Does Fermat's Last Theorem imply the modularity theorem? The Wikipedia article on the proof of Fermat's Last Theorem has this sentence

  If the link identified by Frey could be proven, then in turn, it would mean that a proof or disproof of either of Fermat's Last Theorem or the Taniyama–Shimura–Weil conjecture would simultaneously prove or disprove the other.

This suggests FLT and the modularity theorem are equivalent. While the fact that the modularity theorem implies FLT was a rather important part of Wiles' proof, I wasn't aware the reverse implication was true. Is it?

This discussion seems to me a bit weird. Let us recall the facts:

1) The Shimura-Taniyama-Weil conjecture (now a theorem of Wiles et. al.) states that any elliptic curve $E$ defined over $\mathbf Q$ is "modular". This means roughly that the Hasse-Weil L-function $L_E(s)$ attached to $E$ comes from an L-function L$(f_E,s)$ attached to a certain modular form $f_E$. More precisely, let $f_E$ be the inverse Mellin transform of $(2\pi)^{-s}\Gamma(s)L_E(s)$; then $f_E\in S_2(\Gamma_0 (N))$, where $N$ is the conductor of $E$ and $f_E$ is a Hecke form.

2) Suppose that the Fermat equation $a^p+b^p=c^p$ admits a non trivial solution and consider the elliptic curve $E_{a,b,c}$ defined over $\mathbf Q$ by the equation $y^2=x(x-a^p)(x+b^p)$ (Hellegouarch, 1969). Frey (1986) conjectured that $E_{a,b,c}$ could not be modular (see the appendix). Then (Mazur and) Ribet (1990) proved a deep theorem on modular representations of $Gal(\bar {\mathbf Q}/\mathbf Q)$ which implied Frey's conjecture.

3) FLT follows from 1) and 2). Thus the modularity thm. implies FLT, but the converse  does not hold, for purely logical but also for "intuitive" reasons.

Appendix. I take this opportunity to reproduce a short historical account which I gave as a comment to  A question on FLT and Taniyama Shimura

Frey picked up this curve in Yves Hellegouarch's thesis *Courbes elliptiques et équation de Fermat", Besançon, 1972. Hell. himself has written a nice introductory book on the subject, "Invitation aux math. de Fermat-Wiles", Masson ed., 1997 (I don't know if there is an English translation). At the beginning of the 70s, Hell. had not the necessary coceptual tools to conjecture that his elliptic curve could not be modular, but he was aware that it possessed so many extraordinary properties that he had nicknamed it "Roland's mare" (Roland - Orlando in Italian - was presumably a nephew of emperor Charlemagne, a legendary hero in French mythology, and the central character of Tasso's epic poem "Orlando furioso"), a fabulous animal which had all the qualities in the world except... existence. Only in 1985, with all the theoretical progress made in more than 10 years, Frey was able to state his conjecture that "Roland's mare" could not be modular. See the historical appendix of Hell.'s book.

Yes this reverse is also true
If there were a counterexample for FLT, it would also provide a counterexample of modularity by a specific construction.


Help in understanding search of Vantage-Point tree This is my reference: http://stevehanov.ca/blog/index.php?id=130

A vantage-point tree is a way of organizing a set of points so that finding the n-nearest neighbors is as efficient as possible. It builds this tree by picking a point from a set of points (called the vantage point). The median distance from that point to all other points in space is sort of the boundary that splits off that set into two equally sized subsets. This is repeated recursively for each subset that was made until no more points are left (when all points have been made vantage points).

Now searching a VP-tree is where I need help understanding. Basically, when a search is done, a new point is plotted into that space of points, and you need to find the n-nearest neighbors to that point. I especially have a problem with the "tau." and its purpose in this algorithm.

I think what may be confusing you is that he talks about "the farthest point that we have found" but actually means the farthest point of the up to $n$ candidates we've collected so far. Once we've collected $n$ candidates, we can prune all parts of the tree that contain only points worse than the worst of those candidates. To do that, we need to keep track of the distance to our worst candidate so far. That's $\tau$.

You always search first the subtree where the query point belongs to, on your way down the tree, until you reach a leaf. Every time you return from the search, if the distance to the n-th neighbor found so far intersects the boundary to the other subtree, then you have to search the other subtree as well.

The idea is that the distance to the n-th neighbor will be shrinking as you go down so that you won't have to search within the other subtree (especially the subtrees closer to the root of the tree where each subtree corresponds to a lot of points excluded from the search).

Persuade yourself by drawing a rectangle, as your corpus, and a vantage point as the center of a disk denoting the median distance.

There are two cases for the query point: to fall into the disk or out of it. And two subcases for each case: the disk centered at the query point denoting the distance to its n-th neighbor to remain within the same subspace where the query point is, or to intersect the other subspace as well.


Finding $a_n$ if $a_{n+1}=3a_n-2a_{n-1}$ with $a_1=2$ and $a_2=3$ Given $a_1=2,\; a_2=3,\; a_{n+1}=3a_n-2a_{n-1}$.

How can i prove that:  $a_n=2^{n-1}+1$

I Tried to isolate $a_n$ but it doesn't get me anywhere.

Thanks.

The general method to solve the recursion $$a_{n+1}=\color{green}{\bf 3}a_n-\color{green}{\bf 2}a_{n-1},$$ is to consider the characteristic equation $$r^2=\color{green}{\bf 3}r-\color{green}{\bf 2},$$ whose roots are $r=\color{red}{\bf1}$ and $r=\color{red}{\bf2}$ hence there exists $A$ and $B$ such that, for every $n$, $$a_n=A\cdot\color{red}{\bf2}^n+B\cdot\color{red}{\bf1}^n=A\cdot2^n+B.$$
The initial conditions yield $A$ and $B$.

In full generality, the characteristic equation of the recurrence $$a_{n+1}=\color{green}{\mathbf b_1}a_n+\color{green}{\mathbf b_2}a_{n-1}+\cdots+\color{green}{\mathbf b_k}a_{n-k+1},$$ is the polynomial identity $$r^k=\color{green}{\mathbf b_1}r^{k-1}+\color{green}{\mathbf b_2}r^{k-2}+\cdots+\color{green}{\mathbf b_k}.$$


Differentiability of a Special Function Prove that $f(x)=\sum_{n=1}^\infty 1/2^n(\cos3^nx)$ is continuous but nowhere
differentiable on $\mathbb{R}$.

I have proved the continuity part, but unable to do the second one.
Thanks for any help.

I am not a proper mathematician, so I'll try to give a heuristic argument, why the function is not differentiable at all $x$.

We should start by applying the definition of the derivative to the function:
$$f'(x) = \lim_{\delta \to 0} \frac{f(x + \delta) - f(x)}{\delta}$$

Since the limit and the sum are commutable, we can get the expression:
$$f'(x) = \sum_{n=1}^{\infty} 2^{-n} \lim_{\delta \to 0} {\left( \cos(3^nx) ( \cos(3^n\delta) - 1 ) - \sin(3^nx)\sin(3^n\delta) \right) \over \delta}$$

One can see that no proper limit can be found for this sum, as for large $n$ the value of some of the terms becomes indeterminate when the inequality
$$3^n\delta \ll 1$$
is no longer satisfied. Since some of the terms in the sum don't have a proper limit, the whole sum doesn't have a proper limit.

This is by no means a proper, mathematically rigorous answer, but maybe it helps.

EDIT:

Now let's consider the 'last' terms in the series:
$$\lim_{n \to \infty} 2^{-n} \lim_{\delta \to 0} {\left( \cos(3^nx) ( \cos(3^n\delta) - 1 ) - \sin(3^nx)\sin(3^n\delta) \right) \over \delta}$$

or rather

$$\lim_{n \to \infty} 2^{-n} n{\left( \cos(3^nx) ( \cos(3^n/n) - 1 ) - \sin(3^nx)\sin(3^n/n) \right)}$$

which does not have a proper limit (I hope that the L'Hoptal's theorem can be applied here, but I suspect that is not the case)... Is this not sufficient for mathematicians? :)

Think about the nature of symmetry of the curve here, mainly given an interval $(-a,a)$ (or a similar location about another point than 0) what does the function look like compared to the interval $(\frac{-a}{2}, \frac{a}{2})$, and what would that mean for intervals of the form $(\frac{-a}{2^n}, \frac{a}{2^n})$ as n gets larger?

Another point would be to direct you to look at the Weirstrass function, but that will likely lead to a give away of the answer, so depending on what approach you want to take you can either look there or take the hint I gave and run with it.


First-order logic and Peano Arithmetic paradox I'm working through the Stanford Introduction to Logic course, and seem to have proved that all natural numbers are equal to zero: Ax.(x = 0). Obviously, this is incorrect, but I can't see how. My proof is by induction:

Base case: 0 = 0 by premise
Inductive case: Ax.((x = 0) -> (s(x) = 0))

If we assume that all natural numbers equal zero: x = 0
Then, since all natural numbers equal zero, the successor of any natural number also equals zero: s(x) = 0 by simple substitution

Conclusion: Ax.(x = 0)

I worked this up formally in Stanford's Fitch editor as well:

Can anyone explain my error? Thank you.

When you write “If we assume that all natural numbers equal zero”, you are assuming the thing that you want to prove.

A proof by induction would be as follows:

first you prove that the assertion holds when $x=0$ (as you did);
then you prove that if the statement holds for $x$, then it also holds for $s(x)$.

And it's here where the problem lies, because one of the Peano axioms states that $0$ is never of the form $s(x)$.

David C. Ullrich had the most useful answer, which I'll repeat here: The Universal Introduction on line 3 is invalid because x is free in the active assumption x = 0. Thank you for finding the error!


A concentration problem Consider $N$ variables $X_1,X_2,\ldots,X_N$ with $\Pr(X_i=a_i)=\Pr(X_i=-a_i)=1/2$. Here $a_1,a_2,\ldots,a_N\in [0,1]$.

Does there exist some concentration result about $\sum_{i=1}^N X_i$?

Let $S(a,\epsilon)=\sum_{k=1}^N \epsilon_i a_i$ with $\epsilon=(\epsilon_1,\ldots,\epsilon_N)$ and  $\epsilon_i\in\{\pm 1\}$ independent and uniformly distributed, and $a_i\in [0,1].$ We can actually assume $a_i\leq a_{i+1}$ (since the signs range over all of $\{\pm 1\}^n,$) and take $a_i\in (0,1]$ since otherwise we can reduce the dimension $N$ and reconsider the problem for $N'<N.$

Clearly $S(a,\epsilon)=-S(a,-\epsilon)$ so the distribution of $S(a,\epsilon)$ is symmetric around zero. Let any lower bound $B$ on $$\mathbb{P}\left\{|S((1,\ldots,1),\epsilon)|\leq 1\right\}\geq B$$ for the case $a=(1,1,\ldots,1)$ for all $1\leq i\leq N$ be given. Recall that we can assume $0<a_1\leq a_2\leq \cdots \leq a_n \leq 1.$

Define the all one vector as $a^{(0)}$ and now consider the vector
$$a^{(1)}=(a_{i^{\ast}},\ldots a_{i^{\ast}},1,\ldots,1)$$
where we define $a^{(1)}_i=a_{i^{\ast}}$ for all $i\leq i^{\ast}.$

Lemma: Let $Z$ be any symmetric random variable, with support on the integers. Then $Z+\epsilon a$ where $\epsilon=\pm 1$ with equal probability
satisfies the inequality
$$\mathbb{P}\left\{|Z+\epsilon a|\leq 1\right\}\geq \mathbb{P}\left\{|Z+\epsilon |\leq 1\right\}$$ whenever $a\in (0,1].$ This can be seen by the fact that the probability mass function of $Z+\epsilon a$ is the sum $$\frac{q(z-a)+q(z+a)}{2}$$ while that of $Z+\epsilon$ is the sum $$\frac{q(z-1)+q(z+1)}{2}.$$ Here $q(z)$ is the probability mass function of $Z.$
This is because $a_{i^{\ast}}<1$ and the random sums $S(a^{(1)},\epsilon)$ are ``more central'' than the sums $S(a^{(0)},\epsilon)$ since their distribution result from at least one convolution with dominated random variables on $\{-a_{i^{\ast}}, a_{i^{\ast}}\}$
instead of random variables on $\{-1,1\}.$

The Lemma's argument can then be repeated using induction and applying it to compare
$$S(a^{(0)},\epsilon)\textrm{ with } S(a^{(1)},\epsilon),$$
and then moving from $a^{(1)}$ to $$a^{(2)}=(a_{i^{\ast}-1},\ldots a_{i^{\ast}-1}, a_{i^{\ast}},1,\ldots,1)$$
and applying the argument to obtain that
$$\mathbb{P}\left\{|S(a^{(2)},\epsilon)|\leq 1\right\}\geq B$$
still holds. Then use $a^{(3)},$ etc to finish the argument.

If $a_i$ are deterministic, than you can think about new random variables  $Y_i=\frac{X_i}{a_i}$ and you already have the lower bound for this.


About ordinals and cardinals in structural set theory Among the most important concepts of set theory for mathematical real life applications are ordinal numbers and cardinal numbers. In material set theory, ordinal numbers are defined as transitive sets, well-ordered by the membership relation. It can be shown that every well-ordered set is order isomorphic to a unique ordinal number. In categorical terms, ordinal numbers form a skeleton of the category of well-ordered sets. By the well-ordering theorem, every set admits a well-ordering, whence is isomorphic to the underlying set of an ordinal number. Choosing for each isomorphism class the smallest such ordinal number, we obtain a skeleton of the category of sets given by the cardinal numbers. Note that by a sufficiently strong version of global choice, the existence of skeletons is immediate.

Now the categorical minded reader might feel uncomfortable, since quite a lot "evil" is going on here. First of all, working in structural set theory like SEAR or ETCS, there is no such thing as a membership relation on a set. Secondly, skeletons are not really of use and most often, it is better to just prove theorems for all objects of a given category instead of restricting one's attention to representatives for each isomorphism class. Finally in structural set theory it is not even possible to express equality of abstract sets, hence uniqueness of the ordinal number associated to a well-ordered set.

So the following questions arise: Do we need to know a relatively concrete description of representatives for isomorphism classes in $\mathbf{Set}$ at all? Do we need material properties of those, like being a transitive set in set theory itself or elsewhere? Is it possible to recover the theory of transfinite induction in structural set theory, replacing the partial order of ordinal numbers by the preorder aka thin category of well-ordered sets? Where do the technical details of developing the theory of ordinal numbers in material set theory move? Perhaps most of them disappear or are related to proving that well-ordered sets are "well-preordered" by the initial segment-relation? Are there philosophical issues when interpreting something like $\dim V=\aleph_0$ as "there is a basis of $V$, which is isomorphic to $\mathbb{N}$" rather than "the mathematical object $\dim V$ is equal to the mathematical object $\aleph_0$"? Perhaps the first interpretation is more natural anyway?

To answer the more "philosophical" part: It is very useful to have a skeleton of $\mathsf{Set}$. First recall that natural numbers were invented to form a skeleton of $\mathsf{FinSet}$. Humans wanted to compare finite sets of objects in real life with each other, and it was useful to have a criterion when two finite sets are (what we now call) equipotent. This remains true for not-so-real-life objects in mathematics, and it remains true for non-finite sets. According to your profile ;-), you are more interested in algebra, so let me mention as an example the Ulm invariants which are cardinal numbers which classify certain torsion groups.

(Since you have asked a lot of questions, it could be a good idea to ask them separately, if you are still interested in them.)

In the Homotopy Type Theory Book, a more structural approach to define cardinal and ordinal numbers is presented. See Chapter 10:

  The type of cardinal numbers is the 0-truncation of the type Set of sets:
  $\mathrm {Card} :≡ ||\mathrm{Set}||_0$
  An ordinal is a set $A$ with an extensional well-founded relation which is
  transitive, i.e. satisfies $∀(a, b, c : A).(a < b) → (b < c) → (a < c)$.


Find $f$ such that $(f(x))^2 \ge f(x + y)(f(x) + y), \forall x, y$ Find all $f : (0, \infty) \rightarrow (0, \infty)$ such that $(f(x))^2
 \ge f(x + y)(f(x) + y), \forall x, y \gt 0$

My guess there is no such function but I cannot prove it.

The most obvious idea is to make $y=x$ but it seem to lead nowhere.

Also, it's easy to show $f$ is strictly decreasing

Suppose that such a function $f$ exists.  It is easily seen that $f$ is strictly decreasing.  Let $a_0:=1$ and, for $k=1,2,\ldots$, write
$$a_k:=a_{k-1}+f\left(a_{k-1}\right)\,.$$
From the given inequality, plugging in $x:=a_{k-1}$ and $y:=f\left(a_{k-1}\right)$ yields
$$\big(f\left(a_{k-1}\right)\big)^2\geq f\left(a_{k}\right)\,\big(2\,f\left(a_{k-1}\right)\big)\,,$$
whence
$$f\left(a_k\right)\leq \frac{1}{2}\,f\left(a_{k-1}\right)$$
for all $k=1,2,\ldots$.  By induction on $k=0,1,2,\ldots$,
$$f\left(a_k\right)\leq \frac{b}{2^k}\,,$$
where $b:=f(1)$.  Now, note that
$$a_k-a_{k-1}=f\left(a_{k-1}\right)\leq \frac{b}{2^{k-1}}$$
for each $k=1,2,\ldots$.  We have
$$a_k= a_0+ \sum_{j=1}^k\,\left(a_j-a_{j-1}\right)< 1+\sum_{j=1}^\infty\,\frac{b}{2^{j-1}}=1+2b\,.$$
This shows that
$$f(1+2b)<f\left(a_k\right)\leq \frac{b}{2^k}$$
for all $k=0,1,2,\ldots$.  However, as $k\to\infty$, the upper bound goes to $0$, so that
$$f(1+2b)\leq 0\,.$$
However, this contradicts the requirement that $f(1+2b)>0$.  Ergo, there is no function $f$ with the required property.

Rearranging since $x, y, f > 0$, we have

$\frac{f\left(x+y\right)-f\left(x\right)}{y}\leq\frac{-f\left(x+y\right)}{f\left(x\right)}$

take the limit (if it exists) as $y$ goes to $0$ and $f'(x)\leq-1$.  The function is strictly decreasing (i.e. even where $f'$ doesn't exist) so if $f(1)=a$, then $f(1+a)\leq0$ which is a contradiction.


Rule for factoring out variable in limit I am a bit confused about the proper way to calculate the limit of a function by factoring out the variable that strives for infinity.

If you have this:

$\lim_{x\rightarrow \infty } \sqrt{x} \left(\sqrt{x+1}-\sqrt{x}\right)$

for some reason you cannot do this:

$ = \lim_{x\rightarrow \infty } \sqrt{x} \left(\sqrt{x(1+ \dfrac{1}{x})}-\sqrt{x}\right) = \lim_{x\rightarrow \infty } \sqrt{x} \left(\sqrt{x}-\sqrt{x}\right) = 0$

but rather should do this:

$ = \lim_{x\rightarrow \infty } \sqrt{x} \frac{\left(\sqrt{x+1}-\sqrt{x}\right)\left(\sqrt{x+1}+\sqrt{x}\right)}{\left(\sqrt{x+1}+\sqrt{x}\right)} = \lim_{x\rightarrow \infty } \sqrt{x} \frac{x+1-x}{\left(\sqrt{x+1}+\sqrt{x}\right)} = \lim_{x\rightarrow \infty } \frac{\sqrt{x}}{\sqrt{x(1+\frac{1}{x})}+\sqrt{x}} = \lim_{x\rightarrow \infty } \frac{\sqrt{x}}{2\sqrt{x}} = \frac{1}{2}$

Why is it legit to factor out $x$ in the second approach but not in the first one? Is there a general rule for when you are allowed to do so and when not?

We look at the calculation at the end of the OP that you labelled as legitimate. It is in fact not legitimate, but happens to give the right answer. We should not write
$$\lim_{x\to\infty}\frac{\sqrt{x}}{\sqrt{x\left(1+\frac{1}{x}\right)}+\sqrt{x}}=\lim_{x\to\infty}\frac{\sqrt{x}}{2\sqrt{x}}.\tag{????}$$
For if we do that, we have let $x\to\infty$ in part but not all of our expression. We should instead do the correct algebraic manipulation (cancel the $\sqrt{x}$) and write
$$\lim_{x\to\infty}\frac{\sqrt{x}}{\sqrt{x\left(1+\frac{1}{x}\right)}+\sqrt{x}}=\lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x}}+1}.$$
Now we can safely let $x\to\infty$, with full control over what is going on.

Although $\lim_{x \to \infty} \sqrt{x} (3-3) = \lim_{x \to \infty} \sqrt{x} \cdot 0 =0$, in your case you have $\lim_{x \to \infty} = \sqrt{x}(\sqrt{x} - \sqrt{x}) \neq \lim_x \sqrt{x} \cdot (\lim_x \sqrt{x} - \sqrt{x})$ because $\infty \cdot 0 $ is an indeterminate form. Also $\lim_x (\sqrt{x} - \sqrt{x}) \neq \lim_x \sqrt{x} - \lim_x \sqrt{x}$ because \infty - \infty is an indeterminate form too.


How many people do you need to guarantee that two of them have the same initals? An auditorium has a seating capacity of 800. How many seats must be occupied to guarantee that at least two people seated in the auditorium have the same first and last initials?

I thought $26 \cdot 26$ but the answer key has $26 \cdot 26+1$ Where's the +1 coming from? Is the whole thing about the auditorium a red haring, I don't see how it matters?

$26 \cdot 26$ would count all possible pairs of letters. With $26 \cdot 26$ people it is possible that they all have different initials. The $+1$ ensures there exist at least two people with the same initial.

Maybe it would help to think of a smaller case. If one rolls a dice $6$ times, it's possible to roll a different number each time. However, rolling a dice $6+1=7$ times ensures that at least one number was rolled more than once.

The first two intials can be "$aa,ab,ac,ad,\ldots,zw,zx,zy,zz$".
So, total possibilities for different types of first two intials are $26\times 26 = 676$.
And they asked us to have same first two intials for atleast $2$ children, so we can take any of the $676$ possibilities as a common first two intials.
So, number of ways for at least $2$ students who have the same
first two initials are $676+1 = 677$.


Does span(A) < span(B) imply A < B? Given $A \subseteq B \subseteq V$ for some vector space $V$, we know that $\mathrm{span}(A) \subseteq \operatorname{span}(B) $. We can see that as span of $B$ can make any elements of $A$.

However does the inverse of the statement hold? So if $\operatorname{span}(A) \subseteq \operatorname{span}(B) $, can we say $A \subseteq B$?

No. For example, take $V = \mathbb{R}^2$, $A = \{(2,0)\}$ and $B = \{(1,0), (0,1)\}$. Since $B$ is a basis for $V$, $\operatorname{span}(B) = V$, so $\operatorname{span}(A) \subseteq \operatorname{span}(B)$. However, $A \not\subset B$.

No. In $\mathbb{R}^2$, take $A=\{(1,0)\}$, $B=\{(2,0),(0,1)\}$. Then $\mathrm{Span}\, A=\mathbb{R}\subseteq\mathbb{R}^2$ and $\mathrm{Span}\, B=\mathbb{R}^2$ (so $\mathrm{Span}\, A\subseteq\mathrm{Span}\, B$), but $A\not\subseteq B$, since $(1,0)\notin B$.


What is matrix inequality such as $A>0$ or $A\succ 0$? I am trying to gather here different meanings of the same symbol, inequality symbol or the succ symbol. I find many other use them so many different ways.
Sometimes, $A>0$ means $\bar x^T A \bar x >0$. Sometimes, $A>0$ means element-wise i.e. $a_{i,j}>0$ for all $i,j$.

How do I know which definition of inequality people are meaning?

Examples

[Solved] Meaning of this? Why is it $A\succ 0$?  LMI and example from Boyd's book. $\succ$ means positive-definiteness: check PD with Cholenksy decomposition, positive eigen-value check or Sylvester criteria. Chat.

Strict matrix inequality defined here requiring symmetric matrices.

The curved symbol (in my experience) means that the difference is positive definite (so, with zero it means that the matrix is positive definite). The $>$ depends on the context.

For squared matrix $A$, $A>0$ means that $A$ is positive definite. Also, the real parts of all of $A$s eigenvalues are positive.


Difference between $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}_n$ In every Modern Algebra book I've read, I've seen that the groups $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}_n$ are isomorphic, but not equal. I understand the difference between "isomorphic" and "equal," but this particular example raises a couple of questions for me.

I know the first group consists of cosets $k + n\mathbb{Z}$, and the second group consists of equivalence classes $[k]_n$, but isn't it true that $k + n\mathbb{Z} = [k]_n$ for every integer $k$? (They are both sets containing the same elements from $\mathbb{Z}$.) And if so, then can't we say that $\mathbb{Z}/n\mathbb{Z} = \mathbb{Z}_n$? Thanks in advance!

Suppose we have a surjective homomorphism $h: G \to H$. The distinction, in my mind, between $\Bbb Z/n\Bbb Z$ and $\Bbb Z_n$ is the same as the distinction between $G/\text{ker }h$ and $H$.

Clearly, these are isomorphic groups (they have the same algebraic properties), but they are distinct in their "set structure" (the quotient group elements are elements of the power set of $G$, and the homomorphic image elements are just set elements (singletons) (of $H$)). In some areas of math (cough, topology) it can be crucial to keep track of "which level of set-construction" you are in.

Here is a similar example (from vector spaces) to illustrate the "ontological" versus "abstract" distinction.

The $x$-axis is a (normal) subgroup of the additive (vector addition) group of the Euclidean plane. We can form the coset space of all horizontal lines, which is isomorphic to, but surely not equal to, the $y$-axis.

As algebraic objects, we "abstract away" the particulars of how a given group arises, as we are usually only interested in its properties. However, the group may arise in a particular context we are studying, where its basis of definition may matter. One should not, for example, confuse the congruence class modulo $n$ of the integer $k$, with the integer $k$ itself.

This is the nature of algebraic morhpsims in general-they "filter away" information. In some sense, the quotient object has "more information to keep track of" (in $\Bbb Z/5\Bbb Z$, for example, one has to perform a (simple) mental calculation to resolve $7 + 5\Bbb Z$ to $2 + 5\Bbb Z$), whereas the homomorphic image has discarded the excess baggage.

No , I am suggesting that Z_n is the group multiplicative group of nth roots of unity as I stated above . Surely the books you referred to give a definition for Z_n . How can they mention a group without telling you what it's elements are .If not take my definition ,it's right .Also it is important because is gives a natural connection between the multiplication and addition operations . You can do the same with the additive group of R (real numbers ) and the multiplicative group of the Positive Real numbers via x ---> e^x . Here no quotient group is involved  and they are infinite groups .  Again in your example Z/nZ and Z_n are specific groups with definitions given or known . Of course all this assumes that C (the field of complex numbers and R (real numbers) are fixed throughout ,Of course they have many different constructions better forgotten.(sometimes )


$\lim\limits_{(x,y)\to (3,2)}x^2y^3-4y^2$ I have some problems solving this:

$$\underset{(x,y)\to (3,2)}{\lim}(x^2y^3-4y^2)$$

I know that if $x = 3$ then $$\underset{y\to 2}{\lim}(3^2y^3-4y^2)=\underset{y\to 2}{\lim}(9y^3-4y^2)=56$$

Similarly, when $y = 2$, $$\underset{x\to 3}{\lim}(x^22^3-4(2^2))=\underset{x\to 3}{\lim}(x^28-16)=56$$

But then I get confused - I let $y = mx$ and it becomes $$\underset{x\to 3}{\lim}(x^2(mx)^3-4(mx)^2)=9(27)m^3-4(3)m^2$$. This equation depends on a value of m, let $m = 1$, then $$\underset{x\to 3}{\lim}=9(27)-12$$ which is not equal to 56, therefore I conclude that the limit DNE.

But it seems like this function should be continuous at point (3,2) since it has no restrictions and therefore it should have the limit point.

P.S.: I feel like the situation could be different if I had to check the limit at (0,0), then I could use $y=x^2$ or $y=x$, but this (3,2) bugs me since I don't understand how to check it at this point.

if you want to let y=mx and look at the behavior of the limit as x and y approach in that way, you have to make sure that $y \to 2$ as $x\to 3$. So you problem comes when you set $m=1$, which will make $y \to 3$ as $x\to 3$.

if you set $m=2/3$, then $y \to 2$ as $x\to 3$, and

$\lim_{x\to 3}(x^2(mx)^3-4(mx)^2)=(2/3)^3x^5-4(2x/3)^2= 72-8=56 $

Let me make this clear though, this only shows that you do get the same limit as you use the $y=2x/3$ substitution, but it doesn't show that the limit exist. To show that the limit exist you have to use the continuity argument.

One way you can tell if the limit exists is if you approach the limit from several different paths. For example, if you were to go along the x-axis you would test to see if
\begin{equation}
f(x,y) -> L_{1} for all x = 0
\end{equation}
Then you would test along the y-axis.
\begin{equation}
f(x,y) -> L_{2} for all y = 0
\end{equation}
If $L_1 = L_2$, then you know the limit must exist. In fact, there's a possibility that $L_1$ is your limit.

The always fail-proof way to find any limit is to approximate it by plugging values, very close to the point in question, into your calculator.
In this case, however, I believe you can the squeeze theorem to find it.


Conditions for a map to be $\mathbb{R}$-linear Given a map $f:\mathbb{R}^m\rightarrow \mathbb{R}^n$, is the condition that for all $x,y$ in $\mathbb{R}^m$, $f(x+y)=f(x)+f(y)$ enough for $f$ to be $\mathbb{R}$-linear? I cannot prove it or think of a counter example.

It should also satisfy:
$$f(ax)=af(x).$$


If $f'(x)=0$, is then $f(x+dx)=f(x)$? I am always struggling with infinitesimals, and not sure I'm getting this right. The title basically states the simplest version of my question: If a function has zero slope at some point, is it correct to say that $f(x)=f(x+dx)$ for an infinitesimally small $dx$?
-- If yes, can anyone intuitively explain why I can drop the other terms of the Taylor expansion here?

Edit: It has been pointed out that "infinitesimal" really isn't a well-defined concept in the way that I've used it here, and the answer by Seth explains why. I've clarified what I'm after (without the use of "infinitesimals" in a separate question to avoid mixing too many different problems under one header.

Using the hyperreal number system defined by Robinson, which is a common context in which infinitesimals can be rigorously defined, a function $f$ is differentiable at $x$ if there is a real number $f'(x)$ such that $$f'(x)\approx\frac{f(x+\epsilon)-f(x)}{\epsilon}$$ for all infinitesimals $\epsilon$. Here, $a\approx b$ means that $a$ and $b$ differ by an infinitesimal.

Thus, if $f$ is differentiable at $x$, this implies that $$\frac{f(x +dx)-f(x)}{dx}=f'(x)+\eta$$ for some infinitesimal $\eta$, which implies that $$f(x+dx)-f(x)=dx\,(f'(x)+\eta).$$ Since $dx$ is infinitesimal, the right side is infinitesimal, and therefore, in our alternative notation, $f(x+dx)\approx f(x)$.

Interestingly, this is the definition of continuity: $f$ is continuous at $x$ if $f(x+\epsilon)\approx f(x)$ for all infinitesimals $\epsilon$. Thus the above argument shows that differentiability implies continuity.

The hypothesis $f'(x)=0$ was not used above.  This stronger hypothesis means that $$\frac{f(x+dx)-f(x)}{dx}\approx 0,$$ which means intuitively that $f(x+dx)-f(x)$ is so small that even after dividing it by the infinitesimal $dx$, it is still infinitesimal.

Suppose $f$ continuous differentiable at $x$, as it stated by the $f'(x)=0$. In the case or Riemann integral.

$f'(x)=0\Rightarrow \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=0
\Rightarrow \frac{f(x+dx)-f(x)}{dx}=0\Rightarrow f(x+dx)-f(x)=0$.

Also $dx$ is not a number in the sence of real(in Riemann or differential geometry), it's more like an operator represtation.
In a sense that "it never goes to anywhere". Suppose $m,n\in\mathbb{N}$, $m+dx"="m$, $m+n*dx"="m$ yet $0< m*n*dx<<1$. ($"="$ means "sort of equal" in real, but notice this is very improper.)

For your question, answering in the case of real analysis, I think $f'(x)$ exists, which is differentiable at $x$, already implied that $f$ is continuous at $x$.

Therefore, for anly $\epsilon>0$, there exists $\delta>0$ such that whenever $|x_{new}-x|<\delta$, $|f(x_{new})-f(x)|<\epsilon$. Thus $f("x+dx")=f(x)$.


Homology groups equal when $\cdots \rightarrow 0 \rightarrow H_n(\mathbb{S}^m) \rightarrow H_{n-1}(\mathbb{S}^m) \rightarrow 0 \rightarrow \cdots$ I'm reading a set of notes but I don't understand the following concept. We have a long exact sequence

$\cdots \rightarrow 0 \rightarrow H_n(\mathbb{S}^m) \rightarrow H_{n-1}(\mathbb{S}^m) \rightarrow 0 \rightarrow \cdots$

Apparently this implies that $H_n(\mathbb{S}^m)$ is isomorphic to $H_{n-1}(\mathbb{S}^m)$. Why is this true? I tried to prove it using exactness and rank nullity but all I could show is that rank$(H_n(\mathbb{S}^m)) = $ rank$(H_{n-1}(\mathbb{S}^m))$

It might be clearer if you actually name the maps. You have a short exact sequence within the long exact sequence
$$
0\xrightarrow{f}H_n(S^m)\xrightarrow{g} H_{n-1}(S^m)\xrightarrow{h} 0
$$
where $f$ and $h$ are zero maps. Then by exactness, $0=\operatorname{im}(f)=\ker g$, so $g$ is injective. Also, $\operatorname{im}(g)=\ker h=H_{n-1}(S^m)$ since $h\equiv 0$. Thus $g$ is surjective, hence an isomorphism.

There is no need to look at the ranks of the groups.

If the coefficients in the sequence are in some field, then you have a linear map between two vector spaces of the same rank. By  exactness, the kernel of the map is trivial, so the map is onto. Now from rank-nullity theorem, the image is n-dimensional subspace of n-dimensional space. Alternatively, if not working over a field. then, by exactness ,the kernel of the $0$ map is $H_{n-1}$, i.e., the whole space, which coincides with the image o the map, so the map is onto 1-1 and linear, i.e., an isomorphism.


Covariance of polynomials of random normal variables $\newcommand{\Cov}{\operatorname{Cov}}$If $X$ and $Y$ are random variables with a bivariate normal distribution and:

$X\sim\mathcal{N}(\mu_X,\sigma_X^2)$
$Y\sim\mathcal{N}(\mu_Y,\sigma_Y^2)$
$\Cov(X,Y)\neq0$

May I compute $\Cov(X^m,Y^n)$ for arbitrary positive integers $m$ and $n$?

$\newcommand{\Cov}{\operatorname{Cov}}$Given the new background that the OP provided to his question (most general normal bivariate distribution for variables (X_1, X_2) with non-zero covariance), it is possible to find a general formula for $\Cov(X_1^m, X_2^n)$ as follows:

Consider the generating function $\phi(t_1,t_2)$ for the bivariate distribution given here, in equation (57) (a clean derivation is given for it so I won't repeat it). Then:

$$E(X_1^m X_2^n)=(-i)^{m+n}(\frac{\partial}{\partial t_1})^m(\frac{\partial}{\partial t_2})^n\phi(t_1,t_2)\Big|_{(t_1, t_2)=(0,0)} \tag 1$$

After some tedious and careful algebra we wish to write the generating function $\phi(t_1,t_2)=e^{it_1\mu_1+it_2\mu_2}e^{-\frac{1}{2}(\sigma_1^2t_1^2+2\rho\sigma_1t_1\sigma_2t_2+\sigma_2^2t_2^2)}$ in the form:

$$\phi(t_1,t_2)=e^{-a}e^{-b(t_2+c)^2}e^{-d(t_1+gt_2+h)^2}\tag2$$

which is possible for the values:

$$a=\frac{\mu_1^2}{2\sigma_1^2}+\frac{(\mu_2-\mu_1\rho \frac{\sigma_2}{\sigma_1})^2}{2\sigma^2_2(1-\rho^2)}, \hspace{0.2cm}b=\frac{1}{2}(1-\rho^2)\sigma_{2}^2, \hspace{0.2cm}c=-i\frac{\mu_2-\mu_1\rho \frac{\sigma_2}{\sigma_1}}{2\sigma^2_2(1-\rho^2)}, \hspace{0.2cm}d=\frac{\sigma_1^2}{2},\hspace{0.2cm} g=\frac{\rho\sigma_2}{\sigma_1},\hspace{0.2cm}h=-i\frac{\mu_1}{\sigma_1^2} $$

We change variables
$y_2=\sqrt{b}(t_2+c), y_1=\sqrt{d}(t_1+gt_2+h)$ and we find:

$$\begin{align}\frac{\partial}{\partial t_1}&=\sqrt{d}\frac{\partial}{\partial y_1}\\ \frac{\partial}{\partial t_2}&=g\sqrt{d}\frac{\partial}{\partial y_1}+\sqrt{b}\frac{\partial}{\partial y_2}\end{align}$$

Substitute into (1) for the result:

$$E(X_1^mX_2^n)=(-i)^{m+n}(\sqrt{d}\frac{\partial}{\partial y_1})^m(g\sqrt{d}\frac{\partial}{\partial y_1}+\sqrt{b}\frac{\partial}{\partial y_2})^n e^{-a}e^{-y_1^2}e^{-y^2_2}\Big|_{(y_1, y_2)=(c\sqrt{b},h\sqrt{d})}$$

Expanding the parentheses and using the Rodrigues formula for Hermite polynomials we get:

$$E(X_1^mX_2^n)=e^{-a-bc^2-dh^2}i^{m+n}g^n d^{\frac{m+n}{2}}\sum_{k=0}^n\frac{n!}{k!(n-k)!}\Big(\frac{\sqrt{b}}{g\sqrt{d}}\Big)^{n-k}H_{n-k}(h\sqrt{d})H_{m+k}(c\sqrt{b})$$

The expectation values $E(X_1^m)$ and $E(X_2^n)$ can be calculated by setting $n=0$, $m=0$ in the general formula respectively. Also, the imaginary units in the argument of the Hermite polynomials are precisely cancelled out by the imaginary prefactor, so that the final result is real.

EDIT: According to the analysis by @machfour, there is indeed an error in the expression of $c$, which has been corrected for. Due to this, the formula simplifies to the final expression:

$$\small{E(X_1^mX_2^n)=\Big(\frac{i}{\sqrt{2}}\Big)^{m+n}(\rho\sigma_2)^n \sigma_1^m\sum_{k=0}^n {n\choose k} \Big(\sqrt{\frac{1}{\rho^2}-1}\Big)^{n-k}H_{n-k}\Big(\frac{-i\mu_1}{\sigma_1\sqrt{2}}\Big)H_{m+k}\Big(-i\frac{\mu_2\sigma_1-\mu_1\rho}{\sqrt{2}\sigma_1\sigma_2\sqrt{1-\rho^2}}\Big)}$$

You have said$\newcommand{\Cov}{\operatorname{Cov}}$
\begin{align}
& X\sim N(\mu_X,\sigma^2_X), \\[4pt]
& Y \sim N(\mu_Y,\sigma^2_Y), \\[4pt]
& \Cov(X,Y) = 0.
\end{align}
That falls short of specifying the joint distribution of $(X,Y).$ If it were further specified that $X,Y$ are jointly normally distributed, then the covariance can be $0$ only if $X,Y$ are independent. If $X,Y$ are independent, then so are $X^m,Y^n,$ so their covariance is also $0.$

Here is a simple example: Suppose $X\sim N(\mu_X, \sigma_X^2),$ let $Z = \text{the “z-score''} = (X-\mu_X)/\sigma_X,$ and independently of $X$ you toss a coin. Then let $Y = \mu_Y \pm \sigma_Y Z, $ where the choice between $\text{“}\pm\text{''}$ is determined by the coin toss. Then $X,Y$ have covariance $0$ and have just the distributions you specified in the question, but they are NOT JOINTLY normally distributed and not independent.

But if you assume joint normality, which in the case of two random variables means bivariate normality, then the answer is just as in the first paragraph above.


A definable hyperreal system Reading about hyperreals I learned that a serious problem with such systems is their undefinability.

So I tried to construct a definable system of hyperreals by introducing an infinitismal element $\varepsilon$ such that

$$0^\varepsilon=\frac{1}e$$
by definition.

It seems to me that such system would inevitably include divisors of zero but I am not sure.

I wonder whether such system viable and whether it in fact includes divisors of zero.

I would like to challenge a premise of the original question, namely the assumption that the hyperreal number system is "undefinable".  Most researchers in the field expected this to be the case, but Kanovei and Shelah proved in 2003 the existence of a definable model (see here).

Note. In this recent article (to appear in Journal of Symbolic Logic) we study minimal set-theoretic conditions that enable the construction of a definable hyperreal line.

A hyperreal system with the Transfer Principle (as required in nonstandard analysis) is, indeed, not definable.

But other hyperreal systems are definiable.  By "hyperreal system" I mean an ordered field that properly contains the real numbers.  In particular, these fields have infinite and infinitesimal elements.

Some of the best known of these are:

$\bullet$ rational functions $\mathbb R(X)$ with an ordering where $X > n$ for all $n \in \mathbb N$.

$\bullet$ Levy-Civita numbers

$\bullet$ Hardy fields

$\bullet$ surreal numbers

$\bullet$ transseries


Limit of sequence in complex plane Suppose $Z_{n+1}=\frac{\alpha}{Z_n}+\frac{\beta}{Z_{n-1}}$. where $\alpha$ and $\beta$ and $Z_n$ and $Z_{n-1}$ are all complex numbers.

It is easy to show that $\frac{Z_n}{Z_{n+1}}=\frac{\alpha.\frac{Z_{n-2}}{Z_{n-1}}+\beta}{\alpha.\frac{Z_{n-1}}{Z_{n}}+\beta}$ for all $n$.

From this equation, is it possible to talk about limit of the sequence $\{Z_n\}$?

In general: One can always talk about limits of complex sequences since $\mathbb{C}$ is a metric space where $d(z_1,z_2) = |z_1 - z_2|$.

Regarding your sequence: Let $Z_0 = Z_1 = 1$, $\alpha \neq 1$ and $\beta = 0$. Then
$$Z_{n+1} = \frac{\alpha}{Z_n} ~ , ~ n > 1$$
This means that $(Z_1,Z_2,Z_3,\ldots) = (1, \alpha, 1, \alpha, \ldots)$. Hence, the sequence is not convergent (for this particular choice of $\alpha, \beta$.


Let $F$ be a probability distribution function. Prove that $\int\limits_{-\infty}^{\infty} F(x+a)-F(x)\ dx= a\ \forall a\in\Bbb{R}$ A real-valued map $F$ is called a probability distribution if

$F$ is monotonically increasing
$F$ is right continuous
$\lim\limits_{x\to\infty} F(x)=1$ and $\lim\limits_{x\to-\infty} F(x)=0$

Again I know that for such $F$, there is a probability measure $P$ on borel $\sigma$-field of $\Bbb{R}$. such that $P((a,b])=F(b)-F(a)\ \forall -\infty\le a\le b\le \infty$.
So we have to prove that $\int\limits_{-\infty}^{\infty}P((x,x+a])\ dx=a$
Here $F$ may not be absolutely continuous, so we cannot work with the probability density function.
I'm not getting any idea how to prove this result. Can anyone help me in this regard? Thanks for your help in advance.

Let $\mu$ be the probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$
induced by $F$ (i.e, $\mu(a,b]=F(b)-F(a))$. Let $m$ be the usual
Lebesgue measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Consider
the product measure space $(\mathbb{R}\times\mathbb{R},\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R}),m\times \mu)$.
Fix $a>0$ (the case that $a\leq0$ can be treated similarly). Let
$A=\{(x,y)\in\mathbb{R}^{2}\mid x<y\leq a+x\}$ which is a Borel subset
of $\mathbb{R}^{2}$ (see remark below). By Fubini-Tonelli Theorem,
$$
\int\left(\int1_{A}(s,t)dm(s)\right)d\mu(t)=\int\left(\int_{A}(s,t)d\mu(t)\right)dm(s).
$$
Observe that for each $s$, $\int1_{A}(s,t)d\mu(t)=\mu\left((s,s+a]\right)=F(s+a)-F(s)$.
Therefore $\int\left(\int_{A}(s,t)d\mu(t)\right)dm(s)=\int_{-\infty}^{\infty}\left[F(s+a)-F(s)\right]ds.$
On the other hand, for each $t$, $1_{A}(s,t)=1$ iff $s<t\leq s+a$
iff $s\in[t-a,t)$. Therefore, $\int1_{A}(s,t)dm(s)=m\left([t-a,t)\right)=a$.
It follows that $\int\left(\int1_{A}(s,t)dm(s)\right)d\mu(t)=\int ad\mu(t)=a$.
That is, $\int_{-\infty}^{\infty}\left[F(s+a)-F(s)\right]ds=a$.

Remark: We go to show that $A$ is a Borel set.
Let $A_{1}=\{(x,y)\in\mathbb{R}^{2}\mid x<y\}$ and $A_{2}=\{(x,y)\in\mathbb{R}^{2}\mid y>a+x\}$.
Note that
\begin{eqnarray*}
A_{1} & = & \cup_{r\in\mathbb{Q}}\{(x,y)\mid x<r<y\}\\
 & = & \cup_{r\in\mathbb{Q}}\left((-\infty,r)\times\mathbb{R}\right)\cap\left(\mathbb{\mathbb{R}}\times(r,\infty)\right)
\end{eqnarray*}
which is a countable union of measurable rectangles, so it is measurable.
Similarly, $A_{2}$ is measurable. Finally, $A=A_{1}\setminus A_{2}$,
so $A$ is measurable too.


Decimal expansion of an irrational number not ending in a particular sequence Consider the set $N$ of natural number, $N$={1,2,3,4,5,6,7,8,9,10,11...} and consider the subsequence  {$N_i$} ,$i \in N$

Each $N_i$ consists of elements in ascending sequence greater than $i$

$N_1$={2,3,4,.....}  ,$\quad$$N_2$={3,4,5,.....},$\quad$$N_{11}$={12,13,14,15,...} and so on

0.1234567......., $\quad$ 0.0123456.....,$\quad$ 0.1119012345679810....$\quad$ 0.992018151617181920 $\quad$ are are all irrational numbers in $[0,1] $, but the tail  of each sequence  can be mapped to a subset of {$N_i$}: for example $\quad$ 0.992018151617181920... can be mapped to $\quad$ $N_{15}$={15,16,17,18,19,20,21,....}

My question is can someone give example of irrational number in $[0,1] $ ,such that the tail of its decimal expansion cannot be mapped to any subset of {$N_i$} ?

Of course there are uncountable of such examples. But to construct one is not trivial. You may know the decimal expansion of $\pi$ upto one billion decimal places but that doesn't stops its remaining digits being mapped to {$N_i$}.


A friendly introduction to Hilbert's sixth problem with special focus in its meaning and issues While I was reading the Wikipedia's article dedicated to Hilbert's problems I've known the so-called Hilbert's sixth problem from this page and the corresponding Wikipedia's entry dedicated to this problem that I've read.

  Question. I would like to know a friendly introduction to Hilbert's sixth problem, in which you can do more focus about the meaning of the problem and the issues (I think about the axiomatization of the more relevant theories in physics); isn't required the history of the problem, nor status. Your introduction can be done
  from an informative point of view but adding those remarkable facts, formulas and reasonings if you want to exemplify some point. Many thanks.

Thus I am asking about an explanation from your words and knowledges, in a post as your answer.

If you need/want to enrich your explanation with some reference (preferably also with an informative spirit) add such references and I am going to search and read what I can from the literature.

A poster child would be special relativity and the Lorenz transformation.  Your axioms are that the universe is homogeneous and isotropic, which says the laws of physics are the same in all reference frames, and that the speed of light is independent of the motion of the source.  From those you can derive all of special relativity.  This is done in many texts.

The maximization of entropy can link statistical mechanics to may other subjects.  You can get the  classical black body radiation curve, then introduce photons and get the Planck curve.  You can get the ideal gas law.  You can get a lot of information about phase changes.

I suppose I should recommend my own book,
https://www.amazon.com/Hilberts-Sixth-Problem-Axiomatization-Probability/dp/3330072938
which is "friendly" only in the sense that you should have been either a math major or a physics major.
In particular, I cannot really agree that entropy is "foundational", it is merely a useful approximation under certain limitations, but plays no role in the axioms of the foundations of physics.


Probability - Interview Question Hi I recently got this question in an interview and would like help solving it.

There were 2 boys and 3 girls in a room. A new baby entered the
room whose gender was unknown (but equally likely to be a boy/girl).
A nurse came into the room and picked one baby randomly and it
turned out to be a boy. What’s the probability that the new baby was
a boy?

Let $bg$ denote the event that the new child is a boy, and the randomly chosen child is then a girl. Label other events similarly viz. $P(bb)=P(bg)=\frac{1}{4},\,P(gb)=\frac{1}{6},\,P(gg)=\frac{1}{3}$. Then $$P(\text{new boy}|\text{boy chosen})=\frac{P(bb)}{P(bb)+P(gb)}=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{6}}=\frac{3}{5},$$where to simplify I've multiplied numerator and denominator by $12$.

As an existing answer that reached this value has been downvoted, I double-checked my answer with the Python here. In one run, the conditional probability estimate was $0.602380952381$, which is close enough to $0.6$ for the purposes of a sanity check.

I would like someone to verify this, since my confidence isn't high.
This is a method that helps corroborate chances. If something doesn't quite add up, you'd generally notice, since then one or more of the equations would be wrong.

cube chart

The description explains several events with 4 conditions; the chance a boy or a girl is added, and the chance the nurse picks a boy or a girl.

What you're asking is basically, what are the odds something happens, given that we know that thing already happened. In this case it's "What are the odds a boy was added, given that we know the nurse picked a boy". You can see the events cross together with the answer being $1/4$


The topological conjugacy $h$ guaranteed by the Hartman-Grobman theorem is generally not a lipeomorphism Yesterday I asked a question here Topological conjugacy between linear maps and https://math.stackexchange.com/users/349785/paul-frost answered my question very well ! Thank him. Now I want to ask a very interesting question, at first I prefer to write detail again :
Let $A_{\alpha}: \mathbb{R} \to \mathbb{R}$ denote the linear map
\begin{align}
A_{\alpha}(x)=\alpha x
\end{align}
We proved that if $ 0 < \alpha < 1$ and $ 0 < \beta < 1 $ then $A_{\alpha}$ and $A_{\beta}$ are topologically conjugate.

Notice that two homeomorphisms $f:X \to X$ and $g:X \to X $ are topologically conjugate to each other if there is a homeomorphism $h:X \to X $ such that $hf=gh$

Now the question is this :
What if $\alpha \neq \beta$? In the book of Lan Wen I studied and found out that if $\alpha \neq \beta$ then there is no lipeomorphism $h:\mathbb{R} \to \mathbb{R}$ such that $hA_{\alpha}=A_{\beta}h$. Could anyone help me show this?

Note that We call a homeomorphism $f: E \to E^{'}$ a lipeomorphism if both $f$ and $f^{-1}$ is lipschitz.

Assume that $hA_{\alpha}=A_{\beta}h$ with
$$
c|x-y|\le|h(x)-h(y)|\le d|x-y|.
$$
Then $hA_{\alpha}^n=A_{\beta}^nh$ for all integers $n$ and so
$$
c\alpha^n|x-y|\le|h(A_{\alpha}^nx)-h(A_{\alpha}^ny)|=|A_{\beta}^nh(x)-A_{\beta}^nh(y)|\le d\beta^n |x-y|.
$$
In particular, we have
$$
c\alpha^n|x-y|\le d\beta^n |x-y|.
$$
Now you just need to let $n\to\pm\infty$ to get a contradition.


Given known and and unknown line equation and maximising area of triangle condition. Q) If from a point $P\equiv(4,4)$ perpendiculars to the straight lines
  $-3x+4y+5=0$ and $y=mx+7$ meet at $Q$ and $R$ and area of triangle
  $PQR$ is maximum then $m=__ $ ?

First I drew the known line $-3x+4y+5=0$ and the unknown line $y=mx+7$ then I fixed the perpendicular distance from $P(4,4)$ as a fixed value $b$ .

I also know that area of a triangle $\dfrac{1}2 \times base \times height$ so to maximise this area we need $base=height$

I took $base$ as my assumed constant $b$

Then I found out $b=9/5$

Then I substituted in distance formula and I have no idea how to proceed further

$\frac{9}{5}=\frac{-4m+4+c}{\sqrt{m^{2}+1}}$

Dropping the perpendicular from $P$ onto the given line $g: \>-3x+4y+5=0$ gives the point $Q$. We can consider $PQ$ as base of our triangle and then have to make the corresponding height as large as possible. All lines $h_m:\>y=mx+7$, $m\in{\mathbb R}$, pass through the point $Z=(0,7)$. Since $\angle(PRZ)={\pi\over2}$ the point $R$ has to lie on the Thales circle over the segment $PZ$. The center of this circle is the point $M=\bigl(2,{11\over2}\bigr)$, and its radius computes to ${5\over2}$. The optimal point $R$ is then found as follows: Draw a parallel $g'$ to $g$ through the point $M$ and intersect it with the circle. Take the better of the two points so obtained; it is the left one. Looking at the symmetries of the figure one realizes that $R=(0,4)$. This means that there is a largest triangle $PQR$ satisfying $\angle(PRZ)={\pi\over2}$, but strictly speaking $h_*:=R\vee Z$ does not belong the given family of lines $(h_m)_{m\in{\mathbb R}}$, since $h_*$ is vertical.

First of all your sentence:"I also know that area of a triangle 12base×height so to maximise this area we need base=height" is so wrong cause this triangle is under some constraints and maybe this is not possible. Second, this problem is easy but you have to write it with this hints:
Choose PQ as triangle base and find corresponding height. To find that first write equation of line PR (as a function of m) then find coordinate of point R, now find distance of R from line PQ. So you have a scaler function of m that want to be maximized. Go ahead.


Converting an integral from polar to cartesian Question concerning definite integrals. Lets say we have some integral in cartesian coordinates (like the integral of

$$ \int^{3/4}_0 \sqrt{1+y^2} \space dy$$

I completely understand how to convert the function itself into polar coordinates as well as how to find the indefinite integral of a polar equation. What I do not understand is how to reevaluate the limits of integration. Any help would be appreciated.

Normally an integral in Cartesian coordinates is a two dimensional integral over some region of the plane, not a one dimensional interval as in your example.  If it is a simple rectangular region $\int_a^b\int_c^d f(x,y)dx\ dy$  it is hard to express in polar coordinates.  That is a strong reason to stay in Cartesian coordinates.  If the region is simple in polar coordinates, like a circle at the origin $\int_{-R}^R\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} f(x,y) dx \ dy$ it becomes simple in polar coordinates $\int_0^{2 \pi}\int _0^R f(r,\theta)r\; dr\; d\theta$ where I have abused notation by using $f$ for both versions, irrespective of the arguments.  The second is much simpler for circular regions, so is a candidate when the region is simple in polar coordinates.

@Louie I would say, first of all you have to make it double integral in order to use substitution using polar coordinate. For your case you can assume

$$ I = \int^{3/4}_0 \sqrt{1+y^2} \space dy$$
or it can be
$$ I = \int^{3/4}_0 \sqrt{1+x^2} \space dy$$

both are same as only x variable is replaced by y. Now, multiply both of the above equations, you would get
$$ I^2 = \int^{3/4}_0 \int^{3/4}_0 \sqrt{1+x^2} \sqrt{1+y^2} \space dxdy$$
or
$$ I^2 = \int^{3/4}_0 \int^{3/4}_0 \sqrt{1+x^2+y^2+x^2y^2}  \space dxdy$$

Now use the polar coordinate substitution as

$x^2+y^2=r^2$

and the limit of one integral will be for $\theta$ from $0$ to $2\pi$ and for the other integral it will be from $0$ to $r$.

After evaluation, take sqaure root of $I^2$ to end up with your integral solution which is for

$$ I = \int^{3/4}_0 \sqrt{1+y^2} \space dy$$

I hope it would help you.


Area of parallelogram = Area of square. Shear transform Below the parallelogram is obtained from square by stretching the top side while fixing the bottom.
Since area of parallelogram is base times height, both square and parallelogram have the same area.

This is true no matter how far I stretch the top side.

In below figure it is easy to see why both areas are same.

But it's not that obvious in first two figures. Any help seeing why the area doesn't change in first figure?

A kite is flying into the sky!

In fact, it is what you and Stijn already observed. This drawing summarizes it.


Proof by contradiction confusion I am having confusions on some basic stuff. Could you help me out, please?

Consider the following statement:

"If A then B or C"

If we apply proof by contradiction method to this statement then we assume the contrary of "B or C" which is "not B and not C". Than if we deduce "B or C" by using "A" then does this count as a proof?

Thank you!

Let's say you have to prove
$$A\Longrightarrow B $$
The thing you are looking for is: Assume that $A$ is valid and assume for a contradiction that $B$ is invalid. You would then consider
$$A\land \neg B $$
If that yields a contradiction, then the assumption that $B$ is invalid, is incorrect. Hence $A\Longrightarrow B$

Another way is to prove by showing the contrapositive is valid i.e $\neg B\Longrightarrow \neg A$.

In your case we consider:
$$A\Longrightarrow B\lor C $$
Assume for a contradiction that $\neg B\land \neg C$, then it must be true that:
$$A\land\neg B\land\neg C $$

You had suggested a proof by:
$$A\land\neg B\land\neg C\Longrightarrow B\lor C $$
You will no make progress, for we can apply your method indefinitely and arrive at:
$$A\land\neg B\land\neg C\land A\land\neg B\land\neg C\Longrightarrow B\lor C $$
Which is the exact same thing we started with. Stuck in a circle.

The problem is that you are using the validity of the very thing you try to verify as part of your assumption.

The proof by contradiction in this case would be to show an example where "A" is true but neither "B" nor "C" are true.


Tautologies in Boolean formulas using the bi-conditional connective How can I prove a formula $\phi$ written using only the bi-conditional connective $\leftrightarrow$ (besides variables and parentheses) is a tautology if and only if every variable in $\phi$ occurs an even number of times?

I have attempted to prove this by using the fact that $p \leftrightarrow q \equiv (p \rightarrow q ) \wedge (q \rightarrow p)$ and by generalizing my proof for the fact that $(...((p\rightarrow p)\rightarrow p)\rightarrow...)\rightarrow p$ is a tautology if and only if $p$ occurs an even number of times, but I did not succeed. Would induction be a good choice for this proof?

Show by induction that:

$\iff$ is associative: in an expression built only using variables, "$\iff$," and parentheses, you can rearrange the parentheses however you like. (E.g. "$x\iff (y\iff z)$" is equivalent to "$(x\iff y)\iff z$.")
$\iff$ is commutative: "$x\iff y$" can always be replaced with "$y\iff x$".

Once you have these two facts, then you can show:

  Any $\iff$-formula in which every variable appears an even number of times is equivalent to one of the form $$(x_0\iff x_0)\iff (x_1\iff x_1)\iff . . . \iff (x_n\iff x_n),$$ which is a tautology since "$x\iff x$" is a tautology.

This completes one half of the problem. For the other half:

  If $\varphi$ is a $\iff$-formula in which some variable $x$ occurs an odd number of times, consider the valuation which makes every variable but $x$ true, and makes $x$ false; you can show by induction that this makes $\varphi$ false, so $\varphi$ is not a tautology.

I have a somewhat different approach.

First, we'll take Epp as an axiom (I use Polish notation).  We have four rules of inference:

Wherever any instance of E$\alpha$E$\beta$$\gamma$ appears within a formula, we can replace it with EE$\alpha$$\beta$$\gamma$ and conversely (association).
Wherever any instance of E$\alpha$$\beta$ appears in a formula, we may replace that instance with E$\beta$$\alpha$ (commutation).
We may uniformly substitute any small letter $\lambda$, $\top$, or $\bot$, in any formula $\phi$ only have E's in it with some formula $\psi$, provided that for each instance of the letter $\lambda$, $\top$, or $\bot$, it gets substituted with $\psi$ in $\phi$.
From E$\alpha$$\beta$, as well as $\alpha$, we may infer $\beta$.

Now, with Epp as an axiom, the above four rules allow us to infer all instances of formulas having every variable appearing in a formula $\phi$ an even number of times.  Note that Epp is a tautology, and each of the above four rules qualifies as tautology preserving, so

Meta-Claim 1: This system only produces tautologies.  (Soundness of the above formulation of the equivalential calculus.)

Now, we'll suppose that we can infer some formula $\chi$ having an odd number of at least one variable in $\chi$ to the above.  We proceed as follows:

If there exists two instances of some variable in $\phi$ we use rules 1. and 2. (association and commutation), until it has the form EE$\gamma$$\gamma$$\delta$.  Then we substitute in the axiom Epp, and use rule 4. to detach $\delta$.
If $\delta$ only has variables which appear an odd number of times in $\delta$, then stop.  Otherwise, set $\chi$ to $\delta$ and go to step 1.

Now consider the following cases for $\delta$.

$\delta$ is a propositional variable.  But, if we deduced p, we could substitute it with $\bot$.  By the above, it is not a tautology and not deducible in the above system.
$\delta$ has three symbols.  We can then substitute $\top$ for the first symbol, and $\bot$ for the second symbol.  Similar to case 1., this is not a tautology, and thus not deducible in the above system.
$\delta$ has more then three symbols.  In which case we then substitute $\top$ for the first small letter, $\bot$ for the second small letter, and $\top$ for every other small letter (any letter subscripted by a numeral qualifies as a small letter, as well as any small letter of the ABC's).  Whatever $\delta$ comes as since E$\top$$\bot$ evaluates to $\bot$ and E$\bot$$\top$ evaluates to $\bot$, $\delta$ evaluates to $\bot$.

1., 2., and 3. show that $\delta$ is never a tautology.  Since $\delta$ is never a tautology, $\chi$ is also never a tautology also.  Thus:

Meta-Claim 2: No instances of any formula $\phi$ having some variable appearing an odd number of times occurs in the above system.

Now, it's known that EEpqEErqEpr is a single axiom for all formulas using only the bi-conditional connective... see here for example: http://web.ics.purdue.edu/~dulrich/Single-axioms-for-EQ-page.htm
under rule 3. and rule 4.

I'll use the notation $\nu$ $\alpha$/$\beta$ to indicate that letter $\alpha$ gets uniformly substitute with $\beta$ in formula $\nu$ in the following.

axiom      1 Epp
1 p/EpEqr  2 E EpEqr EpEqr
2, rule 1. 3 EE Epq r EpEqr
3, rule 1. 4 E Epq E r EpEqr
4, rule 1. 5 EEpqEr E Epq r
5, rule 2. 6 EEpqEr EEqpr
6, rule 1. 7 EEpq Er Eq Epr
7, rule 1. 8 EEpq EErq Epr

Thus, EEpqEErqEpr is in the above system.  The above system also has the rule of detachment.  Therefore,

Meta-Claim 3: The above system produces all tautologies having only E in them (completeness).

Since the above system has all tautologies having only E in them (E-tautology hereafter), and since every formula in the above system has it's variables appearing an even number of times, it follows that:

Meta-Claim 4: For all variables in any E-tautology, those variables appear in that tautology an even number of times.

Thus,

If a formula $\phi$ is an E-tautology, then every variable in $\phi$ appears an even number of times.

And the other half of your statement follows also.


Isometry from higher to lower dimensions I was wondering if there was any isometry from $\mathbb{R^2}$ to $\mathbb{R}$, using the Euclidean metric?

no, in the real line, there are not three numbers at a common distance of $1$ apart for each pair. In the plane, there are equilateral triangles

An isometry would be both continuous and 1-1.  So, no way.


Find the sine sign given a pair of 3D vectors I want to find the exact sine between two vectors in 3-dimensional space.

Data:

$x$: vector
$y$: vector
$z = \Vert x \times y \Vert$

I have tried this:

$$\sin \alpha = \frac{\Vert z\Vert}{( \Vert x\Vert\cdot \Vert y\Vert)}$$

but i obtain only the absolute value of the angle, is there a way to obtain even the sign?

Hint: Take the projection of $x$ onto $y$, and check if they are pointing in the same, or opposite directions.

Note that your given formula for $\sin \theta$ is incorrect. It should involve $||z||$.

But if you know a little about inner product then you have that the angle $\,\theta\,$ between both vectors fulfills:

$$\cos\theta=\frac{x\cdot y}{||x||\;||y||}$$

and now simply use the trigonometric identity

$$\sin x=\sqrt{1-\cos^2x}$$

to get what you want. (Warning: the sign above must be carefully chosen depending on the wanted/given domain of $\,\theta\,$ )


Double integral from rectangular region to polar $$\int_{-1}^{1} \int_{0}^{\sqrt{3+2y-y^2}} \cos(x^2 + (y-1)^2)) dy dx $$
How do I go about solving this? I tried substituting $x$ and $y$ for $r\cos(\theta)$ and $r\sin(\theta)$ and $x^2 + y^2 = r^2$. I can't seem to figure it out. So far I believe it's a rectangular region.

$$I=\int_{-1}^{1} \int_{0}^{\sqrt{3+2y-y^2}} \cos(x^2 + (y-1)^2)) dx dy $$

The change of variables in polar coordinates with center $(0,1)$ :
$\quad \begin{cases}
x=r\cos(\theta)\\
y=1+r\sin(\theta)
\end{cases}$ .

$$I=\int_{-\pi/2}^0\int_{0}^{2} \cos(r^2) rdr d\theta $$
$$I=\int_{0}^2\cos(r^2) rdr \int_{-\pi/2}^0 d\theta $$
$$I=\left(\frac12\sin(2^2)\right)\left(\frac{\pi}{2}\right)$$
$$I=\frac{\pi}{4}\sin(4)$$

$$\int_0^{-\frac{3\pi}{2}}\int_0^2\cos\left(r^2\left(\cos^2\theta\right)\ +\left(r\left(\sin\theta\right)-1\right)^2\right)rdrd\theta$$

This will do it!


When does a rational function pass through the horizontal asymptote/how do I know how it behaves? Sorry if this is a bit simple compared to everything here, but I can't really seem to find an answer.

If I have $$f(x) = \frac{(x-2)(x-4)}{x(x-1)}$$
1) When is the horizontal asymptote is crossed? Apparently to check if/where the horizontal asymptote is crossed I solve for f(x) = A, where A is the limit, is this true?

2)After solving for the vertical asymptotes I get x = 0 and x = 1. How do I know how each part behaves? My textbook made us use the behavior of the function as it got closer to the x intercepts, but that was for polynomial functions.

1) If either the $\lim_{x\rightarrow -\infty}f(x)=c_1$ or $\lim_{x\rightarrow +\infty}f(x)=c_2$ exists then you can get the horizontal asymptotes by evaluating $\lim_{x \rightarrow -\infty} f(x)$ and $\lim_{x \rightarrow \infty} f(x)$

If you let $y = f(x)$ then you will see the horizontal asymptotes will occur at the lines $y = c_1$ or $y = c_2$ (provided of course that those limits that defined $c_1,c_2$ exist)

Say you have a function with a horizontal asymptote that you found from evaluating the limit, it's possible (but not guaranteed) that the function has an intercept with the line defined by $y=c_1$. If this is the case (which it is for the f(x) in your question) then you can solve the simultaneous equations defined by $y = f(x)$ and $y=c_1$ or alternatively $f(x) = c_1$ to find the point on the graph that intercepts the horizontal asymptote line. (do the same for $c_2$ if it exists)

2) In a hand wavy way I think the easiest thing to do is to look at what happens on either side of the asymptote.

Start by simplifying the function. This becomes (x^2-6x+8)/(x^2 -x)
The horizontal asymptote is y=1 because both polynomials are the same degree, so it is y=(leading coefficient of numerator)/(leading coefficient of the denominator). Now solve for y=1.
1= (x^2 -6x +8)/(x^2 -x)
 (x^2 -6x +8)=(x^2 -x)
-6x+8=-x
8=5x
x=1.6 (or 8/5)
The asymptote is crossed at (1.6,1)
Use the following rules to find the horizontal asymptote:
If the degree of the top is greater than the degree of the bottom, then the horizontal asymptote does not exist.
If the degree of the top is equal to the degree of the bottom, then the horizontal asymptote is y=(leading coefficient of numerator)/(leading coefficient of the denominator).
If the degree of the top is less than the degree of the bottom, then the horizontal asymptote is y=0 (the x-axis).
Sorry for the format, I am on a phone currently.


Exponential equation for three known points This answer (https://math.stackexchange.com/a/680695/438808) was helpful to me in finding the quadratic equation for three known points.

Is there a similar solution for finding the exponential equation (y = ab^x + c) for three known points?

Thanks in advance!

For certain x-values of points that you're given, you may not be able to solve for the value of $b$ because when you set up a system, a polynomial will be formed with $b$ and polynomials with large degrees aren't necessarily solvable. However, here is a formula for an exponential function passing through any two points $(m,n)$ and $(p,q)$ no matter what the value of $b$ is:

$y=\frac{q-n}{b^p-b^m}(b^x-b^m)+n$

If you can get this through two of your points then you may be able to approximate a value for b.

You can find the base, $b$ by dividing equally separated differences.
if $q - p = n - m$ then
$b^{p-m} = (f(q) - f(p)) / ((f(n)-f(m))$

once you have the base, $b$, you can find the added constant.

$c = (b^{m-n}f(n) - f(m)) / (b^{m-n} -1)$


How find this $\frac{3x^3+125y^3}{x-y}$ minimum value let $x>y>0$,and such $xy=1$, find follow minimum of the value
  $$\dfrac{3x^3+125y^3}{x-y}$$

My idea: let $x=y+t,t>0$
then
$$\dfrac{3x^3+125y^3}{x-y}=\dfrac{3(y+t)^3+125y^3}{t}=3t^2+3yt+3y^2+\dfrac{128y^3}{t}$$
and $$(y+t)y=1$$

I think this can use AM-GM inequality.But I can't.

Thank you very much

First note that $x>1$ and simplify like this:
$$\frac{3x^3+125y^3}{x-y}=\frac{3x^6+125x^3y^3}{x^4-x^3y}=\frac{3x^6+125}{x^4-x^2}$$
Set $t=x^2$. Now we want to find $a\ge0$ such that
$$\frac{3t^3+125}{t^2-t}\ge a\Longleftrightarrow3t^3+at+125\ge at^2$$
for $t>1$ where the equality is possible.

Note that we could use AM-GM inequality on $t^3+t^3+t^3+at+5^3$, but that equality is possible only if all terms are equal, i.e. $t^3=at=5^3\Longrightarrow t=5, a=5^2$.

Luckily this gives us exactly what we wanted:
$$t^3+t^3+t^3+5^2t+5^3\ge5\sqrt[5]{t^{10}5^5}=5^2t^2$$
Therefore $a=25$ is the minimum and it's possible only for $t=5\Longleftrightarrow x=\sqrt5, y=\dfrac1{\sqrt5}$.


Alternating sum of binomial of coefficients up to $\binom{n}{r}$ Question : Evaluate - $$ \binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3} \dots + (-1)^r\binom{n}{r}$$

I tried to solve it but I am able to solve only when $r=n$.In that case I can put $x=-1$ in the expansion of $(1+x)^n$ and the result is $0$.

I know that for asking any question on this forum, I need to show my approach and thoughts but literally I have no idea how to solve up to $r+1$ terms.

Although I tried to add and subtract the terms to reach upto $\binom{n}{n}$, and evaluate the remaining sum, but nothing helpful.

Any help will be appreciable!

You have

$${n \choose k}={n-1 \choose k-1}+{n-1\choose k}$$
meaning that to get the next line in the Pascal's triangle, you sum two numbers above. Now you're computing an alternating sum, so you get a telescoping effect:

$${n\choose 0}-{n\choose 1}+\cdots+(-1)^r{n\choose r}=$$
$$=\bigg(0+{n-1\choose 0}\bigg)-\bigg({n-1\choose 0}+{n-1\choose 1}\bigg)+\cdots + (-1)^r\bigg({n-1 \choose r-1}+{n-1\choose r}\bigg)=$$
$$\textstyle=\big[{n-1\choose 0}-{n-1\choose 0}\big]+\big[{n-1\choose 1}-{n-1\choose 1}\big]+\cdots + \big[{n-1 \choose r-1}-{n-1\choose r-1}\big]+(-1)^r{n-1\choose r}=$$
$$=(-1)^r {n-1\choose r}$$

Short answer: you're counting each number from the upper row of the Pascal's triangle twice with opposite signs, except from the $r$-th one (including the sign).


Injective function proof involving floor function Let $f : \Bbb{Z} \to \Bbb{Z}$ and $g : \Bbb{Z} \to \Bbb{Z}$ be functions defined by $f(x)=3x+1$ and $g(x)=\lfloor\frac{x}{2}\rfloor$. Is $g(f(x))$ one-to-one?

So, $g(f(x)) = \lfloor\frac{3x+1}{2}\rfloor$. I know that $g(f(x))$ is one to one because we are only considering the integers. So for every $x$ integer there is a different $y$ integer. How do I prove this? I'm supposed to prove one-to-one functions by assuming $F(x_1)=F(x_2)$ and then proving $x_1=x_2$, but I don't know how to approach this when the function is a floor function.

Thanks in advance!

One possible Case would be:

Let $[x]=[y]$ in which $[x]=\text{floor}(x)$. If $x\in\mathbb Z$ then $[x]=x$ and so $[y]=x\in\mathbb Z$ so $$y\in [x,x+1)$$ Since, here, $y\in\mathbb Z$ so we necessarily have $y=x$.

Oh for god sake,
$$\lfloor\frac{3(1.1)+1}{2}\rfloor = \lfloor\frac{3(1.2)+1}{2}\rfloor = 2.$$
Thus, $g\circ f$ is not one-to-one.


the function such $p|f(x)+f(y)$,then we have $p|x+y$.find $f(1)$ Question:

  Aussme that the  function $\color{blue}{f:N^{+}\to N^{+}}$,foy any $x\neq y$,if $\color{red}{p|f(x)+f(y)}$,then we have $\color{blue}{p|x+y}$, find $\color{red}{f(1)}$

where $p$ is prime number

Idea:   Assmue $p=2$,if $f(x)+f(y)$ is even,then $x+y$ also even number.

Assmue that $p=3$, if $f(x)+f(y)=3k$,then $x+y=3m,k,m\in N^{+}$.

But How to determine the value $\color{red}{f(1)}$

Note that if $x+y=p^n$, then $p$ is the only allowed prime divisor of $f(x)+f(y)$. $f(x)+f(y)$ can't be $1$, so $p|f(x)+f(y)$.

Now consider the values of $f(1),f(2),...,f(p)$ mod $p$ (with $p>2$). We must have $p|f(1)+f(p-1)$, $p|f(2)+f(p-2)$, etc. If for some $i<j$ both less than or equal to $p$, $f(i)\equiv f(j)$ mod $p$, then $p|f(p-i)+f(j)$ but $p\nmid p-i+j$, a contradiction. Therefore $f(1),f(2),...,f(p)$ are all distinct mod $p$.

Now if $f(p^2-p)\equiv f(i)$ for some $i<p$, then $p|f(p^2-p)+f(p-i)$ but $p\nmid p^2-i$, a contradiction. Therefore $f(p^2-p)\equiv f(p)$ mod $p$.

But by the note above, $p|f(p^2-p)+f(p)$, so $p|2f(p)$. Therefore $p|f(p)$ and $p\nmid f(i)$ whenever $p\nmid i$. In particular, no odd prime divides $f(1)$, $f(2)$, or $f(4)$, so these three values must all be powers of $2$.

$f(1)+f(2)$ and $f(1)+f(4)$ are both odd,. If $f(1)$ is even, both $f(2)$ and $f(4)$ must be odd. Because they are powers of $2$, they would both have to be $1$, which is impossible. Therefore $f(1)$ is odd, so it must be $1$.

(At this point, we know that no prime can ever divide $f(x+1)-f(x)$, so the difference between two consecutive values of $f$ is $\pm1$. However, $f$ can never take the same value twice. The only possible $f$ is the identity function.)

Here's a thought: the identity map, which sends every number to itself, satisfies this property. If you could prove that it is the only such function, you could get the result $f(1) = 1$.
From a meta perspective, we still have the following weaker result:

If this problem has a well-defined answer — meaning that we don't need to know anything more about the function $f$ because every function with these properties assigns the same value to $f(1)$— then the answer must be $f(1)=1$.
The reason is because we know that the identity map has these properties, and it assigns $f(1)=1$.


Find the smallest positive integer $\ell$ such that $3 \cdot \left(4^m + 1\right)$ divides $2^\ell-1$ Find the smallest positive integer $\ell$ such that $3 \cdot \left(4^m + 1\right)$ divides $2^\ell-1$
Hint: The sought $\ell$ is the multiplicative order of $2$ in the ring of integer residues modulo $3\cdot(4^m+1)$.

I am having trouble understanding the hint, are there any "special" characteristics of a ring of integer that might help me?
The answer is supposed to be $4m$ , and I have already managed to show that $3 \cdot \left(4^m + 1\right)$ divides $2^{4m}-1$ , but I do not know how to prove that this is the smallest positive solution.
Link to the question

I may have done this a different way than from suggested in your hint OP.

  Claim: $2^{2m}+1 =4^m+1$ does not divide $2^{\ell}-1$ for $\ell < 4m$.

Proof: $(2^{\ell-2m}-1)(2^{2m}+1) = 2^{\ell}+2^{\ell-2m}-2^{2m}-1 < 2^{\ell}-1$ if $\ell$ satisfies the inequality $\ell -2m < 2m$ or equivalently if $\ell$ satisfies the inequality $\ell < 4m$. On the other hand $2^{\ell-2m} (2^{2m}+1) > 2^{\ell}-1$. [Do you see how this implies that $2^{2m}+1$ indeed does not divide $2^{\ell}-1$? For some integer $a$,
the integer $a(2^{2m}+1)$ is strictly less than $2^{\ell}-1$ yet the integer $(a+1)(2^{2m}+1)$ is strictly greater than $2^{\ell}-1$.]

So $\ell$ as in your problem must be at least $4m$.

However, $\ell=4m$ works; indeed

$$2^{4m}-1 = (2^{2m}-1)(2^{2m}+1) = (2^{2m}-1)(4^m+1).$$

So now it remains to show that $3|(2^{4m}-1)$. However, note that $3|(2^{2m}-1)$ for every integer $m$ [make sure you see why], so $\ell=4m$ indeed works; $3\times(2^{2m}+1)$ divides $(2^{4m}-1)$.

The ring of integers in the hint can be decomposed by the CRT into the product of $A=\Bbb{Z}_3$ by $B=\Bbb{Z}_{4^m+1}$. The order of 2 in each of these two rings is respectively 2 (because in $A$, $2^2=1$), and $4m$ (because in $B$, $2^{2m}=-1$) . Hence the order in the product is $\operatorname{LCM}(2,4m)=4m$.


Deductive proof construction I have for a long time having hard to understand how to create correct deductive proofs. I have been working on trying to solve examples but I don't really know how to proceed.

If we for instance have the argument: (A∧C)′∧C → (B∨C)∧A′
and have a list of equivalences and inference rules, how do I use the list in order to make hypotheses of the argument?

The inference rules are impl, dn, symm, de morgan, mp, mt, conj, add.

Normally you try to see if the statement you would like to prove is equivalent to a tautology. Like the following:

$$\begin{align}
&(A\wedge C)'\wedge C \rightarrow (B\vee C)\wedge A'\\
\Leftrightarrow & [(A\wedge C)'\wedge C]' \vee [(B\vee C)\wedge A']\\
\Leftrightarrow & (A\wedge C)\vee C' \vee [(B\vee C)\wedge A']\\
\Leftrightarrow & (A\wedge C)\vee C' \vee (B\wedge A') \vee (C\wedge A')\\
\Leftrightarrow & [C \wedge(A \vee A')] \vee C' \vee (B\wedge A')\\
\Leftrightarrow & C \vee C' \vee (B\wedge A')\\
\Leftrightarrow & True\\
\end{align}$$

If you would like to write in a better way, write this backward, from $C \vee C' \vee (B\wedge A')$ up to $\Leftrightarrow (A\wedge C)'\wedge C \rightarrow (B\vee C)\wedge A'$.

Here is the OP's question:

  If we for instance have the argument: (A∧C)′∧C → (B∨C)∧A′ and have a list of equivalences and inference rules, how do I use the list in order to make hypotheses of the argument?

Given a goal to derive in propositional logic, perhaps some premises that one may assume, and permitted inference rules the question appears to be: How should one construct a derivation leading to the goal?

One approach is to let a proof checker guide one through the proof by specifying the permitted rules, optionally checking each step, and verifying that the proof is correct when the goal has been reached.

Here is a proof of the example offered by the OP using such a proof checker:

With a proof checker to guide one, this hopefully makes the task of learning how to write such proofs manageable. To see if one understands how this is done, one can try entering the above proof into the proof checker, line-by-line, to see if one gets the same result.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Fall 2019. http://forallx.openlogicproject.org/forallxyyc.pdf


Are the two groups isomorphic? Let $G,H,K$ be three groups. If $G\times H$ is isomorphic to $G\times K$, then is it true for $H\cong K$?

No, take

$$G=\mathbb{Z}\times\mathbb{Z}\times\cdots$$
$$H=\mathbb{Z}$$
$$K=1$$

Note that the thesis holds if $G$ is finite (Hirshon's theorem). See here.

Let $A=\prod\limits_{n=1}^{\infty}G_n$, where $G_n=\mathbb{Z}$ for all $n$.
And then we have $\mathbb{Z}\times A\cong\mathbb{Z}\times\mathbb{Z}\times A$. But $\mathbb{Z}\ncong\mathbb{Z}\times\mathbb{Z}$.


Proving that sequence is bounded Given $|x_n| \leq y_n$ for all $n \in N$, and $(y_n)$ is a convergent sequence, prove that $(x_n)$ is bounded as well.

My attempt:
Let, $\lim(y_n) = a,$ then since $|x_n| \leq y_n$ for all $n,$ we have $|x_n| \leq a.$ Hence, ($|x_n|$) is bounded by $a.$

How do I now go about proving $(x_n)$ is bounded as well?

The statement

" $\lim(y_n) = a$, then since $∣x_n∣ ≤ y_n$ for all $n$, we have $∣x_n∣ ≤ a$ "

is not necessarily true.

You may alter it to

"since  $\lim(y_ n) = a$, we have $ y_n \le M$ for some $M>0$ and since $∣x_n∣ ≤ y_n$ for all $n$, we have $∣x_n∣\le M "$

We know that $- ∣x_n∣\le  x_n \le ∣x_n∣ $, so $x_n$ is bunded.

Try thinking about how $|x_n|\leq y_n$ is the same as $|x_n-0|\leq y_n$.
As $y_n$ is bounded we know there exist some upper bound $U$, we also know that $y_n \ge0$ as $y_n\ge|x_n|$ so $|y_n|=y_n<U$
And $|x_n|\le y_n <U$
At very least $x_n$ is bounded by $+U,-U$


$E(Y^4)$ for exponential distribution. I was doing a question on finding the variance for an exponential distribution and happened on the term $E(Y^4)$. We have the pdf of the exponential distribution as $f(y) = \frac{1}{\beta}e^{\frac{-y}{\beta}}$ where we have $\beta = 10$. So, then we have $$E(Y^4) = \int_0^\infty y^4\frac{1}{10}e^{\frac{-y}{10}} \, dy$$ I was thinking that this looked awfully familiar to the $\Gamma$ function so I set $$E(Y^4) = 10^3\Gamma(5)$$ where $y = \frac{y}{10}$ and $$\Gamma(a) = \int_0^\infty y^{a-1}e^{-y} \, dy$$ But the answer I'm given in the solution manual is $E(Y^4) = 10^4\Gamma(5)$. Could someone please explain why they got the result they did? Why is mine wrong? Thanks.

Note that
$$E(Y^4) = \int_0^\infty y^4e^{\frac{-y}{10}}\frac{1}{10}\,dy
=\int_0^\infty (10u)^4e^{-u}\,du
=10^4\int_0^\infty u^4e^{-u}\,du
=10^4\Gamma(5)
$$
where we made the substituion $u=y/10$.


Expectation and variance of the number of elements of a random non-empty set selected from a finite power set Let $S$ denote a finite set of cardinality $|S| = N$. Select randomly a non-empty subset of $S$. Let $X$ indicate the number of items belonging to this subset.
(a) Describe the probability mass function of $X$ and compute $\textbf{E}(X)$ and $\textbf{Var}(X)$.
(b) Prove that $\displaystyle\lim_{N\rightarrow\infty}\frac{\textbf{E}(X)}{N} = \frac{1}{2}$ and $\displaystyle\lim_{N\rightarrow\infty}\frac{\textbf{V}(X)}{N} = \frac{1}{4}$.

MY ATTEMPT
(a) In the first place, we first notice there are $C(N,x)$ subsets which has $x$ elements, where $1\leq x\leq N$. Therefore we conclude that
\begin{align*}
f_{X}(x) = \frac{1}{2^{N} - 1}{N\choose x}
\end{align*}
From then on we are able to determine $\textbf{E}(X)$ and $\textbf{Var}(X)$. Precisely speaking, we have
\begin{align*}
\sum_{x=1}^{N}{N\choose x}x & = \sum_{x=1}^{N}\frac{N!}{(x-1)!(N-x)!} = \sum_{x=0}^{N-1}\frac{N!}{x!(N-x-1)!}\\
& = N\sum_{x=0}^{N-1}\frac{(N-1)!}{x!(N-x-1)!} = N\sum_{x=0}^{N-1}{N-1\choose x} = N2^{N-1}
\end{align*}
according to the binomial theorem. From whence we conclude that
$$\textbf{E}(X) = \frac{N2^{N-1}}{2^{N}-1}$$
Further, we do also have the following identity
\begin{align*}
\sum_{x=1}^{N}{N\choose x}x^{2} & = \sum_{x=1}^{N}\left[\frac{N!}{(x-1)!(N-x)!}\right]x = \sum_{x=0}^{N-1}\left[\frac{N!}{x!(N-x-1)!}\right](x+1)\\
& = \sum_{x=1}^{N-1}\frac{N!}{(x-1)!(N-x-1)!} + \sum_{x=0}^{N-1}\frac{N!}{x!(N-x-1)!}\\
& = N(N-1)\sum_{x=0}^{N-2}\frac{(N-2)!}{x!(N-x-2)!} + N\sum_{x=0}^{N-1}\frac{(N-1)!}{x!(N-x-1)!}\\
& = N(N-1)\sum_{x=0}^{N-2}{N\choose x} + N\sum_{x=0}^{N-1}{N-1\choose x} = N(N-1)2^{N-2} + N2^{N-1}
\end{align*}
Therefore
$$\textbf{E}(X^{2}) = \frac{N(N-1)2^{N-2} + N2^{N-1}}{2^{N}-1}$$
Finally, we have the variance:
\begin{align*}
\frac{\textbf{Var}(X)}{N} & = \frac{\textbf{E}(X^{2}) - \textbf{E}(X)^{2}}{N}
\end{align*}
(b) As requested, we obtain the desired result
\begin{align*}
\lim_{N\rightarrow\infty}\frac{\textbf{E}(X)}{N} = \lim_{N\rightarrow\infty}\frac{2^{N-1}}{2^{N}-1} = \frac{1}{2}
\end{align*}
However, the same does not apply to the variance when I make use of the results obtained. Am I committing some conceptual mistake? I would be equally grateful if someone provided me a quicker or smarter way to tackle this problem.
EDIT
I edited my answer as suggested by Did. Nonetheless, I am still thinking there is something wrong with my solution. Could someone double-check my calculations? Thanks in advance.

Seem probably simpler if you look at it this way: Say $S$ has $n$ elements. If  $Y$ is the number of elements of a randomly selected element of the power set, including the empty set as a possibility, then $Y$ has the same distribution as $X_1+\dots+X_n$, where the $X_j$ are iid with $P(X_1=0)=P(X_1=1)=1/2$.


Evaluating the Definite Integral $\int_0^{\pi}\cos^{2n} \theta d\theta$ $$\int_0^{\pi}\cos^{2n} \theta d\theta$$

$$u=\cos \theta \implies du= -\sin \theta d\theta \implies d\theta= -\frac{du}{1-u^2} $$

$$\int_{-1}^1 \frac{u^n}{1-u^2} du=\int_{-1}^1 \frac{u^n}{(1-u)(1+u)}du$$

I have no idea what to do next, any guidance is appreciated!

HINT

$\int_0^{\pi}\cos^{2n} \theta d\theta=2\int_0^{\pi/2}\cos^{2n} \theta d\theta$

which is Wallis' integral: integral

See this document (page 5 theorem 2.2): http://www.les-mathematiques.net/phorum/file.php?4,file=57836,filename=507-1088-1-PB.pdf


Different ways to arrange $n$ different groups in $2$ equal parts I was solving a combination problem of arranging N different groups into 2 equal parts where order of elements in the part doesn't matter.
e.g-
We have 3 types of elements/groups A, B and C .
Type A has 1 element, type B has 6 elements and type C has 1 element.

If I arrange them in two equal parts, the possibilities are following :

ABBB | BBBC ,
BBBC | ABBB ,
ACBB | BBBB ,
BBBB | ACBB

Another case,

type X has 2 elements, type Y has 2 elements. the possible arrangements are :

XX | YY ,
XY | XY ,
YY | XX

How to calculate this count ? I was looking forward to generalize this for N types having different frequencies.

Suppose there are $n$ different groups, and the $i^{th}$ group has $a_i$ members total. I will assume that $s=\sum_{i=1}^n a_i$ is even, so that it is possible to equally divide the elements.

Let $x_i$ be the number of members of the $i^{th}$ group that are put into the first part, so that $a_i-x_i$ are put into the second part. The numbers $x_i$ will describe a valid solution if and only if
$$
x_1+x_2+\dots+x_n=s/2,\\
0\le x_i\le a_i\hspace{.6cm}\text{for }1\le i\le n
$$
We just need to count the number of integer solutions to the above equation and inequalities. This can be done using the Principle of Inclusion Exclusion and stars and bars, as described in the answers to this question.

$A$ is in one group.  Either $C$ is in the same group as $A$ or it isn't.

Think about it....


"Simple" rational inequality - How does one deduce this? I am a little puzzled by a past exam question. Part (a) is very easy and part (b) is easy too. However, according to the question, and the mark scheme, I should be able to deduce the answer to part (b) from my answer to part (a) without any further calculation. I am able to find the answer to part (b), but I have to draw some graphs and it is far from straight forward.

Part (a) [6 marks]

Find the set of values of $x$ for which
$$x+4>\frac{2}{x+3}$$
As I said: this is easy. Multiply both sides by $(x+3)^2$, take everything over to one side and then factorise. This gives $(x+2)(x+3)(x+5)>0$ and so $-5<x<-3$ or $x>-2$.

Part (b) [1 mark]

This is where it gets a little odd. The question says: Deduce the values of $x$ for which
$$x+4>\frac{2}{|x+3|}$$
According to the solutions, part (b) should be blindingly obvious. The answer is $x>-2$. However, I don't see how one can deduce this trivially. I drew the graph and got the right answer, but a 1 mark question means it is obvious and there is no need for calculation.

Adding a modulus usually makes a problem more difficult and adds solutions, e.g. $x-1>2 \iff x>3$, while $|x-1|>2 \iff x<-1 \ \ \text{or} \ \ x>3$.

The right-hand-side of (b) is positive. This forces $x > -4$. On the other hand, if $-4 < x < -3$, the left-hand side is less than $1$ but the right-hand-side is greater than $1$. Thus $x > -3$.

This means $|x + 3| = x+3$ so the two inequalities (a) and (b) are the same. Since the solution to (a) is $-5 < x < -3$ or $x > -2$, you must have $x > -2$.

I don't know if it is blindingly obvious, but there you go.

You can see the result from the calc out of a).
but not out of yours.

I multiply first $(x+3)$. Then you have 2 cases.

First case $x+3 < 0$ then you've got $(x+4)(x+3) < 2$ (you have to change the sign, because you multipy by a negative figure)
this case return the solution $−5<x<−3$

And the second case $x+3 \geq 0$ returns $x>−2.$

It IS quite obvious, that if you multiply by $|x+3|$ there is only the second case. So only the solution of the second case is correct.


Integral $\int \arcsin \left(\sqrt{\frac{x}{1-x}}\right)dx$ $$\int \arcsin\left(\sqrt{\frac{x}{1-x}}\right)dx$$

I'm trying to solve this integral from GN Berman's Problems on a course of mathematical analysis (Question number 1845)

I tried substituting $x$ for $t^2$:
$$2\int t\arcsin\left(\frac{t}{\sqrt{1-t^2}}\right) dt$$
And then evaluating it by parts, which made it very complicated.

So how do I go about solving this?

Let $t= \sqrt{\frac{x}{1-x}}$. Then,

$$I = \int \arcsin\left(\sqrt{\frac{x}{1-x}}\right)dx
= -\int \arcsin t\> d\left(\frac{1 }{1+t^2}\right) $$

Integrate by parts,

$$I= -\frac{\arcsin t}{1+t^2}+\int \frac{dt}{(1+t^2)\sqrt{1-t^2}} $$

Let $u^2=\frac{2t^2}{1-t^2}$. Then,
$$\int \frac{dt}{(1+t^2)\sqrt{1-t^2}} = \frac1{\sqrt2}\int \frac{du}{u^2+1}
= \frac1{\sqrt2}\arctan u$$

Thus,

$$I= -\frac{\arcsin t}{1+t^2}+ \frac1{\sqrt2}\arctan \left(\frac{\sqrt2 t}{\sqrt{1-t^2}}\right)+C$$
$$=-(1-x) \arcsin\sqrt{\frac{x}{1-x}}+\frac{1}{\sqrt{2}}\arctan\sqrt{\frac{2x}{1-2x}}+ C$$

This is what you should get. I am very sorry I thought in the denominator its not arcSin, my apologies!!

Please go through this, and let me know if you have concerns or questions.


Max Eigenvalue of A'A A is an $m\times m$ square matrix with (possibly complex) eigenvalues $\lambda_i$. Suppose $\max |\lambda_i(A)|<r$ (for some positive no. $r$ say). Then can we conclude any upper-bound of $\ \lambda_{\max} (A'A)$ (in terms of $r$) ? Thanks in advance.

The question you are asking is deeply related to the relation between spectral norm $\|A\|_2$ and spectral radius $\rho(A)$. If $\lambda_i$'s are eigenvalues of $A$, then spectral radius is defined as:
$$
\rho(A)=\max|\lambda_i|.
$$
Spectral norm is defined as:
$$
\|A\|_2=\max_{\|x\|_2=1}\|Ax\|_2.
$$
See that $\|Ax\|_2^2=x^*(A^*A)x$ and since $A^*A$ is positive semidefinite, we can see:
$$
\|A\|_2^2= \lambda_{\max}(A^*A).
$$
In general $\|A\|_2\geq \rho(A)$, which means that the upper bound that you have does not tell that much about spectral norm. In general the inequality can be strict.

For instance consider $A=\left[\begin{array}{cc}
 0&1\\
0&0 \end{array}
\right]$. The eigenvalue is $0$ which give $\rho(A)=0$. However:
 $$
 A^* A=\left[\begin{array}{cc}
 0&0\\
1&0  \end{array}
\right]
\left[\begin{array}{cc}
 0&1\\
0&0  \end{array}
\right]=
\left[\begin{array}{cc}
 0&0\\
0&1  \end{array}
\right]
$$
which has eigenvalues $0,1$ and hence $\|{{A}}\|_{2}=1$. Hence $\rho(A)<\|A\|_2$.

However for some cases these two are equal. A very general class of these matrices are normal matrices. In this case any upper bound on spectral radius is an upper bound on spectral norm.

To see why $\|A\|_2\geq \rho(A)$, pick a unit norm eigenvector $y$ of $A$ corresponding to the eigenvalue with maximum absolute value. It is easy to see that $\|Ay\|_2=\rho(A)$. From the definition of $\|A\|_2$, we have $\rho(A)\leq \|A\|_2$.

Since $\;\lambda\;$ is an eigenvalue of $\;A\implies \overline\lambda\;$ is an eigenvalue of $\;A^*\;$ , and since $\;|\lambda|=|\overline\lambda|\;$we get that

$$\left|\lambda_\max(A^*A)\right|=|\lambda_\max(A)|^2<r^2$$


What is the difference between normal and perpendicular? What is the difference when a line is said to be normal to another and a line is said to be perpendicular to other?

There are different kind of contexts you use term normal in mathematics. You often use perpendicular in case of two or three dimensional geometry, and in hi-dimensional spaces (e.g. infinite-dimensional) a term orthogonal is more common. On the other hand in the context of vectors, normal usually means also that the vector is of unit length, however this is not a must (but it is for example if you speak about orthonormal base).

There is yet another related meaning in computer graphics, where normal is the direction which you would use to reflect the light. This can be a vector perpendicular to rendered face, but for nice effects you use one perpendicular to the original surface, not its approximation. Also, there is bump mapping technique that lets you change the normal vectors and achieve look of some kind of wrinkles and bumps.

Finally, e.g. in topology or group theory term normal means something completely different, however, I will leave those and others out as I suspect this was not the scope of your question. Hope this helps ;-)

If normal is 90 degree to the surface, that means normal is used in 3D. Perpendicular in that case is more in 2D referring two same entities (line-line) with 90 degree angle between them. Normal in this case could refer 2 different entities (line-surface) making this term valid for 3D case (I definitely remember hearing term "line normal to surface", rather then "line perpendicular to the surface") - line and surface makes 3D case; line and line is one plane - 2D. This I just figured out from one of the answers here which made most sense for me.


Composition of functions that are onto or one-to-one I found part of my answer here: If g(f(x)) is one-to-one (injective) show f(x) is also one-to-one (given that...); however I wanted to flesh out the last two statements I had in a proposition in my notes.

  Proposition:
  Let $f: A \rightarrow B$ and $g: B \rightarrow C$.  Then:

  (i) If $g \circ f$ is one-to-one, then $f$ is one-to-one.

  (ii) If $g \circ f$ is onto, then $g$ is onto.

Proof:
(i)  Suppose $f(x)=f(y)$ for some $x,y$.

Since $g \circ f$ is one-to-one: $$g\circ f(x) = g\circ f(y) \Rightarrow x=y,\forall x,y \in A.$$

Therefore $f$ must be one-to-one.

(ii)  Since $g \circ f (x)$ is onto, then for every $c \in C$ there exists an $a \in A$ such that $c=g(f(a))$.  Then there exists a $b \in B$ with $b=f(a)$ such that $g(b)=c$.  Thus g is onto.

I wanted to confirm that these proofs are both correct for my peace of mind (as they weren't proven in class).

On the first one i would say since g composed of f is one-to-one, if g(f(x))=g(f(y)) for all x,y in A then x=y. By definition of f then there exist some value f(x) and f(y) in B. Then because we already noted x=y. F(x) must equal F(y) by the definition of a function.


Wikipedia's proof that $f : \mathbb{R} \to \mathbb{R}, x \mapsto x^2$ is not uniformly continuous Prove that $f : \mathbb{R} \to \mathbb{R}, x \mapsto x^2$ is not uniformly continuous

Now I actually ask this question, as Wikipedia's proof of this seems wrong to me.

Wikipedia's Proof:

$f$ is uniformly continuous on $\mathbb{R}$ if $\forall \epsilon > 0$, there exists a $\delta > 0$ for all $x_1, x_0 \in \mathbb{R}$ such that

$$|x_1 - x_0| < \delta \implies |f(x_1) -f(x_0)| < \epsilon$$

But $$\begin{aligned}
f(x+ \delta) - f(x) &= x^2 + 2x\delta +\delta^2 - x^2\\
&= 2x\delta + \delta^2\\
&= \delta(2x + \delta)\\
&> \epsilon \ \ \ \text{for suffiently large $x$} \  \ \ \square
\end{aligned}$$

Now what I don't get is that if we set $x_1 = x+ \delta$ and $x_0 = x$, then $|x_1 -x_0| = |x+ \delta - x| = |\delta| \not< \delta$, so everything above should not apply as $|x+ \delta - x| \not < \delta$ in the first place.

Is Wikipedia's proof correct? If so, then why would my argument above be incorrect?

You can replace the $|x - y| < \delta$ with $|x - y | \leq \delta$ in the definition of uniform continuity  and not fundamentally change anything.  However, if it makes you feel comfier, you can fix it up by choosing a $\rho < \delta$ and proceed exactly as shown, but with $f(x+ \rho) - f(x)$.

We can establish the equivalence of the two slightly different definitions as follows:

First, if there exists a $\delta$ such that $|x - y| < \delta \implies |f(x) - f(y)| < \varepsilon$, then there exists a $\rho$ such that $|x - y| \leq \rho \implies |f(x) - f(y)| < \varepsilon$.  For example, $\rho = \delta / 2$.

On the other hand, if there exists a $\delta$ such that $|x - y| \leq \delta \implies |f(x) - f(y)| < \varepsilon$, then of course it's also true that $|x-y| < \delta \implies |f(x) - f(y)| < \varepsilon$.

One can also change the $< \varepsilon$ condition to $\leq \varepsilon$.  To establish this equivalence, first suppose that if we are given any $\varepsilon > 0$, we can find a $\delta$ such that $|x - y| < \delta \implies |f(x) - f(y)|< \varepsilon$.  It follows immediately that $|x-y| < \delta \implies |f(x) - f(y)| \leq \varepsilon$.

On the other hand, suppose that if we are given any $\varepsilon > 0$, we can find a $\delta$ such that $|x-y| < \delta \implies |f(x) - f(y)| \leq \varepsilon$. By choosing a $\delta$ that satisfies this condition for some $\varepsilon' < \varepsilon$, we thus have a $\delta$ such that $|x-y| < \delta \implies |f(x) - f(y)| < \varepsilon$.

I suspect the reason why the strict less-than condition for both $\varepsilon$ and $\delta$ is prefered in analysis texts is because this emphasizes the more general topological definition of continuity.  Namely, the preimage of an open set is open.

If we put

$$f(x)=x^2,$$

$$x_n=\sqrt{n+1},$$

and $$y_n=\sqrt{n}.$$

then we have

$$\lim_{n\to+\infty}(x_n-y_n)=0$$

but

$$|f(x_n)-f(y_n)|=1$$

So, $\;\;f\;\;$is not uniformly continuous at $[0,+\infty)$.

If we take $\epsilon=0.5$ for example,

then $\forall \eta>0$,

we will always find $x_n$ and $  y_n$ such that

$|x_n-y_n|<\eta$ and

$|f(x_n)-f(y_n)|=1>\epsilon$.


What is a simple means of proving that 3 vectors belonging to $\Bbb{R}^2$ are linearly dependent? For my linear algebra class, there is a 2 part problem that asks, given the set {(1 2), (-1 -1), (1 0)}, prove or disprove that it is linearly independent using the definition only AND then prove or disprove that it is linearly independent "with a quicker proof that uses more than just the definition". For the first part I disproved that it is linearly independent by providing a counterexample, but for the second part I don't know what to do. My gut tells me that you can't have more than 2 linearly independent vectors that belong to  $\Bbb{R}^2$ (since the rref of a matrix with these vectors as columns could have at most 2 pivots), but I don't know how to express this in a proof.

If $v_1$ and $v_2$ are linearly dependent, you're done. If $v_1$ and $v_2$ are linearly independent  then $\{v_1,v_2\}$ is  a basis for $R^2$ so there are reals $x,y$ with $v_3=x v_1+y v_2.$

It turns out that what he was looking for was some quick statement based on the following proposition: Let V be an n-dimensional vector space. If {$u_1$, ... , $u_p$} $\subset$ V and p > n, then {$u_1$, ... , $u_p$} is linearly dependent.


Sequence of monotone functions problem Found a problem on "The Way of Analysis" by Strichartz.

Say $f_n$ converges to $f$ (pointwise) and each $f_n$ is increasing.

(a) Must $f$ be increasing?
(b) What happens if each $f_n$ is strictly increasing?

To be honest I have no idea. Taking $x<y$ in the domain, I'm trying to use the definition, finding appropriate $\epsilon$ to somehow deduce $f(x)\le f(y)$ but failed. Any help?

Suppose $x<y$ then $f_n(x) \le f_n(y)$, in particular $f_n(y)-f_n(x) \ge 0$. Taking limits gives $f(y)-f(x) \ge 0$ (the limit of a non negative sequence must itself be non negative, which is essentially the fact that
$[0,\infty)$ is closed) and hence $f(x) \le f(y)$.

Let $f_n(x) = {1 \over n}\arctan x$. Each $f_n$ is strictly increasing, the limit $f(x)  = 0$ is not strictly increasing.

Here's another $f_n(x) = {1 \over n} {x \over 1+|x|}$.

Here's a sketch of a solution:

Choose $x<y$.  We know $f_n(x)\rightarrow f(x)$ and $f_n(y)\rightarrow f(y)$.  This means that for all $\epsilon>0$ there are $N(x,\epsilon)$ and $N(y,\epsilon)$ such that if $n \geq\max\{N(x,\epsilon),N(y,\epsilon)\}$,

$$
f(x)-\epsilon\leq f_n(x)\leq f(x)+\epsilon\\
f(y)-\epsilon\leq f_n(y)\leq f(y)+\epsilon
$$ thus using one side of each of these and the fact that $f_n(x)\leq f_n(y)$ we have

$$
f(x)-\epsilon\leq f_n(x)\leq f_n(y)\leq f(y)+\epsilon
$$ and so

$$
f(x)\leq f(y)+2\epsilon
$$ Since this inequality holds for all $\epsilon>0$, it remains true in the limit $\epsilon\rightarrow 0$.

For the strictly increasing case, you'll end up with a statement like

$$
f(x)<f(y)+2\epsilon
$$ does this imply that the strict inequality holds in the limit $\epsilon\rightarrow 0$?


An inequality with a load of variables: $ (1-a_1)(1-a_2)...(1-a_n) \ge 1/2$ It is known about numbers $$a_1, a_2, ... a_n$$ that $$a_1 + a_2 +...+a_n \le 1/2.$$ Prove that $$ (1-a_1)(1-a_2)...(1-a_n) \ge 1/2$$
I have tried using $a^2 \geq 0$, it led to nothing.
How can I make my inequality look like in the possible duplicate?

Hint: you can consider the inequality for arithmetic and geometric means:

$$\frac{1}{n}\sum_{k=1}^n b_i \geq \sqrt[n]{\prod_{k=1}^n b_i}$$

for any set of non-negative real numbers $\{b_1,\cdots,b_n\}$

Partial attempt:

$$b_i = 1-a_i$$
$$\frac{1}{n}\sum_{k=1}^n (1-a_i) \geq \sqrt[n]{\prod_{k=1}^n (1-a_i)}$$
$$1-\frac{1}{n}\sum a_i \geq \sqrt[n]{\prod_{k=1}^n (1-a_i)}$$

$$lhs > 1-\frac{1}{2n}$$

if product $\leq 1/2$ then rhs not sure to be larger than $\sqrt[n]{\frac{1}{2}}$

Now what to remains is to compare which can be done by calculus comparing a linear function with a power function, treating n as continous variable.

The proof that blue does not overtake red can be left as an exercise.