Mr. Rijan Karkee from Nepal has discoverd his own new theory in physics challenging Einstein and Newton.
his formula states: $ Fg \equal{} \frac {Mc^2} {d}$
for more info. look at: http://www.kantipuronline.com/kolnews.php?&nid=198791
                                 http://wearenepali.blogspot.com/2009/06/nepali-young-scientist-rijan-karki-in.html

I didn't see anything that indicates where this theory comes from.  And here's the thing about Einstein's theory of general relativity: it's been experimentally verified every single time.  In fact, his theory accurately described a slight alteration in the orbit of Mercury and accurately explained the bending of light.  I'm going with Einstein.

I don't think you "discover" a theory.  You verify it, using a LOT of experiments. That's how general relativity is a theory.  Until his "theory" is well verified, then it is not considered a theory.

I think it's best to keep an open mind. Theoretically, it sounds plausible, but I guess we'll have to wait on experimental verification. In the meanwhile, as JRav mentioned, Einstein's theory is more supported by the universe at large, and I think I'll go with that until experiments that prove it wrong and prove this new theory undeniably correct occur.

He was at the conference at NASA. i guess you guys can know more about it from NASA itself. but i heard he derived it experimentally. but a fact that he is only 16 yrs has come up with such a thing without any higher education its worth admiring. Also in a place like Nepal where the middle school education is weak, which i am facing today. people may think twice about this.

I'll remind you that Bohr's theory of the atom was semi-derived from experimentation (he created a theory that explained certain phenomena) and it looked good, until it was tested on other elements besides hydrogen.  Experiment, experiment, experiment!

It sounds really interesting but searching on the net i didn't find any reliable source. All the web pages talking about Rijan Karkee where from Nepal (one blog). There's nothing on the media.

First of all the name  doesn't even appear on the ISDC schedule of conferences (http://www.isdc2009.org//programming). The photo they show in those blogs (http://nepalaawaz.com/images/nepali%20studentinnasa.jpg) doesn't even appear on the ISDC Official web page (http://www.isdc2009.org/) and the only photos that "appear to be from Napal" are of some school students (http://www.avenfoto.com/isdc/content/isdc09_0305_large.html) that where there for, what i read, some kid's contest (http://www.isdc2009.org/education).

So, in my opinion it's fake. I really wanted to know how this kid came up with this idea, because it seemed, at first glance, nonsense:

The Napal's web sites that talked about this kid's thearoy said that this was a replacement for $ E \equal{} mc^2$ Einstein's equation. That can't be possible, how does the inner energy of a mass can possibly depend inversely on the distance to the mass? It doesn't make sense with experimental facts.

I guess he had to be taken in a theoretical physics conference not in NASA. so dosent NASA give its info of the conference in its site?

I web-searched on NASA's homepage and nothing.

The Napal's site calims that he attended at the ISDC conference, but if you search there nothing comes up either.

[i][b][/quote]Rajiv Such a hoax is not uncommon. There have been many nuts in India too who claimed absurd scholarships, or something else which I don't remember (all trash news) from NASA. The news people are hungry for anything sensational, true or false.
In my opinion, it is not easy to beat a brain as that of Einstein.[/quote][/b][/i]
i wanted to ask u if those nuts from india were brought by the media of which country? . and as u talked about scholarship. it is given by a college of our own country. NASA wouldnt have to worry.  And as of beating einstein, who had thought someone could challenge newton and galileo? ok maybe this boy is wrong but it cannot be said that he is wrong because he just challenged einstein? and einstein himself says "The important thing is not to stop questioning.". its admiring he has just come up with some of his results. even ramanujan from ur country, didnt Hardy compare him with polya and hilbert? its glad he got a chance, hadnt hardy and littlewood come to the rescue his theorems would just be papers. and now when someone has got a chance u cannot criticise him.  And he hasnt come to this conclusions just from some grade 9/10 mathematics. he has done it with a exp. i dont know much. we have to wait for a techinal explanation soon.

Ramanujan was right.  Most of his results could be, and were verified.  Big difference.

My comments are not intended to hurt your sentiments or challenge your claim. All I wanted was to point out a possible explanation to this boy, Rijan's news being away from the international media spotlight so far. That, again a possibility I'd like to point out, might be because they are very careful and never publish anything without thorough verification of the authenticity of the said claim.

If you check the dimensions of the left and right side of the equation, you will see that they do not match

@KevinB
Check properly,dimensions are correct.Both have dimensions of force :P

Hello Everybody,
I would like to thanks everybody who gave me such commennts. Brothers and sisters, let me clear you that the news shown in Nepal are just ton attract the readers. I have never said that I have challenged Newton and Einstein, the journalist just understood in that sense and wrote that. I have just said that I have made new thinking on Gravitation force because upto now nowhere is explained about the origin of all, even who kept comment on me may not able to answer where is the  origin? What I have done is that I came with new idea to support the origin of Gravity force.I do support that there is no list my name in the conference because I was not sure to attend the conference because of VISAs and sponsors. I informally requested the members over there where they gave me chance and they have taken my hypothesis.
So, let me tell in brief that I have not challenged Einstein but just added the new idea,
I feel myself as very unlucky guy because when I was 13 , I made the new theory on motion and when I showed to my seniors,he replied me that this is the same theory that Einstein has already explained in his theory of relativity,
even I was right and accordinfg to me I brought new idea but I was wrong , it was already exist, at that age I have even not heard about Theory Of Relativity but what a coincidence Einstein and my thinking became same but it is useless to me.
From this comment , I would like tell everybody that within 10 to 12 years I will be solving dark energy with mathematics because at present I have solved theorically with new idea , now all I have to do is mathematics,
and I am going to make one engine that gonna run cotinuously(perpetual motion)
and if it works, then ............!!!!!!!!!!!
I have already made design and now I need fund.............................
So those who are angry with me I would like to show them..............

Hello RKee.

So you are the one who made this theory eh?

Its very nice to hear that you are thinking of these things, and all i can say is go on..

I am also thinking about these things a lot you know (i dont agree with einstein..).

And an important thing is you shouldnt get stuck with the math..., mathematicians are there to clean that up. The important thing is always the physical thought behind it.

There is a famous saying from a nobel prize winning physicist "In physics, the physics are hard, not the mathematics" so if you keep yourself to that you will be successful. Of course mathematics is very important in physics, just physics is not maths, its very different..

Im sure you have heard about Nicola Tesla, you should make some research on what he did (An exceptional physicist whos legacy is thrown out in the "rubbish bin" by physicists today), and you can get many ideas from him..

*Off-topic: about my greatest countryman, Nikola Tesla

Indeed, Tesla's work has been neglected for a long time. World probably wasn't ready for everything Tesla sensed and discovered. Still, there's a problem today with the over-mistification of his work - this is encouraged by Tesla's non-conventional way of life. But if we clear the occultism out of Tesla's legacy, we can re-discover a lot (both in physics and electrical engineering!).

Yes Hsiljak, thats what we should do..

I already started doing some research on his inventions, and theories (though i look at it more from a physicists few rather than an electrical engineers ... :D ).. And if they could work, that would be amazing.

@Rkee
Can you please explain the idea behind your theory and how did you come to the conclusion?

[quote="Rkee"]
and I am going to make one engine that gonna run cotinuously(perpetual motion)
[/quote]

Now I feel a bit skeptical. Einstein's theory may be wrong (or incomplete), but you claim by this that energy conservation laws and the fundamental laws of thermodynamics are false (which is rather unbelievable, at least for me). I believe you can't reveal details about your idea, but I'd be happy to hear just a bit more about it  .

One cannot say Einstein's theory is wrong unless an experiment is presented which blatantly contradicts his theory.  If however, one conducts an experiment which reveals something new, all it means is that the previous theory is incomplete.

And I don't know how you can possibly conclude that thermodynamics and energy conservation are wrong just because Einstein's theory is not a theory of everything.  And since special and general relativity are rooted in Maxwell's theory, that would imply that E&M theory is wrong, or at least badly misunderstood. :huh:

Actually, JRav, I don't think that Rkee concluded TD and conservation laws are wrong from his analysis of Einstein's theory. It looks to me that he just had an idea about a way how to make a perpetuum mobile*, and he is now trying to find a sponsor to bring this idea to life. Every day people claim to have found perpetuum mobile - but they're wrong. Because of that I'm highly skeptical about this version of it.

________
*and laws of thermodynamics claim that no such machine (perpetuum mobile of first and second kind) is possible.

Rijan Karki told  that two opposite repulsive forces from the two masses is the reason why the universe is expanding.
But if this is the reason then every bit of matter should repel each other. If there exists repulsiveness properties then due to continuous repulsion the matter ,from which stars are born , should repels each other and there is no chance of getting them combined. I am sure that Rijan's theory is wrong.

Can Rijan KArki explain about the two repulsive property?
Then why earth is no repeling every bit of matter.
Why sun is no exerting repulsive property with the repulsive property of the planet?
I am not saying Einstein's Theory is wrong. Einstein Theory is accurate.
But I can't figure it out about Rijan's Theory. He is totally wrong.

[quote]
I feel myself as very unlucky guy because when I was 13 , I made the new theory on motion and when I showed to my seniors,he replied me that this is the same theory that Einstein has already explained in his theory of relativity,

I am going to make one engine that gonna run cotinuously(perpetual motion)
and if it works, then ............!!!!!!!!!!!
I have already made design and now I need fund.............................
So those who are angry with me I would like to show them..............[/quote]

Nice, that you are thinking on these things. I would just like to know what exactly was your theory of motion when you were 13? I would also appreciate if you could elaborate your point regarding your Gravitational theory.

For perpetual mobile, I wish you luck! :?:

Bests.

I am a little sceptical.

The "theory" can be summarised by such a simple formula. Indeed, if $ E \equal{} mc^2$ had not been discovered by Einstein, it certainly would have later discovered experimentally. However, we do know that Einstein's theory is either wrong or incomplete due to it being incompatible with quantum mechanics.

In fact the modern description of general relativity includes a (very small) cosmological constant to account for the expansion of the universe. It has been included to "make things work", although its origins are not clear. It is hypothesised to be due to dark matter in the universe, which has the tendancy to warp space-time in the opposite sense (i.e. causes the space-time manifold to have negative curvature). Bodies in such a space would therefore have a tendency to move apart. His theory may be just a way to describe this curvature - who knows.

The person who wrote the message was probably not the man himself. For instance the phrase "brothers and sisters" is sickeningly stereotypical. Also it seems highly coincidental that he should appear - with his first message on this site - just when people are discrediting him. Unless he has a reason to be on this website, like being in the indian olympiad. Also the claims that he formulated the theory of relativity himself at the age of 13 is almost impossible to believe. Einstein made a leap of thought, although it was only a relative leap - excuse the pun. The inertial frames of relativity were exactly those frames in which Lorenz had shown the Maxwell equations to be invariant. Other evidence that the current theory was inadequite such as the Michaelson Morley experiment, and experiments with light travelling through a moving fluid, were known by Einstein. The building up of relativity from its foundations was a joint effort...for instance Einstein was not the first to find the exact way in which matter cruved space (although it is to be said, Einstein did discover the euqations a few days later independently). This all makes it highly implausible that a 13 year old could come up with relativity himself: and if he was learned enough, then surely he should have stumbled across relativity before hand. Finally, if indeed by some genius much greater than Einstein he did formulate relativity independently, his ignorance of pre-existing physics would be verified, and there would be no reason to trust his theory now.

@^^
Thank you very much for your elaboration.

Therefore, I asked him to elaborate what was exactly his thought on relativity when he was 13.

And I'm afraid, I have a feeling that it is "galilean relativity" which is known since thousands of years and already postulated by Galieo some 400 yrs ago!

I am sorry to say, it seems the understanding of relativity and quantum mechanics is being profoundly missed by the guy.

Considering the fact that "Discovery is a stroke of idea worked in details" I'm yet optimistic to have some details on  the work.  :)

For "perpetual motion machine", I would say, if it happens, we will have to rewrite about the universe again!

-Diwaker


Is that page harder than usual MC problems? :?

Make sure you're going here: http://mathcounts.saab.org/

It is relatively the same.  And it matters what you call usual math problems...

If you mean Warmups/Workouts/Stretches it is a little more difficult (close to what the problems were like in the 80s).

But its pretty close to state.

Those questions tend to resemble some of the ones from previous years.  And in my opinion, mathcounts was a lot harder then then it is now.

 :D

I think some of them are around today's National level.

Yeah...But I have no clue what your exposed to.  I do things from almost any year...(especially old ones).

Well, I've looked at some of the previous MathCounts rounds and they seem a bit harder than the current ones... depending on what you call "normal", these would probably be around high-level States or maybe Nationals.

Yes, when i was using that site last year to study I found most of them to be Nationals level, and some of the easier ones might be the very toughest ones at state, like Treething said I guess.


In $ \Delta ABC$, $ \overline{DE} \parallel \overline{AB}, CD \equal{} 4$ cm, $ DA \equal{} 10$ cm, and $ CE \equal{} 6$ cm. What is the number of centimeters in the length of $ \overline{CB}$?

[asy]pair A,B,C,D,E;
A = (-2,-4); B = (4,-4); C = (0,0); D = A/3; E = B/3;
draw(E--D--C--B--A--D);
label("A",A,W); label("B",B,dir(0)); label("C",C,N); label("D",D,W);
label("E",E,dir(0));[/asy]

Similar triangle occur here $ ( \Delta ABC\sim \Delta DEC)$, so we can set up a proportion:
\[ \frac{CD}{CA}\equal{}\frac{CE}{CB}
\]
Which, by substitution, leads to:
\[ \frac {4}{4 \plus{} 10} \equal{} \frac {6}{CB}\implies\frac {2}{7} \equal{} \frac {6}{CB}
\]
Cross multiplying, we get:
\[ 2CB \equal{} 42\implies CB \equal{} \boxed{21}
\]


integration
limit
calculus
logarithms
real analysis
real analysis unsolved

Define $ a_n \equal{} 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{} \cdots \plus{} \frac{1}{n} \minus{} \int^n_1 \frac{1}{t}dt$. Prove that $ a\equal{}\lim_{n \rightarrow \infty} a_n$ exists and $ 0<a<1$.
Attempt
My issue is the integral in the sequence. I don't know how to take the limit of the sequence with that in there. I think there is a way to break up the integral to take the limit of the sequence easier. I need help with this question. Thanks in advance.

Think about left and right Riemann sums.

As a matter of fact:

$ \int_{1}^{n} \frac {dt}{t} \equal{} \ln n.$

Hence,

$ \ln (n \plus{} 1) \minus{} \ln n \equal{} \int_{n}^{n \plus{} 1} \frac {dt}{t}$

which implies in turn that

$ \frac {1}{n \plus{} 1} \equal{} \int_{n}^{n \plus{} 1} \frac {dt}{n \plus{} 1} < \int_{n}^{n \plus{} 1} \frac {dt}{t} \equal{} \ln (n \plus{} 1) \minus{} \ln n < \int_{n}^{n \plus{} 1} \frac {dt}{n} \equal{} \frac {1}{n}.$

Using this estimate prove now that the sequence is decreasing and bounded below by $ 0$.


Hola! :)

I need the help on seeing whether I know my grammar on spanish right or not. I have an exam coming on Monday and hopefully to A+ it! :lol: Last time I got help from here, I got A+, so I guess this forum really helps! I hope to get same help again!

¿Que piensas regalarle a tu hermana?
Answers:

Answer 1: Voy a darle una planta a mi hermana.
Answer 2: Pienso reglarle una planta a mi hermana.

¿Para quien es el regalo?

Answer: La planta es para mi hermana.

¿Que tipo de regalo buscas?

Answer: Busco unos calcetines para mi hermano.

Perdon, ¿donde esta (should be accent on a) la floreria?

Or another question with same meaning "Perdon, ¿donde queda la floreria?

Esta (should be accent on a) lejos de la jugueteria.

Or another answer: Queda al lado de la zapateria. (Different meaning)

¿Que ropa vas a llevar a la fiesta?

Voy a llevar una camisa con bluejeans.

¿No tienes algo mas formal? (should have accent on a on mas)

Si, pero prefiero llevar ropa comoda.

¿Cual es mas (same case as above) barato --- el vestido de seda o la chaqueta de cuero?

El vestido de seda es menos caro.

¿Son los aretes tan caros como el collar?

Si. Son del mismo precio.
---------------------------

Setting: Para cenar en un restaurante elegante

Para cenar en un restaurante elegante tienes que llevar ropa muy formal.  Puedes llevar un traje y una corbata.

Setting: Para hacer un picnic con unos amigos

Para hacer un picnic con unos amigos puedes llevar unos bluejeans y una camiseta.

Setting: Para ir a clases en el invierno

Para ir a clases en el invierno puedes llevar el sueter de lana.

Thanks again and please be this to be right!!!!!!!!!! :?  :?  :?  :?

In general I do not see any problem with your answers. Just two small comments:

Blue Jeans is used some times, but, at least what I have heard to people of basically al Latin America is: "Pantalones de Mezclilla", or just Jeans.

Don't worry about the lack of accents (I do it all the time when I am writting in Spanish on a computer, because most of the time it is configured not in spanish :D ), because if you think that you were indicated all the accents you should have, well, there are a lot that you are missing (I suppose that it may be important for your assignment).

:D Speaking of accents, I got another question.

Like "esta" without accent and "esta" with accent, if you hear that word, how do you know which one they're talking about? :?

If you hear it on the sentence, you might be able to guess which one they use (since they clearly have different meanings :lol: ) but my spanish teacher always says if you hear the spanish word, you can always know how to spell it because spanish words are pronounced just like how they spell.

The problem is... :( I can't do that.. :? ... And that's what makes spanish hard to me.. Hard for me to catch the words by "ear..."

[quote="Silverfalcon"]Like "esta" without accent and "esta" with accent, if you hear that word, how do you know which one they're talking about? :? [/quote]

Of course, the pitch in the first case is in the first syllable, but in the second case is in the second syllable, at least if you talk about [b]esta[/b] (female this) and [b]está[/b] (it is).


calculus
derivative
geometry
3D geometry
sphere
email
algebra unsolved

Ok the Question was, Find the dimensions of the cone of maximum volume that will fit inside a sphere of readius 10cm so that the top of the cone and the circumference of teh base of the cone all touch the sphere.  You may use any technology at your disposal to arrive at a solution to this qustion.

V=1/3(pi)r^2h

OK, so far, i have derived that i needed to use a right angled triangle to find h, which doesnt really help cos i need to find r also, so therefore ive come up with;
the height = 10+[(100-r^2)^1/2]
subbing that back into equation becomes;
V=1/3(pi)r^2*[10+({100-r^2}^1/2)]

This is about as far as ive gotten and i need to this desperately...Now to find r i have to find the derivative of it using the chain rule,
where f(x)=g[h(x)], and f'(x)=h'(x)*g'[h(x)]

Now if anyone could help me with this it would be greatly appreciated, thanks

PS i also need all the working etc... my email is listoboy89@hotmail.com[/code]

Cut a plane through the center of the sphere and draw the right triangle from the center of the sphere to the base, and to the edge of the base, of the cone.  Then $r^2 + (h - 10)^2 = 100$, where $r$ stands for the radius of the base.  Substitute $r^2 = 20h - h^2$ into $V = \displaystyle\frac{1}{3} (\pi r^2 h)$, and you have $V = \displaystyle\frac{1}{3} (20h^2 - h^3)$.

Straightforward differentiation gives the result as $h = \frac{40}{3}, r^2 = \frac{800}{9}, r = \frac{40}{3}$


topology
real analysis
real analysis unsolved

If $a,b\in\mathbb R^2\setminus A$, the circles which pass through $a,b$ are all disjoint (well, except for the endpoints, $a$ and $b$) and there are uncountably many of them, so there must be one which contains none of the points we eliminated from the plane.

Suppose the contrary. This means there exists $x,y\in\mathbb R^2$ such that any path connecting them contains at least one element of [b][i]A[/i][/b]. Consider the family [b][i]F[/i] [/b] of paths connecting [b][i]x[/i][/b] and [b][i]y[/i] [/b] and filling, for instance, all the ball with center at $\frac{x+y}2$ and radius $\frac{\|x-y\|}2$. By hypothesis each $c\in F$ contains certain $a\in A\Rightarrow$ cardinal number of [b][i]A[/i][/b] is not less than that of [b][i]F[/i][/b]. But [b][i]F[/i][/b] has cardinal of continuum, so we have come to the contradiction with countability of [b][i]A[/i][/b].

So the result is true not only for $\mathbb R^2$ but also for $\mathbb R^n$. Maybe this result is true for any (semi)normed space. If you approve the proof, I propose to solve this more general problem (in the case of (semi)normed space).

[quote="eugene"]Prove that if $A \subset \mathbb{R}^2$ is countable, then $\mathbb{R}^2 -A$ is path connected[/quote]

As a continuation of this problem: if $A$ is a countable set of disjoint closed intervals of a length $1$ in $\mathbb{R}^2$, is it true that $\mathbb{R}^2 -A$ is path connected?

No, it's no longer true. The intervals we're going to use are going to be vertical, and each one is going to lie on a different line $\ell_a$, defined by the equation $x=a$ for some $a\in\mathbb Q$. We place the intervals on the plane inductively:

Let $(D_n)_n$ be a basis of open sets for the topology of the plane. Now, having placed the $n$'th interval on the plane, we place the $n+1$'th such that it lies on a line $\ell_a$ not used by the previous $n$ intervals, and such that its midpoint belongs to $B_{n+1}$. It should be clear that we cannot connect two points in $\mathbb R^2\setminus A$ which lie on different vertical lines by a continuous arc.


In $ \triangle ABC$ the measure of $ \angle B$  is $ 20^{\circ}$ and the measure of $ \angle C$ is $ 40^{\circ}$. Moreover, the length of the bisector of $ \angle A$ is equal to 2. What is the value of $ \overline{BC} \minus{} \overline{AB}$?

Anyone?

Please don't brute force with trig.

[quote="mathemagician1729"]Please don't brute force with trig.[/quote]

[hide="Solution"]

We simply use the Law of Sines. In case you're wondering, this [b]is not brute forcing[/b].

$ \frac {AB}{\sin 100} \equal{} \frac {2}{\sin 20}$

$ \overline{AB} \approx 5.76$

$ \frac {CB}{\sin 120} \equal{} \frac {5.76}{\sin 40}$

$ \overline{BC} \equal{} 7.76$.

$ 7.76 \minus{} 5.76 \equal{} \boxed{2}$

[/hide]

Interestingly, we notice that for this particular case, $ \overline{BC} \minus{} \overline{AB} \equal{} x$, where $ x$ is the length of the bisector.

[b]Extension[/b]: Can we prove for all cases of $ a\minus{}c\equal{}x$ where x is the length of the angle bisector of angle A, and that a,b, and c are the sides of a triangle?

Sorry, but I can't use a calculator on this. What I did was to find the other angle, 120 and then split it using the angles bisector. I drew an adjacent perpendicular and now I have a 30-60-90. I'm stuck now though...

[hide="no trig"]
You might want to draw a picture and label everything in the solution, I'm not too good with geogebra/asymptote, so yeah.
Let the angle bisector of A meet BC at point D.
We know that angle DAC is 60 degrees and angld ACD is 40 degrees, so angle ADC must be 180-60-40=80 degreees.
Construct segment AE such that angle AEB is 80 degrees. We see that angle EAD is 180-80-80=20 degrees, so angle CAD is 60-20=40 degrees. This also means angle BAE is 60+20=80 degrees.
From isosceles triangle ABE, we see that AB=BE.
From isosceles triangle ADE, we see that AD=AE=2, and since ACE is also isosceles, we see that CE=AE=2.
Finally, BC-AB=BC-BE=CE=2, so the answer is 2.
[/hide]

idk if this is brute force but...
[hide="with trig/no calculator"]$ AB \equal{} \frac{2\sin100}{\sin20} \equal{} \frac{2\cos10}{\sin20} \equal{} \frac{1}{\sin10} \equal{} \frac{\sin 40}{\sin10\cdot\sin40}$

$ BC \equal{} \frac{\sin120}{\sin10\cdot\sin40}$

$ BC\minus{}AB \equal{} \frac{\sin120\minus{}\sin40}{\sin10\cdot\sin40} \equal{} \frac{2\cos80\cdot\sin40}{\sin10\cdot\sin40} \equal{} \frac{2\cos80}{\cos80} \equal{} \boxed{2}$.[/hide]

I think you're going the right way mathemagician. Now that you have 30-60-90 triangles I don't think it should be hard to just blow its bejeezez with some law of sines.


superior algebra
superior algebra unsolved

For a ring $ A,$ $ B$ is the same as $ A$, except multiplication as defined by

$ a\circ b \equal{} ba$

Show that $ B$ is a ring and that $ B$ is isomorphic to $ A$ precisely when $ A$ has an antiisomorphism.
When is $ A \equal{} B?$

You will need to use the definition of a ring and put this in. Where do you have problems¿

Another properties:

a) $ B\equal{}R^{\mbox{op}}$ is called the [b]opposite ring[/b] of $ R$.

b) $ R$ has an identity if and only if $ R^{\mbox{op}}$ does.

c) $ R$ is a division ring if and only if $ R^{\mbox{op}}$ is.

d) $ (R^{\mbox{op}})^{\mbox {op}} \equal{}R$.

e) If $ S$ is a ring , then $ R \cong S$ if and only if  $ R^{\mbox{op}} \cong S^{\mbox{op}}$.


inequalities
geometry
analytic geometry
conics
ellipse

If [b]P[/b] is any point on a straight line drawn through the vertex[b]A[/b]
of an isosceles triangle [b]ABC[/b],parallel to the base, prove that
[b]PB+PC>AB+AC[/b][/list]

K, here is my crack at it

[hide]
consider if the triangle were on a plane, since A and P are on the same line then their y values are the same. for their x value we can come up with an inequality statement
$P < A < P$

from now on when i refer to an angle I really mean its x location.$b$ = base and $h$ = height.

we can create equations for the side lengths
$PB = \sqrt{(P-B)^{2}+h^{2}}= \sqrt{(P-C+b)^{2}+h^{2}}$
$PC = \sqrt{(P-C)^{2}+h^{2}}$

$AB = \sqrt{(A-B)^{2}+h^{2}}= \sqrt{(A-C+b)^{2}+h^{2}}$
$AC = \sqrt{(A-C)^{2}+h^{2}}$

So each equation has the same $b, h$ and $C$ values thus we can pick arbitrary values for them which stays the same throughout each equation. Since each equation is under the radical and each has $h^{2}$ within we can eliminate them since those operations/values are constant throughout all 4 equations.

$(P-C+b)^{2}+(P-C)^{2}> (A-C+b)^{2}+(A-C)^{2}$
Lets say the base of our triangle = 4, and the x location of C = 4, Since this is a isosceles triangle the x location of A must equal 2.

So we have two cases, $P < A$ and $P > A$ so we can pick random values of P to substitute in to check our inequality statement as long as they are greater or less than A

$P < A \Rightarrow P < 2, P =-7$
$(-7-4+4)^{2}+(-7-4)^{2}> (2-4+4)^{2}+(2-4)^{2}$
$49+121 > 4+4 \Rightarrow 170 > 8$
So this checks out

$P > A \Rightarrow P > 2, P = 9$
$(9-4+4)^{2}+(9-4)^{2}> (2-4+4)^{2}+(2-4)^{2}$
$81+25 > 4+4 \Rightarrow 106 > 8$
and again this checks out

so it is true that $PB+PC > AB+AC$
[/hide]

Is there no way to solve it by pure geometry without using coordinates? :maybe:

[hide]Let the height of the isosceles triangle be 'h' and let the base have lenth '2a'. Therefore, each of the sides has length sqrt(h^2 + a^2).

Extend BC in both directions. Draw the line perpendicular to AP through P. We can still draw two right triangles, only this time the two sides are: sqrt(h^2 + (a+x)^2) and sqrt(h^2 + (a-x)^2). 2sqrt(h^2 + a^2) < sqrt(h^2 + (a+x)^2) + sqrt(h^2 + (a-x)^2) so [b]PB + PC > AB + AC[/b]. This would work even if x was greater than a, because (a-x)^2 = (x-a)^2.[/hide]

OK.... taking a random value to solve an inequality isn’t a good idea at all. I can give u some example for which the inequality will not hold for random numbers but for an interval.
And wizard would u please explain ur solution?
So i am giving a complete proof:

let h the height and base=2x and a point P on the parallel line and PD be a orthogonal projection on Base such that, DC=y and BD=x-y

using Pythagoras,

BA=AC=sqrt(h^2+x^2)

PB=sqrt(h^2+(2x-y)^2)

PC=sqrt(h^2+y^2)

so the given inequality is, 2sqrt(h^2+x^2)<sqrt(h^2+(2x-y)^2)+sqrt(h^2+y^2)

after a long calculation we get that (if my calculation isn’t wrong!) 4xy<3y^2+2x^2 now we use AM-GM in equality but here is a trick

if we can prove that 2x^2+2y^2>4xy  we will be done... bcz 2x^2+2y^2<2x^2+3y^2

AM-GM>>> (x^2+y^2)/2>sqrt(x^2*y^2) that is x^2+y^2>2xy so our inequality is true. QED

>>>MOON

consider an ellipse with foci at B,C, passing through A

Or consider Minkowki's Inequality. (thanks, t0r!)
Or do some analytic geometry.

Thank u all
But probably i have made a little mistake.After calculation the inequality would be 2xy<=x^2+y^2 so equality,when x=y i mean the triangle is isosceles


there is a problem
it said that i had passed venn diagrams, but when i went back and saw, I was still at 79%

This goes [url=http://www.artofproblemsolving.com/Forum/index.php?f=497]here[/url], so a mod should move it. This problem has been brought up [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=302938]here[/url].


Two stronger inequalities:
\[1.\quad \sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{C}{2}\sin\frac{A}{2}\le\frac{5}{8}+\frac{1}{6}\left(\sin A\sin B+\sin B\sin C+\sin C\sin A\right) \]

\[2.\quad \sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{C}{2}\sin\frac{A}{2}\le\frac{5}{8}+\frac{r}{4R}\]
Open question: $\boxed{\frac{1}{6}\left(\sin A\sin B+\sin B\sin C+\sin C\sin A\right)\le \left(\ge\right) \frac{r}{4R}}\; ?$

[quote="radio"]In triangle $ABC$, show

$\sin{\frac{A}{2}}\cdot\sin{\frac{B}{2}}+\sin{\frac{B}{2}}\cdot\sin{\frac{C}{2}}+\sin{\frac{C}{2}}\cdot\sin{\frac{A}{2}}\leq\frac{3}{4}$.[/quote]

It's true because:
$\sin{\frac{A}{2}}\cdot\sin{\frac{B}{2}}+\sin{\frac{B}{2}}\cdot\sin{\frac{C}{2}}+\sin{\frac{C}{2}}\cdot\sin{\frac{A}{2}}\le\sin^{2}\frac{A}{2}+\sin^{2}\frac{B}{2}+\sin^{2}\frac{C}{2}\le \frac{3}{4}$

[quote="hien"]

It's true because:
$\sin^{2}\frac{A}{2}+\sin^{2}\frac{B}{2}+\sin^{2}\frac{C}{2}\le \frac{3}{4}$[/quote]

Why is this?

No, no i'm wrong. The sign has to reserve.

[quote="April"]Two stronger inequalities:
\[1.\quad \sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{C}{2}\sin\frac{A}{2}\le\frac{5}{8}+\frac{1}{6}\left(\sin A\sin B+\sin B\sin C+\sin C\sin A\right) \]

\[2.\quad \sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{C}{2}\sin\frac{A}{2}\le\frac{5}{8}+\frac{r}{4R}\]
[/quote]
I'm sorry, the first inequality is not correct  :blush: . It must be:
[quote]
\[\sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{C}{2}\sin\frac{A}{2}\le\frac{3}{8}+\frac{1}{6}\left(\sin A\sin B+\sin B\sin C+\sin C\sin A\right) \]
[/quote]
[hide="Proof."] We have:
$\left \{\begin{array}{ll}&\frac{1}{6}\sin A\sin B+\frac{3}{8}\tan\frac{A}{2}\tan\frac{B}{2}\ge\sin\frac{A}{2}\sin\frac{B}{2}\\ &\frac{1}{6}\sin B\sin C+\frac{3}{8}\tan\frac{B}{2}\tan\frac{C}{2}\ge\sin\frac{B}{2}\sin\frac{C}{2}\\ &\frac{1}{6}\sin C\sin A+\frac{3}{8}\tan\frac{C}{2}\tan\frac{A}{2}\ge\sin\frac{C}{2}\sin\frac{A}{2}\\ &\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}=1 \end{array}\right\|\oplus\downarrow\\ \Longrightarrow \sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{C}{2}\sin\frac{A}{2}\le\frac{3}{8}+\frac{1}{6}\left(\sin A\sin B+\sin B\sin C+\sin C\sin A\right)$[/hide]


calculus
integration
trigonometry
calculus computations

Hmm... well making the substitution $u = \frac{1}{x}$ we get

$I = \int_{0}^{\infty}\frac{dx}{a^{4}+\left(x-\frac{1}{x}\right)^{4}}\cdot \frac{1}{x^{2}}$

so we add them together and get

$2I = \int_{0}^{\infty}\frac{dx}{a^{4}+\left(x-\frac{1}{x}\right)^{4}}\cdot \left(1+\frac{1}{x^{2}}\right)$.

Make the substitution $u = x-\frac{1}{x}$ and we find

$2I = \int_{-\infty}^{\infty}\frac{du}{a^{4}+u^{4}}$.

Unfortunately, I cannot find a nice way to evaluate this either.

(I'm assuming paladin8 is right so far; I haven't checked every step.)

One more little substitution: $u=av.$ Now you have

$I=\frac1{2a^{3}}\int_{0}^{\infty}\frac{dv}{1+v^{4}}$

It's a fairly well-known formula (whose slickest proof uses the residue theorem of complex variable theory) that

$\int_{0}^{\infty}\frac{dx}{1+x^{n}}= \frac{\pi}{n\sin\left(\frac{\pi}n\right)}$

Hence we have $I=\frac{\pi}{4\sqrt{2}a^{3}}.$

[quote="didilica"]The integral equals $\frac{\pi}{2\sqrt{2}a^{3}}$.

The substitution $x-\frac{1}{x}=u$ leads to the new limits of integration ${-\infty}$ and $\infty$, but other than that I think it is important the techniques used in the calculations....
Nice methods....
Didi[/quote]

Yeah, my bad.

[quote="Kent Merryfield"]
$ \int_{0}^{\infty}\frac {dx}{1 \plus{} x^{n}} \equal{} \frac {\pi}{n\sin\left(\frac {\pi}n\right)}$[/quote]
How do you show that applyin' real variable tools?


geometry
circumcircle
geometric transformation
reflection
homothety
ratio
geometry proposed

Three circles $\mathcal C_1(O_1)$, $\mathcal C_2(O_2)$ and $\mathcal C_3(O_3)$ share a common point and meet again pairwise at the points $A$, $B$ and $C$. Show that if the points $A$, $B$, $C$ are collinear then the points $Q$, $O_1$, $O_2$ and $O_3$ lie on the same circle.

I presume $Q$ is the common point of the three circles.  Cute problem. :)

Invert about $Q$.  Then, the three circles turn into lines that bound a triangle $A'B'C'$.  Because $A,B,C$ are collinear, $Q$ lies on the circumcircle of $A'B'C'$.  The centers of the three circles are the reflections of $Q$ across $A'B', B'C', C'A'$.  The feet of the perpendiculars from $Q$ to $A'B', B'C', C'A'$ are collinear by Simson's theorem so by a homothecy centered at $Q$ with ratio $2$, the three circle centers are collinear.  Thus, the centers of the three circles  and $Q$ (before inverting) are concyclic, as desired.

Comment.  The problem can actually be strengthened to the following: $A,B,C$ are collinear [b]if and only if[/b] the centers of the three circles and $Q$ are concyclic.  This easily follows from the iff condition of Simson's theorem.


geometry
3D geometry
tetrahedron
geometry solved

Let $ABCD$ be a tetrahedron. Prove that there exists a triangle whose sides are: $AB \cdot CD,$ $AC \cdot BD,$ $AD \cdot BC.$

http://www.mathlinks.ro/Forum/viewtopic.php?t=21209 .

  darij


function
trigonometry
integration
calculus
calculus computations

For the function $ f(x)=\frac {\sin x-x\cos x}{x+\sin x},\ x>0$ ascertain $ \|\begin{array}{c}
A=\int \frac {f(x)}{x}\ \mathrm {dx}\\\\
B=\int \frac {f(x)}{\sin x}\ \mathrm {dx}\end{array}\ .$

First observe that

$ A\equal{}\int\left(\frac{1}{x}\minus{}\frac{\cos(x)\plus{}1}{x\plus{}\sin(x)}\right)dx\equal{}\ln|x|\minus{}\ln|x\plus{}\sin(x)|\plus{}C_{1}$,

and

$ A\plus{}B\equal{}\int f(x)\frac{x\plus{}\sin(x)}{x\sin(x)}dx\equal{}\int\left(\frac{1}{x}\minus{}\frac{\cos(x)}{\sin(x)}\right)dx\equal{}\ln|x|\minus{}\ln|\sin(x)|\plus{}C$.

I hope I made no mistakes.


probability
inequalities
probability and stats

Suppose we keep throwing a die until the sum of the outcomes is 1400 at least. Let B be the probability that more than 420 throws of the die are needed for this. Use the following inequality to find a bound for B:

P(|X - E(X)| ≥ a) ≤ (1/(a^2))variance(X)

Now, I know that
E(X) for rolling the die once = (6+1)/2 = 3.5
and
Var(X) for rolling the die once = (6^2 - 1)/12 = 35/12.

I also know that if you roll the die 420 times then:

E = 420(3.5) = 1470
V = 420(35/12) = 14700/12 = 1225

I guess my trouble lies in finding the actual bound using this information.

The event you want to talk about is the event that it takes more than 420 throws to accumulate 1400. That's a little difficult to deal with, but fortunately, it is identical to the event that the accumulated sum after 420 throws is less than 1400.

What you've done so far is fine. The random variable which is the sum of 420 throws has mean $ 1470$ and variance $ 1225.$ Take the square root of that variance: the standard deviation is $ 35.$

Then $ 1400$ is $ 70$ units or $ 2$ standard deviations below the mean.

By Chebyshev's inequality (which is what you were asked to do), the probability that a random variable is more than $ 2$ standard deviations away from the mean is no greater than $ \frac1{2^2}\equal{}25\%.$ That's the answer you were asked for.

But to use Chebyshev for such a thing is nearly always a gross overestimate. The sum of $ 420$ independent copies of the same nicely-behaved random variable should (by the Central Limit Theorem) be well approximated by a normal distribution of the appropriate mean and variance. The probability that a normal distribution lies more than $ 2$ standard deviations from the mean on one side is approximately $ .023.$ That, which is less than a tenth of the Chebyshev bound, is what we'd estimate this probability as.


Props to Kamior for suggesting math symbols & andy17null for coming up with custom smilies.  I'll be editing this post as we add more symbols.  Keep suggesting more & we'll add the ones we can...

To invoke each of these, type what's in 'code' below between colons, like so

:Code:

Code   -   Symbol   -   Use
^2       -     :^2:     -   Squared
^3       -     :^3: - Cubed
^4       -     :^4: - Fourth power
^5  - :^5: - Fifth power
^o - :^o: - Degree symbol
Sigma - :Sigma: - summation symbol
pi - :pi: - pi
alpha - :alpha: - alpha
stddev - :stddev: - standard deviation symbol (lowercase sigma)
mu - :mu: - mu
pm - :pm: - plus/minus symbol
sqrt - :sqrt: - square root
rt2 - :rt2:
rt3 - :rt3:
rt5 - :rt5:
phi - :phi:
theta - :theta:
inf - :inf:
equiv - :equiv:
approx - :approx:
cong - :cong:
sim - :sim:
le - :le:
ge - :ge:

You can also see all these by clicking 'View more Emoticons' while posting.

Here's another way:

An ampersand before each of these makes:

#176; = °

#177; = ±

#178; = ²

#179; = ³

#181; = µ

And a few more:

#188; = ¼

#189; = ½

#190; = ¾

Thanks for the recognition... How about an approximation sign (double tilda?) And because I'm really picky, why not also add a multiplication dot? Also really useful would be a radical sign. I hate typing out sqrt(x^5) or whatever... More suggestions as I think of them.

-Isaac

How about similar and congruent signs.  It would help with things in geometry.

[color=cyan]How about subscripts?  I just thought of this because of the Fibonacci problem -- you could write the problem using F_1 and F_n, but we had to use function notation.  Actually, we'd have to do that anyways, I guess.  Maybe an nth power, so we can have 1, 2, 3, 4, 5, n?  Also, is it possible to have the math smileys display themselves where the smiling smileys are now?

Joel[/color]

How about an inverse?  Some of the problems being done involve trig, and having a sin^-1 would be nice.

Thanks.

I don't like these mathematical 'emoticons', use LaTeX standard instead.


real analysis
function
limit
integration
search
real analysis unsolved

Let $ (a, b)$ be a finite open interval and $ f$ a Lebesgue integrable function on $ (a, b)$. Show that
\[ \mathop {\lim }\limits_{t \to \infty } \int_a^b {e^{itx} f\left( x \right)dx}  \equal{} 0.
\]

How to solve it?

Name (useful for key-word searching): Riemann-Lebesgue Lemma.

Some related topics here: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=91119]One.[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=131611]Two.[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2069715752&t=26568]Three.[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21933]Four.[/url]

[hide="Hint"]Reduce to the case that $ f$ is a simple function for which the claim is easy to show.[/hide]


trigonometry
inequalities
inequalities unsolved

obvious approach:

---

Use the fact that $\sin A  =  \frac{2\tan \frac{A}{2}}{1 + \tan^2 \frac{A}{2}}$.

Using simple identities like $tan^2 X + 1 = sec^2 X$, it is not hard to get at the equivalent statement $\sum \frac{\tan (A/2) (\cos A + 1)}{\tan (B/2) (\cos B + 1)} \leq \sum \frac{\tan (A/2)}{\tan (B/2)}$,

and after an application of the cosine law, $\sum \frac{\tan (A/2) (a)(b+c-a)}{\tan (B/2) (b)(a+c-b)} \leq \sum \frac{\tan (A/2)}{\tan (B/2)}$

Now, $\tan^2 A/2 = \frac{1- cosA}{1 + cosA}$, so that $\tan (A/2) = \sqrt{\frac{a^2 - (b-c)^2}{(b+c)^2 - a^2}}$, and

$\frac{\tan (A/2)}{\tan (B/2)} = \sqrt{ \frac{a^2 - (b-c)^2}{(b+c)^2 - a^2}} \sqrt{\frac{(a+c)^2 - b^2}{b^2 - (c-a)^2} } = \frac{a-b+c}{b+c-a}$

whereby it remains to prove $\sum \frac{a}{b} \leq \sum \frac{a-b+c}{b+c-a}$

-------

Now, set $a = y+z$ etc., (a standard trick), so that we have $x,y,z$ as positive reals and the inequality to be proved becomes:

$\sum \frac{y+z}{x+z} \leq \sum \frac{y}{x}$

which appears here:

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Belarus&t=47343

(Thanks to billzhao)

[quote="Singular"]

$\sum \frac{y+z}{x+z} \leq \sum \frac{y}{x}$
[/quote]

Oh...Me too  :D  .So i want another solution.This solution is long  :(

[quote="TinTin"]Let a triangle $ABC$
Prove that
[tex]\sum \frac{sinA}{sinB}  \leq   \sum \frac{tg{\frac{A}{2}}}{tg{\frac{B}{2}}}[/tex][/quote]
In$\triangle ABC$,
\[ \frac{sinA}{sinB}+ \frac{sinB}{sinC}+ \frac{sinC}{sinA} \leq \frac{tan{\frac{A}{2}}}{tan{\frac{B}{2}}}+ \frac{tan{\frac{B}{2}}}{tan{\frac{C}{2}}}+ \frac{tan{\frac{C}{2}}}{tan{\frac{A}{2}}}.\]


Together  Joe and Sally can mow the lawn in 40 minutes. If it takes Joe half the time to mow the lawn than it does for Sally. How long would it take for Sally to mow the lawn by herself?

2) Jonathen runs one way 14 mi /h and 10 min later his math runs to him at 16mi/h. How long will it take until they both meet?

1.
[hide]
$40(j+s)=x$
j is rate of Joe
s rate of Sally

$2(j)=s$
$40j+40y=x$
$40j=20s$

$60s=x$

$1hr$
[/hide]

Good job. These problems no longer seem like homework problems. They still may be, but this is good enough.  :D

Keep up the good work!

i'm sry but i didnt get this concept is there another way to go about it

I'm in a bit of a hurry and am not sure about the answer to the second quetion,is it 7/6 hours,sorry i dont know how to hide,i just used relative vel and then did
$2=(7/3)/[(1/6)+t]$
so t comes 1,so total time is 1/6+1=7/6 :huh:

2.

[hide]
xmin(14 m/hr)+10min(14m/hr)=xmin(16m/h)

xmin(14 m/60min)+10min(14m/60min)=xmin(16m/60min)

7m/3min=xmin(2m/60min)

(7m/3min)*(60min/2m)=105 minutes
[/hide]

hmm I get different.... thing....
[hide="Sol"]
$d$ covered in $10$ min=$1.4$mi
Relative velocity,
$2$mi/hour,
So, time=d/s
$1.4/2$
$=0.7$
$=\frac{7}{10}*60$
$=42$mins
[/hide]

1
[hide]$T(\frac{1}{J}+\frac{1}{S})=1$

Where T is time and 1 is basically 1 job.  From the most basic of formulas... Distance=rate*time.  However, in this case, we know that T=40 minutes and that $2J=S$.
So heres how it works out.

$40(\frac{1}{J}+\frac{1}{2J})=1$
Adding the fraction we get
$40(\frac{3J}{2J^2})=1$
Cancel out the Js and divide 40 from both sides.
$(\frac{3}{2J}=\frac{1}{40})$
Cross multiply and get
$2J=120$
$J=60$    Remember we know that $2J=S$
$S=120$
2 Hours.[/hide]

2
[hide]Well we know that Johnathen runs at the rate of 14mi/h and that his math goes at 16 mi/h.

At what point would the math catch up each?  When they both are at the same point or in other words, equal.
Lets set up an equation to get how long each of them have traveled.
T is the time elapsed. Subtracting 10 for the ten minutes elapsed before the math left.
$14(T)=16(T-10)$
Solving we get
$T=80$
Check our answer by plugging it back and verify it.[/hide]

Pakman,ur method is perfect but then why by my method its coming 70 min?

[quote="eminem"]hey i got 7/6 too same as shady. :huh:

about pakman's method , isnt it that u have to convert it into proper units , like u took speed in miles/hr while time in minutes , whereas there should be only  one unit  of time either mins or hrs?[/quote]
 :oops:  I didnt convert $10$mins to hours.
I get the same answer,
$s=d/t$
$2=\frac{\frac{14}{6}}{t}$
$t=\frac{7}{6}$

[quote="g3pilotplaya"]Together  Joe and Sally can mow the lawn in 40 minutes. If it takes Joe half the time to mow the lawn than it does for Sally. How long would it take for Sally to mow the lawn by herself?
[hide]
sally's speed=x

1/(x/2)+1/x=1/40
2/x+1/x=1/40
3/x=1/40
x=120 minutes
[/hide]

[/quote]


Prove that in a triangle $\frac{am_{a}+bm_{b}+cm_{c}}{(a+b+c)(m_{a}+m_{b}+m_{c})}\le \frac{1}{3}$. The symbols have their usual meaning.

Maybe it is solved by Chebyshev inequality when we assume $a\ge b\ge c$
$\rightarrow m_{a}\le m_{b}\le m_{c}$


integration
function
calculus
real analysis
real analysis unsolved

Find f:R->R , f \in C1 such that  \int (0 to y) f^3(x)dx=( \int (0 to y)f(x)dx)^3 for every real y.

f = 0 the function is a solution, from now we are looking
function f not identically 0

There exist x_0 such that f(x_0)=/=0 and the continuity of f we have f(y) =/=0 for any y neighbourhood W of x_0.
Derivation gives f^3(y)=3f(y)(\int_{0 to y }f(x)dx)^2,
For any y in W,  f^2(y) = 3(\int_{0 to y}f(x)dx)^2
2f(y)f'(y) = 6f(y)\int_{0 to y}f(x)dx
f'(y) = 3\int_{0 to y}f(x)dx, derivation again
f''(y) -3f(y)=0 linear differential equation of order two
characteristic equation r-3=0,
(*) f(y)= A.exp( \sqrt 3 y)+B.exp(- \sqrt 3y)
with A,B constants real
if we plug (*) in the integral equation we get conditions on A, B
o \cup \beta  \int  \cap  \cap o  \int

Ok! So f on  W is discovered. But how do I extend to R?

Only one solution : f=0 (the initial conditions are zero)


probability
geometry
geometric transformation
rotation
probability and stats

If $ n$ couples sit in a round table randomly, what is the probability that no wife sits next to her husband?

[hide]If you have n couples the total number of people is 2n, and total permutations is 2n!.  Number the seats 1,2,3,4...(2n), sit the first couple side by side at seats 1 and 2, they could sit in this spot in 2 ways, multiply this by the number of permutations that the other people could take up the remaining chairs, then multiply this by the number of couples, each one of these would be a different permutation, then divide by the total number of permutations, this should make sure that all of the possible permutations with at least one couple sitting next to each other are counted.

2*((2n-2)!)*(n) / 2n![/hide]

Alas, I neither understand the solution, nor believe the answer :(

[quote]If you have n couples the total number of people is 2n, and total permutations is 2n!. Number the seats 1,2,3,4...(2n), sit the first couple side by side at seats 1 and 2, they could sit in this spot in 2 ways, multiply this by the number of permutations that the other people could take up the remaining chairs, then multiply this by the number of couples, each one of these would be a different permutation, then divide by the total number of permutations, this should make sure that all of the possible permutations with at least one couple sitting next to each other are counted.

2*((2n-2)!)*(n) / 2n![/quote]

I was tired when I posted the stuff above and I wasn't paying attention, the equation above should equal the number of permutations that there are at least two couples sitting next to each other, WHEN those couples are sitting in the first 2 seats, divided by the total number of permutations, of course this makes no sense as I realized now that I re-read it.  To fix the equation I subtracted the total number of permutations for a single rotation (2n!/2n) (by rotation I mean that the table would look the same only "rotated", which has to be done since 2*(2n-2)!*n only counts for 1 rotation).

The top half of the fraction should give the number of permutations where no couples sit next to each other for 1 rotation, then divide by the total number of permutations for 1 rotation, and that should be the answer.

((2n!/2n)-(2*(2n-2)!*n)) / (2n!/2n)

For 2 couples the formula ends up with -2/6 so I multiplied -1 by the answer to get the following formula:

(-1*((2n!/2n)-(2*(2n-2)!*n))) / (2n!/2n)

To make it simpler p = number of people: p=2n, then you get the following formula:

(-1*(p!/p)-(p*((p-2)!))) / (p!/p)

p!/p = number of permutations / number of rotations.

I don't know why it ends up negative it doesn't seem to make sense, for some reason (p*(p-2)!) is counting the total number of permutations for a single rotation, plus the number of times when no couples are sitting together, when it should equal the number of permutations for 1 rotation with 1 or more couples sitting next to each other.  But it seems to get the right answers for the ones I can check.


AMC
Polynomials
USAMO
inequalities
lagrange s interpolation

Let $ P$ be the given polynomial, and we're about to prove that there exist $ Q(x),R(x)\in\mathbb{R}[x]$, both having $ n$ real roots, and
\[ R(x) \equal{} 2\cdot P(x) \minus{} Q(x)\]
Let $ x_1< x_2 < \dots < x_n$ be arbitrary real numbers, and suppose that $ a_i \equal{} P(x_i)$, and let's take some real numbers $ b_i$, such that for all odd $ i$:
\[ 2\cdot a_i < b_i, b_i > 0\]
and for all even $ i$ we have:
\[ 2\cdot a_i > b_i, b_i < 0\]
According to the Interpolational Lagrange Formula, there does exist the polynomial, such that $ Q(x_i) \equal{} b_i$. As we know,

$ Q(x_1), Q(x_3), \dots > 0$ and $ Q(x_2),Q(x_4),\dots < 0$, so $ Q$ has at least $ n \minus{} 1$ real roots, consequently it has precisely $ n$ real roots. On the other hand, $ R(x_1) \equal{} 2a_1 \minus{} b_1 < 0$, $ R(x_2) \equal{} 2a_2 \minus{} b_2 > 0$ and so on. Therefore, $ R(x)$ has exactly $ n$ real roots either.

So the only problem left is to both of the polynomials to have leading coefficients equal $ 1$. Note that when using Interpolational Formula, we used only $ n$ points, $ x_1,x_2,\dots,x_n$ so we may choose another point $ x$, such that $ Q$ is monic. Hence, $ R$ is monic either, seeing as $ R(x) \equal{} 2P(x) \minus{} Q(x)$.

P.S I managed to come up with this solution after thinkin' for 30 minutes, so it may contain mistakes.

Let the polynomial be $P(x)$. Let $M$ be any number greater than the maximum value of $|P(x)|$ as $-1 \leq x \leq 1$. Then, let $Q(x) = P(x) + M T_{n-1}(x)$, where $T_{n-1}(x)$ is the $n-1$st Chebyshev polynomial. Then define $x_k = \cos\left(\frac{\pi (n-1-k)}{n-1}\right)$. We get \[Q(x_k) = P(x_k) + M T_{n-1}(x_k) = P(x_k) + M (-1)^{n-1 - k},\] which has the same sign as $(-1)^{n - 1 - k}$.

Additionally, because $Q$ is monic of degree $n$, there is some (large negative) $x < x_0$ such that $Q(x)$ has the same sign as $(-1)^n$, and some (large positive) $X > x_{n-1}$ such that $Q(X)$ has the same sign as $1$ (i.e. is positive).

Since $x < x_0 < x_1 < \cdots < x_{n-1} < X$, by the Intermediate Value Theorem, the number of real roots $N$ of $Q$ is at least the number of pairs of consecutive members in the sequence \[Q(x), Q(x_0), Q(x_1), \cdots, Q(x_{n-1}), Q(X)\] with opposite signs. By our previous results, the members of this sequence have the same signs as the members of the sequence \[(-1)^n, (-1)^{n-1}, (-1)^{n-2},  \cdots, -1, 1, 1.\] Since this sequence has $n$ pairs of consecutive members with opposite sign, $Q$ has at least $n$ real roots. Since it has degree $n$, it has exactly $n$ real roots. By the same argument, $R(x) = P(x) - M T_{n-1}(x)$ has exactly $n$ real roots.

Therefore, $P(x)$ is the average of two polynomials ($Q(x)$ and $R(x)$) of degree $n$ with $n$ real roots.

Let $f(n)$ be the monic polynomial and let $M > 1000\max_{t=1, \dots, n} |f(t)|+1000$.
Then we may select reals $a_1, \dots, a_n$ and $b_1, \dots, b_n$ such that for each $k = 1, \dots, n$, we have
\begin{align*}
	a_k + b_k &= 2f(k) \\
	(-1)^k a_k & > M \\
	(-1)^{k+1} b_k & > M.
\end{align*}
We may interpolate monic polynomials $g$ and $h$ through the $a_k$ and $b_k$ (if the $a_k$, $b_k$ are selected "generically" from each other).
Then one can easily check $f = \tfrac12(g+h)$ works.

Does this work for a solution?

We use induction on $n$. $n=1$ is trivial.
Now suppose it is true for $n-1$. If $n$ is odd or $P(x)$ has a real root, then $P(x)$ has a factor $x-r$ ($r$ real), in which we are done by factoring out $x-r$ and applying the induction hypothesis on $n-1$. Now assume that $n=2m$ is even and $P(x)>0$ for all real $x$.
Let $Q(x)=(x-q_1)(x-q_2)...(x-q_{2m})$ where the $q_i$'s are strictly increasing and positive reals and $q_{2m}$ is arbitrarily large and $R(x)=2P(x)-Q(x)$. Note that $R(q_{2i-1})=2P(q_{2i-1})>0$, since $P$ does not have any real factors.
Also, since $q_{2m}$ is arbitrarily large, $Q((q_{2i-2}+q_{2i-1})/2)>2P(q_{2i-2}+q_{2i-1})/2)$ i.e. $R(q_{2i-2}+q_{2i-1})/2)<0$. Also, $Q(0) < 0$. Thus, there are two roots of $R$ in each of the intervals $(-\infty, q_1), (q_2, q_3), ..., (q_{2m-2}, q_{2m-1})$, so we are done.
I don't know if this is right since $q_1, q_2, ..., q_{2m-1}$ are virtually random. Could someone tell me if this is right? Thank you!

[quote=v_Enhance]Let $f(n)$ be the monic polynomial and let $M > 1000\max_{t=1, \dots, n} |f(t)|+1000$.
[/quote]

Why was the $M > 1000 \ \text{"something"} + 1000$ necessary in the first place? Couldn't we just state that $M$ is arbitrarily large?

[quote=MithsApprentice]Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree $n$ with real coefficients is the average of two monic polynomials of degree $n$ with $n$ real roots.[/quote]

Let's do some thorough analysis and motivation on this problem.

First, the statement of this problem is nice :)
Pretty straightforward if you know two general things about polynomial: the main problem is to $\textbf{construct}$ polynomials.
Now, if we were to find 2 polynomials of degree $n$ with $n$ real roots, the first thing that comes to mind is indeed, [b]Intermediate Value Theorem[/b]. Why do we think of this?
Basically you wanted to construct a polynomial with $n$ real roots, but to do this you just need $n+1$ distinct points such that $f(a_i) < 0$ if $i \le n + 1$ is odd and $f(a_i) > 0$ if $i \le n + 1$ is even. Then, by Intermediate Value Theorem, there would be a root in the interval $(a_i , a_{i + 1})$ for every $1 \le i \le n$, satisfying the hypothesis that we wanted.
----------------
Keeping that in mind, suppose that we have pick $a_i$ such numbers that $g$ passes through, by $\textbf{Lagrange Interpolation Theorem}$, we are pretty much done, since there must exists polynomial of degree at most $n$ that pass these $n + 1$ points.
So, we are pretty much done since we can construct two polynomials $g$ and $h$ that has $n$ roots, and we can indeed choose $2f = g + h$. Formalizing this, choose $g_1, g_2, \dots, g_{n + 1}$ being roots of $g$, and $h_1, h_2, \dots, h_{n + 1}$ being roots of $h$, such that
\[ g_i + h_i = 2f(x_i) \]
for some $x_i$ and $|g_i|, |h_i| > M$ for a large value of $M > 100 \cdot | \max_{x_i, 1 \le i \le n } f(x_i) |^{100} + 200 $ and $g_i$ switch signs depend on parity ($h_i$ as well).
-------
Now, there's just one problem left: the polynomial $g$ is not necessarily monic. But this is easy to fixed. Suppose we have pick $n$ points: $g_1, g_2, \dots, g_n$. This could determine a polynomial of degree $n$ and we could somehow control $g_{n + 1}$ such that this is monic.

Now, if $g$ is monic, we could simply check the leading coefficient of $h = 2f - g$, is monic as well. So we are done.

Cute problem

We will prove this using inducktion on $n$. Our base case is $n = 1$, where the polynomial $P(x) = x-k$, we can just take $x-k, x-k$ as our two monic polynomials.

Now, onto the inducktive step. Assume it is true for $n-1$, we will prove it for polynomials of degree $n$. Let the polynomial be $P(x)$. If $P$ had a real root $r$, then let $P = (x-r)Q$. By our inducktive hypothesis, there exists 2 polynomials $f, g$ with all real roots such that $f+g = 2Q$. Then take $f(x-r), g(x-r)$, which will sum to $2P$, and are both monic and have all real roots, which proves the problem. Thus, assume $P$ has no real roots (which also means $n$ is even), and since $P$ is monic this means $P(x) > 0$ for all real $x$. Now, consider points $(x_{1}, y_{1}), (x_{2}, y_{2}), \ldots (x_{n}, y_{n})$ that satisfy the following conditions:
\[y_{i} > P(x_{i}), x_{i} < x_{i+1}\]
Let $R$ be the degree $n-1$ polynomial that passes through $(x_{i}, -(-1)^{i}y_{i})$ for each $i$. This means
\[R(x_{2k+1}) > P(x_{2k+1}), R(x_{2k+2}) < P(x_{2k+2})\]
and by the intermediate value theorem, this means $P$ and $R$ must intersect at least $n-1$ times. Furthermore, since $R$ is an $n-1$ degree polynomial, which is odd degree, then there exists some $t < x_{1}$ such that $R(t) < P(t)$, but since $R(x_{1}) > P(x_{1})$, they must intersect before $x_{1}$ as well, so $P$ and $R$ intersect $n$ times, which means $P - R$ has $n$ real roots. We can similarly deduce that $P$ and $-R$ intersect $n$ times, so $P+R$ has $n$ real roots. Then, taking $P-R, P+R$ as our two polynomials, they both have $n$ real roots and average to $P$, which completes our inducktive step.

Thus, by inducktion, there exists such polynomials.

[hide=Solution]haha Intermediate Value Theorem go brrr
See attachment[/hide]

[hide=Remarks]At the beginning, this problem seems hard to approach, but the solution's main idea is, at its core, simple. Graphing helped me get a lot of intuition.[/hide]

[quote=MithsApprentice]Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree $n$ with real coefficients is the average of two monic polynomials of degree $n$ with $n$ real roots.[/quote]

In other words,given any monic polynomial $f$ we need to show that $$f \equiv \frac{1}{2}\left[g+h \right]$$
i.e $$2f=g+h$$
By Lagrange interpolation choose points $\{a_1,b_1\},........,\{a_n,b_n\}$ such that \begin{align*}
a_k + b_k &= 2f(k) \\
(-1)^k a_k & > \text{some sufficiently large number} \\
(-1)^{k+1} b_k & >  \text{some sufficiently large number}
\end{align*}
The first condition ensures $2f=g+h$ and the second condition by IVT ensures both $g$ and $h$ have an real root.$\blacksquare$

bruh what is this problem :skull:
too lazy to write-up details lel, just outline
------
Let $P(x)$ be the polynomial. Pick $-1 < x_1 < x_2 \cdots < x_n < 1$; and let $a_1 \cdots a_n$, $b_1 \cdots b_n$ be reals such that $a_i < -1, b_i > 1$ and $a_i + b_i = 2(P(x_i)-x^n)$ for all $1 \le i \le n$.

Use Lagrange Interpolation to let $Q_1(x)$ be the unique polynomial with degree at most n-1 that passes through points $(x_1, a_1), (x_2, b_2), (x_3, a_3), (x_4, b_4), \cdots$ in [i]alternating[/i] order. Similarly define $Q_2(x)$ to be the polynomial that passes through points $(x_1, b_1), (x_2, a_2), (x_3, b_3), (x_4, a_4), \cdots$. Then, we let $P_1(x) = Q_1(x) + x^n$ and $P_2(x) = Q_2(x) + x^n$.

It is easy to see that $Q_1$ and $Q_2$ each have $n-1$ roots by Intermediate Value Theorem, and since $-1 < x^n < 1$ for $-1 < x < 1$, the roots are preserved in $P_1$ and $P_2$. Casework on the parity of $n$ yields that in either case, the end-behavior of $P_1$ and $P_2$ yields another root, giving $n$ roots.

First note that if we have monic $p(x),q(x),r(x)$ such that $2p(x)=q(x)+r(x)$—i.e. the statement holds for $p(x)$—then the statement holds for $p(x)(x-r)$ for any real $r$ as well, as $2p(x)(x-r)=q(x)(x-r)+r(x)(x-r)$, and the two polynomials on the RHS are still monic and have all real roots. Hence it suffices to consider the case where our monic polynomial $P(x)$ (that we want to express as the average) has no real roots. In particular, its degree must be even, so it's always positive (by looking at end behavior).
Define $y_i=P(i)$ for $1 \leq i \leq n$, so $y_i>0$ for all $i$. By Lagrange Interpolation, there exist (not necessarily monic) polynomials $q(x),r(x)$ of degree at most $n-1$ such that
$$q(1)=3y_1,q(2)=-y_2,\ldots,q(n-1)=3y_{n-1},q(n)=-y_n,$$
$$r(1)=-y_1,r(2)=3y_2,\ldots,r(n-1)=-y_{n-1},r(n)=3y_n.$$
Then let $Q(x)=q(x)+(x-1)\ldots(x-n)$ and $R(x)=r(x)+(x-1)\ldots(x-n)$, so $Q,R$ are monic polynomials of degree $n$ which agree with $q,r$ respectively over $\{1,\ldots,n\}$. Then,
$$Q(1)>0,Q(2)<0,\ldots,Q(n-1)>0,Q(n)<0,\lim_{x \to \infty} Q(x)=\infty,$$
hence $Q$ has a root in each of the $n$ intervals $(1,2),\ldots,(n-1,n),(n,\infty)$ by IVT. LIkewise,
$$\lim_{x \to -\infty} R(x)=\infty,R(1)<0,R(2)>0,\ldots,R(n-1)<0,R(n)>0,$$
hence $R$ has a root in each of the $n$ intervals $(1,2),\ldots,(n-1,n),(-\infty,1)$.
Further, since $P,Q,R$ are all monic, $2P(x)-Q(x)-R(x)$ is a polynomial of degree $n-1$ which is evidently zero at $x=1,\ldots,n$, hence it is identically zero and we have $2P(x)=Q(x)+R(x)$ for all $x$ as desired. $\blacksquare$

[b]Remark:[/b] graphing op

I have an email from September 25, 2014 (wow, long time ago) where I describe the solution.

We induct on the degree, it's easy for $n=0,1$.  If $P$ has a real root, WLOG $P(0)=0$.  Then write $P(x)=xQ(x)$, and let $Q(x)=\frac{R(x)+S(x)}{2}$ be such an average.  So $P(x)=\frac{xR(x)+xS(x)}{2}$ is the average of two polynomials with $n-1$ real roots, since $R,S$ have $n-1$ real roots.  Replace $(R,S)$ by $(R+\varepsilon,S-\varepsilon)$ if need be so that $R,S$ do not have a root at $0$.  Then $xR(x), xS(x)$ have $n$ real roots.

If $P$ does not have a real root, it is of even degree $n=2d \ge 2$, and we can take $Q(x)=(x-2)...(x-2d)(x-r)$ for some sufficiently large huge $r$.  Now, $P$ is the average of $Q$ and $R=2P-Q$.  Clearly $Q$ has $n$ real roots.  We claim that $2P-Q$ has two real roots on each of $(-\infty,2), (3,4), (5,6), \cdots, (2d-1,2d)$.  Note that $Q$ is positive on these intervals $(3,4), \cdots, (2d-1,2d)$ but $0$ at their endpoints.  As $r \to \infty$, it blows up on each of these intervals, and so intersects the graph of $2P$ at two points, when $r$ is big enough.  We have $R(2)=2P(2)-Q(2)=2P(2)>0$, and as $R$ is monic of even degree, $\lim_{x \to -\infty}R(x)=\infty$.  However, if $r$ is sufficiently large, $R(1)=2P(1)-Q(1)<0$, since $\lim_{r \to \infty}Q(1)=\lim_{r \to \infty}r(2d-1)!=\infty$.

Let $P(x)$ be the given monic polynomial, and select some random $n$ pairwise distinct $x$-values $x_1<x_2<\dots < x_n$. At each value $x_i$, select two other values $a_i,b_i$ for which $a_i+b_i = 2P(x_i).$ If $i$ is even, make $a_i <0,b_i>0$ and if $i$ is odd, make $a_i>0,b_i<0$. Clearly, if we can find monic polynomials for which $Q(x_i)=a_i$, $R(x_i)=b_i$ then we are done by the Intermediate Value Theorem. Fortunately, we can let \[Q(x)=c_0+c_1x+c_2x^2+\dots + c_{n-1}x^{n-1}+x^n\] and plug in values $x_1, \dots, x_n$ into that, and we get a system of $n$ equations and $n$ variables ($c_0,\dots c_{n-1}.$) Since the coefficients behind the variables have different ratios in each equation, we have a unique solution. We are done.

Feels a bit superficial, easy to do directly.

Pick some very large $N_1, N_2, \dots, N_{n+1} \gg 0$ and real numbers $x_1, x_2, \dots, x_{n+1}$. Let $P(x_i) = y_i$ for each $i$. Set two polynomials $Q, R$ such that $Q(x_i) = N_i$ and $R(x_i) = 2y_i - N_i$ if $i$ is odd, and vice versa if $i$ is even by Lagrange interpolation.

Now, observe that by intermediate value theorem, $Q$ and $R$ both have at least one root in $[x_i, x_{i+1}]$, hence they both have at least $i$ real roots. Furthermore, $P(x_i) = \frac{Q(x_i) + R(x_i)}2$ for $n+1$ values of $i$; thus $P = \frac 12(Q+R)$ exactly, and we're done.

Really silly. The idea is to basically make the initial polynomial irrelevant.

I interpreted $n$ real roots as $n$ distinct real roots, but idk.

Let $P$ be the given polynomial with degree $n$. Let $Q(x) = \prod_{i = 1}^{n - 1} (x-i)$. Let $M$ be a gigantic real number which we will establish later.

Consider the two polynomials
\[ P(x) + MQ(x), P(x) - MQ(x). \]
Note that these two polynomials are monic.

Let's consider $Q$. Fix an arbitrarily small $\varepsilon > 0$. For each $i \in \{ 1, 2, \dots, n-1\}$, choose some $\frac{1}{1000} > \delta_i > 0$ such that $\forall x \in (i - \delta_i, i + \delta_i)$, we have $Q(x) \in (-\varepsilon, \varepsilon )$. This follows from the definition of continuity, and the fact that a subset of a set is in fact a subset of a set [that's craaazy]. Now take the set $U = \bigcup_{i = 1}^{n - 1} (i - \delta_i, i + \delta_i)$, and define
\[ T := \{ t \in \mathbb R | \exists x \in U \text{ s.t. } |P(x)| = t \}. \]
Observe that $T$ is nonempty as $|P(1)|$ is in this set. Thus it makes sense to take $T_M = \sup T$, and choosing $M = \frac{2T_M}{\varepsilon}$, we can guarantee that for $x \in (i - \delta_i, i + \delta_i)$, $ Q(x) - \varepsilon/2 \le \frac{P(x)}{M} + Q(x) \le Q(x) + \varepsilon/2$.It follows then that $P(x) + MQ(x)$ has a root on each of these intervals, and hence has at least $n-1$ distinct roots. Likewise $P(x) - MQ(x)$ has a root on each of these intervals too.

Now we need to show that there's no double roots here. This is really tedious (we just basically remove the contribution of $P'(x)$, by choosing $M$ such that the maximal absolute size of $P'(x)/M$ on any of the delta intervals is bounded by some $c$ such that $0 \not \in [Q'(x) - c, Q'(x) + c]$, also tightening $\varepsilon$ even more such that the derivative is never zero on the interval.) so I'll leave it as an exercise to the reader to work out all the details.

The remaining root can't be complex as $P$ is a real polynomial. Thus the remaining root is real. We're done.

I have a really goofy solution

First, note that every polynomial with real coefficients can be factored into linear and quadratic polynomials with real coefficients, and if the original was monic, all of these linear and quadratic polynomials are monic as well.  Thus, it is sufficient to prove that all linear and quadratic polynomials work.  Linears are trivial as all have 1 real root.  For quadratics, let $x^2+ax+b$ be our quadratic, and say it is the average of $(x+i)(x+j)$ and $(x+k)(x+l)$ for some reals i, j, k, l.  It is sufficient to prove that i, j, k, l exist for all a, b.  Thus, $2a=i+j+k+l$, $2b=ij+kl$.  Say that $i+j=a+n$, $k+l=a-n$ for some arbitrary $n$.  By AM-GM, $ij\leq \frac{(a+n)^2}{4}$, $kl\leq \frac{(a-n)^2}{4}$, thus $ij+kl\leq \frac{a^2+n^2}{2}$, or $4b\leq a^2+n^2$.  If $a^2>4b$, we are simply done.  Otherwise, set $n^2=4b-a^2$, with $i=j=\frac{a+n}{2}$, $k, l=\frac{a-n}{2}$.  We can confirm that this works.  We are thus done.

someone tell me if I am having a skill issue or not

AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

So this is like my first realpoly problem thanks ApraTrip for giving me hint "how to find poly with all real roots"

I guess [b]Intermediate Value Theorem[/b] is like extremely overpowered. Also [b]thinking graphically[/b] actually works I guess.

So basically we just choose $n$ values where it keeps having one polynomial be really big then really small next so it keeps crossing the $x$-axis. Then the last degree of freedom for Lagrange Interpolation is used to make the polynomials monic, gg.


real analysis
limit
integration
number theory
relatively prime
real analysis unsolved

Suppose f(x)f(y)=f(x+y) for all real x and y,
(a) Assuming that f is differentiable and not zero, prove that

$ f(x) \equal{} e^{cx}$

where c is a constant.

(b) Prove the same thing, assuming only that f is continuous.

[b]Attempt at a solution[/b]
(a) The given information implies that $ f(x)^2 \equal{} f(2 x)$ and furthermore that $ f(x)^n \equal{} f(n x)$ for any natural number n. We can differentiate that to obtain  $ f(x)^{n \minus{} 1} f'(x) \equal{} f'(n x)$ for any natural number n.

It is also not hard to show that $ f'(x) f(y) \equal{} f'(y) f(x)$ for all real x and y.

We can take $ y \equal{} 0$ to find that $ f(0) \equal{} 1$ (I have used the fact that f is nonzero here)
(b) Let's do (a) first.

Just use $ f'(x)\equal{}\lim_{h\rightarrow 0} \frac{f(x\plus{}h)\minus{}f(x)}{h}$.

(b). Let's look at what you already have: $ f^n(x)\equal{}f(nx)$ for natural numbers $ n$. Then let's take some rational number $ r\equal{}\frac{p}{q}$, where $ p$ and $ q$ are relatively prime integers, so that $ r\cdot q \equal{} p\cdot 1$. Then \[ f^p(x)\equal{}f(px)\equal{}f(q\cdot rx)\equal{}f^q(rx).\] Raising both sides to the $ q$-th root, \[ f^{p/q}(x)\equal{}f(rx).\] Note that the left hand side is actually $ f^r(x)$. Hence, your identity works for all rational numbers.

Since $ f$ is continuous, we can show that $ f^r(x)\equal{}f(rx)$ actually holds for real $ r$ as well by the sequential characterization of limits: for a sequence $ \{y_n\}$ that converges to $ r$, \[ \lim_{n\rightarrow\infty}f(y_nx)\equal{}f(rx).\] Since the rational numbers are dense in the real numbers, we can pick a sequence of rational numbers $ \{y_n\}$ that converges to $ r$. Then \[ \ln(f(rx))\equal{}\lim_{n\rightarrow\infty}\ln(f(y_nx))\equal{}\lim_{n\rightarrow\infty}\ln(f^{y_n}(x))\equal{}\lim_{n\rightarrow\infty}y_n\ln(f(x))\equal{}r\ln(f(x)).\] Thus, $ f(rx)\equal{}f^r(x)$ for all real numbers $ r$ and $ f(r)\equal{}f^r(1)\equal{}e^{cr}$, where $ e^c\equal{}f(1)$.

[quote="kunny"]Just use $ f'(x) \equal{} \lim_{h\rightarrow 0} \frac {f(x \plus{} h) \minus{} f(x)}{h}$.[/quote]

Here is the complete proof, using only things from Rudin.

$ f'(x) \equal{} \lim_{h\rightarrow 0} \frac {f(x \plus{} h) \minus{} f(x)}{h} \equal{} \lim_{h\rightarrow 0} \frac {f(x)f(h) \minus{} f(x)}{h} \equal{} f(x) f'(0)$

Let c = f'(0).

We know that f is always positive because differentiability implies continuity and f(1)=0 and f is nonzero. Thus, Rudin equation 8.39 tells us that

$ \log(f(x)) \equal{} \int_1^{f(x)} \frac {dx}{x} \equal{} \int_0^{x}\frac {f'(w)dw}{f(w)} \equal{} \int_0^{x} c dw \equal{} xc$

Exponentiating both sides produces the desired result.

There is one small problem with this: I need to show that f is strictly increasing to use the change of variables theorem. Anyone know how to do that?

EDIT: first of all it is not true since f' could be identically 0; otherwise f is strictly increasing or strictly decreasing by f'(x) = f'(0) f(x). This completes the proof.

Can someone just give me A HINT on part b. I don't want to read the post two above since it looks like a complete solution.


pigeonhole principle
induction
combinatorics unsolved
combinatorics

Let $n \in N_0$. Anne write on a line $n$ distinct naturals (differents from $0$). After that, Marc removes some of this numbers (not all and perhaps neither) and put signs $+$ or $-$ in front of the remaining numbers and afterwards does the obtained calculation. If the result is divisible by $2003$, Marc wins. If not, Anne wins. Who has a winning strategy?

Hm... That's not so hard. For $n \leq 10$ Ann wins: she can choose the numbers $1,2,..., 2^{n-1}$. Every result Marc can get is between $-1023$ and $+1023$ and is not equal to 0, since it is of the same sign as the largest of the numbers ($2^j > 2^j - 1 = \sum_{i = 0}^{j-1}2^i$). So Ann wins. For $n \geq 11$, the set $C$ of Annes numbers has $2^n - 1 \geq 2047 > 2003$ non-empty subsets. Therefore, by the Pigeonhole Principle, there exist two of them, say $A$ and $B$, such that the sums of the numbers in $A$ and $B$ are congruent modulo 2003. If Marc puts $+$ in front of the numbers of the set $A \backslash B$, a $-$ in front of the numbers $B \backslash A$ and deletes all remaining numbers of $C$, he wins.

If n>10 there are 2048 subsets of the writen numbers, so there must exist two subsets A and B for that
$\sum_{i\in A} i =\sum_{i\in B}i \mod 2003$. Marc leaves there only numbers that are in exactly one of the sets A an B and he writes + to the numbers from A and - to the numbers of B. The result is divisible by 2003, because
$2003|\sum_{i\in A} i-\sum_{i\in B} i$.  For n=0 Marc wins. For $n\in \{1,2,...,9\}$ Anne wins by writing one of these sequnces:
n=1: 1
n=2: 1, 2
n=3: 1, 2, 4
n=4: 1, 2, 4, 8
By sumation of these numbers Marc obviously can't get 0 (if u don't see it, use induction) and if n<10, sum of all the numbers is lower then 1024 and he can't get multiples of 2003 greater than 0 as well.
I'm not sure about the case n=10. Do u have a tip?  ;)

Yes, I do have a hint: look at my post :D


Suppose that a=(a1,a2,a3) a is a vector

The unit vector is [img]http://img522.imageshack.us/img522/3198/maths.gif[/img]
Is it true??
How can I prove it??[/url]

hello, you must write $ \frac{1}{\sqrt{a_1^2\plus{}a_2^2\plus{}a_3^2}}\left(\begin{array}{c} a_1\\a_2\\a_3\end{array}\right)$.
Sonnhard.

[i]a is equal to the unit in the direction of a scaled to the length of a[/i]
$ \mathbf{a} \equal{} \| \mathbf{a} \| \hat{a}$

[i]Divide by the length of a[/i]
$ \hat{a} \equal{} \frac {1}{\| \mathbf{a} \|} \mathbf{a}$

[i]If length is defined to be the Euclidean norm, then your result follows[/i]


Recently, I've pretty much been doing math every spare second it is physically possible. I myself don't feel too tired, but I don't feel like I'm making much progress.  What you guys think the maximum amount of math someone should do a day before one starts becoming  ineffective?

I'm not sure, but I just do math for fun :D.  I don't really think about improving too much; I think it comes naturally if you're willing to think in new ways and practice.  Every once in a while you get an "eureka moment" and then you know you have improved :).

I don't think there should be any minimum or maximum for the amount of math one should do each day.  I think it varies depending on the person.

For example, at the Math Olympiad Summer Program, the students there do at least 8 to 10 hours of math each day, and it doesn't seem like many of them lose interest or effectiveness.  Also, in his later years, Paul Erdos put in 19 hours of math every day (on 3-4 hours sleep) and was one of the most productive mathematicians in history; he never seemed to get tired.  Of course, not everyone could stand that amount of math.

However, if you have a busy schedule, four hours does seem a good max for the average person.  I would say to do math as much as you feel like it, when you want to.  Overworking oneself can always lead to a dislike of the subject.

As far as making progress, math isn't like athletic sports, where you can feel yourself getting fitter and stronger within a few weeks' training and measure it with split times or weight or other things.  Ability grows very slowly and subtlely, and I notice that the best way to see if I'm making progress is to work consistently for a very long time (a few years perhaps) and then look back at how far I've come.  You'll probably laugh if you look at how you were a few years ago compared to now.   :lol:

I personally am currently doing math every spare second possible too, but that's because AIME and USAMO are coming up...   :o  Speaking of which, better get back to studying.  :)

[quote="ComplexZeta"]I would recommend not doing more than 25 hours of math in any one day.[/quote]
I must say this is very good advice.  I tried such a thing once and failed just one hour short of the goal. :D

From what I've experienced, it's very difficult to do too much math. It's very easy to do not enough math, but not too much... Just do enough so that you are satisfied and/or too tired to do more.

[quote="ComplexZeta"]I would recommend not doing more than 25 hours of math in any one day.[/quote]

Good advice.

By the way, there is a famous story about lawyers for a Wall Street law firm (working on the IBM antitrust case, if I remember right) who billed twenty-seven hours of legal work in one day, because they were working part of that day on an airplane flying from New York to California. Not all firms bill this way.

A great way to see if you've made progress is to take a past AMC/AIME/USAMO, then try one a few months later.  I was very surprised at what I didn't know the first time.

[quote="tokenadult"]Four hours a day.[/quote]

So, we should put our heads down and go to sleep for the last half hour of USAMO?  :P

[quote="MysticTerminator"]hm, I think you're being too harsh, Simon. If a person really wants to do that much, (s)he should be able to. I would recommend not going too much over that though.[/quote]

I'm not at all against more than 25 hours per day. But if you succeed, tell me how so I can try!

[quote="ComplexZeta"]You can do more than 25 hours of math in a day if you take a flight from Europe to the US.[/quote]

nice.. very nice.. you can do almost 48 hours if you cross the international date line right?

[quote="zabelman"][quote="tokenadult"]Four hours a day.[/quote]

So, we should put our heads down and go to sleep for the last half hour of USAMO?  :P [/quote]

The question of the original poster was directed at day-by-day practice. Athletes, musicians, and, yes, math competitors all have to deal with the issue of how much practice is enough to maintain to ensure top performance in competition. I earlier cited a seminal article on the subject of practice, "Expert Performance: Its Structure and Acquisition," in another AoPS thread. (The direct Web link to a .PDF of the article appears to have vanished.) That's the basis for my general recommendation.

The other thought I have is that there is some pretty good biographical information about the previous generation of mathematicians that suggests it is expedient to spend part of your youth reading books, playing sports, hanging out with friends, listening to music, etc. for balanced development and ultimate better long-term productivity as a research mathematician. For example, I rather suspect that Hardy stayed as productive as he did for as long as he did in what he considered "a young man's game" because he took time off to watch cricket matches. A case like that of Paul Erdos might not be a counterexample to a young man who wants to grow up and marry and have children and enjoy forms of recreation other than mathematics.

So how many hours do you put in on a daily basis? Of course I would like to hear the answer to that question from someone who has aced the AMC 12.


Let $ ABC$ be an acute triangle. A circle $ (O)$ pass through $ B$ and $ C$ meet side $ AB$,$ AC$ at $ M$, $ N$, respectively. Construct a circle which touch $ MB$, $ NC$ and in-touch $ (O)$ at $ P,Q,R$, respectively. Let $ T$ is the midpoint of arc $ BC$ which does not contain $ M,N$. Prove that $ PQ,BC,RT$ pass through a common point.

[img]http://i227.photobucket.com/albums/dd72/TPFC/NewImage.png[/img]

My picture???

to: tdl

what about R and T being the same side of BC?!

$ R$ and $ T$ should be same side of $ BC$. this problem is same with $ 8^{th}$ Quato of http://www.mathlinks.ro/viewtopic.php?t=170192. In $ 8^{th}$ Quato of http://www.mathlinks.ro/viewtopic.php?t=170192, $ RP$ bisect $ \angle APC$ is same as $ QP(Q\equal{} NM\cap CD)$ bisect $ \angle DPC$

This problem can be also solved by a solution in this post : http://www.mathlinks.ro/viewtopic.php?search_id=1338641321&t=6086
with some projective lemma.


vector
search
inequalities
Functional Analysis
advanced fields
advanced fields unsolved

i got stuck in the next question in operator theory:

given a self adjoint operator A prove that A+iI is invertible.

can anyone please help me?  :?
thanks

If $A+iI$ is not invertible, then $-i$ belongs to the spectrum of $A$. But the spectrum of a self-adjoint operator is confined to the real line.

by spectrum you mean that -i is an eigenvalue?
why is that true?[/code]

In infinite-dimensional vector spaces, you have to be a little bit careful about throwing around the word "eigenvalue." For instance, consider the "right-shift" operator $R$ on the sequence space $\ell^2.$ Although $R$ is one-to-one, it is not onto and hence is not invertible. Does that make $0$ an eigenvalue of $R$? The problem is that there are no eigenvectors. There is no $x\ne 0$ such that $Rx=0.$ However, it is certain that $0$ is in the spectrum of $R.$

An operator-theory textbook (such as the one you got the problem from?) should have the definition of the spectrum of an operator.

This question appears about hilbert spaces and before banach space and spectrum.
how can i proove it without spectrum?

By direct computation, making full use of the fact that $A$ is self-adjoint:

$||(A+iI)x||^2=||Ax||^2+||x||^2\ge||x||^2.$

Now search what you know among such theorems as the Open Mapping Theorem and the Closed Graph Theorem and their corollaries to see what that norm inequality gets you.

Actually, what you really need to go along with the norm inequality in the previous post is that you need to know that the range of $A+iI$ is dense in the Hilbert space.

Suppose $y$ is in the orthogonal complement of the range of $A+iI.$ Then, $\forall\,x,\,\langle(A+iI)x,y\rangle=0.$ This applies even if $x=y$, so

$\langle(A+iI)y,y\rangle=0$

$\langle Ay,y\rangle+i||y||^2=0$

But $\langle Ay,y\rangle$ is real, so this forces $y=0.$ Hence, the orthogonal complement of the range of $A+iI$ is trivial, which means that the range of $A-iI$ is dense.


geometry
3D geometry
topology
advanced fields
advanced fields solved

Let's consider the half cone,
I know that tip point causes problems. But by definition,if we just flatten the cone onto R^2,
isn't this a chart? And this single chart already covers
the whole half cone.

Or maybe it is the "whole cone" that is not topological manifold?

Can someone give a explanation? thanks.

[quote="mlok"]Half-cone is a topological manifold (with or without boundary). Double cone isn't.[/quote]

I quote from wiki:

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

"Every topological manifold in dimension 1, 2, or 3 has a unique differentiable structure (up to diffeomorphism); thus the concepts of topological and differentiable manifold are distinct only in higher dimensions."

So half cone is a differentiable manifold, even with that tip?

The half-cone is homeomorphic to the plane, and thus it has the same smooth structure if we ignore the $\mathbb{R}^{3}$ picture. It does not embed smoothly into $\mathbb{R}^{3}$; the vertex destroys any chance of smoothness there.

The reason why the double-cone is not a topological manifold is simple. If it was we could find a chart $\varphi:V\to U$ such that $V$ is an open ball say and $U$ is a neighborhood of the vertex $p$. But, $p$ is a cut point, and so ${\varphi:V-\{\varphi^{-1}(p)}\to U-\{p\}$ is a homeomorphism from a connected set to a disconnected set--contradiction.


So, prediction time again.  Gamecocks?  Panthers?  Tigers?  Tar Heels?

Gamecocks: 8-5, including a bowl victory.  With 19 returing (good) players next year, and only 3 starting seniors this year, how about an SEC championship next year?

Panthers: Have you seen their preseason games?  This team will make it to the super bowl again, but probably lose, because the AFC is clearly superior this year.

Tigers: Gamecocks beat them, but they go 8-5 due to a ridiculously easy schedule.

Tar Heels: 0-12, as usual

[quote="Iversonfan2007"]philadelphia eagles[/quote]

Meaning what....I think they'll do well, but definitely not a super bowl berth.

WOW, football obsessed, jk.  Anyways, I say the Panthers.

WOOOOOOH

SPURRIER'S 150TH COLLEGE WIN YESTERDAY!!!!!!

Yep, 15-0 shutout for Tyrone Nix's defense.  We had 2 (I think) seniors starting that gmae for the defense, and we allowed 79 yards.  Great defensive performance; all I can say about the offense is that at one point, our KICKER was leading the team in rushing yards, and our first offensive play of the game resulted in an interception.  But at least we got the job done, and freshman quarterback Smelley had a great outing.


geometry
analytic geometry
linear algebra
matrix
rectangle
vector

Can someone refresh me on the shoelace method for finding the area of a polygon when you know its coordinates?

[quote="hunter34"]http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf[/quote]
Yeah, thats a good site for shoelace.
It's basically listing out all the points of a polygon in clockwise or counterclockwise order, multiplying in a "shoelace" fashion, and half of the difference between the sums of the products.
So say the coordinates were $ (a,b)(c,d)(e,f)(g,h)(i,j)(k,l)$
And they were already in clockwise or counterclockwise order.
You list them
$ (a,b)$
$ (c,d)$
$ (e,f)$
$ (g,h)$
$ (i,j)$
$ (k,l)$
$ (a,b)$
You list the first term again.
Then you multiply diagonally.
Left side:
bc
de
fg
hi
jk
la
Right Side:
ad
cf
eh
gj
il
kb
Then, you add the products of each side up, and find the difference between the two sums.
Then you take half of that distance, and that's the area of your polygon.
I think it works for convex polygons only.

The proof of the theorem is too advanced though

Isn't this similar to Matrix Method?

It sounds like it.

[quote="ZhangPeijin"]What's the matrix method?[/quote]

I think it's basically same thing as shoelace method.. at least it sounds like it.

[quote="Quickster94"]The shoelace method should definetely be remembered.  Ppl before me have already explained it, but it is very useful.  For eg, if u remember 2006 chapter target, there was a problem that asked u to find the area of something with coordinates of like (1,1), (-1,2), (3,3), and (2007,2008).  with the shoelace method, the problem was very very simple, but w/o it was almost impossible.[/quote]
@ZhangPeijin: Quickster94's method is what I was talking about. You simply list the coordinates out with 1's at the end of each (I think this is where people got the shoelace method.) and repeat the last two vertical lines and paste them at the end.

Then you draw a diagonal(southeast, and southwest) thru the numbers and multiply the numbers that you crossed, and add them all up and divide it by 2.

Sorry, if I didn't explain it very well..

[quote="Quickster94"]The shoelace method should definetely be remembered.  Ppl before me have already explained it, but it is very useful.  For eg, if u remember 2006 chapter target, there was a problem that asked u to find the area of something with coordinates of like (1,1), (-1,2), (3,3), and (2007,2008).  with the shoelace method, the problem was very very simple, but w/o it was almost impossible.[/quote]
No, it was very possible. You just make a rectangle and subtract a few things from it, if I remember the problem right.

[quote="alanchou"]Does this work for all polygons?  I originally heard it only works for convex polygons, but it did work for at least one concave one.[/quote]
I am not sure... but the problems usually want you to find the area of triangle with only coordinates given, I think.

How advanced is the proof...exactly? :huh:
And where could I find it?  :P

It involves determinants of matrices.  Very simple actually.  You can probably ask for it in the high school basics.

[quote="Temperal"][quote="alanchou"]Does this work for all polygons?  I originally heard it only works for convex polygons, but it did work for at least one concave one.[/quote]
You can modify it to subtract areas, I think.[/quote]

How can you modify it?  :|

Shoelace is definitely not classroom math

It's proof consists of high-level use of vectors

Well, think about a convex polygon. One can form a concave polygon by subtracting smaller convex polygons from its area.

I get how you subtract stuff--
But how would you modify the formula? Is there a direct way to do that, or would you have to just subtract the other sectors out?  :huh:

You would just have to subtract, but that is a method.

Shoelace Formula is just an easy way to explain the determinants of matrices.

No, the method for evaluating determinants of 3 by 3 matrices is something a bit different.

[quote="Quickster94"]The shoelace method should definetely be remembered.  Ppl before me have already explained it, but it is very useful.  For eg, if u remember 2006 chapter target, there was a problem that asked u to find the area of something with coordinates of like (1,1), (-1,2), (3,3), and (2007,2008).  with the shoelace method, the problem was very very simple, but w/o it was almost impossible.[/quote]
 :blush: I did it out the long way. I made a big rectangle, then subtracted out two triangles from it.


function
algebra
functional equation
algebra proposed

Find all weak monotonic functions $f: \mathbb{R}\rightarrow \mathbb{R}$ so that for all reals $x$
\[ f(f(x))=f(-f(x))=f(x)^2 \]
holds.

(weak monotonic means that for all $a<b$: $f(a) \leq f(b)$ or for all $a<b$: $f(a) \geq f(b)$ or both hold).

Lemma: For every $x\in\mathbb{R}$, if $-|f(x)|\le t\le |f(x)|$, then $f(t)=f(x)^2$.
Proof: If $-|f(x)|\le t\le |f(x)|$ and $f$ is weak monotonic, then $f(-|f(x)|)\le f(t)\le f(|f(x)|)$ or $f(-|f(x)|)\ge f(t)\ge f(|f(x)|)$. But $f(t)$ is trapped in a "sandwich", because $f(-|f(x)|)=f(-|f(x)|)=f(x)^2$.

Solution: There are only two solutions to the functional equation: $f(x)=0$ and $f(x)=1$.
Proof: $-|f(x)|\le 0 \le|f(x)|$ for every $x\in\mathbb{R}$, and then, using the lemma, $f(0)=f(x)^2$, which gives  the constant solution $f(x)=f(0)^{1/2}$. Using $f(0)=f(0)^{1/2}$, we get $f(0)=0$ or $f(0)=1$.

[quote="lordWings"]Lemma: For every $x\in\mathbb{R}$, if $-|f(x)|\le t\le |f(x)|$, then $f(t)=f(x)^2$.
Proof: If $-|f(x)|\le t\le |f(x)|$ and $f$ is weak monotonic, then $f(-|f(x)|)\le f(t)\le f(|f(x)|)$ or $f(-|f(x)|)\ge f(t)\ge f(|f(x)|)$. But $f(t)$ is trapped in a "sandwich", because $f(|f(x)|)=f(-|f(x)|)=f(x)^2$.

Solution: There are only two solutions to the functional equation: $f(x)=0$ and $f(x)=1$.
Proof: $-|f(x)|\le 0 \le|f(x)|$ for every $x\in\mathbb{R}$, and then, using the lemma, $f(0)=f(x)^2$, which gives  the constant solution $f(x)=f(0)^{1/2}$. Using $f(0)=f(0)^{1/2}$, we get $f(0)=0$ or $f(0)=1$.[/quote]

Very nice solution :)

Indeed, your solution is nice but unfortunately it is wrong  :(
Many students fell into the trap that $f(x)=1$ and $f(x)=0$ are the only solutions.

Another possible solution would e.g. be
$f(x)=1$ for $x\geq -1$ and $f(x)=-1$ for $x<-1$.

You have forgotten $\pm f(0)^{1/2}$ in the last line of your solution.

My solution:
[hide]

Obvioulsly, if $f$ is constant, then $f(x)=0$ and $f(x)=1$ are the only solutions, so let us assume that $f$ is not constant.

Case 1: There exist $x,y$ so that $|f(x)|\neq |f(y)|$.
Let $a=min(|f(x)|,|f(y)|)$ and $b=max(|f(x)|,|f(y)|)$.
We have $0\leq a < b$ hence $[-a,a] \subset [-b,b]$.
We have $f(-a) = f(a) = a^2$, hence for all $x\in [-a,a] \rightarrow f(x)=a^2$ due to monotony.
On the other hand we have  $f(-b) = f(b) = b^2$, hence for all $x\in [-b,b] \rightarrow f(x)=b^2$ due to monotony.
But since $[-a,a] \subset [-b,b]$, we have $a^2=b^2$ which is a contradiction.

Case 2: For all $x,y$ we have $|f(x)| = |f(y)|$.
All absolute values are equal, so let $c\in \mathbb{R}^+$ be this value (if $c=0$ then $f$ would be constant).

We have $f(-c)=f(c)=c^2$, so $\pm c = c^2$ hence $c=1$. So $f(x)=\pm 1$ for all $x$.
We have $f(-1)=f(1)=1$.

Case 2.1: $f$ is increasing.
Since $f$ is not constant and $f(x)=\pm 1$, there must somewhere be a jump discontinuity $l$ from $-1$ to $1$.
In other words: $\exists l \in \mathbb{R}: \ f(l)\in\{-1,1\}, \forall x<l\rightarrow f(x)=-1, \forall x>l\rightarrow f(x)=1$.
But since $f(-1)=1$ we have $l<-1$ if $f(l)=-1$ and $l\leq -1$ if $f(l)=1$.
Obviously all functions which satisfy the conditions above are solutions.

Case 2.2: $f$ is decreasing.
Since $f$ is not constant and $f(x)=\pm 1$, there must somewhere be a jump discontinuity $l$ from $1$ to $-1$.
In other words: $\exists l \in \mathbb{R}: \ f(l)\in\{-1,1\}, \forall x<l\rightarrow f(x)=1, \forall x>l\rightarrow f(x)=-1$.
But since $f(1)=1$ we have $l>1$ if $f(l)=-1$ and $l\geq 1$ if $f(l)=1$.
Obviously all functions which satisfy the conditions above are solutions.

[/hide]

:P I was sleeping my siesta when suddenly I woke up with a :idea:: I had forgotten a $\pm$!
Then I corrected my solution after the missing $\pm$, with only a few simple lines:

Either $range(f)=\{0\}$, or $range(f)=\{-1,1\}$.
In the first case there's only one solution:
 a) $f(x)=0$ for every $x$.
In the second case there are three different kind of weak monotonic solutions with $f(t)=1$ for every $-1\le t \le 1$:
 b.1) $f(x)=1$ for every $x$.
 b.2) $f(x)=1$ for every $x<k$ (or $x\le k$) and $f(x)=-1$ otherwise, being $k>1$ (or, respectively, $k\ge 1$).
 b.3) $f(x)=-1$ for every $x<k$ (or $x\le k$) and $f(x)=1$ otherwise, being $k<-1$ (or, respectively, $k\le -1$).

So, there's one good thing and one bad thing about siestas: they clear your head and help you reaching solutions, but when you post them, someone has already beaten you to it!  :lol:

[edit: a sign was missing]

[quote="Yimin Ge"]Indeed, your solution is nice but unfortunately it is wrong  :(
Many students fell into the trap that $f(x)=1$ and $f(x)=0$ are the only solutions.
[/quote]

What a boring problem!
It does not require creativity nor mathematical talent.

Its main intention seems to be to make participants overlook some of the solutions.
Their should be better ways to determine the Austrian IMO team!


Στην Eλλαδα το 2004 πραγματοποιηθηκε η ΔΜΟ (ΙΜΟ στα αγγλικα)
τωρα εαν οι Ι,Μ,Ο ειναι διαφορετικοι θετικοι ακεραιοι και     $ I \cdot M \cdot O\equal{}2004$

Να βρειτε την μεγιτη τιμη του αθροισματος    $ I\plus{}M\plus{}O$

[quote="basted"][hide]$ 2004 \plus{} 1 \plus{} 1 \equal{} 2006$[/hide][/quote]
τα παραπανω γραμματα ειναι [u]διαφορετικοι[/u] θετικοι ακεραιοι
 :wink:

Α οκ. Ε ενταχει αναλυουμε σε πρωτους παραγοντες και περνουμε περιπτωσεις.


f(x)=1/{³√(x²+2x+1)+³√(x²-1)+³√x²-2x+1)}

for all positive integers x, find the value of f(1)+f(3)+f(5)+....+f(997)+f(999)[/hide]

[hide="hint"]

Try to rationalize the denominator

[/hide]

[hide="solution"]
Let $u = \sqrt[3]{x+1}$, and $v = \sqrt[3]{x-1}$.

\begin{eqnarray}\frac{1}{\sqrt[3]{(x+1)^{2}}+\sqrt[3]{x^{2}-1}+\sqrt[3]{(x-1)^{2}}}& = &\frac{1}{u^{2}+uv+v^{2}}\\&= &\frac{u-v}{u^{3}-v^{3}}\\& = &\frac{1}{2}\left(\sqrt[3]{x+1}-\sqrt[3]{x-1}\right)\end{eqnarray}

Then, the sum telescopes, so we end up with $\frac{1}{2}\left(-\sqrt[3]{0}+\sqrt[3]{1000}\right) \rightarrow \boxed{5}$
[/hide]


Let $x , y , z , t$ be positive real numbers .

prove that : \[\left({\frac{x}{y}+\frac{y}{z}+\frac{z}{t}+\frac{t}{x}}\right)^{2}\ge \left({x+y+z+t}\right)\left({\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}}\right)\]

[quote="yongyao"]`The inequality does not hold.`
`a counter-example`
$[1/6,1/103,1/1779,1]$[/quote]

you are right  the inequality does not hold .
thank you yongyao


[quote="4everwise"]Determine how many degrees are in 1 radian.[/quote]

[hide]To convert from radians to degrees: if $r$ is the radians and $d$ is degrees.

$\frac{360^\circ}{2\pi}=\frac{d}{r}$

$\frac{360^\circ}{2\pi}=\frac{d}{1}$

$\frac{360^\circ}{2\pi}=d$

$d=\frac{180^\circ}{\pi}$
[/hide]


2006 takımı belli oldu:
Ahmet Kabakulak
Batuhan Karagöz
Cafer Tayyar yıldırım
Hale Nur Kazaçeşme
Hasan Hüseyin Eruslu
Metehan Özsoy

Bence tek sürpriz isim Cafer oldu.Hepimizi tebrik ederim.İnşallah ilk 10a falan gireriz.

:-) Tebrik ediyorum!!!
Basarilar.
Takimdakilerin hangi okullardan oldugunu ogrenebilirmiyim?

devran

izmir fen lisesi, ankara fen lisesi, özel şehzade mehmet lisesi, samanyolu lisesi ve ilköğretim okulu takımda öğrencisi olan okullar bunlar ama kim hangi okula gidiyo tam hatırlamıyorum

1)Bence Metehan Ozsoy, Ankara fen lisesinde okuyordu, degil mi??
2)Gordugum kadar onlarin hepsi ozel okularda okuyorlar, degil mi??

 Davron

izmir fen lisesi ve ankara fen lisesi devlet okuludur
ahmet kabakulak izmir fen lisesi hasan hüseyin eruslu ve hale nur kazaçeşme şehzade mehmet lisesi

tam olarak:
Ahmet Kabakulak(ben)-İzmir Fen Lisesi(devlet okulu)
Batuhan Karagöz-Ankara Fen Lisesi(devlet okulu)
Halenur Kazaçeşme-Şehzade Mehmet Lisesi(özel okul)
Hasan Hüseyin Eruslu-Şehzade Mehmet Lisesi(özel okul)
Cafer Tayyar Yıldırım-Samanyolu Fen lisesi(özel okul)
Metehan Özsoy-Samanyolu Fen lisesi(özel okul)


Prove the inequality for three nonnegative reals:
\[9(a^{2}+bc)(b^{2}+ca)(c^{2}+ab) \geq 8abc(a+b+c)(a^{2}+b^{2}+c^{2}). \]

Sorry, I forgot the coefficients! Edited.


[quote="idiot"]F[size=75]n[/size]=F[size=75]n-1[/size]+F[size=75]n-2[/size] for n >= 2

prove that 7F[size=75]n+2[/size]:^3:-F[size=75]n[/size]:^3:-F[size=75]n+1[/size]:^3: is divisible by F[size=75]n+3[/size][/quote]

Let $x = F_{n+1}, y = F_n$; then $F_{n+3} = x+2y$.

The problem is to prove that $P(x,y) = 7(x+y)^3 - x^3 - y^3$  is divisible by $x+2y$.
This is the same as to prove that $P(x,y)$ vanishes when $y = -2x$.
Because $P(x,y)$ is homogeneous, this is equivalent to $P(1,-2) = 0$.  That's all.


number theory
greatest common divisor
number theory unsolved

Let $a_1, a_2, ...........$ be a strictly increasing sequence of positive integers such that
 $gcd( a_m, a_n)=a_{gcd(m,n)}$ for all positive integers $m$ and $n$.Let $k$ be the
least positive integer for which there exist positive integers $r<k$ and $s>k$ such that ${a_k}^2=a_r a_s$. Prove that $r$ divides $k$ and that $k$ divides $s$.