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I want to discuss the problem that undermines and ruins music like nothing else does on Earth. At the same time, this problem is still “invisible” for many pianists.

Question : Can a pianist be scared of his/her own ability to play beautifully and suffer because of it ? To help you answer, I would like to remind you that beautiful music is extremely powerful. It can change the minds and hearts of any human being and even of entire nations. A lot of people were executed in my country Russia and elsewhere for performing the “wrong music” or for performing the right music in the wrong way.

 (In spite of deadly jeopardy many Russians enjoyed taking this risk. When serving my term in the Soviet Army, I met a private who amazed his comrades with performances of the first bars of “The International” (Communist sacred anthem). He was extremely lucky that nobody reported his vocal exercises to the Secret Police, because he “sang” them from “a very wrong” part of his body.

Renowned composer Rodion Schedrin wrote the Symphonic Poem “Naughty Tunes”. It was based entirely on the tunes of songs for which the singer was always “awarded” the standard penalty of “10 years in jail”. Schedrin was quite likely to have received those “10 years” if the political climate in USSR had worsened.)

Under the influence of beautiful music, people become better : they do not want to be mean, to participate in injustices, to rob, to kill, to lie, etc. Consider this: what person in power would want this to happen to people under his authority? Who would carry out his plans and orders ? This is the root of the problem. This is why the best performers and composers are kept or driven away from conservatories and universities, why all the favourites of audiences are always “eliminated” in the very first round of any competition, why they are banned directly or indirectly from the opportunity to transmit their thoughts and feelings to others through their beautiful music. I am not talking about myself. I am not the best and my situation is not the worst, but I know plenty of others who are much better musicians than I and who have not had even my opportunities.

Story No 1. After my performance at the All-Russia Competition, an old gentleman (composer Foret) whom I had never met, embraced me, saying “I voted for you because you played like our pianists did in the days of my youth, before communists seized power in 1917. Be careful : they always destroy people like you !”
His words frightened me : : SOME  members of my family who had committed no crimes were executed  simply in accordance with the Communist Party drive to wipe out the “old intelligentsia”. The word “intelligentsia” comes from the original Russian word borrowed by many languages with unfortunately the modified definition : “smart, well-educated, etc.”. Originally, it meant only “ the part of the population that thinks independently”. It signifies that any independently thinking street cleaner or taxi driver is “intelligent” but not necessarily every scientist or professor. No totalitarian regime tolerates this “part of the population”; they always root out any “intelligentsia”. Indeed, hardly any type of political regime in the world likes them.

My head was full of alarming thoughts : Did this fellow say the same things about me during the Board of Adjudicators’ discussion ? Probably yes, because another member of the Board Prof. K.Adjemov formerly friendly towards me, refused to speak to me afterwards and was furiously against me during the discussion, I was told. I remembered that a very old relation of a friend told me that my piano playing reminded him of young Rachmaninov (a personal friend of his) although “Rachmaninov did not play like this in the latter period of his life on his recordings”.

I knew well that Rachmaninov, Prokofiev, Horowitz, Hoffman, Metner, Stravinsky and plenty of other pianists escaped from Russia after 1917 : “these dregs had sold their motherland for a dirty sheaf of dollars”. But how could my playing resemble that of those traitors ? I had received excellent marks for my solid knowledge of textbooks like Yury Kremlev’s “Piano Technique”. I always remembered that “piano playing is a clear political issue”, that “piano technique is a powerful weapon of Soviet propaganda and we must use it in full strength against our ideological enemies”. I felt no joy from that victory ( I won a “pass” to represent Russia in the All-USSR competition) . I realized instead that something was very wrong within me and that I had to do something about it urgently.

Story No 2. Many years later, when I arrived at a very luxurious sanatorium ahead of  a concert I was to give with two singers, a very strange looking man was on the stage where we were supposed be preparing. Obviously unauthorized to be there, he was playing on a huge concert Steinway in just a hospital gown with no trousers. I had to ask him to leave the stage and hall and be outside until we will start our concert. So, I headed towards him.

However, after two steps in his direction, I stopped and was absolutely transfixed to hear his rendition of Tchaikovsky’s Meditation, a piece that I had never much appreciated previously. The two singers had the same reaction. Instead of preparing our own concert, we listened for over an hour to stunning music from this trouser-less pianist. It was very difficult for us to perform on the same stage immediately after this improvised recital of one the of the greatest contemporary pianists : Oleg Boshnyakovich.

Later Boshnyakovich invited us to his own concert which we heard on the same stage and piano… it was a very banal and even a bit boring concert with the same program. “Why did you not play the same way for them that you did for us ?” I asked Boshnyakovich. “Who would ever allow me to play in this fashion ?” was his reply. I knew the bitter reason of these words. For almost half his life, this great pianist was forbidden to play the piano by other “pianists in power” because of his “complete incompatibility with the Soviet style of piano playing.” Instead he performed whistling on stage for many years to make a living.

Story No 3. In St. Petersburg I heard Svyatosolav Richter ask my teacher Vladimir Nielsen about the fingering for Chopin’ study No6 Op25. Later they argued about “Bydlo” from Pictures at an Exhibition by Mussorgsky and discussed Richter’s last recorded version of Rachmininov’s 2nd Concerto. Nielsen told him “Many years ago you played this concerto one hundred times better than Rachmaninov himself, because you were inspired by this music which Rachmaninov was not at the time of his recording. Rachmaninov was obviously rather tired of playing this great work over and over”. Nielsen however did not like Richter’s last recorded version.

I felt myself confident enough to visit Richter at his home where I asked him : why is my teacher still banned from concerts in Moscow despite the end of political repressions ? Nielsen was always on the black list of Soviet censors; even his most famous students were not allowed to mention his name in their professional résumés. Because of someone’s report to the secret police, he was forbidden to play at the Chopin Competition in spite of his “pass No 1” at the All-Union selection, and so on.

Richter told me

TO BE CONTINUED BELOW

CONTINUATION
Richter told me that he was unaware of this situation. He thought that Nielsen simply did not want to play in Moscow. “I will say something to the right people” he promised.

A few weeks later, Nielsen received an “unexpected” invitation and was able to play a concert at the Tchaikovsky Concert Hall in Moscow which was previously a drama theatre with poor acoustics for the piano. The best pianists in Moscow never play there. When I read the article about his concert in “Sovetskaya Musyka” (Soviet Music) written by the Docent of Moscow conservatory Leonid Zhivov, I learned the answer to my question “Why ?”. Zhivov wrote in this article “All pianists in Moscow should learn from Nielsen because his music is “alive”, not “dead”…

However, when I pointed to Nielsen that many of his phrases had “non-standard”, reversed accentuation and asked: “Don’t they make your music alive?”, he only ordered: “Shut up and never said this again anywhere!”
I can tell plenty of stories on this topic, but is it not strange that this very specific aspect of piano technique can cause so much passion and emotion ?

The reply depends on the answer to another question : what is the goal of Art in general ? If we say “Our goal is to reflect the whole world with all its beauty, love, passion, drama, and hope, etc. in our music”, then all possible ways are good except the boring ones.

However, of all the means at our disposal, only one is acceptable : the one which is “approved by the authorities” whose only goal is to make perfect idiots out of normal people. The authorities want obedient robots “ready to carry out any order of ANY government”, who are preoccupied solely with their jobs, watching TV and drinking alcohol and who will die or kill “no questions asked”. From the authorities’ point of view, music is used as a form of brainwashing. To convert people into robots they must be saturated with special “robotic music”. The essential ingredient of robotic art is that “there is no choice”,  “there is no another way”  (to quote from a real song that Soviet people had to repeat and listen to over and over again) other than  the right one provided by the authorities. The pattern “LOUD soft LOUD soft LOUD soft” is absolutely indispensable for this task. This pattern works like a narcotic : look at people sitting in their cars and shaking their heads in time to the beat from their stereo speakers, at mass shows, military parades, etc.

Is the pattern “soft LOUD soft LOUD soft LOUD” any better ? It is not better at all. Everything on Earth is bipolar. There is North and South, left and right, heaven and hell, good and evil, etc. We have two eyes, two ears and nostrils, arms and so on. Imagine that artists were forbidden to draw “anything left” on their pictures: no left legs, no left ears, hands and no clothes with the buttons on the left side; no left-hand traffic portrayed in pictures of countries with motorcars, only right, right and right again.

This is exactly the situation in music. Nowadays, there are more than two generations of pianists who do not suspect nor even imagine that the music Left (one TWO three FOUR) and Right (ONE two THREE four) are equally legal and both absolutely necessary for any normal performance despite official theory in the matter. Listen carefully to any great pianist : no one plays only that “march-like” style; both patterns always support each other. This is what makes their music come alive. But WHAT do we have to teach our students instead ? Only what is written in our textbooks and dictionaries full of false rules and statements. We are only making idiots of our audiences, our students and ourselves when we blindly follow them !

Amazingly enough, some Western musical educators are much more “Soviet” than in Russia. The Russian word “soviet” for many centuries meant “advice” or “council” but since 1917 ”Soviet” has come to be understood as “Totalitarian.” However, no single “Soviet hard liner” hit upon the idea of forcing all musicians in the country to play wrong notes in Mozart or Handel just because some his illiterate editor was unable to print music properly; or of preventing pianists from using pedal; or of appointing one single institution in USSR to take exams from all the students in the country; or of making a list of which Beethoven Sonatas and Bach Fugues are “allowed” and which are “not” at exam performances. In this respect, Western educators have undoubtedly eclipsed the most totalitarian dreams of all Soviet musical evil - doers. They seem to sincerely believe all the written dogma of the so-called “Soviet piano school”, whereas the best Russian performers and teachers always laughed at it (when not in the exam rooms of course) and were great and brilliant not because but in spite of this dictatorship. To paraphrase a Russian saying : “He who knows and can – plays, He who knows but cannot –teaches, He who does not know and cannot – writes instructions for those who teach and play”.

The billions that Communists invested in musical education were not just charity from a bloody dictatorship. It was a battle for minds. If THEY won, we lost. Musical education in the West could not compete with the excellence of Soviet pianists and instead of selectively borrowing the best parts of the Soviet system; it has simply surrendered with discretion and copied only the worst part of this system: stupid dictatorship ( but without Russian inherited resistance to any dictates). Look around and you will realize that you are still living under musical “Soviet” occupation; many Western pianists are still “musical slaves” of a system that died many years ago whether or not they understand the meaning of the word “Soviet”.

In music, many of us are not as free as we would like to believe because we lack courage. We seem to adulate pianists who play faster because fast tempo appears to be the most important thing in music (confer e.g. “examination requirements”). This is nonsensical. In any language (music is only one of them) fast tempo is the LEAST important thing. Are the words “I love you”, “I hate you”, “I am your friend”, “I want to console you” or other words more effective because they are spoken “Presto” rather than “Allegro” or “Andante” ? Do you know any actor on the planet who is admired just because he/she speaks fast ? I doubt it. Fast speech is more of a handicap than an advantage for any actor, politician, priest, etc.

There is no reason to be scared of “robots” – they get high marks only from other “robots”, from their “maintenance people”. Audiences do not like them or their recordings. Robots are rather ridiculous because they can not play even two notes on their own without someone else’s CD downloaded in their head, and they are absolutely unable to play any piece slowly (because CD can not play slowly, of course). Ask any “robot” to play their piece in only ¼ of normal tempo, and robot immediately will “declare bankruptcy”: slow tempo is incompatible with an empty head. As soon as fair and honest competitions are opened to pianists of all styles, all robots will die off quickly.

Would you be unhappy if you were selling normal good pictures while someone else sold “one-eyed”, “one-handed”, “one-eared” pictures ? Of course not. This type of competition would be very useful for your business: people would compare and appreciate your art even more.

However, suppose you were the salesman of this “one-sided” art ? What would you feel about your competitor selling normal art ? What would you want to do to him and his wares ? This is the reason why so many pianists enjoy being sheep-dogs of our worldwide, global “musical concentration camp” packed with “one-sided prisoners”. They are always ready to attack anyone who dares to step off the path determined by our warders. (The standard phrase of escort to prisoners in Soviet Russia was “Should you make a single step to the right or to the left, it will be considered attempted escape and you will be fired upon immediately without warning.” One of our artists was shot in the arm because he raised his hand to touch a blossom on a tree above his head while he was being led in a line of prisoners.) Read a few posts at random in any forum and you will spot these pianists easily.

This situation explains why the “most overrated pianist ever” is detested so much by all “one-sided musicians” which is why he is the best loved pianist in Russia. All Russians have had more than enough of slavery and they appreciate freedom. Van Clyburn, a student of Russian refugees and who obviously played in a pre-1917 fashion, was and still is a symbol of Freedom in music for my country despite anything he lacked in technique.

Listen to any musical phrase of his and you can always hear the living variety of accentuations, just as in normal human speech of any language. No one “uniform accentuation” can compete with this way of “speaking and singing on the piano”. Bankruptcy is guaranteed. Many “kings of all the pianists of the universe” played in Russia after Clyburn and had empty seats in the hall at the end of their concerts, because they were only slightly different models of “robots” in a country “overstocked” with them.

Is there any way to end this “one-sided slavery” and become a normal “two-sided” pianist ? Yes, of course. Anyone can in just a few days. All you need to do is to play for a short period of time each piece of your repertoire in “reversed”, “left” instead of “right” accentuation. It is terribly difficult the first day and especially the first hour, but you will quickly come to appreciate the change. The next step is to play once with “left” then with “right” accentuation, then continue alternating. In this way, you will get a chance to choose the best accentuation for each phrase and store it in your memory. If you manage to play any piece “both ways” easily, then you are FREE for ever. In any of your phrases you will always play the accentuation that you prefer rather than because it is the only one you know
.
A big advantage of this approach is that your performances will always be fresh. You will be amazed to hear that you play differently each time. Your combinations of Right and Left accentuations never will be repeated because they depend on the state of your mind and soul at the time of performance. Each time it will be a kind of improvisation, and you never will be bored with any of your music.

Don’t be scared with this fact; remember words of the greatest pianists: “You have no right to play any piece on a rainy day the same way as you played it on a sunny day”.  This approach will boost your creativity and you will be interested and inspired even when playing your old repertoire. Any audience will immediately feel and appreciate your style: everyone likes freshly cooked meals; no one likes “canned” or stale ones, especially in music.

To support this post, I have recorded a few randomly chosen pairs of absolutely different pieces with “Right” and “Left” accentuations. If somebody would be interested – I will send them for listening. To avoid being influenced by any previous experience, I have chosen the music to record from my students’ repertoire which I rather sight-read. I have never worked on any of it before as seriously as I would normally for myself. Some of my students play better than I because no one is a slave in my class. They are always FREE to choose ANY accentuation they prefer. As for me, I am not completely free yet, because my “slavery” lasted too long. However, I remember the words of Chekhov that the most important thing for any human being is : “to squeeze out the slave from ourselves drop by drop each day”.

 Can ANYONE ELSE do the same and post recordings of the same pieces or others with BOTH types of accentuation: Left and Right? Who will take up my gauntlet? Prove for yourself now, that the title of this post is NOT about you!

I hope you succeed better than I. I wish this sincerely because I do NOT want to feel alone amongst “one-sided” robots. I am sure a lot of “musical humans” are alive out there somewhere. Make yourselves known, please !

In anticipation,

Vladimir Dounin

PS Try to do the same with any of your songs that I did on my "one - sided recordings": simply replace ALL your accents with their "mirrored reflections". (Do you know how great is look of the house with mirrors and how poor it is without any of them on the wall). Not for ever, just for a few times, for a change. It will give you the greatest opportunity of choice. Instead of unreasonable, suggested by our evil-wishers denial you will be always able to take fully INFORMED DECISIONS about the appropriate Left or Right accentuations in each of your phrases in the future.

Play in any group of notes (four eights or sixteenth, bar, phrase etc) the very first SLIGHTLY (read my FUR ELISE) softened and the last one SLIGHTLY stressed. That's all. It is so simple! And you will be amazed: how fresh and new for you will be any of your old pieces.

To encourage you and all the others to try this way, I pasted here below a real quote from the recent letter of my "distant student" from Australia (I never met her in real yet).

Standard HeaderHide Pane
Eleanor Sheldxxx <xxxxbird_22@hotmail.com>
View Saturday, September 16, 2006 8:24:45 PM

Thank you so much for sending me those. I've just been playing with these numbers - and it has an effect like magic on the overall sound when that stressing is applied. It's beautiful.
                                                                                Eleanor

Be positive and logical: why not to look at the reflection? Why your reflection must look necessary worse than yourself? Even more, many of expensive, specially mastered mirrors make your look better than "original you".
(These mirrors make you more high and slim; enhance the colour of your skin etc. It gives you (that is especially important for each lady) confidence that eventually is stronger and works well, better than any beauty.

Good Luck!

Vladimir Dounin

How does this apply to involvement in problem-solving?

I see and point the problem: millions of the people around the globe at least once in their life time tried to learn music. They dropped the music then because they were dissappointed with the result: their music was ugly, not enjoable in spite of the time and money spent and official "Grade - certificates" obtained.

 A lot of people come for the concerts of classical performers and say "never again" in spite of high "official rating" of these performers. These people never will come and for the concerts of better performers in the future as well.

This article analyzes the roots of the problem and gives some ideas for the solution of these problems. This an invitation for discussion on this topic vitally important for the music.

Vladimir Dounin


prove that e^n is asymptotic to lcm(1,2,..,n)

It is not true. But $ ln(lcm(1,2,...,n)\equal{}n(1\plus{}o(1))$. These statment is very small, than your.


k, can some1 explain in simple words how to find each

i get confused in everything

i just know taht critical #s are when first derivative is 0 or undefined

then i put like
                      (-infinity, first crit #)|(first crit #, to 2nd crit #)|(2nd crit #,+infinity)
            test #                                |                                       |
    value in f'(x)

something like that
but can some1 explain it better?
i got online test thingy(that mathxl stuff) and i don't really understand it
and i can't understand it in text, but if some1 explains it better i can
here is the question
[url=http://img463.imageshack.us/my.php?image=math8aj.gif][img]http://img463.imageshack.us/img463/737/math8aj.th.gif[/img][/url]

thx in advance

[url=http://imageshack.us][img]http://img307.imageshack.us/img307/9945/math1ax.gif[/img][/url]

i need help with these two,  what exactly are thy questions asking me?
is it asking for the derivative?
that i can find

i got the second question(s) right
but
[url=http://imageshack.us][img]http://img171.imageshack.us/img171/8199/math3bg.gif[/img][/url]
how do i get this one right?

A relative extremum is (for a differentiable function $f$), the values of $x = x_0$ such that $f'(x)$ changes sign at $x_0$.  So, for example, $f(x) = x^2$ has a relative minimum at $x = 0$, because $f$ changes sign at that point.  However, $f(x) = x^3$ doesn't have any relative extrema, because at the only value where $f'(x) = 0$, $f'(x)$ doesn't change sign.  I hope this helps.

can u explain it simpler  :?

also is this right so far?
[url=http://imageshack.us][img]http://img234.imageshack.us/img234/4282/math4kz.gif[/img][/url]
and then just do quotient with each then multiply? or would i jsut elave it as it was and do the top product then quotient?

nvm, its adding that second problem


i did everything in the AoPSWiki, but it seems that no log file is created when i use the asymptote with pause tool

There isn't a "asymptote with pause tool" in the AoPSWiki. Did you mean the one in the first post at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148557]Reading asymptote error messages[/url] which doesn't create a log file but just pauses the Asymptote window to allow you read any errors?

The version in AoPSWiki at [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Showing_Asymptote_error_messages_in_TeXnicCenter]Showing Asymptote error messages in TeXnicCenter[/url] does create a log file but only shows it if there's an Asymptote error, but doesn't pause. You should only run this [b]after[/b] you have compiled the tex file, since that creates the asy file that is required.

yeah i did mean the pausing asymptote error messages...

i made the batch file that's supposed to pause the error messages but it just flashes with a message. i managed to catch this:
'C:\Program' is not recognized as an internal or external command, operable program or batch file.

Did you forget the quotes around "C:\Program Files\Asymptote\asy.exe" in asymptote.bat? The quotes are needed for any path that involves spaces.

this is what's in the file:
[code]@echo off
"C:\Program Files\Asymptote\asy.exe" -batchView -tex "pdflatex" -align C -f "pdf" %1
pause[/code]

asy.exe is in C:\Program Files\Asymptote?
What happens if you go to Start, Run, cmd, OK so you get a DOS window and then type "C:\Program Files\Asymptote\asy.exe"? Do you get Welcome to Asymptote ...? To get out type exit then you can close the window.

yeah, i get the welcome to asymptote message..

Do you get the same if you do the same but this time you type "C:\Program Files\Asymptote\asymptote.bat"?

yeah, the welcome to asymptote version 1.36 message is there too

That means that asymptote.bat is fine. Did you remember to include the quotes in the Command box in TexnicCenter, Tools, Customize, Tools?


x only divides $ x^2 \plus{} 1$ if x=1 or -1, because $ x \cdot x \equal{}x^2$  :clap2: .

If $ x$ does not equal 1 or -1, then $ x^2\cong \minus{}1 (mod x)$, false.

[quote="Elio (n)"]Can you prove this ?[/quote]

$ x\equiv 0 \mod x$
$ x^2 \equiv 0 \mod x$

So we must have
$ 1\equiv 0 \mod x$
$ \Rightarrow x \equal{} \pm 1$

Also, if $ x$ divides $ x^2 \plus{} 1$, then $ \frac{x^2 \plus{} 1}{x}$ is an integer, and so is $ \frac{x^2 \plus{} 1}{x} \minus{} x \equal{} \frac{1}{x}$. Therefore $ x$ divides 1, so $ x \equal{}\pm 1$


The number 555,555 can be expressed as the product of two 3-digit numbers. Find the smaller of these two numbers.

let's prime factorize that number first

$ 555,555\equal{}3\times5\times7\times11\times13\times37$

now, let's rearrange the numbers into two groups with three numbers each
remember, both products have to be 3-digit numbers

a little hard work reveals that:
$ 3\times7\times37\equal{}777$ and $ 5\times11\times13\equal{}715$

so our answer is $ 715$

let's check our answer
we said the two 3-digit numbers were 777 and 715
let's multiply them and see if the product is 555,555
go ahead, do it yourself

it works!
777 and 715 multiply to 555,555
so our answer is correct

$ 555555 \equal{} 5\cdot 111111 \equal{} 5\cdot 11\cdot 10101 \equal{} 5\cdot 11\cdot 37\cdot 273 \equal{} 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 37 \equal{} 715\cdot 777$

Edit: Beaten


Where Was The 2004 National's Winner From?  Anybody know?

Top Team and Individual were both from Illinois

The national winner was Greg Gauthier from Illinois.
Go Illinois!

I was able to watch a recording from last year's nationals.  It was kind of weird that the final 4 were the 1st, 2nd, 11th, and 12th place winners :huh:

Greg Gauthier went the Nationals in sixth, seventh, and eighth grade if I remember correctly.  Or at least in sixth and eighth.

That's what I thought....He's obviously really strong(in math) and since he won all 3 parts of the National competition....

Good thing he's not there this time (So now I can win!!! haha).

[quote="falconwing64"]No, you were right the first time.
Greg got in the top 30 in sixth grade, 7th in 7th grade, and first in 8th grade.[/quote]

wow, someone told me that he was already taking math courses on the level of college before he went to MathCounts, is that true?

I find it VERY ironic that 3  of the last 5 national winners came from the SAME REGIONAL!

(Joe Jia, Albert Ni, and Greg Gauthier all came from the same Illinois REGIONAL!)

Wow, thats amazing. Well, all four of the people representing Texas at Nationals this year are from the same chapter. Actually, we all have the same coach but one of us (the sixth grader) goes to a different school than the other two mathletes and I. Our coach teaches at two different schools because he used to teach at only one before a new school was built and half of his students had to transfer and he had to follow them.


Yesterday I was 10years old and next year I will be 13 years old. How is this possible.?Whats my birth date?

ez.

ur birthday must be Dec 31, at (example) 11:59. Yesterday ("dec 31") you were 10 'till 11:59. Next year, at Dec 31, 11:59, you will be 13.


number theory
relatively prime
superior algebra
superior algebra unsolved

Let E be the group of permutations of $ \mathbb Z/n$ which has the form $ x\mapsto x^e$
I have that n = 17*19. Then I have to determine E. SHow that teh maximal elemnet order in E is 12. Then I have to find the order of $ x \mapsto x^5.$
Since the maximal order is 12 the order of
$ x \mapsto x^5$ is divisors if 12. And I am having problem with finding the order please help. how can I do it. thanks

This is not very hard to do by brute force, but you need to know that $ x^{\phi(n)\plus{}1} \equiv x\ (\rm{mod}\ n)$, so what we're really studying is the group of units $ (\rm{mod}\ \phi(n))$.

Unfortunately, I think you'll find that there are elements of order $ 24$ in this group, so your problem has a mistake.  Anyway, your question can be rephrased: what is the smallest positive integer $ k$ such that $ 5^k \equiv 1\ (\rm{mod}\ \phi(n))$?

I have a question I know that E is isomorphic to $ (\mathbb Z/144)^*$, then how can I find the order of the permutation $ x \mapsto x^5.$
Should I solve $ 5^k \equiv 1 \, mod(48)$ or should it be $ 5^k \equiv 1 \, mod(48)$? and by the way to find the order why should I solve this, how do you get this
5^k \equiv 1 \, mod(n) .
Thanks.

[quote="peteryellow"]Should I solve $ 5^k \equiv 1 \, mod(48)$[/quote]

These are the same equation.  Anyway, you seem to have noticed my mistake--we don't want to work mod $ \phi(n)\equal{}288$, but mod $ (16,18)\equal{}48$.

As for your last quesiton: what happens when you apply $ x\mapsto x^5$ twice?  Three times?  Etc.?  Now, use the fact that $ x^{48\plus{}1}\equiv x$. (actually, you need a litle more: there exists an element of multiplicative order 48)

Argh, I keep making mistakes.

So we actually need to work mod 144.  The reason is, if we are in fact looking at permutations of $ \mathbb{Z}/n$ rather than $ (\mathbb{Z}/n)^\times$, we need to consider numbers that are not relatively prime to $ n$.  For example, $ 17^{10}\equiv 17$ and $ 19^{9}\equiv 19$.  All in all, the smallest integer $ k$ such that $ x^{k \plus{} 1}\equiv x$ for all $ x$ is $ k \equal{} (48,10 \minus{} 1, 9 \minus{} 1) \equal{} 144$.

It follows that we are looking for the order of $ 5$ mod $ \phi(144)$, which is 24.

On the other hand, if we are really looking at permutations of $ (\mathbb{Z}/n)^\times$, then we can work mod $ \phi(48)$ and the order is 12 as the original problem said.


a problem from a long time ago thats probably made up by mathcounts.

In professional football, it is possible to score 2 points ( for a safety), 3 points (for a field goal), or 6 points (for a touchdown). If a touchdown is scored, it is possible to score 1 additonal point. The score of a football game is 21 to 17. Find the total number of possible combinations of points to acheive this final score.

have fun.  :trampoline:  :ddr:  :heli:

I think I might have counted wrong...
[hide]
There are 17 ways to get 21 points, and 10 ways to get 17 points, so the answer is 170.[/hide]

you miscounted by one brute ;)

I answered it here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1197450497&t=188804


In the picture (attached) $ABCD$ is a parallelogram and $AM=CN$. Prove that the angle bisector of $\angle AEC$ passes through the point $D$.

[color=darkblue]Denote : the points $S\in AB$, $P\in AD$ so that $C\in PS$ and $\overline{CPS}\parallel AM$ ; the point $R\in NP\cap SD$. Then the triangle $CNP$ is isosceles, the bisector of the angle $\widehat{AEC}$ is parallelly with the line $\overline{NPR}$ and the quadrilateral $DENR$ is a parallelogram ... as.o.[/color]

See also here http://www.mathlinks.ro/Forum/viewtopic.php?t=87917

[quote="manunited"]$S(DAM)=S(DCN)$[/quote]
indeed as $BC \parallel AD$ $S(DAM)=S(DCA)$
and as $AB \parallel AD$ $S(DCN)=S(DCA)$


Assume $ a_{i}>0$

If $ \sum_{n\equal{}1}^{\infty}a_{n}$ converges

what about: $ \sum_{n\equal{}1}^{\infty}\frac{a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}}{n}$

whether it converges or not?

That never converges, as long as we have $ a_i>0.$

The second series is greater than $ \sum_{n\equal{}1}^{\infty}\frac{a_1}{n}$ and diverges by comparison to the harmonic series.

Even if we dispense with the positive terms, this would also diverge any time $ \sum_{n\equal{}1}^{\infty}a_n\ne 0.$

[quote="Kent Merryfield"]

Even if we dispense with the positive terms, this would also diverge any time $ \sum_{n \equal{} 1}^{\infty}a_n\ne 0.$[/quote]

[color=red]Do you mean:[/color]

if $ \sum_{n \equal{} 1}^{\infty}a_n\ne 0.$

then  $ \sum_{n \equal{} 1}^{\infty}\frac {a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n}}{n}$ will diverges ?

[color=red]There is a counter-example: [/color]

If we choose some $ a_{i}$, such that : $ S_{1} \equal{} 1,S_{2} \equal{} \minus{} 1,...,S_{n} \equal{} ( \minus{} 1)^{n \minus{} 1},...$

where: $ S_{n} \equal{} a_{1} \plus{} ... \plus{} a_{n}$

so we have: $ \sum_{n \equal{} 1}^{\infty}\frac {S_{n}}{n} \equal{} \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{n \minus{} 1}}{n}$

use Leibniz's Test ,the above series converges :roll:

Ah, but that violates the hypothesis that $ \sum a_n$ converges.


Determine natural numbers n for which there  $ z\in C$ with $ |z| \equal{} 1$ and $ z^n \plus{} z \plus{} 1 \equal{} 0$ .

I think this is true for all n other than n = 3k ie multiples of 3 . i took $ z \equal{} \omega$ .


geometry trigonometry limit function perimeter

I derived a formula to compute the area of a regular polygon of "radius" one (radius being the distance from the center to a corner).  You can do the algebra and see that the answer is that an n-sided polygon with radius 1 has area
[i]n[/i]sin(180/[i]n[/i])cos(180/[i]n[/i])
I put this in my graphing calculator, and then realized that as n approaches infinity, the area approaches pi.  I plugged in 9999999 and it gave me 3.1415926535897932384626433.

Is this a known formula or did I just create it?

One of my old algebra books uses this exact same method to show one way of getting pi.

Yeah, basically, a circle can be viewed as a polygon with infinite sides, so that limit will give you the area of a unit circle, which is $\pi$...

You can also find a formula of an n-gon with apothem 1. Taking the limit as $n\to\infty$, you'll get $\pi$.

[quote="JavaMan"]Which functions did you use to sandwich A?[/quote]

I think he means the standard result

$\lim_{x \rightarrow 0}\frac{\sin{x}}{x}= 1$

is obtained using the Sandwich Theorem.

[quote="t0rajir0u"]It's been known for centuries, but the demonstration is instructive.

First, we convert to radians.  Then our formula becomes

$A = n \sin \frac{\pi}{n}\cos \frac{\pi}{n}= \frac{n}{2}\sin \frac{2\pi}{n}$

Let $m = \frac{1}{n}$.  Then

$A = \frac{\sin 2 \pi m}{2 m}$

Finally, as $n$ becomes large, $m$ approaches zero.  Then

$\lim_{m \to 0}A = \pi$

By Sandwich Theorem.[/quote]

There's a sort of circular logic involved here. The number $\pi$ is BASICALLY defined as the area of a unit circle, but here it's introduced BEFOREHAND in the arguments of trig functions.

If we knew nothing about $\pi$ (which is implicitly understood here), the best we could do is to PROVE that $\lim_{n\to\infty}n\sin{{180^\circ}\over n}\cos{{180^\circ}\over n}$ exists and make a convention that it should be denoted by a greek letter $\pi$. Also, we could prove that if $C$ is that limit (which is later to be denoted by $\pi$), then $\mathrm{Area}=Cr^{2}$ and $\mathrm{Perimeter}=2Cr$

[quote="JavaMan"]Is $\lim_{n \to 0}\frac{\sin mn}{n}= m$? If so, why?[/quote]

Simply because

$\lim_{n \to 0}\frac{ \sin mn}{mn}= 1$

Farenhajt:  I realize that.  It wasn't a proof as much as it was a demonstration, as I said.  In fact, it could be considered an alternate proof that $\lim_{x \to 0}\frac{\sin x}{x}= 1$.

You just multiply and divide by m.. i get it  :oops:

Is it not called the squeeze theorem? not the sanwich theorem...

I think both are acceptable http://en.wikipedia.org/wiki/Sandwich_theorem

[quote="t0rajir0u"]
In fact, it could be considered an alternate proof that $\lim_{x \to 0}\frac{\sin x}{x}= 1$.[/quote]

At best, it is a hint to the proof. For a strict proof you have to have radians introduced (because of the sine argument), and in order to do that you have to have $\pi$ introduced, hence you MUSTN'T use the above limit in deriving the formulae for the area and the perimeter of a circle.


geometry
3D geometry
algebra
polynomial
symmetry

Too long to type out based on my solution.
Hint
[hide]
Subtract the first two equations from one another and get a^2+b^2 = (a^3 - b^3)/3

Sub into the last equation.

Use similar steps for all three to get that the addition of the cubes of any two of the three minus the third variable cubed equals 25.[/hide]

[hide="very brute force"]
Suppose that $a,b,c$ are the roots of the polynomial $x^{3}+px^{2}+qx+r$. We know that $a,b,c$ are distinct. Therefore, we subtract the second equation from the first to get:
$(a-b)(a^{2}+ab+b^{2})=3(b+a)(b-a)$. Since we know $a,b,c$ are distinct, we have:
$a^{2}+ab+b^{2}=-3(a+b)$.  By symmetry, we also have
$b^{2}+bc+c^{2}=-3(b+c)$
$a^{2}+ac+c^{2}=-3(a+c)$
Now, we subtract the second equation from the first to get:
$(b-c)(a+b+c) =-3(b-c)$, so $a+b+c =-3$ Then, $p=3$.
Returning to our system above, we now add the three equations together to get
$2(a^{2}+b^{2}+c^{2})+\sum{ab}=-6(a+b+c)$
$18-3\sum{ab}=18$
$\sum{ab}= 0$. Thus, $q=0$.
Finally, we return to our original set of equations and add them to get
$a^{3}+b^{3}+c^{3}= 6(9)-75 =-21$. Then, we have
$a^{3}+b^{3}+c^{3}= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-ac-bc)+3abc$
$-21 = (-3)(9)+3abc$
$abc = 2$.
[/hide]
I'm pretty sure there is a better solution than that; the three equations look pretty symmetrical, and if we have the polynomial $x^{3}+px^{2}+qx+r=0$,
then, $a^{3}+pa^{2}+qa+r = 0$, and this could match the original set of equations if we repeat for $b$ and $c$ and make some maniupulations...

Solve

$ y^{2} \minus{} x \equal{} 5$

$ \frac {1}{y \minus{} 1} \plus{} \frac {1}{y \plus{} 1} \equal{} \frac {1}{x}$.

Sorry, I meant this to be a separate post; I clicked on reply instead of post a new topic.

hello,plugging $ x\equal{}5\minus{}y^2$ in the second equation simplifying and factoring we get
$ \frac{2y^3\minus{}y^2\minus{}10y\plus{}1}{(y\minus{}1)(y\plus{}1)(y^2\minus{}5)}\equal{}0$
one solution of our polynomial of degree $ 3$ is
$ y\equal{}1/6\,\sqrt [3]{37\plus{}6\,i\sqrt {6267}}\plus{}{\frac {61}{6}}\,{\frac {1}{\sqrt
[3]{37\plus{}6\,i\sqrt {6267}}}}\plus{}1/6$
Sonnhard.

Thanks, Sonnhard. I meant to write the second equation as

$ \frac{1}{y\minus{}1} \minus{} \frac{1}{y\plus{}1}\equal{}\frac{1}{x}$.

hello, oh, in this case you will get by the same procedure the solutions
$ (4,3)$ or $ (4,\minus{}3)$.
Sonnhard.

$ a^3\plus{}3a^2\plus{}25\equal{}3(a^2\plus{}b^2\plus{}c^2)$.

So suppose that we have the system:

$ a^3\plus{}3a^2\plus{}25\equal{}3k$
$ b...$
$ c...$
$ k\equal{}a^2\plus{}b^2\plus{}c^2$

where $ k$ is some fixed number. This may or may not have solutions. Anyways, we look for solutions where $ a,b,c$ are distinct.

Indeed, since this is the case, $ a,b,c$ are the distinct roots of $ p(t)\equal{}t^3\plus{}3t^2\plus{}25\minus{}3k\equal{}0$.

So we have $ a\plus{}b\plus{}c\equal{}\minus{}3$ and $ ab\plus{}bc\plus{}ca\equal{}0$. So $ a^2\plus{}b^2\plus{}c^2\equal{}9$ so $ k\equal{}9$ so $ abc\equal{}2$.

(And we have $ t^3\plus{}3t^2\minus{}2\equal{}0$)


inequalities
integration
real analysis
real analysis unsolved

How to prove that if $L^{p}(\mu)\subset L^{q}(\mu)$ if $\mu (X)<\infty$ for all $1\leq q\leq p \leq \infty$? Does this follows from Jensen inequality?

It also follows from Hölder's inequality.

But you can prove the inclusion using only $x \ge 1 \to x^{q}\le x^{p}$ because $\int_{|f| \le 1}|f|^{p}\le \mu(X)$.

[quote="redkimchi"]It also follows from Hölder's inequality.[/quote]
Right. Because Hölder implies the given inequality and Jensen implies Hölder, then Jensen implies the given inequality. Probably there is some more direct way to prove it by Jensen's inequality.

Since $x \mapsto x^{p/q}$ is convex, it follows that $\left(\int |f|^{q}\right)^{p/q}\le \int |f|^{p}$ provided that $\mu(X) = 1$.


The directions on a can of lemony Lemonade say to use 1 part of Lemony Powder for every 3 parts of water. The committee wants to make 1 gallon of lemonade(128oz).

                  How much water and how much mix will they need?

                 Explain how you solved the problem

By 1 part - 3 part, I assume by volume.

$\frac{128}{4}=32$

$32$  ounces of Lemony Powder and $96$ ounces of water?

EDIT:
After further research, I see you assume fluid ounces.

So measuring Lemony Power in $cm^3$ or $in^3$ would be more applicable.

$32 \cdot 29.57=946.24 cm^3$ (no sig figs) of lemony powder or $32 \cdot 1.8=57.6 in^3$


Okay, anybody have a clue how to do this? 'Inverse variation'. Problems reside here~~

Solve. Assume that y varies inversly as x.

10. $ Suppose y\equal{}6 when x\equal{}2. Find y when x\equal{}3.$

Okay, helpeth please!

$ x\equal{} 2$ then $ y\equal{} 6$ seems to imply that if $ x \equal{} 3$, $ y \equal{} 9$ since I'm supposing $ y \equal{} 3x$ according to the original statement.

Well, that sucked... Um, here~~

 Suppose y = 6 when x = 2. Find y when x = 3.

 Hope that works?

 Sorry 'bout that...

[hide][b]Inverse variation[/b] is xy=k, meaning the product of x and y is always the same.
Prime factorization did [b]direct variation[/b], kx=y, meaning the quotient of x and y is always the same.[/hide]

[hide]To answer your problem:
x=2 and y=6.
2*6=12
12 is k in this example.
You want to find y when x is 3, so you put in x and k into the equation xy=k.
3y=12, and the answer is 4.[/hide]

Oh, okay. Thank you!

 Now, can you help me with a word problem?

[quote="Verridith"]Word Problem~

 The frequency of a musical tone varies inversly as it's wavelength. if a tone has a frequency of 440 vibrations per second, and a wavelength of a tone that has a frequency of 660 vibrations per second.

 Help?[/quote]

Please make use of the edit button; it was created for a reason  :wink:

Also, what are you asking to find?  I don't quite see the question...

If you mean FIND the wavelength of a tone that has a frequency of 660 vibrations per second, then see below.

[hide="below"]The frequency of the frequency of the 660 wave is $ \frac{660}{440}\equal{}\frac32$ as fast as the 440 wave, so it's frequency will be $ \frac1{\frac32}\equal{}\frac23$ of the 440 wave's frequency.[/hide]

Ooops... -Looks- Oh, there IS an edit button! Thank you!

 Yea, it didn't say a /question/.


AMC
AIME
calculus
integration
ratio
geometric sequence
arithmetic sequence

(Me) A sequence of positive integers with $ a_1 \equal{} k$ where $ k$ is a positive integer, and $ a_{200} \equal{} 30300$ is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and in general, for all $ n \geq 1,$ the terms $ a_{2n\minus{}1}, a_{2n}, a_{2n\plus{}1}$ are in geometric progression, and the terms $ a_{2n}, a_{2n\plus{}1},$ and $ a_{2n\plus{}2}$ are in arithmetic progression. Find $ k$.

Level: AIME level

[hide="solution"]
If k=1, the sequence is $ 1,p,p^2,p(2p \minus{} 1),(2p \minus{} 1)^2,(2p \minus{} 1)(3p \minus{} 2)$... where the 2n+2th term is $ ((n \plus{} 1)p \minus{} n)(np \minus{} (n \minus{} 1))$ and the 2n+1th term is $ (np \minus{} (n \minus{} 1))^2$. To get all other values of k, we can multiply the whole sequence by a scalar. If k=1, $ a_{200} \equal{} (100p \minus{} 99)(99p \minus{} 98)$. So $ 30300 \equal{} c(100p \minus{} 99)(99p \minus{} 98) \equal{} 2^2\cdot3\cdot5^2\cdot101$. Solving for the factors, c=3 and p=2 works. [/hide]
Edit: In my post before, I assumed p was 2 on accident.

My solution:

[hide]We know that $ a_{200} \equal{} ky^2x$, where y represents 99 pairs of rates and x represents the last rate, because of the geometric relationship given. If we write the first five terms out as $ a, ar, ar^2, ar^2n, ar^2n^2$, then by using the arithmetic relationship given, $ n \equal{} \dfrac{2r \minus{} 1}{r}$.  If we choose an integral value for r and continue to find rates that satisfy this relation, we realize that many numerators and denominators cancel, so we only have a perfect square for $ y$, but $ x$, the next rate, will leave us with just one composite number that isn't a square.  We now have $ apq$.  Since this is AIME, we prime factorize: $ 30300 \equal{} 2^2*3*5^2*101$.  Choosing various $ r$ shows that p and q are very close together.  Conveniently, when $ r \equal{} 2$, we need something that differs by one, or $ 101$ and $ 2^25^2$.  This implies that $ k \equal{} 3$[/hide]

[quote="calculuslover800"][hide="is there any reason..."]

... that 30300 couldn't be a solution as well? (Aside from the size limit on AIME answers, but this problem was modified)

The geometric sequences have ratio 1 and the arithmetic sequences have difference 0.

You may want to tighten the wording some.

[/hide][/quote]

Okay, I'm not sure if you read my rules on AMC forum because I think I wrote that the answer format in this round WILL BE same as the real AIME so it was "understood" to ignore 30300. Although it is my fault that I don't have a single document outlining all the rules yet, I would really appreciate if people read all the rules before asking, "Well, this problem needs to be tightened..."

Second, this is a proposal problem so the wording really doesn't matter. I'm writing these to give myself some practices on writing (without going so meticulous on grammar and stuffs) and others to practice.

 :)

[quote="123456789"]What does HDMO mean?[/quote]

Click [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=192539]here[/url].


Have any of you qualified through the Group Maths Olympiad (held by CBSE) to write INMO?
If no , have you any idea when the results will be out?

Hi,
I think the results of the CBSE Group Olympiad are out. They have been sent to the various schools. I didnt write it , but a friend of mine has got through it.


Harold tosses a nickel four times. The probability that he gets at least as many heads as tails is

a) $\frac{5}{16}$
b) $\frac{3}{8}$
c) $\frac{1}{2}$
d) $\frac{5}{8}$
e) $\frac{11}{16}$

Source: AMC 8 2002 #21

[hide]
For this question, we need to find the probability that he gets 0 or 1 tails and subtract from 1.
There is 1 chance for the first
There is 4 chances for the second.
There are $16$ total possibilities, so a $\frac{5}{16}$ chance for 1 or 0 tails.
Thus there is a $\frac{11}{16}$ chance for the same number or more heads.[/hide]

[quote="easyas3.14159..."][hide]The probablility of tossing two heads, two tails is 6(1/2*1/2*1/2*1/2)=3/8. Answer B.[/hide][/quote]

You forgot that he can also have 3 heads, 1 tail, or 4 heads (he doesn't have to have the same number)  That makes the answer:

[hide="answer"]2 heads, 2 tails $\Rightarrow{4\choose2}=6\cdot\frac{1}{2^4}=\frac38$
3 heads, 1 tail $\Rightarrow{4\choose3}=3\cdot\frac{1}{2^4}=\frac3{16}$
4 heads, 1 tail $\Rightarrow{4\choose4}=1\cdot\frac{1}{2^4}=\frac1{16}$
$\frac38+\frac3{16}+\frac1{16}=\frac{11}{16}$[/hide]

excuse me, but doesnt [hide]$\frac{3}{8}+\frac{3}{16}+\frac{1}{16}$ [/hide]


Let the sequence is difined by:
\[x_1=1,x_{n+1}=[e(x_n!)]\]
Prove that:
 \[\gcd (x_m,x_n)=x_{\gcd (m,n)}\]

In other words:
\[x_{n+1}=x_n!\left(\frac{1}{0!}+\frac{1}{1!}+...+\frac{1}{x_n!}\right).\]
I don't see any approaches at the moment :(

we have $x_{n+1}=(x_n)!+(x_n)!+\frac{(x_n)!}{2!}+...+1$ by the definition of $e$. now assume that $m<n$ (the case $m=n$ is clear); let's expand everything $\mod x_m$. then $x_{m+1}\equiv 1\mod x_m$ by the above. now, to reach $x_m$ again, we need $n\geq 2m$ and we see that it gets periodic with period $m$. hence, the result.


Consider $ a\equal{}1.1 ^{10}$ and $ b\equal{}1.1 ^{11}$. Compare the numbers $ a^b$ and $ b^a$.

[quote="moldovan"]Consider $ a \equal{} 1.1 ^{10}$ and $ b \equal{} 1.1 ^{11}$. Compare the numbers $ a^b$ and $ b^a$.[/quote]

Are you implicitly asking which among the two numbers is greater, right?

Now, It is slightly obvious that $ b \equal{} 1.1 ^{11}$ is 'bigger' than  $ a \equal{} 1.1 ^{10}$.

The first number  $ a^b$ is an instance where a 'smaller number' has a 'bigger number' as its 'power'. The second number $ b^a$ is an instance where a 'bigger number' has a 'smaller number' as its 'power'.

I had observed that frequently 'real numbers' which when raised to a number greater than themselves (for example: $ a^b$)  tend to be larger than 'real numbers' which are raised to a number lesser than themselves. (for example: $ b^a$)

In the case of integers, this is not the case. ($ 2^4$= $ 4^2$)

So, I'm guessing that the number $ a^b$ is greater than the number $ b^a$.

I'm guessing that the numbers will be really close to being equal-- I haven't done any computation so far, so I think I maybe wrong too.

[quote="mavropnevma"]You may as well take logarithms and notice that $ 11 \equal{} 10\cdot 1.1$.[/quote]

You may as well type out your alternate solution in its entirety
so that we may read it.  Also, I'm not going to give you the benefit
of the doubt that you think you know what you're stating to be
accurate until your steps are shown.

mavropnevma gave a very good hint, which can generalize the problem:

If $ a\equal{}\frac{b}{c}$ and $ b\equal{}c\plus{}1$, then $ (a^b)^{a^c}\equal{}(a^c)^{a^b}$

All we need to prove is $ b\cdot a^c\equal{}c\cdot a^b$, or $ \frac{b}{c}\equal{}a^{b\minus{}c}\equal{}a$ which is the condition

Logarithms? The problem is too simple for using logarithms...

[quote="Bugi"]mavropnevma gave a very good hint, which can generalize the problem:

Logarithms? The problem is too simple for using logarithms...[/quote]

mavropnevma mentioned using logarithms, *so* let's see a solution worked
out with one, then.  No, it's *not" too simple for logarithms as claimed by a
dismissive tone; *you* may have a dislike for doing the problem that way.
Your method can be fine, and there can be plenty of alternate type solutions,
just as mine used manipulation and no logarithms.  It also lends some practice
to working with logarithms, even if this problem is placed too high in a math
forum section for users who would already be fluent with those.

. . . . . . . . .  $ a^b$ versus $ b^a$

$ (1.1^{10})^{(1.1^{11})}$ versus $ (1.1^{11})^{(1.1^{10})}$

$ log(1.1^{10})^{(1.1^{11})}$ versus $ log(1.1^{11})^{(1.1^{10})}$

$ (1.1^{11})log(1.1^{10})$ versus $ (1.1^{10})log(1.1^{11})$

$ (1.1^{11})(10)log(1.1)$ versus $ (1.1^{10})(11)log(1.1)$

$ \frac {(1.1^{11})(10)log(1.1)}{(1.1^{10})log(1.1)}$ versus $ \frac {(1.1^{10})(11)log(1.1)}{(1.1^{10})log(1.1)}$

$ (1.1)(10)$ versus $ 11$

$ 11 \equal{} 11$

So $ a^b \equal{} b^a$.

Relax... And yes, it IS too simple for logarithms...

Compare my solution and logarithm solution... See a huge difference? I say, there are many ways to do this, just logarithms aren't needed... They make the solution pretty much longer...


1.
Show that Sum of (n=1 to infinity) sin(x/n^2) converges uniformly on any closed interval [a, b].
(solved)

2.
Suppose that f[size=75]n[/size] -> f and g[size=75]n[/size] -> g uniformly on E in R.
Show that if f and g are not bounded then {f[size=75]n[/size]g[size=75]n[/size]} does not necessarily converge uniformly on E.
(solved)

3.
If each f[size=75]n[/size]  is continuous on [a, b] and Sum of f[size=75]n[/size]  converges uniformly on [a, b] then Sum of M[size=75]n[/size]  converges, where M[size=75]n[/size]  = max { |f[size=75]n[/size] (x)| a <= x <= b}.

2.

Let $ E \equal{} (0,1)$ and  $ f_n(x) \equal{} \frac {1}{n}$ and $ g_n(x) \equal{} \frac {1}{x}$. Then $ f_n g_n (x) \equal{} \frac {1}{n x}$ can be made arbitrarily large for any $ n$, so it doesn't converge uniformly.


[quote="behemont"]If we fix $k$ and $a_{i}$'s, $i=1,2,...n$, the condition $\sum a_{i}\sum \frac{1}{a_{i}}=k$ holds if we switch $a_{i}$ to $ma_{i}$, $\forall m\in \mathbb{R}$. So there isn't maximum of $P$...

A friend of mine figured this out...

Sorry for this local moment: maks ga je mrknuo.. :wink:[/quote]

I am sorry, the problem is firstlu presented in the following way:

If $x_{1},x_{2},...,x_{n}$ are positive and
\[x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}+\frac{1}{x_{1}^{2}}+...+\frac{1}{x_{n}^{2}}=m\ge 2n, \]
find the minimal value of
\[P=(x_{1}+x_{2}+...+x_{n})\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{n}}\right). \]
I revert it, but it becomes wrong. Sorry, the main problem is the above.

[quote="hungkhtn"] If $ x_{1},x_{2},...,x_{n}$ are positive and
\[ x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}+\frac{1}{x_{1}^{2}}+...+\frac{1}{x_{n}^{2}}=m\ge 2n, \]
find the minimal value of
\[ P=(x_{1}+x_{2}+...+x_{n})\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{n}}\right). \]
[/quote]
The minimal value of $ P$ is $ n^{2}$, which is attained for $ x_{1}=x_{2}=...=x_{n}$. :P

An interesting problem is the following:
If $ a,b,c,d$ are positive numbers such that
$ (a+b+c+d)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac 1{d})=m\ge 16$,
find the maximal value of
$ E=(a^{2}+b^{2}+c^{2}+d^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac 1{d^{2}})$.
I think that $ E\le (m-12)^{2}$, but $ (m-12)^{2}$ is not the maximal value.

Equality holds for $ a=b=1$ and $ cd=1$.
Without loss of generality, we may assume that $ cd=1$. Under this assumption, the inequality becomes

$ \sqrt{(a+b+x+\frac 1{x})(\frac{1}{a}+\frac{1}{b}+x+\frac{1}{x})}\ge 2+\sqrt[4]{(a^{2}+b^{2}+x^{2}+\frac 1{x^{2}})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+x^{2}+\frac{1}{x^{2}})}$,

with equality if and only if $ a=b=1$.


calculus
integration
function
inequalities
linear algebra
matrix
vector

Wow. Could you give any hints as to what to base the proof on? Because right now I have completely no idea :) .

It's based totally on quantum mechanics - the proof which I know. That involves a lot of dirty integrals and other concepts of QM. I was asking whether there exists a simple proof on basic concepts. If nobody else can post, I''l post mine!!

[quote="Rushil"][size=150][b]Prove the Heisenberg Uncertainity principle!!![/b][/size]

i.e. prove that $ \sigma _x \sigma _p \geq \hbar $ where $ \sigma _x $ is the uncertainity(standard deviation in position) and $ \sigma_p$ is the uncertainity in momentum... $ \hbar  = \frac{h}{4 \pi} $

I've got a very dirty proof involving integrals. Any ingenious proof???
Also, is the principle valid if we consider all 3 dimensions??? (Shoul be!!)[/quote]First things first, $\hbar = \frac{h}{2\pi}$.

That said, here goes:

For any operator $\hat{\Omega}$ one can write
$\hat{\Omega}|\psi\rangle =  \langle \Omega \rangle |\psi\rangle + \sigma_{\Omega}|\psi_{\Omega\perp}\rangle$

Likewise $\hat{\Gamma}|\psi\rangle =  \langle \Gamma\rangle |\psi\rangle + \sigma_{\Gamma}|\psi_{\Gamma\perp}\rangle$

Taking the expectation value of the commutator of these two operators:

$\langle [\hat{\Omega},\hat{\Gamma}]\rangle =\langle \psi |[\hat{\Omega},\hat{\Gamma}]|\psi\rangle = \langle \Omega \rangle \langle \Gamma\rangle   -\langle \Gamma\rangle \langle \Omega\rangle + \sigma_{\Omega}\sigma_{\Gamma}(\langle \psi_{\Omega\perp} |\psi_{\Gamma\perp}\rangle-\langle \psi_{\Gamma\perp} |\psi_{\Omega\perp}\rangle) $

$\langle [\hat{\Omega},\hat{\Gamma}]\rangle = \sigma_{\Omega}\sigma_{\Gamma}(\langle \psi_{\Omega\perp} |\psi_{\Gamma\perp}\rangle-\langle \psi_{\Gamma\perp} |\psi_{\Omega\perp}\rangle) $

$\langle [\hat{\Omega},\hat{\Gamma}]\rangle = 2i \;\sigma_{\Omega}\sigma_{\Gamma}Im\{\langle \psi_{\Omega\perp} |\psi_{\Gamma\perp}\rangle\}$

$|\langle [\hat{\Omega},\hat{\Gamma}]\rangle| = 2 |\sigma_{\Omega}\sigma_{\Gamma}||Im\{\langle \psi_{\Omega\perp} |\psi_{\Gamma\perp}\rangle\}|$

Since the wave functions are normalized, $|Im\{\langle \psi_{\Omega\perp} |\psi_{\Gamma\perp}\rangle\}| \le 1$.

That being the case,  $|\langle [\hat{\Omega},\hat{\Gamma}]\rangle| \le 2 |\sigma_{\Omega}\sigma_{\Gamma}|$

Recognizing that $\sigma_{\Omega}$ and $\sigma_{\Gamma}$ are each positive definite we can rearrange the above inequality such that:

$ \sigma_{\Omega}\sigma_{\Gamma} \ge \frac{1}{2}|\langle [\hat{\Omega},\hat{\Gamma}]\rangle|$

When we let $\hat{\Omega} = \hat{X}$ and let $\hat{\Gamma} =\hat{P}$ then $|\langle [\hat{X},\hat{P}]\rangle| = \hbar$ and we get:

$ \sigma_{X}\sigma_{P} \ge \frac{1}{2}\hbar$

That completes the proof.

*Note in most primers on Heisenberg's Uncertainty Principle the factor of $\frac{1}{2}$ is omitted from the RHS because $\hbar$ is the smallest possible action any observable can take, and on that basis the $\frac{1}{2}$ is superfluous.

[quote="Rushil"]UHHHHHH

So you've used some definition I don't know. Anyway, as long as the answer comes!!!
As I said, my solution uses solving all the standard deviation integrals.
Is there a simpler , non too mathematical proof???[/quote]I didn't believe that anyone who asked this question would necessarily have seen the commutator, the Dirac notation or any of the operator algebra involved in the problem, but you did ask if anyone had an "ingenious proof" (not involving integral equations). To my mind, there is no better way to learn advanced concepts than to leap at them full on.

The business of quantum mechanics is mathematics; there is no way to avoid mathematics in proving the HUP, especially the HUP, because the HUP comes from mathematics; not experimentation.

Quantum mechanics is essentially matrix algebra performed upon a Hilbert space (an n-dimensional complex vector space). I exploited that algebra without resorting to integral equations.

If I skipped a step anywhere in my proof, it was that I did not show that the eigenvalue of $\hat{\Omega}|\psi_{\Omega\perp}\rangle$ , i.e. $\sigma_{\Omega}$, is the standard deviation of $\hat{\Omega}$. It is, which is quite easily demonstrated, but I did not wish to clutter a notation already unfamiliar to most in order to prove an obvious intermediate result.

PS To answer your second question, yes, the HUP is valid in all three dimensions, and for energy and time, and angle and angular momentum ... in general it is true of any operators which do not commute.


Show that Beltrami points have pedal triangles similar to those of 3 Apollonian
     circles centers.

     Maj.   Pestich

Inverse in circumcircle of Brocard points.

    Lt.  Pestich


Let $ x;y;z$ be positive real numbers satisfying:
$ 3x^2\plus{}3xy\plus{}y^2\equal{}75$
$ y^2\plus{}3z^2\equal{}27$
$ x^2\plus{}xz\plus{}z^2\equal{}16$
Caculate value of $ P\equal{}xy\plus{}2yz\plus{}3xz$

This problem is in Engel's [i]Problem Solving Strategies[/i] and it is from AUO 1984.

[hide="Hint"] Consider a geometrical interpretation of the problem, and it is unnecessary to find x,y, or z. [/hide]


USAMTS
floor function
function
inequalities
induction
calculus
integration

I'm planning to take USAMTS for 05-06 year.

I think we should start some kind of practice and I want to see how many people will be interested. What I plan to do is basically post some problems that may/may not include proofs but when you post the solution, you must write as you're sending that answer to USAMTS grader.

It would be nice to have someone who are experienced or actual USAMTS grader :D (That would be awesome) to look at the solutions and stuff.

But I want to see how many people will be interested first. Anyone?

After problem, I'll post my problem too so we can see which proof/answer is better.

[size=142]Problem 1[/size]

For any number $n \in \mathbb{Z}$ and $1 \leq r \leq n - 1$ the integer $\binom{n}{r}$ is defined by:

\[\binom{n}{r} = \frac{n(n-1)...(n-r+1)}{r(r-1)...3 \cdot 2 \cdot 1}\]

Show that the greatest common divisior of $\binom{n}{1}, \binom{n}{2},.... \binom{n}{n-1}$ is a prime if $n$ is a power of a prime and is 1 otherwise.

Fix $n$ and consider $\binom {n}{r}$, where $n=p^a$ and $r=p^b$ with $0\leq r\leq a-1$

Setting $b=0$, we see that the greatest common divisor is less than $p^a$ and that it also divides $p^a$

Set $b=a-1$.  It is clear :!: that the greatest power of $p$ dividing $\binom {p^a}{p^{a-1}}$ is $p^1$

Hence the greatest common divisor is less than $p$.

Next, in $\binom {n}{r}$, which is by combinatorial interpretation an integer, consider an enumeration $\{a_i\}$ of $p^a,p^a-1,\ldots,p^a-(r-1)$ which has the property that the greatest power of $p$ dividing $a_i$ is greater than or equal to the greatest power of $p$ dividing $a_j$ if $i>j$, and if the greatest power of $p$ dividing $a_i$ is the same as the greatest power of $p$ dividing $a_j$, then $i>j$ if $a_i>a_j$.  Make a similar enumeration $\{b_i\}$ of $r,r-1,\ldots 1$.

We claim that for each $i$, $a_i\geq b_i$.  For $i\geq 1$, consider the number of $a_i$'s where $p^k$ divides $a_i$, with $1\leq k\leq a-1$.

By a well-known theorem, this is just $\left \lfloor \frac {r-1}{p^k} \right \rfloor + 1$

For the $b_i$'s, the similar result is

$\left \lfloor \frac {r}{p^k} \right \rfloor $

Using a well-known result for the floor function (along the lines of $\lfloor x\rfloor \leq \lfloor x-1\rfloor +1$) and using the fact that strict inequality holds for $i=1$ (that is, the greatest prime dividing $a_1$ is strictly greater than the greatest prime dividing $b_1$) and since there is a bijection from the $a_i$'s to the $b_i$'s , we conclude (there's a subtle, but intuitive, induction step left out there) that when $\binom {p^a}{r}$ is expressed as a rational number in lowest terms, $p$ divides the numerator, and since it is always integral, we conclude that $p$ divides $\binom {p^a}{r}$ for $1\leq r\leq p^a-1$

Here is how you finish the problem (I am not going to prove it).  In the second case, write $n$ as a product of primes $p_1^{r_1}\ldots p_k^{r_k}$, where $k>1$.  For each index $i$, consider

$\binom{n}{p_i^{r_i}}$

By adapting the previous results, it is trivial to show that $p_i$ does not divide $\binom{n}{p_i^{r_i}}$.  By running through all $i$, we may show that the binomial coefficients do not have a common prime factor, and hence have a gcd of $1$.

This marathon will probably move really slowly, since it's USAMTS, but it will be fun!  :coolspeak:

[quote="DPatrick"]I should point out that this problem is somewhat more difficult than the typical USAMTS problem.  In other words, students who may find this problem too difficult should not be discouraged from participating in the USAMTS![/quote]

It's not so much that it's difficult, it's just that a reasonable approach to it requires having seen number theory problems of this sort before (e.g, problems of the sort 'how many zeros does n! end in?').  I don't really follow the USAMTS problems, but it seems to me that you guys do a good job of keeping things self contained so that people with experience don't have a huge advantage over those who don't.

[quote="DPatrick"]I should point out that this problem is somewhat more difficult than the typical USAMTS problem.  In other words, students who may find this problem too difficult should not be discouraged from participating in the USAMTS![/quote]

This was from senior division of math olympiad. Personally, when I first posted the problem, it looked easy (because I understood the problem) but when I actually sat down to prove it, I found it really hard. :?  Thus, I'll choose more appropriate problems.

Thanks to blahblahblah and pbornsztein posting the solutions (I took a look at it but I didn't get any of them).

Anyway, to have new start, today will be Day 1. I'll do same thing with AMC Practices. Send the solution/proof (in this case) to me. I'll take look at it and if there are more than one good solution with different ideas, I'll post all of them.

[size=150]Day 1[/size]

[color=blue]A band of pirates unloaded some number of treasure chests from their ship. The number of pirates was between 60 and 69 (inclusive). Each pirate handled exactly 11 treasure chests, and each treasure chest was handled by exactly 7 pirates. Exactly how many treasure chests were there? Show that your answer is the only solution.[/color]

Note: For those who saw AMC-Practices, these problems are going to be way tougher than they are. Since USAMTS are obviously harder tha most AMC-12 problems and even some of the later AIMEs, if you have hard time, that's okay. Try your best.

[size=142]Day 2[/size]

Let $a,b,c,d,$ and $e$ be the unique rational numbers satisfying the system of equations:

\[a+b+c+d+e = 2\\
16a+8b+4c+2d+e=4\\
81a+27b+9c+3d+e=6\\
256a+64b+16c+4d+e=8\\
625a+125b+25c+5d+e=34\]

Determine the value of $a-b+c-d+e$.

[quote="Silverfalcon"]From today, I will score each answer as the graders do in USAMTS. My grading will be poor compared to them but I'm trying to make this real instead of just giving 1 point for somewhat good answer.

[size=142]Day 3[/size]

Determine the unique pair of real numbers $(x,y)$ that satisfy the equation

\[(4x^2+6x+4)(4y^2-12y+25) = 28\][/quote]
I would recommend you add the condition "prove that this is the only solution," as otherwise a simple demonstration proof would technically warrant full credit (i.e. "here we have $(x_1, y_1)$, substituting into the given equation we see that the relationship is fulfilled, QED").

[quote="zanttrang"][quote="LynnelleYe"]The original question didn't ask that we prove it's the only solution--besides, it's given in the problem that the pair of numbers is unique. Something like "Explain how you found your answer" would be better.[/quote]
Yes, you're right, that wording would probably be better - it doesn't matter how the original question phrased it, though, as Silverfalcon can modify the questions however he wants for this contest.  An even better way of writing the question would be "find all real solutions, etc. etc. and prove that these are the only solutions."  I always feel like the problem telling you there's only one solution is a little bit of a giveaway, so not implying in the problem that there's only one solution makes it a bit more interesting (to me, at least).[/quote]

The problem has exact wording from the contest.

[size=142]Day 4[/size]

Determine, with proof, the rational number $\frac{m}{n}$ that equals:

$\frac{1}{1 \sqrt 2 + 2 \sqrt 1}  + \frac{1}{2 \sqrt 3 + 3 \sqrt 2} + \frac{1}{3 \sqrt 4 + 4 \sqrt 3} + \cdots + \frac{1}{4012008 \sqrt{4012009} + 4012009 \sqrt{4012008}}$.

[size=142]Day 5[/size]

Suppose $\frac{\cos {3x}}{\cos x} = \frac{2}{5}$ for some angle $x$, $0 \leq x \leq \frac{\pi}{2}$. Determine $\frac{\sin {3x}}{\sin x}$ for the same $x$.

[quote="chess64"]I don't remember trig ever being on USAMTS questions. Oh well.[/quote]

There has been trig on USAMTS problems before.  See problem 4, round 4, year 15 [url=http://www.usamts.org/Problems/U_ProblemsPast.php]here[/url].

[quote="Silverfalcon"][size=142]Day 6[/size]

Prove that if $0 < x < \frac{\pi}{2}$, then $\sec^6 x + \csc^6 x + (\sec^6 x)(\csc^6 x) \geq 80$.[/quote]

I'd like to encourage anybody who knows calculus not to use it on this problem.  While using calculus does work, the process is dull and uninteresting, and doesn't provide any real challenge.  USAMTS-level problems (such as this one) never require calculus, and in fact the best problems (the ones that get commended) don't use calculus.  I think you'll be more satisfied if you find another, more elegant route to solving the problem.

[size=142]Day 8[/size]

Compute $1776^{1492!} \mod 2000$; i.e. the remainder when $1776^{1492!}$ is divided by 2000. (As usual, the exclamation point denotes factorial.)

[size=142]Day 9[/size]

Assume that $f(1) = 0$, and that for all integers $m$ and $n$,

\[f(m+n) = f(m) + f(n) + 3(4mn-1)\]

Determine $f(19)$.

[quote="Silverfalcon"][size=142]Day 6[/size]

Prove that if $0 < x < \frac{\pi}{2}$, then $\sec^6 x + \csc^6 x + (\sec^6 x)(\csc^6 x) \geq 80$.[/quote]
[hide]
It's equivalent to

$\cos^6x+\sin^6x+1\geq 80\cos^6x\sin^6x$

By the power-mean inequality,

$\cos^6x+\sin^6x\geq \frac {1}{4}(\sin^2x+\cos^2x)^3$

Hence $\frac {5}{4}\geq 80\cos^6x\sin^6x\iff \frac {1}{64}\geq \cos^6x\sin^6x$

But $\frac {1}{64}\geq \cos^6x\sin^6x\iff \frac {1}{2}\geq \sin x\cos x\iff 1\geq \sin 2x$, which is obvious.[/hide]

Actually, better yet

[hide]Write the LHS as $(\sec^3x-\csc^3x)^2+\sec^3x\csc^3x(\sec^3x\csc^3x+2)$

Since $\sin x\cos x$ has range in $[0,1/2]$ in the given interval, $\sec^3x\csc^3x$ has range in $[8,\infty]$, so that $\sec^3x\csc^3x(\sec^3x\csc^3x+2)\geq 8(8+2)=80$.  Since squares are nonnegative, the conclusion follows.[/hide]

[size=142]Day 10[/size]

Let $a,b,c,d$ be positive real numbers such that

\[a+b+c+d = 12\]

and

\[abcd = 27+ab+ac+ad+bc+bd+cd\].

Find all possible values of $a,b,c,d$ satisyfing these equations.

[size=142]Day 11[/size]

I'm going to separate this problem into two parts which will worth 1 point each. (There is more to the problem but I don't want to grade that part now :D ).

A. If $*$ is $+$ or $-$, determine the appropriate sequence of $+$ and $-$ signs so that:

$2 = \sqrt{6 * \sqrt{6 * \sqrt{6 * \cdots}}}$

B. Find another sequence when 6's are replaced by 7's.

The maximum point in these are 2 points.

I'm going to post Day 12 very soon.

I'm trying to find the problem that I can find the answer with great explanation :lol: .

[size=142]Day 12[/size]

This problem is easy.  :P Try this problem, guys!

Find $x^2+y^2+z^2$ if $x,y$ and $z$ are positive integers such that

$7x^2-3y^2+4z^2 = 8$ and $16x^2-7y^2+9z^2 = -3$.

Just for sidenote, this problem was in same contest where most of the USAMTS problems are from (IMTS).


Find the largest power of $3$ which divides $\displaystyle 10^{3^{1999}}-1$

[hide]
$10^{3^{1999}}$ is a number with a $1$ then $3^{1999}$ zeroes.  Subtracting one gives a number that is just $3^{1999} \text{  } 9\text{'s}$.  This number is divisible by 3, so we can write the largest divisor as $\frac {3^{1999}}{3} = 3^{1998}$
[/hide]


Pythagorean Theorem
geometry
power of a point

Please recheck your problem!
If $ AC<1$ then??? :?:

[quote="Temperal"]Consider the following figure:

[asy]draw(unitcircle); draw((-1,0)--(1,0)--(-1,1)--cycle); label("$A$",(-1,0),(-1,0)); label("$B$",(1,0),(1,0)); label("$C$",(-1,1),(-1,1)); label("$O$",(0,0),(0,-0.1));[/asy]

such that $ \angle CAB$ is $ 90^{\circ}$ and $ AB$ is the diameter of circle $ O$. Now, let there be a point $ D$ on $ O$ such that $ AD$ is an altitude of $ \triangle ABC$.

Show that $ AB^2 \plus{} AD^2 \equal{} BC^2$.[/quote]

That's still false, I think.  $ BC^2 \minus{} AB^2 \equal{} AC^2\ne AD^2$...

Hmm, that's strange, I have a "proof" for it.
Maybe you can help me spot my error?
[hide]
By power of a point, $ CA^2\equal{}CD\cdot DB$.
By the Pythagorean Theorem, $ AC^2\plus{}AB^2\equal{}BC^2$.
By the Geometric Mean Theorem, $ AD\equal{}\sqrt{CD\cdot BD}$. (Actually, I'm not sure about this, I may have remembered the theorem wrong)
Substituting, we get $ AB^2\plus{}CD\cdot DB\equal{}BC^2 \Rightarrow AB^2\plus{}AD^2\equal{}BC^2$.
[/hide]

[quote="Temperal"]$ CA^2 \equal{} CD\cdot DB$.
[/quote]
It isn't correct, it must be $ CA^2\equal{}CD.CB$.
[quote="Temperal"]$ AD \equal{} \sqrt {CD\cdot BD}$. (Actually, I'm not sure about this, I may have remembered the theorem wrong)
[/quote]
It's correct because two triangle $ CDA,ADB$ are similar.

[quote="Temperal"]Consider the following figure:

[asy]draw(unitcircle); draw((-1,0)--(1,0)--(-1,1)--cycle); label("$A$",(-1,0),(-1,0)); label("$B$",(1,0),(1,0)); label("$C$",(-1,1),(-1,1)); label("$O$",(0,0),(0,-0.1));[/asy]

such that $ \angle CAB$ is $ 90^{\circ}$ and $ AB$ is the diameter of circle $ O$. Now, let there be a point $ D$ on $ O$ such that $ AD$ is an altitude of $ \triangle ABC$.

Show that $ AB^2 \plus{} AD^2 \equal{} BC^2$.[/quote]

It must be false because $ BC^2 \equal{} AB^2\plus{}AC^2 > AB^2\plus{}AD^2$.

[quote="tdl"]
It isn't correct, it must be $ CA^2 \equal{} CD.CB$.[/quote]

Ah, so that was the problem. I accidentally used the wrong segments for the power of a point... thanks!


Let $ ABC$ be a triangle.Prove that:
$ -1 <6cosA+3cosB+2cosC<7$

It's false, take $ A=B=\frac{\pi}{6}$.

But it's easy to prove that:
$ -1<6\cos A+3\cos B+2\cos C<8$ (if u want I can show my proof)

[quote="friendlist"]Can you help me?[/quote]
[hide]friendlist, mình nghĩ bạn không nên post bài ở báo THTT khi chưa hết hạn gửi bài
Thân![/hide]


linear algebra
matrix
linear algebra unsolved

Let $A$ be an $n \times n$ matrix with entries in $\{0,1\}$. Denote $\alpha$ for the number of zero's in the matrix and let $\beta$ be the number of one's in the matrix. Is it possible to find relations among $\alpha$ and $\beta$ such that $A$ is invertible (if positions are well chosen)??? It is clear that there are lower and upperbounds for $\alpha$ and $\beta$. Can we relate $\alpha$ and $\beta$ (multiplicatively  :maybe:) ? Would this relation depend on $n$?

Not sure if this question I posed myself has a 'nice' solution. Suggestions, counterexamples, remarks, references are very welcome :)

One condition for $A$ to be invertible is $\beta>n-1$. Afterwards... $\forall \beta \in \{1,2,...,n^{2}\}$ you can find a non invertible matrix by choosing one which has two identical lines or columns .

[quote="Jacobson"]I think we must have $n \leq \beta \leq \frac{n(n-1)}{2}$ for $A$ to be invertible, in general.[/quote]

consider the matrix
$A=\begin{bmatrix}1&1&1\\0&1&1\\1&0&1\end{bmatrix}$
$\mathrm{det}A=1$
but there are 7 1's and 2 0's.
maybe the upper bound is $n^{2}-n+1$.


well...
[hide="hint"]
\[(a+b)^2=a^2+2ab+b^2\]
\[(a-b)^2=a^2-2ab+b^2\]
If you these two, there should be no difficulties with that probleml.
You just follow the steps...
[/hide]

[quote="daleliop"]I saw this in a solution:

$\displaystyle 7 \left(\frac {x_n \pm y_n}{2}\right)^2 + \left(\frac {7x_n \mp y_n}{2}\right)^2 = 2 ( 7x_n^2 + y_n^2 )$

Could you explain how they got this? Thx.[/quote]I can't say for sure exactly "how they got it," but it's not hard to show that it is true. The first step is to realize that the statement is equivalent to the TWO statements:

\[7 \left(\frac {x_n+y_n}{2}\right)^2 + \left(\frac {7x_n-y_n}{2}\right)^2 = 2 ( 7x_n^2 + y_n^2 )\]and\[7 \left(\frac {x_n-y_n}{2}\right)^2 + \left(\frac {7x_n+y_n}{2}\right)^2 = 2 ( 7x_n^2 + y_n^2 ).\]

A little algebra should convince you that these are both true.

OR (this way is much simpler actually)

Just expand and simplify. (I dropped the $n$ subscripts)

\[7 \left(\frac {x_n \pm y_n}{2}\right)^2 + \left(\frac {7x_n \mp y_n}{2}\right)^2 =\frac{7x^2\pm14xy+7y^2+49x^2\mp14xy+y^2}{4}\]\[=\frac{56x^2+8y^2}{4}=2(7x^2+y^2)\]

The question is the first one (on pg.3 under Bulagria) in [url]http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc96-97-01feb.pdf[/url]

The solution is pretty short. Can you explain how to figure out their solution? I am stuck on how to come up with substituting $X = \frac{x_n \pm y_n}{2}$ and $Y = \frac{x_n \mp y_n}{2}$. That's the most important step.


geometry
geometric transformation
reflection
circumcircle
perpendicular bisector
geometry proposed

A triangle $ABC$ has $\angle ACB > \angle ABC$. The internal bisector of $\angle BAC$ meets $BC$ at $D$. The point $E$ on $AB$ is such that $\angle EDB = 90^{\circ}$. The point $F$ on $AC$ is such that $\angle BED = \angle DEF$. Show that $\angle BAD = \angle FDC.$

Of course, this is trivial if you notice that $D$ is an excentre of triangle $AEF$. I was actually looking for some alternative proofs (without excentres) and I decided to reflect $F$ in $DE$ to get a point $G$. It is then sufficient to show that $ADFG$ is cyclic but I can't immediately see why this should be the case. Anyone?

Any alternative solutions?

DE is the perpendicular bisector of FG, which meets the bisector AD of the angle $\angle A$ at D and this must be the midpoint of the arc GF of the circumcircle of the triangle $\triangle AGF$ opposite to the vertex A, so that AGDF is cyclic. And since $GF \parallel BC$, BC is a tangent to this circumcircle at D.

How could I possibly miss that? :D I've been using this little lemma a lot in the past :blush:


algebra
polynomial
induction
algebra solved

Sorry if this is probably a stupid question.

This afternoon I used the well-known Waring formula

$a^n+b^n= (a+b)^n-nab(a+b)^{n-2} +\frac{n(n-3)}{2}a^2b^2(a+b)^{n-4}+.....$

to solve a problem and I asked myself if it exists an analogous formula for 3,4,....variables.

Do you know that?

Since $a^n+b^n+c^n$ is a symmetrical polynomial, there is its representation via $a+b+c$, $ab+bc+ca$ and $abc$. Another question does such representation have so nice look as for $a^n+b^n$.

Yes there is one (very ugly...):
$\sum_{i=1}^k x_i^n =n \cdot \sum_{l_1+2 \cdot l_2+...+k \cdot l_k=n} \frac{(-1)^{n+l_1+l_2+...+l_k}}{l_1+l_2+...+l_k} \cdot (l_1,l_2,...,l_k)!  \cdot (s_1^{l_1} \cdot s_2^{l_2} \cdot ... \cdot s_k^{l_k})$
($s_i$ is the $i$-th elementary symmetrical polynomial)

Don't know at the moment, got it from an algebra book where its only cited...

Edit: possibly works with principle of inclusion-exclusion...

[quote="ZetaX"]Yes there is one (very ugly...):
$\sum_{i=1}^{k}x_{i}^{n}=n \cdot \sum_{l_{1}+2 \cdot l_{2}+...+k \cdot l_{k}=n}\frac{(-1)^{n+l_{1}+l_{2}+...+l_{k}}}{l_{1}+l_{2}+...+l_{k}}\cdot (l_{1},l_{2},...,l_{k})! \cdot (s_{1}^{l_{1}}\cdot s_{2}^{l_{2}}\cdot ... \cdot s_{k}^{l_{k}})$
($s_{i}$ is the $i$-th elementary symmetrical polynomial)[/quote]
This is a problem in my textbook.I have thought of it for almost five days and there is still no solution in my mind.
I think I am going mad.
Is there anyone can solve it?

I suppose it should be possible to prove the equation
by induction on $n$:

For a fixed sequence of $x_{i}$ let $f(k,n) : = \sum_{i=1}^{k}x_{i}^{n}$.
Then (if I am not mistaken) something like the following holds:
\[f(k,n) = \sum_{i = 1}^{k}f(k,n-i)s_{i}\]
This should get the induction working.


geometry
parallelogram
vector
geometry proposed

Let $ ABC$ be a triangle. Points $ D$ and $ E$ lie on sides $ BC$ and $ CA$, respectively, such that $ BD \equal{} AE$. Segments $ AD$ and $ BE$ meet at $ P$. The bisector of angle $ BCA$ meet segments $ AD$ and $ BE$ at $ Q$ and $ R$, respectively. Prove that $ \frac {PQ}{AD} \equal{} \frac {PR}{BE}$.

Hello!

$ \frac {RP}{RE}\cdot\frac {QA}{QP}\cdot\frac {CE}{CA} \equal{} 1\Rightarrow\frac {PR}{PQ} \equal{} \frac {RE}{CE}\cdot\frac {CA}{QA}\Rightarrow\frac {PR}{PQ} \equal{} \frac {BR}{BC}\cdot\frac {CA}{QA}$     (1)

$ \frac {QD}{QA} \equal{} \frac {CD}{CA}\Rightarrow\frac {AD}{QA} \equal{} \frac {CA \plus{} CD}{CA}\Rightarrow\frac {AD}{QA} \equal{} \frac {CA \plus{} BC \minus{} BD}{CA}$     (2)

$ \frac {BR}{RE} \equal{} \frac {BC}{CE}\Rightarrow\frac {BR}{BE} \equal{} \frac {BC}{BC \plus{} CE}\Rightarrow\frac {BR}{BE} \equal{} \frac {BC}{BC \plus{} CA \minus{} AE}$     (3)

(2), (3) $ \Rightarrow\frac {AD}{QA}\cdot\frac {BR}{BE} \equal{} \frac {BC}{CA}\Rightarrow\frac {BR}{BC}\cdot\frac {CA}{QA} \equal{} \frac {BE}{AD}$     (4)

(1), (4) $ \Rightarrow\frac {PR}{PQ} \equal{} \frac {BE}{AD}\Rightarrow\frac {PQ}{AD} \equal{} \frac {PR}{BE}$    :)

Construct the paralellogram EBDG, AEG is an isosceles triangle, AG || CF. Translate QR by the vector PD, get DQ' = PQ and DR' = PR, and the result follows.

Best regards,
sunken rock


Hi, I would like to cite this website

http://www.geocities.com/pluricomplex_amath/ntmid120.html

in one of my documents. However, I don't know how to get LaTeX to treat the underscore as it is, instead of formatting it to create a subscript.

I tried looking around the Guide, but I didn't find anything. I am kind of new to TeX and any help would be greatly appreciated.


calculus
integration
limit
trigonometry
real analysis
real analysis unsolved

The result: $\displaystyle\int_0^{2\pi} e^{e^{i\alpha}}\,d\alpha=2\pi$

$\int_0^{2\pi} e^{e^{i\alpha}}\,d\alpha=\lim \frac{1}{n} \sum_{i=1}^{n}e^{\cos \frac{i}{n}\pi +i \sin \frac{i}{n}\pi}=\lim \frac{1}{n} e^{\cos \frac{1}{n}\pi +\sin \frac{1}{n}\pi} \frac{1-e^{\cos \frac{n+1}{n}\pi +i \sin \frac{n+1}{n}\pi}}{1- e^{\cos \frac{1}{n}\pi +i \sin \frac{1}{n}\pi}}$
who can help me to finish it


Find all functions $f: R \to R$ such that: $|f(x)-f(q)| \leq 5(x-q)^{2}\forall x \in R$ and $\forall q \in Q$

Sorry, I don't understand your step. Where does the n on the outside come from?

[quote="cuopbientcd"]OH!
$\frac{|f(x)-f(y)|}{|x-y|}\leq 5.|x-y|$ therefor $f'(x)=0 \Rightarrow f(x)=a$[/quote]
Actually
$|f(x)-f(y)|\leq |f(x)-f(q)|+|f(q)-f(y)|\leq 5(x-q)^{2}+5(y-q)^{2}\leq 10(x-y)^{2}$
for some $q\in\mathbb{Q}$ between $x\not =y$.(Then $|x-q|,|y-q|\leq |x-y|$)
The rest is correct.

$\lim_{h\rightarrow 0}{f(x+h)-f(x)\over h}$ exists and $=0$ for all $x\in\mathbb{R}$. So $f$ is differentiable with $f'=0$ and $f$ constant.

[quote="cuopbientcd"]no nessecsary because there exist  always $\{q_{n}\}$ such that $\lim q_{n}=x$[/quote]
It is, because $f$ is at first not continuous.

Unless you mean
$|f(x)-f(y)|\leq |f(x)-f(q_{n})|+|f(q_{n})-f(y)|\\ \hspace*{2.5cm}\leq 5(x-q_{n})^{2}+5(y-q_{n})^{2}\mathop{\longrightarrow}_{n\rightarrow\infty}5(x-y)^{2}$
for $q_{n}\mathop{\longrightarrow}_{n\rightarrow\infty}x$, but that's hardly shorter.

rust your solution is not correct or incomplet. because the relation is only for rational. i wil give the complet solution. the only solution is $f=cte$.
let $a$ and $b$ two rational and let $a_{i}=a+i\frac{b-a}{n}$. its clearly that all $a_{i}$ are rational. and we can prove easly that $|f\left(a\right)-f\left(b\right)|\le{5\frac{(b-a)^{2}}{n}}$ it gives when $n\to\infty$.that $f\left(x\right)=cst$ for rational. lets prove the result for reals. for that let $r_{n}$ a sequense converge ver $x$  then we have $|x-r_{n}|\le{\frac{1}{10^{n}}}$ for all $n$. then we have $|f\left(x\right)-c|=|f\left(x\right)-f\left(r_{n}\right)|\le{5(x-r_{n})^{2}}\le{\frac{5}{10^{2n}}}$. then when $n\to\infty$ it gives $\forall{x}\in\mathbb{R}$ $f\left(x\right)=c$.
 :wink:


There's two problems:
[hide="1"]John has 10 balls. 8 are red, 6 are yellow, and 7 are red. How many are red and yellow?[/hide]

[hide="2, reffers to #1"]John puts the balls on 3 shelves. What is the chance that a shelf has all 6 yellow balls?[/hide]

[quote="gauss1181"]Eh isn't there a typo for #1? I mean, isn't there supposed to be a given number of red balls, yellow balls, and red and yellow balls? :maybe:[/quote]

NO, You're supposed to find the number of balls which are BOTH RED AND YELLOW

but it says 8 are red and 7 are red??? Is it me or is there a screwup typo in this? Why are there two different values for the number of red balls?

could there be blue and yellow balls?
or red and blue?

THERE COULD BE ANY KIND! Don't u know venn diagrams?

[quote="jonathanchou711"]Well sometimes tcjy puts some problems that is trying to make US figure it out. In other words, he's not trying to figure out the problem. He's trying to make us figure it out. I've seen him put up some problems that don't make sense and are impossible.[/quote]

well, for this one, i TRIED to make it make sense.  :( and failed.

I feel your pain, my first problem posted didn't make sense AT ALL!


I noticed that when you click on the profile, it gives you the option to change your username.  What happens if you do? will you lose your classes you are enrolled for?

Not sure what will happen generally, but you won't lose the classes you're enrolled in.


geometry
analytic geometry
geometric transformation
reflection
trigonometry

Let $ ABC$ be a right angled triangle of area 1. Let $ A'B'C'$ be the points obtained by reflecting $ A,B,C$ respectively, in their opposite sides. Find the area of $ \triangle A'B'C'.$

[hide="Considering the simplest case"]

I don't think this answer would work as a "proof" but if it was just an open question, I think it should work.

The question's wording is pretty general (just say about area 1) so we can set the triangles $ ABC$ to have sides of $ \sqrt{2}, \sqrt{2},$ and $ 2$. Putting this into the coordinates, we have $ A(0, \sqrt{2}), B(0,0),$ and $ C(\sqrt{2},0)$. Reflecting A over BC, we get $ A'(0,\minus{}\sqrt{2}), B'(\minus{}\sqrt{2},0)$ and $ C'(\sqrt{2},\sqrt{2})$. So, this resulted triangle has sides $ \sqrt{10},\sqrt{10},$ and $ 2$ and area of 3. [/hide]

I'm not sure if I understood the "reflecting" right though.

I am also wondering what do they mean by reflecting  :o any ideas?

is the problem asking if the points were reflected across the side of the triangle that's opposite it? For example A is reflected over side BC?

Hint: What is the height of $ A^\prime B^\prime C^\prime$ taken from $ A^\prime$?

This is what they mean by reflecting:

I'm reviving this because this problem hasn't been fully solved yet.

[hide="Help"]

I'm guessing then from the diagram, $ A'$ is the extension of altitude from $ A$ to $ BC$ when it meets with rays from $ B'$ and $ C'$. For convenience, I decided to use coordinates.

$ A(0,0), B(b,0), C(0, \frac{2}{b})$ since the area of $ ABC$ is 1.

Then:

$ B'(\minus{}b,0), C'(0, \frac{\minus{}2}{b})$

Now, all I need is to find the coordinate of $ A'$ but I'm unsuccessful at it. I found the length of altitude from $ A$, which turns out to be $ \frac{\sqrt{b^4\plus{}4}}{b}$ but this really doesn't get me nowhere (especially when I try to find the length $ CD$ where $ D$ is the point in $ BC$ that altitude from $ A$ touches, I get negative values... from Pythagorean Theorem..). So, hmm, any better idea?

I'm unsure how to find what cosinator asked either.  :oops: [/hide]

You don't have to use coordinates at all.

Since $ A^\prime$ is the reflection of $ A$, $ AA^\prime$ is perpendicular to $ BC$.  So, $ AA^\prime\equal{}2AH$ where $ H$ is the foot of the altitude from $ A$ to $ BC$.  Also, notice that $ AB^\prime\equal{}AB,AC^\prime\equal{}AC,\angle B^\prime AC^\prime\equal{}90$, so $ AH^\prime\equal{}AH$ where $ H^\prime$ is the foot of the altitude from $ A$ to $ B^\prime C^\prime$.  A little angle chasing shows that $ A,H,H^\prime$ are collinear.  So, the height fof $ A^\prime B^\prime C^\prime$ is $ 3AH$.  So, $ [A^\prime B^\prime C^\prime]\equal{}\frac{1}{2}B^\prime C^\prime \cdot3AH\equal{}\frac{3}{2}BC\cdot AH\equal{}3$ since $ \frac{1}{2}BC\cdot AH\equal{}[ABC]\equal{}1$.

Thank you cosinator! Your solution was concise and very clear!

Now I just have one question. I'm not sure why angle chasing was necessary to show $ A, H, H'$ were collinear. To me, it seemed pretty obvious (so what angle chasing was required to verify this then?).  :) I'm not so familiar with collinearity.

Well, you are correct in that it is fairly obvious.  They are collinear because $ \angle CAH\plus{}\angle CAB^\prime\plus{}\angle B^\prime AH^\prime\equal{}\angle C\plus{}\angle A\plus{}\angle B\equal{}180$.

[img]http://www.artofproblemsolving.com/Forum/download/file.php?id=15514&[/img]

Note $[AB'C']=1$.  Let the perpendicular line segment from $A$ to $BC$ equal $h$.  Then $AA'$ is $2h$.  Also, $AC=AC'=b$ and $AB=AB'=c$.  We have: \[[ABC]=\frac{hb\sin{\angle{CBA}}}{2}+\frac{hc\sin{\angle{CAB}}}{2}\] And also: \[[B'AA']+[AA'C']=hb\sin{\angle{CBA}}+hc\sin{\angle{CAB}}\]  Thus $[B'AA']+[AA'C']=2[ABC]=2$ and $\boxed{[A'B'C']=2+1=3}$

$\blacksquare$

Note that I've angle-chased a bit and also used the fact that $\sin{(180-x)}=\sin{x}$.

Let $\triangle ABC$ have a right angle at $B$. Note that points $A,B,C,B'$ are concyclic and let $BB'$ intersect $AC$ at $H$. Note also that $\angle ABB'=\angle C$, $\angle CBB'=\angle A$. Hence $[A'B'C']=[A'BC']+[B'BC']+[B'BC]=\frac{1}{2}ac+\frac{c^{3}a}{b^{2}}+\frac{a^{3}c}{b^{2}}=1+2=3$.

Complete the pentagon $AC'A'CB'$,now $A'C'=AC$,let the feet perpendicular from $B'$ to $AC,C'A'$ are $L,M$,also $M,B,L,B'$ lie on a line.again $3BL=B'M$,so $[A'B'C']=(A'C'.B'M)/2=3$.


Can someone help me with this problem?

Give a hand containing the nine numbered hearts from a deck of cards labeled 1 through 9, what is the probability of drawing 3 cards in sequence from the hand such that each selection results in a number larger than the previous one?

Thanks a lot.

Pretend that you've drawn your three cards, but you forgot the order in which you drew them.  There's 3!=6 ways you could have drawn them.  Of those, only 1 results in them being drawn in increasing order.  So the answer is 1/6.

[quote="breez"]I think it is 7 / (9 x 8 x 7)

So, around 1.38888888... %

There are 7 different permutations that contain card numbers with a common diff of 1. There are a total of 9 x 8 x 7 permutations possible.

The 7 are 3-9 inclusive, 1 and 2 do not count because they are < 3.[/quote]

The question didn't ask for them to be consecutive and increasing, just increasing.  If it was required that they be consecutive (i.e. 1/2/3, 2/3/4, 3/4/5, 4/5/6, 5/6/7, 6/7/8, or 7/8/9) then that'd be right.

I don't know, I did it this way, by sorting out the numbers:

If starting number $1$ and $2$:$1 \cdot 1 \cdot 7$
If $1$ and $3$: $1 \cdot 1 \cdot 6$
If $1$ and $4$: $1 \cdot 1 \cdot 5$, etc.

So if it starts out with $1$, we can have $7+6+5......+1=28$

If starting with $2$ and $3$: $1 \cdot 1 \cdot 6$
If $2$ and $4$: $1 \cdot 1 \cdot 5$, etc

So if starts out with $2$, we have $6+5..+1=21$

Thus we can assume that we have from starting from $1$ through $7$, since $8$ is not possible, we have $28+21+15+10+6+3+1=84$ and since the total number of ways to choose $3$ cards out of $9$ is $9P3=504$

EDIT: Realized my mistake

So I get $\frac{84}{504}=\boxed{\frac{1}{6}}$ :blush:

[quote="keta"]Ok, after 20 minutes of deep thinking, I got it, and thanks to everyone who helped.

The nine values you can draw are 1 through 9. Since no replacements are allowed, any 3 values we draw must be different, and of all those combinations of three, only one way can we make arrange them from least to greatest. So the event has 9C3 ways of happening, but the sample space is 9P3 = 9(8)(7).

So the total probability is (9C3)/(9*8*7) = 1/6, which agrees with Wumbate.[/quote]

Yes, I think I understand now from your explanation, because with $9C3$, order does not matter and even if you choose the numbers $9,8,7$ in that order, since in combinations, order doesn't matter, it can be rearranged to $7,8,9$ to fit the increasing pattern as desired and so it is $\frac{9C3}{9P3}$. Thank you for clearing that up :) But I still can't understand why my method didn't work ;)

So can someone explain to me what went wrong in my method?

[quote="Iversonfan2005"]I don't know, I did it this way, by sorting out the numbers:

If starting number $1$ and $2$:$1 \cdot 1 \cdot 7$
If $1$ and $3$: $1 \cdot 1 \cdot 6$
If $1$ and $4$: $1 \cdot 1 \cdot 5$, etc.

So if it starts out with $1$, we can have $7+6+5......+1=28$

If starting with $2$ and $3$: $1 \cdot 1 \cdot 6$
If $2$ and $4$: $1 \cdot 1 \cdot 5$, etc

So if starts out with $2$, we have $6+5..+1=21$

Thus we can assume that we have from starting from $1$ through $7$, since $8$ is not possible, we have $28+21+15+10+6+1=81$ and since the total number of ways to choose $3$ cards out of $9$ is $9P3=504$

So I get $\frac{81}{504}$, which obviously isn't right :o Can someone explain why my method doesn't work?[/quote]

You're missing the $3$ for if you start with $6$... So you get $84$ possibilities and $\frac{84}{504} = \frac{1}{6}$.


trigonometry
limit
real analysis
real analysis unsolved

we can rewrite product on the left as follows:

$\frac{\sin x}{\cos x}\left( \frac{2\sin x \cos x}{\cos 2x} \right)^{\frac{1}{2}}\left( \frac{4\sin x \cos x \cos 2x}{ \cos 4x} \right)^{\frac{1}{4}}...$

we can see that $(\frac{1}{\cos 2^nx})$ appears once with exponent $\frac{1}{2^n}$ wheres $cos 2^nx$ appears infinitely many times with exponent $\frac{1}{2^{n+1}}+\frac{1}{2^{n+2}}+...=\frac{1}{2^n}$. Hence we get rid of cosinuses.

now we have to prove that :$\prod (\sin x)^\frac{1}{2^n}=(\sin x)^2$ but that's obvious ($1+\frac{1}{2}+\frac{1}{4}+...=2$).
so we're left with proof of:

$\lim_{n \to \infty} \prod _{i=1} ^n \sqrt[2^i]{2^i}=4$ for which I don't have idea for now...  :oops:

well:
$\displaystyle \sum_{i=1}^\infty \frac{i}{2^i} = \sum_{i=1}^{\infty}\sum_{k=i}^\infty \frac1{2^k} = \sum_{i=0}^\infty \frac{1}{2^i} = 2$ ;)

Unfortunately we don't get rid of the cosines.

For all $ x\in\Bbb{C}\setminus \left(\Bbb{Z}\cdot \frac {\pi}{2^{n \plus{} 1}}\right)$

$ |\tan x|\cdot \left|\tan 2x\right|^{1/2} \cdot \left| \tan 4x \right|^{1/4}\cdots \left|\tan 2^n x\right|^{1/2^n}$

$ \equal{} 2^{\sum_{k \equal{} 0}^n k/2^k}\, |\sin x|^{\sum_{k \equal{} 0}^n 1/2^k}\, \left|\frac {1}{\cos x\,\cos 2x\cdots \cos 2^n x}\right|^{1/2^n}$ .

And this is

$ 2^{2 \minus{} (n \plus{} 2)/2^n}\, |\sin x|^{2 \minus{} 1/2^n}\, \left|\frac {2^{n \plus{} 1}\, \sin x}{\sin 2^{n \plus{} 1} x}\right|^{1/2^n} \equal{} \frac {4\, |\sin x|^2}{\left|2\sin 2^{n \plus{} 1} x\right|^{1/2^n}}$ .

Now the question is for which $ x$ does $ \left|\sin 2^n x\right|^{1/2^n}$ tend to 1 for $ n\to\infty$ ?

Maybe there are infinitely many $ n$ such that $ 2^n x$ approximates a multiple of $ \pi$ very well.

John Doe is right.

In fact, I think there is an $ x$ for which if $ |\sin(2^{n}x)|^{1/2^n}$ converges, it converges to $ 0$ and for all $ n, \sin(2^{n}x) \neq 0$.

To construct such an $ x$, we construct the binary representation of $ \beta \equal{} \frac {x}{\pi}$. We construct $ \beta$ as follows:

Pick any positive integer $ i_1$. Make the $ {i_1}^{th}$ digit of the fractional part of $ \beta \equal{} 1$ (fill in zeroes before that). Next pad it with $ 4^{i_1}$ zeroes after that. Now put a 1, which is at the $ {i_2}^{th}$ place, say. Now pad with $ 4^{i_2}$ zeroes, and so on.

Thus we get a sequence of integers $ i_1 < i_2 < ... < i_n < ...$ such that:

the fractional part of $ 2^{i_k}\beta$ is $ < 1/2^{4^{i_k}}$.

Now for $ n \equal{} i_k$, we have that

$ |\sin(2^{n}x)| \equal{} |\sin(2^{n}\pi\beta)| \equal{} |\sin(\pi\alpha)|$, where $ \alpha$ is the fractional part of $ 2^{n}\beta$.

Now,

$ |\sin(\pi\alpha)| < \pi\alpha < \pi/2^{4^n}$.

After raising this to the $ (1/2^n)^{th}$ power, the RHS still is small enough to $ \to 0$.


trigonometry geometry incenter circumcircle inequalities

Let be given a triangle $ABC$. Prove that:
$1.\quad\frac{3}{4}+\frac{1}{4}\sum\cos\frac{B-C}{2}\ge\sum\cos A$
$2.\quad\sum a\cos\frac{B-C}{2}\ge p\left(1+\frac{2r}{R}\right)$

Solution of 1.
Let O ,I ,H denote the circumcentre,incentre and orthocentre respectively.
Let AI cut the circimcircle at Q.

We will use these results from triangle geometry:

1.$\angle OAH$ =$B-C$
2.$AI$bisects $\angle OAH$,therefore $\angle OQA$ =$\angle OAQ$ =$\frac{B-C}{2}$
3.$AI$ =$4R{sin\frac{B}{2}}{sin\frac{C}{2}}$ =$rcosec\frac{A}{2}$
4.$IQ$ =$2Rsin\frac{A}{2}$

Also we will use these two inequalities which are true for any triangle:

5.$\sum{cosec\frac{A}{2}}$ $\geq$ 6
6.$\sum{sin\frac{A}{2}}$ $\geq$ $\sum{cosA}$

In triangle $OAQ$,     $cos\frac{B-C}{2}$ =   ${\frac{AQ}{2}}/R$.   Therefore,
$\frac{3}{4}$ +$\frac{1}{4}$.$\sum{\cos\frac{B-C}{2}}$ $\geq$ $\sum{cosA}$

$\Leftrightarrow$ $3$+$\sum{\frac{AQ}{2R}}$ $\geq$ $4\sum{cosA}$

$\Leftrightarrow$ $6R$ +$\sum{AQ}$ ${\geq}$ $8R(1+\frac{r}{R})$   using $\sum{cosA}$ =$1+\frac{r}{R}$

$\Leftrightarrow$ $\sum{AQ}$ $\geq$ $2R+8r$

$\Leftrightarrow$ $\sum{AI}$ +$\sum{IQ}$ $\geq$ $2R+8r$ (AQ=AI+IQ)

$\Leftrightarrow$ $\sum{rcosec{\frac{A}{2}}}$ +$\sum{2R\sin{\frac{A}{2}}}$ $\geq$ $2R+8r$......$*$ using (3) and (4)

Now using the inequality in( 5) we get $\sum{rcosec{\frac{A}{2}}}$$\geq$ $6r$............(7) and using the inequality in (6) we get

$\sum2R{sin\frac{A}{2}}$ $\geq$ $2R\sum{cosA}$ =$2R.(1+\frac{r}{R})$ =$2R+2r$................(8)

Adding the inequalities (7) and (8) we get the inequality $(*)$

Proof of inequality (5) :We use

$\sum{cosec\frac{A}{2}}$.$\sum{sin\frac{A}{2}}$ $\geq$ $9$  ( AM-GM)

$\sum{sin\frac{A}{2}}$ $\leq$ $3\sin{\frac{A+B+C}{6}}$ ( Jensen)

Proof of inequality (6):

$cosA+cosB$  =$2cos{\frac{A+B}{2}}.cos{\frac{A-B}{2}}$ =$2sin{\frac{C}{2}}.cos{\frac{A-B}{2}}$ $\leq$ $2sin{\frac{C}{2}}$
Add this  inequality to two similar ones obtained by cyclic substitution.

[quote="April"]Let be given a triangle $ABC$. Prove that:
$1.\quad\frac{3}{4}+\frac{1}{4}\sum\cos\frac{B-C}{2}\ge\sum\cos A\quad (*)$
[/quote]
[b]Solution (Unknow author)[/b] :D
\[(*) \Longleftrightarrow 4\sum\cos A\le 3+\sum\cos\frac{B-C}{2}\\ \Longleftrightarrow 4\sum\cos A\le 3\sum\left[\cos\frac{B-C}{2}-\cos \frac{B+C}{2}\right]+\sum\sin\frac{A}{2}\\ \Longleftrightarrow 4\sum\cos A\le 3+2\sum\sin\frac{B}{2}\sin\frac{C}{2}+\sum\sin\frac{A}{2}\]
We know: $\sum\cos A\le\sum\sin\frac{A}{2}$ then we have to prove that:
\[3\sum\cos A\le 3+2\sum\sin\frac{B}{2}\sin\frac{C}{2}\]
But it’s simple because:
\[\sum\sin\frac{B}{2}\sin\frac{C}{2}\ge 3\sqrt[3]{\prod\sin^{2}\frac{A}{2}}=3\prod\sin\frac{A}{2}\cdot\frac{1}{\sqrt[3]{\prod\sin\frac{A}{2}}}\ge6\prod\sin\frac{A}{2}\]

\[\Longrightarrow 12\pro\sin\frac{A}{2}\le 2\sum\sin\frac{B}{2}\sin\frac{C}{2}\]

\[\Longrightarrow 3\left(\sum\cos A-1\right)\le 2\sum\sin\frac{B}{2}\sin\frac{C}{2}\]

\[\Longrightarrow 3\sum\cos A\le 3+2\sum\sin\frac{B}{2}\sin\frac{C}{2}\quad\mathbb{QED}\]