geometry
geometric transformation
invariant
real analysis
real analysis unsolved

suppose that $T: L^{1}(\mathbb{R}) \to L^{1}(\mathbb{R})$ is a bounded linear operator that commutes with translation. Show that there exists a finite measure $\mu$ such that $T(f)=f*\mu$. What about $T: L^{p}\to L^{p}$

[quote="alekk"]What about $T: L^{p}\to L^{p}$[/quote]
A translation invariant operator that has some kind of boundedness on some space should be given by convolution with a tempered distribution - but there's no reason that should be a bounded measure. The classic example is the Hilbert transform, which is given by convolution with (a constant multiple of) the principal value of $\frac1x.$ The Hilbert transform is a bounded operator from $L^{p}$ to $L^{p}$ for $1<p<\infty,$ with an operator norm that is comparable to $\frac1{1-p}$ for $p$ near 1 and to $p$ for large $p.$ That includes that it is an isometry on $L^{2}.$


How many digits are there in 2^1000.
I know that if we show that log[size=75]10[/size] 2 =0.30103... it's easy but how can we show this without using a calculator.

Ravi B wrote:I would use the fact that 210 = 1024 is close to 103 = 1000.  Using that connection, all you have to prove is that 10 < 1.024100 < 100.

[hide]The right inequality is pretty easy to prove.  For example, use the inequality, [tex]\ln(1+x)<x\quad x\ne 0[/tex].

[tex]\ln 1.024<0.024[/tex]

[tex]\ln (1.024^{100})=100\times\ln 1.024[/tex]

[tex]\ln (1.024^{100}) < 2.4[/tex]

[tex]1.024^{100}=e^{\ln(1.024^{100})}<e^{2.4}<3^3<100[/tex]

The left inequality turns out to be much tighter.  But let's give it a go.  We'll use the same technique on the inverse of the middle term of the inequality.

[tex]\displaystyle\frac{1}{1.024^{100}}=\Big(\frac{1}{1.024}\Big)^{100}=0.9765625^{100}[/tex]

[tex]\ln 0.9765625=\ln(1-0.0234375)<-0.0234375[/tex]

[tex]\ln(0.9765625^{100})<-2.34375[/tex]

So, [tex]\ln(1.024^{100})>2.34375[/tex]

If you have memorized [tex]\ln(10)\approx 2.30[/tex], we're done.  But then you probably also would have memorized [tex]\log(2)=0.301^+[/tex], and we would have been done a long time ago.

[tex]1.024^{100}=e^{\ln(1.024^{100})}>e^{2.34375}>^?10[/tex]

From here on, there are many ways to proceed, with many shortcuts.  However, I present a solution that is rather tedious and unpleasant...

[tex]e^{2.34375}>^?10[/tex]

[tex]e^{2.34375}>2.7^{2.34375}>2.7^{7/3}>^?10[/tex]

[tex]2.7^7>^?10^3[/tex]

[tex]27^7>^?10^3\times 10^7[/tex]

[tex]27^7>^?10^{10}[/tex]

[tex]3^{21}>^?10^{10}[/tex]

Do you know [tex]\log 3\approx 0.477[/tex]?  If so, we're done.  How high do you know your powers of 3?  3^21=10460353203.  If not, you can multiply it out right here.  But I continue.

[tex]3^{21}=3\times3^{20}=3\times(3^4)^5=3\times81^5[/tex]

[tex]81^5=(80+1)^5[/tex]

You better know your powers of 2!  I take the first two terms of the binomial expansion (because that's all I need to prove the inequality)...

[tex](80+1)^5>80^5+5\cdot 80^4[/tex]

[tex]80^5+5\cdot 80^4=3276800000+204800000=3481600000[/tex]

Finally,

[tex]3^{21}=3\times(80+1)^5>3\times 3481600000=10444800000[/tex]

[tex]3^{21}>10444800000>10^{10}[/tex]

[/hide]

This just proves you should try to memorize a few of the key logarithms.  Aside from identities, I suggest you start with:  log(2), ln(2), log(5), log(3), ln(10), log(e).


[quote="Ilthigore"]Hi. I don't seem to be able to edit my solutions any more. I think perhaps a board wide 'no editing after a certain time' rule has applied to the contest as well. Any help would be good. I've been sticking all my work in progress up as I go along, so my solutions are far from complete.[/quote]Ooops you're right, I'll look into that (it is the editing rule).


Wow, it's been a long time since I posted here. Anyhow, I'm a sophomore at MIT, majoring in physics. I love it, but I also look back at nostalgia at high school math competitions. I've tried to find classes here that cover similar material/problem-solving, but so far have failed because all my math major friends are doing intense theory-based classes that I'm not as interested in. So I came here, the only other place I thought significant quantities of MIT math undergrads might visit. To you I ask: do you know of any classes that work with problems similar to ARML, AIME, AMC, etc? I miss that kind of math dearly, and would love to take some.

Thanks.

[quote="tokenadult"]I thought MIT had a specific problem-solving course that helps train its Putnam team, but that other students can register for.[/quote]

It does. I've been thinking about joining it. That's the closest you'll get to high school-style competition math. From what I've heard, the difficulty of the problems varies a lot.

Hm...isn't it easier to ask other people around MIT?  That's what I would do if I had a similar question.

Anyway, some of the combinatorics classes might have a little bit of what you're looking for: e.g. 18.062J or 18.314 e.g.  In my experience though, the closest you'll get to what you're talking about is a class that starts off with one or two psets at that level (or a few problems at that level) and then builds up to much more advanced or theoretical levels of the topic.

Hope that helps a little.


integration
trigonometry
geometry
3D geometry
algebra
polynomial
function

$\int sin^{3}x\,dx$

$\int sin^{3}.sinx.dx$

$\int (1-cos^{x})d(-cosx)\,dx$

How did they integrate sinx but in next step going to integrate again?

Bos1234,

     What you wrote in your post was not too clear, but I think you meant something like this:

$\int \sin^{3}x dx = \int (\sin^{2}x) (sin x) dx = \int (1-\cos^{2}x) \sin x dx$
$=\int (1-u^{2}) (-du) =-u+\frac{1}{3}u^{3}=-\cos x+\frac{1}{3}\cos^{3}x$

The justification is as follows:  One can split the cube of the $\sin x$ as a product of $\sin^{2}x$ and $\sin x$ which is clear enough; this is taking place inside of the integrand.   One can then rewrite $\sin^{2}x$ as $1-\cos^{2}x$ using the trig identity; again within the integrand.  One then performs the substitution $u=\cos x$ recognizing that $du=-\sin x$.  Then one integrates the transformed polynomial and then substitutes back in the expression for $u$ to obtain the final answer.

ok similar qns

$\int$ $sec^{3}x$ = $\int$ $secx.sec^{2}x.dx$

$\int$ $secx.d(tanx)dx$
how did they obtain step too so quickly?

-------

$\int$ $cosec^{3}x.dx$ = $\int$ $cosecx.cosec^{2}x$ dx

=$\int$ $cosecx d(-cotx)$ $<---$ ??

$\frac{d}{dx}\tan x = \sec^{2}x$ and $\frac{d}{dx}\sec x = \sec x\tan x$, and so

$P = \int \sec^{3}x\,dx = \int \sec^{2}x \sec x\,dx = \tan x\sec x-\int \tan x \cdot \sec x\tan x\,dx$
$= \tan x\sec x-\int \tan^{2}x \sec x\,dx = \tan x\sec x-\int(\sec^{2}x-1) \sec x\,dx$
$= \tan x\sec x-\int \sec^{3}x\,dx+\int \sec x\,dx = \tan x\sec x-P+\ln|\sec x+\tan x|+C$
$\Rightarrow 2P = \tan x\sec x+\ln|\sec x+\tan x|+C$
$P = \frac{1}{2}(\tan x\sec x+\ln|\sec x+\tan x|)+C$.

For $\int\csc^{3}x\,dx$ proceed in a similar way, taking into acount that

$\frac{d}{dx}\cot x =-\csc^{2}x$.

A bit of notation: for the nicer-looking function names, use a backslash: \sin becomes $\sin$, for example.

Also, the integral signs look better if they're part of the same block of $\LaTeX$ as the following stuff.

[quote="jmerry"]A bit of notation: for the nicer-looking function names, use a backslash: \sin becomes $\sin$, for example.[/quote]
And remember to put a space in. For instance, x\sinx would become $x\sinh$, since there is no command named "sinx", but x\sin x, with the space in there, becomes $x\sin x,$ which is what you probably want.

One other thing: which looks better:

$\int x\sin x dx$ or $\int x\sin x\,dx$ ?

Hold your mouse over them to find out what's different.


function
real analysis
real analysis unsolved

Here is a classic. Find a non-zero infinitely differentiable function $f$ at $0$ so that $f^{(n)}(0)=0$ for all $n\geq 0$ but yet its Taylor series doesnt agree with the function, no matter how small the open interval is.

The example is classic too.
f equals to $e^{-1/x}$ if x>0 and equals to 0 otherwise


Let m, n be positive integers.

Prove, using the method of mathematical induction:

$ m^{2n} \minus{}1$ is divisible by $ m \plus{} 1$ for all positive integers $ n.$

Hint of possible inductive steps of a) what to assume and b) what to show, respectively:
[hide]
a) Let $ P(k) \equal{} m^{2k} \minus{} 1 \equal{} (m \plus{} 1)r,$ for some integer $ r$.
b) Let $ P(k \plus{} 1) \equal{} m^{2(k \plus{} 1)} \minus{} 1 \equal{} (m \plus{} 1)s,$ for some integer $ s$.[/hide]

It's a lot quicker just showing that the remainder when its divided by m+1 is 0 (by plugging in -1 for m)...

He asked for MI proof.

[hide="Solution"]

n=1: not true, maybe you meant for $ n\ge2$

n=2, true

Suppose it's true for n=k.

n=k+1: $ m^{2k \plus{} 2} \minus{} 1 \equal{} m^{2k}\cdot m^2 \minus{} 1\equiv m^2 \minus{} 1\equiv 0\pmod{m \plus{} 1}$

Q.E.D.
[/hide]

Yes, sorry, I thought it was $ m^n$, made a mistake, and then continued with $ m^{2n}$ without correcting the mistake.


If x,y,a,b are naturals

Solve

$ x! \plus{} y! \equal{} 2^{a}\cdot 3^{b}$

[hide="Solution"]
WLOG, let $ x\le y$. If $ x\ge5$, then we can factor out a $ 5$ from $ x!$ and $ y!$, which is impossible. Hence $ 1\le x\le4$.
[b]Case:[/b] $ x\equal{}1$. The LHS is odd if $ y\ge2$, so the only possibility is for $ y\equal{}1$, which makes the LHS not divisible by $ 3$.
[b]Case:[/b] $ x\equal{}2$. If $ y\ge3$, we can take this $ \bmod3$ to get that there are no solutions. If $ y\equal{}2$, it doesn't work.
[b]Case:[/b] $ x\equal{}3$. If $ y\equal{}3$, we get the solution $ (x,y,a,b)\equal{}(3,3,2,1)$. If $ y\ge4$, then we can write this as $ 3!(1\plus{}4\cdot5\cdot\cdots\cdot y)$. Neither $ n\equal{}4$ nor $ n\equal{}5$ work, and for $ n\ge6$, we can take the parentheses $ \bmod6$ to see that the expression is $ 1\pmod6$, and thus can't be divisible by $ 2$ or $ 3$ giving extraneous factors in the LHS which are not $ 2$ or $ 3$, so there are no other solutions in this case.
[b]Case:[/b] $ x\equal{}4$. $ y\equal{}4$ gives the solution $ (x,y,a,b)\equal{}(4,4,4,1)$. $ y\equal{}5$ gives $ (x,y,a,b)\equal{}(4,5,4,2)$. For $ y\ge6$, we can again factor as $ 4!(1\plus{}5\cdot6\cdot\cdots\cdot y)$, and for $ y\ge6$ note that this is $ 1\pmod6$, which is again contradictory.
[/hide]


[quote="Inequalities Master"]Let a,,b,c be positive reals s.t.[/a^2+b^2+c^2=1].Prove that
1/(1-ab)+1/(1-bc)+1/(1-ca) is less or equal than 9/2.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66936
$ \minus{}9a^2b^2c^2\geq\minus{}abc(a\plus{}b\plus{}c)$ it's the Vasc's hint!


I always see those names but when I see the Wiki it full of complex looking things.

Can someone give an overview of each and one easy example in applying it?

Thanking you

[hide="AM-GM"]
For positive numbers $ a_1,a_2,a_3,\ldots,a_n,$ \[ \frac{a_1\plus{}a_2\plus{}a_3\plus{}\ldots\plus{}a_n}{n}\ge\sqrt[n]{a_1a_2a_3\ldots a_n}.\] Equality occurs when $ a_1\equal{}a_2\equal{}a_3\equal{}\ldots\equal{}a_n.$[/hide]

[hide="Cauchy-Schwarz"]
For sequences $ a_1,a_2,a_3,\ldots,a_n$ and $ b_1,b_2,b_3,\ldots,b_n$ of real numbers \[ \left(a_1^2\plus{}a_2^2\plus{}\ldots\plus{}a_n^2\right)\left(b_1^2\plus{}b_2^2\plus{}\ldots\plus{}b_n^2\right)\ge(a_1b_1\plus{}a_2b_2\plus{}\ldots\plus{}a_nb_n)^2.\] Equality occurs when there exist constants $ \alpha,\beta\neq0$ such that for all $ 1\le k\le n,\alpha a_k\equal{}\beta b_k.$[/hide]

AM-GM simply says the arithmetic mean of a bunch of positive numbers is greater than the geometric mean. For example,
$ \frac{3\plus{}7\plus{}8\plus{}9}{4} \ge \sqrt[4] {3 \cdot 7 \cdot 8 \cdot 9}$
If you evaluate each side exactly, you get
$ 6.75 \ge 6.2357$
So for example, AM-GM tells you $ \frac{\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}}{3} \ge \sqrt [3] {1}$, or $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge 3$, if a, b, and c are positive.
For Cauchy, I will just start with an example.
$ (1^2\plus{}3^2\plus{}2^2\plus{}3^2)(1^2\plus{}4^2\plus{}3^2\plus{}2^2) \ge (1 \cdot 1\plus{} 3 \cdot 4 \plus{} 2 \cdot 3 \plus{} 3 \cdot 2)^2$
If you evaluate each side, we get
$ 690 \ge 625$.
With Cauchy, you can have as many numbers in each parentheses as you wan't, I just decided to have 4 in each. But you always have to have 2 as the exponents, or it would be a different inequality called Holder.
Cauchy would tell you, for example, that $ (1\plus{}1)(a^2\plus{}b^2) \ge (a\plus{}b)^2$, or $ 2(a^2\plus{}b^2) \ge (a\plus{}b)^2$

Ok, AM GM is like
$ \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n} {n} \ge \sqrt [n] {a_1a_2...a_n}$. That is, the arithmetic mean of $ n\ge 2$ positive numbers is greater than or equal to the geometric mean, wtih equality hold only when all the terms are equal. Probably the most used version is with 2 terms: $ \frac {a \plus{} b} {2} \ge \sqrt {ab}$. To derive the formula for 2 terms is pretty easy. We have $ (\sqrt {a} \minus{} \sqrt {b})^2 \ge 0$ since no square is negative. Expanding and moving things around gives us the desired result. Im too lazy to type the derivation for 3 terms or more, and you dont really need to know it.
2 Examples
1. $ a_1,a_2,...a_n$ are positive real numbers with a product of 1. Prove that $ (1 \plus{} a_1)(1 \plus{} a_2)...(1 \plus{} a_n) \ge 2^n$.
[hide="solution"]AM-GM gives $ 1 \plus{} a_1 \ge 2\sqrt {1*a_1} \equal{} 2*\sqrt {a_1}$. We do the same to the rest of the terms. Thus, $ (1 \plus{} a_1)(1 \plus{} a_2)...(1 \plus{} a_n)\ge 2\sqrt {a_1}*2\sqrt {a_2}...*2\sqrt {a_n} \equal{} 2^n\sqrt {a_1a_2...a_n} \equal{} 2^n$, so we are done.[/hide]
2. a,b,c are real numbers. Show that $ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} ac \plus{} bc$.
[hide="solution"]We have $ a^2 \plus{} b^2 \ge 2ab$, $ a^2 \plus{} c^2 \ge 2ac$, $ b^2 \plus{} c^2 \ge 2bc$. Adding these together gives us $ 2(a^2 \plus{} b^2 \plus{} c^2) \ge 2(ab \plus{} ac \plus{} bc)$. Dividing by 2 makes us win.
[/hide]

EDIT 2 people beat me to the post

That was amazing thanks a lot guys you can really be very good instructors  :D

I undersdant what are they now it's just need practice to handle them.

Thanking you all  :)


trigonometry calculus function logarithms algorithm LaTeX

Find the number of roots of the equation :

$ sin$ $ x \equal{} log$ $ x$

[hide="Solution-ish"]Drawing a slightly accurate graph (which is not too hard by either plotting a couple of points or using calculus) it appears the answer is $ \boxed{3}$.[/hide]

Thanks a lot. :lol:

But is there another way around it (i.e, without using graphs)?

Can u also help me find the solns???

Using any graphing calculator on internet you could get these intersection points: $ (2.6965, 0.4308), (7.3285, 0.865), (8.2637, 0.9172)$. Note that these are not very accurate solutions as they will fail on 4th or 5th decimal.

I don't know how you would find solutions without graphs.

[hide="Without graphs?"]This is a tad confusing, maybe not rigorous...

Our function is $ f(x)\equal{}\log x \minus{} \sin x$, defined on $ (0,\infty)$, so it is always positive for $ x>k$ when $ \log k > \sin k$ is always true. We can immediately see that $ k\equal{}10$ works (even though it may not be the minimum).

We have zeros when $ \log x \equal{} \sin x$; that is, when $ \sin x$ intersects $ \log x$. On the interval $ (0,10)$, $ \sin x\equal{}\minus{}1$ twice, meaning that it drops below $ \log x$ twice. However, beyond $ x\equal{}10$, $ \sin x < \log x$, so it never crosses over again. Thus, $ \sin x$ crosses $ \log x$ twice during its first drop but only once during its second drop, for a total of $ \boxed{3}$ zeros.

Okay, I admit that I drew a quick sketch of $ \log x$ and $ \sin x$ to help me see this, but I did not have to plot points. So this is another semi-graphy solution.[/hide]

lol, you pretty much just wrote out the steps I took without drawing the graph :)

You can find the roots with algorithms, if you really want to.

hehe sorry... I guess I did. :oops:

What is the algorithm you have in mind?

I was thinking [url=http://en.wikipedia.org/wiki/Newton's_method]Newton's method[/url].

Thanks for the link, grn!

@Adharsh: Okay, we have $ f(x_{n \plus{} 1}) \equal{} 0$, I think.

So basically, Newton's Method finds the x-intercept of a tangent to the curve, finds the tangent at that new x-intercept, finds the x-intercept for that, and so on, nearing the actual zero..

[quote="undefined117"]

@Adharsh: Okay, we have $ f(x_{n \plus{} 1}) \equal{} 0$, I think.

So basically, Newton's Method finds the x-intercept of a tangent to the curve, finds the tangent at that new x-intercept, finds the x-intercept for that, and so on, nearing the actual zero..[/quote]

Can u please explain to me why $ f(x_{n \plus{} 1}) \equal{} 0$ :?:

You want to find roots of the function, so you can set up recursion claiming that $ f(x_{n\plus{}1}) \equal{} 0$ because that's exactly what you want. Then you are looking for $ x_{n\plus{}1}$. Having recursion set up, every $ x_i$ will be approximate value of the root you are looking for.

[quote="Adharsh Raghavan"]Thanks, but in Newton's method, I can't understand this:
[hide]

$ f^'(x_n) = \dfrac{f(x_n) - 0}{x_n - x_n_ + _1}$

why is there a 0 in the numerator?? :?: [/hide][/quote]

Really random $ \text{\LaTeX}$ tip: Don't exponentiate your prime symbol, it just looks weird. Instead, just write f'(x), which looks like $ f'(x)$, which is a lot prettier than $ f^'(x)$.

[quote="Flame"]You want to find roots of the function, so you can set up recursion claiming that $ f(x_{n + 1}) = 0$ because that's exactly what you want. Then you are looking for $ x_{n + 1}$. Having recursion set up, every $ x_i$ will be approximate value of the root you are looking for.[/quote]

I am not good at math, so please tell me this :

if $ f(x_{n + 1}) = 0$ according to our assumption, then what would be the next step in the recursion algorithm?

Wouldn't we have something like this in the next step :

$ f'(x_{n}) =\frac{f(x_{n+1})-0}{x_{n+1}-x_{n}_{+}_{2}}$

which means, in the numerator, $ x_{n + 1} = 0$, so we have $ f'(x_{n}) = 0 - 0 = 0$.
So....help please?? :help:

[quote="Adharsh Raghavan"]Wouldn't we have something like this in the next step :

$ f'(x_{n}) = \frac {f(x_{n + 1}) - 0}{x_{n + 1} - x_{n}_{ + }_{2}}$[/quote]

We have $ f'(x_n) = \frac{f(x_n)-0}{x_n-x_{n+1}}$.

Remember that, like you said, $ f(x_{n+1})=0$ is only an assumption. It's what we're guessing will happen based on the linear approximation near the actual zero. So for the sake of clarity, if we let $ g(x)$ be the tangent to the curve, we have $ f(x_n)=g(x_n)$ and $ f(x_{n+1}) \approx g(x_{n+1}) = 0$ in the equation above.

so , u mean 2 say that

$ f$ and $ g$ $ \rightarrow 0$ as we move closer to the actual root, and the assumption we make is $ f(x_n_+_1) = 0$, where $ x_n_+_1$ is the x-intercept of the tangent after every successive step.


inequalities
geometry
rhombus
rotation
trigonometry
geometry unsolved

Let ABCD be a rhombus with ∠BAD = 60◦. Points S and R inside the
triangles ABD and DBC, respectively, are chosen such that
∠SBR = ∠RDS = 60◦.
Prove that $ SR^2$ ≥ AS · CR.

Nice problem !!!

Lets be $ AS \equal{} x, SR \equal{} y, RC \equal{} z$ . Now, lets rotate $ 120^o$, $ \triangle ASB$ wrt point $ B$. So, $ SB \equal{} BL, AS \equal{} LC \equal{} x$ and $ \angle SAB \equal{} \angle LCB \equal{} a$. Since $ \angle SBR \equal{} \angle LBR \equal{} 60^o$, $ SBLR$ is a kite and $ SR \equal{} RL \equal{} y$.

Similarly, lets rotate $ 120^o$, $ \triangle ASD$ wrt point $ D$. So, $ SD \equal{} DE, AS \equal{} EC \equal{} x$ and $ \angle SAD \equal{} \angle ECD \equal{} b$. Since $ \angle SDR \equal{} \angle EDR \equal{} 60^o$, $ SDER$ is a kite and $ SR \equal{} RE \equal{} y$. So, $ \triangle ECR$ and $ \triangle LCR$ are concurrent (Side - Side - Side) and $ \angle ECR \equal{} \angle LCR$. Since $ \angle DAB \equal{} a \plus{} b \equal{} 60^o$ then $ \angle ECL \equal{} 120^o$. So,  $ \angle ECR \equal{} \angle LCR \equal{} 60^o$. We apply cosine theorem in $ \triangle CRL$:

$ y^2 \equal{} x^2 \plus{} z^2 \minus{} xz$

According to the A-G mean inequality, $ x^2 \plus{} z^2 \ge 2xz$ and therefore $ y^2 \ge xz$

Lokman GÖKÇE (Turkiye)


calculus
integration
limit
function
logarithms
real analysis
real analysis unsolved

Prove that

$ \lim _{k \rightarrow \infty} \int_{\minus{} \infty}^{\infty} [(1\plus{}\frac{z^2}{k})^{\minus{}\frac{k}{2}} \minus{} e^{\minus{}\frac{z^2}{2}}] \mathrm{d}z \equal{} 0$

The second half of your integral is sqrroot(2pi).

Evaluate the first half using residue theory, in terms of k. (It may be useful to take k=2n for this purpose). Then I think the limit can be evaluated.

No need to evaluate the integrals.  $ \left(1 + \frac {z^2}{k}\right)^{ - \frac {k}{2}}\to e^{ - \frac {z^2}{2}}$ essentially by definition of $ e$ and some algebraic manipulation, so for each $ \epsilon > 0$ and $ M > 0$ there is a $ K$ such that for all $ k > K$ we have $ \left \vert \left(1 + \frac {z^2}{k}\right)^{ - \frac {k}{2}} - e^{ - \frac {z^2}{2}}\right \vert < \epsilon$ for each $ |z| < M$.  Hence,
\[ \begin{eqnarray*} \left \vert \int_{ - \infty}^\infty \left(1 + \frac {z^2}{k}\right)^{ - \frac {k}{2}} - e^{ - \frac {z^2}{2}}dz\right \vert & \leq & \left \vert \int_{ - M}^M \left(1 + \frac {z^2}{k}\right)^{ - \frac {k}{2}} - e^{ - \frac {z^2}{2}}dz\right \vert + 2\left \vert \int _M^\infty \left(1 + \frac {z^2}{k}\right)^{ - \frac {k}{2}} - e^{ - \frac {z^2}{2}}dz\right \vert \\
& \leq & 2 \epsilon M + 2\left \vert \int _M^\infty \left(1 + \frac {z^2}{k}\right)^{ - \frac {k}{2}} - e^{ - \frac {z^2}{2}}dz\right \vert
\]
The first term vanishes as $ k\to \infty$ because $ \epsilon$ is arbitrary (and chosen independent of $ M$), and the second term vanishes because the integrand is $ L^1$ (needs to be shown, of course, but not hard) and $ M$ is arbitrary.

Let $ k\geq 2$ then
\[ \left( 1 \plus{} \frac {z^{2}}{k}\right)^{ \minus{} k/2} \leq \frac {1}{1 \plus{} z^{2}/2} \in L^{1}
\]
since

the function $ h(t) \equal{} t \ln (1 \plus{} z^{2}/t)$ is increasing for all $ z$
thus by the Lebesgue dominated convergence theorem $ \lim \int_{\mathbb{R}}\left( 1 \plus{} \frac {z^{2}}{k}\right)^{ \minus{} k/2} dx \equal{} \int_{\mathbb{R}} e^{ \minus{} z^{2}/2}dz$


quadratics
LaTeX
algebra
quadratic formula

The product of two consecutive odd whole numbers is 255.
What is the greater number?

that's probably the easiest way, but you could do it using algebra

let's say we give the value $ x$ to be the number we are looking for, in this case, the larger number
since it is two consecutive odd numbers, the other number must be $ x \minus{} 2$

therefore, we have the equation
$ (x)(x \minus{} 2) \equal{} 255$

so, we solve it

$ x^2 \minus{} 2x \equal{} 255$

we subtract 255 from both sides to get

$ x^2 \minus{} 2x \minus{}255 \equal{} 0$

from then on, use the quadratic equation to find the value of $ x$

i don't know how to write the quadratic equation in latex, but if you figure it out that way, the answer should come out to either $ 17$ or $ \minus{}15$
since it asks for whole numbers, the answer is $ 17$

The quadratic formula is pretty slow to use, though. I might add that since you know that 255 ends in a 5, one of the factors must also end in a 5, which makes the guess/check a lot easier.

very true, but i'm just telling pplz how to solve it if they weren't competing against time


Let $a_1,a_2,...$ be a sequence of real numbers, such that $a_{k-1}+a_{k+1}\geq 2a_k$ for all $k\geq 2$. Prove that for all $n\geq 1$,\[\frac{a_1+a_3+...+a_{2n+1}}{n+1}\geq \frac{a_2+a_4+...+a_{2n}}{n}\]

it is easy to prove when n is odd
but I don't how to do when n is even :(

I don't think it is right for n is even
try n=2

Take $f : \{1; 2; ...; 2n+1\} \rightarrow R$, $i \rightarrow a_i$. The problem states $f$ is convex. Obviouly,
$(2n+1, 2n+1, ... (n\ times) ..., 2n+1, ......, 1, 1, ... (n\ times), 1)$ $\succ$ $(2n, 2n, ... (n+1\ times) ... 2n, ......, 2, 2, ... (n+1\ times), 2)$,
hence Karamata implies the problem.


hi, can you guys help me with the base conversions? i can convert a number of base n to base 10 most of the time. but i find it very confusing to do it the other way. could you guys help?
FOR EXAMPLE:

1) convert $34_7$ to base 10.

2) convert $216_{10}$ to base 6 and 8

if you could explain your work as it goes it would be greatly appreciated. :)

[hide="#1"]Example: 23 in base 10 is $2 \times 10^1 + 3 \times 10^0$.  Following this pattern, $34_7=3 \times 7 + 4 \times 1$ which is 25. [/hide]

[hide="#2"]To convert 216 to base 6, we want to find the nearest power of 6 to 216 without going over it.  That would be 216 (what a coincidence).  216=$6^3$ and we have only one of those, so the number in question is 1000.  To convert 216 to base 8 uses the same idea.  The nearest power to 216 of 8 is 64, but we need 3 of them to be as close to 216 as possible.  After $3 \times 64=192$, we have 216-192=24.  24 can be divided to 3 8's.  So, our number is 330.[/hide]


Anybody play minesweeper?

My personal records are:

Beginner - 2 seconds
Intermediate - 24 seconds
Expert - 83 seconds (though I'm sure I can beat that :))
Sum - 109.

I'm aiming for a sum of 100.

(Edited to update my scores :P)

wow.. :surrender:

I used to play..
my computer says..
beginner: 6 seconds
intermediate: 38 seconds
expert: 93 seconds

whoa, i've never timed my self except for once
It was on the most squares you can get and the most mines you can get.  I did it in about 5 seconds.  I solved 18 of them consecutivly

[quote="math92"]It was on the most squares you can get and the most mines you can get.  I did it in about 5 seconds.  I solved 18 of them consecutivly[/quote]
24 by 30 with 667 mines?
it is (almost) impossible to solve!!

[quote="Geehoon"][quote="math92"]It was on the most squares you can get and the least mines you can get.  I did it in about 5 seconds.  I solved 18 of them consecutivly[/quote]
24 by 30 with 667 mines?
it is (almost) impossible to solve!![/quote]
no mine was like 100 by 100... :?

[quote="Geehoon"][quote="math92"]no mine was like 100 by 100... :?[/quote]
youre not playing on MS Windows?[/quote]
Omg.  THere was only ten mines on 100 by 100

[quote="lightrhee"][quote="no_one_2000"]Anybody play minesweeper?

My personal records are:

Beginner - 2 seconds
Intermediate - 25 seconds
Expert - 83 seconds (though I'm sure I can beat that :))
Sum - 110.

I'm aiming for a sum of 100.[/quote]
 :o
Mine is something like 8, 55, 259 seconds. How can you click so fast!! :o[/quote]

Pretty simple:  I have [u]no[/u] life. :)  That, and I have lots of practice.  The game's incredibly addicting, so I play it all the time, often while I'm waiting for downloads to complete, while I'm waiting for a friend to respond in a chat, while I'm talking on the phone, when I'm bored, etc.

[quote="lightrhee"][quote="alan"]as of february, when i stopped playing...

beginner: 1 (pure luck)
inter: 26
expert: 100[/quote]

1 second!
 :o  :o  :o  :o  :o[/quote]

lol, I dislike the beginner level so I don't think I'm really persistent enough to get 1 second.  It took me forever just to get 2 seconds.  25 seconds (Inter.) was pretty lucky too, though there was once where I would have had 19 if I didn't click the wrong square at the end (that was aggravating).  Luck is required to get high scores, but you still need some speed.

[quote="shobber"]My best record is around 150.
I have tried my best but cannot proceed anymore. Can you please tell me some hints?[/quote]

- Don't overuse the left-right click feature.  Even though it can uncover a lot of cells with one click, there are many situations (usually around the edges) where you can go through much faster by not flagging mines and just using the left mouse button.
- If you're at an impossible situation where you're forced to guess, don't stall.  Just go for it.  There's no use clearing out the rest of the board and coming back to it because you'll still have to guess because if you get it wrong, doing all that extra work didn't help you one bit.
- Drink caffeine ;)

Otherwise, I don't think there's anything I could tell you that you wouldn't already know.  If you've gotten 150 on Expert, you're pretty good already, so you are most likely familiar with all the patterns.

Or watching these videos might give you some tips: http://metanoodle.com/minesweeper/videos.html  I know they've helped me.

I went to click Minesweeper after this topic and look at this...

P.S. This page is in the background ^^

can we calculate the probability of getting 10 mines in one click?

Most games, I don't really think about where I'm placing it. If I have to guess, I do it immediately and go on. I get it wrong quite often, though. :P

My best times are 8 for beginner, 63 for intermediate and 214 for expert :?  How do you guys do it so quickly?
I'm working on bettering my expert time... yesterday I almost had it done in 105 seconds but then I had to guess the last two and guessed incorrectely :wallbash_red:

Practice. :)

I used to play it more than I do now, although I never actually spent a lot of time on it. I just play maybe 20 Expert games a day now, counting the ones where I mess up partway through but not the ones where I don't really get started. Over time, I gradually improve.

7
67
270
given that i only started really playing this game a month or two ago, that is pretty good i guess!

The best time i have ever seen for expert is 41 seconds.  i saw it on the internet, it came with a vid of the guy doing it. my personal records:

3
19
81

Whoa! Mine are 6, 37, 138, give or take a second each. That was from when I was 10 and somewhat smart. Can't even do that now.

Though I'm not sure your two second one was due to skill.  ;)

I rarely play, but these are my times:

21
146
999 :D

Since this topic's been revived, I'll go ahead and post my best times:
B: 12
I:  65
E: 215

Well, since the last time I've posted, my best times have changed to 7, 55 and 186 seconds for beginner, intermediate and expert respectively :)

Sadly, I have yet to beat it on hard. My best time on easy is 9 i think. I randomly clicked a square and it killed like most of them.

Im addicted to minesweeper and rubiks.
my times for minesweeper are 1, 16, and 82

WOOT!  99 sum

[quote]i used to think i was awesome at minesweeper until i got to this post
begginer 9
intermediate 81
expert 214
[/quote]

 lol me too

beginner: 5
intermediate: 47
expert: 164

as of now my times are:
beginner - 7
intermediate - 47
expert - 160


A sequence $\{a_n\}$ is defined by: $a_1 = 1, a_{n+1} = a_n + \dfrac{1}{\sqrt{a_n}}$ for $n = 1, 2, 3, \ldots$. Find all real numbers $q$ such that  the sequence $\{u_n\}$ defined by  $u_n = a_n^q$, $n = 1, 2, 3, \ldots$ has nonzero finite limit when $n$ goes to infinity.

THERE MIGHT BE A TYPO!

[quote="orl"]A sequence $\{a_n\}$ is defined by: $a_1 = 1, a_{n+1} = a_n + \frac{1}{\sqrt{a_n}}$ for $n = 1, 2, 3, \ldots$. Find all real numbers $q$ such that  the sequence $\{u_n\}$ defined by  $u_n = a_n^q$, $n = 1, 2, 3, \ldots$ has nonzero finite limit when $n$ goes to infinity.[/quote]

$a_n$ grows like $C n^{2/3}$ where $C$ is some explicit constant ($C^3 = {3/2}$).   So the intended question is probably about $u_n = {a_n} / {n^q}$.

This is a question that is routine if you know the trick and hard to solve, if you do not.  Strange to see it on a TST.

Here is my solution for this problem
[b]Solution[/b]
It's easy to prove that: $\lim a_n = + \infty$
We have: $\sqrt{a^3_{n + 1}} - \sqrt{a^3_n} = \sqrt{\left(a_n + \dfrac{1}{\sqrt{a_n}} \right)^3} - \sqrt{a^3_n} = \dfrac{\left(a_n + \dfrac{1}{\sqrt{a_n}} \right)^3 - a^3_n}{\sqrt{\left(a_n + \dfrac{1}{\sqrt{a_n}} \right)^3} + \sqrt{a^3_n}} = \dfrac{3a_n \sqrt{a_n} + 3 + \dfrac{3}{a_n \sqrt{a_n}}}{\sqrt{\left(a_n + \dfrac{1}{\sqrt{a_n}} \right)^3} + \sqrt{a^3_n}}$ $= \dfrac{3 + \dfrac{3}{a_n \sqrt{a_n}} + \dfrac{3}{a^3_n}}{\sqrt{\left(1 + \dfrac{1}{a_n\sqrt{a_n}} \right)^3} + 1} \rightarrow \dfrac{3}{2}$
Then by Stolz theorem, we have: $\lim u_n = \lim \dfrac{\sqrt{a^3_n}}{n} = \lim \left(\sqrt{a^3_{n + 1}} - \sqrt{a^3_n} \right) = \dfrac{3}{2}$
So: If $q > \dfrac{3}{2}$ then $\lim u_n = \infty$, if $q < \dfrac{3}{2}$ then $\lim u_n = 0$
Hence: $q = \dfrac{3}{2}$ satisfies the problem


I have a problem on studying mathematics that I'm so slow to learn new methods. How can I learn things and doing problems faster? I mean if I start studying some book, like Neukirch's Algebraic number theory, then it might take some weeks until I have learned Chapter 1. I have found that I can read mathematics text more rapidly than solving exercises. But I feel sometimes that I have no time to read the whole book on my lifetime as some problems might be so difficult. Often as I start studying some book, I can read at most 20 pages until I meet a problem I can't solve.

I took a course of logic given by the department of Philosophy and it helped greatly ... it wasn't about implications and whatnot but rather to know how to know or learn how to learn if I can phrase it that way

Also, you don't have to know the answer to every problem just to master the material.  In an AoPS Textbook, I consider mastery:

70% of regular review problems
30% of regular challenge problems
You don't necessarily have to be able to do the "starred" problems.

ah, but i thing doing 65% regular 35% challenge
and do the stars after.

[quote="Maybach"]Also, you don't have to know the answer to every problem just to master the material.[/quote]
I have always thought I have to. I just thought that if Liu's book has about 600 exercises and he probably knows solutions to all of them, then a reader should also do all of them.

I also thought that as in AOPS there is one problem in one book, would it be useful to write those solutions in one PDF and put on the Internet? For example I can find the solution to exercises of Atiyah and MacDonald in one PDF but for example there is no such file for Liu's or Neukirch's book. But I just realized that my lifetime won't be long enough to solve all of the problems.


charlie ... how is this possible ? :) you haven't tried it yet out?
just have a look at http://www.mathlinks.ro/viewtopic.php?t=3188 and the part 2 at http://www.mathlinks.ro/viewtopic.php?t=3527 :D

[quote="Charlie"]Sorry but what's pingu?[/quote]
In English it is penguin..

I've tried that out but I don't know the its name  :D

if it wasn't this one, what would be the official game then? :D

Btw, Valentin, is this just my comp or has the site become extremely slow at the very moment? :?

from my end is working fine. there are some "rush hours" I might say :) don't worry about it, it will come back at normal speed.

oh well fine no problem

more time to play pingu then :D

Maybe the more relevant question might be: Which is your favourite version of Penguin? (v. 1, 2, 3, or one of the spinoffs like the NASA pic, Greenpeace, finger hurting etc.) :D

[quote="Valentin Vornicu"]so what do you say? I say Yeah! :D[/quote]

The official game, in my humble opinion, should be Unreal Tournament 2004 (free demo  -> Multiplayer).  We could make one large mathlinks squadron and kick the **** out of other world players.

for more information see

http://www.unrealtournament.com/
http://www.unrealtournament.com/ut2004/downloads.php

remember the demo is multiplayer and free!  I've already racked up more hours on this game than I have on pinguin . . . can you believe that?   :D

[quote="integral0"]I've already racked up more hours on this game than I have on pinguin . . . can you believe that?   :D[/quote]

no.

Oldest non-locked thread on Fun Factory :)

WOW that was a lot of clicking

I like the game though

There are a lot of interesting threads if you scroll down a lot


Find the largest real $ \lambda$ such that for every real $ x$ the inequality $ x^{2009}\ge\lambda[x]^{2008}\{x\}$ holds.
(here $ [x]$ is the integer part of $ x$, $ \{x\}\equal{}x\minus{}[x]$ is the fractional part of $ x$.)

is it only me or : for $ x\equal{}\minus{}1$, then $ \minus{}1 \ge 0$ ?

Sorry. :blush:
It should be
[quote]Find the largest real $ \lambda$ such that for every real $ x$ the inequality $ x^{2010}\ge\lambda[x]^{2009}\{x\}$ holds.
(here $ [x]$ is the integer part of $ x$, $ \{x\} \equal{} x \minus{} [x]$ is the fractional part of $ x$.)[/quote]


function
limit
real analysis
real analysis unsolved

Let $f: (0,+\infty)\to\mathbb{R}$ be continuous concave function, $\ds\lim_{x\to+\infty}f(x)=+\infty$, $\ds\lim_{x\to+\infty}\frac{f(x)}{x}=0$. Prove that $\ds\sup_{n\in\mathbb{N}}\ \{f(n)\}=1$, where $\{a\}=a-[a]$ is fractional part of $a$.
(O. Nesterenko)

consider $\Delta_{n}: = f(n+1)-f(n)$. concavity gives $f(\frac{n+2+n}{2}) \ge \frac{f(n+2)+f(n)}{2}$, which is $\Delta_{n+1}\le \Delta_{n}$.
from the hypothesis, you have that $\Delta_{n}\ge 0$ for every $n$, and if you call $\lim_{n\to\infty}\Delta_{n}= a$, you have that $\lim_{n\to\infty}f(n)/n = a$, so you have $a=0$.
from this, it's easy to deduce thesis...

$f(n)$  is a sequence such that $f(n+1)-f(n)$ tends $0$ when n tends +oo

and $f(n)$ tends +oo when n tends +oo

What is the adherence of the set $A=(f(n)-E(f(n)); n in N )$ ?

[quote="Moubinool"]alekk said $f(n+1)-f(n)$ tends to zer when  n tends +infty
Could you prove that ?[/quote]
Let $a_{n}=f(n+1)-f(n)$
We can prove:$a_{n+1}\le a_{n}$(Because $f$ is concave)
And also:$f(x)$ is not decrease.(Because  $f$ is concave,and $\lim_{x\to \infty}f(x)=+\infty$)
Thus,$a_{n}\ge 0$
thus:$\lim_{n \to+\infty}a_{n}$ is exist,note it is L.
then if $L>0$,we will have contradiction with $\lim_{x\to \infty}\frac{f(x)}x=0$
thus we proved  $f(n+1)-f(n)$ tends to zero.
Anyway,Nice problem. :)

Edit:When I posted I have seen the mago's post :)


After James unloaded 1/8 of the cartons in a truck and Jerry unloaded 2/7 of the remaining cartons, 45 cartons were left.  How many cartons were originally in the truck?

The answer is 72 cartons.

How can this be?

Is this the correct set up:

(1/8)x + (2/7)x = 45

Am I suppose to solve for x?

No.

[hide]You're given what's remaining... Therefore...

$\frac{7x}{8}*\frac{5}{7}=45$

$x=72$[/hide]


For $1 \le k \le 1000$, what is the number of values of $k$ for which $3k+2$ is a perfect square?

[hide]$3k+2=n^{2}\implies k=\frac{n^{2}-2}{3}\implies n^{2}\equiv 2 \mod 3$, but $2$ isn't an quadratic residue of modulo $3$, so there are no solutions.[/hide]

[hide]Look at it in $\mod3$ so the $3k$ goes away. The set of quadratic residues $\mod3$ does not include $2$ so $\boxed{0}$[/hide]

[hide="If you mean real values"] For $k=1$, $3k+2=5$. For $k=1000$, $3k+2=3002$. There are $\boxed{52}$ squares between those. [/hide]

[hide="If you mean integers"] $\boxed{0}$ The quadratic residues in mod 3 are 0, 1. $3k+2\equiv 2\pmod3$. [/hide]

Yeah, if you mean all real values and K does not have to be integers, then there are 52 values, but if they are integers, there's 0.

[hide="solution"]It is asking squares in the form 2mod3, but the square root of that is sqrt2mod3, which doesn't exist. Therefore, theer are no solutions.[/hide]


topology
Support
geometry
3D geometry
sphere

Hi.

The Jordan-Brouwer separation theorem states that for every embedding $ f: S^{n\minus{}1} \hookrightarrow S^n$ the set $ S^n \setminus f(S^{n\minus{}1})$ has exactly two components (and the image of $ S^{n\minus{}1}$ is their common boundary).

Is there an elementary proof that there are [i]at least[/i] two components? (let's define "elementary" as "doesn't use algebraic topology")

May be there is no such elementary proof. Maybe one has to use homology. Is it possible to derive the statement from the Eilenberg-Steenrod-axioms (including dimension, infinite additivity and compact support if you want) but without using a concrete homology theory?
It is possible to derive from the axioms that $ S^n \setminus f(S^{n\minus{}1})$ has at most two components. In fact one can show from the axioms that $ \dim H_0(X)$ is an upper bound for the number of components if $ X$ is locally connected.

You can compute the homology of the sphere using just the axioms (in particular you can show that for $ n>1$, $ H_1(S^n)\equal{}0$). Then we can show that any embedding $ S^1\hookrightarrow S^n$ has trivial homology class which means it cuts out some disc.

[quote="LkNsngth"]which means it cuts out some disc.[/quote]
And what do you mean by that? That f can be extended to an embedding of $ D^2$ ? I think, that's wrong in general. I'm not quite sure, but I think that the so called wild arc of Fox and Artin is a counterexample.

And even if it were, why does that imply that $ S^n \setminus f(S^{n\minus{}1})$ has (at least) two components?


Γειά σας, είμαι νέος στο forum.
Μπορείτε να με βοηθήσετε στην άσκηση :
177...7955...52-533...328 =..?
--\  ν /--\ ν  /-----\ν-1/---

(ν 7αρια, ν 5αρια, (ν-1) 3αρια

Ευχαριστώ

ΥΣ(Ξέρω να χρησιμοποιώ το 10^ν -1/9 , αλλά σε πιο απλές ασκήσεις. Παρακαλώ οποιος μπορεί ας με βοηθήσει )

Γειασου Κωστα!Αποδεικνιεται επαγωγικα οτι το αποτελεσμα της διαφορας ισουται με:
$ 177..7422...24$ με ν 7-αρια και ν 2-αρια.

Ευχαριστώ φίλε ..ειμαστε και συγχωριανοί
σ΄'αυτον τον τρόπο δεν είχε παει το μυαλό μου. Πολύ απλό.

Χαρηκα!Εισαι ο τριτος Καλαματιανος της παρεας τοτε! :)

Χάρηκα κι εγώ. Ποιός είναι ο τρίτος?

Ειναι ο kostas ενας φιλος μου αλλα εχει καιρο να εμφανιστει...καλυτερα να τα πουμε σε p.m(πατας στο ονομα μου και υστερα στον φακελο που θα εμφανιστει...)Σορρυ guys... :wink:


Write each point given in rectangular to polar form.

(1) (0, 2)

(2) (sqrt{3}, 1)


Prove that in any triangle, a line passing through the incenter halves the perimeter of the triangle iff it halves the area of the triangle.

I didn't quite understnad the problem or the hint!!!??? :(


To the second one:
From f(f(n))=n you have this: For distinct x,y integers f(x)<>f(y). Suppose it's not true, so f(x)=f(y) for some x,y. Then x=f(f(x))=f(f(y))=y which is not true, contradiction (from now we can use f(a)=f(b) => a=b). We also know that f(f(0))=0 so f(1)=0. We can show by induction that f(2k+1) for k positive integer (also 0) is -2k. It's true for k=0. Suppose it's true for some k. Then set n=2k+1:
f(f(2k+1))=f(f(2k+3)+2)
From this we have
f(2k+3)=f(2k+1)-2=-2(k+1)
so we proved it. Very similary you can prove that f(2k)=-(2k-1) for k>=0.
You can do similar things with negative integers. Let's prove that
f(-(2k+1))=2k+2
Set n=-1. Then
f(f(-1))=f(f(1)+2) => f(-1)=2
By setting n=-3,-5,... (induction) you can prove it for other values. It's similar also for even negative numbers.
Sorry for writing so many times word similar, but it's similar and I am lazy to write it completely :)

To the first one: I've seen a solution but not going to post immediately :)


Prove that

$ \frac{F_{2n}}{F_{n}}= L_{n}$

Where $ F_{n}$ denotes the nth Fibonacci number and $ L_{n}$ the nth Lucas number.

This is a classic property. Just plug in the explicit forumlas for both sequences and the result follows directly [by difference of squares...].


una pregunta

alguien tiene en su poder algun apunte de desigualdades asi como para empezar desde el principio a aprenderlas?

mas que nada apuntes en PDF para aprender desigualdades desde cero, eso de de la media aritmetica, geometrica, armonica, cuadratica, desigualdades importantes.

se agradece cualquier aporte.

[color=blue]El año pasado hice apuntes de algunas teorías de desigualdades, pero están en un cuaderno.

¿Es necesario que se envíe en un archivo .pdf? Lastimosamente, no tengo aún medios para crear uno. ¿Cómo proceden para crearlos? (Especialmente pregunto a Carlos Bravo)[/color]

en realidad, es muy facil teniendo el adobe acrobat 7.0 professional (o una version mas nueva), dicho programa trae incluida la impresora PDF Maker, y con eso, se pueden traspasar cosas hechas en word por ejemplo, a un .pdf.

seria muy util si pudieses darme dicho material en esa extension por favor.

[color=blue]Aparte de la desigualdad cuadrática - Aritmética - Geométrica - Armónica, cuáles más querrías tener en el mismo material?[/color]

Hola como estas LINE,

AKI tienes unos pdf q he colgado en la comunidad,
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58478
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61334

pero hay mucha info en la web, aunq la mayoria en ingles..

-como en los cursos de verano en Princeton: http://www.princeton.edu/~ploh/olympiad.shtml

-el forum de inecuaciones de mathlinks: http://www.mathlinks.ro/Forum/index.php?f=54

-e inclusive en los directorios de los websites olimpicos:
http://olympiads.win.tue.nl/imo/

Espero te sea de utilidad...

CARLOS BRAVO
Lima - PERU


abstract algebra
group theory
superior algebra
superior algebra unsolved

Hello,
I was given the following problem:
Let $ G$ be an abelian group of order n, an let m be an integer. Show that the elements $ g \in G$ with $ o(g)|m$ form a subgroup of $ G$.

I came up with the following solution.
Let ${ K=\{ g \in G : o(g)|m}$.
Let $ e$ be the identity in $ G$. Since $ o(e)=1|m$, then the identity lives in $ G$.
Let $ g_1,g_2 \in G$. Then, $ {{g_1}^m}{{g_2}^m}=e.e=e$. Thus, $ ({g_1}{g_2})^m=e$, since G is abelian, and so $ {g_1}{g_2}$ lives in $ G$.
Moreover, for any $ g\in G$, $ g^m=e$, thus, $ (g^m)^{-1}=e$ and so $ (g^{-1})^m=e$, which means that the inverses live in $ G$. Thus, those elements form a subgroup of $ G$.

Is this argument correct? Is there any way to prove this using direct products of cyclic groups? Please let me know.
Thanks

I think it's really ok.

Referring your second question, do you know the prime factor analysis of m?

Yes, i do, but i cannot see how to use this result. Actually the book was giving this as a hint, and i could't discover how to use it. That's why i came up with this quite elementary solution. Do you have any further ideas for any proof using the fact that any finite abelian group is isomorphic to a direct product of cyclic groups?
Thanks

If you let G_m = { g^m = 1 : g in G }, then (G×H)_m = G_m × H_m.  Then you could prove that (G×H)_m was a subgroup because G_m and H_m are subgroups.

I prefer your original solution since "(G×H)_m = G_m × H_m" has almost exactly the same proof, and I don't think it is too much easier to see that G_m is a subgroup when G is a cyclic group of prime power order (versus an arbitrary, possibly infinite, abelian group).

Guess I'll never forget [url=http://www.mathlinks.ro/viewtopic.php?t=184800]that[/url] one line proof due to -oo-.


Let $ n>m>1$ be odd integers, let $ f(x)\equal{}x^n\plus{}x^m\plus{}x\plus{}1$. Prove that $ f(x)$ can't be expressed as the product of two polynomials having integer coefficients and positive degrees.

It is not true. If $ n\minus{}m$ is odd $ f(\minus{}1)\equal{}0$, therefore $ f(x)\equal{}(x\plus{}1)g(x)$ with $ g(x)\equal{}x^m(x^{n\minus{}m\minus{}1}\minus{}...\plus{}1)\plus{}1$.

[quote="Rust"]It is not true. If $ n \minus{} m$ is odd $ f( \minus{} 1) \equal{} 0$, therefore $ f(x) \equal{} (x \plus{} 1)g(x)$ with $ g(x) \equal{} x^m(x^{n \minus{} m \minus{} 1} \minus{} ... \plus{} 1) \plus{} 1$.[/quote]
A/Q $ n,m$ both odd. $ n\minus{}m$ must be even.

I'm blocked here, anyone could continue in this way or propose another solution?

See Ljunggren W, On the irreducibility of certain trinomials and quadrinomials, Math.Scand. 8 (1960)

Main idea is if $ f$ is not irreducible, then there exist its divisor $ g(x)$ such that $ g(x)\equal{}x^{deg(g)}\cdot g(x^{\minus{}1})$. This leads $ f(x)$ has a root $ a$ such that $ a^{\minus{}1}$ is its another root. After some calculation, $ a$ must be even and odd degree root of $ \minus{}1$, that leads to contradiction.

Proof of the mentioned result is quite long and boring. I doubt it can be invented during TST. I would like to see something simple in case $ m$ and $ n$ are odd.

Here's an explicit write up for this case.

Suppose that FTSOC $f(x) = g(x)h(x)$.
Define $\tilde{f}$ as the reciprocal polynomial.
Then if $k = g \tilde{h}$, $k \ne \pm f, \pm \tilde{f}$ and \[ k\tilde{k} = f\tilde{f} = (x^n + x^m + x + 1)(x^n + x^{n-1} + x^{n-m} + 1) \]
By considering the coefficient of $x^n$, the sum of square terms in $k$ is the same as $f$ and thus $4$. WLOG suppose that $k$ is monic.
Let $k = x^n \pm x^{u} \pm x^{v} \pm 1$ for $u, v < n$. Then, all coefficients must be positive since $f\tilde{f}$ has $16$ positive coefficients.
As such, \[ (x^n + x^m + x + 1)(x^n + x^{n-1} + x^{n-m} + 1) = (x^n + x^{u} + x^{v} + 1)(x^n + x^{n-v} + x^{n-u} + 1) \]
It follows that $u + v = m + 1$ must hold by considering the sum of exponents.
Now, consider the $x^{2n - 1}$ term which appears in the LHS. This only holds if one of $u, v$ is either $n - 1$ or $1$.
This implies that $k$ is either $f$ or $\tilde{f}$, contradiction.


True Or False :

If $ (f_\tau)$ is a collection of measurable functions on $ [0,1]$ such that $ \tau_1 < \tau_2 \ \Longrightarrow f(\tau_1)\leq f(\tau_2) , \forall x\in [0,1]$ , then $ \underline{lim} (f_\tau)$ and $ \overline{lim} (f_\tau)$  are measurable .

[hide]
Surely , it's not true. since we can work on uncountable collection of measurable functions such that lim sup and lim inf is not measurable.

My thinking is to work on the set $ P$ (Vitali set ) by making an Ascending chain of subsets of $ P$ each one of subsets is finite set . So, we can define the sequence of simple functions on these sets . Thus the sup of them it will be on $ P$ which is not measurable function .

Are my thinking true ??[/hide]

 :)

Actually this is true because the limits can be taken over the rational values only (the monotonicity condition is crucial for showing that these restricted limits are the same as the limits over $ [0,1]$).

Do u mean the statement is true ? No , that's is not true .

The statement is false .

Take :

$ E_{\tau_k} \equal{} \{x_1,x_2,\cdots,x_k\}\subset P$ , define $ f_{\tau_k} \equal{} 1_{E_{\tau_k}}$ then :

the limit sup on these collection of measurables functions it will be on $ P$ . which is not measurable functions.

What do u see??
 :)

I see that your family of functions is (probably) not monotone in $ \tau$ to start with ("probably", because I had to guess the exact meaning of what you wrote), and that you are a bit sloppy in your writings to end with: $ \lim$ as $ \tau$ tends to what?, How are $ x_k$ related to $ \tau_k$? (remember, you need to define $ E_\tau$ for all $ \tau\in[0,1]$ and you just did it for some strange subsequence about which you said nothing). :P

The English language is not my native language at all. Don't blame me on my writing  :mad: . I'm still learning English. :P

I don't said this is my solution in my 1st post , I said that's my thinking  only  :P  :D

I need to know how to write that sequence only .

I'm not blaming you for English (which is not my mother tongue either) at all. Your writing is faulty in purely mathematical sense :).

You're right :-)

Since I don't understand this question very well yet.

I need to know how to write the answer of it.

Thx dude  :)


How do I typeset square-bracket binomial coefficients?  They should look like $ \left[ \begin{array}{c} n \\ k \end{array} \right]$ but not take up so much space so as to resemble $ {n \choose k}$ more closely.

Use \genfrac on which \binom{n}{k} $ \binom{n}{k}$ is based: $ \genfrac{[}{]}{0pt}{}{n}{k}$
In a document you would use \newcommand{\[i]yourname[/i]}[2]{\genfrac{[}{]}{0pt}{}{#1}{#2}} so as to be able to use \yourname{n}{k}.
For more on \genfrac see [url=http://tug.ctan.org/tex-archive/info/math/voss/mathmode/Mathmode.pdf]Math Mode[/url] page 58.


Let x, y be reals satisfying:

sin x+cos y=1
sin y+cos x=-1

Prove cos 2x=cos 2y

[hide="Alternately"] Let $z = \frac{\pi}{2}-y$.  Then

$\sin x+\sin z = 1$
$\cos x+\cos z =-1$

And we want to prove

$\cos 2x = \cos (\pi-2z) =-\cos (2z)$

Well, let $u = e^{i x}, v = e^{i z}$.  Then

$u+v = i-1$

Of course, geometrically, we can only have two cases (there are only two right triangles with side lengths $1, 1, \sqrt{2}$ such that the hypotenuse is the line from $(0, 0)$ to $(-1, 1)$):

$u = i, v =-1$

Or

$u =-1, v = i$

Both of which satisfy our condition.  QED. [/hide]

[quote="junggi"][quote="D.P.L"]Let x, y be reals satisfying:

$\sin x+\cos y=1$
$\sin y+\cos x=-1$

Prove $\cos 2x=\cos 2y$[/quote]
[hide]square $\sin x+\cos y=1$
$\sin^{2}x+2\sin x\cos y+\cos^{2}y=1$
square $\sin y+\cos x=-1$
$\sin^{2}y+2\sin y\cos x+\cos^{2}x=1$
add the two and simplify
$\sin (x+y)=0$
subtract the two and simplify
$\sin^{2}x-\cos^{2}x+\cos^{2}y-\sin^{2}y+2\sin x\cos y-2\sin y\cos x=0$
$-\cos2x+\cos2y+2(\sin(x-y))=0$(1)
multiply two original equations together
$\sin x\sin y+\cos x\cos y+\sin x\cos y+\sin y\cos x=1$
$\cos(x-y)+\sin(x+y)=1$
$\cos(x-y)=1$
so $\sin(x-y)=0$
then (1) becomes
$\cos2y-\cos2x=0$
$\cos2y=\cos2x$ AWD[/hide][/quote]

[hide]From $\sin(x+y) = 0$, you immediately get $x+y = k\pi$, so $y = k\pi-x$. Therefore, \[\cos 2y = \cos(2k\pi-2x) = \cos(-2x) = \cos(2x)\] [/hide]


let  $x,y,z$ be three positive real numbers . Prove that :
 $\sum \frac{1}{x^{5}+1}\geq \frac{12}{(x+1)(y+1)(z+1)}$
    question :  ineqs true or false with
$\sum \frac{1}{x^{5}+1}\geq \frac{12}{(\sqrt[3]{xyz}+1)^{3}}$

[quote="tranthanhnam"]let  $x,y,z$ be three positive real numbers . Prove that :
 $\sum \frac{1}{x^{5}+1}\geq \frac{12}{(x+1)(y+1)(z+1)}$[/quote]
There is a mistake. For $x=y=z>9$ it is false.


I have a question with number 1c, When it says how many words have exactly the same letters, wouldn't all of the words have the same letters because they're just variations of the same thing?

Yes. But the ones in a) are different subsets {m, a, t} may be one and {t, a, m} is another in a). They ask how many subsets in a) have the same letters in set b). While {m, a, t} = {m, t, a} in set b), they are different in subset a) , thus you already have two subsets of a) that have the same letters as a subset of b). There are more - hence the problem asks you how many there are total.

where in the world is question number 1c?
i'm lost fubu.

i found it (i haven't been doing the problem sets)
its on problem set 2 of the AMC/MATHCOUNTS Counting Class.
Mr. Moderator should move it there.


AMC
USA(J)MO
USAMO
Harvard
college
MIT
analytic geometry

Okay, so just to give you a backgroup of myself. I'm an azn guy that lives in Pennsylvania, and I'm currently a junior. I'm also currently valedictorian and president of math club, physics league, and Tri-M music club. I almost certainly qualified for the USAMO this year, and I'm basically putting all my time into studying so that I can make it to MOSP. The Engel Book takes up the majority of my time, and even at the pace I'm going, I'll probably only get through half of it.

However, I just got back my SAT scores, with a wretched 1490. As an azn guy hoping to get into Harvard or MIT, I'm debating whether to retake or not. However, the next SAT is May 1, just a few days after the USAMO. Should I spend some time studying for the SATs? Or should I spend all my time preparing for the USAMO? From my personal experiences, math and english do not mix at all...

If I make it into MOSP, I think that'll help me get into Harvard a lot more easily than a 1600 on the SATs. What do you guys think I should do? I'd hate to study for both exams at the same time...

Please give me your opinion here.

Your SAT score is better than mine was. Why don't you put it away and prepare for the USAMO? You'll get a lot more personal satisfaction out of doing well on the USAMO than you will out of 1600ing the SAT I would imagine.

My advice to ANYONE who has a shot at USAMO this year is to leave the SAT for the fall. One reason I give this advice is that training for the USAMO and a shot at MOP will be good SAT training anyhow, after you digest the math you learn over the summer. Another reason is that MOP is a MUCH more impressive credential--especially to the schools that care about math credentials at all--than a higher score on the SAT taken as a junior. You have time to take an SAT that will reach a college admissions office in time, and you have a strategy to do something much more impressive than merely score high on the SAT: go for it.

I just saw my son's March 27 SAT I scores online. I thought the SAT was a cakewalk back when I took it, but I only took it once, in the autumn of my senior year (because of an earlier grade skip, that was at age sixteen). I have been telling my son, and he is beginning to believe my advice  :D that the key to the SAT has been and always will be to READ well and reason carefully about what the questions are asking. For anyone on this thread who, like my son, has a non-English language speaking in his head, and who has more math time in his free time activities than reading time, my advice given in earlier threads stands: READ, READ, READ, and READ. He is pulling up his verbal scores dramatically on standardized tests over the past year, and he is avoiding a lot of "stupid mistakes" now on math tests because he is a more attentive, thoughtful reader. Definitely spend your free time this summer reading interesting books and building reading experience. I have some [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8674]reading suggestions about math[/url] in an earlier thread. Don't worry one bit about retaking the SAT right away: you have time to do something much more important.

Good luck, and have fun.

[quote="mathnovice"]I wish I could have your "problems," masamune.[/quote]

I was going to say something similar, but probably more rudely :).

I would agree that you should study for the USAMO.  If your trying to get into Harvard or MIT MOP would probably look a lot better than a 1600 SAT.  But I'm not entirely sure.  In my opinion MOP and a 1490 would give you a good shot at being accepted.  My only other advice is to definitely retake the SAT next year because a 1490 is good but it s not very reassuring if your trying to get into Harvard.  If you get into MOP and get over 1550 on the SAT I believe you should have no problem getting into Harvard/MIT.
Good Luck

Masamune, what was your breakdown? 800-690 is not too bad if you're applying engineering, but if you made a few mistakes on math...=/.

Anyway, a 1490 is not a bad score. If you just made a bunch of careless mistakes, it's okay. What I would do is glance over a vocabulary list like Wordsmart (I at least, II if you're feeling up to it--I like the series). I trust you have no problems with the math section in terms of ability; just try not to pull a 2+3=7 :).

I'm not sure you can really "study" a lot for USAMO. Sure, there's a lot of practice to be done, but I think you've already been studying for USAMO, indirectly or directly, for perhaps years. Combine that with your valedictorianship and whatnot, and you'd probably make Harvard/MIT anyway.

So...hmm, I lost track of what I was trying to say. When I took the SAT in January, I basically winged it, but I looked at the above mentioned vocabulary lists. They help a lot. If your problem is critical reading, though, I don't know what to say. CR skill comes with practice and time.

[quote="JBL"]I wish people would get over the idea that anyone can say, "Well, given your credentials, you're all set to get in to <school>."  At the percentage of acceptances that we're talking about, there is a very sizeable dose of random involved.[/quote]

I was agreeing with you in adopting the term "randomness" in the last thread that raised this point, but another Ivy League grad seemed to think that it's [b]mostly[/b] merit that matters, so that the word "random" isn't fitting. I think you have the truth of it. What I have read is that if one put together an alternative entering class from among all the entrants who were denied admission to [name of top university], the alternative entering class would be every bit as strong on the basis of the usual credentials as the admitted class. I think it's fair to describe such situations as involving a certain degree of unavoidable randomness.

I'm deeply skeptical that any applicant can do more than increase chances for admission, which is NEVER a sure thing. And Asian-American males have an especially steep hill to climb in increasing chances for admission, the way most "affirmative action" programs are working these days. Anyway, everyone ought to have a plan B, and a Plan C, and no one should think it is the end of the world if the most-desired school says no.

What I would wonder about the application of the person who opened this thread is:

1) what activities do you engage in to help the less fortunate?

and

2) are you motivated by genuine intellectual curiosity to study the things you study, or are you just trying to check off items on a check list for college admission?

Admissions officers will want to know the answers to both of those questions for most applicants, at most top schools.

[quote]If you have a 3.0 grade average, it may be another situation.[/quote]

I'm pretty sure he has a higher GPA than that given that he is his school's valedictorian.  Unless his school is much different than the majority of other schools in the nation he probably has a 4.0 or higher(depending on how the classes are weighed).

I didn't mean for my previous comment to come across that if he has accomplished such and such he [/u]will[u] get into a certain school.  I meant that his chances would be greatly increased.  Please excuse me for making that impression because I know that nothing in this world is a given. And I also know that your a fool if think something is.[/u]

[quote="tokenadult"]
What I would wonder about the application of the person who opened this thread is:

1) what activities do you engage in to help the less fortunate?

and

2) are you motivated by genuine intellectual curiosity to study the things you study, or are you just trying to check off items on a check list for college admission?

Admissions officers will want to know the answers to both of those questions for most applicants, at most top schools.[/quote]

I've participated and helped coordinate both Relay for Life (cancer fundraiser) and Habitat for Humanity (constructing homes in Philadelphia for the poor). However, other than that, I haven't done much else...

And of course, I am genuinely interested in mathematics and wish to persue some actuarial major. It is this whole college appication process that sickens me...the unfairness of it all...

I can tell you that English is certainly not my forte. I've studied very long for the SAT I and I still did not do that well.

However, my GPA is at a good 4.42 out of 4.6 (A+)

you can take the sat again.. not the usamo

Well, USAMO is in late April I think. This means that if he studies for USAMO, he won't be able to adequately prepare for the May SAT. And if he's doing ARML, he can't do the June SAT either. If that's the case, he can't take the SAT's until early next year, and that makes it kind of uncomfortable for ED/EA.

The October SAT is plenty timely for early action. Even the November SAT is timely for early action at Harvard, according to Harvard reps whom I saw speaking in Minneapolis last year. The harder time crunch for early action comes from setting up alumni interviews.

Thanks everyone for the advice. Now I have a decent idea of what to do: focus on the upcoming USAMO and take the SAT I next fall.

Unfortuantely, that doesn't take any pressure off my econ AP exam, stat AP exam (I have to learn all of stat next month, considering I'm not taking the class), SAT II Writing, final exams...etc. But I'm sure most of the juniors at AoPS are in the same boat as me.

Good luck everyone!

My advice is just to take the SAT as many times as you can, and not study for it unless you are memorizing vocab words.  A kid at my school took the SAT 5 times and went from a 13something to a high 15something, and he was accepted to millions of good places.  Colleges don't care how many times you take it, just your best score, and they take maths and verbals from different test days.  I swear.  People will tell you differently, but they are wrong.  Most likely, some person who is not an adcom prepares your file, w/your best math and verbal, best 3 SAT iis, and the rest of your file -- then gives it to an adcom for review.  So USAMO is way more important for you, for personal achievement (and college apps -- MIT will def. care about this).  But, you can still probably get a high 15something by taking it a few times.  Don't stress out, just think -- "I can take it again, and no worries if I do worse or anything like that, b/c I already have a 1490."  Stress and the whole "this test decides my life" mentality will make you do worse, the SAT is pretty hard to study for so your not prepping hard enough probably didn't cause this "bad" score (hey if its any consolation I would have loved to have a 1490).  It is not a test that you study for like the USAMO.  It is an aptitude (though you can argue) test, NOT an acheivement test.  There is a difference.  As for your verbal woes, here's some things that I would suggest in studying:

(1) Memorize vocab.  Get a CD that Kaplan has that has flashcards on it, this really maximizes your studying and remembering, and has all the good words on it.  Whatever you do, do not get the Barrons 3500 word list.  LOL that is a huge waste of time.  Kaplan lists are the best in my opinion.

(2) Practice doing analogies, get a Kaplan or PR book and do them from that.  There are some strategies to solving them that are pretty straightforward and sortof like a math formula (well as close as icky verbal can get to it...)  Anyway, w/practice, these are the easiest to prep for.

(3) Don't guess randomly on vocab questions.

(4) On critical reading, make sure you are not just reading it, but fully UNDERSTANDING the material.  If you are reading and don't understand a paragraph, then go back and read it again.  Don't freak out about time on the CR sections, it might seem like you don't have a lot of time left, but reading it and UNDERSTANDING takes a lot longer than answering those multiple choice questions.  If you fully understand the passage, then the only ones you will have to go back and look at in the passage are those "what does this word mean in context" questions.

(5) Buy 10 Real SATs.  Don't do tests from other books b/c those are not the real thing.  Use the real thing if you can.

Remember that you can check around w/people on this board who are already in college, and (correct me if im wrong) those who go to programs like USAMO or MOSP or stuff like that usually go to Harvard/MIT.

So basically to sum up this response: you can have both.
-- Take the SAT a bunch of times to reach your max score and do a little prepping for verbal.  Remember that it really dosent matter how many times you take it.
-- Prep for the USAMO, this is a much bigger deal, more fun, and a greater achievement than the lousy SAT.

=)

Hope my advice is helpful...

Hahaha. I've exhausted the 10 Real SAT's book. Both editions. :D

[quote="JS1527"]Stress and the whole "this test decides my life" mentality will make you do worse . . . .

So basically to sum up this response: you can have both.
-- Take the SAT a bunch of times to reach your max score and do a little prepping for verbal.

Hope my advice is helpful.[/quote]

The first quoted point made above is VERY helpful, and ought to be taken seriously by everyone who visits this site. The single best way to prepare for a test is not to be too worried about the consequences of one or another score. Don't fill your mind with worry about your future; just deal with the problems in front of your face with full concentration, and what happens, happens.

My advice for those prepping for the verbal section of the SAT, which I can't repeat enough, it seems, is READ, READ, READ, and READ. Reduce TV-watching time to read more. DEFINITELY eliminate video game-playing time to read more. Read while you're waiting for the bus. Read every chance you get. Read what interests you, and read what is supposedly good for you. Read a lot, and you will gain reading skill.

I grew up with a lot of boys in the 1970s who all thought that the SAT verbal section is a cakewalk, too easy to be useful for deciding who should get into college. The new generation of young people, at least the boys, is much less likely to read as much as my generation did, but reading is still the royal road to a higher SAT score. If you go to MOP in the summer time, bring some books along. If you're working at a job and you have break times at your workplace, read during your breaks. You get the idea. It's a lot more important to read lots and lots of connected text than to spend time memorizing word lists or looking at flash cards.

Yea as tokenadult said, read.  I dunno maybe my advice is sortof biased b/c I already read a lot and didnt have trouble w/CR when I took the SAT.  Wordlists def. help though.  If you want to read a lot and forget the wordlists, when you see a word you don't know when your reading then write it down and define it or something like that.

Thanks again!

Haha. I just noticed that the poll is 80% to 19%. What happened to the other 1%?? :-p

I think it should be 81%

17/21 * 100 = 80.95%

yikes!

The mean score at the vast, vast, majority of colleges for the SAT I is below 1490. For reference, Swarthmore's 25-75% is something like 2(690-770) and Caltech's is ~2(740-790). So, a 1490 is probably not going to get you kicked out of anywhere. If you want to take it again, then due so, but I wouldn't kill myself over it, as it's not a big deal. I would not waste time studying for something I'm not interested in.

You'll get into any college you want to, beyond any shadow of a doubt.

With that said, you should never do anything for the sole purpose of getting into college.  Do what you'll enjoy more, which, unless I am thouroughly mistaken, will be USAMO.

[quote="Magnara"]You'll get into any college you want to, beyond any shadow of a doubt.[/quote]

No -- it is true that there is very little he can do to increase his chances of getting in, however.  In particular, retaking the SAT's is unlikely to help him particularly.

[quote="Magnara"]You'll get into any college you want to, beyond any shadow of a doubt.[/quote]

I don't think applicants should ever think they have a lock on admissions. There is just enough unpredictability in the process of admissions decision making at any of the top schools that the best one can hope for is avoiding a no-brainer rejection. The person who opened the thread certainly is doing the things that are usually advised for applicants to the top schools (other than reading quite as much as I would advise) but admission is not a sure thing. I have already given my advice that making a good effort on USAMO would be much more meaningful than taking the June SAT, but that is not to say that following my advice will guarantee admission. I am trying to help my own children do the right kind of preparing for college, but one kind of preparing they all have to do is preparing for being rejected for inexplicable reasons by their dream school. Everyone needs a plan B, and a Plan C. There isn't any sure formula for admission to the top schools.

hahaha.. no one voted for SAT...
ya... around 1000 people get 1600 on sat i..
now how many people go to the mosp? you want something that'll make you stand out..


function
logarithms
inequalities unsolved
inequalities

for a,b,c >0 and abc=1:

prove or disprove this ineq:

$\frac{a-1}{a^2(a^4+1)}+\frac{b-1}{b^2(b^4+1)}+\frac{c-1}{c^2(c^4+1)}\leq0$ :D  :?  :ninja:

tell me if you know  it's true or false, please :oops:

Im new in math science but if a+b+c>=3 then this ineq is false
I could write the solving of that problem but dont know how to do it correct  :blush: - not to look like (A^2)((b^c)^q) or something ... :roll:

sorry, what's your meaning??
If you think this ineq is wrong, can you give a counterexample?

Dont know if my solving of the problem is correct but... here it is:
f((1/n)(x1+x2+..+xn))<=(1/n)(f(x1)+f(x2)+..+f(xn)) - a formula I will use!
suppose that f(x)=(x-1)/x^2(x^4 + 1)
then
f(a)+f(b)+f(c)>=(1/3)(f(a)+f(b)+f(c))>=f(1/3(a+b+c))
(1/3)(a+b+c) = t
f(t)>=0 if t>=1

[quote="Gohan"]f((1/n)(x1+x2+..+xn))<=(1/n)(f(x1)+f(x2)+..+f(xn)) - a formula I will use!
[/quote]
this is not true for all functions -- it's only true for convex functions and the f you set is neither convex/concave all the time.

i think karamata is useful here :D

considering this funtion $e^x$

with $x=ln(\frac{1}{a^2(a^4+1)})$

hei , no one want to solve it :blush:

its results is really pretty :P
try it please :oops:

It suffices to show that
$\frac{2(a-1)}{a^2(a^4+1)}\le \ln{a}$.

with all my respect, many thanks, vasc :)

and its result is
with abc=1 and a,b,c>0
$\sum\frac{b+c}{a^3+bc}\leq\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$


So where does this quadratic residue identity come from?

The generalization is this: if $p$ is an odd prime, $S$ is the set of the $\frac{p-1}{2}$ nonzero squares modulo $p$, and $z$ is a primitive $p$th root of 1, then
\[\left(1+2\sum_{k\in S}z^{k}\right)^{2}=p\cdot (-1)^\frac{p-1}{2}\]
I forget the proof, but it goes something like this: [hide]define $f$ so that $f(0)=0$, $f(a)=1$ if $a$ is in $S$, and $f(a)=-1$ otherwise. After multiplying out the left hand side, which is equal to
\[\left( \sum_{k=0}^{p-1}z^{k}f(k) \right)^{2}=\sum_{k=0}^{p-1}\left( z^{k}\sum_{j=0}^{p-1}f(j)f(k-j) \right)\]
(and the index of 0 in the right hand sum can be replaced by 1 since $f(0)f(k)=0$), the expression $f(j)f(k-j)$ for $j\neq 0, k\neq 0$ is rewritten in some clever way [size=150][b](...)[/b][/size] (using things such as $f(a)f(b)=f(ab)$ and $a\neq 0 \Rightarrow f(a^{2})=1$) to make it clear that
\[\sum_{j=1}^{p-1}f(j)f(k-j)=-f(-1)=-(-1)^\frac{p-1}{2}\]
for $k\neq 0$. For $k=0,j\neq 0$, $f(j)f(k-j)=f(-1)=(-1)^\frac{p-1}{2}$, which would make the sum equal to
\[(p-1)(-1)^{p-1}{2}-(z+z^{2}+\cdots+z^{p-1})(-1)^\frac{p-1}{2}=p\cdot (-1)^\frac{p-1}{2}\][/hide]
Now the fix the hole where the [size=150][b](...)[/b][/size] is...

[hide="An crazy solution"] Let $\alpha=\frac{\pi}{11}$. We have to prove that:
\begin{eqnarray*}\tan{3\alpha}+4\sin{2\alpha}&=&\sqrt{11}\\ \Longleftrightarrow\frac{\sin{3\alpha}}{\cos{3\alpha}}+4\sin{2\alpha}&=&\sqrt{11}\\ \Longleftrightarrow(4\sin 2\alpha.\cos 3\alpha+\sin 3\alpha)^{2}&=&11\cos^{2}3\alpha\\ \Longleftrightarrow\sin^{2}3\alpha+8\sin 2\alpha.\sin 3\alpha.\cos 3\alpha+16\sin^{2}2\alpha.\cos^{2}3\alpha&=&11\cos^{2}3\alpha\\ \Longleftrightarrow\frac{1-\cos 6\alpha}{2}+4\sin 2\alpha.\sin 6\alpha+4(1-\cos 4\alpha)(1+\cos 6\alpha)&=&\frac{11}{2}(1+\cos 6\alpha)\\ \Longleftrightarrow 2(\cos 4\alpha-\cos 8\alpha)+4(1-\cos 4\alpha+\cos 6\alpha-\cos 4\alpha.\cos 6\alpha)&=&5+6\cos 6\alpha\\ \Longleftrightarrow 2(\cos 2\alpha+\cos 4\alpha+\cos 6\alpha+\cos 8\alpha+\cos 10\alpha)&=&-1 \end{eqnarray*}
We have: $\text{LHS}.\sin\alpha=\sin 11\alpha-\sin\alpha=\sin\pi-\sin\alpha=-\sin\alpha\Longrightarrow\text{LHS}=-1\quad\mathbb{QED}$[/hide]


function
algebra
polynomial
group theory
combinatorics proposed
combinatorics

Consider the set $ M = \{1,2, \ldots ,2008\}$. Paint every number in the set $ M$ with one of the three colors blue, yellow, red such that each color is utilized to paint at least one number. Define two sets:

$ S_1=\{(x,y,z)\in M^3\ \mid\ x,y,z\text{ have the same color and }2008 | (x + y + z)\}$;
$ S_2=\{(x,y,z)\in M^3\ \mid\ x,y,z\text{ have three pairwisely different colors and }2008 | (x + y + z)\}$.

Prove that $ 2|S_1| > |S_2|$ (where $ |X|$ denotes the number of elements in a set $ X$).

Coz no one posts a solution ,I'm posting my one.Hope that I haven't made any idiotic misake!!Personally,I adore this one!
The residue(I mean zero here!) is just meaningless.In other words,one can prove the same comparation for $ x\plus{}y\plus{}z\equal{}d(\mod n)$(I put $ 2008\equal{}n$ for generality)
  So assume three generating functions$ : A(x) \equal{} \sum_{a panted red}x^{a \minus{} d/3}$ ,$ B(x) \equal{} \sum_{b panted blue}x^{b \minus{} d/3}$ and $ C(x) \equal{} \sum_{c panted yellow}x^{c \minus{} d/3}$.
So all we need to prove is as follows:
$ 2\times\frac {1}{n}(\sum_{x|x^n \equal{} 1}A(x)^3 \plus{} B(x)^3 \plus{} C(x)^3) > \frac {6}{n}(\sum_{x|x^n \equal{} 1}A(x)\times B(x)\times C(x))$
  It follows directly from Cauchy-Schwartz inequality(easily to check that the equality won't occur)!

I heard of on fact that this one was taken from a article due to Titu Andresscu on generating functions.So the contest wasn't fair completely,right?  :|

[quote="ricardo4"]any solutions without generating function ?[/quote]

Yes. There are. We have two such solutions.

The first one base on the facts:
Let M = {1, 2, ..., 2008}. R, G, J be red, green, jelow sets respectively.
Let $ N(ABC) \equal{} {(a,b,c) \in M^3| a \in A, b \in B, c \in C   and  2008 | a \plus{} b \plus{} c }|$
Then:
1) N(RGJ) = N(RJG) = R(GJR) = ....
2) N(A.B.M) = |A|.|B| for any A, B

The second one use polynomials, actually:

    $ f(x) \equal{} \sum_{a \in R} x^a, g(x) \equal{} \sum_{a \in G} x^a, h(x) \equal{} \sum_{a \in J} x^a$

In fact, this problem is closer to group theory problem rather that generation function.

[i]Can anybody post a clearer solution than what Namdung said  :maybe: [/i]

My solution:

Lemma:
$ n$ is a positive integer. Consider the set $ S=\{ 1,2,\cdots, n\}$. Color every number in the set $ S$ with either red or blue. Denote sets $ T_1=\{(x,y,z)\in S^3\ \mid\ x,y,z\text{ have the same color and }n | (x + y + z)\}$ and $ T_2=\{(x,y,z)\in S^3\ \mid\ x,y,z\text{ don't have the same color and }n | (x + y + z)\}$. Given that there are $ a$ red numbers and $ b$ blue numbers ($ a+b=n$), then $ |T_1|=a^2-ab+b^2$ and $ |T_2|=3ab$.

Lemma's proof:Choose a red number $ x$ and a blue number $ y$, then $ x\not= y$. Assume $ z$ is the unique number in $ S$ such that $ n|x+y+z$, w.l.o.g. we assume that $ z$ is red. Consider following two cases:

Case 1: $ z$ is neither $ x$ nor $ y$. That means $ x,y,z$ are distinct. Consider the triples in set $ T_2$ that contains both $ x$ and $ y$.There are six such triples:$ (x,y,z),(x,z,y),(y,z,x),(y,x,z),(z,x,y),(z,y,x)$. Now it is the key idea. Corresponde the unordered pair $ (x,y)$ with exactly three elements in set $ T_2$:$ (x,y,z),(x,z,y),(y,x,z)$.(Choose these three triple for $ x$ is in front of $ z$ in them). According to the same definition we corresponde the unordered pair $ (z,y)$ with exactly three elements in set $ T_2$:$ (y,z,x),(z,x,y),(z,y,x)$. So in case 1 each unordered pair $ (x,y)$ ($ x$ and $ y$ are of different color) correspondes exactly three elements in set $ T_2$.

Case 2: $ z$ equals either $ x$ or $ y$. w.l.o.g. we assume that $ z=x$. Then there are also three triples in set $ T_2$ that contains both $ x$ and $ y$:$ (x,x,y),(x,y,x),(y,x,x)$. Of course corresponde the unordered pair $ (x,y)$ with them.

So in both case every unordered pair $ (x,y)$ ($ x$ and $ y$ are of different color) correspondes exactly three elements in set $ T_2$ and each element in set $ T_2$ is used exactly once. Therefore we get that $ |T_2|$ is 3 times of the number of unordered pairs $ (x,y)$ ($ x$ and $ y$ are of different color). Finally the number of unordered pair $ (x,y)$ ($ x$ and $ y$ are of different color) is $ ab$ obviously, so $ |T_2|=3ab$ is proved. For given $ x$ and $ y$ the number $ z$ in set $ S$ satisfying $ n|x+y+z$ is unique. And both $ x$ and $ y$ has $ n$ choices, so $ |T_1\cup T_2|=n^2=(a+b)^2$. Since $ |T_1\cap T_2|=0$, then $ |T_1|=a^2-ab+b^2$.
Our lemma is proved.

Back to this problem, assume there $ a,b,c$ blue, yellow and red numbers, respectively.($ a+b+c=2008$)
Denote set $ A=\{(x,y,z)\in M^3|x,y,z \text{ are all blue and }2008|x+y+z\}$. Sets $ B$ and $ C$ are denoted similarly.
Denote set
$ AB=\{(x,y,z)\in M^3|\text{ All colors of }x,y,z \text{ are blue and yellow and }2008|x+y+z\}$.
Sets $ BC$ and $ AC$ are denoted similarly.
Denote set
$ ABC=\{(x,y,z)\in M^3|\text{ All colors of }x,y,z \text{ are blue, yellow and red and }2008|x+y+z\}$

Now consider blue and yellow the same color, applying the lemma we have that
$ |A|+|B|+|C|+|AB|=(a+b)^2-c(a+b)+c^2$ and $ |ABC|+|AC|+|BC|=3(a+b)c$
Similarly we have
$ |A|+|B|+|C|+|BC|=(b+c)^2-a(b+c)+a^2$ and $ |ABC|+|AB|+|AC|=3(b+c)a$
$ |A|+|B|+|C|+|AC|=(a+c)^2-b(a+c)+b^2$ and $ |ABC|+|AB|+|BC|=3(a+c)b$

Take the sum of both sides. The sum of left minus the sum of right we have the following equality:
$ 6(|A|+|B|+|C|)-3|ABC|=3[(a-b)^2+(b-c)^2+(c-a)^2]$
i.e. $ 2|S_1|-|S_2|=[(a-b)^2+(b-c)^2+(c-a)^2]>0$(Assume $ (a-b)^2+(b-c)^2+(c-a)^2=0$, then $ a=b=c=\frac {2008}{3}$, a contradiction)
Q.E.D.

by the way: I think that the lemma is useful.It can help to solve this problem in IMO-SL 2007:
http://www.mathlinks.ro/viewtopic.php?p=1187179#1187179

Nice solution, thank you very much, linboll.

[quote="Namdung"]
In fact, this problem is closer to group theory problem rather that generation function.[/quote]

Could you please elaborate?


trigonometry
inequalities
inequalities unsolved

[quote="quocbao153"]1. $ \sqrt {\sin x} \plus{} \sqrt {\cos x} \le \sqrt [4]{8},x \in (0;\frac {\pi }{2})$
2. $ |\frac {{(1 \minus{} x^2 )\sin a \plus{} 2x\cos a}}{{1 \plus{} x^2 }}| \le 1,\forall a,x$[/quote]
Use Cachy schawrt inequality and $ sin^2x\plus{}cos^2x\equal{}1$ for all $ x$,we have immediately your inequality
I think they should be put at pre-olympiad box ,They are too easy for here


geometry
LaTeX
trigonometry
congruent triangles
AMC

Regular hexagon $ ABCDEF$ has vertices $ A$ and $ C$ at $ (0,0)$ and $ (7,1)$, respectively.  What is its area?

$ \textbf{(A) } 20\sqrt {3} \qquad \textbf{(B) } 22\sqrt {3} \qquad \textbf{(C) } 25\sqrt {3} \qquad \textbf{(D) } 27\sqrt {3} \qquad \textbf{(E) } 50$

[hide="Answer"]$AC=\sqrt{7^2+1^2}=5\sqrt{2}$
Call $M$ the midpoint of $AC$. $AM=CM=\frac{5\sqrt{2}}{2}$, and since we are dealing with a regular hexagon, these are each the long legs of 30-60-90 triangles $AMB$ and $CMB$, respectively. Therefore, $|\triangle ABC|=|\triangle DEF|=\frac{5\sqrt{2}}{2}\left(\frac{5\sqrt{2}}{2\sqrt{3}}\right)=\frac{25\sqrt{3}}{6}$, so $|\triangle ABC|+|\triangle DEF|=\frac{25\sqrt{3}}{3}$. Also, we find that the length of each side of the hexagon is $\frac{5\sqrt{6}}{3}$. Therefore, $|ACDF|=\frac{5\sqrt{6}}{3}(5\sqrt{2})=\frac{50\sqrt{3}}{3}$. Adding the two, we have $25\sqrt{3}\Rightarrow \boxed{C}$.[/hide]

It's not necessary to find the side length of the hexagon.

The area of the hexagon is 2 times the area of the equilateral triangle ACE.

Right, I knew that... I just wanted to do it a harder way... :whistling:

[quote="mysmartmouth"]Hmm...if we examine these two ways of finding the area of a hexagon, we can derive a neat little formula for the area of a hexagon using side length $s$.

[hide="Formula"]
Area of a hexagon: $\frac{3s^2\sqrt{3}}{2}$
[/hide][/quote]

Eh, two ways? Just remember that equilateral triangle's area * 6 = hexagon. So, $6 \cdot \frac{s^2 \sqrt{3}}{4} = \frac{3 s^2 \sqrt{3}}{2}$

[hide]Area of $ABCDEF$ is twice the area of the equilateral triangle with side length AC. $AC=\sqrt{50}$.
$ABCDEF=2(\frac{50\sqrt{3}}{4})=25\sqrt{3}$. So $\boxed{C}$.[/hide]

[quote="asdf4"]Actually there's a neat little formula that I made for [i]s[/i] number of sides and [i]n[/i] which is the measure of each side.

[hide]tan((180(s-2))/(2s))*sn^2/2[/hide][/quote]

I had to latex this so I could decipher it...

$\tan\frac{180(s-2}{2s})*\frac{sn^2}{2}$

(Is that right?)

Since this is for regular polygons, we can connect the center of the polygon to the endpoints of one of the sides to create a triangle.  Since connecting the center to all of the vertices would create $s$ congruent triangles, this triangle's area is $\frac 1s$ the area of the polygon.  Draw in the altitude from the center to the side of the polygon (a.k.a. the apothem).  You now have a right triangle with one of its angles equal to $180(s-2)/(2s)$.  Using the definition of $\tan$ and the fact that $A=\frac 12bh$, you get that formula.


ARML
analytic geometry
graphing lines
slope
conics
parabola

Find the equation of the circle with [b]smallest[/b] radius that is tangent to the two parabolae:
\[y= x^{2}-4 \]
and
\[y= 9-(x-9)^{2}\]

[quote="cincodemayo5590"]Find the equation of the circle with largest radius that is tangent to the two parabolae:
\[y= x^{2}-4 \]
and
\[y= 9-(x-9)^{2}\]
[/quote]

noob...you better be doing the arml hw yourself...

and you might mean inside...otherwise it is just 2 maximization problems instead of 1

Rofl, alex... that's not the arml homework.

noob. of course I'm doing (*have basically finished) the arml hw by myself.  :P

[quote="cincodemayo5590"][quote="Altheman"]if the circle is on the inside, then the center lies on the line connecting the intersections

otherwise, the center lies on x=0 for tangent above

or x=4 for tangent below

if it is tangent outside, then it is obviously unbounded [also on the line connecting the centers[/quote]

but they don't intersect...[/quote]

oh man thats lame then...there is such an awesome way to find the slope of that

then obviously the radius is unbounded since we don't need to worry about the radius of curvature for the parabolas anymore

to find a circle,

take a point on a parabola

take the tangent, the center lies on the normal line to the tangent

move the center "away" from the point of tangency, to the "outside" of the parabola

obviously, there is a circle that intersects the other parabola once, and intersects the other parabola two times

by geometric continuity, it also can be tangent

--

the problem is better if they intersect

or you want minimum radius

[quote="Altheman"]
or you want minimum radius[/quote]

Lol, sorry that's what I meant. I was thinking about english hw while writing this...  :P


It's an old USAMO or Putnam or something.  It's also appeared on the forum before.


First class assignment due tomorrow and i want to reassure myself of it before i hand in the work so here we go :)

You have 6 large empty folders and you select a number of them and into
those selected you insert 4 medium folders. You then select a number of the
medium folders and into each of them you insert 4 small folders.
(i) At this point you have have 36 empty folders, how many full ones do
you have?
(ii) Is it possible to nish with 35 empty folders? Justify your answer.
In solving this problem you must abstract and model the problem, and
identify a suitable invariant. Marks will be awarded for appropriate notation
and structure in your solution.

I have my own way of doing it and i understand it fully my problem is how to write it down i have it down atleast 3 different ways so if anyone could help me out or just give opinions it would be great.

Thanks

[hide="Solution"]We note that we originally have $ 6$ empty folders, and by adding a medium folder, we add $ 3$ empty folders, and by adding a small folder, we also add $ 3$ empty folders. Therefore, we need to add $ 10$ folders, either medium or large. Therefore, we fill $ \boxed{10}$ folders.

We cannot have $ 35$ folders, because the number of empty folders is always divisible by $ 3$.[/hide]


While on the forum I saw a problem very similar to this somewhere....anyway,
[hide]Give each man a number, 1-100.  1-85 are married. 86-100 have a telephone, and 1-55 have a telephone as well.  56-100 own a car, and 1-30 own a car as well.  Last, 31-100 own a home, and 1-10 do as well.

Numbers 1-10  had all 4 things.  So the answer is $\displaystyle{\boxed{10}}$.  :D [/hide]

[quote="ch1n353ch3s54a1l"]In a small town of 100 men, 85 are married, 70 have a telephone, 75 own a car, and 80 own their own home. On this basis, what is the smallest number of men who are married, have their own telephone, own their own car, and own their own home?[/quote]

I just drew a picture and got 10.

$10$

[hide]
Add them up for a total of 310. You are looking for the smallest number of men that can have 4. Since 100*3 is 300, 10 must saisfy all 4.[/hide]


logarithms
inequalities
search
function
geometric sequence

(1) The value of $ \log_{3}2$ is $ 0.631$, rounded to the nearest thousandth.  Compute the smallest positive integer $ n$ for which $ 3^n>2^{102}$.

(2) Solve $ 2^{41}\equal{}2\plus{}\sum_{k\equal{}1}^{39}\log_{10}x^{(2^k)}$ for all real value(s) of $ x$.

Thanks in advance!

[b]1.[/b] Use the fact that for $ n\in\mathbb{N}$ the inequality $ 3^n>2^{102}$ is equivalent with $ n>\log _32^{102}\equal{}102\log _32$ and use the approximation $ \log _32\approx 0.631$.

[b]2.[/b] There is not possible to solve the given equation for all real $ x$. Namely $ \log x$ is not defined for $ x\equal{}0$. Because $ 2^k$, $ k\in\mathbb{N}$, is even we can search the solution in the set $ \mathbb{R}\setminus\{ 0\}$. You obtain simply
$ 2\plus{}\sum_{k\equal{}1}^{39}\log x^{2^k}\equal{}2\plus{}\sum_{k\equal{}1}^{39}2^k\log |x|\equal{}2\plus{}\log |x|\cdot\sum_{k\equal{}1}^{39}2^k$
which can be simplified by the finite summation of terms which are the members of the geometric sequence $ \{ 2^k\}_k$.

[quote="alkjash"][hide="1"]
So $ 3^{102*.631} \equal{} 2^{102} \implies 3^{64.36...} \equal{} 2^{102} \implies 3^{\boxed{65}} > 2^{102}$.
[/hide]

[hide="2"]
We have
$ \sum_{k \equal{} 1}^{39}\log_{10}x^{(2^k)} \equal{} \sum_{k \equal{} 1}^{39}2^k\log_{10}|x|$
$ \equal{} \log_{10}|x|\sum_{k \equal{} 1}^{39}2^k$
$ \equal{} \log_{10}|x|(2^{40} \minus{} 1)$

Thus, $ \log_{10}|x| \equal{} 2$ is clearly a solution, and since $ \log_{10}x$ is an increasing function, $ x \equal{} \boxed{\pm100}$ are the only real solutions.[/hide][/quote]

Could you explain your second solution a bit more, please?  I don't quite understand the way that you use the summations.  Thanks!

[b]Identity 1[/b]: $ \log a^n \equal{} n \log a$.

In this case, in the summation we have $ \log_{10} x^{2^k}$ and applying the identity above, we get $ 2^k \log_{10} x$.

But these two expressions are not exactly equal yet. As you can see the original expression is defined for negative $ x$, but the new expression isn't. So we modify the new expression to be $ 2^k \log_{10} |x|$ to account for negative values of $ x$.

Next, putting the summation back in, it is important to notice that the index of summation is $ k$. Therefore, $ x$ is a "constant" with respect to the summation so we can pull that out to get $ \log_{10} |x| \sum_{k \equal{} 1}^{39} 2^k$.

Time for another identity.

[b]Identity 2[/b]: $ \sum_{k \equal{} 1}^n 2^k \equal{} 2^{n \plus{} 1} \minus{} 1$

So the summation boils down to $ \log_{10} |x| (2^{40} \minus{} 1)$, from which point it is easy to pick up.


I have on my desktop an image with a problem. I have forgotten the source.Before I delete it , I send it  for a last attemt to gother nice solutions.I searched for a while but I did not find it again.
 Babis

[quote="silouan"]See here

http://www.mathlinks.ro/viewtopic.php?p=118682#118682

There is also a link there with three more solutions .[/quote]

Thank you Silouan  :) !

 [color=red][b]Good luck to you and to the whole greek team at the  Balkan Math Olympiad .[/b][/color]

[u] Babis[/u]


function
inequalities
combinatorics proposed
combinatorics

We have volleball competition which has $ n$ teams participating.Any 2 teams have only one match, the result of every match is win or lose. Assume that  the number of teams win any 2 teams are less than $ T$. Prove that $ n\le 4T \plus{} 3$

You can replace "less than $ T$" by "less or equal to $ T$" and the problem still remains true. Before I solve it, I rewrite it in a more mathematical way:

[color=blue][b]Theorem 1.[/b] Let $ X$ be a finite set, and let $ T$ be an integer. Let $ >$ be a binary relation on the set $ X$ (written the infix way, i. e. we write $ a>b$ for "the pair $ (a,b)$ satisfies the relation $ >$") such that for any two elements $ a$ and $ b$ of $ X$, exactly one of the three assertions $ a>b$, $ b>a$ and $ a=b$ holds. (This relation $ >$ needs not be transitive, however.) Assume that for any two distinct elements $ a$ and $ b$ of $ X$, we have

[/color][color=green][b](1)[/b][/color][color=blue] $ |\{c\in X\mid c>a\text{ and }c>b\}|\leq T$.

Then, $ |X|\leq 4T+3$.[/color]

[i]Proof of Theorem 1.[/i] We assume that $ |X|-1>0$ (else, $ |X|\leq 1$ and everything is obvious). We have

[color=green][b](2)[/b][/color]       $ |\{(a,b,c)\in X^3\mid a\neq b\text{ and }c>a\text{ and }c>b\}|$
$ =\sum_{\substack{a,\ b\text{ distinct}\\\text{elements of }X}}\underbrace{|\{c\in X\mid c>a\text{ and }c>b\}|}_{\leq T\text{ by (1)}}\leq \sum_{\substack{a,\ b\text{ distinct}\\\text{elements of }X}}T$
$ =T\cdot(\text{number of pairs of distinct elements }a,\ b\text{ of }X)=T\cdot|X|\cdot(|X|-1)$.

On the other hand, for any $ c\in X$, let us denote by $ L_c$ the set of all $ d\in X$ such that $ c>d$. Then,

$ \sum_{c\in X}|L_c|=\sum_{c\in X}|\{d\in X\mid c>d\}|$
$ =|\{(c,d)\in X^2\mid c>d\}|$
$ =\frac12\cdot(|\{(c,d)\in X^2\mid c>d\}|+|\{(c,d)\in X^2\mid c>d\}|)$
$ =\frac12\cdot(|\{(c,d)\in X^2\mid c>d\}|+|\{(c,d)\in X^2\mid d>c\}|)$ (we have just substituted $ (c,d)$ for $ (d,c)$ in the second set)
$ =\frac12\cdot|\{(c,d)\in X^2\mid c>d\text{ or }d>c\}$ (since there is no $ (c,d)\in X^2$ which satisfies $ c>d$ and $ d>c$ simultaneously, because exactly one of the three assertions $ c>d$, $ d>c$ and $ c=d$ holds)
$ =\frac12\cdot|\{(c,d)\in X^2\mid c\neq d\}$ (in fact, $ c>d\text{ or }d>c$ is equivalent to $ c\neq d$, because for any $ c$ and $ d$, exactly one of the three assertions $ c>d$, $ d>c$ and $ c=d$ holds)
$ =\frac12\cdot|X|\cdot(|X|-1)=\frac{|X|\cdot(|X|-1)}{2}$,

so that

$ \frac{\sum_{c\in X}|L_c|}{|X|}=\frac{|X|-1}{2}$.

Besides,

$ |\{(a,b,c)\in X^3\mid a\neq b\text{ and }c>a\text{ and }c>b\}|$
$ =\sum_{c\in X}|\{(a,b)\in X^2\mid a\neq b\text{ and }c>a\text{ and }c>b\}|$
$ =\sum_{c\in X}|\{(a,b)\in X^2\mid a\neq b\text{ and }a\in L_c\text{ and }b\in L_c\}|$
$ =\sum_{c\in X}|\{(a,b)\in L_c^2\mid a\neq b\}|=\sum_{c\in X}|L_c|\cdot(|L_c|-1)$.

Thus, [color=green][b](2)[/b][/color] becomes

$ \sum_{c\in X}|L_c|\cdot(|L_c|-1)\leq T\cdot|X|\cdot(|X|-1)$.

In other words,

[color=green][b](3)[/b][/color] $ \frac{\sum_{c\in X}|L_c|\cdot(|L_c|-1)}{|X|}\leq T\cdot(|X|-1)$.

Now, since the function $ f: \mathbb{R}\to\mathbb{R}$ defined by $ f(x)=x(x-1)$ is convex, the Jensen inequality yields

$ \frac{\sum_{c\in X}f(|L_c|)}{|X|}\geq f(\frac{\sum_{c\in X}|L_c|}{|X|})$,

so that

$ \frac{\sum_{c\in X}|L_c|\cdot(|L_c|-1)}{|X|}\geq \frac{\sum_{c\in X}|L_c|}{|X|}\cdot(\frac{\sum_{c\in X}|L_c|}{|X|}-1)$
$ =\frac{|X|-1}{2}\cdot(\frac{|X|-1}{2}-1)=\frac{|X|-1}{2}\cdot\frac{|X|-3}{2}$.

Combined with [color=green][b](3)[/b][/color], this becomes

$ T\cdot(|X|-1)\geq\frac{|X|-1}{2}\cdot\frac{|X|-3}{2}$.

Thus, $ 4T\geq|X|-3$ (we have divided by $ |X|-1$, what is okay since $ |X|-1>0$), so that $ |X|\leq 4T+3$, and Theorem 1 is proven.

  Darij

A nice problem and a nice solution too  :D


Find the minimum of $\log_{\frac{3}{2}x}{(\frac12-\frac{1}{36x^{2}})}$
when $x\in (\frac{1}{4},\frac{2}{3})$
It comes from Moldova 2006


trigonometry
logarithms
function
integration
limit
floor function
complex numbers

show that the sum $sigma$ $from 1 to$$infinity$
 $sin(ln(n))/n$ diverge

$S = \sum_{n=1}^{\infty}\frac{\sin(\ln n)}{n}= \sum_{n=1}^{\infty}\frac{e^{i\ln n}-e^{-i\ln n}}{2in}= \sum_{n=1}^{\infty}\frac{n^{i}-n^{-i}}{2in}$
$= \frac{1}{2i}\sum_{n=1}^{\infty}\left(\frac{1}{n^{1-i}}-\frac{1}{n^{1+i}}\right)$.

The first series diverges and the second converges. As the sum of a divergent series with a convergent one is always a divergent series, so S diverges.

Erm..., aren't $\frac 1{n^{1+i}}$ and $\frac 1{n^{1-i}}$ just complex conjugate to each other? If so, how can it happen that the sequence of numbers (partial sums of the first series) doesn't have a limit but the sequence of their conjugates does?

for series like $\sum_{n \geq 1}f(n)$ where $f$ is a function wich does not oscillate too much in the sense where $\int_{1}^{\infty}|f'(x)| dx < \infty$, it is a good idea to compare the sum with $\int_{1}^{\infty}f(x) dx$

Ok, so according to Mathmanman,

$n^{i}-(n+1)^{i}= \frac{\sin(\ln n)}{n}-\frac{i \cos(\ln n)}{n}+O\left(\frac{1}{n^{2}}\right)$
$\Rightarrow  \frac{\sin(\ln n)}{n}= Re[n^{i}-(n+1)^{i}]-O\left(\frac{1}{n^{2}}\right)$.

But $\sum_{n=2}^{+\infty}[n^{i}-(n+1)^{i}] = 2^{i}-\lim_{n\to\infty}n^{i}= \infty$

and so we can conclude that the series $\sum_{n=2}^{+\infty}\frac{\sin(\ln n)}{n}$ diverges. Is this correct?

Here's a different approach. Break the sum into

$\sum_{k}\sum_{n=\lfloor e^{k\pi}\rfloor+1}^{\lfloor e^{(k+1)\pi}\rfloor}\frac{\sin(\ln n)}n=\sum_{k}b_{k}$

I claim that there exists a constant $c>0$ such that $|b_{k}|>c$ for large enough $k.$ That would then show that the partial sums of the original series do not form a Cauchy sequence.

Note that all of the terms in the sum for $b_{k}$ are of the same sign, so no cancellation occurs within that sum. Suppose $e^{\left(k+\frac16\right)\pi}<n<e^{\left(k+\frac56\right)\pi}.$ Then $|\sin(\ln n)|>\frac12$ and $\left|\frac{\sin(\ln n)}{n}\right|>\frac1{2n}>\frac1{2e^{(k+1)\pi}}=\frac1{c_{2}e^{k\pi}}.$ The number of values of $n$ for which this is true is greater than $e^{k\pi}\left(e^{\frac{2\pi}3}-1\right)=c_{1}e^{k\pi}.$

These estimates can be combined to produce $|b_{k}|>\frac{c_{1}e^{k\pi}}{c_{2}e^{k\pi}}=c.$


superior algebra
superior algebra unsolved

find a group that containselements a and b such that |a| = |b| = 2
and
(i)  |ab| =3
(ii) |ab| = 4
(iii) |ab| = 5

relationship between |a|, |b| and |ab|?

Thanks.

You can construct all of these with dihedral groups (taking $a$ and $b$ as reflections).


What is the sum of the reciprocals of the first three positive even integers? Express your answer as a common fraction.

The first three positive even integers are $ 2, 4, 6$. The reciprocal of a number $ a$ is simply $ \frac{1}{a}$. So all we need to do is add:

$ \frac{1}{2} \plus{} \frac{1}{4} \plus{} \frac{1}{6} \implies \frac{6}{12} \plus{} \frac{3}{12} \plus{} \frac{2}{12} \implies \boxed{\frac{11}{12}}$

EDIT: Beaten to it.


[b]Andrew Wiles Last Conjecture[/b] - Once you prove a Last Theorem, it becomes your last theorem.

As of 2009, Andrew Wiles is unable to prove his last conjecture, which is evidence that it is correct.

I am currently searching for a counter-example, but if I end up proving it instead, I'll be forced to give up on my dream of being a mathematician :(

Is there any point to this post?  And it doesn't make any sense, so please give your posts meaning or stop posting.

It was a joke sheesh... I was poking fun at the fact that once you've proven Fermat's Last Theorem, anything else you do will pale in comparison.

I would say spam is a more accurate description.

[quote="BenDoverBooks"]spam usually involves more than trying to get a laugh... didn't get enough hugs as a child or what?[/quote]I think you've got it in reverse.