IFRAME SYNC
September 2019
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19:48

$AH$ is a altitude of $\triangle ABC$, $AD$ is the bisector of $\angle CAB$ and $BC$ meets the incircle of $\triangle ABC$ at point $E$ and $(D \in (ABC), H \in BC)$. Let the intersections between $DE$, $DH$ and $(ABC)$ are respectively $F$, $L$ $(L \not\equiv D \not\equiv F)$. $AF$ and $FL$ cut $BC$ correspondingly at $K$ and $J$. Prove that the tangent of $(ABC)$ at point $F$ passes through the midpoint of $KJ$.

enter image description here

Let the incenter of $\triangle ABC$ be $I$.

We would have that $IF \perp AK$ and $H \in (AIK)$

$\implies \angle KFI = \angle KIA = 90^\circ \implies \angle FAD = \angle FIK$

But that's all.



from Hot Weekly Questions - Mathematics Stack Exchange
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03:06

The paper “On the crest factor for dissipative partial differential equations“, by Michele Bartuccelli, has been published in Proceedings A of the Royal Society of London. In the paper Michele proves new results on the “crest factor” (defined in the figure left), and argues that it is a fundamental criterion for discerning between large and small space scales, and hence a useful diagnostic for turbulence. A link to the published paper is here.



from Surrey Mathematics Research Blog https://ift.tt/2oycipm
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23:25

Senior at a commuter school here, and I have a test for Real Analysis I in about 14 hours. I genuinely feel such a strong hatred for math at the moment, and maybe i'm just being dramatic, but I really feel as if I haven't learned anything since my intro to proof class. I used to feel so passionate about mathematics, but even when I do understand a proof (which is actually quite often lately) the most it draws out of me is a "that's kinda cool i guess". Maybe it's an academic culture type of thing that I don't mesh well with, but am I just feeling like that because i'm an idiot, or is the whole "rigor or die" thing real because at this point, i don't even feel like i'm struggling productively like i did in Intro to Proofs & Linear Algebra, but just for the sake of struggling. Sorry if this is against the rules, I will delete if required.

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23:23

Rational Limit Theorem.

For $f(x, y) =\frac{|x|^a|y|^b}{|x|^c+|y|^d}$, with $a, b, c, d$ positive,

$$\lim_{(x,y)→(0,0)}f(x, y) \text{ exists and equals zero}\Leftrightarrow \frac{a}{c}+\frac{b}{d}>1.$$

We will break this down into three parts: first proving one direction by our techniques to show limits don’t exist, then using a famous inequality to help prove the other direction. As a guideline, each proof can be written in two or three lines.

$1.$ Show that if $\frac{a}{c}+\frac{b}{d}≤1$, then the limit does not exist.

For the next two problems, you are free to use the following inequality:

Young’s Theorem. For positive real numbers $w$, $z$ and any $0≤t≤1$,

$$w^tz^{1−t}≤tw+ (1−t)z$$

$2.$ Show that if $\frac{a}{c}+\frac{b}{d}= 1$, where $a, b, c, d$ are all positive, then $f(x, y)≤1$ for all $(x, y)∈\mathbb{R^2}\backslash\{(0,0)\}$.

$3.$ Show that if $\frac{a}{c}+\frac{b}{d}>1$, then $$\lim_{(x,y)→(0,0)}f(x, y) = 0.$$

Before I start to prove this, i'm thinking why $1-3$ together implies

$$\forall a,b,c,d>0,\lim_{(x,y)→(0,0)}\frac{|x|^a|y|^b}{|x|^c+|y|^d} \text{ exists and equals zero}\Leftrightarrow \frac{a}{c}+\frac{b}{d}>1$$

"$1.$" looks like the contrapositive of direction "$\Rightarrow$", actually, "$1.$" implies "$\Rightarrow$", but "$\Rightarrow$" doesn't implies "$1.$", this makes the statement stronger, which is good.

To show "$1.$" implies "$\Rightarrow$"

Let $f(x,y)=\frac{|x|^a|y|^b}{|x|^c+|y|^d}\text{ and },a,b,c,d > 0$, assume "$1.$" we have

$$\frac{a}{c}+\frac{b}{d}\le1\rightarrow \lim_{(x,y)→(0,0)} f(x,y) \text{ not exists}$$ $$\Rightarrow(\frac{a}{c}+\frac{b}{d}\le1 \wedge \lim_{(x,y)→(0,0)} f(x,y)=0)\rightarrow \lim_{(x,y)→(0,0)} f(x,y) \text{ not exists}$$ Since $((a \wedge b)\rightarrow c)\Leftrightarrow(a\rightarrow(\neg b \vee c))$ we have: $$\Leftrightarrow\frac{a}{c}+\frac{b}{d}\le1\rightarrow (\lim_{(x,y)→(0,0)} f(x,y) \text{ not exists} \vee \lim_{(x,y)→(0,0)} f(x,y)\neq0)$$

Which is the contrapositive of "$\Rightarrow$", so this make sense $\dots$ logically. $\tag*{$\square$}$

Then I suppose "$2.$" and "$3.$" together should implies direction "$\Leftarrow$".

"$2.$" states the following: $$\forall a,b,c,d>0, \frac{a}{c}+\frac{b}{d}=1\rightarrow \forall (x,y)\in\mathbb{R^2}\backslash\{(0,0)\},f(x,y)\le 1$$

"$3.$" says that:

$$\forall a,b,c,d>0,\frac{a}{c}+\frac{b}{d}>1\rightarrow\lim_{(x,y)→(0,0)}f(x, y) = 0.$$

And together should implies "$\Rightarrow$":

$$\forall a,b,c,d>0,\frac{a}{c}+\frac{b}{d}>1\rightarrow\lim_{(x,y)→(0,0)} f(x,y) \text{ exists} \wedge \lim_{(x,y)→(0,0)} f(x,y)=0$$

Proof.

Assume "$3.$" have:

$$\forall a,b,c,d>0,\frac{a}{c}+\frac{b}{d}>1\rightarrow\lim_{(x,y)→(0,0)}f(x, y) = 0.$$

Since $$\lim_{(x,y)→(0,0)} f(x,y)=0\rightarrow \lim_{(x,y)→(0,0)} f(x,y) \text{ exists}$$

Directly implies "$\Rightarrow$"

$$\forall a,b,c,d>0,\frac{a}{c}+\frac{b}{d}>1\rightarrow\lim_{(x,y)→(0,0)} f(x,y) \text{ exists} \wedge \lim_{(x,y)→(0,0)} f(x,y)=0\tag*{$\square$}$$

$1.$

$$\text{WTS }\forall a,b,c,d>0,\frac{a}{c}+\frac{b}{d}\le1\rightarrow \lim_{(x,y)→(0,0)} f(x,y) \text{ not exists}$$

Proof.

Let $a,b,c,d\in\mathbb(0,\infty)\cap{\mathbb{R}}, S=\mathbb{R^2}\backslash\{(0,0)\}$

Assume

$$\frac{a}{c}+\frac{b}{d}\le1$$

Show

$$\forall L\in\mathbb{R}, \neg(∀ε>0,\exists δ>0, s.t. ((x,y)\in S \wedge 0<|(x,y)−(0,0)|<δ)\rightarrow |f(x,y)−L|<ε)$$

$$\Leftrightarrow\forall L\in\mathbb{R}, \exists ε>0,\forall δ>0, s.t. (x,y)\in S \wedge 0<|(x,y)−(0,0)|<δ\wedge |f(x,y)−L|\ge ε$$

Let $L\in\mathbb{R}$

Basicly I need to pick a $ε$ that

$$\forall \delta>0, (x,y)\in \mathbb{R^2}\backslash\{(0,0)\} \wedge 0<|(x,y)|<δ)$$

$$\wedge 0<|(x,y)|<δ)\wedge |\frac{|x|^a|y|^b}{|x|^c+|y|^d}−L|\ge ε$$

$\dots$ okay, forget epsilon-delta, let's parameterize $x$ and $y$

$\color{orange}{\text{Updates: (parameterize $x$ and $y$)}}$

Let $x=t^\frac{1}{c}, y=mt^\frac{1}{d}$ where $m\ge0$, we have the following

$$\lim_{(x,y)→(0,0)}f(x,y)=\lim_{t→0}\frac{|t^\frac{1}{c}|^a|mt^\frac{1}{d}|^b}{|t^\frac{1}{c}|^c+|mt^\frac{1}{d}|^d}$$

Try approch $t$ from right side, so everything is positive, then we have:

$$\lim_{t→0^+}\frac{mt^{\frac{a}{c}+\frac{b}{d}}}{(m+1)t} =\lim_{t→0^+}\frac{1}{m+1}t^{\frac{a}{c}+\frac{b}{d}-1}$$

Consider two cases:

Case 1:$\frac{a}{c}+\frac{b}{d}=1$

Have $$\lim_{t→0^+}\frac{1}{m+1}t^{0}=\frac{1}{m+1}$$

The limit depend on the value of $m$, that implies limit d.n.e

Case 2:$\frac{a}{c}+\frac{b}{d}<1$

Have $\frac{a}{c}+\frac{b}{d}-1$ is negative, implies limit diviges, that

$$\lim_{t→0^+}\frac{1}{m+1}t^{\frac{a}{c}+\frac{b}{d}-1}=\infty$$

Where $\frac{a}{c}+\frac{b}{d}-1$ is negative, implies limit diviges, that

Therefore in both cases limit d.n.e$\tag*{$\square$}$

$\color{orange}{\text{Please check this proof}}$

$2.$

$$\text{WTS }\forall a,b,c,d>0, \frac{a}{c}+\frac{b}{d}=1\rightarrow \forall (x,y)\in\mathbb{R^2}\backslash\{(0,0)\},f(x,y)=\frac{|x|^a|y|^b}{|x|^c+|y|^d}\le 1$$

Maybe I can use Young's Theorem here

$$\forall w,z\in\mathbb{R},t\in[0,1]\cap\mathbb{R}, w^tz^{1−t}≤tw+ (1−t)z$$

Let $w = |x|^c, z = |y|^d, t = \frac{a}{c}$ then we have

$$(|x|^c)^{\frac{a}{c}}(|y|^d)^{1−\frac{a}{c}}≤\frac{a}{c}|x|^c+ (1−\frac{a}{c})|y|^d$$

($\color{green}{\text{here i'm trying to follow brain's hint, but where is $b$, can anyone explain this in detail$\dots$}}$)

$3.$

$$\forall a,b,c,d>0,\frac{a}{c}+\frac{b}{d}>1\rightarrow\lim_{(x,y)→(0,0)}f(x, y) = 0.$$

Proof.

Let $a,b,c,d\in\mathbb(0,\infty)\cap{\mathbb{R}}, S=\mathbb{R^2}\backslash\{(0,0)\}$

Assume $\frac{a}{c}+\frac{b}{d}>1$

Show $∀ε>0,\exists δ>0, s.t. ((x,y)\in S \wedge 0<|(x,y)|<δ)\rightarrow |f(x,y)|<ε$

Yet I don't know where to start, still trying to find some examples $\dots$

Any help or hint or suggestion would be appreciated.



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23:03

Let $S\subseteq \mathbb Z^+$ be a set of positive odd numbers. I am asked to prove that there exists a sequence $(x_n)$ such that for any positive integer $k$, $$ \sum_{n=1}^\infty x_n^k $$ converges iff $k\in S$. I have no idea where to start. Even in the special case $S=\{1\}$ I don't know if any sequence would work.

Any hints?

If $S=\{1\}$, then we have to find a sequence such that $\sum x_n$ converges but $\sum x_n^k$ doesn't for $k\geq 2$. However, for a positive sequence, if $\sum x_n^k$ converges then $\sum x_n^{k+1}$ converges, so we must not choose $(x_n)$ to be a sequence of positive terms. But what can I do next?



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22:23

The usual axioms I've seen for a group are: associativity; existence of two-sided identity; existence of two-sided inverses for all elements.

$$\forall a,b,c\in G: a\left(bc\right)=\left(ab\right)c$$ $$\exists e\in G, \forall a\in G: ae=a=ea$$ $$\forall a\in G \exists a'\in G: aa'=e=a'a$$

I recently came across a different axiomatisation, and there were no proofs of equivalence. They were: associativity; existence of left-identity; existence of left-inverses.

$$\forall a,b,c\in G: a\left(bc\right)=\left(ab\right)c$$ $$\exists e\in G, \forall a\in G: ea=a$$ $$\forall a\in G, \exists a'\in G: a'a=e$$

Are these equivalent? I kind of doubt it, since we have associative semigroups with left but not right identities, but maybe the left-inverses part changes things.

There was a proof that given these axioms, a left-inverse is a right-inverse, and hence that the original inverses axiom is proven, but what about right-identity?

Proof: Let $g\in G$ then $g$ has a left inverse, call it $g'\in G$ and this too has a left inverse, call it $g''\in G$. Then, $g'g=e$, $g''g'=e$ and so $$gg'=egg'=g''g'gg'=g''g'=e$$ so $g'$ is the right-inverse of $g$ also.



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03:37

For what values of n is the expresion $2^{2n} -1$ is divisible by $4n+1$.

I have checked using a computer and the values of 2n I get are 8,20,36,44,48,56,68,96,116,120,128,140,156,168,170,176 $\cdots$etc [For $2n<200$] .

I don't seem to find a relation between these values.I am curious to see if we can find a relation by applying simple number theory.



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02:37

Let $g:\mathbb{R}\to\mathbb{R}$ be strictly increasing, continuous, and satisfy $g(x) > x$ and $\lim\limits_{x\rightarrow -\infty}g(x) = -\infty $. Is there a continuous function $f :\mathbb{R} \rightarrow \mathbb{R} $ such that $f(f(x)) = g(x)$?

I have noted a few properties about a candidate $f$; one must have

(1) $f$ is injective

(2) $f$ is strictly increasing or strictly decreasing

(3) $f$ is surjective

We also have that $g$ has a continuous inverse $g^{-1}$.

The following maybe useful;

We can define an equivalence relation $\sim$ on $\mathbb{R}$ where $x \sim y$ if $y =g_k(x)$ (here $g_{m+1}(x) = g(g_{m}(x))$ , $g_{m-1}(x) = g^{-1}(g_m(x))$, $g_{0}(x) = x$ ).

One can also observe that for a candidate $f$ that if $f(a) = b$ then $a \nsim b$ as if $f(a) = g_k(a)$ then $f(f(a)) = f(g_k(a))$ or $g_1(a) = f(g_k(a))$; hence $f(g_1(a)) = g_{k+1}(a)$; thus by two-sided induction we must have for $u \in \mathbb{Z} $ that $f(g_u(a)) = g_{k+u}(a)$; hence $f(g_k(a)) = g_{2k}(a)$ but from above $f(g_k(a)) = f(f(a)) = g(a)$; hence $g(a) = g_{2k}(a)$ which is impossible as $g$ has no fixed points.



from Hot Weekly Questions - Mathematics Stack Exchange
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02:05

Do teams that spend a lot win a lot? (just updated!) -  Students compare team wins with team salaries in the four major North American sports ... the NBA, NFL, MLB and NHL.  Students use data from recent seasons to create scatter plots comparing wins and salaries for each sport.  5.G.1, 5.G.2, 6.SP.2, 6.SP.5, 8.SP.1, 8.SP.2, 8.SP.3, HSS.ID.6, HSS.ID.C.8, HSS.ID.C.9

4th down - It's 4th down.  Should you go for it?  Punt?  Kick a field goal?  Instead of using your gut, use research, data and math to make the right call .   7.NS.1 , 7.NS.2 , 7.NS.3 , 7.SP.C.7 , HSS.MD.A.2 , HSS.MD.B.5 , HSS.MD.B.6 , HSS.MD.B.7

Home team advantage - In this activity students use an infographic to compare NFL team home and away wins. Students consider the best home team, the best away team and consider if NFL teams really do seem to have a home field advantage. 6.RP.1, 6.RP.2, 6.RP.3, 7.SP.4

Typical Super Bowl scores - An activity on predicting Super Bowl scores after being given the historical data. Which of the central measurements gives the most accurate picture of the scores?   6.SP.2, 6.SP.5, 7.SP.4, S-ID.2

Cost of Super Bowl Ads - How has the cost of a 30-second ad changed in the last 54 years?  What will next year's ads cost? In what way is the cost graph changing ... linearly? exponentially? Or what?  8.F.3, 8.F.5, 8.SP.1, HSS.ID.B.6a, HSF.LE.A.1, HSF.IF.C.8b

Benjamin-Watson

Watson saves - Benjamin Watson stopped a touchdown in the last instant of a game by running diagonally across the entire field.  How far did he run? Teddy Braschi said that he must have run 120 yards.  Did he? Pythagorean Theorem or scale. 8.G.7, G-SRT.8

Extra point or 2-point conversion? - The NFL changed the extra point kick distance to 33 yards and now teams have a tough decision to make, go for one or two points?  The one point extra point is no longer a given with a success rate of about 94%.  Teams can go for two by trying to get the ball in the end zone from the two yard line. Teams have been successful at two-point conversions in the past about 50% of the time and 53% of the time after the first two weeks of the NFL season.  So we ask students, which does it make more sense to do? 7.SP.6, 7.SP.7, HSS.MD.B.5, HSS.MD.B.6, HSS.MD.B.7

Running&Passing

Rush or Pass - Are teams better off if they have a great passing game or a great rushing game? We have four scatter plots to be made and analyzed in this activity.  8.SP.1, HSS.ID.B.6, HSS.ID.C.8, HSS.ID.C.9, HSS.IC.A.1, HSS.IC.B.6

Clean close shave In this lesson students compare NFL QB rating of quarterbacks with and without facial hair.  This task can be used to explore a number of math ideas such as sampling, causation versus correlation, misleading data, measures of central tendency, variability and data visualization. 6.SP.1, 6.SP.3, 6.SP.4, 7.SP.1, 7.SP.2, 7.SP.3, 7.SP.4, HSS.ID.A.2, HSS.ID.C.9

Fantasy football In the midst of another NFL season, we introduce students to Fantasy Football. Students first calculate football points given touchdowns, yardage gains and interceptions. They are then challenged to generalize an equation for that gives a player's total fantasy points. Students solve equations as they try to find the number of passes, touchdowns, or interceptions that yield given point totals. 6.EE.1, 6.EE.2, 7.NS.1, 7.NS.2, 7.EE.3, 7.EE.4, HSA.SSE.A.1, HSA.CED.A.4, HSA.CED.A.2, HSA.REI.B.3deflated-football

Deflate-gate The deflated ball controversy makes an interesting segue into negative number line understanding.6.NS.5, 6.NS.6, 6.NS.7, 6.EE.5, 7.NS.1,HSA.CED.A.1, HSA.CED.A.3

How many wings did the Bills send the Bengals? - In 2018 the Bills sent a thank-you gift of Duff's famous chicken wings and sides to the Bengals. How many wings, pounds of carrots, and gallons of sauce, do you think that they sent to fill-up the whole Bengals team? MP1, 5.NBT.5, 6.RP.3

Big Bucks - Students look at various sport's contracts for highly paid athletes.  How much do they earn per year? per season? per game?  Students judge from slope which athlete receives the largest rate of pay.  6.RP.2, 6.RP.3, 7.RP.2, 8.EE.5

NFL draft picks - Every spring, roughly 250 college football players are drafted by NFL teams through the NFL draft. How much does where they are drafted effect their earnings? 6.EE.9, 8.F.3, HSF.LE.A.1

Losing team in the playoffs -  How often do teams with losing records qualify for the playoffs?  Who was the worst team ever to make the playoffs?  In this activity students explore teams in the three major US sports who have made their playoffs with a losing record. How will they determine which team had the worst record. 4.NF.2 , 6.RP.1 , 6.RP.2 , 6.RP.3 , 7.RP.3



from Yummy Math
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13:02

Let $n,u,m\in \mathbb{N}$

$n_{u,m}$ is a number defined as

$$n_{u,m}= n^m+(n+1)^m+(n+2)^m+...+(n+u)^m$$

$$= \sum_{i=0}^{u}(n+i)^m$$

example: $3_{2,4}=3^4+(3+1)^4+(3+2)^4=962$

Question: Is the following claim true?

Show that $2^t$ cannot be written in $n_{u,m}$

$$n_{u,m}\ne 2^t \ \ \ \ \ \forall n,u,m,t\in \mathbb{N}$$

I proved for $n_{u,1}$ and $n_{u,2}$ never equals a power of two

Proof for $n_{u,1}\ne 2^t$

Proof

Let suppose $$n_{u,1} = n+(n+1)+...+(n+u)$$

$$=\frac{(u+1)(2n+u)}{2}= 2^t$$

So $$ (u+1)(2n+u)= 2^{t+1}$$

Case$1$: $u$ is $odd$

Then $u+1= even$ and $2n+u = odd$ it implies $ even×odd \ne 2^{t+1}$ because $ 2^{t+1}$ content only $even$ multiples except $1$ and $2n+u>1$.

Case$2$: $u$ is $even$

Then $u+1= odd$ and $2n+u = even$ it implies $odd×even \ne 2^{t+1}$ similarly as case1

So both cases shows complete proof for $n_{u,1} \ne 2^t$

Note

By using Newton's interpolation method, we can calculate formula for $n_{u,m}$ for any $m$.

So $$ n_{u,2}=n^2(u+1)+(2n+1)\frac{(u+1)u}{2} +\frac{(u+1)u(u-1)}{3} \ \ \ \ \ \ ...eq(1)$$

Proof for $n_{u,2}\ne 2^t$

Proof

Let suppose $n_{u,2} = 2^t$

We can write $eq(1)$ as

$$ (u+1)(6n^2+3(2n+1)u+2u(u-1))= 3×2^{t+1} \ \ \ \ ...eq(2)$$

Case$1$: $u =even$

$\implies u+1 = odd$

$\implies u+1=3$ $\ \ \ $ By $eq(2)$

$\implies 3n^2+3(2n+1)+2=2^{t}=even$

But we know, if $n$ is $even$ then $3n^2+3(2n+1)+2\ne even$

and if $n$ is $odd$ then $3n^2+3(2n+1)+2\ne even$

Hence it implies $3n^2+3(2n+1)+2\ne2^{t}$

Case$2$: $u =odd$

$\implies u+1=even=2^x$ for some $x$.

$\implies 6n^2+3(2n+1)u+2u(u-1)= even=3×2^y$ for some $y$.

Where $2^x2^y=2^{t+1}$

$\implies 2n+1= even$, which is not true.

Hence both cases shows complete proof for $n_{u,2}\ne 2^t$

I may not have tried much that you could reject using counter example



from Hot Weekly Questions - Mathematics Stack Exchange
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07:07

Say $A$ is a square matrix over an algebraically closed field. Say $m$ is the minimal polynomial and $p$ is the characteristic polynomial.

Of course C-H implies that $m|p$. Conversely, if we can show $m|p$ then C-H follows; the question is whether one can give a "simple", "elementary" or "straightforward" proof that $m|p$.

Note. What I really want is a proof such that I feel I actually understand the whole thing. Hence in particular no Jordan form allowed.

Edit. When I posted this is was an honest question that I didn't know the answer to. I think I got it; if anyone wants to say they believe the argument below (or not) that would be great.

First, it's clear that linear factors of $m$ must divide $p$:

If $m(\lambda)=0$ then $p(\lambda)=0$.

Because $m(t)=(t-\lambda)r(t)$, so $(A-\lambda)r(A)=0$. Minimality of $m$ shows that $r(A)\ne0$, hence $A-\lambda$ is not invertible, hence $p(\lambda)=0$.

If we could show that $(t-\lambda)^k|m$ implies $(t-\lambda)^k|p$ we'd be set. Some possible progress on that, first restricted to a simple special case:

If $t^2|m(t)$ then $\dim(\ker(A^2))\ge 2$.

Proof: Say $X=K^n$ is the underlying vector space. Say $m(t)=t^2q(t)$. Let $$Y=q(A)X,$$ $$B=A|_Y.$$ Then $Y\subset\ker(A^2)$. Say $d=\dim(Y)$.

Now $B^2=0$, and it follows easily that $B^d=0$. But $B\ne0$, hence $d\ge2$.

Similarly

If $(t-\lambda)^k|m$ then $\dim(\ker(A-\lambda)^k)\ge k$.

So we only need

If $\dim(\ker(A-\lambda)^k)\ge k$ then $(t-\lambda)^k|p$.

Which I gather is true, but only by hearsay; I'm sort of missing what it "really means" to say $t^2|p$.

Wait, I think I got it. Say $$m(t)=(t-\lambda)^kq(t),$$ $$q(\lambda)\ne0.$$ The "kernel lemma" shows that $$X=\ker((A-\lambda)^k)\oplus\ker(q(A))=X_1\oplus X_2.$$

Each $X_j$ is $A$-invariant, so we can define $$B_j=A|_{X_j}.$$Since similar matrices have the same determinant we can use any basis we like in calculating the determinant $p(t)$; if we use a basis compatible with the decomposition $X=X_1\oplus X_2$ it's clear that $$p_A=p_{B_1}p_{B_2},$$so we need only show that $$p_(t)=(t-\lambda)^k.$$ In fact it's actually enough to show $(t-\lambda)^k|p_{B_1}$, and that's clear:

Lemma. If $B$ is a $d\times d$ nilpotent matrix then $p_B(t)=t^d$.

Proof: We're still assuming $K$ is algebraically closed; $B$ cannot have a non-zero eigenvalue.

So if $d=\dim(\ker((A-\lambda)^k)$ then $$p_{B_1}(t)=(t-\lambda)^d;$$we've already shown that $d\ge k$, so $(t-\lambda)^k|p$.



from Hot Weekly Questions - Mathematics Stack Exchange
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05:57

Double palindrome:

  • ...is a number nontrivially palindromic in two consecutive bases $b,b\pm1$

  • Let $d_1,d_2$ be numbers of digits in the two bases: nontrivially means $d_1,d_2\gt 1$.

  • Let $d=\max\{d_1,d_2\}$ be called the degree of a double palindrome.

  • Example: $10$ is palindromic in bases $(b,b-1)=(4,3)$ with $(d_1,d_2)=(2,3)$ digits: $$10=(1,0)_{10} =(2,2)_4=(1,0,1)_3$$

Theorem 1. If $d$ is even, there are no examples.

  • Even length (amount of digits) palindromes in base $b$ are divisible by $b+1$.
  • Thus, such palindrome will end in $0$ in the other base, and can't be a double palindrome.

From now on, assume we have an odd degree $d=2l+1,l\in \mathbb N$.

Theorem 2. If $d_1=d_2$, there are infinitely many double palindromes for every fixed $d$.

  • Example: $(b^{2l}-1)/(b+1)$ is palindromic in $(b,b+1)$ for all $b\gt \binom{2l}{l}$, with $d=2l-1$.

  • the above result was discussed and proven in my other question.

Conjecture. If $d_1\ne d_2$, there are finitely many double palindromes, for every fixed $d$.

Question. Is there any hope in proving this conjecture?

Results on small cases of $d$ via brute force search:

  • If $d=3$, it can be shown the only solution is $10$ in bases $3,4$, as:

$$(1,0)_{10}=(1,0,1)_3=(2,2)_4$$

  • For $d=5$, the following should be all of the solutions: $$ 130=(1, 1, 2, 1, 1)_{3}=(2, 0, 0, 2)_{4}\\ 651=(1, 0, 1, 0, 1)_{5}=(3, 0, 0, 3)_{6}\\ 2997=(1, 1, 5, 1, 1)_{7}=(5, 6, 6, 5)_{8}\\ 6886=(1, 0, 4, 0, 1)_{9}=(6, 8, 8, 6)_{10} $$

  • For $d=7$, the following should be all of the solutions: $$ 9222=(2, 1, 0, 0, 0, 1, 2)_{4}=(2, 4, 3, 3, 4, 2)_{5}\\ 26691=(1, 3, 2, 3, 2, 3, 1)_{5}=(3, 2, 3, 3, 2, 3)_{6}\\ 27741=(1, 3, 4, 1, 4, 3, 1)_{5}=(3, 3, 2, 2, 3, 3)_{6}\\ 626626=(1, 1, 5, 4, 5, 1, 1)_{9}=(6, 2, 6, 6, 2, 6)_{10}\\ 1798303=(1, 0, 1, 9, 1, 0, 1)_{11}=(7, 2, 8, 8, 2, 7)_{12}\\ 1817179=(1, 0, 3, 1, 3, 0, 1)_{11}=(7, 3, 7, 7, 3, 7)_{12} $$

And so on. For every $d$, solutions seem to only exits in relatively small bases.

For a general fixed $d=2l+1,l\in\mathbb N$, is it possible to set upper bounds on base $b$, after which solutions can't exits? - to prove the conjecture?

That is, how to show that double palindromes can't exist in (arbitrarily large) number bases $(b,b\pm1)$, when $b\gt b_0$, for some value $b_0:=b_0(d)$, if degree $d$ is fixed, and $d_1\ne d_2$?

Given $d=2l+1$ digits and bases $b,b+1$, then:

I have following data: digits [number of terms] (last b with terms / last b checked) {terms}

3  [1]    (3/100)  {10} 
5  [4]    (9/100)  {130, 651, 2997, 6886} 
7  [6]    (11/100) {9222, 26691, 27741, 626626, 1798303, 1817179} 
9  [12?]  (17/50)  {11950, 76449, 193662, 704396, 723296, 755846, 883407, 4252232, 10453500, 65666656, 2829757066, 7064428133} 
11 [14?]  (21/30)  {175850, 2347506, 2593206, 48616624, 160258808, 630832428, 5435665345, 8901111098, 9565335659, 37180386607, 131349132446, 746852269786, 7056481741113, 17985535104496} 
13 [>32?] (25/25)  {6643, 749470, 1181729, 17099841, 17402241, 25651017, 32317933, 295979506, 297379006, 402327541, 9689802280, 54453459798, 54464523606, 55027793502, 827362263728, 2909278729092, 2926072706292, 4036309890977, 7448647872250, 8013269088838, 17901027912530, 34577567573550, 34811609537160, 35194041720930, 54489277730565, 54768340178775, 55150772362545, 142077571662616, 682765460591464, 683230317449824, 733909097713709, 59777562308125626, ...} 
15 [>19]  (15/15)  {11435450, 203509031168, 204191148800, 231773764784, 321015775216, 3741580511478, 19404342621340, 41275222257214, 42143900934124, 218053292350812, 218210353012812, 218254595452812, 251569181965152, 259799383997952, 3338546970154550, 3617178283518590, 23044579418585216, 26926823266016368, 38322172687372936, ...}
17 [>21]  (12/12)  {16516113567, 16619231967, 198522549056, 204185363456, 240971251611, 246467321391, 303520083621, 330347455102, 341225573632, 4102350269485, 12262956787888, 13267882222408, 68995850733945, 1366179755723700, 1767662936108630, 4782537117352874, 5987078778707895, 140538057123815013, 388816019726293166, 396289206590671310, 411924791551509530, ...}
19 [>15]  (9/9)    {916821671, 956613659, 1136307905, 155784877126, 4262839618051, 126532386891655, 6615812399178042, 6622944330543930, 6641481107049786, 10688365729164780, 81877825421774500, 120168724989001390, 190076027720670091, 194216405504612491, 547906983389609745, ...}
21 [>9]   (6/6)    {1422773, 2806999337418, 3101308506654, 275956595195822, 451853066660344, 6116904274791985, 6875219172190387, 10229280954883514, 10231408608585002, ...}
23 [>8]   (5/5)    {5415589, 46746179770, 77887660577, 37004798195346, 47470618709562, 48517516968462, 3099677168429681, 9779924118261554, ...}
25 [>2]   (4/4)    {635913760790, 383478037564629, ...}
27 [>1]   (4/4)    {5892002867556037, ...}
...

That is, the conjecture is: How to prove that each row in this table will be finite?



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02:47

Remark: I assume the equality symbol „$=$“ with its standard semantics to always be included in first-order logic.

Suppose we want to formalize some first-order theory. For our first-order language, we need to fix some set of variables and some set of contstants, e.g. one constant in group theory and Peano arithmetic, no constant at all in ZFC. The choice of a set of constants is most often done quite canonical.

Also, the cardinality of the set of constants can have a direct influence on semantics. For if the set of constants has cardinality $\kappa$ and we add for any two distinct constants $c,d$ the axiom $c\neq d$, each model must have at least cardinality $\kappa$.

What consequences does the choice of the set of variables have?

I can think of these two cases:

  • One usually takes an at least infinite set of variables to be able to nest arbitrary many quantifiers. Example: Let $P$ be a 1-ary predicate and $F$ be a 2-ary function, then the formula $$ \forall x_1 \forall x_2 \forall x_3 \dots \forall x_n : P(F(x_1,F(x_2,F(x_3, \dots ,F(x_{n-1},x_n)))))$$ does only make sense, if one has at least $n$ distinct variables.
  • If one has at least $n$ distinct variables, one can add axioms like $$ \exists^{\ge n} x : P(x) \qquad \text{or} \qquad \exists^{\le n} x : P(x)$$ which imply each model to contain at least $n$ or at most $n$ elements.
  • If one wants to enumerate the set of all strings and do some gödelian reasoning, one must only take a countable set of variables.

Do you know any more cases, where the cardinality of the set of variables does matter? Does it, as long as it is infinite, affect the class of models for our formal theory?



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23:32

The instructor in our Differential Equations class gave us the following to solve: $$ \frac{dy}{dx} = \frac{x - y}{xy} $$

It was an item under separable differential equations. I have gotten as far as $ \frac{dy}{dx} = \frac{1}{y} - \frac{1}{x} $ which to me doesn't really seem much. I don't even know if it really is a separable equation.

I tried treating it as a homogeneous equation, multiplying both sides with $y$ to get (Do note that I just did the following for what it's worth)... $$ y\frac{dy}{dx} = 1 - \frac{y}{x} $$ $$ vx (v + x \frac{dv}{dx}) = 1 - v $$ $$ v^2x + vx^2 \frac{dv}{dx} = 1 - v $$ $$ vx^2 \frac{dv}{dx} = 1 - v - v^2x$$

I am unsure how to proceed at this point.

What should I first do to solve the given differential equation?



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04:12
Let be the proposition “You are first,” and let be the proposition “You are last.” Translate the logical statement . This matches the famous catchphrase from “Talladega Nights: The Ballad of Ricky Bobby.” Context: Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can […]

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01:47

The question is as follows:

Let $G$ be a finite group of order $n$ having a proper normal subgroup $N$ that is not contained in the center of $G$. Prove that $G$ has a proper subgroup $H$ with $|H|>\sqrt{n}$.

I believe I have most (if not all) the pieces of an argument, but I'm struggling to put them together. I know that since there exists $N\triangleleft G$ such that $N\not\subset Z(G)$, then $|Z(G)|<\frac{n}{2}$. Also, by application of the orbit-stabilizer theorem and Lagrange's theorem, the size of the conjugacy class of an element $g\in G$ is $n$ divided by the centralizer of $g$. That is, $$|cl_g|=\frac{n}{|C_G(g)|}.$$ So I think I'm trying to show that the order of the centralizer of some element $h\in N$ is at least $\sqrt{n}$, but I'm not sure.



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01:02

The least integral value of a for which all the roots of the equation $x^4-4x^3 -8x^2 +a=0$ are real.

let $f(x) =x^4-4x^3 -8x^2 +a=0$

$ f'(x) = 4x^3 -12x^2 -16x $

Put $f'(x) = 0 $ we get x =0, -1, 4. How to proceed further, I am not getting any idea on this. Please guide thanks.



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21:22

I have been starting at the following integral for the past 2 hours and can not see where I have gone wrong. Can anyone please assist in isolating where I've made a mistake.

My sanity thanks you in advance,

\begin{equation*} I = \int \frac{\sqrt{x^2 + 1}}{x}\:dx \end{equation*} Let $x = \tan(s)$: \begin{equation*} \frac{dx}{ds} = \sec^2(s) \rightarrow dx = \sec^2(s)\:ds \end{equation*} Thus, \begin{align*} I &= \int \frac{\sqrt{x^2 + 1}}{x}\:dx = \int \frac{\sqrt{\tan^2(s) + 1}}{\tan(s)}\cdot \sec^2(s)\:ds = \int \frac{\sec(s)}{\tan(s)}\sec^2(s)\:ds = \int \frac{\sec^3(s)}{\tan(s)}\:ds \\ &= \int \frac{\frac{1}{\cos^3(s)}}{\frac{\sin(s)}{\cos(s)}}\:ds = \int \frac{1}{\cos^3(s)} \cdot \frac{\cos(s)}{\sin(s)}\:ds = \int \frac{1}{\cos^2(s)\sin(s)}\:ds = \int \frac{1}{\left(1 - \sin^2(s)\right)\sin(s)}\:ds \\ &= \int \frac{1}{\left(1 + \sin(s)\right)\left(1 - \sin(s)\right)\sin(s)}\:ds \end{align*} Applying a Partial Fraction Decomposition we see: \begin{equation*} \frac{1}{\left(1 + \sin(s)\right)\left(1 - \sin(s)\right)\sin(s)} = \frac{1}{\sin(s)} - \frac{1}{2}\cdot \frac{1}{1 + \sin(s)} - \frac{1}{2}\cdot \frac{1}{1 - \sin(s)} \end{equation*} Thus, \begin{align*} I &= \int \frac{1}{\left(1 + \sin(s)\right)\left(1 + \sin(s)\right)\sin(s)}\:ds = \int \left[\frac{1}{\sin(s)} - \frac{1}{2}\cdot \frac{1}{1 + \sin(s)} - \frac{1}{2}\cdot \frac{1}{1 - \sin(s)} \right]\:ds \\ &= \int \frac{1}{\sin(s)}\:ds - \frac{1}{2} \int \frac{1}{1 + \sin(s)}\:ds - \frac{1}{2} \int \frac{1}{1 - \sin(s)}\:ds \end{align*} We now employ the Weierstrass Substitution $t = \tan\left(\frac{s}{2} \right)$: \begin{equation*} ds = \frac{2}{1 + t^2}\:dt, \quad \sin(s) = \frac{2t}{1 + t^2} \end{equation*} Thus, \begin{align*} I&= \int \frac{1}{\sin(s)}\:ds - \frac{1}{2} \int \frac{1}{1 + \sin(s)}\:ds - \frac{1}{2} \int \frac{1}{1 - \sin(s)}\:ds \\ &= \int \frac{1}{\frac{2t}{1 + t^2}} \cdot \frac{2}{1 + t^2}\:dt - \frac{1}{2} \int \frac{1}{1 + \frac{2t}{1 + t^2}}\cdot \frac{2}{1 + t^2}\:dt - \frac{1}{2} \int \frac{1}{1 - \frac{2t}{1 + t^2}}\cdot \frac{2}{1 + t^2}\:dt \\ &= \int \frac{1}{t}\:dt - \int \frac{1}{1 + t^2 + 2t}\:dt - \int \frac{1}{1 + t^2 - 2t}\:dt \\ &= \int \frac{1}{t}\:dt - \int \frac{1}{\left(t + 1\right)^2}\:dt - \int \frac{1}{\left(t - 1\right)^2}\:dt = \ln\left|t\right| - -\frac{1}{t + 1} - - \frac{1}{t - 1} + C \\ &= \ln\left|t\right| +\frac{1}{t + 1} + \frac{1}{t - 1} + C = \ln\left|t\right| +\frac{2t}{t^2 - 1} + C \end{align*} Where $C$ is the constant of integration. Here: \begin{equation*} t = \tan\left(\frac{s}{2} \right) = \tan\left(\frac{\arctan(x)}{2} \right) = \frac{\sqrt{x^2 + 1} - 1}{x} \end{equation*} Thus, \begin{align*} I&= \ln\left|t\right| +\frac{2t}{t^2 - 1} + C = \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| +\frac{2\left(\frac{\sqrt{x^2 + 1} - 1}{x}\right)}{\left(\frac{\sqrt{x^2 + 1} - 1}{x} \right)^2 - 1} + C \\ &= \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| + 2 \cdot \frac{\sqrt{x^2 + 1} - 1}{x} \cdot \frac{1}{\left(\frac{x^2 + 1 - 2\sqrt{x^2 + 1} + 1}{x^2} \right) - 1} + C \\ &= \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| + 2 \cdot \frac{\sqrt{x^2 + 1} - 1}{x} \cdot \frac{x^2}{\left(x^2 + 1 - 2\sqrt{x^2 + 1} + 1\right) - x^2} + C \\ &= \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| + 2 \cdot \frac{\sqrt{x^2 + 1} - 1}{x} \cdot \frac{x^2}{2\left(1 - \sqrt{x^2 + 1} \right)} + C = \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| - x + C \\ \end{align*}



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21:02

I'm trying to prove that $\forall b \in \mathbb{N}, b>2,$ $b^{3} = 6\binom{b} {3} +6 \binom{b}{2} +b$ without just using algebra. The idea I've thought about is that there are $b^{3}$ ways to choose a triple of numbers from $\{1,\ldots, b\}$, and there are obviously $b$ ways to choose one element, the issue is I'm not sure how the $\binom{b}{3}$ and $\binom{b}{2}$ terms contribute. Any help would be greatly appreciated.



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20:02

We call a number "holy" if it contains no $666$ in its decimal expansion, and "unholy" otherwise. For instance, $12366621$ and $666609$ are unholy, while $7777$ and $66166266366$ are holy.

Question: Is the set $$\{2^n \ | \ n \in \mathbb N, 2^n \text{ is holy}\}$$ infinite?

Of course, tons of similar questions can be asked by changing the number $666$, the base $2$, and the base for extension (we asked for decimal, so the default was $10$). I do not feel that I am the first one asking this, and I appreciate it if someone gives me references if applicable.

But my thought is the following:

Conjecture: No.

I will share my reasoning at the end of the post, but let us first see some facts:

Smallest unholy instances: $$ \begin{aligned} 2^{157} &= 182687704\color{magenta}{666}362864775460604089535377456991567872\\ 2^{192} &= 6277101735386680763835789423207\color{magenta}{666}416102355444464034512896 \end{aligned} $$

Then, we witnessed a cluster of unholy powers: $2^{218}, 2^{220}, 2^{222}, 2^{224}, 2^{226}$, and then kept holy for a while, until we hit the unholy $2^{243}$.

Largest holy instances: I did not throw in a lot of CPU time to pursue holy numbers, nor did I try hard enough to optimize my programs, but among the $3715$ holy powers of $2$, the largest of them are $$2^{25357}, 2^{25896}, 2^{26051}, 2^{26667}, 2^{29784}.$$

I tested up to around $2^{110000}$, but that is all I got. It probably will be reasonable for an average computer to test up to say $2^{10^6}$ or $2^{10^7}$, but I will be surprised to see a new holy number.

Statistics: For an integer $n$, let $H(n)$ be the number of holy powers of $2$ up to $2^n$.

n      | H(n)     || n      | H(n)     || n      | H(n)
 1000  |  875     || 11000  | 3567     || 21000  | 3700
 2000  | 1560     || 12000  | 3602     || 22000  | 3703
 3000  | 2059     || 13000  | 3621     || 23000  | 3705
 4000  | 2442     || 14000  | 3645     || 24000  | 3707
 5000  | 2747     || 15000  | 3655     || 25000  | 3709
 6000  | 2984     || 16000  | 3670     || 26000  | 3712
 7000  | 3171     || 17000  | 3682     || 27000  | 3714
 8000  | 3332     || 18000  | 3689     || 28000  | 3714
 9000  | 3440     || 19000  | 3693     || 29000  | 3714
10000  | 3514     || 20000  | 3695     || 30000  | 3715

A plot of this: enter image description here

The heuristics of the conjecture:

This is definitely not close to a proof at all, and I still hope if rigorous arguments exists:

The idea is that we want to estimate, for an integer $n$, the probability $P(n)$ that $2^n$ is holy, and then compute $\sum_{n=1}^\infty P(n)$.

We know that $2^n$ has $O(n\ln 2)$ decimal digits, so there are $O(n\ln 2)$ groups of three. For each group there is a $1-10^{-3}$ chance to be not $666$, so very roughly $$ P(n) = (1-10^{-3})^{n\ln 2} \approx e^{-10^{-3}\ n\ln 2}. $$

And note that $$ \sum_{n=1}^\infty P(n) \approx \int_{n=0}^\infty e^{-10^{-3}\ x\ln 2} dx < \infty. $$

The red "estimation line" in the figure above follows from this integral.

Of course, one can easily argue the properness of the heuristic above:

  • The distribution of the digits close to the left are not uniform; they are affected by the growth of logarithmic functions.
  • The distribution of the digits close to the right are not uniform; they are affected by the pattern of $2^n \pmod{10^k}$.
  • $P(n)$ and $P(n+i)$ are not independent, partially because of the awful choice of the number $666$: $6\cdots 6 \times 2^2 = 26\cdots 64$.

Any thoughts are appreciated.



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19:52

I'd like to proof the following inequality for $d,D \in \mathbb{N}$:

$$ \sum_{k=0}^{D} \binom{D}{k}\binom{d+D-k-1}{D-1} \leq 2D d^{D-1}. $$

In case $B>A$ we define $\binom{A}{B}:=0$.

This upper bound is exact in the case $d=1,2$. I'm not quite sure how tackle this problem in a somewhat elegant way.



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04:04

The paper “Approximate filtering of conditional intensity process for Poisson count data: application to urban crime”, co-authored by Naratip Santitissadeekorn, David Lloyd, Martin Short (Georgia Tech), and Sylvain Delahaies, has been accepted for publication in Computational Statistics and Data Analysis, one of the leading journals on Data Science. A link to the final form manuscript is here. The picture below shows Figure 3 from the paper.



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04:04

Anne Skeldon is in Vancouver Canada this week (20-25 September) to attend the 2019 World Sleep Congress. It is the largest international conference devoted to research into sleep, and is expected to attract over 2500 participants. Anne is presenting a poster on “What determines our light exposure patterns and how do we quantify them?“.



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02:23

My Attempt:

$1.$ $G$ is abelian if and only if the mapping $g\mapsto g^{-1}$ is an isomorphism on the group $G$.

$2.$If $G$ is finite and every irreducible character is linear then $G$ is abelian.

$3.$If $\operatorname{Aut}(G)$ acts on the set $G-\{e\}$ transitively then $G$ is abelian.

$4.$If $\mathbb Z_2$ acts by automorphism on a finite group $G$ fixed point freely then $G$ is abelian.

$5.$ If $\forall a,b\in G$ $ ab=ba$ then $G$ is Abelian.

My Question:

The above are the things which I already use to show a group will be Abelian.

  • Is/are there any other way(s) to show a group $G$ to be Abelian?


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02:13

$\displaystyle\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+1}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+n}}\right)^{n}$

left=$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+1}}}\right)^{n}}=e^{\displaystyle n \ln{\frac{n}{\sqrt{n^2+1}}} }=e^{0}=1$

right=$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+n}}}\right)^{n}}=e^{\displaystyle n \ln{\frac{n}{\sqrt{n^2+n}}} }=e^{-\frac{1}{2}}$

left $\ne$ right ,what to do next?



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01:33

I want to know if the group $G=\mathbb Z \times \mathbb Z$ can be written as union of finitely many proper subgroups of it ?

It is clear that $\mathbb Z$ can't be written as union of finitely many proper subgroups as the subgroups are of the form $n \mathbb Z$ for some integer $n$ and there are infinitely many primes in $\mathbb Z.$

My way to think: If possible $G= H_1 \cup \cdots\cup H_r $ where $r>1$ and $H_i's$ are proper subgroups of $G.$ Now considering the projection maps $\pi_1$ and $\pi _2$ on $G$ there exist $i$ and $j$ such that $\mathbb Z=\pi_1(H_i)$ and $\mathbb Z=\pi_2(H_j).$ I can't complete my arguments after that. Any helps will be appreciated. Thanks.



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01:23

I am slightly confused about the interchangeability of the terms isometry and biholomorphism. This confusion is rooted in the following statement, which I have seen more than once in some discussions on equivalent versions of the Uniformization Theorem.

Suppose $(M,g)$ is a complete, simply connected, $2$-dimensional Riemannian manifold with constant curvature. By the Killing-Hopf theorem, it is isometric to either $\mathbb{R}^2$, $S^2$, or $\mathbb{H}^2$. Identify $M$ as a simply connected Riemann surface and identify $\mathbb{R}^2$, $S^2$, and $\mathbb{H}^2$ as, respectively, $\mathbb{C}$, $\hat{\mathbb{C}}$, and $D_1$. Then $M$ is biholomorphically equivalent to $\mathbb{C}$, $\hat{\mathbb{C}}$, or $D_1$.

To me, an isometry between two Riemannian manifolds $(M,g)$ and $(M',g')$ is a diffeomorphism $f:M\longrightarrow M'$ such that $g=f^*g'$. Thus, it preserves angles and can be described as a conformal mapping.

On the other hand, given two Riemann surfaces $M$ and $N$, a biholomorphism is a bijective map $f:M \longrightarrow N$ such that both $f$ and $f^{-1}$ are holomorphic.

So, in the context of the italicized statement, does it really follow that the isometry turns into a biholomorphism just because we switched from Riemannian manifolds to Riemann surfaces, and because both isometries and biholomorphisms preserve angles? Or is there something more subtle going on here?



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01:03

I flip a coin until I get $m$ more heads than tails, or $n$ more tails than heads. Let the expected number of flips of the coin before stopping be f(m,n).

I obtained $f(m,n)=mn$ from the recursion $f(m,n)=1+\frac{f(m-1,n+1)+f(m+1,n-1)}2$ with $f(k,0)=f(0,k)=0$ for all $k$.

Other than going through this recursion (and either solving by inspection or by writing as linear recurrence in single variable and solving brute force), is there an intuitive reason you should expect this process to take $mn$ flips? I was thinking about the more general problem with probability $p$ of getting heads and was struck by how simple the formula became when handling, what turned out to be a special case (general formula broke down) of $p=\frac12$.



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00:13

How can I simplify this trigonometric expression?

$$ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$$

I used

$$\sin \frac{\pi}{15}=2 \sin \frac{\pi}{30} \cos \frac{\pi}{30}$$ and $$\cos\frac{\pi}{15}=2\cos^2\frac{\pi}{30}-1$$

But these give me more complicated expressions.



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23:43

Given positives $x_1, x_2, \cdots, x_{n - 1}, x_n$ such that $$\large \sum_{k = 1}^n\frac{1}{x_k + m} = \frac{1}{m}$$. Prove that $$\large \frac{\displaystyle \sqrt[n]{\prod_{k = 1}^nx_n}}{m} \ge n - 1$$

We have that $$\sum_{k = 1}^n\frac{1}{x_k + m} = \frac{1}{m} \iff \sum_{k = 1}^n\frac{m}{x_k + m} = 1 \iff \sum_{k = 1}^n\frac{x_k}{x_k + m} = n - 1$$

Let $x_1 \ge x_2 \ge \cdots \ge x_{n - 1} \ge x_m$, using the Hölder's inequality, it is seen that $$\left(\sum_{k = 1}^nx_k\right) \cdot \left(\sum_{k = 1}^n\frac{1}{x_k + m}\right) \ge m \cdot \sum_{k = 1}^n\frac{x_k}{x_k + m} \implies \left(\sum_{k = 1}^nx_k\right) \cdot \frac{1}{m} \ge m(n - 1)$$

Unfortunately, I can't go for more.



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22:01

Everyone loves a good pun! They make people laugh, they make people smile, and if they're really good, they might even make people groan! With that in mind, we thought we'd give you a few math jokes to make the mathematician in your life smile! (Note: the mathematician in your life will probably also groan at a couple of these as well)

Why couldn't the number stay still?
It was a Roamin' Numeral!

Why shouldn't you talk to zero?
It's pointless!

But why shouldn't you talk to pi?
It will just go on forever!

If you ever need help with something, don't ask an obtuse triangle.
It will never be right!

We know 6 was afraid of 7 because 7 8 9, but why did 7 eat 9?
You're supposed to eat 3 squared meals a day!

How do mathematicians travel short distances?
They take a Rhom-bus!

How do mathematicians travel long distances?
They take an x-y plane!

Why are math lectures so long?
The teachers like going on tangents!

How do you stay warm in a cold room?
Stand in the corner - they're always 90 degrees!

Did you see the math professors with the graph paper?
They must be plotting something!

Did you hear about the retired calculus teacher?
He's doing alright - just having some trouble dealing with the aftermath.

Infinitely many mathematicians walk into a bar.
The first mathematician says "I'd like one beer, please."
The second mathematician says "I'd like half a beer, please."
The third mathematician says "I'd like a quarter of a beer, please."
The bartender pours two drinks.
"Come on," he says, "you guys should know your limit!"

from The Center of Math Blog https://ift.tt/2noH4At
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19:48

Let $\varphi(x)$ be the Euler totient function. For $k = 1,2,3,\ldots$ I calculated the number of solutions of $\varphi(x) = \varphi(x+k)$. I observed that we have very few solutions when $k = 3,9,15,21,\ldots$ and a very large number of solutions when $k = 6,12,18,24,\ldots$ as shown below for $n \le 10^7$.

enter image description here

Question 1: Why do we have a very few solutions when $k$ is an odd multiple of $3$ and a very large number of solutions when $k$ is an even multiple of $3$?

Question 2: Is the following claim true?

For any six consecutive positive integer values of $k$, and a sufficiently large $x$, the number of integers $n \le x$ such that $\varphi(n) = \varphi(n+k)$, is the lowest when $k \equiv 3 \text{ (mod 6)}$ and is the highest when $k \equiv 0 \text{ (mod 6)}$.

Related question: Gap between integers having the same number of co-primes



from Hot Weekly Questions - Mathematics Stack Exchange
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18:18

for $x \in \mathbb R ,|x|< \frac12$ define:

$A_n(x)$ solution of the relation:

$A_{n+1}(x) = \frac{x^2}{1-A_n(x)}$

$A_0(x) = 0$

(I found that $A_n(x) = 2x^2 \cdot \frac{(1+\sqrt{1-4x^2})^n - (1 - \sqrt{1-4x^2})^n}{(1+\sqrt{1-4x^2})^{n+1} - (1 - \sqrt{1-4x^2})^{n+1}}$)

prove that the sequence:

$p_0(x)=1$

$p_{n+1}(x) = p_n(x) \cdot (1-x^2 - A_n(x))$

consists of integer polynomials in $x$



from Hot Weekly Questions - Mathematics Stack Exchange
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02:04

The paper “Existence of stationary fronts in a system of two coupled wave equations with spatial inhomogeneity” co-authored by Jacob Brooks, Gianne Derks, and David Lloyd, has been accepted for publication in Nonlinearity. The picture left is from Figure 13 in the paper. An electronic version of the final-form preprint is available for downloading here.



from Surrey Mathematics Research Blog https://ift.tt/2mz83sz
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01:43

I had a long discussion in chat over what seems like a simple question:

Is $\sin \infty$ an indeterminate form?

What do you think? :)

I'm labelling this as (soft-question) to be safe, but it should have a clear answer depending on the definition of indeterminate form used. The term "indeterminate form" is seldom used in post-calculus mathematics, but I believe it has one or more accepted definitions, which are either informal or formal. So any answer which takes a standard definition and argues the case would be interesting to me.

How to make a good answer:

  1. State the definition of indeterminate form, either from an online or textbook source, or a definition you came up with on your own.

  2. Determine, using your stated definition, whether $\sin \infty$ is indeterminate or not.

EDIT: What do I mean by $\sin (\infty)$?

It's not a well-defined expression, but neither are any of the other indeterminate forms: $\frac{0}{0}$ doesn't exist, $1^\infty$, doesn't exist, and so on. So the question is whether this expression -- which is not well-defined, just like any other indeterminate form -- is an indeterminate form.



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