calculus
integration
limit
function
real analysis
trigonometry
vector

[quote="Beginner"]I need for help.
Is it true or not if in evaluating inprper integral of this type
$\int_{- \infty}^{\infty} f(x) dx$ I transform it to
$\lim_{a \rightarrow \infty} \lim_{b \rightarrow - \infty} \int_{- \infty}^{\infty} f(x) dx$ As far as I know tihs called iteration limit.
Hmmm :?  Maybe its different from
\[ \lim_{(x,y) \rightarrow (0,0)} f(x,y)\]

Thanks in Advance.[/quote]Yes, but you're complicating the situation more than is necessary.

$\int_{- \infty}^{\infty} f(x) dx \equiv \lim_{h \rightarrow \infty} \int_{- h}^{h} f(x) dx$

An example of a divergent integral with a principal value: $\int_{-\infty}^{\infty}\frac{x}{x^2+1}\,dx$.

[quote="liyi"][quote="Dr. No"]
$\int_{- \infty}^{\infty} f(x) dx \equiv \lim_{h \rightarrow \infty} \int_{- h}^{h} f(x) dx$[/quote]
Er... I don't think so.
$\lim_{h \rightarrow \infty} \int_{- h}^{h} f(x) dx$ is just Cauchy's Principal Value...
If the integral does not exist, the principal value may exist.[/quote]I was just pointing out that $\lim_{h \rightarrow \infty} \int_{- h}^{h} f(x) dx$ is how $\int_{- \infty}^{\infty} f(x) dx$ is defined; I wasn't suggesting that such a limit always exists.

[quote="blahblahblah"]No, liyi was pointing out that you are wrong.  You let the limits go to infinity one at a time.[/quote]You won't be wrong if you consider them one at a time. Nevertheless, it's not necessary to do so.

$\lim_{a \rightarrow -\infty} f(a) = \lim_{h \rightarrow \infty} f(-h)$

It is a hair-splitting argument to insist that such limits must be considered separately.

[quote="blahblahblah"]No, it isn't.

By your definition, $\int^{\infty}_{-\infty} xdx=0$.  However, this is not in $L^1$ (obviously), and the iterated limits don't exist.  In analysis, sometimes it is crucial that both positive and negative parts of a function be well behaved - your definition only requires some weighted average of positive and negative parts to be well-behaved.[/quote]Your counterexample is fatally flawed. $f(x) = x$ is an odd function. The integral of any odd function over a symmetric interval is zero. (To assume $x \not \in L^1$, is quirky, argumentative, and beyond the scope of the original poster's question.)

[quote="blahblahblah"]The only 'part' of being in $L^1$ that I'm considering is just the 'size' bit; ${\int_{\mathbb{R}} |x|}=\infty$.

I really don't get what you're saying.  First of all, your original definition is not correct.  Liyi stated that your integral is what's referred to as a principal value.  Jmerry provided an example of a function that has a principal value, but which grows fast enough (even though it tends to zero) that its improper integral doesn't exist.  My example was similar, although admittedly more ridiculous.

It's nice to take principal values, especially when you're using the residue theorem or something similar, but you have to say that you are taking a principal value, because it is not contained in the defintions of the Riemann or Lebesgue integrals.[/quote]Let's back up about a half step.

It was evident to me that Beginner, the original poster, was asking a first semester integral calculus question; namely, how to evaluate $\int_{-\infty}^{\infty} f(x) \mathrm{d}x$, for $f: \mathbb{R}^1\rightarrow\mathbb{R}^1$. In such a case, the Cauchy Principal Value is the value of the integral, if the Cauchy Principal Value exists for the integral in question. (It might not, which is a point I don't dispute.)

I'll weigh in as someone who frequently teaches this material. (For us, it's at the very beginning of the second semester of calculus.)

From the beginning, I always teach the official definition of improper integral. I tend to discuss it this way: for how many different reasons is this integral improper?

$\int_0^3\frac{dx}{\sqrt{x}}$ is improper for one reason - the integrand is unbounded as $x\to0^+.$

$\int_0^{\infty}x^2e^{-x}\,dx$ is improper for one reason: $+\infty.$

$\int_{-\infty}^{\infty}\frac{x}{1+x^2}\,dx$ is improper for two reasons: $-\infty$ and $+\infty.$

$\int_0^{\infty}\frac{\sin x}x\,dx$ might be improper for two reasons: we know it's improper as $x\to +\infty$ and we're suspicious of it as $x\to 0^+.$ Even though it turns out that it's actually a proper integral around $0^+,$ our suspicions were appropriate. We may very well break it into two pieces when we try to prove that it does converge.

$\int_{-2}^3\frac{dx}x$ is improper for two reasons: $0^-$ and $0^+.$

And the message for all of this is that whenever we have an integral improper for more than one reason, we need to break it up into pieces such that on each piece it's improper for no more than one reason.

$\int_{-\infty}^{\infty}\frac{x}{1+x^2}\,dx= \int_{-\infty}^0\frac{x}{1+x^2}\,dx+ \int_0^{\infty}\frac{x}{1+x^2}\,dx=\infty-\infty$ and that's not arithmetically defined.

In particular, that's not an inderminate case, in which we say $\infty-\infty$ but mean "the limit of very large minus very large": this is literally infinity minus literally infinity, and we're not going to try to define that.

$\int_{-2}^3\frac{dx}x=\int_{-2}^0\frac{dx}x+\int_0^3\frac{dx}x=-\infty+\infty.$ Same thing; not convergent.

This is really the way I talk in that course. One goal is to make sure everyone recognizes the following "calculation" for the nonsense that it is:

$\int_{-2}^3\frac{dx}{x^2}=-\frac1x|_{-2}^3=-\frac13- (-\frac1{-2})=-\frac56.$

[quote="blahblahblah"]Not quite.  There's a classic example from most calculus II courses where it is inappropriate to take a principal value - take a symmetric interval around zero, and integrate $f(x)=1/x$ :)[/quote]This topic is hardly worthy of getting into a semantic food fight. You whelps in short pants can argue all you wish about the particulars of specific cases; I'm not interested. As far as I am concerned improper integrals are defined in terms of limits. If some particular integral diverges then said limits do not exist. All the crap, about $L^1$ vector spaces, and special cases, e.g. $\int_{-\infty}^{+\infty} \frac{\mathrm{d}x}{x}$, is just that -- crap.

I said in the beginning that $\int_{-\infty}^{+\infty} = \lim_{h\rightarrow \infty} \int_{-h}^{h} f(x)\; \mathrm{d}x$, [b] only if[/b] the limit exists.  And under that constraint, I can prove my statement.

[quote="Dr. No"]I was just pointing out that $\lim_{h \rightarrow \infty} \int_{- h}^{h} f(x) dx$ is how $\int_{- \infty}^{\infty} f(x) dx$ is defined; I wasn't suggesting that such a limit always exists.[/quote]

No, the definition for improper Riemann integral should be $\lim_{a\to \infty\atop b\to-\infty} \int_a^b f(x)dx$. In $\epsilon-X$ language, it reads, for any $\epsilon>0$, there exists $X$, such that for all $a>X$ and $b<-X$, it holds $\left|\int_a^b f(x)dx- I\right|<\epsilon$. Then we say $\int_{-\infty}^\infty f(x)dx = I$

[quote="blahblahblah"]You can define things any way you want them to, but if someone is asking for a definition, then I would prefer to give him the proper definition.  However, continue to insist whatever you will.[/quote]

 :wallbash:

From Anton (1980), pp. 537-539,

[i]If the two improper integrals $\int_{-\infty}^{0} f(x) \mathrm{d}x$ and $\int_{0}^{+\infty} f(x) \mathrm{d}x$ both converge then we say $\int_{-\infty}^{+\infty} f(x) \mathrm{d}x$ [b]converges[/b] and we define:

$\boxed{\int_{-\infty}^{+\infty} f(x) \mathrm{d}x = \int_{-\infty}^{0} f(x) \mathrm{d}x + \int_{0}^{+\infty} f(x) \mathrm{d}x}$[/i]

Immediately prior to that statement Anton had written:

[i]If f is continuous on the interval $\left[ a, +\infty\right)$ , then we define the improper integral $\int_{a}^{+\infty} f(x) \mathrm{d}x$ as the limit: $\int_{a}^{+\infty} f(x) \mathrm{d}x = \lim_{l\rightarrow +\infty}\int_{a}^{l} f(x) \mathrm{d}x$[/i]

And the corresponding statement:

[i]If f is continuous on the interval $\left( -\infty, b\right]$ , then we define the improper integral $\int_{-\infty}^{b} f(x) \mathrm{d}x$ to be the limit: $\int_{-\infty}^{b} f(x) \mathrm{d}x = \lim_{l\rightarrow -\infty}\int_{l}^{b} f(x) \mathrm{d}x$[/i]

Substituting h or - h for l, as appropriate, in the above statements leads to the result $\int_{-\infty}^{+\infty} f(x) \mathrm{d}x = \lim_{h\rightarrow \infty}\int_{-h}^{h} f(x) \mathrm{d}x$.

But then I've only been doing calculus for 25 years, so what do I know?

[quote="Kent Merryfield"]That quote from the Anton textbook (and it's not just Anton - all the other textbooks would say the same thing) agrees precisely with what I was saying in #13 and with everything that blahblahblah and liyi have been saying.

You can say this - and be careful of the direction of implication:

[b]IF[/b] $\int_0^{\infty}f(x)\,dx$ and $\int_{-\infty}^0f(x)\,dx$ both converge [b]then[/b] $\int_{-\infty}^{\infty}f(x)\,dx=\lim_{h\to\infty}\int_{-h}^hf(x)\,dx$

But it would be equally valid to say that if both integrals converge, then

$\int_{-\infty}^{\infty}f(x)\,dx=\lim_{h\to\infty}\int_{-\ln h}^{e^h}f(x)\,dx$

There nothing special about that balanced (Cauchy principal value) limit in this formulation. You can, as I just have, pair the sides up in a wildly unbalanced fashion. And the existence of the principal value - or any particular unbalanced analogue - does not imply the convergence of the integral.

Anton (and most of the other textbook writers) do feel that it is necessary to give the correct definition of convergence for an integral that is improper for two or more reasons, and to give that correct definition the first time it is taught. Presenting the principal value, or anything like it, as if it were the definition is misleading.[/quote]First, I understand that Anton and all others agree. The reason I quoted from Anton is because I happen to have Anton at hand. (It was my own freshman text, and is well annotated I assure you.) Second, to supply wonky assymetric limits to an improper integral, as in your example, is unnecessary and misleading. The integral between $\pm \infty$ will either converge or not. If not, one still has a problem regardless of whether one is dealing with $L^1, L^2,$ a Hilbert space, or a scalar function of $\mathbb{R}^1$.

[quote]Second, to supply wonky assymetric limits to an improper integral, as in your example, is unnecessary and misleading.[/quote]
That was precisely as unneccessary and precisely as misleading as supplying the symmetric limits of the principal value. My point was that neither one [b]is[/b] the improper integral.

Way back at the top of this topic you wrote "Yes, but you're complicating the situation more than is necessary. [Improper integral] = [formula for principal value]."

That was wrong. [b]That[/b] was unneccessary and misleading. That's not the way calculus should be taught. It's not "hairsplitting" to insist on the correct definition of an improper integral. And that's why both liyi and blahblahblah felt compelled to respond to you.

[quote="Kent Merryfield"]That was precisely as unneccessary and precisely as misleading as supplying the symmetric limits of the principal value. My point was that neither one [b]is[/b] the improper integral.

Way back at the top of this topic you wrote "Yes, but you're complicating the situation more than is necessary. [Improper integral] = [formula for principal value]."

That was wrong. [b]That[/b] was unneccessary and misleading. That's not the way calculus should be taught. It's not "hairsplitting" to insist on the correct definition of an improper integral. And that's why both liyi and blahblahblah felt compelled to respond to you.[/quote] Fine. But I've also added the qualifier (repeatedly) [b]only if[/b] the limit (as h becomes arbitrarily large) exists. Refusal to acknowledge that qualifier is hairsplitting. Nevertheless, it's not worth arguing about any further.


x,y,z are positive real numbers such that xy+yz+zx+2xyz=1.Prove that x+y+z≥2(xy+yz+zx)

Denote $ x \equal{} \frac {a}{b \plus{} c}$ and similarly and  follows Schur Inequality


geometry
advanced fields
advanced fields unsolved

Hi,

 For every degree $ d >0$, give the equation of a non singular curve of degree $ d$ in $ \mathbf{P}^{2}$ over a field of characteristic 0.

Thanks for the help.

Examples are given by the so-called Fermat curves. Can you guess their equation from their name? :D


let a sequence of integers be defined by: $(n-1)a_{n+1}=(n+1)a_n-2(n-1)$, si $2000$ divide a $a_{1999}$ find the smallest $m$ such that $2000$ divides  $a_m$.


It's not realistic, but here's a strip detailing exactly how Windows Vista screws over even the best of computers:

(P.S. It also has liquid cooling.)

[quote="gogators"]nice...

your handwriting is really messy but really neat in a way....

thats a compliment...

 :lol:
 :rotfl:[/quote]

Yeah..that is so true..

His handwriting is neat, and messy....
(And no..thats not a paradox)

really appreciable , creative, outside the box.....

and true too...lolzzz.. :wink:


topology
real analysis
real analysis theorems

I decided to learn Topology on my own. So I got myself Munkres' book. I want to do a lot of exercises from the book.

One of my exercises goes like this:

1. Let X be a topological space with topology $ T$; let A be a subset of X. Suppose that for each $ x\in A$ there is an open set U containing x such that $ U \subset A$. Show that A is open in X.

Proof from previous knowledge: Choose $ x\in A$. Then an open set U exists with $ x \in U$.
Since U is open, there exists $ \delta > 0: \left(x \minus{} \delta, x \plus{} \delta\right) \subset U$.
Then $ U\subset A \implies \left(x \minus{} \delta, x \plus{} \delta\right) \subset A$.
Thus, A is open.

I am supposed to use the definition of openness which says A is open if A belongs to the collection $ T$. I guess I am looking for meaning of being belong to the collection $ T$.

Hint: by the definition of the topological space, the union of open sets is an open set. Show that $ A$ is the union of open sets.

Pf: Choose $ x\in A$.
Then $ \exists$ an index $ \alpha: x\in U_{\alpha} \implies x\in \bigcup U_{\alpha} \implies A\subset \bigcup U_{\alpha}$.

If $ y$ belongs to $ \bigcup U_{\alpha}$, then there is at least one index $ \alpha: y\in U_{\alpha}$. But $ U\subset A \implies y\in A$.
Hence $ A \equal{} \bigcup U_{\alpha}$.

Correct?

Yes, the proof is correct except that you don't define what $ U_{\alpha}$ and $ \alpha$ are; also when you write $ U\subset A$, do you mean $ U_{\alpha}\subset A$? Whether $ U_{\alpha}\subset A$ or not depends on your definition of $ U_{\alpha}$.


algorithm
combinatorics unsolved
combinatorics

We are given a connected, undirected graph $ G \equal{} (V, E)$ with positive edge lengths {$ \ell_e$}, and we want to find a spanning subgraph of it. Now suppose we are willing to settle for a subgraph $ H \equal{} (V, F)$ that is "denser" than a tree, and we are interested in guaranteeing that, for each pair of vertices $ u, v \in V$, the length of the shortest $ u\minus{}v$ path in $ H$ is not much longer than the length of the shortest $ u\minus{}v$ path in $ G$. By the [i]length[/i] of a path $ P$ here, we mean the sum of $ \ell_e$ over all edges $ e \in P$.

Here’s a variant of Kruskal's Algorithm designed to produce such a subgraph.

[list][*]First we sort all the edges in order of increasing length. (You may assume all edge lengths are distinct.)
[*]We then construct a subgraph $ H \equal{} (V, F)$ by considering each edge in order.
[*]When we come to edge $ e \equal{} (u, v)$, we add e to the subgraph $ H$ if there is currently no $ u\minus{}v$ path in $ H$. (This is what Kruskal's Algorithm would do as well.) On the other hand, if there is a $ u\minus{}v$ path in $ H$, we let $ d_{uv}$ denote the length of the shortest such path; again, length is with respect to the values {$ \ell_e$}. We add $ e$ to $ H$ if $ 3\ell_e < d_{uv}$.[/list]

In other words, we add an edge even when $ u$ and $ v$ are already in the same connected component, provided that the addition of the edge reduces their shortest-path distance by a sufficient amount.

Let $ H \equal{} (V, F)$ be the, subgraph of $ G$ returned by the algorithm

a) Prove that for every pair of nodes $ u, v \in V$, the length of the shortest $ u\minus{}v$ path in $ H$ is at most three times the length of the shortest $ u\minus{}v$ path in $ G$.

b) Show that the maximum cycle that $ H$ contains is a 4-cycle.

[quote="modulo"]b) Show that the maximum cycle that $ H$ contains is a 4-cycle.[/quote]  Either I misunderstand something or the problem is wrong: suppose that $ G$ has vertex set $ \{1, 2, 3, 4, 5\}$ with edge-lengths $ \ell(12) \equal{} 7$, $ \ell(23) \equal{} 8$, $ \ell(34) \equal{} 9$, $ \ell(45) \equal{} 10$ and $ \ell(51) \equal{} 11$.  Then $ H \equal{} G$ is a pentagon.

(Part (a) is cute.  I believe we can induct on the number of edges in the path: first, if $ u$ and $ v$ are connected in $ G$ then the path between them can't grow by a factor of more than 3 or else the edge would have been added in.  Now we should be able to build up for longer paths.)

Edit: Part (b) is probably meant to be, "Show that all cycles in $ H$ have length 5 or longer, i.e., $ H$ contains no squares or triangles."  This follows immediately because you add the edges in increasing order of length, since $ 3d < a \plus{} b \plus{} c$ implies $ d < \max\{a, b, c\}$.

First of all thank you very much for your time. :)
True, I should have said "all cycles in $ H$ have length 5 or longer".

I have to show that:

b) A connected, undirected graph with $ n$ vertexes and $ \omega\left(n^{3/2}\right)$ edges contains a cycle of length $ \leq 4$. [i]Note:[/i] Generally, each connected and undirected graph with $ n$ vertexes and $ \omega\left(n^{1 \plus{} 1/k}\right)$ edges contains a cycle of length $ \leq 2k$.

c) Show that the number of edges in $ H$ is $ \mathcal{O}\left(n^{3/2}\right)$, where $ n \equal{} |V|$

That's why I have to show that "all cycles in $ H$ have length 5 or longer".

Proving (c) was easy in the end, but how could we prove the lemma in (b)? :maybe:


group theory
abstract algebra
superior algebra
superior algebra unsolved

hello,

Supposed that G is finite group such that any two elements of the order p are conjugate for any prime divisor p of |G|. Then 1) Sylow 2-subgroup of G is cyclic or generlized quaternion group or not?2) Sylow p-subgroup is normal cyclic for odd prime number p  or not?

Thanks


1) Find all real functions for $\large{xf(x)-yf(y)\,=\,(x-y)f(x-y)}$ .

hmm... is it
[hide]$f(x) = kx, k\in\mathbb{R}$   ?[/hide]

Edit: nevermind that's wrong....

[hide="Hint"]Define $g(x) = x f(x)$.  :) [/hide]

I don't have time now, but I'm on my way to getting it. I will post asap if I'm able to access my laptop and the internet.

ok well going on t0rajir0u's hint we have[hide]
$x \cdot f(x)=g(x)$
$g(x)-g(y)=g(x-y)$
from this we can get $g(x)=kx,  k \in \mathbb{R} \implies \boxed{ f(x)=k,  k \in \mathbb{R} }$. Im not sure if this is the only solution, but higher order functions dont seem to work. If anyone wants to help, go for it.[/hide]

I proved it`s right for $\large{Z}$ using telescoping , but how prove it for reals ?

It still only shows for integers.  Centy is right, we need another condition.

[quote="Centy"]The problem is that although the Cauchy functional equation $f(x + y) = f(x) + f(y)$ only has a solution $f(x) = kx$ for $f : \mathbb{Q} \mapsto \mathbb{Q}$, when we consider $f : \mathbb{R} \mapsto \mathbb{R}$ we get wild solutions which are nowhere bounded.

To avoid this, we need another weak condition which bounds our function.[/quote]

According to [url=http://mathworld.wolfram.com/CauchyFunctionalEquation.html]mathworld[/url], it looks like Cauchy's functional equation applies for $g: \mathbb{R} \mapsto \mathbb{R}$, but we still have the problem that we are dealing with $g(x-y) = g(x) - g(y)$ and not Cauchy's equation.  However if we assume that $g(-x) = - g(x)$ then we can show that $g(x) = kx$ is the only solution because $g(x-y)=g(x) - g(y) \implies g(x + y) = g(x) + g(y)$ if we make that assumption.  (here, $g(x) = x*f(x)$)

[quote="calculuslover800"]However if we assume that $g(-x) = - g(x)$ [/quote]

There is no need to [b]assume[/b] it!

$g(0) = g(x) - g(x) \implies g(0) = 0$
$g(0 - (-x) ) = g(0) - g(-x) \implies g(x) = - g(-x)$

:)

[quote="shobber"]$f$ must be continuous.[/quote]shobber is correct, continuity keeps $f$ bounded.


In a tangle  $ABC$  we take the midpoint $M$ of side $AB$ and let $BD$ the bisector of angle $\angle B$ . If $MD$  is perpendicular to $BD$ , prove that $AB = 3 BC$ .

[u] Babis [/u]

Cute :)

Let P be the midpoint of MB.  Note that PD = PM = PB so in particular, we have

< DPM = 2 < DBM = < ABC, which implies NM, PD, and BC are parallel.

Thus, D is the midpoint of NC and by the angle bisector theorem, AB/BC = AD/DC = 3.

This problem is very nicely, but also is very easily.
It is a good exercise for the beginners ![hide="Here are three methods collected from the my students"] [u]The first method ([b]George[/b]).[/u] I note the middlepoint $N$ of the segment $[BM]$.

$\widehat {ABD}\equiv \widehat {CBD}\Longrightarrow \frac{DA}{DC}=\frac ca;\ DB\perp DM\Longrightarrow NB=NM=ND=\frac c4$

and $ND\parallel BC\Longrightarrow 3=\frac{NA}{NB}=\frac{DA}{DC}=\frac ca\Longrightarrow c=3a.$

[u]The second method ([b]Daniel[/b]).[/u] $DB\perp DM\Longleftrightarrow l_b=\frac c2\cos \frac B2\Longleftrightarrow$

$\frac{2ac}{a+c}\cos \frac B2=\frac c2\cos \frac B2\Longleftrightarrow 4a=a+c\Longleftrightarrow c=3a.$

[u]The third method ([b]Silviu[/b]).[/u] I note $E\in BC\cap MD$ ($DM=DE$) and

I apply the Menelaus's theorem to the transversals $\overline {ADC}$, $\overline {EDM}$

for the triangles $BME$, $ABC$ respectively $\left(\frac{DA}{DC}=\frac ca\right)$:

$\frac{AM}{AB}\cdot \frac{CB}{CE}\cdot \frac{DE}{DM}=1\Longrightarrow \frac{CB}{2}=\boxed {\frac{CE}{1}=\frac{BE}{3}}\ \ (1).$

$\frac{EC}{EB}\cdot \frac{MB}{MA}\cdot \frac{DA}{DC}=1\Longrightarrow \boxed {\frac{EC}{a}=\frac{EB}{c}}=\frac{a}{c-a}\ \ (2).$

From the relations $(1)$ and $(2)$ results $\frac a1=\frac c3$, i.e. $c=3a.$[/hide]


for how many positive real values of K will (2+Ki)cubed be a real number?

[quote="Ubemaya"][hide]
None.
[/hide][/quote]

Not so.

[hide="Correct answer"]One value.[/hide]

$0$ is not a positive value.

[hide]Since 2 is positive and K is positive, the number, when written in polar format, must have a direction between $0$ and $\pi/2$. We need an angle such that $3\theta=n\pi$, where $n$ is a positive integer.

The only such value we can achieve is $\pi$, since $2\pi/3>\pi/2$.

$3\theta=\pi \implies \theta=\pi/3 \implies k/2=\sqrt{3}\implies \boxed{k=2\sqrt{3}}$

This is the only value of $k$. Checking, $(2+2\sqrt{3}i)^{3}=-64$.
[/hide]

or you could just know that $(\frac{1}{2}+\frac{\sqrt3i}{2})^{3}=1$, so $(x+x\sqrt3i)^{3}$ must be real if $x$ is real

All you want is $\Im[(2+ki)^{3}]=0\implies k=2\sqrt{3}$.

[quote="pianoforte"][hide]If it is real, the imgainary part is 0. The imaginary part $=4k-k^{3}=k(2+k)(2-k)=0$. The positive roots are $\pm2$, so $2$.[/hide][/quote]

Pianoforte, I think you forgot to multiply by 3. The imaginary part is actually $12k-k^{2}$, which gives you a positive root of $2\sqrt{3}$.


ceiling function
logarithms
real analysis
real analysis unsolved

If $ k$ is positive integer,  prove that the number of 1's in the binary expansion of $ k$ is at most $ \lceil \log_2 (k) \rceil$.

Thanks

[quote="JRav"][hide="Hint"]
If there is a 1 in the $ n$th place of the binary expansion, what does that mean?  Now what if every space is a 1?  What does that tell you about the number?[/hide][/quote]

I see , thanks


calculus integration derivative vector trigonometry LaTeX MATHCOUNTS

see question
i'm going into calculus next year, btw.

A TI 89 can be [i]very[/i] useful in calculus. It does definite and indefinite integrals and a whole bunch of other stuff that an 83 can't.

It's not necessary, but if you're like me, and you like to check your work, and you are able to buy one, I'd recommend getting it.

89 is excellent, especially in calculus.
I highly recommend it.

The TI-89 is amazing. Especially since they let you use it on the AP Test :P. It really just does your calculus homework for you.

looks like i know what i'll be asking for my birthday.  :lol:

The TI 89 is very useful, but you have to learn a completely new system.

Then again, I presume that committing time to learning how to use a calculator is not a problem for you.

[quote="mcalderbank"]Then again, I presume that committing time to learning how to use a calculator is not a problem for you.[/quote]

not at all  :D

[quote="paladin8"]The TI-89 is amazing. Especially since they let you use it on the AP Test :P. [/quote]

They do?  Awesome!

On the original topic, yes, an 89 is very useful.  Symbolic integration and differentiation is excellent for checking your work (or if you are very, very lazy).  I highly recommend getting one, if the price and time required to learn to use it are not problems.

It's not very fast though.  A simple task like evaluating a big number could take one minute.

The 89 is great! I was hoping to get another one at nats last year, but they changed to 83's. owell, variety is good i guess.

hehe went two years got an 89 and 82

anyways more on the topic, as said before, especially in a course like calculus, the 89 is incredibly useful. i just got an 89-titanium (my old 89 broke down for some reason) and i'm too sure what the difference is, but my guess is they probably improved the speed somewhat, probly still not as fast as an 83 plus, but it works fine

I'm pretty sure the -89 is faster than any -83, but even if it isn't, the -89 is still well worth the extra 50 bucks.  Here's a short list of reasons why the -89 is better than the -83 or -86:

-Differential Equations
-Solves almost any equation algebraically
-3-D Graphs (& implicit plots)
-Indefinite integrals & derivatives
-Infinity button (allows you to do limits to infinity, improper -integrals, etc.)
-Extensive programming capabilities
-Smaller screen resolution
-Leaves things in fractions (not decimals all the time, unless you want it to)
-Units (very, very useful in physics, chemistry, thermophysics, thermochemistry, and more!)
-Vectors (rectangular, polar, cylindrical, and spherical)
-Tab menus
-Countless fun characters, including greek
-Organize variables & programs in folders
-Make any variable name up to 8 characters long
-Split Screen Capabilities
-Complex Numbers (in rectangular and polar)
-Displays things how you would write them on paper - i.e. $ \frac{x^2-9 \frac 1 y}{\sqrt{x^{2x+1}-1}}$ instead of
(x^2-9*(1/y))/( :sqrt: (x^(2x+1)-1))
-Taylor series
-Custom menus
-Numbers in base 2,10, or 16
-Hyperbolic trig
-factor & expand algebraic, complex, and trig expressions
-factor numbers
-Plus everything the -83 and -86 do!

Are you sold yet? :D

For a little extra money, I recommend a TI-89T.

so it alleviates the need for a bunch of messy parentheses? or no?

[quote="adidasty"]The 89 writes like $\LaTeX$????????[/quote]

did anyone say that? i dont think it does...but it would really cool if it did

I think adidasty was referring to the pretty print which makes expressions look, well, pretty!

Get a TI-89! It saved my life a bunch of times on tests in Adv. Analysis II and Number Theory class....

I agree with everyone here that the 89T is awesome.  However, just be careful that you don't become too dependent on it since most teachers know what it is capable of and won't allow it on tests.  (Also the AP exam only allows calculators on half of the free responce part.  Otherwise your on your own).

you mean there is a calculator that tastes better than the 83 silver....wow, i'm gonna be addicted.

I ran into one of those at a circuit city...It was on display and someone got it off of the metal string that tied it to the wall so people wouldn't steal it.  I walked around playing with all the functions...

I still regret that I didn't just walk out with it... :(

[quote="Farazy"]I still regret that I didn't just walk out with it... :([/quote]
but that would be stealing and just by chance some employee would see you and hit the alarm buttom. then the police would come and arrest you. you would go to jail and make many new friends. you would have lots of fun playing cards and basketball. and once a year you would get to play the guards in a football game (*waiting for the longest yard to come to the movies*)

[quote="miraculouspostmaster"]Just be careful.  The TI-89 tastes a lot better than the 83, so the temptation to eat it is a lot stronger.  I might recommend coating it with some flavor you don't like so it's easier to resist.

TI-89 on wheat with thousand island...mmmm <drooling>[/quote]
weirdest analogy ever

I have the 83-plus, but I hope they give out 89T's at MathCounts nats!

Yes, the one in Mathgeek2006's avatar.  The 89 Titanium is pretty cool; I played around with one at some store.   It has an organizer, a spreadsheet (I think) thing and a CLOCK.


IBO 2k14
analytic geometry
parallelogram

Let $AC$ be a line segment in the plane and $B$ a points between $A$ and $C$. Construct isosceles triangles $PAB$ and $QAC$ on one side of the segment $AC$ such that $\angle APB = \angle BQC = 120^{\circ}$ and an isosceles triangle $RAC$ on the other side of $AC$ such that $\angle ARC = 120^{\circ}.$ Show that $PQR$ is an equilateral triangle.

By the law of cosine: $RP^2=AP^2+AR^2-2 \cdot AP \cdot AR \cdot cos{\angle{PAR}}$. Since $\angle{PAR}=\angle{RAC}+\angle{PAB}=30^o+30^o=60^o$, so $RP^2=AP^2+AR^2-AP \cdot AR$. Similarly we have $RQ^2=CQ^2+CR^2-CQ \cdot CR$. Since $AR=CR$, so:

$RP^2-RQ^2=AP^2-CQ^2+AR \cdot (CQ-AP)$
$=(AP+CQ)(AP-CQ)+AR \cdot (CQ-AP)$
$=(AP-CQ)(AP+CQ-AR)$.

Because: $AP \cdot \sin{30^o}+ CQ \cdot \sin{30^o}=\frac{1}{2}AC=AR \sin{30^o}$, so $AP+CQ=AR$, thus $AP+CQ-AR=0$, thus $PR^2=QR^2$, $PR=QR$.

$PQ^2=PA^2+QC^2-2 \cdot PA \cdot QC \cdot \cos{120^o}=PA^2+QC^2+PA \cdot QC$.
Let $RP^2-PQ^2$, we can get:
$RP^2-PQ^2=(CQ+AR)(CQ-AR+AP)=0$

Therefore $PR=QP$, thus $PR=RQ=QP$, so $\triangle{PQR}$ is an equilateral triangle.

Another interesting thing is that the circumcircles of $\triangle{APB}$, $\triangle{BQC}$ and $\triangle{ARC}$ meet at one point.

let P1,Q1,R1 be the points symmetrical to P,Q,R with respect to AC. Then PAP1, QAQ1 are equilateral , and PP1R,QQ1R,PQr1 are equal. So PR=RQ=QR.

[quote="Altheman"]isn't this just napoleon's theorem, except the triangle is degenerate
(the one relating to the first fermat point?)[/quote]
Yes. If we construct equilateral triangles with side $AB$, $BC$, $AC$, then $P$, $Q$, $R$ are the centres of the triangles, thus $\triangle{PQR}$ is equilateral due to Napoleon's Theorem.

But I don't know how it is related the Fermat point...?

the circumcircles of the equilateral triangles drawn on the sides of the triangle in napolean's theorem concur at the first fermat point, only for acute triangles though, so you are correct that it would not be the case for this triangle, but that is how i remember it

just curious, if you wrote that on the actual exam, would you get full credit, your solution would be like a sentence  :lol:

[quote="Altheman"]just curious, if you wrote that on the actual exam, would you get full credit, your solution would be like a sentence  :lol:[/quote]
I'd still use the law of cosine. I don't think that Napolean's theorem on a straight line is acceptable...

Indeed, the Altheman's remark is nicely: this problem is a particular case (just on the limit) of the well-known Napoleon's problem.

[b]A equivalent enunciation.[/b] Let $ABCD$ be a rhombus with $m(\widehat {ABC})=60^{\circ}$. For a point $M\in (BD)$ denote the points $N\in (AB)$, $P\in (AD)$ so that $MN\parallel AD$ and $MP\parallel AB$. Then the triangle $CNP$ is equilaterally.

[hide="Proof using the complex numbers."] Denote $\omega =\cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}$, i.e. $\omega ^3=1$, $\omega ^2=-1-\omega$. Choose the origin $M(0)$ and the line $BD$ as the axis $Ox$. Thus, $B(-1)$, $D(d)$, $A(a)$, $C(c)$, $N(n)$, $P(p)$, where $d\in R$, $d>0$ and $\{a,c,n,p\}\subset \Im\setminus \Re$.

$\blacktriangleright NB=NM,\ m(\widehat {BNM})=120^{\circ}\Longrightarrow 0-n=(-1-n)\omega\Longrightarrow n=\frac{\omega}{1-\omega }.$
$\blacktriangleright PM=PD,\ m(\widehat {DPM})=120^{\circ}\Longrightarrow d-p=(0-p)\omega \Longrightarrow p=\frac{d}{1-\omega }.$
$\blacktriangleright CB=CD,\ m(\widehat {BCD})=120^{\circ}\Longrightarrow -1-c=(d-c)\omega\Longrightarrow c=\frac{-1-d\omega}{1-\omega}.$

$\bigstar \ \ \triangle CNP$ is equilateral $\Longleftrightarrow c^2+n^2+p^2=cn+np+pc\Longleftrightarrow$

$\omega^2 +d^2+(1+d\omega )^2=d\omega -\omega (1+d\omega )-d(1+d\omega )\Longleftrightarrow$

$(d^2+d+1)(\omega ^2+\omega +1)=0$, what is true.[/hide]

[b]A generalization of the proposed problem.[/b]

If $PA=PQ$, $QB=QC$, $RA=RC,\ m(\widehat {APB})=m(\widehat {BQC})=m(\widehat {ARC})=\phi$,
then $PR=PQ\Longleftrightarrow RP=RQ\Longleftrightarrow QP=QR\Longleftrightarrow \phi=120^{\circ}$
and in this case the triangle $PQR$ is equilaterally.

Brute forced with coordinate geometry.
I didn't realise it was a degenerated form of Napoleon's theorem (although in the rmo, I don't think it would be possible to simply state this)

$\{D\}\in AP\cap CQ$; see that $ARCD$ is a rhombus, with $m(\widehat{RAD})=60^\circ$, $PBQD$ a parallelogram, so $AP=DQ, DR=AR, \angle DAR=\angle CDR$, hence $\triangle PAR\equiv \triangle QDR$ and, a $60^\circ$ rotation around $R$ will map each other, i.e. $PR=RQ, m(\widehat{PRQ})=60^\circ$, done.

Best regards,
sunken rock

Let $AP=PB=a$ and $BQ=QC=b$ and $AR=RC=c$.

In $\Delta APB$, by sine rule $AB=\dfrac{AP\sin{120^\circ}}{\sin{30^\circ}}=a\sqrt{3}$
Similarly, $BC=b\sqrt{3}$ and $AC=c\sqrt{3}$.
But $AB+BC=AC$ which gives $a+b=c...(1)$ or $a^2+c^2-ac=b^2+c^2-bc...(2)$

In $\Delta QRC$ by cosine rule $QR^2=b^2+c^2-bc$ and in $\Delta PRA$ $PR^2=a^2+c^2-ac$. Hence from (2) $PR=QR$ and $PR^2=QR^2=a^2+b^2+ab$

Now by applying sine rule and cosine rule in $\Delta QRC$ and $\Delta PRA$ for angles $QRC$ and $ARP$ respectively we(after a lot of calculations) we find $\sin{(\angle{QRC}+\angle{ARP})}=\sqrt{3}/2$ which gives $\angle{PRQ}=60^\circ$ and we are done.

Complex numbers, $p=\frac{a-b\epsilon}{1-\epsilon}, q=\frac{b-c\epsilon}{1-\epsilon}, r=\frac{c-a\epsilon}{1-\epsilon}$, so $r+p\epsilon+q\epsilon^2=0$...

We can start off by labelling the angles. As $\triangle{APB}$ and $\triangle{BQC}$ are isosceles triangles, $\angle{PAB} = \angle{PBA} = 30^{\circ}$ and $\angle{QBC} = \angle{QCB} = 30^{\circ}$. Similarly, $\triangle{RAC}$ is isosceles. This means that $\angle{RAC} = \angle{RCA} = 30^{\circ}$.

Now, extend line $\overline{PA}$ to point $D$ where $D$ is in line with point $C$. Using the fact that angles on a line add up to $180^{\circ}$, $\angle{BPD} = \angle{BQD} = 60^{\circ}$ and $\angle{PBQ} = 120^{\circ}$. Since the angles of a quadrilateral add up to $360^{\circ}$, $\angle{D} = 120^{\circ}$. This implies that $DPQB$ is a parallelogram.

Let $PD = x$ and $PB = y$. As $DPQB$ is a parallelogram, $DQ = y$ and $BQ = x$. Since $\triangle{APB}$ and $\triangle{BQC}$ are isosceles triangles, $PA = y$ and $QC = x$. We know that $\angle{DAR} = \angle{DCR}$ and $\angle{CDA} = \angle{ARC}$. This means that $ADCR$ is a parallelogram. As $DA = DC = x + y$, $CR = AR = x + y$.

Using Law of Cosines on $\triangle{PDQ}$ (to find side PQ), we see that:
$$PQ^2 = x^2 + y^2 - 2xy\cos{120^{\circ}} = x^2 + y^2 + xy$$
Therefore, $PQ = \sqrt{x^2 + y^2 + xy}$.

We can use the Law of Cosines on triangles $PAR$ and $QCR$ as well (to find sides PR and QR).
$$PR^2 = y^2 + (x + y)^2 - 2y(x + y)\cos{60^{\circ}} = x^2 + y^2 + xy$$
Therefore, $PR = \sqrt{x^2 + y^2 + xy}$.

$$QR^2 = x^2 + (x + y)^2 - 2x(x + y)\cos{60^{\circ}} = x^2 + y^2 + xy$$
Therefore, $QR = \sqrt{x^2 + y^2 + xy}$.

All the sides of triangle $PQR$ are equal, implying that it is equilateral. $\blacksquare$

Consider the composition $$\Gamma  = \text{Rot}(R, 120^{\circ}) \  \circ \  \text{Rot}(Q,120^{\circ})  \ \circ \  \text{Rot}(P, 120^{\circ})$$ $\Gamma$ has to be a translation, since the angles add to $360^\circ$, and it takes $A \rightarrow B \rightarrow C \rightarrow A$ so it has a fixed point $A$. Thus $\Gamma$ must be the identity transformation.

Now, note that the inverse function of the rotation $\text{Rot}(A,\alpha)$ is simply the rotation $\text{Rot}(A,-\alpha)$, therefore, composing $\Gamma$ on the left with inverse of $\text{Rot}(R,120^{\circ})$, we get that (since $\Gamma$ is just the identity), that
\[\text{Rot}(Q,120^{\circ}) \ \circ \text{Rot}(P,120^{\circ}) = \text{Rot}(R,-120^{\circ})\]
Therefore, by the[url=https://rijulsaini.wordpress.com/2012/04/12/the-magic-of-rotations/] composition law of rotations[/url], $R$ must be the point of intersection of the lines which make an angle of $60^{\circ}$ with $PQ$ at $P,Q$. Hence, $\angle RPQ = \angle RQP = 60^{\circ}$, and we’re through. $\square$


this is an example of a telescoping series, a series in which the terms have a common form and may be rewritten so that many terms cancel out

for example this sequence becomes
$ (\frac11\minus{}\frac12)\plus{}(\frac12\minus{}\frac13)\plus{}(\frac13\minus{}\frac14)\plus{} \ldots$
and when you write it out like this, it is apparent that many terms cancel leaving $ \frac11\minus{}\frac{1}{100}\equal{}\frac{99}{100}$


Find all real values of  $0\leq x\leq \frac{\pi}{2}$  such that:  $sin^3x-cos^3x=0,008$.

Did you use Cardano's formula :D  :D  :D .

[quote="hungkhtn"]Let $t=sinX-cosX$,we have:

$sin^3X-cos^3X=t(t^2+3\frac{1-t^2}{2})=\frac{t(3-2t^2)}{2}$

And we need to solve

$t(3-2t^2)=0,016$ With $-1\le t \le 1$

I think It's not so difficult.[/quote] I try to understand this equality. I don't find her :?  :( :blush:

[quote="kunny"][quote="hungkhtn"]Let $t=sinX-cosX$,we have:

$sin^3X-cos^3X=t(t^2+3\frac{1-t^2}{2})=\frac{t(3-2t^2)}{2}$[/quote]

You are wrong,hungkhtn.[/quote] Finally, I think that too :P  ;) !

Ok,here's my solution:
First observe that the function $f(x)=sin^3x-cos^3x$ is stricktly increasing.Also $f(0)=-1,f(\frac{\pi}{2})=1$ hence the equation  $f(x)=p,0<p<1$
has a unique root.Next denote $sinx-cosx=2sin\phi$ then you'll see that $f(x)=sin(3\phi)$ So $\phi=\frac{\arcsin p}{3}$.Also from the equation $sinx-cosx=2sin\phi \Leftrightarrow sin(2x)=cos(2\phi)-1$ we'll get $x=\frac{arcsin(cos(2\frac{\arcsin p}{3})-1)}{2}$


The 40 unit squares of the 9 9-table (see below) are labeled. The horizontal or vertical row of 9 unit squares is good if it has more labeled unit squares than unlabeled ones. How many good (horizontal and vertical) rows totally could have the table?

answer is 16, (Example is very simple). assume we can do more - at least 17. Then we have at least $ (17 \cdot 5)/2\equal{}85/2\equal{}42.5$
labeled unit squares. Contradiction.


inequalities
probability
inequalities unsolved

Let $m$ ,$n$ natural numbers with $m,n \geq 2$ and  $x$, $y$ real numbers with $0 <x,y <1$, $x+y=1$. Show that \[\left(1-x^{m}\right)^{n}+\left(1-y^{n}\right)^{m}>1\]


limit
trigonometry
logarithms
calculus
calculus computations

$ a\not\in [\minus{}1,1]$
find
$ \lim_{x\to\infty}\frac{a\plus{}\sin [\ln (x\plus{}b)]}{a\plus{}\sin (\ln x)}$

$ \lim_{x\to\infty}\frac{a\plus{}\sin [\ln (x\plus{}b)]}{a\plus{}\sin (\ln x)} \equal{}\lim_{x\to\infty}(1\plus{}\frac{\sin [\ln (x\plus{}b)]\minus{}\sin (\ln x)}{a\plus{}\sin (\ln x)})\equal{}\\
\equal{}1\plus{}\lim_{x\to\infty}\frac{2\sin \frac{\ln (x\plus{}b)\minus{}\ln x}{2}\cos\frac{\ln (x\plus{}b)\plus{}\ln x}{2}}{a\plus{}\sin (\ln x)}$
Because $ 2\sin \frac{\ln (x\plus{}b)\minus{}\ln x}{2}\to 0$, and $ \left|\frac{\cos\frac{\ln (x\plus{}b)\plus{}\ln x}{2}}{a\plus{}\sin (\ln x)}\right|\le\frac{1}{|a|\minus{}1}$, the limit is equal to 1.


trigonometry
ratio
geometry
geometric transformation
reflection
function
group theory

How can we construct a 27-degree-angle using compasses and ruler?

[hide]
Bisect a $ 60^{\circ}$ angle twice to get a $ 15^{\circ}$ angle. Construct $ \dfrac{\sqrt{5}-1}}{4}$, which is $ \sin{18^{\circ}}$, to get $ 18^{\circ}$. Subtract them and copy the $ 3^{\circ}$ angle 9 times to get $ 27^{\circ}$.
[/hide]

Well, actually, we just need to construct a 18-degree-angle and then bisect it to get a 9-degree-angle and copy it three times.  :)

Well, actually I was being stupid :|

But my way can construct all multiples of 3 :D

[quote="bibobeo"][color=darkred]How can we construct an angle with the measure $ 27^{\circ}$ using compasses and ruler ?[/color][/quote]
[color=darkblue]This problem is equivalently with the following remarkable construct problems

using only the compasses and the ruler :[/color]

[quote="Virgil Nicula"][color=darkred]$ 1\blacktriangleright$ Find $ X\in (AB)$ so that  $ AX^2 \equal{} AB\cdot XB$ ([b]medium and extreme ratio[/b]).

$ 2\blacktriangleright$ Let $ w$ be a circle with the diameter $ [AB]$. Construct the position

of $ X\in w$ for which $ XA \equal{} YB$, where $ Y$ is the projection of $ X$ to $ AB\ .$[/color][/quote]
[color=darkblue]Look for the remarkable book (two tomes - plane and space)[/color]

[color=darkred][b]" [u]Jacques Hadamard[/u] - Lecons de Geometrie elementaire, 1949 "[/b] [/color]

Or construct a regular pentagon, say ABCDE, then construct AC and AE.  Then CAE=108, so then bisect it twice.

[quote="bibobeo"][color=darkred]How can we construct an angle with the measure $ 27^{\circ}$ using compasses and ruler ?[/color][/quote]
[color=darkblue]This problem is equivalently with the following remarkable construct problems :[/color]

[color=darkred]$ 1\blacktriangleright$ Find $ X\in (AB)$ so that  $ AX^2 \equal{} AB\cdot XB$ ([b]medium and extreme ratio[/b]).

$ 2\blacktriangleright$ Let $ w$ be a circle with the diameter $ [AB]$. Construct the position

of $ X\in w$ for which $ XA \equal{} YB$, where $ Y$ is the projection of $ X$ to $ AB\ .$

$ 3\blacktriangleright$ The construct of a regular pentagon inscribed in a given circle ([b]Mathgeniuse^ln x[/b]).
.[/color]

[color=darkblue]Here are three proofs for the first construct problem.[/color]

[color=darkblue][b]Proof I[/b] (from the [b]Hadamard[/b]'s book). Construct the points $ A'$, $ B'$ so that $ AA'\perp AB$, $ BB'\perp AB$, $ AA' \equal{} BB' \equal{} AB$ and the line $ AB$ separates the points $ A'$, $ B'$. The circle $ C(O)$ with the diameter $ [A'B']$ cuts the line $ AB$ in the point $ D$ so that $ B\in (AD)$. Construct the point $ X\in (AB)$ for which $ AX \equal{} BD$. Indeed, for $ AB \equal{} 2$, obtain $ OA \equal{} OB \equal{} 1$, $ AA' \equal{}$ $ BB' \equal{} 2$, $ OD \equal{}$ $ OB' \equal{}$ $ \sqrt 5$, $ XA \equal{}$ $ BD \equal{}$ $ \sqrt 5 \minus{} 1$, $ XB \equal{} AB \minus{} XA \equal{}$ $ 3 \minus{} \sqrt 5$ and $ XA^2 \equal{} AB\cdot XB$ $ \Longleftrightarrow$ $ \left(\sqrt 5 \minus{} 1\right)^2 \equal{} 2\left(3 \minus{} \sqrt 5\right)$, what is truly.
[b]Proof II[/b] ([u]my former teacher[/u]). Construct : the middlepoint $ O$ of the segment $ [AB]$ ; the reflection $ C$ of the point $ O$ w.r.t. the point $ A$ ; the point $ B'$ for which $ BB'\perp AB$ and $ BB' \equal{} AB$ ; the point $ X\in (AB)$ for which $ CX \equal{} OB'$. Prove easily that $ AX^2 \equal{} AB\cdot XB$. Indeed, $ CA \equal{} AO \equal{} OB \equal{} 1$, $ BB' \equal{} 2$, $ CX \equal{} OB' \equal{} \sqrt 5$, $ XB \equal{} 3 \minus{} \sqrt 5$, $ AX \equal{} \sqrt 5 \minus{} 1$. Thus, $ XA^2 \equal{} AB\cdot XB$.[/color]

Uhm, we can construct an angle by ruler and compass if it is a multiplication of 3 just by using $ cos 36^o\equal{} \frac{1\plus{} \sqrt{5}}{4}$ . So can you construct an 1-degree angle?

[quote="borislav_mirchev"]May we constuct by compass and ruller angle with dimension $ 1^{o}$ ?[/quote]

No.  In particular, regular 9-gons are not constructible.  If we could construct a $ 1^\circ$ angle, we could construct a $ 40^\circ$ angle and then easily construct a nonagon.

[quote="borislav_mirchev"]May we constuct by compass and ruller angle with dimension $ 1^{o}$ ?[/quote][color=darkblue]Maybe you, I cann't. Here is an another problem. Do you [b]really[/b] mean it ?!
Or you have a [b]small bell[/b] in the your head ? [b]Sorry for these words[/b] !
Here is peace and stillness. [b]Good luck and ... healthfulness ![/b][/color]

I think it is really interesing questions as well as: "Are 5 degree equations are really not sovable in elementary functions?". I know it is proved by Abel. I don't know the proof of it, but I think there are many open questions in math. As well as construction of angle with dimension of 1 degree with compas and ruller. I found that: http://planetmath.org/?op=getobj&from=objects&id=5728 confirms your answers - I don't really believe to this proof but I don't want to die too yang and beautiful reading some proofs.
Virgil Nicula, you are too emotional. "No take the life in serious - no one is not left this world alive". Don't forget also - math is only my hobby so stupid questions are accepted from time to time for me.

[quote="borislav_mirchev"]... [b]stupid questions are accepted[/b] from time to time for me.[/quote]
If it was proved to be stupidity, then sorry, I don't.

[color=darkblue][b]Proof III[/b] ([u]mine[/u]). Let $ w \equal{} C(O)$ be the circle with the diameter $ [AB]$. Construct the point $ C\in AA$ for which $ AC \equal{} AB$ and the points $ \{E,F\} \equal{} OC\cap w$, where $ E\in (OC)$. Construct the point $ X\in (AB)$ for which $ AX \equal{} CE$. Then the required point is $ X$. Indeed, I"ll use the power of the point $ C$ w.r.t. the circle $ w$ : $ CA^2 \equal{} CE\cdot CF$ $ \Longrightarrow$ $ AB^2 \equal{} CE\cdot (CE \plus{} EF)$ $ \Longrightarrow$ $ AB^2 \equal{} AX\cdot (AX \plus{} AB)$ $ \Longrightarrow$ $ AB\cdot (AB \minus{} AX) \equal{} AX^2$ $ \Longrightarrow$ $ AX^2 \equal{} AB\cdot XB$.[/color]

The easiest way to show that a $ 1^{\circ}$ angle is not constructible is to consider the field of constructible points (subfield of the complex numbers). This is an extension over the rationals and a point is constructible iff there exists a sequence of degree 2 field extensions such that the last one contains the point. A $ 1^{\circ}$ angle would require an extension of degree 3, so it cannot be constructed.

Showing that you cannot solve a general quintic is simply showing that $ S_5$ is not a solvable group and that there exist a quintic such that $ S_5$ is the Galois group of the polynomial.

Both of these proofs are well beyond pre-olympiad level.

Thank you. I see there are lots of very skilled people in mathlinks.


Triangle ABC has median AM. A line $ d$ through M and perpendicular to AM. $ B',C'$ are the projection of $ B,C$ on $ d$. A point N moves on $ d$. The perpendicular bisectors of $ B'N, C'N$ meets $ AB,AC$ at $ P,Q$. The line $ d'$ through N and perpendicular to PQ. Prove that $ d'$ passes through a fixed point when N moves on d.

[b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ M$ be, the midpoint of its side-segment $ BC.$ We draw the line $ (d)$ through $ M$ and perpendicular to $ AM$ and let $ D,\ E,$ two arbitrary points on $ (d).$ We denote as $ P,$ the intersection point of the lines through $ D,\ E$ and perpendicular to $ AC,\ AB,$ respectively. Prove that $ PN\perp BC,$ where $ N$ is the midpoint of the segment $ DE.$[/color][/size][/b]

$ \bullet$ Let $ D,\ E$ be, the midpoints of the segments $ NB',\ NC'$ respectively and we denote as $ D',\ E',\ M',$ the orthogonal projections of $ D,\ E,\ M$ respectively, on the line segments $ AC,\ AB,\ PQ.$

It is well known that the line segments $ DD',\ EE',\ MM',$ ae concurrent at one point so be it $ K,$ as the [b][size=100]orthopole[/size][/b] of the triangle $ \bigtriangleup APQ,$ with respect to the line $ (d)$ $ ($ easy to prove by [b][size=100]Carnot theorem[/size][/b] $ ).$

Because of now, $ DE \equal{} \frac {B'C'}{2}$ and $ \angle KDE \equal{} \angle MAC$ and $ \angle KED \equal{} \angle MAB,$ we conclude that  $ \bigtriangleup KDE,$ is a constant triangle, when the point $ N,$ is variable on the line $ (d)$ and then, the segment $ KL,$ where $ L$ is the midpoint of the segment $ DE,$ is constant too.

$ \bullet$ Baesd on the above [b][size=100]Lemma[/size][/b], we have that $ KL\perp BC.$

Because of now, the points $ M,\ D,\ E,\ L,$ are the midpoints of the segments $ B'C',\ NB',\ NC',\ DE$ respectively, it is easy to show that $ ML \equal{} LN$ $ ,(1)$

The line segment $ KN,$ intersects the midperpendicular of $ BC,$ at one point so be it $ Z$ and because of $ (1),$ we have $ KN \equal{} KZ$ $ ($ from $ KL\parallel MZ$ $ ).$

It is easy to show that the point $ Z,$ is the [b][size=100]orthopole[/size][/b] of the given triangle $ \bigtriangleup ABC,$ with respect to the line $ (d),$ because of the lines through the points $ B',\ C'$ and perpendicular to $ AC,\ AB$ respectively, pass through the point $ Z$ $ ($ from $ B'D \equal{} DN$ and $ C'E \equal{} EN$ $ ).$

So, we conclude that the line $ (d'),$ through the point $ N$ and perpendicular to $ PQ,$ as parallel to $ MK,$ intersects the line segment $ MZ,$ as the midperpendicular of $ BC,$ at one point so be it $ T,$ such that $ MT \equal{} MZ \equal{} 2\cdot KL \equal{} constant$ $ ,(2)$

As an interesting remark in our configuration, we can say that the point $ T,$ is the [b][size=100]orthopole[/size][/b] of the triangle $ \bigtriangleup A'BC,$ where $ MA \equal{} MA',$ with respect to the line $ (d)$ ( easy to prove ).

Hence, from $ (2)$ we conclude that the line $ (d'),$ as the problem states, passes through the fixed point $ T$ and the proof is completed.

$ \bullet$ This proof is dedicated to [b][size=100]Fang Jianhua[/size][/b].

Kostas Vittas.

PS. I will post here later, the proof of the Lemma I have in mind.

[quote="vittasko"][quote="vittasko"][b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ M$ be, the midpoint of its side-segment $ BC.$ We draw the line $ (d)$ through $ M$ and perpendicular to $ AM$ and let $ D,\ E,$ two arbitrary points on $ (d).$ We denote as $ P,$ the intersection point of the lines through $ D,\ E$ and perpendicular to $ AC,\ AB,$ respectively. Prove that $ PN\perp BC,$ where $ N$ is the midpoint of the segment $ DE.$[/color][/size][/b][/quote]
[b][size=100]PROOF[/size][/b]. - Let $ D',\ E'$ be, the orthogonal projections of $ D,\ E$ respectively, on $ AC,\ AB.$

We draw the line through the point $ D$ and perpendicular to $ AB,$ which intersects the line segments $ PN,\ AB,$ at points $ K,\ K',$ respectively.

Because of $ DK\parallel PE$ and $ ND \equal{} NE$ $ \Longrightarrow$ $ PN \equal{} NK$ $ ,(1)$

Because of the cyclic quadrilaterals $ ADMD'$ and $ AMDK',$ we have that $ \angle PDN \equal{} \angle MAC$ $ ,(2)$ and $ \angle KDN \equal{} \angle MAB$ $ ,(3)$

$ \bullet$ Because of now, the triangles $ \bigtriangleup ABC$ and $ \bigtriangleup DKP,$ have equal the angles formed by their medians $ AM,\ DN,$ through their homologous vertices $ A,\ D$ respectively, with respect to their homologous sidelines $ AB,\ DK$ and $ AC,\ DP,$ we can say that these triangles are similar.

This is a not well argument, but it is not difficult to prove it. It is enough to think that for every pair of points $ B',\ C',$ on $ AB,\ AC$ respectively, such that $ B'C'\parallel BC,$ the line segment $ AM,$ bisects the segment $ B'C'.$ Inversely, if the line segment $ AM$ bisects the segment $ B'C',$ the conclusion of $ B'C'\parallel BC,$ it holds.

So, because of the similarity of the triangles $ \bigtriangleup ABC$ and $ \bigtriangleup DKP,$ we conclude that $ \angle DPK \equal{} \angle ACB$ and so, the quadrilateral $ PD'CL,$ where $ L\equiv BC\cap PK,$ is cyclic.

Hence, we conclude that $ \angle PLC \equal{} \angle PD'C \equal{} 90^{o}$ $ \Longrightarrow$ $ PL\equiv PN\perp BC$ and the proof is completed.

Kostas Vittas.[/quote]

Thank Kostas Vittas very much.


1. For any quadrilateral $ ABCD$ with perimeter $ p$, prove that \[ p>AC\plus{}BD>\frac{p}{2}\]

2. For any real numbers $ a$ and $ b$, prove that $ |a\plus{}b|\plus{}|a\minus{}b|\ge |a|\plus{}|b|$. When does equality hold?

[quote="nayel"]1. For any quadrilateral $ ABCD$ with perimeter $ p$, prove that
\[ p>AC\plus{}BD>\frac{p}{2}\]
[/quote]

Let $ AC\cap BD\equal{}O$
$ AB\plus{}BC>AD$
$ BC\plus{}CD>BD$
$ CD\plus{}DA>AC$
$ DA\plus{}AB>BD$
Adding the inequalities we have $ 2p>2(AC\plus{}BD)\Rightarrow p>AC\plus{}BD$

$ AO\plus{}OB>AB$
$ BO\plus{}OC>BC$
$ CO\plus{}OD>CD$
$ DO\plus{}OA>DA$
Adding the inequalities we have $ 2(AC\plus{}BD)>p\Rightarrow AC\plus{}BD>\frac{p}{2}$

[quote="nayel"]
2. For any real numbers $ a$ and $ b$, prove that $ |a\plus{}b|\plus{}|a\minus{}b|\ge |a|\plus{}|b|$. When does equality hold?[/quote]
2 cases:
1.for $ |a|\geq |b|$
$ |a\plus{}b|\plus{}|a\minus{}b|\geq 2|a|\geq |a|\plus{}|b|$
2.for $ |b|\geq |a|$
$ |a\plus{}b|\plus{}|\minus{}(b\minus{}a)| \equal{} |a\plus{}b|\plus{}|b\minus{}a|\geq 2|b|\geq |a|\plus{}|b|$

so equality case is same as triangle inequality.that is $ ab\geq 0$
[hide="secret"]feeling sleepy so something may be wrong :huh: [/hide]

[quote="Moonmathpi496"]so equality case is same as triangle inequality.that is $ ab\geq 0$[/quote]

You're wrong! :D  For example, take $ a \equal{} 3,b \equal{} 2$. :wink:

[hide="Correct version"]If $ |a|\ge |b|$, we have $ |a\plus{}b|\plus{}|a\minus{}b|\ge |a\plus{}b\plus{}a\minus{}b| \equal{} |2a| \equal{} 2|a|\ge |a|\plus{}|b|$. Else, $ |a\plus{}b|\plus{}|a\minus{}b| \equal{} |a\plus{}b|\plus{}|b\minus{}a|\ge |2b|\ge |a|\plus{}|b|$.

Now to determine the equality cases, consider the two inequalities we used:

$ 1.\ |a\plus{}b|\plus{}|a\minus{}b|\ge 2|a|$ (or, $ |a\plus{}b|\plus{}|a\minus{}b|\ge 2|b|$)

$ 2.\ 2|a|\ge |a|\plus{}|b|$ (or, $ 2|b|\ge |a|\plus{}|b|$)

In the first inequality, equality holds if $ a\plus{}b$ and $ a\minus{}b$ (resp. $ b\minus{}a$) has the same sign. And in the second one equality holds for $ |a|\equal{}|b|$. First suppose that $ a\plus{}b$ and $ a\minus{}b$ (resp. $ b\minus{}a$) are both positive. These clearly imply $ a\ge b$ and $ a\ge\minus{}b$. Hence $ |a|\ge |b|$. This is our second inequality, in which equality will hold if $ |a| \equal{} |b|$. Combining both cases, it's immediate that equality holds only if $ |a| \equal{} |b|$. Now if $ a\plus{}b$ and $ a\minus{}b$ are both negative, the argument is similar. Only in this case we will have $ a\le b$ and $ a\le\minus{}b$. So we conclude that equality holds if and only if $ \boxed{a \equal{}\pm b}$.[/hide]

[quote="nayel"][quote="Moonmathpi496"]so equality case is same as triangle inequality.that is $ ab\geq 0$[/quote]

You're wrong! :D  For example, take $ a \equal{} 3,b \equal{} 2$. :wink:

[/quote]

huh..this is why i should not do math (and send solution) at 12:43 AM :( .As i was sleepy i thought that something might be wrong.Whatever...thank u for this nice problem.

The second one's easy if you simply recognize that

$ a^{2}\plus{}2|ab|\plus{}b^{2}\geq a^{2}\plus{}2ab\plus{}b^{2}$

Then all that is necessary is factorization and you have your inequality. A similar approach also holds for $ |a\minus{}b|\geq |a|\minus{}|b|$


geometry
Triangle
midpoint
incircle
concurrency
IMO
IMO 1982

A non-isosceles triangle $A_{1}A_{2}A_{3}$ has sides $a_{1}$, $a_{2}$, $a_{3}$ with the side $a_{i}$ lying opposite to the vertex $A_{i}$. Let $M_{i}$ be the midpoint of the side $a_{i}$, and let $T_{i}$ be the point where the inscribed circle of triangle $A_{1}A_{2}A_{3}$ touches the side $a_{i}$. Denote by $S_{i}$ the reflection of the point $T_{i}$ in the interior angle bisector of the angle $A_{i}$. Prove that the lines $M_{1}S_{1}$, $M_{2}S_{2}$ and $M_{3}S_{3}$ are concurrent.

This is the most difficult problem of IMO1982. It is problem 2

[i]The following table shows the distribution of scores on each individual question.

Score        0   1   2  3   4   5   6   7

Question 1 04 18 08 11 20 00 04 54
Question 2 80 09 08 02 01 03 08 08
Question 3 31 01 36 08 09 02 05 27
Question 4 36 04 02 14 08 06 05 44
Question 5 17 03 03 01 03 03 07 82
Question 6 46 12 11 09 08 07 06 20[/i]

http://www.srcf.ucam.org/~jsm28/imo-scores/1982/

I remember reading somewhere that this was the most difficult, but I didn't want to scare people off :D  :D It took me some time to solve it (meaning thet it took me A LOT of time), but I don't think it's THAT nasty...

Hint: I think a problem posted on this forum before might help (I'm talking about [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=764]the proof for Feuerbach's theorem, which Lagrangia posted in the 'Geometry' section[/url]; it might be in the 'Solved Problems' section now )

[b]EDIT:[/b] Sketch of a solution: An angle chase will show that $S_iS_j\|a_k$ for all permutations $(i,j,k)$ of $(1,2,3)$, so the triangles $S_1S_2S_3,M_1M_2M_3$ are homothetic, meaning that the lines connecting the corresponding points, namely $S_iM_i$, must be concurrent.

P.S.

The concurrence point is the Feuerbach point of $ABC$, i.e. the point where the nine-point circle touches the incircle.

The Additional Theorem in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118923#118923]http://www.mathlinks.ro/Forum/viewtopic.php?t=14927 post #2[/url] is stronger than the problem: it shows that the lines $M_{1}S_{1}$, $M_{2}S_{2}$ and $M_{3}S_{3}$ pass through the Feuerbach point of triangle $A_{1}A_{2}A_{3}$.

  Darij

Dear Mathlinkers,
you can see a synthetic proof on
http://perso.orange.fr/jl.ayme  vol. 9 Le triangle reflechi p. 10.
Sincerely
Jean-Louis

[u][b]PROOF SKETCH:[/b][/u]

[b]1.[/b]$S_1,S_2,S_3$ all lie on the incircle.

[b]2.[/b]$S_1S_2 \parallel M_1M_2,S_2S_3 \parallel M_2M_3,S_3S_1 \parallel M_3M_1$.

[b]3.[/b]$\triangle{M_1M_2M_3} \sim \triangle{S_1S_2S_3}$.

[b]4.[/b]$M_1S_1,M_2S_2,M_3S_3$ concur at the centre of homothety.

For the proof of Sayantan da in a bit detail.
See http://www.math.ust.hk/excalibur/v9_n4.pdf

Rename the vertices so that $A_1=A,A_2=B$ and $A_3=C$. Let $I$ be the incenter of $\triangle ABC$.

[b]Claim 1:[/b] $S_1, S_2, S_3$ all lie on the incircle.
[i] Proof: [/i]By construction, $IS_1$ is the reflection of $IT_1$ with respect to the $A$ angle bisector since $I$ lies on the angle bisector of $A$. So, $IS_1=IT_1$ so $S_1$ lies on the incircle.  Similarly, $S_2$ and $S_3$ lie on the incircle. $\Box$

Note that $I$ is the circumcenter of $\triangle S_1S_2S_3$ because of equal radii due to Claim 1.

[b]Claim 2: [/b]$\angle S_1IS_2=2\angle C.$
[i] Proof: [/i]In cyclic quadrilateral $T_1IT_2C$, $\angle T_1IT_2=180^{\circ}-\angle C$. Let $D$ be the point where the $A$ angle bisector intersects $BC$. Due to the construction of $S_1$, $\angle S_1IT_1=2\angle DIT_1. $ But, $\angle ADC=180^{\circ}-\angle C-\frac{1}{2}\angle A$ so $\angle DIT_1=\frac{1}{2}\angle A+\angle C-90^{\circ}.$ Thus, $$\angle S_1IT_1=\angle A+2\angle C-180^{\circ}.$$ Similarly, $\angle S_2IT_2=\angle B+2\angle C-180^{\circ}.$ Therefore, $\angle S_1IS_2=\angle S_1IT_1+\angle T_1IT_2+\angle T_2IS_2=2\angle C.$$\Box$

Hence, we use Claim 2 to get $\angle S_1S_3S_2=\frac{1}{2}\angle S_1IS_2=\angle C$ and we similarly get $\angle S_3S_1S_2=\angle A$ and $\angle S_1S_2S_3=\angle B.$

[b]Claim 3: [/b]Let $X_1$ be the midpoint of $S_2S_3.$ Then, $T_1IX_1$ is collinear.
[i]Proof: [/i]We already derived that $\angle T_1IT_2=180^{\circ}-\angle C$ and $\angle S_2IT_2=\angle B+2\angle C-180^{\circ}.$ Recall that a radius is perpendicular to the midpoint of a chord so $S_2S_3\perp IX_1.$ Then, $\angle X_1IS_2=90^{\circ}-\angle X_1S_2I=90^{\circ}-(90^{\circ}-\angle A)=\angle A.$ Thus, $$\angle T_1IX_1=\angle X_1IS_2+\angle S_2IT_2+\angle T_2IT_1=\angle A+\angle B+2\angle C-180^{\circ}+180^{\circ}-\angle C=180^{\circ},$$ as desired.$\Box$

Then, $T_1IX_1\perp S_2S_3$ because a diameter is perpendicular to a chord at its midpoint and because $T_1IX_1$ is part of a diameter, by Claim 3.  Hence, $S_2S_3\parallel BC.$ But, $M_2M_3\parallel BC$ so $S_2S_3\parallel M_2M_3$.  Similarly, we get $S_1S_2\parallel M_1M_2$ and $S_1S_3\parallel M_1M_3$ so $\triangle M_1M_2M_3\sim \triangle S_1S_2S_3$ with corresponding parallel sides. Thus, there exists a homothety that maps $M_1M_2M_3$ to $S_1S_2S_3$ and $M_1S_1, M_2S_2,$ and $M_3S_3$ intersect at the center of the homothety. $\blacksquare$

Here's a proof I found.
The distance from $S_2$ to ${A_2}{A_3}$ is equal to the distance from $T_2$ to ${A_1}{A_3}$ which can be written as ${A_1}{T_2}\centerdot\sin\angle{A_2}{A_1}{A_3}$
Similarly, The distance from $S_3$ to ${A_2}{A_3}$ can be written as ${A_1}{T_3}\centerdot\sin\angle{A_2}{A_1}{A_3}$
And we have ${A_1}{T_2}={A_1}{T_3}$
Therefore ${S_2}{S_3}\parallel{A_2}{A_3}\parallel{M_3}{M_2}$
So $\triangle{M_3}{M_2}{M_1}$ is homothetic to $\triangle{S_3}{S_2}{S_1}$, which implies the concurrency of ${M_1}{S_1},{M_2}{S_2},{M_3}{S_3}$

An interesting fact about this proof is that we don't even need to prove that ${S_I}$ lies on the incircle.


integration
calculus
calculus computations

Find the volume of the region in $ \mathbb R^3$ bounded by the paraboloids with equations $ z\equal{}3\minus{}x^2\minus{}y^2$ and $ z\equal{}x^2\plus{}y^2\minus{}5$.

The paraboloids intersection when $ 3 \minus{} x^2 \minus{} y^2 \equal{} x^2 \plus{} y^2 \minus{} 5\Rightarrow x^2 \plus{} y^2\equal{}4$ so the volume is

\[ V\equal{}\iint_{x^2 \plus{} y^2\le 4}\left[(3 \minus{} x^2 \minus{} y^2 )\minus{}(x^2 \plus{} y^2 \minus{} 5)\right]\, dA\]
\[ \equal{} \int_{0}^{2\pi}\int_{0}^{2}(8\minus{}2r^2) r \,dr\, d\theta \equal{}2\pi \left[4r^2\minus{}\frac{1}{2}r^4\right]_{0}^{2} \equal{}16\pi\]


let $n$ be positive integer ($n\ge2$) find the maximum value of
$\sum_{1\le i<j\le n}^{} {x_i{x_j}}({x_i}+x_j)$
where $x_1,x_2,.....x_n\ge0 $and$ x_1+x_2+...+x_n=1$.

I think you want the maximal value. Yet, it is strange to see it was on a such early shortlist, I remember to have seen something very similar in IMO 1999.  :?

Ok, I have now a nice solution. Observe that the sum is in fact $(\sum {x_i})\cdot (\sum{x_i^2})-\sum{x_i^3}=\sum{x_i^2(1-x_i)}\leq\frac{1}{4}\cdot\sum{x_i}=\frac{1}{4}$. I used that $ 4a(1-a)\leq 1$. Thus the maximal value is $1/4$ attained when two variables are equal to $1/2$ and the rest are 0.


Let $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}1$.

Prove, that $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\geq\frac{25}{1\plus{}48abc}.$

[quote="ggggg"]Let $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$.

Prove, that $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\geq\frac {25}{1 \plus{} 48abc}.$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119301

[quote="ggggg"]Let $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$.

Prove, that $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\geq\frac {25}{1 \plus{} 48abc}.$[/quote]

setting $ F(a,b,c) \equal{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \minus{} \frac {25}{1 \plus{} 48abc}$

we have $ F(a,b,c) \minus{} F(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2}) \ge 0 \leftrightarrow 3a\ge 0$ (right)

$ \rightarrow F(a,b,c)\ge F(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2})$

we need prove $ F(a,b,c)\ge 0$ true with $ a \equal{} 1 \minus{} 2t$ and $ b \equal{} c \equal{} t$

$ \leftrightarrow F(1 \minus{} 2t,t,t)\ge 0$ (ineq that easy)

done


abstract algebra
superior algebra
superior algebra unsolved

Let $ A$ be a ring, $ I$ directed set and let $ (M_i)_{i\in I}$ be a family of $ A$-modules indexed by $ I$. For each pair $ i,j\in I$ such that $ i\leq j$, let $ \mu_{i,j}: M_i\to M_j$ be an $ A$-homomorphism, and suppose that the following axions are satisfied:
1. $ \mu_{i,i}$ is the identity mapping of $ M_i$, for all $ i\in I$.
2. $ \mu_{i,k}\equal{}\mu_{j,k}\circ \mu_{i,j}$ whenever $ i\leq j\leq k$.
Let $ C$ be the direct sum of the $ M_i$, and identify each module $ M_i$ with its canonical image in $ C$. Let $ D$ be the submodule of $ C$ generated by all elements of the form $ x_i\minus{}\mu_{i,j}(x_i)$ where $ i\leq j$ and $ x_i\in M_i$. Let $ M\equal{}C/D$, let $ \mu : C\to M$ be the projection and let $ \mu_i$ be the restriction of $ \mu$ to $ M_i$.

1. Show that every element of $ m$ can be written in the form $ \mu_i(x_i)$ for some $ x_i\in M_i$.
2. Show that if $ \mu_i(x_i)\equal{}0$ then there exists $ j\geq i$ such that $ \mu_{i,j}(x_i)\equal{}0$ in $ M_j$.

for the lazy of us: http://www.math.rochester.edu/people/grads/towsley/AMSols.pdf


That was a great movie but, Great Debators FTW!!!

[quote="suma_milli"][quote="mewto55555"]I'm a huge fan of the Barney and Teletubbies movies........anything with Sesame Street is good too.[/quote]

Omg, what a coincidence?

Fellow barney and teletubbie fans unite!

... :P[/quote]
 :P YAY!!!!

I saw a really funny movie recently. What was it called,  "An Inconvient Truth" I think....


Evaluate the limit:

$ lim_{n\minus{}>\infty}cos\frac{\pi}{2^2}cos\frac{\pi}{2^3}...cos\frac{\pi}{2^{2n}}$

[quote="violetcraze"]Evaluate the limit:

$ lim_{n \minus{} > \infty}cos\frac {\pi}{2^2}cos\frac {\pi}{2^3}...cos\frac {\pi}{2^{2n}}$[/quote]
[hide="Solution"]
We note that $ \sin \frac {\pi}{2} \equal{} 2\sin \frac {\pi}{4}\cos \frac {\pi}{4} \equal{} 2\cos \frac {\pi}{4}(2\sin \frac {\pi}{8} \cos \frac {\pi}{8}) \equal{} 4 \cos \frac {\pi}{4} \cos \frac {\pi}{8}(2\sin \frac {\pi}{16}\cos \frac {\pi}{16})...$  If we keep on doing this for $ n$ times, we will find that $ \frac {\sin \frac {\pi}{2}}{2^{2n \minus{} 1}} \equal{} \cos \frac {\pi}{4} \cos \frac {\pi}{8} \cos \frac {\pi}{16}...\cos \frac {\pi}{2^{2n}}\sin \frac {\pi}{2^{2n}} \equal{} \cos \frac {\pi}{4} \cos \frac {\pi}{8} \cos \frac {\pi}{16}...\cos \frac {\pi}{2^{2n}}$ since $ \sin \frac {\pi}{2^{2n}} \equal{} 1$ as $ n$ goes to infinity.  So, our answer is $ \boxed{\frac {1}{2^{2n\minus{}1}}}$ [/hide]

[quote="t0rajir0u"]Not quite.  You messed up the last step there; see [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=170350]my latest blog entry[/url].  This is Vieta's identity :)[/quote]
Oh, sorry, t0rajir0u... yeah, I did mess up...
By the way, I got the solution idea from your blog (actually, I completely forgot that you actually had the problem and the solution on your blog...) I just made a mistake at the end because I though it was $ \cos \frac {\pi}{2^{2n}}$...
[hide="Revised"]
$ \sin \frac {\pi}{2^{2n}} \equal{} \frac {\pi}{2^{2n}}$ when $ n$ nears infinity.  So, we have that $ \cos \frac {\pi}{4}\cos \frac {\pi}{8}...\cos \frac {\pi}{2^{2n}}(\frac {\pi}{2^{2n}}) \equal{} \frac {1}{2^{2n \minus{} 1}}$.  Thus, $ \cos \frac {\pi}{4}\cos \frac {\pi}{8}...\cos \frac {\pi}{2^{2n}} \equal{} \frac {2}{\pi}$. [/hide]


Για τους φοιτητές μας , που τόσο δυναμικά συμμετέχουν στο φόρουμ ,  έφτιαξα τον δικό τους χώρο!
 Εδώ μπορείτε να προτείνετε , να λύνετε και να συζητάτε προβλήματα που αφορούν τον Πανεπιστημιακό - Πολυτεχνικό χώρο .
  Δεν θα ήταν άσχημα να μαζεύετε σιγά - σιγά και θέματα εξεταστικών περιόδων , κάτι που θα μαζέψει περισσότερο κόσμο και η επικοινωνία θα απλωθεί πιο ευχάριστα !

[color=red][b] Καλή δύναμη -  Καλές εμπνεύσεις και Καλές σπουδές !!![/b][/color]

Από την ανισότητα $ (x \plus{} 1)^y \ge 1 \plus{} xy, y \ge \minus{} 1$ πέρνουμε:

$ A \equal{} a^b \plus{} b^a \equal{} (a \minus{} 1 \plus{} 1)^b \plus{} (b \minus{} 1 \plus{} 1)^a \ge 1 \plus{} b(a \minus{} 1) \plus{} 1 \plus{} a(b \minus{} 1) \equal{} 2 \plus{} 2ab \minus{} a \minus{} b \equal{} B$

έστω $ b \equal{} a \plus{} c$ τότε (προφανώς) $ \minus{} 1 \le c < 1(I)$ άρα

$ B > 1 \Leftrightarrow 1 \plus{} 2a(a \plus{} c) \minus{} a \minus{} (a \plus{} c) \ge 0 \Leftrightarrow 2a^2 \minus{} 2a(1 \minus{} c) \plus{} (1 \minus{} c) > 0$

αλλά η διακρίνουσα ως προς a είναι $ D_a \equal{} 4(1 \minus{} c)^2 \minus{} 8(1 \minus{} c) \equal{} 4(1 \minus{} c)(1 \minus{} c \minus{} 2) \equal{} \minus{} 4(1 \minus{} c)(1 \plus{} c)$ άρα
(λόγω $ (I)$) είναι $ D_a < 0$ άρα $ 2(2a^2 \minus{} 2a(1 \minus{} c) \plus{} (1 \minus{} c)) > 1 \Leftrightarrow 2a^2 \minus{} 2a(1 \minus{} c) \plus{} (1 \minus{} c) > 0$

άρα πράγματι $ B > 1 \Rightarrow A > 1 \Leftrightarrow a^b \plus{} b^a > 1$ q.e.d. :lol:

(Δυσκολούτσικη ήταν!)

[quote="paulos18"]Πολύ ωραία λύση...τι να πω τώρα που το βλέπω δεν ήταν και τόσο δύσκολο ήταν τελικά  :P
Υπάρχει καμιά λύση και με ανώτερα μαθηματικά;Μας την έβαλε ένας καθηγητής για αυτό το σκοπό...[/quote]

Τι ακριβώς σημαίνει "ανώτερα μαθηματικά"? ;)

jk :)

Cheerio,

Durandal 1707

Ίσως λίγο λάθος έκφραση>>>>Εννούσα παραγώγους κτλ. και όχι στοιχειώδης τρόπο όπως εδώ :P
H έκφραση βεβαια είναι θέμα συνήθειας γιατί "ανώτερα μαθηματικά" δεν υπάρχουν απλά συνηθίζεται να ονομάζουν έτσι τα μη στοιχειώδη... :wink:

Σκεφτόμουν ξανά τη λύση αυτή και παρατήρησα μόλις τώρα ότι η ανισότητα $ (1 \plus{} x)^y\geq1 \plus{} xy$ δεν ισχύει για $ y\geq \minus{} 1$ αλλά για $ y\geq1$ ή $ y\leq0$ και $ x > \minus{} 1$ ενώ δεν ισχύει για $ 0 < y < 1$ και $ x < 0$ αφού π.χ για $ x \equal{} \minus{} \frac {1}{2},y \equal{} \frac {1}{2}$ γίνεται $ \frac {1}{2}^{\frac {1}{2}}\geq\frac {3}{4}\Rightarrow \frac {4}{3}\geq \sqrt2$ άτοπο. Άρα η λύση είναι λάθος αφού δεν θα ισχύει $ (a \minus{} 1 \plus{} 1)^b \ge 1 \plus{} b(a \minus{} 1)$ γιατί $ 0 < b < 1$ και $ a\minus{}1<0$ :maybe:
Καμιά άλλη πρόταση; :maybe:

Λοιπον, η γενικευμενη ανισοτητα Bernoulli λεει οτι

Αν $ \minus{} 1 < x\neq 0$ τοτε

$ (1 \plus{} x)^a > 1 \plus{} ax$, αν $ a > 1$ ή $ a < 0$

και

$ (1 \plus{} x)^a < 1 \plus{} ax$, αν $ 0 < a < 1$

___________________________________________

Εμεις εχουμε $ \minus{} 1 < a \minus{} 1 < 0$ και $ 0 < b < 1$ αρα

$ (a \minus{} 1 \plus{} 1)^b < 1 \plus{} b(a \minus{} 1)$

Οποτε maniopa η λυση σου ειναι λαθος...

Ισως να λυνεται με αναλυση. Αν παρεις την συναρτηση $ f(x)\equal{}x^b\plus{}b^x\minus{}1$ παραγωγιζοντας την ευκολα διχνεις οτι $ f'(x)>0$ για καθε $ x\in (0,1)$ και επομενως $ f(x)>f(0)\equal{}0$  :maybe:

Το έχω προσπαθήσει αλλά πώς ακριβώς βγάζεις $ f'(x)\equal{}bx^{b\minus{}1}\plus{}b^xlnb>0$? :maybe:

:oops: εχεις δικιο δεν μπορεις να το δειξεις.

Δίκιο έχετε, η λύση μου είναι λάθος.

Αλλά νομίζω ότι αρχίζω να τα χάνω τελείως.  :(

Δείτε το παρακάτω και πείτε μου πού είναι το λάθος:

έστω $ f(x)\equal{} e^x \minus{} \frac{1}{2}$ τότε

$ f'(x) \equal{} e^x >0$ άρα $ f(x) \equal{}$ αύξουσα

άρα (επειδή $ b < 1 \Leftrightarrow lnb < 0 \Leftrightarrow alnb < 0$)

$ f(a lnb) < f(0) \equal{} \frac{1}{2}$

ομοίως
$ f(b lna) < \frac{1}{2}$

άρα (από πρόσθεση)

$ f(a lnb) \plus{} f(b lna) < 1 \Leftrightarrow e^{a lnb} \plus{} e^{b lna} < 1 \Leftrightarrow b^a \plus{} a^b < 1$

[quote="paulos18"][color=red]Να δειχθεί ότι α^β+β^α>1 για κάθε 0<α,β<1 και α<>β[/color]
Ίσως σας φανεί εύκολη ανισότητα δεν ξέρω,πάντως εμένα με παιδεύει πολύ και δεν μπορώ να τη βγάλω.  :o[/quote]

Οταν γραφεις 0<α,β<1 εννοεις οτι $ 0<a<1$ και $ 0<b<1$ ;  :huh:

Τα -1/2 που πήγανε στην τελευταία ανισότητα????

[quote="maniopas"]
$ f(a lnb) \plus{} f(b lna) < 1 \Leftrightarrow e^{a lnb} \plus{} e^{b lna} < 1 \Leftrightarrow b^a \plus{} a^b < 1$[/quote]

maniopa, βρηκα το λαθος σου.

$ f(a lnb) \plus{} f(b lna) \equal{} b^a \plus{} a^b \minus{} 1$ οποτε αυτο που εδειξες ειναι οτι

$ b^a \plus{} a^b < 2$  το οποιο ειναι προφανες απο το $ a<1$ και $ b<1$ (αν ισχυει)

ΥΓ. bouzoukman με προλαβες   :roll:

[quote="mailo"][quote="paulos18"][color=red]Να δειχθεί ότι α^β+β^α>1 για κάθε 0<α,β<1 και α<>β[/color]
Ίσως σας φανεί εύκολη ανισότητα δεν ξέρω,πάντως εμένα με παιδεύει πολύ και δεν μπορώ να τη βγάλω.  :o[/quote]

Οταν γραφεις 0<α,β<1 εννοεις οτι $ 0 < a < 1$ και $ 0 < b < 1$ ;  :huh:[/quote]
Ακριβώς αυτό εννοώ :wink:

Τελικα δεν ειναι και τοσο ευκολη οσο φενεται. Νομιζω οτι πιο καλα θα ηταν να φτιαξουμε καινουργιο thread με την ανισοτητα αυτη.


probability
geometry
3D geometry
symmetry
pyramid
calculus
limit

Three real numbers are chosen at random between $ 0$ and $ 1$.  What is the probability that they form the side lengths of a triangle? [i] Prove your answer.[/i]

[hide]The answer is $ \dfrac{1}{2}$.

Let the three numbers be $ x,y,z$. Note that any ordered triple $ (x,y,z)$ corresponds to a point in 3-D space within the unit cube bounded by the 8 binary strings of 0's and 1's: $ (0,0,0),(0,0,1)$, etc. Note that this has volume 1. We want the volume of the set of points such that $ x\plus{}y>z$, $ y\plus{}z>x$, and $ z\plus{}x>y$. Equivalently, we can take the volume of the set of points such that either $ x\plus{}y<z$, $ y\plus{}z<x$, or $ z\plus{}x<y$, and subtract it from the volume of the whole cube. Note that we can disregard cases such as $ x\plus{}y\equal{}z$ because these represent planes, not contributing anything to the volumes.

Note that it is impossible for any point $ (x,y,z)$ to belong to more than one of the regions $ x\plus{}y<z,y\plus{}z<x,z\plus{}x<y$, since if, for example, $ x\plus{}y<z$, since $ x,y,z>0$, we have $ z>y,x$. This implies that $ y\plus{}z>x$ and $ z\plus{}x>y$, so $ (x,y,z)$ can only be in one of these three regions. Thus we can find the volumes of each of the regions separately. By symmetry, we only need to consider $ x\plus{}y<z$. $ x\plus{}y\equal{}z$ represents a plane, so we want the volume of the set of points on one side of the plane, within the unit cube. Note that $ (0,0,0)$, $ (0,1,1)$, and $ (1,0,1)$ all satisfy $ x\plus{}y\equal{}z$, so these points are on the plane. Now, $ (0,0,1)$ satisfies $ x\plus{}y<z$, so we want the volume of the part of the cube on the same side of the plane as $ (0,0,1)$. We see that this is simply the pyramid with apex $ (1,0,1)$ and base vertices $ (0,0,1),(0,0,0),(0,1,1)$. The height is clearly 1 (the distance between $ (1,0,1)$ and $ (0,0,1)$, since this length is perpendicular to the base), and the base has area $ 1/2$, so the volume is $ \dfrac{1}{3}\cdot\dfrac{1}{2}\cdot1\equal{}1/6$.

Therefore, the total volume of the region we don't want to count is $ 3\cdot\dfrac{1}{6}\equal{}\dfrac{1}{2}$, and thus the region we want is $ 1\minus{}\dfrac{1}{2}\equal{}\dfrac{1}{2}$. The volume of the cube is 1, giving a probability of $ \dfrac{1}{2}$, as desired.[/hide]

I decided to be lame and use calculus on this one in my solutions, but here's another solution:

[hide="Another Solution"]Consider the set $ \{\frac1n, \frac2n, \cdots 1 \}$. Let $ P_n$ represent the probability that a randomly chosen subset $ \{a, b, c\}$ of three not necessarily distinct elements of the set form a triangle.

We here find $ 1\minus{}P_n$, i.e. the probability that they don't form the sides of a triangle. For simplicity let $ r\equal{}na, s\equal{}nb, t\equal{}nc$. Note that the condition that $ r, s, t$ don't form a triangle is equivalent to exactly one of $ r\plus{}s \leq t$, $ s\plus{}t \leq r$, and $ t\plus{}r \leq s$ being true. WLOG let $ r\plus{}s \leq t$. Then, by a standard argument, there exists some positive integer $ m$ such that $ r\plus{}s\plus{}m\equal{}t\plus{}1$. For each value of $ t$, by Balls and Urns, there are $ \binom{t}{2}$ ways to select $ r, s, m$. Since $ t \leq n$, there are then $ \sum^n_{t\equal{}1} \binom{t}{2}$ total ways to choose $ r, s, t$, which by Hockey stick is equal to $ \binom{n\plus{}1}{3}$. Thus, because of our assumption that $ r\plus{}s \leq t$, there are $ 3\binom{n\plus{}1}{3}$ ways to choose $ r, s, t$ so that they don't form a triangle. Thus, since there are $ n^3$ ways to choose $ r, s, t$, $ P_n\equal{}\frac{n^3\minus{}3\binom{n\plus{}1}{3}}{n^3}\equal{}\frac{n^2\plus{}1}{2n^2}$.

Now, as $ n$ gets large, the discrete case we've been looking at approaches the continuous case the problem asks for. Therefore, $ \lim_{n \to \infty} P_n\equal{}\frac12$, which is our answer.[/hide]

[quote="perfect628"]
Now, as $ n$ gets large, the discrete case we've been looking at approaches the continuous case the problem asks for. Therefore, $ \lim_{n \to \infty} P_n \equal{} \frac12$, which is our answer.
[/quote]

Well, even as $ n$ gets large, $ P_n$ is only a set of rational numbers.  I don't see a quick way to adapt the argument to the real line without starting over.  (Then again, there probably is one, so please prove me wrong.)

By the way, I'm with you on the calculus thing.  I mentioned geometry, but I do not consider "geometric probability" to be a legitimate proof.  Besides, with the calculus it took absolutely no work.  I didn't even do the calculation by hand; I just used Maxima.

Blech there I am not paying attention--I had actually seen this problem before, and that's where I got the argument above. Well, I thought I had--Now that I look back it specified rationals there. Luckily I used the boring integration on the quiz!

I don't understand why the limit wouldn't work - the rationals are uniformly distributed among the reals, are they not?

Intuitively the limit thing makes sense to me, but I don't have enough mathematical knowledge (or the right angle on the question) to make it rigorous.  Could you please explain what you mean in a precise mathematical sense by "the rationals are uniformly distributed among the reals"?  The rationals are dense in the reals, but the entire rational line is of measure zero, so what does a "distribution" of the rationals among the reals entail?

And then if you want to be really nice you can pamper me and give a full justification of the "limit" argument bridging the rational argument with the real one.

Well, the probability function perfect describes is bounded on the interval we're considering, so the integral and the limit are interchangeable.

As for uniform distribution, essentially it means that on any closed subinterval, the expected value of of a randomly chosen rational and a randomly chosen real are the same.

[hide="Very simple solution"]
Without loss of generality, assume $ c$ is the largest of $ a,b,c$. Divide each length by $ c$. We now must find the probability that $ a \plus{} b > 1$, which is obviously $ \frac {1}{2}$. (Just draw a unit square -- we want the probability the probability $ (a,b)$ is above the diagonal.)

You would want to do a little more work to show that all values in the range $ (0,1)$ are still equiprobable, which is not hard -- just treat it as if you picked $ c$ first.[/hide]

Thank you to Josh Alman, who, by sheer coincidence, shared the similar problem with a near-identical solution method "Randomly pick $ a,b,c$ from (0,1); find the probability they form an acute triangle" just a few days before the qualifying quiz was posted.

Unfortunately, I haven't had calculus yet (self-studying with Spivak right now, so it should take a while, too >_> I guess I'll study it concurrently with my ACTUAL calculus class, heh), so I did it Catalyst's way. I wasn't quite sure it was rigorous either, and I'd actually quite like to know whether geometric probability is or is not a valid proof method.

I see no reason why a geometric solution wouldn't be rigorous. Really it's the same thing as the calculus solution, except we're finding the volume of the region by basic geometry instead of integration.

Definitely.  I mean, I suppose both the $ P_n$ thing and the geometry could both be justifiable, but at the root of it, the only way I see to justify them is to set up the calculus and then, once that is established, show that your alternate solution is equivalent to the calculus.

And Zeph, I've been trying to get my hands on Spivak's book for the longest time.  What I used this year (in addition to an AP calculus class -- a great deal of good that has done) was a short old real analysis text that I borrowed from my teacher.  Everyone raves about Spivak, but I haven't been able to find it in a library.

A lot of this comes down to how you define probability, which can be a bit of a deep subject. In this situation it would be reasonable to define the probability of picking a point in some set $ E$, out of a set of possible points $ \Omega$ to be:
$ P(E) \equal{} \frac{m(E)}{m(\Omega)}$
(where $ m(X)$ represents the [i]measure[/i] (volume) of a set $ X$).

In this case either integration or geometry would be valid ways to find $ m(E)$.

Now you're getting beyond me.  But that does seem reasonable.  Defining probability in terms of geometry never really sat well with me, but it would work because of the link between synthetic geometry and Euclidean space which I find to be very nontrivial.  Anyway, even in the geometry, I ended up with a couple of tetrahedra, and I don't know how to find the volume of one of those without calculus.  At every step of the way it is clear that the analytic and the analytic-geometric methods are the same.  What discomforts me personally about it is that I'm basically saying,

[quote]
If we wish to find the probability that three real numbers $ a_1,a_2,a_3$ chosen independently and randomly from $ (0,1)$ with uniform distribution satisfy the so-called "triangle" inequalities $ a_i \plus{} a_j > a_k$, then it is sufficient to consider three mutually perpendicular lines in Euclidean three-space, and then label each point in the space with coordinates, and consider the locus of points whose coordinates satisfy the inequalities.  We can prove that this is a region bounded by this plane, this plane and that plane, which takes the form of an octahedron.  In this space, we define "volume" to mean such-and-such, and we define probability to be the ratio of volumes.  Oh, by the way, all this is independent of the choice of mutually perpendicular lines.  Incidentally, this definition can be shown to coincide with other definitions based on axioms that did not originate in 300 B.C.
[/quote]

To someone who is more experienced than me in this, that's probably trivial.  But personally I try to avoid using geometry to prove things that in my mind "are not geometry" because I feel that I'm BS-ing it, for the sole reason that if someone sat me down in a room and asked me to justify everything I had said down to symbols, I would not be able to do so.  Yes, I am a strange person.

I actually usually feel the opposite way. For things like this I try to see spacial arrangements because when I'm messing with symbols it seems like they do the thinking for me and I end up not actually understanding what I'm doing.

Anyways, in this case I also initially did this with calc. But when I saw the answer I looked for a better solution and imagined a three dimensional graph, had one dimension be the long side, and realized that the other two had to form a 45 45 90 triangle cross section with linearly increasing sides to contain all the volume that would not form a triangle with the three lengths. Since one third of a cube would just be the same thing with a square instead of the triangle, 1/2 of possibilities form triangles.

Yeah, I'm really not describing this in the right way.  Let me try again.  I consider a proof to be a proof, and any other consideration of mathematics might be BS or someone might claim that you have failed to justify it.  (even if you're obviously right, there are some people like that no I am not saying the mathcamp people are like that)  Maybe it would help to clarify if I mentioned that I really, really wanted to get into this program and I had serious doubts that I would make it, so I was VERY careful to make sure nothing went unsaid in my solutions.  They were eleven pages long Lol.  Plus diagrams.

You're not alone in that. My solutions were 24 pages last year (although I always started a new problem on a new page). Still, I did things such as proving that, if you convert the sequence of powers and sums of powers of three into ternary, order will be preserved, and overall just tried to let nothing go unstated.

I know some others did similarly, such as one girl who mentioned that she wrote her proofs as if "talking to a three year-old."

My proofs weren't too verbose, in my opinion, and didn't do anything unnecessary, and ended up being 14 pages long. To be fair, two of the pages only had two lines, but still...

@Darmani LOL.  It's because the people at Mathcamp are so stupid and have so little experience in math.  We need to make sure they can digest our complex ideas.  And it worked, didn't it?  We got in.

@perfect628 Why did you sometimes have two lines to a page?  Did you start each solution on a new page?

Yes I started each solution on a new page...Did others not?

I didn't.  I don't think they really cared, but the template on their site (which I used) did not build in a line break between "solution" environments.


search
limit
set theory
combinatorics proposed
combinatorics

Prove that there exist positive constants $ c$ and $ n_0$ with the following property. If $ A$ is a finite set of integers, $ |A| \equal{} n > n_0$, then
\[ |A \minus{} A| \minus{} |A \plus{} A| \leq n^2 \minus{} c n^{8/5}.\]

What is $ |A\minus{}A|$ and $ |A\plus{}A|$?

[quote="mavropnevma"]The following are called Minkowski sum and difference of sets:

$ A \plus{} A : \equal{} \{a \plus{} a' \ \mid \ a, a' \in A\}$
$ A \minus{} A : \equal{} \{a \minus{} a' \ \mid \ a, a' \in A\}$

The other notation is the cardinality $ |X|$ of a set $ X$.[/quote]

Thank your Mr.[b]mavropnevma[/b]. Could you refer us to some documents about those Minkowski sum and difference of sets, especially the $ A \plus{} A$ and $ A \minus{} A$ mentioned in the problem? I have no idea to find $ c$ :blush:

[quote="mavropnevma"]An easy search on Google yields [url]http://en.wikipedia.org/wiki/Minkowski_addition[/url][/quote]

Of couse I Googled it before question. But I found out that they majorly talk about the Minkowski sum of a Euclidean figure in the plan, not about that in the side of set theory and Combinatorics. Do you have any reference about Minkowsky sum in Combinatoris, in particular, the number of elements $ |A\plus{}A|$ and $ |A\minus{}A|$?

That is just a name and a notation. There is a large literature on "sumsets", with some links following

[url]http://en.wikipedia.org/wiki/Restricted_sumset[/url]
[url]http://math.nju.edu.cn/~zwsun/csz.htm[/url]
[url]http://www.emis.de/journals/INTEGERS/papers/b2/b2.pdf[/url]

Names of mathematicians involved: Noga Alon, Imre Rusza, etc.

I have an idea.

Let $ A \equal{} \{a_1,a_2,...,a_n\}$ where $ a_1 < a_2 < \cdots < a_n$, we have $ a_1 \plus{} a_2 < a_1 \plus{} a_3 < \cdots < a_1 \plus{} a_n$ are $ n \minus{} 1$ pairwise distinct elements of $ A \plus{} A$, thus $ |A \plus{} A| \geq n \minus{} 1$. On the other hand, $ |A \minus{} A| \leq {n \choose 2} \equal{} \frac {n^2 \minus{} n}{2}$, the equality holds when we choose every $ a_{i \plus{} 1}$ large enough compared to $ a_i$ for $ i \equal{} 1,2,\cdots,n \minus{} 1$. Thus: $ |A \minus{} A| \minus{} |A \plus{} A| \leq \frac {n^2 \minus{} n}{2} \minus{} (n \minus{} 1)$. Hence, it is enough to find $ c$ such that $ \frac {n^2 \minus{} n}{2} \minus{} (n \minus{} 1) < n^2 \minus{} cn^{\frac {8}{5}}$. But this is equivalent to $ c < \frac {\frac {n^2}{2} \plus{} \frac {3n}{2} \minus{} 1}{n^{\frac {8}{5}}}$. Now the existence of $ c$ is followed because $ \lim_{n\to \plus{} \infty}\frac {\frac {n^2}{2} \plus{} \frac {3n}{2} \minus{} 1}{n^{\frac {8}{5}}} \equal{} \plus{} \infty$. QED.

@Mr.[b]mavropnevma[/b]: Could you please tell me how can Combinatorial Nullstellensatz involve here?

It does not (those links were just to show diverse methods around the subject). In my language there is a saying "not all that flies can be eaten"  :)

It is not necessarily true that $ |A\minus{}A| \leq \binom{n}{2}$, because subtraction is not commutative.

If $ A\equal{}\{1, 3, \dots 3^{n\minus{}1}\}$, then $ |A|\equal{}n$ but $ |A\minus{}A|\equal{}n(n\minus{}1)\plus{}1$.

Here's something that does [b]not[/b] go all the way to proving the problem statement, but feels like it should be improvable somehow.

Let $ a_1, \dots, a_k$ be the elements of $ A \minus{} A$, and $ b_1, \dots, b_\ell$ be the elements of $ A \plus{} A$.

Let $ x_i$ be the number of ordered pairs of elements $ (c, d)$ of $ A$ solving $ c \minus{} d \equal{} a_i$, and let $ y_i$ be the number of ordered pairs solving $ c \plus{} d \equal{} b_i$.  We know that  $ 1 \leq x_i, y_i \leq n$.  We also know
\[ \sum_{i \equal{} 1}^k x_i \equal{} \sum_{j \equal{} 1}^\ell y_i \equal{} n^2,\]
and I claim
\[ \sum_{i \equal{} 1}^k x_i^2 \equal{} \sum_{j \equal{} 1}^\ell y_i^2: \equal{} m\]
for some unknown $ m$.  This last equation stems from the observation that an ordered quadruple $ (c_1,c_2,c_3,c_4)$ satisfies $ c_1 \plus{} c_2 \equal{} c_3 \plus{} c_4$ iff it satisfies $ c_1 \minus{} c_3 \equal{} c_4 \minus{} c_2$, and that the sum of the squares of the $ x_i$ and $ y_i$ count the number of such solutions.

By Cauchy-Schwarz we have $ n^4 \leq m k$ and $ n^4 \leq m \ell$.   To get an upper bound on $ m$, note that $ \sum_{i\equal{}1}^k x_i^2$ is maximized relative to the constraints $ 1 \leq x_i \leq n$ and $ \sum x_i \equal{} n^2$ when all of the $ x_i$ are either $ 1$ or $ n$.  Solving for how many $ x_i$ need to be $ 1$, we see
\[ m \leq \frac{n^2\minus{}k\plus{}1}{n} n^2 \plus{} k\]
Comparing this with our Cauchy-Schwarz bound, we see
\[ \ell \geq \frac{n^4}{m} \geq \frac{n^4}{n(n^2\minus{}k\plus{}1) \plus{} k }\]
Letting $ z\equal{}n^2\minus{}k\plus{}1$, we see that
\[ |A\minus{}A|\minus{}|A\plus{}A| \leq n^2\plus{}1\minus{}z \minus{} \frac{n^4}{n z \plus{} n^2\minus{}1\minus{}z}.\]
Unfortunately, we have no control over $ z$, so can only assume the worst case, which is $ z$ of order $ n^{3/2}$.  Doing so gives a bound of $ |A\minus{}A|\minus{}|A\plus{}A| \leq n^2\minus{}Cn^{3/2},$ not quite strong enough.

kevinatcausa, your solution may be finished by the following easy observation. Assume that some difference $ c$ is attained at least $ k\geq Cn^{4/5}$ times. Then $ a_i\minus{}b_i\equal{}c$ for $ i\equal{}1,2,\dots,k$, $ a_i$, $ b_i$ $ \in A$.

The claim. There exists a set $ V$ of at least $ k/3$ elements $ v_1$, $ v_2$, $ \dots$, $ v_l$, $ l\geq k/3$ of $ A$ such that each difference $ v_i\minus{}v_j$ is also realized as $ p\minus{}q$, where $ p,q\in A\setminus V$.

Proof of the claim. Consider all elements $ a_i$ and $ b_i$. They form few disjoint arithmetic progressions with difference $ c$, each progression has at least 2 members. Define $ V$ as teh set of members of these progressions with even indecies (say, if we have $ x,x\plus{}c,x\plus{}2c$, then take $ x\plus{}c$, if we have $ x,x\plus{}c,x\plus{}2c,x\plus{}3c$, take $ x\plus{}c$ and $ x\plus{}3c$ and so on). Easy to see that such $ V$ satisfy our claim.

Then \[ |A\minus{}A|\leq |(A\setminus V)\minus{}V|\plus{}|V\minus{}(A\setminus V)|\plus{}|(A\setminus V)\minus{}()A\setminus V|\leq 2l(n\minus{}l)\plus{}(n\minus{}l)^2\equal{}n^2\minus{}l^2,\]
and we are done.

So, $ x_i$'s are bounded from above by $ O(n^{4/5})$ and we may improve your upper estimate of $ \sum x_i^2$ so that the solution works.


calculus
integration
calculus computations


Prove that if one chooses more that $ n $ numbers from the set $ (1,2,...,2n) $ , then the one number is a multiple of another . Can this be avoided with exacly $ n $ numbers ?

  The first question is famous . I'm thinking about the second one

You are right this is a really standard problem.
For the second question, the answer is no, as proved by choosing $n+1,n+2, \cdots, 2n.$

Pierre.


$n$ is a natural number and $n\geq 2$.suppose that $p(x)=a_0+a_1x+...+a_nx^n$
and $0<a<-\sum_{k=1}^{[\frac{n}{2}]}\frac{1}{2k+1}a_{2k}$.
prove that there exist $q \in [-1,1]$ such that $p(q)=0$.

[quote="conceive"][quote="jensen"]$n$ is a natural number and $n\geq 2$.suppose that $p(x)=a_0+a_1x+...+a_nx^n$
and $0<a<-\sun_{k=1}^{[\frac{n}{2}]\frac{1}{2k+1}a_2k}$.
prove that there exist $q \in [-1,1]$ such that $p(q)=0$.[/quote]

Did you mean:

$n$ is a natural number and $n\geq 2$.suppose that $p(x)=a_0+a_1x+...+a_nx^n$
and $0<a<\sum_{k=1}^{[\frac{n}{2}]}\frac{1}{2k+1}a_{2k}$.
prove that there exist $q \in [-1,1]$ such that $p(q)=0$

  :?:[/quote]
yes exactly. :D

May I ask what $a$ is supposed to represent?

[quote="jensen"][quote="conceive"][quote="jensen"]$n$ is a natural number and $n\geq 2$.suppose that $p(x)=a_{0}+a_{1}x+...+a_{n}x^{n}$
and $0<a<-\sun_{k=1}^{[\frac{n}{2}]\frac{1}{2k+1}a_{2}k}$.
prove that there exist $q \in [-1,1]$ such that $p(q)=0$.[/quote]

Did you mean:

$n$ is a natural number and $n\geq 2$.suppose that $p(x)=a_{0}+a_{1}x+...+a_{n}x^{n}$
and $0<a<\sum_{k=1}^{[\frac{n}{2}]}\frac{1}{2k+1}a_{2k}$.
prove that there exist $q \in [-1,1]$ such that $p(q)=0$

  :?:[/quote]
yes exactly. :D[/quote]
If that is exactly what you meant, then howcome you have a negative sign in front of the summation? I'm not sure what the problem is, especially what $a$ is supposed to be.  Could you please clarify? :)

Use \[\int_{-1}^{1}p(x)\,dx=2\sum_{k=0}^{[n/2]}\frac{a_{2k}}{2k+1}\] Thus, if $a$ means $a_{0}$ the 1st post was right.
It means that p(0)>0 and $\int_{-1}^{1}p(x)\,dx<0.$
Hense the statement is obvious.


1. Let a, b, and n be positive integers satisfying:

1997/1998 + 1999/n = a/b, with a/b in reduced form.
Given that n is the smallest positive integer for which 1000 divides a, what is the sum of the digits of n?

2. Let a_2, a_3,..., a_100 be defiined as follows:
a_100 = 100
a_99 = 99^a_100 = 99^100
a_98 = 98^a_99 = 98^99^100
.
.

Thus in general a_n = n^a_(n+1), for n = 2,3,...,100. Calculate the units digit of a_2.

For #2, the units digit of 2^x depends on the residue of x mod 4. Thus, consider a_n mod 4. a_100 will be 100 :equiv: 0. a_99 will be (I'm doing this in mod 4) 3^0 :equiv: 1. a_98 will be 2^1 :equiv: 2. a_97 will be 1^2 :equiv: 1. a_96 will be 0^something :equiv: 0. You can see it cycles period 4. Thus, a_3 mod 4 will be 1. Bumping out of mod 4, the units digit of 2^(something :equiv: 1 mod 4) will be 2.

Am I right?  :shock:  :shock:

Heres an answer to the first one.

[hide]Multiplying out and factorizing gives b(1997n + 1997.1998)=a(1998n).

Now we are assuming a is divisible by 1000 and b is neither even nor divisible by 5.

Since b is not even, n must be as the right hand side is. Let n=2m, now we get b(1997m+1999.999) = a(1998m).

Taking both sides mod 1000, we get b(-3m+1) :equiv: 0 (mod 1000) and thus 3m :equiv: 1 (mod 1000). This means m = 1000c + 667 for some c (can get this just by letting 3m=1000c+1 and considering mod 3 or something like that).

Now we just check the cases. m=667 gives us

b(1997.667 + 1999.999) = a(1998.667). Now this means that (1997.667 + 1999.999) is divisible by 1000 and also by an extra factor of 2 that comes from the 998. So 1997.667 + 1999.999 must be divisible by 2000. Mod 2000 gives us that sum is 1000, so this does not work.

m=1667 similarly needs 1997.1667 + 1999.999 divisible by 2000, and mod 2000 shows this is true.

So the smallest n is 2x1667 = 3334 whose sum of digits is 13.

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This isn't quite as elegant, but for #1, I rearranged the left hand side as (1997+1998*1999/n)/1998. I figured I would just working on the numerator being a multiple of 1000 and forget about reducing. Anyway, I took the numberator mod 1000, and set 1998*1999/n :equiv: 3 mod 1000 (bc 1997+3 :equiv: 0, which is what I want). Multiplying both sides by n, and simplifying 1998*1999, I get 2 :equiv: 3n mod 1000 which means n :equiv: 334 mod 1000. Just trying the first few n, I got n=3334 as the first n that worked. sorry, wasn't quite as elegant.

MysticTerminator wrote:gauss202 wrote:(spoiler):

[hide]Yes. Your answer is right, but consider this.. did you really need to consider the powers all the way up near 100, 99, 98 etc? What if those digits were replaced by something else, would the answer still be the same?[/hide][/hide]

No... I didn't really; I could have said a_4 :equiv: 0 (duh), and then worked down from there, but it was pretty much the same thing as I just showed it was a cycle of 4. Unless you're talking about some totally different radical way to do it, the two methods are pretty much the same.

Right, that's what I was referring to. I just thought it was interesting how it doesn't matter what the higher exponents are. It only depends on the last 3 recursions.

Ok. good just checking. Yes, that is rather interesting. You could come up with a table of the last digit based on n mod 20 pretty easily, couldn't you? I get 20 as LCM of 4 and 10. (10 determines the last number, and 4 determines the stage in the cycle of powers as all powers cycle in 1, 2, or 4 I think)


Campbell's says:

"MHC proteins also play a key role in the development of T-cell self tolerance. During the development of T cells in the thymus, developing T cells interact with other thymic cells, which have high levels of both class I MHC and class II MHC molecules. Only T cells bearing receptors with affinity for self MHC proteins reach maturity."

I'm not entirely sure how self-tolerance develops in T-cell. Could someone explain it to me? Particularly, I don't get why "bearing receptors with affinity for self MHC proteins" results in self-tolerance. Thanks!

t-cells themselves hav MHC proteins.those who hav receptors of these protein are recognised as self and r not destroyed
. :maybe:


Let $ a\neq b, b \neq c, c \neq a$ and $ \frac {a}{b \minus{} c} \plus{} \frac {b}{c \minus{} a} \plus{} \frac {c}{a \minus{} b} \equal{} 0$.

Prove that: $ \boxed {\frac {a}{(b \minus{} c)^2} \plus{} \frac {b}{(c \minus{} a)^2} \plus{} \frac {c}{(a \minus{} b)^2} \equal{} 0}$

[quote="thanhnam2902"]Let $ a\neq b, b \neq c, c \neq a$ and $ \frac {a}{b \minus{} c} \plus{} \frac {b}{c \minus{} a} \plus{} \frac {c}{a \minus{} b} \equal{} 0$.
[/quote]

Well, I may be dumb but is it possible?
We have $ (\frac {a}{b \minus{} c} \plus{} \frac {b}{c \minus{} a} \plus{} \frac {c}{a \minus{} b} \equal{} 0)\iff (\sum_{cyc}a(a\minus{}b)(a\minus{}c)\equal{}0)$
But by schur we have $ \sum_{cyc}a(a\minus{}b)(b\minus{}c)\geq 0$ with equality that cannot occur. So, where is my mistake?

Schurr works only for nonnegative numbers - check the proof. Here, example $ (a,b,c)$ are $ (1,0,\minus{}1)$.

Aaah, sorry. I mechanically assumed $ a,b,c\in R_{\plus{}}$  :blush:

[hide="hint"]
Expand out $ \left( \frac{a}{b\minus{}c}\plus{}\frac{b}{c\minus{}a}\plus{}\frac{c}{a\minus{}b}\right)\left( \frac{1}{b\minus{}c}\plus{}\frac{1}{c\minus{}a}\plus{}\frac{1}{a\minus{}b}\right)$ and deal with the other terms.
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Thank [b]scorpius119[/b]. Can you post full solution.

The expansion is our desired identity plus $ \frac{(a \plus{} b)(a \minus{} b) \plus{} (a \plus{} c)(a \minus{} c) \plus{} (b \plus{} c)(b \minus{} c)}{(a \minus{} b)(b \minus{} c)(c \minus{} a)} \equal{} 0$.