Is this :$\displaystyle\lim_{x\to \infty} f(x) =f(\infty) $ true? I would like to know the difference between limit and image of function. For example, if I define the function by $f(x)= x^2$ , the image of the function by $x=2$ is $f(2)=2^2=4$ and if i would like to calculate $\displaystyle\lim f(x) ,x\to 2$ we w'll obtain $4 =f(2)$ in this case Image has the same mathematical meaning with limit . My humble question here is :

  Question:
  What is the mathematical difference between limit and Image of function and have they the same meaning for $x=\infty$ ?.

Note: I know only they have the same meaning if $f$ is a continuous function .

Thank you for any help

Just like any other value, $\lim_{x \to \infty} f(x) = f(\infty)$ holds precisely when $f$ is continuous at $\infty$.

The use $\infty$ and $-\infty$ in calculus can be made into actual values one can compute with; adding these two points to the ends of the real line gives something we call the "extended real numbers".

We usually, we only bother to define the value of a function at $\infty$ when doing so makes the function continuous — e.g. we define $\arctan(\infty) = \frac{\pi}{2}$ and $\arctan(-\infty) = -\frac{\pi}{2}$, but we leave $\sin(\infty)$ undefined.

In other words, the limit at and the value at $\infty$ always the same is the result of convention. If we don't stick to that convention, then we have to be prepared for situations where $f(\infty)$ exists, but $\lim_{x \to \infty} f(x)$ has a different value, or doesn't exist at all.

This is almost correct if interpreted in the context of a hyperreal extension $\mathbb R\hookrightarrow{}^\ast\!\mathbb R$.  Namely, if $H$ denotes an infinite hyperreal then the limit $\lim_{x\to\infty}f(x)$ will exist if and only if for every $H$ the value of $f(H)$ is infinitely close to a suitable real $L$ (independent of the choice of $H$).  Then
$$
\lim_{x\to\infty}f(x)\approx f(H)
$$ where $\approx$ stands for the relation of infinite proximity.


Show that $H \leq C_G(C_G(H))$. Show that $H \leq C_G(C_G(H))$.

We have that $C_G(H) = \left\{ g \in G | gh = hg \; \forall h \in H \right\}$. So then $C_G(C_G(H))$ is the $g$'s that commute with all $g$'s that commute with $h$. This would mean $$C_G(C_G(H)) = \left\{ g \in G | gg'h = hg'g, \; \forall g' \in G, \; \forall h \in H \right\}$$
But since $gg' \in G$ this is just some element of $G$, so doesn't this become the same as $C_G(H)$?

I feel like what I have is wrong but I am not sure why.

Pick $h\in H$. We wish to show that $hg=gh$ for any $g\in C_G(H)$. But this is tautological from the definition of $C_G(H)$, so $H\subset C_G(C_G(H))$.

in the definition $ C_G(C_G(H)) = \left\{ g \in G | gg'h = hg'g, \; \forall g' \in G, \; \forall h \in H \right\}$, is not $\forall g' \in G$, but $\forall g' \in C_G(H)$.

I believe that the question is wrong, because:

If $h_1\in C_G(C_G(H))$ and $h_1 \in H$, we have $$ h_1g'h_2=h_2g'h_1  ,\forall g' \in C_G(H), \forall h_2 \in H  $$
It's clear that the identity $e \in C_G(H)$, so $$h_1h_2=h_2h_1, \forall h_2 \in H $$, then $H \subset C_G(C_G(H))$ only if H is abelian. But $H \leq C_G(C_G(H))$ also can indicate relation of order.


Are these infinite groups decomposable? I have been asked to:

Decide whether the following groups are decomposable:
(a) - $(\mathbb{R^*}, \cdot)$
(b) - $(\mathbb{C}, +)$
(c) - $(\mathbb{Q^*}, \cdot)$
(d) - $(\mathbb{Q}, +)$

I would like a hint for item (a). I believe I was able to do itens (b), (c) and (d).
Regarding item (a), I tried to decompose $\mathbb{R^*}$ in rationals and irrationals (but this failed, since the irrationals are not a subgroup) or into algebraic and transcendental numbers (which also fails, since the transcendental numbers are not a subgroup). I also thought about showing that if $\mathbb{R^*} = A \times B$ then $A$ and $B$ do not intersect trivially (thus showing that the group is indecomposable), but I couldn't prove this idea.
Regarding item (b), I decomposed $\mathbb{C}$ into $\mathbb{R}$ and
$i\mathbb{R} = \{iy \ | \ y \in \mathbb{R} \} $.
Regarding item (c), I wrote that $\mathbb{Q^*} = \langle \ p \ | \ p \  \text{is a prime} \rangle = \langle 2 \rangle \ \oplus \ \langle \ p \ | \ p \  \text{is an odd prime} \rangle $.
EDIT: As pointed in the comments, this decomposition is for the multiplicative group of positive rational numbers. A correct decomposition would be, for instance, $\mathbb{Q^*} = \langle 2, -1 \rangle \ \oplus \ \langle \ p \ | \ p \  \text{is an odd prime} \rangle $.
Regarding item (d), I proved that the group is indecomposable by proving that two non-trivial subgroups don't intersect trivially. My reasoning was the same as in: Why is the additive group of rational numbers indecomposable?.
Can anyone give me a hint for item (a)? Thanks in advance.

For item (a), you can decompose where one factor is the sign and the other is the absolute value. Notice that you probably should do (c) the same way. Your decomposition is for the multiplicative group of positive rational numbers.

It turns out $(\mathbb R, +) $ is decomposable as well, but it's not quite as easy to see this.

$(\mathbb R^*, \cdot) = \left(\mathbb R^+, \cdot\right)\times \{1, -1\}$, because (i) $\left(\mathbb R^+, \cdot\right)$ and $\{1, -1\}$ are clearly normal subgroups of $(\mathbb R^*, \cdot)$, (ii) common element of $\left(\mathbb R^+ , \cdot\right)$ and $\{1, -1\}$ is the identity $1$ only, and clearly (iii) $\mathbb R^* = \left(\mathbb R^+\right)\{1, -1\}$. Here $\mathbb R^*$ is the set of all non-zero real numbers and $\mathbb R^+$ is the set of all positive real numbers.


Does markovian property imply independence? A stochastic process has the Markov property if the conditional probability distribution of future states of the process (conditional on both past and present states) depends only upon the present state, not on the sequence of events that preceded it. A process with this property is called a Markov process.

Is the google definition.

My book defines:

$Pr(X_{i+1}<x | X_i = x_i,X_{i-1}=x_{i-1},...,X_0=x+0) = Pr(X_{i+1}<x | X_i = x_i)$

Does this mean that:

X_{i+1} is independent of $X_{i-1}$ etc.
or
does this mean that $X_{i}$ contains all the information of $X_{i-1}$,$X_{i-2}$ etc?

$$\mathsf P(X_{i+1}<x \mid X_i=x_i,X_{i-1}=x_{i-1},\ldots, X_0=x_0) ~=~\mathsf P(X_{i+1}<x\mid X_i=x_i)$$

For events $F,C,P$, the expression $\Pr(F\mid C, P)=\Pr(F\mid C)$ means that $F,P$ are conditionally independent given $C$.

In this example you have events for the future state, current state, and all prior states:

$F = \{X_{i+1}<x\}$
$C = \{X_i=x_i\}$
$P = \{X_{i-1}=x_{i-1},\ldots, X_0=x_0\}$

So, it follows from the Markov property: an event for a furture state is conditionally independent of values for all prior states when given a value for the current state.

The Google definition states that the probability that the $i+1$ random variable has a value less than some $x$ given all the previous values is equal to the probability given only the single previous value.


Number of 3 element subsets of ${1,2,...,n}$ with a certain property. Given the set $\{1,2,\dots,n\}$, I want to count the number of three-element subsets $\{x,y,z\}$ such that $|x-y|=1$ or $|x-z| = 1$ or $|y-z|=1$. I'm not familiar with how to deal with the "inclusive or" in a counting problem.

As someone pointed out, we can equivalently count the number of triples $\{x,y,z\}$ such that $|x-y|>1$ and $|x-z|>1$ and $|y-z|>1$.

You need a pair of successive integers and an arbitrary one. And then subtract those three integers that have two successive ones. That likes Inc-Exc Principle. You have $n-1$ choices for successive ones and then have $n-2$ choices for that arbitrary one. For subtracting, we know there successive integers have two successive ones that there is $n-2$ ones. So the answer is:
$$(n-1)(n-2) - (n-2) = (n-2)^2$$

I think your first have to built a two elements subset of two consecutive integers. There is $n - 1$ ways to do. Then, you chose the last element between the $n - 2$ elements that remain. Finally there is $(n - 1)(n - 2)$ such subsets.


How many equivalence relations $S$ on $A$ are there for which $R⊆S$ ($R$ is an equivalence relation on a set $A$, with $4$ equivalence classes) Suppose $R$ is an equivalence relation on a set $A$, with four equivalence classes.
How many different equivalence relations $S$ on $A$ are there for which $R⊆S$?
Thanks in advance

Let's imagine $R$ is an equivalence relation on $A$ with four equivalence classes exactly. But for $R$ to have exactly four equivalence classes then $A$ must be a four-element set. For example

$A=\{a,b,c,d\}$

and $R=\{(a,a),(b,b),(c,c),(d,d)\}$

Then the four equivalence classes of $R$ are

$$[a] = \{a\}$$
$$[b] = \{b\}$$
$$[c] = \{c\}$$
$$[d] = \{d\}$$

Now, the reason your book says there are 15 total equivalence relations (which is the correct answer) is because $R$ is a subset of all possible equivalence relations (since each equivalence relation must be reflexive and will have all ordered pairs found in $R$).

It's easy to see that
$R=\{(a,a),(b,b),(c,c),(d,d)\}$

and that $R$ a subset of $R_1$ through $R_{15}$ below which is the total number of equivalence relations $S$ on $A$.

Thus you could list out all $15$ equivalence relations:

$$R_1=\{(a,a),(b,b),(c,c),(d,d),(a,b),(b,a)\}\\
R_1=\{(a,a),(b,b),(c,c),(d,d),(a,c),(c,a)\}\\
R_2=\{(a,a),(b,b),(c,c),(d,d),(a,d),(d,a)\}\\
R_3=\{(a,a),(b,b),(c,c),(d,d),(b,c),(c,b)\}\\
R_4=\{(a,a),(b,b),(c,c),(d,d),(b,d),(d,b)\}\\
R_5=\{(a,a),(b,b),(c,c),(d,d),(c,d),(d,c)\}\\
R_6=\{(a,a),(b,b),(c,c),(d,d),(a,b),(b,a),(a,c),(c,a),(b,c),(c,b)\}\\
\vdots\\
R_{15}\{(a,a),(b,b),(c,c),(d,d),(a,b),(b,a),(a,c),(c,a),(b,c),(c,b),\cdots,(d,a),(a,d)\}$$

But this would take a long time!

It is easier to recognize that the total number of equivalence relations possible on a set is equal to the number of possible equivalence classes on a set. Knowing the number of equivalence classes can be determined by using Bell numbers. https://en.wikipedia.org/wiki/Bell_number

Since this is the bell number $B_{4}=15$ then $15$ is the answer.

Since R is a subset of S the equivalence classes can be the same as R or we can merge a few together. We can have S=R (1 case) or two equivalence classes merged(6 cases), or two merged and another  two merged (6 cases) or three merged(4 cases) or all of them merged together (1 case).


Show that no group of order 48 is simple Show that no group of order 48 is simple

I was wondering if I was allowed to do something along this line of thinking:

Let $n_2$ be the number of $2$-Sylow groups.

$n_2$ is limited to $1$ and $3$ since these are the only divisors of 48 that are equivalent to $1 \mod 2$.

$n_2=3$ (since if $n_2=1$ the group is definitely not simple)

Each $n_2$ subgroup contains 1 distinct element and there are 3 of these subgroups hence there are 3 distinct elements.

We have $48-3=45$ elements to account for.

At this point, can I assume that these 45 elements form a subgroup and then solve this proof by proving a group of 45 elements form a p-Sylow normal subgroup?

A Sylow $2$-subgroup contains fifteen non-identity elements. Two distinct
Sylow $2$-subgroups could conceivably meet in an order $8$ subgroup, so it
is difficult to count how many elements are contained in the union of the Sylow
$2$-subgroups.

But there are either $1$ or $3$ Sylow $2$-subgroups. In the latter case
$G$ acts on them transitively by conjugation, so there is a non-trivial
homomorphism from $G$ to $S_3$.

I obviously did not understand what a $2$-Sylow subgroup encompasses. I've attached my corrected answer, comments are welcome. It's a bit different from the answer above but it's the same gist I think.

$48=3 \times 2^4$

$n_2=1,3$ since $1, 3$ are the only divisors of $48$ which are equivalent to $1\mod 2$.

Consider the $2$-Sylow subgroup.

$n_2\neq 1$ or else the $n_2$ subgroup will be normal to itself so $n_2=3$.

Each $2$-Sylow subgroup has $2^4$ elements and the intersection of these $2$-Sylow subgroups (assuming they're unique) is $\{e\}$ the identity.

So there are $15$ distinct element in each $2$-Sylow subgroup.

$3 \times 15=45$.

So there are $3$ distinct elements not accounted for but one of them is $e$, the identity, so there are actually only $2$ distinct elements not accounted for.

Consider the $3$-Sylow subgroup.

$n_3$ must be $1$ since there are only $3$ elements remaining, one of which is $e$.

So the $3$-Sylow subgroup conjugates to itself so it must be normal.


Limit $\frac{0}{0}$ which tends to $\frac{\pi}{2}$ I'm trying to evaluate the following limit:
$$\lim_{x\rightarrow\pi/2}\frac{\cos(x)}{(1-\sin(x))^{2/3}}$$
The limit has the form $\frac{0}{0}$, I've tried using L'Hopital's rule but I can't resolve it.
Any idea?

For a simple approach, we can use the following identities:

$1 - \sin 2x = (\cos x - \sin x)^2$

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

Be careful when you consider the limits on either side.


How to factor $a^2+3b^2+4ab+2ac+6bc-4b+4c-4$? Can someone please explain in the simplest language they can manage strategies (best if one that applies to most polynomial factorization) to complete the formula above? And how in general to improve factorizing? I'm only fourteen with limited knowledge (duh) so I might not understand what you're saying but say it anyways cause then at least I'll know where to look/ get the gist of it. Thanks in advance!

Usually it amounts to noticing differences of squares and such consider this
$$a^2+4ab+3b^2=(a+2b)^2-b^2\\(a+2b)^2-b^2+2ac+6bc-4b+4c-4\\-b^2-4b-4=-(b+2)^2\\(a+2b)^2-(b+2)^2+2ac+6bc+4c\\2ac+6bc+4c=2c(a+3b+2)\\(a+2b)^2-(b+2)^2+2c(a+3b+2)=\\(a+2b+b+2)(a+2b-b-2)+2c(a+3b+2)=\\(a+3b+2)(a+b-2)+2c(a+3b+2)=\\(a+3b+2)(a+b-2+2c)$$
Though this amounts to trial and error,usually completing to square and noticing well known identities.


what does "a set of sets that are not members of themselves" of Russell’s Paradox mean Russell’s Paradox begins with a statement of "Let $R$ be the set of sets that are not members of themselves", i.e. $R=\{S\mid S\notin S\}$.

I'm a little bit confused  with the statement, for example, let  $S=\{1,\{2,3\}\}$, of course $S\notin S$ since $S$ doesn't have an element $\{1,\{2,3\}\}$, $S$ only have two elements which are $1$ and $\{2,3\}$, So $R$ only have one element i.e. $R=\{\{1,\{2,3\}\}\}$, and again, of course $R\notin R$, I must have something wrong here but I don't know where I go wrong.

Sets are collections of mathematical objects which are themselves mathematical objects. That means that in principle, a set can be a member of itself. We can therefore ask which sets are not members of themselves, and that is a valid question from a mathematical point of view.

Russell's paradox shows us that the collection of sets which are not members of themselves is not a set. It is a collection we can define, but it cannot be a set. And that was the main purpose of the paradox, to dispel the notion that every collection we can define forms a set.

Your example is not good, though, because the paradox applies to the entire mathematical universe. It encompasses all the sets out there. Not just that specific $S$.

Formally speaking, modern set theory has an axiom which prevents a set from being a member of itself. But it is possible to replace that axiom by one of several axioms which guarantee the existence of sets which are members of themselves (e.g. $x=\{x\}$, and more), and these new axioms do not introduce new contradictions to mathematics.

The problem arises from the late 1800's when mathematicians were trying to formalize the notion of what "set" means, what are numbers, what is logic? Georg Cantor proposed Naive set theory where "a set is described as a well-defined collection of objects. These objects are called the elements or members of the set. Objects can be anything: numbers, people, other sets, etc."

Anything that could be described in words was a set, for example the Universal Set, U, meaning the set that contains ALL sets. Since U is a set it must also be a member of the Universal Set so $U \in U$. That was a bit disturbing, even pathological but hey, we can describe it so U must be a set in Naive Set Theory. And since U contains itself, there might be other sets that also that contain themselves.

In 1901 Bertrand Russell proposed that we look at the set $A$ that contains only the "non-pathological" sets, $A = \{ S : S \notin S\}$ That is "the set of all sets that don't contain themselves." $A$ would be a great set to base mathematics upon except for one problem, is $A \in A$ ? If it is, then by the definition of $A$, $A \notin A$. We also get a contradiction if we assume $A \notin A$. So $A$, and by extension, Naive Set Theory, is logically inconsistent.

Because of Russell's Paradox, mathematicians realized they needed a new theory to restrict what a "formal" "set" means in such a way to exclude $U$ and $A$ from being mathematical "sets". Of the several theories created, the most popular today is Zermelo-Fraenkel set theory or ZF. But that's a topic for a different question.


Proof that $\lim_{x\to0}\frac{\sin x}x=1$ Is there any way to prove that
$$\lim_{x\to0}\frac{\sin x}x=1$$
only multiplying both numerator and denominator by some expression?
I know how to find this limit using derivatives, L'Hopital's rule, Taylor series and inequalities. The reason I tried to find it using only multiplying both numerator and denominator and then canceling out indeterminate terms is because the most other limits can be solved using this method.This is an example:
$$\begin{align}\lim_{x\to1}\frac{\sqrt{x+3}-2}{x^2-1}=&\lim_{x\to1}\frac{x+3-4}{\left(x^2-1\right)\left(\sqrt{x+3}+2\right)}\\=&\lim_{x\to1}\frac{x-1}{(x+1)(x-1)\left(\sqrt{x+3}+2\right)}\\=&\lim_{x\to1}\frac{1}{(x+1)\left(\sqrt{x+3}+2\right)}\\=&\frac{1}{(1+1)\left(\sqrt{1+3}+2\right)}\\=&\frac18\end{align}$$
It is obvious that we firtly multiplied numerator and denominator by $\sqrt{x+3}+2$ and then canceled out $x-1$. So, in this example, we can avoid indeterminate form multiplying numerator and denominator by
$$\frac{\sqrt{x+3}+2}{x-1}$$
My question is can we do the same thing with $\frac{\sin x}x$ at $x\to0$? I tried many times, but I failed every time. I searched on the internet for something like this, but the only thing I found is geometrical approach and proof using inequalities and derivatives.
Edit
I have read this question before asking my own. The reason is because in contrast of that question, I do not want to prove the limit using geometrical way or inequalities.

If I remember correctly, it is not possible to solve this limit algebraically, as therefore the Numerator $f(x)=\sin{x}$ and the Denominator $g(x)=\frac{1}x$ must have both an existing limit, so one can use the rule
$$\lim_{x\to x_0}\frac{f(x)}{g(x)}=\frac{c}{d}$$

The right and lefthand side limits of the Denominator $g(x)=\frac{1}x$ are
$$\lim_{x\to 0+}\frac{1}x=\infty$$ and
$$\lim_{x\to 0-}\frac{1}x=-\infty$$

So you cannot solve this limit algebraically, as the limit is not defined at $x=0$.

Hope this helps!


Prove that a directed graph with no cycles has at least one node of indegree zero How would I show this? I know a directed graph with no cycles has at least one node of outdegree zero (because a graph where every node has outdegree one contains a cycle), but do not know where to go from here.

Suppose that there exists a graph with no cycles and there are no nodes of indegree $0$. Then each node has indegree $1$ or higher. Pick any node, since its indegree is $1$ or higher we can go to its parent node. This node has also indegree $1$ or higher and so we can keep doing this procedure until we arrive at the node we already visited. This will prove that there exists a cycle which contradicts our initial assumption. So we proved that every directed graph with no cycles has at least one node of indegree zero.

This proof will prove that DAGs (Directed Acyclic Graphs) have at least one node of indegree $ 0 $ and one node of outdegree $ 0 $ as well:
A DAG will contain all paths of finite length (Because of absence of cycles). Let's consider WLOG, the longest path $ P $ of the graph in consideration from a vertex $ u $ (our source) to vertex $ v $ (our destination). Since this path is the longest path from $ u $ to $ v $, there would not be any incoming edge on $ u $ from any other vertex $ u ' $ (if that had been the case, then our longest path would have started from $ u ' $ itself or any other vertex but not $ u $ for sure) and similarly there will not be any outgoing edge from $ v $ to any vertex $ v ' $ (if that had been the case, then our longest path should not have ended at $ v $ but at $ v ' $ or some other vertex, but definitely not $ v $). So this proves that $ u $ has indegree $ 0 $ and $ v $ has outdegree $ 0 $ and this proof includes your answer.


remainder question for big number Every 2 digit positive integer is repeated 5 times and strung together (ex. 1818181818). Find the remainder when all of these integers are added together and divided by 505

Not sure best way to do it -

If you repeat the two digit number $n=10a+b$ (ie digits $a,b$) five times you get $ababababab$, which is $n\times 101010101$. It is easy to check that $101010101=1\bmod505$. So the sum of all these 10 digit numbers mod 505 is the same as the sum of all two digit numbers mod 505.

$10+\dots+99$ is an AP with sum $45\times109$ (as usual pair the numbers $10+99=109,11+98=109$ etc and multiply by half the number of terms).

It is easy to check that $45\times109=360\bmod505$.

A mechanistic way to write your initial sum is
$$
S = \sum_{n=10}^{99} \sum_{k=0}^{4} n10^k
$$
Then your question reduces to
$$
S \equiv x \mod{505}
$$

Can you solve this relationship?


The space $x^3-y^2=0$ Consider $\{(x,y)\in\mathbf{R}^2 \ | \ x^3-y^2=0\}$ as a subspace of $\mathbf{R}^2$. Intuitvely I understand that this is not supposed to be a differentiable manifold because it has a cusp at $0$. But how to prove this formally, is there a clean way to do this? Is this even a topological manifold? Thanks in advance.

a) Yes your curve $C=\{(x,y)\in\mathbb R^2 \ | \ x^3-y^2=0\}$ is a topological manifold because it is homeomorphic to $\mathbb R$ by the homeomorphism $\mathbb R\to C :t\mapsto (t^2,t^3) $, whose inverse is $C\to \mathbb R: (x,y)\mapsto \sqrt [3] y$.

b) No, $C$ is not a differential submanifold of $\mathbb R^2$:

Consider any  differentiable map $\gamma:\mathbb R\to C:t\mapsto (u(t),v(t))$ with $\gamma(0)=(0,0)$ and of course $$u^3-v^2=0 \;(\bigstar )$$  Since $u$ has a minimum at $t=0$ (all points of $C$ satisfy $x\geq0$ !) we have $u'(0)=0$.
On the other hand since $u(0)=0, v(0)=0$ we can write $u=tU, v=tV$  with $U,V$ differentiable and satisfying $u'(0)=U(0), v'(0)=V(0)$.
Substituting into $(\bigstar)$ and cancelling $t^2$ we  see that $V^2-tU^3=0$, so that $V^2(0)=0$, hence  $V(0)=v'(0)=0$ and thus finally $\gamma'(0)=(u'(0),v'(0))=(0,0)$.
This proves that $C$ is not a manifold because if it were we could find such a differentiable  $\gamma$ with $\gamma(0)=(0,0)$ but $\gamma'(0)\neq (0,0)$.

The map $(x,y)\mapsto (x^3-y^2,y)$ is a homeomorphism of $\mathbb{R}^2$ with inverse $(a,b)\mapsto ( (a+b^2)^{\frac{1}{3}}, b)$ and takes $C$ to the vertical $a=0$, so $C$ is a topological manifold.
We'll give a proof that $C$ does not have a regular parametrization around $(0,0)$ by a $C^1$ function ($(0,0)$ is not a regular point of $C$). First notice that if $O$ is a regular point on a curve then for $P$, $Q$ near $O$ we have $|P,Q| \simeq \text{length}(P,Q)$, the length of the curve from $P$ to $Q$, that is
$$\frac{PQ}{\text{length}PQ}\to 1$$
as $P$, $Q$ approach $O$ on the curve.
. Now take two points $P_1$, $P_2$ on $C$, close to $O=(0,0)$ and symmetric. We have

$$\frac{P_1P_2}{\text{length}(P_1,P_2)}=\frac{P_1P_2}{\text{length}(P_1,O)+\text{length}(O,P_2) }\le \frac{P_1P_2}{P_1O+OP_2}\to 0 $$
as $P_1$, $P_2 \to O = (0,0)$ (look at the cusp). Therefore $O$ is not a regular point of $C$.

In fact, for regular points $O$ and $P_1$, $P_2$ on both sides of $O$ and close to $O$, the points $P_1$, $O$, $P_2$ are almost in line, that is $\frac{P_1P_2}{P_1 O + OP_2} \to 1$, and that is  not true for cusps.

Or, for regular points $O$ and $P_1$, $P_2$ on both sides of $O$ and close to $O$, the directions of the vectors $P_2 0, OP_1$ are almost the same. For this curve they become almost opposite.


Are there natural numbers that are not the descendant of 0? Based on the Peano Axioms (wich are a way to correctly absolutely define the set of natural numbers - correct me if i'm wrong) it is possible to construct a set of symbols that doesn't quite look the way i imagine the natural numbers:
If there is a circle of other symbols next to the infinite row of known natural numbers, doesn't this also fit all the requirements?
So are there multiple unequal sets of natural numbers?

The induction axiom ensures that $\Bbb N$ cannot contain a cycle like your $a,b,c$ cycle. It says that if

$0\in A$, and
for each $n\in\Bbb N$, $n\in A$ implies that $n+1\in A$,

then $A=\Bbb N$. Take $A$ to be everything in your diagram except $a,b$, and $c$; this $A$ satisfies both of these requirements, yet it’s not the whole set shown in your diagram. Thus, the set in your diagram doesn’t satisfy the Peano axioms, and indeed they characterize $\Bbb N$.

However, the induction axiom cannot be expressed in first-order logic, and there are structures other than $\Bbb N$ that satisfy the first-order counterpart of the Peano axioms, though they still don’t contain cycles. All of them are linearly ordered and consist of a copy of the standard $\Bbb N$ followed by copies of $\Bbb Z$ (so that everything except $0$ has a unique immediate predecessor). There are restrictions on how these copies of $\Bbb Z$ can be ordered relative to one another. For instance, the only possibility for a countable non-standard model looks like $\Bbb N$ followed by $\Bbb Q\times\Bbb Z$ ordered lexicographically.

Going by wikipedia.  https://en.wikipedia.org/wiki/Peano_axioms

The first 8 axioms (particularly 8): For every natural number n, S(n) = 0 is false. That is, there is no natural number whose successor is 0.) seem to assure that the is the "number line" 0->1->2->3.... and that it is infinite and not circular.  But they don't say any thing about there not being any a->b->c->d.... or oom<->pah. which are completely distinct and separate from this set of number lines.

They do state that they must be distinct though.  If S(S(S(....(a)))) = a then a $\ne $ S(S(.....(0))))).  Or if b $\not \in$ 0->1->2->3.... then no predecessor or successor of b is either.

But 9) the induction axiom does imply there are not distinct threads.

If K is a set such that:

0 is in K, and
for every natural number n, n being in K implies that S(n) is in K,

then K contains every natural number.

As K = 0->1->2->3 ... is such a set we have $\mathbb N \subset K \subset \mathbb N$ and 0->1->2->3.... are precisely the natural numbers and nothing else are.

This is not just theoretical.  1/2 ->1 1/2 -> 2 1/2 -> 3 1/2 $\not \subset \mathbb N$ is just such a separate string that does exist but are not the natural numbers.

I like to think (and I'm probably abusing some theory and I could be wrong) that 1-8 define the smallest thing the natural numbers can be.  And the induction axiom restricts it to the smallest it can be.  I may be naively abusing math though.


Functional Equation Problem: Strictly Increasing $ f $ satisfying $ f \big( f ( n ) \big) = n + 2 $ If $ \mathbb N $ denotes all positive integers, then find all functions $ f : \mathbb N \to \mathbb N $ which are strictly increasing and such for all positive integers $ n $, we have:
$$ f \big( f ( n ) \big) = n + 2 $$

So far I know that $f(n)$ is greater than $n$ because the function is strictly increasing, but I'm not sure how to use this in order to solve the equation.

As kindly explained below strictly increasing means $f(a) > f(b)$ for $a >b.$
Let's consider $f(0)$.

$f(0) = 0$ is not possible because then $f^2(0) = 0 \neq 2.$
$f(0) = 1$ is possible.
$f(0) = m > 1$ is not possible, since $f^2(0) = f(m),$ and since $f$ is strictly increasing we have $f(m) > f(0) > 1$ which implies $f(m) \ge m + 2 > 2$ in contradiction with the assumption $f^2(0) = 2.$

Hence $f(0) = 1.$ Now suppose there is a $k$ such that $f(k) = k+1.$ What can you say about $f(k+1)$?

Hint: Set
$$f(n)=an + b$$
Viz,
\begin{align}
f(f(n)) &= a(an+b)+b \\
&= a^2n+ b(a+1)
\end{align}
Thus $a= \pm 1$ and
$$
 b(a+1) = \begin{cases}
      0, & \text{if}\ a=-1 \\
      2b, & \text{if}\ a= +1
    \end{cases}
$$


Empty Set $\{\}$ is the Only Basis of the Zero Vector Space $\{0\}$ Question

Suppose we want to find a basis for the vector space $\{0\}$.

I know that the answer is that the only basis is the empty set.

Is this answer a definition itself or it is a result of the definitions for linearly independent/dependent sets and Spanning/Generating sets? If it is a result then would you mind mentioning the definitions of bold items which based on them this answer can be deduced.

Useful Links

These are the links that I found useful for answering this question. It needs some elementary background form mathematical logic. You can learn it by spending a few hours on this Wikipedia page.

Link 1, Link 2, Link 3, Link 4, Link 5, Link 6

The standard definition of basis in vector spaces is:

$\mathcal B$ is a basis of a space $X$ if:

$\mathcal B$ is linearly independent.
The span of $\mathcal B$ is $X$.

You can easily show both of these statements are true when $X=\{0\}$ and $\mathcal B= \{\}$. Again, you have to look at the definitions:

Is $\{\}$ linearly independent? Well, a set $A$ is linearly independent if, for every nonempty finite subset $\{a_1,a_2\dots, a_n\}$, we have that if $$\alpha_1a_1 + \dots + \alpha_n a_n=0,$$ then $\alpha_i=0$ for all $i$. This condition is satisfied automaticall in the case of an empty set (everything follows from a false statement). This part may be difficult to understand, but since there is no nonempty finite collection of vectors from $\{\}$, any statement you say about nonempty finite collections of vectors from $\{\}$ must be true (because any such statement includes an assumption that a nonempty finite collection exists. It does not, meaning that any such statement is of the type $F\to A$ and is automatically true). This means $\{\}$ is linearly independent.
Is the span of $\{\}$ equal to $\{0\}$? Well, the span of a set $A\subseteq X$ is defined as the smallest vector subspace of $X$ that contains $A$. Since all vector subspaces contain $\{\}$, it is clear that $\{0\}$, which is the smallest vector subspace at all, must be the span of $\{\}$.

Alternatively, the span of $A$ is the intersection of all vector subspaces that contain $A$. Again, it should be obvious that this implies that the span of $\{\}$ is $\{0\}$.

To answer the question of why $\text{Span}\{\}=\{0\}$ is true, I considered the following argument for myself.

I think all the things draw back to the operation of addition.

Addition is a map defined as follows

$$
\begin{align}
(\cdot+\cdot): & V \times V \to V \\
& (u,v) \mapsto (u+v)
\end{align}$$

with the commutative and associative properties

$$
\begin{align}
(u+v) &= (v+u) \\
((u+v)+w) &= (u+(v+w))
\end{align}
$$

so according to this definition, whenever we are talking about addition we should provide two inputs to get one output.

From a programming point of view it is useful to have outputs in case when we have one or no inputs (see the detail of Plus function in wolfram language). It also turns out to be useful in proofs like the ones using induction. So what is the most useful definitions to make for such cases? Experience shows that these are

$$
\begin{align}
(u+\text{null}) &= u\\
(\text{null}+u) &= u \\
(\text{null}+\text{null}) &= 0
\end{align}
\tag{1}$$

where you can think of null meaning that no argument is provided!

Now the following definition can be easily interpreted in the special cases when we have a set with one element or no element.

  Linear Combination and Span.
  A linear combination of a set $A=\{v_1,v_2,...,v_m\} \subseteq V$ is a vector $v$ defined by $v=\sum_{j=1}^{m}a_jv_j$. The set of all linear combinations of $A$ is called the span of $A$ denoted by $\text{Span}A$.

Now, if we make the convention that $m=0$ means $A=\{\}$ and when $m=1$ then $A=\{v_1\}$, according to $(1)$, we can interpret the definition as follows

$$
v=\sum_{j=1}^{m}a_jv_j:=s_m, \qquad
s_i =
\begin{cases}
0,  & i=0 \\
(a_1v_1+\text{null}),  & i=1 \\
(a_1v_1+a_2v_2),  & i=2 \\
(s_{i-1}+a_{i}v_{i}), & \text{otherwise} \\
\end{cases}
, \qquad 0 \le i \le m
$$

So, we can see that $\text{Span}\{\}=\{0\}$. Note that this is a result of our own convention for the addition operation and the definition of the span. I think that the vacuous truth argument has no advantage over this one! However, it keeps repeating in many other examples! So it is good to learn it once and for all!


Prove or refute that $\log_5(0.7-2^x)=\log_2(0.7-5^x)$ if and only if $x=-1$ I have been trying to prove that the following two functions intersect at a single point $(-1,-1)$:

$f(x)=\log_5(0.7-2^x)$
$g(x)=\log_2(0.7-5^x)$

It is sufficient to prove that:

If $x>-1$, then $g(x)>f(x)$
If $x<-1$, then $g(x)<f(x)$

I believe it should be relatively easy to prove that they don't have more than two intersection points (though I fail to see how this might help in proving that they only have a single intersection point).

Can anyone help proving or refuting that?

Here is the graph of both functions:

It is sufficient to prove that $f'<g'$:

\begin{align}f'(x)<g'(x)&\iff-\frac{2^x}{.7-2^x}\frac{\log2}{\log5}<-\frac{5^x}{.7-5^x}\frac{\log5}{\log2}\\
&\iff\frac{2^x}{.7-2^x}\frac{.7-5^x}{5^x}>\frac{\log^25}{\log^22}=:k\\
&\iff\frac{.7\cdot5^{-x}-1}{.7\cdot2^{-x}-1}>k\\
&\iff(.7\cdot5^{-x}-1)>(.7\cdot2^{-x}-1)k\\
&\iff2^{-x}k-5^{-x}<\frac{10}7(k-1)\\
\end{align}

Thus we are showing that $h(x):=2^{-x}k-5^{-x}$ has a particular upper bound. Since it is a difference of exponentials, we already know it has one root and is upper bounded. The derivative $h'(x)=2^{-x}k\log 2-5^{-x}\log 5$, which is zero when $x=x^\circ:=-\log_{5/2}\log_25$. Then $$h(x)\le 2^{-x^\circ}k-5^{-x^\circ}\approx5.805<6.273\approx\frac{10}7(k-1).$$

(Sorry I had to resort to numerical evaluations, but this is a rigorous proof, since with a good enough evaluation of the two constants you can bound one above the other.)

Let $f(x) = \log_5(0.7-2^x)-\log_2(0.7-5^x)$; $f(x)$ is continuous and differentiable for $x<\log_2(0.7)\approx-0.51$, so you can apply Rolle's Theorem. By absurd, suppose there exist two numbers $x_1$ and $x_2$, such that $F(x_1)=f(x_2)=0$, then by the Rolle's Theorem, there exists a number $c$ in $]x_1;x_2[$, such that $f'(c)=\frac{2^c\ln(2)}{(2^c-0.7)\ln(5)}-\frac{5^c\ln(5)}{(5^c-0.7)\ln(2)}=0 \Rightarrow \frac{2^c\ln(2)}{(2^c-0.7)\ln(5)}=\frac{5^c\ln(5)}{(5^c-0.7)\ln(2)}$ but this equation has no real solution! Absurd conclusion $\to$ there is only 1 solution $(x=-1)$.


$(f(1 - f(x)) = 1 - x^9$, $f(1) = 0$ and $f'(1) < 0$, then where is the real number $r$ such that $f(r) = r^{99}$? If $f(1 - f(x)) = 1 - x^9$, $f$: R $\to$ R is differentiable, $f(1) = 0$ and $f'(1) < 0$, how to show there is a real number $r$ such that $$f(r) = r^{99}?$$

Edit: Taylor Theorem makes no use. I try to take $a = 1 - {1 \over n}$, where n is also a real no. Then as $f'(1) < 0$, $f$ is decreasing at $x = 1$ and $f(a) > f(1) = 0$. By mean-value theorem, $$f(a) - f(1) = f'(a_0)(a - 1), a < a_0 < 1$$

Let $g(x) = f(x) - x^{99}$, then clearly $g(1) < 0$ and $g(r) = 0$.

How about showing $g(a) > 0$? Or to take $a$ as something else?

After almost 4 years I finally think of these. Let me know anywhere wrong.

Firstly if $f(1 - f(x)) = 1 - x^9$, then let $a \in \mathbb{R}$ so that $f(a) = 1$, hence \begin{align} f(0) = f(1 - f(a)) = 1 - a^9. (1) \end{align}

Also we rearrange the terms in expressing $f(1 - f(x))$ to be $f(x) = 1 - f^{-1}(1 - x^9)$, then \begin{align} f(0) = 1 - f^{-1}(1 - 0) = 1 - a. (2) \end{align}

Combining $(1)$ & $(2)$, we have $a^9 - a = 0$, which $a = 1$ (rejected, or $f(1)$ has $2$ values), $0$ or $-1$ (rejected).

If $a = -1$, i.e. $f(-1) = 1$, then $f(1 - f(-1)) = f(0) = 1 - (-1)^9 = 2$. If we let $g(x)$ as I tried above, $g(0) = 2 > 0$. However, by differentiating $f(1 - f(x))$ against $x$, we have $f'(1 - f(x))\cdot -f'(x) = -9x^8$.

\begin{align} \Rightarrow f'(1 - f(-1)) \cdot -f'(-1) = f'(0)f'(-1) &= -9; \\ f'(1 - f(0)) \cdot -f'(0) = f'(-1)f'(0) &= 0 \end{align}

produces contradiction, so we have to reject this possibility.

If $a = 0$, i.e. $f(0) = 1$, then $g(0) = 1 > 0$ as well without the confusion on (the product of) the derivatives. Now the mean-value theorem becomes useful that

differentiabilty of $f$ (so does of $g$) guarantees continuity on at least $[0, 1]$.
$g(0) = 1 > 0 > -1 = g(1)$.

So $\exists \text{ } r \in [0, 1] \Rightarrow g(r) = 0$, i.e. $f(r) = r^{99}$. $\Box$


Geometric locus of points. The vertices of triangles having a common base and congruent altitudes.
I came up with the picture below.

I we consider the segment AB as our common base, and all altitudes length congruent the segment CB, then the geometric locus of points is the line HC and IJ. Is this correct? I came up with two sets of points which are the two lines. But I thought the geometric locus of points is a unique set of points that satisfy some condition. But the two lines I got satisfy the given condition but not unique.

André already pointed out the key aspect in his comments, but I'd like to get this question out of the unanswered queue.

The locus is the set of all points satisfying given constraints. In your case, any point on either of the two (infinite) lines will satisfy the requirements, as you already pointed out yourself. Therefore the locus is the union of all these points, i.e. the union of the two lines you described. So the locus consists of both lines taken together. As such it is unique.

Andre and MvG already answered this question very well. But still I want to add something from the Master Euclid:

If you study the Proposition 37 of Euclid's Book 1, it directly relates to this question. His proposition states that "Triangles which are on the same bases and between the same parallels (lines) are equal to one another."

You may be worried about the converse of this theorem. But its converse really holds true. If you like to further investigate the converse or related theorems, please see this https://mathematicshub.wordpress.com/2015/08/29/proposition-37-euclids-elements/.

Also, this is a question, that I saw in the Kiselev's Geometry Book 1 as question number 199. I've solved a lot of questions from this book. I also want someone who is interested in matching the answers of the questions as the author never wrote any solution manual.


Topology - multiplication of real valued functions $fg$ is continuous. Problem: If $f,g: X \rightarrow \mathbb{R}$ are continuous real valued functions on a topological space $X$, then so is $fg$.

I could show that, by standard consequences of product topology, that the function $f\times g: X \times Y \rightarrow X' \times Y'$ is continuous, given continuous functions $f:X \rightarrow X'$ and $g:Y\rightarrow Y'$.

Similarly, I could also show that the 'diagonal' map, $\Delta: X \rightarrow X \times X$ defined by $\Delta (x) = (x,x)$ is continuous.

Now $fg = m \circ (f \times g) \circ \Delta$, where
$m: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ is the multiplication of real numbers. It suffices to show $m$ is continuous.

It seems to me that $m$ must be continuous, but I could not really understand how: if $U$ is open $\mathbb{R}$ then $m^{-1}(U)$ is open in $\mathbb{R} \times \mathbb{R}$. How should I approach?

Let $x$ be in $U$ then there is an $\varepsilon>0$ such that $]x-\varepsilon, x+\varepsilon [\ \subset U$.

For $m((y,z))=x$ choose $\delta < \frac{1}{3} \inf \left\{ \frac{\varepsilon}{y}, \frac{\varepsilon}{z}, \sqrt{\varepsilon}  \right\}$, then for $(y',z')\in \ ]y- \delta, y+\delta[\ \times \ ]z-\delta , z+\delta[$

$$ y'z' \in \  ]x-\varepsilon, x+\varepsilon [ .$$

This means any point $(y,z)$ in $m^{-1}(U)$ the open ball $]y- \delta, y+\delta[ \ \times\ ]z-\delta , z+\delta[$ is contained in $ m^{-1}(U)$.

Since $\mathbb{R}$ is metric space，how about using sequential criterion for continuity? That is $x_n\to x,y_n\to y$, we have $x_ny_n\to xy$.

Or use the definition of continuity in metric space, that is
$\forall (x_1,y_1)\in \mathbb{R}\times \mathbb{R},\forall \epsilon > 0,\exists \delta>0$, such that for all $(x_2,y_2)\in \mathbb{R}\times \mathbb{R}$, with $d((x_1,y_1), (x_2,y_2)) <\delta$, we have $|x_2y_2-x_1y_1|<\epsilon$.


For each n, find an example of a critical 2-connected graph with a vertex of degree at least n. Definition:
Let G be a critical 2-connected graph; this means that G is 2-connected but no graph G − e for e ∈ E(G) is 2-connected.

Question
For each n, find an example of a critical 2-connected graph with a vertex of degree at least n.

I am pretty new to this, would you give me some examples or hint?

My first thought is a tree, which is minimally connected. Say n = 5, I have a tree T with only 4 edges.Removing any 1 edge, T is disconnected. Is this what the question is looking for? would any tree with order n work for this example? Please help me to understand this question better.

2-connected means the graph is connected, and if you remove any edge it is still connected. In practice this means the graph must contain a connected simple subgraph in which every vertex must have a degree of at least 2. (We'll just consider simple graphs because anything else is just redundant)

If you have a tree, the vertices on the "ends" of the graph only have one edge, so it doesn't work.

The easiest example of a graph that's definitely critically 2-connected is a cycle graph. This is where (if you label your vertices from 1 to k), there's an edge between vertex 1 and vertex 2, vertex 2 and vertex 3, ..., vertex k-1 and vertex k, and vertex k and vertex 1. Now if you remove any edge, say the edge between vertex 1 and vertex 2, then vertex 1 has degree 1 and vertex 2 has degree 1, so they are no longer 2-connected. Good. However you had an extra condition - for each n, you need an example where there is a vertex of degree at least n. For a cycle graph, every vertex has degree 2, so this example only works for n = 1, 2.

Intuitively you need to have one vertex with many edges coming off it, but each of these edges can only connect to a vertex with one other edge. Otherwise if you removed an edge adjacent to the "vertex with degree at least n", it will still be 2-connected and hence you graph wasn't critical. Well the key is to combine this idea with cycle graphs. (All you need are 3-cycles.)

Visualise $G_1$ to be a graph with 3 vertices, where each of these verticies are connected to the other two. You end up with a cycle-graph that looks like a triangle, critically 2-connected. Now make a copy of this graph which shares a vertex with $G_1$. You have a "central vertex" with degree 4, and four verticies around it which all have degree 2. It looks like a bow-tie. This is still critically 2-connected because every edge connects to a vertex with only one other edge. However, we now have a graph for n = 1, 2, 3, and 4.

In the same way, for any n, you:

need to consider any even k greater than it
have k edges radiating out from one vertex
add a vertex to the end of each of these edges
add an edge between every pair of these vertices.

Here's an example of a graph which is critically 2-connected with a vertex of at least degree 8.


Limit of $\sqrt[n]{(x+1)...(x+n)} - x$ as $x \to +\infty$ Let $n \in \mathbb{N}^{\ast}$. I want to determine the following limit :

$$ \lim \limits_{x \to +\infty} \sqrt[n]{(x+1)\ldots(x+n)} - x.$$

Let $x = \frac{1}{t}$ with $t \to 0$. It is equivalent to the following limit :

$$ \lim \limits_{t \to 0} \sqrt[n]{\displaystyle \frac{(t+1)\ldots(t+n)}{t^{n}}} - \frac{1}{t}. $$

$$
\begin{align*}
\sqrt[n]{\displaystyle \frac{(t+1)\ldots(t+n)}{t^{n}}} - \frac{1}{t} &= {} \frac{1}{t} \sqrt[n]{(t+1)\ldots(t+n)} - \frac{1}{t} \\[2mm]
 &= \frac{1}{t} \sqrt[n]{ n! + u(t) } - \frac{1}{t} \\[2mm]
\end{align*}
$$

where $u(0)=0$ and $\displaystyle \lim \limits_{t \to 0} u(t) = 0$. Since $\displaystyle \sqrt[n]{1+t} = 1 + \frac{t}{n} + o(t)$ as $t \to 0$, $$\displaystyle \sqrt[n]{n! + u(t)} = \sqrt[n]{n!} \; \sqrt[n]{\displaystyle 1 + \frac{u(t)}{n!} } = \sqrt[n]{n!} \bigg( 1 + \frac{u'(0) t}{n(n!)} + o(t) \bigg).$$

I do not see how to go on from there !

There is a very simple trick that involves bounding such products by matching terms of opposite sizes. This idea motivates considering the following inequalities.

$(x+1)(x+n) \le (x+k)(x+n+1-k) \le (x+\frac{n+1}{2})(x+\frac{n+1}{2})$ for any $x \ge 0$ and $k \in [1..n]$.

The above can be proven by many different methods. After this, it is trivial to bound the expression under the limit by $\sqrt{(x+1)(x+n)} - x$ and $(x+\frac{n+1}{2}) - x$, and then taking limits for both yield $\frac{n+1}{2}$ as $x \to \infty$, and hence by squeeze theorem we are done.


Prove using contour integration that $\int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$ Prove using contour integration that $\displaystyle \int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$

I am at a loss at how to start this problem and which contour to pick. I have been trying to get the sector with angle $2\pi/3$ to work with a bump around the pole at $e^{i2\pi/3}$ and the origin, but I am getting 5 or 6 different integrals and it is not really getting me anywhere.

Actually, we do need to worry about the pole at $x=1$ if we intend to use contour integration, for reasons that are a bit subtle.  I will demonstrate below.

The standard way to treat integrals of rational functions times logs over $[0,\infty)$ in complex analysis is to consider a keyhole contour, and an integral over that contour of the next higher power of log.  In this case, the integral is

$$\oint_C dz \frac{\log^2{z}}{z^3-1}$$

$C$, however, is a modified keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$.  The modification lies on small semicircular bumps above and below $z=1$ of radius $\epsilon$, and we will consider the limits as $\epsilon \to 0$ and $R\to\infty$.

Let's evaluate this integral over the contours.  There are $8$ pieces to evaluate, as follows:

$$\int_{\epsilon}^{1-\epsilon} dx \frac{\log^2{x}}{x^3-1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log^2{\left (1+\epsilon e^{i \phi}\right )}}{(1+\epsilon e^{i \phi})^3-1} \\ + \int_{1+\epsilon}^R   dx \frac{\log^2{x}}{x^3-1} + i R \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{\log^2{\left (R e^{i \theta}\right )}}{R^3 e^{i 3 \theta}-1} \\ + \int_R^{1+\epsilon} dx \frac{(\log{x}+i 2 \pi)^2}{x^3-1} + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{(\log{\left (1+\epsilon e^{i \phi}\right )}+i 2 \pi)^2}{(1+\epsilon e^{i \phi})^3-1} \\ + \int_{1-\epsilon}^{\epsilon} dx \frac{(\log{x}+i 2 \pi)^2}{x^3-1} + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\log^2{\left (\epsilon e^{i \phi}\right )}}{\epsilon^3 e^{i 3 \phi}-1} $$

(To see this, draw the contour out, including the bumps about $z=1$.)

As $R \to \infty$, the fourth integral vanishes as $\log^2{R}/R^2$.  As $\epsilon \to 0$, the second integral vanishes as it is $O(\epsilon^3)$, while the eighth integral vanishes as $\epsilon \log^2{\epsilon}$.  This leaves the first, third, fifth, sixth and seventh integrals, which in the above limits, become

$$PV \int_0^{\infty} dx \frac{\log^2{x} - (\log{x}+i 2 \pi)^2}{x^3-1} + i \frac{4 \pi^3}{3}$$

EDIT

It should be appreciated that, in the fifth, sixth, and seventh integrals, the $i 2 \pi $ factor appears because, on the lower branch of the real axis, we write $z=x \, e^{i 2 \pi}$.  In the sixth integral, in fact, $z = e^{i 2 \pi} + \epsilon \, e^{i \phi + 2 \pi}$.

END EDIT

The $PV$ denotes the Cauchy principal value of the integral.  As it stands, the integral does not actually converge.  Nevertheless, we are not actually considering the integral straight through the pole at $z=1$, but a very small detour around the pole.  Thus, in the limit, we get the Cauchy PV.  A little rearranging cancels the $\log^2$ term, and we now have two integrals to evaluate:

$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^3-1} + 4 \pi^2 PV \int_0^{\infty} \frac{dx}{x^3-1} + i \frac{4 \pi^3}{3}$$

Note we could remove the $PV$ on the first integral because the pole is a removable singularity.

The contour integral is equal to $i 2 \pi$ times the sum of the residues at the poles. The poles here are at $z=e^{i 2 \pi/3}$ and $z=e^{i 4 \pi/3}$. Note that the pole at $z=1$ is not inside the contour $C$ because of the detour around that "pole".  It should be appreciated that the poles must have their arguments between $[0,2 \pi]$ because of the way we defined $C$.

In any case, we now have that the above 1D integrals over the positive real line are equal to

$$i 2 \pi \left [\frac{-4 \pi^2/9}{3 e^{i 4 \pi/3}} + \frac{-16 \pi^2/9}{3 e^{i 8 \pi/3}}  \right ] = -\frac{4 \pi ^3}{3 \sqrt{3}}+i \frac{20 \pi ^3}{27} $$

Equating real and imaginary parts, we find that

$$ \int_0^{\infty} dx \frac{\log{x}}{x^3-1} = \frac{4 \pi^2}{27} $$

$$ PV \int_0^{\infty} \frac{dx}{x^3-1} = -\frac{\pi}{3 \sqrt{3}} $$

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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\large\tt Alternative:\ A\  "\Re eal"\ Integration,}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln\pars{x} \over x^{3} - 1}
\,\dd x} =
\left.\bracks{\nu^{1}}\int_{0}^{\infty}{1 - x^{\nu} \over
1 - x^{3}}\,\dd x\,
\right\vert_{\,\Re\pars{\nu}\ \in\ \pars{-1,2}}
\\[5mm] = &\
{1 \over 3}\bracks{\nu^{1}}
\int_{0}^{\infty}{x^{-2/3} - x^{\nu/3 - 2/3}\,\,\, \over 1 - x}
\,\dd x
\\[5mm] = &\
{1 \over 3}\bracks{\nu^{1}}
\int_{0}^{\infty}{x^{\color{red}{1/3} - 1}\ -\
x^{\color{red}{\nu/3 + 1/3} - 1}\,\,\, \over 1 - x}\,\dd x
\end{align}
However,$\ds{{1 \over 1 - x} = \sum_{k = 0}^{\infty}
\color{red}{\Gamma\pars{1 + k}\expo{\ic\pi k}}\,
{\pars{-x}^{k} \over k!}}$
Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln\pars{x} \over x^{3} - 1}
\,\dd x}
\\ = &\
-\,{1 \over 3}\bracks{\nu^{1}}
\\ &
\overbrace{%
\Gamma\pars{\!%
\color{red}{{\nu \over 3} + {1 \over 3}}\!}
\!\Gamma\pars{\!1 -
\bracks{\color{red}{{\nu \over 3} + {1 \over 3}}}\!}
\!\exp\pars{\!%
-\ic\pi\bracks{\color{red}{{\nu \over 3} + {1 \over 3}}}\!}}
^{\substack{\ds{Ramanujan's} \\[0.5mm] \ds{Master}\\[0.5mm] \ds{Theorem}}}
\\[5mm] = &\
-\,{1 \over 3}\bracks{\nu^{1}}
\bracks{{\pi \over \sin\pars{\pi\nu/3 + \pi/3}}\,
\exp\pars{-\ic\pi\bracks{{\nu \over 3} + {1 \over 3}}}}
\\[5mm] = &\
-\,{\pi \over 3}\bracks{\nu^{1}}
\cot\pars{{\pi\nu \over 3} + {\pi \over 3}} =
\left.-\,{\pi \over 3}
\totald{\bracks{\cot\pars{{\pi\nu \over 3} + {\pi \over 3}}}}{\nu}\,\right\vert_{\ \nu\ =\ 0}
\\[5mm] = &\
-\,{\pi \over 3}\bracks{-\,{\pi \over 3}
\csc^{2}\pars{\pi \over 3}} =
\bbx{4\pi^{2} \over 27} \approx 1.4622 \\ &
\end{align}


Product of two independent, random draws is usually less than the square of a random draw. Why? Suppose a uniform distribution, D over positive real numbers. On average, the product of two independent draws from D is less than the square of a random draw from D. I wrote a MWE in R code that simulates the above. My question is, why?

n_sim = 1000000
min_val = 1
max_val = 10000000000
a = runif(n   = n_sim,
          min = min_val, max = max_val)
b = runif(n   = n_sim,
          min = min_val, max = max_val)
c = runif(n   = n_sim,
          min = min_val, max = max_val)
table(a<b)
table(a*b < c^2)

Your finding is true for any distribution, positive or not, uniform or not. Let $X$ and $Y$ be independent, each with the same distribution as $Z$. Your observation is that
$$E(XY -Z^2)\le0.\tag1$$
Since $X$ and $Y$ are independent, the LHS of (1) equals $E(X)E(Y)=[E(Z)]^2$ so asserting (1) is the same as asserting
$$
[E(Z)]^2-E(Z^2)\le0.\tag2
$$
You can rearrange (2) to the form
$$
E(Z^2)-[E(Z)]^2\ge0,\tag3$$
which is a true statement for any $Z$ because the LHS of (3) equals $E[(Z-E(Z))^2]$, which is the expectation of a non-negative random variable (it's also the variance of $Z$).


Taylor Series of an arctangent function The question is what is the Taylor series expansion of $\tan^{-1}(\frac{2x}{1-x^2})$ at $x=0$? I know that $\tan^{-1}(x)=x-x^3/3+x^5/5+...$

This is complicated to replace $x$ with $2x/1-x^2$ is there a neater way to compute this?

Notice that after simplifications,

$$\left( \arctan \frac {2x} {1-x^2} \right) ' = \frac 1 {1 + \frac {4x^2} {(1-x^2)^2}} \frac {2(1-x^2) - 2x(-2x)} {(1-x^2)^2} = \frac 2 {1+x^2} = (2 \arctan x)' ,$$

whence it follows that

$$\arctan \frac {2x} {1-x^2} - 2 \arctan x = C$$

with $C \in \Bbb R$ a constant. To find out $C$ just evaluate the equality at $x=0$, to get

$$\arctan 0 - 2 \arctan 0 = C ,$$

whence $C=0$ and thus

$$\arctan \frac {2x} {1-x^2} = 2 \arctan x$$

and the required series is obvious now if you now the series of $\arctan$.

Alternatively, use the formula

$$\tan (u+v) = \frac {\tan u + \tan v} {1 - \tan u \tan v}$$

with $u = v$ to get

$$\tan (2u) = \frac {2 \tan u} {1 - \tan^2 u} ,$$

whence

$$2u = \arctan \frac {2 \tan u} {1 - \tan^2 u}$$

whence, denoting $x = \tan u$, we get

$$2 \arctan x = \arctan \frac {2x} {1-x^2} .$$

These trigonometric computations are a bit formal, since I haven't payed attention to the domain of definition of the above functions, but they can be made perfectly rigorous with just a bit of supplementary work.

You could try using the Taylor series expansion of $\dfrac{2x}{1-x^2}$:
$$\dfrac{2x}{1-x^2}=2x\left(1+x^2+x^4+\cdots\right)=2x+2x^3+2x^5+2x^7+\cdots$$
which is valid if $|x|<1$.


When is $\sqrt[3]{a+\sqrt b}+\sqrt[3]{a-\sqrt b}$ an integer? I saw a Youtube video in which it was shown that
$$(7+50^{1/2})^{1/3}+(7-50^{1/2})^{1/3}=2$$
Since there are multiple values we can choose for the $3$rd root of a number, it would also make more sense to declare the value of this expression to be one of $2, 1 + \sqrt{-6},$ or $1 - \sqrt{-6}$

We may examine this more generally. If we declare $x$ such that
$$x=(a+b^{1/2})^{1/3}+(a-b^{1/2})^{1/3}$$
$$\text{(supposing } a \text{ and } b \text{ to be integers here)}$$
one can show that
$$x^3+3(b-a^2)^{1/3}x-2a=0$$
Which indeed has $3$ roots.

We now ask

  For what integer values of $a$ and $b$ is this polynomial solved by an integer?

I attempted this by assuming that $n$ is a root of the polynomial. We then have
$$x^3+3(b-a^2)^{1/3}x-2a$$
$$||$$
$$(x-n)(x^2+cx+d)$$
$$||$$
$$x^3+(c-n)x^2+(d-nc)x-nd$$
Since $(c-n)x^2=0$ we conclude that $c=n$ and we have
$$x^3+3(b-a^2)^{1/3}x-2a=x^3+(d-c^2)x-cd$$
And - to continue our chain of conclusions - we conclude that
$$3(b-a^2)^{1/3}=d-c^2 \quad\text{and}\quad 2a=cd$$
At this point I tried creating a single equation and got
$$108b=4d^3+15c^2d^2+12c^4d-c^6$$
This is as far as I went.

Use $$x^3+y^3=(x+y)(x^2-xy+y^2)$$ identity. Things will cancel out.


Is spending a long time to understand logic very well worth it to learn math better? My goal is to become a computer/electrical/mechanical engineer. To become one i have to read a lot of math based books (physics, operation research, etc). I know some real analysis, linear algebra and more already, but in college we go over topics too fast and i can't grasp everything so well. So I want to start studying math from the beginning on the side. I discovered that there exist a "foundation of mathematics" which is pretty much set theory and logic if I'm correct (plus a few more) and I was wondering how important logic is to understand more complicated topics later on (let's say, multivariable calculus, multilinear algebra, differential geometry etc). Is it worth spending a "long time" studying logic (prepositional and predicate logic), since sometimes definitions can be confusing (vacuous truths for example) or written symbolically, or it would just be a waste of time?

I would say that it is absolutely not helpful ... in that particular way. Logic is super valuable in other contexts and interesting in its own right, but that's a separate issue.

Actually, if anything you should go the other way around: although "foundationally" logic comes first, in terms of pedagogy one's not generally prepared to really get the most out of logic until one has some decent grounding in some fairly abstract mathematics - in particular, I would recommend abstract algebra. I personally don't see the point in tackling logic until after you're quite comfortable with something on the level of abstract groups.

(Computability theory is a bit of an exception here, except $(i)$ it's not the kind of logic you're talking about and $(ii)$ even for it I think having some abstract algebra is very useful.)

CAVEATS:

In my opinion it is worth spending some time learning the "jargon," with quantifiers ($\forall$ and $\exists$) in particular. But even this isn't a make-or-break point: I know serious mathematicians who can't read quantifiers fluidly and prefer to avoid them.
There's also the pedagogical value of knowing that it's there - that is, knowing that the math you see (which is often presented semi-formally) can actually be grounded in a completely formal system, and that you can look up the details later when you have the time. But one's mileage may vary with regards to this point: some people don't care in the first place, and others aren't satisfied until they see the details.

I think it's worth to look at a basic level logic principles at first. The problem is that most of us learn better first with examples. Take for instance real analysis; then you would like to continue with perhaps functional analysis or measure theory (depending on the subfield you are going to specialize). You will find that the proofs in functional analysis are "generalizations" of those in real analysis, but the underlying principles are the same.

In this sense, more than logic I would say that what really matters in modern math is set theory. Take for example Jech's book, you can find many proofs given in real analysis there.

In short, sometimes is better to climb the stairs not from the beginning, but from the middle and the going up and down as the matter requieres.

As a side effect, learning logic will teach you to structure your ideas and make them coherent and correct. This is important in writting thesis or research papers, or mathematical proofs for that matter...


Chances of getting exactly one heads in two coin flips? I know the chances of getting exactly one "heads" in two coin flips is 50%, as there are only four possible outcomes: HH, HT, TH, TT; and only two of those four meet the restraint I've placed. To figure this out, I think you can use a binomial distribution: 2C1 times 1/2 times 1/2 gives the probability of 1 in 2 that I'm looking for.

But why doesn't the formula $$P(A \text{ or }B) = P(A) + P(B) - P(A \text{ and } B)$$ (where A and B are heads on the first and second attempts, respectively) work? I get 75% (1/2 + 1/2 -1/4) which would work if I wanted the restraint "at least one 'heads,'" or simply what are the chances of getting one "heads."

Moreover, if you were to find the chances of getting "heads," wouldn't you take 1 minus the probability of getting two "tails" (25%) to get the 75% you wanted? So when would you use P(A or B)?

The symmetric difference, of exclusive or, is the probability of getting exactly one head. What you were calculating was the union, which is the probability of at least one head.

$\newcommand{\P}{\mathbb{P}}$In mathematics, "or" means "either $A$ is true, $B$ is true, or both $A$ and $B$ are true. See this Wikipedia page which has nice diagrams: https://en.wikipedia.org/wiki/Logical_disjunction.

In other words $\mathbb{P}(A\text{ or }B)= \mathbb{P}(A\text{ only})+ \P(B\text{ only})+ \P(\text{both }A\text{ and }B)$.

If I am interpreting your question correctly, $A = \{HH, HT \}$ and $B = \{ HH, TH\}$.

This means that $$A\text{ or }B = (A \text{ is true}) \lor (B \text{ is true}) = A \cup B = \{HH, HT, TH  \}$$

Therefore $\P(A \text{ or }B) = \frac{3}{4}$ and the formula which you claim does not work does work. $$\P(A)=\frac{1}{2}, \quad \P(B)=\frac{1}{2}, \quad P(A \cap B)=\frac{1}{4}, \quad \frac{1}{2}+\frac{1}{2} - \frac{1}{4} = \frac{3}{4}$$

The word "or" is often used in English to mean "exclusive or" or "XOR", which means "either $A$ only, or $B$ only, but not both". This corresponds to the set operation of symmetric difference, rather than union. However, in probability, it is always union, and not symmetric difference, that is meant when the term "or" is used. This makes most mathematical expressions and statements much easier to state and formulate.

Here is a picture below of symmetric difference; contrast it with that of union you see above:

This means that $$ A \text{ xor }B = [(A\text{ is true})\lor(B\text{ is true})]\land ((A\text{ and }B)\text{ is not true}) \\= A \bigtriangleup B = \{HT, TH \} = A \cup B \setminus A \cap B = \{HH, HT, TH \} \setminus \{HH \}$$

No. See what. You look seperatly. If you define A as "the probability that only the first toss is head" its outcome probability will be 1/4, rather than 1/2.  If you want to look the way you mentioned, than we will have the following $P(A)=P(the first is head)=\frac{1}{2}$ ; $P(B)=P(the second is head)=\frac{1}{2}$ , and $P(A&B) = P(both the first and th second tosses give head) =\frac{1}{4}$ (as it is one case of four). And we will have $P(A or B) = \frac{1}{2}+\frac{1}{2} - \frac{1}{4}$.  But $P(A or B) = P( either first or second (or both) toss is head)$. See however that this does not exclude the situation when both are head. More precisely "(A or B) = (A and not B) + (B and not A)".


Example of hermitian semidefinite positive Could someone show me an example of a $3\times 3$ hermitian semidefinite positive matrix? (I mean not only symmetric, but hermitian, with complex coefficients)

There is a lot of work on hermitian semidefinite positive matrices, but it seems that people talk about them and don't take the time to exhibit examples.

Note that a Hermitian matrix will be positive definite if and only if it has positive eigenvalues.  A good "recipe" to quickly build a positive definite matrix is to select diagonal elements that are sufficiently big, since selecting large positive diagonal entries will guarantee positive eigenvalues (positive, since the eigenvalues of a Hermitian matrix are necessarily real).  An example of a more precise (and useful) description of this phenomenon is given by the Gershgorin circle theorem.

So with that being said: without computing eigenvalues explicitly, I can guarantee that the Hermitian matrix
$$
\pmatrix{100 & 1 - 2i & 3 + 5i\\
1 + 2i & 100 & -9 + i\\
3 - 5i & -9 - i & 100}
$$
is positive definite.

Hint: The identity matrix is Hermitian and psd.


non-cartesian set product? Foremost, this question is asked from a point of a computer scientist undergrad, so please don't nag me for inconsistent notation, or lack of proper vocabulary.

Is there a concept in mathematics for a non-cartesian product of sets, i.e. a one which maps to a set of products of all elements from two sets? Let me define more formally:

$$
for\quad { S }_{ 1 },{ S }_{ 2 },{ S }_{ 3 }\subseteq \mathbb{R},\quad let\quad { f }: ({ S }_{ 1 },{ S }_{ 2 })\mapsto { S }_{ 3 }\\ so\quad that\quad {f}({ S }_{ 1 },{ S }_{ 2 })=\{ x|\exists a\in { S }_{ 1 },\exists b\in { S }_{ 2 },(ab=x)\}
$$
For example: let ${ S }_{ 1 }=\{ 1, 2, 3\}$ and ${ S }_{ 2 }=\{ 4, 5\}$.
Therefore ${ S }_{ 1 }.{ S }_{ 2 }=\{ 1*4,\quad 1*5,\quad 2*4,\quad 2*5,\quad 3*4,\quad 3*5\} =\{ 4,5,8,10,12,15\}$

It would seem to me that this concept would be useful and widely used in mathematics, yet I have never stumbled upon it. Does this concept exist? If yes, where and how is it used? Or is it something that I inherently misunderstood about sets that prevents such concept to be useful?

The common notation for the element-wise product of two sets (which must both be subsets of some given magma, i.e., a set with an operation) is $AB$ or $A\cdot B$ or, if the operation is addition, $A+B$ (or any other symbol for the operation) and is defined as the set $\{a+b\mid a\in A,b\in B\}$ (suitably adapted to use the relevant operation symbol). It is commonly used throughout mathematics whenever convenient.

It's called a binary operation. A common application is Graph Theory, esp. the graph product.

Its similar to a Cartesian product, but this returns an ordered pair $(a,b)$ without defining an operation on $a$ and $b$.


Is ZFC without Axiom of Infinity consistent? The incompleteness theorem states that one cannot prove whether ZF or ZFC is consistent, but what about ZFC withouth Axiom of infinity? (Assuming the empty set exists)

Furthermore, let $M$ be a consistent model not invoking infinity and $A,B$ be statements invoking infinity such that $A$ contradicts $B$. Then, let's assume both $M+A$ and $M+B$ are consistent. If statements $\phi_A$ and $\phi_B$ invoking infinity are provable in $M+A$ and $M+B$ relatively, then are finite pieces of $\phi_A$ and $\phi_B$ both provable in $M$?

The corollary from the incompleteness theorems is that you cannot prove the consistency of $\sf ZFC$ from $\sf ZFC$ itself. You have to have a stronger theory.

For example, in $\sf ZFC+\text{There exists an inaccessible cardinal}$, you can in fact prove the consistency of $\sf ZFC$ because this is a stronger theory.

Similarly this is the case of $\sf ZF_{fin}$ ($\sf ZF$ without infinity). The theory itself cannot prove its own consistency. However $\sf ZFC$ is a strictly stronger theory, and it proves the consistency of $\sf ZF_{fin}$. It does so by exhibiting a set which is a model of the theory, $V_\omega$ - the set of the hereditarily finite sets.

Large cardinal axioms are often called "strong infinity axioms" because they mimic the axiom of infinity, in the sense that they make a stronger theory by describing that a certain set of ordinals exists.

Usally "ZFC without Axiom of Infinity" is considered consistent, since $\sf ZFC \setminus  AOI + \neg AOI$ has the natural numbers as a model via Ackermann Coding. If $\sf ZFC \setminus AOI + \neg AOI$ has a model, so does $\sf ZFC \setminus AOI$. This is a model like the Hereditary Finite Sets:

  Hereditarily finite set
  https://en.wikipedia.org/wiki/Hereditarily_finite_set

The Ackermann Coding was shown here:

  Ackermann, Wilhelm (1937), "Die Widerspruchsfreiheit
  der allgemeinen Mengenlehre", Mathematische Annalen,
  114 (1): 305–315

On the other hand, to formally prove it, "ZFC with Axiom of Infinity, the usual ZFC" should do, it would show that $V_\omega$ is a model of $\sf ZFC \setminus AOI + \neg AOI$. So you would need to first accept $\sf ZFC$.

Remark: For the rest of your question concerning $M+A$, I have no answer. How do you combine a model $M$ and a formula $A$?


Find the Derivative of $f(x)=\frac{7}{\sqrt {x}}$ using the definition. I get that $$\frac{d}{dx}\left(7\times\dfrac{1}{\sqrt{x}}\right)=\frac{d}{dx}(7x^{.5})=\dfrac{7}{2}x^{-.5}$$ is the derivative, but I can't ever use $\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.

If someone or anyone could go step by step and do the problem, I would be eternally grateful.

Hint:

Use binomial series expansion of $(x+h)^{\frac{-1}{2}}$


Show that $a^2$ congruent 0 (mod 4) or $a^2$ congruent 1 (mod 4). Proof: Suppose for the sake of contradiction, there is an integer a such that $a^2$ congruent $0 \pmod {4}$ and $a^2$ congruent $1 \pmod {4}$. Then $(a^2)-0=4k$ and $(a^2)-1=4l$ for some $k,l$ members of the integers.
Now,
$$a^2-0=4k$$
$$a^2=4k$$
Thus,
$$a^2-1=4l$$
$$((4k)^2)-1=4l$$
$$(16k^2)-1=4l$$
$$(16k^2)-4l=1$$
$$2((8k^2)-2l)=1$$

$2m=1$ for some integer m where $m=(8k^2)-4l$
By the definition of an even integer $1$ is even, however we know $1$ is odd, therefore a contradiction.

I feel the answers provided aren't clear or complete enough, so I'm going to put my two cents in. The proof is done by cases.

Case $1$: $a = 2k$ is even

By the definition of divisibility and congruence, $$a^2 \equiv 0\pmod4\impliedby 4\mid(a^2-0)\impliedby 4\mid(2k)^2\impliedby4\mid4k^2$$ and since $\exists\, c\hspace{0.1cm}: 4k^2 =4\cdot c$ with $c=k^2$, the statement is true.

Case $2$: $a=2k+1$ is odd

The second case is done almost exactly, except you're proving that $a^2 \equiv 1\pmod4$ holds for odd numbers, so instead of inputting $n$ as $2k$, you input it as $2k+1$. You will find that the witness $c$ for this case is $c = k^2+k$. You have proven $a^2$ is congruent to either $0$ or $1\pmod4$ for all $a$, thus the proof is considered complete. $\square$

You only have to make a table of squares modulo $4$, which can be done readily since there are only $4$ congruence classes:
$$\begin{array}{c|ccccc}a&0&1&2&3\\\hline
a^2&0&1&0&1\end{array}$$


Prove that $\varphi(2n)=2\varphi(n)$ if $n$ is even. So, far, I have the following which I have seen the basic idea in other similar proofs:

Let $n$ be an even, positive integer. Then $n=2^{s}(p_{1}^{x_{1}}\cdot p_{2}^{x_{2}} \cdot\cdot\cdot p_{r}^{x_{r}})$ since $n$ is even, where $p_{r}$ is an odd prime number and $x$ is an positive integer. Then $\frac{n}{2^{s}}$ is odd, and so it is relatively prime to $2^{s}$ which is even. But $n=2^{s}\cdot \frac{n}{2^{s}}$, which implies that $\varphi(n)=\varphi(2^{s}) \cdot \varphi(\frac{n}{2^{s}})$.

I am hoping to somehow get $2$ out of $\varphi(2^{s})$ and combine the leftovers with the term $\varphi(\frac{n}{2^{s}})$, but not sure if it's possible? It seems like I keep end up proving the statement false, when it should be true...is there a clear way to finish this, or is there an easier way to do it?

Write $n = 2^k\cdot m$ where $m$ is odd. Then
$$\varphi(2n) = \varphi(2^{k+1} \cdot m) = \varphi(2^{k+1})\varphi(m) = 2^k\varphi(m) = 2\varphi(2^k)\varphi(m) = 2\varphi(n).$$

Since $n$ is odd, $\gcd(2,n)=1$.

  $$\varphi(2n) = \varphi(2)\varphi(n) = 1\varphi(n)$$

Since $n$ is even, $n = 2^km$

  $$\varphi(2n) = \varphi(2\cdot 2^km) = \varphi(2^{k+1}m) = \varphi(2^{k+1})\varphi(m)= 2^k\varphi(m)= $$$$2\cdot 2^{k-1}\varphi(m)=2\cdot \varphi(2^k)\varphi(m)=2\cdot \varphi(2^km) = 2\varphi(n)$$


How to calculate the parameters of a line through two points with homogeneous coordinates $p_1, p_2$? So in my notes it says it is $l=p_1 \times p_2$. I don't quite understand that, for example, suppose the points are $p_1=(1,1,1)$, $p_2=(2,2,1)$. Then the cross product would give a line perpendicular to these points, in other words perpendicular to the plane in orange as seen in the image below:

In 2D, the equation of any line $\ell$ has the form $a x + by + c = 0$.

Similar to one can use homogeneous coordinate $\vec{p} = (x,y,1)$ to represent a point $p = (x,y)$ in 2D, one can use homogeneous coordinates $\vec{\ell} = (a,b,c)$ to represent a line $\ell$ in 2D. The condition that $p$ lies on $\ell$ (i.e. $a x + by + c = 0$ ) can be rewritten as $\vec{\ell} \cdot \vec{p} = 0$.
In short,

  point $p$ lies on line $\ell$ in 2D $\quad\iff\quad$ $\vec{p}$
  perpendicular to $\vec{\ell}$  in 3D.

A consequence of this is if $\ell$ is a line passing through 2 points $p_1, p_2$, we will have

$$\vec{\ell}\cdot\vec{p_1} = 0\quad\text{ and }\quad
  \vec{\ell}\cdot\vec{p_2} = 0\quad\implies\quad \vec{\ell} \propto \vec{p_1} \times \vec{p_2}$$

i.e the determination of a line by 2 points in 2D corresponds to cross product of corresponding 3D vectors of homogeneous coordinates.

Another interesting consequence is if $p$ is the intersection of two lines $\ell_1$ and $\ell_2$, then $\vec{p} \propto \vec{\ell}_1 \times \vec{\ell}_2$.

It is this sort of correspondences between geometric operations in 2D and vector
operations in 3D that make homogeneous coordinates useful in analysis of 2D problems.

The cross product $\vec{w}=\vec{u} $ x $ \vec{v}$ is the vector which is orthogonal to $\vec{u}$ and to $\vec{v}$ or in other words, the vector which is orthogonal to the plane defined by the vectors $\vec{u}$ and $ \vec{v}$.


Distributional Laplacian of logarithm and the Dirac delta distribution Consider $S(\mathbb{R}^2)$ the space of rapidly decreasing functions.

Consider $F(x) = \displaystyle\frac{1}{2 \pi} \ln|x| , x \in \mathbb{R}^2 - \{ 0 \}$.

I want to prove this  : $\Delta F = \delta $ in $S^{'}(\mathbb{R}^2).$

My try: For every $\varphi \in S({\mathbb{R}}^2)$ we have

$$\langle \Delta F, \varphi\rangle  = \langle F , \Delta \varphi\rangle  = \displaystyle\int_{\mathbb{R}^2} F(x) \Delta\varphi(x) \ dx = \displaystyle\lim_{r \rightarrow + \infty} \displaystyle\lim_{\epsilon \rightarrow 0^{+} } \displaystyle\int_{\epsilon \leq |x| \leq R} F(x) \Delta \varphi(x) \ dx $$

I don't know how to prove this :$\displaystyle\lim_{r \rightarrow + \infty} \displaystyle\lim_{\epsilon \rightarrow 0^{+} } \displaystyle\int_{\epsilon \leq |x| \leq R} F(x) \Delta\varphi(x) \ dx = \varphi (0) = \delta(\varphi)$

I don't know what to do from here.

(Indeed, $F$ is not in $\mathscr{S}$; it is any rate more convenient to consider $\mathscr{D}=C^\infty_0$.) Let $\varepsilon>0$ and $\Omega_\varepsilon:=\mathbf{R}^2-B_\varepsilon(0)$; now
$$\left\langle F,\Delta\phi\right\rangle=\lim_{\varepsilon\downarrow 0}\int_{\Omega_\varepsilon} F\Delta\phi \ dx,$$
and by Green's identity
$$\int_{\Omega_\varepsilon} F\Delta\phi \ dx=\int_{\Omega_\varepsilon} \phi\Delta F \ dx+\int_{\partial\Omega_\varepsilon} F\frac{\partial\phi}{\partial n}ds-\int_{\partial\Omega_\varepsilon}\phi\frac{\partial F}{\partial n} \ ds.$$
In the RHS the first integral vanishes as $\Delta F=0$ on $\mathbf{R}^2-\left\lbrace 0\right\rbrace$; the second integral can be estimated in absolute value to be at most
$$C\int_{\partial\Omega_\varepsilon} Fds=\frac{C}{2\pi}2\pi\varepsilon\log \varepsilon$$
which tends to $0$ as $\varepsilon\downarrow 0$ ($C$ is an upper bound for $\partial\phi/\partial n$). Now $\partial F/\partial n=-1/2\pi|x|$ (the normal to $\partial\Omega_\varepsilon$ is $-x/|x|$) so that we are left with
$$\left\langle F,\Delta\phi\right\rangle=\lim_{\varepsilon\downarrow 0}\frac{1}{2\pi\varepsilon}\int_{\partial\Omega_\varepsilon}\phi\ ds=\phi(0),$$
by a standard argument using, say, polar coordinates.

F(x) appears to be an increasing function! Nonetheless, I think the proposition is true (but with a minus sign), and it can easily be proved using Gauss's theorem: http://en.wikipedia.org/wiki/Divergence_theorem .

Take the gradient of F to obtain a vector field. The Laplacian of F is the divergence of this vector field. The field points everywhere radially outwards, and has magnitude 1/(2*pi*|x|). The flux of the field across a sphere (circle) of radius r is therefore 1. The integral of the divergence of the field inside the sphere is therefore 1. Since this is true for all r, the divergence must be confined to the origin.


Compute $I(U), I(V),I(U\cap V)$ and $I(U)+I(V)$. Let $U=Z(x^2-y)$ and $V=Z(y)$ in $k[x,y]$ where $k$ is a field. I recall that $$Z(I)=\{(x,y)\in k^2\mid \forall f\in I, f(x,y)=0\}$$
and $$I(U)=\{f\in k[x,y]\mid \forall (x,y)\in U, f(x,y)=0\}.$$

For $I(V)$, I have that $$f\in I(V)\iff f(x,0)=0\iff f\in (y).$$
We have that $I(U)=(x^2-y)$ but I don't understand why. Indeed,
$$(x,y)\in Z(x^2-y)\iff \forall f\in (x^2-y), f(x,y)=0\iff x^2-y=0\iff y=x^2,$$
therefore, $$Z(x^2-y)=\{(x,x^2)\mid x\in k\},$$
and thus $$f\in I(U)\iff f(x,x^2)=0$$
but how can I continue ?

For the rest, I think it's fine.

If $f(x,x^2)=0$ then $y=x^2$ is a root of $g(y)=f(x,y)\in F[x][y]$, therefore, $g(y)=(y-x^2)k(y)\in F[x][y]$ and thus $f(x,y)\in (y-x^2)$.

I wish to help you. we have theorem: $k[U]\cong k[x,y]/I(U) $, we have $k[U]=k[x,x^2]=k[x]$, then $k[x]\cong k[x,y]/I(U)\Rightarrow I(U)=(y-x^2).$ In similar way we have that $I(V)=(y)$.  and $I(U\cap V)=\sqrt{I(U)+I(V)}$.
In your question $I(U\cap V)=I((0,0))=(x,y)$, and $I(U)+I(V)=(x^2-y,y)=(x^2,y)\Rightarrow \sqrt{I(U)+I(V)}=(x,y)$. we note that $I(U\cap V)\ne I(U)+I(V)$.


Do eigenvalues depend smoothly on the matrix elements of a diagonalizable matrix? Suppose I have a matrix $M(t)$ whose matrix elements depend smoothly on a real parameter $t$. I also know that this matrix is diagonalizable for the $t$s I'm interested in. Can I say that its eigenvalues depend smoothly on $t$?

Thanks very much!

The answer is NO; more pecisely, even for the real symmetric matrices, a $C^{\infty}$ parmetrization $M(t)$ gives birth to (only) a continuous paremetrization of the eigenvalues.

Of course, if, for every $t$, the  eigenvalues of $M(t)\in M_n$ are simple, then there are smooth local parametrizations of the spectrum: $\lambda_1(t),\cdots,\lambda_n(t)$. More generally, this property stands when the mutiplicy of the eigenvalues are locally constant. When the mutiplicity of some eigenvalue varies, then the storm is threatening...

Now we assume that we work in $S_n$, the set of symmetric real matrices of dimension $n$ and that the eigenvalues may be multiple.

For the $C^{\infty}$ parametrization, the standard counterexample is due to Rellich, but I have not it on hand; you can make one by using the function $\exp(-1/t^2)$.

There exists a precise result when the symmetric matrix depends analytically on one parameter.

$\textbf{Proposition.}$ Assume that $t\in\mathbb{R}\rightarrow M_t\in S_n$ is analytic. Then the eigenvalues and a basis of (unit length) eigenvectors of $M_t$ are globally analytically parametrizable (even if the eigenvalues present some mutiplicities; note that the natural ordering of the eigenvalues is not met).

$\textbf{Remark.}$ That works also when $t\in\mathbb{R}\rightarrow M_t\in S_n$ is smooth; we must add the condition that two continuous curves $(t\rightarrow \lambda_i(t),t\rightarrow \lambda_j(t))$ (where $(\lambda_i(t),\lambda_j(t))$ are any couple of eigenvalues of $M_t$) are the same or intersect only a finite number of times.

Yes. The characteristic equation has is monic, i.e. its leading coefficient is 1, so its roots, which are the eigenvalues of the matrix, are functionts,continuously differentiable to all orders, of the coefficients , which are themselves, up t0 parity, the sums of the principal minors,i.e. the determinants of symmetrically formed sub-matrices of appropriate size of the matrix and determinant functions are just polynomials. So the eigenvaluesare smooth functions of the matrix elements and thus smooth functions of your parameter.


Calculate the value of sum for any given n $2n\choose0$ - $2n\choose2$ + $2n\choose4$ - ... + $(-1)^n$$2n\choose2n$

Q: What's the value for any given n $\in$ $N$?

Task is quite easy for odd n and I completed it. You just have to notice that the first number and the last one are opposite etc. I have a problem with even numbers. I noticed it will be $2^n$, but not always positive. Any ideas?

The given expression is the real part of $(1+i)^{2n}$.
\begin{align*}
(1+i)^{2n} &= \left(\sqrt{2}(\cos \pi/4 + i \sin\pi/4)\right)^{2n} \\
&= 2^n (\cos n\pi/2 + i\sin n\pi/2)
\end{align*}
Thus when $n$ is odd, the sum is zero and when $n = 2k$, sum equals $(-1)^k 2^{2k}$

Precalculation + OEIS
http://codepad.org/tjwaQvuT
https://oeis.org/search?q=-4%2C16%2C-64%2C256%2C-1024%2C4096%2C-16384%2C65536%2C-262144%2C1048576%2C

Answer maybe is (-4)^n


How is a decibel a tenth of a Bel? By definition, one Bel is log10(P2/P1) (in the power example) and one dB is 10log10(P2/P1), so it appears to me that one dB is ten Bels, not one-tenth of a Bel.  Yet, many online publications, university publications, and others all seem to agree that one dB is one-tenth of a Bel.  How do they figure?  (And please don't say anything about the definition of the deci- prefix, which only clouds the explanation.)

Thanks!
Noji Ratzlaff
Orem, Utah

One Bel is not $\log_{10}(P_2/P_2)$ (especially as this depends on $P_1$ and $P_2$). Rather, the number of Bels by which $P_2$ differes from $P_1$ is obtained by computing $\log_{10}(P_2/P_2)$. And because one dB is one tenth of one B, the number of decibels is ten times as large.

This is the same as that the number of decimeters is obtained by multiplying the number of meters that make a distance by $10$: I am $1.85$ meters tall, so I am $18.5$ decimeters tall (or $185$ centimeters, or $0.00185$ kilometers)

A bel is defined by using the formula: $$\log (x÷y)$$
So, if we compare 2 power ratios... Say 10 and 1; then our Bel = log (10÷1) = log (10) = 1
Similarly, if we compare 100 and 1; then our Bel = log (100÷1) = log (100) = 2
A Decibel is 1/10th of a Bel. Which means, there are 10 Decibels in 1 Bel.
So for the same ratios...
10:1 = 1 Bel OR 10 decibels;
100:1 = 2 Bels OR 20 decibels; and
1000:1 = 3 Bels OR 30 decibels.
As you see, for every Bel, we get an equivalent value of 10 Decibels.
Meaning…
If our Bel value = 1, our dB value = 10.
If our Bel value = 2, our dB value = 10 x 2 = 20.
If our Bel value = 3, our dB value = 10 x 3 = 30.
Which means, in other for us to get our Decibel value, we need to multiply our Bel value by 10.
So effectively, if we were to compute a formula for decibels directly, we can say that...
dB = 10 x log (x÷y)


Prove Fibonacci Sequence Property: $x^2_n + x^2_{n+1}=x_{2n+1}$ For Fibonacci Sequence, We know that the recursive difference equation is:

$$
x_{n+2} = x_n + x_{n+1}\ \ \ \ n\ \geq 0
$$

And that the closed form solution is:

$$
x_n = \frac{1}{\sqrt{5}}\left[ \left(\frac{1+\sqrt{5}}{2}\right)^{n} - \left(\frac{1-\sqrt{5}}{2}\right)^{n}  \right]\ \ \ \  n \geq 0
$$

But how to prove that this fibonacci property is true using this information?:

$$x^2_n + x^2_{n+1}=x_{2n+1}$$

Hint: substitute the closed form expression for the fibonacci sequence into difference equation and verify that its true.

well...I've tried this at least 3 different ways of doing exactly as hinted without much luck...the equation always blows up on me.  Any ideas how to prove  it?

The 'standard proof' uses strong induction, and I'll leave you to figure that out.

Let $\varphi$ and $\psi$ represent the two roots of $x^2-x-1=0$ where $\varphi > \psi$ (i.e. $\varphi,\psi=\frac{1\pm \sqrt{5}}{2}$). The given closed form solution is $$F_n=\frac{\varphi ^n-\psi ^n}{\sqrt{5}}$$ so we know $$F_n^2=\frac{\varphi ^{2n}-2(\varphi \psi)^n+\psi ^{2n} }{5}$$ and $$F_{n+1}^2=\frac{\varphi ^{2n+2}-2(\varphi \psi )^{n+1}+\psi ^{2n+2}}{5}$$ so $$F_n^2+F_{n+1}^2=\frac{\varphi ^{2n}-2(\varphi \psi)^n+\psi ^{2n}+\varphi ^{2n+2}-2(\varphi \psi )^{n+1}+\psi ^{2n+2} }{5}$$But we know that $\varphi \psi =-1$, so either $(\varphi\psi)^n=1$ and $(\varphi\psi)^{n+1}=-1$ or $(\varphi\psi)^n=-1$ and $\varphi\psi)^{n+1}=1$. Either way, the two terms cancel each other, so$$F_n^2+F_{n+1}^2=\frac{\varphi^{2n}(1+\varphi ^2)+\psi ^{2n}(1+\psi^2)}{5}$$It can be easily seen that $\frac{1+\varphi^2}{\sqrt{5}}=\varphi$ and $\frac{1+\psi^2}{\sqrt{5}}=-\psi$ so the above becomes \begin{align*}F_n^2+F_{n+1}^2&=\frac{\varphi^{2n}(1+\varphi ^2)+\psi ^{2n}(1+\psi^2)}{5}\\ &=\frac{\varphi^{2n}\cdot \varphi\sqrt{5} + \psi^{2n}\cdot (-\psi\sqrt{5})}{5} \\ &=\frac{\sqrt{5}(\varphi^{2n+1}-\psi^{2n+1})}{5} \\ &=\frac{\varphi^{2n+1}-\psi^{2n+1}}{\sqrt{5}} \\ &= F_{2n+1}\end{align*}completing the proof.

Based on the title of the topic and the hint, you need to plug the Closed form expression into your equality

$$x_{n}^2+x_{n+1}^2-x_{2n+1}$$

and make sure it equals zero identically.

Best way is to taky any CAS and plug the figures. For Wolfram Mathematica

  x[n_] := 1/Sqrt[5]*(((1 + Sqrt[5])/2)^n - ((1 - Sqrt[5])/2)^n);

  x[a]^2 + x[a + 1]^2 - x[2*a + 1]

  Simplify[Out[]]

Then you get something like this

$$\frac{1}{5} \left(\left(\frac{2}{\sqrt{5}+1}\right)^{-a-1}-\left(\frac{1}{2} \left(1-\sqrt{5}\right)\right)^{a+1}\right)^2+\frac{1}{5} \left(\left(\frac{1}{2} \left(\sqrt{5}+1\right)\right)^a-\left(\frac{1}{2} \left(1-\sqrt{5}\right)\right)^a\right)^2-\frac{\left(\frac{2}{\sqrt{5}+1}\right)^{-2 a-1}-\left(\frac{1}{2} \left(1-\sqrt{5}\right)\right)^{2 a+1}}{\sqrt{5}}$$

You need to carefully open up all brackets, cancel out the terms to get zero.


Borel $\sigma$-algebra = monotone class = Dynkin system Let $G_n = \{U \subset \mathbb R^n \mid U \text{ open }\}$.

Why does it hold true that $\sigma(G_n) = m(G_n) = \delta(G_n)$, where $\sigma$ is the Borel $\sigma$-algebra, $m$ the smallest monotone class created by the open sets and $\delta$ the smallest Dynkin-system produced by the open sets?

$1)$ $G_n$ is a family of sets that is closed under finite intersections,so by a known theorem we have that $\sigma(G_n)=\delta(G_n)$

Also note that every closed set in $\Bbb{R}^n$ can be expresed  as a  countable intersection decreasing open  of open sets.

Indeed if $F$ is open then $F=\bigcap_n\{x \in \Bbb{R}^n:d(x,O)<\frac{1}{n}\}$

$m(G_n)$ contains all open and closed sets.

So $m(G_n)$ contains the algebra generated by the open sets.

$2)$ Note that every Dynkin class is a monotone class.

$3)$ Also note that $m(G) \subseteq \delta(G)$, whre $G$ is any family of subsets of $\Bbb{R}^n$

From $1),2),3)$ you have the conclusion.


How Do Integrals Behave In Higher Dimensions? As I started looking into Linear Algebra and reading Flatlands: A Romance of Many Dimensions after learning Calculus, I wondered how integrals act in higher dimensions. So here I am on Stack Exchange. How do integrals behave in higher dimensions?

So far you've been taking integrals with respect to one variable, whether that be $x$ or $y$. An integral's main purpose is to provide the area under the curve.

Of course in higher dimensions, for example, 3$D$, that area is known as "volume".

Now you actually have 2 axis, the $x$ and $y$ axis, and so you will encounter what's known as a double integral, where you integrate with respect to both axes.

Right now you are fimilar with single integrals, or integrals pertaining to only one axes.

Here is a visual representation.

Here you can integrate via $x$ or $y$, and you will get the area of the blue region.

In 3D, you will have two set of axes, and your output will be on the third axes (usually $z$)

Notice how here the axes are $x$ and $y$ and your output is in the $z$ axis. Here you have volumn between two sets of points.

Anyways, that's the basic idea. Integrals.

How bout this? Your velocity as a function of time is $<2t,t^2>$. Then your location is $\int <2t,t^2>\, dx=<t^2,t^3/3>+C$.


Approximating a rational function by a rational of exponentials Given

$G(s)=\frac{\displaystyle\sum\limits_{k=0}^{m-1} b_ks^k}{\displaystyle\sum\limits_{j=0}^{n-1} a_js^j}$

with known positive integers $m,n$ and known real coefficients $b_k, a_j$, and moreover

$H(s)=\frac{\displaystyle\sum\limits_{k=0}^{p-1} c_ke^{ksT}}{\displaystyle\sum\limits_{j=0}^{q-1} d_je^{jsT}}$

with known positive integers $p,q$, known real value T, and unknown real coefficients $c_k, d_j$. Is it possible to find a least-square approximation of G(s) by H(s) on a compact interval of the imaginary axis ($s=i\omega$)

$\int\limits_{\omega_1}^{\omega_2} |G(i\omega)-H(i\omega)|^2 d\omega= min$

with a finite algorithm (i.e. does it reduce to a linear equation system by means of some clever trick), or is it a general nonlinear problem, all with iterative solution, ambiguity and such?

More specifically if $G(s)$ is sampled on an interval of the imaginary axis (finite Fourier series, from measurements) and least-square approximation is demanded only for the samples, does this simplify anything?

As far as I can see, the latter could also be stated slightly differently: is it possible to approximate a Fourier series by the quotient of two Fourier series with presumably much lower orders, by means of some finite algorithm?

After thinking about this a little deeper, I have found a solution - not of the original problem, to be fair, but of a problem that is similar with respect to the underlying approximation goal.

If $G(s)$ is approximated by

$H(s)=\frac{U(s)}{V(s)}$

(on the imaginary axis) then for many practical cases (I wish I had the time and mathematical discipline to clarify which cases these are...) will also $V(s)\cdot G(s)$ be approximated by $U(s)$. So if I modify the original problem to finding a minimum of the functional

$S[U,V]:=\displaystyle\int_{-\omega_0}^{\omega_0} |V(i\omega)G(i\omega)-U(i\omega)|^2 d\omega$

then setting the partial derivatives of $S$ to zero reduces to a usual linear least square problem, the solution of which is equivalent to an inhomogeneous matrix equation. The only thing that is needed to guarantee the uniqueness of the solution is to normalize the fraction that is $H$ by setting one coefficient to a known constant, for example $d_0=1$.

It turns out that the coefficients of the matrix equation are just one sort or the other of a discrete Fourier transform of $G(s)$ and $|G(s)|^2$. These amount to the impulse response of the "transfer function" $G(s)$ and its autocorrelation function.

I already did some numeric testing and the method (I don't believe that I am the first one to discover it) seems pretty stable for usual filter theory transfer functions (like Butterworth) and physical transfer functions (like one mass spring damper).

The only problem I have identified is, that the polynomials in $H(s)$ do sometimes contain roots that make the digital filter $H(s)$ unstable although the analog filter $G(s)$ is stable. That is the general problem of the sensitivity of roots from the polynomial coefficients. Since the least square method constrains the polynomial coefficients and not the roots, the roots can be pretty off, numerically.

It was impressive for me to see that relative coefficient changes of 1e-12 can result in changes of root magnitudes by several % (from slightly below 1 to slightly above 1). For the general topic of root instability, see https://en.wikipedia.org/wiki/Wilkinson%27s_polynomial

Sorry for my colloquial style, but I just wanted to quickly share the principle with those who are interested in the original question. The mathematics behind all that is really trivial for the most part.


Large degree polynomial function How does one solve for $x$ in a polynomial of the form
$ax^{n} + bx^{\left(n - 1\right)} + c = 0$, given that $n$ is a larger number ?. For example:

$\displaystyle x^{100} - 3x^{99} + 1 = 0$


Correct Notation of Mean Value Theorem for Vector-Valued Function Let ${\bf f}({\bf x})$ be a function ${\bf f}: \mathbb{R}^n \to \mathbb{R}^m$ with continuous derivatives ${\bf H}({\bf x})$. We wish to approximate ${\bf f}({\bf x}_0)$ by ${\bf f}({\bf x})$.

It is well known that for $m > 1$, we cannot guarantee the existence of a vector $\bf \tilde{x}$ between ${\bf x}_0$ and ${\bf x}$ such that
$${\bf f}({\bf x}) = {\bf f}({\bf x}_0) + {\bf H}({\bf \tilde{x}})({\bf x}-{\bf x}_0) $$

Hence, the MVT cannot be directly applied to vector-valued function. However, a straightforward modification to the MVT can yield fruitful results, but seems ignored by many sources (including this very site).
We can apply the MVT to each of the $m$ components of ${\bf f}$ separately, and thus we can write
$${f_k}({\bf x}) = f_k({\bf x}_0) + {\bf h}_k({\bf \tilde{x}}_k)({\bf x}-{\bf x}_0) , k = 1,...,m$$
I have seen some authors combine the $m$ results and write

$${\bf f}({\bf x}) = {\bf f}({\bf x}_0) + {\bf H}({\bf {x}_*})({\bf x}-{\bf x}_0) $$
where ${\bf {x}_*} =[{\bf \tilde{x}}_1, ...,{\bf \tilde{x}}_m]^T$ is now a $m \times n$ matrix, and ${\bf H}({\bf {x}_*}) = [{\bf h}_1({\bf \tilde{x}}_1),...,{\bf h}_m({\bf \tilde{x}}_m)]^T$.

Although I remember seeing this notation multiple times, I can't remember how exactly how it was expressed, or in which article I viewed it.
Could someone kindly reminds me how to properly use this notation for vector-valued functions?

The following might be what you are looking for:

Given such an ${\bf f}$ with open domain $\Omega\subset{\mathbb R}^n$, let $\tilde\Omega:=\bigl\{({\bf x},{\bf y})\bigm| [{\bf x},{\bf y}]\subset\Omega\bigr\}$, whereby $[{\bf x},{\bf y}]$ denotes the segment connecting ${\bf x}$ with ${\bf y}$. Then there is a continuous function $L:\>\tilde\Omega\to{\cal L}({\mathbb R}^n,{\mathbb R}^m)$, the linear maps from ${\mathbb R}^n$ to ${\mathbb R}^m$, such that
$${\bf f}({\bf y})-{\bf f}({\bf x})=L({\bf x},{\bf y}).({\bf y}-{\bf x})\qquad\forall\>({\bf x},{\bf y})\in\tilde\Omega\ .$$
Furthermore $$L({\bf x},{\bf x})=d{\bf f}({\bf x})\qquad\forall\,{\bf x}\in\Omega\ .$$
This $L({\bf x},{\bf y})$ is exactly what you denote by ${\bf H}({\bf x_*})$. But neither your ${\bf x}_*$ is a point, nor ${\bf H}$ is a function of one vector variable.

Proof. Fix $({\bf x},{\bf y})\in\tilde\Omega$, and consider the auxiliary function
$$\phi(t):={\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr)\qquad(0\leq t\leq1)$$
with derivative $\phi'(t)=d{\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr).({\bf y}-{\bf x})$. One then has
$${\bf f}({\bf y})-{\bf f}({\bf x})=\phi(1)-\phi(0)=\int_0^1\phi'(t)\>dt =\int_0^1 d{\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr).({\bf y}-{\bf x})\>dt=\left(\int_0^1 d{\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr)\>dt\right).({\bf y}-{\bf x})\ .$$
It follows that
$$L({\bf x},{\bf y}):=\int_0^1 d{\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr)\>dt$$
has the stated properties.


Number of solutions $X$ to $AX=XB$ in $\mathbb F_2$ It is a well-known theorem that in an arbitrary field $F$, if $A$ is an $m\times m$ square matrix and $B$ is an $n\times n$ square matrix, then there is a unique $m\times n$ solution $X$ to the equation
$$AX=XB$$
if and only if $A$ and $B$ share no eigenvalues.

My question is this: what can be said about the number of solutions $X\in \mathbb F_2$ to the above equation? Is the number of nonzero solutions equal to the number of common eigenvalues? If so, how can one prove this?

$X$ maps each eigenvector of $B$ to either $0$ or an eigenvector of the same eigenvalue of $A$, since if $Bv = \lambda v$, $AX v = X B v = \lambda X v$.

Let's suppose for simplicity that $B$ is diagonalizable, with a basis $b_j$ of eigenvectors corresponding to eigenvalues $\lambda_j$ that may or may not be eigenvalues of $A$.

For each $j$, we can choose $X b_j$ to be any member of the eigenspace of $A$ for $\lambda_j$, and once we have done that we have specified a solution $X$.  Thus the number of solutions for $X$ is the product over $j$ of the cardinalities of the eigenspaces of $A$ for $\lambda_j$ (note that this is $1$ if $\lambda_j$ is not an eigenvalue of $A$).

Suppose $X_1$ and $X_2$ are both solutions. Then $A(X_1+X_2)=AX_1+AX_2=X_1B+X_2B=(X_1+X_2)B$. Thus, $X_1+X_2$ is a solution. So the number of solutions can be $0$, $1$, or infinity. This is a common situation in linear algebra: you either have no solutions, a unique solution, or infinite solutions.


In a finite dimensional vector space a positive operator can have infinitely many square roots "In a finite dimensional vector space a identity operator (which is positive) can have infinitely many square roots". I see only $2 ^ {C(n,2)}$ when dimension is $n (>1)$ by swapping two basis elements. Where am I wrong ?

Let $X: \mathbb{R} \to \mathbb{R}^{2 \times 2}$,

$$X(t) := \begin{bmatrix} 1 & t \\ 0 & -1 \end{bmatrix}.$$

Then

$$(X(t))^2 := \begin{bmatrix} 1 & t \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & t \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$

for all $t \in \mathbb{R}$. I leave it to you to generalize this example to higher dimension spaces.

Note that these are merely all triangular matrices $X$ such that $X^2 = I$. There are many other, nontriangular ones, with the same property.

First of all, a positive operator and an identity operator are much different.

Second of all, the square root of an identity is an identity, and so forth.


Find nearest integer with certain prime factors Trying to answer this question, I have sketched a strategy:

Find an integer $N$ "near" to $10^9$ with factors $5$ and $7$, that is, $5^n\cdot7^m$
Conpute $k=N-\varphi(N)$ ($\varphi$ is the totient function). This is the number of multiples of $5$ or $7$ lesser than $N+1$. Think in $k$ as an estimation.
If $N$ is really near from $10^9$ we can count by hand to fix $k$.

My question is: is there some efficient method to achieve the first step? I have written $10^9$ in base $35$ (it is $19.01.13.21.18.20$), but I'm stuck.

Generally speaking: given a positive integer $N$ and some primes $p_1,\ldots,p_k$ and a tolerance $T$, is there any efficient method to find nonnegative integers $n_1,\ldots,n_k$ such that
$$|p_1^{n_1}\cdot\ldots\cdot p_k^{n_k}-N|\leq T$$
if they exist, or prove that they don't if that is the case?

For two factors there is an efficient algorithm.
The following program will find the requested approximation using $O(log(n))$ operation.
It is modeled on  the well known but unnamed algorithm that checks two sorted lists if there is a number in each list such that they sum up to a given number (here and here)

def find_min(n,factor1,factor2):
    product=factor1
    while product<n:
        product*=factor2
    min_diff=product-n
    min_diff_at=product
    while True:
        if product>n:
            if product%(factor2*factor2)!=0:
                break
            product//=factor2
        elif product<n:
            product*=factor1
        else:
            break
        if abs(product-n)<min_diff:
            min_diff_at=product
            min_diff=abs(product-n)
    return(min_diff_at)

The program generates the following output.

print(find_min(1000000000,5,7))
1008840175

This algorithm can be extended to more factors using two sorted lists where the first list contains all products smaller than the target number that are products of powers of the first half of the given factors and the second list contains the analogous numbers created by the second half of the factors but I don't think that such an algorithm will make sense.
Either the length of the list is rather small and the probability that the approximation is very near to the target number is small or the lists are very large (about the square root of the target number) and scanning the whole list will take to long.
Here a version of the program that outputs more details.

Sketch of an plan:  Using $log$ base $10$, write

$$n \log 5 +m \log 7 \approx 9.$$

Now we need to approximate $9$ with linear combinations of
two irrational numbers.  The method here seems to be just
the ticket:

https://mathoverflow.net/questions/70035/