image or projection of a universally measurable set Suppose $A \subset [0,1]\times[0,1]$ is universally measurable. Is it true that its projection to the first coordinate is a universally measurable subset of $[0,1]$?

What is known is that the projection of any analytic subset (and hence also any Borel-measurable subset) of $[0,1]\times[0,1]$ is analytic and hence universally measurable, which is proved by first showing that (Lusin's theorem) the collection of universally measurable sets is closed under Suslin operations, but that doesn't seem to give any hint as to what can be said about projections of arbitrary universally measurable sets.

This is consistently false: Under $V = L$, there are $\Delta^1_2$ non measurable sets of reals. I don't see how to construct a ZFC counterexample though.


Is the function $F(x)=\int_{0}^{x} f(t) dt$ differentiable at 0? Let
$$
f(x)=\left\{\begin{matrix}
\cos\frac{1}{x} & x\neq0\\
0, & x=0
\end{matrix}\right.$$

I think the answer is not differentiable at 0 because of the chain rule, but I don't know how to prove it.

It is indeed differentiable with derivative zero: Of course, $\int_0^h \cos(1/x)\;\mathrm dx\to 0$, and it happens to go very quickly. This can be seen by a change of variables as follows.
$$
\int_0^h \cos(1/x)\;\mathrm dx=\int_{1/h}^\infty \frac{\cos(x)}{x^2}\mathrm dx
$$
which is easier to analyze. Namely, through an integration by parts
$$
\int_{1/h}^\infty \frac{\cos(x)}{x^2}\mathrm dx=
-h^2\sin(1/h)+\int_{1/h}^\infty \frac{2}{x^3}\sin(x)\;\mathrm dx
$$
so,
$$
\left|\int_{1/h}^\infty \frac{\cos(x)}{x^2}\mathrm dx\right|=
\left|-h^2\sin(1/h)+\int_{1/h}^\infty \frac{2}{x^3}\sin(x)\;\mathrm dx \right|\\
\leq
h^2|\sin(1/h)|+\int_{1/h}^\infty \frac{2|\sin(x)|}{x^3}\;\mathrm dx
$$
by the triangle inequality. Evaluating the integral and using the obvious bound on $\sin$ we have
$$
\left|\int_{1/h}^\infty \frac{\cos(x)}{x^2}\mathrm dx\right|\leq
2h^2
$$
Using the above in the difference quotient:
$$
\left|\frac{1}{h}\int_0^h \cos(1/x)\;\mathrm dx\right|=
\left|\frac{1}{h}\int_{1/h}^\infty \frac{\cos(x)}{x^2}\mathrm dx\right|\leq
2|h|\to 0
$$

Remark: If you keep playing the integration by parts game, you'll see it actually vanishes to the order of $O(h^n)$ for any $n\in \mathbb{N}$, i.e. faster than any polynomial.

If the function is not continuous at a point, the function at that point is not differentiable, too.

To check the continuity of the function at $x=0$, we must compute it's L.H.L and R.H.L

$$L.H.L=\lim_{x\to0^-}cos(\frac{1}{x})$$
(Something oscillating between $[-1,1]$)

$$R.H.L=\lim_{x\to0^+}cos(\frac{1}{x})$$
(Something oscillating between $[-1,1]$)

Now, neither L.H.L nor R.H.L is defined. So, we can conclude that limit at $x=0$ doesn't exist and$f(x)$ is not continuous at $x=0$. Hence, $f(x)$ is not differentiable at $x=0$ .


Why is $\Bbb Z_2 \times \Bbb Z_4$ not isomorphic to $\Bbb Z_8$? $\Bbb Z_2 \times \Bbb Z_4$ has 8 elements and obviously so does $\Bbb Z_8$. How do you conclude they are not isomorphic then? They essentially have a $1-1$ relationship.

Indeed, they do have a one-to-one correspondence between, them (in fact, there are $8!$ such correspondences). The problem is that they don't have the "same structure," so none of these correspondences will be homomorphisms, and so no isomorphism exists. But it's tedious in the extreme to check so many functions to see if any of them is a homomorphism, so we instead make use of one of many properties preserved under isomorphisms. Two such (related) properties are that:

isomorphisms preserve orders of elements (meaning that any given element will be mapped to an element of the same order), and
isomorphic images of cyclic groups are again cyclic groups.

Those two properties are probably the simplest way to show that these two groups are not isomorphic, since (in particular) there are four different order $8$ elements of $\Bbb Z_8,$ but no such elements in $\Bbb Z_2\times\Bbb Z_4.$ So, we can directly conclude that they are not isomorphic by fact 1. Alternately, we can note that this means $\Bbb Z_8$ is cyclic and $\Bbb Z_2\times\Bbb Z_4$ is not, so can indirectly conclude that they are not isomorphic by fact 2.

There are many other such properties that we can use in other cases. It's good to keep several of them in mind. As Mark points out in the comments, though, the first property is a necessary condition for a function to be an isomorphism, but not sufficient. However, two cyclic groups of the same order will necessarily be isomorphic. Unfortunately, most groups aren't cyclic, so this property won't usually be useful.

The range of every homomorphism $f:\mathbb{Z_2 \times Z_4 \rightarrow Z_8}$ is {0,2,4,6} or {0,4}.


Requirements of the edges of a tetrahedron What are the requirements of the six edges $a,b,c,d,e,$ and $f$ of a tetrahedron?

I know for each face of the tetrahedron, the three edges must satisfy the triangle condition. I also know by just satisfying all four faces does not meet the requirement of forming a tetrahedron. The requirements looks complicated. Can somebody helps me out?

The condition is indeed quite complicated, too much so for me to be tempted to write it down here, but if you have access to Mathematica or  similar software, here is a simple method to deduce it in a few minutes.  We suppose that the tetrahedron is ABCD with $a$, $b$, $c$, $d$, $e$ and $f$ the lengths of $AB$, $BC$, $CA$, $AD$, $BD$ and $CD$ in that order.  By a simple argument, one can choose coordinates so that the vertices are $(0,0,0)$, $(a,0,0)$, $(p,q,0)$, and $(r,s,t)$.  Then if you use the command

Solve[{(p-a)^2+q^2==b^2,p^2+q^2==c^2,r^2+s^2+t^2==d^2,(r-a)^2+s^2+t^2==e^2,(r-p)^2+(s-q)^2+t^2==f^2},{p,q,r,s,t}]

this will not only solve your problem but it will give you coordinates for the required tetrahedron.  The condition for existence is simply the positivity of the terms which occur under a square root sign in the solution.

The answer is beautifully given in the paper "Edge lengths determining tetrahedrons" by Wirth and Dreiding, Elemente der Mathematik, 64, 160 (2009).
Roughly speaking, the Cayley-Menger determinant must be positive.


What is the most expensive item I could buy with £50? I was set the following question during the discrete mathematics module of my degree and despite my instructor explaining his working to me I still disagree with the answer he says is correct.

Can someone please help me either understand where my mistake is or help me prove that my instructor's answer is incorrect?

  It is the Christmas festive season again and your manager is very pleased with your performance and gives you a £50 Amazon gift card as a Christmas bonus. Value added tax (VAT) or sales tax is currently 20%. Determine the price of the most expensive taxable item you can buy with the gift card. Show your working and not just the answer.

It's a pretty horribly worded question! My gut feeling was £50, as all UK retail prices are already inclusive of VAT.

My Answer:

Let the total price inclusive of VAT $=x=$ £50

Let the rate of VAT $=y=0.2\ (20\%)$

Let the price exclusive of VAT $= z$

$x = z + zy$

$50 = z + 0.2z$

$50 = 1.2z$

$50 / 1.2 = z$

$z = 41.666...$

Instructor's Answer:

Let the price of the most expensive taxable item be $= x$

Let the 20% VAT on £50 $= y = (20/100)*50 = 10$

Our equation can be written as:

$50 = x + y$

$x = 50-y$

$x = 50-10$

$x = 40$

Update from instructor:

  It looks like you and I are going to have some interesting discussions
  during the course of this module. I see where you are “going wrong”
  for want of a better phrase. You are assuming that the £50 includes
  VAT, but that is the wrong assumption. Sometimes easy to make that
  automatic jump or connection to real life scenario, but this question
  has nothing to do with the actual UK VAT laws. Maybe it could have
  been phrased differently, but the £50 is SUBJECT to a 20% VAT which
  implies that VAT is not included and has to be deducted from the £50.
  Forget the UK law for now and you will see why it’s £40. The point
  really is not even about how much the item is, but it is about
  rearranging an equation and solving for x.

In my opinion there are so many errors in his logic that it's not worth me pushing this point any further as he will be my teacher for the next 3 months anyway.

Your answer looks right.   The instructor made the mistake of assuming the taxable part would be 20% of the total,  which is a common mistake with percentages, to forget about what percentages mean.   You can see that $40$ is right out,   by simply taking $40+.2*40=48$, and that's not 50.

This is totally out of the realm of mathematics, but still...

The most expensive item that you can buy with a £50 gift card costs £50. Unless Amazon doesn't actually have any product that costs £50, in which case the most expensive item might cost £49.99 for example.

Now your instructor seems to interpret this as "what could be the most expensive item pre-tax that you can buy for £50". In the UK, most items have 20% VAT. His calculation then is totally bogus. He subtracts 20% from (item price + VAT) and gets £40. But you only pay £8 VAT on a £40 item. As others have calculated, you can buy an item costing £41.67 + £8.33 rounded VAT with your £50 gift card.

But since your instructor is determined to not be corrected, I'll just note that there are plenty of items that are VAT free, like baby wear, children's clothing and children's footwear, books, newspapers, and magazines, and other items with a 5% VAT rate. If he doesn't accept VAT free items as an answer, then you can most definitely buy a children's car set at Amazon for £49.99 including 5% VAT, which would be £47.61 + VAT.


Find a surjection $(\mathbb Z/m)^\times\to (\mathbb Z/n)^\times$ Assume $n|m$ and suppose further that $m$ and $n$ have the same prime factors.

How can we construct a surjective map
$$(\mathbb Z/m)^\times\to (\mathbb Z/n)^\times$$
which preserves the group multiplication?

I tried the map $a+m\Bbb Z\mapsto a+n\Bbb Z$ but I can't prove that it's surjective. I don't know how to use the fact that $m$ and $n$ have the same prime factors.

Your map is fine. Here's a proof that may help you.

Suppose $a\not\in(\mathbb{Z}/n)^\times$ and $a\in(\mathbb{Z}/m)^\times$. Then $\exists b\in(\mathbb{Z}/m)^\times : a b \equiv 1 \mod{m}$.

It holds that $a\not\in(\mathbb{Z}/n)^\times$ so it does not have an inverse. Since $n$ divides $m$, $1 \mod{m} \equiv 1 \mod{n}$. But in that case $a b \equiv 1\mod{n}$. This is a contradiction.

Therefore, if $a\not\in(\mathbb{Z}/n)^\times$ then $a\not\in(\mathbb{Z}/m)^\times$.

Use the negation of that:
$$
¬ (a\not\in(\mathbb{Z}/n)^\times \Rightarrow a\not\in(\mathbb{Z}/m)^\times) \Leftrightarrow  (a\in(\mathbb{Z}/m)^\times \Rightarrow a\in(\mathbb{Z}/n)^\times)
$$

$a\in(\mathbb{Z}/m)^\times \Rightarrow a\in(\mathbb{Z}/n)^\times$ is the clue to what you want to prove.

This shows that your map is well-defined. Every element from $\mathbb{Z}/m^\times$ maps to an element from $\mathbb{Z}/n^\times$. The map is also a surjection because the image of every element is never empty.

I hope this helps :)

Suppose

$$m=\prod_{k=1}^s p_k^{a_k}\;,\;\;n=\prod_{k=1}^s p_k^{b_k}\;,\;1\le b_k\le a_k\;,\;\;p_k\;\;\text{primes}$$

From the above is clear now that if $\;x\le n\;$ , then $\;gcd(x,n)=1\iff gcd(x,m)=1\;$ and there you have your surjectivity proof.


Equations of Motion to follow an trajectory There are two points $\vec{x}(t)$ and $\vec{u}(t)$ that change over time. We are interested in the time interval $t \in [0, T]$. The positions and velocities of both points at time $t=0$ are known. The entire trajectory of the second point $\vec{u}(t)$ is known in advance. It is also known that the acceleration of either point can not exceed a constant $a_{max}$ at any time.

The goal is to construct the trajectory $\vec{x}(t)$ that minimizes the average squared distance $S = \frac{1}{T}\int_{t=0}^{T}|x(t)-u(t)|^2dt$ while fulfilling the above acceleration constraint.

What is an aproppriate formalism to address this question, and, if possible, what are the equations that can be solved for the trajectory.

I have attempted to use Euler-Lagrange to solve this problem, but it does not seem to give satisfactory results. For example, a naive Lagrangian

$$L = \frac{mv^2}{2} - \frac{k|x(t)-u(t)|^2}{2}$$

has several problems

Even if $u(t)$ is a constant, $x(t)$ would oscillate around it instead of converging
The trajectory of $x(t)$ does not take any advantage of the knowledge of future values of $u(t)$, it only uses the current value
The acceleration of $x(t)$ can exceed $a_{max}$ if the points are sufficiently far away. I am not aware of a way to include the acceleration constraint into Lagrangian, as it typically only depends on position and velocity.

Perhaps Euler-Lagrange is not the correct way to approach this problem? Any hints are useful

Well, FWIW, in principle we could pick a Lagrangian

$$ L(\vec{x},\vec{v},\vec{a},t)~=~\frac{1}{2}|\vec{x}-\vec{u}(t)|^2 + {\infty} 1_{]a_{\max},\infty[}(|\vec{a}|), $$

where we implicitly assume the rule $\infty\cdot 0=0$.


Proof by well ordering: Every positive integer greater than one can be factored as a product of primes. I am reading the book Mathematics for Computer Science and have reached the chapter 2.3 Factoring into Primes where there is an explanation how Well Ordering can be used to prove the following:

  Theorem 2.3.1. Every positive integer greater than one can be factored as a product of primes.

Here is the excerpt of the proof:

  Let C be the set of all integers greater than one that cannot be
  factored as a product of primes. We assume C is not empty and derive a
  contradiction.
  If C is not empty, there is a least element n ∈ C by
  well ordering. The n can’t be prime, because a prime by itself is
  considered a (length one) product of primes and no such products are
  in C.
  So n must be a product of two integers a and b where 1 < a; b <
  n. Since a and b are smaller than the smallest element in C, we know
  that a; b ∉ C. In other words, a can be written as a product of primes
  p1p2...pk and b as a product of primes q1...ql.
  Therefore, n =
  p1....pkq1....ql can be written as a product of primes, contradicting
  the claim that n ∈ C. Our assumption that C is not empty must
  therefore be false.

I've managed to understand the proof, but I'm stuck on the following statement:

  So n must be a product of two integers

Why is this a 'must'?

Since $n$ is in $C$, it cannot be factored into a product of primes.  $n$ cannot itself be prime, otherwise it would have the trivial factorization.  Since $n$ is not prime, it has proper integer divisors.

This proof is appallingly convoluted! Here's a formulation that cleanly distinguishes the conceptually important parts, and is furthermore constructive. Your question is indirectly answered by the key lemma.

Lemma. Given any number $n$, the smallest factor $p$ of $n$ such that $p>1$ is prime.

Proof. Given a number $n$, let $C$ be the set of factors $a$ of $n$ with $a>1$. Since any factor $a$ of $n$ satisfies $1\leq a\leq n$, if $n=1$ then $C$ is empty and the statement is vacuous.

Otherwise, if $1<n$, then $C$ is non-empty since $n$ is a factor of itself; hence by well-ordering $C$ has a least element $p$. Since $1<p$ by definition of $C$, to show that $p$ is prime it suffices to show that the only factor $a$ of $p$ such that $1<a$ is $a=p$. But if $a$ is factor of $p$ with $1<a\leq p$, we also have that $a$ is a factor of $n$ such that $1<a$ (since $a$ is a factor of $p$ which is a factor of $n$). This means $a$ is in $C$, and since $p$ is the least element of $C$ we must have $p\leq a$. But since $a\leq p$ already it follows that $p=a$.

Proposition. Any number $n>1$ has a prime factorization.

Proof. Let $C$ be the set of quotients of $n$ by products of primes, i.e. let $C$ consist of numbers $m$ such that $n=p_1\cdots p_k\cdot m$ for some sequence of (not necessarily distinct) primes $p_1,\dots,p_k$. The set $C$ is not empty when $n>1$ since by the lemma $n$ has a prime factor $p$ (given by the smallest factor $p$ of $n$ such that $p>1$), in which case the quotient $m$ in $n=pm$ is an element of $C$.

It follows by well-ordering that $C$ has a minimal element $m$ so that $n=p_1\cdots p_k\cdot m$ for sequence of (not necessarily distinct) primes $p_1,\dots,p_k$. If $m>1$, then by the lemma $m=pq$ for some prime $p$, hence $n=p_1\cdots p_k\cdot p\cdot q$ implies $q$ is in $C$. But $1\leq q<m$ since $q$ is a quotient of $m$, which contradicts $m$ being minimal in $C$. Therefore, it cannot be the case that $m>1$, so we must have $m=1$. This shows that $n=p_1\cdots p_k$ for some sequence of primes $p_1,\dots,p_k$. Furthermore, the sequence may be constructed by successively taking smallest factors $p$ such that $p>1$.


How to find the derivative with respect to the transformed co-ordinates. I am stuck with something very simple , would be glad to get help .
Suppose if i have a transformation matrix J , how do i find the derivative with respect to new co-ordinates , and derivative of function with respect to the transformed co-ordinates.
For example i have a transformation

My interest is to find $dx_i'$ and $\frac{\partial f'}{\partial x_i'}$

$\mathbf{X}'=\mathbf{JX}\Rightarrow x'=x\cosh\theta+y\sinh\theta\Rightarrow dx'=\cosh\theta dx+\sinh\theta dy$

$\frac{\partial f'}{\partial x'}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial x'}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial y'}=\frac{f_{x}}{\sinh\theta}+\frac{f_{y}}{\cosh\theta}$

the same for $y$.


direct sum of image and kernel in a infinitedimensional space Is it true that in an infinitdimensional Hilbert space the formula $$\text{im} S\oplus \ker S =H$$holds, where $S:H\rightarrow H$ ?
I know it is true for finitely many dimensions but I'm not so sure about infinitely many. Would it be true under some additional assumption, like assuming that the rank of $S$ is finite ?

Assuming you intended to ask what I proposed in a comment, the answer is yes, this also holds for infinite-dimensional vector spaces, assuming the axiom of choice. Take a basis of $\ker S$ and extend it to all of $S$. The additional basis vectors induce a basis of $S/\ker S$, which by the first isomorphism theorem is isomorphic to $\operatorname{im} S$. This established a linear bijection between $S$ and $\operatorname{im} S\oplus\ker S$. Note that there is no canonical choice for the basis of $S/\ker S$, and you need the axiom of choice to get one.

Consider $l_2$, and the translation operator $T:~e_n\mapsto e_{n+1}$, it's injective but not surjective. So that $ker~T=0,im~T\neq l_2,ker~T\oplus im~T\neq l_2$.

If $rank~im T$ is finite, i remember i have learned that the equality holds in some book. (but it's vague to me now, so take care)

You can try to consider the coordinate function on the finite dimensional space $im~T$ in this case.

Other kinds of additional assumptions may concern with

Projection operator
Compact operator
Fredholm operator

Try your best


How to prove $\ln x<x$? How can we prove that the inequality
$$\ln~x<x$$

It is trivial in the cases $0<x\le 1$. I couldn't do anything for $x>1$.

This is one way to see it among others.

The function $x\mapsto\ln(1+x)$ is a concave function (it's twice differentiable and its second derivative is strictly negative). Thus it's below all its tangents.

The tangent at the point $(0,0)$ is the line $y=x$. Hence
$$\forall x>0,\, \ln(1+x)\leq x$$

We deduce from this that
$$\forall x>0,\, \ln x<x$$

Well, graphically, you can easily prove that. Plotting $\color {blue}{y = e^x}$ vs $\color{red}{y = x}$ graph, we can see:

$e^x > x$

Take log on both sides (Considering only +ve values of $x$).

$x > \ln x$


Solve the equation $10x^3-6x^2-12x-8=0$ Solve the equation $10x^3-6x^2-12x-8=0$
I was doing this question but got stuck and hence looked at the solution, which went as follows $(x+2)^3=x^3+6x^2+12x+8$ and hence the equation is written as $11x^3=(x+2)^3$ from which we have that $\frac{x+2}{x}=\sqrt[3]{11}$ and hence $x=\frac{2}{\sqrt[3]{11}-1}$.
Could you please explain to me how they thought of using the fact that $(x+2)^3=x^3+6x^2+12x+8$ and if there is a more intuitive approach, could you please explain it to me?

I imagine the first thing is to try to factor.  That will fail as this has not rational roots but with $8 = 2^3$ as the last term of a third degree polynomial they think in terms of $(ax \pm 2)^3 = a^3 \pm 6a^2 + 12 a \pm 3$.  That fails as there is no such possible $a$ but the coefficients of $6,12, 8$ are just too convenient to ignore.
$(x\pm 2)^3 = x^3 \pm 6x^2 + 12x \pm 8$ and $P(x) = 10x^3 - 6x^2 -12x -8$.  So if we subtracted those then we'd wipe out every coefficient and something useful may come of it.  We'd get, if nothing else, a shorter equation.
So $P(x) + (x+2)^3 = (10x^3 - 6x^2 -12x -8) + (x^3 + 6x + 12x + 8) = 11x^3$.
Well that's a clean expression $P(x) + (x+2)^3 = 11x^3$ but what can we do with it?

I'll be honest my first thought is not what the book did, but to set $P(x) = 11x^3 - (x+2)^3$ and factor knowing $m^3 - n^3 = (m-n)(m^2 + mn + n)$.  But then I'd realize $11$ doesn't factor nicely.  But... I suppose I must be honest.  I didn't solve this before reading the books answer so if I hadn't read the solution .... I'd probably plow ahead.

$11x^3 - (x+2)^3 = (\sqrt[3]{11}x - (x+2))[(\sqrt[3]{11}^2x^2 + \sqrt[3]{11}x(x+2) +(x+2)^2]$ so either $\sqrt[3]{11}x - (x+2)=0$ and $x=\frac 2{\sqrt[3]{11}-1}$ (the book's solution)

or $(\sqrt[3]{11}^2x^2 + \sqrt[3]{11}x(x+2) +(x+2)^2=(\sqrt[3]{11}^2+\sqrt[3]{11} + 1)x^2  +(2\sqrt[3]{11}+4)x + 4=0$

And by quadratic equation: $x = \frac{-(2\sqrt[3]{11}+4) \pm \sqrt{(2\sqrt[3]{11}+4)^2 - 16(\sqrt[3]{11}^2+\sqrt[3]{11} + 1)}}{2(\sqrt[3]{11}^2+\sqrt[3]{11} + 1)}$.  But that has a negative determinate so no solution.

But the book made a simple observation that would have saved me a lot of time and headache
Well, we are attempting to solve $P(x) = 0$ so we need to solve $(x+2)^3 = 11x^3$ and now we are doing what the text did.
Maybe I'd do it differently $x+2 = \sqrt[3]{11} x$ so $x-\sqrt[3]{11}x = -2$ so $x = \frac {-2}{1-\sqrt[3]{11}} = \frac 2{\sqrt[3]{11}-1}$.


prove that $\lim_{n\to\infty}\int_{-\infty}^\infty \frac n {\sqrt {\pi}}e^{-(nx)^2} f(x) dx = f(0)$ prove that for any integrable function $f$ defined on $(-\infty,\infty)$ (or maybe square integrable function.. my professor didn't gave exact imformation about $f$.)
$$\lim_{n\to\infty}\int_{-\infty}^\infty \frac n {\sqrt {\pi}}e^{-(nx)^2} f(x) dx = f(0)$$

I think I have to get some $\delta>0$ and divide integral into three terms.

$$\int_{-\infty}^{-\delta} f*g_n +\int_{-\delta}^{\delta}f*g_n + \int_\delta ^\infty f*g_n$$
($g_n(x)=\frac n {\sqrt {\pi}}e^{-(nx)^2}$)

since $f*g_n$ converges uniformly to $0$ if $x\neq0$, first and third term vanish.
so what I have to do is to take $N$ and $\delta$ (from the given value $\epsilon>0$) so that if $n>N$  $$\left|\int_{-\delta}^{\delta}\frac n {\sqrt {\pi}}e^{-(nx)^2} f(x)dx - f(0) \right|<\epsilon$$

I can't go any further.

Hints: 1. Since we could redefine $f$ at the one point $0$ and not change any of the integrals, we need further conditions on $f$ for this to be true. For example, continuity of $f$ at $0.$ 2. This is not a DCT problem. 3.This problem is but one example of the following situation: We have a sequence integrable functions $g_n$ on $\mathbb R$ such that i)$\int_{\mathbb R}g_n = 1$ for all $n;$ ii)for all $r>0,$  $g_n(x)\to 0$ uniformly on $\{|x|>r\}.$

If we have that, then given any integrable $f$ on $\mathbb R$ that is continuous at $0,$

$$\lim_{n\to \infty} \int_{-\infty}^\infty g_n\,f = f(0).$$

Generalizing like this actually makes the proof easier I think.

This is my try. The continuity at $0$ and the boundedness of $f$ should be enough.

Rewrite the LHS as:
$$\int_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi\frac{1}{2n^2}}} e^{-\frac{x^2}{2\frac{1}{2n^2}}} f(x)dx$$

This is the expectation of function $f(X_n)$, where $X_n \sim N(0, \frac{1}{2n^2})$. We need to show that: $E(f(X_n)) \to f(0)$.

We can show that $X_n \overset{a.s.}{\longrightarrow} 0$. As $f$ is continuous at $0$, we have: $f(X_n) \overset{a.s.}{\longrightarrow} f(0)$. $f(X_n)$ is bounded, so we can use DCT to conclude.


Proof of $\|P\| \leq \|I-P\|$ for L2 norm in Hilbert space when $P$ is a projection Knowing that $P$ is a projection in Hilbert space $\mathcal{H} \rightarrow \mathcal{H}$, not necessarily orthogonal and that neither $\mathcal{R(P)}$ nor $\mathcal{N(P)}$ equals $\mathcal{H}$, how can we prove the following inequality:

$$\|P\| \leq \| I - P \| $$

for L2 norm using its supremum definition?

$$ ||A|| = \sup_{||x||=1}{||Ax||}$$

I ended up with:
$$
\begin{align*}\
 \|I-P\| &\geq \|I\| - \|P\|   \\
 \| I - P \| &\geq 1 - \|P\|   \\
 0 \leq \| P \| &\leq 1.
\end{align*}
$$

Am I missing some obvious property?

[EDIT: This answer was posted (long) before OP changed the conditions to $P\neq I$ and $P\neq 0$.]

The claim is false. $P = I$ is a projection.

Moreover, $0\le\|P\|\le 1$ is only true for orthogonal projections. In fact, from $P^2 = P$ it follows that $\|P\| = \|P^2\|\le\|P\|^2$ and thus $\|P\|\ge 1$.

$$1=\|I\|=\|I-P+P\| \le \|I-P\|+\|P\|$$

If $P$ is a projection in a Hilbert Space, then using $x=(I-P)x+Px$ one sees that $\|x\|^2=\|(I-P)x\|^2+\|Px\|^2$. Taking supremum over all $x \in H$ with $\|x\|=1$, we see that $1=\|I-P\|+\|P\|$.

Like amsmath mentions, $P=I$ is a counterexample to your claim.


Is every ideal in $\hat{A}$ extended? Let $A$ be a Noetherian ring, $I\subset A$ an ideal, $\hat{A}$ the $I$-adic completion. Is it true that every ideal of $\hat{A}$ is of the form $\hat{J}$ for some ideal $J\subset A$?

The answer is no. Consider $A=k[x,y]$ over a finite or countable field $k$, $I$ the maximal ideal $(x,y)$. Then $\hat{A}=k[[x,y]]$. Let $f(y)$ be a power series transcendental over $k(y)$ with $f(0)=0$ (such power series exists by cardinality arguments). Now consider the ideal $J'=(x-f(y))k[[x,y]]$. I claim that $J'\cap k[x,y]=0$, which implies that $J'$ can not be a $\hat{J}$.

Let $P(x,y)\in J'\cap k[x,y]$. Then $P(x,y)=(x-f(y))g(x,y)$ for some $g(x,y)\in k[[x,y]]$. Replacing in this equality $x$ with $f(y)$ (this is legal becase $f(y)\in yk[[y]]$), we get $P(f(y),y)=0$. As $f(y)$ is trenscendental over $k(y)$, this implies that $P(x,y)=0$.

Update An explicit power series which is transcendental. Consider $e^z\in \mathbb C[[z]]$. If it were algebraic, we would have an integral relation
$$e^{dz}+f_{d-1}(z)e^{(d-1)z}+\dots+f_0(z)=0$$
with a positive integer $d$ and $f_i(z)\in \mathbb C(z)$. There exists a positive integer $m$ such that $P_i(z):=z^{m(d-i)}f_i(z)\in\mathbb C[z]$ for all $i\le d-1$. Then
$$(z^me^z)^{d}+P_{d-1}(z)(z^me^z)^{d-1}+\dots+P_0(z)=0$$
in $\mathbb C[[z]]$.This implies the same equality in the ring of entire functions. But this is clearly impossible because when the real part of $z$ tends to $+\infty$, $e^{z}$ grows much faster than any polynomial function.

Note that $e^z\in\mathbb Q[[z]]$ is then transcendental over $\mathbb Q((z))$. In fact this holds for any field of characteristic $0$.

That should indeed be true. To prove this take an ideal $\hat{J}$ in $\hat{A}$ and prove that the set of numerators of elements in $\hat{J}$ forms an ideal in $A$.

Edit: Sorry I did not read careful enough. This prove only holds for the localization not the completion.


How does the L.Hopital rule work when numerator $\neq$ $\infty$ L.Hopital rule can be used to find the limits of the form

$\lim_{x \to a} \dfrac{f(x)}{g(x)}$ when $\lim_{x \to a}f(x)=\lim_{x \to a}g(x)=\infty$

Today I saw page claiming that $\lim_{x \to a} f(x)=\infty$ is not necessary (When $\lim _{x \to a}g(x)=\infty$) . Is it true? Can it be proved that this condition isn't necessary.

Edit: To clarify, I am solving a limit problem with L.Hopital where $f(x)\neq \infty$[This doesn't prove it, I am trying to clear what I am trying to ask]

$\lim_{x \to \infty}\dfrac{1}{x^2}$ (as $\lim_{x \to \infty} x^2 =\infty$, limit of numerator need not ne $\infty$)

Applying L.Hopital once:

$\implies\dfrac{0}{2x}$ (as $\lim_{x \to \infty}2x=\infty$,limit of numerator need not ne $\infty$)

Applying L.Hopital again:

$\implies \dfrac{0}{2}=0$

Edit:

I think what you are looking for is the proof found in here using the Cesaro-Stolz Theorem. http://www.imomath.com/index.php?options=686

One of the cases is when $\lim_{x\to a}g(x)=\infty$ but makes no hypothesis on $f$.

Another link that may help you is  this one:

In this link a proof is provided that seems to match what your picture wants.

if you need to find $\lim_{x\rightarrow a} \frac{f(x)}{g(x)}$ and you have $\lim_{x\rightarrow a} f(x) \neq \infty$ and $\lim_{x\rightarrow a} g(x) = \infty$, then l'hopital's rule doesn't apply: limiting to (something finite)/$\infty$ gives 0, as opposed to something indeterminate.

Even worse, trying to use L'Hopital's rule when it's inappropriate can get you into trouble: $$\lim_{x\rightarrow 0}\frac{\sqrt[3]{x}}{\csc(x)}=\frac{0}{\infty}=0$$
But if you apply l'hopital's rule this gives $$\frac{1}{3}\frac{x^{-\frac{2}{3}}}{\csc x \cot x} = \frac{\infty}{\infty}$$
and repeated applications of the rule don't get rid of this indeterminate form.


What is really meant by "letting a variable tends to" I have a certain need of rigorously justify some steps of proofs that use limits (maybe because I don't want to rely on intuition about limits that I have from first courses of calculus or somewhat), but I'm not sure what to think when I read "let $x\to a$".

I not sure if I'll make myself clear. I have done basic courses of calculus and I'm actually studying undergraduate level Real Analysis. I know basic properties of limits and know how to prove limits using epsilons and deltas. I'm interested in whether I can use the term "tending to" as it is used, for example, in the following proof of Fermat's Theorem (this proof is given on theorem 5.8 of Rudin's Principles of Mathematical Analysis):

  Theorem: Let $f$ be defined on $[a,b]$; if $f$ has a local maximum at a point $x\in(a,b)$, and if $f'(x)$ exists, then $f'(x)=0$.

Beginning of the proof: Choose $\delta>0$ so that $f(x)$ is a local maximum, that is, $f(x)\ge f(t)$ for all $t\in (x-\delta, x+\delta)$. Then take $x-\delta<t<x$ then $$\frac{f(t)-f(x)}{t-x}\ge 0.$$ Letting $t\to x$, we have $f'(x)\ge 0$.

I do know I can assume that $f'(x)$ is negative and choose $\epsilon = -f'(x)$ so that the quotient will also be negative, getting a contradiction hence we must have $f'(x)\ge 0.$ But this is quite non-practical as the term "tending to" suggests. Any answer that helps me "translate" the term into rigorous thought would be appreciated.

Essentially, it means taking the limit of $t\to x$. So if we have

$$\frac{f(t)-f(x)}{t-x}\ge 0$$

and we let $t\to x$, then we mean

$$\lim_{t\to x}\frac{f(t)-f(x)}{t-x}\ge 0$$

or $f'(0)\geq 0$.

I also suspect you've made a typo there since $x+\delta<t<x$ is impossible if $\delta>0$. I expect they mean $t$ is larger than $x$ and we let $t\to x$ from above.

For $t<x$ we have $\frac{f(t)-f(x)}{t-x}\ge 0.$  Since $f'(x)= \lim_{t \to x}\frac{f(t)-f(x)}{t-x}$, it follows that $f'(x) \ge 0$.


Transformation of beta distribution into gamma distribution How can I convert a Beta Distribution to a Gamma Distribution?
Strictly speaking, I want to transform parameters of a Beta Distribution to parameters of the corresponding Gamma Distribution. I  have mean value, alpha and beta parameters of a Beta Distribution and I want to transform them to those of a Gamma Distribution.

Let $X_a$ and $X_b$ denote independent gamma random variables with respective parameters $(a,c)$ and $(b,c)$, for some nonzero $c$. Then $\dfrac{X_a}{X_a+X_b}$ is a beta random variable with parameter $(a,b)$.

if x denotes beta distribution of 1st kind with parameters @ and 1 then -logx will follow gamma distribution with parameters @ and 1


Discreteness of $\{ a+b\sqrt2 \mid a,b \in \Bbb N \}$ in $\Bbb R$ How to prove that $\{ a+b\sqrt2 \mid a,b \in \Bbb N \}$ is discrete in $\Bbb R$?

If I sum over $\Bbb Z$ instead of over $\Bbb N$, it becomes dense, which is quite confusing to me.

Also, when I plot the points, they appear to become denser as I go to the right, which leads me to wonder if the set is really discrete.

It is discrete, as for every $N$ there are only finitely many elements from the set less than $N$, because for every $a+b\sqrt2 < N$, we know tha $a+b<N$, but there are only at most $N(N+1)/2$ possible pairs of $(a,b)$ with $a+b < N$, so the number of elements from the set that are less than $N$ is also finite.


Understanding the proof for Lebesgue integrability criterion I found a somewhat rigorous proof of Lebesgue's integrability criterion, where $\Delta x_k = x_k - x_{k-1}$, and
$$M_k = \sup_{x \in [x_{k-1},x_k]} f(x),m_k = \inf_{x \in [x_{k-1},x_k]} f(x)
$$
However, how is the proposition
$$ D_i \subseteq \bigcup_{k \in T} [x_{k-1},x_k]
$$
justified?

We read "$T$ is the set of those $k$ with $D_i\cap (x_{k-1},x_k)\neq \emptyset$."  Furthermore, the $[x_{k-1},x_k]$ intervals collectively cover $[a,b]$ per the statement that $a = x_0<x_1< \cdots < x_n=b$.  We are also told that $[a,b]$ is the domain of $f$, and $\omega_f$ is defined on that domain of definition, so $D_i$, defined as the set of points $x$ where $\omega_f(x)$ exceeds a certain value, must be a subset of that domain.

We thus have $D_i\subset \bigcup_{k=1}^n [x_{k-1},x_k]$.  From the latter union, we may omit all those $[x_{k-1},x_k]$ where $D_i$ does not intersect $(x_{k-1},x_k)$ without affecting the correctness of the inclusion, a conclusion which is along the lines of "if $X\subset A\cup B\cup C$ and $X\cap C=\emptyset$, then $X\subset A\cup B$."  By the definition of $T$, those are precisely the intervals which are omitted in the union $\bigcup_{k\in T} [x_{k-1},x_k]$.  We thus have $D_i \subseteq \bigcup_{k \in T} [x_{k-1},x_k]$ as desired.


Steinhaus theorem for topological groups $G$ is a locally compact Hausdorff topological group, $m$ is a (left) Haar measure on $X$, $A$ and $B$ are two finite positive measure in $G$, that is $m(A)>0$, $m(B)>0$.

My question is:

Can we conclude that $AB= \{ab, a\in A, b\in B\}$ contains some non-empty open set of G?

Is this question right?
Or is this right just for $G=R^n$, $R^n$ is the Euclid space, and $m$ is the Lebesgue measure on $R^n$. If so, how to prove it?

Thanks a lot.

Here is another proof, using regularity of the measure instead of convolution.

  Claim: The result holds when $B=A^{-1}$.

Proof: By regularity there is a compact set $K$ and an open set $U$ such that $K\subset A\subset U$ and such that $m(U)<2m(K)$. The multiplication map sends $\{1\}\times K$ into $U$, so by continuity of multiplication and compactness of $K$ there is a neighbourhood $V$ of $1$ such that multiplication sends $V\times K$ into $U$. But then if $x\in V$ the sets $K$ and $xK$ are each more than half of $U$, so $K\cap xK$ is nonempty, so $x\in KK^{-1}$. Thus $KK^{-1}$ contains a neighbourhood $V$ of $1$. $\square$

  Claim: The result holds in general.

Proof: By regularity we may assume both $A$ and $B$ are compact. For $x$ running over $G$ we have
$$\int m(A\cap xB^{-1}) \,dx = \int\int 1_A(y) 1_B(y^{-1}x)  \,dy\,dx = m(A)m(B)>0$$
by Fubini's theorem, so there is some $x$ such that $m(A\cap xB^{-1})>0$. Now apply the previous result to $A\cap xB^{-1}$. Since
$$(A\cap xB^{-1})(A\cap xB^{-1})^{-1} \subset ABx^{-1},$$
we deduce that $AB$ contains a neighbourhood of $x$. $\square$

You can take this example if the topological group $G$ has finite Haar measure, then you can construct a compact set $K$ satisfies for every $x \in G$, you have $K \cap xK \neq \phi$ see proposition 1.4.5 in the book of principles of Harmonic analysis of Deitmar and Echterhoof. Then you can easily show that $G=KK^{-1}$ then you can take $A=K$ and $B=K^{-1}$.


A simple fundamental group I think I have this one, but I want to make sure:

Using Van Kampen's theorem,

Find the fundamental group of two disjoint spheres with each north pole identified, and each south pole identified.

The open sets I chose are sphere 1 minus the north pole, sphere 2 minus the south pole. The intersection is nonempty with trivial fundamental group since poles are identified, so we get trivial*trivial=trivial. Correct?

Let $X$ be your space. Another possibility, not using van Kampen theorem (explicitely), is to glue a 3-cell inside each sphere. Because the fundamental group depends only on the 2-skeleton, $\pi_1(X)$ is isomorphic to the fundamental group of our new space $\tilde{X}$. Now, $\tilde{X}$ clearly retracts by deformation on a circle, hence $$\pi_1(X) \simeq \pi_1(\tilde{X}) \simeq \pi_1(\mathbb{S}^1) \simeq \mathbb{Z}.$$

First of all, we may picture the two spheres as a torus pinched in two points, namely North and South points of the two spheres.
Doing this, it's clear that our space $X$ is path connected and can be covered by two open sets: $A = \left[\left(S^2 \amalg S^2 \right)/ \sim \right] \setminus \left[\,\{N\}\,\right]$ where $\sim$ is the relation with North and South poles identified and $\left[\,\{N\}\,\right]$ is the class of equivalence of North poles of the two spheres; and $B$, the first half of the 2-pinched torus, namely $\left[ S^2 \amalg \{S\} \right]$.

$A$ and $B$ are obviously open sets and path connected; $A$ is homeomorphic to $S^2 \vee S^2$, $B$ is homeomorphic to $S^2$. Their intersection $A\cap B$ is homeomorphic to an infinite cylinder and thus its fundamental group is $\mathbb{Z}$.

Now, Van Kampen says that $\pi(X) = \pi(A) \times \pi(B)$ if $A \cap B$ is simply connected. In a more general version, it says that $\pi(X) = \left[ \pi(A) \times \pi(B) \right] \star \pi(A\cap B)$, where $\star$ is the free product. Now we know that $\pi(A)$ is trivial (again for Van Kampen, if you desire) and $\pi(B)$ is trivial. So, $\pi(X) = \pi(A\cap B) = \mathbb{Z}$.


Trick to solve an improper integral $$ \int_{-\infty}^{\infty} e^{-i a x} e^{-b x^2} \,\mathrm d x$$

Is there any good trick to solve this type integral? I guess there's a way to make that form like $\int e^{-x^2} \,\mathrm d x$. So I could get answer in terms of $\pi$.  Is there any good trick that I can solve this  easily?

If we can differentiate under the integration sign,

$$\frac{\partial I}{\partial a}=-i\int x\,e^{-iax-bx^2/2}dx.$$ (Notice the $/2$ for convenience.)

Then we can write
$$-iaI+\frac bi\frac{\partial I}{\partial a}=\int(-ia-bx)e^{-iax-bx^2/2}dx=\left.e^{-iax-bx^2/2}dx\right|_{-\infty}^\infty=0.$$

Hence the differential equation

$$\frac{\partial I}{\partial a}+\frac abI=0$$

that is separable and solved by

$$I=Ce^{-a^2/2b}.$$

The integration constant can be determined by setting $a=0$,

$$C=\int_{-\infty}^\infty e^{-bx^2/2}dx=\frac1{\sqrt b}\int_{-\infty}^\infty e^{-x^2/2}dx.$$

You have $$\int_{-\infty}^{\infty}{\frac{1}{e^{bx^2+axi}}dx}$$
Try substituting out $u=bx^2+axi$


Class of directed graph for which there is only one path to a given parent? From a nomenclature standpoint, I am wondering if there is a name for a class of directed graph that has only one path to any given parent.

I can visualize this shape as an upside down tree that may or may not have a root, kind of like this:

In looking at this, I can see it is not really an upside down tree.

Is there a name for this type of graph? Again the rules are:

graph is directed
there are no cycles
nodes can share children
no node can have multiple paths to the same parent

Note that these rules work the same in both directions. For example, if we take the graph shown above and reverse the direction of all the arrows, then the rules are still satisfied. What is the name of this thing?

An orientation such that for every $u-v$ path there exists a $v-u$ path is called a strong orientation. Conversely, an orientation that only has either a $u-v$ or a $v-u$ path is a weak orientation.


Find point(s) of intersection between a line and a circle whose radius is parameterized by the same variable as the line Let's assume we have a line:

$$\begin{align}
x&: x_0 + v_xt, \\
y&: y_0 + v_yt \end{align}$$

and a circle

$$\begin{align}
x&: X_0 + kt\cos(s), \\
y&: Y_0 + kt\sin(s).\end{align}$$

Where does the line and the circle intersect?

Set $x_0+v_xt=X_0+kt\cos(s)$ and rearrange to make t the subject.

Set $y_0+v_yt=Y_0+kt\sin(s)$ and rearrange to make t the subject.

Make equal to each other and solve for s ... like this:

$x_0-X_0=kt\cos(s)-v_xt$

$t= \frac {x_0-X_0}{k\cos(s)-v_x}$

$y_0-Y_0=kt\sin(s)-v_yt$

$t= \frac {y_0-Y_0}{k\sin(s)-v_y}$

$\frac {x_0-X_0}{k\cos(s)-v_x}= \frac {y_0-Y_0}{k\sin(s)-v_y}$

$a \cos(s)-b\sin(s)=c$ where $a=k(y_0-Y_0)$ and $b=k(x_0-X_0)$ and $c=v_x(y_0-Y_0)-v_y(x_0-X_0)$

$rcos(s+\alpha)=c$ where $\alpha=\tan^{-1}(\frac b a)$ and $r=\sqrt{a^2+b^2}$

$s=\cos^{-1}(\frac c r)-\alpha$

Hint: convert the cirlce to the form $(x-a)^2+(y-b)^2=r^2$ and the line to $y=mx+b$.
substitute to the cirlce, this would lead a quadratic equation and we know how to solve!


Continuity, derivatives of rational function I'm studying the continuity of a function and its derivatives checking if the function is continuous, differentiable and calculating some derivatives.
The function is

\begin{cases}
\dfrac{x^2y}{x^2+y^2}& \text{if }\, (x,y)\neq 0\\
0& \text{if}\,\, (x,y)=0
\end{cases}

I'm asked to study the continuity in $(0,0)$.

My work:

I started checking the limit of the function for $(x,y) \rightarrow (0,0)$ and considered the trasformation in polar coordinates to get a easier limit:

$$\lim_{\rho\rightarrow 0} \frac{\rho^3\cos^2\theta \sin\theta}{\rho^2}=\lim_{\rho\rightarrow 0} \rho \cos^2\theta \sin\theta=0$$
So I got the continuity.

Then I've been asked to study the value of partial derivatives in $(0,0)$ and the directional derivatives in $(0,0)$ calculated on $y=x$.

I calculated the derivatives:

$f_x=\dfrac{2xy^3}{(x^2+y^2)^2}$,
this is not defined in $(0,0)$ so I considered the limit using polar coordinates again, and I got:
$$\lim_{\rho\rightarrow 0} \frac{2\rho^4 \cos\theta \sin^3\theta}{\rho^4}=2\cos\theta \sin^3\theta$$

Is it right? What can I say now about this limit?
The value of the derivatives in this point is supposed to be $0$ according to the solution given but I'm stuck in this point.

Besides  $f_y=\dfrac{x^4-x^2y^2}{(x^2+y^2)^2}$, from which I get to the same kind of problem (again the values is supposed to be $0$)

Finally the directional derivatives: I used the definition and got:

the vector of the direction is $(\sqrt2/2;\sqrt2/2)$ and $f(0,0)=0$ so

$$\lim_{t\rightarrow 0} \frac{f(\frac{\sqrt2}{2}t;\frac{\sqrt2}{2}t)-0}{t}=\lim_{t\rightarrow 0} \frac{\frac{1}{2}t^2\cdot\frac{\sqrt2}{2}t}{\frac{1}{2}t^2+\frac{1}{2}t^2}\cdot\frac{1}{t}=\frac{\frac{\sqrt2}{4}t^3}{t^2}\cdot\frac{1}{t}=\frac{\sqrt2}{4}$$

I have no idea how to finish the discussion of the two first derivatives and so how to check if the whole function is differentiable in the point $(0,0)$

To work out $\mathrm{f}_x$ you need to use the limit definition:

$$\left.\frac{\partial\mathrm{f}}{\partial x}\right|_{x=y=0} = \lim_{h \to 0} \ \frac{\mathrm{f}(h,0) - \mathrm{f}(0,0)}{h} = \lim_{h \to 0}\ \frac{0-0}{h} = 0$$

To find the partial derivative $\mathrm{f}_y(0,0)$ you need to find the limit

$$\left.\frac{\partial\mathrm{f}}{\partial y}\right|_{x=y=0} = \lim_{h \to 0} \frac{\mathrm{f}(0,h)-\mathrm{f}(0,0)}{h} = \lim_{h \to 0} \ \frac{0-0}{h} = 0$$

To see if $\mathrm{f}$ is differentiable at $x=y=0$ then we need to show that $z=\mathrm{f}(x,y)$ is well approximated by a plane close to $x=y=0$. Since $\mathrm{f}(0,0)=\mathrm{f}_x(0,0) = \mathrm{f}_y(0,0) = 0$ then the only candidate for this plane is $z=0$. To see if $z=\mathrm{f}(x,y)$ is approximated by the plane $z=0$ we need to show that

$$\lim_{(x,y)\to(0,0)} \frac{\mathrm{f}(x,y)-(0x+0y)}{\sqrt{x^2+y^2}} = 0$$

This is the ideal time to use polar coordinates: $x=r\cos\theta$ and $y=r\sin\theta$. We need to show

$$\lim_{r \to 0} \ \frac{\mathrm{f}(r\cos\theta,r\sin\theta)-(0x+0y)}{\sqrt{r^2\cos^2\theta+r^2\sin^2\theta}} = 0$$

We can simplify this quite considerably:

$$\lim_{r \to 0} \ \frac{\mathrm{f}(r\cos\theta,r\sin\theta)}{r} = \lim_{r \to 0} \frac{r\cos^2\theta\sin\theta}{r} = \cos^2\theta\sin\theta$$

This limit depends on $\theta$, which means it is not unique. Since the limit is not well-defined, it follows that $\mathrm{f}$ is not differentiable at $x=y=0$. I've attached a picture below which shows the surface $z=\mathrm{f}(x,y)$ and the plane $z=0$. Notice what bad of a job the plane does at being tangent to the surface at the origin. The blue line is the line $z=\mathrm{f}(0,y)$, the green line is $z=\mathrm{f}(x,0)$ and the red line is $z=\mathrm{f}(t,t)$.

I know it has been past two years when this question was asked, but I study that example and I think I can show you some new idea that hasn't been shown so future reader can learn something.
 The main idea is to use Squeeze theorem and AG inequality.
$$\frac{x+y}{2}>=\sqrt{xy}$$
$$\frac{x^2+2xy+y^2}{4}>=xy$$
$$\frac{x^2+y^2}{1}>=2xy$$
$$\frac{xy}{x^2+y^2}>=\frac{1}{2}$$
$$\frac{xy*|x|}{x^2+y^2}>=\frac{1}{2}*|x|$$
So we can squeeze our limit and use absolute of that:
$$\lim_{(x,y)\rightarrow 0,0}|\frac{x^2*y}{x^2+y^2}|$$
$$0=<|\frac{x^2*y}{x^2+y^2}|<=\frac{1}{2}*|x|$$
SO by the squeeze theorem this limit is equal to zero and function is continous at $(0,0).$


$\ell_0$ Minimization (Minimizing the support of a vector) I have been looking into the problem $\min:\|x\|_0$ subject to:$Ax=b$. $\|x\|_0$ is not a linear function and can't be solved as a linear (or integer) program in its current form. Most of my time has been spent looking for a representation different from the one above (formed as a linear/integer program). I know there are approximation methods (Basis Pursuit, Matching Pursuit, the $\ell_1$ problem), but I haven't found an exact formulation in any of my searching and sparse representation literature. I have developed a formulation for the problem, but I would love to compare with anything else that is available. Does anyone know of such a formulation?

Thanks in advance, Clark

P.S.
The support of a vector $s=supp(x)$ is a vector $x$ whose zero elements have been removed. The size of the support $|s|=\|x\|_0$ is the number of elements in the vector $s$.

P.P.S.
I'm aware that the $\|x\|_0$ problem is NP-hard, and as such, probably will not yield an exact formulation as an LP (unless P=NP). I was more referring to an exact formulation or an LP relaxation.

Consider the following two problems
$$ \min:\|x\|_0 \text{ subject to } Ax = b \tag{P0} $$
$$ \min:\|x\|_1 \text{ subject to } Ax = b \tag{P1} $$
The theory of compressed assert that the optimal solution to the linear program $(P1)$ is an optimal solution to $(P0)$ i.e., the sparsest vector, given the following conditions on $A.$ Let $B = (b_{j,k})$ be an $n \times n$ orthogonal matrix (but not necessarily orthonormal). Let the coherence of $B$ denoted by $$\mu = \max_{j,k} | b_{j,k} |.$$ Let $A$ be the $m \times n$ matrix formed by taking any random $m$ rows of $B.$ If $m \in O(\mu^2 |s| \log n)$ then $(P1)$ is equivalent to $(P0).$ More in the papers referenced in the Wikipedia article on compressed sensing.

the problem can be modeled by an integer formulation: introduce binary variables y and implications $x > 0 \longrightarrow y = 1$ (for example with $M$ constraints $x\le My$). Then minimize the sum of y


What do finite, infinite, countable, not countable, countably infinite mean? Possible Duplicate:
  Different kinds of infinities?

Those terms keep confusing me. Could somebody give a clear and unambiguous definition? How do we prove a set is countable, finite, ...? I know that there are more real numbers than there are integers. How do we prove this? What about other sets like $\mathbb{Z, Q, C}$?

The definitions are:

We say that a set $A$ is finite if and only if there exists some $k\in\mathbb N$ such that there exists $f\colon A\to\{n\in\mathbb N\mid n<k\}$ such that $f$ is a bijection.

Namely, $A$ is finite if we can write it as $\{a_0,\ldots,a_{k-1}\}$.
We say that a set $A$ is countable if and only if there exists $f\colon A\to\mathbb N$ which is injective. Clearly every finite set is countable, but also some infinite sets are countable.

Note that some places define countable as infinite and the above definition. In such cases we say that finite sets are "at most countable".
We say that a set $A$ is uncountable if and only if it is not countable.

Now to show that $\mathbb{N,Z,Q}$ are countable we do the following:

Note that $\mathbb N$ is countable by definition (the identity function witnesses that);
The set $\mathbb Z$ is countable since we can consider the following function: $$f(m)=\begin{cases}2m& m\geq 0\\ -2m+1& m<0\end{cases}$$ We can show that this is an injective function (in fact a bijection) by considering the two cases.
To show that $\mathbb Q$ is countable we use the following facts:

If $A$ is countable then $A\times A$ is countable.
If $A$ is infinite and countable and $B$ is a set such that there is $f\colon A\to B$ which is onto $B$, then $B$ is countable, and vice versa: if $B$ is countable then such surjective function exists.

Now we have that since $\mathbb Z$ is countable so is $\mathbb Z\times\mathbb Z$; we consider the function $f\colon\mathbb Z\times\mathbb Z\to\mathbb Q$ defined as: $$f(m,k)=\begin{cases}\frac{m}{k}&k\neq 0\\ 0 &k=0\end{cases}$$ It is not hard to see that this is a surjective function, since every rational number can be written as a fraction $\frac{m}{k}$. Therefore $\mathbb Q$ is countable too.
Lastly, $\mathbb R$ is uncountable. For this we use Cantor's theorem which states that if $A$ is a set then there is no surjective function from $A$ onto $\mathcal P(A)$ (the set of all subsets of $A$).

We can use the above theorem to show that $\mathbb R$ is in fact with bijection with $\mathcal P(\mathbb N)$, and therefore $\mathbb R$ is not countable.
Since the above shows that $\mathbb R$ is uncountable, and $\mathbb R\subseteq\mathbb C$ we have that the complex numbers are also uncountable.

If there is a bijection from $\mathbb{N}$ to the set $A$ then we say that $A$ is infinite countable. If there is no such bijection, then we say that $A$ is not countable or uncountable.

If there is a bijection from a finite subset of $\mathbb{N}$ to the set $A$ then we say that $A$ is finite countable.


Quick Check on Bayes Rule in Terms of Sets Recently, I took a quiz with a question asking me to find the conditional probability of X given Y in terms of sets.
I solved the question as:
$$Pr(X|Y) = \frac{Pr(Y|X)Pr(X)}{Pr(Y)}$$
$$Pr(X|Y) = \frac{Pr(X \cap Y)}{Pr(Y)}$$
The answer was revealed to be "the probability of the intersection of X and Y, divided by the probability of X." Have I made a mistake somewhere? Any way I think about it, $Pr(Y)$ should be the denominator.

Let's take definition of conditional probability  of X given Y, when $P(Y) \gt 0$
$$P(X|Y)=\frac{P(X \cap Y)}{P(Y)}\quad (1)$$
Obviously, when $P(X) \gt 0$ we have
$$P(Y|X)=\frac{P(Y \cap X)}{P(X)}\quad (2)$$
Now solving $P(Y \cap X)$ from second $P(Y \cap X) =P(Y|X) \cdot P(X)$ and putting it in (1), using $Y \cap X = X \cap Y$, gives
$$P(X|Y)=\frac{P(X \cap Y)}{P(Y)}=\frac{P(Y|X) \cdot P(X)}{P(Y)}\quad (3)$$
As to the your last question in comments, then generally, when $P(X) \ne P(Y)$ and $P(X \cap Y) \ne 0$, then of course $P(X|Y) \ne P(Y|X)$.


What algebra is generated by $\mathrm{O}(2)$? The unit complex numbers can be identified with the $2 \times 2$ special orthogonal matrices $\mathrm{SO}(2)$. The problem with $\mathrm{SO}(2)$, however, is that its not closed under $\mathbb{R}$-linear combinations. To rectify this, we can consider the $\mathbb{R}$-algebra generated by $\mathrm{SO}(2)$, namely $\mathbb{R}[\mathrm{SO}(2)]$, which turns out to be isomorphic to $\mathbb{C}$. It seems natural to play the same trick with $\mathrm{O}(2)$, and see what we get.

  Question. What is $\mathbb{R}[\mathrm{O}(2)]$?

If its not too much to ask, I'd also like to know what $\mathbb{R}[\mathrm{SO}(3)]$ is. I'll ask another question if this isn't answered here.

Well, as you said, any "complex number" $\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$ is a linear combinations of rotations. This implies that any matrix of the form $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} a & b \\ -b & a \end{pmatrix} = \begin{pmatrix} a & b \\ b & -a \end{pmatrix}$ is a linear combinations of products of $\mathrm{diag}(-1,1)$ and a rotation, so belongs to the vector space generated by $O(2)$ (which is the same thing as $\mathbb R[O(2)]$). With these two kinds of matrices, it's easy to get every $2 \times 2$-matrix, so $\mathbb R[O(2)] = M_2(\mathbb R)$.

EDIT: the question about $\mathbb R[SO(n)]$ has actually already be answered on MSE. It turns out that for $n \geq 3$, the answer is a bit disappointing (from your point of view): you get all of $M_n(\mathbb R)$.

You have a split short exact sequence
$$
0 \to \mathbb{R}[SO(2)] \to \mathbb{R}[O(2)] \to \mathbb{Z}/2 \to 0
$$
where the action of the generator of $\mathbb{Z}/2$ on $\mathbb{C}$ would be by complex conjugation. I'm not sure if there is a better description of this, or if this suffices to slake your curiosity.

(I'm not sure about $\mathbb{R}[SO(3)]$)


How to determine if a triangle can be drawn with the given points. Given $3$ points
$$(x_1, y_1), (x_2, y_2), (x_3, y_3),$$ how does one determine whether they are vertices of a triangle?

Thanks.

1: Find area of triangle formed by 3 points. if not zero they can form triangle.

2: Find line equation of 2 vertices and check if the 3rd vertices is present on this line. if not they can form triangle. Similarly Find the slope of the line joining points A and B,then find the slope of the line joining A and C.If they are same,then you can't draw a triangle.(@rah4927)

A triangle can be drawn with three points iff these three points are three different points.

EDIT: in order to be more precise in the core of my answer, and not in comment: three distincts colinear points form a degenerate triangle, but a triangle nonetheless. It is a geometric figure with three angles (they can be calculated, and not in the case where there are only one or two disctinct points) and their sum is 180°...


Is $\operatorname{span}\{(0,1)\}$ a point in $\mathbb{R}^2$? Is $S_1 = \operatorname{span}\{(0,1)\}$  a point in $\mathbb{R}^2$?
Does $0$ vector belong to $S_1$?

Is $S_2=\operatorname{span} \{(0,1),(0,2)\}$ a line in $\mathbb{R}^2$?
Does $0$ vector belong to $S_2$?

How to determine whether $0$ belongs to the span or not?

The definition of the span of a set $S$, denoted by $\operatorname{span}(S)$, is the set of all finite linear combinations of the elements of $S$.

In your case of $S_1$, we therefore have that $\operatorname{span}\{(0,1)\} = c \cdot (0,1) = (0,c)$ for scalars $c \in \mathbb{R}$. The scalar can be anything — $c = 1, c = 3.1$ or anything else as long as it is in $\mathbb{R}$. So yes, $0$ is in $S_1$ as we can choose $c = 0$ to get the $0$ vector. So $\operatorname{span}\{(0,1)\}$ is much more than just a single point. If you choose different values for $c$ and plot them you will see that you'll end up with a line going through the origin. As a matter of fact, it will be the $y$-axis.

Now for $S_2$, notice that $\operatorname{span}\{(0,1)\}$ already contains the point $(0,2)$ as if we were to choose $c = 2$ we get the point $2 \cdot (0,1) = (0,2)$. So the point $(0,2)$ in $S_2 = \operatorname{span}\{(0,1), (0,2)\}$ is redundant. However, lets formally prove that $S_1 = S_2$:

By definition, $S_2 = a \cdot(0,1) + b \cdot(0,2)$ for scalars $a,b \in \mathbb{R}$. Note that $$a\cdot(0,1) + b\cdot(0,2) =  (0,a) + (0,2b) = (0,a + 2b) = (a + 2b) \cdot (0,1).$$

So we have again that $S_2$ is the set of all linear combinations of $(0,1)$ and therefore, $S_1 = S_2$.

Can you see that again $0$ is in $S_2$? Can you see also that in general $0$ is always in the span of a set?

Hope this helps!

$S_1$ is a line, connecting the origin to the given point $[0,1]$, and $S_2$ is the same line, because it already contains $[0,2]$, too.

${\rm span\,}(v)=\{\lambda\,v\,\mid\,\lambda\in\Bbb R\}$.

So, $[0,0]$ and $[0,2]$ both are in ${\rm span\,}([0,1])$.


Swapping and handshakes with neighbours in a circle $N$ people are at a party and decide to play a cooperative game. They begin by standing in a circle. The game proceeds in turns. In each turn, one person is chosen to perform one of the following actions:

Shake hands with someone adjacent to them
Swap positions with someone adjacent to them

The game ends when every pair of players has shaken hands, and the aim of the game is to minimize the number of swaps required. What is the optimal strategy?

I've tried working through some small examples (e.g. $N = 3,4,5,6$ require $0,1,3,5$ swaps minimum but I can't really find a pattern in the strategy).

Suppose $\{a_1, a_2, ...,a_n\}$ people play this game. The pairs of neighbours are

$$ \{a_1,a_2\} $$
$$ \{a_2, a_3\} $$
$$ \vdots $$
$$ \{a_n, a_1\} $$

Which gives $n$ pairs of neighbours. Swapping with a neighbour perturbs the set by

$$ \{a_1, a_2, ... a_{n-k+1}, a_{n-k}, ..., a_n\} $$

So that, now, $a_{n-k+1}$ can shake with $a_{n-k}$ (which presumably they already have) and (more importantly) shake with $a_{n-k-1}$

In order to shake with everyone, the person originally  at location $n-k$ must migrate (by swapping) around the ring to position $n-k+3$ (and I assume this person shakes with everyone along the way).

We need a total of $\frac{n(n-1)}{2}$ handshakes, and each swap allows us to shake hands with 2 new people (one with the person who swapped, and one with the person who got swapped).

I will propose that $\left\lceil \frac{n(n-1)}{4} \right\rceil $ swaps will guarantee everyone has shaken hands, but I have not come up with a minimum expression.

Perhaps someone might take over.


Is there a general formula for solving 4th degree equations (quartic)? There is a general formula for solving quadratic equations, namely the Quadratic Formula:

$$ x = \frac{ -b \pm \sqrt{ b^2 - 4ac } }{ 2a } $$

For third degree equations of the form $ax^3+bx^2+cx+d=0$, there is a set of three equations: one for each root.
Is there a general formula for solving equations of the following form?
$$ ax^4 + bx^3 + cx^2 + dx + e = 0 $$
How about for higher degrees? If not, why not?

There is, in fact, a general formula for solving quartic (4th degree polynomial) equations.  As the cubic formula is significantly more complex than the quadratic formula, the quartic formula is significantly more complex than the cubic formula.  Wikipedia's article on quartic functions has a lengthy process by which to get the solutions, but does not give an explicit formula.

Beware that in the cubic and quartic formulas, depending on how the formula is expressed, the correctness of the answers likely depends on a particular choice of definition of principal roots for nonreal complex numbers and there are two different ways to define such a principal root.

There cannot be explicit algebraic formulas for the general solutions to higher-degree polynomials, but proving this requires mathematics beyond precalculus (it is typically proved with Galois Theory now, though it was originally proved with other methods).  This fact is known as the Abel-Ruffini theorem.

Also of note, Wolfram sells a poster that discusses the solvability of polynomial equations, focusing particularly on techniques to solve a quintic (5th degree polynomial) equation.  This poster gives explicit formulas for the solutions to quadratic, cubic, and quartic equations.

edit:  I believe that the formula given below gives the correct solutions for x to $ax^4+bx^3+cx^2+d+e=0$ for all complex a, b, c, d, and e, under the assumption that $w=\sqrt{z}$ is the complex number such that $w^2=z$ and $\arg(w)\in(-\frac{\pi}{2},\frac{\pi}{2}]$ and $w=\sqrt[3]{z}$ is the complex number such that $w^3=z$ and $\arg(w)\in(-\frac{\pi}{3},\frac{\pi}{3}]$ (these are typically how computer algebra systems and calculators define the principal roots).  Some intermediate parameters $p_k$ are defined to keep the formula simple and to help in keeping the choices of roots consistent.

Let: \begin{align*}
p_1&=2c^3-9bcd+27ad^2+27b^2e-72ace
\\\\
p_2&=p_1+\sqrt{-4(c^2-3bd+12ae)^3+p_1^2}
\\\\
p_3&=\frac{c^2-3bd+12ae}{3a\sqrt[3]{\frac{p_2}{2}}}+\frac{\sqrt[3]{\frac{p_2}{2}}}{3a}
\end{align*}
$\quad\quad\quad\quad$

\begin{align*}
p_4&=\sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+p_3}
\\\\
p_5&=\frac{b^2}{2a^2}-\frac{4c}{3a}-p_3
\\\\
p_6&=\frac{-\frac{b^3}{a^3}+\frac{4bc}{a^2}-\frac{8d}{a}}{4p_4}
\end{align*}

Then: $$\begin{align}
x&=-\frac{b}{4a}-\frac{p_4}{2}-\frac{\sqrt{p_5-p_6}}{2}
\\\\
\mathrm{or\ }x&=-\frac{b}{4a}-\frac{p_4}{2}+\frac{\sqrt{p_5-p_6}}{2}
\\\\
\mathrm{or\ }x&=-\frac{b}{4a}+\frac{p_4}{2}-\frac{\sqrt{p_5+p_6}}{2}
\\\\
\mathrm{or\ }x&=-\frac{b}{4a}+\frac{p_4}{2}+\frac{\sqrt{p_5+p_6}}{2}
\end{align}$$

(These came from having Mathematica explicitly solve the quartic, then seeing what common bits could be pulled from the horrifically-messy formula into parameters to make it readable/useable.)

If anyone is interested, there is a quartic equation calculator with a method and complete steps to solving the roots in the link here.


Showing $1+p$ is an element of order $p^{n-1}$ in $(\mathbb{Z}/p^n\mathbb{Z})^\times$ I'm trying to work through Dummit & Foote, but I've gotten stuck on the following question:

  Let $p$ be an odd prime and let $n$ be a positive integer. Use the
  binomial theorem to show that $(1+p)^{p^{n-1}} \equiv 1\bmod{p^n}$ but
  $(1+p)^{p^{n-2}} \not \equiv 1\bmod{p^n}$. Deduce that $1+p$ is an
  element of order $p^{n-1}$ in the multiplicative group
  $(\mathbb{Z}/p^n\mathbb{Z})^\times$.

The trouble I'm having is mostly with respect to the first implication, since I'm not completely confident with what I've done so far.  I started by letting $m = p^{n-1}$ and $a_i = {m\choose i}/m$. I then ended up with
\begin{align*}
(1+p)^m = \sum_{k=0}^{m} {m\choose k}p^k &= 1 + {m \choose 1}p + {m \choose 2}p^2 + \cdots + {m \choose m-1}p^{m-1}
+ p^m\\
&= 1 + mp + a_2mp^2 + a_3mp^3 + \cdots + a_{m-1}mp^{m-1}+ p^m\\
& = 1 + p^n + a_2p^{n+1} + a_3p^{n+2} + \cdots + p^m\\
&= 1 + p^n(1+ a_2p + a_3p^2+ \cdots + p^{m-n})\\
&\equiv 1 \bmod{p^n}
\end{align*}

...but based on some numerical tests I did, the $a_i$'s aren't necessarily integers, so this doesn't work. Would someone please point out what I'm missing here?

For the second implication, I can see a clean application of Euler's theorem starts things off. To finish, I'd like to show that no power of $(1+p)$ less than $(1+p)^{p^{n-1}}$ is congruent to $1\bmod{p^n}$, but I'm not sure how knowing the first part plays into this.

Any pointers would be appreciated.

It's easier to think of this as the result of repeated raising to the power $p$.  Start with a general number of the form $1+kp^r$ with $r\ge 1$ and $p \nmid k$ and see what you can deduce about $(1+kp^r)^p$.

For the second part, if an element $x \in (\mathbb Z/p^n \mathbb Z)^\times$ satisfies $x^m = 1$, what can you say about the order of $x$?  (You can say something quite a bit stronger than "it could be anywhere from $1$ to $m$".)

$(1+p)^{p^r}=1+\binom {p^r}1 p+\binom {p^r}2 p^2+\cdots+p^{p^r}$
$=1+p^{r+1}+p^{r+2}(\cdots)\equiv1+p^{r+1}\pmod{p^{r+2}}$ as $p^r\ge r+2$ for $p\ge3,r\ge 1$

$\implies (1+p)^{p^r}\not\equiv1\pmod{p^{r+2}}$

and  $(1+p)^{p^r}\equiv1\pmod{p^{r+1}}\implies (1+p)^{p^{r+1}}\equiv1\pmod{p^{r+2}}$

So, $ord_{p^{r+2}} (1+p)\mid p^{r+1}-->(1),$ but $ord_{p^{r+2}} (1+p)\not\mid p^r-->(2)$

$(1)\implies ord_{p^{r+2}} (1+p)$ must be $p^s$ where $s\le r+1$

$(2)\implies  s\not\le r$ or $s>r$

So, $(1),(2)$ together $\implies  s=r+1,$ i.r., $ ord_{p^{r+2}} (1+p)=p^{r+1}$

More generally, we can prove using Binomial Theorem,

if $ord_{p^s}a=d$ then $ord_{p^{s+1}}a=d, \space or \space pd--->(1)$

if $ord_{(p^s)}(a)=d$ and $ord_{p^{(s+1)}}(a)=pd,$ then $ord_{p^{(s+2)}}(a)=p^2d--->(2)$

Here, $1+p\equiv 1\pmod p,$ more generally $1+k\cdot p^r \equiv 1\pmod {p^r}$
if $p\not\mid k$

and $1+p\not\equiv 1\pmod{p^2},$ more generally $1+k\cdot p^r \not\equiv 1\pmod {p^{r+1}}$  if $p\not\mid k$

So, using $(1), ord_{p^{r+1}}(1+k\cdot p^r)=p\cdot 1=p$

So, using $(2), ord_{p^{r+2}}(1+k\cdot p^r)=p\cdot p=p^2$

Using $(2)$ repeatedly we can derive $ ord_{p^{r+s}}(1+k\cdot p^r)=p^s$


If a triangle's area equals its perimeter, what special condition must its angles satisfy? If we have a triangle where the Perimeter >0 and the Area >0 , and Area=Perimeter, what special condition must the angles of this triangle satisfy for this to happen?

I've done a bit of research and have found information on Heron's forumla and on integral triangles but I'm at a loss at how to figure this problem out.

Through research I have found that if the lengths of the sides of the triangle are either {6,8,10),(5,12,13),(9,10,17),(7,15,20), or(6,25,29) we get a triangle that fits the requirements (im not sure if this information is relevant) but the question is about the "special condition" that the angles must meet to create such a triangle not the sides.

I'm looking for both an answer and an explanation, any help is greatly appreciated. Please let me know if I left anything out (that's all the information I was given) or if something is unclear.

Hint: If you scale the triangle by a factor of $\alpha$, the perimeter is scaled by $\alpha$ and the area is scaled by $\alpha^2$.

Hence, given any triangle, there is a unique triangle that is similar to it, which satisfies perimeter = area.

This a Pythagorean triple treatment of area to perimeter ratios where $\angle AB$ is right and $A^2+B^2=C^2$.

We can find Pythagorean triples, if they exist, for any ratio $R$ of area/perimeter by finding the $m,n$(s) that represent them using the following formula which includes a defined finite search for values of $m$. Whenever the $m$ $R$ combination yields a positive integer for $n$, we have the $m,n$ for a triple.
We begin by solving the area/perimeter equation for $n$ where area=$D$ so as not to confuse it with $A,B,C$.

$$D=\frac{AB}{2}=\frac{(m^2-n^2)*2mn}{2}=mn(m^2-n^2)\quad P=(m^2-n^2)+2mn+(m^2+n^2)=2m^2+2mn$$

$$\frac{D}{P}=\frac{mn(m^2-n^2)}{2m^2+2mn}=\frac{mn(m-n)(m+n)}{2m(m+n)}=\frac{n(m-n)}{2}=R\qquad n^2-mn+2R=0$$

$$n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{m\pm\sqrt{m^2-4*1*2R}}{2*1}$$

$$n=\frac{m\pm \sqrt{m^2-8R}}{2}\text{ where }\lceil\sqrt{8R}\space \rceil \le m \le 2R+1$$

Example: For $R=0.5\quad \sqrt{8*0.5}=2\le m \le 2*0.5+1=2$

$$n=\frac{2\pm \sqrt{2^2-8*0.5}}{2}=1\qquad f(2,1)=(3,4,5)$$
But you are asking about $R=1$ and there only two such Pythagorean triples to be found.

$$R=1\rightarrow 3\le m \le 3\quad f(3,2)=(5,12,13)\quad f(3,1)=(8,6,10)$$
For other ratios:
$$R=1.5\rightarrow 4\le m \le 4\quad f(4,3)=(7,24,25)\quad f(4,1)=(15,8,17)$$

$$R=2\rightarrow 4\le m \le 5\quad f(4,2)=(12,16,20)\quad f(5,4)=(9,40,41)\quad f(5,1)=(24,10,26)$$

$$R=2.5\rightarrow 4\le m \le 6\quad f(6,5)=(11,60,61)\quad f(6,1)=(35,12,37)$$

$$R=3\rightarrow 4\le m \le 7\quad f(5,3)=(16,30,34)\quad f(5,2)=(21,20,29)\quad f(7,6)=(13,84,85)\quad f(7,1)=(48,14,50)$$

$$R=18\rightarrow 12\le m \le 37\quad f(12,6)=(108,144,180)\quad f(13,9)=(88,234,250)\quad f(13,4)=(153,104,185)\quad f(15,12)=(81,360,369)\quad f(15,3)=(216,90,234)\quad f(20,18)=(76,720,724)\quad f(20,2)=(396,80,404)\quad f(37,36)=(73,2664,2665)\quad f(37,1)=(1368,74,1370)$$

Aside: you can always find one primitive triple for a given $R$ if you let $(m,n)=(2R+1,2R)$.


A closed form for $\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx$? I would like some help to find a closed form for the following integral:$$\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx $$ I was told it could be calculated in a closed form. I've already proved that $$\int_0^1 \frac{\log (1+x)}{x}dx  = \frac{\pi^2}{12}$$ using power series expansion.

Thank you.

do you know this results ?

$\int_0^1 \frac{\left(\log (1+x)\right)^2}{x}dx=(ζ(3))/4$


How to prove that $(z^{-1})^{-1} = z$ and $(zw)^{-1} = z^{-1}w^{-1}$? I need to prove that $(z^{-1})^{-1} = z$ and $(zw)^{-1} = z^{-1}w^{-1}$ but the only thing I can think about is to consider

$$z = a+bi, w = c+di$$

and then prove it algebraically using laws of multiplication for complex numbers. Any ideas to prove it structuraly?

You know that $\mathbb{C}\setminus\left\{0\right\},\cdot$ is a group. Hence the inverse of an element is unique. Since $zz^{-1}=1=z^{-1}z$, we conclude that $z$ is the inverse of $z^{-1}$, this element is denoted by $(z^{-1})^{-1}$. Hence $(z^{-1})^{-1}=z$.

In the same fashion we can conclude that $(zw)^{-1}=w^{-1}z^{-1}$.

Generally, $(z^a)^b=z^{ab}$, so $(z^{-1})^{-1}=z$.

We have $(zw)(zw)^{-1}=1.$ Pre-multiply by $z^{-1}$ and then $w^{-1}$ to get the result.


Run away from lions in a cage I came across an interesting problem:

  There is a round cage and you are in it. Also two lions are in this cage
  too. The start position is that the distance between you and both lions is
  the diameter of the circle (you are on opposite sides of the cage).
  The speed of the lion is 1 m/s.

And the question is:

  What is the minimal speed you need to have to always run away from lions and never be caught.

Probably it is a bit simpler to search the maximal speed when the lions will catch you. And the result of the original task will be the upper limit of that value.

I think the radius of the cage doesn't matter - it is only a scale problem. The only important thing is that the cage is round.

There is a similar problem here for one lion. But the answer links to buy some book and I couldn't find where to download it for free =).

And also I wonder if there is a solution for the generalized task with $N$ lions. That looks too complicated but I think the idea is the same - the lions should build a line when you can't run between any two of them and two lions on the ends of a chain will behave like the ones in the two-lions problem.

It seems so that this is related to the optimization theory problem The Lion and Man and is in general a pursuit problem.

Peter

First, i don't get why two lions are there. But supose they will move always directly to you. If you would move same speed as them around the cage, they would be coming closer and closer to you until they would finnaly cacht you. Or not? I thing they would be still-closing but never able to catch you, since it will end up that they will move 0.0000001mm just behind you. So the conclusion is that you need to move any larger speed than theirs.


Trying to find what proof works best with this question? If  $ (n+1)^2   $ is even then   $ n $    is odd
The work part isn't the issue for me it's finding what is the best and most simple proof to use on this question
My three options are Direct proof, Indirect proof(Contrapositive) or proof by Contradiction
I know n is odd if $N=2k+1$
N is even if $n=2k$
I want to say solving it by Contrapositive is the best method because you flip n to Even which will make it easier

I would approach this by proving the contrapostive, as you suggest.  The contrapostive of the statement, which is equivalent to the statement, is:

If $n$ is even, then $(n+1)^2$ is odd.

Assume $n$ is even.  So $n= 2k$, for some integer $k$.  Then $(n+1)^2 = n^2+ 2n + 1 =  (2k)^2 + 2(2k) + 1 =  4k^2 + 4k + 1= 2(2k^2+ 2k) +1 = 2m+1$,  which is an odd integer, (noting that we can represent $m = 2k^2+2k$, another integer, given that $k$ is an integer.)

Direct requires proving that if $(n+1)^2$ is even then $n+1$ is even.  And the only real way I see to do that is be contrapositive.  If $n+1=2k$ is even then $(n+1)^2 = 4k^2$ is even but if $(n+1) =2k + 1$ is odd then $(n+1)^2=4k^2 + 4k + 1= 2(2k^2 + 2k) + 1$ is odd.  Or by claiming by Euclid's Lemma that $2$ is prime so if $2|(n+1)^2$ then $2|n+1$ and $n+1$ is even.  Then once we know $n+1$ is even then $n+1 = 2k$ so $n = 2k-1$ is odd.
That's not hard but it's more work.
COntrapositive is fairly.... er, dirct.....  If $n=2k$ is even then $n+1 = 2k+1$ is odd, and $(n+1)^2 = 4k^2+4k + 1$ is odd.  So contrapositively if $(n+1)^2$ is not odd then $n$ is not even.  Done.
Proof by contradiction sets up unnecessary conditions:
Suppose $(n+1)^2$ even and $n$ is even.  Then $n+1$ is odd and $(n+1)^2$ is odd so that's a contradcition.
Fine, but why did we need to suppose $(n+1)^2$ was even.
I'd go with contrapositive.
But they are all relatively easy and straightforward.


(Product) Lebesgue measure on infinite dimensional spaces? I am trying to understand measure construction procedures on infinite-dimensional spaces. Why is it not possible in general to construct Lebesgue measure on $\mathbb{R}^\mathbb{N}$ or $\mathbb{R}^\mathbb{R}$?

I am not quite sure what you mean. But the construction of Lebesgue measure on $X$ I learnt is by using the Riesz representation theorem to identify Lebesgue measure as one continuous linear functional over $C(X)$. But this requires the $X$ to be locally compact, and if $X$ is a vector space, the only locally compact ones are $\mathbb{R}^n$.

Feynman has done something like this when he defines path integral. It is infinite dimensional measure.

$\int_{\mathbb{R}^\mathbb{N}}d^{\infty}x \mathcal{Dx}e^{i\mathcal{S}[x]}$

Is such beautiful formula but many mathematician have problem justifying? Answer: Is not infinite dimensional measure.


Error in the proof for Let $A$ and $B$ be sets, then $A^{\complement}\cup B^{\complement}\subseteq (A \cap B)^{\complement}$ I have a proof that I know has a fundamental error in it, but I have no idea where the error is. Here is the proof:

  Proof: Suppose that $x \in A^{\complement} \cup B^{\complement}$. This implies that $x \in A^{\complement}$ or $x \in B^{\complement}$. This disjunction will be true if $x \in A^{\complement}$. In this case, we have that $x \notin A$. But if $x \notin A$ then $x \notin A \cap B$, since $A \cap B$ is the set of all elements that are in both $A$ and $B$. And if $x \notin A \cap B$, then $x \in (A \cap B)^{\complement}$. Therefore, $A^{\complement}\cup B^{\complement}\subseteq (A \cap B)^{\complement}$ as desired.

There is no fundamental error in it.

You proved legally that $x\in A^{\complement}\implies x\in(A\cap B)^{\complement}$ so that $A^{\complement}\subseteq(A\cap B)^{\complement}$.

What lacks somehow is a statement as:

"...likewise we prove that $B^{\complement}\subseteq(A\cap B)^{\complement}$ so that $A^{\complement}\cup B^{\complement}\subseteq(A\cap B)^{\complement}$".

You have not taken into account the fact that you can have both $x\in  A^{\complement}\cup B^{\complement}$ AND $x\in A$...

Since $ A^{\complement} \cup B^{\complement}$ is the addition of $A^{\complement}$ and $B^{\complement}$, so you can add back elements that were excluded from $A$ but are in $B^{\complement}$.


Matrix of arguments and amplitudes of cosinus with software (MatLab, Maple, Mathematica...) My problem is that, I have a big matrix with $6\times 6$ elements, with each term of the matrix is a big function of sum of cosines where I need to work only with the arguments and the amplitudes separately.

Simplifying the task is, I have a function like:

$$f(\phi) = A_{1} \cos \left(\phi -\frac{2 \pi }{3}\right)+
A_{2} \cos \left(2 \phi +\frac{\pi }{3}\right)+
A_{3} \cos \left(3 \phi -\frac{\pi }{2}\right);$$

and with any software like Matlab, Maple, Mathematica or any software. Input the matrix $f(\phi)$ and have the output a vector of amplitudes and a vector of arguments:

$$A=\left[ \begin{array}{ccc} A1 & A2 & A3 \end{array} \right] \text{and}
\arg= \left[ \begin{array}{ccc} \phi -\frac{2 \pi }{3} &  2 \phi +\frac{\pi }{3} & 3 \phi -\frac{\pi }{2} \end{array} \right]$$

I have tried with Regexp of MatLab but it is not possible to choose only the argument of the cosine.

If anyone can help me please,
Thank you very much

Mathematica handles automatically provides these reductions
$$
\begin{align}
%
  \cos \left(\phi -\frac{2 \pi }{3}\right) &= -\sin \left(\frac{\pi }{6}-\phi \right) \\
%
\cos \left(2 \phi +\frac{\pi }{3}\right) &= \sin \left(\frac{\pi }{6}-2 \phi \right) \\
%
 \cos \left(3 \phi -\frac{\pi }{2}\right) &= \sin (3 \phi )
%
\end{align}
$$

The TrigReduce expression will yield
$$
\text{TrigReduce}\left[A_{1} \cos \left(\phi -\frac{2 \pi }{3}\right)+A_{2} \cos \left(2 \phi +\frac{\pi }{3}\right)+A_{3} \cos \left(3 \phi -\frac{\pi }{2}\right)\right] =
-A_{1} \sin \left(\frac{\pi }{6}-\phi \right)+A_{2} \sin \left(\frac{\pi }{6}-2 \phi \right)+A_{3} \sin (3 \phi )
$$

Using TrigExpand
$$
\text{TrigExpand}\left[A_{1} \cos \left(\phi -\frac{2 \pi }{3}\right)+A_{2} \cos \left(2 \phi +\frac{\pi }{3}\right)+A_{3} \cos \left(3 \phi -\frac{\pi }{2}\right)\right] = \frac{1}{2} \sqrt{3} A_{1} \sin \phi -\frac{1}{2} A_{1} \cos \phi -\frac{1}{2} A_{2} \sin ^2(\phi )+\frac{1}{2} A_{2} \cos^{2}\phi -\sqrt{3} A_{2} \sin \phi  \cos \phi -A_{3} \sin ^3(\phi )+3 A_{3} \sin \phi  \cos^{2}\phi
$$

Using Simplify or FullSimplify:
$$
\text{Simplify}\left[A_{1} \cos \left(\phi -\frac{2 \pi }{3}\right)+A_{2} \cos \left(2 \phi +\frac{\pi }{3}\right)+A_{3} \cos \left(3 \phi -\frac{\pi }{2}\right)\right]
=
-A_{1} \sin \left(\frac{1}{6} (\pi -6 \phi )\right)+A_{2} \cos \left(\frac{1}{3} (6 \phi +\pi )\right)+A_{3} \sin (3 \phi )
$$

TrigFactor
$$
\text{TrigFactor}\left[A_{1} \cos \left(\phi -\frac{2 \pi }{3}\right)+A_{2} \cos \left(2 \phi +\frac{\pi }{3}\right)+A_{3} \cos \left(3 \phi -\frac{\pi }{2}\right)\right]
=
-\frac{1}{2} \sqrt[3]{-1} \left(\left((-1)^{5/6} A_{1}+i A_{1}\right) \sin (\phi )+\left(A_{1}-\sqrt[3]{-1} A_{1}\right) \cos (\phi )+\left(-(-1)^{5/6} A_{2}-i A_{2}\right) \sin (2 \phi )+\left(\sqrt[3]{-1} A_{2}-A_{2}\right) \cos (2 \phi )+2 (-1)^{2/3} A_{3} \sin (3 \phi )\right)
$$

The trick is to decide which of these forms to work with and use FullForm. For example
$$
\text{FullForm}\left[A_{1} Cos\left[\phi - \frac{2}{3} \pi\right]\right]
=
\text{Times}[\underbrace{A_{1}}_{amplitude},\text{Cos}[\underbrace{\text{Plus}[\phi ,\text{Times}[-1,\text{Times}[\text{Times}[2,\text{Power}[3,-1]],\text{Pi}]]]}_{angle}]]
$$
Extract

This problem seems better suited for the crowd at [Mathematica Stack Exchange][2]

Solution


Why is $\infty \cdot 0$ not clearly equal to $0$? I did a bit of math at school and it seems like an easy one - what am I missing?

$$n\times m = \underbrace{n+n+\cdots +n}_{m\text{ times}}$$

$$\quad n\times 0 = \underbrace{0 + 0 + \cdots+ 0}_{n\text{ times}} = 0$$

(i.e add $0$ to $0$ as many times as you like, result is $0$)

So I thought an infinite number of $0$'s cannot be anything but $0$? But someone claims different but couldn't offer a reasonable explanation why. Google results seemed a bit iffy on the subject - hopefully this question will change that.

The problem is that the laws of addition and multiplication you are using hold for natural numbers, but infinity is not a natural number, so these laws do not apply. If they did, you could use a similar argument that multiplying anything by infinity, no matter how small, gives infinity, thus $\infty \times 0 = \infty$. More sophisticated arguments can also be made, like $\infty \times 0 = \lim_{x \to \infty} (x \times 1/x) = 1$. Clearly all these different values for $\infty \times 0$ mean that $\infty$ cannot be treated like other numbers.

In order to work with infinity, you must first define it. You may think you know what infinity is, but really you don't have a concrete definition. In fact, there are many different definitions of infinity that you could use, each of which result in different behaviors. For example, the real projective line has a concept of infinity such that $1/\infty = 0$, while when talking about infinite sets one uses cardinal numbers (another type of infinity) to represent the sizes of these sets. You must make it clear what infinity you are talking about in order to work with it.

In summary, the expression $\infty \times 0$ using multiplication defined for the natural numbers does not have any meaning, so it cannot be said to be equal to $0$.

$$
\underbrace{\frac 1 n + \cdots + \frac 1 n}_\text{$5n$ terms} = 5
$$
As $n$ approaches $\infty$, each of these terms approaches $0$, but their sum remains equal to $5$, so its limit is $5$.


graphing and functions one of my problem says that the average slope formula is: $\dfrac{f(x_2)-f(x_1)}{x_2-x_1}$
It tells us the average slope over the interval from $x_1$ to $x_2$
Then it says to take the average slope of the function $x^2+3x+1$ over one unit intervals from $0$ to $10$.

can i use a particular function function to solve this problem?

As André said, what they want is the 10 numbers
$$\frac{f(1)-f(0)}{1-0}, \frac{f(2)-f(1)}{2-1}, \ldots, \frac{f(10)-f(9)}{10-9}$$
-- that is, with $x_1$ increasing from $0$ to $9$ while $x_2$ increases from $1$ to $10$. Then plot them against the $x_1$ values for each of them, and look for a pattern.

$f(x)=x^2+3x+1, x_1=0, x_2=10$
so the average slope on the interval [0,10] is $\frac{f(10)-f(0)}{10-0}=\frac{131-1}{10}=13$.

If you want the average slope over all 1 unit intervals $[a, a+1]$ for $0\le a\le 9$ then this is the symbolic expression $\frac{f(a+1)-f(a)}{a+1-a}$ which simplifies to $2a+4$.

For example with $a=4.5$ we also get average slope of 13, but for $a=9$ we get average slope of 22.


Characteristic Function of a Conditioned Random Varaible Let $(X_j)$ be iid random variables and $N \sim \mathrm{Poisson}(\lambda)$, independent of $X_j\, \forall j$. Define $S_n:=\sum_j^n\,X_j$ and consider $S_N$.

Find the characteristic function of $S_N$. Then find $\mathbb{E}S_N$ and $\operatorname{Var}(S_N)$.

I can find $\mathbb{E}S_N$ and $\operatorname{Var}(S_N)$ from the first and second moments, once I have the characteristic function. The last bit is what is tripping me up. Here is what I have so far:

\begin{align*}
\varphi_{S_N}(t) &= \varphi_{\sum_j^N\,X_j}(t)\\
  &= \mathbb{E}e^{it\sum_j^N\,X_j}\\
  &= \mathbb{E}e^{it\sum_j^N\,X_j}\sum_j^N\mathbb{1}_{\{N=n\}}\\
  &= \sum_{n=0}^\infty\,\mathbb{E}e^{it\sum_{j-1}^N\,X_j}\mathbb{P}(N=n)\\
  &= \sum_{n=0}^\infty\,(\varphi_{X_1}(t))^n\lambda^n\frac{e^{-\lambda}}{n!}\\
  &= e^{-\lambda}\sum_{n=0}^\infty\,(\varphi_{X_1}(t))^n\frac{\lambda^n}{n!}\\
  &= e^{-\lambda} + \sum_{n=1}^\infty\,(\varphi_{X_1}(t))^n\frac{\lambda^n}{n!}
\end{align*}

Have I made an error, or gone off in the wrong direction? Thanks in advance.

I think earlier in your calculation some things got confusing. You should just find:

$$\phi_S(t)=E[e^{it \sum_{j=1}^N X_j}]=\sum_{n=0}^\infty E \left [e^{it \sum_{j=1}^N X_j} \mid N=n \right ] P[N=n] \\
= \sum_{n=0}^\infty E \left [ e^{it \sum_{j=1}^n X_j} \right ] P[N=n] \\
= \sum_{n=0}^\infty \phi_X(t)^n P[N=n]$$

using the total expectation formula and the fact that $X_j$ are iid.

Now the point is to make the result look more familiar:

$$\phi_S(t)=\sum_{n=0}^\infty \frac{e^{-\lambda} \phi_X(t)^n \lambda^n}{n!} = e^{-\lambda} \sum_{n=0}^\infty \frac{(\phi_X(t) \lambda)^n}{n!}.$$

You should recognize sums like that one.

You derailed on the third line.   Backtrack and try this:

$$\begin{align}
\varphi_{S_N}(t) = &~ \mathsf E(\mathscr e^{\imath t S_N})
\\[1ex]=&~ \mathsf E(\mathscr e^{\imath t \sum_{k=1}^N X_k})
\\[1ex]=&~ \mathsf E(\prod_{k=1}^N\mathscr e^{\imath t X_k}) & \because~e^{a+b+c+\ldots}=e^a\cdot e^b\cdot e^c\cdots
\\[1ex]=&~ \mathsf E(\mathsf E(\prod_{k=1}^N \mathscr e^{\imath t X_k}\mid N)) & \textsf{Law of Iterated Expectation}
\end{align}$$

Then just apply independence and identicallity of your random variables and everything should fall into place.


What is a condition for two real functions $f,g$ to "commute", so $f(g(x))=g(f(x))$? Say I'm given two functions $f,g$. Can I tell if they "commute" without actually trying them in the formula
$f(g(x))=g(f(x))$?
And given a function $f$, is there a way to find all functions $g$ such that
$f(g(x))=g(f(x))$?
I've tried using derivatives and the chain rule but haven't got anything interesting yet.

There is, in general, no characterization other than the evaluation of $f(g(x))$ and $g(f(x))$.
Also, for fixed $f$ there is no an easy way to find all the $g$ such that $f \circ g = g \circ f$.
Special cases can be studied. For example, a theorem of J. Lipiński says that if $f$ is a continuous and strictly monotonic function on an interval $I$ such that $f(I) \subseteq I$ then there is an infinity of solutions $g$ (depending on an arbitrary function).

If $f(x) = g^{-1}(x)$, then $f(g(x)) = g(f(x))$ but the converse is not true.
If $f(x) = g(x)$, then $f(g(x)) = g(f(x))$.


Best Intermediate/Advanced Computer Science book I'm very interested in Computer Science (computational complexity, etc.). I've already finished a University course in the subject (using Sipser's "Introduction to the Theory of Computation").

I know the basics, i.e. Turing Machines, Computability (Halting problem and related reductions), Complexity classes (time and space, P/NP, L/NL, a little about BPP).

Now, I'm looking for a good book to learn about some more advanced concepts. Any ideas?

The Art of Computer Programming
(Donald Knuth)
The legendary book (of multiple volumes, still incomplete) can't go without mention. For learning about algorithms and their complexities, there is no rival. It's written with practicality in mind, though from a largely theoretical perspective.

(source: wikimedia.org)

The Mythical Man-Month more for software engineering but absolutely one of the best books ever written about programing IMO.


4 cards are shuffled and placed face down. Hidden faces display 4 elements: earth, wind, fire, water. You turn over cards until win or lose. Question: 4 cards are shuffled and placed face down in front of you. Their hidden faces display 4 elements: water, earth, wind, fire. You turn over cards until win or lose. You win if you turn over water and earth. You lose if you turn over fire. What is the probability that you win?

I understand that wind is effectively absent from the sample space. Does not affect your chances of winning or losing. I also know that 1/3 (because we removed wind), you can pick fire where you lose the game.

Since there are 4 cards, there are 4! = 24 possible orderings of the cards.

You lose in the following cases:

1) Fire is the first card. There are 3! = 6 ways for this to occur.
2) Wind is the first card and Fire is the second. There are 2! = 2 ways for this to occur.

Hence, the probablity of losing = $\frac{6+2}{24}=\frac{1}{3}$
So your answer is correct.

Probablity of losing is not 1/3. It should be 2/3.

As Below,
He wins if,

First three cards are Water/Earth/Wind = 3!
First two cards are Water/Earth and third is Fire = 2

so the total = 6 + 2 = 6

Probability of winning  = 8/24 = 1/3 , so loosing 2/3.


3-connected multigraph and parallel edge In Diestel book, 5th edition, page 62, he says in the footnote "Recall also that 3-connected multigraphs cannot have multiple edges". Also, in the proof for Lemma 3.2.1, he uses "the fact" that "$G\,\dot{-}\,e$ graph has no parallel edges."

Isn't it obviously false? I can start with any 3-connected graph and start adding arbitrary edges and parallel edges and it will still remain 3-connected. What am I missing here?

As is often the case, the problem is cleared up when we see the part of the context that you didn't tell us about. Immediately before the sentence that you quote, in that same footnote, he suggests to “See Chapter $1.10$ for the formal definition of suppressing vertices in a multigraph”. Chapter $1.10$ contains the following explanation:

  A few differences, however, should be pointed out. [...] The ends of loops and parallel edges in a multigraph $G$ are considered as separating that edge from the rest of $G$. [...] Thus, a multigraph with a loop is never $2$-connected, and any $3$-connected multigraph is in fact a graph.


Hypothesis test (p-value) Suppose I conduct a hypothesis test and $H_0: \mu = 7$ and $H_A: \mu \neq 7$. Let's say the $p$-value is 0.03. Would a 95% confidence interval contain the value $7$? I think it would since alpha (which is 0.05) is higher than the $p$-value.

However, the question that is really bothering me is would I have enough information from this hypothesis test (if the only information we have is the $p$-value and hypotheses) to conclude if the value $6$ is contained in this 95% confidence interval?

I don't think I would since I have no information on the sample mean and variance.

This appears to be a two-sided test.  Since the $p$-value $0.03$ is the probability of an observation as extreme as or more extreme than the observed sample mean $\hat{\mu}$, then assuming normal distributions and large samples, $\hat{\mu} = 7 \pm 2.17\sigma$, that is,
the observed sample mean $\hat{\mu}$ is larger (or smaller) than the mean $7$ under the null hypothesis by $2.17$ standard deviations.  Note that $P\{Z > 2.17\}\approx 0.015$ for a standard normal random variable $Z$.  But, under the assumption that the data is a large sample of independent identically distributed normal random variables, the sample mean is normal with standard deviation $\sigma$, its observed value $\hat{\mu}$ is the
maximim-likelihood estimator
for the mean, and the $95\%$ confidence interval for the
mean is the interval $(\hat{\mu} - 1.96\sigma,  \hat{\mu} + 1.96\sigma)$ and does not include $7 = \hat{\mu} \pm 2.17\sigma$.

Note that this answer makes several assumptions that might not be applicable in this problem.  Other assumptions e.g. small samples, non-normal data, different method
for construction of the confidence interval, disbelief in frequentist hypothesis
testing and confidence intervals, etc. can lead to different conclusions as to
whether the confidence interval contains $7$ or not.


How I can prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}$ converges to 2? Prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}},  \sqrt{2\sqrt{2\sqrt{2}}} \ $ converges to $2$.

My attempt

I proved that the sequence is increasing and bounded by $2$, can anyone help me show that the sequence converges to $2$?
Thanks for your help.

We have

$$\sqrt{2}=2^{\frac{1}{2}}.$$
Exciting, no?
We also have
$$\sqrt{2\sqrt{2}}=2^{\frac{1}{2}+\frac{1}{4}},$$
and
$$\sqrt{2\sqrt{2\sqrt{2}}}=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}.$$
For the next term, we take the previous term, multiply by $2$, getting exponent of $2$ equal to $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$, then take the square root, getting exponent of $2$ equal to $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}$.
The pattern continues, since the "next term" is always obtained by the same process.

It is well-known that the series $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$ converges to $1$. Since the function $f(x)=2^x$ is continuous, our original sequence converges to $2^1$.

You can also observe that

$$\sqrt{2\sqrt{2\sqrt{2 ...\sqrt{2}}}} \cdot \sqrt{\sqrt{\sqrt{...\sqrt{2}}}}=2$$

thus

$$\sqrt{2\sqrt{2\sqrt{2 ...\sqrt{2}}}} =\frac{2}{\sqrt[2^n]{2}}$$


Taking modulo of both sides On the following page at the bottom there is an algorithm for calculating modular inverses. In the proof I am confused with the line 'Taking both sides modulo $m$'. How does that work getting a congruence from the above equation?

$j = k\,\Rightarrow\, \color{#0a0}{j\bmod m} = k\bmod m\,$ by $\,f(x) = x\bmod m\,$ is a function so  $\,j=k\,\Rightarrow\, f(j) = f(k)$.

Your special case has $\, \color{#c00}0 \le \underbrace{m\bmod i}_{\Large\color{#0a0} j} < i < \color{#c00}m\,$ so $\,\color{#0a0}{j\bmod m = j}\,$ is already reduced $\!\bmod \color{#c00}m$

Remark $ $ Note that the modular inversion algorithm discussed there is the same as the algorithm I discussed in your prior question, i.e. Gauss's modular inversion algorithm (in non-fractional form).

If $x=y$ then $x \equiv y (mod m)$, so far, so good.

Now, let's say we have what I think you may be struggling with $ax + bm = y$

Then $ax + bm \equiv y\space (mod m)$,

but $bm \equiv 0 \space (mod m)$, so that's why we get $ax \equiv y \space (mod m)$ To get there, you may want to subtract $bm$ from both sides


Find an angle in a given triangle $\triangle ABC$ has sides $AC = BC$ and $\angle ACB = 96^\circ$. $D$ is a point in $\triangle ABC$ such that $\angle DAB = 18^\circ$ and $\angle DBA = 30^\circ$. What is the measure (in degrees) of $\angle ACD$?

In $\triangle ADB,\angle ADB=(180-18-30)^\circ=132^\circ$

Applying sine law in $\triangle ADB,$ $$\frac{AB}{\sin 132^\circ}=\frac{AD}{\sin30^\circ}\implies AD=\frac{AB}{2\sin48^\circ}$$ as $\sin132^\circ=\sin(180-132)^\circ=\sin48^\circ$

$\angle ABC=\angle BAC=\frac{180^\circ-96^\circ}2=42^\circ$

Applying sine law in $\triangle ABC,$ $$\frac{AC}{\sin 42^\circ}=\frac{AB}{\sin 96^\circ}\implies \frac{AC}{AB}=\frac{\sin 42^\circ}{\sin 96^\circ}=\frac{\cos 48^\circ}{2\sin 48^\circ \cos 48^\circ}$$ (applying $\sin 2A=2\sin A\cos A$)

So, $$AC=\frac{AB}{2\sin48^\circ} \implies AC=AD$$

So, $\angle ACD=\angle ADC=\frac{(180-24)^\circ}2=78^\circ$

Hint:

Have you drawn your picture?

I drew the picture out and I got an isosceles triangle, with base angles being ∠A and ∠B. Then I placed point D inside the triangle and connected that point to the 3 vertices of the triangle.

The first statement tells us that we are in an isosceles triangle ( meaning base angles are the congruent ). Therefore we can find the measure of the base angles since we know that ∠ACB=96∘. Each base angle turns out to be 42∘ each.

Consider triangle ABD. Can you find all angles of that triangle? And can you then find the angles of triangle BDC? and ADC?

I hope this helped! If you are still stuck let us know!


$p$ is a $3$-digit prime, then there always exist $p$ consecutive composite numbers. True/False
If $p$ is a $3$-digit prime, then there always exist $p$ consecutive composite numbers.

How to approach this?

$P_k =(s + 1)! + k$ for $k = 2$ to $(s + 1)$

are $s$ consecutive positive integers

as $P_k$ is always divisible by $k$, hence the statement is true

The statement is false!
The first 3 digit prime is 101, the next prime is 103; difference is 2 not 101!
Take another 3 digit prime, 131 the next prime is 137; difference is 6 not 131!
You can look up any of the 3 digit primes and by inspection the difference is well under 100.
  Ron


If a triangle $PQR$ is constructed such that All sides are integers ; and , $PQ= 37 , QR= M$ , and $M <37$ , then the possible values of $PR$ : If a triangle $PQR$ is constructed such that All sides are integers ; and , $PQ= 37  , QR= M$  , and  $M <37$ , then the possible values of $PR$  :

A) $2M-2$

B)$2M-1$

C)$2M$

D)$2M+1$

I tried to apply triangle inequality

$37-M <PR <M+37$
$1 <PR <2M+1$         since Max (m)=$36$

But i can't reject except  $D$  ; what should I do?

Your question is confused, but from $PR<M+37$ you can't conclude $PR<2M+1$, you can only say $PR<max(M)+37$


Conditional expectation $E(X | X<Y)$ when $X$ and $Y$ are independent I'm stuck calculating the following expectation:

  Given a (discrete) RV $X$ with $P(X=0)=P(X=1)=P(X=2)=1/3$ and $Y \sim \text{Unif}(0,2)$ what is $E(X | X<Y)$?

Now what I did was conditioning on $Y$, i.e.
\begin{align}
E(X | X<Y)
&=\int_0^2\frac{1}{2}E(X | X<y)dy \\
&=\int_0^1\frac{1}{2}E(X | X<y)dy+\int_1^2\frac{1}{2}E(X | X<y)dy \\
&=\frac{1}{2}\cdot 0 + \frac{1}{2} \cdot 1=\frac{1}{2}.
\end{align}
To confirm this I simulated the situation with R. I simulated an instance of $X$ and $Y$ 10,000 times and if $X < Y$ I remembered $X$. However the mean of this list was $\frac{1}{3}$ and not $\frac{1}{2}$ as expected.

Can someone tell me what I did wrong? Thanks!

The answer is $1/3$.  Given that $X<Y$, there are three possible situations, depending on the value of $X$, but they're not all equally likely:
$X=0, Y>0$ (probability $2/3$); $X=1,Y>1$ (probability $1/3$), and $X=2,Y>2$ (probability $0$).  Thus $E[X | X<Y] = 0(2/3) + 1(1/3) + 2(0) = 1/3$.

Another way to look at the problem (and to check your result)  is to approach from this angle: $$E(X|X<Y)=\sum_{i=0}^2 i \cdot P(X=i \ | \ Y>X)$$
and $$P(Y>X) = \sum_{i=0}^2 P(Y > X \ | \ X=i) \cdot P(X=i) $$

I took advantage of this approach and determined that the expected value is in fact $\frac{1}{3}$


Fundamental Theorem of Trigonometry This is a pretty open ended question and I apologize, in advance, if this is not the place for it.  But what do you recommend should be given the title of the Fundamental Theorem of Trigonometry and why?  Should we have to restrict ourselves to the planar case... I think so.

$$\frac{\sin x}{\cos x} = \tan x$$
$$sin^2 x + cos^2x = 1$$