In how many bit strings of length 10 are there with no 3 0s adjacent it is hard for me dealing with more than 2 not adjacency problems, I need some straightforward approaches for these kinda problems, I have searched most of the community looking for similar problems but I got sorta more confused.
 it would be very nice of you to put me in the right direction. The title is my question A
But for B:
B)if at least 3 0s are not adjacent?

A) My notion is, 2 0's are allowed,
So at first as we got only 3 0s to place in,  I get the total by $10\choose3$ =$120$
And for 3 adjacent 0s:
0001111111,1000111111,1100011111,1110001111,1111000111,1111100011,1111110001,1111111000, indicating we have only 8 locations to place our consecutive 0s, subtracting this from the total , $120 - 8 = 112$ , was this a correct approach?

For B)  it means we can place in more 0s than just 3 right? If so , non consequtive 3 0s,4 0s, ... , 10 0s
Summing them all up we have 36, hence 120 - 36 = 84 .  ?
Help please..

if the question is "How many bitstrings of length 10 exist with three 0's, but non-adjacent?" Your solution to A. of 112 is right.

For B, you'd have to write a recurrence relation.

You can compose any acceptable string by combining $1, 01, 001$.
The number of words you can create like this satisfies $w_n = w_{n-1} + w_{n-2} + w_{n-3}$. Since you can also finish with 0, 1, or 2 trailing 0's, you are interested in $w_8 +w_9 + w_{10}$ which happens to be $w_{11}$.

You have the tribonacci numbers, with an offset, so I think the answer is 504.
http://oeis.org/A000073

For all who might come across my problem and would like to know about the combinatorial solution:

We start with 10 1s and 0 0.
$10\choose0$ = 0
9 1's, means we got 10 gaps to place one 0,
$10\choose1$ = 10
8 1's,  we got 9 gaps left and we should choose 2 locations to place 0, $or$ choose one single location to place $00$
$9\choose2$ + $9\choose1 $
7 1's,  we got 8 gaps left and we should choose 3 locations to place 0, $or$ choose 2 locations to place 00 and 0, with $\frac{2!}{1!}$ ways for their arrangement.
$8\choose3$ + $8\choose2 $ $\times$ $\frac{2!}{1!}$
The string includes 6 1's, leaving us with 7 gaps in between, either we should choose 4 locations to place 0, $or$ choose 3 locations to place 00 and 0,0 with $\frac{3!}{2!}$ ways for their arrangement.. $or$ 2 locations to place 00,00
$7\choose4$ + $7\choose3 $ $\times$ $\frac{3!}{3-1!}$ + $7\choose2$
The string includes 5 1's, leaving us with 6 gaps in between, either we should choose 5 locations to fill with 0's, $or$ choose 4 locations to place 00 and 0, 0, 0 with $\frac{4!}{3!}$ ways for their arrangement.. $or$ 3 locations to place 00,00 ,0
$6\choose5$ + $6\choose4 $ $\times$ $\frac{4!}{4-1!}$ + $6\choose3$ $\times$ $\frac{3}{3-1}$
The string includes 4 1's, leaving us with 5 gaps in between, either we should choose 5 locations to place 00,0,0,0,0 with $\frac{5!}{4!}$ ways for their arrangement $or$ choose 4 locations to place 00 and 00, 0, 0 with $\frac{4!}{2!2!}$ ways to arrange them. $or$ we can choose 3 locations to place 00,00 ,00
$5\choose5$ $\times$ $\frac{5!}{5-1!}$ + $5\choose4 $ $\times$ $\frac{4!}{4-2!4-2!}$ + $5\choose3$
The string includes 3 1's, leaving us with 4 gaps
we should choose 4 locations to place 00,00,00,0 with $\frac{4!}{3!1!}$ ways for their arrangement
$4\choose4$ $\times$ $\frac{4!}{3!1!}$

Summing them all up, we get 504...


Kalman filter: state estimate Assume we have a linear state-space model:
$$
z_{k} = Hx_{k} + v_{k}\\
x_{k} = F x_{k-1} + w_{k}
$$

If I understand correctly, having observations $z_{0}, \dots, z_{k}$, in filtering problem the goal is to get the estimate $\hat{x}_{k|k} $ of $E[x_{k}|z_{k}, \dots, z_{0}]$.

The question: why do we call $\hat{x}_{k|k}$ the state estimate if state is itself random?

In wiki, it is even more confusing:

In what follows, the notation $\hat{\mathbf{x}}_{n|m}$ represents the estimate of $x$ at time $n$ given observations up to and including at time $m \leq n$.

Each measurement is associated with a certain time index and this then also fixes the associated estimate of the state with respect to an index. For example $\hat{x}_{k|k} = E[x_{k}|z_{k}, \dots, z_{0}]$ is a different problem compared to $\hat{x}_{k+1|k} = E[x_{k+1}|z_{k}, \dots, z_{0}]$. Namely, in that case the measurement $z_{k+1}$ is not used, so during that time step one can only predict. For example if $w_k\sim\mathcal{N}(0,W)$ then the expected value of the prediction would yield $\hat{x}_{k+1|k} = F\,\hat{x}_{k|k}$.

Also note that the state itself is a stochastic process, which is does not have to be completely "random" since it can be modeled by $x_{k} = F\,x_{k-1} + w_{k}$. Only if $F=0$ would the state inherit all the stochastic properties from $w_k$.

The notation $\hat{ x}_{n|m}$ means $E(x_n|z_m,\cdots,z_0)$. When $m=n$, estimating $x_t$ is the filtering problem, when $m>n$, it is the prediction problem, and it is the smoothing problem for the cases where $n<m$.


Using chain rule take the derivative of $x^{\sin(x)}$ How do I use specifically the chain rule for $x^{\sin(x)}$?

In your application of the chain rule you have defined

$$ v(u) = x^u, \quad \text{and} \quad u(x)=\sin(x), $$

and noticed that it follows that

$$ v(u(x)) = x^{\sin(x)}, $$

in order to apply the rule to $v(u(x))$.

The problem is that $v(u)$ as you defined it also depends on $x$, which is the argument of the "inside function", $u(x)$. This invalidates the application of the rule.

Treating the exponent as a constant, the derivative is $\sin(x)\cdot x^{\sin(x)-1}$, treating the basis as a constant the derivative is $x^{\sin(x)}\cdot\ln(x)\cdot\cos(x)$. Adding both gives the derivative of $x^{\sin(x)}$.  (This is not a coincidence.)


On the homotopy type of $\mathbb R P^3$ and $\mathbb R P^2 \vee S^3$ I want to prove $\mathbb R P^3$ and $\mathbb R P^2 \vee S^3$ are not homotopy equivalent. After some work I've shown that they have the same fundamental group and homology, and $X = S^3 \vee S^2 \vee S^3$ is a universal covering for $\mathbb R P^2 \vee S^3$ (just attach one copy of $S^3$ to each one of the poles in $S^2$ and you get a $2$-fold covering).

I know this question has already been answered here before, but everyone I ask uses some variation of an argument based on the cohomology ring of these spaces, and I don't know nothing about cohomology or cup products yet (neither has the text book said anything about it when this exercise was proposed). Apparently the second homotopy group of $\mathbb R P^2 \vee S^3$ is $\mathbb Z$ while we know $\pi_2(\mathbb R P^3)$ is trivial, so that's  the way to go. Unfortunately, I do know how to compute the homotopy groups of the wedge sum $X = S^3 \vee S^2 \vee S^3$ as I can't use van Kampen, so I'm honestly quite lost here.

Could someone show me how does one prove they are not homotopy equivalent using just the homology and maybe homotopy and the covering spaces?

As you observed, $$\pi_2(\Bbb{RP}^3) = \pi_2(\widetilde{\Bbb{RP}^3}) = \pi_2(S^3) = 0.$$ On the other hand, the wedge of two spaces $X \vee Y$ retracts onto either of the summands, by sending every point in $Y$ (or $X$, depending) to the common basepoint. In particular, $\pi_2(X) \to \pi_2(X \vee Y)$ is injective, because composing it with the map $\pi_2(X \vee Y) \to \pi_2(X)$ induced by the retraction gives the identity. So $\Bbb Z = \pi_2(\Bbb{RP}^2) \hookrightarrow\pi_2(\Bbb{RP}^2 \vee S^3)$, and that's enough to get a contradiction.

Hint: look the multiplicative structure of the cohomology groups. $H^*(\mathbb{R}P^3,\mathbb{Z}/2)$ is generated by an element of degree $1$, it is not the case of $H^*(\mathbb{R}P^2\vee S^3,\mathbb{Z}/2)$.


3, 5, 7 are the only three consecutive odd natural numbers that are prime. Can someone please help me proof this theorem. So far I assumed towards contradiction that there exist three consecutive odd natural numbers that are primes P, P+2, and P+4.

Theorem: 3, 5, 7 are the only three consecutive odd natural numbers that are prime.

All odd primes, except for $3$, are of the form $6n \pm 1$.

So, if two odd primes bigger than $3$ exists and differ by $2$,  they must be of the form

$6n-1, 6n+1$ for some integer $n$.

The number that are $2$ before and $2$ after will show up as

$6n-3, 6n-1, 6n+1, 6n+3$

If $n=1$, you get $3,5,7,9$ and the first three numbers are prime.

For any $n>1$, only the middle two numbers might be prime.

A simpler proof.

Let $(x, x+2, x+4)$ be three consecutive odd numbers.

Modulo 3, those three numbers will have to look like one of

(0, 2, 1)  or  (1, 0, 2)  or (2, 1, 0)

Hence at least one of them will be a multiple of three.

Since the number 3 is a prime number, we need to look at those cases where one of those three numbers is 3.

(-1, 1, 3)  or
(1, 3, 5)  or
(3, 5, 7)

So (3, 5, 7) is the only sequence of three consecutive odd prime numbers.

Here $nk$ represents $10n +k$

We start by saying ,$$$$leaving 2 all prime numbers have odd units place , $$$$possible combinations for 3 consecutive prime numbers will be.

$$n1,n3,n5$$
$$n3,n5,n7$$
$$n5,n7,n9$$
$$n7,n9,(n+1)1$$
$$n9,(n+1)1,(n+1)3$$
As all the numbers $\gt 5$ with units place 5 are divisible by 5, so by eliminating cases with n5 we are left with,

$$n7,n9,(n+1)1$$
$$n9,(n+1)1,(n+1)3$$

  Now if we see 3's table
  $$9,39,69,\cdots,(3n)9$$
  Are divisible by 3,and
  $$21,51,81,\cdots,(3n+2)1$$
  Are divisible by 3,
  $$27,57,87,\cdots,(3k+2)7$$
  Are divisible by 3

So on our possible cases if $n=3k$

  $$(3k)7,(3k)9,(3k+1)1$$
  $$(3k)9,(3k+1)1,(3k+1)3$$
  As  $(3n)9$ is divisible by 3, $n\not= 3k$ for three consecutive prime numbers to exist.

For $n=3k +1$

  $$(3k+1)7,(3k+1)9,(3k+2)1$$
  $$(3k+1)9,(3k+2)1,(3k+2)3$$
  As (3k+2)1 is divisible by 3,
  $$n\not= 3k+1$$ for three consecutive prime no to exist

For $n=3k+2$

  $$(3k+2)7,(3k+2)9,(3(k+1))1$$
  $$(3k+2)9,(3(k+1))1,(3(k+1))3$$
  As (3k+2)7 is divisible by 3 ,
  $$n\not= 3k+2$$for three consecutive prime numbers.

We all cases are disapproved we thus say there are no three consecutive prime numbers $\gt 5$.


The number of circles that will fit inside the area of larger circle? Let's say circle $\omega_1$ has a diameter $X$. Let $X>Y$; $Y\in \mathbf{R}^{+}$. How many circles with diameter $Y$  will fit inside $\omega_1$?

Is there a formula for this?


Why is the projection of a closed polytope closed? In general, projection of a closed set into a subspace does not result in a closed set. However, I was able to prove that in $\mathbb{R}^n$, the projection of a closed polytope (intersection of finitely many closed half spaces) onto a 1 dimensional subspace is closed. However the proof involved induction on dimension and using the theory of linear optimization.

Is there a short topological proof of this fact? And perhaps a generalization that projection onto an m-dimensional subspace preserves closure? Intuitively imagining the shadow of a polytope, I feel it's true.

This is a property of linear maps in $\mathbb{R}^n$, not just projections.

The key result here is that the class of polyhedral sets (intersection of a finite number of closed half-spaces) is the same as the class of finitely generated sets (Rockafellar, "Convex analysis", Theorem 19.1).

To quote Rockafellar, "This classical result is an outstanding example of a fact with is completely obvious to geometric intuition, but which wields important algebraic content and is not trivial to prove".

Suppose $P$ is a polyhedral set, then there are vectors $v_1,...,v_l, d_1,....,d_m$ ($l$ or $m$ may be zero with appropriate adjustments, the $v_i$ are 'points', the $d_j$ are 'directions') such that
$P = \{ \sum_i \lambda_i v_i + \sum_j \mu_j d_j | \sum_i \lambda_i =1, \lambda_i \ge 0, \mu_i \ge 0 \}$.

Suppose $A$ is a linear operator. Then
$AP = \{\sum_i \lambda_i A v_i + \sum_j \mu_j A d_j | \sum_i \lambda_i =1, \lambda_i \ge 0, \mu_i \ge 0 \}$, hence $AP$ is finitely generated, and so is a polyhedral set, hence closed.

Projections are continuous functions, and continuous functions preserve limits. If a set contains all of its limit points, then the image of that set under a continuous function also contains all of its limit points. Not sure off the top of my head what the conditions are for this to hold in a general topological setting, but it sure should hold in Euclidean space.


Inequality proof $ay + bz + cx < 1$ with conditions Question: Given that for all $a,b,c,x,y,z\in\mathbb{R}^+$ and $a+x = b + y = c + z = 1$, show that $ay + bz +cx < 1$.

Attempts: Using AM-GM and squaring all sides of the given equality and rearranging is something I feel maybe be important, although I have been unsuccessful in doing so. Another idea was to consider cases, although that is certainly not going to work as there is definitely a more elegant way of going about it.
I managed to get to a point where $ay + xy + b = bz + yz + c = cx + zx + a = 1$ although I believe this is also a dead-end.
Any help or guidance would be greatly appreciated!

By AM-GM (!), $$bz\le b^az^x\le ab+xz$$
Then $$ay+bz+cx\le ay+ab+xz+xc=a+x=1$$

Substitute $a=1-x$. Then the expression $ay+bz+cx$ becomes linear in $x\in(0,1)$, hence you only have to check that the (non-strict) inequality is valid for extremal values of $x$, i.e. for $x=0$ and $x=1$. Argue similarly with $y$ and $z$. Then you are left with checking a finite number of cases.


Suppose $n\in\mathbb{Z}$. If $3 \nmid n^2$, then $3 \nmid n$. Suppose $n\in\mathbb{Z}$. If $3 \nmid n^2$, then $3 \nmid n$.

Assume $3|n$, that is $n=3m$ for some $m\in\mathbb{Z}$, so $n^2=(3m)^2=9m^2=3(3m^2)$. Let $3m^2=x$, then we have  $n^2=(3m)^2=9m^2=3(3m^2)=3x$, it means that $3|n^2$, so we are done.

Can you check my proof? Thankss...

Also, for any prime $p$
and positive integer $a$,
if
$p^a | n$
then
$n = p^am$
so
$n^2
= p^{2a}m^2
= p^a(p^{a}m^2)
$
so
$p^{a} | n^2$.

Therefore,
if
$p^{a} \not\mid n^2$
then
$p^{a} \not\mid n$.

Also see this question of mine
for a further generalization:

If $n \mid a^2 $, what is the largest $m$ for which $m \mid a$?


Functions that are their own inverse. What are the functions that are their own inverse?

(thus functions where $ f(f(x)) = x $ for a large domain)

I always thought there were only 4:

$f(x) = x , f(x) = -x , f(x) = \frac {1}{x} $ and $ f(x) = \frac {-1}{x} $

Later I heard about a fifth one

$$f(x) = \ln\left(\frac {e^x+1}{e^x-1}\right) $$ (for $x > 0$)

This made me wonder are there more? What are conditions that apply to all these functions to get more, etc.

These are called involutions.

If you don't require continuity, then the set of involutions $\Bbb R\to\Bbb R$ corresponds to the set of ways to partition $\Bbb R$ into at most $2$-element subsets. This train of thought follows from thinking about the cycle decomposition of an involution, considered as a permutation of an infinite set.

This is massively infinite, with $2^{\frak c}$-many elements, where ${\frak c}=|\Bbb R|$. Indeed, if we take for granted the fact that every infinite set has a decomposition into $2$-sets, or equivalently every infinite set has a fixed-point-free involution (which I am sure follows from some level of choice), then a lower bound for the number of involutions would be the number of coinfinite subsets of $\Bbb R$ chosen to be the fixed-point sets, and a lower bound for the number of coinfinite subsets is the number of subsets of say $\Bbb R\setminus[0,1]$ of which there are $2^{\frak c}$. An upper bound is given by the number of functions $\Bbb R\to\Bbb R$, which is also $2^{\frak c}$ - see here.

If you do require continuity there are ${\frak c}=|\Bbb R|$-many continuous involutions. As an upper bound, there are $\frak c$ continuous real-valued functions - see here. As a lower bound, $\log_a\left(\frac{a^x+1}{a^x-1}\right)$ is a distinct continuous involution for every $a>0$, of which there are $\frak c$-many.

To see where these ideas are coming from, we employ two ideas: conjugation from group theory and linear fractional transformations from complex analysis. Given any involution $\sigma$ and any homeomorphism $f:\Bbb R\to\Bbb R$ (continuous function with continuous inverse) there is a conjugate function $f\circ\sigma\circ f^{-1}$ which is also an involution. Of course, this can still equal $\sigma$ in general.

A linear fractional transformation is a map $[\begin{smallmatrix} a & b \\ c & d\end{smallmatrix}]z:=\frac{az+b}{cz+d}$. This creates a group action of the group of invertible matrices. That is, if $A$ and $B$ are matrices, then $A(Bz)=(AB)z$ in this notation, where $AB$ is the product of $A$ and $B$. Usually we have them acting on the complex plane or the Riemann sphere (or the upper half plane in hyperbolic geometry), but we can certainly consider matrices with real coefficients acting on the real number line $\Bbb R$. Note that scalar multiples of the identity matrix act as the identity function.

Note $[\begin{smallmatrix}1&1\\1&-1\end{smallmatrix}]^2$ is proportional to the identity matrix, and conjugating the corresponding linear fractional transformation $\frac{u+1}{u-1}$ by $a^x$ yields the aforementioned family of examples. Technically $a^x$ is a homeomorphism $\Bbb R\to(0,\infty)$, but this is fine since $\frac{u+1}{u-1}$ is well-defined on $(0,\infty)$.

I'll give you an example of a nonlinear involution, that is a transcendental function, and a composition of $e^x$ and $\frac{1}{\log(x)}.$ Composing one way gives $1/x$ (an involution) and composing the other way gives a different involution, whose reciprocal is an involution, and whose fixed points are reciprocals of one another: $f(x)=\exp(\frac{1}{\log(x)}).$


identity theorem to prove/disprove the existence of a holomorphic function Does there exist a holomorphic function $f$ the unit disk $\{z :|z|<1 \}$ such that
$f(\frac{1}{n}) \leq \frac{1}{2^n} $ for all $ n \in \mathbb{N}$?
I know by identity theorem we can prove that $f$ not exists with $f(\frac{1}{n}) = \frac{1}{2^n} $ but not sure if this inequality can change the conclusion.

I guess the conjecture of @mjw is right, here a tentative proof. Let me now if there are mistakes.
The zero-function satisfies the conditions, and looking for a contradiction,
let $f$ be a non-zero holomorphic function satisfying the condition too. The function $f$must have a zero of order $m\in\mathbb N_{>0}$ in $z=0$, hence
$g(z):=f(z)/z^m$ is also holomorphic and $g(0)\neq 0$. On the other hand, it holds
$$|g(0)|= \lim_{n\to \infty}|g(1/n)|=\lim_{n\to \infty} |n^m f(1/n)|\leq \lim_{n\to \infty}\frac{n^m}{2^n}=0$$
a contradiction.

Conjecture:  The function $f(z)=0$ suggested by @Desperado is the unique holomorphic function that satisfies $$|f(1/n)| \le \frac{1}{2^n}.$$


The definition of a subspace in linear algebra I'm trying to learn linear algebra on my own but I am stuck on the definition of a linear subspace.

Let's assume I want to find out if $S$ is a subspace of $\mathbb{R}^2$, where $ S = [X_1 , X_2]  $ and $ X_1 > 0 $.

$S$ would not be a subspace since it violates rule 3 of the definition. A negative one times the entire matrix would produce a vector that is not in the subspace.

But I would say it is quite obvious that the area where X is positive is a SUBSPACE of $\mathbb{R}^2$? It seems intuitive.

Why did mathematicians choose the three rules they chose? What's the motivation?

The definition of a subspace is a subset that itself is a vector space. The "rules" you know to be a subspace I'm guessing are

1) non-empty (or equivalently, containing the zero vector)

2) closure under addition

3) closure under scalar multiplication

These were not chosen arbitrarily. If you look through the definition of a vector space you'll notice that every condition will automatically be satisfied by vectors in any subset except these three.

What I mean is, for instance, if $U$ is a subset of $V$ and $V$ is a vector space, then I already know that $u_1+u_2=u_2+u_1$ for any $u_1,u_2\in U$ because they are also elements of $V$, and this property holds for elements of $V$.

Therefore what you are thinking of as some random "rules" to be a subspace are really just the minimal requirements for a subset of $V$ to itself be a vector space.

That can't be right because the set {(x1,x2):x1>0}{(x1,x2):x1>0}  does not contain the 0 vector due to the condition.


how long on average would it take for a person to randomly step on any point $n$ times. how long on average would it take for a person to randomly step on any point $n$ times.
This person walks in a random way every odd step: he can choose to go north or south $1$ meter.
Every even step, he walks 1 meter east or west.
My question is, on average, how many steps will it take to go to any point $n $ times?
I would like to learn how to solve problems like this. I'm a 9th grader hoping to learn probability.

Your walk is essentially two independent random walks, so let's just consider one of them, i.e. a (symmetric) random walk on the integers.
You are asking about the expected time of the $n$th visit to any arbitrary point. Let's first try to answer the problem, but with the point being the starting point.
It turns out that you can show that such a walk is recurrent (the probability of return is 1), but not positive-recurrent (the expected (1st) return time is infinite).
You can learn more by consulting these lecture notes which i found nice http://www.statslab.cam.ac.uk/~rrw1/markov/M.pdf


Is $\sum_{n=1}^{\infty}\arctan{\frac{1}{n}}$ finite? As I have seen Showing that $ \sum_{n=1}^{\infty} \arctan \left( \frac{2}{n^2} \right) =\frac{3\pi}{4}$

Is $$\sum_{n=1}^{\infty}\arctan{\frac{1}{n}}<\infty?$$

With Stolz-Cesaro and L'hopital rule:

$$L=\lim_{n\to \infty}\sum_{k=1}^{n}arc\tan\frac{1}{k}=\lim_{n\to \infty}\frac{\sum_{k=1}^nn \: arc\tan\frac{1}{k}}{n}=\lim_{n\to \infty}\frac{narc\tan\frac{1}{n}}{n-(n-1)}=\lim_{n\to \infty} narc\tan\frac{1}{n}=\lim_{x\to 0}\frac{arc\tan x}{x}=\lim_{x\to 0}\frac{1}{1+x^2}=1$$

Result: $L=1$


How do you respond to "I was always bad at math"? Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

To put a slightly different spin on this (before @Zev votes to close :) ). When folks ask me what I teach, I ask them to name their least favorite subject. Occasionally I do get English or history or science. But 90% of the time I get math ... to which I reply "bingo." :) I then engage them in a discussion about how math (generically) and history (for me and many others) are ruined for the masses by secondary teachers who just don't get it and who teach relying almost totally on rote memorization/manipulation. They concur 99% of the time on this. At some point, I reassure them that there are some really good teachers out there (and I claim to be one) :P

Finding some practical application for math can get people quite a lot more interested; helping them to feel like they have a mastery of the numbers; and thus of the world around them.

That most likely sounded like a paragraph of BS though, so what I'm really trying to advertise is the LOGO Programming Language; the story of Seymour Papert's Lisp-like language for small children is filled with numerous instances of kids who were "never good at math" finding that they had really just been keeping away from it due to adverse learning experiences. Here's the book that I did my final term paper on. Here you have kids below 3rd grade, learning systems of geometry, degrees, and even programming idioms like subroutines, simply through the ability to experiment with them.

Applying those kind of ideas universally in a school system would be a huge undertaking, but it could apply very well in a private learning environment. Also: LOGO is old as disco, but the basic principles mentioned in the book are available in the 'turtle' library of Python.


AB = I and BC = I then what is the relation between A and C? B = matrix given below. I is identity matrix.

 [1 2 3 4]
 [3 2 4 3]
 [1 3 2 4]
 [5 4 3 7]

So What will be the relation between the matrices A and C if AB = I and BC = I?

I think that A = C because both AB and BC have B in common and both of their product is an identity matrix but I'm not sure.

Suppose $AB=I$ and $BC=I$; then
$$
A=AI=A(BC)=(AB)C=IC=C
$$

Hint : Multiply $AB=I$ from right with $B^{-1}$ and $BC=I$ from left with $B^{-1}$. What do you get ?


How to solve in integers the equation, $(x^2-y^2)^2=1+16y$? How to solve in integers the equation,
$$(x^2-y^2)^2=1+16y$$

By observation we can take $y=3,5$

Is there another method to solve the equation?

Notice that $y\ge 0$, $y\neq 1$ and $x \neq 0$. If $(x, y)$ is a solution then $(-x, y)$ is a solution too. Hence W.L.O.G $x>0$.

It seems that $x$ and $y$ cannot be too far from each other, otherwise LHS would become larger than RHS. We'll find out this. Suppose that $z=|x-y|$. Notice that $z\neq 0$. Now, we have

$$z^2(x+y)^2=|x-y|^2 (x+y)^2=(x^2-y^2)^2=16y+1$$

And since $y\ge 0$, $y\neq 1$ and $x > 0$, then

$$z^2=\frac{16y+1}{(x+y)^2}<\frac{4(1+y)^2}{(1+y)^2}=4  \ \ \Longrightarrow \ \ z<2$$Therefore $z=1$. It follows that $x=y \pm 1$ and equation turns into a simple quadratic for $y$.

The integers solution I found are: $(\pm1,0),(\pm4,3),(\pm4,5)$.
Here's how I did it:
\begin{eqnarray}
(x^2-y^2)^2&=&1+16y\\ &=& 1\textrm{ mod }16 \qquad \textrm{(means when you divide a number by 16, remainder is 1)}\\x^2-y^2 &=& \pm\sqrt{1\textrm{ mod }16},
\end{eqnarray}
The only values of $y$ that give an integer square root for the above are 0,3 and 5.
For example, $y=3$ works because, $16\times3 + 1 = 49, \pm\sqrt{49} = \pm7$.
Now, $x^2-3^2 = \pm 7 \implies x^2 = \pm7+9 \implies x = \pm4$ or $x = \pm \sqrt{2}$. But since we are ol,y interested in integer solutions we only take the integer value of $x$. Hence, one possible integer solution is $x = \pm 4, y = 3$.


Speeding up energy costs. Am I dumb? I have maybe very easy question but my friend keep telling me I am stupid.

Let's have 10kg mass in vacuum without gravity forces. To speed up this mass to 10m/s I need 500J.

Now I want to speed up this mass from 10m/s to 20m/s. It needs surely more energy for same amount of speed up right? I am again speeding up just 10m/s but I need 1500J now right?

My friend keep telling me its not true and it needs same amount of energy to speed up mass from 0-10m/s as when for example 1000-1010m/s

Where is the truth?

If you apply a constant force to the mass, its speed will increase linearly, and so does the power (force times speed).

The work increases quadratically because the space travelled increases quadratically. And accordingly, the kinetic energy increases quadratically. If the body is in free fall, this energy is drawn from the loss of potential energy, proportional to the space, also quadratic.

You are correct.

You put energy into something by doing work on it. You do work by applying a force through a distance.

This works no matter how you apply the force, but the math is easier to do when the force is constant, so let's assume it is.
Then the acceleration of the object is uniform, that is, the speed initially increases linearly from zero to $10\ \mathrm{m/s}$. Suppose it takes exactly one second to reach that speed; it takes a total force of $100\ \mathrm{N}$ in the direction of travel to do this.

In that first second of motion, the object's average speed is $5\ \mathrm{m/s}$.
So it travels $5\ \mathrm{m}$ in that time.
The work done by a force of $100\ \mathrm{N}$ applied through a distance of
$5\ \mathrm{m}$ is $100\ \mathrm{N} \times 5\ \mathrm{m} = 500\ \mathrm{J}.$

Now keep applying that same force for one second longer. The object speeds up from
$10\ \mathrm{m/s}$ to $20\ \mathrm{m/s}$ during that time.
The average speed is $15\ \mathrm{m/s}$ and the distance traveled during that second is $15\ \mathrm{m/s}$.
The work done by a force of $100\ \mathrm{N}$ applied through a distance of
$15\ \mathrm{m}$ is $100\ \mathrm{N} \times 15\ \mathrm{m} = 1500\ \mathrm{J}.$

There are a lot of things that are proportional when a constant total force is applied to an object moving in a straight line.
Speed is proportional to the time since the speed was zero.
Work done (and therefore energy put into the object) is proportional to the distance traveled.
But distance traveled is not proportional to time, and therefore the things that are proportional to distance (such as energy) are not proportional to the things that are proportional to time (such as speed).

A non-constant force is more complicated but it still turns out that the kinetic energy of the object is still proportional to the square of its speed, not simply proportional to speed.


Proving $\frac{1}{2}(n+1)<\frac{n^{n+\frac{1}{2}}}{e^{n-1}}$ without induction I want to show that $\displaystyle\frac{1}{2}(n+1)<\frac{n^{n+\frac{1}{2}}}{e^{n-1}}$. But except induction, I do not know how I could prove this?

If $n\ge3$, $\frac12(n+1)<n$ and $n\ge\mathrm{e}$ hence $n^{n-1}\ge\mathrm{e}^{n-1}$ and it is enough to prove that $n<n^{3/2}$, which is obviously true. Then you can check manually the cases $n=1$ (for which the strict inequality is false, by the way) and $n=2$.

This is equivalent to showing $$n \leq 2 \frac{n^{n+\frac{1}{2}}}{e^{n-1}}-1 =  2\left(\frac{n^{n}n^{1/2}e}{e^n}\right)-1 $$

Let $$f(n) = 2\left(\frac{n^{n}n^{1/2}e}{e^n}\right)-n-1$$

Then $f(n) \geq 0$ (i.e. find critical points and use first derivative test).


Solve the logarithmic equation by $x$ Solve the eqation for all real $x$:
$\log_2(x^2+7)+\log_3(x+6)=6$.

What I tried:
$\log_2(x^2+7)=a$ and $\log_3(x+6)=b$, then $a+b=6$ and $2^a=3^{2b}-4\cdot3^{b+1}+43$.
But the problem is $a$ and $b$ don't have to be integers...

It is easy to show that for $x>3$ and $3>x\ge 0$ there are no soultions. Which leaves us only with the interval $0>x>-6$.

Take derivative and consider where left part increases and decreases, where are extrema (at least approximately). After proving that it has a single root, we can guess that it is $x=3$.

Hint: $x$ is a natural number. Also $log_2(x^2+7)$ and $log_3(x+6)$ are natural numbers. You will end up with $4+2=6$


Categories without a tensor product What is an example of a category that is useful in geometry and which does not have a tensor product, i.e. a category which we do not know how to turn into a monoidal category?

(I am excluding cartesian monoidal categories too.)

In any monoidal category $(C, \otimes, I)$, the unit object $I$ has the property that its endomorphism monoid $\text{End}(I)$ is commutative, by the Eckmann-Hilton argument. So if $C$ is a category in which every endomorphism monoid is noncommutative, then it cannot have any monoidal structure.

A reasonably natural geometric example is given by the category of closed connected surfaces. It's a cute exercise to show that every endomorphism monoid here is noncommutative.

Would you be interested categories with too many tensor products? (I'd argue that having too many choices can sometimes be similar to having no choices at all, depending on your purpose.)

Our varieties are smooth. The derived category of a coherent sheaves of variety has a tensor product, but monoidally derived equivalent varieties are isomorphic, by Balmer reconstruction. So if you want to consider varieties up to derived equivalence (and have this be a weakening of isomorphism) then you want to forget the natural monoidal structure. So, this category has a tensor product, in fact many tensor products, but frequently we don't want them.

Here is the paper: https://arxiv.org/abs/math/0111049

(In the smooth case, all complexes are perfect, so the bounded derived category of perfect complexes is the same as the usual bounded derived category.)

I'm not expert on this though (I'm not an expert on anything, but this stuff especially so), so make up your own mind.


How can I solve this system of equations? Here is a system of equations:

$$\begin{cases}
x^2 + 10y = 41\\
y^2-2z = 23\\
z^2-6x = 17
\end{cases} $$

What's the value of $x$ and $y$ and $z$?

It is straightforward to determine the solutions over $\mathbb{C}$. We obtain indeed a polynomial of degree $8$, hence $8$ complex roots. More precisely, we obtain
$$
x=\frac{z^2 - 17}{6},\; y=\frac{ - z^4 + 34z^2 + 1187}{360},
$$
where $z$ is a root of the polynomial
$$
t^8 - 68t^6 - 1218t^4 + 37516t^2 - 837431.
$$
This may not look nice, but it is a valid answer.

Is it possible that the system has a typo ? If you, say, consider the equations
\begin{align*}
x^2+10y & = 41, \\
y^2-2z & = 32, \\
z^2-6x & = 10,
\end{align*}
then $(x,y,z)=(9,-4,-8)$ is a nice solution.

You can do this is with substitution.

$$y = \frac{41-x^2}{10}$$

$$z = \frac{20-y^2}{-2}$$

Now plug this $z$ into the third equation. This gives you an equation in terms of $x$ and $y$. Substitute the above equation into $y$, and you have a polynomial in $x$. Solve the polynomial equation and find $x$.


Give an example of a non-ellipsoid ovalid The compact, connected surfaces of $\mathbb{R}^3$ for which the Gaussian curvature is always positive are called ovaloids.

So, ellipsoids are ovaloids.

Is there an ovaloid that is not an ellipsoid? Some literature?

How about the surface with equation $x^4+y^4+z^4=1$?

I happen to have been acquainted with Robert Greene from my days at ucla... from geometry seminars etc...  and I happen to remember there is a paper by him, entitled "ovaloids" or something like that,  and the reference is in the bibliography of one of the volumes of Spivak's comprehensive introduction to differential geometry.

Perhaps look it up...  (it's back there somewhere... )


Suppose $I\oplus K$ is a free module, then "$KI\subseteq K\cap I$" I have been staring at this proof in the Proceedings of the AMS, and I don't follow the author's logic.

Here's the setup:

$I$ is an ideal in a ring $R$, and it is projective as a right $R$ module. Therefore it is a summand in a free right $R$ module, $F=I\oplus K$.

Now, the line of reasoning continues (verbatim except for symbol changes):

  Suppose $K\neq 0$. Since $F$ is isomorphic to a direct sum of copies of $R$, it has canonical multiplication. Let $\operatorname{Ann}_F(I)$ be the annihilator of $I$ in $F$. Then $KI\subseteq K\cap I=0$, so $\operatorname{Ann}_F(I)\neq 0$.

Now if $K$ was a right ideal of $R$, then $KI\subseteq K\cap I$ would be easy to understand. The only set $K$ and $I$ are comparable in $F$, so $(I\oplus 0)\cap( 0\oplus K)=0\oplus 0$ in the direct sum.  But the left side is apparently multiplying $K$ through the $F$ module action, so that we're actually talking about $(0\oplus K)I$. Why would one say that's a subset of $I\oplus 0$?

Sure, every element of $(0\oplus K)I$, when expressed as a tuple in $F$ has entries in $I$, but as far as I can see, that doesn't mean anything about its membership in $I\oplus 0$.

I should also say that the ring $R$ is right self-injective and the ideal $I$ has zero left annihilator in $R$, but I'm not sure that it makes a difference. The author appeals to neither property in the line of thought above. In fact, that $I$ has zero left annihilator immediately allows you to say that $Ann_F(I)=0$, but since the whole point of this is to derive a contradiction, I need to see if there's any merit in what the author has claimed.

I haven't managed to cook up a counterexample yet, mostly because I have a hard time realizing projective ideals as summands in free modules. Am I missing some observation or is my doubt justified? I have a sneaking suspicion that a cognitive error occurred on the author's part.

I think your doubt is justified.

Here’s a very simple example (where the ring is right self-injective and the ideal $I$ has zero left annihilator in $R$).

Let $R=k$, a field, and let $I=k$.

$I$ can be regarded as the (right $k$-module) summand of $F=k\oplus k$ spanned by $(1,1)$, with complement $K$ spanned by $(0,1)$.

Then the annihilator of $I$ in $F$ with the “canonical multiplication” (which I presume means coordinate-wise) is zero.

Of course, in this example you could make a different choice of the embedding of $I$ into $F$ so that it had non-zero annihilator. But the “canonical” multiplication, and hence the annihilator, depend on the choices you make.

Hint: look carefully the proof of the existence of $K$, $K$ is the kernel of the surjective morphism $F\rightarrow P$.


Help me get my head around this simple linear regression problem Given a very simple linear projection model $Y=X\beta+e$ with $E(e|X)=0$ and $X$ a scalar. Notice that this is a simple linear model with no intercept.
Then $\beta = E(XY)/E(X^2)$ from the least squares formula.
On the other hand, if I take expectation on both sides of the original model, I get $E(Y)=\beta E(X)+E(e)=\beta E(X)$. Therefore, $\beta=E(Y)/E(X)$. (Of course, suppose $E(X)\neq 0$.)
I just don't see how the two quantities could be identical. Did I do something wrong?


$\lim_ {x\rightarrow−2} \frac{1}{x + 1} = −1$ Been repeatedly stumped by this limits, epsilon-delta question:
$\lim_
{x\rightarrow−2}
\frac{1}{x + 1} = −1$.
The answer in the back of Anton Transcendentals is: Min$(\frac{1}{2}, \frac{\epsilon}{2})$
I have tried manipulating it every which way, but cannot get my numbers to come out right, any help greatly appreciated.

If we set $f(x)=\frac{1}{x+1}$, then
$$f(-2)=\frac{1}{-2+1}=-1\in\mathbb{R}$$

Hence,
$$\lim_{x\to-2}f(x)=f(-2)=-1$$

No need to manipulate the numbers .


How to calculate the lower and upper bound of the error in estimating a ratio with low sample size? I have the following problem,
I am trying to estimate the conversion rate of product sales online, my data is simple,
I have the number of clicks and the number of sales for each product.
Sales are sparse so there are few sales per product. how can I estimate the upper and lower bound of the error on each product?
For example, if I have 21 clicks and 1 sale, the conversion rate is 1/21=0.07, I want to be able to add a lower and upper bound on the error of this estimation.
Thank you

If you are trying to get a confidence interval for the 'population' conversion rate based on $1$ sale among $21$ clicks, you can view this as a confidence interval for
a binomial proportion. The relevant Wikipedia page, of @tomi's Comment,
discusses several styles of confidence intervals.
In R, the procedure prop.test gives an 'exact' 95% CI $(0.0066, 0.1828)$ based on binomial CDFs (with no normal approximation) as shown below. [The suffix $conf.int shows just the CI.]
prop.test(1, 27, cor=F)$conf.int
[1] 0.006568146 0.182834659
attr(,"conf.level")
[1] 0.95

A Bayesian interval estimate, based on the Jeffries noninformative prior $\mathsf{Beta}(.5,.5),$ is often used as a frequentist CI.
For your data, this 95% CI $(0.0052,0.2018)$
uses quantiles $.025$ and $.975$ of the Bayesian
posterior distribution $\mathsf{Beta}(.5+x, .5+n-x)
= \mathsf{Beta}(1.5, 20.5).$
qbeta(c(.025, .975), 1.5, 20.5)
[1] 0.005187043 0.201755968

The Wald CI of the form $\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}},$ where $\hat p = x/n.$ This is an asymptotic interval meant for use with large $n$ and so
one does not expect reliable results with $n = 21.$
For your data the Wald CI is taken as $(0, 0.1327),$
suppressing the impossible negative lower limit.
n = 21;  x = 1;  p.h = x/n
CI.wald = p.h + qnorm(c(.025,.975))*sqrt(p.h*(1-p.h)/n)
CI.wald
[1] -0.04346329  0.13870138

Agresti and Coull (1998) proposed a modification of Wald's CI,
using point estimate $\tilde p = (x+2)/(n+4)$ to obtain the interval
$\tilde p \pm 1.96\sqrt{\frac{\tilde p(1-\tilde p)}{n+4}},$ which has a more reliable 95% coverage probability for small $n$ than the Wald CI.
The Agresti
interval is currently used in many elementary and intermediate level statistics texts because it gives reasonably good good results and can be computed without specialized software (e.g.,  using just a mobile phone calculator). For your data, this 95% CI amounts to
$(0, 0.2474).$
n = 21;  x = 1;  p.e = (x+2)/(n+4)
CI.ac = p.e + qnorm(c(.025,.975))*sqrt(p.e*(1-p.e)/(n+4))
CI.ac
[1] -0.007382581  0.247382581

Note: I have shown endpoints of 95% CIs to four
decimal places to make clear the differences among
types of intervals. In practice, with small $n,$ it might be appropriate to show two-place accuracy. To one place: 'Exact', Jeffries, and Agresti intervals are all
$(0.0, 0.2).$

Trivially, the lower bound is zero and the upper bound is given by the total market of your product. Any other bounds assigned based on distributions, will depend on assumptions on their parameters, which is not better than taking a guess.
Build your bounds based on an economic argument, using data on industry sales, firm advertising and seasonality.


Poisson Integral is equal to 1 Show
  $$
\int_{-\pi}^{\pi}P(r, \theta)d\theta = 1
$$

Let $\alpha(r) = \frac{r^2 - 1}{2r}$ and $\gamma(r) = -\big(\frac{r^2 + 1}{2r}\big)$.
  Then
  $$
    \frac{1}{2\pi}
    \int_{-\pi}^{\pi}\frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}d\theta
    = \frac{\alpha}{2\pi}
       \int_{-\pi}^{\pi}\frac{1}{\cos(\theta) + \gamma}d\theta
$$
where
    $$
    \frac{r^2 - 1}{2r}\frac{1}{\cos(\theta) - \frac{1}{2r} - \frac{r^2}{2r}} = \frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}
    $$
Next, let $z = e^{i\theta}$. Then $d\theta = \frac{-i}{z}dz$. Since $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$, $\cos(\theta) = \frac{z + z^{-1}}{2}$.
$$
    = \frac{-i\alpha}{\pi}\int_C\frac{1}{z^2 + 2z\gamma + 1}dz
$$
and $C$ is the contour oriented counter clockwise with simple poles at $z = -\gamma\pm\sqrt{\gamma^2 - 1}$. Let $f(z) = \frac{1}{z^2 + 2z\gamma + 1}$. Then
$$
    = \Big(\frac{-i\alpha}{\pi}\Big)2\pi i\sum\text{Res}_{z = z_j}f(z)
       \tag{1}
$$
The only pole in $\lvert z\rvert < 1$ is $z_j = -\gamma + \sqrt{\gamma^2 - 1}$.
  Then
  $$
  2\pi i\lim_{z\to z_j}\Bigg[\big(z + \gamma - \sqrt{\gamma^2 - 1}\big)
  \frac{1}{\big(z_j + \gamma + \sqrt{\gamma^2 - 1}\big)
    \big(z + \gamma - \sqrt{\gamma^2 - 1}\big)}\Bigg] =
  \frac{\pi i}{\sqrt{\gamma^2 - 1}}
  $$
Now, we can substitute $\frac{\pi i}{\sqrt{\gamma^2 - 1}}$ for $2\pi i\sum\text{Res}$ in equation (1).
  \begin{align*}
    \Big(\frac{-i\alpha}{\pi}\Big)2\pi i\sum\text{Res}_{z = z_j}f(z)
    &= \frac{\alpha}{\sqrt{\gamma^2 - 1}}\\
    &= \frac{r^2 - 1}{2r\sqrt{\frac{(r^2 + 1)^2}{4r^2} - 1}}\\
    &= \frac{r^2 - 1}{\sqrt{(r^2 + 1)^2 - 4r^2}}\\
    &= \frac{r^2 - 1}{\sqrt{r^4 - 2r^2 + 1}}\\
    &= \frac{(r - 1)(1 + r)}{(r - 1)(r + 1)}\\
    &= 1
  \end{align*}

  I have been unable to convince myself that the only pole in $\lvert z\rvert < 1$ is $z_j = -\gamma + \sqrt{\gamma^2 - 1}$.  I know it is the case because if I use the other pole, the integral becomes $-1$

For $\boldsymbol{r\in(-1,1)}$
$$
\begin{align}
1
&=\frac1{2\pi i}\oint\frac1{z-r}\mathrm{d}z\tag{1}\\
&=\frac1{2\pi i}\int_{-\pi}^\pi\frac1{e^{i\theta}-r}\mathrm{d}e^{i\theta}\tag{2}\\
&=\frac1{2\pi}\int_{-\pi}^\pi\frac{1-re^{i\theta}}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta\tag{3}\\
&=\frac1{2\pi}\int_{-\pi}^\pi\frac{1-r\cos(\theta)}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta\tag{4}\\
&=\frac12+\frac1{2\pi}\int_{-\pi}^\pi\frac{\frac12-\frac12r^2}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta\tag{5}\\
1
&=\frac1{2\pi}\int_{-\pi}^\pi\frac{1-r^2}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta\tag{6}
\end{align}
$$
Explanation:
$(1)$: $\frac1{z-r}$ has residue $1$ at $z=r$ (inside the unit circle)
$(2)$: parametrize the unit circle
$(3)$: multiply integrand by $\frac{e^{-i\theta}-r}{e^{-i\theta}-r}$
$(4)$: take the real part of both sides
$(5)$: add $\frac12$ to and subtract $\frac12$ from the integral
$(6)$: multiply both sides by $2$ and subtract $1$

For $\boldsymbol{r\not\in(-1,1)}$

If $r^2=1$, the integrand is $0$, so assume $r\not\in[-1,1]$.

If $r\not\in[-1,1]$, then the left side of $(1)$ starts at $0$ and step $(6)$ changes the left side to $-1$.

Alternatively, if $r\not\in[-1,1]$, then $\frac1r\in(-1,1)$, and therefore, $(6)$ says
$$
\begin{align}
\frac1{2\pi}\int_{-\pi}^\pi\frac{1-r^2}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta
&=\frac1{2\pi}\int_{-\pi}^\pi\frac{\frac1{r^2}-1}{\frac1{r^2}-\frac2r\cos(\theta)+1}\mathrm{d}\theta\\
&=-\frac1{2\pi}\int_{-\pi}^\pi\frac{1-\frac1{r^2}}{1-\frac2r\cos(\theta)+\frac1{r^2}}\mathrm{d}\theta\\
&=-1\tag{7}
\end{align}
$$

From your definition, $\gamma = -\frac{r^2+1}{2r}$ so $-\gamma>1$ for all positive $r$.  So $-\gamma+\sqrt{\gamma^2-1}>-\gamma>1$ so that root cannot possibly lie in $|z|<1$.  Are you sure you do not have a sign error somewhere between lines 3 and 4 of your derivation?


Does (Riemann) integrability of a function on an interval imply its integrability on every subinterval? For example, if $f$ is integrable on $[0,3]$, is it also integrable on $[1,2]$? I tried thinking of a counterexample but couldn't, since I've only learned what implies integrability but not what integrability implies.

First note that if $a \leq a'$ and $b' \leq b$ and $g: [a,b] \to \mathbb R$ is $0$ outside of $[a',b']$, then $\int_a^b g(x)\, dx = \int_{a'}^{b'} g(x)\,dx$. This is easily shown using the definition of the Darboux (or Riemann, if you prefer that definition) integral using partitions (tagged partitions for the Riemann definition).

The product of Riemann–Darboux integrable functions are Riemann–Darboux integrable, so in particular you have (in your concrete example) that $\int_1^2 f(x)\, dx = \int_0^3 f(x) \chi_{[1,2]}(x)\, dx$, where $\chi_{[1,2]}$ denotes the characteristic function of the interval $[1,2]$, i.e.

$\displaystyle\qquad \chi_A(x) = \begin{cases} 1 & \text{if } x \in A \\ 0 & \text{if } x \notin A\end{cases}$

If $f$ is integrable on $[0,3]$ then there is a $F:[0,3]\to \Bbb R$ :$F'=f$ on  $[0,3]$.The restriction of $F$ on $[1,2]$ let's name it $G$ means that $G'=f$ on $[1,2]$. So if $f$ is integrable on $[0,3]$ is integrable on every connected subset(<=>here is equal to every subinterval)
 of $[0,3]$


Specific integral with continuous function Let $f: [0,1] \times [0,1] \to \mathbb{R}$ be a continuous function. How can I compute:

$$ \lim_{n \to \infty} \left( \frac{(2n+1)!}{(n!)^2} \right)^2 \int_{0}^{1} \int_{0}^{1} (xy(1-x)(1-y))^n f(x,y)\, dx\, dy$$
?

$$\iint_{[0,1]^2} x^n (1-x)^n y^n (1-y)^n \,dx\,dy = \left(\int_{0}^{1}z^{n}(1-z)^n\,dz\right)^2=\left(\frac{n!^2}{(2n+1)!}\right)^2$$
hence the limit, by convergence in distribution, is just $\color{red}{f\left(\frac{1}{2},\frac{1}{2}\right)}.$

HINT: Expand $f(x,y)$ in a power series $f(x,y)= \sum_{i,j=0}^\infty a_{ij}x^iy^j$. Then use the Definition of the Beta function.

For computing the Limit $n \rightarrow \infty$ you must use $\Gamma(n)=(n-1)!$ for natural number $n$.


Finding the value of the supremum of a set Consider the set $S = \{x \in \mathbb R: x < \frac2x\}$. Determine the value of $sup$ $S$ (if it exists).

Here is my attempt:
Firstly, $S = \{x \in \mathbb R: 0 < x < \sqrt 2$ $\lor$ $x < -\sqrt2 \}$.
Since $1 \in S$, $S \ne \emptyset$.
Also, $\forall x \in S$, $x < \sqrt2$ and so $S$ is bounded above.
By the least upper bound property of $\mathbb R$, $sup$ $S$ exists.
Suppose $\exists y \in \mathbb R: y < \sqrt2 \land y$ is an upper bound of $S$.
Since $1 \in S$, $1 \le y < \sqrt2 \Rightarrow y \in S$.
...

How do I show that $sup$ $S$ $= \sqrt 2$?

To find $S$ the equation $x<\frac2{x}$ or equivalently $\frac{(x-\sqrt2)(x+\sqrt2)}{x}<0$ must be solved.

This leads to $S=(-\infty,-\sqrt2)\cup(0,\sqrt2)$.

Here $\sqrt2$ is an upper bound of $S$ and proving that $\sup S=\sqrt2$ comes to showing that every $y<\sqrt2$ is not an upper bound of $S$. That means: showing that $y<s$ is true for some $s\in S$.

You shall prove that the supremum is not greater nor less than $\sqrt 2$. ie by supposing so in both cases and getting a contradiction.


What is the difference between these two sets of symbols? I am considering two sets of symbols. Are these two sets different or the same? Why?

\begin{align}(1)\qquad&\lim_{n\to\infty}\sum_{j=1}^n\\
(2)\qquad&\sum_{j=1}^\infty\end{align}

The series (2) in the form
  $\sum_{j=1}^{\infty}a_n$
  admits two different interpretations.

  It can be seen as sequence  of partial sums
  \begin{align*}
\sum_{j=1}^{\infty}a_j:=\left(a_1,a_1+a_2,a_1+a_2+a_3,\ldots\right)=\left(\sum_{j=1}^na_j\right)_{n\geq 1}
\end{align*}
  and it can be seen as limit of the sequence of partial sums
  \begin{align*}
\sum_{j=1}^{\infty}a_j:=\lim_{n\to\infty}\sum_{j=1}^na_j
\end{align*}

  In the second case the symbols (1) and (2) have the same meaning whereby (2) is defined by (1).

Note: You can find this line of argumentation e.g. in K. Knopps Theory and application of infinite series.

I guess in some formal setting the sum $$\sum_{j=1}^{\infty}()$$
is defined to be $$\lim_{n \to \infty} \sum_{j=1}^{n}()$$, how else can we take infinite sums or prove results? With the limit. Indeed if you "believe" in infinity or not the results can be positively guaranteed.


Proving a certain map on the closed unit disc must be the identity Bounty expired. Will gladly re-create one if a satisfactory answer is posted in the future.

Prove: Let $f$ be a continuous function on the closed unit disc with two properties:

1. $f$ is the identity on the boundary, i.e., on the unit circle. That is, if $|z| = 1$, then $f(z) = z$.

2. $f^2$ is the identity, i.e., for all $z$ in the closed unit disc, we have $f(f(z)) = z$.

Then $f$ must be the identity function.

Motivation: I came across this question for the closed unit sphere here on MathOverflow. It seemed to me like considering the two-dimensional case might be a good place to start in trying to tackle the problem there. Ultimately, the MO question was resolved using some nontrivial results. I am wondering whether there is a proof of the question here, for the closed unit disc, which uses methods that don't extend beyond those of basic point-set topology or a first course in real analysis.

Remark: Please note that I am looking for a fundamentally different proof of the proposition above, that is, not a modification of either of the responses given at the MathOverflow link into more digestable language.

In response to a comment: If you have thought up a proof and are concerned the level of presentation is too high, then I hope you will still post it as an answer.

As pointed out in the comments on the MO question, this follows immediately from a 1931 theorem of M.H.A. Newman, which is Theorem 2 in this paper. Notice that Theorem 2 is basically a lemma (used to prove the main theorem 1) whose proof takes two pages, and uses nothing other than definition of manifold.

Consider the fact that $f(f(z))=z$. This shows that $f$ is its own inverse. Since $f$ is invertible, it is a bijection; a bijection from the compact space closed disk into the Hausdorff  space of the closed disk, so that it is a homeomorphism (continuous bijection from compact to hausdorff is a homeomorphism). Since the closed unit disk is contractible, any two maps into it are homotopic, so $f$ is homotopic to the identity, so that , from the conditions on the boundary,$f(z)=z$ or $f(z)=-z$. From the conditions given, we must have $f(z)=z$


Harmonic functions on $S^2$ Consider the sphere $S^2 = \lbrace (x,y,z) :\ x^2 + y^2 + z^2 = 1 \rbrace$. This is a smooth manifold in $\mathbb{R}^3$, and for a given point $s \in S^2$, one can consider its coordinate neighborhood.

There are many ways to put a smooth structure on $S^2$, but all require at least two coordinate neighborhoods. One simple way is to make use of the local coordinates $\theta, \phi$ (azimuthal angle and inclination angle). For $s \in U = S^2 \setminus {(0,0,1), (0,0,-1)}$, define a coordinate map by

$ g : (0,2\pi) \times (0,\pi) \rightarrow U $

with formula given by

$ g(\theta, \phi) = (\sin(\phi) \cos(\theta), \sin(\phi) \sin(\theta), \cos(\phi)) $

This map is one-to-one, onto, and bicontinuous, so it is a homeomorphism. One can construct a similar map onto $V = S^2 \setminus {(1,0,0),(-1,0,0)}$, so the whole sphere is covered. The fact that this defines a differentiable structure is easy to work out.

The question I want to ask is, are there any non-constant solutions to the equation

$ \frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2} = 0 $

where $f : S^2 \rightarrow R$ is a smooth function.

Furthermore, what sorts of boundary conditions are necessary in order to ensure uniqueness.

If I understand your question correctly, you are aware that $\frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2}$ is not actually the Laplacian on a sphere, but you want to know whether there are any nontrivial smooth functions on $S^2$ which are harmonic with respect to this modified Laplacian. The answer is no.

A function $f(\theta,\phi)$ which satisfies $\frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2} = 0$ is really a harmonic function on the rectangle $[0,2\pi] \times [0,\pi]$, which is mapped onto the sphere via $g^{-1}$. Continuity on the sphere requires that $f(\theta,0) = \text{const}$ (continuity at the north pole), $f(\theta,\pi) = \text{const}$ (continuity at the south pole), and $f(0,\phi) = f(2\pi,\phi) = h(\phi)$ for some $h:[0,\pi]\rightarrow \mathbb{R}$ (continuity across the prime meridian). For any $h$, the boundary conditions are independent of $\theta$, so $f$ is a function of $\phi$ alone. But then $\frac{\partial^2 f}{\partial \phi^2} = 0$ implies that $f$ is linear in $\phi$. If the slope is nonzero, $f$ on the sphere fails to be smooth at the poles. Therefore, for $f$ to be smooth on the sphere and satisfy $\frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2} = 0$, it has to be constant.

Partial differentiation in Laplacian does't works at all in all curved surfaces. SO the answer is No. However I can't say anything about $S^n$ in an continuous series


Proving that $ \chi(G) = \omega(G) $ if $ \bar{G} $ is bipartite. I know that $ \chi \! \left( \bar{G} \right) = 2 $ and that $ \chi(G) \geq \alpha \! \left( \bar{G} \right) $, but how can I conclude that $ \chi(G) = \omega(G) $?

First, it’s always true that $\chi(G)\ge\omega(G)$, so your real problem is to show that $\chi(G)\le\omega(G)$ if $\overline{G}$ is bipartite. Note that it need not be true that $\chi(\overline{G})=2$: the chromatic number of a bipartite graph is $1$ if the graph has no edges. In that case $G$ is a complete graph, and the result is trivial, so I’ll assume that $G$ is not complete.

Suppose that $\overline{G}$ is bipartite with parts $V$ and $W$. Without loss of generality we may assume that all isolated vertices of $\overline{G}$ belong to $V$, and that $|V|\ge|W|$. Clearly $V$ is a clique in $G$. Color each vertex of $V$ with a different color. Now observe that for each $w\in W$ there is at least one $v\in V$ such that $vw$ is an edge of $\overline{G}$ and hence such that $wv$ is not an edge of $G$, so that $w$ can be given the same color as $v$. Thus, $\chi(G)\le|V|\le\omega(G)$.

suppose if $w \in W$ is adjacent to every vertex of V in G then  we cannot color that w with that and  cardinality of V and W are less tahn or equal to clique of G complement not exactly equal to . @gcolucci\in


The number $n^4 + 4$ is never prime for $n>1$ I am taking a basic algebra course, and one of the proposed problems asks to prove that  $n^4 + 4$ is never a prime number for $n>1$.

I am able to prove it in some particular cases, but I am not able to do it when $n$ is an odd multiple of $5$.


Prove $(\forall x, y \in E)(\exists ! z\in Z)$ s. t. $x + z = y$. I am currently studying Introduction to Hilbert Spaces with Applications, the third edition, by Debnath and Mikusinski. Chapter 1, exercise 1, is as follows:

Prove that for every $x, y \in E$ there exists a unique $z \in E$ such that $x + z = y$.

The definition of vector space is given as follows:

By a vector space we mean a nonempty set $E$ with two operations:
$(x, y) \mapsto x + y$ from $E \times E$ into $E$ called addition.
$(\lambda, x) \mapsto \lambda x$ from $\mathbb{F} \times E$ into $E$ multiplication by scalars,
such that the following conditions are satisfied for all $x, y, z \in E$ and $\alpha, \beta \in \mathbb{F}$:
(a) $x + y = y + x$;
(b) $(x + y) + z = x + (y + z)$
(c) For every $x, y \in E$ there exists a $z \in E$ such that $x + z = y$;
(d) $\alpha (\beta x) = (\alpha \beta) x$;
(e) $(\alpha + \beta)x = \alpha x + \beta x$;
(f) $\alpha(x + y) = \alpha x + \alpha y$
(g) $1x = x$
Elements of $E$ are called vectors. If $\mathbb{F} = \mathbb{R}$, then $E$ is called a real vector space, and if $\mathbb{F} = \mathbb{C}$, $E$ is called a complex vector space.

Chapter 1 already contains a proof that it seems would prove this (modified by me):

Let $x, y \in E$. By (c), there exists a $w \in E$ such that $x + w = y$. If $x + z_x = x$ for some $z_x \in E$, then, by (a) and (b),
$$y + z_x = (x + w) + z_x = (x + z_x) + w = x + w = y.$$
This proves that $z$ exists for every $x, y \in E$.
Now let $z_1, z_2 \in E$ such that $x + z_1 = x$ and $y + z_2 = y$ for all $x, y \in E$.
Therefore, we have that
$$z_1 + z_2 = z_1$$
and
$$z_2 + z_1 = z_2$$
Therefore, we can conclude that $z_1 = z_2$.

I would greatly appreciate it if people would please take the time to review this proof for correctness.

Existence is guaranteed by (c). You just need to prove uniqueness. Suppose there exists $z_1,z_2 \in E$ such that $$x+z_1=y, x+z_2=y.$$ Then $$x+z_1=x+z_2.\tag{1}$$ By (c), corresponding to $x \in E$ and $0 \in E$ (whose existence and uniqueness are proved here), there exists $x' \in E$ such that $x+x'=0.$
Therefore by $(1),$ $$x'+(x+z_1)=x'+(x+z_2)$$ and finally (b) implies $z_1=z_2.$

Your solution actually shows the uniqueness of additive identity.

Let $z, z^\prime, w \in E$.

From (c), it follows that there exists a $w \in E$ such that $z + w = z^\prime$.

Let $x + z = y = x + z^\prime$ and $z + w = z^\prime$.

Then

$$y = x + z^\prime = x + z + w = y + w.$$

Therefore, we must have that $w = 0$.

And so, we have that

$$z^\prime = z + w = z.$$


Prove two Ideal are equal Consider ideals of the ring $\frac{F[x]}{x^n-1}$ where $F$ is an arbitrary finite field.

Suppose $g(x) | (x^n-1)$ and $f(x) | (x^n-1)$ . Also suppose $A  = \left <g(x)\right>$ be the ideal that is generated by $g(x)$ and $B  =  \left<f(x)\right>$ be the ideal that is generated by $f(x)$ and $C =\left <g(x)\right> + \left<f(x)\right>$. Now find $h(x)$ such that $C= \left<h(x)\right>$.

hint = I think $h(x) = f(x) +g(x)+f(x)g(x)$ and I proved $\left<f(x) +g(x)+f(x)g(x)\right> \subseteq C$

If $h(x) = gcd (f(x),g(x))$ and $c \in < f(x) > +< g(x) > $ we have $c \in < h(x) > $ because $$c= a(x)f(x)+b(x)g(x)$$ for some $a(x)$ and $b(x)$ and then exist a $k(x)$ such that $c= k(x)h(x)$. And also $< h(x) > \subset < f(x) > +< g(x) >$ because exist a linear combination of $ f(x), g(x) $ such that $h(x)= t(x)f(x)  +s(x)g(x) $ so if $c \in h(x)$ then exist $l(x)$ such that $c= l(x)t(x)f(x)+l(x)s(x)g(x)$ and so we have $$< h(x) > = < f(x) > +< g(x) >$$


Is $ \pi $ definable in $(\Bbb R,0,1,+,×, <,\exp) $? Is there a first-order formula $\phi(x) $ with exactly one free variable $ x $ in the language of ordered fields together with the unary function symbol $ \exp $ such that in the standard interpretation of this language in $\Bbb R $ (where $ \exp $ is interpreted as the exponential function $ x \mapsto e^x $), $\phi (x) $ holds iff $ x=\pi $?

EDIT: As Levon pointed out, a negative answer to this problem would imply that $π$ and $e$ (and $e^e$, $(2e)^{3e^2}$, and so on) are algebraically independent over $\Bbb Q$, which is an unsolved problem. So, if you think that a definition of $\pi$ is impossible, I would be pleased if you could show something like, that it is possible to reduce $\phi$ to a formula which contains no terms involving bound variables inside exponential functions, which would reduce the problem more or less to a question on algebraical independece.
However because there are such intricate connections between exponential and trigonometric functions, I don't think that $\pi$ should be undefinable.

Todd Trimble provided the answer to this question on MO:

Assuming Shanuel's conjecture, this treatise about exponential rings proves that (see Theorem 2.5.1) the exponential ring generated by $\pi$ looks just as the exponential ring generated by nearly every other real number, which implies that there is no defining relation of $\pi$ over $\Bbb R_{\exp}$.

Using that $\int_{-\infty}^\infty e^{-x^2} = \sqrt{\pi}$

$$
\phi(x) := (\forall_y \psi(y) \implies y^2 = x)
$$

Where
$$
\psi(y) := \forall_{\epsilon > 0} \exists_N \forall_{n > N} \forall_x s(n, x) \implies |x - y| < \epsilon
$$
The formula $\psi(y)$ says that $y$ is a limit of $x_n$ such that $s(n, x_n)$ holds. Now, $s(n, x)$ says that $\int_{-n}^n e^{-t^2} = x$:

$$
s(k, x) = \forall_{\epsilon > 0} \exists_N \forall_{n > N} \forall_y p(n, k, y) \implies |x - y| < \epsilon
$$

Now $p(n, k, y)$ says that $y$ is a value of k-th Riemann sum of integral $\int_{-n}^n e^{-t^2}$. I'll leave the details.


Showing two stochastic process are independent Let $X=\{X_t, t \ge 0\}$ and  $Y=\{Y_t, t \ge 0\}$ are two stochastic processes defined on common probability space. Then two show X and Y are independent, does it suffice to show for any $t$ and $s$, $X_t$ and $Y_s$ are independent? Or do we need stronger conditions?

You need to show that the sigma-algebras $\sigma(\{X_t \mid t \ge 0\})$ and $\sigma(\{Y_t \mid t \ge 0\})$ are independent, it isn't sufficient that all pairs $X_t, Y_t$ are independent.

This can easily be seen by the following example: Let $X, Y$ be two independent random variables with nontrivial distribution. Then the processes
$$\begin{align*}X_t &= X \cdot I\{t = 0\} + Y \cdot I\{t = 1\} \\
Y_t &= X \cdot I\{t = 1\} + Y \cdot I\{t = 0\}\end{align*}$$
are not independent, although all pairs $X_t, Y_t$ are independent.

Edit: The question was changed since I've answered it. To answer the new question, consider the standard example of random variables $X_1, X_2, X_3$ that are pairwise independent, but not jointly independent. Now consider the processes
$$\begin{align*}Y_t &= X_1 \cdot I\{t = 0\} + X_2 \cdot I\{t = 1\} \\
Z_t &= X_3 \cdot I\{t = 0\}\end{align*}.$$


$\sum^{\infty}_{n=1}x_n$ converges $\Rightarrow$ $\sum^{\infty}_{n=1}x_n^3$ converges also? Let the series $\sum^{\infty}_{n=1}x_n$ converge. $\{x_n\}_{n=1}^{\infty} \in \mathbb R$

  Does $\sum^{\infty}_{n=1}x_n^3$ converge too?

I tried to find some counter-examples but found none. I tried to prove this also, but I can't... I don't know even if it's true or not.

Let $a= \exp(2 \pi i/3)$ (such that $1+a+a^2=0$ and $a^3=a^6=1$)

Take the series $$x_{3n} = \frac{1}{n^{1/3}},\quad x_{3n+1} = ax_{3n},\quad x_{3n+2}=a^2x_{3n}.$$

With such choice of $x$, the series $\sum_n x_n$ converges, but the series $\sum_n x_n^3$ does not.

edit

Choose real numbers $a$, $b$, $c$ such that $$a+b+c=0\\a^3+b^3+c^3>0.$$
For example, $a=-1$, $b=-2$, $c=3$. Then take

$$x_{3n} = \frac{a}{n^{1/3}},\quad x_{3n+1} = \frac{b}{n^{1/3}},\quad x_{3n+2}=\frac{c}{n^{1/3}}.$$

With such choice of $x$, the series $\sum_n x_n$ converges, but the series $\sum_n x_n^3$ does not.

Hint : justify that there exists $N$ such that for $n\geq N$ :

$$(x_n)^3\leq x_n $$

and conclude.


Why can't you flatten a sphere? It's a well-known fact that you can't flatten a sphere without tearing or deforming it.  How can I explain why this is so to a 10 year old?

As soon as an explanation starts using terms like "Gaussian curvature", it's going too far for the audience in question.

A great explanation would work just as well for a hemisphere as for a sphere, as both have positive curvature.

Experimentation is fun!

Give him a globe, some string, a ruler, a compass, some paper, and a pencil.  Then have him try to map 4 different cities from the globe onto the paper, keeping ALL the distances between them the same as on the globe.

You can, just imagine a perfect sphere, being put into a piece of paper that breaks away as it enters, then take a picture of it. Perfect sphere, the reason you can't physically flatten a sphere, is because all of that third dimension of space has to go SOMEWHERE, so it's either going to fall apart, and split the third dimension perfectly (EXTREMELY unlikely) or, it'll stay together, and become deformed.

A perfect sphere is merely a circle anyway, it's just a filled in surface of the spheres diameter.

This is a complete shot in the dark, and I hope my answer helps, but it's obvious so it probably won't :/


Find remainder of division of $x^3$ by $x^2-x+1$ I am stuck at my exam practice here.

  The remainder of the division of $x^3$ by $x^2-x+1$ is ..... and that of $x^{2007}$ by $x^2-x+1$ is .....

I tried the polynomial remainder theorem but I am not sure if I did it correctly.

By factor theorem definition, provided by Wikipedia,

  the remainder of the division of a polynomial $f(x)$ by a linear polynomial  $x-r$ is equal to $f(r)$.

So I attempted to find $r$ by factorizing $x^2-x+1$ first but I got the complex form $x=\frac{1\pm\sqrt{3}i}{2}=r$.

$f(r)$ is then $(\frac{1+\sqrt{3}i}{2})^3$ or $(\frac{1-\sqrt{3}i}{2})^3$ which do not sound right.

However, the answer key provided is $-1$ for the first question and also $-1$ for the second one. Please help.


Reasoning that $ \sin2x=2 \sin x \cos x$ In mathcounts teacher told us to use the formula $ \sin2x=2 \sin x \cos x$.
What's the math behind this formula that made it true?  Can someone explain?

$$\color{red}{\sin 2x}=\color{blue}{2\sin x}\cos x$$

Let's start with the power series definitions:
$$\sin x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!} -+ \dots, $$
$$\cos x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-+ \dots. $$
If you expand the product $$\sin x \cos x= \left( x-\frac{x^3}{3!}+\frac{x^5}{5!} -+ \dots \right) \left( 1-\frac{x^2}{2!}+\frac{x^4}{4!}-+ \dots \right), $$
and collect terms, you will find that only odd powers of $x$ appear, and the coefficient of $x^{2n+1}$ is $$(-1)^n \left(\frac{1}{(2n+1)!0!}+\frac{1}{(2n-1)! 2!}+ \dots +\frac{1}{1!(2n)!} \right).$$
You can prove (using induction for example) that the last expression equals $$(-1)^n \frac{2^{2n}}{(2n+1)!}=(-1)^n \frac{1}{2} \frac{2^{2n+1}}{(2n+1)!}.$$
Thus we have $$\sin x \cos x=\frac{1}{2} \sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!}, $$ which is precisely $\frac{1}{2} \sin 2x$.


Solving Coupled Eigenvalue equations I wish to solve the following set of coupled eigenvalue equations. How should I do it?

For real matrices $A$,$B$,$D$ and vectors $x \in R^m$, $y \in R^n$
$$
A x + B y = \lambda x
$$
$$
B^T x + D y = \mu y
$$
where, $A$ and $D$ are symmetric.

Background:

I am trying to solve the following optimization problem:

$$
\min x^T A x + y^T D y + x^T By
$$
$$
\textrm{such that } \, x^T x = 1 , y^T y = 1
$$
This leads to the above eigenvalue problem. $\lambda$ and $\mu$ are the Langrangian parameters for the constraints.

EDIT: I need to actually find numerical solutions for these equations given $A$, $B$, $D$

If you write $I^m$ for an $m \times m$ identity matrix, you can make a big matrix equation out of it $$\left(\begin {array}{c  c}A-\lambda I^m & B\\B^T & D-\mu I^n \end {array}\right)\left(\begin {array}{c}x \\ y \end {array}\right)=0$$    which is a classic eigenvalue problem.  Chapter 11 of Numerical Recipes (free for the obsolete versions) or any numerical analysis text can help you.

Perhaps adding the two equations in order to reeduce it to one equation with eqienvalue (lambda + mu) and Eqienvector x+y is the trick. Then find the inverse using a calculator or standard methods. But there is probably a trick involving the fact that the matrices satisfies the properties they do.


Is there any situation where the joint probability is equal to the marginal probability? Is there any situation in that joint probability $p(x,y)$ equals to marginal probability $p(x)$? What is the interpretation of this situation?

Assume these are the distributions of $(X,Y)$ and $X$, where $X$ and $Y$ are random variables with values in $\mathbb X$ and $\mathbb Y$ respectively, and let (C) denote the condition that [For every $(x,y)$ in $\mathbb X\times\mathbb Y$, $p(x,y)=p(x)$].

First setting: If $\mathbb X$ and $\mathbb Y$ are discrete spaces, (C) is impossible unless $\mathbb Y$ is a singleton, and then (C) is always met. In particular, if $Y$ is almost surely constant but with values in $\mathbb Y$ not a singleton, then (C) is not met.

Second setting: Assume that $\mathbb X$ and $\mathbb Y$ are Borel subsets of some Euclidean spaces and that $p$ and $p$ are densities with respect to some measures $\mu\otimes\nu$ and $\mu$ on $\mathbb X\times\mathbb Y$ and $\mathbb X$ respectively. Then, (C) is never met unless $\nu(\mathbb Y)=1$, and then (C) means that $Y$ has distributon $\nu$ and is independent of $X$.

If $f(x,y)=f(x)$, then $y$ could be $y=x$.


Generally, $T_i$ implies $T_j$ for $j < i$. (Although some care is needed, because some people would not necessarily understand a $T_i$ space for $i > 0$ to be $T_0$, too)

So, there are two settings here: $T_0$ spaces with 3 elements, and $T_1$ spaces with 3 elements. Any finite $T_1$ space is already discrete, as open sets are closed under finite intersections. Thus, there is only one $T_1$ topology on $\{1,2,3\}$.

For $T_0$, we can try to construct a minimal $T_0$ topology up to permutation of the elements. It is ok if one element has only $X$ as neighborhood, wlog let that one be 3. Both 1 and 2 need smaller neighborhoods, in particular this implies that $\{1, 2\}$ is open. Then again, for one point this can be the smallest neigherborhood, wlog let that be 2. Then, $\{1\}$ is open, too.

So, the $T_0$ topologies are those that (after some permutation) include the open sets $\{1\}$ and $\{1,2\}$.


Show that $ x< \sin(\frac{\pi x}2)\forall x\in (0,1)$ I don't know how to prove this inequality, some help would be appreciated, thank you!

$x< \sin(\frac{\pi x}2)\quad\quad \forall x \in (0,1)$

Let we have to prove $$t<\sin \left(\frac{\pi t}{2}\right)\;\forall t \in (0,1)$$

Now put $\displaystyle \frac{\pi t}{2} = x\Rightarrow t = \frac{2x}{\pi}$

So we have to prove $$\frac{2x}{\pi}<\sin x\;\forall x \in \left(0,\frac{\pi}{2}\right)$$

Now Slope of $$\displaystyle \bf{OB} = m_{0B} = \frac{\sin x-0}{x-0} = \frac{\sin x}{x}.$$

and Slope of $$\displaystyle \bf{OA} = m_{0A} = \frac{1-0}{\frac{\pi}{2}-0} = \frac{2}{\pi}.$$

So Here in above graph $$\displaystyle \bf{m_{OB}>m_{OA}}$$

So we get $$\displaystyle \frac{\sin x}{x}>\frac{2}{\pi}\Rightarrow \frac{2x}{\pi}<\sin x\;\forall x\in \left(0,\frac{\pi}{2}\right)$$

For $x\in(0,1)$ the function $\sin( \frac\pi2x) $ is an increasing function and we have
$$1< \frac\pi2x$$
so $1< \sin (\frac\pi2x) $


Why the Equality Axioms are considered as logical axioms? This question is from Introduction to Mathematical Logic by Elliot Mendelson , page 94-95 , First order theories with equality.There is a line which is kind of confusing for me.

The numbering (A6) and (A7) is a continuation of the numbering of the logical axioms.

Here , (A6) and (A7) are the following axioms:
$(A6)$ $(\forall x_1)x_1=x_1$
$(A7)$ $x=y \to (\mathscr B(x,x) \to \mathscr B(x,y))$
My question is , why they are considered logical axioms ? Shouldn't they be in the category of proper (or non-logical) axioms?

What is your criterion of what counts as a "logical constant"? The connectives and first-order quantifiers count as logical notions -- but why?
There are various stories on the market for demarcating the logical constants.  Depending on the details, some count the identity predicate as a logical constant, some don't. For an accessible discussion of some of the options here, see e.g. https://plato.stanford.edu/entries/logical-constants/
But certainly, some standard accounts do treat the notion of identity as logical: it is topic neutral and general enough; it is governed by introduction and elimination rules in the right sort of way, and so on. Mendelson is not at all going out on a limb in treating the identity laws as logical.

I think Mendelson did not even say that $(A6)$ and $(A7)$ are logical axioms.I completely misinterpreted the above line.
@James Hanson Already answered my question.They are indid non-logical because they are not logically valid.An axiom (generated from a schema) is a logical axiom iff it is logically valid.
Second , The author meant to say that he would number the non logical equality axioms according to the continuation of the numbering of the logical axioms (There are 5 logical axioms in this case).Not that the equality axioms themselves are logical axioms.


Prove that if $deg(u)+deg(v)\geq n$ for every pair $u.v$ of non adjacent vertices of $G$, then $G$ is non-separable. Prove that if $G$ is a graph of order $n \geq 3$ with property that $deg(u)+deg(v)\geq n$ for every pair $u.v$ of non adjacent vertices of $G$, then $G$ is non-separable.

I'm trying to show that $G$ has no cut-vertex, but I can't see how the property  $deg(u)+deg(v)\geq n$ can help me. I would appreciate any hint.

Suppose that $G$ is separable, and let $a$ be the cut node, $H_1$ and $H_2$ the nodes in the two distinct parts of the graph, so that $|H_1|+|H_2|+1=r+s+1=n$.

For all couple of nodes $(u,v)$ such that $u\in H_1$, $v\in H_2$, $deg(u)+deg(v)\geq n$ holds, because they're not adjacent.

So

$$
rsn\le \sum_{u\in H_1, v\in H_2}[deg(u)+deg(v)]=\\=s\sum_{u\in H_1}deg(u)+r\sum_{v\in H_2}deg(v)\le sr(r-1)+rs(s-1)=rs(r+s-2)
$$

And this is absurd

$$
r+s+1=n\le r+s-2
$$

Simpler proof can be:

If we assume our graph to have $\mathrm{deg}(u)+\mathrm{deg}(v) \ge n$ for every pair of non adjacent vertices $u$ and $v$, our graph is therefore Hamiltonian.

And by definition, a Hamiltonian graph can not have a cut vertex, therefore making it nonseparable.


how to differentiate skew symmetric matrix $[\mathbf{v}]_{\times}$ with respect to $\mathbf{v}$ can anyone explain how to differentiate a skew-symmetric matrix $\mathbf{v}_{\times}$ with respect to $\mathbf{v}$ i.e. $\frac{\partial{[\mathbf{v}]_{\times}}}{\partial \mathbf{v}}$ where $\mathbf{v}\in\mathbb{R}^3$?

In addition, how to derive $\frac{\partial [\mathbf{v}]^2_{\times}}{\partial \mathbf{v}}$?

Thank you!

Using the Levi-Civita (aka Permutation) tensor $\varepsilon$ the problem has a straightforward approach $$\eqalign{
 [v]_\times &= -\varepsilon\cdot v \cr\cr
d[v]_\times &= -\varepsilon\cdot dv \cr\cr
\frac{\partial [v]_\times}{\partial v} &= -\varepsilon \cr
\cr
}$$
This is the same as what you derived, but the notation is more standard.

Update
Thinking about the second part of your question, your initial expansion is incorrect
$$\eqalign{
[v]^2_\times &= vv^T - (v^Tv)\,I \,\,\,\neq vv^T \cr\cr
}$$
Update #2
Here's a partial result for the question that you linked in your comment.
Given the vector $w$ and the cross-matrix $$W=[w]_\times$$ you generate a rotation matrix $$R=\exp W$$ which you'd like to differentiate.

First, define a scalar $\theta$ representing the length of vector and take its differential
$$\eqalign{
 \theta^2 &= w\cdot w = \frac{1}{2}W:W \cr
 2\theta\,d\theta &= W:dW \cr
 d\theta &= \frac{W:dW}{2\theta} \cr
\cr
}$$
Next expand $R$ via Rodrigues' formula
$$\eqalign{
 R &= I + \frac{\sin\theta}{\theta}W + \frac{1-\cos\theta}{\theta^2}W^2 \cr
  &= I + \alpha W + \beta W^2 \cr
\cr
}$$
Now let me denote the $4^{th}$ order isotropic tensor by ${\mathcal A}$ with components
$${\mathcal A}_{ijkl}=\delta_{ik}\,\delta_{jl}$$
and the dyadic ($\star$) product $C = A\star B$ with components
$$C_{ijkl} = A_{ij}\,B_{kl}$$
and a colon to  denote the double-dot product, i.e.
$$A:B = A_{ijkl}\,B_{klmn}$$

Finally we're ready to differentiate the rotation matrix
$$\eqalign{
  dR &= \alpha dW + \beta(W\cdot dW + dW\cdot W) + W\alpha^\prime d\theta + W^2\beta^\prime d\theta \cr
 &= \Big[\alpha{\mathcal A} + \beta(W\cdot{\mathcal A} + {\mathcal A}\cdot W^T) + \frac{\alpha^\prime}{2\theta}\,W\star W + \frac{\beta^\prime}{2\theta}\,W^2\star W\Big]:dW \cr
 &= -\Big[\alpha{\mathcal A} + \beta(W\cdot{\mathcal A} + {\mathcal A}\cdot W^T) + \Big(\frac{\alpha^\prime W}{2\theta}+\frac{\beta^\prime W^2}{2\theta}\Big)\star W\Big]:\varepsilon\cdot dw \cr\cr
\frac{\partial R}{\partial w} &= -\Big[\alpha{\mathcal A} + \beta(W\cdot{\mathcal A} + {\mathcal A}\cdot W^T) + \Big(\frac{\alpha^\prime W}{2\theta}+\frac{\beta^\prime W^2}{2\theta}\Big)\star W\Big]:\varepsilon \cr
}$$where
$$\alpha^\prime=\frac{d\alpha}{d\theta},\,\,\,\,\beta^\prime=\frac{d\beta}{d\theta}$$

You can rearrange this result to better suit your tastes if you keep in mind a few things.

${\mathcal A}$ is the identity for the double-dot product
$${\mathcal A}:X = X:{\mathcal A} = X$$
and $$W:\varepsilon = \varepsilon:W = -2\,w$$

Update #3
Looking more closely at your linked question, you appear to be interested in functions of the $W$ matrix of the form
$$\eqalign{
 F &= I + \alpha W + \beta W^2 \cr
 \alpha &= \alpha(\theta),\,\,\,\beta = \beta(\theta) \cr
}$$
The case $F=e^W$ was analyzed above. The good news is that all of that analysis carries over to the case $F=\frac{e^W-I}{W}\,\,$ just using different functions for the scalar coefficients
$$\eqalign{
 \alpha &= \frac{1-\cos(\theta)}{\theta^2} \cr
 \beta &= \frac{\theta-\sin(\theta)}{\theta^3} \cr
}$$
Therefore
$$\eqalign{
\frac{\partial F}{\partial w} &= -\Big[\alpha{\mathcal A} + \beta(W\cdot{\mathcal A} + {\mathcal A}\cdot W^T) + \Big(\frac{\alpha^\prime W}{2\theta}+\frac{\beta^\prime W^2}{2\theta}\Big)\star W\Big]:\varepsilon \cr
\cr
 &= \Big(\frac{\alpha^\prime W}{\theta}+\frac{\beta^\prime W^2}{\theta}\Big)\star w - \alpha\,\varepsilon - \beta\,W\cdot\varepsilon - (\beta{\mathcal A}\cdot W^T):\varepsilon \cr
}$$

I derive a solution using the notations of linear algebra:
$$
\frac{\partial [\mathbf{v}]_\times}{\partial \mathbf{v}} = \frac{\partial \begin{bmatrix}
0 & -v_3  & v_2\\
v_3 & 0 & -v_1 \\
-v_2 & v_1 & 0
\end{bmatrix}}{\partial \begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}}
$$
using $v_1$ as an example:
$$
\frac{\partial [\mathbf{v}]_\times}{\partial v_1} = \begin{bmatrix}
0 & 0  & 0\\
0 & 0 & -1 \\
0 & 1 & 0
\end{bmatrix} = [e_1]_\times
$$
where $e_1$ is the first column of an identity matrix $\mathbf{I} \in \mathbb{R}^{3\times3}$. Then, we have:
$$
\frac{\partial [\mathbf{v}]_\times}{\partial \mathbf{v}} = \begin{bmatrix}
[e_1]_\times\\
[e_2]_\times\\
[e_3]_\times
\end{bmatrix}
$$

since we have $[\mathbf{v}]_\times^2=\mathbf{v}\mathbf{v}^\top$, we could derive that (thank to greg's help, I made the following correction.)
$$
\frac{\partial [\mathbf{v}]_\times^2}{\partial v_i} = \frac{\partial \mathbf{v}\mathbf{v}^\top-\mathbf{v}^\top\mathbf{v}\mathbf{I}}{\partial v_i} = \frac{\partial \mathbf{v}}{\partial v_i}\mathbf{v}^\top + \mathbf{v} [\frac{\partial \mathbf{v}}{\partial v_i}]^\top-\frac{\partial (\mathbf{v}^\top\mathbf{v})}{\partial v_i}\mathbf{I}=e_i\mathbf{v}^\top+\mathbf{v}e_i^\top-2v_i\mathbf{I}
$$
where $e_i$ is the i-th column of the $\mathbf{I}$. Correct me if anyone see any problem with this.


limits of integration and derivative I have an integral that gives
$$\left[\frac{d}{dx}[f(x)]\right]_a^b$$

is it possible in general to claim that this is equal to
$$\frac{d}{dx}\left(\left[f(x)\right]_a^b\right)\text{ ?}$$

If not in general, are there any sufficient conditions to claim this?

Thanks!!

The first expression is

$$
f'(x)|^b_a = f'(b) - f'(a)
$$
whereas the second one, since neither $a$ nor $b$ are functions of $x$, means
$$
\frac{d}{dx} \left[f(b)-f(a)\right] = 0
$$

So necessary and sufficient condition for equality is $f'(b) = f'(a)$, and one sufficient condition is if $f'$ is constant (so $f$ is linear).

On the left hand side we have: $\left[\dfrac{\mathrm d}{\mathrm d x} f(x)\right]_a^b = \dfrac {\mathrm d f(b)}{\mathrm d b}-\dfrac{\mathrm d f(a)}{\mathrm d a}$.

On the right hand side we have: $\dfrac{\mathrm d}{\mathrm d x}{\big[f(x)\big]}^b_a = \dfrac{\mathrm d}{\mathrm d x}\big(f(b)-f(a)\big) = 0$

In general, the left hand side is not necessarily equal to the right hand side.

One counter example is simply, $f(x)=x^2, a=0, b=1$.  Then $\left[\dfrac{\mathrm d x^2}{\mathrm d x} \right]_0^1 = 2$.


Let p, q be prime; if p | q + 1 then also p | q^(q + 1) + q, proof? Let $p$, $q$ be prime. if $p | q + 1$ then $p | q^{q + 1} + q$
Any elementary proof will be appreciated!

Let us add the condition that $q$ is odd. We are told that $q\equiv -1\pmod{p}$. Since $q+1$ is even, it follows that $q^{q+1}\equiv 1\pmod{p}$, and the result follows.

Remark: We could write that $q+1=kp$ for some integer $k$, and therefore $q=kp-1$. Then we could imagine expanding
$$(kp-1)^{q+1} +(kp-1).$$
Since $q+1$ is even, the term of $(-1)^{q+1}$ in the expansion is $1$, which cancels the $-1$ from the $kp-1$ term. All the rest of the terms are divisible by $p$.

We mention this way of doing it to point out how much more efficiently the congruence machinery does the job.

if neither p nor q is not 2 then $oddnumber∤evennumber$ so the first term will be false.
if only p is 2 and q is an odd prime number , then q+1 is even too and q^(q+1) is odd so q^(q+1)+q is even and then p | q^(q + 1) + q
if only q is 2 and p is an odd prime number , then only if p=3 the first term will be true. but in this case we will get to 3|10 which is false!
if both q and p are two then the first term will be false.

so based on number 3 it can't be prooved.


How would you solve this system if linear equations in 3 variables? The equations are-
$$
a+b+c=6
$$
$$
ab+bc+ca=11
$$
$$
abc=6
$$
Also,is there a way to generalise this to a formula for varying values of the constants?

Hint: $a,b,c$ are roots of the cubic equation
$$
x^3-6x^2+11x-6=0.
$$

An elementary computation (to get to the same end result as in the same answers):

Use the first equation to find that
$$
a=6-b-c.
$$

Substitute this into the second equation and solve for $b$ using the quadratic formula to get
$$
b=-\frac{1}{2}c+3\pm\frac{1}{2}\sqrt{-3c^2+12c-8}
$$

Substituting this into the third equation results in
$$
c^3-6c^2+11c=6
$$
(which then must be solved).


The inclusion function $Haus\hookrightarrow Top$ has a left adjoint. Can someone help me with my proof that the inclusion functor $Haus\hookrightarrow Top$ has a left adjoint? The inclusion functor is continuous. However, I am failing to prove that the inclusion functor satisfies the solution set condition.

Let $X$ be any topological space. Since $\textbf{Haus}$ is a full subcategory of $\textbf{Top}$, it is enough to provide a set $\{X_i\}_{i \in I}$ of Hausdorff spaces such that every continuous map $f: X \to Y$, where $Y$ is Hausdorff, factors through some $X_i$. We can take $\{X_i\}_{i \in I}$ to be the set of Hausdorff quotients of $X$.

Let $f: X \to Y$ be continuous, with $Y$ Hausdorff. Define an equivalence relation $R$ on $X$ by $x R x'$ iff $f(x) = f(x')$, and let $\pi: X \to X / R$ be the quotient map. Then $f$ factors as $\bar{f} \pi$, for some continuous $\bar{f}: X/R \to Y$. So we are left to show that $X / R$ is Hausdorff. Let $\pi(x), \pi(x') \in X/R$ be distinct, then by definition of $R$ we have
$$
\bar{f}(\pi(x)) = f(x) \neq f(x') = \bar{f}(\pi(x')).
$$
So by Hausdorffness of $Y$, we find opens $\bar{f}(\pi(x)) \in V$ and $\bar{f}(\pi(x')) \in V'$, such that $V \cap V' = \emptyset$. Then $\bar{f}^{-1}(V)$ and $\bar{f}^{-1}(V')$ are opens in $X / R$, containing $\pi(x)$ and $\pi(x')$ respectively, while having empty intersection. So indeed, $X / R$ is Hausdorff.

Edit: as mentioned in the comments by Jeremy, a more elegant solution would be to take $\{X_i\}_{i \in I}$ to be the set of all Hausdorff spaces with a surjective continuous map from $X$. Since every continuous map factors as a surjective map (to a subspace) followed by an injective map (the inclusion of that subspace), this would work as well.

Define the relation $R$ on every topological $X$ such that $xRy$ iff every open subset which contains $x$ contains $y$ and $S$ the equivalence relation generated by $R$.  We denote by $X_S$ the set of equivalence classes with the quotient topology. The map $X\rightarrow X_S$ is the right adjoint.


$\frac{2019(x^{4}+2)}{x^{2}y+1} \in \mathbb{N^*}$ Find all pair $(x;y)$ of positive integers such that $k=\frac{2019(x^{4}+2)}{x^{2}y+1} \in \mathbb{N^*}$
$\Leftrightarrow 2019x^4-kyx^2+4038-k=0$.Hence $\Delta\geqslant 0$
$\Leftrightarrow k^2y^2+4 \cdot 2019(k-4038)\geqslant 0\Leftrightarrow y^2\geqslant \frac{4 \cdot 2019(4038-k)}{k^2}$
$$x^2y+1 \mid 2019(x^2-2y)$$
I don't know what I should do next. Help me, please?


"Existence and Uniqueness" Theorems for first order IVP: two or just one? Let's say that I've got the following IVP:

$\frac{dy}{dx} = f(x,y)$

$y(x_0) = y_0$

And I want conditions that guarantee existence and uniqueness of its solution.

On the one hand I've got the Picard–Lindelöf theorem. It asks that there exists a rectangle $R = [a,b] \times [c,d]$, containing $(x_0, y_0)$ as an interior point, where $f$ is continuous in $x$ and Lipschitz continuous in $y$.

On the other hand I've got a theorem, which I've encountered in many undergraduate text books, that requires $f$ and $\frac{\partial f}{\partial y}$ to be continuous in the aforementioned rectangle.

Are these two different theorems? It seems to me that the hypotheses of the first one are implied by those of the second one. But in that case, why would some authors prefer this more restrictive form of the theorem? Could it be just so that students don't need to learn the concept of Lipschitz continuity?

Suppose $\frac{\partial f}{\partial y}$ is continuous. Then $|\frac{\partial f}{\partial y}|$ has a greatest value $K$ on the rectangle in question (by the extreme value theorem). The mean value theorem for derivatives says that
$$
f(x,y_2)-f(x,y_1) = \frac{\partial f}{\partial y}(x,\eta) \, (y_2-y_1)
$$
for some $\eta$ between $y_1$ and $y_2$, which implies
$$
|f(x,y_2)-f(x,y_1)| \le K |y_2-y_1|
,
$$
so $f$ is Lipschitz continuous with respect to $y$ on that rectangle, with Lipschitz constant $K$.

Hence the assumptions that $f$ and $\frac{\partial f}{\partial y}$ are continuous are just (somewhat weaker) replacements for the “proper” assumptions about Lipschitz continuity, with the advantages that they are often very easy to verify, and that you don't have to explain to your readers what Lipschitz continuity means.

The first, that requires that the partial derivative of f with respect to y exist, is an "if and only if" theorem.  The second, that requires "Lindelof" in y, says "if" but not "only if".  Any function that is differentiable with respect to y is "Lindelof" but the converse is not true.


Why Dirichlet's energy uses a **squared** norm? $E = \int_{\Omega}\left \| \nabla u(x)\right \|^2 dx$

So, Dirichlet's energy measures the integral of the squared norm of the gradient. Why squared norm? What would we get if we use just a norm? It's still going to be non-negative.

If I calculated a $u(x)$ that minimizes $E = \int_{\Omega}\left \| \nabla u(x)\right \| dx$ (not squared) would it be worse than Dirichlet? My goal is exactly what I specify: the gradient at any point should have minimum length.

Is the reason for the squared norm minimisation equivalent to the role of the square in the least squares fit? (here)

I think the square comes from physical modelling the bending energy.

But then, the square of a Hilbert space norm has several nice properties: it is uniformly convex and twice continuously differentiable with constant second derivative.

The Dirichlet problem
$$
\min \int_\Omega \frac12|\nabla u|^2 - uf \ dx
$$
is of course not equivalent to
$$
\min \int_\Omega |\nabla u| - uf \ dx.
$$
The latter problem is much harder to solve and analyze.

For instance, it is much more difficult to prove that the second problem has a solution with integrable gradient.

The minimizer of the Dirichlet energy satisfies
$$
-\Delta u = f,
$$
whereas the minimizer of the second problem satiesfies
$$
-\nabla\cdot\left( \frac{\nabla u}{\|u\|} \right) =f.
$$
To have a finite-dimensional analogue, consider the minimization of
$$
\sum_{i=1}^n \frac12|a_i|^2 - a^Tc
$$
and of
$$
\sum_{i=1}^n |b_i| - b^Tc.
$$
The first one has solution $a_i=c_i$, the second has a solution only if $|c_i|\le 1$ for all $i$ (which is then $b_i=c_i$).

I think in this form, the Dirichlet's energy is a strictly convex quadratic function and should have an UNIQUE minimum for reasonable boundary conditions .
This also holds for the least squared norm minimisation.


A little exercise about free modules Let $R$ be a non-zero ring with unity. Let $a$ be a proper ideal of $R$. I wish to prove the following statement:

$R/a$ is a free left module over $R$ iff $a=0$

where $R/a$ has the usual scalar multiplication induced by considering the left regular module $_{R}R$

I'm having little trouble proving that if $R/a$ is free over $R$ then one must have $a=0$. Here is what I've done so far:

Suppose that $R/a$ is free over $R$. Let $X$ be a basis of $R/a$ over $R$. Since $a$ is a proper ideal, we have that $R/a \neq 0$ and so $X \neq \emptyset$ as a generating set. Let $x + a \in X$. Since $X$ is linearly independent over $R$, we have in particular that $ \{x+a\}$ must be linearly independent over $R$. This means that $\forall r\in R$, $r(x+a)=a \implies r=0$ or equivalently $\forall r \in R, rx \in a \implies r=0$.

Now if one could take $x=1$ the result would follow immediately. Can anyone help me from the last step forward? Or provide a different way of looking at the problem? Thank you in advice.

If $R$ is commutative, this is true. Noncommutative rings like the ring of endomorphisms of an infinite-dimensional vector space are counterexamples in general. I'll give a relatively inexplicit proof-this could be reduced to manipulating elements.

So, assume $R$ is commutative. If $R/a$ were isomorphic to $R$, then the map $R\to R/a\cong R$ would split, making $R\cong R\oplus M$. We want to show $M=0$, for which it's enough to suppose $R$ is local: $M=0$ if and only if $M_{\mathfrak p}=0$ for every prime ideal $\mathfrak p$. Then if $R$ is local with maximal ideal $\mathfrak m$ and residue field $\kappa,$ we have $\kappa\cong \kappa\oplus M/\mathfrak m M$, so $M/\mathfrak m M=0$. But then by Nakayama's lemma, $M=0$, as was to be shown.

EDIT: A simpler answer which doesn't use commutativity was posted and deleted, so I'll reproduce it here for now. If $R$ is a unital ring, not necessarily commutative, $a$ is two-sided and $f:R\to R/a$, then for $x\in a, xy\in a$ for every $y\in R$. Thus $x[f(1)]=[xf(1)]=[f(x)]=[0]$ in $R/a$, and $f$ cannot be injective.

To be quite clear about how this fails in case $a$ is only a left ideal, take $R$ the endomorphism ring of an infinite-dimensional vector space and $a$ the left ideal of endomorphisms killing the supspace generated by $e_{2k}$ the standard basis vectors with even index. Then $f:R\cong R/a$ as left modules by sending the identity to the endomorphism $T:e_{2k-1}\mapsto 0, e_{2k}\mapsto e_k$. Since $T$ is surjective, $f$ is injective, in contradiction to the two-sided case.

Suppose that $R/a$ is free over $R$. Let $X$ be a basis of $R/a$ over $R$. Since $a$ is properly contained in $R$, we have that $R/a \neq 0$ and so $X \neq \emptyset$ as a generating set. Let $x + a \in X$. Since $X$ is linearly independent over $R$, we have in particular that $ \{x+a\}$ must be linearly independent over $R$. This means that $\forall r\in R$, $r(x+a)=a \implies r=0$ or equivalently $\forall r \in R, rx \in a \implies r=0$. But since $a$ in particular a right ideal, one has that $\forall r \in a, rx \in a$. So for $r \in a$, it follows that $r=0$ by the previous linear independence argument, and thus $a=0$.


Let $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits. Let $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits.

My attempt is:
Since the sequence is bounded , there exists $M>0$ such  that $x_n \in [-M,M]$ for all $n \in \mathbb{N}$. Since the sequence does not converge to $x$, there exists $\epsilon_0>0$ such that $ \forall N \in \mathbb{N}$, there exists $n \geq N$ such that $|x_n-x| \geq \epsilon_0$.

Then we have $x_n \in [-M,x-\epsilon_0] \cup x_n \in [x+\epsilon_0,M]$. By Bolzano-weierstrass theorem, there exists a convergent subsequence in the two intervals.

Is my proof valid? ${}{}$

Use Bolzano-Weierstrass to extract a subsequence $x_{i_1}, x_{i_2}, \dotsc$ that converges to some $a$. Since $x_1, x_2, \dotsc$ does not converge to $a$, there exists some $\varepsilon > 0$ such that for each positive integer $N$, there exists some $j(N) > N$ such that $|x_{j(N)} - a| \geq \varepsilon$. Use Bolzano-Weierstrass to extract from $x_{j(1)}, x_{j^2(1)}, x_{j^3(1)}, \dotsc$ a subsequence $x_{k_1}, x_{k_2}, \dotsc$ that converges to some $b$. Clearly, $a \neq b$.
By the way, this method works for any $x_1, x_2, \dotsc$ in $\mathbb{R}^n$, which goes beyond what the OP intended.

Yes, valid, but could write it even clearer.

For each $N$ there is an $n$ such that $|x_n-x|\ge \epsilon_0$. Let such an $n$ be chosen for all $N$ and denote it by $k(N)$, for example. Thus we get a subsequence $(x_{k(1)},x_{k(2)},\dots)$ of $(x_n)$, which lies in $[-M,x-\epsilon_0]\cup [x+\epsilon_0,M]$, so this subsequence has a convergent subsequence, and the limit of that cannot be $x$.