frac{1}{v} - \frac{v}{1+v^2}]dv$$

$$=\ln|v|-\frac{1}{2}\ln(1+v^2)+C$$

$$\log x -\log(1+x^2)/2$$

does it, I think.


Convert $y^2 = 4(x + 1)$ to a polar equation I'm trying to convert the rectangular cartesian equation

$$
y^2 = 4(x + 1)
$$

to a polar equation. After replacing $y = r \sin \theta$ and $x = r \cos \theta$, I get

$$
r^2 \sin^2 \theta = 4(r \cos \theta + 1)
$$

After replacing $\sin^2 \theta = 1 - \cos^2 \theta$ and rearranging, I get

$$
r^2 - r^2 \cos^2 \theta - 4r \cos \theta -4 = 0
$$

That's where I'm stuck and I can't solve the equation in terms of $r$.

Basic approach. Rewrite your equation as

$$
(\sin^2\theta) r^2 - (4\cos\theta) r - 4 = 0
$$

and use the quadratic formula for $r$.


Multilateration of Sound in 3D Space TL:DR - How can you find the 3D coordinates of a emitter than transmits an impulse signal?

STORY:

I'm working on something to improve my bird-watching.  I've got a camera that can take pictures of the birds when I'm not around, but currently it has to be zoomed out all the way to guarantee they're in frame.  This doesn't make for good pictures, so here's what I've done:

Mounted camera on a motor so it can rotate, zoomed in enough that the pictures will be better quality, and attempted multilateration to make the camera turn.

ATTEMPTED SOLUTION:

My multilateration is simple.  4 microphones listen for sound.  When an impulse (such as a chirp) is created from an emitter (bird), the microphones can detect the impulse, and my microcontroller can calculate the time differences between all 4 mics receiving the impulse.

My microcontroller then uses a home brew program that converts these time differences and the known locations of the microphones relative to each other into matrix form.

Once the program has the matrices, it can solve for the distance from each microphone to the bird's origin, which then can be used to figure out the coordinates of the bird relative to the microphones.

PROBLEM:

The problem with this, is that it needs to be really precise.  I'm talking ~10 nanoseconds of difference in reception time between mics in theoretical math space will cause the program to miscalculate where the bird is.

I've muddled with the code to see if implementing more mics will lessen the need for precision, but I can't find a way to achieve a tolerance greater than ~±25ns.

With my setup, I can only calculate a reception time difference on the level of 10-5 seconds, so it's not possible for me to guarantee the level of precision that this type of math needs.

Can anyone think of a way for me to improve my setup so that it works?  Are there other ways to accomplish multilateration?  How else could I find where the bird is when it's chirping?

Also, I think my question is different from this one.

Thanks guys, you're always awesome!!!

EDIT:

I have written out the mathematical process I have used for this problem.  Pictures of that, an excel sheet for generating initial conditions, and Matlab code for handling the maths can be found here.

A mayor factor you didn't mention is the distance between your microphones!
Also you didn't mention how precise you want to locate the bird.
But you can only relax timing needs if you have greater distance between the mics.

    ...it can solve for the distance from each microphone to the bird's origin..
    I don't understand your calculations. What you should get is a horizontal and a vertical angle of the chirp's origin, relativ to the base line. That needs no matrix calculation - its only geometric, isn't it?

As far as my practical experience in electronics precision measurement goes, I even don't trust your 10µs resolution. There are many error sources, from microphone differences (frequency response) to reflections in trees, air turbulences, other sounds and overall white noise level...
10ns resolution from <20kHz sound is practically absolutely impossible.
But:
10µs deviation for a 1 meter base line amounts to an angle variance of
0,2 degree.
Is that not sufficient???


repeating number, when n is multiple of 3. If we take any integer  multiple of $3$, and add up the cubes of its decimal digits, then take
the result and sum the cubes of its digits, and so on, we invariably end up with $153.$

Why the above result is true for only an integer multiple of $3?$ can we prove the reasons for it? Is there any such numbers will invariably end with any number and not multiple of $3?$ If yes, how to find them.

A number is only divisible by three if the sum of its digits is a multiple of three. Since $a^3\equiv a\mod 3$, we see that after applying the 'step' (the function $s$) to a number divisible by three will still be divisible by three. Also, a number not divisible by $3$ will never be divisible by three after the step (because everything here works in two directions). Since $3|153$ (because $3|1+5+3=9$), only numbers divisible by $3$ can end in $153$.

In general, because $9^3<1000$, $s(n)<n$ when $n>10000$ and $s(n)<10000$ when $n\leq 10000$ (these are not a sharp bounds). This implies that the sequence $\left(s^k(n)\right)_{k\geq 0}$ will be decreasing until it falls below $10000$. Because of that, the function will eventually get in a finite loop. Thus, there is an $l$ such that, for a sufficient large $K$, $s^{k+l}(n)=s^k(n)$ for all $k>K$. It is possible to calculate all fixpoints $n$ of $s$ for which $s(n)=n$. I think this only applies to $153$, since it is given in this problem, but maybe there are some other numbers with this property.

EDIT As user44197 points out, the only other fixpoints (I assume) are $370$, $371$ and $407$. And of course $0$ and $1$, but they are trivial. As you can see, none of these numbers is divisible by $3$, so they can only end up in $153$ or in a cycle, but there aren't any cycles apparently. For other numbers then $3$, I don't know whether or not multiples will always end in the same number, but I think not. The only number that may be of interest is $11$, because a number is a multiple of $11$ when the alternating sum of its digits is a multiple of $11$. Problem is that $2^3=8\neq \pm 2\mod11$, so divisibility it not conserved. For any other interesting numbers, you can search for $n$ s.t. $i^3\equiv \pm i\mod n$ for all $1\leq i\leq n$. Thus, $2^3=8\equiv \pm 2\mod n$. This is true for $n=2$, $n=3$, $n=5$, $n=6$ and $n=10$. $n=2$ doesn't work, because $s(12)=1+8=9$ and $9$ is odd. Same for $n=10$: $s(10)=1+0=1$. For $n=6$, we get $s(12)=9$. For $n=5$, we get $s(10)=1$, so this won't work either. Thus, only multiples of $3$ have the property that they are still multiples of $3$ after applying $s$.


Solve $\sin(x)=-\frac{1}{2}$ I have to solve the following equation for $x$

$$\cos(2x)+\sin(x)=0$$

After simplification i got

$\sin(x)=1$ or $\sin(x)=-\frac{1}{2}$

$\Rightarrow x=90^0$

But don't know how to solve for $x$ for $\sin(x)=-\frac{1}{2}$ ?

If $\sin x=-1/2$, then $x=210^{\circ}$ or $x=330^{\circ}$ are solutions. In the third and fourth quadrants sine becomes negative. Alternatively you can say $x=-30^{\circ}$ or $x=-150^{\circ}$.

The trick here is to have a mental picture of what the sine function is doing. Also the values are the same when you add or subtract $360^{\circ}$.

The way I was taught to remember which one is positive was remembering a mnemonic for the letters A, S, T and C.

A = All

S = sine

T = tangent

C = cosine

If $\space sin(x)=-1/2\space $ then $\space x=\pi n+\frac{\pi}{6}\cdot(-1)^{n+1}$

Edit: $\space n\in Z$


Linear Algebra Wizardry I am reading a textbook which features results on the multivariate normal distribution and the author writes out results with no justification, which take me a page to verify, and they end up being right. These claims don't seem obvious to me. I'll give an example.

Here's the setup:

Let $y_1,y_2,...,y_n, \mu \in \mathbb{R}^d$, let $A \in \mathbb{R}^{d \times d}$

Let $Y$ be the $n \times d$ matrix, formed by taking row $i$ to be $y_i$.

Let $e \in \mathbb{R}^d$ be a vector of 1's.

An example of a claim:

$\sum_{i}(y_i-\mu)^{T}A(y_i-\mu)=\text{trace}(A(Y-e\mu^{T})^{T}(Y-e\mu^{T}))$

This claim took me a long time to verify, but the author included zero justification and just followed it on from his last line of work. There are more examples of claims he makes which would take me a while to verify, but I feel like I might be missing a trick given the frequency of this happening.

Can anyone explain how either I can get better at doing these, or at least some verification that I'm not crazy and it's not super trivial?

Many thanks.

Here is a short derivation of the identity in question.

$$
\sum_{i}(y_i-\mu)^{T}A(y_i-\mu) =
\sum_{i}\operatorname{tr}[(y_i-\mu)^{T}A(y_i-\mu)]\\
= \sum_i [\operatorname{tr}A(y_i-\mu)(y_i-\mu)^{T}]
= \operatorname{tr}\left[ A\sum_i(y_i-\mu)(y_i-\mu)^{T}\right].
$$
Now, note that
$$
\sum_i(y_i-\mu)(y_i-\mu)^{T} = \pmatrix{y_1 - \mu & \cdots & y_n - \mu} \pmatrix{(y_1 - \mu)^T \\ \vdots \\ (y_n - \mu)^T} = (Y - e\mu^T)^T(Y - e\mu^T).
$$

Can anyone explain how either I can get better at doing these

The same way you get better at doing anything else - by practice.

  or at least some verification that I'm not crazy and it's not super trivial?

It's not trivial at the slightest, but that's not the criterion. It's not about "is it trivial", it's about "is it interesting enough".

Let me explain.

The book you read was not about linear algebra, it was about something else, in your case, it sounds like probability. When writing a book about one subject, it is usually best if the author includes as little of other subjects as possible, while still keeping the book understandable. The derivation of the claim you cite is a purely algebraic manipulation that the author did not consider interesting enough to include.


How to prove that inverse Fourier transform of "1" is delta funstion? $\mathscr{F}\{\delta(t)\}=1$, so this means inverse fourier transform of 1 is dirac delta function so I tried to prove it by solving the integral but I got something which doesn't converge.

In the following $\langle f, \cdot \rangle$ denotes the linear functional on Schwartz space induced by $f$ and $f^\lor$ stands for the inverse Fourier transform of $f$. By definition, for any Schwartz function $\varphi$
\begin{align*}
\langle 1^\lor, \varphi \rangle=\langle 1, \varphi^\lor \rangle=&\int_\mathbb{R} \left(\int_\mathbb{R} e^{2\pi ixy}\varphi(y) dy\right)dx
=\lim_{M\to\infty}\int_{-M}^M \left(\int_\mathbb{R} e^{2\pi ixy}\varphi(y) dy\right)dx.
\end{align*}
By Fubini's theorem we have
\begin{align*}
\int_{-M}^M \left(\int_\mathbb{R} e^{2\pi ixy}\varphi(y) dy\right)dx=&\int_\mathbb{R} \varphi(y)\left(\int_{-M}^M e^{2\pi ixy}dx\right) dy
=\pi^{-1}\int_\mathbb{R} \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy.
\end{align*}
Since $\varphi$ is differentiable at $y=0$, we have $|\varphi(y)-\varphi(0)|\le C|y|$ for some constant $C$. Thus
$\left(\varphi(y)-\varphi(0)\right)/y\in L^1_{loc}(\mathbb{R}).$
Then by Riemann-Lebesgue Lemma we have
\begin{align*}
\lim_{M\to\infty}\int_{-1}^1 \left(\varphi(y)-\varphi(0)\right)\frac{\sin (2\pi My)}{y} dy=0,
\end{align*}
which means
\begin{align*}
\lim_{M\to\infty}\int_{-1}^1 \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy=&\varphi(0)\lim_{M\to\infty}\int_{-1}^1\frac{\sin (2\pi My)}{y} dy
\\
=&\varphi(0) \int_{-\infty}^\infty \frac{\sin y}{y}dy=\pi\varphi(0).
\end{align*}
Note that $\varphi\left(y\right)/y$ is integrable on $\mathbb{R}\setminus [-1,1]$. Thus by Riemann-Lebesgue Lemma we have
\begin{align*}
\lim_{M\to\infty}\int_{1}^\infty \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy=\lim_{M\to\infty}\int_{-\infty}^{-1} \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy=0.
\end{align*}
To sum up the above argument we have for all Schwartz functions $\varphi$,
\begin{align*}
\langle 1^\lor, \varphi \rangle=\pi^{-1}\lim_{M\to\infty}\int_\mathbb{R} \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy=\varphi(0)=\langle \delta, \varphi \rangle.
\end{align*}
Therefore $1^\lor=\delta$.

$$\mathscr{F^{-1}}\{1\}=\int_{-\infty}^{\infty}e^{2\pi ixy}dy=\lim_{M\to\infty}\frac{\sin{2\pi Mx}}{\pi x}$$
Now we need to consider 2 cases:

1) $x=0$, then $\lim_{M\to\infty}\frac{\sin{2\pi Mx}}{\pi x}=\infty$

2) $x\ne0$, then $\lim_{M\to\infty}\frac{\sin{2\pi Mx}}{\pi x}=0$

Hence combining these 2 cases, we obtain

$$\mathscr{F^{-1}}\{1\}=\delta(x)$$


Simplify sum of angles expression (Trigonometry) I'm stuck on this.I need to simplify this given expression
$$\cot(\fracπ4+\theta)(\cos^2\theta-\sin^2\theta)$$
So far I simplified the first parentheses to $\frac{\cos\theta-\sin\theta}{\cos\theta\sin\theta}$ which I dont if this is correct but now I dont know how to multiply this by the second parenthesis

Remember this:

$$\cos^2(x) - \sin^2(x) = \cos(2x)$$

and $$\cot\left(\frac{\pi}{4} - x \right) = \tan\left(\frac{\pi}{4} + x\right)$$


exponential of second derivative, not positive? I'm interested in whether the following function $h(x,y)=e^{\partial_x\partial_y}f(x)g(y)$, is positive in the sense that $h(x,y)\geq 0$, whenever $f(x)\geq 0$ and $g(y)\geq 0$ (positive and real). I would have thought that it was positive because the derivatives are in the exponential, but it appears not to be. Any idea why it isn't positive? Maybe because ${\partial_x\partial_y}$ is not a normal operator? But I think it is even self-adjoint (or at least symmetric), so should have real eigenvalues.

To see that it is not positive one can expand the exponential $h(x,y)=f(x)g(y)+f'(x)g'(y)+\frac{1}{2}f''(x)g''(y)+...$ and take (for example) $f(x)=x^2$ and $g(y)=y^2$. Then the series terminates at second order and is $(xy+2)^2-2$, which is negative at $x=-\sqrt{2}$, $y=\sqrt{2}$ etc.

To see that $\partial_x\partial_y$ is symmetric we have $\int dx dy \psi^*(x,y)\partial_x\partial_y \phi(x,y)=\int dx dy \partial_x\partial_y \psi^*(x,y) \phi(x,y)$ after integrating by parts twice, plus a boundary term. So for the boundary term to vanish we should have that $\psi^*(x,y)$ is zero at infinity, so maybe  $e^{\partial_x\partial_y}f(x)g(y)$ is positive on functions which decay to zero  at infinity? This is not the case with polynomials which are the examples I know that give negative values.

We will assume Clairaut's Theorem holds, such that $\partial_x\partial_y=\partial_y\partial_x.$ Using the generalized Leibnitz Rule for the higher-order derivatives of a product, we have
\begin{align*}
e^{\partial_{x}\partial_{y}}\{fg\}&=\sum_{n=0}^{\infty}\left[\frac{1}{n!}\left(\frac{\partial^2}{\partial x \, \partial y}\right)^{\!n}(fg)\right] \\
&=\sum_{n=0}^{\infty}\left[\frac{1}{n!}\frac{\partial^n}{\partial x^n}\frac{\partial^n}{\partial y^n}(fg)\right] \\
&=\sum_{n=0}^{\infty}\left[\frac{1}{n!}\frac{\partial^n}{\partial x^n}\left(\sum_{i=0}^{n}\binom{n}{i}\left\{\frac{\partial^{n-i}}{\partial y^{n-i}}\,f\right\}\left\{\frac{\partial^i}{\partial y^i}\,g\right\}\right)\right] \\
&=\sum_{n=0}^{\infty}\left[\frac{1}{n!}\sum_{j=0}^{n}\sum_{i=0}^{n}\binom{n}{i}\binom{n}{j}\left\{\frac{\partial^{n-j}}{\partial x^{n-j}}\frac{\partial^{n-i}}{\partial y^{n-i}}\,f\right\}\left\{\frac{\partial^j}{\partial x^j}\frac{\partial^i}{\partial y^i}\,g\right\}\right].
\end{align*}
Because you're dealing with the derivatives of functions that could be decreasing, even if they are positive, there is no guarantee that the result here will be positive.

An operator being normal just means it's diagonalizable (infinite-dimensional space: spectral theorem holds). There's still no guarantee that its spectrum is real, which is what you'd need here, I think. Unitary operators are normal, and their eigenvalues are definitely not guaranteed to be real. If you're exponentiating an Hermitian operator, then I think you'd have yourself a positive operator coming out.

I presume you mean $e^{(\partial_x f(x))(\partial_y g(y))}= e^{f_x(x)g_y(y)}$.  For given x and y, assuming they are real numbers, whether the partial derivatives are positive or not, the exponential is always positive.


Can we rely on Confidence Intervals? Suppose the mean is in (7.6,8.4) with 95% confidence. I understand that this means 95% of the confidence intervals from different samples will contain the population mean. But, what is the significance of this particular interval on its own. Since I am sure that 95% of sampling intervals will contain the mean can I be somewhat certain that this interval is one of them? If not, how is this interval useful at all to me?
In other words, how sure can I be that mean is in (7.6,8.4) and if I can't be sure then what's the use of this?

Only one specific $95 \%$ confidence interval $(7.6, 8.4)$ is usually not sufficient to derive the wanted information about the statistical parameter of interest. Nevertheless it provides more information than simply stating that the mean is approximately $8.0$. One aspect is the resulting precision of the CI.
Precision of CI: A confidence interval respects random statistical fluctuations due to sampling variations. The resulted precision of the interval $(7.6, 8.4)$ is given by its length $0.8$ and we can think about if this is appropriate for our needs or not.
It might indicate that we should try to get a larger sample size  to obtain a smaller CI. It could also indicate to weaken the confidence level in order to reduce the size of the interval.
So, this specific CI can help to fine-tune the process and improve our model of the population under analysis.
Notes:

In real life just one CI is by far not enough to derive any conclusions. A population in  real life  is influenced by many different aspects and a statistical model is typically a rough and simplified version of the real life situation.
Crucial for the validity of the confidence interval is also the selection mechanism of the sample(s).
In order to overcome these difficulties we need to learn from real life by repeating (if possible) the tests, getting new samples and calculate more and more CIs this way and analyse the situation each time. We can use them to improve our knowledge of the unknown parameter of interest and to increase our confidence in the so-derived region of the true mean.

I'd like to mention in this context Statistical Intervals - A Guide for Practitioners by G.J. Hahn and W.Q. Meeker, which provides helpful information and examples to calculate confidence intervals, prediction intervals and tolerance intervals following different distributions.
Regarding precision of CIs the authors state:
... We wish to reiterate that the issue of data quantity is often secondary to that of the quality of the data. In particular, in making a statistical estimate or constructing a statistical interval, one assumes that the available data were obtained by using a random sample from a defined population of process of interest. As stated previously, when this is not the case, all bets are off. Just increasing the sample size - without broadening the scope of the investigation - does not compensate for lack of randomness; all it does is allow one to obtain a (possibly) biased estimate with greater precision. Putting it another way, increasing the sample size per se usually improves the precision of an estimate, but not necessarily its accuracy.

In other words, how sure can I be that mean is in (7.6,8.4) and if I
can't be sure then what's the use of this?

You can be $95\%$ sure that the mean is in your confidence interval, in a probabilistic sense: as you said, if you repeated the sampling $100$ times, $95$ times you would be facing a confidence interval that contains the real mean. While looking at a single sampling, someone will say that the mean is or is not in the interval, so you cannot talk about probability. But if you had to place a bet about the mean being in your interval, then... you'd just have $95\%$ chances of winning.