geometry
geometric transformation
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articles
reflection

Hello,

First than all, if I say something that is not totally correct, I confess: [b]I know basically nothing about asian languages[/b]. But one of my office mates and my advisor are chineses (and some other friends too), and they answer some of my questions. Now, from the next quote comes from [url=http://www2.ignatius.edu/faculty/turner/languages.htm]This Webpage[/url]:

[quote]The Summer Institute for Linguistics (SIL) Ethnologue Survey (1999) lists the following as the top languages by population:
(number of native speakers in parentheses)

   1. Chinese* (937,132,000)
   2. Spanish (332,000,000)
   3. English (322,000,000)
   4. Bengali (189,000,000)
   5. Hindi/Urdu (182,000,000)
   6. Arabic* (174,950,000)
   7. Portuguese (170,000,000)
   8. Russian (170,000,000)
   9. Japanese (125,000,000)
  10. German (98,000,000)
  11. French* (79,572,000)

* The totals given for Chinese, Arabic, and French include more than one SIL variety[/quote]
And latter on they make the aclaration
[quote]* "Chinese" for this chart is composed of the total of the following SIL designated languages:
	Mandarin 		836,000,000
	Xiang 		36,015,000
	Hakka 		34,000,000
	Gan 		20,580,000
	Minbei 		10,537,000 	[/quote]
Well, if you consider that Mandarin is [b]spoken[/b] by 836 millions, then, it would be without any doubt the principal native language of the world. Nevertheless chineses that I know (including my advisor) say that Mandarin Speakers that comes from different places of China, do not understand each other if they are speaking (even when they understand each other perfectly if they are writting), but just locally they really can [b]speak[/b] their Mandarin. Therefore, depending on your definition of speakers, it would not be the most widely speaking mother language, but the most widely writing mother language.

The interesting part is that, as a native english speaker, and as a second-language english speaker, I can say that in both of them, even when you can spend some time trying to [i]get the accent[/i] (sometimes I have problem understanding Australians, but my spoken english is very American), nevertheless I can speak fluently in spanish with basically every native spanish speaker, and I claim I could do the same in english.

Hence, would you say that Mandarin (or Chinese) is really [u]the world's most widely spoken native language[/u]?

I would like to hear the opinion of other people in the forum, and I am craving to read specifically the opinion of chinese people, because as I already said, I know basically nothing about chinese, and everything I am saying about it is second hand information.

[b]Remark[/b]: I honestly spect that chinese MathLinkers don't feel offended by anything I said here, if so, I appologize, I tried to express everything in the more neutral way, as a sincere question!

Having moved to the US early, I have lost a lot of Chinese fluency, but while there are very different accents, it is still often comprehensible.  Also, many Chinese can speak both standard Mandarin and their local dialects (possibly similar to Mandarin).  Furthermore, Mandarin is becoming more and more standard.  I'm not sure about specific numbers, but I think the impression I get from your post is somewhat inaccurate.  However, there certainly are many different accents and dialects, so if you wanted the language to be as consistent as English, your number might become something like 550 million.  Of course this is just an estimate--someone living in China could give you better information.

Well yea duh, China has the largest population in the world, and Mandarin is the most common Chinese dialet. When most people say "Chinese", most of the time they refer to "Mandarin", or "Cantonese". I'm surprised Cantonese didn't appear on the chart.

I've never heard of the last four types of Chinese dialets on the second chart. ;) So yea I guess Mandarin is still the most widely spoken language in the world, because like almost all Chinese can speak Mandarin, and that's a whole lot of people. It's like the basic language you have to learn in China.

[quote="shobber"]Well, i wonder how do you define Madarin??
I think that Chinese [b]is[/b] the most widely spoken native language. There are only some differences in the pronunciation like british english and amerian english, but they are the same language![/quote]
Yes, but my point is that an American can easely understand a British and viseversa, nevertheless according to some chineses I know, in China it is almost impossible to understand other people if they are from far away of your place!

many chinese speaks more than 1 dialect. my dad speaks 3

[quote="sheepwarrior"]i'm surprsied the number for "chinese" is only 900,000,000 actually... but neways
i think u should count mandarin as number 1 language because even though there are many dialects, as probability1.01 said they are still based on the mandarin language. when he said they can speak mandarin and their local dialects i think thats sorta true but there are also people whose mandarin just sounds different i dont know what he meant by mandarin and local dialects because the local dialects are mandarin just with different sounds and pronunciations
im no expert on this and sorry for being confusing but id still say mandarin is #1 language because number 2 on the list was 1/3 of number 1 anyways  :D[/quote]
I speak both Mandarin and Shanghainese, and knowing the former is not near sufficient to understand the latter. Shanghainese is probably derived from Mandarin, but it's too different to be counted as Mandarin.

[quote="shobber"]Well, i wonder how do you define Madarin??
I think that Chinese [b]is[/b] the most widely spoken native language. There are only some differences in the pronunciation like british english and amerian english, but they are the same language![/quote]
Well see in Mandarin there are four accents (1, 2, 3, 4) while in Cantonese there are nine accents.
So I suppose that's a way to define Mandarin.

[quote="djimenez"] Nevertheless chineses that I know (including my advisor) say that Mandarin Speakers that comes from different places of China, do not understand each other if they are speaking (even when they understand each other perfectly if they are writting), but just locally they really can [b]speak[/b] their Mandarin. Therefore, depending on your definition of speakers, it would not be the most widely speaking mother language, but the most widely writing mother language.
[/quote]
We can understand each other in Mandarin (offcially called Pu Tong Hua) and we also have accents. But we do not understand their dialects. People from South China have stronger accents, esp from Fujian and Guangdong. I am from Fuzhou, Fujian. And I do not speak Mandarin correctly for all sounds (that is an accent), but they can understand me well.

Okay, this stuff with the "dialects" being counted as different languages and whatnot is just wrong. It's kind of like saying we should split English into "American English" and "British English" because the Brits use words like buggy. And furthermore, we should create different categories for "New York English", "Texas English", "California English", and etc to represent every region that employs its own colloquialisms.

All of the regional dialects that are under discussion are merely ways of accenting "Pu Tong Hua" that have been adopted by those areas. And frankly, no matter how different two dialects of Mandarin are, anyone who has a solid foundation in the language (a decade or more) should be able to be the general gist of another speaker's message. A lot of adults don't know what "top dawg" or "sup" means, but that doesn't mean they speak a different language. I know northerners that have never heard anyone say "ya'll" but that doesn't alienate them from the English language. There's less than a half billion fluent English speakers and we already have different accents. Imagine what would happen if we doubled (or, probably more appropriately, tripled) that number.

Cantonese and Taiwanese (which isn't really a separate language at all, btw) may be arguably a greater departure from mainstream Mandarin, but even those are by no means separate languages.

Imagine what Madrid would think if Mexico City all of a sudden said they speak Mexican instead of Spanish.

Anyway, first, let's correct a few misconceptions:
-Mandarin has four tones plus one neutral tone. So that means there are five tones total.
-Cantonese is reported to have six tones. (Personally, I don't know how true that is because although I speak the language fluently, I could never identify only six distinct tones. Some people report that there are nine tones. So I guess the number of tones range from six to nine, depending on whom you ask.)

I really don't think that understanding people who speak Mandarin with different accents is a problem. That's why it's called "Pu Tong Hua," or literally, the "Common Language." (It's also known as "Guo Yu", or the "National Language," but personally, I have some issues with that name.  :P )

Now, onto the dialect issue.  :lol:

This is actually quite tricky to answer. Each dialect is distinctly different, and there is actually no guarantee that knowing one will allow you to even partially understand another.

This next analogy is a stretch, but I'm going to try it anyway. Think of all the Romantic languages that are in existence today. One could argue that if you spoke Latin, then although you would not be able to converse in each of the Romantic languages, you would be able to understand a lot of it and get the main idea. As you know, this isn't really true, but your knowledge of Latin certainly doesn't hurt.

The various Chinese dialects are like this in a sense. Each dialect carries is own character and flair. If you are fluent in more than one dialect, then you'll notice that each time you speak another dialect, you assume a new persona and attitude. What's more, even people who spent their life speaking one dialect of Chinese (so they are quite fluent) may not necessarily understand other dialects. This is most evident in the elderly generations who had their language training before the Cultural Revolution.

In the past, the various Chinese dialects were united under a strong leadership, and thus, for political and historical reasons, all Chinese dialects are [i]written[/i] in the same way. Still, they are spoken very differently. For example, in Cantonese, you speak in patois and would usually not speak as you would write.

Today, the Chinese dialects are united under technology. For example, when movies are filmed today, actors and actresses speak their native tongue. Later, the editing crew would then record the appropriate voice over in the desired language. Personally, this drives me crazy when I watch the films. But *shrugs* what can I do?

Hey guys,

Thank you very much for your opinions and comments. I have another question. I was talking to a friend, a chinese girl, one of these days, and talking to here I told her how [b]China[/b] is pronounced in Spanish, and she told me that it sounds like an offensive word in chinese, but she didn't want to tell me what exactly mean. We don't pronounce the [b]i[/b] in China as in english, it is more like the vowel sound in [b]green[/b] or in [b]Teena[/b], you could probably say that we pronounce [b]Cheena[/b]. Could someone explain me what does it mean?

[/quote]
I speak both Mandarin and Shanghainese, and knowing the former is not near sufficient to understand the latter. Shanghainese is probably derived from Mandarin, but it's too different to be counted as Mandarin.[/quote]

Shanghainese is not derived from mandarin. i believe, i may be wrong, some of it is derived from european languages.  Shanghainese is totally different from mandarin.

so where do u actually live....California...???

Yeah I live in California. Born in California too.

I was wondering if any of you guys read the China's Century NEWSWEEK.  Inside it there was an article about how more and more American schools are offering Chinese. Pretty nice...

yeah, I have.  But I've heard that Chinese is an impossibly difficult language to learn, unlike Spanish, which is ridiculously easy to learn (at least for me).  I would still choose Spanish over Chinese if I had the choice.

In addition, Chinese isn't that useful yet for us to learn because in our daily experience, we won't use Chinese too often.  The only time we'd use is when we're in China.  On the other hand, there are 17 million + spanish-speaking people in the U.S., so Spanish is very useful.  And if you're on vacation in Mexico, Spain, Latin America...

However, nothing you learn in school is useful.  I don't think that's the point of school--if it were, we should go back to the apprentice system of the 1700s.  School is to stretch your mind and grow--education for education's sake, so in that case, it'd probably be better to learn Chinese.

But it's not about now but the future, soon China will become a major force (in fact it is already) in world economics and politics so knowing Chinese will definitely give you an edge in society.

[quote="fedwinri"][quote="ambierona"]Yeah it's a Cuban flag. I'm half Cuban and half Chinese, so I keep changing my flag from Cuba to Chinese.. maybe I should put the US flag there.. but I put California as my location so that's cool.[/quote]Cool.  Just a thought: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=239316#p239316]How do you feel about this?:[/url]  "Bush got 566..."  j/k.
"One reason why I wonder what it means to be American is because I'm not Caucasian. ...  Now, I'm proud of my ethnic heritage but I'm a little bothered by the fact that no matter how many generations I and my descendants live in the U.S., we may still be considered less than full-blooded Americans."
It's an interesting remark (and understandable).

I will say that this statement strikes me differently knowing you're (a) Asian--sorry if that sounds reminiscient of the other (linked above) thread--(b) in America.  What else would one learn--besides English (over Spanish??) when approaching an age like two?[quote="ambierona"]... I wish I could speak Chinese.. grr I was stupid though and wanted to learn English when I was 2.[/quote][/quote]

Well, about that remark thing.. I don't think there are any "full-blooded Americans." So yeah.. the closest you can get is a Native American or Indian or whatever we call that race to be politically correct. And America is a nation of all these mixed races, so I think I'm pretty American.

Uh, I was actually speaking Chinese (Cantonese) and Spanish until I was 2.. because my parents spoke that to me. But then my older brother got into preschool without knowing English, and people made fun of him, he came home crying, and learned English. And I wanted to be just like him, so I learned English and forgot Chinese and Spanish. But it would have been nice to keep up three languages, so I could communicate with my grandparents. (that's what I would use Chinese and Spanish for in the real world, communicating with relatives... but yeah, here in California, Spanish is more useful to know for like going to Mexico or something)

[quote="tokenadult"][quote="juicybooty911"]mandarin and taiwanese are actually very very similar[/quote]

English and German are more similar, but I don't see many English speakers counting themselves as speakers of German just because of that similarity.[/quote]

I would like to tell you, mandarin and taiwanese may be a little different in accent, but, the grammar is the same. Well, I think the grammar of English and German might be totally different(just guess).

Just wondering, why you have almost every opinion on the opposite side of China in almost every topic? such as, you insist on to find the difference between Mandarin and taiwanese--a branch of Chinese.

Besides, I can tell you, I am from south Jiangsu, if you would like difference between Mandarin and other branches of Chinese, hmm, I would like to tell you: personally, I think my native tongue(similar to shanghainese, called Wu)  has greater difference between mandarin than taiwanese(although my hometown is nearer to centre China than taiwan) or cartoonese, many people say that it has similar accent with janpanese.

[quote="tokenadult"][quote="Zratonx"]no matter wut dialect ppl in China speak they write in the same exact words.[/quote]

That is a factually incorrect statement. A lot of visitors to Hong Kong who look carefully at local writing in Han characters will see Han characters used there that aren't used anywhere in China where Cantonese is not spoken. The same is true, but less frequent because of the pervasiveness of Mandarin as a second language, in Taiwan.

There has always been a tendency in the writing of people who are literate at all in Chinese (which was always a small minority of Chinese people until very recently) to write in patterns that reflect the standard language, even if that standard language is not the normal pattern of speech for that speaker. Thus, Cantonese speakers and Taiwanese speakers and Wu speakers are implicitly learning Mandarin when they learn to write in Han characters. But even at that most speakers of Sinitic languages write as they talk--as all human beings do when they write--and that means there are various forms of "mistaken" usage (from the point of view of the standard national language) that creep into everyday writing. I have had to deal with that as a translator on many occasions.[/quote]

I speak Wu.
I can ensure that Wu and Mandarin only have difference in accent.
Give an example:
English: Hello, what is your name?
Mandarin: Ni3 hao3, ni1 shi4 she2 me0 ming2 zi4?
Wu:          Ni  ho   ,  ni   si    se    me   min    si?  (wu doesn't have tones)

besides, word "what" , in mandarin is she2 me0, in wu is se me, but it could also be accented as "so ge" or "sa ge"
or "dia gu" in Wu.

[quote="Alexander Lenin"][quote="tokenadult"][quote="juicybooty911"]mandarin and taiwanese are actually very very similar[/quote]

English and German are more similar, but I don't see many English speakers counting themselves as speakers of German just because of that similarity.[/quote]

I would like to tell you, mandarin and taiwanese may be a little different in accent, but, the grammar is the same. Well, I think the grammar of English and German might be totally different(just guess).

[b]Just wondering, why you have almost every opinion on the opposite side of China in almost every topic? such as, you insist on to find the difference between Mandarin and taiwanese--a branch of Chinese.[/b]

Besides, I can tell you, I am from south Jiangsu, if you would like difference between Mandarin and other branches of Chinese, hmm, I would like to tell you: personally, I think my native tongue(similar to shanghainese, called Wu)  has greater difference between mandarin than taiwanese(although my hometown is nearer to centre China than taiwan) or cartoonese, many people say that it has similar accent with janpanese.[/quote]

First time poster here,

I don't think there is any political intentions behind his posts.
You are reading too much into his posts.

As a Taiwanese,  i can tell you Taiwanese(Minnan) is in fact very different from Mandarin.
Just look at the origination of each dialect, one (Minnan/Taiwanese) originated from southern coast of china  whereas the other (Mandarin) originated from the very northeastern portion of China (Manchuria, bordering north Korea). Heck, mandarin and Manchurians were considered as both barbaric and barbarians until the establishment of the Qing Dynasty.

And  take my professor for my Chinese Film and Literature  class for example, She is a native taiwanese, as in born in Taiwan. She graduated from NTU(Natiaonal Taiwan University) and received her Ph.D from Stanford.  She speaks perfect Mandarin, yet, she told us can neither speak nor understand Taiwanese.

Heck, just look at me, I was not taught Taiwanese as a kid growing up.
Even though i hear it spoken around me all the time, I can not speak Taiwanese or understand it when spoken to me.

And the Taiwanese "writting system" is somewhat different from the standard Chinese writting system.
For example, i was watching cable the other day, and a Taiwanese song came on.  I was reading the lyrics, and the usage of some of the characters made me go "eh?""

[quote="Zratonx"]manchurian was barbaric until the Qing Dynasty cause they ruled China in the Qing Dynasty, how the heck is mandarin barbaric come on ever since the Zhou Dynasty the Han have been the predominant force and gained total control after the Qin Dynasty.....

the character differences r prob just traditional and simplied differences....[/quote]

Mandarin is the language of the Manchurians who like you said, were considered as barbarians until they established the Qing dynasty.

And no, Mandarin was spoken  during the Han dynasty. If anything else, the common dialect spoken in the courts of the the Han dynasty is closer to your souther coastal dialects such as fujiannese.

Another example, dialect of courts of the Tang dynasty is closer to Cantonese. Some tang poems will not rhyme when read in mandarin, however, it will if you read it using cantonese.

thanks very mucgh about the language lesson about mandarin...i really had no clue....

 :blush:

 but still mandarin is spoken but most ppl today so it is...well no longer barbaric...

[quote="Valentin Vornicu"]Actually the English and German grammar are quite different, rather than similar. They have some common words, but that's it. You might say that the French and English grammars are similar :)[/quote]I'll say that English actually sounds similar to German in some respects and gets a lot of things from it--enough to be considered non-romantic.  It is true, when comparing structure (except for our idea of conjugating verbs--English) and roots of many words, English ties more closely to the Latin than anything else; i.e., yay, Christian church and intercom. Btween France and also Mexico, etc.  ~~
None of that jaunt to take away from your good Chinese discussion forum (only that I have nothing to say for any kind of Chinese--or not-Chinese).


limit
integration
induction
real analysis
real analysis solved

proove that:
a) the sequence x_n=\sum{k=1,n}(1/(n+k)) is monotonous.
b)there exists a sequence (a_n) \in {0,1} s.t.
lim{n->oo}\sum{k=1,n}(a_k/(n+k)).

I don't think the statement for (b) is complete. We're supposed to show that there is a sequence (an)n>=1 with values 0 or 1 s.t. that limit Kuba mentioned is 1/2.

Statement b) is not complete , I just looked over the subjects in 2001 and the problem goes like so:

Prove that there exits a sequence (an)n>=1 with values of 0 or 1 so that

lim(n-->) (a1/(n+1)+...+an/(n+n))=1/2.
(Radu Gologan)

OKay... lets post a solution:

try to construct this sequence using induction, so that we have:

1/2<=sum(k=1,n) ak/(n+k)<=1/2+1/(2n)

because we have 1/2<=1/3+1/4<=1/2+1/4 we can choose a1=1 and a2=1. Now lets consider that all numbers a1,..,an were chosen so that we have: 1/2<=sum(k=1,n) ak/(n+k)<=1/2+1/(2n)
now we need to consider two cases:
i) if a1/(n+2)+..+an/(2n+1)>=1/2 then we take a_(n+1)=0
ii) if a1/(n+2)+..+an/(2n+1)<1/2 then we take a_(n+1)=1
(see what happens in both cases)

then the limit is 1/2.

cheers! :D :D


calculus
integration
real analysis
real analysis unsolved

Calculate

$ \int{\frac{t^2}{\sqrt{t^4\plus{}t^2\plus{}1}}}$

[quote="xxxxtt"]but can it be computed??[/quote]

It's an elliptic integral.  So, yes, it can be computed in terms of elliptic integrals E and F.
The zeros in the denominator are all on the unit circle, so this may be a Crystoffel-type integral which maps the circle to a certain polygon.  I'd have to look it up to be sure.


Find the three roots such that the equation $ x^3\minus{}x\plus{}k\equal{}0\ (k>0)$ has imaginary root with its absolute value 1.

[hide="Solution"]Let $ a\plus{}bi$, $ a\minus{}bi$, and $ r$ be the roots. Since the absolute value of the imaginary roots is $ 1$, we know that $ a^{2}\plus{}b^{2}\equal{}1$.

Then, we have:
$ (x\minus{}(a\plus{}bi))(x\minus{}(a\minus{}bi))(x\minus{}r)\equal{}(x^{2}\minus{}2ax\plus{}(a^{2}\plus{}b^{2}))(x\minus{}r)\equal{}$
$ x^{3}\minus{}(2a\plus{}r)x^{2}\plus{}(a^{2}\plus{}b^{2}\plus{}2ar)x\minus{}r(a^{2}\plus{}b^{2})\equal{}x^{3}\minus{}(2a\plus{}r)x^{2}\plus{}(1\plus{}2ar)x\minus{}r$

Comparing to the given equation, we get, $ 2a\plus{}r\equal{}0$ and $ 1\plus{}2ar\equal{}\minus{}1$. Solving, we get $ r\equal{}\pm \sqrt{2}, a\equal{}\mp \frac{\sqrt{2}}{2}$. Since $ k > 0$, we can eliminate $ r\equal{}\sqrt{2}$ as a solution. So, we have $ r\equal{}\minus{} \sqrt{2}$, $ a\equal{}\frac{\sqrt{2}}{2}$, and $ b\equal{}\frac{\sqrt{2}}{2}$. The roots are $ \boxed{\minus{}\sqrt{2}, \frac{\sqrt{2}}{2}\plus{}\frac{\sqrt{2}}{2}i, \frac{\sqrt{2}}{2}\minus{}\frac{\sqrt{2}}{2}i}$.[/hide]

That's correct answer.
You can use directly Vieta's fomula.


inequalities
integration
geometry
calculus
derivative
complex analysis
complex analysis unsolved

Suppose $f$ is analytic on a set that contains the closed disk centered at $a$ of radius $R.$ Prove that

$|f(a)|^2\le\frac1{\pi R^2}\int_0^R\int_0^{2\pi}|f(a+re^{i\theta}|^2r\,d\theta\,dr$

The challenge: can you find at leat two [i]different[/i] proofs of this?

$f(z)=p(z)+iq(z)$ where $p,q: \mathbb{C}\to \mathbb{R}$ are harmonic functions. Hence:
$p(a)=\frac{1}{\pi R^2} \int \int_D p(z)dz$, $p(a)^2 \leq \frac{1}{\pi R^2} \int \int_D p(z)^2dz$.
Similarly:
$q(a)^2 \leq \frac{1}{\pi R^2} \int \int_D q(z)^2dz$.
This is why:
$|f(a)|^2=p(a)^2+q(a)^2\leq \frac{1}{\pi R^2} \int \int_D p(z)^2+p(z)^2 dz$,
$|f(a)|^2 \leq \frac{1}{\pi R^2} \int \int_D |f(z)|^2 dz$,

$D$ the unit open disk of $\mathbb{C}$
$f: D \to  \mathbb{C}$ an injective analytic function.
Prove that the area $A$ of $f(D)$ is greater than $\pi |f'(0)|^2$.

That should be $\ge$; we don't want a problem when $f$ is multiplication by a constant.

The proof is an easy application of the theorem already discussed and the change-of-variables formula for multiple integrals. $\int\int_{f(D)}1\,dA=\int\int_{f(D)}\det f'(f^{-1}(x,y))\cdot \det(f^{-1})'(x,y)\,dx\,dy=\int\int_D \det f'(x,y)\,dx\,dy$.
Those derivatives are $2\times 2$ matrices; the determinant of $f'$ in the $\mathbb{R}^2$ sense is $|f'|^2$ in the complex sense.

The Fourier series argument:

$f(z)=\sum_{n=0}^{\infty}c_n(z-a)^n,$ with radius of convergence at least $R.$

Thus, $f(a+re^{i\theta})=\sum_{n=0}^{\infty}c_nr^ne^{in\theta}$

So $\int_0^{2\pi}|f(a+re^{i\theta}|^2\,d\theta=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty} \int_0^{2\pi}c_n\overline{c_m}\,r^nr^me^{in\theta}e^{-im\theta}\,d\theta$

$=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_n\overline{c_m}\,r^{n+m}2\pi\delta_{nm} =\sum_{n=0}^{\infty}|c_n|^22\pi r^{2n}$

So $\frac1{\pi R^2}\int_0^R\int_0^{2\pi}|f(a+re^{i\theta}|^2r\,d\theta\,dr= \sum_{n=0}^{\infty}\frac{2|c_n|^2}{R^2}\int_0^Rr^{2n+1}\,dr$

$=\sum_{n=0}^{\infty}|c_n|^2\frac{R^{2n}}{n+1}\ge |c_0|^2=|f(a)|^2.$


Mel Colly was found dead in his library. His forehead was resting on the keys of a tape recorder, a pool of blood lay on the desk beside him. From the door, Shadow could see that Mel's right hand covered the gun and that a bullet had entered his right temple. Shadow removed the tape recorder from under Mel's head and pressed the play button. A voice said,"This is Mel Colly speaking. I am not going to go on with a long winded farewell. I'm tired of life and all its anguish. I'll see you on the other side, wherever that may be." There was a gunshot and the sound of his head hitting the tape recorder keys, followed by a click indicating the tape recorder had been shut off. Shadow was certain that someone else had imitated Mel's voice and made this recording to sound like a suicide. Why does Shadow suspect murder?

Feel free to ask questions.

[hide="classic..."]Someone else must have shut the tape recorder off after Mel died.[/hide]

[hide="Something Else"]His left hand could've landed on a switch or something that turned it off.[/hide]

When his head hit the tape recorder it turned off...

was there someone else present in the room at that time?
or...
shadow is teh murderer!! lol, no probably not...
hmm....tricky, tricky, tricky.....

[hide]If Mel Colly shut off the tape recorder when he died by hitting it with his forhead then the recording should've left off at the point after he died.

But when Shadow played the recording it started from the beginning, but if Mel Colly really killed himself Shadow would've had to rewind the recorder to listen to it.

So another person (murderer) would have had to rewound the tape recorder so that whoever picks it up can hear the fake message.[/hide]

am i right?


Putnam
number theory unsolved
number theory

This problem may have been already posted in this forum.I am sorry if it is the case.
Find all real numbers $ x$ such that for all integer $ n$, $ n^x$ is an integer.

$ x$ should be integer or $ n\equal{}a^{x}$ where $ a$ is integer,and $ x$ is Real number.

[url]http://www.mathlinks.ro/viewtopic.php?t=24042[/url] (Putnam problem)
I am just posting a post there, asking some clarification...


inequalities
geometry
geometric transformation
reflection
inequalities proposed

For $ a,b,c$ positive reals and $ k > 1$ prove that

$ \sum \sqrt[k] {\frac{a^2\plus{}b^2}{c^2\plus{}ab}} \leq \frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}}$

[quote="manlio"]Dear can_hang2007,

have you found a solution for this very nice and very hard inequality?

Thank you very much. :)[/quote]
Thanks Manlio. I haven't seen this problem yet. I'll try.

[quote="manlio"]For $ a,b,c$ positive reals and $ k > 1$ prove that

$ \sum \sqrt [k] {\frac {a^2 \plus{} b^2}{c^2 \plus{} ab}} \leq \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}}$[/quote]
The inequality is not true for $ 2 \ge k \ge 1$ (for example $ a\equal{}1.4,b\equal{}c\equal{}1$), for $ k \ge 3$, I got the following solution

[quote="manlio"]For $ a,b,c$ positive reals and $ k > 1$ prove that

$ \sum \sqrt [k] {\frac {a^2 \plus{} b^2}{c^2 \plus{} ab}} \leq \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}}$[/quote]
Dear Manlio,
I don't recall that I posted the problem in this form; more specifically, I don't know where the $ k\ge 1$ is from. But my original post can be seen here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=113060[/url]
Since then, a more or less direct proof of it can be found in Ho Phu Thai's solution to Math. Reflections S27 [url]http://reflections.awesomemath.org/2006_6/2006_6_solutions.pdf[/url]

Dear Ductrung,

sorry for my misprint  :blush: ; I copied it from VIF(Vietnam Inequality Forum -- http://batdangthuc.net/forum/showthread.php?t=125.) and I supposed $ k \geq 1$ but I was wrong. Sorry again.

Dear Can_hang,

thank you very much for your solution.

[quote="manlio"]Dear can_hang2007,

in your solution you say: "....According to my lemma at Mathematical Reflection."

but I don't remember you published it. Do you refer at ABC theorem?[/quote]
I don't know what's ABC theorem at all, the lemma I told is in the article "On a class three variable inequality".
As in the theorem, we have $ \frac {1}{27}(p \plus{} q)^2(p \minus{} 2q) \le r \le \frac {1}{27} (p \minus{} q)^2(p \plus{} 2q)$, now if we denote $ x \equal{} \frac {p \minus{} q}{3} \ge 0, y \equal{} \frac {p \plus{} 2q}{3} \ge 0$ then $ r\le x^2y$.  Moreover, it's easy for us to check that
\[ p \equal{} a \plus{} b \plus{} c \equal{} 2x \plus{} y
\]

\[ ab \plus{} bc \plus{} ca \equal{} x^2 \plus{} 2xy \equal{} \frac {p^2 \minus{} q^2}{3}
\]
That's what I want to talk about. :)

[quote="can_hang2007"]I don't know what's ABC theorem at all, the lemma I told is in the article "On a class three variable inequality".
As in the theorem, we have $ \frac {1}{27}(p \plus{} q)^2(p \minus{} 2q) \le r \le \frac {1}{27} (p \minus{} q)^2(p \plus{} 2q)$, now if we denote $ x \equal{} \frac {p \minus{} q}{3} \ge 0, y \equal{} \frac {p \plus{} 2q}{3} \ge 0$ then $ r\le x^2y$.  Moreover, it's easy for us to check that
\[ p \equal{} a \plus{} b \plus{} c \equal{} 2x \plus{} y
\]

\[ ab \plus{} bc \plus{} ca \equal{} x^2 \plus{} 2xy \equal{} \frac {p^2 \minus{} q^2}{3}
\]
That's what I want to talk about. :)[/quote]

Sorry if I am so stupid, but I don't understand why, from $ r\leq x^2y$ you can deduce that it suffices to consider the expression in your proof only for $ b \equal{} c \equal{} 1$  :oops:

$ \frac{(\sum a)^3}{\sum a^2} \plus{} 6\frac{(\sum a^2)^2}{\sum a^3 \plus{} 3abc} \geq 6\sum a$

[quote="manlio"][quote="can_hang2007"]I don't know what's ABC theorem at all, the lemma I told is in the article "On a class three variable inequality".
As in the theorem, we have $ \frac {1}{27}(p \plus{} q)^2(p \minus{} 2q) \le r \le \frac {1}{27} (p \minus{} q)^2(p \plus{} 2q)$, now if we denote $ x \equal{} \frac {p \minus{} q}{3} \ge 0, y \equal{} \frac {p \plus{} 2q}{3} \ge 0$ then $ r\le x^2y$.  Moreover, it's easy for us to check that
\[ p \equal{} a \plus{} b \plus{} c \equal{} 2x \plus{} y
\]

\[ ab \plus{} bc \plus{} ca \equal{} x^2 \plus{} 2xy \equal{} \frac {p^2 \minus{} q^2}{3}
\]
That's what I want to talk about. :)[/quote]

Sorry if I am so stupid, but I don't understand why, from $ r\leq x^2y$ you can deduce that it suffices to consider the expression in your proof only for $ b \equal{} c \equal{} 1$  :oops:

$ \frac {(\sum a)^3}{\sum a^2} \plus{} 6\frac {(\sum a^2)^2}{\sum a^3 \plus{} 3abc} \geq 6\sum a$[/quote]
Well, if you change the inequality into $ p,t,r$ ($ t\equal{}ab\plus{}bc\plus{}ca$), then it becomes
\[ \frac{p^3}{p^2\minus{}2t}\plus{}\frac{6(p^2\minus{}2t)^2}{p^3\minus{}3pt\plus{}6r} \ge 6p
\]
Now,
\[ \frac{p^3}{p^2\minus{}2t}\plus{}\frac{6(p^2\minus{}2t)^2}{p^3\minus{}3pt\plus{}6r} \ge \frac{p^3}{p^2\minus{}2t}\plus{}\frac{6(p^2\minus{}2t)^2}{p^3\minus{}3pt\plus{}6x^2y}
\]
It suffices to prove
\[ \frac{p^3}{p^2\minus{}2t}\plus{}\frac{6(p^2\minus{}2t)^2}{p^3\minus{}3pt\plus{}6x^2y} \ge 6p
\]
Now, replace $ p,t$ by $ 2x\plus{}y,x^2\plus{}2xy$, respectively, it becomes
\[ \frac{(x\plus{}2y)^3}{x^2\plus{}2y^2} \plus{}\frac{6(x^2\plus{}2y^2)^2}{x^3\plus{}3xy^2\plus{}2y^3} \ge 6(x\plus{}2y)
\]
The inequality is homogenous, then we can assume $ y\equal{}1$ :)

Many thanks again  :)

Your proof is very nice and very interesting  :wink:


function
floor function
logarithms
AMC
AIME

If $\lfloor {x} \rfloor$ is the greatest integer less than or equal to $x$, then

\[ \sum^{1024}_{N=1} \lfloor {\log_2 N} \rfloor =  \]

$\text{(A)} \ 8192 \qquad \text{(B)} \ 8204 \qquad \text{(C)} \ 9218 \qquad \text{(D)} \ \lfloor {\log_2 (1024!)} \rfloor \qquad \text{(E)} \ \text{none of these}$

I've never gotten the sum function...Could someone explain it to me?

You basically just apply the the thing on the right for each number, starting with the bottom number, to the top number.  For example, in this problem, you would add up all the values of ${\lfloor {\log_2 N} \rfloor}$, where N is all the numbers from 1 to 1024.

So $\sum^{5}_{N=1} N = $ would be 15.

I posted a thread concerning the use of summation a while ago.

Here it is:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Summation&t=22820

Hope this helps you on the understanding the $\sum$!

Thanks!!  That's helped me to understand a lot more of the sum function.  I've seen it a lot before, just all I knew was that it meant sum.  There's another greek symbol for product (It is an uppercase $\pi$: I believe), would it be appropriate to ask what that means here?

$\prod$ works same as $\sum$.

Only difference is instead of $\sum^{5}_{i=1} i = 1+2+3+4+5 = 15$, $\prod^{5}_{i=5} i = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 5! = 120$.

Does this answer your question?

Yes, it does.  Thanks for all your help!!!

[hide]For $N=1$, $\lfloor \log_2 N \rfloor = 0$.
For $2 \le N \le 3$, $\lfloor \log_2 N \rfloor = 1$.
For $4 \le N \le 7$, $\lfloor \log_2 N \rfloor = 2$.
...
For $512 \le N \le 1023$, $\lfloor \log_2 N \rfloor = 9$.
For $N=1024$, $\lfloor \log_2 N \rfloor = 10$.

So $0 + 1(2) + 2(4) + 3(8) + \cdots + 9(512) + 10 = 10 + 2 + 8 + 24 + \cdots + 4608$

Don't know how to evaluate this though. :blush: [/hide]

[quote="chess64"]So $0 + 1(2) + 2(4) + 3(8) + \cdots + 9(512) + 10 = 10 + 2 + 8 + 24 + \cdots + 4608$

Don't know how to evaluate this though. :blush:[/quote]

Take out your calculator (or your computer program)  :roll:

[quote="chess64"]Oh. But no calculators were allowed on AHSME :( Guess you'll have to do it by hand :P[/quote]

They weren't?  :o I guess I was thinking the AMC 12  :P
I cheated  :blush:

[quote="chess64"]Oh. But no calculators were allowed on AHSME :( Guess you'll have to do it by hand :P[/quote]Yeah, there was a problem like this on AIME one year.  The only way is just do it by hand, I think.


AMC
AMC 10
AMC 12
AIME
MIT
college
USA(J)MO

I have heard a ton of people talking about the point system for next year's AMC; everyone is giving me different numbers for the amount of points gained when you leave a question blank, so does anyone know how the point system will work next year?

6 points for a correct answer.
1.5 points for no answer.
0 points for an incorrect answer.

25 questions still. Generally, if you're one of those people who purposely does 17 questions on the AMC10 or 11 questions on the AMC12 to qualify, this change is to disrupt you and prevent you from qualifying for the AIME easily like this.

[quote="krustyteklown"]rofl; wikipedia?

I think it was announced in the manual(for this year's AMC)

also, AMC Director has made posts on AoPS about the change[/quote]

This is correct. There was a thread that asked whether 1.5 point scoring is true or not and I think Mr.Dunbar told people that it is true. I think he also mentioned that the main reason for this was to prevent some "random" students who would just solve exactly 11 problems and qualify without looking at the later problems (which are much better and exciting). By new scoring, it won't be that easy I think.

I do not really like wikipedia and never recommend this for any type of serious information as this case. It is because that anyone can edits it and who knows, someone could've put something like "If you do well on AoPS's Mock AMC E, you'll qualify to AIME next year :D "

ahh, 14 questions. I got 13 questions right this year.  :blush:

The first I heard about it was in the 2005 awards booklet.  As for the information on wikipedia, I'm the one who put it there, and it came from the booklet.

[quote="Silverfalcon"]
...It is because that anyone can edits it...[/quote]

Point in case. :-)

i didnt think people actually updated those online encyclopedias that are available for anyone to change lol...i guess i was wrong. well, thanks!

Wow I actually really like Wikipedia. It gives lots of information about things that other online encyclopedias don't. And usually if you edit in something incorrect, it is changed within seconds. Try it - I've done it once. But whatev, maybe it's just because I'm the only one who doesn't have an encyclopedia at home... :blush:

Wikipedia pretty much rules all!

Check out its entries for some of the prestigious universities around the country, I found the MIT entry to pretty amusing (and accurate, on the whole)

Additionally, Wikipedia is a nice place to get a quick dose of information about famous mathematicians.

It's not like Wikipedia isn't nice if you want to look up song lyrics or something but you really shouldn't use it if you're doing formal research; it's not a credible course.

I am rather happy about the changes, it filters out all the borderline people. Not that I'm great at math or anything, but I did qualify decently this year for AIME (125.5 on AMC 12B)
But if USAMO # of qualifiers expands it seems natural for all the numbers for AIME and AMC12 to expand too... unless the AMC people want to save money on those who won't be moving on at all, which is a very natural thing to do :)

Many of us on this site have been used to a rather elitist outlook on math competitions. AIME qualification seems to come naturally for almost all the users who browse this site on a normal basis, so it doesn't mean all that much anymore. To many people, making the AIME is what making USAMO or MOP is to most of us here. The traditional 12,000 qualifiers, while certainly not a top-level prestige, does make a statement about an individual's problem solving abilities. And that, I think, is a level of achievement that should be preserved. A change in the point system is a great way to do so.

Heck, even mathletes not enamored by AIME qualification should be glad that this move more fairly calculates a person's true problem solving ability by cancelling some of the USAMO index inflation created by people simply leaving many answers blank.

Heh, I'd like to point out that I got 3 on this year's AIME  :rotfl: and I'm a junior, so time is running out  :o

But that's ok, Mith, you're absolutely right!

[quote="randomdragoon"][quote="hockeyadi23"][quote="zmli"]This year, 28,000 qualified for AMC12.  Among them, 17,000 scored between 100 - 110.  I guess you won't see these people in AIME next year with the change.[/quote]

I think it makes as much of an impact on the AMC 10. TONS of people qualify with scores like 121.5 (ahem, me last year) and like 123. They probably wouldn't get in now, since you need 19 right/6 blank or better to get in. This probably also explains why the first few questions are SOO easy in both tests...[/quote]
Does it really? I suppose it does explain.[/quote]

Well a lot of AMC 10-ers who make the AIMEs get most of the first 20, then just bomb the last five, or at least like four of them. Given that it's wicked easy to make a silly mistake on the AMC10 (cough cough number sixteen :wallbash_red: surprisingly the least answered question), we can assume this hits the 16-20 question right range. That leaves a lot of people on the boarderline, forcing at least 19 right (while before 17 right, 8 blank gets in), and that's if you leave the rest blank. A lot of middle schoolers and freshmen thus get knocked out. I don't exactly know why I took the AMC 10 (maybe to get a perfect? :lol:) but I was one of only three in my school, while 45 (!!!) made it on the AMC 12. Now my score was clearly in, but at least one of the other two may not have made it with the new system.

I think they're going to lower the difficulty of the questions on the AMC's because of the new scoring system, so scores should be about the same.

That sounds like a positive change, IMO......

I could never do the borderline slip by approach anyhow, since I am almost guarenteed to miss 1 or more problems that I am convinced are right when I'm doing them.  Last 2 years I slipped by, since I got 4 questions wrong both times.  I am the LORD of stupid mistakes.  Something I should think about/work on for next year, since I really can't afford it if I want to do the USAMO next year.


A group of 25 friends were discussing a large positive integer. "It can be divided by 1," said the first friend. "It can be divided by 2," said the second friend. "And by 3," said the third friend. "And by 4," added the fourth friend. This continued until everyone had made such a comment. If exactly two friends were incorrect, and those two friends said consecutive numbers, what was the least possible integer they were discussing?

I think that the two numbers were 16 and 17 so the answer is 787386600.

Do you multiply this all out by hand, do you need to memorize factorials up to $25$, or would this be a calculator question?

no, the two numbers were the two largest prime numbers, 19, 23

I'm not sure if we have to memorize the factorials up to 25 (I don't think so) I think we just do this out by hand or of this part of the test we were aloud to use a calc


calculus
integration
trigonometry
calculus computations

yello !  :D

$ \text{(i)}$
note that signs alternate, and the denominators are composed of [i]odd[/i] numbers that are [u]not[/u] divisible by $ 3$, i.e.

$ 1 \minus{} \dfrac{1}{5^3} \plus{} \dfrac{1}{7^3} \minus{} \dfrac{1}{11^3} \plus{} \dfrac{1}{13^3} \minus{} \dfrac{1}{17^3} \plus{} \dfrac{1}{19^3} \minus{} \dfrac{1}{23^3}\;\; \plus{} \dfrac{1}{25^3} \minus{} \dfrac{1}{29^3}\;\; \plus{} \;\; \minus{} \;\;\ldots$

$ \text{(ii)}$
note that signs alternate, and the denominators are composed of [i]odd[/i] numbers that are [u]not[/u] divisible by $ 7$, i.e.

$ 1 \minus{} \dfrac{1}{3^3} \plus{} \dfrac{1}{5^3} \minus{} \dfrac{1}{9^3} \plus{} \dfrac{1}{11^3} \minus{} \dfrac{1}{13^3} \plus{} \dfrac{1}{15^3} \minus{} \dfrac{1}{17^3}\;\; \plus{} \dfrac{1}{19^3} \minus{} \dfrac{1}{23^3}\;\; \plus{} \;\; \minus{} \;\;\ldots$

do [b]not[/b] use complex variables or fourier series. those are things that are taught at later stages in an university. instead, try and use techniques that a 7th or 8th grader will be likely to have, i.e. general integral calculus, trigonometry, etc. i know the answers and a rather elementary method for solving both, but i want to see if anyone here knows of any simpler method...

[quote="Misan"]that a 7th or 8th grader will be likely to have, i.e. general integral calculus[/quote]

Again, a general 7th or 8th grader is not supposed to know calculus. You deal with special students, don't forget. Also, for your genius students complex variables and Fourier series should be elementary, right? Aren't you exagerating a little already?


ok so tell me how farr youve made it in Mathcounts.....and how many years it took you

im bored...might as well find out

if anyof you are wondering...last year was my first year and i came 6th in state

I joined when I was 9, and I competed when I was 10, 11, and 12. I got 29th at nats when I was 12(going on 13 very soon then)

I got 18th in state in 6th grade and then 16th at Nationals in 7th grade.

I got 19th in chapter in 6th grade and 5th at state in 7th grade (sounds much more of an improvement than it really is).

Havn't gone yet going next year  :lol:

I've competed at age 9,10,11 in ARML, PuMAC, AMCs (GO 25 ON 8!), AIME, NYSML, HMMT, BCA math competition, ATL-PAF, AMCNJ, Math Contests, Four-by-Four, Fall Startup event.

I think.

Good luck to everyone at MC (Unless, of course, you're facing me in CD)

I made Nats ~130th in 8th grade, the first year I'd even heard of Mathcounts.

6th: 1st district, 5th state
7th: 1st district, 2nd state, 65th nats
8th: tie 1st school, ___ district, ___ state, ___nats

6th: 8th chapter, 22nd state (in Michigan)
7th: (moved to Ohio) 1st chapter, 4th state, 92nd nats

6th grade: didn't make school team  :(
7th grade: 3rd chapter, 45th state

8th grade: 1st state (MN), my first nats, T5th. Good luck to everyone else...

I wish I could have been smarter earlier in my lifetime... (like the younger people who won last year :P)

6th grade: Alternate...  :o
7th grade: hasn't happened yet... definitely made the team though...
8th grade: 2010  :P

6th grade: Got lucky and was an individual.  Rank in Chapter: Not Placed ( I got 9 right in the sprint)
7th Grade: chapter: 13th place and alternate for states
8th grade: Chapter 0th place, state: 0th place, nats: 0th place

0= TBA

6th grade: 6th chapter. Didn't make state. School sucks.
7th grade: 4th chapter, went to state as an individual. 6th place state.
8th grade: Hoping for 1st chapter and 1st state, top 20 (?) nationals

Our school didn't have mathcounts when I was in 6th grade.
7th: This awesome teacher comes, our team makes state as a wild card (although we had the 2nd highest chapter score in the state, we came 2nd in chapter, lost by a quarter of a point :P) and gets 4th place there. Individually, I got 3rd at chapter, and 16th at state.
8th: Our school got 1st chapter, and 4th state again. I got 1st chapter, 1st state, T3 nationals.

Good luck to everyone who's preparing for it this year!

someone really needs to make a thread or forem for all of these surveys...
this is getting nuts...

6th grade, 3rd regional, blown out of water in cd by two people who ended up making nats (they had 1-2-3 play cd in first round!!!). At state that year, I was tied for first after sprint round with a guy who would go on to come in 12th at nats. Then I epic failed target and dropped to 11th!

7th grade I made nats coming in a tie for 69th. Too many potatoes that morning  :rotfl:


function calculus derivative algebra domain integration Putnam

[quote="Originally"]Assume that $ |f(x)| \leq 1$ and $ |f'(x)| \leq 1$ for all x on an interval of length at least 2. Show that $ |f''(x)| \leq 2$ on the interval.[/quote]

[u]Corrected statement of the problem:[/u] Assume that $ |f(x)| \leq 1$ and $ |f''(x)| \leq 1$ for all x on an interval of length at least 2. Show that $ |f'(x)| \leq 2$ on the interval.

This is indeed a challenge! If we start with $ f'(x) \equal{} \frac {|x|}{x}$, by smoothing at $ x \equal{} 0$ we can make $ f''(0)$ arbitrary large.

I would assume that the intended problem was that $ |f(x)|\le1$ and $ |f''(x)|\le1,$ from which we seek a bound on $ |f'(x)|.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123244]This[/url] isn't the same problem but it is related.

I did indeed mistype the problem. It is corrected above. Sorry for the mistake.

[hide="Kent:"]I saw that other thread and I do suspect that the solution involves Taylor's Polynomial/Mean Value Theorem. But since this problem does not specify the derivatives at any particular point in the interval, I'm having trouble pinning it down.[/hide]

There was a similar problem in Rudin's Principles of Mathematical Analysis (Ch. 5 #15)

[hide="Solution?"]We use Taylor's Theorem and we have $ f(x\plus{}2)\equal{}f(x)\plus{}2f'(x)\plus{}2f''(c)$ where $ c\in (x, x\plus{}2)$. We solve for $ f'(x)$ and we have $ f'(x)\equal{}\frac{1}{2}[f(x\plus{}2)\minus{}f(x)]\minus{}f''(c)$. We now notice that $ |f'(x)| \leq \frac{1}{2}[|f(x\plus{}2)|\plus{}|f(x)|]\plus{}|f''(c)| \leq 2$ as desired.[/hide]

[quote="1/(ln x)"]There was a similar problem in Rudin's Principles of Mathematical Analysis (Ch. 5 #15)

[hide="Solution?"]We use Taylor's Theorem and we have $ f(x \plus{} 2) \equal{} f(x) \plus{} 2f'(x) \plus{} 2f''(c)$ where $ c\in (x, x \plus{} 2)$. We solve for $ f'(x)$ and we have $ f'(x) \equal{} \frac {1}{2}[f(x \plus{} 2) \minus{} f(x)] \minus{} f''(c)$. We now notice that $ |f'(x)| \leq \frac {1}{2}[|f(x \plus{} 2)| \plus{} |f(x)|] \plus{} |f''(c)| \leq 2$ as desired.[/hide][/quote]
I think this works only if $ x \pm 2$ is in the domain of $ f$.

[hide="Proof by Picture"]WLOG say $ f$ is defined on $ [\minus{}1,1]$, and let $ s \in [\minus{}1,1]$. Using $ |f''| \le 1$ and the MVT we get $ f'(x) \ge f'(s) \minus{} (s\minus{}x)$ for $ x \le s$ and $ f'(x) \ge f'(s) \minus{} (x\minus{}s)$ for $ x \ge s$. Then
\[ 2 \ge f(1) \minus{} f(\minus{}1) \equal{} \int_{\minus{}1}^s f'(x)dx \plus{} \int_s^1 f'(x)dx \\
\ge \int_{\minus{}1}^s f'(s) \minus{} (s\minus{}x) dx \plus{} \int_s^1 f'(s) \minus{} (x\minus{}s) dx \\
\equal{} 2f'(s) \minus{} s^2 \minus{} 1 \ge 2(f'(s)\minus{}1).\]
This shows $ f'(s) \le 2$, and we can similarly get $ f'(s) \ge \minus{}2$.[/hide]

When $ x\geq 0$
We have
$ f(x)\minus{}f(0)\equal{}\int_{0}^xf'(x)dx\equal{}xf'(x)\minus{}\int_{0}^xxf''(x)dx$

$ |xf'(x)|\equal{}|\int_{0}^xxf''(x)dx\plus{}f(x)\minus{}f(0)|$

              $ \leq |f(x)|\plus{}|f(0)|\plus{}|\int_{0}^xxf''(x)dx|$

              $ \leq |f(x)|\plus{}|f(0)|\plus{}\int_{0}^xx|f''(x)|dx$

              $ \leq 1\plus{}1\plus{}\int_{0}^xxdx$

$ |xf'(x)|\leq 2\plus{}\frac{x^2}{2}$

$ x^2\minus{}2x|f'(x)|\plus{}4\geq 0$ which is true for all $ x\in R$

Discriminant $ \leq 0$

$ 4|f'(x)|^2\leq 16$

$ |f'(x)|\leq 2$

This solution was on this website and I had copied it long back.

Thats bad notation all the way Pankaj :wink:


complex numbers
geometry
perpendicular bisector

Two fixed points $ Z_{1}, Z_{2}$ and a variable point $ Z$ represent the complex numbers $ z_{1},z_{2}, z$ respectively.

Give the locus of Z if $ |z\minus{}z_{1}| \equal{} |z\minus{}z_{2}|$

Why does Z have to be on the perpendicular bisector? Why can't it remain where it is??

Well the question stated that $ z$ is a variable point which means it's not fixed........

So by definition your locus represents the set point points $ z$ which are equi-distance from $ z_{1}$ and $ z_{2}$. So it is a perpendicular bisector.

Can't Z be a variable point at the vertex and moving so that the 2 arms remain equal??

Instead of being the perendicualr bisector?

[url=http://img459.imageshack.us/my.php?image=81210114qt5.png][img]http://img459.imageshack.us/img459/3755/81210114qt5.th.png[/img][/url]

bos1234...yes but any point at the vertix of this trinagle will lie on the perpendicular bisector...

Proof

Using your diagram, let the line $ zx$ be perpendicular to $ z_{1}z_{2}$ with $ x$ lying on the side $ z_{1}z_{2}$.
By using pythagoras' theorem

$ {z_{1}z}^{2}\equal{}{z_{1}x}^{2}\plus{}{xz}^{2}$ and $ {z_{2}z}^{2}\equal{}{z_{2}x}^{2}\plus{}{xz}^{2}$

But since $ z_{1}z \equal{} z_{2}z$ then $ {z_{1}z}^{2}\equal{}{z_{2}z}^{2}$

$ \Rightarrow{z_{1}x}^{2}\plus{}{xz}^{2}\equal{}{z_{2}x}^{2}\plus{}{xz}^{2}$

$ \Rightarrow{z_{1}x}^{2}\equal{}{z_{2}x}^{2}$

$ \Rightarrow z_{1}x \equal{} z_{2}x$

Hence $ x$ is the midpoint of $ z_{1}z_{2}$. But since $ xz$ is perpendicular to $ z_{1}z_{2}$, then $ zx$ is the perpendicular bisector of $ z_{1}z_{2}$.

Hence $ z$ lies on the perpendicular bisector.

wonderful proof. They should include this in the books

Now I understand thoroughly
THANKS!!


Has anyone played .hack/infection here? What did you guys think of it??? Being only able to play few video games with my little time, I have found it to be pretty interesting. My only regret was that there was no bonus round where you could kill/hunt grunty's... that would have been pretty tite.

Enjoy.
IW.OAV

I have no clue what it is or where to get it. Can you post some more info? Thanks.

-Isaac

[color=white]Same here.  What platform is it for?  It sounds like Banjo-Kazooie...grunty's?[/color]

I've never played it, having spent too much time on this forum.  But I think grunty's are referring to grunts, as in common, usual soldiers.  Dunno.  Just a guess.

[color=white]Wow, mathfanatic...you're really gaining on ComplexZeta on posts.  Syntax Error isn't going to be happy.[/color]

And guess what... every time you post something, I post something in reply...so it's all YOUR fault. :lol:   But thanks.  #2 right now.

[color=white]...

Well, when I post, I get a point as well.  So we're helping each other.  Seeing that you're not too far away, from ComplexZeta, some faraway day in the future you may be #1.[/color]

Oooh, jojo's gunning in on 100 posts and a Poincare Conjecture.  Eek!  He's coming up FAST too.  I'd better watch out.

P.S. Maybe if I have another "hyper mouse session" and post this 5 times, I'll get credit for 5 posts.  <kidding>

Can you help me get posting points too?

-Isaac

You guys can find about it here.. it's a freakin' awesome tastic game!

http://www.dothack.com/game/index.html

BTW it's one of the best RPGs out next to the FF series.

Me going away for a couple of weeks ruined my post ranking :( oh well, I'm trying to come back..

Gosh...it seemed like yesterday that Mathfanatic passed 210, but now he's in the 240s???

-interesting_move

yah, I need to make a comeback :P

-Isaac

[color=white]He he, I'm catching up, Issac...and I passed IW.OAV a long time ago.[/color]

the sequel to infection is already out. and more is to come. i have so many games that are just sitting there to be played on my ps2. like xenosaga. it rocks. but i dunno. i havent been in the mood to play. .hack is also a show on cartoon network.

I think "nope, I don't play this game" and I agree.


Determine all triples of positive integers $(a, m, n)$ such that $a^m +1$ divides $(a+1)^n$.

[quote="Peter"]Determine all triples of positive integers $ (a, m, n)$ such that $ a^m \plus{} 1$ divides $ (a \plus{} 1)^n$.[/quote]
http://www.kalva.demon.co.uk/short/soln/sh00n4.html

If $ a,m > 1$ and $ (a,m) \neq (2,3)$, by Zsigmondy's theorem there is a prime $ p$ dividing $ a^m \plus{} 1$ but not $ a \plus{} 1$. Thus there are no solutions.
Trivially checking the other cases, we get that all other cases always are solutions, regardless what $ n$ is, the only exception being $ (a,m) \equal{} (2,3)$ where we need $ n \geq 2$.


Prove that $ x^{\sin^2 a} \cdot y^{\cos^2 a}<x\plus{}y$ , $ x,y>0, a \in R$.

[quote="balan razvan"]Prove that $ x^{\sin^2 a} \cdot y^{\cos^2 a} < x \plus{} y$ , $ x,y > 0, a \in R$.[/quote]
Assume that $ y>x$ ( similarly when $ y<x$ )
We have
$ x^{\sin^2 a} \cdot y^{\cos^2 a} < x \plus{} y \Leftrightarrow x^{1\minus{}cos^2 a}y^{cos^2 a}<x\plus{}y \Leftrightarrow \left(\frac{y}{x}\right)^{cos^2 a}<1\plus{}\frac{y}{x}$
Which is clearly true because $ \left(\frac{y}{x}\right)^{cos^2 a}<\frac{y}{x} \forall y>x$

For $ x\geq y>0,\ \left(\frac {x}{y}\right)^{\sin ^ 2 a} < \frac {x}{y} < \frac {x}{y} \plus{} 1$.

Since the maximum value of $ sin^2a$ is 1, $ x^{sin^2a}\leq x$ and $ y^{sin^2a}\leq y$.
Multiplying them together we get that the LHS is less than or equal to xy, which is less than x+y because by AM-GM, 2xy<=x+y, and so xy<x+y since x and y are positive reals.
Does that work?

$ x^{\sin ^ 2 a}\leq x$ holds only when $ x\geq 1$, then my solution isn't perfect. :maybe:


AMC
AIME
USA(J)MO
USAMO
geometry
AMC 12
AMC 10

I predict 18000 students taking AIME in 2006, with a floor value of 6 -- on 2 killer-AIME tests that compensate for the 4 easy-to-qualify-for-AIME AMC tests.  :D

[quote="Silverfalcon"]Okay, using our smart math skills, let's guess what the floor value will be.. I'm guessing 8 since the tests seem to be little bit harder than last year.

What do you guys think?[/quote]

Hmm did you take account of the expansion??

Predicting 7/ 210 index.

7/210 index here too

actually, bump it up to 218 : | just to motivate myself to do more on AIME

i'm predicting a lot more AIME qualifiers this year, as the first 15 or so questions on both contests seemed easier than normal.  so the index needs to be higher to get the top 375 from the expanded pool.  my guess: index between 220 and 229.5, with a floor of 8.

nah, man those people who got in on the first 15 won't even have a nice fat AMC score to factor into the index first 15 perfect is what, 115? they'd need 10-11 on the aime, highly unlikely if they only got the first "easy" 15

USAMO index: 200
AIME floor: 7
Like krustyteklown said, USAMO qualifiers almost always get at least 120 on the AMC 12, and the 12B in particular made it hard to get too far past that. Plus, since the AMC's and the last AIME were comparatively easy, I predict this AIME will be pretty hard. Then there's the expansion...

7/200, since the 12 was harder to get a perfect score on this time, just look at the last 7-1(excluding 22)!

I don't think the AMC's were [i]that[/i] hard. The influence of AoPS certainly will combat the the qualifier inflation. And I really mean that. The 50% increase will have no effect at all, because AoPS itself has raised the index 15 points (look at all our members at the AMC 130+)

Various indices:
210/6 (AIME extremely hard, like 1994)
219/7 (hard, like 2004 I)
228/8 (average difficulty, like AIME 2003 I)
233/9 (easy, like the AIMEs in the 80's or last year's)

215/8

That means I need 11 on the AIME...

In 2002 it was 6, in 2003 it was 8, in 2004 it was 7, in 2005 it was 9, so if we continue the pattern, it should be 8 this year.  But because of the expansion, it's probably go down by 1, so it's probably 7 this year.

i would say
9/220
this is because i think the AIME questions will be something like last year...

For sophmores and under the amc does not matter. So people that barely qualify still have a chance.

I'm hoping for a killer AIME, just so there are fun problems.

Err... the floor is set by the top 120 people or something like that. :rotfl:

[quote="krustyteklown"][quote="calc rulz"]If nobody qualifies for AIME in a US territory or state, then who gets to take the USAMO?[/quote]

I'm sure someone from your state will qualify, [i]most definitely[/i] someone from your school too[/quote]

My state yes. However, nobody qualified in Guam I'm quite sure. (I don't need to know. I'm just curious.)

I believe the top scorer (index) in each state is guaranteed into the USAMO. I'm not too sure about territories though.

Well if nobody in a state(like Wyoming, hopefully this year they'll have someone qualifying since it's much easier to qualify) qualifies for AIME then nobody will be invited to take the USAMO.

[url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/05usamoqualstate.html#wi]www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/05usamoqualstate.html#wi[/url]

But it's so awesome to live in Wyoming, if you qualify for AIME you are pretty much guaranteed to qualify for USAMO  I wonder why nobody moved there yet :lol:

[quote="JesusFreak197"][quote="beta"]But it's so awesome to live in Wyoming, if you qualify for AIME you are pretty much guaranteed to qualify for USAMO  I wonder why nobody moved there yet :lol:[/quote]

Probably because that would mean that qualifying for USAMO is the most important thing in your life, and hopefully it isn't for anyone.[/quote]

It was a joke incase you didn't know...

[quote="WarpedKlown1335"]Really?  I don't know, after all considering it was from beta I thought it might've been awfully serious.

LoL jk.[/quote]

Not knowing beta that well, I'm not exactly sure.  From most people, considering the smilie and the end, I would say it was a joke.  You might be right though.

Yeah it was a joke! But then good math people moving to Wyoming is actually good because then the good math people will bring math to Wyoming and Wyoming's general math performance will improve, which is the whole purpose of AMC/AIME/USAMO :lol:

Yeah, but Wyoming has those big-horned sheep. And bears that fight people like...

http://www.youtube.com/watch?v=UymHE0-3mEA&search=salmon%20commercial

Here's what I wanted. From the AMC website (http://www.unl.edu/amc/e-exams/e8-usamo/factoidsusamo.html) [participation selection - 5]

[i]The student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO. [/i]


algebra
polynomial
superior algebra
superior algebra theorems

Name all criterias and tricks you know about irreducible polynomials. Also some useful polynomial identities.

I want to hear your tricks.
Not the usual theorems.


Heyfam omad ino nafrestam
Agar $ a,b,c$>=0 va dashte bashim $ a\plus{}b\plus{}c\equal{}3$ sabet konid
\[ \frac{a}{1\plus{}b^2}\plus{}\frac{b}{1\plus{}c^2}\plus{}\frac{c}{1\plus{}a^2}\geq \frac{3}{2}\]

[quote="m bahador"]be rahati ba estefade az raveshe lagranj tabe $ L$ ra tarif mikonim:
\[ \begin{array}{l} L \equal{} \frac {a}{{1 \plus{} b^2 }} \plus{} \frac {b}{{1 \plus{} c^2 }} \plus{} \frac {c}{{1 \plus{} a^2 }} \minus{} \lambda (a \plus{} b \plus{} c \minus{} 3) \\
\Rightarrow \frac {{\partial L}}{{\partial a}} \equal{} \frac {1}{{(1 \plus{} b)^3 }} \\
\frac {{\partial L}}{{\partial b}} \equal{} \frac {1}{{(1 \plus{} c)^3 }} \\
\frac {{\partial L}}{{\partial c}} \equal{} \frac {1}{{(1 \plus{} a)^3 }} \\
\to a \equal{} b \equal{} c \equal{} 1 \\
 \\
\end{array}
\]
pas ekstermom tabe
\[ f \equal{} \frac {a}{{1 \plus{} b^2 }} \plus{} \frac {b}{{1 \plus{} c^2 }} \plus{} \frac {c}{{1 \plus{} a^2 }}
\]
dar noghteye $ (a,b,c) \equal{} (1,1,1)$ ast va choon tabe mohadabast pas in ekstermom , minimum ast.va be ezaye in noghte masale hal ast[/quote]
 :huh:
heife soale be in ghashangi nis ke ba in raha kharabesh konim?

$ x(x \minus{} 1)^2\geq 0\Rightarrow \frac 1{1 \plus{} x^2}\geq\frac {2 \minus{} x}2$

baraye har $ x\in\mathbb{R}^ \plus{}$,pas darim:

$ \frac {a}{1 \plus{} b^2}\geq\frac {a(2 \minus{} b)}2$

pas darim:

$ \sum\frac {a}{1 \plus{} b^2}\geq\sum\frac {a(2 \minus{} b)}2$

pas kafie sabet konim:

$ \sum\frac {a(2 \minus{} b)}2\geq\frac 32\iff 3\geq ab \plus{} bc \plus{} ca$

ke inam vazehe cherake $ a \plus{} b \plus{} c \equal{} 3$ va $ (a \plus{} b \plus{} c)^2\geq 3(ab \plus{} bc \plus{} ca)$

are bavar kon ke ghalban ham razi naboodam in raho post konam vali choon ke too emtehanaeem va vaght kafi baraye fek kardan nadaram majboor shodam ,kholase bebakhshid  :oops:

akh akh gofti emtehana yade parsal oftadam,cheghad hal midad. :D


function logarithms algebra polynomial limit induction calculus

Consider a differentiable function f:R---->R which satisfies $ f^2(\frac{x}{\sqrt{2}})\equal{}f(x)$ for all x belonging to R and f(1)=2
Find f(x)

Note that $ f(t)$ on $ \left[1, x\right]$ satisfies the hypotheses of the Mean Value Theorem.
So there exists $ \theta \in \left(1, x\right)$ such that we have:

$ \frac {f(x) \minus{} f(1)}{x \minus{} 1} \equal{} f'(\theta) \implies f(x) \equal{} \left(x \minus{} 1 \right)\cdot f'(\theta) \plus{} f(1)$ *

The condition that $ f^{2}\left(\frac {x}{\sqrt {2}}\right) \equal{} f(x) \implies \sqrt {2} \cdot f\left(\frac {x}{\sqrt {2}}\right) \cdot f'\left(\frac {x}{\sqrt {2}}\right) \equal{} f'(x)$

$ \implies \sqrt {2} \cdot f(1) \cdot f'(1) \equal{} f'(\sqrt {2})$

Now to find $ f(x)$ explicitly you need to know the values assigned to $ f(1)$ and $ f'(1)$ and then the * will give us the result by setting $ \theta \equal{} \sqrt {2}$.

To akech:
Note, that in your formula * you have that $ \theta \equal{} \theta(x)$ depends on $ x$ and,hence, you cannot take $ \theta \equal{} \sqrt {2}$ for arbitrary $ x$ :)

Probably the point is that the condition that $ f$ is differentiable [b]at 0[/b] is the crucial thing that rules out all other functions.

The obvious solution is $ f(x) \equal{} 2^{(x^2)}$

If I'm not mistaken, there should be some extra conditions on our function or on its derivative(if not, then there exist answers,which are not very  ''simple").But maybe I've done some stupid mistake.I'll check it later :)

it's easy to note that $ f(0) \equal{} 1$ and $ f > 0$ for all $ x$
consider $ g(x) \equal{} \ln f(x)$ then we have $ g(1) \equal{} \ln 2$ and $ 2g(x/\sqrt {2}) \equal{} g(x)$
so if we suppose that $ g$ is twise differentiable function then it's easy to show that $ g$ is polinomial  for example by this way :
$ g''(x/\sqrt {2}) \equal{} g''(x)$ so for any point $ z$ we have $ g''(z) \equal{} g''(z/2^{n/2}) \to g''(0)$ thus $ g''(z) \equal{} g''(0)$
so $ g$ is polynomial hence it's equal to $ x^{2} \ln 2$ since $ g(1) \equal{} \ln 2$
I think it's not hard to solve this problem without assumption twise differentiability of $ g$.
for example consider $ y(x) \equal{} \frac {g(x)}{x^{2}}$ then condition $ 2g(x/\sqrt {2}) \equal{} g(x)$ becames $ y(x/\sqrt {2}) \equal{} y(x)$ thus similary for any point $ z$ we have $ y(z) \equal{} y(z/2^{n/2}) \to y(0)$ thus we have $ g(x) \equal{} y(0)x^{2}$ so $ g(x) \equal{} x^{2} \ln 2$
so you just need to explain why $ y(x)$ is continous at point $ 0$ or why there exist limit $ \lim _{0}\frac {\ln f(x)}{x^{2}}$ , does it follows from the differentiability of $ f$ ?

No, it does not follow from the differentiability,in general :wink:

Note that $ f(x)\ge 0$ for all $ x$. It is easy to prove using induction that for all naturals $ n$,
$ f(x)\equal{}f^{2^n}\left(\dfrac{x}{(\sqrt{2})^n}\right)\qquad(1)$
where $ f^k(x)$ means $ (f(x))^k$. This means
$ \sqrt[2^{n}]{f(x)}\equal{}f\left(\dfrac{x}{(\sqrt{2})^n}\right)$
As such
$ \sqrt[2^{n}]{f(1)}\equal{}f\left(\dfrac{1}{(\sqrt{2})^n}\right)\qquad \Rightarrow\ \sqrt[2^{n}]{2}\equal{}f\left(\dfrac{1}{(\sqrt{2})^n}\right)$
Taking limit as $ n\to \infty$, we obtain $ 1\equal{}f(0)$.
Next, if for some $ x_0\neq 0$, $ f(x_0)\equal{}0$, then that means that at $ x\equal{}x_0/\sqrt{2}$ also $ f\equal{}0$ which further means that at $ x\equal{}x_0/(\sqrt{2})^2$, $ f\equal{}0$ and so on and in general at  $ x\equal{}x_0/(\sqrt{2})^n$, $ f$ is zero. But again, in the limit, that will imply that $ f(0)\equal{}0$ --- a contradiction. So $ f$ is not zero for any $ x$. Hence, $ f>0$ for all $ x$.

By setting $ F(x)\equal{}\log_af(x)$, for some positive $ a$, the given functional equation can be written as
$ F\left(\dfrac{x}{\sqrt{2}}\right)\equal{}\dfrac{1}{2}\,F(x)$
which is the so called $ \mathrm{Schr\ddot{o}der}$ equation. Either, from this equation or by taking log of the equation (1), we get that $ F(x)$ satisfies for all $ x$ the following condition for all natural $ n$:
$ F(x)\equal{}2^n F\left(\dfrac{x}{(\sqrt{2})^n}\right)$
It is easy to see that $ F(x)\equal{}x^2$ satisfies the above equation for all $ x$ (I haven't been able to prove so far the uniqueness of this assertion). As such
$ f(x)\equal{}a^{x^2}$
By the condition $ f(1)\equal{}2$, we get $ a\equal{}2$. As such, we get
$ f(x)\equal{}2^{x^2}$

Ok.I'll try to briefly explain the idea of my counterexample (maybe somebody explain to me where I'm wrong :maybe: ).
First note, that if we have differentialble function $ g: R\to R$ for which $ g(\frac {x}{\sqrt {2}}) \equal{} \frac {1}{2}g(x)$ and $ g(1) \equal{} 1$ then we can take $ f(x) \equal{} 2^{g(x)}$ and get the function that satisfies our functional equation.
Now choose the 'small' interval $ (a,b)$ for $ a < 1 < b$ and $ a\sqrt {2} > b$ and define nonconstant  function $ h(x)$ that is differentiable on $ \mathbb {R}$ and $ h(1) \equal{} 1$, $ h(x) \equal{} 0$ for $ x\notin (a,b)$ (this is a classical example). Consider the collection of intervals $ (a(\sqrt {2})^n, b(\sqrt {2})^n)$, $ n\in\mathbb {Z}$ and define our function $ g$ recursively using our equation  $ g(\frac {x}{\sqrt {2}}) \equal{} \frac {1}{2}g(x)$ and $ h(x) \equal{} g(x)$ on $ [a,b]$, $ g(0) \equal{} 0$. Then because our intervals do not intresect each other the only thing that we should check is the differentiability in point $ 0$. But this is obvious because if $ x < \varepsilon$ then
f(x)=0 if $ x\ne \frac{c}{(\sqrt {2})^n}$ for some $ c\in (a,b)$ or $ f(x) \equal{} \frac {f(c)}{2^n}$ if  $ x \equal{} \frac {c}{(\sqrt {2})^n}$. But then $ |\frac {f(x)}{x}| < 2^{ \minus{} \frac {n}{2}}|\frac {f(c)}{c}|$.This  bound provides differentiabiliry in point $ 0$. It's easy to extend oyr function for $ x < 0$ in the same way.

You're right, of course, but let's understand the word "differentable" as "has as many derivatives as you wish" (one is certainly not enough, but, I guess, two or three would do the trick). ;)

i didn't understand both of the above solutions.
 :(

If we suppose that the function $ h(x)\equal{}log_2 f(x)$ has two dirivatives, and second derivative is continous in point $ 0$ then,of course, everything will be fine.
Indeed, for function  $ h(x)$ we have the equation $ h(\frac{x}{\sqrt{2}})\equal{}\frac{1}{2} h(x)$. After differentiation two times we'll get $ h^{''}(x)\equal{}h^{''}(\frac{x}{\sqrt{2}})$ or $ h(x)\equal{}lim_{n\to\infty}(h(\frac{x}{2^\frac{n}{2}}))\equal{}h(0)$.


Let $f: [a,b]\to\mathbb{R}$ a derivable on $[a,b]$ and continous on $(a,b)$. prove that there is
 $c\in (a,b)$ such that: $(b-a)f(c)=(c-a)f(a)+(b-c)f(c)$ ;)

Manifestly that's false ! :D ! Could you make a correct statement  ?
 :cool:

The obvious fix is to make it more symmetric and replace $(b-c)f(c)$ with $(b-c)f(b)$. The problem is, that isn't true either.

when $f=constant$ then $\forall c\in(a,b) :$ $(b-a)f(c)=(c-a)f(a)+(b-c)f(c)$


Given a function $f: X \to Y$
We have
1. $f(f^{-1}(B)) \subset B \textrm{ for } B \subset Y$ and
2. $A \subset f^{-1}(f(A)) \textrm{ for } A \subset X$.
Equality need not hold in either case.

Why? Could someone explain that? thanks

Two simple counterxamples. In #1, we need a function that does not map $X$ onto $Y.$ In #2 we need a function that is not one-to-one.

#1. Define $f: \mathbb{R}\mapsto\mathbb{R}$ by $f(x)=e^x$ and let $B=[-1,1].$

$f^{-1}(B)=\{x: e^x\in[-1,1]\}=(-\infty,0].$

$f(f^{-1}(B))=f((-\infty,0])=(0,1]$ which is not $B.$

#2. Define $f: \mathbb{R}\mapsto\mathbb{R}$ by $f(x)=x^2$ and let $A=[1,2].$

$f(A)=[1,4].$

$f^{-1}(f(A))=\{x: x^2\in[1,4]\}=[-2,-1]\cup[1,2].$


Consider the sequence of primes $ p_{1}, p_{2},\ldots$ such that for each $ i\geq 2$, either $ p_{i}\equal{} 2p_{i\minus{}1}\minus{}1$ or $ p_{i}\equal{} 2p_{i\minus{}i}\plus{}1$. An example is the sequence $ 2, 5, 11, 23, 47$. Show that such a sequence has a finite number of terms.


function
algebra
functional equation
algebra proposed

Find all functions f defined on the positive reals with values in the same set such that f(1+xf(y))=yf(x+y).

The only significant thing I managed to get is f(1)=1... :D I also got that f takes all values in (0,1], if that is somehow helpful. That's not much, is it? :D  :D

Well, thank you very much, as if it were extremely difficult to find that f(x)=1/x is a solution. Many IMO problems can be "solved" in this way , observing that f(x)=x verifies the conditions. I would give 0 points for such a solution. So, I'm waiting for a solution.

My solution doesn't use the fact that f(1)=1 and that f takes all the values from (0,1], but this may be helpful. Nobody says there is only one solution to this problem. In fact, I would be very happy to get another solution.

I'm still here and noone solves me! I will start crying if you let me alone!

harazi! If you have solution what do you still want? What is the source of this problem?

A quick answer:
 MORE SOLUTIONS
 VRANCEANU-PROCOPIU CONTEST, BACAU, 2003

Do you, by any chance, know the grade for which this was proposed?

There was a kind of TST for seniors and there was the problem proposed. Of course, nobody managed to finish it and that took my attention.

to be more precise: a districtual contest in Romania (of no big importance whatsoever) makes a simulation of TST (at least that's what they call it). Of course it's never such difficult, but this year this problem was difficult enough. But please do not underestimate the Romanian Team :D Andreis solved this problem in 20 minutes, and a nice one too. I will ask him to post it.

Only two students from the ones that have participated last year at RoMOP (first 25 in country) went to this contest :)

So the problem itself is not that hard :)

You say the problem is not that hard. I really don't have the same opinion. Could you please post your solution?

And you should not concider that it is easy only because someone solved it in 20 minutes. Just consider the fact that it stayed for quite a long time on this forum with no solution received. If you consider it "not so difficult", then why didn't you post any solution?

well it's not MY solution, because Andreis solved it. I said above that I will ask him to post it. Unfortunately at this moment he's on his way to Singapore (in the plane) so it might take a while for him to get near a computer.

Also do not feel upset: I said that the problem is not that difficult, not that it is easy. You said that at a Romanian TST nobody could solve it, but I have explained you that only 2 (and 2 of the youngest!) components of the big team (25 students) could not solve it, and that is was not a TST. Andreis is one of the first 6 and he solved it pretty quickly. :D

I said that it was a kind of romanian TST, not a TST. I'm convinced that the members of the Romanian team would have solved the problem, but I'm really not convinced that more than half of the first 25 people you mentioned would have solved it. I find it very difficult and I gave it to lots of people who couldn't do absolutely anything with it.
  And what do you want to say:"it's not that difficult". In my opinion, when you say something like this, you must have a solution-otherwise the problem is difficult. Could you post it?

I already know Andreis' solution and I will let him post it. It's rather hard to think about another solution :) once you know one.

You are probably right about that half :) but then again didn't you solve the problem? If this is a very diffcult problem as you say, how do you call a functional equation that you can't solve? (considering that you already have a solution for this :D) an extremely difficult problem? :) it's a matter of choosing the right adjectives I think. :)

It depends on how much time you have spent for finding that solution. Taking into account hat I have managed to solve all the functional equations given in any IMO, except problem 6 proposed by Bulgaria one year, I can say that there aren't so many functional equations given in IMO's that could be called "extremely difficult". When I said this one is very difficult, I compared the time spent for finding a solution to it with the time spent for finding solutions to many other problems. And just another thought: If someone solves a problem, that doesn't become automatically easy. Reid Burton or Ciprian Manolescu solved extremely difficult problems and I don't think that you could say: well, if you solved it, the problem is  easy.

I've found your interesting disscusion only yesterday. I don't think that this problem is very simple (30 min) but I don't think that it's extremelly
hard (30 min).

Sol.
Answer: f(x)=1/x.

Let g(x)=1/f(x),  then for g we get an equation g(x+y)=y g(1+x/g(y)).
Putting x=a g(y) we get that

1. g(ag(y)+y)=y g(1+a)

2. Now we can see, that if g(y_1)=g(y_2) than y_1=y_2.

2'. putting y=1 in 1 we get, that g(1)=1.

3. For each t>0 there exist y such that g(y)=t:  y=a g(t/g(a+1)) +t/g(a+1)

4. If y_2>y_1 than g(y_2)>g(y_1). If (y_1-y_2)/(g(y_1)-g(y_2))<0 then
let a=-(y_1-y_2)/(g(y_1)-g(y_2)) >0. Then a g(y_1)+y_1=a g(y_2)+y_2, then
y_1g(1+a)=y_2g(1+a), so y_1=y_2.

5. From 4&3 we can get, that g is continious, and we can define th limit
g(0),  so we can put a=0 in the equation 1. So g(y)=y.

Excellent solution, Fedor. Only now I have realised that the problem is not very difficult. Well, I just hope now that everyone has this feeling at least once in his life. My solution is much more complicated and made me believe that the problem is so difficult: I WAS WRONG!  Anyway, I still believe it isn't trivial. I saw far less difficult problems given in important contests. MY solution:
   First of all, we prove that f(x)<=1 for x>1 and x>=1 for x<1. This follows easy if we suppose for example that for an y>1 we have f(y)>1. Take x=(y-1)/(f(y)-1).
   Thus, f(x+y)=f(1+xf(y))/y<=1/y for all x and y, from where f(x)<=1/x.
   Then, yf(x+y)=f(1+xf(y))=1/xf(y)*f(1+f(xf(y)) and so xyf(y)f(x+y)=f(1+f(xf(y)). Take x=1/f(y) and find that f(x)>=(1-k)/(kx) for all x, where k=1-f(1+f(1))>0.
   We prove now the injectivity:If f(a)=f(b), then for all x we have af(x+a)=bf(x+b) and so a^nf(na)=b^nf(bn). Let a<b then for n large we have f(nb)<(a/b)^n, which contradicts the fact that f(nb)>=(1-k)/(knb) for all n.
    Finally, let y>1 and x>0. Then yf(x+y)=f(1+xf(y))=f((1+xf(y)-1/y)+1/y)=yf(1+f(1/y)(1+xf(y)-1/y)). From injectivity we deduce that
 x+y=1+f(1/y)(1+xf(y)-1/y) from where f(1/y)f(y)=1 and y=1+f(1/y)-1/yf(1/y). From here we find that for y>1 we have f(y)=1/y and since f(1/x)=1/f(x) we must have f(x)=1/x for x<>1. It's easy to see that f(1)=1.

hi, Fedor
I don't know you how to get this?  can you ......?thank!

2.Now we can see, that if g(y_1)=g(y_2) than y_1=y_2


geometry
3D geometry
algebra
function
domain
greatest common divisor
number theory unsolved

Find all $x,y\in Z$ such that
$x^2+2=y^3$.

[quote="ZetaX"]I'm not sure if it was posted before, but I think yes...

Looking $\mod 4$ gives us that $x$ is odd.
Factor it as $(x+ \sqrt{-2})(x-\sqrt{-2}) = y^3$. Now $\gcd(x+ \sqrt{-2} , x-\sqrt{-2}) =1$, so the numbers are coprime.
Thus (using unique factorisation in $\mathbb{Z}{\sqrt{-2}]}$) $x+\sqrt{-2}$ is a cube, so $x+\sqrt{-2} = (a+\sqrt{-2}b)^3 = a^3 + 3 \sqrt{-2} \cdot a^2b - 6 \cdot ab^2 - 2 \sqrt{-2}\cdot b^3$.
Comparing imaginary components gives $1= b \cdot (3 \cdot a^2 - 2\cdot b^2)$, thus $b = \pm 1 = 3 \cdot a^2 - 2\cdot b^2 = 3 \cdot a^2 -2$, thus $a=\pm 1 , b=1$, giving the solutions $x=\pm 5 , y=3$.[/quote]
I am sorry $ZetaX$,but I can't understand what means $\gcd(x+ \sqrt{-2} , x-\sqrt{-2}) =1$ :? .

As ever, it is the greatest common divisor, but this times in another ring as $\mathbb{Z}$, here $\mathbb{Z}[\sqrt{-2}]$:
When $a|b$ (in the ring $R$), then call $a \leq b$ as a definition. By this, the term "greatest" is defined. What remains to show is that there is (up to units) exactly one greatest divisor. That possibly sounds rather obvious, but isn't. You will need a unique factorisation domain (or some other property) to show this.
At least, the known rule for $\gcd$'s applies: $\gcd(a,b) = \gcd(a,b+ka)$ for all $k$. This follows directly from the definition the same way as for the standard $\gcd$.

For this special case, it is
$(x-\sqrt{-2}) \cdot (x+\sqrt{-2}) =x^2+2=4n^2+4n+3$ (odd) together with
$(x+\sqrt{-2})+(x-\sqrt{-2})=2x=4n+2$ (even, coprime as normal integer to the above) gives after using
$c \cdot (4n^2+4n+3) + d \cdot (4n+2) =1$ from Bezeout finally
$(c(x-\sqrt{-2})+d) \cdot (x+\sqrt{-2})+d \cdot(x-\sqrt{-2})=1$, showing the desired property.
Well, looks, ugly, so another way:
$\gcd(x+\sqrt{-2},x-\sqrt{-2}) = \gcd(2\sqrt{-2},x-\sqrt{-2})$, dividing
$\gcd(2\sqrt{-2},2x-2\sqrt{-2})=\gcd(2\sqrt{-2},2x)=2$.
But $\sqrt{2}$ doesn't divide $x+\sqrt{2}$, so the $\gcd$ is $1$.

There is also another definition with ideals:
Let $\mathfrak{a} , \mathfrak{b}$ be ideals of the ring $R$, then their $\gcd$ can be defined as $\mathfrak{a} + \mathfrak{b} : = \{ a+b | a\in \mathfrak{a} , \ b \in \mathfrak{b} \}$.

PS: could some moderator please move it to Number Theory¿

[quote="ZetaX"]I'm not sure if it was posted before, but I think yes...
[/quote]
You are right :) : http://www.mathlinks.ro/Forum/viewtopic.php?t=15408

On a side note, I've learnt this technique with Euclidian rings not long ago, and I've to admit that I'm now a fan of it!  :love:

Thanks $ZetaX$,but I will be glad to learn more about this theory,can you advise any books in the net?


number theory
greatest common divisor
geometry
3D geometry

Hello, can anyone please give me a hint on how I would solve the following problems?
1.P.T. the equation $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 1969^2$ does not have any solutions for $ x,y,z\in\mathbb{N}$.
2. Let $ S$ denote the set of all 6-tuples $ (a,b,c,d,e,f)$ of natural numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2 \equal{} f^2$. Consider the set
                   $ T \equal{} [abcdef: (a,b,c,d,e,f)]\in S$
   Find the gcd of all the members of $ T$
3.Prove that the # of 5-tuples of positive integers $ (a,b,c,d,e)$ satisfying
                 $ abcde \equal{} 5(bcde \plus{} acde \plus{} abde \plus{} abce \plus{} abcd)$
IS AN ODD INTEGER.
4.Find all $ n\in\mathbb{N} ; n\geq3$ such that
        $ n_{C_0} \plus{} n_{C_1} \plus{} n_{C_2} \plus{} n_{C_3} | 2^{2001}$

Please help me solve these problems ASAP! :|
Thanks!!

[hide="3"]The expression is symmetric, so consider possible permutations of any solution.[/hide]

I think the first problem can be done by considering mod 7.
The right hand side is 4 mod 7 whereas any cube can only be 0,1,6 mod 7

Any permutation of 0,1,6 for the right hand side can never give 4 mod 7.
So we are done.

[hide="hint for number four"]

Pascal's Identity: $ \binom{n}{k} \plus{} \binom{n}{k\plus{}1} \equal{} \binom{n\plus{}1}{k\plus{}1}$[/hide]

[quote="befuddlers"]Any permutation of 0,1,6 for the right hand side can never give 4 mod 7.[/quote]

Except $ (6,6,6)$, you mean. ;)

Sorry guys mod 7 does not work.

But mod 9 does.

Any cube is only 0,1,-1 mod 9.

But $ 1969^2$ is 4 mod 9.

Any permutation of 0,1,-1 never yields 4 mod 9.

I have got quite a bit far in no. [b]2[/b]::
First of all, all of $ a, b, c, d, e$ cannot be odd. As a square is congruent to 1(mod8)
Hence, if $ a^2\equiv b^2\equiv c^2\equiv d^2\equiv e^2\equiv1(mod8)$, then it implies that $ f^2\equiv5(mod8)$ which, of course, is impossible.
Thus at least one of $ a, b, c, d, e$ have to be even.
Also I have observed that if none of $ a, b, c, d, e$ are divisible by $ 3$,i.e. if
    $ a^2 \equiv b^2\equiv c^2\equiv d^2\equiv e^2\equiv1(mod 3)$, then it would imply that $ f^2\equiv5(mod 3)\equiv2(mod 3)$ which is impossible for any square no. :)

        BUT I DON'T KNOW HOW TO CONTINUE FROM NOW ON!! :(
        Can anyone help me?
                                  Thanks

I have found no.[b]1[/b]!! :idea:
I read in a number theory book that $ x^3 \plus{} y^3 \plus{} z^3\equiv x \plus{} y \plus{} z\equiv1(mod3)$
Well, if $ x, y, z$ would have been in a complete residue system then it $ \Rightarrow$ $ x \plus{} y \plus{} z\equiv0(mod3)$ :)
Hence we may assume that $ x\equiv y(mod3)$
We have three cases(obvious):
[b][u]CASE I:[/u][/b] $ x \equal{} y \equal{} 0,z \equal{} 1 \Longrightarrow x^3 \plus{} y^3 \plus{} z^3\equiv 1(mod9)$
[b][u]CASE II:[/u][/b] $ x \equal{} y \equal{} 1, z \equal{} 2 \Longrightarrow x^3 \plus{} y^3 \plus{} z^3\equiv 1(mod9)$
[b][u]CASE III:[/u][/b] $ x \equal{} y \equal{} 2, z \equal{} 0 \Longrightarrow x^3 \plus{} y^3 \plus{} z^3\equiv 7(mod9)$

But, as $ 1962\equiv \minus{} 2(mod9)$,
          $ \Longrightarrow 1962^2 \equiv 4(mod9)$

But as observed none of our three cases yield the fact above and hence there are no solutions to the given equation! :!:  :idea:  :lol:

BUT, is it correct??? :maybe:

hello, I have found the answer to no 2. I continue from the place where I had written how far I got...
[hide="answer"]$ 24$ is the gcd[/hide]
[hide="continuation"]I found out that the smallest such set a,b,c,d,e,f is:(such that the product $ abcdef$ is smallest)::
$ (a,b,c,d,e,f)\equal{}(1,1,1,2,3,4)$ as $ 1^2\plus{}1^2\plus{}1^2\plus{}2^2\plus{}3^2\equal{}16\equal{}4^2$!
Then with the help of a few examples I saw that in all cases $ 2^3$ is the highest power of $ 2$ dividing $ abcdef$
[hide="request"] Can you please help me to prove the fact that $ 8$ really is the highest power of $ 2$ dividing $ abcdef$?[/hide]
[/hide]


Let be given a triangle $ABC$. Prove that: $R^{2}+r^{2}\ge\frac{5S}{3\sqrt3}$

Let F denote the area of the triangle and s the semiperimeter.

To prove the given inequality we use these facts:

1.  $\sqrt(27)R$ $\geq$ 2s
(which we will use in the form $\frac{R^{2}}_{4}$ $\geq$ $\frac{s^{2}}_{27}$)

2.  $s^{2}$ $\geq$ $\sqrt(27)F$
(which we will use in the form $s^{6}$ $\geq$ $\sqrt27)^{3}F^{3}$)

3.   rs=F

Now $\frac{R^{2}+r^{2}}_{5}$ =  $\frac{1}{5}$($\frac{R^{2}}_{4}$ +$\frac{R^{2}}_{4}$ +$\frac{R^{2}}_{4}$ +$\frac{R^{2}}_{4}$ +$r^{2}$) $\geq$ $\frac{1}{5}$($\frac{s^{2}}_{27}$+$\frac{s^{2}}_{27}$+$\frac{s^{2}}_{27}$+$\frac{s^{2}}_{27}$+$r^{2}$)   using the inequality in 1 above.
                     $\geq$ ${\frac{s^{2}.s^{2}.s^{2}.s^{2}.r^{2}}_{27^{4}}}}^{\frac{1}{5}}$ by the arithmetic geometric mean inequality.
=${\frac{s^{6}.F^{2}}_{27^{4}}}^{\frac{1}{5}}$  after  using equation 3
$\geq$ ${\frac{\sqrt27)^{3}F^{5}}_{27^{4}}}^{\frac{1}{5}}$ after using the inequality 2 above
=$\frac{F}{\sqrt27}$

 Concerning inequalities 1. and 2.

The first is equivalent to SinA +SinB +SinC $\leq$ $\frac{\sqrt27}{2}$ and the proof is left as an exercise.

For the second apply AM-GM to s-a,  s-b  ,s-c  and use F in the form $\sqrt{s(s-a)(s-b)(s-c)}$

[quote="April"][color=darkred]Let be given a triangle $ABC$. Prove that :  $\frac{5}{3\sqrt3}\cdot S\le R^{2}+r^{2}\ \ (*)$[/color][/quote]

[color=darkblue][b]Proof.[/b] Denote $2p=a+b+c\ .$ Therefore,

$\left\{\begin{array}{c}p\le\frac{3\sqrt 3}{2}\cdot R\Longrightarrow S=pr\le \frac{3\sqrt 3}{2}\cdot Rr\Longrightarrow\frac{5}{3\sqrt 3}\cdot S\le \frac{5}{2}\cdot Rr\\\\ 2r\le R\Longrightarrow 0\le (R-2r)(2R-r)\Longrightarrow\frac{5}{2}\cdot Rr\le R^{2}+r^{2}\end{array}\right\|$ $\Longrightarrow$ $(*)\ .$[/color]


Please Post any Video Game Systems you have (even ones not on the poll but still relevant)

Me:
Super Nintendo
Gameboy Color
Gameboy Advance SP
N64
GameCube
Wii*
PSP
TI-83+ Siver Edition
TI-89 Titanium
Motorola Cellphone
Ipod

Edit:
If you have more than one of the systems on the poll
Please choose the one you like better

Edit2:
What happened to the poll?
oh well
just post your systems
and
put an * next to your favorite

I chose PC because it's the only one I have on the poll. :D

Me:
Xbox*
PS2
Game Boy COLOR
Game Boy Advance (doesn't work)
Had a Game Boy Advance SP (little brother lost it  :mad: )
Nintendo 64
Nokia Cell Phone
iPod (can play breakout and parachute)
TI-84 Plus Silver Edition  :D (currently has ChuChu Rocket, Donkey Kong, Dr. Mario, Super Mario Bros., Phantom Star, Phoenix and Tetris)
Getting an Xbox 360 after Nationals...

Sega Genisis(got boring)
Computer
iPod
TI-84 Plus Silver Edition
my imagination.

Whoa, u people have so many game systems...
Wii*
PS1 (never play)
GameBoy SP (Stupid L button broke)
IPod
TI-84 Silver

Coming Soon:
Cell phone
Nintendo DS

Gameboy Color
Gameboy Advance
Gameboy SP
Nintendo DS
N64
Gamecube
Wii*
PS2
PC
TI-83
TI-84
Ipod
Cellphone

hey perfect628
and
ra5249

do you want to pm me your wii number?
I'll tell you mine

Xbox 360*
Game Boy

No Live  :(   and I want to buy the Elite.

PC*
GBA
Ipod Nano (parachute & solitaire)
TI-84+SE (mostly games I build; the most entertaining other than the computer)

[quote="ra5249"]whats a wii number? where can i find it?[/quote]

Point at the lower right hand corner of your screen
there should be a message board type thing there
there a bunch of buttons in the lower right hand corner
click on one of those
I can't remember exactly
but the rest you should be able to find by yourself (or with the manual)

[quote="alkjash"]Wow. how do you guys have time to play on so many systems?

PC
TI 84 Plus Silver[/quote]

Well I play the Super Nintendo when i get nostalgic
Never play Gameboy Color
Sometimes on roadtrips or vacations I play the Gameboy Advance SP or even at school sometimes
N64 I gave away recently
GameCube I don't play anymore (the Wii can play GameCube games)
Wii I play the most during free time at home
PSP the most when I'm out of my house and sometimes at school (those days that are like free days aka last day before "holiday")
I only play tetris on my calculators at school
I rarely (if ever) play on my cellphone
I play on my Ipod when listening to music

So Ones I play frequently:
Wii
PSP
TI-89 Titanium
Ipod

Edit:
PC once in a while

Wow... everyone has a LOT of game consoles. :o