trigonometry
integration
geometry
rectangle
real analysis
real analysis unsolved

For $ 0 < x < \pi$ and $ \alpha > 0$ show

$ \begin{aligned}&\sum_{k\in\Bbb{Z}} \frac {\cos 2kx}{k^4 + \frac {\alpha^4}{4}} = \frac {2\pi}{\alpha^3} \biggl(\sin \alpha x \,\cosh \alpha x - \cos\alpha x\,\sinh\alpha x + \biggr.
\\
&\left.\frac {(\sinh\alpha\pi + \sin\alpha\pi)\cos\alpha x\,\cosh\alpha x - (\sinh\alpha\pi - \sin\alpha \pi)\sin\alpha x\,\sinh\alpha x}{\cosh\alpha\pi - \cos\alpha\pi}\right)\end{aligned}$

it's easy.
consider :
$ \sum \frac {e^{2ikx}}{k^{4} \plus{} \frac {\alpha^{4}}{4}} \equal{} \sum \frac {( \minus{} 1)^{k}e^{ik(2x \minus{} \pi)}}{k^{4} \plus{} \frac {\alpha^{4}}{4}}$
thus:
$ 2x \minus{} \pi \in ( \minus{} \pi,\pi)$
hence consider
$ \int_{C}\frac {e^{iz(2x \minus{} \pi)}}{\sin(\pi z)(z^{4} \plus{} \frac {\alpha^{4}}{4})}dz$
prove that:
if $ C$ is big rectangle then
$ \sum_{k} \frac {( \minus{} 1)^{k}e^{ik(2x \minus{} \pi)}}{k^{4} \plus{} \frac {\alpha^{4}}{4}} \equal{} \minus{} \pi\sum_{res (z^{4} \plus{} \frac {\alpha^{4}}{4})}\frac {e^{iz(2x \minus{} \pi)}}{\sin(\pi z)(z^{4} \plus{} \frac {\alpha^{4}}{4})}$
just take  $ Re \sum \frac {e^{2ikx}}{k^{4} \plus{} \frac {\alpha^{4}}{4}}$ .Done

More general we can write $ \frac {1}{k^n \plus{} \alpha^n}$ as $ \frac {1}{n\, \alpha^{n \minus{} 1}} \sum_{j \equal{} 0}^{n \minus{} 1} \frac {e^{\frac {i\pi}{n}(2j \plus{} 1) }}{\alpha\, e^{\frac {i\pi}{n}(2j \plus{} 1)} \minus{} k}$

and use the formula $ \sum_{k\in\Bbb{Z}} \frac {( \minus{} 1)^k \cos kx}{\beta \minus{} k} \equal{} \pi \, \frac {\cos\beta x}{\sin\beta \pi}$ to obtain

$ \sum_{k\in\Bbb{Z}} \frac {( \minus{} 1)^k \cos kx}{k^n \plus{} \alpha^n} \equal{} \frac {\pi}{n\, \alpha^{n \minus{} 1}} \sum_{j \equal{} 0}^{n \minus{} 1} \frac {\cos\left(\alpha\, e^{\frac {i\pi}{n}(2j \plus{} 1)}x\right) \, e^{\frac {i\pi}{n}(2j \plus{} 1)}}{\sin\left(\alpha\, e^{\frac {i\pi}{n}(2j \plus{} 1)}\pi\right) }$

[quote="Doe John"]More general we can write ...[/quote]
.. by same idea just write :
$ \sum_{k} \frac {( \minus{} 1)^{k}e^{ik(x \minus{} \pi)}}{k^{n} \plus{} \alpha^{n}} \equal{} \minus{} \pi\sum_{res (z^{n} \plus{} \alpha^{n})}\frac {e^{iz(x \minus{} \pi)}}{\sin(\pi z)(z^{n} \plus{} \alpha^{n})}$


Find positive integers $ a,b$ if for every $ x,y\in[a,b]$, $ \frac1x\plus{}\frac1y\in[a,b]$.

[hide="solution"]
condition:$ a\le b$.
it's necessary that $ \frac{1}{x}\plus{}\frac{1}{y}\ge a$,then let $ x\equal{}y\equal{}a$,we have $ a^2\le 2$ i.e.$ a\equal{}1$.
Let $ x\equal{}y\equal{}b$,we have $ b\le 2$ and we get $ b\equal{}1$or$ b\equal{}2$.

1)when $ b\equal{}1$
when $ x\equal{}y\equal{}1$, $ \frac1x \plus{} \frac1y\equal{}2>1\equal{}b$,ruled out

2)when $ b\equal{}2$
if $ x,y\in[1,2]$ then $ \frac1x \plus{} \frac1y\ge \frac12\plus{}\frac12\equal{}1$ and $ \frac1x \plus{} \frac1y\le \frac11\plus{}\frac11\equal{}2$ works.

so $ (a,b)\equal{}(1,2)$.

[/hide]


I'm having a hard time with the simplifying issues.  Can someone please tell how to simplify the following expression:

2(x^2-xy)-2(2x^2+1)+xy+3x

Thank you, any help would be very much appreciated.

okay, well i have the answer:
-2x^2 +3x-xy-2
now, you can factor that also. Do you want me to?
All you have to do is take it one number at a time. Do distributive property and then simplify. Don't forget to put your answer in descending order. Hope this helped. If you have any more questions, just ask me. Good luck!!

Click Mr. Green for the answer!
[hide=" :D "]2(x^2-xy)-2(2x^2+1)+xy+3x
2x^2-2xy-4x^2-2+xy+3x
-2x^2-xy+3x+xy[/hide]

Wow jeanna, I bet that really helped him he knew what to do =p

[hide]
2(x^2-xy)-2(2x^2+1)+xy+3x
2(x^2-xy)-2(2x^2+1)+xy+3x
2(x-y)-2(2x^2+1)+xy+3x
(x-y)-(2x^2+1)+xy+3x/4
(-y)-(2x+1)+xy+3x/4
(-y)+-(2x+1)+xy+3x/4
(-y)+-(x+1)+xy+3x/8
(-y)+(-x)+xy+3x/7
(-y+-x)+(xy)+3x/7
3x/7
x/2.3 repeating

Wow I really think I messed this up. If mine is different from anyone else that has a lot more stars than me, completely ignore me![/hide]

Please don't revive topics from a long time ago (meaning more than one month old and outdated).  Thanks.  :)


algebra
polynomial
real analysis
real analysis unsolved

[quote="flip2004"]For $n\in{\mathbb N}$ consider the series $\sigma(n,x): =\sum_{k=0}^{\infty}\frac{k^{n}}{k!}x^{k}.$ If $x^{n}=\sum_{k=0}^{n}S(n,k)x(x-1)\cdots (x-k+1)\; ,\; \forall x \in{\mathbb R},$ let
 $T(n,m;x): =\sigma(n,x)\sum_{k=0}^{m}S(m,k)x^{k}.$ [b] Prove that for any positive integers $n,m$ and all ${x > 0}$
\[T(n,m;x)=T(m,n;x)\; . \]
[/b][/quote]
Yes, but what is $S(n,k)$?

[quote="Extremal"] Yes, but what is $S(n,k)$?[/quote]
The polynomials $\phi_{0}(x): =1\; ,\; \phi_{k}(x): =x(x-1)\cdots (x-k+1) \; ,(k\ge 1),$ is a basis in the linear space $\Pi_{n}$ of all real polynomials of degree $\le n .$
 The reason is that [b]degree[/b]$(\phi_{k}) = k\; , k\in \{0,1,\ldots ,n\}\; .$
 Therefore each polynomial from $\Pi_{n}$ (e.g. $x^{n}$) may be written in an uniques way as a linear combinations of $\{\phi_{0}(x),...,\phi_{n}(x)\}.$
   As you may observe, $\parallel S(n,k)\parallel $  are the coefficients of this linear  combinations .

In fact these numbers are well-known: i.e. are the [b]Stirling numbers of second kind[/b].


logarithms
arithmetic sequence
number theory open
number theory

A Niven number is a natural number divisible by the sum of its digits. Does there exists an infinite arithmetic progresion free of Niven numbers?
 I have no idea, but I think the problem is very beatiful. Don't hesitate to think about it!

Sorry, mathmanman, I promised you to publish my solution one year ago, but since I have no time to develope my forum I will publish it here :(

Consider an arbitrary arithmetic progression of natural numbers $an+b$.

1) There is a natural number of form $99...90..0$ which is divisible by $a$. So we may WLOG assume $a=99...90..0$ (just consider subsequence).
Let $A$ be the decimal representation of $a$ and $B$ be the decimal representation of $b$. Let $s(m)$ is the sum of digits of natural number $m$.

2) We will find $k$ s.t. $(s(ak+b),s(a))=q$, where $q=1$ if $3\nmid b$, $q=3$ if $3\mid b$ and $9\nmid b$, $q=9$ if $9\mid b$. Indeed, if $(s(a),s(b+a\cdot 0))\neq q$, then consider $11...1\cdot a$ (number of '1' excedes number of '9' in $A$ by one). It is easy to show $s(11...1a)=9+s(a)$. So if we choose $k=11...1\cdot 10^{h}$, where $h$ is greater than number of digits in $B$, then $(s(ak+b),s(a))=(s(b)+s(a)+9,s(a))=(s(b)+9,s(a))$. It means we may achive $(s(ak+b),s(a))=(s(b)+9r,s(a))$ for appropriate $k$. Thus just take $s$ satisfying $(s(b)+9r,s(a))=q$ (note that $s(a)=9m$). Denote $ak+b$ as $b$.

3) Numbers $\frac{s(b)}{q}$ and $\frac{s(a)}{q}$ are coprime. It follows there is sufficient large prime $p$ (e.g. $p>10a$) s.t. $pq=d\cdot s(a)+s(b)$ (by Dirichlet theorem).

4) We will show there are a number in progression of form \[A0..0A0..0A...A0..0B=X\] ($d$ symbols $A$) which is divisible by $p$ (it clear it is divisible by $q$). Note that remainder of $A\cdot 10^{s}$ modulo $p$ depends only on remainder of $s$ modulo $p-1$. Moreover $10^{s}$ modulo $p$ gives at least $\log_{10}p$ distinct remainders. So each $A$ in $X$ gives us a freedom of at least $\log_{10}p$ remainders. By Cauchy-Devenport theorem we conclude $X$ can take at least \[\min(p,d\cdot (\log_{10}p-1)+1)\] remainders. But $d \approx \frac{pq}{s(a)}$ for large $p$, therefore $d\cdot (\log_{10}p-1)+1>p$ for large $p$. It means $X$ takes remainder 0 modulo $p$ and $s(X)=pq$.

Hope it is correct.


Hey I was wondering if there's an approach to finding nontrivial solutions to problems such as $ x\minus{}\cos{x}\equal{}0$ without a calculator? I really hate having to use the thing.  :maybe:

[hide="example"]Let $ f(x)\equal{}\frac{\cos{x}}{x}$. Find the equation of the line that is tangent to $ f$ at exactly two distinct points.[/hide]
Thanks!

You mean an algebraic one?  No, not really.  Most "transcendental equations" don't have closed-form solutions, and there's no reason to expect one.


articles Harvard college calculus geometry

My next article for my college paper will be on exchange rates. I am trying to do a little research on both fixed and flexible exchange rates. Any opinions here?

I guess that there aren't a whole lot of AoPSers interested in Economics yet.  It has always seemed to me that math people got into Economics mostly after getting to college if they ever do and most of the users on the site are not yet in college.

I think the exchange rates question is fairly obvious, but I'm sure some would disagree.  Pegging one currency to another masks the opinions of economic participants on the relative value of each.  One of three things must eventually happen -- (1) demand for the overvalued currency will dry up completely resulting in an inability to carry on international trade (and even domestic trade with partners who have international interests), (2) an arbitrager like George Soros will arbitrage the relative value so dramatically that it will cost the taxpayers from the nation with the currency peg billions of dollars, or (3) the peg will be removed resulting in an economic shock to the holders of the currency that is instantly devalued by the supply/demand effects of the marketplace.

I keep holding my breath that the Chinese currency peg will be lucky enough to escape international disaster only because its true value will move back into the direction of the current peg.  I guess I should have mentioned this as a fourth possibility except that exactly hitting a peg is like throwing a dart from the moon and hitting a dartboard in Cleveland.  There will almost surely be some economic fallout.

I guess I am always showing that I am illogical but can you explain to me the following in even more straightforward logic? I was trying to find a lot of literature on exchange rate regimes and a bit on the depreciation on the dollar, but I could not understand too much of the articles, I still do not have a great grasp of why they are good or bad.

(1) demand for the overvalued currency will dry up completely resulting in an inability to carry on international trade (and even domestic trade with partners who have international interests), (2) an arbitrager like George Soros will arbitrage the relative value so dramatically that it will cost the taxpayers from the nation with the currency peg billions of dollars, or (3) the peg will be removed resulting in an economic shock to the holders

[quote="hello"]
(1) demand for the overvalued currency will dry up completely resulting in an inability to carry on international trade (and even domestic trade with partners who have international interests), (2) an arbitrager like George Soros will arbitrage the relative value so dramatically that it will cost the taxpayers from the nation with the currency peg billions of dollars, or (3) the peg will be removed resulting in an economic shock to the holders[/quote]

A peg requires that the government continually take one side of a currency trade.  This does not actually deter the market from believing the relative value is different, so the government simply trades and trades and trades, but the capacity to take one side of a trade is finite and risk levels grow very high with such a one-sided trade.  Meanwhile, holders of the currency that should be more highly valued are wary about trading away their currency and holders of the currency that is overvalued find fewer and fewer people will to take their currency in the marketplace because people educated in finance don't believe in the value of the currency.

I don't want to go into Soros' arbitrage of the peg of the pound, but I'm sure you can Google it.

Any time a currency is dramatically devalued all at once, economic instability results for a number of reasons.  I don't want to write a paper about this right now, but Google on the results of the devaluation of the Russian ruble in 1998 and you'll see what I'm talking about.

For that matter, the Soros attack on a currency peg I mentioned above is a more moderate example that was paid for by taxpayers.  Taking a billion or two out of an economic in one fell swoop doesn't make life easier for a lot of people.  That said, it was better sooner than later.

I am not sure I follow the following, sorry that I dont understand anything.

A peg requires that the government continually take one side of a currency trade. This does not actually deter the market from believing the relative value is different, so the government simply trades and trades and trades, but the capacity to take one side of a trade is finite and risk levels grow very high with such a one-sided trade. Meanwhile, holders of the currency that should be more highly valued are wary about trading away their currency and holders of the currency that is overvalued find fewer and fewer people will to take their currency in the marketplace because people educated in finance don't believe in the value of the currency.

hello,

If you asked more specific questions, MCrawford would probably be better able to answer in a manner that would help you understand, for instance...

"Could you explain what sides of a currency trade are?"

"What is relative value?"

"Why is 'one sided trading' finite and why does the risk grow?"

or maybe something like...

"Could you maybe provide a live example of this following situation that might help me understand this, 'Meanwhile, holders of the currency that should be more highly valued are wary about trading away their currency and holders of the currency that is overvalued find fewer and fewer people will to take their currency in the marketplace because people educated in finance don't believe in the value of the currency.'"

And then of course... maybe a thank you? :)

OK, I want to thank Mathew for all the help he has provided in all my econ forums. Thanks for the suggesstion to you Spoon also. Well, I am not familiar with all the sentences that you copying from Mathew, so I guess I might need an explanation with that. Sorry about not specifying what I need help with exactly, that is because I understand like nothing from the whole paragraph, so the only way I could have written it was how I did.

What I understand about exchange rates (maybe that will help explain what I dont understand) is that flexible exchange rates have a market in which they are traded, like stocks and bonds, so people can buy them, right? In fixed exchange rates, the currency is fixed to another value, for example 8.28 rambini (what is that by the way, Chinese currency or what?) equals $1. So that the Chinese currency is fixed and is not exchanged on the market. I know that a flexible exchange rate causes inflation, because monetary policy has to monitor it or something like that (I would be glad for a more thorough explanation) and that is why it is not always adopted, yet a fixed exchange rate slows economic growth but manages to hold down inflation. I know that most countries are in the flexible exchange system these days but some like China are not, the US government want them to become flexible. I looked at some academic literature to understand the topic but I understood very little I think. I saw that some economists were not for completely flexible exchange rates like this harvard economist Frankel suggested pegging the dollar to the export price or somthing like that. So that is all I understood, or at least most of what I understood, I am trying to find the logic of the rest of what is going on.

Also, I am interested in the value of the dollar, if it is good or bad that its declining and how much longer will it decline? I looked at an interview with economists that had different opinions. I understand that it is good from the perspective that we can get more exports but obviously there is a negative side like prices going up. So I wanted to understand the logic of this and understand it in furthur detail.

I would like to thank Mathew once again for all his help and anyone else who replies to my questions.

Unfortunately I do not have the time to write complete explanations of my thoughts on all these topics.  My hope was to provide some ideas to think about and research.

In regards to the falling relative value of the dollar, I don't regard the changing of the relative values of currencies as good or bad overall unless it's dramatic because volatility of value eats wealth (a la declining Sharpe ratios).  Otherwise, the shifting of values appropriately rewards successful economic activity.

ok, thanks. I will try todo research even though I did some I will try again. I tried to find most of my info in academic papers, maybe that is why I did not find a lot of success? Is it too advanced to be reading academic papers when completing only 2  introductory econ classes? Should I just read news type articles? Thanks

I personally never favored reading a lot of academic articles (at least for the first two or three years) in order to learn economics.  I think a better education comes from practical game theory, playing economics-based strategy games (like poker), working in an environment that involves economic decision making, and reading just enough to enable you to gain academic context for everything you've learned.

It doesn't hurt to study enough mathematics such that you're always one step ahead of your economics education in terms of understanding the kinds of processes you are examining and exploring.

I think in my whole undergraduate program at my college, there is no calculus based courses for econ which is really strange considering that top colleges have them almost since the beginning of their undergrad career, of course they are more advanced, but nevertheless.

I never really understood why a trade deficit means debt. Doesnt it just mean that there is much more coming into the country? What if all the money that is being paid for the products coming in is paid right away? Isnt that not a problem of debt?

my article regarding this is published, u can take a look and comment on it, nobody comnes to this forum, darn:

http://www.thetriangle.org/media/paper689/news/2005/07/22/EdOp/U.Dollar.Threatened.In.Global.Market.Bad.Policies-963860.shtml

Give it time, and get the word out on other forums and this place might take off, hopefully. There aren't many alternatives.

The Mundell-Fleming is useful in determining the relative merits of exchange rate regimes. In this model, monetary policy (interest rate/ money suppy changes) effect national output under flexible exchange rates but not fixed. The opposite is true of fiscal policy (taxes, government spending). Output does not rise with a flexible x-rate regime due to 'crowding out'.

There are many scenarios aside from purely flexible and fixed x-rate regimes. I live in the 'ultimate', a common currency union of european nations. This evolved from a target-zone regime of semi-fixed rates.

Regarding national debt and trade deficits, consider:
(S)aving = (I)nvestment in a closed economy (eg the World)
If a country saves more than it invests, it is borrowing to the rest of the world (net capital outflows). This is accompanied by a trade surplus (exports > imports) as foreign countries have more of your money than you of theirs.
It is an identity (necessarily true) that
S-I = X-M
Paul Krumgan's 'pop internationalism' is highly readable

I really cannot think of any "advantage" for more than one currency.
Currency trade is nothing more than shear gamble: we are gambling the on the exchange rate, which has no contribution to the production output.
I demand if anyone have better idea on it :?:

The nice thing about competing currencies is that they give us a very objective measure of the relative health between economies and with that a gauge of the relative strength of various economic policies.

Currency trading could be viewed as gambling, but so could any capital investment.  However, that description is a little misleading.  I do believe that some people have better insight into relative values between forms of capital than others.  To the common investor, it might be throwing darts at boards.  But there are people with great skill at playing darts.

[quote="hello"]Bchuchi, so you think that exchange rates on markets are determined randomly, is that what you mean?[/quote]

I don't think that's what he's saying.  There is a standard criticism about many markets, currencies in particular, that the relative value movements are random.

This is true in some sense as any knowledge of future moves that are well understood by the marketplace should be priced in and therefore movements should be stochastic (not exactly, but close enough -- market crashes tend to be more one-way than booms, so downward moves are on average larger, though less frequent, for any security that is evaluated against a "high standard" currency).

However, we can't assume that the relative values of currencies are well understood by the market.  Indeed, if they were, we would not see discrete shifts of value -- and we do.

Mathew, I am not sure what you mean by understood by relative currencies being understood by the market?

I believe what Matthew means is that the rates fluctuate showing that the relative value of the currencies is understood by the participants of the marketplace.
Currencies are harder to trade and to forecast primarily due to the sheer amount of geopolitical factors that feed into the market and influence the participants. It is an area which is very interconnected in nature which is the reason why you have actions from one currency precipitating down to others. For example a good example of this is last week when the chinese officials raised the range of traded between the RMB and the dollar. This resulted in a series of currency moves which caused sudden spikes and then reverted back to where they were. USDJPY had a spike down (RMB strength is favorable to the Japanese a big trading partner) while the EURUSD had a spike upwards. In both these cases prices reverted back to the normal range after 5 minutes or so.

Whether "the exchange rate is a Brownian motion" requires a large sample set of data to be verfiied. The problem is: "what is the good of having more than one currencies in one unified globalized market?", possibly could we look consider the problem in terms of "transaction cost"? Could currency be merely consuming
transaction cost but with no added value to the total production of the world economy?  :idea:

It might also be mentioned that something could appear to behave with Brownian motion and still be arbitragable.

The notion of randomness has been much better understood in recent years, especially in regards to market behavior.  It's often best to view randomness as compression of information to the point where we can't predict it's behavior.  Of course, information is different for everybody, so what might be random to one person might not be random to another who has a more complete information set.


I have a question about problem No.456 in the September Issue.
I'm not quite sure what is counted as one move. Am i supposed to count the two that  moves Token k to cup m and  the other  which moves
Token k+1 to cup k? or just regard both as one move?

PS: I'm not discussing the solution but if it's not supposed to talk about, please delete it. Thanks

This is from a problem set that is not due until October 31, so it should not be discussed until then. The topic is locked.


trigonometry
inequalities
inequalities proposed

In every triangle there is a relation: $ \cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma+2\cos\alpha\cos\beta\cos\gamma = 1$ and it is also true that if $ x, y, z$ satisfy the condition $ x^{2}+y^{2}+z^{2}+2xyz=1$ then there are cosines in some triangle.
Using this for $ \sqrt{ab}, \sqrt{bc}, \sqrt{ca}$ we are left with proving that $ cos\alpha+\cos\beta+cos\gamma \geq \frac{3}{2}$ but it is well known and easy with Jensen.

Just a little note: The sign of the inequality is reversed ;)
(it should be $ \cos\alpha+\cos\beta+\cos\gamma\leq \frac32$)

[quote="Litlle 1000t"]If $ ab+bc+ac+2abc=1$ and $ a,b,c>0$ Prove that
$ \sqrt{ab}+\sqrt{bc}+\sqrt{ac}\ge{\frac{3}{2}}$[/quote]
It is wrong. Counterexample, $ a\to 0$, $ b=c=1$.

for the edited problem you can also use $ x=\sin{\frac{A}{2}},y=\sin{\frac{B}{2}},z={\frac{C}{2}}$
take a look at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=149752[/url] posts #$ 11,14$

At first a clarification: meaning inequality must be reversed!
Now my solution:
With the substitutions $  a=\ \frac {a} {2} $,     $  b=\ \frac {b} {2} $,  $  c=\ \frac {c} {2} $ inequality is rewritten as follows:
1) $ \sqrt(ab)+\sqrt(bc)+\sqrt(ca)\le 3  $
if $ a $,$ b $,$ c $ are positive numbers such that $ ab + bc + ac + abc = 4 $
The substitutions $ a=\ \frac{yz}{x} $, $  b=\ \frac{zx}{y} $, $ c=\ \frac{xy}{z} $ transform (1) in the following equivalent relationship:
1')$ x+y+z \le  3   $
valid if $ x $, $ y $,$ z $are positive numbers such that $ x ^ 2 + y ^ 2 + z ^ 2 + xyz = 4 $
The result of (1 ') is well-known! Not insist!
______________________________________
Sandu Marin, Bucuresti, Romania


geometry
parallelogram
ratio
analytic geometry

A parallelogram $ ABCD$ is given where $ P,Q,R,S$ are mid points of $ AB,CD,DA,BC$ respectively.Find the ratio of areas of the shaded region to that of the parallelogram.

For those who don't want to download:
[asy]size(1.5inch); pair A,B,C,D,P,Q,R,S; A=(0,0);B=(3,0);C=(4,3);D=(1,3);P=A/2+B/2;Q=C/2+D/2;R=D/2+A/2;S=B/2+C/2; draw(A--B--C--D--cycle);draw(A--S);draw(B--Q);draw(C--R);draw(D--P); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,NE);label("$D$",D,NW); label("$P$",P,SE);label("$Q$",Q,NE);label("$R$",R,SW);label("$S$",S,NE);[/asy]
The shaded region to which the question refers are the 4 smallest triangles. I would shade them, but I don't want to compute the coordinates of the vertices not on the parallelogram boundary.

[hide="extremely non-rigorous"]WLOG, ABCD is a square. After that, it's area chasing time.[/hide]

Just skew the diagram so it is a square. This transformation preserves length ratios [and hence area ratios]

I think it is 1 to 5.

[hide="hint"]$ RACS$ is a parallelogram because one pair of opposite sides are $ \cong$ and $ \parallel $ so $ \angle{RAS}\cong\angle{RCS}$. So similarly $ \angle{DPA}\cong\angle{BQC}$ So by ASA we get that the two triangles are congruent. So from there on by vertical angles and we are done. [/hide]


Hey everyone,
I seem to get distracted by a lot of things (fantasy baseball news, NBA boxscores, checking collegeconfidential, AIM, facebook, checking AOPS and other message boards, etc). Throughout my last four years of high school, I often procrastinated and waited until the absolute last minute to do everything. Many times, I would not get certain tasks done, and I have missed out on many wonderful opportunities as a result. I was wondering what methods fellow AOPSers use to keep their focus on the tasks at hand? There are so many smart and hardworking people on this board, so I was hoping that I can learn from you guys! Thanks in advance  :)

PS. I definitely have OCD, but I'm not sure if I have ADD.

lol definately do not procrastinate, especially in high school. In middle school it might be allowed but in high school grades actually count in your future.

2.) always plan your "homework" time ahead of time...for example if you have an after school activity and lots of homework, force yourself to not get on the computer until your homeowkr is done

I'm actually already a senior in high school, unfortunately  :oops: . I was wondering though, do most people just eventually develop the habit of not procrastinating if they stick to a set schedule for a long enough time (such as going home, doing all of their homework, then doing anything else)? I'm hoping to erase my procrastinating ways before I go off to college next year, where the classes will be harder, and the consequences for waiting until the last minute to do everything will probably be much more severe.

I have quite a strict anti-procrastination policy, and it's always worked well for me. Whenever I do start to slack, I always regret it.

Be mildly autistic.   :P

I have a lot of trouble with this, too.  I haven't figured out a good solution yet.  The real problem is that none of my homework is interesting.

[quote="7h3.D3m0n.117"]force yourself to not get on the computer until your homeowkr is done[/quote]

When you have to type all of your homework because your teachers refuse to read your handwriting, this could get tricky  :oops:

[quote="t0rajir0u"][quote="7h3.D3m0n.117"]force yourself to not get on the computer until your homeowkr is done[/quote]

When you have to type all of your homework because your teachers refuse to read your handwriting, this could get tricky  :oops:[/quote]

lol touche =)

well don't get on homework time consuming sites (like AoPS, yes this has happened before and i had to do an english essay the last minute....darn you AoPS! lol)...also try to avoid using AIM because of annoying ppl that might IM you but sometimes if im relaly busy I just turn it to away.

how about:

PRocrastinate!

Some people do well under pressure, possibly even better.

i would start a new topic, but this is similar.

my problem is, if theres something i dont want to do, i just dont.

like if i have hw, and its boring, i dont do it. and i cant help it, and my grades suck because of that.

help?

[quote="neelnal"]get a friend on AIM that has the same hw with you.
race against him to finish it with quality or better, set a time where you guys have to confirm that each other finished the work.

another suggestion to slowly cut down is to pick a favorite activity, say visiting aops. say you have a 6 paragraph essay for hw and each paragraph takes you 20 minutes. then after each paragraph, allow yourself a 5 minute break on aops or every other paragrpah. depnding on ure liking. might take longer to type the essay (3 hours instead of 2) but its better than procrastinating.[/quote]

Yes, these are very good suggestions. The second idea might even improve efficiency, because sometimes you run into blocks when working full-time on a single task (for essay-type stuff).

[quote="7h3.D3m0n.117"]Okay here is as tip I would rarely give:

sometimes if there is absolutely no way of getting something done just do it the period before it is due...[/quote]

Heh... There is a rule among Murphy's laws stating something like
\[(\text{time left for the project}) \cdot (\text{effort invested in completing the project}) = \text{const.}\]
And a corollary: if there wasn't for the last moment, nobody would have gotten anything done ;)

Funny how everyday stuff supports Murphy's laws...

But on topic, I think that you just have to make yourself do everything on time. No tricks -- just don't let yourself get up and do anything other than homework.

[quote="now a ranger"]what if there is a test that takes up the whole period for everyone in the period before your homework is due one day ?[/quote]

you rush the test and rush the homework...

 idont' really ever recall a situation where a test required a full hour

This is one area in which having dial-up internet is really handy.  If I don't need the Internet for homework, I stay disconnected.  I also play music to distract me from all of the other distractions.  It sounds sort of weird and illogical, but it seems to work.

[quote="randomdragoon"][quote="Elemennop"][quote="7h3.D3m0n.117"]Okay here is as tip I would rarely give:

sometimes if there is absolutely no way of getting something done just do it the period before it is due...i do that alot but you have to careful not to get caught...also beware of teachers that walk around too much and can see you doing it...that would suck =)[/quote]

This isn't a tip, this is my life. I have elaborate systems where I do homework for every class in the class immediately preceding it; it's high pressure, but it works and I haven't done homework at home in a couple months.[/quote]
I have a different system -- I do the homework during the class and day that it is assigned, and if I am feeling tired I do it during the next period as well.[/quote]

i do that in math but my teacher caught me...now its harder to do it as she walks around too much and i have to hide the math book every other problem  :(

[quote="7h3.D3m0n.117"]
i do that in math but my teacher caught me...now its harder to do it as she walks around too much and i have to hide the math book every other problem  :([/quote]

Hold up, your math teacher stops you from working more advanced topics in class even though you're clearly bored with the work being done? Blasphemous!

I agree and I would love to skip that class into Geometry but my parents wont let me take the Algebra II test.

To maintain one's concentration, I leave everything off to the near end (like 7h3.d3m0n.117 said) and read a book at the same time. It always helps. For me of course. Only problem is that you might write something like, "He did that," on ur math homework.


This is a very beautiful problem:

Given $n>3$ prove that there exist $2^{n-1}$ points in the plane, no
three collinear, s.t. no $2n$ form a convex polygon.

I already know two solutions: one on http://www.kalva.demon.co.uk/short/soln/sh94c7.html and another my own.
Can you find another solutions?
Can you prove an upper bound? (this must be harder)

It is a famous Erdos-Szekeres problem.

I know very nice article devoted to this problem in the Bulletin of the AMS, October 2000, volume 37, number 4.

In this article following result is proved:
[i]there are $2^{n-2}$ points, no three collinear, s.t. no $n$ form convex polygon.[/i]

The famous Erdos-Szekeres conjecture that for any $2^{n-2}+1$ points we can find $n$, which form convex polygon.

myth,  is there are version of this theorem for polytopes?

Rrrr... [b]Myth[/b]

Yes, there is high-dimensional version of the problem.

Just little problem:

[b]Statement.[/b]
[i]Let $N_d(n)$ be minimal number s.t. it is always possible to choose convex $n$-top from given $N_d(n)$ points in general position in $\mathbb{R}^n$.
Prove that $N_d(n)\leq N_{d-1}(n)\leq...\leq N_2(n)\leq {2n-5 \choose n-3}+2$.[/i]

Note there are many different estimates for $N_d(n)$.

Actually this is Klein-Erdos-Szekeres Problem or Happy end problem. And you can find something romance about it.
You can see at :
 http://mathworld.wolfram.com/HappyEndProblem.html
The reason why it was called "Happy end problem" ... Go and see
http://martin.winkler.net/happy_end.html

Sorry  :D :P

Hi all,

just to remember you guys I am working on the ISL/ILL project and I kindly requested you to tell me that you post any of these problems, either per PM or per posting in the ISL year you took the problem from.  :) Otherwise I don't hesitate to repost problems and if I find yours by chance guess what will happen to it...  :no:


function
quadratics
algebra
special factorizations
algebra proposed

$ a)$ Determine the functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that:

$ 3 \cdot f(x) \cdot f(y) \minus{} 5 \cdot f(x)\plus{}7 \cdot f(y)\plus{}4\equal{}xy, \forall x,y \in \mathbb{R}$

$ b)$ Consider $ f: \mathbb{R} \rightarrow \mathbb{R}$, $ f(x) \equal{}
\begin{cases}
  x\minus{}4,  & x<2m \\
  3x\minus{}2, & x \ge 2m
\end{cases}$, $ m \in \mathbb{R}$.

Determine $ m \in \mathbb{R}$ such that:

$ f(2m\minus{}1)\plus{}f(m^2\plus{}m\plus{}1)\plus{}f(2m\plus{}2)\equal{}m f(2m\minus{}2)$

a) Can it be for all $ x,y \in \mathbb{R}$? Just put $ x\equal{}y\equal{}0$ we get a quadratic equation that has no real solution.

b)We have $ m^2\minus{}m\plus{}1>0$ by completing the square
$ \Leftrightarrow m^2\plus{}m\plus{}1>2m$

So the equation becomes $ (2m\minus{}5)\plus{}(3m^2\plus{}3m\plus{}1)\plus{}(6m\plus{}4)\equal{}m(2m\minus{}6)$
$ \Leftrightarrow m^2\plus{}17m\equal{}0$, so $ m\equal{}0$ or $ \minus{}17$


[hide]Let $m^2=a$ and $n^2=b$.  Then we want to find $3ab$.  We know $a+3ab=30b+517$.  Using Simon's factoring trick (subtract $10+30b$ on both sides, factor out the $a$, then factor out $10$, and then factor out $1+3b$) we have $(3b+1)(a-10)=507$.  Factoring $507=13^2\cdot 3$.  Since both $a$ and $b$ are perfect squares we find $a-10=39$, so $a=49$ and $3b+1=13$, so $b=4$.  We want the value $3ab=3\cdot 49\cdot 4=588$.[/hide]


Let $a_n$ the number of digits greater than or equal to 5 in $n$. Prove that $\sum_{n\geq 1}{\frac{a_{2^n}}{2^n}}=\frac{2}{9}$.

This is another cute problem by harazi I would like to give a second life to. I should confess that, at first, I did not believe it and even tried to disprove the statement using a computer. But when I got $0.222222222...$ for the sum of the first $40$ terms, I decided that it might be true after all, though it took me some time after that to figure the problem out. The only hint I can give without ruining the problem is that, unlike many other problems by harazi, it is [i]very simple[/i].

As usual, if nobody figures it out in a couple of weeks, I'll post a solution. :)

We can use the relation:

$s(2^n) = 2s(2^{n-1}) - 9a_{2^{n-1}}$
[try to prove this :)]

where $s(x)$ is the sum of digits of x

Divide by $2^n$, sum and get : $s(2^n)/2^n - 1 = -(9/2)x_n$ where $x_n$ is the partial sum of our serie.

From here the result is clear because $s(2^n)/2^n \to  0$ [$s(2^n) \leq l(2^n) = O(n)$] where $l(x)$ = nr of digits of x

do you know the proof of the relation you used ?

[Edit] ok, I've got it. ;)

nice solution,mareleG;
was you solution the same, fedja ?

[quote="eugene"]
was you solution the same, fedja ?[/quote]
Yes. In a sense it is [b]the[/b] solution. At least, I'll be extremely amazed if somebody finds a really different approach.

i see these $O(n)$ these days looking on mathlinks
what does it mean ?
is it related to asymptotic notation with big tetha that stuff introductory in cormen ?

[quote="spx2"]i see these $O(n)$ these days looking on mathlinks
what does it mean ?[/quote]
Read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31517]Kent's stickie[/url] in Computations and Tutorials.


let n is nature and n>2. Prove that $(n^{n^{n^{n}}}-n^{n^{n}})\vdots 9$ :lol:

What is it exactly that we have to prove?

maybe he wants us to prove that $9|n^{n^{n^{n}}}-n^{n^{n}}$

I suppose that you could let $n{}^n{}^n=x$, where you know that $n|x$.  From there, you have
$n^x-x$
And then...mods?
 :o
And...
Uh...
 :huh:
:?
nooooooooo
 :(


ARML
geometry
geometric transformation
reflection
trigonometry
algebra
polynomial

Hello to any participants of the ARML forum... You may or may not know about the previous ARML Practice Series (located at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30264]here[/url] (Solutions/Results [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32273]here[/url]) and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86168]Here[/url] (Solutions/Results [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86171]here[/url])), but I (along with my partner in crime, pythag011) am ready and willing to revive another! That is, if there is interest. Please PM me or post here if you are interested!

If we get enough responses, we'll start up a ARML Practice Series 3 thread, along with a companion Scoreboard/Results thread. There will be a new problem each day, and a format likely similar to the previous Practice Series (different point values, cumulative contest, and possible mini-"reward" (mmm, food) for winners who go to the same site as me. (Sorry, people not going to Las Vegas...))
If not, pythag and I will probably still post some problems, but in either the AMC or Pre-Olympiad forum, and probably not as many...

For now, here is a sample of problems that at least somewhat reflect the problems that will be present in the series...

1. Let $ x,y$ be two reals such that $ x^3 \plus{} y^3 \equal{} 2xy \minus{} \frac {8}{27}$. Let $ a$ and $ b$ be the maximum and minimum possible values of $ x \plus{} y$, respectively. Find $ a \minus{} b$. [Source: Brut3Forc3]

2. For positive integer $ k$, let $ f_k(x) \equal{} \frac {1}{k}(\sin^k x \plus{} \cos^k x)$ Find the minimum possible value of $ f_4(x) \minus{} f_6(x)$. [Source: 103 Trigonometry Problems]

3. Prove that there are infinitely many triples of positive integers $ (a,b,c)$ such that $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca \plus{} a \plus{} b \plus{} c}$ is a positive integer. [Source: pythag011]

(Note: As anyone with prior knowledge of myself and pythag011 would know, my problems will tend to the easier side, whereas pythag's will tend to the harder side. Most likely, this problem will be dealt with by adjusting the point value of corresponding problems.)

Once again, please post here or PM me if you are interested in participating! (Please? :P)

Okay, I've received several responses, so this is a go! I think I'll start it on Friday, at around 8 PM PST, with problems going up at around that time every day after.

(But signups are still open!)

[quote="appillai"]Hey, I'm trying out for ARML chicago, but I'm not sure if I'm ready. I think this would be great practice! Please start this series. Thanks![/quote]
Try solving these 3 problems above first! They are all nice problems!
Here's my solution for the 1st one:
We have: $ (x\plus{}y)^3\plus{}8/27\minus{}3xy(x\plus{}y\minus{}2/3)\equal{}0$.
Factorize this one we get: $ (x\plus{}y\plus{}2/3)[(x\plus{}y)^2\minus{}2(x\plus{}y)/3\plus{}4/9\minus{}3xy]\equal{}0$
We either have $ x\plus{}y\equal{}\minus{}2/3 or (x\plus{}y)^2\minus{}2(x\plus{}y)/3\plus{}4/9\equal{}3xy$
Since $ (x\plus{}y)^2/4 \geq xy$, we have: $ 0 \geq (x\plus{}y)^2\minus{}3(x\plus{}y)^2/4\plus{}4/9\minus{}2(x\plus{}y)/3$, which means $ 0 \geq [(x\plus{}y)/2\minus{} \minus{}2/3]^2$.
The equality occurs when $ x\plus{}y\equal{}4/3$.
Therefore, summing from both cases, we get max$ (x\plus{}y)\equal{}4/3$. The equality occurs at $ x\equal{}y\equal{}2/3$

[quote="ghjk"][quote="appillai"]Hey, I'm trying out for ARML chicago, but I'm not sure if I'm ready. I think this would be great practice! Please start this series. Thanks![/quote]
Try solving these 3 problems above first! They are all nice problems!
Here's my solution for the 1st one:
We have: $ (x \plus{} y)^3 \plus{} 8/27 \minus{} 3xy(x \plus{} y \minus{} 2/3) \equal{} 0$.
Factorize this one we get: $ (x \plus{} y \plus{} 2/3)[(x \plus{} y)^2 \minus{} 2(x \plus{} y)/3 \plus{} 4/9 \minus{} 3xy] \equal{} 0$
We either have $ x \plus{} y \equal{} \minus{} 2/3 or (x \plus{} y)^2 \minus{} 2(x \plus{} y)/3 \plus{} 4/9 \equal{} 3xy$
Since $ (x \plus{} y)^2/4 \geq xy$, we have: $ 0 \geq (x \plus{} y)^2 \minus{} 3(x \plus{} y)^2/4 \plus{} 4/9 \minus{} 2(x \plus{} y)/3$, which means $ 0 \geq [(x \plus{} y)/2 \minus{} \minus{} 2/3]^2$.
The equality occurs when $ x \plus{} y \equal{} 4/3$.
Therefore, summing from both cases, we get max$ (x \plus{} y) \equal{} 4/3$. The equality occurs at $ x \equal{} y \equal{} 2/3$[/quote]
I have changed the problem a bit since you saw it, and it now includes finding the min. (But it's basically the same.) Good solution!

For the other two, 2 is fairly easy, whereas 3 is much harder, and not likely to be seen on an ARML.

I think even 1 is too hard for an ARML problem.

I'll post solutions to the other 2.

[hide="3"]
The problem is just a simple application of backwards Vieta jumping.

Note that (1,2,4) is a solution to $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca \plus{} a \plus{} b \plus{} c} \equal{} 1.$ We claim there are infinitely many solution. Assume that (a,b,c) is a root, with $ a \le b \le c$ and a,b,c positive. Multiply out to get $ a^2 \plus{} b^2 \plus{} c^2 \equal{} ab \plus{} bc \plus{} ca \plus{} a \plus{} b \plus{} c,$ or $ a^2 \minus{} (b \plus{} c)a \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} b \minus{} c \equal{} 0.$ Consider the polynomial $ x^2 \minus{} (b \plus{} c)x \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} b \minus{} c.$ By the above, one of its roots is a. Vieta gives us that b+c-a is also a root, and that it is greater than c, so (b+c-a,b,c) and (b,c,b+c-a) are solutions. We can keep doing this to give us infinitely many solutions.[/hide]

[hide="2"]Note that $ sin^6 \plus{} cos^6 \equal{} (sin^2 \plus{} cos^2)(sin^4 \minus{} sin^2cos^2 \plus{} cos^4) \equal{} sin^4 \minus{} sin^2cos^2 \plus{} cos^4.$ Thus, the expression equals $ \frac {sin^4 \plus{} cos^4}{4} \minus{} \frac {sin^4 \minus{} sin^2cos^2 \plus{} cos^4}{6} \equal{} \frac {sin^4}{12} \plus{} \frac {sin^2cos^2}{6} \plus{} \frac {cos^4}{12} \equal{} \frac {sin^4 \plus{} 2sin^2cos^2 \plus{} cos^4}{12} \equal{} \frac {(sin^2 \plus{} cos^2)^2}{12} \equal{} \frac {1}{12},$ so it always equals 1/12.[/hide]

[quote="Differ"]Free problems are free problems. Who doesn't like free problems? :P

I'd definitely be interested.[/quote]

That is why I'll do it and i need that practice in general.

I'm interested, although I probably won't do much because difficulty/schoolwork. I promise I'll look at them though :P

These are some lovely problems. I will definitely be doing this.

[quote="pythag011"]I'll post solutions to the other 2.

[hide="3"]
The problem is just a simple application of backwards Vieta jumping.

Note that (1,2,4) is a solution to $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca \plus{} a \plus{} b \plus{} c} \equal{} 1.$ We claim there are infinitely many solution. Assume that (a,b,c) is a root, with $ a \le b \le c$ and a,b,c positive. Multiply out to get $ a^2 \plus{} b^2 \plus{} c^2 \equal{} ab \plus{} bc \plus{} ca \plus{} a \plus{} b \plus{} c,$ or $ a^2 \minus{} (b \plus{} c)a \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} b \minus{} c \equal{} 0.$ Consider the polynomial $ x^2 \minus{} (b \plus{} c)x \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} b \minus{} c.$ By the above, one of its roots is a. Vieta gives us that b+c-a is also a root, and that it is greater than c, so (b+c-a,b,c) and (b,c,b+c-a) are solutions. We can keep doing this to give us infinitely many solutions.[/hide][/hide][/quote]
Just checked over this, there's a slight error:
[hide]You said "Multiply out to get $ a^2 \plus{} b^2 \plus{} c^2 \equal{} ab \plus{} bc \plus{} ca \plus{} a \plus{} b \plus{} c,$ or $ a^2 \minus{} (b \plus{} c)a \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} b \minus{} c \equal{} 0.$", but the polynomial is actually $ a^2 \minus{} (b \plus{} c\plus{}1)a \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} b \minus{} c \equal{} 0$.
This means that $ (a,b,c)$ being a root actually implies that $ (b,c,b\plus{}c\minus{}a\plus{}1)$ is a root. However, that doesn't invalidate the proof at all.[/hide]

I will look at them but i dont think i will have the time to do all of them... srry... :(


compute $ (\sqrt\frac {1}{2} \plus{} \sqrt\frac {i}{2})^ {12}$ without using a calculator.

[hide]
What is so nice about this question anyway?  It's tedious and not very interesting, in my opinion.  Looks like a homework problem.
[hide]
$ \left(\sqrt{\frac{1}{2}} \plus{} \sqrt{\frac{i}{2}}\right)^{12} \equal{} \left(\frac{1}{2} \plus{} 2\sqrt{\frac{i}{4}} \plus{} \frac{i}{2}\right)^6 \equal{} \left(\frac{1\plus{}i}{2} \plus{} \sqrt{i}\right)^6$.

Since

$ \sqrt{i} \equal{} \left(e^{i \pi/2}\right)^{1/2} \equal{} e^{i \pi/4} \equal{} \frac{1\plus{}i}{\sqrt{2}}$,

we have

$ \left(\frac{1\plus{}i}{2} \plus{} \sqrt{i}\right)^6 \equal{} \left(\frac{1}{\sqrt{2}} \plus{} 1 \right)^6 (e^{i\pi/4})^6 \equal{} \left(\frac{1}{\sqrt{2}} \plus{} 1\right)^6 (\minus{}i) \equal{} \frac{\minus{}i}{8} (1\plus{}\sqrt{2})^6$.

Now all that remains is to expand the real factor in parentheses:

$ (1\plus{}\sqrt{2})^6 \equal{} (3 \plus{} 2\sqrt{2})^3 \equal{} 3^3 \plus{} 3(3^2)(2\sqrt{2}) \plus{} 3(3)(2\sqrt{2})^2 \plus{} (2\sqrt{2})^3 \equal{} 99\plus{}70\sqrt{2}$.

Hence the answer is $ \frac{\minus{}i}{8}(99 \plus{} 70 \sqrt{2})$.
[/hide][/hide]

I kind of had a more interesting solution in mind but oh well.....

Sorry to double post but why do all my posts get moved here when i post them in HSB?


The sequence $(x_{n})$ is defined by $x_{1}= a \in R$ and
$x_{n}+1 = 3x_{n}^{3}-7x_{n}^{2}+5x_{n}$for n = 1, 2, 3,
  Find all $a$ for which $(x_{n})$ has a finite limit and determine
that limit.

I think your problem is as follows.
Please edit your post.
$x_{n+1}=3x_{n}^{3}-7x_{n}^{2}+5x_{n}\ (n=1,2,3,\cdots).$


linear algebra
matrix
vector
algebra
polynomial

compute the determinant of the matrix attached...

Thank you

my hint:
det(A2)=(a+b)^2-b^2=a^2+2ab
det(A3)=(a+b)^3-b^2(a+b)+2b^3=a^3+3a^2b
.
.
det(An)=a^n+nba^(n-1)  is it ok??

Here's how I went about solving the problem.

Explicit calculations show that the determinant is $ a^2 \plus{} 2ab \equal{} a(a \plus{} 2b)$ in the case where the matrix is two by two, and $ a^3 \plus{} 3a^2b \equal{} a^2(a \plus{} 3b)$ in the case where the matrix is three by three.  This pattern seems likely to persist, so we probably know the answer already, even if we can't prove it yet.

I tried using row reduction and didn't get anywhere.  So, I have to force myself to try something different, which is difficult because I tend to want to keep trying the same thing, hoping I can finally get it to work.  This matrix is a "circulant matrix", and I remember hearing something about eigenvalues and eigenvectors of a circulant matrix, and that it is easy to guess an eigenvector, or something like that.  This reminds me that the determinant is the product of the eigenvalues.  So maybe I can compute the determinant by finding all the eigenvalues (and maybe I can find all the eigenvalues by explicitly guessing eigenvectors).

A simple guess for an eigenvector (which I vaguely remember from discussions of circulant matrices) is a vector consisting of all $ 1$s.  In the three by three case, this vector is indeed an eigenvector with eigenvalue $ a \plus{} 3b$.  It seems likely that the other eigenvalue is $ a$, with a multiplicity of two.

It turns out to be pretty easy to guess two linearly independent eigenvectors with eigenvalue $ a$.  They are $ [1,\minus{}1,0]^T$ and $ [1,0,\minus{}1]^T$.  So in the three by three case we have found all the eigenvalues, so we know the determinant.  Moreover, you can give a similar solution in the general case.

An alternative approach.

1.  Let's suppose that instead every entry in the matrix was $ b$.   What is the rank of this new matrix?  What are the eigenvalues of this new matrix?

2. To get to the original matrix from the modified matrix, we add $ a$ times the identity matrix.  What does this do to each eigenvalue, and what are the eigenvalues of our original matrix?

The matrix $ J$ which consists of $ 1$s in every position satisfies the relation $ J^2\equal{}nJ.$ If a matrix satisfies a polynomial equation, then all of its eigenvalues also satisfy that same polynomial equation. Hence the only possible eigenvalues of $ J$ are $ 0$ and $ n.$ But since the trace of $ J$ is $ n$ and the eigenvalues add up to the trace, we see that the eigenvalues must be $ n\minus{}1$ copies of $ 0$ and one copy of $ n.$ We would come to the same conclusion by noting that $ J$ has rank $ 1,$ hence its null space (the eigenspace associated to the eigenvalue $ 0$) has dimension $ n\minus{}1.$

However we arrive at it, we conclude that the characteristic polynomial of $ J$ is $ x^{n\minus{}1}(x\minus{}n).$

What about the matrix $ bJ?$ This satisfies the relation $ (bJ)^2\equal{}b^2nJ\equal{}(bn)(bJ).$ We conclude that the eigenvalues are $ 0$ and $ bn$ and that the characteristic polynomial is $ x^{n\minus{}1}(x\minus{}bn).$ That is, $ \det(bJ\minus{}xI)\equal{}(\minus{}1)^nx^{n\minus{}1}(x\minus{}bn).$

In this problem, we're asked to find $ \det(bJ\plus{}aI).$ All we have to do is let $ x\equal{}\minus{}a.$

$ \det(bJ\plus{}aI)\equal{}(\minus{}1)^n(\minus{}a)^{n\minus{}1}(\minus{}a\minus{}bn)\equal{}a^{n\minus{}1}(a\plus{}bn).$

----

Three small tips for studying linear algebra:

1. Be aware of possible polynomial identities for matrices; the principle that an eigenvalue must satisfy any polynomial that the matrix does (or, phrased another way, every root of the minimal polynomial is a root of the characteristic polynomial) is useful.

2. Be aware of the trace of a matrix. It is easy to compute, conserved under similarity, and hence the sum of the eigenvalues.

3. The particular matrix $ J$ should be familiar - an old friend of a matrix, whose properties you know.


Prove for every prime p the following equation: \sum^{p-1}_(k=1) li(k^3/p) = 1/4*(p-2)*(p-1)*(p+1).

Note that li(x) denotes the greatest integer n with n \leq x.

First I will use [x] instead of li(x).
Now, note that if x is not an integer then [-x] = -[x] - 1.  (1)
Moreover, for k =1,...,p-1, the number k 3 /p is not an integer, since p is prime.

Let S(p) be the LHS of the desired identity.
Clearly, S(2) = 0, so that the identity holds in this case thus, from now, we assume that p = 2q+1 is odd.
For k = 1,...,q, we have :
(p-k) 3 / p = p 2  - 3pk + 3k 2  - k 3  / p.
Thus, from (1), we deduce that :
[(p-k) 3 / p ] + [ k 3  / p] = p 2  - 3pk + 3k 2 - 1.
Then :
S(p) =  \sum {1 \leq k \leq q} ( [(p-k) 3 / p ] + [ k 3  / p] )
= \sum {1 \leq k \leq q} (p 2  - 3pk + 3k 2 - 1)
= (p 2  - 1)q - 3pq(q+1) + q(q+1)(2q+1)/2
= (p 2  - 1)(p-2)/4, as desired.

Pierre.


Hi all, I have a "problem" of sorts that I do not know how to solve. On my laptop, if I build a PDF document in TexnicCenter, then view it, then make some changes and recompile and attempt to build again, I get 3 errors along the lines of:

- I can't write on file 'test.pdf'

- Emergency Stop!

- Fatal error occurred, the output PDF file is not finished!

Consequently, I must always close the currently opened PDF file (created from a previous build) before I can recompile my source code. On my other computer, this is not necessary as it always just overwrites the current PDF file, so I'm wondering if there is some setting or something else that is causing this to happen, or something I can do?

Thanks.

In TeXnicCenter, go to the "Define Output Profiles..." option of the Build menu.

Make sure that "Latex => PDF" is selected, and click on the Viewer tab.

Then the one at the bottom -- "Close document before running (La)TeX" -- the "DDE command" button needs to be selected, and the command is

[DocClose("%bm.pdf")]

Compare the settings on this page on your laptop with those on your desktop, and see what's different.  If it still doesn't work, click on the "Wizard" button and try to reset the output profiles.

On the other hand, every once in a while TeXnicCenter will no longer talk to Adobe Reader in the way that it's suppose to.  This often necessitates closing both TeXnicCenter and Reader and then restarting them, or in some cases rebooting your PC.

Ah ok I got it now. I did have those close document commands previously which was confusing me. However, for some reason under "View Project's Output" and "Forward Search", it was using Command line arguments instead of DDE Commands.

For both "View project's output" and "Forward search", DDE command must be selected, the command should be

[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]

and for Server, enter "Acroview" (without the quotes), and for Topic, enter "Control" (without the quotes). Also make sure those are the entries for Server and Topic under the close document part.

Wow, that was annoying.

EDIT: Oh wow, check this out:

http://en.wikipedia.org/wiki/TeXnicCenter

It has all the information you want as well as more goodies, such as making it so your PDF documents go back to the page you were originally on!

hmm, i tried to set texniccenter so that its reopens the file after a recompilation at the same location as it was just before by copying the lines from wikipedia

but when i try to compile, its gives an error that says it cannot excecute the commands...
even after a restart...

 :(

Note that if you follow the Wikipedia instructions for sending a pdf back to where it was previously opened, the pdf must have already been opened, or you will get an error message. The error message does not actually affect the compilation, and you'll get it the first time you compile and open a pdf after starting TexnicCenter. Afterward, every build and recompile will send you to whatever page you had previously been looking at.

nope, even after a pc reboot it still doesn't work.

(i already mentioned that i first needed to open it but that didn't help either

That's strange, are you sure you have Server: Acroview, and Topic: Control?

hmm, now i know that it can be done i begon to miss it, so here's what i have

i'm using texniccenter 1 beta 6,31 'Firenze'
and acrobat reader   Versie 7.0.5.2005092300

in attachment you can find how i set the viewer part in texniccenter
the command i used were these i copied from
any solution on this?? [url=http://en.wikipedia.org/wiki/Texniccenter]wikipedia[/url]

Hmm, that is exactly identical to what I have. I have no idea what the problem is. I believe I am using beta version 7.00, but otherwise there is no difference.

hm, even after an update to texniccenter 7 it still doesn't work  :(

[quote="Watson"]Windows can't handle files properly. Only one process at a time can open them. So if you have Adobe Acrobat open, and LaTeX attempts to write, you have an error.[/quote]Not quite. I have a file opened with Acrobat, and I can re-write it using WinEdT.


AMC
USA(J)MO
USAMO
search
group theory
linear algebra
superior algebra

Hi,

Does anyone know of any book/website which has examples of combinatorics or other problems which can be solved by elementary group theory? Can anyone give an example of a combinatoric problems which can be solved by group theory?

Thanks

See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1116182#1116182 for an application of group theory to USAMO 2008 #6.

I've actually also been interested in exactly the same thing, i.e. finding more examples of applications of group theory to olympiad problems.

There are, to my mind, two big ways group theory comes up in Olympiad-level (especially combinatorics) problems:  [url=http://en.wikipedia.org/wiki/Burnside%27s_lemma]Burnside's lemma[/url] and linear algebra over finite fields (especially $ \mathbb{F}_2$).  See, for example, Gowers' discussion of the latter [url=http://gowers.wordpress.com/2008/07/31/dimension-arguments-in-combinatorics/]here[/url].

Edit:  I also love citing [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=272516508&t=215429]IMO 2008 #5[/url]; the solution I gave implicitly uses the discrete Fourier transform on $ (\mathbb{Z}/2\mathbb{Z})^n$, although I didn't recognize it at the time, but I guess I wouldn't call this purely a "group-theoretic" technique:  you know it as the roots of unity filter.


integration
calculus
logarithms
function
trigonometry
calculus computations

The indefinite integral is not elementary.

[quote="FOURRIER"]$ \int e^x ln(x) dx$

Its sufficent to compute :$ \int ln(ln(x)) dx$[/quote]

$ \int e^x \ln x dx \& = \int \ln x d \left( e^x \right) = e^x \ln x - \int \frac {e^x}{x} dx$

Recall : $ e^x = \sum_{\tiny{k = 0}}^{\tiny \infty} \frac {x^k}{k!} \;\; \Rightarrow \;\; \frac {e^x}{x} = \sum_{\tiny{k = 0}}^{\tiny \infty} \frac {x^{\tiny{k - 1}}}{k!}$

So :

\begin{align} \int e^x \ln x dx & = e^x \ln x - \int \frac {e^x}{x} dx \\
& = e^x \ln x - \int \left( \frac{1}{x} + \sum_{\tiny{k = 1}}^{\tiny \infty} \frac {x^{\tiny{k - 1}}}{k!} \right) dx \\
& = e^x \ln x - \ln x + \sum_{\tiny{k = 1}}^{\tiny \infty} \int \frac {x^{\tiny{k - 1}}}{k!} dx \\
& = e^x \ln x -  \ln x + \sum_{\tiny{k = 1}}^{\tiny \infty} \frac {x^{\tiny k}}{k.k!} \end{align}

You can do certain definite integrals, but the indefinite integral is impossible. (I suppose you could express it in terms of Ei(x) if you want)
$ \int_{ \minus{} \infty}^{0} e^{x}ln|x|dx \equal{} \minus{} \gamma$

[quote="FOURRIER"]do you mean it can not be computed?[/quote]

I mean it cannot be expressed in terms of "elementary functions" (unless you consider the [url=http://en.wikipedia.org/wiki/Exponential_integral]exponential integral[/url] an elementary function).

[quote="FOURRIER"]$ \int e^x ln(x) dx$[/quote]

rather than just staring at the ceiling and coming up with these integrals and posting them on the forum, do some preliminary work yourself . for example, when you   see a definite integral, first   convince yourself   that the integral   converges ( and don't   blindly use a software for that).  if you find that it converges, proceed to find out a closed form....... failing that, post it on the forum saying what you actually did in your work and where you may have gotten stuck  so others can try to point you in the right direction.

don't just blurt out some indefinite integral like   :D

$ \int \; \ln\,\left(\dfrac{\tan^{ \minus{} 1}\;\sqrt {\; 1 \plus{} x^3\; }}{e^x \minus{} 2\cos^{ \minus{} 1}\,x^3 \plus{} \sin\, \sqrt x}\right) \;\textbf dx$

and expect someone to solve it. it doesn't work that way, unfortunately.

[quote="FOURRIER"]
Its sufficent to compute :$ \int ln(ln(x)) dx$[/quote]

same thing with your vague   hints...  :|  care to explain what you mean by "it's sufficient" ... i am interested to know...

[quote="misan"]
[quote="FOURRIER"]
Its sufficent to compute :$ \int ln(ln(x)) dx$[/quote]

same thing with your vague   hints...  :|  care to explain what you mean by "it's sufficient" ... i am interested to know...[/quote]

Put $ t \equal{} e^x$

[quote="misan"][quote="FOURRIER"]$ \int e^x ln(x) dx$[/quote]

rather than just staring at the ceiling and coming up with these integrals and posting them on the forum, do some preliminary work yourself . for example, when you   see a definite integral, first   convince yourself   that the integral   converges ( and don't   blindly use a software for that).  if you find that it converges, proceed to find out a closed form....... failing that, post it on the forum saying what you actually did in your work and where you may have gotten stuck  so others can try to point you in the right direction.
[/quote]

what do you mean by convergence of an integral??

misan is talking about improper definite integrals.  Often a problem like this (where the integrand does not have an elementary antiderivative) is phrased as a definite rather than indefinite integral, and the answer turns out to be expressible in terms of well-known constants (rather than functions).


Prove that

$ h_a\plus{}h_b\plus{}h_c\geq\frac{9[ABC]}{s}$

Rewrite the inequality as : $ \frac{2[ABC]}{a}\plus{}\frac{2[ABC]}{b}\plus{}\frac{2[ABC]}{c}\ge \frac{9[ABC]}{s}$

$ \Longleftrightarrow\ \frac 1a\plus{}\frac 1b\plus{}\frac 1c\ge\frac{9}{a\plus{}b\plus{}c}$ , which is true by [b]Cauchy-Schwarz[/b] .


inequalities
algebra
binomial theorem
number theory unsolved
number theory

For every natural number $ n$ prove the inequality
\[ (n\minus{}1)^n\plus{}2n^n\leq (n\plus{}1)^n\leq 2(n\minus{}1)^n\plus{}2n^n.\]

Dividing by $ n^n$ yields:

$ (1\minus{}\frac{1}{n})^n\plus{}2 \le (1\plus{}\frac{1}{n})^n \le 2(1\minus{}\frac{1}{n})^n\plus{}2$

Just looking at the right side and binomial theorem you only have to show that

$ \binom{n}{2k}\frac{1}{n^{2k}}\minus{}3\binom{n}{2k\plus{}1}\frac{1}{n^{2k\plus{}1}} \ge 0$

that should be straight forward...

lhs should solve analogously

[quote="wauwau"]

Just looking at the right side and binomial theorem you only have to show that

$ \binom{n}{2k}\frac {1}{n^{2k}} \minus{} 3\binom{n}{2k \plus{} 1}\frac {1}{n^{2k \plus{} 1}} \ge 0$

[/quote]

I don't think it's just this...


Find the exact values of:

$\sum_{n=1}^{\infty}\frac n{2^n}$

$\sum_{n=1}^{\infty}\frac{n^2}{2^n}$

$\sum_{n=1}^{\infty}\frac{n^4}{2^n}$

First one is equal with $2$.hope I'm not mistaken. :?

#2

let P be the original sum

(1/2)*P shifts all the term to the right

P - 1/2P =1/2P gives  the first term from the original summation +  sum n=1 to inf [ ((n+1)^2-n^2)/2^(n+1)]

the new summation can be calculated using the technique used in #1 . we get 1/2P = 1/2 + 5/2

P = our desired sum = 6

i believe #3 can be solved using the same method. i'll try to use latex


Does anyone have gregx007's old USNCO pdf notes?
Could you post them by any chance? The wayBackMachine (internet archive) hath failed me, and, as far as I can tell gregx007's account is inactive.

here it is. saved it from a while ago.

enjoy :)


$ \sum_{n} \frac{(\minus{}1)^{[\sqrt{n}]}}{n}$

[color=darkblue]where $ [x]$ denotes the biggest integer less than $ x$.[/color]

hello, your sum does not converge.
Sonnhard.

The series converges: http://www.mathlinks.ro/Forum/viewtopic.php?p=708340#708340

No, Sonnhard's not right. Like him, you make the mistake of putting too much trust in computer algebra systems. This depends on putting the terms into blocks of the same sign and estimating their size, which computer tools are not clever enough to notice.


geometry
projective geometry
linear algebra

[quote="mita2002"]I must now learn a subject affine geometry but I haven't any good ebook for this , so if you have it , up it here for me . Thanks :)[/quote]
Please reply me , I don't wanna this but I haven't got any good book  for this , more it's hard to find one

Nobody replied this topic , so Can you let me know the name and author of the book about Affine geometry you think it's best

I don't know if it's the "best" (as it's the only book on the subject I've ever read), but I like [url=http://www.amazon.com/Affine-Projective-Geometry-M-Bennett/dp/0471113158/ref=sr_1_1?ie=UTF8&s=books&qid=1242330269&sr=8-1]Affine and Projective Geometry[/url] by M. K. Bennett. It's approach is very much tied to the subjects of Linear Algebra and Abstract Algebra (particularly Division Rings and Fields), so a good knowledge of those subjects is necessary for a thorough understanding of this book.


when i come here i see all these topics about quants but there are hardly any (if any at all) about actuaries.  I was wondering if someone could provide me with more information on this field of work.

Think of actuaries as the "engineers" of the insurance business (which is broad enough to include pension and benefit programs). They work for insurance companies, pension funds, and as consultants. Some also work for the government agencies that regulate insurance companies or oversee government pension programs.  Twenty years ago, most of the top officers of insurance companies were trained as actuaries (and most of the top officers of oil companies were trained as engineers or geologists). That may not be true any more in either case.

Actuaries have a very defined career path and structure, with a series of nationally-set exams being the centerpiece. The first exam involve calculus, linear algebra, probability and statistics. Some of the later exams are mathematical in nature, but others involve finance, law, or regulations. Becuase of the mathematical nature of some of the early exams, it's easer for college math majors or minors to break into the field than for others, but it's not an absolute requirement.

The descriptions of "quants" that have appeared in this forum make it appear to be a high-risk, high-reward, high-stress job with an emphasis on creativity and dynamic problem-solving. The actuarial career path is nearly the opposite - a stable-income, regular-hours job in which knowledge and experience count for a lot.

The actuarial societies' own propaganda site would be a place to start: http://www.beanactuary.org .

Interesting.
Sounds familiar but I haven' heard of the term actuaries before.

I have been an actuary for 30 plus years.

We are applied mathematicians specializing in insurance.  Generally we determine the insurance premium to charge for some sort of insurance deal (life, health, pension, property, casualty), or we determine the amount of funds to set aside for reserves for the eventual claims for such deals.  Actuaries do many other things, and some are president of their companies.

We have a series of actuarial exams to pass to be fully credentialed.  The first exam has changed from what is described above.  We probably do not have the same kind of work hours and pressures of financial engineers, but we are not necessarily 9 to 5 ers either.  We have the solvency of our firms in our hands.  We also have a code of ethics, and can potentially lose our credentials, like doctors or lawyers - many other professionals have nothing like that.

A friend of mine has a good intro page to actuaries called Actuarial Advice:  http://users.aol.com/fcas/advice.html


inequalities
logarithms
function
integration
calculus
calculus computations

Show that $ \frac{1}{x} > \ln (1\plus{}x) \minus{} \ln x > \frac{1}{1\plus{}x}$ for $ x>0$

Deduce that, for $ x>0$, $ (1\plus{} \frac{1}{x})^{x}$ increases and $ (1\plus{}\frac{1}{x})^{x\plus{}1}$ decreases as $ x$ increases.

[quote="AndrewTom"]Show that $ \frac {1}{x} > \ln (1 \plus{} x) \minus{} \ln x > \frac {1}{1 \plus{} x}$ for $ x > 0$

Deduce that, for $ x > 0$, $ (1 \plus{} \frac {1}{x})^{x}$ increases and $ (1 \plus{} \frac {1}{x})^{x \plus{} 1}$ decreases as $ x$ increases.[/quote]
Use the mean value theorem ..  :cool:

Since $ y \equal{} \frac {1}{t}\ (t > 0)$ is an decreasing and convex function,

$ x\leq t\leq x \plus{} 1\ (x > 0)\Longrightarrow \frac {1}{x \plus{} 1}\leq t\leq \frac {1}{x}$

$ \therefore \int_x^{x \plus{} 1} \frac {1}{x \plus{} 1}\ dt < \int_x^{x \plus{} 1} \frac {1}{t}\ dt < \int_x^{x \plus{} 1} \frac {1}{x}\ dt$

$ \therefore \frac {1}{x \plus{} 1} < \ln (1 \plus{} x) \minus{} \ln x < \frac {1}{x}$

A nice proof of the given inequality(often called Napier's ineq) is in the book [Proof without Words] by American Mathematical Association.


take $ a\equal{}\minus{}b ,c\equal{}\minus{}d$, $ abcd>0$
then $ LHS\equal{}0$,$ RHS>0$
so your statement is wrong  :wink:

[quote="∫FaiLurE∮"]take $ a \equal{} \minus{} b ,c \equal{} \minus{} d$, $ abcd > 0$
then $ LHS \equal{} 0$,$ RHS > 0$
so your statement is wrong  :wink:[/quote]

sorry
the question is
$ a, b,c ,d$are real positive numbers


Guess the word palindrome.
Pronounced as one letter but written with three, only two different letters are used to make me.
I'm double, I'm single I'm black, blue, and gray.
I'm read from both ends and the same either way.

[hide="answer"]Eye[/hide]

Just go through the letters...

There is no reason to revive a 5-year old thread, especially when you are not adding anything to the discussion.


Asymptote
LaTeX
function
geometry
rectangle
analytic geometry
conics

how do you draw semi-figures with asy?
I know how to draw regular figures, but couldn't figure out how to draw semi-figures..

Uh?

draw((x,y)--(x2,y2)-- [etc.] );

will draw about anything. Consult asymptote.pdf inside your asymptote download for a full tutorial.

To use asymptote on the forums:

[asy]size(200); pair A,B,C,D,P; A=(0,0); B=(200,0); C=(200,150); D=(0,150); P=(124,85); draw(A--B--C--D--cycle); label("A",A,(-1,0)); dot(A); label("B",B,(0,-1)); dot(B); label("C",C,(1,0)); dot(C); label("D",D,(0,1)); dot(D); dot(P); label("P",P,(1,1)); draw((0,85)--(200,85)); draw((124,0)--(124,150)); label("w",(124,0),(0,-1)); label("x",(200,85),(1,0)); label("y",(124,150),(0,1)); label("z",(0,85),(-1,0)); dot((124,0)); dot((200,85)); dot((124,150)); dot((0,85));[/asy]

for example.

<asy></asy> is for the wiki.

And there's a specific forum from asymptote and latex questions: http://www.artofproblemsolving.com/Forum/index.php?f=123

EDIT: Okay, moved.

[color=red]Unnecessary quote removed by moderator[/color]

I know how to do those...
What I want to know is how to draw semi-circles and other semi-stuffs.

Semi-circles can be drawn with arc command. For instance,
draw(arc((0,0),2,0,180));
produces
[asy]draw(arc((0,0),2,0,180));[/asy]
As to semi-stuff, that depends on what is that stuff. There is no command "semi()" with picture argument in asy (for obvious reasons: not everything can be divided into two equal halves). If you tell exactly what other semi-stuff you want to draw, somebody will show you how to do it :)

Oh, semi?

Well, half of a square is just a rectangle. Unless you're splitting it down the diagonal?

Well, anyways, you could write a function to draw a rectangle given it's square;'s coordinates, double, I suppose. You could repeat the process for any other polygon, you wouldn't need anything except draw commands and a few conditionals.

Semi circles and ellipses are trickier...

Well, how do you close the bottom?

EDIT: Never mind.. I figured it out.
[asy]draw((-2,0)--(2,0)); draw(arc((0,0),2,0,180));[/asy]


sorry I counldn't find a website but try going to mathcounts.org and seeing if they have anything
also, i think the registration deadline was sometime in the past few days, so you might want to hurry up

The deadline is over, and I couldn't find a Maryland website either :(...

[quote="TheDeity"]Hi, do you know if there is even an official Maryland Mathcounts website? I believe there should be since 3 Marylanders placed in the 2009 Nationals, but I can't find it, so I can't register.[/quote]

Completely unrelated, but if you are referring to like national countdown, that would be 3 Marylanders in 2008 Denver.

umm, i'm pretty sure that there is a maryland mathcounts website, but it hasn't been updated for several years. anyway, I think you're supposed to register on the official MathCounts website.

yea. But I think the deadline's already over. Our teacher proabably already registered or something

no problem. Did you get a chance to register though?


Find the eighteen digit in decimal of numbers $ \sqrt[3]{37}$

Super title... Use titles that have more content, see http://www.mathlinks.ro/Forum/viewtopic.php?t=135914 .
And I don't see the fun of calculating digits ;)

Your calculator can do it if you're tricky.  :)


induction
combinatorics unsolved
combinatorics

Hello, I have the following problem:

You have the numbers from 1 to k and you take n elements of this set.
Now I want to prove that there are two distinct subsets of this set which have the same sum of their elements.
Interesting for me are the following cases:

a) k=2^(n-1)
b) k=2^n-1

I would be pleased if you could find a solution (positive or negative) for this problem.

Naphthalin

ok assume that $X=\{1,...,k\},A=\{x_1,x_2,...,x_i\},B=\{y_1,y_2,...,y_j\}$
and also $A \bigcup B=X$ such that $\sum_{r=1}^i x_r=\sum_{s=1}^j y_s$
we have:$1+2+...+k=\frac{k(k+1)}{2}$
but if we can decompose the set into two equal subsets then the sum of the elements of $X$ must be an even number
there for $4|k(k+1)$ there for $k \equiv 0,-1 (\bmod 4)$
now we prove that for every $k$ such that $k \equiv 0,-1 (\bmod 4)$ we can decompose $X$ into two equal subsets
assume that $k=4n$ or $k=4n-1$
now use induction on $n$ for $n=1$ the problem is solved
now for $k=4n$ assume that we have solved the problem for $k=4n$ now we want to solve it for $k=4n+4$
put $A'=A \bigcup \{4n+1,4n+4\},B'=B \bigcup \{4n+2,4n+3\}$
by the induction hypothesis we have $|A|=|B|$ but $|A'|=|A|+4n+1+4n+4,|B'|=|B|+4n+2+4n+3$ there for $|A'|=|B'|$
now for $k=4n-1$ assume that we have solved the problem for $k=4n-1$ now we want tpo solve it for $k=4n+3$
we know that $A \bigcup B=X$ there for number $1$ should be in $A$ or $B$ assume that $1 \in B$ now move $1$ from $B$ to $A$
now put $A'=A \bigcup \{4n-1,4n+3\},B'=B \bigcup \{4n+1,4n+2\}$
by the induction hypothesis we have $|A|=|B|$ but $|A'|=|A|+1+4n-1+4n+3,|B'|=|B|-1+4n+1+4n+2$
there for $|A'|=|B'|$
and the problem is solved for every $k \equiv 0,-1 (\bmod 4)$
now for $n>2$ we know that $k=2^{n-1} \equiv 0 (\bmod 4)$
and for $n\geq 2$ we know that $k=2^n-1 \equiv -1 (\bmod 4)$
there for we can do it for part $a,b$ with the above conditions
and the problem goes false for part $a$ for $n\leq 2$ and it goes false for part $b$ for $n<2$

Thanks for your answer, but I think you understood it in a wrong way because I didn't describe my problem exactly enough.
My problem is: I have the set $Y=\{1,\dots ,k\}$ and I take a subset $X$ with $n$ different elements of the set $Y$. Now I want to take two distinct subsets $M$ and $N$ of $X$ which have the same sum of their elements.
Interesting for me are the two following cases:
Can I find two distinct subsets $M$ and $N$ of $X$ with the same sum of their elements for every subset $X$ of $Y$ for
a) $k=2^{n-1}$
b) $k=2^{n}-1$ ?

Naphthalin

ooooooooooooooooooooopps  :oops:
sorry
Im really sorry I`ll try this one thanks for the advice

no problem bcause i calcuted it and the maximum $k$ for different $n$ is:
$k=13$ for $n=4$ instead of $k=15$, $k=24$ for $n=5$ instead of $k=31$, $k=46$ for $n=6$ instead of $k=65$ and $k=85$ for $n=7$ instead of $k=127$.
now i want to have the maximum $k$ for each $n$.

Naphthalin


we call a L-tromino a figure with 3 squares joined together in an "L" shape.

a square courtyard with side length $2^n$ units is divided into unit squares by lines parallel to its sides. one of the unit quares is reserved for a drainage outlet.

prove that the rest of the courtyard can be tiled with unborken, non-overlapping l-tromino shaped tiles.

A bit hard to explain without a figure.

Cut the courtyard in 4 parts. The part (let's consider it is the upper right) with the missing square is coverable by induction hypothesis.

Lets remove the bottom right square of the upper left part and cover it with 'L'
Lets remove the upper right square of the bottom left part and cover it with 'L'
Lets remove the bottom left square of the botom rigth part and cover it with 'L'

.........X|...........
---------l---------
.........X|X..........

There is a small 'L'-shaped hole that can be covered with an 'L' between the 3 parts.

QED  :P


$1$ The sum of several consecutive positive integers is $171966464$. what are the numbers?

$2$ In how many ways can $608608$ be expressed as the sum of several consecutive numbers?

$3$ Given $N$, in how many ways can $N$ be expressed as the sum of consecutive positive integers?

[quote="t0rajir0u"][hide="Hint"] A sum of $2k-1$ consecutive integers is divisible by $2k-1$.  [/hide][/quote]

Can you give some more details?

[hide="Explanation of hint"] The $2k-1$ numbers

$(a-(k-1))+(a-(k-2))+...+(a-1)+a+(a+1)+...+(a+(k-2)+(a+(k-1))$

Sum by Gaussian pairing to

$(2k-1)a$

Because $a-(k-1) = b$ must be positive this is better written as

$(2k-1)(b+(k-1))$ [/hide]
[hide="Second hint"] A sum of $2k$ integers is congruent to $k \bmod 2k$. [/hide]

My homework over break was something a lot like this.  It says "Call T(n) the number of ways to represent a positive integer n as the sum of consecutive positive integers."  Then it asks you to prove all of the following:
a.  T(prime) = 2
b.  T(power of 2) = 1
c.  T(prime raised to the power of a) = a+1
d.  T((p^a)(q^b)) = (T(p^a))(T(q^b)) if p and q are distinct odd primes and a and b are positive integers.

(Any time it says "prime", 2 is NOT included.)


integration
vector
linear algebra
matrix
algebra
polynomial
real analysis

Prove that there exist real coefficients not all zero, such that:

$ (a_0 \plus{} ... \plus{} a_{n\minus{}1})^2 \equal{} n^2 \int_0^1{(a_0 \plus{} a_1 x \plus{} ... \plus{}a_{n\minus{}1}x^{n\minus{}1})^2 dx}$

where n is integer at least 2 ( for example in case n = 2 we can find vector $ a_0,a_1 \equal{} \minus{}3a_0$ )

Let's rephrase this as a linear algebra problem.

Index vectors in $ \mathbb{R}^n$ and $ n\times n$ matrices so that the indices run from $ 0$ to $ n\minus{}1.$ Let $ H$ be the $ n\times n$ matrix whose $ (i,j)$ entry is $ \frac1{i\plus{}j\minus{}1}.$ Let $ J$ be the $ n\times n$ matrix with every entry equal to $ 1.$ We're asked to show that for some nonzero vector $ a,$
\[ a^T\left(n^2H\minus{}J\right)a\equal{}0\]
This requires precisely that the matrix $ n^2H\minus{}J$ be singular.

For the $ n\equal{}2$ example that mareleG gave, we have that
\[ n^2H\minus{}J\equal{}\begin{bmatrix}3&1\\1&\frac13\end{bmatrix}\]
and it is clear that that is a singular matrix and that $ a\equal{}\begin{bmatrix}1&\minus{}3\end{bmatrix}^T$ lies in its null space.

For $ n\equal{}3,$ $ n^2H\minus{}J\equal{}\begin{bmatrix}8&\frac72&2\\ \frac72&2&\frac54\\ 2&\frac54&\frac45\end{bmatrix}.$

A brute force calculation shows that this is also a singular matrix.

Now, as for the general case ...

this was my approach too, the next step is to prove singularity of that matrix.
hint: same technique used for Cauchy matrix, with elements $ 1/(i \plus{} j)$ or $ 1/(x_i \plus{} y_j)$
using this problem one can find a solution for problem 4 - SEEMOUS 2008, different from official solution

That $ n^2$ really reminds me of the result [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=146373]here[/url] that $ j^T H^{ \minus{} 1}j \equal{} n^2$, where $ j$ is the $ n\times 1$ vector of all ones.

OK, what if $ a \equal{} H^{ \minus{} 1}j$? We have
$ a^T(n^2H \minus{} J)a \equal{} j^T(H^{ \minus{} 1})^T(n^2H \minus{} jj^T)H^{ \minus{} 1}j$
$ \equal{} n^2 j^T H^{ \minus{} 1} H H^{ \minus{} 1}j \minus{} j^T H^{ \minus{} 1}j j^T H^{ \minus{} 1}j$
$ \equal{} n^2 j^T H^{ \minus{} 1}j \minus{} (j^T H^{ \minus{} 1}j)^2 \equal{} n^4 \minus{} n^4 \equal{} 0$

Second thought: the connection is even closer than I thought at first. The appropriate polynomial $ A(x)\equal{}a_0\plus{}a_1x\plus{}\cdots\plus{}a_{n\minus{}1}x^{n\minus{}1}$ has the property that $ \langle A(x),x^k\rangle\equal{}1$ for each $ k\in \{0,1,\cdots,n\minus{}1\}$ (or it's a multiple of that one)

It seems that Colombian Olympiad 06 problem is the same as SEEMOUS problem :)
Hilbert matrix is a particular case of Cauchy matrix $ 1/(x_i \plus{} y_j)$
My solution relies on elementary proving that matrix $ 1/(x_i \plus{} y_j) \minus{} 1$ is singular when $ x_1 \plus{} ... \plus{} x_n \plus{} y_1 \plus{} ... \plus{} y_n \equal{} 1$ and choose $ x_i \equal{} i/n^2, y_j \equal{} (j \minus{} 1)/n^2, i,j \equal{} 1,...,n$


Prove that if x^2 + xy + y^2 is divisible by 10 it's divisible by 100.

[quote="miyomiyo"]How does this prove it?[/quote]
Since $10|x$, $100|x^{2}$.

Also, $100|y^{2}$.

So $x^{2}+xy+y^{2}=100(\text{blah})$.

[quote="rnwang2"][quote="miyomiyo"]Why is $x$ divisible by $10$?[/quote]
Since $10|(x^{2}+xy+y^{2})$, it divides each term of the sum.[/quote]

um, that's not true with 3,7,13,19,31,37,43,61... (primes that are 1 mod 6). You have to show that it's actually true for 2 and 5.

Here's a proof with two cases.

Case 1: $x=y$. Then $3x^{2}\equiv 0 \pmod{10}$, so we conclude that $x \equiv 0 \pmod{100}$ and the result follows from there.

Case 2: $x \neq y$. Rewrite the expresion as $\frac{x^{3}-y^{3}}{x-y}\equiv 0 \pmod{10}$. Thus $\frac{x^{3}-y^{3}}{x-y}=10k$ for some integer $k$, or $x^{3}-y^{3}=(x-y)10k$, from which it follows that $x^{3}\equiv y^{3}\pmod{10}$.
This last congruence implies $x \equiv y \pmod{10}$, so our original equation becomes $3x^{2}\equiv 0 \pmod{10}$, from which we get $x^{2}\equiv 0 \pmod{10}$ and then $x \equiv 0 \pmod{10}$.
Thus we actually have $x^{2}\equiv 0 \pmod{100}$, so that we can write $x^{2}+xy+y^{2}\equiv 3x^{2}\equiv 0 \pmod{100}$.

I must have been hallucinating when I wrote those posts.  :blush:

[quote="GlenthemanN"]I cant find it anywhere, but what does the | mean when you post things such as $10|x$ etc..

thanks  :oops:[/quote]

It means that 10 divides x, or x is divisible by 10.


This is to my friend, pvthuan, but all are welcome to submit solution.
For any positive real numbers $a,b$ and $c$,
\[\frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ca}+\frac{1}{c^{2}+ab}\le\frac{a+b+c}{ab+bc+ca}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right) \]

Here's my proof.

Using Jensen's ineq on $f(x) = \frac{1}{x}$.

We get $ \sum_{cyclic}\frac{1}{a^{2}+bc}\leq \frac{9}{\sum_{cyclic}a^{2}+\sum_{cyclic}ab}$.

and $ (a+b+c)\sum_{cyclic}\frac{1}{a+b}= 3+\sum_{cyclic}\frac{c}{a+b}\geq \frac{9}{2}$ (nebit's ineq)

So, it is enought to prove that

$\displaystyle ab+bc+ca \leq a^{2}+b^{2}+c^{2}$.

This is self-evident.

Q.E.D.

[quote="Chang Woo-JIn"]Here's my proof.

Using Jensen's ineq on $f(x) = \frac{1}{x}$.

We get $\sum_{cyclic}\frac{1}{a^{2}+bc}\leq \frac{9}{\sum_{cyclic}a^{2}+\sum_{cyclic}ab}$.

[/quote]

I think it's invalid. the sign whould be reversed.

[quote="Chang Woo-JIn"]
We get $\sum_{cyclic}\frac{1}{a^{2}+bc}\leq \frac{9}{\sum_{cyclic}a^{2}+\sum_{cyclic}ab}$.
[/quote]
You have mistake here. ;)

My proof involves very little computation and no expansion ;-)
Note that this inequality is stronger than
\[\frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ca}+\frac{1}{c^{2}+ab}\le \frac{3(a+b+c)^{2}}{2(ab+bc+ca)^{2}}\]
which is mentioned at http://www.mathlinks.ro/Forum/viewtopic.php?t=116853


Hi guys, I've got a project to write on a maths topic, and am looking to LaTeX to write it. It's not my first encounter with it, but I'm no pro by any stretch of the imagination :s
Firstly, I have a title page - fine. Before moving onto the contents page, I want a page with just a couple of quotes, to set the scene a bit, you know. So what I want is a page with a couple of quotes which are roughly near the middle of the page vertically. I tried using the command \vspace(2.5in) and then the quote environments etc, but the space doesn't appear. I tried preceding it with \text{ } which worked, so I guess the vspace needs something (in my case at the top of the page) away from which to set my quote. Ok so what I did works, but it hardly seems "correct" or neat. Is there a less clumsy way to do this?
The next trouble I have (which I haven't yet been able to solve at all) is in page numbering. Ideally I want no page numbering on the title page, then start counting in Roman numerals starting with said quotes page as page i, and then when I get to the "meat" of the project start counting in Arabic from 1. Does that make sense, what I'm trying to do? Can anyone help me do this?

Many thanks, and if it helps, you're all such lovely people ;)

Hi, I found this post from Google because I was looking for a solution to the vspace not showing up at the top of a page. It's been a few weeks since you posted, so you probably found what you need, or it's too late now, but I can help you with page numbering if you haven't found it yet.

To avoid numbering on your title page, try enclosing it in a \begin{titlepage} ... \end{titlepage} environment.

To get roman numerals starting at i on the page after this, you just need \pagenumbering{roman}.

Then what you want is \pagenumbering{arabic} \setcounter{page}{1}.

I hope this helps.


I noticed that a user named [u]GAMEBOT[/u] is posting all the Alcumus Problems and Video Discussions. Why? Is anyone else allowed to post some? ? ? ?

GameBot is a bot that was made to do exactly that, when someone hits "Discuss Problem on Forum" or "Discuss Video on Forum".


algebra
polynomial
Vieta
quadratics
quadratic formula

hello.  this is an easy problem.

Let $ P(x)\equal{}x^3\minus{}8x^2\plus{}mx\plus{}n$ for some $ m,n\in\mathbb{Z}$.  It is known that $ P(x)$ has three distinct positive roots and that exactly one is an integer, which is the sum of the other two roots.  How many values of $ n$ are possible ?

[hide]
Let the roots be $ a,b,c$ with $ a,b \not\in \mathbb{Z}$ and $ c, ab, (a\plus{}b) \in \mathbb{Z}$ and $ a\plus{}b\equal{}c$

Using Vieta's, and the facts that integers are closed under addition

$ a\plus{}b\plus{}c \equal{} 8 \Longrightarrow a\plus{}b\equal{}c\equal{} 4$

$ b\equal{}4\minus{}a \Rightarrow a(4\minus{}a) \in \mathbb{Z}$

Since $ a(4\minus{}a) \equal{} 4\minus{}(a\minus{}2)^2$ we know $ (a\minus{}2)^2$ is an integer, but $ (a\minus{}2)$ is not

So $ a\equal{} 2 \plus{} \sqrt{k}$ where $ k \in \mathbb{N}$ and is not a surd.

Hence $ n \equal{}\minus{}abc\equal{} \minus{}4(2\plus{}\sqrt{k})(2\minus{}\sqrt{k}) \equal{} 4(k\minus{}4)$ which has infinite solutions[/hide]

[hide="solution"]
WLOG let $ x_1>x_2>x_3>0$.
From Vieta's we have:

$ x_1\plus{}x_2\plus{}x_3\equal{}8$

$ x_1\equal{}x_2\plus{}x_3 \implies x_1\equal{}4$

$ x_1x_2x_3\equal{}\minus{}n \implies 4x_2x_3\equal{}\minus{}n$

Substituting $ x_3\equal{}4\minus{}x_2$ gives us $ 4x_2^2\minus{}16x_2\minus{}n\equal{}0$ (we would get same quadratic if we substituted $ x_2$) thus

$ x_{2,3}\equal{}\frac{4\pm \sqrt{16\plus{}n}}{2}$

Since roots are positive $ 4\minus{}\sqrt{16\plus{}n}>0 \implies n<0$.

Similarly $ 16\plus{}n \geq 0 \implies \minus{}16 \leq n<0$.

Since $ x_2$ and $ x_3$ can't be integers we have to take out $ n\equal{}\minus{}12$ and $ n\equal{}\minus{}16$ (which would give $ x_2\equal{}\{3,2\}, x_3\equal{}\{1,2\}$).

If we return to starting polynomial and plug in $ x\equal{}4$ we get $ 4m\plus{}n\equal{}64$ or $ m\equal{}16\minus{}\frac{n}{4}$ so $ n$ must be multiple of $ 4$ which gives us $ n\equal{}\{\minus{}4,\minus{}8\}$ for total of $ \boxed{2}$ values.[/hide]

[quote="maths486"]hello.

Let $ P(x) \equal{} x^3 \minus{} 8x^2 \plus{} mx \plus{} n$ for some $ m,n\in\mathbb{Z}$.  It is known that $ P(x)$ has three distinct positive roots and that exactly one is an integer, which is the sum of the other two roots.  How many values of $ n$ are possible ?[/quote]

EDIT: Solution not consistent with known solutions.

My answer doesn't agree with the other conclusions.

[hide]Let a, b be integers
Let n = ab
Because all the roots are positive, $ n \ne 0$
Then a $ \ne$ 0, b $ \ne 0.$

Let x = a be the only positive integer root of P(x) = 0.

Divide P(x) by the corresponding factor (x - a) and get the quotient
$ x^2 \plus{} (a \minus{} 8)x \plus{} m \plus{} a(a \minus{} 8)$ and a remainder of $ ab \plus{} am \plus{} a^2(a \minus{} 8)$.

Set the remainder equal to zero for this to have its three factors and solve for m:
$ m \equal{} \minus{} a^2 \plus{} 8a \minus{} b$.

Suppose you want to factor the quadratic in x, and solve for its roots:

$ x \equal{} \frac {(8 \minus{} a) \pm \sqrt {(a \minus{} 8)^2 \minus{} 4(1)[m \plus{} a^2 \minus{} 8a]}}{2a}$

$ x \equal{} \frac {(8 \minus{} a) \pm \sqrt {a^2 \minus{} 16a \plus{} 64 \minus{} 4m \minus{} 4a^2 \plus{} 32a}}{2a}$

$ x \equal{} \frac {(8 \minus{} a) \pm \sqrt { \minus{} 3a^2 \minus{} 16a \plus{} 64 \minus{} 4m}}{2a}$

Substitute $ m \equal{} \minus{} a^2 \plus{} 8a \minus{} b$ and simplify:

$ x \equal{} \frac {(8 \minus{} a) \pm \sqrt {(a \minus{} 8)^2 \plus{} 4b}}{2a}$  . . . (**)

Besides x = a (the integer root), these second and third roots from
this quadratic formula (**) add up to

$ \frac {8 \minus{} a}{a} \equal{} \minus{} 1 \plus{} \frac {8}{a}$

What integer values may we choose for a?

Potentially, the variable a = 1, 2, 4, 8, as it is a positive integer, and it
divides 8.

Look at their placements into (**).

When a = 8, one of the x-values becomes $ \frac { \minus{} \sqrt {4b}}{16}$, and so
it violates having three positive roots.

In (**), if a = 1, 2, or 4, then with suitable negative b-values, both x-values
are positive along with their respective a-values.

I am getting three n values, then, for my answer.

EDIT:  This solution is not consistent with the known soution.
[/hide]

Please check my work.

[quote="Arrange your tan"]
Please check my work.[/quote]
But you didn't bother to check ours..?

Anyway I missed that the roots were positive,

$ (x\minus{}4)(x\minus{}2\plus{}\sqrt{k})(x\minus{}2\minus{}\sqrt{k}) \equal{} x^3 \minus{} 8x^2 \plus{} (20\minus{}k)x \minus{} 4(4\minus{}k)$

So $ n\equal{}\minus{}4(4\minus{}k)$ with $ k$ an integer, but not a surd

Since roots are positive $ 2\minus{}\sqrt{k}$ is positive, therefore $ k\le 4$,
 but $ 1, 4$ are a perfect square, and $ k\equal{}0$ implies two roots aren't distinct, so $ k\equal{}2,3$

And $ n\equal{}\minus{}4, \minus{}8$

hello

[hide]

I think we have the correct answer by Flame.  We get $ 2$ possible values for $ n$ since the problem also restricts $ m$ to be an integer, and also since all the roots should be positive.  We will get $ a\equal{}4$ for the integer root and we must therefore have $ 2\plus{}\sqrt{b}$ and $ 2\minus{}\sqrt{b}$ as roots also, where $ b\in\mathbb{Z}$

$ 2\minus{}\sqrt{b}>0$ only when $ b\equal{}2,3$.  We can't have $ b\equal{}0,1$ since we only have one rational (integer) root.  There are $ 2$ values of $ b$ which will correspond to two values for $ n$.

[/hide]

[quote="maths486"]  this is an easy problem.

[/quote]
In the future, leave off subjective descriptions such as these
to try to entice problem solvers.


linear algebra
matrix
linear algebra unsolved

Let $A,B$ two real matrix of order 3, such that $rank(A)=rank(B)=1$.

  Prove that $det(A+B)=0$. ;)

$Ker(A)$ and $Ker B$ are two planes. So $Ker(A) \cap Ker(B) \neq \{0\}$. Hence there exists $v$ such that $Av=Bv=0$. Conclusion follows ?

$\text{rank}(A+B)\le\text{rank}(A)+\text{rank}(B).$

This is easy to see: the union of a basis for the columns of $A$ and a basis for the columns of $B$ must span the column space of $A+B.$

So $\text{rank}(A+B)\le2$ and hence $A+B$ is singular.

This is basically the same argument that alekk gave, just stated a little differently.

[quote="alekk"]$Ker(A)$ and $Ker B$ are two planes. So $Ker(A) \cap Ker(B) \neq \{0\}$[/quote]

why these two plane are not parallel ?

Since $ 2+2>3$, Moubi! Just apply Grassman theorem.


Let $ a,b,c\ge 0$ s.t. $ ab \plus{} bc \plus{} ca \equal{} 3$. Prove that:

$ (a \plus{} b \plus{} c)(a^2b \plus{} b^2c \plus{} c^2a)\ge 9$

Edit: oops, a bit inattentive. I see it's too easy. Please don't write more replies in this topic.

[quote="Inequalities Master"]Let $ a,b,c\ge 0$ s.t. $ ab \plus{} bc \plus{} ca \equal{} 3$. Prove that:

$ (a \plus{} b \plus{} c)(a^2b \plus{} b^2c \plus{} c^2a)\ge 9$[/quote]
But it's Cauchy-Schwartz. :wink:


$ a_{0} \equal{}5000$, $ a_{n\plus{}1} \equal{} \frac {a_{n}^2 } {a_{n} \plus{}1}$

Then find the value of $ [a_{2007}]$.

I think it is easy for you. Please solve it!

from the condition we have $ a_{n \plus{} 1} \equal{} \frac {1}{a_n \plus{} \frac {1}{a_n}}$ which means $ \frac {1}{a_{n \plus{} 1}} \equal{} \frac {1}{a_n} \plus{} a_n$
now let {$ u_k$}={$ \frac {1}{a_k}$} we get the same problem as the problem you 've posted http://www.mathlinks.ro/viewtopic.php?t=205618


Ανοίγω αυτό το topic,ώστε οι παλαιότεροι να βοηθήσετε στη λύση προβλημάτων διαχείρησης των λογαριασμών μας και ό,τι άλλο προκύψει.Ιδίως για μας τους καινούργιους...
Και ξεκινάω με τα πρώτα ερωτήματα:
1)πώς θα εμφανίζεται η υπογραφή μου;
2)γιατί όταν στέλνω pm εδώ και κάτι μέρες μου τα βγάζει στo outbox και όχι στo sentbox;Περιττό να σημειώσω ότι δεν έχω πάρει απάντηση σε κανένα!(Σημ.:στην αρχή μου τα έβγαζε στο sentbox και επικοινωνούσα κανονικά)

Τα μηνύματα που εμφανίζονται στο outbox ειναι μηνύματα που δεν έχουν φτάσει ακόμη στον προορισμό τους, είτε γιατί ο μελλοντικός παραλήπτης δεν έχει ακόμη συνδεθεί είτε γιατί για κάποιο λόγο δεν έχει ακόμη διαβάσει το μήνυμα. Μόλις παραλάβει το μήνυμα και το διαβάσει θα μεταφερθεί αυτόματα σε σένα στον φάκελο sentbox.
Αν τώρα είσαι σίγουρος ότι ο παραλήπτης έχει λάβει και έχει διαβάσει το μήνυμά σου και παρ'όλα αυτά συνεχίζει να εμφανίζεται στο outbox, τότε μάλλον υπάρχει κάποιο πρόβλημα με το λογαριασμό σου.

Για να εμφανιστεί τώρα η υπογραφή σου πηγαίνεις πάνω δεξιά εκεί που λέει profile. Από τις επιλογές στα αριστερά πηγαίνεις Preferences-->Posting a message και στο Always attach my signature τικάρεις το Yes (...αυτά εφόσον βέβαια έχεις βάλει μια υπογραφή από το Profile-->Signature control panel) .

Όλα θέμα συνήθειας είναι και λίγο ψαξίματος. :)

[quote]Για να εμφανιστεί τώρα η υπογραφή σου πηγαίνεις πάνω δεξιά εκεί που λέει profile. Από τις επιλογές στα αριστερά πηγαίνεις Preferences-->Posting a message και στο Always attach my signature τικάρεις το Yes (...αυτά εφόσον βέβαια έχεις βάλει μια υπογραφή από το Profile-->Signature control panel) .[/quote]

 :blush: Το είχα ψάξει,το είχα φτιάξει αλλά έφυγα απ' το παράθυρο χωρίς submit!Τhx,anyway(και για τα pm).
Σημ.:Ο kostas έφταιγε που τα διάβαζε τόσο γρήγορα και νόμιζα ότι πήγαιναν απευθείας στο sendbox :lol:

ετσι ε??? :D


This is problem E2 from the AMY packet.  It should be "Easy," but I'm not seeing it:
Is there a function from the real numbers, excluding zero, to the real numbers excluding zero such that:
f(f(x)) = 1/x

for all x in the domain?

I don't have said packet and I didn't attend the program, but

[hide]Yes. Define $ f(1) \equal{} \minus{}1, f(\minus{}1) \equal{} 1$ and for all $ x > 1$,
$ f(x) \equal{} \minus{}x$
$ f(\minus{}x) \equal{} 1/x$
$ f(1/x) \equal{} \minus{}1/x$
$ f(\minus{}1/x) \equal{} x$[/hide]

you might want to think of it as making a piecewise function that breaks up the intervals like this:

$ ( \minus{} \infty, \minus{} 1)$
$ \{ \minus{} 1\}$
$ ( \minus{} 1,0)$
$ (0,1)$
$ \{1\}$
$ (1, \infty)$