Power set of set containing empty set, sets of empty set, and mixes of the former The title might be sort of confusing. The set is an infinite set like
$$
A = \{
\emptyset, \{ \emptyset \}, \{ \{ \emptyset \} \}, ...
\{ \emptyset, \{ \emptyset \} \}, \{ \{ \emptyset \}, \{ \{ \emptyset \} \} \}, ...
...
\}
$$
And it could be defined by the following rules

#1: ∅ ∈ A
#2: if a ∈ A, {a} ∈ A
#3: if a ∈ A and b ∈ A, a ⋃ b ∈ A

Consider a subset of 2 elements of A, like {x, y} ⊂ A, say, x ∈ A and y ∈ A, so {x} ∈ A and {y} ∈ A (#2), then {x} ⋃ {y} ∈ A (#3), as a result {x, y} is also an element of A.

Similar deductions could be applied to subsets of 3 or even more elements. But that should have no chance according to Cantor's theorem. What mistakes I've made and what is the power set of A?

I understand that the problem is that the "set" A appears to contain all subsets of itself and therefore P(A) ⊂ A  so that |P(A)| <= |A| which would certainly upset Cantor.

Taking ZFC as the axiomatic framework to work in, then I don't think that your "set" is a valid contruction. The ZFC axioms were laid down to exclude pathological cases created by arbitrary definition, like Russell's "set" R of all sets which are not members of themseleves (is R ∈ R ?????): by the axioms of ZFC Russell's R is not a set, and nor do I think is yours.

You would have to demonstrate the construction of A quoting from the 10 ZFC axioms as you go along why each step is vaild.

1) ∅ the empty set exists by the axiom of the empty set.

2) {∅} esists by axiom of pairing

3) {∅, {∅}} exists by axiom of paring

4) there exists a set containing ∅ and the "successor" of each of it's elements by the axiom of infinity where the successor of a is {a}

Bit of a problem though to find axioms that will support your third rule. You could  construct a set which contains the union of two specific elements or even a finite number of such, but not I think an infinite collection that satisfied this condition.

On the other hand, if you start from the powerset (which exists by axiom of the powerset)  P(A) then your rule three is valid, but not rule 2: the powerset does not contain the sucessor of all of its elements.


Let$\ p_n$ be the$\ n$-th prime. Is$\ \lim_{n\to\infty} \log \log n \prod_{i=1}^{\lfloor \log n \rfloor} \frac{p_i-1}{p_i}>0$? I'm less than a novice in analysis, I don't even know how to approach this. Thanks in advance.

The limit is $e^{-\gamma}$ where $\gamma$ is the Euler–Mascheroni constant.
This is a direct consequence of

  Mertens' $3^{rd}$ theorem
  $$\lim_{x\to\infty} \log x \prod_{\substack{p\,\le\,x\\p \text{ prime}} }\left(1-\frac{1}{p}\right) = e^{-\gamma}$$

By Prime number theorem, we know $p_n \sim n\log n$, so
$$\lim_{n\to\infty}\frac{\log n}{\log p_n} = 1$$

If we substitute $x$ by $p_n$ in Mertens' theorem and multiply its LHS by $\displaystyle\;\frac{\log n}{\log p_n}$, we get

$$
\lim_{n\to\infty} \log n \prod_{i=1}^n \left(1 - \frac{1}{p_i}\right)
= \lim_{n\to\infty} \frac{\log n}{\log p_n} \left[ \log p_n \prod_{i=1}^n \left(1 - \frac{1}{p_i}\right)\right]\\
= \lim_{n\to\infty} \log p_n \prod_{\substack{p\,\le\,p_n\\p\text{ prime}}}\left(1 - \frac{1}{p}\right) =
 e^{-\gamma}$$

For a proof of Mertens' theorem, see

G.H Hardy's and E.M. Wright,
Chapter XXII - The Series of Primes (3),
An Introduction to the Theory of Numbers

It is theorem $429$ there.

No, because as $n \to \infty$, we have that $x_n=\prod_{i=1}^{\log n} \frac{p_i-1}{p_i}=\prod_{i=1}^{k} \frac{p_i-1}{p_i}$ diverges*, where $k=\log n \to \infty$.  Hence, we have:

$$\displaystyle \lim_{n \to \infty}\log \log nx_n=\log \log \infty=\infty$$

*(by your other question Let$\ p_n$ be the$\ n$-th prime. Can you give me a proof for$\ \prod_{i=1}^\infty \frac{p_i-1}{p_i}=P\approx \frac{1}{11.0453}$?)


Scale invariance and the Mellin transform? The Mellin transform of a function is given by:

$$\mathcal{M}[f](s) = \int_0^{\infty}x^{s-1}f(x)dx$$

Supposedly, the magnitude of the Mellin transform is invariant to scaling, analogous to how the magnitude of the Fourier transform is invariant to translation. Yet, when I try to derive this I get:

$g(x) = f(kx)$

$|\mathcal{M}[g](s)| = |\int_0^{\infty}x^{s-1}f(kx)dx|=|\int_0^{\infty}(\frac{x'}{k})^{s-1}f(x')\frac{dx'}{k}|=|k^{-s}||\mathcal{M}[f](s)|$

which is not invariant, though it does have a simple relation between them. Where did I go wrong?

$k^{-s}$ is proportional to $s$ coefficient so you can extract and compare $k$ by $\log \dfrac{g(s)}{f(s)}$ for each integral of $s$ like you do with shift on fourier.


Cover a cicular hole with planks A friend of mine asked me the following question.

  Whats the minimum number of rectangular planks of unit
  width (and infinite length) needed to cover a circular hole with diameter $n$?

Obviously, $n$ is an upper bound because the hole can be covered by putting $n$ planks in parallel next to each other. Also it is easy to see that for small $n$, in particular $n=2$ and $n=3$ one cannot make do with fewer than $n$.

Is there a proof that in the general case $n$ planks are needed?

This is a theorem of Bang from 1950-51, solving Tarski's Plank Problem. See Wikipedia  for description and links.

How many pieces of unit width planks and infinite length are needed to cover a hole of diameter n?

Easy!

One piece :)

Just cut up each bit as required from that infinite length piece :)


Any open subset of $\Bbb R$ is a countable union of disjoint open intervals Let $U$ be an open set in $\mathbb R$. Then $U$ is a countable union of disjoint intervals.

This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.

Here’s one to get things started.

Let $U$ be a non-empty open subset of $\Bbb R$. For $x,y\in U$ define $x\sim y$ iff $\big[\min\{x,y\},\max\{x,y\}\big]\subseteq U$. It’s easily checked that $\sim$ is an equivalence relation on $U$ whose equivalence classes are pairwise disjoint open intervals in $\Bbb R$. (The term interval here includes unbounded intervals, i.e., rays.) Let $\mathscr{I}$ be the set of $\sim$-classes. Clearly $U=\bigcup_{I \in \mathscr{I}} I$. For each $I\in\mathscr{I}$ choose a rational $q_I\in I$; the map $\mathscr{I}\to\Bbb Q:I\mapsto q_I$ is injective, so $\mathscr{I}$ is countable.

A variant of the same basic idea is to let $\mathscr{I}$ be the set of open intervals that are subsets of $U$. For $I,J\in\mathscr{I}$ define $I\sim J$ iff there are $I_0=I,I_1,\dots,I_n=J\in\mathscr{I}$ such that $I_k\cap I_{k+1}\ne\varnothing$ for $k=0,\dots,n-1$. Then $\sim$ is an equivalence relation on $\mathscr{I}$. For $I\in\mathscr{I}$ let $[I]$ be the $\sim$-class of $I$. Then $\left\{\bigcup[I]:I\in\mathscr{I}\right\}$ is a decomposition of $U$ into pairwise disjoint open intervals.

Both of these arguments generalize to any LOTS (= Linearly Ordered Topological Space), i.e., any linearly ordered set $\langle X,\le\rangle$ with the topology generated by the subbase of open rays $(\leftarrow,x)$ and $(x,\to)$: if $U$ is a non-empty open subset of $X$, then $U$ is the union of a family of pairwise disjoint open intervals. In general the family need not be countable, of course.

The following is certainly not the quickest approach to a proof, but when this question was first posed to me in class, my first intuition was to use some elementary graph theory:

Let $U$ be an open set of $\mathbb{R}$. As we know, $\mathbb{R}$ has a countable basis $\mathcal{B}$ comprised of connected open sets and so we may write $U=\bigcup_{n\in I} U_n$, where for each $n$ we have $U_n\in\mathcal{B}$ and $I$ is some countable index set.

Let $G$ be the intersection graph of $\{U_n\}$. That is to say, the vertex set of $G$ is simply $\{U_n\}$ and there is an edge between $U_i$ and $U_j$ iff they have nonempty intersection. It's easy to convince yourself that:

This graph must have countably many graphically-connected components (otherwise we'd have uncountably many vertices which is impossible).
The intersection graph of $A\subseteq\{U_n\}$ is graphically-connected iff for any two $V,W\in A$ there is a sequence $V=U_{n_1},U_{n_2},\ldots,U_{n_k}=W$ such that $U_{n_i}\cap U_{n_{i+1}}\neq\varnothing$.
The union $\bigcup A$ is a connected set of $\mathbb{R}$ whenever the intersection graph of $A$ is graphically-connected.

Thus, when we take the union of all the vertices within a graphically-connected component, for every component, we obtain countably-many connected open sets. The union of these sets is of course $U$ itself. Since the connected open sets of $\mathbb{R}$ are intervals (including rays), we're done.

Side Note: This would also work in $\mathbb{R}^n$ or in general, any topological space $X$ that has a countable basis comprised of connected sets. Well, so long as we replace countable union of disjoint open intervals with countable union of disjoint open connected sets.


How do we know that we found all solutions of a differential equation? I hope that's not an extremely stupid question, but it' been in my mind since I was taught how to solve differential equations in secondary school, and I've never been able to find an answer.

For example, take the simple differential equation $$y'=y.$$

While I understand that $y=C\cdot e^x$ satisfies this equation, I do not understand why there can't be other solutions as well. So, how do we know this?

When you study the topic in University, you will probably encounter uniqueness theorems that show that for certain ODE's, not only does a solution exist, but this solution is also unique - there are no other solutions.

For first order ODE's such as that in your example, the relevant theorem is the Picard–Lindelöf theorem, which shows that your solution is indeed unique around a given interval. These intervals can then be strung together to show this is the only solution across the entire domain.

I'll add that for some other forms of differential equations it is not at all clear that solutions are unique. The most famous example is probably the Navier-Stokes equations in three dimensions, which govern the motion of gases and liquids all around us, and yet no proof of uniqueness in all cases is known.

In the case of y'=y...
Move along the x axis, follow the ordinate line up to the curve where the slope ='s the y-ordinate. In this way you trace out the curve, being only one curve save for translation.


How to show that $T$ is linearly independent? This question came in my exam, and is by no mean a homework.

Let $\{v_1,v_2\}$ be a linearly independent subset of a vector space $V$ and let $w$ belong to $V$ but not to $\operatorname{Span}(\{v_1,v_2\})$.
We need to show that the set $T= \{v_1 + w, v_2 + w\}$ is linearly independent.

My reasoning so far:

$\{v_1,v_2\}$ are linearly independent, thus there exist $c_1$ and $c_2$ such that
$c_1v_1 + c_2v_2 = 0$, $c_1=c_2=0$.

$w$ is not in $\operatorname{Span}(\{v_1,v_2\})$, thus the set $\{v_1,v_2,w\}$ is linearly independent.

Thus there exist another $c_1,c_2,c_3$ such that $c_1v_1+c_2v_2+c_3w = 0$.

$c_1v_1+c_2v_2+c_3w = 0 \implies$

$c_3w = -c_1v_1 - c_2v_2 \implies  w = (-c_1/c_3)v_1 + (-c_2/c_3)v_3$

$T= \{v_1 + w, v_2 + w\}$. Let there be $d_1$ and $d_2$ scalar such that

$d_1(v_1 + w) + d_2(v_2 + w)= 0$

$\implies  d_1(v_1 + (-c_1/c_3)v_1) + d_2(v_2 + (-c_2/c_3)v_3) = 0$

$\implies v_1(d_1 + (-c_1/c_3)d_1) + v_2 (d_2 + (-c_2/c_3)d_2) = 0$,

but $v_1$ and $v_2$ are linearly independent.

Thus, $d_1 + (-c_1/c_3)d_1 = d_2 + (-c_2/c_3)d_2 = 0$

We conclude that the set $T$ is linearly independent.

Is this correct?

EDIT:

$c_1v_1+c_2v_2+c_3w = 0$

T linearly independent, there exit $d_1(v_1 + w) + d_2(v_2 + w)= 0$

$\implies d_1.v_1 + d_1.w + d_2.v_2 + d_2.w = 0$

$\implies (d_1.v_1 + d_2.v_2) + (d_1.w + d_2.w )= 0$

But $ d_1.v_1 + d_2.v_2 = 0 $ since $v_1$ and $v_2$ are linearly independent

thus $d_1 = d_2 = 0$

$\implies  d_1(v_1 + w) + d_2(v_2 + w)= 0$  with $d_1 = d_2 = 0$

T is linearly independent

Note that $v_1,v_2,\ldots,v_k$ are linear independant does not mean (as you write) that there exist $c_1,c_2,\ldots,c_k$ such that $c_1v_1+\ldots +c_kv_k=0$ and $c_1=\ldots =c_k=0$.
Rather it menas that for all $c_1,c_2,\ldots,c_k$ with $c_1v_1+\ldots +c_kv_k=0$  we have that $c_1=\ldots =c_k=0$.
However, in your reasoning you seem to make use of the correct form.

Nevertheless, you divide by $c_3$ without knowing whether $c_3\ne 0$. Can you modify your calculation to avoid this division?

Apart from this a remark about notation: I am always unhappy when people talk about a set of vectors being linearly independant. For example (not related to the problem statement at hand), assume $\{v_1,v_2\}\subset V$ is linearly indpendant and let $w=v_1+2v_2$. Can it happen that the set $\{v_1,v_2,w\}$ is linearly independent? Maybe surprisingly the answer is: Yes, it can! (Can you find an example for this?)

you have shown that $\{v_1, v_2, w\}$ is linearly independent. you want to show $\{v_1+w, v_2 + w\}$ is linearly independent.

suppose $$c_1(v_1+w) + c_2(v_2 + w) = 0.\tag 1$$ we need to show that $$c_1 = 0, c_2 = 0 $$

rewrite $(1)$ as $$c_1v_1 + c_2v_2 + (c_1+c_2)w = 0.$$ use the fact that $\{v_1, v_2, w\}$ is linearly independent to conclude $c_1 = 0, c_2 = 0.$ that concludes the proof.


Showing that a certain formula of second order logic with full semantic is true in all and only non-standard model of arithmetic. Showing that a certain formula of second order logic with full semantic is true in all and only non-standard model of arithmetic.

$\exists X(\exists x Xx \wedge \forall x\forall y((Xx \wedge (x=0 \vee x=y+1)) \rightarrow (x\not=y \wedge Xy)))$

How should I approach this problem?

This should be the right answer to the question.

Suppose that the formula is true in an arbitrarily standard model $M$, then we get that there is a property $X$, such that if there is an $x$ that has that property, for every number, if $Xx$ and $x\not=0$ then also the predecessor of $x$, namely $y$ has the property $X$. Now we can iterate this reasoning, given that $Xy$ and the if $y\not=0$ then also the predecessor of $y$ has the property $X$, until we get to the case in which $x=0$. In this case the antecedent of the implication would be true, given that the left disjunct in the antecedent is true, but the consequent would be false having $0\not=0$. This is so because we know that if the successor of 0 has the property X, then also 0 must have the property X. Also we have $\forall y$ in the formula, hence we will have the case in which $y=0$. Therefore we get the implication false, and this contradicts our assumption. By the arbitrariness of the standard model $M$ we can conclude that the formula has no standard model.

Now if we assume that the formula is true in a full semantic model, then we get that it must exists a set of all the non-standard numbers, given that the interpretation of $X$ range over all the possible subsets of the power set of the domain. That is our property $X$. So it is clear that the formula must be true for all the non-standard numbers. This is so because whenever we pick a non-standard number $n^*$, it must have a predecessor, say $n_{i}^*$ (because that non-standard number cannot be 0). Now given that every successor of a standard number is standard, it follows that also $n_{i}^*$ is non-standard. By applying this reasoning over and over we get an infinite descending sequence of non-standard numbers. By a very similar reasoning this allows us to say that the formula is valid in such a model. Whenever $x$ has the property $X$ ($x$ is a non standard number), we can always find a predecessor with the property of being non-standard without ever coming to a stop. Now by arbitrariness of the model, we have that (given that no standard model satisfy the formula), the formula is true in all and only non standard models.


What is the rule of equating exponents called? For example:

$$2^{2n-1} = 2^{n+2}
\Rightarrow 2n - 1 - n - 2 = 0 \Rightarrow n = 3$$

I couldn't find this rule in properties of exponents i.e when the bases are equal, the exponents can be equated. What is this rule called?

In the end, it comes down to the fact that the function
$$x\longmapsto 2^x$$
is "one-to-one"; that is, that different inputs yield different outputs. That is, that the graph of $y=2^x$ passes the so-called "horizontal line test": a horizontal line intersects the graph at most once.

It's the same reason that we can go from $a^3 = b^3$ to $a=b$: because the function $x\longmapsto x^3$ is one-to-one; and why we cannot go from $a^2=b^2$ to $a=b$: because $x\longmapsto x^2$ is not one-to-one (different inputs may give the same output; e.g., $(-1)^2 = 1^2$ even though $-1\neq 1$).

When a function is one-to-one, it has an inverse; and applying the inverse "undoes" what the original function does. That's what taking "logarithm base 2" is: the inverse of the exponential base 2.

If $f(x)$ is monotone (a slightly more restrictive condition than $1-1$ but satisfied by $f(x) = 2^x$) and $f(y) = f(x)$,  then $y = x$.

Most of the tables of "rules of exponents" I found were given names like "Rule 1" or "Rule 2", so I see the complaints about failing on giving a "name" somewhat arbitrary. If you wanted to use a "rule of exponents", then try combining the rule that might be called: "ratios imply subtraction of powers" and the "rule" that says if 2^x=1 that x=0 :

 a^m/a^n = a^(m-n) combined with the "rule" that if 2^m= 1 then m=0

You can then divide both sides 2n-1 and simplify.


suppose that $f:A \rightarrow B$ and $g:B \rightarrow C$ are functions. Show that if $g\circ f$ is onto then $g$ is onto. $\forall x \in A ,\forall g\circ f(x) \in C$ (since $g\circ f$ is onto)
$\implies \forall g[f(x)] \in C$ (since $g\circ f(x) = g[f(x)]$ )
$\implies \forall f(x) \in B$ (since $g:B\rightarrow C$ and $B$ is the domain of the $g$)
Therefore if $\forall x \in A,\forall g\circ f(x) \in C, then \forall f(x) \in B,\forall g\circ f(x) \in C$.
So if $g\circ f$ is onto then $g$ is onto.
Is this correct and complete? Is there another way to prove it?

Your proof has just no sense. What you have to prove is :
$$\forall y\in C, \exists x\in B: y=g(x).$$

Let $y\in C$. Since $g\circ f$ is onto, there is $x\in A$ s.t. $y=g\circ f(x)$. Set $u=f(x)\in B$. Then $y=g(u)$, and thus $g$ is onto.

To be honest I can't understand the terminology of your question
We need to use the definitions for this purpose.
Given that $g\circ f$ is onto and the range of the function $g$ is $C$ whereas its domain is $B$. Also the range of the function $f$ is $B$. Let $x \in A$. This implies $f(x) \in B$. Now consider $g\circ f (x)$ or $g\big(f(x)\big) \in C$ (because range of function $g$ is $C$). Now we need to prove that for every $y\in C$, there exists an $f(x)\in B$ such that $f=g\big(f(x)\big)$. If $g\circ f$ is onto (given) then for every $g\big(f(x)\big)$, there exists an $f(x)\in B$ such that $g\big(f(x)\big)\in C$. Let $f(x) = k\in B$. and let image of $f(x)$ under $g$ be $y$. Hence $y=g(k)$. This shows that for all $y\in C$ there exists a $k\in B$ such that $y=g(k)$ which is the definition of onto function.


Linear Algebra Question from Mock GRE exam. Suppose $V$ and $W$ are finite dimensional vector spaces, and that $f:V\mapsto W$ is a linear map. Suppose $\{e_1,...,e_n\}\subset V$ and that $\{f(e_1),...,f(e_n)\}$ is a basis of $W$. Then which of the following are true?

I. $\{e_1,...,e_n\}$ is a basis of $V$

II. There exists a linear map $g:W\mapsto V$ such that $g\circ f=Id_V$

III. There exists a linear map $g:W\mapsto V$ such that $f\circ g=Id_W$

(A) I only

(B) II only

(C) III only

(D) I and III

(E) II and III

The solutions manual says it is (B), but I think it should be (C) since we can regard $W$ to be isomorphic to some subspace of $V$?

Thanks in advance!

Consider III. Suppose it is true.
 $$w \in W \implies w=fg(w)=f(g(w))\\\implies w \in \;\;\text{range}\; f$$ Hence $ \text{range}\; f=W$, concluding $f$ is onto.

Is $f$ onto ?

For sure the I. it's false infact V could be $\mathbb{R^{n+1}}$ and $f(e_{n+1})=0$ and the hypothesis still true. The II. could be false infact if as in the first example $f(e_{n+1})=0$ we have that $g(f(e_{n+1}))=g(0)=0$ so the rank of $g(f(v))$ is $n$ and not $n+1$. So the right one is the III. infact if you choose $g$ such that $g(f(e_n))=e_n$ we have that $f(g(v))$ on $W$ in the base ${f(e_1),f(e_2),...,f(e_n)}$ is the identity.

If the answer it's wrong i will fix it tomorrow.


How to determine countability of $\Bbb N \times \Bbb N \times \Bbb N$? Is $\Bbb N \times\Bbb N \times\Bbb N$ countable or not, where $\Bbb N$ is the infinite set of all natural numbers ?

Please explain how it is done.

Define $f:\mathbb{N}\times\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ by $f(m,n,l)=2^m3^n5^l$ and observe that $f$ is injective by the Fundamental Theorem of Arithmetic.

Notice how this can be extended to show that any finite number of copies of $\mathbb{N}$ is countable.

Hint: Let $(x,y,z) \in \mathbf{N}\times\mathbf{N}\times\mathbf{N}$, let $w=x+y+z.$ What do you know about each $w \geq 3$ in relation to $x,y,z$?


Prove if $|z-1| \le \frac{1}{2}$ then $|\frac{z}{|z|} - 1| \le |z - 1|\sqrt2$ for z complex As the title, problem is to prove if $|z-1| \le \frac{1}{2}$ then $|\frac{z}{|z|} - 1| \le |z - 1|\sqrt2$ for complex $z$.

I noticed using the condition we can get $|\frac{z}{|z|} - 1| \le \frac{\sqrt2}{2}$.

Then can square both sides to get $2(1 - \frac{\operatorname{Re}z}{|z|}$) $\le \frac{1}{2}$.

From here I try further manipulations but am stuck and unable to proceed.

How would you solve this problem?

Let $z=re^{i\theta}$,  and then you need to prove $|e^{i\theta}-1|^2\leq2|re^{i\theta}- 1|^2$. Here $1/2\leq r\leq 3/2,-\pi/6\leq \theta\leq\pi/6$.
$$LHS=2-2\cos\theta;$$
$$RHS=2(r^2-2r\cos\theta+1);$$
$$\frac{1}{2}(RSH-LHS)=r^2-2r\cos\theta+\cos\theta\geq 0.$$

Let $\arg z=\theta.$

Thus, it's enough  to show that:
$$|\cos\theta+i\sin\theta-1|\leq\frac{1}{\sqrt2}$$ or
$$2-2\cos\theta\leq\frac{1}{2}$$ or
$$\cos\theta\geq\frac{3}{4},$$ which is
$$0^{\circ}\leq \theta\leq\arccos\frac{3}{4}$$ or
$$360^{\circ}-\arccos\frac{3}{4}\leq \theta<360^{\circ},$$
which is true because by the given
$$0^{\circ}\leq \theta\leq30^{\circ}$$ or
$$330^{\circ}\leq \theta<360^{\circ}.$$


nontrivial 'sub' Banach spaces I'm looking for a pair of Banach spaces $(X,\left\|\right\|_X)$ and $(Y,\left\|\right\|_Y)$ where $Y\subset X$, $\left\|\right\|_Y \leq \left\|\right\|_X$ and $Y$ is dense in $X$ for the $X$-norm, yet $Y\neq X$. I suppose such a pair exists, but have some trouble finding one.

The puzzling property of such a pair is of course that every $\left\|\right\|_X$-Cauchy-sequence in $Y$ is necessarily also a $\left\|\right\|_Y$-Cauchy sequence. Hence any $\left\|\right\|_X$-Cauchy sequence in $Y$ converging to some $x \in X\setminus Y$ (A sequence which exists by assumption.) must converge to some different $y \in Y$ for the $Y$-norm.

I suppose this question may be quite standard, so apologies in advance if the answer can be found in simple places (which I have not found until now.)

In order for $Y\subset X$ to be complete, $Y$ must be closed as a subset of $X$, so that $Y=\overline Y$. However, $Y$ being dense in $X$ means that $\overline Y=X$.

So, the only dense "Banach subspace" of $X$ is $X$, if the norm on $Y$ is inherited (or topologically equivalent to the inherited norm) from $X$.

On the other hand, considering $(\ell^1,\|\cdot\|_1)$ as a subset of $(\ell^2,\|\cdot\|_2)$ gives you the desired result.


How to solve system of 4 equations with four unknowns? I have got system of 4 equations as shown below and I am considering if there is any other method than brute force to solve them.

B + C + D = S1
A + C + D = S2
A + B + D = S3
A + B + C = S4

Values of S1-S4 are given:

S1 = 70
S2 = 75
S3 = 80
S4 = 75

I tried to to solve that, but always I have infinite equation what means to solve A I need B, to solve B I need C, to solve C I need D and to solve D I need.. A. I know the answer, but need the solution how to solve it. Answer:

A = 30
B = 25
C = 20
D = 25

Use Gaussian elimination/method:

Your equation becomes:

$$\left[\begin{array}{cccc|c}
      1 & 1 & 1 & 0 & 75         \\[0.55ex]
      1 & 1 & 0 & 1 & 80        \\[0.55ex]
      1 & 0 & 1 & 1 & 75\\[0.55ex]
      0 & 1 & 1 & 1 & 70
\end{array}\right]$$

You can multiply rows and add/subtract them from another row, which will not affect the solution. So you can subtract the first row from the second and the thirdrow and the matrix will become:

$$\left[\begin{array}{cccc|c}
      1 & 1 & 1 & 0 & 75         \\[0.55ex]
      0 & 0 & -1 & 1 & 5        \\[0.55ex]
      0 & -1 & 0 & 1 & 0\\[0.55ex]
      0 & 1 & 1 & 1 & 70
\end{array}\right]$$

Now add the third row to the fourth:

$$\left[\begin{array}{cccc|c}
      1 & 1 & 1 & 0 & 75         \\[0.55ex]
      0 & 0 & -1 & 1 & 5        \\[0.55ex]
      0 & -1 & 0 & 1 & 0\\[0.55ex]
      0 & 0 & 1 & 2 & 70
\end{array}\right]$$

And at last add the second row to the fourth row:

$$\left[\begin{array}{cccc|c}
      1 & 1 & 1 & 0 & 75         \\[0.55ex]
      0 & 0 & -1 & 1 & 5        \\[0.55ex]
      0 & -1 & 0 & 1 & 0\\[0.55ex]
      0 & 0 & 0 & 3 & 75
\end{array}\right]$$

Now from the last row we get that $3D = 75 \implies D=25$

From the third row we get that $D-B = 0 \implies D=B \implies B=25$

From the second row we get that $D-C = 5 \implies C=25-5 \implies C=20$

And from the first row we get that: $A+B+C = 75 \implies A=30$

B=D
So use the first equation b+c+d=b+c+b=70
and the second a+c+d=75
and the third a+b+d=80
so b-c=5
(b+c+b+5) divided by 3 equals b and d
(70+5) divided by 3 equals 25(b and d)
25-5=20(c)
(a+c+b)-(b+c+b)=5(a-b)
25+5=30(a)
a=30
b=25
c=20
d=25


How to prove this property of the Wronskian? I couldn't understand why it says the linear system has a nontrivial solution since the number of rows is :$1$(the original function）$+(n-1)$ (take the derivative $n-1$ times)$=n$. Also there are $n$ columns or unknowns. Why the system is not an independent one?

First of all consider the $n$ functions $f_1,...,f_n \in C^{n-1}(\mathbb{R})$ and suppose there exists $k_1,...k_n$ real numbers such that

$$g=\sum_{i=1}^n k_if_i(x)=0$$

for every $x\in\mathbb{R}$; that is $g$ is the zero-function.
The set $C^{n-1}(\mathbb{R})$ is a real vector space of infinite dimension, so $g$ is still a function belong to it. Consider now the first derivative of $g$

$$g'(x)=\sum_{i=1}^n k_if_i'(x), $$

because the derivative of the sum is the sum of the derivative, on the other hand $g\equiv0$, so $g'(x)\equiv0$; then$$g'(x)=\sum_{i=1}^n k_if_i'(x)=0. $$

This means that if $f_1,...,f_n$ are linearity dependent, then $f'_1,...,f_n'$ needs to be linearity dependent in $C^{n-1}(\mathbb{R})$. If you repeat for all order of derivation, you obtain the same result

$f_1,...,f_n$ are linearity dependent iff $f_1^{(i)},...,f_n^{(i)}$ are linearity dependent for every $i=1,...,n-1$.

Finally, consider the functions $f_1,...,f_n \in C^{n-1}(\mathbb{R})$ and take the following matrix
$$A=\begin{pmatrix}
f_1(x)&  ... & f_n(x)\\
... & ... & ...\\
f_1^{(n-1)} & ... & f_n^{(n-1)}
\end{pmatrix}$$

the Wronskian $W(x)$ of $A$ is its the determinant, which clearly it depend on $x$. If there exists a non-zero vector column of real numbers such that

$$A=\begin{pmatrix}
f_1(x)&  ... & f_n(x)\\
... & ... & ...\\
f_1^{(n-1)} & ... & f_n^{(n-1)}
\end{pmatrix}
\begin{pmatrix}
k_1\\
...\\
k_n
\end{pmatrix}=
\begin{pmatrix}
0\\
..\\
0
\end{pmatrix},$$

for every $x\in\mathbb{R}$, this mean that the function are linearity dependent by the discussion above. Viceversa if there exists a real number $\overline{x}$ such that $W(\overline{x})$ is not zero, then the function are linearity independent.

I use [1] as the basis of my proof. As in [1], I employ a proof by contradiction [2].
Step 0. Define the proposition to be proved
By $P$, I denote the proposition to be proved, which is

$P := $ Given $n$ functions $f_1(x), f_2(x), \ldots, f_n(x)$ that are elements of the set $C^{n-1}{(I)}$ [3]  (i.e., the space of continuous functions that have continuous first $n-1$ derivatives in the interval $I$), if the Wronskian is not equal to zero for any value $c$ in the interval $I$, then the $n$ functions are linearly independent in the interval $I$ [4].

Step 1. The proposition to be proved, $P$, is assumed to be false
I assume that given $n$ functions $f_1(x), f_2(x), \ldots, f_n(x)$ that are elements of the set $C^{n-1}{(I)}$, if the Wronskian is not equal to zero for any value $c$ in the interval $I$, then the $n$ functions are  linearly dependent in the interval $I$.
Step 2. I derive a contradiction
I know that a sequence of functions $f_{1}{(x)},f_{2}{(x)},\dots ,f_{n}{(x)}$  from a function space are said to be linearly dependent, if there exist constant scalars $a_{1},a_{2},\dots ,a_{n}$, which are not all zero, such that
$$a_{1}\,f_{1}{(x)}+a_{2}\,f_{2}{(x)}+\cdots +a_{k}\,f_{k}{(x)}= 0_f,\quad\text{for all}~x\in  I ;\tag{1}$$
where $ 0_f$ denotes the zero function [5].  Next, I take the derivative of Equation $1$ $n-1$ times. I obtain the following system of $n$ equations in $n$ unknowns:
$$
\begin{bmatrix}
f_1{(x)} &f_2{(x)} & \cdots & f_n{(x)}
\\
f^{'}_1{(x)} &f^{'}_2{(x)} & \cdots & f^{'}_n{(x)}
\\
\vdots & \vdots & \vdots & \vdots
\\
f^{(n-1)}_1{(x)} &f^{(n-1)}_2{(x)} & \cdots & f^{(n-1)}_n{(x)}
\end{bmatrix}
\,
\begin{bmatrix}
a_1\\a_2\\\vdots\\a_n
\end{bmatrix}
=
\begin{bmatrix}
0\\0\\\vdots\\0
\end{bmatrix},\quad\text{for all}~x\in  I.
$$
This may be re-written in terms of any constant $c\in I$.
$$
\begin{bmatrix}
f_1{(c)} &f_2{(c)} & \cdots & f_n{(c)}
\\
f^{'}_1{(c)} &f^{'}_2{(c)} & \cdots & f^{'}_n{(c)}
\\
\vdots & \vdots & \vdots & \vdots
\\
f^{(n-1)}_1{(c)} &f^{(n-1)}_2{(c)} & \cdots & f^{(n-1)}_n{(c)}
\end{bmatrix}
\,
\begin{bmatrix}
a_1\\a_2\\\vdots\\a_n
\end{bmatrix}
=
\begin{bmatrix}
0\\0\\\vdots\\0
\end{bmatrix}\qquad~\text{for any}~c\in I.
$$
According to my proposition, I have
$$
\begin{vmatrix}
f_1{(c)} &f_2{(c)} & \cdots & f_n{(c)}
\\
f^{'}_1{(c)} &f^{'}_2{(c)} & \cdots & f^{'}_n{(c)}
\\
\vdots & \vdots & \vdots & \vdots
\\
f^{(n-1)}_1{(c)} &f^{(n-1)}_2{(c)} & \cdots & f^{(n-1)}_n{(c)}
\end{vmatrix}
\neq 0\qquad~\text{for any}~c\in I.
$$
I can therefore invert the square matrix of coefficients. I obtain a solution for my unknowns (i.e., $a_1, a_2,\ldots, a_n$). I obtain that
$$
\begin{bmatrix}
a_1\\a_2\\\vdots\\a_n
\end{bmatrix}
=
\begin{bmatrix}
f_1{(c)} &f_2{(c)} & \cdots & f_n{(c)}
\\
f^{'}_1{(c)} &f^{'}_2{(c)} & \cdots & f^{'}_n{(c)}
\\
\vdots & \vdots & \vdots & \vdots
\\
f^{(n-1)}_1{(c)} &f^{(n-1)}_2{(c)} & \cdots & f^{(n-1)}_n{(c)}
\end{bmatrix}^{-1}
\begin{bmatrix}
0\\0\\\vdots\\0
\end{bmatrix}\qquad~\text{for any}~c\in I.
$$
Upon matrix mulitplication, I find
$$a_1 = a_2=\cdots = a_n=0.$$
These functions were said to be linearly dependent since there existed scalars $a_1, a_2, \ldots, a_n$ that are not all zero. Yet, I just arrived at the contradictory statement that the scalars $a_1, a_2, \ldots, a_n$ are all zero.
Step 3. Conclusion
Since the statements that, the scalars $a_1, a_2, \ldots, a_n$ are not all zero, and the statement that, the scalars $a_1, a_2, \ldots, a_n$ are all zero, cannot both be true, the assumption that $P$ is false must be wrong.  I conclude that $P$ must be true.
BIBLIOGRAPHY
[1] https://math.vanderbilt.edu/sapirmv/msapir/prwronskian.html
[2] https://en.wikipedia.org/wiki/Proof_by_contradiction#:~:text=In%20logic%20and%20mathematics%2C%20proof,false%20leads%20to%20a%20contradiction.
[3] https://en.wikipedia.org/wiki/Function_space
[4] https://mathworld.wolfram.com/LinearlyDependentFunctions.html
[5] https://en.wikipedia.org/wiki/Linear_independence#Definition


prove conditional probability inequality Prove that if $P(A | B) > 0$ then $P ( B | A) \geq P(B)$. I've a couple of different things but I didn't get any closer to solution. Any suggestions would be appreciated.
I know that $ P (B | A) = \frac{P(A \cap B)}{P(A)}$. The fact that $P(A | B) > 0$ basically tells me that $A$ and $B$ intersect somewhere and their intersection is non zero, correct? Sadly I don't have any idea on how to continue.

The inequality seems false indeed let consider for example
$$A=\{1,2,3,4\} \quad B=\{3,4,5,6\}$$
then

$P(A)=\frac23$
$P(B)=\frac23$
$P(A \cap B)=\frac13$
$P (A | B)=P (B | A)=\frac12$


Borel-Cantelli lemma and set-theoretic limits Let $A_n$ be a sequence of events in probability space.

The event $A(i.o.)$=$\{A_n$ happens infinitely often $\}$ is formally stated as $\lim \sup A_n$ (in other words $\cap_{k=1}^\infty \cup_{n=k}^\infty A_n$).

This is because if an element belongs to $\lim \sup A_n$, it can be shown that it belongs to infinitely many $A_n$ (because it belongs to all sets $\cup_{n=k}^\infty A_n$).

In the proof of Borel-Cantelli lemma on page 2, it says:

$P(A(i.o.)) = \lim_{k \to \infty}P(\cup_{n=k}^\infty A_n)$.

According to the definition above, it's exactly the same as

$$P(\cap_{k=1}^\infty \cup_{n=k}^\infty A_n)= \lim_{k \to \infty}P(\cup_{n=k}^\infty A_n)$$

Could anyone prove this equality?

Suppose that $B_1,B_2,\dots$ are a decreasing sequence of events. Then

$$
P\left(\bigcap_{n\ge 1} B_n\right) = \lim_{n\to\infty} P(B_n)
$$

To see this define $E_{n}=B_1 \setminus B_n$. $E_n$ are measurable (i.e events) so we can assign them probability. Note also since $B_n$ is decreasing $E_n$ is increasing. We then have
$
B_1=E_n \cup B_n
$

and as $E_n$ and $B_n$ are disjoint

$$
P(B_1)=P(E_n)+P(B_n) \implies P(E_n)=P(B_1)-P(B_n)
$$

By De Morgan's Law we also have $\bigcup E_n = B_1\setminus \bigcap B_n$. Therefore we have

$$
P\left(\bigcup E_n\right) = P(B_1)-P\left(\bigcap B_n \right)
$$

Since $E_n$ is increasing, we have $P\left(\bigcup E_n \right)=\lim_{n\to\infty} P(E_n)$ (I entrust you to show this since it's a little easier, use the same technique - write the union as union of disjoint events)

Therefore

\begin{align*}
P(B_1)-P\left(\bigcap B_n \right)&=\lim_{n\to\infty} P(E_n)\\
&=\lim_{n\to\infty} (P(B_1) - P(B_n))\\
&=P(B_1)-\lim_{n\to\infty} P(B_n)
\end{align*}

(This result is standard for any measure space, as long as we are given that measure of the set $B_1$ is finite. But in probability space this is true for any event $B_1$)

Now apply this result to $B_k=\bigcup_{n\ge k} A_n$

The equality follows by the continuity of probability $(*)$. See here for a proof of BCL1 and half of the continuity of probability.

$(*)$ If events $B_1, B_2, ...$ increase or decrease to some event $B$, then the $P(B_n)$'s increase or decrease, respectively, to $P(B)$.

In this case we have decreasing events $B_1, B_2, ...$ given by $B_k = \bigcup_{n=k}^{\infty} A_k$


The expected payoff of a dice game There's a question in my Olympiad questions book which I can't seem to solve:

  You have the option to throw a die up to three times.  You will earn
  the face value of the die.  You have the option to stop after each
  throw and walk away with the money earned. The earnings are not additive. What is the expected payoff of this game?

I found a solution here but I don't understand it.

Let's suppose we have only 1 roll. What is the expected payoff? Each roll is equally likely, so it will show $1,2,3,4,5,6$ with equal probability. Thus their average of $3.5$ is the expected payoff.

Now let's suppose we have 2 rolls. If on the first roll, I roll a $6$, I would not continue. The next throw would only maintain my winnings of $6$ (with $1/6$ chance) or make me lose. Similarly, if I threw a $5$ or a $4$ on the first roll, I would not continue, because my expected payoff on the last throw would be a $3.5$. However, if I threw a $1,2$ of $3$, I would take that second round. This is again because I expect to win $3.5$.

So in the 2 roll game, if I roll a $4,5,6$, I keep those rolls, but if I throw a $1,2,3$, I reroll. Thus I have a $1/2$ chance of keeping a $4,5,6$, or a $1/2$ chance of rerolling. Rerolling has an expected return of $3.5$. As the $4,5,6$ are equally likely, rolling a $4,5$ or $6$ has expected return $5$. Thus my expected payout on 2 rolls is $.5(5) + .5(3.5) = 4.25$.

Now we go to the 3 roll game. If I roll a $5$ or $6$, I keep my roll. But now, even a $4$ is undesirable, because by rerolling, I'd be playing the 2 roll game, which has expected payout of $4.25$. So now the expected payout is $\frac{1}{3}(5.5) + \frac{2}{3}(4.25) = 4.\overline{66}$.

Does that make sense?

Can someone explain to me why the expectation value of the outcome is not the same as the markov solution?

Imagine you were playing such that you cash out after 500 tries (instead of 2 or 3 treis).  On your first roll, you roll a 4.  Would you stop at the first roll because it's higher than 3.5?  I think not.  Your odds of getting a six are almost certain.

What is the expectation value of rolling a die twice, while only keeping the highest value?  We can compute this as follows:

How many ways can I roll a 1?  The answer is 1.  1 on my first roll. (1)

How many ways can I roll a 2?  The answer is 2.  Either on the first roll, (2) or on the second roll, assuming my first roll was a 1 (1,2).

How many ways can I roll a 3?  The answer is 3.  Either on the first roll (3) or on one of the second rolls, assuming my first roll was a 1,2 or 3 (1,3) (2,3).

And so on...

Therefore, there is only 1 outcome where I'd make \$1, two outcomes where I'd make \$2, 3 outcomes where I'd make \$3 dollars, up to 6 outcomes where I'd make \$6.

The total number of outcomes, then, is (1+2+3+4+5+6) = 21.

The expectation value is then:

$<E> = \$1 \frac{1}{21} + \$2\frac{2}{21} + \$3\frac{3}{21}$

$<E> = \frac{1}{21} \sum_{n=1}^{6} n^2$

$<E> = \$91/21 $

$<E> = \$4.3\bar{3}$

I guess I'm trying to answer the question "how much would you pay to play this game"?  I'd pay up to $4.333.

The other solutions look more sophisticated, so maybe I'm blowing smoke.


Is $\sum\limits_{n=0}^{\infty} \frac{n}{n+1} (-1)^{n}$ convergent? I will like to know whether $\sum\limits_{n=0}^{\infty} \frac{n}{n+1} x^{n}, \: x \in \mathbb{R}$ is convergent for $x=\pm1$ or not.
I found the radius of convergence to be $r=1$. So I have to check each ($x=1$ and $x=-1$) case separately. $x=1$ is easy. $x=-1$ is not so easy. I have used the limit test, M-test, root test, Leibniz test, comparison tes. But none if them work. I do not think I have other tests in my arsenal.

Any help is welcome.

You've already noted @Tom explained that, if $\sum_{n\ge0}a_n$ converges, the triangle inequality implies $\lim_{n\to\infty}a_n=0$. Whatever name that test has, it proves your series diverges. Indeed$$a_n:=\frac{(-1)^nn}{n+1}\implies\lim_{n\to\infty}|a_n|=1\implies\lim_{n\to\infty}a_n\ne0.$$In fact $\lim_{n\to\infty}a_n$ doesn't exist, because the subsequences for even (odd) $n$ have respective limits $1$ ($-1$), which differ. (That test appears to have a name too.)

OK, note that

$$\frac{-n}{n+1} + \frac{n+1}{n+2} = \frac{1}{(n+1)(n+2)}.$$

So if you pair up the terms, you get a series with terms less than $1/n^2$.  So the series converges when $x=-1$.

Now if you pair them up the other way, you'll get a series that also converges to a different number.  So now what?


Is there another way of graphing a line $y = mx + c$ rather than letting $m = 0, 1$? I was graphing the rule $y = 2x - 3$. Since this rule is of the form $y = mx + c$, we have substituted $m = 2$ and $c = -3$ when $x = 0$. This means the gradient (slope, steepness, etc) of the line $y$ we want to graph will be $2$ since $m$ denotes the gradient. We can mark our first co-ordinate on the graph by knowing that $x = 0$ and $c = -3$, forming the ordered pair $(0, -3)$ which is a point on our graph. Since every point is of the form $(x, y)$ then $c$ is equal to the $y$-intercept (the point where the line crosses the $y$ axis).
Now I have a question. My teacher asked me to let $x = 1$ after I set it equal to $0$, and according to the rule, $y = 2\cdot 1 - 3 = 2 - 3 = -1$ so we form another point $P(1, -1)$ where we write $P$ in front of the ordered pair to make it clear that it is a point on the graph. Since the rule is of the form $y = mx + c$, we are forming a linear graph where our line is always straight (of course since the measured gradient of the line at each of its points must be the same; a constant gradient). So from here, we can just draw a straight line through the two points we have plotted on the graph and Bob's your uncle.
However, I figured out something interesting. We could re-write $y = 2x -3$ as $y + 1 = 2(x - 1)$ and here we have constants $1, -1$ which are exactly the $x$ and $y$ coordinates of the point on our line for which $m = 1$. Is this a coincidence?
I am a little eager and impatient, and I do not want to wait for my next maths class to discuss this.
Thank you in advance.

Edit:
I have found my answer here $\longrightarrow$ How to Graph Equations $-$ Linear, Quadratic, Cubic, Radical, & Rational Functions
But thanks for the help anyway :)

Once you have recognized that the graph will be a straight line, you just need two different points on it to actually draw it with a ruler.

For $y=mx+c$, setting $x=0$ and $x=1$  is a particularly  convenient way to get those two points (because those two numbers are both trivially easy to multiply by $m$, so it takes only a single addition to find all the numbers you need). But you don't have to do it that way if you have a different way to choose two points on the line you'd rather use.

If, instead, you have the equation on the form $y-b=m(x-a) $ like you rewrote it to, it makes perfect sense to let $(a, b) $ be one of your two points.

to graph the line $$y=mx+c$$ we will need two different points:
if $$x=0$$ we get $$y=c$$ this is the first point:
$$P_1(0,c)$$
second: $$y=0$$ we get $$x=-\frac{c}{m}$$ we get
$$P_2\left(\frac{-c}{m},0\right)$$


Gap between the induced norm of a matrix and largest Eigenvalue? Are there any known results on how big the gap between the absolute value of the largest Eigen  value of matrix and the induced norm can be?

More formally, let the induced norm of A is given by $\|A\| = max_{\|x\| = 1}\|Ax\|$ and let $\lambda_{max}$ denote the largest Eigenvalue. I am interested in the quantity $\|A\| - |\lambda_{max}|$. Since the norm bounds the absolute value of the Eigenvalues, the quantity $\|A\| - |\lambda_{max}|$ is always positive. I also know that for positive-definite matrices, the quantity $\|A\| - |\lambda_{max}|$ is zero. But are there any results known for a generic matrix $A$?

p.s: I am working on a problem where I am trying to compute a bound on the norm of a matrix. I have bounds on the Eigenvalues of my matrix via Gresghorin's circle theorem and I am trying to see whether I can use that in some way to obtain a bound on the matrix norm....

Edit: To clarify, A is a square matrix over the field of reals and I am using the standard 2-norm on $R^n$

Presumably you are talking about the induced 2-norm, i.e. the largest singular value. When the spectral radius of $A$ is fixed, the 2-norm can be unbounded, as illustrated by the example $A=\pmatrix{1&n\\ 0&1}$. We have $\rho(A)=1$ but $\|A\|_2$ (and in turn the gap $\|A\|_2-\rho(A)$) approaches infinity when $n\to\infty$.

Your two notions of norm are equal. If the maximum absolute value eigenvalue of $A$ is $\lambda$ and you take any vector $w$ of length $1$, then you will have $||Aw|| \leq |\lambda| ||w|| = |\lambda|$, and you will have equality when you take $w$ to be an eigenvector of $A$ with eigenvalue $\lambda$.


SVD with rectangular matrix U I have a matrix $A_{3\times2}$ , $U_{3\times2}$ , $V_{2\times2}$.

I was asked to find the singular values, which I did with by multiplying $U'AV$.

I got the $U'$ by following this rule: Since $U$ is an orthogonal matrix, the inverse of $U$ is equal to its transpose.

But now the question:

  If I didn't follow the rule of transposing $U$, then I would have a
  matrix multiplication of dimension $(3\times2)(3\times2)$ which
  doesn't work.

Any suggestions, tips on what am I missing here?

$$A=U\Sigma V'\implies AV=U\Sigma\implies U'AV=\Sigma$$.

Sizes - $A_{m\times n}, U_{m\times m},\Sigma_{m\times n},V_{n\times n}$.

All multiplications are valid, and you got the singular values on the diagonal of $\Sigma$

When you get a non square matrix $U$, you need to "complete" it to be a square matrix. How? Just add random (almost) vectors to it. Just be sure they keep your matrix columns orthonormal (Gram Shcmidt to the rescue). Now we need to fix $\Sigma$. Just add rows of zeros to it (yes, it is that simple) and you are done. As the comment suggests, this is just for mathmatical equality, and you dont have to do this in order to complete the SVD decomposition.


help with $\lim\limits_{(x,y) \to (0,0)} f(x,y) = {\cos(x) -1 - {x^2/2} \over x^4 + y^4}$ $\lim\limits_{(x,y) \to (0,0)} f(x,y) = \dfrac{\cos(x) -1 - {x^2 \over 2}}{x^4 + y^4}$

Is the following approach correct?

If we approach the origin from $y$ , that is $x = 0$:

$\lim\limits_{(x=0,y) \to (0,0)} f(0,y) = {0 \over y^4} = 0$

Now we approach the origin from $x$ and use $\cos(x) \sim_{0} 1 - {x^2\over 2}$

$\lim\limits_{(x,y=0) \to (0,0)} f(0,y) = { -x^2 \over x^4} =- {1 \over x^2}=-\infty$

Then we can conclude that the limit does not exist.

My teacher took another approach where he uses the $\cos (2x)$ formula and ends up using $\lim\limits_{u\to0}\dfrac{\sin u}{u}=1$ so I'm wondering if my solution is OK or not.

Use the straight line $y=mx$ for $m \in \mathbb{R}$:
$$
\lim_{x \to 0} \frac{\cos x -1 -\frac{x^2}{2}}{(1+m^2)x^2} = \lim_{x \to 0} \frac{1-\frac{x^2}{2}+\frac{c^4}{4!}-1-\frac{x^2}{2}}{(1+m^2)x^2}=\lim_{x \to 0} \frac{-x^2+\frac{c^4}{4!}}{(1+m^2)x^2}
$$
for some $c \in (0,x)$. Therefore
$$
\lim_{x \to 0} \frac{-x^2+\frac{c^4}{4!}}{(1+m^2)x^2} = -\frac{1}{1+m^2},
$$
which clearly depends on $m$. The original limit cannot exist.


Integral involving square root of sine and cosine Is there any closed formula for
$$
\int_{0}^{\pi/2} \dfrac{e^{-x}\sqrt{\cos x}\ dx}{\sqrt{\cos x} + \sqrt{\sin x}}
$$
I know
$$
\int_{0}^{\pi/2} \dfrac{\sqrt{\cos x}\ dx}{\sqrt{\cos x} + \sqrt{\sin x}} = \dfrac{\pi}{4}
$$
replacing $x$ by $\pi/2 - y$. Thanks for any help.

There exists no closed form for your proposed Integral. However if we generalize the problem as $$I(\alpha ) = \int\limits_0^{\frac{\pi }{2}} {\frac{{{e^{ - \alpha x}}\sqrt {\cos x} }}{{\sqrt {\cos x}  + \sqrt {\sin x} }}dx}$$ we could obtain the variation of the integral with respect to $\alpha$ (numerically) as


Help with finding the $\lim_{x\to 0} \tan x \csc (2x)$ So I am trying to figure out the limit

$$\lim_{x\to 0} \tan x  \csc (2x)$$

I am not sure what action needs to be done to solve this and would appreciate any help to solving this.

Hint: We have $\csc(2x)=\frac{1}{\sin(2x)}$ and $\sin(2x)=2\sin x\cos x$.

$$
\lim_{x\to 0} \tan x  \csc (2x)= \lim_{x\to 0} \dfrac{\sin x}{2\cos^2x\sin x} = 1/2
$$


Power set ordered by sum and Dijkstra shortest path I've needed to enumerate the power set ordered by the sum of elements in each subset. Luckily I've found a nice solution here: Algorithm wanted: Enumerate all subsets of a set in order of increasing sums.

But I'm still wondering if this is an equivalent of Dijkstra shortest path algorithm.

Empty set is the source, target is the whole set, and each subset's distance from the source is the sum of the elements in them.

The graph which is enumerated is the lattice of subsets or $(2^N,\cup, \cap)$, where $N$ is the whole set.

For example, let $N=\{1,4,5,9\}$, then:

$d(\emptyset,\{1\})=1$ and $d(\emptyset,\{1,4\})=5$, as defined above and  therfore $d(\{1\},\{1,4\})=d(\emptyset,\{1,4\})-d(\emptyset,\{1\})=4$.

No, the algorithm described in the other post is not quite the same as Dijkstra's algorithm executed on the lattice.  It uses an additional property of the lattice to "prune" the search that Dijkstra's algorithm does at each iteration.  The goal is to tweak Dijkstra's to make it more efficient.  (But as I note below it is not clear that it achieves this goal.)
To explain, I'll describe what Dijkstra's algorithm on the lattice would do, and then describe how the other algorithm can be viewed as a modification of that.
Dijkstra's algorithm maintains the collection $K$ of all subsets of $N$ enumerated ("known") so far.  At each iteration it does the following. It adds to $K$ the cheapest subset, say, $S'$, not yet in $K$.  To do this efficiently, Dijkstra's algorithm maintains, for each set $S\not\in K$, a value $D[S]$ which is the cheapest way to make $S$ by adding just one element to some set already in $K$. It finds $S'$ by choosing the subset $S$ with minimum $D[S]$.
Dijkstra's stores the values $D[S]$ for $S\not\in K$ in a min-heap so that the desired minimum can be found quickly (in time logarithmic in the number of values). However, note that this number of values can be exponential, because there is one for every subset not yet in $K$.  (As an aside, because heap operations take time logarithmic in the number of keys stored, I think Dijkstra's on the lattice does actually have polynomial delay for this problem.)
The linked-to algorithm uses the following observation.  For each element $i\in N$, it maintains the cheapest set $S_i$ in $K$ that does not contain $i$.  Then, it only needs to consider, in each iteration, the sets of the form $S_i\cup\{i\}$ for each $i$.  This is polynomially many, instead of exponential, and one of the sets of this form will be the set that Dijkstra's would choose (the next cheapest set).
On the other hand, Step 4 of that algorithm, which updates $S_i$ when $S_i\cup\{i\}$ is added to $K$, is not clearly polynomial, because it has to scan through $K$ to find the next value for $S_i$.  So, because of its Step 4, unlike Dijkstra's algorithm, that algorithm does not clearly have polynomial delay between enumerated sets.
EDIT 1: It seems this would be easy to fix by maintaining for each $i$ the subset $K_i$ of $K$ of sets that don't contain $i$ (ordered by cost).  When a set $S$ is added to $K$, it would be appended to each $K_i$ such that $i\not \in S$.  Then the update to $S_i$ would take constant time, so the whole algorithm would have polynomial delay between sets.
EDIT 2: By the way, a poly-delay algorithm for a more general problem is described in an answer to this related question: https://cstheory.stackexchange.com/questions/47023/can-we-efficiently-enumerate-the-words-accepted-by-a-dfa-by-order-of-increasing

Well these two are not same. Consider the first step in Dijkstra. In Dijkstra, we start with the source, iterate over all the edges starting from source and select the edge with the lowest weight among them . So even though there exists a edge with a lesser wight between two vertices where none of them are the source, Dijkstra will not select that first.

Also, there are $2^{|E|}$ subsets of edges for a graph with |E|-edge and Dijkstra algorithm does not enumerate all of them, being a polynomial time algorithm.

Also it reduces the distance between two vertices after "fixing" any vertex, if possible. That means, it does not always, produce a subset of edge (present in the path) in the order or increasing sum.

So they have lots of differences.


Find least number of radial-subgraph of a graph Background:
Here is a group G of a people, one maybe another's friend. How to select least number of people to be a leader of a subgroup, so that everyone in the group G has a friend as a leader?

Translate:
find least number of radial-subgraph of a graph. By radial-subgraph, it means a subgraph which has at least one point which connects all the other point in the subgraph.

This is actually Dominating set problem, check in wikipedia https://en.wikipedia.org/wiki/Dominating_set

I think the application is, how to select least number of people in group to tell a message, so that all of the people in the group can get that message via friends.


The definition of $\oplus$ I would like to understand why the books give two different concepts to $\oplus$ between vector spaces:

See:

Concept 1: $W=V_1\oplus V_2=\{(v_1,v_2) \mid v_1\in V_1, v_2\in V_2\}$.

Concept 2: $W=V_1\oplus V_2=\{v_1+v_2 \mid v_1\in V_1, v_2\in V_2\}$, where $V_1\cap V_2=\{0\}$.

Are they equivalent?

I'm thinking to prove this is an isomorphism $(v_1,v_2)\mapsto v_1+v_2$.

Thanks

The two concepts are indistinguishable from the point of view of algebra.

The mathematical formulation of this is:
$$
V_1\cap V_2 = \{0\}\implies
\{(v_1 +v_2 |v_1 ∈V_1 ,v_2 ∈V_2 \} \cong\{
(v_1,v_2) |v_1 ∈V_1 ,v_2 ∈V_2
\}
$$

I let you find the isomorphism.

In addition to MattS's excellent answer, we can add that the two notions, Cartesian products and external direct sum, differ when considering infinitely many vectors spaces.

We consider a family $(V_j)_{j\in J}$ of vector spaces, indexed by a set $J$. Then:

The product $\displaystyle{\prod_{j\in J}V_j}$ is, as a set, the Cartesian product of the $V_j$ with the following operations: if $v=(v_j)_{j\in J}$ and $w=(w_j)_{j\in J}$, $\lambda v=(\lambda v_j)_{j\in J}$ and $v+w=(v_j+w_j)_{j\in J}$. It is easily checked that these two operations give to $\displaystyle{\prod_{j\in J}V_j}$ a structure of vector spaces.
The external direct sum $\displaystyle{\bigoplus_{j\in J}V_j}$ is the subspace of $\displaystyle{\prod_{j\in J}V_j}$ of all vectors $v=(v_j)_{j\in J}$ such that all component $v_j$, except a finite number of them, are zero.

For a finite number of spaces $V_j$, these two spaces are the same.

These definition are not actually restricted to vector spaces but can be applied to many algebraic structures (groups, modules, rings,...), with the suitable operations.

Edit: after re-reading the OP and the comments, I feel this answer is more a complement to Pedro Arbizu's comment and Mookid's answer.


$|2-|1-x||=|x^2-1|$ equation, incorrect solution When I solved this equation I got $x_1 = -1$, $x_2 = 0$ and $x_3 = \frac{-1+\sqrt{17}}{2}$
I solved it like this: I first sketched a graph and I could see that the first two solutions were trivial, I got the third one by solving $x^2-1 = -(x-3)$ for $x > 1$, since I could see from my graph that there was an intersection.
Wolfram alpha only displays the first two solutions and doesn't mention the third one, is my solution to the whole problem incorrect?
I also have a question, how do you view if someone solves a problem by sketching a graph and solves it from there and doesn't do all the required calculation work to show that the expression inside every absolute value can be negative and what happens if it is positive and does everything from there.
Thank you for your answers.

Alpha does all computations in the complex and it turns out that the solution includes a whole locus with a terribly complicated expression (because the equation is equivalent to a quartic one). But that locus certainly includes the third real root.
Now the following Alpha request will give you the three real roots: https://www.wolframalpha.com/input/?i=real+roots+of+abs%282-abs%28x-1%29%29-abs%28x%5E2-1%29

Your method is very good but it is not considered a rigorous proof. You can use the graph to turn this into a rigorous proof. We see on the graph that there are only three cases to study: $x\lt -1$ and  $-1\le x \le 1$ and $1 \le x$. Each of these three cases is very easy to study.


How prove this $|1+x|^a\ge 1+ax+\dfrac{1}{1000}|x|^a$ let  $2\le a\le 13,a\in R$,and $x\in R$,show that:
  $$|1+x|^a\ge 1+ax+\dfrac{1}{1000}|x|^a\tag{1}$$

My try: let
$$f(x)=|1+x|^a-1-ax-\dfrac{1}{1000}|x|^a$$

and since if $x>-1$,then  $$|1+x|^a=(1+x)^a=1+ax+\dfrac{a(a-1)}{2}x^2+\cdots+x^a$$
and I fell this is nice reslut.because it is well konw this follow Bernoulli inequality

$$(1+x)^a\ge 1+ax,x>-1,a>1$$
But my inequality is strong than this .and  I use computer test found this inequality $(1)$ is true.and I can't
 prove it.

BY the way I found in china  book have this

Thank you for you help

Here is a partial answer : I show below that inequality (1) holds
when $x\geq 0$ or $x \leq c_1=-\frac{1}{1-\big(\frac{1}{1000}\big)^{\frac{1}{13}}}$
(note that $c_1 \approx -2.42 \ldots$).

Let $g(x)=(1+x)^a-1-ax-x^a$ for $x\geq 0$. Then $g'(x)=a(1+x)^{a-1}-a-ax^{a-1}$,
$g''(x)=a(a-1)\big((1+x)^{a-2}-x^{a-2}\big)$, so $g'$ is increasing, and hence
$g'(x) \geq g'(0)=a(a-1) >0$, so $g$ is increasing, and hence
$g(x) \geq g(0)=0$. So when $x\geq 0$ we have
$|1+x|^a \geq 1+ax+|x|^a$, which is stronger than (1).

Next, if $x\leq c_1$ then $|x| \geq |c_1|$,
$1-\frac{1}{|x|}\geq 1-\frac{1}{|c_1|}=\big(\frac{1}{1000}\big)^{\frac{1}{13}}$,
so $\bigg(1-\frac{1}{|x|}\bigg)^a \geq \frac{1}{1000}$. It follows that
$|1+x|^a \geq \frac{1}{1000}|x|^a$, which is also stronger than (1).

The value of the function in 0 is 0 for every a. The function is continuous, so you could do the derivative and use binomial theorem to prove that for every x > 0 is positive and for every x < 0 is negative.

EDIT: http://i.imgur.com/CZcJutJ.jpg


What's a polynomial witness for this TSP variation? I was thinking about the Travelling Salesman Problem when I thought of a possible variation, the minimum salesman path(MSP).

In the original TSP, the question is to decide whether a path exists through the weighted graph such that the sum of the weights in the path be less than, or equal to some integer, $n$. A polynomial-time witness for this is such a given path.

In the minimum salesman path variation, I ask whether the length of the smallest path in the graph is equal to some integer, $m$ (The path satisfies the TSP route through the nodes).

If we restrict ourselves to non-negative integer values for the weights on the paths, MSP is Turing reducible to TSP. The reduction is achieved thus :

Given a weighted graph, and the value, $m$, we use an algorithm for TSP to decide whether there are any paths with length $\le m$. If yes, then we run TSP algorithm again to check whether there are any paths with length $\le m-1$.

So, MSP Turing reduces to TSP, and is a decision problem. Hence, it should be in the class $NP$, or $Co-NP$
But I am unable to find any polynomial-time witness for either acceptance, or rejection.

Can someone provide me with any?

EDIT : changed $\lt$ to $\le$ in the text.

Your problem is known as EXACT-TSP and lives in the complexity class DP. EXACT-TSP was proven DP-complete over thirty years ago in the paper "The complexity of facets (and some facets of complexity)" by C. H. Papadimitriou and M. Yannakakis (see Theorem 2).  As such, there is no polynomial-time witness for this problem unless the polynomial hierarchy collapses to the first level.

TSP is in $NP$ because there are polynomial-time witnesses for path length $\le m$.  But it is not known to be in Co-$NP$, because there are no known polynomial-time witnesses for length $\ge m$.

EDIT: A polynomial time witness for minimal path length $=m$ would also be a witness for minimal path length $\ge m$.  Thus MSP is not known to be in $NP$.


Finding primes $p$ and $q$ such that $p$ divides $q^2 -4$ and $q$ divides $p^2-1$ Finding all sets of primes $p$ and $q$ such that $p$ divides $q^2 -4$ and $q$ divides $p^2-1$.

Hint :

$p$ divides $q+2$ or $q-2$ and $q$ divides $p+1$ or $p-1$

Consider one by one case :

$p=l(q+2),q=m(p+1)\Rightarrow p=l(m(p+1)+2)=lmp+lm+2l\Rightarrow p= ??$

try other cases...

$p=l(q+2),q=m(p-1)$
$p=l(q-2),q=m(p-1)$
$p=l(q-2),q=m(p+1)$

Four cases would be formed and only two would be left .

$p+1 \equiv 0 \pmod q$

$q+2 \equiv 0 \pmod p$

After some estimation it would be found that only $p=5$ and $q=3$ can be the solution.


The deep reason why $\int \frac{1}{x}\operatorname{d}x$ is a transcendental function ($\log$) In general, the indefinite integral of $x^n$  has power $n+1$.  This is the standard power rule.  Why does it "break" for $n=-1$?  In other words, the derivative rule $$\frac{d}{dx} x^{n} = nx^{n-1}$$ fails to hold for $n=0$.
Is there some deep reason for this discontinuity?

I'll try to give a soft answer to what I see as the spirit of the question, which is not why you get exactly log, but why the behaviour is different when integrating $x^{k}$ for $k=-1$.

The way I see it is that the logarithm would actually be there for other powers of $x$ too, but it's "hidden" by the fact that in a geometric series one term "gobbles up" all the others together, except in the very special case when all terms are equal. Put in other words, the intuition is very close to the reason why $\sum_{i=a}^b c^i$ can be reasonably approximated by just the first term of the sum ($c^a$) if $c<1$, and just by the last ($c^b$) if $c>1$, regardless of how large $b-a$ is, i.e. of how many terms you have in the sum. But if $c=1$ no single term dominates all the others, and that's when you have to count them all, and you end up seeing the $b-a$ term "emerge" in $\sum_{i=a}^b 1^i=b-a+1$.

Informally, you can see how this applies to the case at hand writing $\int_{x_0}^{x_f} x^k dx$ as $\int_{x_0}^{2 x_0} x^k dx + \int_{2x_0}^{4x_0} x^k dx + ...$. Each of the $\approx \log_2 \frac{x_f}{x_0}$ terms has the same weight as the others if, and only if, $k$ has a very specific value (which?), in which case $\int_{x_0}^{x_f} x^k dx$ equals $\approx \log_2 \frac{x_f}{x_0}$ times $\int_{x_0}^{2 x_0} x^k dx$. If it's just an $\epsilon$ smaller, or larger, you get the sum within a constant factor of either $\int_{x_0}^{2 x_0} x^k dx$ or of $\int_{\frac{1}{2}x_f}^{x_f} x^k dx$, respectively, regardless of $\frac{x_f}{x_0}$. Note that instead of using $2$ as a base, we could have used $3$, or $e$, or $7.24$, or $\pi$, and the critical value of $k$ would have remained the same: it's the value that ensures that if you integrate $x^k$ over an interval $7.24$ longer, but with a starting point $7.24$ times larger, the integral does not change, $k=-1$.

This is actually a phenomenon that I've seen pop up really really often in math, physics, and computer science. You often have a sum of many terms, and a parameter that, for small values, makes the first term of the sum dominate all the others put together, and for large values, makes the last term dominate all the others put together, regardless of how many terms you have in the sum. But when the parameter equals exactly the critical value at which the transition between the two regimes occurs, all sort of strange quantities related to the number of terms in the sum appear in any approximation of it.

For me, the key observation has always been that $\int \frac{dx}{x}$ is "scale-free" in the following sense.  Suppose that you perform a scaling operation on $x$ by defining $u=ax$ for some constant $a$.  How does the integral change?  You get:
$$ \int \frac{dx}{x} = \int \frac{\frac{du}{a}}{\frac{u}{a}} = \int \frac{du}{u} $$
It doesn't change!  This would not be true for any other power of $x$, since the numerator and denominator would not cancel out perfectly.
This is fundamentally the same observation as the highest-voted answer by @Anonymous above, and there are hints of it in the other answers as well.  But I think that this way of explaining it is much more concise.
In physical terms, applying a scaling factor could represent measuring $x$ in feet rather than meters, for example.  It's unusual that the result does not depend at all on the units that you use, and this is ultimately the reason why the logarithm has all of its interesting properties, such as $\log ab = \log a + \log b$.
(For those who are interested, the term "scale-free" is a buzzword in statistical physics to describe some similar behavior in physical systems near a phase transition.  I have intentionally misused it here, because in physics you typically don't care about creating a constant factor when doing the scaling, so all power laws in physics are "scale-free," but here we are talking about strict equality.)


In how many ways we can put 12 different balls to 4 equal boxes We put 12 different balls to 4 equal boxes. How many configurations are possible?

If the boxes were distinguishable it would be simply $4^{12}$, but I do not know how to consider the fact that they are distinguishable.

When the balls are indistinguishable, you will have the following formula:

\begin{equation}
{n+k-1 \choose n}
\end{equation}

The formula counts the number of ways in which n indistinguishable balls can be distributed into k distinguishable boxes. In the following link, you can find a good explanation of the reasoning behind it. Number of ways of distributing balls into boxes


Prove the transcendence of the number $e$ How to prove that the number $e=2.718281...$ is a transcendental number?   The truth is I have no idea how to do it.

If I can recommend a book or reference on this topic thank you.

There are many tests on the transcendence of $ e $?

I'd read several shows on the transcendence of $ e $

Try Michael Spivak's Calculus. I find it amusing that he would prove the transcendence of $ e $ in a calculus textbook.

I realize that amWhy already gave one answer: the Lindemann-Weierstrass theorem.  I don't know how to prove this... to be frank it is over my head.  But in general it states that in the equation:
$$e^a = b$$
If $a\ne 0$ is algebraic then $b$ is transcendental. The contrapositive is also true: if $b$ is algebraic then $a$ must be transcendental. Obviously $a=0,b=1$ has to be an exception.

Take that on faith for a moment. Like I said, I cannot prove the L-W theorem but I welcome you to research it.

If you ask whether or not $\ln(2)$ is algebraic or transcendental, take note that $e^{\ln(2)}=2$ is algebraic, and you can conclude that $\ln(2)$ is transcendental.

Similarly, you asked if $e$ itself was algebraic or transcendental. That is easy. $e^1 = e$ by definition. The power is algebraic therefore $e$ is transcendental. This is not circular reasoning. It rests squarely on the truth of the Lindemann-Weierstrass theorem: for any non-zero algebraic power, $e^a$, whatever that may be, is transcendental.

If you ever see a proof for the transcendence of $e$ more complicated than the one I just gave, I assure you it is just a proof for a special case of the L-W theorem. The general case proof would suit you better.

Likewise, take Euler's Identity: $e^{i\pi} = -1$.  The result is algebraic, therefore the power $i\pi$ is transcendental. But we already know $i$ to be algebraic, so it is the $\pi$ that is transcendental. This is how you prove the transcendence of $\pi$.

I would also like to mention a theorem that hasn't been mentioned yet: the Gelfond-Schneider theorem.  It states that:
$$a^b$$
is always transcendental if:

$a$ is algebraic and $a\ne 0,1$, and
$b$ is algebraic and irrational

Numbers such as $2^{\sqrt{2}}$ are proven transcendental in this way.

We can also prove that $\sqrt{2}^{\sqrt{2}}$ is transcendental as a consequence of the G-S theorem, because it is simply the square root of $2^{\sqrt{2}}$. The square root of a transcendental number is still transcendental.

I, personally, do not know of any other helpful theorems. But then I am new to transcendental number theory.  Good luck.


What is meaning of 2n factors? Can any one explain what does 2n factors mean and how it's been written n factors in next step. I tried to figure out but couldn't make out.

It means that there are 2n terms, each term being a factor of the form $(1+\omega^2)$. In the second step, they've grouped two terms together, for eg. $[(1+\omega)(1+\omega^2)]$ is one term and so, there are n such factors.

It means number of our factors and there are even number of factors like 8 factors or 10 factors whatever
let me solve it to understand your answer why and how we final get the result= 1
your question prove that: $(1+ω)(1+ω^2)(1+ω^4)$ etc.......to 2n factors $=1$
so they are factors multiply each other and they equal to 1
how we multiply  "product" factors
let me show you how we multiply factors
now let's take like number 12 what is factors of "12"
$1\times12$ ,$2\times6$ , $3\times4$
so factor will equal ={1,2,3,4,6,12}
then how we multiply factors ? by the awesome formula
$N^{no.of factors/2}$
n=12 the number repeat by product of two numbers
{$1\times12$ , $2\times6$ , $3\times4$}
number of factors{1,2,3,4,6,12} =6
by using the formula $12^{6/2} =1728$
Getting back on our question
note $ω^3=1$ , $ω+ω^2+1=0$
in the third bracket we will find
$(1+ω^4)=(1+ω\timesω^3)=(1+ω)$
the first two brackets are repeat$=n$
and number of our factors $=2n$
let's use our formula
${((1+ω)(1+ω^2))}^{2n/2}$.two will go with two
${((1+ω)(1+ω^2))}^{n}$.multiply two bracket.${(1+ω^2+ω+ω^3)}^{n}={(1+ω^2+ω+1)}^{n}={(1+0)}^{n}=1^n=1$

i think that help you to understand the first answer .$\color{gold}{Yahia}$ $\color{gold}{Kamal}$ $\color{red}{♥♥}$
refrences:
https://www.youtube.com/watch?v=g4zTQveQuuY


Find infimum of $\{\frac{1}{n^2} | n \in \Bbb N\}$ in $\mathbb{R}$ So I need to find the infimum of$\{\frac{1}{n^2} | n \in \Bbb N\}$

I know that this means that I need to find some $x$ where $x < \frac{1}{n^2} \forall n \in \Bbb N$.

By intuition I know that $\lim_{n \to \infty}$, is $0$, which means that the infimum should be $0$, but is there a more formal way of proving that without using limits?

Maybe this can help. Let us write $$A=\bigg\{\frac{1}{n^2}:n\in\Bbb N\bigg\}.$$
Clearly, $0$ is a lower bound of $A$. So, the set $A$  is a non-empty set bounded below. Thus, we can find a real number $w$ such that $$\inf A=w.$$ Then, $$w\geq 0.$$
Let $\epsilon >0$. Then, using the Archimedean Property, we can find $n\in\Bbb N$ such that $$\frac{1}{n}<\sqrt{\epsilon}.$$ Since $w$ is a lower bound of $A$ and $\frac{1}{n^2}\in A$, we get
$$w\leq\frac{1}{n^2}<\epsilon.$$
Hence,
$$w< \epsilon\quad \forall \epsilon>0.$$
Thus, $$w\leq 0$$ and hence $$w=0.$$

There is no inf
that is in the set,
because all values are positive
but for any value,
there is a smaller one.

However,
there is a lim inf,
and it is zero,
because,
for any $c> 0$,
all values beyond a $n$
that depends on $c$
are within $c$ of zero.


Easy way to compute $\int_{-\pi}^\pi \cos^4(x)dx$. Is there an easy way to compute $$\int_{-\pi}^\pi \cos^4(x)dx\ \ ?$$

What I did is using $$\cos^4(x)+1-1=1+(\cos^2(x)-1)(\cos^2(x)+1)=-\sin^2(x)(1+\cos^2(x))$$

and then using formula $$\cos^2(x)=\frac{\cos(2x)+1}{2}$$
and $$\sin^2(x)=\frac{1-\cos(2x)}{2},$$
but at the end, I have to compute $$...+\int_{-\pi}^\pi\cos^2(2x)dx.$$
Using finally that $$\cos^2(2x)=\frac{1+\cos(4x)}{2},$$
I concluded (and found $\frac{3\pi}{4}$ which is correct). But is there an other (shorter) method ?

We have that
$$ \int_{-\pi}^{\pi}e^{ni\theta}\,d\theta = 2\pi\cdot \delta(n) \tag{1}$$
and since $\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$, by the binomial theorem it follows that
$$ \int_{-\pi}^{\pi}\cos^4(x)\,dx = \frac{1}{16}\int_{-\pi}^{\pi}\binom{4}{2}\,d\theta = \color{red}{\frac{3\pi}{4}}.\tag{2}$$

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\color{#f00}{\int_{-\pi}^{\pi}\cos^{4}\pars{x}\,\dd x} =
2\int_{0}^{\pi}\
\overbrace{\bracks{{3 \over 8} + \half\,\cos\pars{2x} + {1 \over 8}\,\cos\pars{4x}}}
^{\ds{\cos^{4}\pars{x}}}\ \,\dd x = 2\times{3 \over 8}\times\pi=
\color{#f00}{{3 \over 4}\,\pi}
\end{align}

  Note that
  $\ds{\left.\int_{0}^{\pi}\cos\pars{nx}\,\dd x
\,\right\vert_{\ n\ \in\ \mathbb{Z}\,\backslash\,\braces{0}}\;\;\;\; =\ 0}$


A class consists of 15 boys of whom 5 are prefects. How many committees of 8 can be formed if each consists of at least 2 prefects? A class consists of $15$ boys of whom $5$ are prefects. How many committees of 8 can be formed if each consists of at least $2$ prefects?

I had the solution following solution in mind but that seems to be incorrect. can anyone please explain to me why?

First, I select $2$ students from the perfect category using $\binom{5}{2}$ and then I select remaining $6$ students combined from both the perfect and unperfect category i.e. $\binom{13}{6}$. So the answer becomes $\binom{5}{2}\times \binom{13}{6} = 17160$.
But, the correct answer is $5790$. Can you please help me?

There should be minimum $2$ perfects in the committee, there is no limit for maximum of perfects. So, The number of ways can be counted as:

$1$. $2$ perfect and $6$ non perfect..... Number of ways become ${{5}\choose{2}}\times{{10}\choose{6}}$.

$2$. $3$ perfect and $5$ non perfect....... Number of ways become ${{5}\choose{3}}\times{{10}\choose{5}}$.

$3$. $4$ perfect and $4$ non perfect.....Number of ways become ${{5}\choose{4}}\times{{10}\choose{4}}$.

$4$.  $5$ perfect and $3$ non perfect.....Number of ways become ${{5}\choose{5}}\times{{10}\choose{3}}$.

Add all the ways and get the answer.

15C8-(10C8 + 10C7*5C1) = 5790

At least two means, total number of ways - number of ways to make committee without any prefect + number of ways to make committee with one prefect and 7 non prefects.

15C8 - total number of ways to choose 8 people committee.
10C8 - number of ways not to choose any prefects.
10C7*5C1 - number of ways to choose one prefect and 7 non prefects


Finding the largest constant $C$ such that $|\ln x−\ln y| \geq C|x−y|$ for all $x, y \in (0, 1]$ Find the greatest value of C such that
$|\ln x−\ln y|≥C|x−y|$
for any $x,y∈(0,1]$. What should my approach be? I can't think of much options.

To find the largest $C$ so that $|\log(x)-\log(y)|\ge C|x-y|$, note that, by the Mean Value Theorem, for some $t$ strictly between $x$ and $y$
$$
\frac{\log(x)-\log(y)}{x-y}=\frac1t
$$
The minimum of $\frac1t$ on $(0,1]$ is $1$. Therefore, the largest $C$ can be is $1$.

$f(b)- f(a) = f'(\xi) (b - a),\, \xi \in (a, b)$. When $x, y$ near $0$, the constant $C$ will become very large.


Why is there never a proof that extending the reals to the complex numbers will not cause contradictions? The number $i$ is essentially defined for a property we would like to have - to then lead to taking square roots of negative reals, to solve any polynomial, etc. But there is never a proof this cannot give rise to contradictions, and this bothers me.

For instance, one would probably like to define division by zero, since undefined values are simply annoying. We can introduce the number "$\infty$" if we so choose, but by doing so, we can argue contradictory statements with it, such as $1=2$ and so on that you've doubtlessly seen before.

So since the definition of an extra number to have certain properties that previously did not exist may cause contradictions, why aren't we so rigourous with the definition of $i$?

edit: I claim we aren't, simply because no matter how long and hard I look, I never find anything close to what I'm looking for. Just a definition that we assume is compatible with everything we already know.

There are several ways to introduce the complex numbers rigorously, but simply postulating the properties of $i$ isn't one of them. (At least not unless accompanied by some general theory of when such postulations are harmless).

The most elementary way to do it is to look at the set $\mathbb R^2$ of pairs of real numbers and then study the two functions $f,g:\mathbb R^2\times \mathbb R^2\to\mathbb R^2$:

$$ f((a,b),(c,d)) = (a+c, b+d) \qquad g((a,b),(c,d))=(ac-bd,ad+bc) $$

It is then straightforward to check that

$(\mathbb R^2,f,g)$ satisfies the axioms for a field, with $(0,0)$ being the "zero" of the field and $(1,0)$ being the "one" of the field.
the subset of pairs with second component being $0$ is a subfield that's isomorphic to $\mathbb R$,
the square of $(0,1)$ is $(-1,0)$, which we've just identified with the real number $-1$, so let's call $(0,1)$ $i$, and
every element of $\mathbb R^2$ can be written as $a+bi$ with real $a$ and $b$ in exactly one way, namely $(a,b)=(a,0)+(b,0)(0,1)$.

With this construction in mind, if we ever find a contradiction involving complex number arithmetic, this contradiction can be translated to a contradiction involving plain old (pairs of) real numbers. Since we believe that the real numbers are contradiction-free, so are the complex numbers.

It became obvious early on that the real numbers were not sufficient to proceed with building mathematics.  The equation x2 + 1 = 0 tells you that right away.  Okay, so we need some more numbers.  We can define anything we want, but it might not be particularly useful.  By defining the complex numbers in the rigorous ways described above we get an enormous amount of usefulness out of a simple definition.

It is consistent as an extension of the real numbers, because it preserves addition and multiplication of the real numbers.  The study of functions of a complex variable have led to immense insights into all sorts of things; and straightened out some apparent anomalies on the real line.

Although the complex numbers are closed and complete, I could hypothesize that we may need yet another new extension of the reals. (Yes, there are already many extensions).  The reason for this thought is that certain problems, which can be easily stated, are very difficult to prove.  As one example, problems about the distribution of the prime numbers have proved quite intractable.

Now sometimes a hard problem is just hard, but we have a history of simplifying hard problems and making them easier by changing our point of view and/or adding more tools to work with.  I keep thinking that a different set of numbers, maybe not based on the buildup from integers thru rationals thru reals thru complex, might place a different light on many of these difficult problems, and allow of straightforward solutions.

This is all theorizing, of course.

Would my imagined new number system be "consistent"?  If it is to be of any use, it must be consistent in some sense.


Prove $x^{n}-5x+7=0$ has no rational roots This question arises in STEP 2011 Paper III, question 2. The paper can be found here.

The first part of the question requires us to prove the result that if the polynomial
$$x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_{0}$$
where each of the $a_{n}$ are integers, has a rational roots if and only if that root is an integer. It does not give a name for this result.
EDIT: user69810 has pointed out that this is in fact the rational root theorem.

We are then to use this result to prove that the polynomial $$x^{n}-5x+7=0$$
has no rational solutions for $n\ge 2$. My argument was the following:
If there exists a rational root, then it must be an integer.
If there exists an integer root, then $$x^{n}=5x-7$$
for some integers $x$ and $n\ge 2$. Then the LHS is divisible by $x$, meaning that $7$ must be divisible by $x$. Therefore $x \in \{-7,-1,1,7\}$. Checking each of these shows that there is no rational root.

Is this solution correct? It isn't the one in the solutions, but if it's right then I think it's more elegant than theirs.

You have identified the possible rational roots correctly and shown that none of them work.  It is a fine application of the rational root theorem.

I think it's not correct, because you are assuming that if $x|5x-7$ then $x|7$. This is not true in general.Think for example to $2+4$. Of course $6|2+4$ but $6$ doesn't divide $2$ nor $4$. What you can say is that $x(x^{n-1}-5)=7$, but $7$ is prime, then you must have $x=1$ and $x^{n-1}-5=7$ or $x=7$ and $x^{n-1}-5=1$. But both are absurd because $x=1$ and $x=7$ doesn't solve the equation. As you can see the result is quite the same, but the procedure is quite different for what you assumed at the beginning. Just a remark, what you called Eisenstein criterion is Gauss Lemma, Eisenstein criterion is about irriducibility in $\mathbb{Q}[x]$.


Probability distribution and density of Y=g(X) $Y=g(X)$ as shown below.

Find $f_Y(y)$ and $F_Y(y)$ in function of $f_X(x)$.

I began with writing $Y=g(X)$ as the following piecewise function: $
 Y =
  \begin{cases}
   \ -b & \text{if } x < -a \\
   bx/a       & \text{if } -a \leq x \leq a \\
   b       & \text{if } x > a
  \end{cases}
$

For the middle part I think it's simply $f_Y(y)= f_X(y)$ and $F_Y(y)=F_X(y)$.

For the constant parts of the functions I had the following: $f_Y(y)=\delta(y+b)$ when $x<-a$ and $f_Y(y)=\delta(y-b)$ when $x>a$. Is this correct? And how should I put these parts together to express the overall functions $F_Y(y)$ and $f_Y(y)$?

If $b\leq y$ then $g\left(x\right)\leq y$ is true for every $x$ so
that: $$F_{Y}\left(y\right)=P\left\{ g\left(X\right)\leq y\right\} =1$$

If $-b\leq y<b$ then $g\left(x\right)\leq y\iff x\leq\frac{a}{b}y$
so that: $$F_{Y}\left(y\right)=P\left\{ g\left(X\right)\leq y\right\} =P\left\{ X\leq\frac{a}{b}y\right\} =F_{X}\left(\frac{a}{b}y\right)$$

If $y<-b$ then $g\left(x\right)\leq y$ is not true for any $x$
so that: $$F_{Y}\left(y\right)=P\left\{ g\left(X\right)\leq y\right\} =0$$

It is not sure that $Y$ has a density. For this we need $P\{X\in[-b,b]\}=1$, and in that case: $$f_{Y}\left(y\right)=\frac{a}{b}f_{X}\left(\frac{a}{b}y\right)$$

From the graph, $Y$ cannot be less than $-b$ nor greater than $b$.   $$\Pr(Y<-b)=0\\ \Pr(Y>b)=0$$

However, at the points of inflection, we know $Y$ has a probability mass: $$\Pr(Y=-b)=\Pr(X\leq -a) \\ \Pr(Y=b)=\Pr(X\geq a)$$

In the interval between there is a linear relation between $X$ and $Y$ so $$\Pr(Y\le y)=\Pr(X\le ay/b) , \forall y\in (-b, b)$$

Putting it together,
$$F_Y(y) = \begin{cases} 0 & y < -b \\ F_X(-a) & y=-b\\ F_X(ay/b) & y\in (-b, b) \\ 1 & y\ge b\end{cases}$$
By derivation,
$$f_Y(y) = \begin{cases} 0 & y < -b \\ F_X(-a)\delta(y+b) & y=-b \\ a f_X(ay/b)/b & y\in (-b, b)\\ (1-F_X(a))\delta(y-b) & y=b \\ 0 & y> b\end{cases}$$
Thus:
 $$f_Y(y) = F_X(-a)\delta(y+b)+(1-F_X(a))\delta(y-b) + \frac{a f_X(ay/b)}{b}\operatorname{\bf 1}_{(-b,b)}(y)$$

$$F_Y(y) = F_X(ay/b)\operatorname{\bf 1}_{[-b,b)}(y)+\operatorname{\bf 1}_{[b,\infty)}(y)$$


Solution to the equation of a polynomial raised to the power of a polynomial. The problem at hand is, find the solutions of $x$ in the following equation:

$$ (x^2−7x+11)^{x^2−7x+6}=1 $$

My friend who gave me this questions, told me that you can find $6$ solutions without needing to graph the equation.

My approach was this: Use factoring and the fact that $z^0=1$ for $z≠0$ and $1^z=1$ for any $z$.

Factorising the exponent, we have:

$$ x^{2}-7x+6 = (x-1)(x-6) $$

Therefore, by making the exponent = 0, we have possible solutions as $x \in \{1,6\} $

Making the base of the exponent = $1$, we get $$ x^2-7x+10 = 0 $$
$$ (x-2)(x-5)$$

Hence we can say $x \in \{2, 5\} $.

However, I am unable to compute the last two solutions. Could anyone shed some light on how to proceed?

Denote $a=x^2-7x+11.$ The equation becomes $a^{a-5}=1,$ or equivalently* $$a^a=a^5,$$ which has in $\mathbb{R}$ the solutions $a\in \{ {5,1,-1}\}.$ Solving the corresponding quadratic equations we get the solutions $x\in \{1,6,2,5,3,4\}.$

*Note added: $a=0$ is excluded in both equations.

If this is just a casual riddle, then I can agree with the accepted answer. However, if we want to be mathematically strict, I claim that $3$ and $4$ are not solutions of the equation because they lie outside the domain.

Disclaimer: in this post I only consider real exponentiation. It is not my intention to dive into the complex numbers.

What we mean by solving an equation like $f(x) = g(x)$ is finding all $x$ such that both sides make sense and evaluate equal. Hence the first step is always determining the intersection of the domains of $f$ and $g$ - that is, the set of all $x$ such that both sides make sense. Let's consider what that would be in your case.

The left side of your equation naturally decomposes as follows:

$$(x^2-7x+11)^{x^2-7x+6} = p(q_1(x), q_2(x))$$

where

$$\begin{align*}
q_1(x) & = x^2-7x+11 \\
q_2(x) & = x^2-7x+6 \\
p(a, b) & = a^b
\end{align*}$$

So we have to determine the set of all $(a, b)$ such that $a^b$ makes sense (that is, the domain of exponentiation) and then find the set of all $x$ such that the pair $(q_1(x), q_2(x))$ belongs to this set.

And here is the problem.

There is no uniform way to define both $(-1)^5$ and $3^{\sqrt{2}}$. These are different kinds of exponentiation - the first one is obtained as repeated multiplication, the second one is the result of some limit process, and neither definition works for the other side. So we have a choice: if we allow zero and negative numbers as bases, the exponent must be a non-negative integer, so the domain is $\mathbb{R} \times \mathbb{N}$. If we exclude $0$ as a base, we can use negative exponents, which makes the domain $(\mathbb{R} \setminus \{ 0 \}) \times \mathbb{Z}$. If we go further and exclude negative numbers as bases, we can use limits to pass to real exponents, so the domain becomes $(0, \infty) \times \mathbb{R}$.

One could argue that since the three kinds of exponentiation pairwise agree on the intersections of their domains, we could glue them, i.e. consider the exponentiation on $\mathbb{R} \times \mathbb{N} \cup (\mathbb{R} \setminus \{0\}) \times \mathbb{Z} \cup (0, \infty) \times \mathbb{R}$. But that would be unnatural, useless and - in my opinion - ugly.

Now: which exponentiation does the original equation involve? If the one with natural or integral exponents, then we would have to restrict the domain to those $x$ for which $x^2-7x+6$ is an integer. That doesn't seem right.

Hence we are left with the third, which means we should not consider those $x$ for which $x^2-7x+11$ is negative. This rules out $3$ and $4$ as potential solutions*.

Of course, if we just substitute $x=2$, we get

$$1^{-4} = 1$$

which we know is true and if we substitute $x=3$, we get

$$(-1)^{-6} = 1$$

which we know is equally true, leading to an illusion that both solutions are on equal terms. But that illusion results from using the same notation $a^b$ for two different kinds of exponentiation and it does not stand passing to a strict setting.

*Note: I chose the type of exponentiation that seemed to me to fit in better with the equation. In fact, that choice is an inseparable part of the problem, so it should be disambiguated by the author of the equation (and stated alongside it).


Studying the function $f(x) = x^4-6x^2$ using derivatives: minima, maxima, inflection, concavity (I know this is my second question today, but I'm explaining what I'm doing so I hope it's okay)

Consider the graph of $f(x) = x^4-6x^2$.

  a) Find the relative maxima and minima (both x and y coordinates).

I actually forgot how to do this.. not the point of my question though.

  b) Find the coordinates of the point(s) of inflection.

So for this, you take the derivative of the derivative, correct?

$f'(x) = 4x^3-12x$

$f''(x) = 12x^2-12.$

$f''(x)=0 \Leftrightarrow 12x^2 = 12 \Leftrightarrow x = -1, 1.$

So the points are $(-1,5)$ and $(1,5)$ (y coordinate from plugging back into the original)?

  c) Determine the interval(s) on which the function is increasing.

When I set the derivative equal to zero to find the potential turning points, I get 0, rad 3, and negative rad 3. Forgot what to do from here. Not important.

  d) Determine the interval(s) on which the function is concave up.

I get $(-\infty,-1) \cup (1,\infty)$

*Answer only what you want. Any help is appreciated!

Source: http://online.math.uh.edu/apcalculus/exams/AB_SECTION_II_version_1.pdf - #4

a) Min/Max: calculate the derivative of the function and equal it to zero to find the x-coordinate of the extreme points of the function. Then calculate the second derivative and evaluate it on these extreme values of $x$. If the 2nd derivative is positive, then the slope of the tangent is growing and thus we have a minimum at $(x, f(x))$. if the 2nd derivative is negative; we have a maximum $(x, f(x))$

b) That's right.

c) function opens up when $f''(x)>0$

after that find $f'(x)=0$ here you get two points.

Then evaluate these two points in $f''(x)$ for each of these two points.

then you get one negative -ve(local maximum),put put this value in f(x).

then other you will get +ve positive (local minimum) value after putting in f''(x). put this value in f(x).

Note: put the x-values in f(x), not that positive and negative value.


Solution for $x = -c_1  e ^ x + c_2  e ^{-x}$ How can I solve this equation,
$$x = -c_1 e ^ x + c_2e ^{-x}, \;\;\; 0 < c_1, c_2 < 1$$
We can use $t = e^x$ which will result in,
$$t \ln(t) + c_1 t ^ 2 - c_2 = 0, \;\;\; 0 < c_1, c_2 < 1$$
but how can I solve this one then?

Let's find first derivative of the both sides of the equation:

$x'=(-C_1e^x)'+(C_2e^{-x})'$

$1=-C_1e^x -  C_2e^{-x}$ and now let's find first derivative of the left and right side:

$(1)'=(-C_1e^x)' -  (C_2e^{-x})'$

$0=-C_1e^x + C_2e^{-x} \Rightarrow C_1e^x=C_2e^{-x}$ , which means that:

$x=-C_1e^x + C_1e^x$

$x=0$


Find probability distribution given constraints? I am looking at the following problem. I have a function $f(x)$ with support on $[0, \infty)$. Furthermore, $f(x)$ is bounded between 0 and 1, monotonically increasing and concave everywhere.

Separately, there is a r.v. $X$ with some unknown density $p(x)$ which is also defined on $[0, \infty)$. We know that $E[X]=m$.

I would like to maximize $E[f(X)]$. Clearly it is bounded by $f(m)$ regardless of the distribution of $X$ because of Jensen’s inequality. But can I do better (obtain a lower upper bound) by finding a specific pdf for $X$ which maximizes $E[f(X)]$ subject to the constraints on $X$? Something along the lines of how the exponential distribution has maximum entropy for the class of distributions with support on the positive real line and known mean, except instead of entropy I am maximizing the expectation of $f(x)$. Any ideas or direction would be much appreciated. Thanks!

Wouldn't the distribution that just puts all the mass at $m$ give you $E[f(X)] = f(m)$?  (Or if you don't want to have a mass point, you could come arbitrarily close to this by putting almost all the mass arbitrarily close to $m$.)


How to solve a logarithmic inequality? I mean
 $$\left(\log _{\frac{3 x^2}{2}-\frac{4 x}{3}+\frac{5}{6}}\left(4 x^2+1\right) \log _{\frac{3 x^2}{2}-\frac{4 x}{3}+\frac{5}{6}}\left(1-4 x^2\right)+1\right) \log _{1-16 x^4}\left(\frac{3 x^2}{2}-\frac{4 x}{3}+\frac{5}{6}\right)\geq 1. $$
My (unsuccessful) try was to replace $\log _{1-16 x^4}\left(\frac{3 x^2}{2}-\frac{4 x}{3}+\frac{5}{6}\right) $ by $\frac 1 {\log _{\frac{3 x^2}{2}-\frac{4 x}{3}+\frac{5}{6}}\left(1-16 x^4\right)} $ and to determine the domains of the logarithms. My Mathematica 12.0 is running for hours without any result, solving it.

Let $$f=\frac{3 x^2}{2}-\frac{4 x}{3}+\frac{5}{6},\quad g=4x^2+1,\quad h=1-4x^2$$

Then, your inequality is

$$\left(\log _{f}\left(g\right) \log _{f}\left(h\right)+1\right) \log _{gh}\left(f\right)\geq 1, $$
i.e.
$$\left(\log _{f}\left(g\right) \log _{f}\left(h\right)+1\right)\frac{1}{\log _{f}(gh)}\geq 1\tag1$$

By the way, we have to have
$$f\gt 0,f\not=1,g\gt 0,h\gt 0,gh\not=1,$$
i.e.
$$x\in \left(-\frac{1}{2},-\frac 19\right)\cup \left(-\frac 19,0\right)\cup\left(0,\frac{1}{2}\right)\tag2$$

Now, let us separate it into two cases :

Case 1 : If $\log_f(gh)\gt 0$, then
$$\begin{align}(1)&\iff \log _{f}\left(g\right) \log _{f}\left(h\right)+1\ge \log_f(g)+\log_f(h)\\\\&\iff (\log _{f}\left(g\right)-1)(\log _{f}\left(h\right)-1)\ge 0\tag3\end{align}$$

Here, noting that $gh=1-16x^2\lt 1$ under $(2)$, we have
$$\log_f(gh)\gt 0\iff f\lt 1\iff x\in\left(-\frac 19,0\right)\cup\left(0,\frac{1}{2}\right)\tag4$$

Also, noting that $f-g=-\frac{1}{6}(5x+1)(3x+1)\lt 0$ under $(4)$, we have $\log_f(g)-1\lt 0$.

So, we have
$$(3)\iff \log _{f}\left(h\right)-1\le 0\iff f-h=\frac 16(3x-1)(11x+1)\le 0$$

So, in this case, we have
$$x\in \bigg[-\frac{1}{11},0\bigg)\cup\bigg(0,\frac 13\bigg]$$

Case 2 : If $\log_f(gh)\lt 0$, then
$$(1)\iff (\log _{f}\left(g\right)-1)(\log _{f}\left(h\right)-1)\le 0\tag5$$

Since $gh\lt 1$, we have
$$\log_f(gh)\lt 0\iff f\gt 1\iff x\in\left(-\frac{1}{2},-\frac 19\right)\tag6$$
Also, noting that $f-h=\frac 16(3x-1)(11x+1)\gt 0$ under $(6)$, we have $\log_f(h)-1\lt 0$.

So, we have
$$(5)\iff \log_f(g)-1\ge 0\iff f-g=-\frac{1}{6}(5x+1)(3x+1)\le 0$$

So, in this case, we have
$$x\in \bigg(-\frac{1}{2},-\frac 13\bigg]\cup\bigg[-\frac 15,-\frac 19\bigg)$$

Therefore, the answer is
$$\color{red}{x\in \bigg(-\frac{1}{2},-\frac 13\bigg]\cup\bigg[-\frac 15,-\frac 19\bigg)\cup\bigg[-\frac{1}{11},0\bigg)\cup\bigg(0,\frac 13\bigg]}$$


What is the linear preserving criterion? I have a question regarding the linearity preserving criterion:

From an article I read, is said that the linearity preserving criterion required that the discretization scheme is exact whenever the solution is piecewise linear and the diffusion tensor is piecewise constant.

When writing "exact", does this mean that the scheme is an exact representation of the real world, or is this referring to something else?

When writing that it is exact whenever the solution is piecewise linear, does this mean that the solution gives is piecewise linear, it follows that this solution is exact?

I understant the linearity preserving criterion is very important when we have distorted mesh, but as you can see I have a problem understanding what the linearity preserving criterion actually is.

It means that you are assuming that for a given degree of freedon on your mesh, the diffusion phenomenon behaves linearly. In other words, for a 1D problem, Integral of Grad(F) = 0 will give you a straight line. This will be valid for each degree of freedom, since you are assuming that your physics behaves linearly at that point. Wich is fairly reasonable. If you zoom in the continuum, the solution, if your derivatives are continuous, then you´d see a straight line for a delta x as small as possible.

I hope it will help.